HEAT AND MASS TRANSFER
Problems and Solutions
About the Author
Dhirendra Kumar Dixit was until recently on the faculty of Mechanical
Engineering at Visvesvaraya National Institute of Technology, Nagpur, and
has more than 35 years of industrial, teaching and research experience. He has
taught almost all subjects and guided projects to the undergraduate mechanical
engineering students and to postgraduate students of heat power engineering.
Besides being an accomplished teacher and a prolific writer, he is well versed
in diversely different fields like Journalism, Management, Astrology, Yoga and
Spiritualism.
He has travelled worldwide and visited several countries including the USA, the UK, the Philippines,
the Netherlands, Luxumberg, Belgium, Singapore, Hong Kong, Thailand, etc. He has also published and
presented more than 150 research papers both nationally and internationally. He has published more
than 2000 articles in English, Hindi and Marathi in almost all leading newspapers and magazines. He
has organized many workshops and seminars and was the convener of All India Seminar on Energy
Management organized by the Institution of Engineers (India). Prof. Dixit has been frequently invited by
the Akashwani and Doordarshan for topical talks and scientific programmes.
Based on his doctoral research at IIT Bombay, he was awarded the prestigious best paper prize by the
Institution of Engineers (India) and is the recipient of KF Antia Award. He also won the Nagpur Times
Gold Medal and the Austen Wingate Nazareth memorial prize for journalism. Prof. Dixit is a Life Fellow
of the Institution of Engineers, a Life Member oflnternational Solar Energy Society, a Life Member of the
Indian Society for Heat and Mass Transfer and Life Member of Indian Society for Technical Education.
He has been a visiting faculty at many institutes like the Department of Business Management, Nagpur
University, Bhavan's College of Communication and Management, Department of Mass Communication,
Nagpur University, National Power Training Institute, Nagpur, etc. His areas of interest and specialization
are Energy Management, and Renewable Energy sources, Refrigeration and Air-Conditioning Thermal
Engineering, Heat Transfer and Marketing Management.
HEAT AND MASS TRANSFER
Problems and Solutions
DK Dixit
Retired Professor
Department of Mechanical Engineering
Visvesvaraya National Institute of Technology (VNIT) Nagpur
Maharashtra
Tata McGraw Hill Education Private Limited
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Heat and Mass Transfer
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Contents
Preface ........................................................................................................................................... xi
Nomenclature ............................................................................................................................... xiii
1.
FUNDAMENTAL CONCEPTS OF HEAT TRANSFER
1.1
1.2
1.3
2.
1.1-1.39
Introduction .................................................................................................................................. 1.1
Modes of Heat Transfer ................................................................................................................ 1.1
Energy Balance ............................................................................................................................. 1.3
Solved Examples ........................................................................................................................... 1.4
Multiple Choice Questions ......................................................................................................... 1.36
True/False ................................................................................................................................... 1.38
Fill in the Blanks ........................................................................................................................ 1.38
Exercises ..................................................................................................................................... 1.38
STEADY-STATE HEAT CONDUCTION-ONE DIMENSION
2.1-2.128
2 .1 Introduction .................................................................................................................................. 2.1
2.2 General Heat-Conduction Equation ............................................................................................. 2.1
2.3 Initial and Boundary Conditions .................................................................................................. 2.2
2.4 Steady-State Heat Conduction-One Dimension ........................................................................ 2.3
2.5 Composite Systems ...................................................................................................................... 2.4
2.6 Critical Radius of Insulation ........................................................................................................ 2. 7
2.7 Variable Thermal Conductivity ..................................................................................................... 2.8
Solved Examples ........................................................................................................................... 2.8
Multiple Choice Questions ....................................................................................................... 2.119
True/False ................................................................................................................................. 2.122
Fill in the Blanks ...................................................................................................................... 2.122
Exercises ................................................................................................................................... 2.122
3.
STEADY-STATE, ONE-DIMENSIONAL
HEAT CONDUCTION WITH INTERNAL HEAT GENERATION
3 .1
3.2
3.3
3.4
3.5
3.6
3.7
3.1-3.85
Introduction .................................................................................................................................. 3.1
Plane Wall with Uniform Heat Generation ................................................................................... 3.1
Long Solid Cylinder with Uniform Heat Generation ................................................................... 3.3
Hollow Cylinder with Uniform Heat Generation ......................................................................... 3.5
Solid Sphere ................................................................................................................................. 3. 7
Heat Generation in Nuclear Fuel Element ................................................................................... 3. 7
Heat Generation in Nuclear Fuel Element with Cladding ............................................................ 3.8
Solved Examples........................................................................................................................... 3.9
Multiple Choice Questions ......................................................................................................... 3.80
True/False ................................................................................................................................... 3.82
Fill in the Blanks ........................................................................................................................ 3.82
Exercises ..................................................................................................................................... 3.83
vi ,
4.
Contents
HEAT TRANSFER FROM EXTENDED SURFACES
4 .1
4.2
4.3
4.4
4.5
4.1--4.78
Introduction .................................................................................................................................. 4.1
Fins of Uniform Cross Section ..................................................................................................... 4.1
Fin Efficiency and Fin Effectiveness ............................................................................................ 4.4
Fins of Non-Uniform Cross-Section ............................................................................................ 4.5
Fin Arrays ..................................................................................................................................... 4.6
Solved Examples ........................................................................................................................... 4.8
Multiple Choice Questions ......................................................................................................... 4. 73
True/False ................................................................................................................................... 4. 75
Fill in the Blanks ........................................................................................................................ 4. 75
Exercises ..................................................................................................................................... 4. 75
5.
MULTIDIMENSIONAL HEAT CONDUCTION
5 .1
5.2
5.1-5.46
Introduction .................................................................................................................................. 5.1
Two-Dimensional Steady-State Heat Conduction ........................................................................ 5.2
Solved Examples ........................................................................................................................... 5.4
Multiple Choice Questions ......................................................................................................... 5.44
True/False ................................................................................................................................... 5.45
Fill in the Blanks ........................................................................................................................ 5.45
Exercises ..................................................................................................................................... 5.45
6.
TRANSIENT HEAT CONDUCTION
6.1
6.2
6.3
6.4
6.5
6.6
6. 7
6.1-6.110
Introduction .................................................................................................................................. 6.1
Lumped Capacity Formulation ..................................................................................................... 6.1
Finite Internal Resistance: One-Dimensional Systems ................................................................ 6.3
Semi-Infinite Medium .................................................................................................................. 6.6
Multidimensional Systems: Product Solutions ............................................................................ 6.8
Heat Transfer in Multi-Dimensional Systems .............................................................................. 6.9
Periodic Variation of Surface Temperature ................................................................................. 6.10
Solved Examples ......................................................................................................................... 6.11
Multiple Choice Questions ....................................................................................................... 6.103
True/False ................................................................................................................................. 6.106
Fill in the Blanks ...................................................................................................................... 6.106
Exercises ................................................................................................................................... 6.106
7.
BASICS OF CONVECTION HEAT TRANSFER
7.1-7.55
7 .1 Introduction ................................................................................................................................. 7.1
7.2 Heat-Transfer Coefficient ............................................................................................................ 7.1
7.3 External Flows: Concept of Velocity and Boundary Layer ......................................................... 7.2
7.4 Thermal Boundary Layer ............................................................................................................ 7.3
7.5 Analogy between Momentum and Heat Transfer ........................................................................ 7.4
7.6 Internal Flow ............................................................................................................................... 7.4
7. 7 Dimensional Analysis .................................................................................................................. 7. 6
Contents
r
vii
Solved Examples ........................................................................................................................... 7. 6
Multiple Choice Questions ......................................................................................................... 7.52
True/False ................................................................................................................................... 7.54
Fill in the Blanks ........................................................................................................................ 7.55
Exercises ..................................................................................................................................... 7.55
8.
FORCED CONVECTION: EXTERNAL FLOW
8 .1
8.2
8.3
8.4
8.5
9.
10.
Introduction ................................................................................................................................. 8.1
Parallel Flow Over Flat Plate ...................................................................................................... 8.1
Flow Across a Cylinder ............................................................................................................... 8.4
Flow Over Spheres ...................................................................................................................... 8.6
Flow Across Banks of Tubes ....................................................................................................... 8.6
Solved Examples ........................................................................................................................... 8.8
Multiple Choice Questions ......................................................................................................... 8. 76
True/False ................................................................................................................................... 8. 77
Fill in the Blanks ........................................................................................................................ 8. 77
Exercises ..................................................................................................................................... 8. 78
FORCED CONVECTION: INTERNAL FLOW
9 .1
9.2
9.3
9.4
9.5
9.6
9.7
9 .8
9.9
8.1-8.84
9.1-9.95
Introduction ................................................................................................................................. 9.1
Parallel Flow between Two Plates ............................................................................................... 9.1
Average Velocity and Temperature .............................................................................................. 9.1
Laminar and Turbulent Flow ....................................................................................................... 9.2
The Entrance Region ................................................................................................................... 9.2
Constant Heat Flux and Constant Wall Temperature Conditions ................................................ 9.3
Laminar Flow in a Tube .............................................................................................................. 9.5
Turbulent Flow Heat Transfer in Circular Tubes ......................................................................... 9. 6
Liquid-Metal Heat Transfer ......................................................................................................... 9. 7
Solved Examples ........................................................................................................................... 9.8
Multiple Choice Questions ......................................................................................................... 9.90
True/False ................................................................................................................................... 9.91
Fill in the Blanks ........................................................................................................................ 9.91
Exercises ..................................................................................................................................... 9.92
NATURAL (FREE) CONVECTION
10.1-10.109
10 .1 Introduction ............................................................................................................................. 10.1
10.2 Vertical Flat Plate ..................................................................................................................... 10.1
10.3 Horizontal Plates ..................................................................................................................... 10.4
10.4 Horizontal Cylinders ............................................................................................................... 10.5
10.5 Spheres .................................................................................................................................... 10.6
10.6 Simplified Correlations for Air.. .............................................................................................. 10.6
10.7 Enclosures ................................................................................................................................ 10.6
10.8 Concentric Cylinders ............................................................................................................... 10.8
10.9 Concentric Spheres .................................................................................................................. 10.8
viii ,
Contents
10.10
10.11
11.
HEAT TRANSFER WITH PHASE CHANGE
11.1
11.2
11.3
11.4
11.5
11.6
12.
Turbine Rotors/Rotating Cylinders/Disks/Spheres ................................................................ 10.9
Combined Free and Forced Convection ................................................................................ 10.9
Solved Examples .................................................................................................................. 10.10
Multiple Choice Questions ................................................................................................ 10.101
True/False .......................................................................................................................... 10.102
Fill in the Blanks ................................................................................................................ 10.102
Exercises ............................................................................................................................ 10.102
11.1-11.78
Introduction ............................................................................................................................. 11.1
Boiling Heat Transfer .............................................................................................................. 11.1
Boiling Heat-Transfer Correlations ......................................................................................... 11.2
Condensation Heat Transfer .................................................................................................... 11. 5
Film Condensation .................................................................................................................. 11.5
Heat-Transfer Correlations for Film Condensation ................................................................ 11. 6
Solved Examples ................................................................................................................... 11.10
Multiple Choice Questions .................................................................................................... 11. 72
True/False .............................................................................................................................. 11. 73
Fill in the Blanks ................................................................................................................... 11. 73
Exercises ............................................................................................................................... 11. 74
HEAT EXCHANGERS
12.1-12.113
12.1 Introduction ............................................................................................................................. 12.1
12.2 Overall Heat-Transfer Coefficient ........................................................................................... 12.J
12.3 Analysis of Heat Exchangers .................................................................................................. 12.3
12.4 Selection Criteria ..................................................................................................................... 12. 7
Solved Examples ..................................................................................................................... 12. 7
Multiple Choice Questions .................................................................................................. 12.102
True/False ............................................................................................................................ 12.105
Fill in the Blanks ................................................................................................................. 12.105
Exercises ............................................................................................................................. 12.105
13.
RADIATIVE PROPERTIES AND PROCESSES
13.1-13.55
13 .1 Introduction ........................................................................................................................... 13 .1
13.2 Thermal Radiation ................................................................................................................. 13.J
13.3 Black-Body Radiation ........................................................................................................... 13.2
13.4 Laws of Radiation ................................................................................................................. 13.2
13.5 Band Emission ...................................................................................................................... 13.4
13.6 Emission, Irradiation and Radiosity ...................................................................................... 13.4
13.7 Radiative Properties .............................................................................................................. 13.5
13. 8 So lid Angle and Radiation Intensity ..................................................................................... 13. 6
13.9 Kirchhoff's Law .................................................................................................................... 13.8
13.10 Atmospheric and Solar Radiation .......................................................................................... 13. 7
Contents
r
ix
Solved Examples ..................................................................................................................... 13.8
Multiple Choice Questions .................................................................................................... 13.50
True/False .............................................................................................................................. 13.52
Fill in the Blanks ................................................................................................................... 13.52
Exercises ............................................................................................................................... 13.53
14.
RADIATION HEAT EXCHANGE BETWEEN SURFACES
14.1
14.2
14.3
14.4
14.5
14.6
14.1-14.124
The Shape Factor ..................................................................................................................... 14.l
Radiation Heat Exchange between Black Surfaces ................................................................ 14.3
Radiation Heat Exchange between Diffuse-Gray Surfaces ..................................................... 14.3
Radiation Shields .................................................................................................................... 14.5
Radiation Effect on Temperature Measurement ...................................................................... 14.6
Radiant Heat Exchange with Emitting and Absorbing Gases ............................................................. 14.6
Solved Examples ..................................................................................................................... 14. 7
Multiple Choice Questions .................................................................................................. 14.112
True/False ............................................................................................................................ 14.113
Fill in the Blanks ................................................................................................................. 14.113
Exercises ............................................................................................................................. 14.114
15. MASS TRANSFER
15.1-15.65
15 .1 Introduction ........................................................................................................................... 15.1
15.2 Mass Diffusion ...................................................................................................................... 15.l
15.3 Binary Diffusion Coefficient ................................................................................................ 15.2
15.4 Boundary Conditions ............................................................................................................ 15.3
15.5 Steady State Mass Diffusion in a Stationary Medium .......................................................... 15.3
15.6 Diffusion in a Moving Medium ............................................................................................ 15.4
15.7 Equimolar Counter Diffusion ............................................................................................... 15.4
15.8 Diffusion of Vapour through a Stagnant Gas ........................................................................ 15.5
15.9 Transient Mass Diffusion ...................................................................................................... 15.6
15.10 Convective Mass Transfer ..................................................................................................... 15.6
15 .11 Analogy between Heat and Mass Transfer ........................................................................... 15. 7
15.12 Simultaneous Heat and Mass Transfer ................................................................................. 15.8
Solved Examples ................................................................................................................... 15.8
Multiple Choice Questions .................................................................................................. 15. 63
True/False ............................................................................................................................ 15. 64
Fill in the Blanks ................................................................................................................. 15. 64
Exercises ............................................................................................................................. 15. 65
Preface
Heat transfer along with Thermodynamics and Fluid Mechanics constitutes the trinity of the Mechanical
Engineering discipline. A sound knowledge of fundamentals is the essential prerequisite to master these
subjects. The present book on Heat Transfer does precisely that. The target audience is essentially the
undergraduate students of Mechanical Engineering, Chemical Engineering, and Aerospace Engineering.
The book is also tailored to meet the requirements of the candidates aspiring to take IES, GATE, AMIE
and other competitive examinations.
The USP of the book is clear, coherent and cogent presentation of the relevant theory in a nutshell
followed by an incredibly large number of solved examples with different degrees of difficulty. The
focus is on presenting a refreshing approach with a unique problem-solving methodology coupled with
an emphasis on consistency and compatibility of units, so sadly neglected by the student fraternity.
Salient Features
• Provides complete review of all the fundamental concepts of Heat and Mass Transfer
• Pedagogical features developed in conformation with the requirements of examination pattern of
universities and competitive exams
• Tutorial approach of problem solving employed for all the solved examples
• More than 1500 solved examples, problems, and objective type questions with answers:
• Solved Examples: 600
• Problems: 500
• Objective Type Questions:
• Multiple Choice Questions: 300
• True/False: 75
• Fill in the Blanks: 75
Chapter Organisation
The content is spread over 15 chapters covering topics like Fundamental Concepts of Heat Transfer,
Steady-State Conduction-One Dimension, Steady State Conduction With Heat Generation, Steady State
Heat Transfer From Extended Surfaces, Multidimensional Steady State Heat Conduction, Unsteady State
Heat Conduction, Convection Principles and Dimensional Analysis, Forced Convection: External Flow,
Forced Convection: Internal Flow, Free Convection, Heat Transfer with Change of Phase, Heat Exchangers,
Radiation-Properties and Processes, Radiation Exchange Between Surfaces and Mass Transfer.
The end-of-chapter problems will be useful in gaining confidence acquired with constant practice. I am
more than confident that the book either used independently or as a companion volume to a standard text
on the subject, will be well received by students and teachers alike.
xii , Preface
Acknowledgements
I am grateful to my wife, Chandrakanta, for her forbearance and fortitude during the course of preparation
of this work. My thanks are also due to Mrs Vaishali Temburne for her painstaking typing of the
manuscript and flawless drawing of Figures and illustrations. At this stage, I would also like to thank
the reviewers who went through the manuscript and provided valuable suggestions. Their names are
given below.
Akhilesh Gupta
Indian Institute of Technology (IIT) Roorkee, Uttarakhand
Sunil Punjabi
Government Engineering College, Ujjain, Madhya Pradesh
Amarnath Mullick
National Institute of Technology (NIT), Durgapur, Paschimbanga
Samir Kumar Saha
Jadavpur University, Kolkata
Kanchan Chatterjee
Dr. B C Roy Engineering College, Durgapur, Paschimbanga
MK Bhatt
Sardar Vallabhbhai National Institute of Technology (SVNIT), Surat, Gujarat
AS Dhoble
Vivesvaraya National Institute of Technology (VNIT), Nagpur, Maharashtra
SM Joshi
Pillai College of Engineering, Mumbai
K R Balasubramanian
National Institute of Technology (NIT), Thiruchirapalli, Tamil Nadu
B Sudheer Prem Kumar
Jawaharlal Nehru Technological University(JNTU) College of Engineering, Hyderabad
Finally, I must appreciate the keen interest and involvement of the Tata McGraw-Hill publishing team,
especially Ms. Harsha Singh, for ensuring expeditious publication of the book.
DK Dixit
Nomenclature
Many symbols have more than one meaning. The context will indicate the specific meaning.
a
Speed of sound, mis
Biot number
Specific heat, kJ/kg °C or kJ/kg K or J/kg K
Thermal capacity, W /°C
Speed of light, mis
ct
Friction coefficient
d, D
Diameter, m
Dh
E,E,._
Hydraulic diameter, m
Emissive power, W/m 2 (total) or W/m2 µm (spectral)
Est
erf
Total energy stored, J
erfc
Complementary error function
FiJ
Shape factor between surface i and surface j
f
Frequency, Hz
f
Friction factor
f,.
Fraction of blackbody radiation between wavelengths O and ;\,
Fo
Fourier number
g
Acceleration due to gravity, m/s 2
q
Heat generation rate, W/m 3
G
Irradiation, W/m2
Gr
Grashof number
Gz
Graetz number
h
Heat transfer coefficient, W/m2 °C or W/m 2 K
H
Nondimensional heat-generation parameter
h
Planck's constant, J s
Error function
Radiation heat transfer coefficient, W/m 2 °C or W/m2 K
Latent heat of melting/solidification, kJ/kg
xiv , Nomenclature
lo, 11
Modified Bessel function of order zero and order one, respectively
I
Radiation intensity, W/m 2 sr (total) or W/m 2 sr µm (monochromatic)
J
Radiosity, W/m 2
k
Boltzmann constant, kJ/kmol K
k
Thermal conductivity, W/m °C or W/m K
L
Length, m
LC
Characteristic length, m,
LMTD
Logarithmic mean temperature difference, °C or K
Mean beam length, m
m
Fin parameter for a uniform area fin, m- 1
M
Mach number
m
Mass, kg
NTU
Number of transfer units
Nu
p
Nusselt number
p
Pressure, atm
Pe
Peclet number
Pr
Prandtl number
q
Heat flux, W/m 2
Q
Heat-transfer rate, W
Re
Electrical resistance, {l
R
Radial coordinate or radius, m
R
Radius, m
R
Ratio of heat capacities in heat exchangers
R
Thermal resistance, °C/W or m2 °C/W
Ra
Rayleigh number
Re
Reynolds number
Rf
Fouling resistance, °C/W or m2 °C/W
s
Conduction shape factor
Perimeter, m
Solar constant, W/m 2
Surface area, m2
Diagonal pitch, m
Longitudinal pitch, m
Stanton number
Nomenclature
Sr
Transverse pitch, m
T
Temperature, °C or K
T
Time, s
't
Transmissivity or transmittance
U
Free stream velocity, mis
U
Overall heat-transfer coefficient, W/m2 °C or W/m2 K
u
x-component of velocity, mis
v
y-component of velocity, mis
¥
Volume, m3
x
x-coordinate, m
y
y-coordinate, m
z
z-coordinate, m
00
Greek Symbols
a
Absorptivity
<l
Thermal diffusivity, m2/s
~
Isobaric volumetric expansion coefficient, K- 1
o
o
Boundary layer thickness, m
Depth of penetration, m
81
Thermal boundary layer thickness, m
</>
Phase angle, rad
E
Eddy viscosity
E
Effectiveness of a fin array
E
Emissivity
E
Heat exchanger effectiveness
eH
Eddy diffusivity of heat
TJ
Blasius similarity variable
TJ
Fin efficiency
K:
Absorption coefficient, m- 1
µ
Cosine of angle with respect to normal
v
Kinematic viscosity, m2/s
v
Photon frequency, Hz
0
Angle, degree or radian
0
Nondimensional temperature
r
xv
xvi , Nomenclature
p
Density, kg/m 3
p
Radius ratio
p
Reflectivity
p
Resistivity, Q-m
CJ
Stefan-Boltzmann constant, W/m 2 K4
't
Shear stress, Pa
't
Time constant, s
't
Transmissivity
co
Circular frequency, rad/s
(0
Solid angle, sr
~
Similarity variable
Subscripts
f
Fluid
00
Pertaining to the ambient
rad
Pertaining to radiation
r
Radial
sur
Pertaining to surroundings
1, 2 etc Pertaining to a specific position
1
FUNDAMENTAL CONCEPTS OF
HEAT TRANSFER
Concept Review
INTRODUCTION
• I. I
•
Heat is the energy in transit that can be transferred from one place to another as a result of temperature
difference. The science of heat transfer has evolved from thermodynamics as a means to quantify heat
exchange and temperature. The amount of heat transferred per unit time is called heat-transfer rate and
is denoted by Q . The rate of heat transfer per unit area is called heat flux, q.
MODES OF HEAT TRANSFER
• 1.2
•
Heat can be transferred in three different modes: conduction, convection and radiation, which can, and
often do, act together.
1.2.1
Conduction
Conduction takes place in a stationary medium, and although present in fluids, it is usually more
significant in solids. Conduction is that mode of heat transfer in which energy exchange takes place
from the high-temperature region to the low-temperature region by the kinetic motion or direct impact
of molecules, as in the case of a fluid at rest, and by the drift of electrons, as in the case of metals. In
a solid which is a good electric conductor, a large number of free electrons move about in the lattice;
hence materials that are good electric conductors are also generally good heat conductors (i.e., copper,
silver, etc.).
Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent
less energetic ones as a result of interactions between the particles. The rate of conduction heat transfer
is expressed by Fourier's law of heat conduction as
(l.l)
where k is the thermal conductivity of the material, Ac is the area normal to the direction of heat transfer,
and dT/dx is the temperature gradient. The magnitude of the rate of heat conduction across a plane wall
of thickness L is given by
(1.2)
where AT is the temperature difference across the wall equals to (T1 - T2), as illustrated in Figure l. l.
1.2 ,
Heat and Mass Transfer
1.2.2
Convection
Convection is possible only in a fluid medium.
Convection requires a moving medium; the motion may
be generated by buoyancy force (free convection) or
through external means (forced convection). Thermal,
or infrared radiation, is part of the electromagnetic
spectrum; all bodies with a temperature above absolute
zero emit thermal radiation.
Convection is the mode of heat transfer between
a solid surface and the adjacent liquid or gas that
is in motion, and involves the combined effects of
conduction and fluid motion. The rate of convection
heat transfer is expressed by Newton 's law of cooling as
(1.3)
T
~-----<~--~- ~ x
L
0
Fig. 1.1
Conduction heat transfer through a solid or
stationary medium
where h is the convection heat transfer coefficient
in W/m 2 °C, As is the surface area through which convection heat transfer takes place, Ts is the surface
temperature, and T= is the temperature of the fluid sufficiently far from the surface, as indicated in
Figure 1.2.
~--+----+- u
~----a>----+- T
Temperature
/profile
Airflow
Hot block
Fig. I .2
Convection heat transfer from a hot surface to the moving air
It is noteworthy that the thermal conductivity is a thermophysical property of matter, the heat transfer
coefficient is not; it depends on the velocity of the fluid, the presence of turbulence, surface roughness,
geometry and the various thermophysical fluid properties.
1.2.3
Radiation
Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result
of the changes in the electronic configurations of the atoms or molecules. The maximum rate of radiation
that can be emitted from a surface at an absolute temperature Ts is given by the Stefan- Boltzmann law
as Qemit max = <JAsT/,
where <J = 5.67 x 10-s W/m 2 K4 is the Stefan- Boltzmann constant.
Fundamental Concepts of Heat Transfer
1.3
When a surface of emissivity£ and surface area
As at an absolute temperature Ts is completely
enclosed by a much larger ( or black) surface at
absolute temperature Tsur separated by a gas (such
as air) that does not intervene with radiation, as
shown in Figure 1.3, the net rate of radiation heat
transfer between these two surfaces is given by
IQrad
= ecrA. (T,4 -T.~,)1
(1.4)
Surface
As, Es, Ts
Fig. 1.3 Radiation heat transfer from a small surface to
In this case, the emissivity and the surface
the large surroundings
area of the surrounding surface do not have any
effect on the net radiation heat transfer.
The rate at which a surface absorbs radiation is determined from Qabsorbed = ex Qincident where Qincident is
the rate at which radiation is incident on the surface and ex is the absorptivity of the surface.
ENERGY BALANCE
• 1.3
•
A system with prescribed quantity of matter is called a closed system and a system that has an identifiable
region in space and involves transfer of mass across its boundaries is called an open system or control
volume. The first law of thermodynamics or the energy balance for any system undergoing any process
can be expressed as
(1.5)
In the rate form, the control volume energy balance, as illustrated in Figure 1.4, can be written as
(1.6)
where £gen is the _rate of ~olumetric internal heat generation and £51 is the rate of thermal energy (internal)
storage system. Ein and Eout are the rates of energy transfer (heat/work interactions) entering and leaving
the control volume.
Fig. 1.4
Control volume energy balance
When a stationary closed system involves heat transfer only and no work interactions across its
boundary, the energy balance relation reduces to
( 1.7)
1.4 ,
Heat and Mass Transfer
where Q is the amount of net heat transfer to or from the system. When heat is transferred at a constant
rate of Q, the amount of heat transfer during a time interval M can be determined from
(1.8)
Under steady conditions and in the absence of any work interactions, the conservation of energy
relation for a control volume with one inlet and one exit with negligible changes in kinetic and potential
energies can be expressed as
(1.9)
where m= p AcV is the mass flow rate and Q is the rate of net heat transfer into or out of the control volume.
Control surface energy balance gives
(1.10)
This is ii:respectiye of the presence or absence of £gen and Est terms which are volumetric phenomena
as against Ein and £out which are surface phenomena.
In the heat-transfer analysis, one should be able to identify the separate modes of heat transfer and,
especially, the relative contributions made by convection and radiation.
Solved Examples
(A) CONDUCTION
•
•
Example 1.1 ~ Calculate the heat transfer rate across a plane wall, 15 cm thick, with a cross-sectional area
of 5 m2, and of thermal conductivity 9.5 W/m K. The steady-state end surface temperatures
are I 20°C and 30°C. Also, determine the temperature gradient in the direction of heat flow.
Known: Heat flo'Y across a plane wall.
dT
Find: Heat rate, Q(W) and temperature gradient, -(Kim).
dx
Schematic:
Ac=5m 2
\-\eat i\oW
k=9.5W/m K
---
-L-x
Fig. I.S
Fundamental Concepts of Heat Transfer
1.5
Assumptions: (1) Steady-state, one-dimensional conduction. (2) No internal heat generation. (3) Constant
thermal conductivity. (4) Constant cross-sectional area.
Analysis: The rate of heat transfer in the x-direction is
(7; -7;)
dT
-Q· =-kA -=kA
c dx
c
L
where dT is the temperature gradient.
dx
Substituting proper values,
Q = (9 .5W /m K) (5m2/120-30)oC or K]
0.15 m
= 28.5 x 103 W
or
28.5 kW
(Ans)
Temperature gradient,
dT =-_il.._=28.5xI03 W
dx
9.5W/ mKx5m 2
=-600K/m
(Ans)
Since k and Ac are constant, and Q is constant in steady-state with no heat generation, the temperature
gradient or slope of the temperature distribution T(x) is constant.
Alternatively,
-7;
dT = T(x=L)-T(x=O) = T2
dx
(x=L)-(x=O)
L-0
= (30-120)°C or K = _ 600 Kim
0.15m
Example 1.2 ~ Consider a slab of thickness L = 0.25 m. One surface is kept at I00°C and the other surface
at 0°C. Determine the heat flux across this slab if the slab is made from (a) pure copper,
(b) pure aluminium, (c) pure iron, (d) building brick, (e) cement, and (f) loosely packed asbestos.
Known: Thickness of slab with end temperatures.
Find: Heat flux, q (W!m 2) for (a) copper, (b) aluminium, (c) iron, (d) building brick, (e) cement, and
(f) loosely packed asbestos.
Schematic:
r 1= 100°c
T2 = 0°C
----<f---------+--Q/A
j-
L=O.OSm
Fig. 1.6
-j
1.6 ,
Heat and Mass Transfer
Assumptions: ( 1) Steady-state, one-dimensional conduction. (2) Constant thermal conductivity.
Properties:
k (Yv Im °C)
Pure copper
Pure aluminium
Pure iron
Building brick
Cement
Loosely packed asbestos
401
237
80.2
0.69
0.29
0.158
Analysis: Fourier's rate equation
Q=-kAdT
dx
=kA(1,-J;)
L
Heat flux,
q = Q = k(li -J;) = k(W/m 0 C) (l00-0)°C = 400 k(W/m2 )
A
L
0.25m
(a) Pure copper (k= 401 W/m 0 C):
q = 400 X 401 = 160.4 X 103 W/m 2
(Ans) (a)
(b) Pure aluminium (k= 237 W/m 0 C):
q = 400 X 237 = 94.8 X 103 W/m 2
(Ans) (b)
q = 400 X 80.2 = 32.08 X 103 W/m 2
(Ans) (c)
(c) Pure iron (k= 80.2 W/m 0 C):
(d) Building brick (k= 0.69 W/m 0 C):
q = 400 X 0.69 = 276 W/m 2
(Ans) (d)
q = 400 X 0.29 = 116 W/m 2
(Ans) (e)
(e) Cement (k= 0.29 W/m 0 C):
(f) Loosely packed asbestos (k= 0.158 W/m 0 C):
q = 400 X 0.158 = 63.2 W/m 2
(Ans) (f)
Comment: Heat flux being proportional to thermal conductivity of the slab material, we find maximum
heat flux with copper and minimum heat flux with asbestos. The heat fluxes in the six cases are in
decreasing order.
Example 1.3
L=
The variation in temperature with distance (at a given instant) in a 6 cm thick concrete
(k 0.8 W/m 0 C) wall is as follows:
T(x) = 44 - 27x + xl (0 C)
where x is in cm. The temperature at the left face where x = 0 is 44°C, the right face is at x = 6 cm. Determine
the location and magnitude of the maximum temperature in the wall, and the magnitude and direction of the heat
flux (heat rate per unit area) at each face and at the midplane of the wall. Sketch the temperature profile, T(x).
Fundamental Concepts of Heat Transfer
I •7
Known: Temperature distribution in a concrete wall of prescribed thickness and conductivity.
Find: TmaxC 0 C) and x(cm); q(x=O)• q(x=L/2), q(x=L)
Schematic:
x=O
I cm
2cm
4cm
5cm
l
x=L
=6cm
x=L/2
=3cm
Fig. 1.7
Assumptions: (1) One-dimensional heat conduction. (2) Constant thermal conductivity.
Analysis: The prescribed temperature profile is
T(x) = 44 - 27 x + x 3
where x is in cm and T is in °C.
At x = 0 (left face):
T1 = T(O) = 44°C
Temperature gradient,
d
=-[44-27
x+x 3 ] I
-dTI
dx x=O
dx
x=O
=(-27+3x 2 ) lx=O =-27°C/cm
heat flux at the left face is
dTI
q, =-kdx x=O
=
(o.s~)
m°C
(27°C)(lOOcm),
cm
Im
= +2160W /m 2 (~)
(Ans)
1.8 ,
Heat and Mass Transfer
Temperature gradient at the midplane is
-dTI
=-27+3x2 1
dx x=L/2
x=3 cm
= -27+(3x32 )=0°C/cm
Heat flux at the midplane,
dTI
q =-kc
dx x=3cm
=0W/m2
(Ans)
Temperature gradient at the right face is
-dTI
=-27+3x2 1
dx x=L
x=6cm
= -27 + 3 x 62 = + 81°C/cm
Heat flux at the right face is
dTI
q2 =-kdx x=6cm
= (-0.8W/m 0 C) (+81°C/cm) ( 101°;m)
= - 6480 W/m 2(+-)
(Ans)
To find minimum/maximum temperature in the wall, we differentiate T(x) with respect to x and equate
the resulting derivative to zero.
dT
- = 0 =-27+3x2
dx
x = ~27 /3 = 3 cm
d2 T
--=0+6x, i.e., +ve
dx2
As the second derivative is positive for positive values of x, x = 3 cm is the location of minimum
temperature whose magnitude is
Tmin=T(x=3 cm)=44-(27x3)+(3) 3
=-10°C
To find location and magnitude, let us plot the temperature distribution by calculating the temperatures
T( 0 C) at a few values of distance x(cm).
Fundamental Concepts of Heat Transfer
T (x = l cm) = l 8°C
I •9
T (x = 2 cm) = - 2°C
T (x = 3 cm) = -l 0°C
T (x = 4 cm) = 0°C
T( x = 5 cm) = 34°C
T (x = 6 cm) = 98°C
Tmax = 98°C at x = 6 cm, i.e., at the right face.
The temperature profile is sketched in the schematic.
Example 1.4 ~ The heat flux on the 30° diagonal surface of the Bakelite (k = 1.4 W/m 0 C) wedge shown
in Fig. I .8(a) is 2000 W/m 2 in the direction shown. Calculate the heat flux and temperature
gradient in the x- and y-directions.
y
30°
~-~--~-~x
Fig. l.8(a)
Known: Heat flux on the diagonal surface is specified.
Find: Heat flux and temperature gradient in the x- and y-directions.
Schematic:
y
Bakelite wedge
(k = 1.4 W/m K)
0=30°
~~-----~---x
Fig. I .8(b)
Assumptions: (1) The steady-state heat flux is constant over the entire wedged surface.
Analysis: The wedge angle 0 = 30°, the angle between the x-axis and the direction of heat flux is
<I> = 90 + 0 = I 20°
qx= qn cos <I>= (2000 W/m 2) cos 120° = -1000 W/m 2
(Ans)
qy= qn sin <I>= (2000 W/m 2) sin 120° = +1732 W/m 2
(Ans)
1.10 ,
Heat and Mass Transfer
Fourier's rate equation:
Qn =q =-k dT
A
dn
n
temperature gradient in the x-direction is
dT = _ qx = -( -1000W/m2 )
dx
k
1.4 W/mK
+ 714.3 K/m
(Ans)
and, the temperature gradient in the y-direction is
dT =- qy =-1732W/m 2 = _ 1237 K/m
dy
k
l.4W/mK
(Ans)
Comment: Positive heat flux is associated with negative temperature gradient. Remember that heat flux
is a vector quantity.
(B) CONVECTION
•
•
Example I .S ~ A flat plate of dimensions I00 x 50 mm and negligible thickness, maintained at 46°C, loses 1.2 W
from both sides by free convection to ambient air at 30°C. Determine the heat transfer coefficient.
Known: Both sides of a flat plate exposed to convective conditions.
Find: Heat transfer coefficient, h.
Schematic:
Still air
r~= Jo c
0
::·r--l
_---;../_ _ As= 2 LW
/
.
=2xO.I mxO.OSm
=0.01 m2
--j
L= 100mm
Fig. 1.9
Assumptions: (1) Steady-state conditions. (2) Plate thickness is negligible. (3) Radiation effects not considered.
Analysis: By Newton's law of cooling,
where Q is the total heat transferred from the two exposed surfaces of the plate = 1.2 W
Ts is the surface temperature = 46°C
T~ is the ambient air temperature= 30°C
and
As is the total surface area losing heat= 0.0 l m 2
Substituting values, the natural convection heat transfer coefficient is
Q
1.2W
h -~~-=-------A. (T.-T~ ) O.Olm2 x(46-30)°C
= 7.5 W/m 2 °C
(Ans)
Fundamental Concepts of Heat Transfer
Example 1.6
1.1 I
~ The sun heats the top of a car to about 70°C on a calm summer day when the air
temperature is 26°C. If the average heat transfer coefficient at 48 km/h is 85 W/m 2 °C, what
is the initial cooling rate per unit area on the roof!
Known: Heat is lost to ambient air from a heated surface under forced convection conditions.
Find: Initial coo ling rate per unit area, Q I A(W !m 2 ).
Schematic:
Top roof of a car
T5 =70°C
Air
T==26°C
V=48kmh
h = 85 W/m2°C - - - - -
Fig. 1.10
Assumptions: (1) Constant and uniform heat transfer coefficient. (2) Convection is the only mode of
heat transfer.
Analysis: Initial cooling rate per unit area by forced convection is determined from
Q
A=h(A,-T=)
= (85 W /m 2 C) (70 - 26)°C
0
=3740W/m 2
Example 1.7
(Ans)
~ The surface of a small ceramic kiln is at 65°C when the kiln is in operation. The kiln is of
cubical shape and is supported in air at 25°C with negligibly small legs, with a heat transfer
coefficient of 15 W/m 2°C (assumed to be an average for all surfaces). If 900 watts are required to keep the
kiln in steady-state operation, determine the size of the cubical kiln. What type of heat transfer phenomenon
is involved!
Known: A cubical kiln at a specified temperature is exposed to air. Heat rate and heat transfer coefficient
are prescribed.
Find: Size of cubical kiln.
Schematic:
Air
T5 = 65°C
a
T =25°C
15W/m2°C
h=
a
Q=900W-
-Q
-----------------------------------'
·····-.......
a
·--._
··-...
Fig. I.I I
1.12 , Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Uniform heat transfer coefficient and surface
temperature. (3) Conduction through small legs is negligible.
Analysis: In steady-state, the heat supplied to the cubical small kiln must equal the rate of heat dissipation
from the kiln to the surrounding air. Clearly, convection is the dominant mode of heat transfer in this
case because of the surface-air temperature difference.
(Ans)
For convective heat transfer, we have
It follows that
A.=
900W
Q
h(T,-T.J
(15W /m2 0 C)(65-25)°C
= 1.5m2
The total surface area of the cube of each side a is
A.= 6a2 = l.5m2
Therefore, the size of the cube is
S1"de a=
Example 1.8
V~
6 m- = 0.5m or 50cm
(Ans)
~ A vertical heated plate of 1-m height dissipates heat by natural convection to the ambient still
air at 35°C. The plate surface temperature and the heat flux are known to very according
to the following expressions:
T,(x) = 50x+60
and
q(x) = 100x2 +250
where T, is the temperature of the surface in °C, and q is the heat flux in W/m 2• The distance x (in m) is
measured from the bottom edge of the plate. Determine the local heat transfer coefficient hx at x = 0 (bottom),
0.5 (midway) and I (top) metre.
Known: Surface temperature and heat flux from a heated vertical plate to the surrounding air are functions
of height x from the bottom edge at x = 0, 0.5 and 1 m.
Find: hx at x = 0, 0.5 and 1 m.
Schematic:
Vertical flat plate
x=L= Im
Still air
r.. =35°c
x=0.5m
Q
-JV\/'+- q= -
A
x=Om
Fig. 1.12
=IOOx2+25°
Fundamental Concepts of Heat Transfer
1.13
Analysis: Local heat transfer coefficient,
h
=
x
q(x)(W/m2) (W/m2 oq
[T.(x)-T=](oC)
t At x = 0 : Ts = 50 (0) + 60 = 60°C
f:.T= Ts= 50(0) + 60 = 60°C
q = 100(0) 2 + 250 = 250 W/m 2
h
_ 250W!m 2 _
20
-10 WIm C
250 C
(Ans)
x(bottomJ -
t At x = 0.5 m : Ts= 50 (0 .5) + 60 = 85°C
q = 100(0.5 2) + 250 = 275 W/m 2
t:.T = 85 - 35 = 50°C
q
hx (middle) = f:.T =
275W!m 2
500 C
5.5 W/m 2 °C
(Ans)
t At x = 1 m : Ts = 50 (I) + 60 = 110°C
q = 100(12)+250 = 350 W/m 2
t:.T = 110-35 = 75°C
Hence,
_ 350 W /m2 _
20
-4.67 W/m C
750 C
(Ans)
hx (top) -
Example I. 9
~ The temperature profile of a flat plate exposed to water at 60°C is given by T(y) = 50 + 640 y +
0.1 y2 where T is in °C. Calculate the heat-transfer coefficient if the thermal conductivity of
the water is 0.65W/m 0 C.
Known: Temperature profile for a hot plate exposed to convective environment.
Find: Heat-transfer coefficient, h.
Schematic:
y
Water
Temperature distribution,
T(y) = 50+ 640y+ O. ly2
Plate
Fig. 1.13
1.14 ,
Heat and Mass Transfer
Assumptions: (1) Constant properties. (2) Isothermal plate.
Analysis: The heat flux at the surface of the plate is
q=h(T,,-T-)
=-k1(!:l
where
k1 . . . thermal conductivity of the fluid
T. . . . temperature of surface
T- ... temperature of fluid far removed from surface.
(ar /a y). ... fluid temperature gradient, measured at the surface in the direction y normal to the surface.
Hence, the heat transfer coefficient is given by
Surface temperature, Ts= T (y = 0) = 50°C
= [O+ 640-0.2 Y]y=O
= 640°C/m
It follows that
h = -(0.65 W/moC) 640oC/m
(50-60)°C
= 41.6 W/m2 °C
(Ans)
(C) RADIATION
•
•
Example I. IO ~ A flat plate is irradiated on the top surface at the rate of 1892 W/m 2 and on the bottom
surface at the rate of 1261 W/m 2• At the time of observation, the plate is at such temperature
that the top and bottom surface each emits 3 ISW/m 2• The absorptivity and reflectivity of the plate are 0.3 and
0.4 respectively. Determine the radiosity at the top and bottom surfaces. Is the plate temperature increasing or
decreasing?
Known: A flat plate of prescribed absorptivity and reflectivity receives radiant energy on both top and
bottom surfaces while emitting energy from the two surfaces.
Find: Radiosities, J 1 and J2 (W/m2).
Fundamental Concepts of Heat Transfer
I.IS
Schematic:
E1 =E2
. .i .... L
,\
····· aGj
a.Gi_
Top surface (I)
.........
pG2
Bottom surface (2)
tG1
"-~--~y
h
Fig. 1.14
Assumptions: (1) Steady operating conditions exist. (2) Emitted energy from both top and bottom is
same. (3) Plate is isothermal.
Analysis: The quantity G is irradiation which is the sum of all radiation striking the surface from all
possible sources. A part of it is reflected, a portion of it is absorbed and the balance is transmitted. Thus,
G = G,.r+ Gabs+ Gt,an = pG + cxG+ 1:G
or
cx+p+1:=l
Let G 1 be the irradiation striking surface ( l) and irradiation G2 striking surface (2). At a specific
instant, the plate is at an absolute temperature T such that the emissive power, £ 1 = £ 2 .
The three components of energy leaving the surfaces (l) and (2) are reflected component, transmitted
component and emitted component. The sum total of these components is defined as radiosity, J. Thus,
J 1 =pG 1 +1:G2 +E 1
and
J2=PG2+1:G, +E2
Transmissivity, 't of the plate surface is
1: = l - {p +ex)= l - (0.4 + 0.3) = 0.3
Noting that £ 1 = £ 2 = 315 W/m 2, and substituting appropriate numerical values, one gets
J, = (0.4) (1892) + (0.3) (1261) + 315
= 1450 W/m 2
J2 = (0.4)(1261) + (0 .3)(1892) + 315
= 1387W/m2
Now we can observe that
G 1 + G 2 = 1892 + 1261 = 3153 W/m 2
and
J, + J 2 = 1450 + 1387 = 2837 W/m 2
(Ans)
(Ans)
I .16 ,
Heat and Mass Transfer
Since incoming energy ( G 1 + G2) > outgoing energy (J1 + J 2 ), the plate temperature T is increasing
with respect to time.
(Ans)
Example 1.11
~ Two very large parallel planes separated by vacuum having surface conditions idealised as
a blackbody are maintained at I 200°C and 550°C. Calculate the net radiation heat transfer
per unit area between the two planes.
Known: Radiant heat exchange between two large parallel black surfaces.
Find: Heat-transfer per unit area.
Schematic:
T2=550°C
or 823.15 K
r,=1200°c
or 1473.ISK
)
__,,...__ _ _ ___,____ QIA
Fig. I.IS
Assumptions: (1) Vacuum is transparent to radiation. (2) Planes have black surfaces.
Analysis: Heat-transfer rate per unit area by radiation is
Q = (j ( 7;4 _ ½4 )
A
=5 .67 x I0-8 W/m 2 K 4 [1473.15 4 -823.15 4 ]K 4
= 241 x 103 W/m 2
or
241 kW/m 2
(Ans)
Example 1.12 ~ The outside surface of a spacecraft in space has an emissivity of 0.8 and an absorptivity of
0.25 for solar radiation. If solar radiation is incident on the spacecraft at a rate of I 200W/m 2,
calculate the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Known: A spacecraft of given emissivity and solar absorptivity is exposed to incident solar radiation.
Find: Surface temperature, Ts.
Schematic:
\ \ i s = 1200W/m2
qabsorbed
Spacecraft
asolar = 0.25
c=0.8
- - - qemitted
Fig. 1.16
Fundamental Concepts of Heat Transfer
I .17
Assumptions: ( l) Diffuse-gray surface.
Analysis: Radiation emitted = Solar energy absorbed
qemitted = qabsorbed
i.e.,
or
f,(J
T/ = (Jsolar Gsolar
Surface temperature of the spacecraft is
T =[O"solar Gsolar]l / 4 =[
ea
s
= 285.18 K
or
0.25xl200W/m2
]0.25
0.8x5.67xl0-8 W/m 2 K 4
12°c
(Ans)
(D) ENERGY BALANCE
•
•
Example 1.13 ~ A 4-m diameter thin-shelled spherical tank containing liquid nitrogen at I atm and -I 96°C
is exposed to large surroundings at a temperature of 20°C. The surface emissivity of the
tank is 0.1. At I atm, the density and latent heat of vaporization of liquid nitrogen are 810 kg/m 3 and 198 kJ/kg
respectively. Neglecting convection, determine the rate of evaporation of the liquid nitrogen in the tank in litres
per hour as a result of net radiative heat exchange.
Known: A liquid nitrogen spherical container is placed in a large enclosure.
Find: Rate of evaporation of liquid nitrogen.
Schematic:
Tsur= 200c
e 5 =0.I
D=4m
Liquid nitrogen
(p=810kg/m 3, h1g= 198kJ/kg)
Fig. 1.17
Assumptions: ( l) Steady-state conditions prevail. (2) Being thin-shelled, the outside surface temperature
of the spherical tank is almost the same as the temperature of the liquid nitrogen inside. (3) Convection
heat transfer is neglected.
Analysis: Energy balance
I .18 , Heat and Mass Transfer
i.e.,
(Net heat transfer by radiation from the surroundings)= (Heat lost by evaporation)
to the liquid nitrogen vessel
of liquid nitrogen
Net radiative heat transfer to liquid nitrogen is
Qrad = A. es <1 [I'.~r -T.4]
= (1tD2)(es)a[T.~r -T.4]
= (1tx 42) m2 xO. lx(5.67x 10-8 W / m 2 K 4) x [(20+ 273.15) 4 -(-196+273 .15) 4] K4
=2094.7 W
Hence,
Rate of boil-off of liquid nitrogen is
m= Q,ad =
hfg
2094.71 / s (3600s)=38k /h
198xl03 J / kg
lh
g
m=p¥,
Since
Rate of evaporation,
¥
=; =(3s; )( s:~~g)( \o;;)
= 47 litres per hour
(Ans)
Example 1.14 ~ Water at a temperature of 70.6°C is to be evaporated slowly in a vessel. The water is in
a low-pressure container which is surrounded by steam. The steam is condensing at 103°C.
The overall heat transfer coefficient between the water and the steam is 1200 W/m 2 0 C. Determine the surface
area of the container which would be required to evaporate water at a rate of 0.01 kg/s.
Known: Water is evaporated at a steady rate in a steam jacketed vessel under prescribed conditions.
Find: Surface area of container.
Schematic:
r:=:=====l
I
Vapour (0.01 kg/s)
-
Condensate out
(103°C)
-
Saturated steam in
(103°C)
-_-_-_-_-_-_-_-_-_-·----=-=-=-=-=-=-=-= Steam-jacketed vessel
--=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Overall heat transfer coefficient
:=:=:-Ez-0:6-":CJ=:=:===:=:=:=:=:=== u = 1200 W/m2 °c
- -
-
- - -
-
-- - -
- -
- -- - -
-
-
Fig. 1.18
Assumptions: (I) Pressure in the vessel is maintained constant at the saturation pressure corresponding
to 70.6°C by suitable evacuation means. (2) Steady-state conditions exist.
Fundamental Concepts of Heat Transfer
I .19
Analysis: Energy balance under steady operating conditions
Heat transferred from steam to water = Heating rate required to evaporate water
Rate of evaporation of water at saturation temperature of 70.6°C (Saturation pressure, P = 0.32 bar) =
0.01 kg/s.
Latent heat of vaporisation of water at 70.6°C (or 0.32 bar) is
hfg
= 2332.4 kJ/kg
(from steam tables)
Heat rate,
.
103 JI
Q=mh
1g = (O.Olkg/s)(2332.4kJ/kg) 1lkJ
= 23 324 J/s
or W
Steam temperature, Tsteam = 103°C
Water temperature, Twater = 70.6°C
Overall heat transfer coefficient, U = 1200 W/m 2 °C
Hence, the surface area of the vessel required is
A=
Q
~ U ( 'f.team - Twater)
23 324W
- - - - - - - - - - - - = 0.60m2
(1200 W /m 2 0 C) (103- 70.6)°C
(Ans)
Example 1.1 S ~ A 500 W iron is left on the iron board with its base exposed to the ambient air at 20°C.
The convection heat transfer coefficient between the base surface and the air is 31 W/m 2 K.
The emissivity of the base is 0.7 and its surface area is 0.020 m2 . Calculate (a) the temperature of the base of
the iron, and (b) the radiation heat transfer coefficient.
Known: An iron of prescribed rating dissipates heat by both convection and radiation to the surrounding air.
Find: Iron base temperature, T(°C).
Schematic:
Q=SOOW
Base
T5 (°C),e 5 =0.7
A=0.02m2
Fig. l.19(a)
Air
=
-T==20°C
h=JI W/m2°C
1.20 ,
Heat and Mass Transfer
1/hA
~;,OT-~Q=SOOW
I/h,.A
Thermal circuit
Fig. I.I 9(b)
Assumptions: ( l) Steady operating conditions. (2) Diffuse-gray base surface. (3) The iron can be treated
as a small body in a large enclosure.
Analysis: Power supplied to the iron, £, = Rate of heat transfer by convection and radiation = Q = 500 W
Q = Qconv + Qrad
Heat rate,
= h A,(T, -T=)+ A, E, a(T,4 - I'.~,)
= A.[h(T,-T=)+e, a(T,4 -To!)]
Substituting numerical values and taking care that all temperatures are expressed in absolute units,
we have
500 W = [(0.02 m2 )(3 l W /m 2 K)(T, -293.15 K)]
+[(0.02 m 2 )(0.7)(5.67 x 10-8 W /m 2 K4 ) x {(T,,K) 4 -(293.15 K) 4 }]
A trial and error solution is necessary.
T5 (°C)
450
460
462
RHS (W)
= Qconv + Qrad
266.6 + 211.2 = 477 .8
272.8 + 223.5 = 496 .3
274 + 226 = 500
LHS (W)
Difference
(LHS - RHS) (W)
500
22.2
500
3.7
500
0
Therefore, base surface temperature, T, = 462 °C
Combined convection plus radiation heat transfer coefficient can be evaluated from
(Ans) (a)
Q= hcombinedA,(T,-T=)
Q
500W
hcombined = A,(T,-T=) = 0.02m 2 x(462-20)°CorK
= 56.56 W/m 2 K = h + h,
Radiation heat transfer coefficient,
h,= hcombined -h = 56.56 - 31 = 25.56 W/m 2 K
l
l
Comment: The convection and radiation resistances are - - and -h- respectively.
h A,
r A,
(Ans) (b)
Fundamental Concepts of Heat Transfer
1.21
They are connected in parallel as shown in the thermal circuit. The heat rate,
Q=Ll ¾verall / R,otal
Rtotal - [-}
Rconv
+ _1
i-1 --[hAs+hrAs]-I -- - - - 1
(h+h,)As
R,ad
hcombined As
so that
Example 1.16 ~ A metal plate is placed on a driveway and receives 950 W/m 2 of incident radiant energy from
the sun. The plate absorbs 80% of the incident solar energy and has an emissivity of 0.05.
Consider the lower surface of the plate to be thermally insulated from the driveway. If the air temperature is
20°C and the natural convection heat transfer coefficient between the plate's surface and the surrounding air
is 10 W/m 2 K, estimate the temperature of the plate.
Known: A metal plate with lower surface
Find: Plate temperature, T5 ( 0 C). Insulated, and subjected to solar flux, is exposed to convection and
radiation.
Schematic:
Tsur= 200c
95
qs= ~0W/ml
~Metalplate(T5 )
~ - - - ........----+~---~, £=0.05
a=0.8
/
h= IOW/m2K
(qconv +qrad)
,
T= =20°C
i==:====.====':/
L
1nsulated
t t t
Fig. 1.20
Assumptions: (1) Steady state conditions. (2) The bottom surface of the plate is insulated. (3) The
surroundings temperature equals the ambient air temperature.
Analysis: From energy balance.
.
.
E;n - Eout +
.
0
Ege
(noh t
eration)
. 0
=~
(ste
ystate)
Hence
or
a qs = %onv + qrad
Expressing all temperatures in K, one can write
(0.8)(950 W /m 2 ) = (10 W/m 2 K)[(I'. - 293 .15)K]
+ (5.67 x I0-8 W/m 2 K 4 )(0.05)(I'.4 -293.15K 4 )
1.22 ,
Heat and Mass Transfer
Ts can be obtained by trial and error.
760 = lO(T, _ 293.15). 0.2835 [
c~J- l
73.85
Ts (K)
LHS
RHS
LHS- RHS
350
760
568 .5 + 2 1.6 = 590. 1
169 .9
370
760
768 .5 + 74.1 = 842.6
- 82.6
366.15
760
730 +30 = 760
0
By linear interpolation,
T.s -350 = 0-169.9 =0.673
370-350 -82.6-169.9
plate surface temperature is estimated to be
Ts= 350 + 0.673(370 - 350)
= 363.46 K
Check:
RHS = 703+ 28.5 = 31.5 W/m 2
Since this is still less than 760 W/m 2, let
1'.s = 366.15K i.e., 93°C
RHS = 730+ 30 = 760 W/m 2 = LHS
Hence,
(Ans)
T, =93°C
Example 1.17 ~ A closed container (Figure 1.21 a) filled with hot coffee is placed in a room whose air and walls
are at a fixed temperature. Identify and list down all the heat transfer processes that contribute
to the cooling of coffee. Also comment on the features that would contribute to a superior container design.
Hot
coffee
Cover
Air space
Plastic flask
Fig. l.21(a)
Fundamental Concepts of Heat Transfer
1.23
Known: Hot coffee gets cooled by heat-transfer through a plastic flask, an air space and a cover to the
colder surroundings.
Find: All appropriate heat-transfer processes.
Schematic:
Air space
Surroundings
/Q7
--+-+--Q6
~
Plastic flask
-
Qa
Cover
Fig. 1.21 (b)
Analysis: The various heat-transfer processes that can be easily identified (and indicated in the schematic)
are listed below:
Q1 • • • From the coffee (fluid) to the flask (surface) ~ Natural convection
Q2 . . . Through the stationary medium (flask) ~ Conduction
Q3 . . . From the flask (surface) to the still air (fluid) ~ Natural convection
Q4
Net heat exchange between the outer surface of the flask and the inner surface of the cover
Q5
~ Radiation
From the quiescent air (fluid) to the cover (surface) ~ Natural convection
{1 .... Through a stationary medium (cover) ~ Conduction
Q7 .... From the cover (surface) to the ambient air (fluid) ~ Natural convection
Q8 ... Net heat exchange between the external surface of the cover and the surroundings ~
Radiation
To ensure superior container design, one can evacuate the air space to eliminate natural convection
and use low emissivity material (like aluminised surface) for the flask and cover to curtail radiation heat
exchange. These measures can go a long way to reduce the heat transfer from the hot coffee to cooler
surroundings so that the hot coffee remains 'hot' for a longer period.
Example 1.18 ~ A 25 cm thick wall has a steady-state temperature distribution given by T(x) = 385 - 3600 x2
where T(x) is in °C and x is in metres. The thermal conductivity of the material of the wall
is l.2W/m K. The ambient air temperature is 10°C. Determine (a) the location and magnitude of the maximum
temperature in the wall, (b) the heat flux at the left and right faces of the wall, (c) the surface heat-transfer
coefficient, (d) the rate of volumetric heat generation, and (e) the average wall temperature.
Known: Temperature distribution in a plane wall exposed to ambient air.
Find: (a) x and Tmax; (b) q(O) and q(L); (c) h; (d) q; (e) Tav·
1.24 , Heat and Mass Transfer
Schematic:
!
Wall (k= l.2W/m K)
Air
T(x) = 385-3600x2
iii
h, r .. = 10°c
Fig. 1.22
Assumptions: (1) Steady-state conditions. (2) One-dimensional conduction. (3) Uniform heat-transfer
coefficient.
Analysis: (a) Maximum temperature will occur at a location x where dT = 0.
dx
The steady-state temperature distribution is expressed as
T = 385 - 3600 x 2 (°C)
Differentiating with respect to x, we obtain the temperature gradient.
n = 0-(3600)(2)x = -7200x (~)
-
-
dx
m
Since dT = - 7200 x, it will be zero at x = 0, i.e., at the left face and its magnitude will be
dx
Tmax = T( x = 0) = 385 - (3600) (0) = 385°C
(Ans) (a)
(b) At the left face (x = 0):
Heat flux,
dTI
q(O) =-kdx x=O
= (-1.2 W /mK)(-7200 X 0) = 0
(Ans) (b)
At the right face (x + L)
Heat flux,
q (L) = q (0.25 m)
= {-1.2 W /mK) {-7200°C/mx0.25m)
(Ans) (b)
=2160W/m 2
(c) At the outer surface of the wall (x = L = 0.25 m), we have from the surface energy balance:
qcond
= qconv
Fundamental Concepts of Heat Transfer
r
1.25
i.e.,
With Tw = T(x = L) = 385-3600 (0.25) 2 = 160°C, we have
-(l.2W/mK)(-7200x0.25) °C =h(W/m2 K)[160-10]°C or K
m
Surface heat-transfer coefficient,
h = 1.2 X 7200 X 0.25
150
=14.4W/m2 K
(Ans) (c)
(d) Applying the energy balance:
0
0
hee.ring
st.&.tate
FL -£out +£gen= E/
the wall
Therefore,
Rate of volumetric heat generation,
_ hA(Tw-T=) h(Tw-T=)
q=
AL
=
L
q(L)
L
2160W/m2
0.25m
(Ans) (d)
=8640W/m 3
(e) Average temperature of the wall is
l L
T,,v = - JT(x) dx
Lo
l L
= - J(385 - 3600 x 2 )dx
Lo
3JL = 385 L-1200.V
= _!_[ 385 x_ 3600 x
L
3
= 385 -
0
L
1200 L2
= 385-(1200 X 0.252 )= 310°C
(Ans) (e)
1.26 ,
Heat and Mass Transfer
Example 1.19 ~ A cylindrical rod of IO cm diameter (k = 42 W/m K) has a steady-state radial temperature
profile given by
T(r) = 275-1.8 x I04 r2
where T is in °C and r in metres. The ambient fluid temperature is 30°C. Determine
(a) Maximum temperature in the rod
(b) Rate of volumetric heat generation
(c) Convection coefficient at the rod's outer surface
(d) Average temperature of the rod
Known: Specified temperature distribution in a solid cylinder.
Find: (a) Tmax (0 C); (b) q; (c) h; (d) Tav·
Schematic:
k=42W/mK
--------------·····
t
T(r)=275-l.8x I04r2
-vW'- f. gen
Fig. 1.23
Assumptions: (I) Steady-state conditions. (2) Uniform volumetric heat generation.
Analysis: (a) The temperature distribution is
T(r) = 275- 1.8 x 10 4 r2
Maximum temperature will occur where dT = O
dr
dT
-
dr
= 0 = 0 -(1 .8 XI 04 )(2) r
::::} Ir=
O! i.e., at the centreline
The maximum temperature is
Tmax = T(r = 0) = 275°C
Control volume energy balance gives
0
/! - +
Eout
0
£,gen
Eout = £,gen =
i.e.,
%
=
q¥
.
dTI
Q. = Eout
= -k(2rcRL)dr r=R
(Ans) (a)
Fundamental Concepts of Heat Transfer
r
1.27
Tw = 1i.r=R) = 275 -1.8 X104 R2
= 275-1.8 X10 4 X(0.05) 2
= 230°c
-dTI
=-3.6x10 4 R
dr r=R
= -3.6x10 4 x0.05 = -1800°C/m
Qconv = £out lr=R = h(2 TC R L )(Tw - T-)
.I
=E
dTI
= -k (2rcRL)-
dr r=R
m r=R
Q = -(42W /mK)(2rcx0.05mxlm)(-1800°C/m)
=23 750 W
Heat generation rate per unit volume,
=
23750 W
=3.024xl0 6 W/m 3
rcx0.052 m2 xlm
(Ans) (b)
Convection coefficient,
=
-J?SW/m2 K
23750W
(2 rc x 0.05 x l)m2 x (230-30)K
Average temperature in the cylindrical rod is
2 R
I'av = -J
T(r) r dr
R2
0
2 R
= -J(275-1.8x 104 r 2 ) r dr
R2
0
= ; 2 [( 275x
~ )-(l.8xlo {r;
= ~[137.5R2 -4500R4 ]
R2
4
)I
(Ans) (c)
1.28 ,
Heat and Mass Transfer
= 275 - 9000R 2
= 275-9000 X (0.05) 2 = 252.5 °C
(Ans) (d)
Example 1.20 ~ The steady-state radial temperature variation in a IO cm diameter solid sphere (k = 15 W/m 0 C)
known to be T(r) = 125 - 1.4 x I 0 3 r2 where T is in °C and r in metres. The sphere is exposed
to a convective environment with an ambient at 34°C. Calculate (a) the maximum temperature in the sphere,
(b) the convection heat transfer coefficient at the outer surface, (c) the rate of volumetric internal thermal
energy generation, if any, and (d) the average sphere temperature.
Known: A solid sphere with prescribed temperature profile is placed in a convective atmosphere.
Find: (a) Tmax; (b) h ; (c) q; (d) Tav
Schematic:
Sphere
_ _ _ (k= 15W/m C)
0
T(r)=235-l.4x 10,2
~ qin
Fig. 1.24
Assumptions: (1) Steady-state conditions. (2) Constant thermal conductivity. (3) Uniform convection
coefficient and volumetric heat generation.
Analysis: (a) We note that
T(r) = 125-1.4 x 103 r2
Maximum temperature occurs at a location where
dT/dr= 0
Differentiating with respect to r,
dT
-=0=0-2xl.4xl0 3 r
dr
Hence,
Tmax = T(r = 0) = 125 °C
(b) At the sphere's exposed outer surface (r = R)
qin,cond
= qout,conv
(Ans) (a)
Fundamental Concepts o( Heat Transfer
r
1.29
- kdTI
= h(Tw -T_)
dr r=R
- kdTI
=-2xl.4x10 3R
dr r=R
where
= -2xl.4x10 3 x0.05
=-140 °am
Also
T(r= R) = Tw
= 125 -1.4 X 103R2
= 125-(l.4x103 x0.052 ) = 12l.5°C
It follows that
q = -(15W Im 0 C)(-140°C/m) = h(W /m2 0 C) x(121.5-34)°C
Convective heat-transfer coefficient
h=
15x140
(121.5-34)
(Ans) (b)
= 24.0W /m2 °C
(c) Heat-transfer rate= Rate of thermal energy generation
Rate of volumetric internal heat generation
_ Q
q(41tR2 )
q = ¥- = (4/3)1tR3
3q 3x15x140
=R =
0.05
= 1.26 x 105 W/m3
(Ans) (c)
(d) Average sphere temperature,
3 R
2 dr
T..v = -JT(r)r
R3
0
3 R
=
-f
(125-1.4 x 103 r r dr
R3
2)
2
0
= l._[125 R3 -280R5]
R3
3
= 125-(3 x280 x0.05 2 )
= 122.9 °C
(Ans) (d)
1.30 ,
Heat and Mass Transfer
Example 1.21
~ Consider a cubical cavity with 0.25 m side and I-cm-thick wall of thermal conductivity
0.06 W/m 0 C. Ice at the saturation temperature 0°C with density 920 kg/m 3 and latent heat
of fusion of ice 335 kJ/kg is enclosed in the container. The outer surface temperature of the wall is 25°C. Find
the time required to melt the ice completely.
Known: Melting of ice contained in a cavity.
Find: Time required to melt the ice, tm
Schematic:
Cubical cavity
Side
a=25m
Ice
A
Tr=o c
0
Fig. I .2S(a)
I-
a=0.25m
Containing wall
k=0.06W/m°C
L=lcm
Section A-A
Fig. l.2S(b)
Assumptions: ( 1) Constant properties. (2) The inner surface temperature of the wall is at the melting
temperature of the ice. (3) One-dimensional conduction.
Analysis: Control volume energy balance gives
or
Fundamental Concepts of Heat Transfer
I .3 I
Mass of ice,
m = p-¥- = p[a-2L]3 = (920 kg / m 3 )(0.25-0.02)3m 3
= 11.2kg
With area, A= 6a 2,
0
2 (25-0)°C
Ein -(0.06 W / m C) 6 (0.25m) ---Xtm(s)-56.25tm
O.Olm
(J)
11 Es, = m \ r = (11.2 kg) (335 X 10 3 J/kg)
= 3.75 X 106
(J)
It follows that
56.25 tm = 3.75 X 106
Time needed to melt the ice is
1~1
t = 3.75 X 106 S
= 18.5 h
m
56.25
3600 S
(Ans)
Example 1.22 ~ (a) Consider a long, thin copper wire of diameter D and length L which has an electrical
resistivity Pe· The wire (density p and specific heat Cp) is initially in steady state in a room
The convection heat transfer coefficient is h and the
with the surroundings (Tsur) and the ambient air (T
surface emissivity is £. The wire temperature is uniform at any instant of time. After an electric current / is
passed through the wire, the wire temperature increases. Develop an expression for the rate of temperature
change with time, dT!dt, during passage of the current.
(b) What will be the maximum operating current through the conductor if the permissible safe temperature
is 65°C? The 1.5 mm diameter wire of electrical resistivity 70 µil-cm and emissivity 0.85 is exposed to the air
and the surroundings at 25°C with a convection coefficient of 27.5 W/m 2 0 C.
~>·
Known: Temperature of wire increases with time due to passage of an electric current.
Find: Expression for dT/dt.
Schematic:
Air
Copper wire
[
_,
T(t)• £, p, CP
()
L
Fig. 1.26
Assumptions: (1) Uniform wire surface temperature at any time t. (2) Constant properties (p, CP and£).
(3) Uniform internal heat generation due to Ohmic heating. (4) Radiant heat exchange between the wire
surface and the surroundings is between a small body and a large enclosure.
1.32 ,
Heat and Mass Transfer
Analysis: Applying the control volume energy balance on the rate basis:
0
ft- £out + £gen = Est
where £gen = rate of heat generation = 12 Re = 12 Pe
L
(rc/4)D2
Est = rate of thermal energy storage
re 2 LC )
d
d
=-(mC
T) =-(p-¥-C
T) = -dT( p-D
dt
P
dt
P
dt
4
P
£out= rate of heat dissipation from the wire surface by convection and radiation
= Q.,onv + Qrad = h(rcD L)(T-T-)+ oe(rcD L)(T 4
-T.!r)
Substituting for £out, £gen and Est into the energy balance, one gets
12
L
Pe (re/ 4)D2
= C L(rcD2 ) dT
p
4
P
dt
Hence, the variation of the wire temperature with time is given by
12 Pe (
dT
dt
=
dT
or
dt
~ )-rcDL[h(T-T-)+eo(T
4 -
T.!r)]
pCP L(rcD2 !4)
4 1 2 Pe/rcD2 - rcD [ h (T -T-)+ ea (T 4 - T.!r )]
=-----------------
pCP(rcD2 /4)
(b) Under thermal equilibrium conditions, dT = O. Then
dt
412 Pe= rc2 D 3 [h (T-T-)+ eo(T 4 -T.!r)
Current passing through the wire is then determined from
(Ans) (a)
Fundamental Concepts of Heat Transfer
1.33
Substituting numerical values,
/ -[
1t2 (1.5 X 10-3 m) 3
4x 70 x 10-6 x 10-2 ohm m
x {(27.5W/m2 0 C)(65-25)°C+ (0.85)(5.67 x 10-8 W/m2 K 4 )(338.15 4 -298.15 4 ) K 4
= 4.0 A
}f
(Ans) (b)
(E) COMBINED HEAT TRANSFER MECHANISMS
Example 1.23 ~ The following data were obtained from a single glass cover flat-plate solar heat collector:
Mean plate temperature = 70°C
Ambient and sky temperature = I0°C
Back insulation thickness = 5 cm
Insulation thermal conductivity= 0.05 W/m °C
Coefficient of heat transfer by convection from plate to cover= 3 W/m 2 °C
Equivalent coefficient of heat transfer by radiation from plate to cover= 6 W/m 2 °C
Coefficient of convective heat transfer from cover to ambient air= 25 W/m 2 °C
Equivalent coefficient for radiant heat transfer from cover to sky= 5 W/m 2 °C
Compute the total heat loss per m2 collector area. Assume that resistance to heat flow by convection at the
back surface of the insulation is negligible.
Known: Configuration of a flat-plate solar collector.
Find: Total heat loss per unit collector area.
Schematic:
h(c-a) = 2.5 }
h,(c-a) = 5.0
_
h(p-c)- 3.0
h,(p-c)=6.0
W/m 2 °C
}
W/m2 oC
~
r
Fluid passage
o
7sky -Ta- IO C
Ab(~~7~o~ t
Transparent glass cover
;h:~~~:,~:~~n
--- ------------------------------------------- -------
T
L=0.05m
~ - - - + - - - - - - - - - . - - - - - - ~ l_
Casing
Fluid tubes
Fig. I.27(a)
Rconv, 2
Rrad, I
~Q11
Fig. l.27(b)
Rrad, 2
1.34 ,
Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Constant thermal conductivity. (3) Uniform convection
and radiation heat transfer coefficients. (4) Negligible convective resistance at the back surface of
insulation.
Analysis: The thermal resistance network is shown in the schematic. Heat will flow from the absorber
plate to ambient air through insulation by conduction as well as from the plate to the top cover and the
air by convection and radiation.
On per unit area basis,
"fl
i'cond
= !:.. =
k
0.05 m = 1m2 oc;w
0.05 W Im °C
Convection and radiation resistances are in parallel. Hence, between plate and cover, the equivalent
resistance is
Similarly, between cover and ambient, the equivalent thermal resistance is
-[h.
R •q2 -
·,c-a)
+hr(c-a) ]
-1
1
1 m 2 °C
- - -W
- 25+5 - 30
Total thermal resistance,
Heat loss per m2,
.
TP -T;,
{70-10)°C
Qu =-R--= {13/90)m2 °C/W
total
= 415.4W
Heat loss per m2,
.
TP -T;, {70-10)°C
QI = - - = - - - = 6 0 W
Rcond
lm2 oc;w
Hence, the total heat loss per m2 solar collector area is
Q =QI+ QII = (415.4 + 60)W
=475.4 W
(Ans)
Fundamental Concepts of Heat Transfer
1.35
Example 1.24 ~ Electronic power devices are mounted on a heat sink having an exposed surface area of
450 cm 2 and an emissivity of 0.85. When the devices dissipate a total power of 20 W and
the air and surroundings are at 27°C, the average sink temperature is 45°C. What average temperature will
the heat sink reach when the devices dissipate 30 W for the same environment conditionr
Known: Power dissipation with average sink temperature, sink surface area and emissivity. Ambient air
and surroundings temperature.
Find: Sink temperature, T,,* for power dissipation of 30 W.
Schematic:
Tsur = T= = 27°C or 300.15 K
sa= 20W
,.........~_~_...._-_-~...........
Heat sink C:s=45~C or
.: : : : : : : : : : : : , : :~318.15 K with f,J- 20 W,
'
·
,A5 =450cm2, c=0.85]
./
:
:~.
Qrad
:
'
Qconv
· · c · · · ---·
Control volume
Fig. 1.28
Assumptions: ( l) Steady operating conditions. (2) Heat sink is at constant temperature. (3) Sink is small
compared to large surroundings. (4) All the power dissipated by the device is transferred to the sink.
Analysis: We identify the control volume around the heat sink as indicated by the dotted line in the schematic.
0
0
f}t. =JI~
Energy balance: Ein - Eout +
-
or
f.J=
Heat transferred
to sink from the device
Qconv
Ein
= Eout
+
-
..._,,.....,
Heat
lost
by convection
Q,ad
Radiation
heat exchange
or
Convection heat-transfer coefficient,
Substituting the numerical values, we have
[(20 W /0.045 m2 )-0.85 x 5.67 x 10-8 W/m 2 K4 (318.15 4 - 300.154 ) K4 ]
h=~--------------------------~
(45-27)°C or K
= 19.0W/m2 K
1.36 ,
Heat and Mass Transfer
When the power level is increased from 20 W to 30 W, the average sink temperature T8 is determined
from the same energy balance with h = 19 W/m2 K calculated above. It follows that
30 W = (19W/m2 K)(0.045m2 ){T,* -300.15) K
or
+(0.85)(0.045m2 ) (5.67 x 10-8 W/m2 K 4 )(T.*4 -300.15 4 )
30= 0.855 (T.* -300.15)+2.1688x10-9 (T.* 4 -300.15 4 )
or
Solving by trial and error, with r,• = 326.85 K (or 53.7°C)
LHS = 30
and
RHS "" 30
Hence, the average sink temperature is
T_* = 53.7°C
(Ans)
1
MULTIPLE CHOICE QUESTIONS
Mark the wrong statement in respect of conduction in a homogeneous medium:
(a) No heat flows along an isotherm.
(b) Heat transfer takes place between a region of high temperature and a region of low temperature.
(c) Heat flows along the direction of largest temperature range and hence normal to an isotherm.
(d) Conduction heat transfer can take place in a medium both at rest and in motion.
1.2 The rate of heat transfer per unit area per unit thickness of wall when a unit temperature difference
is maintained across the opposite faces of the wall is called
(a) thermal conductance
(b) thermal conductivity
(c) thermal resistance
(d) heat flux
1.3 Match the properties with their respective units:
1.1
Property
A. Thermal conductivity
B. Heat-flow rate
C. Heat-transfer coefficient
D. Heat flux
Codes:
(a)
(b)
(c)
(d)
A
p
s
u
u
B
R
Q
R
p
Units
(P) Wis
(Q) N/m2
(R) W
(S) W/m 2
(T)W/m2 K
(U)W/m K
C
T
D
u
T
T
T
u
s
s
Fundamental Concepts o( Heat Transfer
r
1.37
1.4 Natural convection differs from forced convection in that
(a) the two velocity profiles are different
(b) one does not obey Newton's law of cooling
(c) in forced convection, the fluid flow is caused by temperature-induced density changes
(d) in natural convection, the fluid flow can be caused by pressure differences perpendicular to the
gravitational force
1.5 Consider the following phenomena:
(1) Boiling
(2) Free convection in air
(3) Forced convection
(4) Conduction in air
Their correct sequence in the increasing order of heat transfer is:
(a) 4, 2, 3, 1
(b) 4, 1, 3, 2
(c) 4, 3, 2, 1
(d) 3, 4, 1, 2
1.6 Using thermal-electrical analogy in heat transfer, match List I (Electrical quantities) with List II
(Thermal quantities) and select the correct answer using the codes given below the lists:
List I
(Electrical quantities)
A. Voltage
B. Current
C. Resistance
D. Capacitance
Codes:
(a)
(b)
(c)
(d)
A
2
4
2
4
B
3
1
1
3
List II
(Thermal quantities)
1. Thermal resistance
2. Thermal capacity
3. Heat flow
4. Temperature
C
1
3
3
1
D
4
2
4
2
1.7 The control volume energy balance is expressed as £in - £out + £gen =Est. Which of the terms in
the above equation are volumetric phenomena?
(a) £in and £out
(b) £gen and Est
(c) £gen alone
(d) £in and Est
1.8 A wire of 3 mm diameter, 0.5 m long at 130°C is submerged in boiling water at 1 atm. The electric
power consumed is 6.80 kW. The boiling heat transfer coefficient is
(a) 48 100 W/m2 K
(b) 17 200 W/m2 K
(c) 43 500 W/m2 K
(d) 80 000 W/m2 K
1.9 A pipe carrying steam at 250°C has a surface emissivity of0.8 and the temperature of the surroundings
temperature is 25°C. The radiation heat transfer coefficient is
(b) 0.787 W/m2 K
(a) 177.22 W/m2 K
(b) 25.0 W/m 2 K
(d) 13.5 W/m2 K
1.10 A steel plate of thermal conductivity 50 W/m Kand 10 cm thickness passes a heat flux by conduction
of25 kW/m2 • If the temperature of the hot surface of the plate is 100°C, then what is the temperature
of the cooler side of the plate?
(a) 30°C
(b) 40°C
(c) 50°C
(d) 60°C
1.38 ,
Heat and Mass Transfer
1
TRUE/FALSE
1.1 In the radiator, radiation is the dominant mode of heat transfer.
1.2 In conduction and convection problems, we deal with linear temperature differences and any
consistent temperature scale (Celsius or Kelvin) may be used.
1.3 Heat transfer takes place in accordance with the third law of thermodynamics.
1.4 Surface energy balance is valid only under steady operating conditions.
1.5 The convection heat transfer coefficient, unlike thermal conductivity, is not a property of the fluid.
1
FILL IN THE BLANKS
1.1 The driving force for heat transfer is the - - - - - gradient.
1.2 A woman modelled as a cylinder, 0.2-m-diameter and 1.7-m-height with top and bottom surfaces
insulated, and the side surface at an average temperature of 35°C is exposed to an atmosphere
at 47°C. The surface heat transfer coefficient is about 20 Wlm 2 0 C. The rate of heat gain is
1.3 SI units of thermal conductivity and thermal diffusivity are respectively _ _ _ or _ __
and _ __
1.4 Flow of cigarette smoke in a still room is an example of _ _ _ convection while the mechanism
of heat flow in a car radiator is _ _ _ convection.
1.5 If Tm is the mean temperature of the surface and the surroundings in K, E the emissivity of the surface,
and, CJ the Stefan-Boltzmann constant, the radiation heat transfer coefficient, hr is approximately
1
EXERCISES
1.1 A glass window (k = 1.4 W Im 0 C), 1.2 m high and 2 m wide, is 7 mm thick with its inner and
outer surface temperatures maintained at 25 and 15°C. Determine the heat loss through the
window.
[4.8 kW]
1.2 A commercial heat flux meter uses thermocouple junctions to measure the temperature difference
across a thin layer of vermiculite (k = 0.068 W Im 0 C) which is 0.005 mm thick. The temperature
difference is observed to be 3°C. What is the heat flux?
[40.8 kW/m2]
1.3 A 100 W incandescent lamp has the glass bulb of 80 mm diameter and its filament is 0.5 mm
diameter and 50 mm long. Calculate (a) the heat flux on the surface of the glass bulb, (b) the heat
flux on the surface of the filament, and (c) the yearly electricity bill if the bulb is kept on for 8 hours
a day and the cost of electricity consumption is ~ 5 per unit.
[(a) 5000 W/m2 ; (b) 1.273 x 106 W/m2 ; (c)Rs 1460 per year]
1.4 A standing student can be modelled, for heat transfer purposes, as a 30 cm diameter and 168 cm
long vertical cylinder with both top and bottom surfaces insulated and the side surface at a mean
temperature of 35°C. Estimate the convective heat loss from the student's body to surrounding air
at 14°C if the heat transfer coefficient is 12 Wlm 2 °C.
[399 W]
Fundamental Concepts o( Heat Transfer
r
1.39
1.5 Hot gas at 100°C flowing over one side of a flat plate with dimensions 10 cm by 50 cm is estimated
to be 135 W when the plate surface is maintained at 25°C. Evaluate the convection heat transfer
coefficient between the plate and the gas.
[36 W/m2 0 C]
1.6 Pressurized water at uniform temperature of 60°C flows through a 40 mm diameter, 2 m long
pipe with the wall temperature maintained at 150°C. If the heat transfer coefficient between
the water and the pipe is 2100 W /m 2 K, compute the rate of heat transfer from the pipe to the
water.
[47.5 kW]
1.7 Estimate the radiation heat transfer from a 6 cm diameter spherical lamp with a surface temperature
of 90°C, when the surroundings are at 26°C. Assume blackbody behaviour.
[6.0 W]
1.8 A refrigerator stands in a kitchen where the air temperature is 22°C. The outside surface temperature
is 18°C, the sides are 30 cm thick and have an equivalent thermal conductivity of 0.10 W/m °C. The
outside convection coefficient is 10 W lm2 °C. Calculate the net heat flux and the surface temperature
on the inside.
[40 W/m2, 6 °C]
1.9 A horizontal small thin plate receives 950 Wlm 2 of incident radiant energy from the sun. The plate
absorbs 80% of the incident solar energy and has an emissivity of 0.05. The lower surface of the
plate is kept thermally insulated. The ambient air temperature is 20°C and the free convection heat
transfer coefficient is 10 Wlm 2 K. Estimate the equilibrium temperature of the plate.
[93 °C]
1.10 Two large parallel plates, 10 mm apart, are maintained at 100°C and 0°C under steady state
conditions. Assuming both surfaces to be black, calculate the rate of heat transfer per unit area if
the gap between the plates is
(a) Filled with atmospheric air (k = 0.0283 W Im 0 C)
(b) Evacuated
(c) Filled with mineral wool insulation (k = 0.05 WIm 0 C)
(d) Filled with superinsulation (k = 0.000017 WIm 0 C)
[(a) 1066.7 W, (b) 783.7 W, (c) 500 W, (d) 0.17 W]
1.11 The combined heat-transfer coefficient for convection and radiation is 170 W lm 2 °C for the outer
surface of a pipe in a large enclosure. The pipe surface is at 200°C and is black. Calculate the
radiation heat-transfer coefficient and convection heat transfer coefficient if the walls of the
enclosure are at 100°C.
[17 .4 and 152.6 W /m2 0 C]
1
ANSWER KEY
Multiple Choice Questions
1.1 (d)
1.2 (b)
1.3 (c)
1.4 (a)
1.7 (b)
1.8 (a)
1.9 (d)
1.10 (c)
1.2 T
1.3 F
1.4 F
1.5 (a)
True/False
1.1 F
1.5 T
Fill in the Blanks
1.1 temperature
1.2 346 W
1.4 Free, forced
1.5 4e<T~
1.3 Wlm°C, Wlm K, m 2ls
1.6 (d)
2
STEADY-STATE HEAT
CONDUCTION-ONE DIMENSION
Concept Review
INTRODUCTION
• 2.1
•
Heat conduction in a medium is said to be steady when the temperature does not vary with time and
unsteady or transient when it does. Heat conduction in a medium is said to be one-dimensional when
conduction is significant in one dimension only and negligible, in the other two dimensions. It is said to
be two-dimensional when conduction in the third dimension is negligible, and three-dimensional when
conduction in all dimensions is significant. In heat-transfer analysis, the conversion of electrical, chemical,
or nuclear energy into heat (or thermal) energy is characterised as heat generation.
GENERAL HEAT-CONDUCTION EQUATION
• 2.2
•
Heat flux is a vector quantity with both magnitude and direction. Thus, for a three-dimensional temperature
field, one can write t?e h~at fl~x vector (q) in terms of its x-, y-, and z-components, (qx, qy, q.J,
respectively, as q=qxi+qyj+q2 k. Partial differentials are used to describe the heat flux in the three
directions to obtain the Fourier's law as follows:
lq
X
= -k dT
dX ,
q
y
= -k dT
dy ,
q = -k dT I
z
(2.1)
dz
The general heat-conduction equation for a homogeneous solid in the three coordinate systems
(Figure 2.1) is given by
z
z
z
z
T (x, y, z)
y
~---+~~~y
T (r, <j>, z)
• T (r, <j>, 0)
~----;--~y
x~---~
X
X
(b) Cylindrical (polar)
(a) Cartesian
Fig. 2.1
(c) Spherical
(a) Cartesian (b) Cylindrical (polar) (c) Spherical spatial coordinates
Cartesian or Rectangular Coordinates
~(k dT) + ~(k dT) + _i_(k dT) + q = pC dT
dx
dx
dy
dy
dz
dz
P dt
(2.2)
2.2 , Heat and Mass Transfer
Cylindrical or Polar Coordinates
(2.3)
Spherical Coordinates
1ci ( kr 2 -ciT) + - 1- - ci ( kciT)
. ciT) +q=pC
_
- + -1
- - ( ksm0-ciT
r 2 cir
cir
r 2 sin 2 0 ci<I>
ci<I>
r 2 sin 0
ci0
p cit
(2.4)
Both steady and nonsteady (transient) problems can be solved with these equations to obtain a
temperature distribution. Using the temperature distribution (temperature field) and Fourier's law, one
can determine the heat flux at any location.
The one-dimensional heat-conduction equations in Cartesian, cylindrical, and spherical coordinate
systems for the case of constant thermal conductivities are expressed as
ci2T
cix2
q
1 ciT
k
a cit
-+-=--
(plane wall)
(2.5)
(long cylinder)
(2.6)
(sphere)
(2.7)
where the property a= k/pC P is the thermal diffusivity of the material.
INITIAL AND BOUNDARY CONDITIONS
• 2.3
•
The solution of a heat-conduction problem depends on the conditions at the surfaces, and the mathematical
expressions for the thermal conditions at the boundaries are called the boundary conditions. The solution
of transient heat-conduction problems also depends on the condition of the medium at the beginning of
the heat-conduction process. Such a condition, which is usually specified at time t = 0, is called the initial
condition, which is a mathematical expression for the temperature distribution of the medium initially.
The three boundary conditions that commonly occur in heat-transfer analysis are presented in Table 2.1.
Table 2.1
1
Boundary conditions for one-dimensional system
Constant wall (surface) temperature
T(O, t) = T,
Ts~
T(x,t)
(Continued)
Steady-State Heat Conduction-One Dimension
Table 2.1
2
2.3
(Continued)
Constant wall (surface) heat flux
(a) Specified heat flux
-koTI
. = qs
OX
T (x,t)
x=O
(b) Insulated or adiabatic surface
oTI
-o
OX x=O
T (x,t)
3
Convective surface condition
oTI
-k-
OX x=O
Fluid
= h[T= -T(O , t)]
h, T=
T(O,t)
ttt
• 2.4
T(x,t)
STEADY-STATE HEAT CONDUCTION-ONE DIMENSION
•
One-dimensional heat transfer through a simple or composite body exposed to convection from both sides
to mediums at temperatures T~ 1 and T~ 2 can be expressed as
.
T~i -T~2
Q = - - (W)
(2.8)
R,ota/
where R101• 1 is the total thermal resistance between the two mediums. For a plane wall (infinite slab) exposed
to convection on both sides, as shown in Figure 2.2, the total thermal resistance can be expressed as
1
L
1
+ Rcond + Rconv '2 = -h A + -kA + ,
h A
Rtotal = Rconv l
1
(2.9)
2
This relation can be extended to plane walls that consist of two or more layers by adding an additional
resistance for each additional layer. The elementary thermal resistance relations can be expressed as
follows:
+ Conduction resistance (plane wall):
+ Conduction resistance (cylinder):
L
R wall
= kA
ln(r2 /r1 )
R cyl
=--2nLk
(2 .10)
(2 .11)
2.4
,
Heat and Mass Transfer
L
h1T=1
lll
111
x=O
x=L
---------Q
L
kA
Fig. 2.2 A plane wall exposed to convection on both sides
(2.12)
+ Conduction resistance (sphere):
1
R conv =hA
-
+ Convection resistance:
+ Radiation resistance:
R
(2.13)
1
(2.14)
--h A
rad -
rad
+ Interface (contact) resistance:
1
Rinterface =
h A=
RC
A
(2.15)
C
where he is the thermal contact conductance, Re is the thermal contact resistance, and the radiation
heat-transfer coefficient is defined as
(2 .16)
Once the rate of heat transfer is available, the temperature drop across any layer can be determined from
(2.17)
The thermal resistance concept can also be used to solve steady heat transfer problems involving
parallel layers or combined series- parallel arrangements, as discussed in the next section.
COMPOSITE SYSTEMS
• 2.5
2.5.1
Multilayered Wall
•
Consider a series composite wall and the equivalent thermal resistance network shown in Figure 2.3.
2.5
Steady-State Heat Conduction-One Dimension
r=,1 - -......
Area,A
Cold fluid
h2, T=, 2
lll
Lx
-------------------Q
lll
I
I
hj"A
h2 A
Fig. 2.3
Composite wall and the equivalent thermal circuit
Heat-transfer rate,
.
Overal temperature difference, ~Toverall
Q=------------
(2 .18)
Total thermal resistance, R101• 1
T
Q=
= ,I
-T
= ,2
[{I/h1A) +(LA /kA A)+ (L 8 /k8 A) + (Lc /kcA) + (1/h2 A)]
(2 .19)
=UA(T=,l -T=,2 )
where U is the overall heat transfer coefficient (W/m 2 0 C) given by
(2.20)
For a series- parallel composite wall, the schematic and thermal circuit are shown in Figure 2.4.
Heat-transfer rate,
·
~Toverall
Q=---
(2.21)
Rtotal
(W)
(2.22)
2.6 , Heat and Mass Transfer
Area, A
@
.~
ko
ls/ks (A/2)
Le.Ike (A/2)
Fig. 2.4
2.5.2
Equivalent thermal circuit for a series-parallel composite wall
Composite Cylinder
Consider radial condition through a composite cylinder with convective boundaries as shown in Figure 2.5.
Neglecting interface (contact) resistances, the heat-transfer rate can be expressed as
(2.23)
Note that
(2.24)
.
1
ln(r2 I r1) ln(r3 I r2 ) ln(r4 I r3 )
Q = ---- = ------------ = ---h1 (21Cr1L)
2nkAL 2nk 8 L 2nkcL
h2 (2nr4 L)
2.5.3
Composite Sphere
Spherical composites can be treated in much the same way as composite cylinders.
We have, on similar lines, for a three-layered composite spherical shell with convection on both sides,
using appropriate forms of thermal resistances:
(2.25)
Steady-State Heat Conduction-One Dimension
2. 7
Heat-transfer rate,
·
LlToverall
Q=---1
(2.26)
Rtotal
Cold
fluid
Hot
fluid
111
T=,l• h1
T=,2• h2
lll
T=,I
Fig. 2.5
Composite cylinder and its equivalent thermal circuit
CRITICAL RADIUS OF INSULATION
• 2.6
•
Adding insulation to a cylindrical pipe or a spherical shell will increase the rate of heat transfer if the
outer radius of the insulation is less than the critical radius of insulation, defined as
kins
rcr,cylinder
=h
rcr,sphere
= -h-
(2.27)
2kins
(2.28)
R-Value The effectiveness of an insulation is often given in terms of its R-value, the thermal resistance
of the material per unit surface area, expressed as
L
R-value = k
(flat insulation)
where L is the thickness and k is the thermal conductivity of the material.
(2.29)
2.8 , Heat and Mass Transfer
VARIABLE THERMAL CONDUCTIVITY
• 2.7
•
+ For a Plane Wall If we need to consider thermal conductivity to vary with temperature and can
use a linear approximation
k(T) = k 0 [1 ± ~T], k0 and ~ being constants,
(2.30)
mean thermal conductivity,
km =k
or
0
[1±(~)]=k (l±~Tm)
0
(2.31)
IQ =-k(T)A~I
(2.32)
.
(Ti_ -T2)
AT
Q=kA--=m
L
Rwall
(2.33)
L
R wan-k A
(2.34)
where
m
+ For a Long Hollow Cylinder
8;J
whore
R
- _Zn_(,:~2/~1j~)
cyl -
(2.35)
2rckm L
+ For a Hollow Spherical Shell
G
where
Rsph
-(_!_ -_!_)
=- 1
4rckm
lj
1i
(2.36)
Solved Examples
(A*) DIFFERENTIAL EQUATIONS AND BOUNDARY CONDITIONS
Example 2.1 ~ A large plane wall of thickness b and thermal conductivity k has its left surface subjected to a
heat flux q0 while the temperature at this surface is T1• Assuming no heat generation within
the wall and constant thermal conductivity, express the differential equation and boundary conditions for steadystate, one-dimensional heat conduction and get the temperature distribution within the wall.
Known: A plane wall has its one surface subjected to a heat flux and held at a constant temperature.
Find: Differential equation, boundary conditions, temperature variation.
Steady-State Heat Conduction-One Dimension
Schematic:
2. 9
Plane
' " " . - - - - - - - + - - , wall (k)
~-------~---x
x=b
b
Fig. 2.6
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) No heat generation. (3) Constant thermal
conductivity.
Analysis: The differential equation can be expressed as
Thus, T= f(x) and we have
(Ans)
This is a second-order differential equation and clearly, two boundary conditions must be prescribed.
The boundary conditions are
•
At x = 0 : % = const
(Constant su,face heat flux condition)
•
At x = 0 : 'Ii = const
( Constant surface temperature condition)
(Ans)
Integrating twice the differential equation, we have
dT = C1 and T(x) = C1 x+C2
dx
Control volume energy balance gives:
or
{Assuming A(x =0) = A (x = b)}
Therefore, the heat flux at the left face equals that at the right face and is constant throughout
q=~=%
dT
dT
q
%
With q = - k - the temperature gradient, - = - - = - - = C1
dx'
dx
k
k
2.10 ,
Heat and Mass Transfer
~ !C2 = 'Ii I
Also, T(x = 0) ='Ii= C1 xO+C2
Substituting for C 1 and C2, the temperature distribution T(x) is given by
~
1~-T(x)=-¥1
(Ans)
dT
The slope is represented by - = C1 = constant. For constant slope, we have a linear temperature
dx
profile as shown in the schematic.
Example 2.2 ~ Consider the following geometry bounded by two quarter circle arcs. The corner may be
assumed to extend to infinity perpendicular to the plane of paper. The thermal conductivity
of the material is 1.7 W/m 0 C. Find the temperature distribution and the rate of heat transfer from T1 to T2,
assuming unit depth normal to the paper.
J- im-J
Fig. 2.7a
Known: Region between two concentric quarter-circles.
Find: Temperature distribution, heat-transfer rate.
Schematic:
0=0
k= l.7W/m°C
""'-------'------'- e= rc/2
T2 = 0 °C
f-- Im
Im
Fig. 2.7b
--j
Steady-State Heat Conduction-One Dimension
2.1 I
Fig. 2.7c
Assumptions: ( l) Heat transfer in the r-and z-directions is negligible. (2) Constant thermal conductivity.
(3) No internal heat generation. (4) Steady-state conditions.
Analysis: The general differential equation in cylindrical coordinates with constant material properties is
given by
i
no heat generation
no heat transfer in
the z-directon
insulated over the
entire boundary
f
+
=~
steady state
Hence, the equation reduces to
l rJ2T
---=0
r2 ae2
~
d 2T
-=0
d0 2
(since T =f (0) only, the partial derivative becomes total derivative)
Integrating with respect to tangential coordinate 0 twice, we have
Boundary conditions required to determine the two constants of integration are:
B.C.I: At 0 = 0,
T= T1 = 100 °C
~
T 1= C2 or C2 = 100
B.C.I1:At0=rcl2,
or
T=T2 =0°C
~
O=c,(%)+100
le,=- 2~01
The temperature distribution is then given by
T(0) = 100- 200 0
1C
2.12 ,
Heat and Mass Transfer
Heat-transfer rate in the 0-direction is
Q9 =-kAa[ dT]=-~(drdz) .!!_[100- 200
r
d0
d0
1t
e]
= ~ 200 (d rd z)
r 1t
For unit depth,
Heat-transfer rate per unit depth from T1 to T2 is
Q = 200 k
f dr 200k I n
=
1t'ir
r
r,_
1j
= 200(1.7 W/m C)(lm) In( 2m)
0
lm
1t
= 75 W
(Ans)
Example 2.3 ~ A hollow cylinder of a material that has a thermal conductivity of 3.46 W/m °C has an inner
radius of 7.5 cm and an outer radius of 15 cm. The temperature on the inner surface is
I 50°C and the temperature profile is given by
T(r)=150-73/n(-' )( 0 c)
7.5
where r is in cm. Calculate the temperature at the outer face and the heat transfer rate per unit length at the
inner and outer faces. Do steady-state conditions exist? Sketch the temperature distribution.
Known: Temperature profile of a hollow cylinder.
Find: T(r2); Q( 'i) and
L
Q('2)
L
Schematic:
Hollow cylinder
k = 3.46 W/m °C
T(r) = 150- 73 /n[...!'._J
7.5
T(r) = 150- 73 /n
Fig. 2.8a
Steady-State Heat Conduction-One Dimension
2. 1l
Assumptions: (1) One-dimensional (radial) heat conduction. (2) Constant thermal conductivity.
Analysis: The temperature profile is given by
T(r) = 150-73 ln(r I 7.5)
[0 C]
At the inner face (r = r 1 = 7.5 cm):
Ti = 150 - 73 In 1 = 150°C
(given)
At the outer face (r = r2 = 15 cm):
Ji= 150-73
/n(~)
7.5
= 99.4°C
(Ans)
Temperature gradient,
dT
d
-=-[150-73{/n r-ln 7.5}]
dr dr
Heat-transfer rate per unit length:
At the left face,
Qri
= - k (2 7t 'i L) dT I
dr r=ri
J
= (-3.46 W/m °C){21tx0.075mxlm) [ -73
- -°C
7.5 m
(Ans)
=15.87 W
At the right face,
Qr2 = - k (2 1C 'i L) dT I
dr r=r,
0
= (-3.46 W/m °C) {21tx0.15 mxlm) [ -73]
- -c
15 m
= 15.87 W
Since, Qr.I = Qr.,2 steady-state conditions exist.
(Ans)
(Ans)
2.14 ,
Heat and Mass Transfer
The temperature distribution is sketched below:
.-.
u
~100°- - - - - - - - + - - - - - - - - - + -~ - - - - -----I
I-
0+---------+---------+-----------1
0
0.05
0.1
0.15
Fig. 2.8b
(A) PLANE WALL WITH SPECIFIED BOUNDARY TEMPERATURES
Example 2.4 ~ A TY advertisement by a reputed insulation manufacturer states: it is not the thickness of the
insulating material that counts, it's the R-value (defined as Uk where L (cm) is the thickness
of the insulation and k (W cm/m 2 K) is the thermal conductivity of the material). The ad then goes on to
show that to obtain an R-value of 3.35 you need 5.5 m of rock, 37.5 cm of wood or just 15 cm of fibreglass
insulation. Is this ad technically reasonable?
Known: Different thicknesses of three insulating materials.
Find: R-value of the three insulating materials.
Properties:
Material
Thermal conductivity, W/m K
Rock
2.0
Wood
0.15
Fibreglass
0.05
Analysis: Heat-transfer rate, per unit area is
k(J;-7;)
L
i;-i;
R
q = ---- = - -
•
where
R-value =
L(cm)
Rock: k= 2 W/m K
5.5
2
R-value = - = 2.75 m K/W
2
•
Wood: k= 0.15 W/m K
R-value =
37 5xI0-2
·
0.15
.
m2 K
(
) , 1.e., -WkW cm/m2 K
= 2.75 m2 K/W
Steady-State Heat Conduction-One Dimension
•
2.1 S
Fibreglass: k = 0.05 W Im K
15 X 10-2
R-value = - - - = 3.0 m2 K/W
0.05
the value of 3.35 therefore seems to be reasonable.
Example 2.5
(Ans)
~ The inner and outer surface temperatures of a freezer compartment in the form of a cubical
cavity of 2.5-m side are 30°C and -I 5°C, respectively. The steady-state heat load to be
maintained is 850 W . The bottom of the compartment is perfectly insulated. Neglecting heat loss from the
corners, calculate the thickness of the insulation of thermal conductivity W/m 0 C.
Known: Geometry and surface temperatures of the freezer compartment.
Find: Insulation thickness, L.
Schematic:
~ T0 =30°C
Insulation -__r;;===::=========:;-i
(k=0.035W/m C)
'-.. T=-IS°C
0
1
~1
I
Cubical cavity
L-
2.5 m
Q=850W -VV,,
·::::::•.
Ac=SH 2
There are only 5 surfaces
for heat transfer
::;:;:;:;:;:;:::;:;:;:;:;: ;:;:;:;:;:::
Fig. 2.9
Assumptions: Steady-state, one-dimensional conduction. (2) Constant thermal conductivity. (3) Heat loss
from the corners is neglected.
Analysis: From Fourier's law
Heat load,
.
.
Qn =-Q =
-k Ac('I'; - I;,)
L
Insulation thickness is calculated from
(0.035 W/m°C)( 5 x 2.5 2 m2 )[(30-(-15)]°C
L=-----------=------=850W
= 0.058 m or 58.0 mm
(Ans)
2.16 ,
Heat and Mass Transfer
(B) PLANE WALL BOUNDED BY SPECIFIED FLUID TEMPERATURE
Example 2.6
~ It is required to reduce heat loss from a slab by doubling the thickness of brick work.
The temperature of inner surface of brick work is 500°C and ambient air is at 30°C. The
temperature of outer surface of initial brick work was 200°C. Calculate the percentage reduction in heat loss
per m2 because of doubling the thickness.
Known: Brick work thickness is doubled to reduce the heat loss.
Find: % reduction in heat flux.
Schematic:
- - - + - - - - - + - - - QI
Brick
~ - - - ~ T2 = 200 °C
111
I-- L - I
Fig. 2.IOa
Finally
ooc\-L
~
Brick
work
I
L------1
Brick
work
Q11
Fig. 2.IOb
Equivalent thermal circuit
Case I
qi_
T1=500°C
•
Vl/v
Rcond = Uk
T2 =200°C
•
Vl/v
Rconv = I/h
T~=200°C
• -
(WI
QI
m
2)
Fig. 2.16c
Case II
Q1-
T1 =S00°C
T =30°C
....---Vvv-----~Vvv,....--~-.
Rconv= I /h
Fig. 2.IOd
-
qu(W/m 2)
Steady-State Heat Conduction-One Dimension
2. I 7
Assumptions: (1) Steady-state, one-dimensional heat conduction. (2) Constant properties. (3) No internal
heat generation.
Analysis:
Q
Initially: Heat flux, q1 =-A=
T{-T=
T;-J;
=
(LI k+l/ h)
LI k
(1)
.
T{ -T=
Finally: Heat flux, q[[ = (2 LI k + 11 h)
(2)
LI k+l/ h = T{-T=
Llk
r;-i;
From equation (1):
l+~= 500-30 = 470°c = 1. 5667
h L 500 - 200
300°C
or
~=0.5667
hL
or
1 0.5667 L
or
h
k
From equation (1 ),
=
q1
500-30
=300k!L
(0.5667+1)L/k
And, from equation (2):
q =
500-30
= 183. lk/L
[[ 2 L +0.5667~
k
k
Percentage reduction in heat loss
= (300-183.l)k/LxlOO
300k/L
=39%
(Ans)
2.18 ,
Heat and Mass Transfer
Example 2. 7
~ The wind chill, which is experienced on a cold, windy day, is related to increased heat
transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty
tissue that is 3-mm thick and whose interior surface is maintained at a temperature of 36°C. On a calm day,
the convection heat-transfer coefficient at the outer surface is 25 W/m 2 K but with 30 km/h winds it reaches
65 W/m 2 K. In both cases, the ambient air temperature is -I 5°C.
(i) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day?
(ii) What will be the skin's outer surface temperature for the calm day and for the windy day?
(iii) What temperature would the air have to assume on the calm day to produce the same heat loss occurring
with the air temperature at -I 5°C on the windy day? Take k = 0.2 W/m K (for tissue fatty layer).
Known: Heat transfer from human skin to atmosphere in cold season on a calm day and on a windy day.
Find: (i) %aim / qwindy; (ii) ½,calm•
½,windy; (iii) T=,calm if %aim= qwindy with T= =-15°C on windy day.
Schematic:
- + - - - - q windy
T1 =36 ' C
- + - - - - qwindy = qcalm
T1 =36 ' C
T1=36 ' C
ih=2SW/m 2 K
T_ =-IS'C
T_=-IS ' C
k
IL=0.003 m, k=0.2W/mK!
Fig. 2.1 la
Assumptions: (I) Steady-state conditions. (2) One-dimensional heat conduction. (3) Uniform heat transfer
coefficient. (4) Constant thermal conductivity.
Analysis: (i) Calm day
q---
---q
2Uk
1/h
Fig. 2.llb
The heat flux, i.e., heat loss per unit area for this case is
[36-(-15)]°C or K
T.,-T=
%aim= L/k +I / h
(
0.003 m ) (
I
)
0.2W / mK + 25W !m 2 K
5 IK
= 927.27 W /m 2
(0.015+0.04)[ m 2 K/W]
Windy day
The convective thermal resistance alone will change to 1/h = 1/65 = 0.0154 m 2 K/W.
Steady-State Heat Conduction-One Dimension
2. I 9
Hence, the rate of heat loss per unit area will be
51K
qwi ndY = (0.015+0.0154)m2 K/W
= 1678.48 W /m2
Ratio of heat losses is
%aim
qwindy
= 927.27 = 0.552
1678.48
(Ans) (i)
Heat loss on a windy day is 1/0.552 or 1.81 times more than that on a calm day.
(ii) Outer skin temperature, T2
Heat loss per unit area,
q = h (T2-T=)
Hence,
Ji =T=+!i
h
y: = _ 150C+ 927.27W/m2
2
25W/m2 K
On a calm day,
= 22.1°c
(Ans) (ii)
y: = _ 150C+ 1678.48W/m2
2
65W/m2 K
On a windy day,
= 10.8°C
(Ans) (ii)
(iii) Heat loss on a windy day with T= = -15°C = 1678.48 W/m2
= Heat loss on a calm day=
36 - T=,calm
'
= ----LI k+l/ h,calm 0.015+0.04
J;_ -T= calm
Ambient air temperature required,
T=,calm = 36°C -
(1678.48 W/m 2) (0.055 m2 K/W)
=-56.3°C
Example 2.8
(Ans) (iii)
~ A steel plate of thickness L and thermal conductivity k is subjected to a constant heat
flux q0N /m2) at the boundary surface at x = 0. From the other boundary surface, the
heat is dissipated by convection into a fluid at temperature T_ with a heat-transfer coefficient h. Develop the
expressions for the surface temperatures T1 and T2 at the surfaces x = 0 and x = L, respectively. For the following
data, calculate the surface temperatures T1 and T2 if L = 2 cm, k = 20 W/m K, q = I 05 W /m2, T= = 50°C and
h = 500 W/m 2K.
Known: Steel plate subjected to constant heat flux at one boundary surface and at the other exposed to
convective environment,
Find: T1 and Ti( 0 C).
2.20 , Heat and Mass Transfer
Schematic:
\ - L = 2 cm
r1
q=IOS W/m2
------I
!! !
k=20 W/m K
h=SOO W/m2 K
T= =S0°C
-
r=
Steel plate
x=O
x=L
Fig. 2.12a
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Constant thermal conductivity. (3) Uniform
heat-transfer coefficient.
Analysis: Energy balance
£in -£out +
= ....,,_,
Est
£gen
.....,,....,
no heat generation
steady state
Hence,
That is
Per unit cross sectional area, the steady-state heat flux is
q(x = 0) = q(x = L) = constant
Using the electrical analogy,
7; -T2 = iJ Ri.
where R 1 is the thermal conduction resistance
r; - T= = iJ R2
and
Adding these two expressions,
where R 2 is the convective resistance
7; - T= = q( Ri_ + R2 )
L (m)
where
Ri = k(W/m oq
l (m2 °C)
and
Ri = h (W)
Hence, the boundary surface temperature at x = 0, is
Ir.= r_ +;;r,+¼JI
and
(Ans)
Substituting the known values, one gets
T. =50oc+1os~x[ 0.02m
1
m2
= 350°c
20W/mK
+
l
]
500W/m2 K
(Ans)
Steady-State Heat Conduction-One Dimension
2.21
T = 5ooc+ 10sw1m2
2
500W/m 2 K
and
= 250°C
(Ans)
The thermal circuit and temperature profile are given below:
1-- L = 0.02 m --1
--------~ q
r 1= 350°c
Too
i~= 250°c
Uk
1/h
q
+ - - - - - - - - - t........_
Fig. 2.12b
Example 2.9
~ Hot gases at 980°C flow past the upper surface of the blade of a gas turbine and the lower
surface is cooled by air bled off the compressor. The convective heat-transfer coefficient at
the upper and lower surfaces are estimated to be 2830 and 1415 W/m 2 °C respectively. The blade material
has a thermal conductivity of 11.6 W/m 0 C. If the temperature of the blade is limited to 870°C, work out the
temperature of the cooling air. Consider the blade as a 0.115-cm thick flat plate and presume that steady-state
conditions have been reached.
Known: A gas turbine blade is exposed to hot gases on one side and the cold air on the other.
Find: Cooling air temperature, T=i( 0 C).
Schematic:
Rconv, I
k = 11.6 W/m °C
T= I = Tgas=980 °C
r,
h 1= 2830 W/m2 °C
(870°C)
~---T1=870°C
,1----:::::=::::""?"""-----::==i.-=:::'
___l_ L = 0.115cm
h2 = 1415 W/m2 °C
Rconv,2
T=2 = Tair=?
Q
Fig. 2.13
2.22 , Heat and Mass Transfer
Assumptions: (1) Steady-state conditions. (2) Gas turbine blade is approximated as a flat plate.
(3) Convection coefficients are uniform. (4) Constant thermal conductivity.
Analysis: The equivalent thermal circuit is shown in the schematic. The individual thermal resistances
based on per unit area are the following:
+
Rconv,l (from the hot gases to the upper swface of the blade)
1
"1A
+
Rcond (through the blade)
L
kA
+
- - - -1- - - = 0.3533x1Q-3 °C/W
2830W/m2 °Cxlm2
0.115x10-2 m
------=0.09914x1Q-3 °C/W
11.6W/m °Cxlm2
Rconv,2 (from the lower swface of the blade to the cooling air)
1
hiA
- - -1- - - = 0.7067x1Q-3 °C/W
1415W/m2 °Cxlm2
The steady-state heat transfer rate,
J?.:ond
+ J?.:onv,2
Therefore, the cooling air temperature is
= 8700C-( 980 _ 870 ) 0C (0.09914+0.7067)(10-3 )°C/W
0.3533xl0-3 °C/W
= 619.14 °C
(Ans)
Example 2.10 ~ An ice chest with outer dimensions of 50 by 70 by 90 cm and wall thickness of 5 cm is
constructed of an insulating material (Styrofoam) with a thermal conductivity of 0.033 W/m K.
The outside surface of the ice chest is exposed to surrounding air at 32°C with a convection coefficient of I0
W/m2 K. Neglecting any heat transfer from the 60 cm x 80 cm base of the ice chest, determine the time for
the ice to completely melt. Initially, the chest is filled with 60 kg of ice at 0°C and the inner surfaces of the
chest may be considered at 0°C at all times. The latent heat of fusion of ice at 0°C may be taken as 333.7 kJ/kg.
Known: An insulated ice chest is exposed to convective environment.
Find: Time required to melt the ice in the chest completely.
Steady-State Heat Conduction-One Dimension
Schematic:
2.23
Outside dimensions:
50cm x 70cm x 90cm
Air
Insulation
h= IOW/m2 K
T==J2°C
Base (insulated)
(60x 80cm)
Fig 2.14a
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Constant thermal conductivity. (3) Uniform
convection coefficient.
Analysis: The thermal circuit for heat flow from the ambient air to the inner surfaces of the ice chest is
shown below:
Ti =0°C
Q-
T = 25°C
T2
----'VV\r------'V\/1,----=
_ L
Rcond - kA
_ I
Rconv - hA
-Q
Fig. 2.14b
Total effective inside surface area (excluding the base area) is
A= L 2 L 3+2(L 1L 3+L 1L 2 )
= {0.6 X 0.8)+ 2[{0.4 X 0.8)+ {0.4 X 0.6)] = 1.6 m 2
Also
k = 0.033 W/m K, L = 0.05 m
Heat transfer rate from the ambient air to the ice inside,
.
T= - I;
A ( T= - I; )
Q = [~+ - 1 ] = {LI k)+(l/ h)
kA
hA
l.6m 2 (32-0)°C orK
=--------------(0.05m / 0.033 W/mK)+(l/lOW/m2 K)
=31.7 W
Energy balance yields:
(
Heat transfer
into the chest
l
= (~ass of ice) x (He~t of fusion
m the chest
of ice
l
2.24 , Heat and Mass Transfer
Heat transfer into the chest = Mass of ice in the chest x Heat of fusion of ice
31.7W ={60kg){333.7x10 3 J/kg)/t(s)
Time required to completely melt the ice is
t= 60x333.7x10 3 sl____!.!:_11 dayl
3600s 24h
31.7
= 7.31 days
(Ans)
(C) COMPOSITE WALL WITH PRESCRIBED BOUNDARY TEMPERATURES
Example 2.1 I ~ An exterior wall of a house may be approximated by a I0-cm layer of common brick [k = 0.7
W/m K] followed by a 4.0-cm layer of gypsum plaster [k = 0.48 W/m K]. What thickness
of loosely packed rock-wool insulation [k = 0.065 W/m K] should be added to reduce the heat loss (or gain)
through the wall by 80 per cent?
Known: Composite slab consisting of two materials of known dimensions and thermal conductivities.
Find: Thickness of additional third layer of given conductivity to ensure 80 percent decrease in heat
transfer rate.
..........
.. .. . .. .
.. .. .......... .. .. .... .. .......... ..
.........
Schematic:
.........
. .. . . . .. .
Brick
Plaster
:::::i~.~~::::
Brick
Qw = 0.20 C2wo
Plaster · · · · · · · · ·
:/W.QQ]:'.:::
.........
.........
.. .... .... .... .... .... .... .... .... ..
kb=0.70W/m K
kp=0.48W/m K
krw=0.065W/m K
Fig. 2.15
Assumptions: (1) Steady-state, one-dimensional heat flow. (2) Constant properties.
Analysis: The overall heat transfer is given by
Q = Ai:iverall
I:J?i
.Qwith insulation = 0 _20 = I: Ri without insulation
Q without insulation
I:i?i with insulation
Steady-State Heat Conduction-One Dimension
2.25
For unit area, the thermal resistances for brick and plaster are:
4,
O.lm
Rt, =-=---=0.143m2 K/W
kb
0.7W/mK
~=I., = 0.0 4 m = 0.083 m2 K/W
kP 0.48W/mK
Thus, I.R1 ( without insulation) = Rb + RP
= 0.143 + 0.083 = 0.226 m2 K/W
0.226
m2 K
Then, I.R 1 ( with insulation) = - - = 1.13 - 0 .2
W
This is equal to Rb+ RP+ R,w where R,w is the thermal resistance due to rock-wool insulation.
m2 K
Rrw = 1.13-0.226 = 0.904 "°w
R,w = 4w ,
krw
As
thickness of insulation,
L,w = Rrwkrw = 0.904 X 0.065
= 0.0588 m or 58.8 mm
(Ans)
Example 2.1 2 ~ A 30-cm thick wall of size Sm x 3m is made of red bricks (k = 0.35W/mK). It is covered on
both sides by layers of plaster, 2-cm thick k = 0.6W/mK. The wall has a window of size of
Im x 2m. The window door is made of glass, 12-mm thick having thermal conductivity of 1.2 W/mK. Estimate
the rate of heat flow through the wall. Inner and outer surface temperatures are I0°C and 40°C respectively.
Known: A composite wall with specified boundary surface temperatures.
Find: Heat-transfer rate through the wall, Q{W).
Schematic:
Outer surface
T0 =40°C
Inner surface
T;= I0°C
Fig. 2. 16
2.26 , Heat and Mass Transfer
Assumptions: (1)
Steady-state conditions. (2) One-dimensional conduction. (3) Constant thermal
conductivity.
Analysis: The total thermal resistance is
Rtot
=[
1
Rwindow
+-l-]-1
Rwall
Individual resistances are
_ _12_x_l_0_-3_m
_ _ = 0 005 K/W
1.2W/mKx(lx2)m2
·
For wall resistances, the effective area,
AP= Ab= [(5x3)-(2xl)Jm2 = 13 m2
Rwall
= Rplaster + Rbrick + Rplaster
LP
Lb
LP
=--+--+-kPAP
kPAP
kbAb
0.02m
0.30m
0.02m
0.6W/mKx{13m2)
0.35W/mKx{13m2)
0.6W/mKx{13m2)
=--------+--------+--------
= 0.07106 K/W
1
1
R101 = [ - - + - - 0.005 0.07106
]-I
=0.00467 K/W
Heat-transfer rate,
Q = _A_T = _T;,_-_I; = _(4_0_-_10_)_C_o_r_K_
0
Rtot
Rtot
0.00467 K/W
= 6422 W
Example 2.13
(Ans)
~ A composite insulating wall has three layers of material held together by a 3-cm diameter
aluminium rivet per 0.1 m2 of surface. The layers of material consist of I0-cm-thick brick
with hot surface at 200°C, I-cm-thick timber with cold surface at I0°C. These two layers are interposed by a
third layer of insulating material 25-cm-thick. The conductivities of the materials are
k (brick)=0.93 W/m K
k (insu/ation)=0.12 W/m K
k (wood) =0.175 W/m K
k (aluminium)= 204 W/m K
Assuming one-dimensional heat flow, calculate the percentage increase in heat transfer rate due to rivet.
Known: A three-layered composite wall is held together by a rivet.
Find: Percentage increase in heat transfer due to rivet.
Steady-State Heat Conduction-One Dimension
Schematic:
2.27
T;=200°C
Brick
t ..................
d=Jcm
J ...............
Lr=Lb+L;+4
Fig. 2. 17a
Assumptions: ( l) Steady-state conditions. (2) One-dimensional heat conduction. (3) Negligible contact
resistance. (4) Constant properties.
Analysis: Without rivets
Qwo
(200-10)°C orK
190K
=~------~---=---A
{0.10 0.01 0.25} m
2 _248 m2 K
W
0.93 + 0.175 + 0.12 W/ mK
or
Q = 190Kx0.lm2 = 8 _45 W
wo
(Ans)
2.248m 2 K/W
With rivets
r ; 1
Rb
•
R;
•
•
Rt
Fig. 2.17b
Q = ~¾verall
w
I.Rt w
~=-4-=_L~b+_4~+_L~1 =_(_0_.l_O_+_O_.O_l+_0_._25_)_m_
k,A,
k,A,
204 W/ mK x ~(0.03)2 m2
4
= 2.4966K /W
2.28 , Heat and Mass Transfer
"f.R
= 2.248 = 2.248 = 22.48 K / W
I.W O
A WO
O• l
=22.64 K /W
The three resistances in series are in parallel with the rivet resistance R,. Hence,
= [ - l- + -l 22.64 2.4966
Q.
w
J-l = 2.2486 K/W
= (200-10)°C = 84 .SW
2.2486K/W
Percentage increase in heat transfer due to rivets,
= f2w ~f2w o xlOO
Qwo
= 84.5-8.45 xlOO
8.45
(Ans)
=900%
(D) COMPOSITE WALL BOUNDED BY FIXED FLUID TEMPERATURES
Example 2.14
~ In a manufacturing process, a transparent film is being bonded to a substrate as
shown in the sketch. To cure the bond at a temperature T0 , a radiant source is
used to provide a heat flux q0 fN/m 2), all of which is absorbed at the bonded surface. The back of
the substrate is maintained at T1 while the free surface of the film is exposed to air at T= and a
convection heat-transfer coefficient h.
4=0.25mm
- =
ti)00
kr= 0.025 W/m K
Ls=I.Omm
Film
,h
T
Lt
Bond T0
Substrate
Ls
l
ks= 0.05 W/m K
T1
Fig. 2.18
Steady-State Heat Conduction-One Dimension
2.29
(a) Show the thermal circuit representing the steady-state heat-transfer situation.
(b) Assume the following conditions: T~ = 20°C, h = SO W/m 2 K and T1 = 30°C. Calculate the heat flux q0 that
is required to maintain the bonded surface at T0 = 60°C.
Known: Transparent film cured by radiant heating with substrate and film surface subjected to prescribed
thermal conditions.
Find: (a) Thermal circuit for this situation, (b) Radiant heat flux, q0 to maintain bond at curing temperature, T0 •
Schematic:
___
T = 20 °c
qo
L~~
• h:S0Wlm'k:=0.02SW1mK
,,Y:2smm
f
Bond, T0 =60 C - ..
k5 =0.0SW/m K
4 = 1.0mm
J_
Fig. 2.19a
Assumptions: (1) Steady-state, one-dimensional heat conduction. (2) All the radiant heat flux % is
absorbed at the bond. (3) Contact resistance is negligible. (4) Constant properties.
Analysis: (a) The thermal circuit for this heat transfer situation is shown below, with all terms expressed
on per unit area basis:
Thermal circuit
qo
(W/m 2)
4
k,
h
q2
(W/m2)
T=
i
Ls
k.
q,
To
T1
(W/m2)
Fig. 2.19b
(b) Performing an energy balance on the film-substrate interface, and using the thermal circuit given above,
Substituting numerical values, we have
60-20
60-30
40
30
% = _!_+ 0.00025 + 0.001 = (0.02+0.0l) + 0.02
50
0.025
0.05
2.30 , Heat and Mass Transfer
Hence, the heat flux, % = (133 3 + 1500) W/m2
=2833 W/m 2
(Ans)
Example 2. 15 ~ The inner surface of a high temperature reactor will operate at 1623 K. The wall of
the reactor will have an overall thickness of 350 mm and is to be made up of an
inner layer of firebrick material (k, = 0.86 W/m K) covered with a layer of insulation (k1 = 0.16 W/m K).
This insulating material has a maximum operating temperature of 1473 K. The ambient temperature will
be 293 K and it is estimated that the heat transfer coefficient at the exposed surface of the insulation
will be IO W/m 2 K. Calculate the thickness of refractory and insulation which gives minimum heat loss
and the magnitude of this loss in W /m2 . Also calculate the surface temperature of the insulation. If the
calculated loss is unacceptable, would the addition of another layer of insulation be a satisfactory solution?
Give your reasons.
Known: A composite wall comprising two layers of prescribed dimensions and properties with its outer
surface exposed to convective environment.
Find: Minimum heat loss per unit area, and temperature of outer surface of insulation.
Schematic:
r---- Lr
T, =1473K
-----,.--+-
lll
L1 ------j
h= IOW/m2K
T,,1= 1623 K
Refractory
Insulation
kr= 0.86 W/m K
k1=0.16W/m K
i-----
L= 350mm - - - - . - i
Fig. 2.20
Assumptions: ( l) Steady-state conditions. (2) One-dimensional heat conduction. (3) Constant properties.
T -T
Analysis: Heat loss per unit area, q = - '- '- '
4 / k,
={162 3 - 1473)K 1[4(m) / 0.86W/mK]= l 50x0.86
4
Thus,
129 W
q=--
(1)
4 m2
Also
(1623-293)K
[4 / k, +L, I k; +l / h]
q =[~ + (0.35-4) + _!_] m2 K
0.86
0.16
10 W
= 1330 / (2.2875 - 5.0872 L,)
(2)
Steady-State Heat Conduction-One Dimension
2.3 I
Equating (1) and (2)
1330
129
2.2875-5.08724
4
1330 L, = (129 X 2.2875) - (129 X 5.0872 L,)
or
or
1986.25 L, = 296.0875
L,= 0.1486 m =150 mm
Thickness of refractory, material, L, = 150 mm
Thickness of insulation, Li= 350 - 150 = 200 mm
Heat loss can be obtained from equation (1)
(Ans)
(Ans)
q = 129 = 129 = 860W /ml
L,
0.15
(Ans)
Outer surface temperature of insulation,
l
¾o =T= +qf\onv =293K+-x860W/m2 =379K
(Ans)
10
Addition of another layer of insulation will add to the total thermal resistance, thereby reducing the
heat loss further. As the area normal to heat flow is constant in the case of a composite or multilayered
wall, there is no critical thickness of insulation.
Example 2. 16 ~ A square plate heater ( 15 cm x 15 cm) is inserted between two slabs. Slab A is 2-cm -thick
(k = 50 W/m K) and Slab B is I-cm thick (k = 0.2 W/m K). The outside heat transfer
coefficients for slabs A and 8 are 200 and 50 W/m 2 K. The surrounding air temperature is 25°C. If the rating
of the heater is I kW, determine
(a) the maximum temperature in the system
(b) the outer surface temperatures of the two slabs
Draw the equivalent thermal circuit of the system.
Known: Heater of given rating sandwiched between two slabs of specified thicknesses and thermal
conductivities. Convection coefficients on both sides and ambient air temperature are given.
Find: Maximum temperature in the composite system and outer surface temperatures of the two slabs.
Schematic:
Heater (Q= ! kW)
Air
Air
Tmax =?
T= 1=25°C
h 1= 200W/m2K
LI
®
®
kA=50W/mK
Fig. 2.21
kB=0.2
W/mK
T=2 =25°C
h2=50W/m 2 K
2.32 , Heat and Mass Transfer
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Constant thermal conductivities and
uniform heat-transfer coefficients.
Analysis: The equivalent thermal circuit is given below:
T=I
Rconvl
r.1
Rcondl
•
Wv
•
Wv
Tmax
•
Wv
t
Q1~
r.2
Rcond2
•
Rconv2
T=2
Wv
•
...f\/Vv- Q2
Q
Fig. 2.22
_½
J?.:ond2 -
k A
B
1
J?.:onvl
= h,_A
1
J?.:onv2
= hi.A
Heater rating, Q = 1000 W
Area of heater, A= (0.15 x 0.15) x 2
= 0.045 m2
[The heater has two swfaces exposed to slabs A and BJ
The heat flux,
q= Q
A
1000W
0.045m2
= 22 222.2 W/m2
or 22.222 kW/m2
Now
or
{Tmax-25)K
{Tmax-25) K
22222 x 103 (W/m2 ) = -0-.0~2-m_ _ _l_m_2_K-+-0-.0~1-m--1-m_2_K_
----+- - - -----+- - 50 W/mK 200 W
0.2 W/mK 50 W
- (T
-
max
-25) [
1
+
1
{4 X lQ-4 )+{5 X10-3) {0.05+ 0.02)
= (Tmax -25)[185.1852+ 14.2857]
= 199.47 Tmax -(25X 199.47)
l
2.ll
Steady-State Heat Conduction-One Dimension
maximum temperature in the system,
T
max
= 22 222.2+4986.77
199.47
= 136.4°C
(Ans) (a)
Heat flux,
q, = (136.4-25) (185.1852)
= 20 630.8 W/m2
Surface temperature,
1
(20630.8)
T.1 =T=I +q,. ,;;-=25+
200
= 128.15°C
(Ans) (b)
Surface temperature,
where
q2 = (136.4 - 25) (14.2857) = 1591.5 W/m 2
T = 25 +(1591.5)
s2
50
= 56.83°C
(Ans) (b)
Check
q = q 1 + q2 = 20630.8 + 1591.5 = 22222.3 W/m 2
Example 2.17
Hence OK
~ The masonry wall of a building consists of an outer layer of facing brick (k = 1.3 W/m C),
0
I0-cm thick, followed by a 15-cm thick layer of common brick (k = 0.72 W/m 0 C), followed
by a 1.25-cm layer of gypsum plaster (k = 0.22 W/m 0 C). The inside air is characterized by 22°C temperature
and a film heat-transfer coefficient of 12 W/m 20 C. In cold climatic conditions, the outside air is characterised
by a film heat transfer coefficient of 30 W/m 2°C and a diurnal varying temperature of the form
0 ~ t ~ 12h
12 ~ t ~ 24h
The total surface area of the wall is 150 m2• Neglecting changes in energy storage within the wall, estimate
the daily heat loss through the wall.
Known: A composite wall with prescribed convection processes at inner and outer surfaces.
2.34 , Heat and Mass Transfer
Find: Daily heat loss for specified day- night cyclic variation in ambient air temperature.
Schematic:
A=ISOm 2
----Q
Inside
Ambient
air
air
®
ttt
T=,;=22°C
ttt
h0 = 30W/m2°C
h;= 12W/m2°C
-+j+.t-- 15cm -J f-1.2Scm
L=6cm~1 ......,------...
T=,o (°C) = 4 sin[~:~
0 :5 t :5 12 h
~
12 :5 t:5 24 h
T=,0 (°C) = IOsin [~:
Fig. 2.23
Assumptions: ( 1) One-dimensional, steady-state conduction. (2) Negligible contact resistance.
Analysis: The heat loss through the wall can be estimated as
Q=
24h T
f =,i -T=,0 dt
o
l?iotal
Total thermal resistance is determined from
l?i
ota
I=_!_[_!_+!!_+½+½+_!_]
A h; kl k2 k3 ho
1
[l
0.10
0.15
l]- 1 m2 °C
0.0125
= 150m2 12+1.3+ 0.72 + ~ + 30
~
= 0.00306°C /W
It follows that
Q=
f Qdt= f T=,I. -T=,0 df
24h
24h
O
O
f
l?iotal
12
2 } dt+
= -l - [ h{ 22-4sin~t
l?iotal
O
24
f
24
2
h{ 22-lOsin~t } dt
12
24
1
= 0.00306°C/W
[{22t-(4)(~)(-cos 2 1t t)}]
2n
24
12
0
l
+{22t-(lo)(~)(-cos 2 1t t)}
2n
24
24
12
Steady-State Heat Conduction-One Dimension
2.35
=327W/°C [{22(12-0)+ :(cos(~x12)-coso)}
!
+ {22{24-12)+ 1 0 (cos~x24-cos~x12)} 0 c h]
J
48
120
=327 [ (22xl2)+~{-l-1)+(22xl2)+--;-{l+l)
Wh
= (327) [573.84] Wh
MJI
= 187.635 kWh 13·6
= 675.5 MJ
lkWh
(Ans)
Example 2.18 ~ An iron plate of thickness L and thermal conductivity k is subjected to a constant heat flux
q0 W/m 2 at the boundary surface at x = 0. From the other boundary surface at x = L, the
heat is dissipated by convection into a fluid at temperature T~ with a heat transfer coefficient h. Develop the
expressions for the surface temperatures T1 and T2 at the surfaces x = 0 and x = L respectively.
For the following data, calculate the surface temperatures T1 and T2 if L = 2 cm, k = 20W/m °C, q0 = I0 5
W/m 2, T~= S0°C and h = 500 W/m 2 0 C.
Known: A plate with one boundary surface subjected to uniform heat flux and the other exposed to
convective cooling.
Find: Surface temperatures, T1 (x = 0) and T2 (x = L).
Schematic:
Iron plate
k=20W/m°C
Heatflux,q0= I05W/m2 -
---
h=SOOW/m2°C
----T==S0°C
0-------------x
/-
L=0.02m
r,
qo
•
-J
T2
'\M
L
k
Thermal circuit
•
T=
'\M
I
•
h
Fig. 2.24
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Constant and uniform heat flux at the
inner surface. (3) No heat generation. (4) Negligible radiation effects. (5) Constant thermal conductivity.
Analysis: The governing differential equation for this problem is
d 2T
-=0
dx 2
2.36 , Heat and Mass Transfer
The general solution is obtained by two successive integrations to be
dT -C
I
dx
and
where C1 and C2 are arbitrary constants.
Applying the first boundary condition (at x = 0):
-kd:~0)=%
~ -kC,=% ~ le,= -:ol
The application of the second boundary condition gives
-k dT(L) = h [T(L)-T=]
dx
~ -kc,= h [(c,L+C2)-r=J
Substituting C1 = - ~ and solving for C2 , one has
C =T + % + %L
2
= h
k
Substituting C1 and C2 in equation (a) gives
T(x)= T= +%[¾+ (L;x)]
Surface temperatures at x = 0 and x = L are evaluated to be
and
The thermal circuit representation is given in the schematic.
Now after putting specified data, the inside and outside surface temperatures are calculated to be
T, =50oc+1os~[
1
+ 0.02m ]
m2 500W/m2 °C 20W/m°C
1
=350°C
and
(Ans)
T- = so 0 c+(10s ~x
2
m2
=250°C
l
)
500W/m2 °C
(Ans)
Steady-State Heat Conduction-One Dimension
2.37
Alternatively
In the absence of internal heat generation, it is much easier to calculate the surface temperatures by
using the thermal resistance concept.
From the equivalent thermal circuit:
Ti-7;=%[f]
Ti - r_ = % [ ¾J
and
7; = T_ +(%/h) = 50°C+[10 W/ m /500W / m °C]
Hence,
5
2
2
= 2so c
0
Ti= J; +%LI K =250°C + [10 5 W/m2 x0.02m /20W /m2 °C]
and
(Ans)
=350°C
Example 2.19 ~ A steam-to-liquid heat exchanger surface of 3200 cm 2 face area is constructed of 0.7-cm
nickel with a 0.2-cm plating of copper on the steam side. The resistivity of a water scale
deposit on the steam side is 0.0017 m2 K/W and the steam and liquid surface conductances are 5465 W/m 2 K
and 580 W/m 2 K respectively. The heated steam is at 111 °C and the heated liquid is at 75°C. Calculate:
(a) Overall steam-to-liquid heat-transfer coefficient
(b) Temperature drop across the scale deposit
(c) Temperature at the copper-nickel interface
Take k(Copper) = 384 W/m K and k(Nickel) = 58 W/m K.
Known: Heat exchanger surface with copper and nickel plating on the steam side. Heat transfer coefficients
and fouling resistance on steam side.
Find: (a) Overall heat-transfer coefficient. (b) Temperature drop across scale deposit. (c) Temperature at
the Cu- Ni interface.
Schematic:
Steam
= 111 °c
Liquid
r_2=75°c
A=3200 cm2
r_,
CD
~
h 1= 5465 -W
m2K
Copper
Nickel
k I = 384
W/mK
k2=58W/mK
(
->+----+--------+-Q
Scale deposit
R, 1=0.0017m2K.JW
L =
1
-I 0.2cm
1-- L2=0.7cm - I
Fig. 2.25a
W
m2K
h2=580-
2.38 , Heat and Mass Transfer
Q
q =A -
T..2
T.. 1
.___,VV\.-------'VV\r---+---'\I\J'\r---__,\1\/\,_ _ _ _-'V\l\r--....
R.:onv, I
Rr1
R.:ond, I
R.:ond,2
Rconv,2
Fig. 2.2Sb
Assumptions: (1) Steady operating conditions. (2) One-dimensional conduction. (3) No fouling on liquid
side. (4) No thermal contact resistance between copper and nickel.
Analysis: Referring to the schematic, we have
Heat transfer rate,
Total thermal resistance per unit area:
L
~h =[ R.:onv,l +
J?n + R.:ondl + R.:ond2 + R.:onv2] ( m~K)
=_!_+Rt:+ Li+½.+_!_
'"1.k2hi
h,_
_
o.2x10-2 o.1x10-2
1
- - - +0. 0017 + - - - - + - - - - + 5465
384
58
580
= 0.003 733 W/m2 K
Overall steam-to-liquid heat transfer coefficient is
1
U= [ -Ll?ih
]-I
A
1
=---=267.88 W/m2 K
0.003733
(Ans) (a)
Heat-transfer rate per unit area,
q = Q = U ( T..1 - T..2 )
A
= (267.88 W/m2 K) (111 - 75) °C or K
= 9643.67 W/m2
Temperature drop across the scale deposit,
7; -7; =(QI A)J?n = (9643.67::) ( 0.0017 m~K)
= 16.4°C
(Ans) (b)
Temperature at the copper-nickel interface is
½ = T..2 + (R.:onv2 + R.:ond2) (Q f A)
= 75oc+[-l-+ 0.007] m2 K (9643.67)J!..._
580
58
W
m2
= 92.79°C
(Ans) (c)
Steady-State Heat Conduction-One Dimension
2.39
Else
½ =T=l -(QI A)[ l\onvl +Rn +Ri]
= l l 1°C-(9643.67W/m 2) [ -1-+0.0017+ 0.002 ] m 2 K
5465
384
W
= 92.79°c
Comment: Face area A was, in fact, not necessary in our calculation.
Example 2.20 ~ A composite wall is made up of asbestos and
bakelite sheets bolted together by four brass
bolts as shown in the adjoining sketch:
Determine the rate of heat flow across the wall. Asbestos plate
(150 mm x 150 mm) :Thickness= 6 mm, kA = 0.113 W/m K. Bakelite
plate ( 150 mm x 150 mm) : Thickness= 6 mm, k8 = 0.233 W/m K,
kc= 111 W/m K.
Brass bolt (4 Nos) : I 0-mm diameter x 12-mm long
Heat-transfer coefficient on the asbestos surface= 35 W/m 2 K
Heat-transfer coefficient on the bakelite surface= 15 W/m 2 K
Hot fluid (adjoining exposed surface A) temperature= 200°C
Cold fluid (adjoining exposed surface B) temperature= 50°C
Fig. 2.26
Known: A composite slab consisting of square plates of asbestos and bakelite exposed to convective
environments and fastened together by four brass bolts.
Find: Rate of heat flow through the composite system, Q[W]
Schematic:
®
®
J__
111
D= 10mm
h 1=35W/m2K
T
Four
bolts
r= 1=200°c
©
rrr
1--1--+-- LA= Ls= 6 mm
hi= 15W / m 2K
T= 2 = 50°C
[4:: = LA + 4i = 12 mm]
7t
2 =1t xO.Ol2m2
Ac =4 x -D
4
AA =Ag =A=0 .15m x0 .15m=0.0225m 2
Fig. 2.27a
2.40 , Heat and Mass Transfer
-
T=I
R1
T=2
Q ---'\/\1\r--
Rs
--~\l\/\r--- -
Q
Fig. 2.27b
Assumptions: (1) Steady-state, one-dimensional conduction. (2) No heat generation. (3) Isothermal
surfaces, constant properties and uniform convection coefficients.
Analysis: The heat-transfer rate across a composite wall is
Q = Ai:iverall
Rtotal
The thermal circuit is shown. We note that the resistance offered by the four bolts is in parallel with
the series combination of the two resistances due to asbestos and bakelite layers.
Besides, there are two convective resistances R 1 and R 2 •
Total thermal resistance is
Rtotal= Ri. + [
-]-l
l
+-1
RA +Rs
Re
+ R2 = Ri_ + Req + Rz
The individual resistances are now evaluated:
1
1
Ri_ = = - - - - - - - = 1.27°C/W
2
h1A 35W/m °Cx0.0225m2
1
1
R =-=-------=2.963°C/W
2
h2 A l5W!m2 °Cx0.0225m2
R = LA =
0.006m
= 2360CJW
A
kAA 0.113W/m°Cx0.0225m2
Rs = Ls =
0.006 m
= 1.144oc;w
k8 A 0.233W/m°Cx0.0225m2
Re,= _!:s;__ =
kcAc
Req
0.012m
= o.344oc;w
lllW/m°Cx1txO.Ol2m2
1
1
1
=---+-=----+--=3.192
RA +Rs Re (2.36+1.144) 0.344
Req = 0.3132°C/W
Therefore
Riotal = 1.27 + 2.963 + 0.3132 = 4.546°C/W
The heat flow rate is
(Ans)
Steady-State Heat Conduction-One Dimension
Example 2.21
2.41
~ A leading manufacturer of household appliances is proposing a self-cleaning oven design
that involves use of a composite window separating the oven cavity from the room air.
The composite is to consist of two high-temperature plastics (A and B) of thicknesses LA= 2 Ls and thermal
conductivities kA = 0.1 S W/m K and ks= 0.08 W/m K. During the self-cleaning process, the oven wall and air
temperatures, Tw and r., are 400°C, while the room air temperature T~ is 25°C. The inside convection and
radiation heat transfer coefficient h; and h,, as well as the outside convection coefficient h0 , are each approximately
25 W/m 2 K. What is the minimum window thickness, L =LA+ Ls, needed to ensure a temperature that is S0°C
or less at the outer surface of the window?
Known: Composite self-cleaning oven window with dimensions and properties of the materials and
conditions of operation.
Find: Composite window thickness for safe self-cleaning operation.
Schematic:
Ts;
Tw=400 °C
h,=2SW/m2K
Composite window
L=LA+Ls
Oven cavity
Air
kA=O. ISW/mK
A B ks=O.OSW/mK
iii
h;=2SW/m2K
T. =400 °C
Air
lll
h0 =2W/m 2 K
Too=25°C
Fig. 2.28a
Assumptions: ( l) Steady state one-dimensional conduction. (2) Constant properties. (3) Negligible contact
resistance. (4) Radiation exchange negligible.
Analysis: The thermal circuit for the heat transfer situation and the resistances involved are represented
as follows:
Tw
h,A
-
LA
kAA
T. =Tw
-
LB
h0 A
ksA
Too - Q
Ts,i
Ts,o
1/h;A
Tw
Fig. 2.28b
The required window thickness may be obtained by applying an energy balance at the outer surface
of the window which is prescribed.
2.42 , Heat and Mass Transfer
The total thermal resistance between the oven cavity and the outer surface of the window includes an
effective resistance associated with convection and radiation, which act in parallel at the inner surface of
the window, and the conduction resistances of the window materials.
Hence
But
Substituting into the energy balance, it follows that
Solving for LA> we have
[
1
1 ]
T,,, -T.o
1
LA kA +2kB =h0 (T., 0 -·T=)-(hi+hr)
or
or
L (m) [-1-+
'A
1 ] mK _
(400-50)°C
0.15 2x0.08 W - (25W/m2 K)(50-25)°C
1
m2 K
(25+25) W
12.917 LA= 0.56 - 0.02 = 0.54
L
'A
= 0.5 4 m =~mm
12.917
= 0.0418 m
12.917
or 41.8 mm
LA
Ls =-=20.9mm
2
Therefore, total thickness of the composite window,
L = LA +Ls= 41.8+20.9 = 62.7 mm
(Ans)
Steady-State Heat Conduction-One Dimension
2.43
Example 2.22 ~ The inside temperature of a furnace wall, 200-mm thick, is I 350°C. The mean thermal
conductivity of the wall material is 1.35 W/m 0 C. The heat-transfer coefficient of the outside
surface is a function of temperature difference and is given by
h = 7.85 + 0.08 L1T
where '1T is the temperature difference between outside wall surface and surroundings. Determine the rate of
heat transfer per unit area if the surroundings temperature is 40°C
Known: A furnace wall with its outer surface exposed to convection environment.
Find: Heat-transfer rate per unit area,
A(W I m2 ).
Q/
f- L=0.2m
Schematic:
--,
T1=1350 °C
T.. =40°C
Furnace wall
T2
k= I .35W/m °C
h=7.85+0.08 (T2- T.. )
Q/A
QIA-
r,
r..
T2
R1=L/K
lll
R2= 1/h
Fig. 2.29
Assumptions: ( l) Steady operating conditions. (2) One-dimensional conduction without heat generation.
(3) Constant properties.
Analysis: Heat-transfer rate per unit area,
(T;-T.. )/h
Q
A
T. -T = - - - ~ 2
(L/k)+(l/h)
(Llk)+(l/h)
00
Heat-transfer coefficient,
h = 7.85+0.08(7;-T.. )
= 7.85+0.08(7i-T.. >j[1+ \L]
= 7.85 + 0.08(1350-40)
1+(~)h
1.35
2.44 , Heat and Mass Transfer
h+0.14815 h2 =7.85+(7.85)(0.14815)h+(0.08)(1310)
or
0.14815 h2 -0.16296 h-112.65=0
or
~
h2 -l.l h-760.4=0
Solving this quadratic equation, we get
-(-1.1) + ~(-1.1) 2 - 4(1) (-760.4)
h = - - - - - - - - - - - = 2 8 . 1 3 W/m 2 °C
2
[Fo,ax'+bx+c=O, x= -b±~:a-•ac]
It follows that
Q
A
=
1350-40
[C~;
2
5)+Cs~13)]
= 7131.3 W/m2
(Ans)
(E) THERMAL CONTACT RESISTANCE
•
•
Example 2.23 ~ Two instrumented duralumin (k = 165 W/m 0 C) bars are placed in an apparatus to measure
contact resistance. The bars are 6 cm 2 and are oriented as shown in Figure 2.45.
c~I
pressure
r
r,
-kA
f
r2
} I\ r3l
rt
-ke
rt
l
r4
l~ct
pressure
Fig. 2.30
The temperatures r, = 60°C and r2= 55°C were measured 5-cm apart as were r3= 50°C and r4= 45°C. The
temperatures T{ and lj* were measured 0.5 cm either side of the contact plane of the two bars. Determine
the contact resistance under these conditions.
Known: Two duralumin bars placed in contact under specified conditions.
Find: Contact resistance, R,,c (m 2 °C/W).
Schematic:
Magnified view of contact cross-section
T
Fig. 2.Jla
Steady-State Heat Conduction-One Dimension
T1
I
i~
T2
i~ i
T3
T4
j
j
2.45
I
Contact
®
q,
Contact
pressu~ ~--k_A_ _ _ _ _ _~+-+--®-B_k_B_ _ _~ pressure
ri
r;
Fig. 2.Jlb
T(°C)~----------------~
(60) T1
(55) T2
r;
(50.5) r;
(54.5)
(50) T3
/B
Fig. 2.Jlc
Assumptions: ( l) Steady-state conditions. (2) Constant thermal conductivity.
Analysis: From energy balance:
l
where he is unit area contact conductance (W/m 2 °C) and thermal contact resistance, l?i,c = h.
C
Hence,
kA (z;-z;)
h =---c
LA (r:2 -T-*)
3
Also
z;• = 54.5 °C and y;• = 50.5 °C
h = 165 W/ m°C
C
0.05m
(by graphical extrapolation)
{60-55)°C
(54.5-50.5)°C
= 4125 W/m 2 °C
:. thermal contact resistance,
R, c = 2.424 x 10-4 m2 °C/W
(Ans)
2.46 ,
Heat and Mass Transfer
Example 2.24 ~ Heat losses from the windows are to be reduced by covering them from inside with a
polystyrene insulation (kins= 0.027 W/m K). Consider application of 25-mm thick insulation
panels to 6-mm thick windows (ky, = 1.4 W/m K). The contact resistance between the glass and the insulation
may be approximated as (l\.c = 0.02 m1 K/W), while the convection coefficient at the outside surface of the
window is nominally losing heat (h 0 = 20 W/m 1 K). With the insulation, the convection coefficient at the inner
surface is h; = 2 W/m1 K, without the insulation it is h; = S W/m 1 K.
(a) What is the percentage reduction in heat loss associated with the use of insulation?
(b) If the total surface area of the windows is A5 = 12 m1, what are the heat losses associated with insulated
and uninsulated windows for interior and exterior temperatures of T-.; = 20°C and T-.o = - I 2°C?
(c) If the home is heated by gas furnace operating at an efficiency of llt = 0.80 and the natural gas is priced at
Cg= Re I per MJ, what is the daily saving associated with covering windows for 12 hours?
Known: Temperatures and convection coefficients associated with fluids at inner and outer surfaces of a
composite wall. Contact resistance, dimensions, and thermal conductivities associated with wall materials.
Find: (a) Percentage reduction in heat loss due to insulation
(b) Heat loss with and without insulation
(c) Daily saving due to insulation.
Schematic:
Rt,c = 0.02 m2 K/W
A5 =12m 2
®
®
Insulation
Window
(Glass)
kA =0.027W/m K
ks= 1.4
W/mK
T=,;=20°C
h-,,w =2W/m2K
h-1,wo =SW/m2K
T=,o =-12°C
h0 =20W/m 2 K
-+-------------------<--Q
f---
LA= 25 mm
---j Ls= 6 mm f--
Fig. 2.32
Assumptions: ( l) Steady-state conditions. (2) One-dimensional heat transfer. (3) Negligible radiation.
(4) Constant properties.
Analysis: (a) Heat loss with insulation,
Q = t:.Toverall =
w
R,01,w
(r -T )K
00
,1
00
,0
[ Rconv,i+Rcond,A +Rcond,B+R,,c+Rconv,o ] {K/W)
Heat loss without insulation,
Q.
wo
= t:.Toverall =
R tot,wo
Percentage reduction in heat loss
(T= ,1 -T=,o
)K
J
[ Rconv,i+ Rcond,B+ Rconv,o {K/W)
Steady-State Heat Conduction-One Dimension
2.47
Total thermal resistance without insulation is
Rtot,wo
=
~ [f--+: +; ]
S
8
1,WO
0
=-1 [.!..+ 6xl0-3 +_!_]=0.2543/As
A. 5
1.4
20
Total thermal resistance with insulation is
R
tot,w
=_!_[_l_+LA+R +LB+_!_]
A
h,
1,w
s
k
A
t,c
k
B
l,
"o
25xl0-3
6x10-3 1]
=1- [1-+---+0.02+---+A. 2
0.027
1.4
20
= 1.500/As
:. percentage reduction in heat transfer
= 1- 0 ·2543 / /4 = 0.8305
1.500 I A.
(Ans) (a)
=83.05%
(b)
AJ;,verall =[20-(-12)]°C or K =32 K
A.= 12 m 2
Q = 32x12 = 256 W
w
and
Q.
WO
1.500
= 32 X 12 = 1510 W
0.2543
(Ans) (b)
(Ans) (b)
(c) Reduction in heat loss = Qwo - Qw
=(15I0-265)w = 1254 wl21 3600 sl
lWs 1 h
=4.515x106 J/h
Heat supplied by the gas furnace of efficiency TJr= 0.80 to compensate for this heat loss
= 4.515xl06 J/h = 5.6435MJ/h
0.80
Daily saving by covering the windows with insulation for 12 hours is
=5.6435MJl12 hi Re 11
h lday MJ
= Rs 67.72
(Ans)(c)
2.48 , Heat and Mass Transfer
(F) VARIABLE AREA OF CROSS-SECTION
•
•
Example 2.25 ~ A truncated solid cone of circular cross-section has its diameter related to axial coordinate by
D = ax312 where a= I m - 112 and the distance xis measured from the apex. At x 1 = 100 mm,
T1 = 250°C and at x2 = 250 mm, T2 = S0°C. The sides are well insulated. Assume one-dimensional model and
k = 43 W/m K. Calculate the temperature distribution and the rate of heat transfer.
Known: Geometry and surface conditions of a truncated solid cone.
Find: Temperature distribution and rate of heat transfer.
Schematic:
X
x, = 100mm
_l_
;
x2=250mm
i
;
i
I
!
k=43!W/m
K
;
i
Fig. 2.lla
Assumptions: ( 1) Steady-state conditions. (2) One-dimensional (axial) conduction. (3) Constant thermal
conductivity.
Analysis: (a) From Fourier's law
.
dT
1t
dT
Q = -k Ac(x)- = -k-D 2 (x)dx
4
dx
= - 1tk ( a x312
4
)2 dT = - 1t ka2 x3 d T
dx
4dx
Separating the variables and integrating,
4Q x dx
T
-f---fdT
1tka2 x 3
I;
x,
or
or
]x
4Q [x-3+1
- - - - =I;-T
1tka2 -2 x,
Steady-State Heat Conduction-One Dimension
2.49
Therefore,
T(x)
=Ji-~[_!__
__!__]
rck a xf x
2
(Ans) (a)
2
(b) Replacing x with x 2 and T with T2 one gets
It follows that
. rcka 2
Q=
(z;-i;)
2 [(llx?)-(llxi}]
Substituting values, the heat rate is
.
rcx43W/mKx(lm-li 2}2x(250-50)°C or
K
Q=----------2x[(o. t21m2 )-(0.25~m2 )]
.-. 150 + -- ~ ,- - - + - - - - - + - - - - - - - ,
u
._.
I- I00 +-- - - - - + - -' ~ - - + - - - - - . . . ,
0
= 160.8 W
(Ans) (b)
Substituting for T1, Q, k, a and x 1 in the expression
for T(x), we have
IT(x) = 250-2 .38[ 100-x-2
0+-----+-----+--------,
0.1
0.15
0.2
0.25
x(m)
]I
The temperature profile is sketched below.
Fig. 2.llb
Example 2.26 ~ A tapered stainless steel rod (k= 17 W/m 0 C), 20-cm long, with end diameters of 12cm
and 6cm has its lateral surface effectively insulated. The smaller end surface is maintained
at 50°C and the rate of heat loss is 75 W. Derive an expression for temperature distribution and sketch the
temperature profile.
Calculate
(a) the temperature of the larger end surface,
(b) the temperature of the rod at 12.5cm from the smaller end, and
(c) the temperature gradient at the two extreme ends of the rod.
Known: Heat loss from a tapered rod of given dimensions with adiabatic curved surface.
dT
Find: (a) T1( 0 C); (b) T(x=7.5cm) (°C); (c) at x=O and x=L. Temperature distribution.
dx
2.50 , Heat and Mass Transfer
Schematic:
Q-
D 1= 12 cm ------------ -StaTnTess-ste-efrocf___ - D2 = 6cm -
1
(k=l7W/m C)
0
Q= 75 W
.1
_
1
L x
Insulated
x=O
T=T1
x = L = 20 cm
T=T2=S0°C
Fig. 2.34a
Assumptions: (l) Steady operating conditions. (2) One-dimensional (axial) conduction without heat
generation. (3) Constant thermal conductivity.
Analysis: Energy balance
0
~
Ein-Eout+
no heat generation
'
steady state
or
Consider a differential element of thickness dx at a distance x from the larger end. It follows that
!2cond,x
.
But,
Qcond,x+dx
.
= !2cond,x+dx
.
d [.
J
= Qcond,x
+ dx Qcond ,x dx
.
d ( .
Qcond,x+dx -Qcond,x = dx Qcond,x
)
dx = 0
Area of cross-section, in this case is a function of x and is given by
1t
Ac ( X) = - D 2 ( X)
4
dT] dx=O
-d [ -kA,,(x)dx
dx
or
Integrating,
dT
-kA,, (x)-=C1
dx
1t
dT
-k-D 2 (x)-=C
4
dx
I
or
where
C1 = Q =constant= 75 W
To express D(x) in terms of x, let us consider similar triangles OAB and OEF.
OE
OA
EF
AB
Steady-State Heat Conduction-One Dimension
L-x
2.5 I
[D(x)-Di)/2
or
L = [D, - D2 ]12
or
D(x) = D2 +(D1 -n2
{1-f]
= D1 -~(D1 - DJ
L
X
= 0.12 m---(0.12-0.06)m
0.2m
= 0.12-0.3x[m]
4C1 dx
dT=---rck D(x) 2
I dx
JdT-- 4C
rck (0.12-0.3x)
1
or
2
Let (0.12 - 0.3 x) be equal to t.
Then
And
or
-0.3 dx = dt
4C1 J dt
4C1 1
T(x) = - rck -0.3t2 =- rck 0.3t +C2
4 QI0.3
+C2
rck[0.12-0.3x]
T(x) = -
- - -4Q
- - - + C2
0.3x0.3rck(0.4-x)
4x75W
T(x) = - - - - - - - - - - + C2
0.09rcxl 7W/m°C[0.4- x]m
The temperature distribution is given by
T(x) = - 62.414(0.4-x)- 1 +C2
To find C2, the boundary condition: T(x = L = 0.2 m) = 50°C can be applied.
C2 = 50 + 62.414(0.4 - 0.2)- 1
or
C2 = 362.07
T(x) = 362.07- 62 •414
(0.4-x)
(Ans)
(a) Temperature of larger end surface is
T(x = 0) = T. = 362.07 - 62 .4 14
I
(0.4-0)
=206°C
(Ans) (a)
2.52 ,
Heat and Mass Transfer
(b) Temperature at 12.5cm from smaller end, i.e., at x=(20-12 .5)cm=7.5cm is
T(x = 0.075m) = 362.07-
62 .4 14
(0.4-0.075)
= 110°c
(Ans) (b)
(c) Temperature gradient,
250
4C1
-=----'---=
dT
dx
1C k D ( x )2
4x75
1C X 17 X D 2
206
170
.-. 150
=-5.617 / D 2
At
-----------
u
._.
x=O,
---- ~!'-......
fT(x)
0
~'
I- 100
dT =- 5.617 =- 390 , 10 C/m
dx
0.12 2
and,
(Ans) (c)
50
~
at x =L,
0
dT =- 5·617 =-1560.3°C/m (Ans) (c)
dx
0.06 2
0
0.05 0.075 0.1
0. 15
0.2
X (m)
Fig. 2.34b
The temperature profile is sketched below:
Example 2.27 ~ Develop an expression for the steady state heat flow rate through an insulated rod sketched
below:
_ _____. Q
.,.j I-____
(
Insulated
T
--'-------------' l
k_1 __ _
r--
L1
Thermal contact
conductance, he
Fig. 2.35
Known: An insulated composite rod of prescribed dimensions, configuration and thermal conductivities.
Contact resistance. Convective processes at the two ends.
Find: Expression for heat-flow rate, Q.
Schematic:
...---L1--~------L2-----+-1
111
Fig. 2.36a
Steady-State Heat Conduction-One Dimension
2.53
r~.1
I
heh
Fig. 2.36b
Assumptions: ( l) Steady-state, one-dimensional heat conduction. (2) No internal heat generation.
(3) Constant properties.
Analysis: Applying energy balance to the rod as control volume, we note that
i i
=
Ein-Eout+
No heat
generation
£in = £out
or
Steady state
Qin = !2out
Q = const.
The lateral (curved) surface of the rod is insulated. Hence, the heat transfer is one-dimensional (axial).
However, one part of the rod is tapered (of varying area of cross-section) and the other of uniform crosssectional area.
The thermal circuit is shown in the schematic. And, the heat flow rate in the axial direction is
Q = Ll T,,verall
Rtotal
T~, - T~,
I.Rth
Let us first identify and evaluated the different thermal resistances.
(J) Convective resistance at the small end of the rod
_1t 2
l
With
Ai - -Di, Rconv i - 4
'
hiAi_
4
1th,D?
---
(2) Conduction resistance for the segment of the rod with variable area of cross-section
Invoking Fourier's law of heat conduction,
.
dT
Q=-kiA(x)dx
1tD2
dT
=-ki-4-dx
where D is a function of x while k 1 is constant.
Then, separating the variables and integrating between
A(x)
Ts
r--~
x~~ID
~ T
D1
__L
4Q dx =-dT
1tki D2
Lx
From the given geometry, we note that
x-0
(D-Di)12
x
½-0
(D2 -Di)12
½
- - = ~---- ~ -
dx=
½
D2-Di
[dD-0]
D-Di
= ---
D2 -Di
dx
x=O, T=T1
-----Q
Fig. 2.36c
1
2.54 , Heat and Mass Transfer
It follows that
or
or
or
or
Thermal conduction resistance,
~D2~D2
4
4
I
2
(3) Conduction resistance for the segment of the rod with constant area of cross-section
R
-~-~
"2A2 - re kiD}
cond 2 -
(4) Convection resistance at the large end
R
=-1-=
4
re h2Di
L1
L2
h2A2
conv,2
Therefore, the total thermal resistance is
R,otal
=-
I
h1A1
+
I
,.-;-;- + - - + - -
k1 \J A1A2
k 2A2 h2A2
4LI
4
4L2
4
=--+-~-+--+-re h,D? re k 1D1D2 re k 2D} re h2 D}
and
Heat transfer rate, Q =
4
[
T -T
I
~,
L
~,
L
I
]
1
2
- - -+- -+-+re fZiD? k1D1D2 "2 D} h2 D}
or
(Ans)
Steady-State Heat Conduction-One Dimension
2.55
(G) VARIABLE THERMAL CONDUCTIVITY
•
•
Example 2.28 ~ A plane wall built from fire bricks is 200-mm thick. The temperatures of the inner and outer
surfaces of the wall are 90()<>C and I00°C respectively. The thermal conductivity of the wall
material varies linearly with temperature as k = k0 ( I + aT), where k0 is 0.8 W/m K and a is 0.0005 per 0 C.
Determine the temperatures at points 50 and 100mm away from the higher temperature surface. What would be
the error in the temperatures calculated on the basis of an average thermal conductivity?
Known: A plane wall with variable thermal conductivity and end surface temperatures.
Find: T(x = 50mm) and T(x = 100mm), error in temperature computed based on km.
Schematic:
T(x = 0.05 m) !
T(x=10.I m)
,---+--- k = k 0 [ I + a T]
=0.8[1 +0.0005T] (W/m C)
0
L-.x
---L=Q2m---·I
Fig. 2.37
Assumptions: (I) Steady-state conditions. (2) Isothermal inner and outer surfaces. (3) One-dimensional
conduction.
Analysis: According to Fourier's law of heat conduction,
Q = -k(T)A dT
dx
Steady-state heat flux,
Q
dT
q =A= -ko (I+cxT) dx
qdx = -k0 (I+aT)dT
or
Integrating between limits x = 0, T= T 1 and x = L, T= T2,
L
f
q dx = -k0
0
or
T2
f(I+aT)dT
r.
q L = ko [ ( I; - I;) +
%(7;2 - 7;2 )]
2.56 , Heat and Mass Transfer
f ( 1i { a ( i;: 1i)l
or
q=
or
q = ko (7; - Ji ) (1+ a Tm )
L
1+
i; -
(A)
Integration between limits x = 0, T= T1 and x = x, T= T yields
q=
~ [ ( 7; - T )+ %(7;2 - T2 )]
(B)
Equating (A) and (B),
x
a~1 aP
-(J;-Ji)(I+aTm)=(J;-T)+--
L
2
2
2
T
-a 2T- + T + { x 7; - Ji) (1+ a Tm)- 7; ( 1+ a T. )} = 0
or
L(
This is a quadratic equation like a x2+bx+ c = 0
where a =
f, b
= l, c = {
Z(7; - Ji) (1+ aTm)- 7; ( 1+ a; )}
The solution is
-b±~b2-4ac
x=-~--2a
Therefore, taking only the positive sign, the temperature distribution in the wall is given by
-1+ 1-4(%){z(Ti-Ji) (l+aTm)-7; (1+
«:)}
T=---------------2x(a/2)
X
= - -1 + [ - 1 + 2 a x - 1 { T. + -a T. 2 - -(T.
a
a2
a2 I 2 I L I - r:2 ) (1 + a 1'.m )
1
2
= _ _!_+ ( -T, )(1 + a'l'. )
a
a2 + 3_
a T,I + T.I ) _3_
a X ~(T.
L I 2
m
1
1) -i(Ti-J;)(l+aTm)
2
T(x)=-«+VI/1( J;+a
With
2
Tm = _!_(900 + 100) = 500°C
2
}]1/
2
Steady-State Heat Conduction-One Dimension
2.57
L = 0.2 m, a= 0.0005°C- 1, T1 = 900°C and T2 = 100°C, one has
1 (900+().()()()5
1) - {2
0.0005 x 0.2 X(900-100)(1+0.0005x500)
2
X
T(x) =- 0.0005 +
}
= -2000+~8.41 X 106 -20 X 106 X
= -2000 + 1000~8.41- 20 X
+ For x = 0.05 m,
T(x = 50 mm)= -2000+ 1000~8.41-(20x0.05)
= 122.1°c
(Ans)
+ For x= 0.1 m,
T(x = 100mm) = -2000+ 1000 ~8.41-(20x0.1)
= 531.8°C
(Ans)
Average (mean) thermal conductivity,
km= k0 (l+aTm)
= 0.8(1+ 0.0005 x500) = 1.0 W/m °C
Heat flux,
km(J;-T)
km(Ti-J;)
L
q=---=--X
~(J;-J;)=J;-T
or
L
T(x) = J; -~(J; -7;)
L
.
x
50 mm
200 mm
+ With -=---=0.25,
L
T(50mm) = 900-0.25(900-100)
= 100°c
Error = ( 722 ·l-?OO) (100) = 3.06%
722.1
(Ans)
+ With~= lOOmm = 0.5,
L
200 mm
T(lOO mm)= 900- 0.5 x 800 = 500°C
Error= ( 53 1. 8 - 500 )(100) = 5.98%
531.8
(Ans)
2.58 , Heat and Mass Transfer
Example 2.29 ~ A composite slab has two layers of 5-cm and I0-cm thickness. The thermal conductivities of
the materials of these layers are temperature-dependent and are prescribed by the relations:
k 1 = 0.05 (I + 0.006 T) WI m °C
k2 = 0.04 (I + 0.007 T) WI m °C
where T = temperature in degree centigrade. The inside and outside surface temperatures of the slab are
maintained at 500°C and 200°C. Determine the steady-state heat flux through the composite, and the interface
temperature.
Known: A composite slab comprises two materials of variable thermal conductivity.
Find: Steady-state heat flux, q (W/m2); Interface temperature, ~{ 0 C).
Schematic:
0
CD
0°C I
q (Wlm2)
T; ( C) = 1
0
8
8
, T2= 200°c
Lx
Fig. 2.38
Assumptions: (1) Steady-state, one-dimensional heat conduction. (2) Isothermal surfaces. (3) Temperaturedependent thermal conductivity.
Analysis: Fourier's rate equation is
Q. =-kA dT
C dx
Q
dT
dT
Steady-state heat flux, -=q=-k(T)-=-k0 (1+aT)Ac
dx
dx
qd x = -ko (1 +aT)dT
or
For Slab (1)
q
x=0.05m
T=T,
x=O
T=500°C
f d x = -0.05 f (1 +0.006T)dT
500°C
or
q= 0.0 5
0.05
f (l+0.006T)dT
T;
Steady-State Heat Conduction-One Dimension
q = [ T + 0.~0 6 T2
or
Io
= (500-7;) + 0.003(500 2 - 7;2)
lq = -0.003 7;2 -7; + 12501
or
2.59
(a)
For Slab (2)
x=0.15
q
T=200°C
J dx = -0.04 J
or
(l+0.007T)dT
T=T,
x=0.05
J
q = 0 ·04 (l+0.007T)dT
O.lO 200
= o.4[r+ o.001 r2Jr.
2
200
= 0.4(7;-200)+ 0.4 x0.007(7;2 -2002)
2
lq = + 0.0014 7;2 +0.47; -1361
or
(b)
Equating (a) and (b), one gets
0.0014 7;2 +0.47; -136 = -0.003 7;2 -7; + 1250
or
or
0.0044 7;2 + 1.47; -1386 = 0
11;2 + 318.187; - 315 ooo = ol
Solving this quadratic equation, the interface temperature is
I Ti = 424.27°C I
(Ans)
Substituting this value of Ji in either (a) or (b), one finds steady-state heat flux,
q = 0.0014( 424.27) 2 + 0.4( 424.27)-136
lq = 285.7W / m I
2
(Ans)
Example 2.30 ~ A steam boiler furnace is made of a layer of 12.5-cm thick fire clay and a layer of red
50-cm thick brick. If the wall temperature inside the boiler furnace is I 100°C and that on
the outside wall is 50°C, determine the amount of heat loss per m2 of the furnace wall.
It is desired to reduce the thickness of the red brick layer in this furnace to half by filling in the space
between the two layers by diatomite whose k = [0.113 + 0.00023 T] W/m K. Calculate the thickness of the
filling to ensure an identical loss of heat for the same outside and inside temperatures.
Assume: k (for fire clay)= 0.533 W/m K
k (for red brick)= 0.7 W/m K
Known: Boiler furnace comprises fireclay and red brick layers bounded by fixed surface temperatures.
Find: Heat flux. Thickness of diatomite layer sandwiched between fireclay and red brick layer required
to reduce red brick layer thickness by half for the same heat flux.
2.60 , Heat and Mass Transfer
Schematic:
T;= I I00 °C
r-
L1 =
12.5cm
Case I
Fig. 2.39a
T;= I I00 °C
Fire clay
T0 =50°C
k1 =0.533
-+----1'7,7,'1777.~?57.:'1777.ffl,;~e,e~~e-e,e-i~ q
Case II
Fig. 2.39b
T;
-
T1
T0
T;
q ~
T1
T2
T0
--'\JV\r-----'\N\,--+---'V\1\,--- -
L1 I k 1
q
*
L2lkm
'--v---'
'--v---'
Fire clay
Diatomite
Fig. 2.39c
Assumptions: (1) Steady operating conditions. (2) One-dimensional conduction.
Q
Analysis: Case I : Heat flux, q = - =
A
t:.T
overall
I.Rth
Q
I;-T;,
q=A= LI ½
- +kl
k2
100-50)°C or
[(~:~!!)+(0/10)] W/:K
(l
K
=---------q = 1106.65 W / m2
(Ans)
2.61
Steady-State Heat Conduction-One Dimension
Case II
L 1 = 12.5cm
L 2 = ?,
L3 = 25cm.
1106.65 = ll00-7;
( 0.125)
0.533
T. =ll00-(1106.65x0.125)
0.533
I
= 840.47°C
(Ans)
As the heat loss of 1106.65 W/m2 remains the same, the temperature at the interface between the two
layers of diatomite and red brick can be found as follows.
Thus
J: -50
2
1106.65 = (
)
0.25
0.7
1i = 5o+(l 106.65x0.25) = 445.23cc
0.7
Average thermal conductivity of diatomite (km) at the average temperature of
km=~= 0.113+ 0.00023
T.+J:) would be
(T
[S40.47~44523 ] = 0.261 W/mK
To find L;, we can write
1106.65 = (1100- 445.2.3)
(~:~~~)+( 0~~1)
L* = 0.261[(1100-445.23)-(0.125)] = 0.0932m
or
2
1106.65
0.533
or 9.32cm
Diatomite layer thickness is
C,,=9.32cm
(Ans)
(H) LONG CYLINDERS BOUNDED BY FIXED SURFACE TEMPERATURES
Example 2.31
~ A wire whose resistance per cm length is 0.1 ohm is embedded along the axis of a cylindrical
cement tube of 0.06-cm radii and 1.2cm. A current of 6 A flows in the wire to keep a
steady temperature difference of I 30°C between the inner and outer surfaces. Calculate the conductivity of
cement and the amount of heat supplied per cm length.
Known: Heat is supplied electrically to a hollow cement cylindrical tube.
2.62 , Heat and Mass Transfer
Find: Thermal conductivity of cement.
Schematic:
Cylindrical cement tube
·- ·- · - · - · -·- ·- ·- · - · - · - ,-·~,,--
Wire
.'
' '
'' ''
' ' ' -~
·-·--- -- ·- ·- ·- ·- ·- ·- ·- ·-·-·- ·- ·- ·-·-·,-·
·-·
'
'I
'
k=?
L
Fig. 2.40
Assumptions: (I) Steady-state conditions. (2) One-dimensional (radial) conduction. (3) Constant thermal
conductivity.
Analysis: Re= 0.1 Q/cm; T;- T0 = 130°C,
ri=0 .06cm; r0 = 1.2cm, 1=6 A,
Heat generated in wire = Heat supplied to the hollow cement cylinder
Heat supplied per cm length,
~ =/2 ( :·) = {6)2 {0.1)= 3.6 W
(Ans)
Also, the heat rate for a hollow cylinder is given by
. 21tkL(I;-I;,)
Q=-~-'------'-
ln(,;, 1,;)
Thermal conductivity of cement,
k = (2 In(,;, I ,;)
L21t(I;-I;,)
= ( 3 _6 "J!.._) /n{l.2 / 0.06) [ 100 cm]
cm 21t{l30)K
Im
= 1.32 W /mK
(Ans)
Example 2.32 ~ A hollow cone of inconel (k = 15 W/m K), 0.5-mm thick connects a 3-cm pipe in diameter with
the exterior metal sheath of an insulated vessel. The base of the cone is IO cm in diameter and its
length measured along the cone is 12 cm. If the pipe is at 2()()<>C and the base of the cone at OOC, what is the rate of
heat loss through the cone? If the pipe were connected direcdy with the outer sheath, what will be the heat-loss rate?
Known: A hollow cone connects a pipe with the external metal sheath of an insulated vessel.
Find: Heat-loss rate.
Steady-State Heat Conduction-One Dimension
Schematic:
2.63
Hollow cone as sector of circle
k= ISW/m K
T0 =0°c--Base of cone
I'.----Thickness, L = 0.5 mm
p.. .fraction of a circle
Fig. 2.41
Assumptions: ( l) Steady-state, one-dimensional (radial) conduction. (2) Constant thermal conductivity.
Analysis: The hollow cone is frequently used to connect the outlet pipe of a vessel containing very hot
or very cold liquids with a base or surface at the room temperature. It can be looked upon as made from
the sector of a circle.
One can use the expression for radial heat conduction through a cylinder with some modification:
.
21tpkL(I;-T,,)
Q=---/n (,;, /1;)
(1)
where L is the cone thickness and the radii r; and r0 are to be found as follows. Let p represent the
fraction of the circle as the cone is equivalent to sector of a circle.
21t r; p = 1t x 3 cm
and
21t r0 p = 1t x 10cm
Also
10
- r. - r.
3
I
I
= 12
12 x3
"=--=5.14cm
I
7
or
and
Now
p = ~ = J.2_ = 0.292
21t,; 5.14
2.64 , Heat and Mass Transfer
Substitution of these values in Eq. (I) gives
.
2rcpkL(T;-I;,)
Q=----'-----'ln(,;, !,; )
= (2rc)(0.292)(15 W/m K)(0.5 x 10-3 m)(200-0)K = 2 3
17.14
(Ans)
' W
/n - -
5.14
If the pipe is directly connected with the outer sheath as the centre of a 10-cm diameter circle of the
same metal, 0.5 mm thick, assuming the circumference of the circle at 0°C, the heat-loss rate would be
. (2rc)(l5W/m K)(0.5xl0-3 m)(200K)
Q = - - - - - - - - - - - = 7.83W
In 10
3
(Ans)
This shows the efficacy of the hollow cone in reducing the heat loss.
(I) LONG CYLINDER BOUNDED BY FIXED FLUID TEMPERATURES
Example 2.33
~ A long. hollow cylinder with constant thermal conductivity (k = 77 W/m C) has its inner surface
0
(r; = 4cm) subjected to constant heat flux qi = I 05 W/m2 while the heat is dissipated from the outer
surface (r0 = 8cm) by convection into a fluid at T- = 50°C with a heat transfer coefficient h = 200 W/m 2 0 C. Develop an
expression for the steady state temperature distribution in the cylinder and calculate the surface temperatures T; and T0 •
Known: A hollow cylinder with prescribed heat flux at its inner surface and outer surface exposed to
convection condition.
Find: Temperature distribution, T(r). Surface temperatures T. and T0 .
Schematic:
q;= I05W/m2}
r;=0.04m
r0 =0.08m
Fluid
h=200W/m 2 °C
T==S0°C
Q=q;(21tr;L)
-
T;
T0
T=
~
/n(r0 /r;)
I
21tkL
h(21tr0 L)
Fig. 2.42
Assumptions: (I) Steady-state, one-dimensional (radial) conduction. (2) No internal heat generation.
(3) Constant thermal conductivity. (4) Negligible radiation effects.
Analysis: The governing differential equation is
!~(krdT)=o
r dr
dr
Steady-State Heat Conduction-One Dimension
2.65
Successive integration twice gives
~ dT = ..s_
k r dT = c,
dr
and
dr
kr
C
T(r)=_!_ zn r+C2
(a)
k
where C1 and C2 are arbitrary integration constants.
The boundary conditions are
BCI:
At r= ri,
i.e.,
BCII
[_s_]
or
h [ c, Zn r + C -T ] = - k
k
O
2
=
k,;,
or
C =T ----Zn r
2
= h,;, k
0
c, c,
With
Substituting for C1 and C2 in equation (a),
(Ans)
Surface temperatures are determined from
(r )
q1. r-.1 q.1 r-.1
(
) =I;, =T-+Tr=,;,
+-- Zn ..E..
h,;,
k
,;,
(10 5 W/m2 )(0.04 m)
=5 0 ° C + - - - - - - - (200 W/m2 0 c){0.08 m)
= 300°C
(Ans)
2.66 , Heat and Mass Transfer
(10 5 W/m2)(0.04 m) (0.08 m)
=300°C+------~ln - (77 W/m 0 c)
0.04 m
= 336°C
(Ans)
Alternatively
We can solve this problem by using the thermal resistance concept because no heat generation is involved.
From the thermal circuit shown in the schematic, we can write
In(,;,/,;)
T-T = q. (2rcr, L ) x - ~ ~
1
o
1
1
2rckL
and,
Therefore
T =T +
0
-
(qi,;)
h,;,
=50oc+ 10sw;m2 x(4cm)
200W/m2 °C
8 cm
= 300°C
and
(Ans)
(r )
q
T = T + - 1.- r,1 In ...£.
1
0
k
,;
(105 W/m2){0.04 m) (8 cm)
=300°C+-'------'--'---'-ln
-77W/m °C
4 cm
=336°C
(Ans)
0) COMPOSITE CYLINDERS BOUNDED BY FIXED BOUNDARY TEMPERATURES
Example 2.34 ~ A steam pipe, 170/ I 60 mm in diameter, is covered with two layers of insulation. The
thickness o1 of the first layer is 30mm and that of the second layer o,_ is 50mm. The thermal
conductivities k 1, k2 and k3 of the pipe and the insulating layers are 50, 0.15 and 0.08 W/m K, respectively. The
temperature of the inner surface of the steam pipe is 300°C and that of the outer surface of the insulation layer
is 50°C. Determine the quantity of heat lost per metre length of steam pipe and the layer contact temperatures.
Steady-State Heat Conduction-One Dimension
2.6 7
Known: A steam pipe is equipped with two layers of insulation.
Find: Rate of heat loss per metre length. Interface temperatures.
Schematic:
r4
·- - -
r1 =80mm
k1=SOW/mK
r2=8Smm
k2=0.ISW/mK
r3 =115mm k3 = 0.08 W/m K
- ·-
T0 = S0°C r4= 165mm
T;=300 °C
Fig. 2.43a
T;
T1
To
T2
R2
To---- - -
R3
Fig. 2.43b
Assumptions: (I) Steady operating conditions exist. (2) One-dimensional conduction. (3) No contact
resistance. (4) Constant properties.
Analysis: Steady heat-transfer rate through the three-layered composite cylinder of length L is expressed as
Q = _~_T_ov_era_l_l = __T_i-_T;,__
Riotal
Ri + R2 + R3
Total thermal resistance is
Steam pipe
ID= 2 r 1 = 160mm
or
r 1 = 80mm
OD= 2 r 2 = 170mm
or
r 2 = 85mm
Insulation 1 Thickness 8 1 = 30 mm
'3 = 'i + <>1 = (85 + 30) mm = 115 mm
Insulation 2
Thickness 82 = 50 mm
r4 = 1j +82 = (115+50) mm= 165 mm
2.68 , Heat and Mass Transfer
Per metre length
R =
1
l
Zn 85
21t{50W/mK){lm) 80
=0.193x10-3 K/W
R =
2
l
Zn~
21t(0.15W /mK)(lm)
85
= 320.7 x10-3 K/W
R =
3
l
/n 165
21t(0.08W/mK)(lm) 115
= 718.2 X 10-3 K/W
Total resistance, ~otal = 1.039 K/W
Overall temperature difference, AJ;,verall = I; - I;,
= 300 - 50 = 250°C
or
K
.
250K
Heat loss per m length, Q = - - - 1.039 K/W
=240.6 W
(Ans)
Layer contact temperatures (T1 and T,)
From the thermal circuit, we have
i;-I;=QRi
J; = J; -Q J?i = 300°C-(240.6 W)(0.193 X 10-3 KJW)
=299.95 °C
(Ans)
Furthermore,
I;-J;,=QR3
I;=J;,+QR3
= 50°C + (240.6 W) (718.2 x 10-3 K/W)
=222.8 °C
Example 2.35
(Ans)
~ A steel pipeline (k = 50 W/m K) of I00-mm ID and 110 mm OD is to be covered with two
layers of insulation, each having a thickness of 50mm. The thermal conductivity of the first
insulation material is 0.06 W/m K and that of the second is 0.12 W/m K.
Calculate the loss of heat per metre length of the pipe and the interface temperature between the two layers
of the insulation when the temperature of the inside tube surface is 250°C and that of the outside surface of
the insulation is 50°C.
Known: A composite pipeline with two layers of insulation with specified boundary surface temperatures
on both sides.
Find: Heat rate Q; Interface temperature, T3 •
Steady-State Heat Conduction-One Dimension
2.69
Schematic:
(<\
- -- · - · - ·
T1 T2
T3
T4
Fig. 2.44a
Tl
(OC)
250
T4 (OC)
k 1 (W/m 0C)
k2 (W/m °C)
k3 (Wlm 0C)
50
50
0.06
0.12
ri(mm) r2 (mm) r3 (mm) r4 (mm)
50
55
105
155
Assumptions: ( l) Steady state, radial conduction. (2) Constant conductivities and heat transfer coefficients.
(3) No contact resistance.
Analysis: The thermal resistance network for this problem is shown below:
Ts
I
'2
I
"ti
27tk 1L1
'•
27tki '",;
Fig. 2.44b
The individual thermal resistances (per metre length) are
t
_
l
r2 _
l
55 _
-4
R1 - - - I n - ( )() In- - 3.0338x 10 K/ W
2rc'G_L 'i 2rc 50 l
50
1- / n 1j =
1
/n l0 5 = 1.71523 K/W
2rck2 L 72 2rc(0.06){1) 55
• Js = -
+ R = -1-!n 14 =
l
ln~=0.51654 K/W
3
2rcfsL '3 2rc(0.12){1) 105
Total resistance, Riotal = R 1+ R 2 + R3
= 2.23238 K/W
Overall temperature difference, ~~verall = 'Ii - T4
={250-50)°C=200°C
or
K
Hence, the heat transfer (per m length) is
Q = ~~verall =
Riotal
=89.6 W
200 K
2.23238 K/W
(Ans)
2. 70 , Heat and Mass Transfer
Interface temperature between two layers of insulation, T3 can be found from the expression
T;-T4 = QR3
I;= 50°C+(89.6W)(0.51654K/W)
= 96.28°C
(Ans)
(K) COMPOSITE CYLINDERS BOUNDED BY FLUID TEMPERATURES
Example 2.36 ~ Brine at -7 °C and 0.05 m3/min is carried by a 20-mm ID and 25-mm OD copper pipe. The
ambient air temperature is 20°C and has a dew point of I0°C. How much insulation with
a thermal conductivity of 0.002 W/m °C is required to prevent condensation on the outside of the insulation
if the outside surface heat transfer coefficient is 11.2 W/m 2 0 C.
Known: A copper pipe carrying brine is wrapped with insulation to prevent condensation of water vapour
on the outer surface of insulation.
Find: Thickness of insulation.
Schematic:
T3(pin) IO °C = Dew point
Air
lll
h= I l.2W/m 2 °C
T=2 =20W/m2°C
r1 =r2= IOmm=
12.5mm
t
Copper
1=.--;nsulation (k2 = 0.002 W/m 0 C)
Fig. 2.45
T=2
---'\f\,f\r-------...J\l\l\r---e-----'\J'\l\r-----'\I\I\,--- -
Q
I
r2
21tk I L In 'I
Fig. 2.46
Assumptions: ( l) Steady, one-dimensional conduction. (2) Constant properties. (3) Negligible inside
convection and pipe conduction resistances.
Analysis: The thermal circuit is shown in the schematic.
Since the values of h 1 and k 1 are not specified, the resistance due to convective film on inside pipe
surface as well as the conduction resistance due to pipe wall are considered negligible.
Then
Steady-State Heat Conduction-One Dimension
2. 71
Essentially, we are left with only two thermal resistances.
'3
1
Rcond = - - - l n 21t 10_L 12
where L is the pipe length, and k2 the thermal conductivity of insulation. Thickness of insulation is
R
=
I
conv "2 ( 2 1t '3L)
Also
Heat-transfer rate from ambient air to brine is
Q = A J;,verall =
T=2 - T=l
Riotal
Rcond + Rconv
(a)
One can also write
(b)
Equating (a) and (b), we have
or
_T;,,~2~-_T_=~l=_R~co=nd~+_R~c=on~v=l+-~-on_d
T=2 - ½
Rconv
Rconv
or
_20_-_(-_7) = l + ( - 1-ln '3) ( "2 ( 21t 13L ))
20-10
21t10_L 12
1
]
27
[ -ln-x"213
1 '3
-=1+
10
10_ 12
or
or
or
or
11.2W /m2 °C x [ r3 (mm) x 10-3 Jm [ r (mm)]
1.7 = - - - - - ~ ~ - - - - ~ - l n3 ~ - 0.002 W /m °C
12 (mm)
'3 ( mm ) 1n
13(mm)
12 (mm)
r,,
or
Trial and error solution is called for.
Trial# 1
LHS
0.5099
r3 = 13mm
13 ln-3-
12.5
=
1.7 x 0.002
11.2 x 10-3
= 0.30357
2. 72 , Heat and Mass Transfer
Trial# 2
r3 = 12.8mm
0.30357 = RHS
Hence, r 3 = 12.8mm
thickness of insulation, t = '3 - 'i
= 12.8 - 12.5 = 0.3mm
(Ans)
Example 2.37 ~ A hot gas at 600°C flows through a long metal pipe of 15cm outer diameter and 5-mm
thick. From the standpoint of safety and reducing the heat loss from the pipe surface,
mineral wool insulation (k = 0.02 W/m K) is wrapped around so that the exposed surface of the insulation is at
a temperature of 600C. Calculate the thickness of insulation required to achieve this temperature if the inside
and outside surface heat transfer coefficients are 50 and 20 W/m 2 K and the surrounding air temperature is
28°C. Also, find the corresponding heat loss if the pipe length is 3 m.
Known: A metal pipe of prescribed geometry provided with insulation. Gas flowing inside the pipe while
the insulation exposed to ambient air. Insulated surface (outer) temperature specified.
Find: Insulation thickness and heat loss.
Schematic:
ttt
k1 (not given)
k2 = 0.02 W/m K
L=3m
h;=50W/m2K
15 or 7.5cm
, 2= 2
T=;=600°C
r1 = r2-t1=7.5-t1
T=0 =28°C
= 7.5-0.5 = 7 cm
t 1= thickness of insulation
T3= 60°C
= r3-r2
h0 =20W/m2K
Fig. 2.47
Assumptions: ( l) Steady-state radial (one-dimensional) conduction. (2) Constant properties. (3) Negligible
metal thermal resistance
( km etaI >> kinsulation
and
'i / 'i "' l) ·
Analysis: The equivalent thermal circuit is as under:
Q T=I
-
Too0
- - ~ · \ / \ / \ ~ - - - - - < 1 - - ~ \ l \ l \ ~ - - - - ~ \ l ' \ l \ r - - - - - - - ' \ I \ .I \ ~ - -
Rconv.l
Rcond, I
Rcond,2
Rconv,2
Steady-State Heat Conduction-One Dimension
2. 73
1
'i
Metal: Rcondl = - - - / n - "" 0
2nlciL lj
.
1
13
Insulation: Rcond 2 = - - - l n 2nlci L 1i
Gas-side: Rconvl =
. •,1
A lr-Slue:
Rconv
2
( l )
Ji.. 21t1j L
1
= h (21r13L )
0
½ -T=o
.
T · -I;
Heat rate Q = --=-1- - Rconvl
+ Rcond2
Rconv2
600 - 60
or
1
27rlj LJi..
60-28
1
+--1-/n 13
2nlciL 1i
{540)(2nL)
or
+ _l_/n 13 (cm)
1
(0.07 X 50)
0.02
13(cm)x10-2 x20
7.5
32
540
or
0.2857 + 50/n(13 /7.5)
540
or
_
0.2857 + 50/n(13 /7.5)
4
·
- 6 13
By trial and error:
13 (cm) I LHS IRHS
9.0
57.45 57.6
As LHS "" RHS, the thickness of insulation is
t2 = 9 - 7.5 = 1.5 cm
(Ans)
Heat loss,
Q=h0 (21r13L)(J;-T= 0 )
= 20 X 21t X 0.09 X 3 X ( 60 - 28)
= 1085.7W
(Ans)
Example 2.38 ~ A steel pipe having I0-cm bore and 12-cm-outside diameter carries hot water at 80°C when
the surrounding temperature is I 5°C. The thermal conductivity of pipe material is 54 W/m K
and inner and outer heat transfer coefficients are I kW/m 2 K and 9 W/m 2 K, respectively. Calculate the heat
loss per metre length of the pipe and the surface temperatures. Also, calculate the heat loss and the surface
temperatures when the pipe is covered with a 4-cm thick insulation having thermal conductivity of 0.048 W/m K
with the outer surface heat transfer coefficient reduced to 7 W/m 2 K.
2. 7 4 , Heat and Mass Transfer
Known: Dimensions, conductivities, heat transfer coefficients and temperatures of a pipe with and without
insulation.
Find: Heat loss and surface temperatures without and with insulation.
Schematic:
/
Pipe wall
.///T =80°C
(k
=54
W/m K)
/
~= I kW/m2
K _
.,__/
____
_h_____
____
I
1
.
/
1+-- - r 1= 5 cm - --+1
i+-- - - r 1= 6cm - - --+1
With insulation
Fig. 2.48
Assumption: (1) Steady operating conditions. (2) One-dimensional conduction. (3) Constant properties.
Analysis: Without insulation
Heat loss Q.
'
WO
= t:.ToveraII
Rr.ot,wo
117;,verall = T=I -
T=2 = 80 - 15 = 65°C
Total thermal resistance, based on unit length of pipe,
R,01,wo
= Rconv, l + Rcond, pipe + Rconv,2
=
l
/; (21r 1 L)
ln('i / 1 )
l
2n lciL
h2 (2n riL)
+---+---l
ln(6cm/ 5cm)
l
(103 W/ m2 K)(21t x 0.05m x lm)
21t x 54 W/ mK x lm
(9W/ m2 K)(21t x 0.06m x lm)
=-,-------,--------+--~---~-+-,-----,--------
= 3.183 X 10-3 + 0.5374 X 10-3 + 0.29473
= 0.29845 K/W or °C/W
Hence,
650 c
QWO -- 0.29845
oc;w -- 217 .8 W
Surface temperatures
T2 = T=2 + Q Rconct,pipe = 15°C + (217.8W)(0.29473 °C/W)
= 79.2°c
(Ans)
2. 7 5
Steady-State Heat Conduction-One Dimension
1;_ = ½ + Q Rcond,pipe = 79.2°C + (217.8W)(0.5374 x 10-3 °C/W)
and
= 79.3°C
With insulation
Q = A J;,verall
w
~otal,w
~ot,w
= Rconv,l + Rcond,pipe + Rcond,ins + R~onv,2
zn(1icc:)
1
21t X 0.048W/mK X lm
(7W/m2 K){21t X 0.10m X lm)
= 3.183 X IQ-3 + 0.5374 X IQ-3 + - - - - - - - - + - - - - - - - - - = 0.003183 + 0.0005374 + 1.6938 + 0.22736
= 1.9248 °C/W
. 65oc
- 33 8 w
•
Heat loss, Qw - 1_92480 C/W -
(Ans)
Suiface temperatures:
½ = T~2 + Qw R~onv,2 = 15°C + {33.8 W){0.22736 °C/W)
(Ans)
=22.7°C
½ = ½ + Qw Rcond,ins = 22.7°C + {33.8W){l.6938 °C/W)
= 79.9°C
1;_ = ½ + Qw Rcond,pipe = 79.9°C + {33.8W){0.0005374 °C/W)
= 79.95 °C
(Ans)
(Ans)
Comment: The outer surface temperature of the insulated pipe drops dramatically to 22.7°C from
79.2°C in the case of the bare pipe. The rate of heat loss is also reduced by 84.5% from 217.8 W
to 33.8 W.
Example 2.39 ~ A 30-mm diameter and 1-m-long pipe carrying steam at 150°C is to be provided with two
insulating layers A and 8 to minimise heat loss.
The exposed surface of the insulated pipe is in contact with room air at I 5°C with an associated heat
transfer coefficient of 15 W/m 20 C. Details of insulating materials A and 8 are tabulated below:
Material
A
B
Quantity (kg)
Density (kg!m 3)
Thermal conductivity (Wlm 0 q
0.27
0.505
190
160
0.046
0.0017
Calculate the rate of heat loss when (a) the insulation A is next to the pipe, and (b) the insulation 8 is next
to the pipe. Which arrangement is more effectivel
2. 7 6 , Heat and Mass Transfer
Known: A pipe with two insulations A and B exposed to convective environment.
Find: Better option (layer A near the pipe or layer B near the pipe) to minimise the heat loss.
Schematic:
......-·
kA = 0.046 W/m °C
Air
kB=0.0017Wlm °C
mA=0.27kg
111
mB=O.SOSkg
®
,,,.-·
Case I
Fig. 2.49a
Air
1 11
PA= 190 kg/m 3
PB= 160 kg/m 3
h= ISW/m2°C
T= = IS°C
20mm
Case II
Fig. 2.49b
Assumptions: (1) Steady operating conditions. (2) One-dimensional (radial) conduction. (3) Constant
thermal conductivities. (4) Uniform heat-transfer coefficient.
Analysis: Case I: Insulation A next to the pipe
Mass A, mA =pA 1t('i2 -1j2 )L
mA
0.27 kg
Volume of material A - - - - - - PA
190kg / m3
-J'i = 1.421 X I0-3 m 3
= 1t( r}-0.015 2 )m2 x lm
Steady-State Heat Conduction-One Dimension
2. 77
Then, the radius r.z = ~0.0152 +(-Ji/re)
= 0.0152 +(1. 421 : 10- 3 ) =0.026m
.
mB
0.505 kg
Volume of matenal B = - = - - - PB 160 kg/m3
¥s = 3.156 x 10-3 m3 = rc(,f - ,r)L
=rc(,f-0.0262 )(1)
Hence,
Rate of heat loss,
Q.I -_
AT,,verall
RA + Rs + Rconv
Total thermal resistance is
Riotal
= RA + Rs + Rconv
=-
1-zn r.z + - 1-zn '3 +
l
1j
2rckBL r.z h(21t13L)
2rckAL
=-1-[_l_zn 0.026 +-1-ln 0.041 +
1
]
2rcxl 0.046 0.015 0.0017 0.026 15x0.041
=44.8 °C/W
.
{150 -15)°C
QI= 44.8 oc;w = 3.0lW
Case II: Insulation B next to the pipe
¥s =rc(,r-,f)L=rc(,r-0.0152 )(1)=3.156x10-3 m3
r.z = 0.0152 + (
3.156x10-3 )
TC
= 0.035 m
Also,
rc(,r - ,r)L = -Ji
or
rc(,f - 0.0352 )(1) = 1.421 x 10-3 m3
'3 = ~0.0352 + (1.421 X IQ-3 /re
=0.041 m
(Ans)
2. 78 , Heat and Mass Transfer
Riotal
= ~ + RA + Rconv
1 [ -ln-+-ln-+1 'i
1 13
1 ]
=-21tL k8 1j kA 1i h13
=-1-[_l_ln 0.035 +-1-ln 0.041 +
1
]
21t X 1 0.0017 0.015 0.046 0.035 15X0.041
=83.14 °C/W
Rate of heat loss,
Q.II -_ Ai:iveran -_ (150 -15)°C
Riotal
83.14 oc / W
= 1.62 W
(Ans)
Thus, we find that the better insulating material with lower thermal conductivity (insulation B) must
be placed next to the pipe (as in Case II) to minimise the heat loss.
Percentage reduction in heat transfer when insulations A and B exchange each other is determined from
QI~ Qn xlOO = (3.01 -1.62)(100)
QI
3.01
=46%
Comment: We note that the convective resistance is the same in both cases. While r 1 and r 3 are same,
i.e., 15 mm and 41 mm, the intermediate radius r2 is different in the two arrangements. In Case I,
the thickness of layer A next to the pipe is ('i -lj) = 11 mm and that of layer B is (13 -'i) = 15 mm.
However, in Case II, layer B next to the pipe is 20-mm thick whereas layer A is now only 6-mm thick.
(L) CRITICAL RADIUS OF INSULATION (CYLINDER)
•
•
Example 2.40 ~ A heat exchanger shell of 15-cm outside radius is to be insulated with glass wool of thermal
conductivity 0.0825 W/m K. The temperature at the surface of the shell is 280°C and it can be
assumed to remain constant after the layer of insulation has been applied to the shell. The convective film coefficient
between the outside surface of slag wool (insulation material) and the surrounding air is 8 W/ m2 K. It is specified that
the temperature at the outer surface of insulation must not exceed 30°C and the loss of heat per metre length of
the shell should not be greater than 200 W. Would the slag wool serve the intended purpose of restricting the heat
loss. If yes, what should be thickness of the insulating material to suit the prescribed conditionsl
Known: Cylindrical insulated shell exposed to ambient air.
Find: Thickness of insulation.
Steady-State Heat Conduction-One Dimension
Schematic:
2. 79
k = 0.0825 W/m K
Ambient
air
lll
h=8W/m2 K
Too=30°C
r-lnsulation----j
Shell surface temperature,
thickness
Ts=280°C
Fig. 2.SOa
-
Q
T;
•
T0
•
VVv
'o
I
2nkL /nr;"
Fig. 2.SOb
Assumptions: ( l) Steady operating conditions exist. (2) One-dimensional heat flow in radial direction.
(3) Constant thermal conductivity and convective film coefficient.
Analysis: Note: There is an apparent mistake in the problem statement. The temperature 30°C refers to
ambient air and not the outer surface of insulation.
We recognise that for reducing heat dissipation in radial systems, the bare (uninsulated) radius must
be more than the critical radius.
Let us first determine the critical radius of insulation for a cylindrical system.
k 0.0825W/mK
,;,c,=-=------=0.0103m or
'
h
8W /m 2 K
1.03cm
Since the shell radius is 15 cm ( >> 1.03 cm), the insulation provided will certainly decrease heat loss.
Referring to the thermal resistance network shown in the schematic, one can write
Q = ~~verall =
J; - Too
I.Rth
_ l _ zn,;, + - - - 2 rc k ,; h(2rc,;,L)
Q = 2rc(J; -T
_!_Zn,;, + - 1-
00 )
or
k
or
,;
h,;,
_!_zn,;, + _l_ = 2 rc(J; .- T
00 )
k
,;
h,;,
Q
2.80 , Heat and Mass Transfer
Substituting proper values,
_1_zn2__+ _I_= 21t{280 - 30) = 7 _854
0.0825 0.15 8r,,
200
Clearly, r0 can be determined by trial and error.
I
r0 (cm)
LHS
0.20
4.11 2
(< 7.85 4)
0.30
8.818
(> 7.854)
0.275
7.802
(< 7.854)
10.216 11 1.844
I I,,, 7.854 1
Thus, the radius of the insulated shell is
r 0 = 0.276 m
or
27.6cm
Thickness of insulation= ,;, - ,; = (27.6- 15)cm
= 12.6 cm
Example 2.41
(Ans)
~ Consider a 50-m-long copper pipe of 15-mm outside diameter and I-mm wall thickness
used to convey water at 70°C. Calculate the heat loss from this length of pipe, with a
15-mm radial thickness of insulation and compare this to the value without insulation. Take the ambient air
temperature as 20°C and the respective values of inside and outside heat transfer coefficients as I00 W/m 2 K
and IO W/m 2 K. The thermal conductivity of copper is 400 W/m K and the thermal conductivity of the
insulation is 0.05 W/m K.
Known: A copper pipe equipped with insulation carries water.
Find: Rate of heat loss with and without insulation.
Schematic:
Copper pipe (k 1 = 400 W/m K, L = 50 m)
Insulation (k 2 = 0.05 W/m K)
r 1 =6.5mm
r2=7.5mm
Air
r=,2 =20°c
Fig. 2.51
Steady-State Heat Conduction-One Dimension
2.81
Assumptions: (1) One-dimensional steady-state radial conduction with constant thermal properties.
(2) Negligible heat transfer by radiation. (3) The outside heat-transfer coefficient is independent of the
pipe's outside diameter. (4) Perfect thermal contact between adjacent layers.
Analysis: Let T1 and T2 be the surface temperatures on the copper at r = r 1 and r = r2 , respectively, and T3
be the temperature on the outside of the insulation at r = r 3• Let the ambient air temperature be T= and
the water temperature be T=,I. The thermal conductivity of the copper is k 1, that of the insulation k2 , and
the inside and outside heat transfer coefficients, h 1, h2 •
With insulation, since the heat flow, Q, is constant
. 21tki L(Ji - 7;)
Q=----
(a)
ln(rilri)
_ 21t"2 L(J; -7;)
-
(b)
ln('3!72)
= 21tLh2'3 (7; - T=,2)
(c)
= 21tLh17i (T=,I - Ji)
(d)
After eliminating the unknown surface temperatures by adding (a), (b), (c) and (d), this can be
rearranged to the form
q= Q =UAT
A
where
~ = { ( ;'i) + ( 13 ln~ I ri)) + ( 13 ln~ I 72)) + ( ~)}
AT= (T=,I -T=,2 )
A= 2 1t r 3 L
Substituting values for the thermal conductivities, heat-transfer coefficients and radii gives
_!_=(
U
0.0225 )+(0.0225 ln(0.0075/0.0065))
0.0075 X 100
400
0.0225 ln(0.0225 I 0.0075))
0.05
1
10
+ ( - - - ~ - - - - ~ +=0.03+(8x10-<:i)+0.494+0.1=0.6244 m2 K/W
U = 1.60 W/m 2 K
A= 21t x 0.0225 m x 50 m = 7.0686 m2
Hence, with insulation the heat loss,
Q = 1.60 W / m2 K x 7.0686m2 x (70 - 20) °C or K
=566 W
(Ans)
2.82 , Heat and Mass Transfer
For a bare pipe, the term r3 /n (r/r2)/k 2 is absent from the overall heat transfer coefficient. Since the
most significant of the remaining terms is l/h 2, the overall heat transfer coefficient can be said to be
dominated, or governed by the external heat transfer coefficient and U "' h 2, from which
Q = 10X(21t X 0.0075 X 50) X 50 = 1178 W
(Ans)
Example 2.42 ~ A 3-mm-diameter and 5-m-long electric wire is tighdy wrapped with a 2-mm-thick plastic cover
whose thermal conductivity is 0.15 W/m 0 C. Electrical measurements indicate that a current of
IO A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed
to a medium at 30 °C with a convective heat transfer coefficient of 12 W/m 2 °C, determine the temperature at
the interface of the wire and the plastic cover in steady operation.Also determine whether doubling the thickness
of the plastic cover will increase or decrease this interface temperature.
Known: Insulated wire dissipating heat to a convective medium.
Find: Interface temperature, T 1. Effect of doubling the insulation thickness.
Schematic:
Q
r1=1 .Smm,L=5m, T1
lll
r ==30°c
h= 12W/m2°C
r2 =r1+t
= ( 1.5+ 2) mm= 3.5 mm, T2
Plastic cover (k = 0.15 W/m 0 C)
~
~
~
Q .., ---Vvv------~Vvv--/n (r2/r1)
Rconv
I
h (21tr2L)
2nkL
Fig. 2.52a
Assumptions: (1) Steady operating conditions. (2) Constant thermal conductivity. (3) One-dimensional
conduction. (4) Uniform heat-transfer coefficient.
Analysis: Rate of heat generation as a result of resistance heating in the wire is
£gen= VI= (8V) (IOA)= 80 W
Energy balance gives
l!,
no heat inflow
-£out +£gen=
J£
steady-state
Steady-State Heat Conduction-One Dimension
2.83
Therefore, the rate of heat dissipation from the insulated electric wire is
Q. = Eout = Egen = 80 W = ~ l)~verall
"total
The thermal resistance network is shown in the schematic. Note that the wire is excluded since it
involves internal heat generation.
Total thermal resistance,
_Zn ('2-/'i)+ - -I- - -_ -I- [ln('2
- - -/'i)
-+ - I
Riotal -
2rckL
h(2rcr,_L)
2rcL
k
hr,_
l
I
[ ln(3.5/ 1.5)
I
]
= 2rc(5m) 0.15 W / m°C + 12W / m2 °C x 3.5 x I0-3 m
= - 1-[5.64865 + 23 .8095]
!Ore
= 0.9377°C
Then
~~veran = I; - T= =
Q Rtotal = (80 W) (0.9377°C I W)
=75°C
Interface temperature between the wire and its plastic cover is
z; = (30 + 75) c = 1os c
0
(Ans)
0
Effect of doubling the cover thickness
In this case, 12· = 'i + 2 t = 1.5 + (2 x 2) = 5 .5 mm
Critical radius of insulation for a cylindrical wire is
r. =!= 0.15W/ m °C(10 3 mm)=l 2 _5 mm
c
h
12W / m2 °C
lm
Since 12· <,;,,the heat transfer rate will increase until r,_* =,;, corresponding to maximum heat dissipation
rate, provided the wire surface temperature remains unchanged. Now,
Q(W)
Qmax= 113.26
-- -- - - -- - - --- - -- - --~-- -.---.
IAToverall= con st I
98.94
80
12.5
(re)
Fig. 2.52b
r0 (mm)
2.84 ,
Heat and Mass Transfer
In (r; I ,i)
Conduction resistance, R~ond = - ~ - ~
2rckL
= In (5 .5!1. 5) =0.2757°C/W
2rc X 0.15 X 5
R*
.
.
C onvectlon
resistance, conv =
=
1
(
• )
h 2nri L
l
= 0.4823°C / W
(12)(2rc X 5.5 X lQ-3 X 5)
R;otal = (0.2757
+ 0.4823) °C/W = 0.7580°C/W
Since Q is held constant,
AT.i:eran = QR;01a1 = (80 W) (0. 7580°C I W)
= 60.64 °C
Hence, the interface temperature will be
T/ = (30 + 60.64)°C = 90.64°C
(Ans)
Comment: If T1 were to be held constant, the new heat transfer rate would have been
Q* = A T.iverall = 105 - 30 = 98.94 W
R;otal
0.758
Also, with ,;, = ,;,r = 12.5 mm, Riotal,cr = (0.45 + 0.2122) K /W = 0.6622 K /W
and
Q.
max
= 105 - 30 = 113.26 w
0.6622
(M) VARIABLE THERMAL CONDUCTIVITY (CYLINDER)
Example 2.43 ~ A reinforced concrete smoke stack with an inner diameter of 80cm and an outer diameter
of 130cm is to be lined with a refractory on the inside. Determine the thickness of the
refractory lining and the temperature of the outer surface of the smoke stack if the heat loss from the outer
surface of the smoke stack does not exceed 2 kW per metre length of the stack and the temperature of the
inner surface of the reinforced concrete smoke stack does not exceed 200°C. The temperature of the inner
surface of the lining is 425°C.
k for refractory= (0.84 + 0.0006T) W/m K, where T is in °C.
k for reinforced concrete= 1.1 W/m K
Known: Reinforced concrete smoke stack of given ID and OD, inner surface temperature, provided with
refractory lining. Heat loss from stack, and lining's inner surface temperature. Materials' properties.
Steady-State Heat Conduction-One Dimension
Find: Refractory lining thickness and stack outer surface temperature.
Schematic:
Q=2000W/m
CD
k1 (T)
I ..... Refractory
2..... Reinforced concrete
r2=40cm
r3= 65cm
k 1 = (0.84 + 0.0006 T) W Im K
k2= I.I W/m K
Fig. 2.53
Assumptions: (I) Steady-state, one-dimensional conduction. (2) Negligible radiation.
Analysis: Heat transfer rate through the smoke stack with refractory lining,
·
LlToverall
Tl - T3
Q=---=---------I.Rth
_ l_ /n r2 + __I _ In r3
2rck 1L r 1 2rc k 2 L r 2
The equivalent thermal circuit is
Also,
T1 -T2 (425 - 200)K
R 1 =--.-=-----=0.1125mK/W
Q
2000W/m
I
r2
R 1 =--In-= 0.1125 m K/W
2rck 1 r1
or
/n2 = 2rc x 0.1125 x [ 0.84 + 0.0006Tm]
rl
where
k 1 = mean thermal conductivity, km
=a+ bTm = 0.84 + 0.0006Tm
2.85
2.86 , Heat and Mass Transfer
As
Tm =½(r, +T2
)= 425 ; 200 =312.5°C
/n( ~) = (2<}(0.1125}(0.84+0.0006X312.5)
= 0.7263
r2 = 2.0674
r,
As
r2 = 0.40 m,
40cm
r1 = - - - = 19.35 cm
2.0674
Thickness of refractory lining,
40 - 19.35 = 20.65cm
Also,
.
(Ans)
Q
r3
T2 -T3 =Q.R2 = - - l n 2rck2L r2
Outer surface temperature,
65
2000
= 200 - - - - / n - = 59.5°C
2rc X 1.1 40
(Ans)
Example 2.44 ~ The inner and outer radii of a hollow cylinder are 5 and I0cm, respectively. The inside
surface is maintained at 3000C, and the outside surface at I 00°C. The thermal conductivity
varies with temperature over the range of I 00 < T < 300 °C as k(D = 0.5 ( I + I o-3
where T is in degrees
Celsius. Determine (a) the heat flow rate per 1-m length of cylinder, (b) the temperature at a radius of 7.5cm
(mid-thickness of the shell), and (c) the radius at which the temperature is 250°C.
n.
Known: Radii and temperatures of hollow cylinder. Variable thermal conductivity.
Find: (a) Heat flow rate, Q; (b) Temperature, T (r= 7.5cm); (c) Radius at which, T= 250°C.
Schematic:
k=0.5(1 +O.OOIT)
L= Im
Fig. 2.54
Steady-State Heat Conduction-One Dimension
2.87
Assumptions: (1) Steady state, one-dimensional conduction. (2) Linear variation of thermal conductivity
with temperature.
Analysis: (a) The heat-flow rate,
. 21tkL(T, -T2 )
Q = ---,-~-,-~
lnhJ,i)
Fork (T) = k0 (1 + bT), k should be replaced by mean thermal conductivity, km= k0 (1 + bTm), Tm being
equal to
½( T + T
1
2)
With
1
T2 = 100°C, Tm= -(300 + 100) = 200°C,
and
and
2
km= 0.5{1+0.001x200) = 0.6W/m°C
Heat-transfer rate per metre length is
. 21t X 0.6 W /m °C X lm X {300 -100°C)
Q=
/n(l0/5)
= 1087.8 W
(Ans) (a)
(b) Starting from the fundamentals,
Q = - k(T)Ac ~:
dT
= - k (l+bT)(21trL)0
dr
or
or
2]
bT1
bT2
= T, +-2- -(T+-2-)
[
bT2
{-QI n -r - [ T1 +
bT/
-+T+
-]} =0
2
21tk0 L r1
2
or
or
or
1Q-3 T 2 + T + {
2
1087.8 /n 0.075 -( 300 + 0.001 x 300 2 ) } = O
21t X 0.5 X 1 0.05
2
0.0005T2 + T- 204.6 = 0
Solving this quadratic equation, we have
T = 187.1°C
(Ans) (b)
2.88 ,
Heat and Mass Transfer
(c) To find the radius r for T= 250°C, we proceed as follows:
__g_
In_.'.:._ = (T
21tk L r
1 -
0
T) +
1
~ (T/ - T 2 )
2
=(T1 -r)(1+%(T1
+T))
= (300 - 250)[ 1 + (0.005 X 550)]
= 63.75
tn!.. = 63.75 x 2rc x 0.5 x l
or
r1
1087.8
= 0.1841
_.'.:._ = 1.202
or
r1
radius r = 1.202 x 5 cm = 6.01 cm
(Ans) (c)
Example 2.45 ~ A hollow cylinder, r 1 ~ r ~ r2, has its boundary surfaces at r = r 1 and r = r2 maintained at
temperatures T1 and T2, respectively. The thermal conductivity of the material varies with
temperature in the form k (1) = k0 (I + ~T2). Develop an expression for the heat flow through a unit length of cylinder.
Known: A long hollow cylinder with prescribed variation in temperature-dependent thermal conductivity.
Find: (a) Expression for heat flow per unit length (2) Thermal resistance (c) Heat-transfer rate.
Schematic:
---+----e--- , 2 = IO cm
T2 =0°C
~ - - - - k = k 0 =60W/m°C
and ~=0.25 x IQ-4°c-2
Fig. 2.55
Assumptions: ( l) Steady-state, one-dimensional (radial) conduction. (2) No internal heat generation.
Analysis: (a) Fourier's rate equation is
.
dT
Q = -k(T)A(r)dr
where A (r) = (2 re r L)
dT
Q. = -k0 ( 1 + ~T 2 ) (2rcrL)dr
Steady-State Heat Conduction-One Dimension
2.89
Separating the variables and integrating,
Q r2
T3]T'
[T+f3-ln-=k
or
27rL
r1
°
3
T
2
or
Heat flow through the cylinder per unit length is
(Ans) (a)
where
ln{rijri)
(b) Thermal resistance, Rth = - - ~
2rckmL
=
/n(lO IS)
= 1.38 x 10-3 °C /W
2rc X 80W / m°C X lm
(Ans) (b)
(c) Heat-transfer rate per unit length is
Q = 1i - i; =
~h
(200 - 0) °C
1.38 X 1Q3 oC/W
= 145 x 103 W
or
145 kW
(Ans) (c)
(N) SPHERICAL SHELL WITH FIXED SURFACE TEMPERATURES
Example 2.46 ~ A 60 W spherical lamp is buried in soil, k = 0.0084 J/s cm °C at 0°C and burned till steady
state is reached. Find the temperature 30 cm away, if I W lamp produces I J/s heat.
Known: A spherical lamp buried in soil attains steady state after it is burnt.
Find: Temperature (at r = 0.3 m), Ti (0 C).
2. 90 ,
Heat and Mass Transfer
Schematic:
~ Qsph = 60 W
Soil
k = 0.0084 J/s cm
0
c
Fig. 2.56
Assumptions: (1) Steady-state conditions. (2) One-dimensional conduction. (3) Constant properties.
Analysis: Heat-transfer rate for a hollow sphere is
. 41tk(J; - I;,)
Q=---
[¾- ~l
As
r0 ~ 00 , l/r0 = 0, and T0 = 0°C, we have
. 41tkT
Q=--' =41tkrT
1/rj
I
I
Hence, the temperature at ,; = 0.3m is
Q
T=-1
41tk,;
60 J/s
41t (0.00841/scm C) (100cm/lm)(0.3m)
0
= 18.95 °C
(Ans)
Example 2.47 ~ Consider three geometries: (a) plane wall, (b) cylindrical annulus, and (c) spherical shell made
up of the same material of the same thickness and the same surface temperatures. Plot
the temperature distribution for the three cases. The inner and outer surface temperatures are T1 = 100°C
and T2 = 25°C. The thermal conductivity of the material, k = 15 W/m 0 C. For the plane wall, the thickness,
L = 5 cm. For the cylindrical and spherical walls: r 1 = 5 cm and , 2 = IO cm [thickness, (r2 - r 1) = 5 cm].
Also, calculate the heat-transfer rates and the temperatures at the mid-thickness in the above three cases.
For plane wall, assume the area of cross-section, A = I m2 and for the cylindrical annulus, the length L = I m.
Known: A plane wall, a cylindrical annulus, and a spherical shell of the same thermal conductivity, same
surface temperatures and the same thickness.
Find: Temperature distribution, Heat-transfer rate, and mid-plane temperature in the three cases.
Steady-State Heat Conduction-One Dimension
Schematic:
r
2. 91
Plane wall (k = 15 W/m 0 C)
k= 15W/m°C
r2 = IOcm
T2 =25°C
1-- L=Scm --I
(a)
(b)
Spherical shell
r
1= 100°c
r1 =Scm
k= 15W/m°C
(c)
Fig. 2.57
Assumptions: ( l) Steady state, one-dimensional conduction without heat generation. (2) Constant thermal
conductivity.
Analysis:
• Plane Wall
Fourier's rate equation is
Q =-kAdT
dx
Temperature gradient,
dT
Q
=- = constant C
dx
kA
' 1
-
Integrating with respect to x, the temperature distribution is
2. 92 , Heat and Mass Transfer
T (x = 0) = T1 = C1 (0) + C2
T~=~=~=~L+~
IT(x) = i; -(i;
~
~
C2 = T1
½ -i;
q=-L
-Ti)zl
T(x) = lOOoC _ (100 - 25) °C Xx (m)
0.05m
or
I T(x) = 100 - 1500 x I
(Ans) (a)
Temperature at the mid-thickness,
= 62.5 °C
(Ans) (a)
+ Cylindrical Annulus
Fourier's rate equation is
.
(
)dT
Q=-k2rcrLdr
Temperature gradient,
!
dT = -Q
= - C1
dr 2rckL r
r
where C 1 is a positive constant
Integrating with respect to r, the temperature variation is given by
T(r) = -C1 In r+ C2
T (r= r 1) = T1:
T1 =-C1 ln r 1 + C2
T (r= r2) = T2 :
T2 = -C1 In r2 + C2
C1 = 108.202
C2 = 7; + C1 In r1
= 100 + 108.202 In 0.05
I C2 = -224.145 I
!T(r) = -108.202 In r - 224.145!
(Ans) (b)
1
At mid-thickness (r = -(5 + 10) cm= 7.5cm),
2
Tm= -108.202 Zn 0.075 - 224.145 = 56.13°C
(Ans) (b)
Steady-State Heat Conduction-One Dimension
2.93
• Spherical Shell
Fourier's rate equation is
.
dT
Q =- k(41tr 2 ) dr
Temperature gradient,
Q
dT =-J]____-= Ci
dr
41tk r 2
r2
where C = - - 1
41tk
~
I; =--+C2
Temperature profile is given by
T(r=r1 )=I;
c1
rt
and
.- (¼-tr
C __
I;-T;
__
(100-25)
(1 / 0.05)-(l / 0.1)
1cl =-7.51
Also
The temperature profile is given by
Ci
T(r)=--+C 2
r
IT(r)=+
7!- -sol
(Ans) (c)
At mid-thickness,
7.5
Hence T = - - - 50 = 50°C
' m
0.075
(Ans) (c)
2.94 ,
Heat and Mass Transfer
The temperature profiles for these three cases are shown below:
Cyliindrical annulus (b)
:Spherical shell (c)
'i - -- -- --- ----- -- -- -- --- --- -- - ------ - --- - ------ - -- - -- -- --- --- 62.5°C
················································ 56.l3°C
....... ................................................ 50°C
T2 =25°C
L------------'---------=-.J
r2= IOcm
rm=7.5cm
r 1=5cm
---Radius, r
x=O
x=L=5cm
Fig. 2.58
Heat-transfer rates
(a)
_kA(T.-T)-15W/ m°Cxlm 2 (100-25)°C
L I
2 0.0 5 m
.
Qplane wall -
= 22.5 x 103 W
.
2rckL(T; -1;)
(b) Qcylind•r=
ln(r2 / r1)
or
=
22.5 kW
(Ans) (a)
2rc{l5W/m°C){im)(!OO- 25)°C
/n(I0 / 5)
= 10.2 x 10 3 W = 10.2 kW
(Ans) (b)
4rck(T; -T2 )
4rc(15 W/m°C){IOO - 25)°C
(c) Qsphere = (l / r1)-(l/r2 ) =
[(110.05)-(1/0.l)]m-I
.
= 1.41 x 10-3 W
or
1.41 kW
(Ans) (c)
Example 2.48 ~ A spherical thin-walled vessel of 16-m-diameter contains saturated liquid ammonia at 516 kPa
(5°C). The vessel is insulated with a I 0-cm thick PUF (polyurethane foam) of thermal conductivity
0.025 W/m K. The outer surface temperature of the insulation is 33°C. Calculate the heat rate and the required
refrigeration capacity to maintain the sphere temperature in tons of refrigeration (TR). [ I TR= 211 kJ/min]
Know: An insulated sphere with prescribed temperatures and dimensions.
Find: Heat-transfer rate and tonnage of refrigeration.
Schematic:
Insulation (k = 0.025 W/m K)
r;=Sm
Spherical
container
/&@~~~~~~~ f-i-.-r = 8.1 m
0
T0 = 33°C
Fig. 2.59
2.95
Steady-State Heat Conduction-One Dimension
Assumptions: ( l) Steady-state, radial (one-dimensional) conduction. (2) Constant thermal conductivity.
(3) Thin-walled sphere (thermal resistance of the sphere wall is negligible).
Analysis: The heat rate,
.
41tk(J;-T0 )
41t(0.025W / mK){5-33)°C
or
K
Q-~-~--=--~--=----~-=-~---- [{1/ 71 )- (1/,;,)J[(1/8) - (1/8. l)] {l / m)
= -5700 W (heat loss)
or 5.7 kW (heat gain)
(Ans)
Tonnage of refrigeration,
_ 5.7 x 60 kJ/min _
- - - - - - - 1 .62
211 kJ/min/TR
(Ans)
(0) SPHERICAL SHELL BOUNDED BY KNOWN FLUID TEMPERATURES
Example 2.49 ~ A spherical, thin-walled, metallic container is used to store liquid nitrogen at -196°C. The
container has a diameter of 45cm and is covered with an insulation that is 2.5cm thick. Its
outer surface is exposed to ambient air at 30°C. The heat-transfer coefficient is 20 W/m 2 K. The latent heat
of vaporization and density of liquid nitrogen are 200 kJ/kg and 800 kg/m 3 respectively.
Determine (a) the rate of heat transfer to the liquid nitrogen, and (b) the liquid boil-off rate. Take
k (insulation)= 0.0017 W/m K. What is the loss of the cryogenic fluid per day in litres?
Known: An insulated spherical container filled with liquid nitrogen is subjected to convection heat transfer
at its outer surface.
Find: (a) Rate of heat transfer. (b) Rate of evaporation of liquid nitrogen.
Schematic:
Air
T=, 0 =J0°C
h=20 W/m2 K
Q T= ,I·=-196°C
h(41tr~)
Fig. 2.60
2. 96 , Heat and Mass Transfer
Assumptions: (1) Steady-state, one-dimensional (radial) conduction without internal heat generation.
(2) Constant thermal properties and heat transfer coefficient. (3) Convective resistance at inside surface
and conduction resistance due to container wall are negligible.
Analysis: The thermal resistance in this case is both due to conduction as well as convection.
Rt,cond
R
and
= 4 l k [_!_ - _!_]
t,conv
1t
=
ro
,;
(
1
h 41tr02
)
The steady rate of heat transfer to the liquid nitrogen,
.
AT
T.. o -T..i
Q = --- = -----Rt,overall
Rt,cond
+ Rt,conv
= -----,:---~,-----4-!-k [:. -
l
~ + -41t-~-r-;
[30 -(-196)]°C
=----------,:-----=---------==------------1
[
1
1 ]
1
41t(O.OOl W/mK) 0.225m - 0.25m + 41t(20 W/m2 K)(0.25m) 2
=
226cc
= 10.83 w
20.868(K/W)
(Ans)
Also, the boil-off rate,
m=JL=
hfg
10.83W (3600s)(1J/s)=0.2 kg/h
200 X 103J/kg lm
lW
Loss of cryogenic fluid per day
m 0.2 kg/h X (24 h/d)
p
800 kg/m3
= 0.006 m3/day
or
6 liters/day
(Ans)
Example 2.50 ~ A hollow sphere made of a material of constant thermal conductivity (k = 40 W/m 0 C) has its
inner boundary surface (r; = 3cm) subjected to uniform heat flux (q; = 1.5 x 105 W/m 2). The
outer boundary surface (r0 = 5cm) is exposed to convective conditions characterised by ambient fluid (T.. = 35°C)
and the convection coefficient (h = 180 W/m 2 0 C). Derive an expression for temperature distribution T(r) and
determine the boundary surface temperatures T; and T0 •
Known: A hollow sphere has its inner surface subjected to constant heat flux. Heat is dissipated from
its outer surface by convection.
Find: Temperature distribution, T(r). Boundary surface temperatures, Ti and T0 •
Steady-State Heat Conduction-One Dimension
2. 97
Schematic:
111
h= 180W/m2°C
Too=JS°C
k=40Wlm°C
Fig. 2.61
Assumptions: ( l) Steady state, one-dimensional (radial) conduction. (2) Constant thermal conductivity
and uniform heat flux .
Analysis: The governing differential equation is
_!_~[kr 2 dTJ = O
r2 dr
dr
Successively integrating w. r. t. r, we get
k r 2 dT -- C I
dr
~
dr
kr 2
(a)
The temperature distribution in the radial direction is
C1
T(r)=--+C 2
kr
(b)
where C1 and C2 are arbitrary constants to be evaluated from the following two boundary conditions.
B.C.I Inner boundary surface: -k(dT)
= q.
dr r=r;
I
From equation (a):
or
B.C.II:
or
or
Outer boundary surface:
2.98 , Heat and Mass Transfer
qr,
2
qr,
2
C -T + - 1 _ 1 _ _1 _ 1
2-00
h2
k
ro
ro
or
Substitution for C1 and C2 in equation (b) gives the temperature distribution.
T(r)= T + qi r;2 - qi r;2 - (-qi ,;2)
oo
hr;
k ro
kr
q.
r,
q. r,
1
1
h
,;,
k
r
r0
T(r) =Too+ ( _!_ ) ( ...!...) 2 +-1_1_2 [ - - - ]
or
(Ans)
The boundary surface temperatures are determined to be
T, = T(<= ~)= T_ + ( ~)[~
J
= 350c + (1.5 x 105 W/m2 )(3 cm)2
180W/m2 °C
5cm
= 335°c
(Ans)
and
= I;, + qi r/ [_!_ _ _!_]
k
r; ,;,
= 3350C + (1.5 X 105 W/m2)(0.03m}2 (-1- _ _1 _)
40W/m°C
0.03m
= 380°C
0.05m
(Ans)
Alternatively
Thermal resistance concept enables us to calculate the boundary surface temperatures fairly easily when
there is no internal heat generation. Referring to the thermal circuit shown in the schematic, Ti and T0
are found to be
(1.5 x 10 5 W/m2)(0.03m)2
(1sow1m20c)(o.05m)2
= 35°c + ---,--------,-~--=335°C
(Ans)
Steady-State Heat Conduction-One Dimension
2. 99
and
= 335 oc + (1.5 x l0 5 W/m2 )(0.03m)2(0.05 x 0.03)m
(40W/m20C)(0.03m x 0.05m)
(Ans)
(P) COMPOSITE SPHERE WITH FIXED BOUNDARY TEMPERATURES
Example 2.51
~ A spherical vessel of 50-cm radius contains a liquefied gas at 187°c. It has two jackets of
lagging each I0-cm thick. The thermal conductivities of the inner and outer layers are in
the ratio of 2:3. The temperature of the outermost surface is Is0c. If the heat leakage rate into the liquid gas
is 225 W, find the thermal conductivities of the tagging material.
Known: Spherical vessel with two layers of insulation.
Find: Thermal conductivities of the two insulations.
Schematic:
Q=22SW
R1 =50cm
Fig. 2.62
Assumptions: (I) Steady operating conditions. (2) Thermal resistance of the sphere material is negligible
so that T=i = Ti . (3) One-dimensional conduction. (4) Constant properties.
Analysis: We note that
R 1 = 50cm = 0.50 m
R 2 = (50 + 10) cm= 60cm = 0.60m
R 3 = (60 + 10) cm= 70cm = 0.70m
2.100, Heat and Mass Transfer
and
k1 2
-=-
or
k1 = -k2
3
k2
3
2
For the given composite sphere, we have
or
41t[15-(-187)] 0 c or K
225 = 225 W = - - - - - - - - - - - - - - - - 1
[(0.6 - 0.5)]
1 [(0.7 - 0.6)]{ 1
m}
(2/3)k 2 (0.5)(0.6) +~ (0.6)(0.7) W/mK · m2
or
i_!_(0.10) + _!_(0.10) = (202)(41t)
k2 0.42
225
2 k2 0.30
or
;2 [(¾)(½) + :~] = 81:;
or
1
lei
8081t
(225)(0.738)
=-----
k 2 = <225 )(0. 738) = 0.0654 W /m K
8081t
and,
2
2
k1 = 3k 2 = 3(0.0654) = 0.0436 W /m K
(Ans)
(Ans)
Example 2.52 ~ A cylindrical tank with hemispherical ends is used to store liquid oxygen at -I 83°C. The diameter
of the tank is 1.5 m and the total length is 8 m. The tank is covered with a I0-cm thick layer of
insulation. Determine the thermal conductivity of the insulation so that the boil-off rate does not exceed 10.Skg/h.
The latent heat of vaporisation of liquid oxygen is 214 kfkg. Assume that the outer surface temperature of the
insulation is 27°C and that the thermal resistance of the wall of the tank is negligible.
Known: Rate of evaporation of liquid nitrogen stored in a cylindrical tank with hemispherical ends and
provided with insulation.
Find: Thermal conductivity of insulation, k.
Steady-State Heat Conduction-One Dimension
Schematic:
2.101
Q
Insulation (k = ?)
__L
T
T
_L
J_
D0 =D;+2t
D;= I.Sm
= 1.5 + 2 (0.1)
= 1.7m
L=6.5m
T;=-183°C
Q----
-
-- Q
R.ph
Fig. 2.63
Assumptions: ( l) Steady-state conditions. (2) One-dimensional conduction (3) Thermal resistance of the
tank wall is negligible.
Analysis: Cylinder length, L = Ltotal - 2 X ri = 8 m - 2 X O. 7 5 m = 6 .5 m
Two hemispherical ends together with insulation make up a hollow spherical shell.
We note that total heat gain by liquid oxygen due to conduction equals heat lost due to evaporation, i.e.,
Qin = mevap hfg
To find Qin• let us first find the total thermal resistance. We note that Rcylinderand Rsphere are in parallel.
Equivalent thermal resistance is given by
l
l
l
2nkL
4nkr0 ~
Rtotal
Rcyl
Rsph
In( ro I ri)
( ro - ri)
--=--+--=---+--2n x 6.5 + 4n x 0.85 x 0.75]
0.85 - 0.75
0.75
= k [ In 0.85
= k[326.5+ 80.11] = 406.4l(m) k [W/m°C]
Heat-transfer rate (inwards),
Q=
t:.T
T -T
overall = - 0- -1 = {406.41 k) W /°C(27 + 83)°C
Rtotal
Rtotal
2.102 ,
Heat and Mass Transfer
Since Qin = m evaphfg •
Thermal conductivity of insulation is
k = (10.8 I 3600)kg I s(214 x 103 )J/kg
(406.41 m)(210°C)
= 0.00752 W/m °C
(Ans)
(Q) COMPOSITE SPHERE BOUNDED BY KNOWN FLUID TEMPERATURES
Example 2.53
~ Iced water at 0°C is stored in a spherical tank with 5-m ID and 25-mm thickness made
of stainless steel (k = 15 W/m 0 C). The emissivity of the outer surface of the tank is 0.95.
The ambient air and surroundings temperatures are both 35°C. The heat-transfer coefficient between the inner
surface of the tank and the iced water is 90 W/m2 °C and that between the surroundings and the outer
surface of the tank is 15 W/m2 °C. Determine the amount of ice at 0°C that melts per day. The latent heat
of ice is 334 kJ/kg.
Known: A spherical tank containing iced water and made of steel has its inner surface with a convective
boundary and the outer surface exposed to both convection and radiation environment.
Find: Amount of ice melted in a day.
Schematic:
Q
Tsur= 350c
Air
T= 2 = 35°C
h2 = 15 W/m 2 °C
c =0.95
Stainless steel (k = 15 W/m 0 C)
Fig. 2.64
Assumptions: (I) Steady-state prevails. (2) One-dimensional heat conduction. (3) Constant properties.
Analysis: Energy balance
Heat transferred into the tank (Qin )(W)
= Rate of melting of ice, m(kg/s) x Latent heat of fusion of ice, h r(J/kg)
5
The thermal circuit is shown in the schematic. We note that at the outer surface, Rconv,2 and R,ad are
connected in parallel.
Steady-State Heat Conduction-One Dimension
l
total thermal resistance, Riot = Rconv,I + Rcond + [ R
l
+ -R--
rad
2.103
]-I
conv,2
Radiation heat transfer coefficient,
hr = rn (7;2 + T.~ ) (½ + T.ur )
But T2 is presently unknown. It must lie between 35°C and 0°C and is likely to be near 0°C. Let
T2 = 7°C. Then
hr= (0.95) (5.67 x 10-8 W/m 2 K4) (308.15 2 + 280.15 2) K 2 (308.15 + 280.15)K
= 5.5 W/m2 K
With
Rrad
=
1
andR
2
hr(41tr2 )
1
1
Equivalent resistance, l\q = [ , ; - + -R--]
.L'rad
1
conv,2 - h2(4 1tr;)
-I
2
= [ ( hr + h2)( 41tr2 )
conv,2
1
-I
J
1
=------=-------------hcombined (41tri) {(15 + 5.5)W/m2 °C}( 41t X 2.525 2)m2
= 0.61 X 10-3 °C/W
= 0.1415 X 10-3 °C/W
1(1
1)
1
[1
1]1
Rco nd = 41tk r1 - r2 = 41t(15 W/m 0 C) 2.5 - 2.525 m
= 21x lo-6 °C/W
or 0.021x 10-3 0 C/W
Riotal = Rconv,I + Rcond + Req
= [(0.1415 X lQ-3) + (0.021 X lQ-3) + (0.61 X lQ-3)] °C/W
= 0.7725 X 10-3 °C/W
Hence, the rate of heat inflow is
Q = T=2 - T=I =
Riotal
(35 - 0)°C
0.7725 X 10-3 oC/W
= 45.3 x 10-3 W
or 45.3 k.J/s
2.1 04 ,
Heat and Mass Transfer
Amount of ice that melts in 24 h or one day is
m=fl_= 45.3 kJ/s 13600sll24 hllltonnel
hsf
334 kJ/kg
lh
!day 103 kg
= 11.7 tonnes per day
(Ans)
Check
T= 2 -T2 = QReq = (0.61 x 10-3 °C/W) = (45.3 x 10 3 W) = 27.6°C
T2 = 35 - 27.6 = 7.4°C
We had assumed T2 = 7°C. The difference is too small to warrant repeat calculation.
Example 2.54 ~ A hollow sphere of 20-cm inner diameter and 40-cm outer diameter (k = IO W/m 0 C). The
inside surface is maintained at a uniform temperature of 230°C and the outside surface dissipates
heat by convection with a convective heat-transfer coefficient 20 W/m 2 °C into an ambient air at 30°C. Determine
the thickness of asbestos insulation (k = 0.5 W/m 0 C) required to reduce the heat loss by 50%.
Known: A hollow sphere is provided with insulation exposed to convective environment.
Find: Insulation thickness to reduce heat loss by 50 %.
Schematic:
1
!!
h=20W/m2°C
T==J0°C
- - - - " , / \ l ' , ~ - - - - - - ' \ . 1 \ / \ ~ - - - - - J \ / \ / \ ~ - -... - + -
<2w= 0.5 <2wo
I
c;:;---;:;-I J
I [ I
47tk2
41trfh
Fig. 2.65
Assumptions: (I) Steady-state conditions. (2) One-dimensional (radial) conduction. (3) Constant properties.
(4) Uniform heat-transfer coefficient.
Analysis: Heat-dissipation rate without insulation is
~¾verall
I [1 r1] 41t,r1h
- - - - - +- - 41tk1 1j
2
Steady-State Heat Conduction-One Dimension
41t AJ;,verall
2.1 05
(1)
=------
_!_(_!__l)+-1]G_ 1j 12
h r.J
Heat loss rate with insulation is
= - - - - -AJ;,verall
-------1 (1 1)
1 (1 1)
1
41t]G_ 1j 12 + 41tk2 12 13 + 41t,f h
or
0.5 Q =
WO
41tAJ;,verall
1(1
(2)
1(1 1) 1
'i 12 + Is 12 '3 + h,f
/G_
1)
Dividing (1) by (2), we have
1(1 1) 1(1 1) 1
'i 12 +Is 12 '3 +h,f
2=---------_!_ (_!_ - _!_) + _1_
]G_ 1j 12
h r.J
/G_
or
1(1
1) 1(1 1)
1
10 0.1 0.2 + 0.5 0.2 '3 + 20,f
2 =--1-(_1__1_)___1_ __
10 OJ- 0.2 + 20x(0.2)2
or
or
3.5 =0.5 + 10-(~) +(
o;
5)
7 = 3__ 0.05
'3
,f
7,f-273 + 0.05 = 0
or
Solving this quadratic equation, we get
r 3 = 0.258 m or
25.8cm
Hence, the thickness of the insulation is
r 3 - r3 = 25.8 -20 = 5.8cm
(Ans)
2.1 06 ,
Heat and Mass Transfer
Example 2.55 ~ A spherical container with 40-cm-lD consists of 4-cm of lead (k =35.3 W/m 0 C) covered by I-cm
of stainless steel (k = 14.2 W/m 0 C) and 5-mm of concrete (k = 1.1 W/m 0 C). The container is
filled with nuclear wastes that generate heat at a rate of q(W/m 3 ). The sphere is placed in the deep ocean where
outside surface temperature is 8°C and the heat transfer coefficient is 300 W/m 2 0 C. If the lead temperature is to
remain IOOoC less than its melting point of 601 K, what is the maximum allowable heat generation rate!
Known: A composite spherical shell containing radioactive wastes is placed in oceanic waters.
Find: Maximum permissible heat generation rate, qmax (W /m 3 ) .
Schematic:
111
r==s c
0
h=300W/m2°C
Tpb.max = 60 I - I 00 = 50 I K
= Ti .max
, I , C' I• •I
i - - - - 2 0 c m - - - - - - - 4cm
0.5 cm
I cm
Fig. 2.66a
R3 = Rconcrete
Q= <2gen
Q
T3
Fig. 2.66b
Assumption: (1) Steady operating conditions exist. (2) Constant properties and uniform convection heat-transfer
coefficient. (3) Nuclear (radioactive) waste heat generation. (4) One-dimensional (radial) heat conduction.
Analysis: Energy balance
0
0
¢-£out +£gen=¢
£gen= £out
Hence,
Qgen = Qout = Q(r) or Q
or
or
From the equivalent thermal circuit,
·
LlToverall
Tl -
T=
Q = ---= ------Riot
Ri + R2 + R3 + R4
Steady-State Heat Conduction-One Dimension
2.1 07
T,,max = Melting point of lead - 100°C
= (601 - 273)°C - 100°c = 228°C
and
T .. = 8°C
AToveran = 228 - 8 = 220°C
The thermal resistances are determined from
~ - ~)
R, = Rlead = 4~kl (
= [1/(41t)(35.3W/moc)J[ 0.2~m - 0.2~m]
= 0.00188°C/W
= [1/(41t){14.2W/m 0 c)] [ 0 _2~m - 0 _2~m]
= 0.000934 C/W
1 -1)
R3 -R
- -I- ( concrete - 41tk3 r3
r4
= [I/( 41t){1.IW/m°C)] [ 0.2~m - 0.2;5m]
= 0.005674 °C/W
R4 = Rconcrete = (
2)
41tr4
h
= [1/(41t){0.255m)2 (300W/m 2 0 c)]
= 0.00408 °C/W
Therefore
R101 = [0.001 88 + 0.000 934 + 0.005 674 + 0.00408]°C/W = 0.012 568 °C/W
Heat-transfer rate, Q = Heat generated by nuclear wastes, Qgen = AT0veran/R101
= { 220° c/(0.012 568 °C!W)}[ 1~::]
= 17.5 kW
Rate of internal heat generation per unit volume
_ Q /{( 4 / 3)
q=
gen
=522 kW/m3
3}
17.5 kW
m, =(4/3)1tx0.203 m3
(Ans)
2.1 08 ,
Heat and Mass Transfer
Example 2.56 ~ Determine the rate of heat flow through a spherical boiler wall which is 2 m in diameter
and 2-cm thick steel (k = 58 W/m K). The outside surface of the boiler wall is covered with
asbestos (k = 0.116 W/m K), 5-mm thick. The temperature of the outer surface and that of the fluid inside are
50°C and 300°C, respectively. Take inner film resistance as 0.0023 K/W.
Known: Composite spherical shell with convective conditions inside and fixed outer surface temperature.
Find: Heat flow rate, Q(W).
Schematic:
Spherical boiler wall (Ts,o = 50 °C)
Asbestos (ks= 0.116 W/m K)
Steel
(kA = 58W/m K)
Q
r 1= Im
___,__-,---
r2 = I + 0.02 = 1.02 m
~~~cJ------1-.i-- r3
= 1.02 + 0.005 = I .025 m
Inner fluid
(T==J00 °C)
- I
Rmm- h-A
I
I
=0.0023K/W
Fig. 2.67a
Assumptions: ( 1) Steady-state, radial heat conduction. (2) Constant thermal conductivities. (3) No thermal
contact resistance.
Analysis: The thermal resistance network for the two-layered spherical wall is shown below:
T==J00°C
•
-----Q
VV'v--~---~VVv~--•--~VVv
r., =50°C
•
0
Fig. 2.67b
R film -A- R
-k- (1
---- --ll R asbestos ll
- h1
- steel - 4 1
- 4 1k (1
i i
1t A rt r2
1t B r2 r3
The rate of heat loss through the composite sphere is
·
LlToverall
T= - 'F.,o
Q=---=--------Rtotal
Rfilm + Rsteel + Rasbestos
(300 - 50)°C
or
K
=---------------------------------(0.0023K IW) + [ -41t_ X_5_
8~/ m- K
- n- -l.~-2}m-t] + [ -41t_ x _O-. l-:6- W
- /m- K ]{-1.~-2 - -l.0-l2_5}m-t
=
°
25 K
= 44.58 x 103 W
0.005608 K/W
=44.58 kW
(Ans)
Steady-State Heat Conduction-One Dimension
2.109
Example 2.57 ~ A small hemispherical oven is built of inner layer of insulating firebrick 12.5-cm-thick, and
an outer covering of 85% 40-mm thick magnesia. The inner surface of the oven is at 800°C
and the heat-transfer coefficient for the outer surface is IO W/m 2 °C; the room temperature is 20°C. Calculate
(a) the rate of heat loss through the hemisphere, (b) the interface temperatures, and (c) the temperature at
the mid-thickness of the layer of firebrick if the inside diameter is 1.2 m. Take the thermal conductivities of
firebrick and 85% magnesia as 0.31 and 0.05 W/m °C, respectively.
Known: A hemispherical oven with two layers of insulation is exposed to convective environment.
Find: (a) Q(W); (b) T2 and T3 (DC); (c) T (r= 66.25cm) (DC).
Schematic:
iii
h= IOW/m 2 °C
r ~ =20°C
85% Magnesia
(k2 = 0.05 W/m C)
0
Firebrick
(k 1= 0.05 W/m2 °C)
r, =800 °C
Fig. 2.68a
-
Q
---VVv----~VVv~-----'VV'v--...
R2
R3
Fig. 2.68b
Assumptions: (1) Steady operating conditions. (2) Constant thermal conductivities. (3) Uniform heat
transfer coefficient.
Analysis: (a) From Fourier's rate equation for hemispherical shell, one can write
.
(21tr 2 )dT
Q=-
dr
Separating the variables and integrating,
or
or
Q[- {-!}'•]
1
21tk
r r,
= T' -TD
. [ - 1 {1 -1 }'·] -T -T
Q 21tk ri
rD r, - i
D
2.1 I O ,
Heat and Mass Transfer
Since AT = QR1h, the thermal resistance of a hemisphere by conduction is
Referring to the thermal circuit shown in the schematic,
Q = AToveran /L Rth
AToveran = 7; - T= = {800 - 20)°C = 780°C
and
"'I,.Rth =R1 +R2 +R 3
=
2:, {:. - :,
h:k,{:, -:,}h(2~,')
+
= 21t/ci ;/m°C {
0.:
m - 0.7;5 m} + 21t x O.O~W/m°C { 0.7;5 m - 0.7:5 m}
1
10W/m2 °C{21t X 0.765 2 m2)
+---------= 0.1475 + 0.2296 + 0.0272 = 0.4043°C/W
The heat loss rate is
Q=
780 °C
= 1930 W
0.4043°C/W
or 1.93 kW
(Ans)
(b) Interface temperatures
.
With
i;-i;
Q=-R-,
I
= 800°C- (1930 W) (0.1475 °C/W)
= s1s.3°c
and
(Ans)(b)
J; = T= + QR 3
= 20 + (1930 W) (0.0272 °C/W)
(Ans) (b)
(c) At mid-thickness of firebrick layer,
r= 1j + (r2 ; 1j) = [ 0.6
°·~
25 ]m = 0.6625 m
Steady-State Heat Conduction-One Dimension
2.111
Temperature at r = 0.6625 m is
T = T. l
Q[-1
21tk1
(__!__ -
=800°C-1930
rt
_!_)]
r
w[---
1- - - 1- } ]
1 --{21tx0.3IW/ m°C 0.6m 0.6625m
(Ans) (c)
Example 2.58 ~ Consider a 2-m diameter thin-shelled, insulated liquid oxygen at I atm and -I 83°C, initially
filled to 80% of its capacity. The tank is exposed to ambient air at 22°C, with a combined
convection and radiation heat-transfer coefficient of 35 W/m 2 K. The tank is being transported by road and
the amount of the liquid oxygen ( p = 1140 kg/m 3, h1g = 213 kJ/kg) evaporated during a journey of 2.5 h should
not exceed 1%. (a) Determine the thickness of the fibre-glass insulation (k = 0.035 W/m K) provided. (b) If the
spherical tank contains liquid nitrogen at I atm,-196 °C (p=810kg/m 3, h1g= 198 kJ/kg), what would be the
insulation thickness!
Known: An insulated spherical liquid oxygen tank exposed to surrounding air.
Find: Insulation thickness.
Schematic:
k=O.Q35 W/mK
!!!
h=35W/m2K
T =22°C
00
T1
T2
T
00
~ -
R1 = r2-r1
41tkr1r1
Q
R2= __I _
41tr22h2
Fig. 2.69
Assumptions: ( 1) Steady operating conditions. (2) The inside surface of the spherical tank is at the
temperature of the liquid oxygen. (3) Constant thermal conductivity of the insulation. (4) Uniform heattransfer coefficient.
Analysis: (a) We note that, Q = mevap fzrg
where mevap is the rate of evaporation (kg/s) and hrg is the latent heat of vaporisation (J/kg).
The heat transferred from the outside air to the liquid oxygen inside the spherical vessel is
.
T
00
-7i
41t[22-(-183)]°C
or
K
Q------------------------- R1 + R2 - {
( r2 - 1) m
1
}
(0.035 W/m K){lm){ri )(m) + (35 W/m 2 K)r} (m 2 )
Volume of liquid oxygen initially in the tank is
¥- = 1tD3 X 0.80
6
(A)
2.1 12 ,
Heat and Mass Transfer
Mass of liquid oxygen is
m = p¥- = (1140kg/ m3 )( ~ X 23 m3 }o.80) = 3820 kg
Rate of evaporation of liquid oxygen is
m
evap
= O.Ol x ~ = 0.01 x 3820kg
At
2.5 X 3600 S
= 4.244 X lQ-3 kg/s
Hence, Q = (4.244 x 10-3 kg/s)(213 x 103 J/kg) = 904.1 W
(B)
Equating (A) and (B), one gets
41t X 205
904 l =
. 28.5714 _ 28.5714 + 0.02857
2
r2
r2
or
285714 _ 28.5714 + 0.02:57 = 41t X 205
r2
r2
904.1
or
25.722r2 - 28.5714r2 + 0.02857 = 0
2
Solving the quadratic equation, one has
Required thickness of insulation is
r2 -r1 =(1.ll-l)m=O.llm
or
11cm
(Ans) (a)
(b) With liquid nitrogen,
T 1 = - 196 °C,
p = 810 kg/m 3
and
hrg = 198 kJ/kg
Then
41t X [22-{-196)]K
----------=
28.5714 _ 28.5714 + 0.02:57
r2
810 X !!_ X 23 X 0.8 X 0.01 X 198 X 103
6
=597.154
2.5 X 3600
r2
or
41t X 218 = 28 14 _ 28.5714 0.02857
.57
---+
2
597.154
r2
r2
or
23.984r2 - 28.5714r2 + 0.02857 = 0
2
Solving for r2, one finds
r2 = 1.19 m
Insulation thickness is
r2 - r 1 = (1.19 - 1.0)m = 0.19 m
or
19 cm
(Ans) (b)
Steady-State Heat Conduction-One Dimension
2.113
Example 2.59 ~ Consider a composite spherical shell comprising an insulating layer of unknown thermal
conductivity sandwiched between two layers of steel of thermal conductivity 45 W/m K.
The inner shell has I00-mm inner diameter and a wall thickness of 5 mm. The outer shell has 150-mm inner
diameter and a wall thickness of 2mm. The inner surface of the inner spherical shell was experimentally found
to be I20°C while the outer surface of the outer spherical shell was at 67°C. The ambient air was at 30°C
with the heat-transfer coefficient at the outer surface estimated to be 6 W/m 2 K. (a) Determine the thermal
conductivity of the insulating material. (b) The experiment was repeated on another day when the ambient air
temperature was 13 °C. The inner surface temperature of the inner shell and the convection coefficient remained
unchanged. Calculate the temperature of the outer surface of the outer shell in this case.
Known: A composite spherical shell with an insulating shell between two steel shells and the outer surface
exposed to the convection process.
Find: (a) kinsulation and (b) Touter surface for changed ambient temperature.
Schematic:
Air
r3 =75mm
h=6W/m2 K
T= =30°C
r:= 1J c
0
T1 = 120°C
Insulation (ks= 1)
Fig. 2.70
Assumptions: (I) Steady operating conditions. (2) Constant thermal conductivities. (3) Uniform heat-
transfer coefficient. (4) Thermal contact resistance is negligible. (5) One-dimensional (radial) conduction.
Analysis: Case (a): Ambient temperature, T= = 30 °C
The thermal circuit for the composite spherical system is given below:
The heat-transfer rate is
The individual thermal resistances are determined to be
• Roo,a.A = 4•~A [:, -
:,]=
4•(45 ~ Im
K)[o~s - o~s}-•
= 3.21525 x I0-3 K/W
t
1 [1 1]
1[1
1J
Rco nct,B = 41tk 8 r2 - r3 = 41t( k 8 ) 0.055 - 0.075
= 0.38583/k8
2.1 14 ,
•
Heat and Mass Transfer
Rc0nd·c = 47t~c [~ -
~] = 41tt45)[0.~75 - 0.~77]
= 6.1243 X lQ-4 K/W
•
R
=
conv
l
=
l
= 2.23696 K/W
h(41tri) (6W/m2 K)(41tx0.0772 m2 )
Heat loss rate is
l
R,0 m1 = LR,, = [ 3 .21525 X JO-S + 0.3:: 83 + 6.1243 X I 04 + 2.23696
= 2.2408 X (0.38583/ks)
T4 -T- _ (67-30)°C or K
Q. -_ ---'--- -'-------'---- = 16.54 W
Rconv
2.23696 K/W
We note that
Rtotal
2.2408 + (
or
= AToverall / Q
0.38583)- 90
k
- -B
16.54
0.38583 = 3.2005
kB
Thermal conductivity of the insulating material is
k = 0.3 8583 = 0.12 WI mK
B
3.2005
Case (b) Ambient air temperature T:, = I 3°C
R
cond,B
= 0.3 8583 = 3.2005 K / W
O. l 2
Total thermal resistance,
+ Rcond,B + Rcond,C + Rconv = 2.2408 + Rcons,B
= 2.2408 + 3.2005 = 5.4413 K/W
Riotal = Rcond,A
Else
T1 - T-
Rtotal =
~=
(120 - 30)K
16.54W
= 5.4413 K/W
Heat-transfer rate,
. • 7; -T~ {120 -13)K
Q =--=--Rtotal
5.4413 K / W
= 19.664 W
(Ans) (a)
Steady-State Heat Conduction-One Dimension
2.115
But
Hence, the outer surface temperature of the composite shell is
r,.* = r:, + Q* Rconv
= 13°C + (19.664 W) (2.23696 K/W)
=57°C
(Ans) (b)
(S) VARIABLE THERMAL CONDUCTIVITY (SPHERE)
Example 2.60 ~ Develop an expression for steady state, radial heat flow across a spherical wall whose thermal
conductivity varies with temperature as follows:
k = k0(1 + ~T2) where k0 and ~ are specified constants. A hollow sphere of 0.5-m ID and 25-mm thick
has its inner and outer surfaces maintained at I 50°C and 30°C respectively. The sphere is made up of
an insulating material whose thermal conductivity varies with temperature according to the expression:
k (T) = 0.06( I + 0.0008 T2) where T is in °C. Determine the rate of heat transfer.
Known: Inner and outer temperatures, and dimensions of a hollow sphere of a material with variable
thermal conductivity.
Find: Rate of radial heat flow, Q(W)
Schematic:
r 1 =0.25m
r 1=150°c
ff-----t--.i--
r2 = 0.25 + 0.025 = 0.275 m
r 1=Jo c
0
k = 0.06 [ I + 8 x I Q-4 T2]
Fig. 2.71
Assumption: ( 1) Steady operating conditions. (2) One-dimensional (radial) conduction. (3) No internal
heat generation. (4) Isothermal surfaces.
Analysis: For steady-state, one-dimensional heat conduction through a spherical wall, the Fourier's rate
equation can be written as
.
dT
Q = -k(T)A(r)dr
= -k0 (1 + ~r2 ){41tr2) : :
Separating the variables and integrating, we have
.
r=r,
f
T=T,
f
dr= - Q
(l+~T 2 )dT
2
41tk 0
r
r=r,
T=T,
2.1 16 ,
Heat and Mass Transfer
__{1_[-.!.Jr2 = -[T- 13!2JT2
or
41tk0
I
Q=(~J-(:,J
·
or
3
r r
41t(7i-J;) {
(
k,
~[
2
T
I
2J)}
!+3,; +T,T,+T,
· 41tke(Ti. - J;)7i7i
Q=----'i - 1i
or
where
Substituting the numerical values, the effective (or equivalent) thermal conductivity of the insulating
material is
k. = 0.06[ 1 + g \ 104 {1502 + (150)(30) + 302 }]
= 0.5064 W/m °C
or
W/m K
Hence, the rate of radial heat flow is
Q = 41t X 0.5064W/m°C X (150 - 30)°C X 0.25m X 0.275 m
(0.275-0.25) m
=2100 W
Example 2.61
or 2.1 kW
(Ans)
~ A hollow sphere of inside and outside radii 'r 1' and 'r2' respectively is heated such that its inner
and outer surfaces are maintained at uniform temperatures 'T 1' and 'Ti'- If the material of which
the sphere is composed has a thermal conductivity which varies with temperature according to the expression
find the heat-flow rate through the sphere.
Known: A hollow sphere made of material of variable thermal conductivity.
Find: Expression for heat-flow rate.
Steady-State Heat Conduction-One Dimension
2.1 17
Schematic:
/ ..-·
,,.,..-··
,,....--··
/ ..-··
.,.,.--··
Hollow sphere
ki-k1
k(T)=k1+-- (T-T1)
T2-T1
Fig. 2.72
Assumptions: ( l) Steady-state, one-dimensional (radial) heat conduction. (2) No internal heat generation.
(3) Isothermal surfaces.
Analysis: Fourier's rate equation in the radial r-direction is
Q. = - k ( T ) Ac -dT
dr
Ac = 41tr2
where
Substituting for k(T), rearranging and integrating, we have
or
Jf( 7i) dT
· [
k
k T,
i1_
_ _!_]'2 = kifT, dT + ( ~
T-
41t
r rI
T
2
½ 'li
T
2
or
or
k2-k1J(T;-7i)]
= (7i - ½)[( kl + 2 - ½ - 1i
2.118 ,
Heat and Mass Transfer
Heat flow rate,
. 4.irr1 r2 (i;_ -i;)(k1 +k 2 )
Q = ---~--',------'-I
(Ans)
2(r2 - r1 )
Example 2.62 ~ The inside and outside surfaces of a hollow sphere, a < r < b, at r = a and at r = b are
maintained at uniform temperatures T1 and T2, respectively. The thermal conductivity varies
with temperature as
k(T) = k0 (1+ aT + ~T2)
(a) Develop an expression for the total heat flow rate Q through the sphere.
(b) Develop a relation for the thermal resistance of the hollow sphere.
Known: Variable thermal conductivity in a hollow sphere.
Find: (a) Expression for Q, and (b) Thermal resistance, Rth·
Schematic:
r=a
Q
k=k 0 (I +aT+~T2)
Fig. 2.73
Analysis: ( l) Steady operating conditions exist. (2) One-dimensional conduction. (3) No internal heat
generation.
Analysis: For a differential element at a radius rand of thickness dr, we have from the Fourier's rate equation:
Q = -k(T)A, : :
Q. = -k0 (l + <J.T + ~T 2 ) ( 41tr2 )dT
-
or
dr
Boundary conditions: At r = a, T = T1, and at r = b, T = T2.
Separating the variables and integrating, we get
b
f
Q r12 dr = -41tk
a
T,
0
f (l + <J.T + ~T2)dT
T1
Steady-State Heat Conduction-One Dimension
2. I I 9
Hence, the heat-transfer rate is
. 4rck 0 ab
[
a
~(
)]
Q = --(Ti -7;) l +-(Ti+ Ti)+ - 2 +Ti.Ti+ 7;2
(b-a)
2
3
r.
(Ans)
From electrical analogy,
.
AT
T. -T,
Rth
Q
Q=- ~ Thermal resistance, Rth = - 1- .-2
Substituting for Q, one gets
1
Rth = - - - - - - - - - - - - - - -
4rck0(¾-¼)[1+ aTm +~(Ti. +Ti.I; +7;
2
2 )]
1
where, Tm= -(7; + 7;)
2
1
MULTIPLE CHOICE QUESTIONS
2.1
Match List I with List II and mark the correct answer using the codes given below:
List I
List II
A. V 2T+i = _!_ aT
k a at
1. Laplace equation
B. V 2T+.i = o
k
2. Fourier equation
C. V2T = _!_ aT
a at
D. V 2T=O
3. General heat-conduction equation
4. Poisson equation
Codes:
A
B
C
D
(a)
4
3
2
1
(b)
2
4
1
3
(c)
1
3
2
4
(d)
3
4
2
1
2.2 Match List I and List II according to the code given below:
List I (Statement)
A. T (0, t) = 100°C
List II (Description)
1. Initial condition
(Ans)
2.120 ,
Heat and Mass Transfer
B
'
C.
(dT)
dx
= 1000
-k (dT)
dx
2. Constant surface temperature
k
x=L
= h(T. - T=)
3. Constant surface heat flux
x=O
D. T(r, 0) = 25°C
4. Convective boundary
Codes:
A
B
C
(a)
2
3
4
(b)
2
3
4
(c)
3
2
4
D
(d)
4
2
3
In descending order of magnitude, the thermal conductivity of (a) pure iron, (b) liquid water,
(c) saturated water vapour, and (d) pure aluminium can be arranged as
(a)abcd
(b)bcad
(c)dabc
(d)dcba
2.4 A plate of thickness L and thermal conductivity k is exposed to a fluid at temperature T= 1 with the
convection coefficient h 1 at the inner surface, and at the outer surface the fluid temperature is
2,
with convection coefficient h2 . The positive x-direction is from the inner surface to the outer surface.
The convective boundary condition is given by
2.3
T=
2.5
Heat flows through a composite slab, as shown
below. The depth of the slab is 1 m. The k values
are in W/m K. The overall thermal resistance in
K/Wis
(a) 17.2
(c) 28.6
(b) 21.9
(d) 39.2
k=0.02
q~
k=0.10
k=0.04
- - - O.Sm
0.2Sm-
1
A composite hollow sphere with steady internal heating is made of two layers of materials of equal
thickness with thermal conductivities in the ratio of 1:2 for inner and outer layers. The ratio of
inside to outside diameters is 0.8. What is the ratio of the temperature drops across the inner and
outer layers?
(a) 0.4
(b) 1.5
(c) 2 /n 0.8
(d) 2.5
2.7 The schematic of a cylindrical tank with spherical ends is shown below:
2.6
Ei:~_E:O_K___ :_:_E_29~:K:_________ ~; ::@0.025 W/m K
------7m-----Fig. 2.74
Steady-State Heat Conduction-One Dimension
2.8
2.9
2.10
2.11
2.121
The rate of heat transfer into the tank is
(a) 23 W
(b) 380 W
(c) 0.38 W
(d) 120 W
What is the heat lost per hour across a wall 4-m high, 10-m long, and 115-mm-thick, if the inside
wall temperature is 30°C and the outside ambient temperature is 10 °C? Conductivity of brick wall
is 1.15 W/m K, heat transfer coefficient for the inside wall is 2.5 W/m 2 K, and that for the outside
wall is 4 W/m 2 K.
(a) 3635 kJ
(b) 3750 kJ
(c) 3840 kJ
(d) 3920 kJ
The two surfaces of a 1.25-m thick plane wall are at 200°C and 50°C. The thermal conductivity of
the wall material is k = 0.25 (l + 0.04 T) where k is in W/m °C and T is in °C. The steady-state
heat flux is
(b) 120 W/m 2
(c) 150 W/m 2
(d) 180 W/m 2
(a) 90 W/m 2
For the three-dimensional object shown in Figure 2.127, five faces are insulated. The sixth face
(PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer
coefficient of 10 W/m 2 K . The ambient temperature is 30°C. Heat is uniformly generated inside the
object at the rate of 100 W/m 3. Assuming the face PQRS to be at uniform temperature, its steadystate temperature is
(a) 10°c
(b) 20°c
(c) 30°C
(d) 40°c
Consider steady-state heat conduction across the thickness in a plane composite wall (as shown in
Figure 2.129) exposed to convection conditions on both sides.
Given:
h; = 20 W/m 2 K
h0 = 50 W/m 2 K
T= ,I. =20°C
T= ,o =-2°C
!G_
= 20 W/m K
k2 =50 W/m K
L 2 = 0.15 m
L1 =0.30 m
Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in
0 C), of the two walls will be
h;, r=,i
CD
lll
lll
i ~ L1 ----I~ L2 ~j
Fig. 2.75
(a) -0.50
(b) 2.75
(c) 3.75
(d) 4.50
2.122 ,
Heat and Mass Transfer
1
TRUE/FALSE
Wood has directionally dependent thermal conductivity.
Steady-state heat flux is constant in a long hollow cylinder in the radial direction.
Critical thickness of insulation is relevant only in radial systems.
In Figure 2.134, the dotted lines are isotherms and dark continuous lines are for heat flux. (Assume
2-D heat transfer only by conduction).
2.5 Absolute vacuum is the ultimate insulator.
2.6 It is worth using the asbestos (k = 0.1 W/m K) as an insulating material on a 20-mm-OD pipe
exposed to air with a convection coefficient of 5 W/m 2 K.
2.1
2.2
2.3
2.4
1
FILL IN THE BLANKS
2.1
2.2
2.3
2.4
2.5
Imperfect thermal contact between mating surfaces is accounted for by incorporating __ resistance
with units - - ·
A material with kx = ky = k2 = k is known as - - ·
V 2 T = 0 is called the __ equation.
__ mean area is used in the case of a hollow cylinder and __ mean area in the case of a hollow
sphere.
A 6-cm diameter sphere at 125°C is provided with insulation (k = 1 W/m 0 C) exposed to air at
25°C with h = 20 W/m 2 °C. For maximum heat loss, the thickness of insulation is __ cm.
1
EXERCISES
2.1 A long copper bar of rectangular cross-section, whose width W is much greater than its thickness
L, is maintained in contact with an ice bath at its lower surface, and the temperature throughout
the bar is equal to that of the ice, T0 • Suddenly an electric current is passed through the bar and
an air stream at a temperature T= is passed over the top surface, while the bottom continues to
be maintained at T0 by the ice bath. Draw an appropriate figure and write down the mathematical
formulation of the problem. List any assumptions made.
2.2 A spherical metal ball of radius R, thermal conductivity k(T) and surface emissivity E is heated to
a temperature Ti throughout and then allowed to cool in ambient air at T=. The average temperature
of the surrounding surfaces is Tsur- The convection coefficient on the outer surface of the ball is h.
Obtain the mathematical formulation of this problem. Do not solve.
[-k aT~=• =
t)
h[T(R)- T=] + O't:[T(R)4 -T4.ur]
J
2.3 A lake surface is covered by 16-cm layer of ice, the ambient air temperature being 22.5°C.
A thermocouple embedded near the upper surface of the ice layer indicates a temperature of -10°C.
Assuming steady state conduction through ice and no liquid cooling, calculate the heat loss from
the lake per km square of lake surface. Also calculate the conductance of the ice layer and the
convection coefficient. Take kice = 2.0 W/m K.
[10 W/m 2K]
Steady-State Heat Conduction-One Dimension
2.123
2.4 A 2-cm-thick wall is to be constructed from a material which has an average thermal conductivity
of 1.3 W/m °C. The wall is to be insulated with a material having an average thermal conductivity
of 0.35 W/m °C so that the heat loss per square metre will not exceed 1830 W. Assuming that
the inner and outer surface temperatures of the insulated wall are 1300 and 30°C. Calculate the
thickness of the insulation required.
[23.75 cm]
2.5 An oven wall measures 1.2m x 0.75m x 8mm. The wall is covered by a layer of insulation that
is 50-mm thick. The thermal conductivity of the wall material is 12.5W/mK and the thermal
conductivity of the insulation is 0.25 W/m K. The inside wall temperature is 300°C, and the
temperature at the outside of the insulation is 30°C. Calculate the amount of heat lost in kJ/h and
the interface temperature.
[299.2°C]
2.6 A furnace wall is composed of 220 mm of firebrick (k = 4 kJ/m h 0 C), 150-mm of common brick
(k = 2.8 kJ/m h 0 C), 50-mm of 85% magnesia (k = 0.24 kJ/m h 0 C) and 3-mm of steel plate
(k = 240 kJ/m h 0 C) on the outside. The inside and outside surface temperatures are 1500°C and
90°C respectively. Estimate the temperatures between layers and heat loss in kJ/hm 2 .
[l017°C]
2.7 An insulated wall is to be constructed of common brick, 20-cm thick and plaster 2.5-cm thick,
with the intermediate layer of packed rock wool. The outer surface of the brick and the plaster are
maintained at 600°C and 50°C respectively. Calculate the thickness of the rock wool required in
order that the heat loss per m2 of the surface area should not exceed 600 W. The conductivities
of brick, rock wool and plaster are 0.32, 0.045 and 0.7 W/m K respectively. Also find the contact
surface temperatures.
[71.43°C]
2.8 (a) Consider a composite plane wall consisting of four different materials as shown in figure. The
upper and lower surfaces are insulated. The heat flow is assumed to be one-dimensional. For
a temperature difference of 300°C, determine the heat-transfer rate if the area is 50 000 cm 2.
Also, find the overall heat-transfer coefficient.
Ti
'\__
CD
k= 10
W/mK
r
@
Im
k=70W/m K
k=SW/mK
1
r
Im
@)
k=70W/mK
,_/
@
1
f- 3 cm--+----- 9cm _ ___,_..._ 4cm -j
Fig. 2.76
2.9
2.10
[81.4 W /m 2 K]
Determine the thickness of the insulating material of thermal conductivity 0.104 W/m °C necessary
to reduce the heat loss from a hot water tank to 25 per cent of the unlagged loss. Assume the
combined convection and radiation heat transfer coefficient to be 8.5 W/m2 °C for the uninsulated
tank and 6.5 W/m2 °C for the exposed surface of the lagging.
[33 mm]
Determine the rate of heat flow per unit area through a boiler wall made of 20-mm thick
steel (k = 58 W/m K). The outer surface of boiler wall is covered with asbestos insulation
(k = 0.116 W/m K) 5-mm thick. The temperature of the outer surface and that of the fluid inside
are 50°C and 300°C respectively. The inner film resistance is 0.0023 K/W.
[5465 W]
2.124 , Heat and Mass Transfer
2.11
2.12
2.13
2.14
2.15
2.16
2.17
A heater element of 1 kW rating and an area of 0.02 m 2 is protected on the backside with an
insulation (60-mm thick and k = 1.4 Wlm K) and on the front side by a plate (20-mm thick and
k = 55 Wlm K). The backside is exposed to surrounding air at 10°C with a convective heat-transfer
coefficient of 15 Wlm 2 K. The front side is exposed to ambient air at 25°C with a combined
convection and radiation heat-transfer coefficient of 200 Wlm 2 K. Calculate (a) the heater element
temperature, and (b) the two extreme surface temperatures.
[262. 7°C]
A furnace has a composite wall-a refractory material for the inside layer and an insulating
material on the outside-to prevent the heat leakage. The total wall thickness is limited to 60 cm.
The temperature of the gases in the furnace is 850°C and the atmospheric air temperature is 30°C.
The temperature at the interface of the two materials of the furnace wall is limited to 500°C. Take
the following data:
k 1 (conductivity of refractory material)= 2 Wlm K
k2 (conductivity of insulating material)= 0.2 Wlm K
hi (heat transfer coefficient between gas and wall)= 200 Wlm 2 K
h0 (heat transfer coefficient between wall and air)= 40 Wlm 2 K
Find the required thickness of each material and the heat loss to the atmosphere in Wlm2
[1291W/m2 ]
A cold storage room has walls made of 0.23 m of brick on the outside, 0.08 m of plastic foam
and finally 1.5 cm of wood on the inside. The outside and inside air temperatures are 22 and -2°C
respectively. If the inside and outside heat transfer coefficients are 29 and 12 Wlm2 K respectively
and the thermal conductivities ofbrick, foam and wood are 0.98, 0.02 and 0.17 Wlm K, respectively.
Determine
(a) The rate of heat removal by refrigeration, if the total wall area is 90 m2
(b) The temperature of the inside surface of the brick
[20.28°C]
Two stainless-steel plates, each 10-mm thick, are subjected to a contact pressure of 1 bar under
vacuum conditions for which the contact conductance is approximately 666 Wlm 2 K. The thermal
conductivity of the stainless steel may be taken as 15.2 Wlm K. (a) Find the rate of heat transfer
across the plates for a temperature difference of 100°C per square metre of the surface. (b) Also,
calculate the contact plane temperature drop.
[53.3°C]
A truncated cone 30-cm high is constructed of copper (k = 385 W Im °C). The diameter at the top
is 7.5 cm, and the diameter at the bottom is 15 cm. The top surface is maintained at 180°C and
the bottom surface at 65°C. The lateral surface is effectively insulated. Determine the rate of heat
transfer. State clearly the assumptions made.
[1304.0 W]
Heat is conducted through a uniformly tapered 40-cm long rod. The temperature of one end is
500°C and the diameter is 2 cm. At the other end, the temperature is 100°C and the diameter is
7 cm. The thermal conductivity of the material is 20 WIm K. Assuming that the heat is conducted
only along the length of the rod, fmd (a) the rate of heat conduction, and (b) the temperatures at
points 10, 20 and 30 cm from the hot end.
At T = 500 - 89.6(6.25 - 3.846) = 284.6°C
X = 0.2m
T = 500 - 89.6(6.25 - 2.778) = 188.9°C
At x = 0.3m
T= 500 - 89.6(6.25 - 2.174) = 134.8°C
Derive an expression for the rate of heat flow through an infmite slab of thickness L, if its thermal
conductivity varies with temperature according to the law
k=k0
(l+ ;)
Steady-State Heat Conduction-One Dimension
2.125
Sketch the temperature distribution when (a) is positive(> 0), (b) a is negative(< 0), and (c) a= 0.
Q = kA(I; - T2 ) + ~n(I; / T2 )
r
1
L(l+ - )
T
2.18
2.19
Derive the temperature distribution equation for an infinite plane slab of thickness L whose thermal
conductivity varies linearly with temperature as k = k0 (1 + cT) where k and c are constants. Sketch
the temperature distribution when c is positive and when c is negative.
The thickness of a large slab is 18 cm, the inner and outer surfaces are at 310°C and 40°C
respectively. Find the heat lost per m 2 of the surface if k = (0.4 + T 2 x 10--{i) where Tis in °C
and k is in W/m K.
[655 W]
Find the steady-state heat flux flowing through the infinite slab shown below. The thermal
conductivity of the material of the slab varies linearly with temperature in the following manner:
k= 0.05 (1 + 0.008 T) W/m °C where Tis in °C.
r==Jo c
0
h= IOW/m2k
T
5cm
_l_
600°c
Fig. 2.77
[1684 W/m 2 ]
2.20 The thermal conductivity of a plane wall of thickness L varies with temperature as
k(T) = k 0 (1 + ~7'3)
2.21
2.22
where k0 and ~ are constants. The boundary surfaces at x = 0 and x =Lare maintained at temperatures
T1 and T2, respectively.
(a) Develop an expression for the heat flow rate across an area A of the wall.
(b) Calculate the heat transfer rate for A = 0.1 m2, L = 0.4 m, T 1 = 200 °C, T2 = 0 °C, k0 = 60 WIm °C,
and ~ = 0.25 10-4 °c-3.
[153 KW]
Calculate the heat loss per 100-m length of pipe when the pipe is insulated by corrugated asbestos
for which the thermal conductivity is 0.1 W/m K. The pipe temperature may be taken as 120°C and
its outside diameter as 40 mm. The thickness of the lagging is 50 mm, and its external temperature
is 55°C.
[3260 W]
A steam pipe of 75-mm OD and 30-m length conveys 1000kg of steam per hour at a pressure
of 2 MN/ m 2 . The steam enters the pipe with a dryness fraction of 0.98 and is to leave the other
end of the pipe with a minimum dryness fraction of 0.96. This is to be accomplished by suitably
insulating the pipe, the thermal conductivity of insulating material being 0.19 WIm K. Neglecting the
temperature drop along the steam pipe, determine the necessary conditions. Take the temperature of
outside surface of insulation as 27°C. For steam at 2 MN/m 2, Ts= 212.4°C and hrg = 1888.6 kJ/kg.
[33.l mm]
2.126 ,
Heat and Mass Transfer
2.23 A copper tube of 20-mm outer diameter, 1-mm thickness and 20-m long (thermal
conductivity= 400 W/m K) is carrying saturated steam at 150°C (convective heat transfer
coefficient= 150 W/m 2 K). The tube is exposed to an ambient temperature of 27°C. The
convective heat-transfer coefficient of air is 5 W/m 2 K. Glass wool is used for insulation (thermal
conductivity= 0.075 W/m K). If the thickness of the insulation used is 4mm higher than the
critical thickness of insulation, calculate the rate of heat lost by the steam and the rate of steam
condensation in kg/h (Enthalpy of condensation of steam = 2230 kJ/kg).
2.24 A thick-walled tube of stainless steel with 2-cm ID and 4-cm OD is covered with a 3-cm thick layer
of asbestos insulation (k = 0.2 W/m 0 C). If the inside wall temperature of the pipe is maintained
at 600°C and the outside of the insulation at 1000°C, calculate the heat loss per metre length.
[-543.lW/m]
2.25 Calculate the overall heat-transfer coefficient (based on inner diameter) for a steel pipe covered
with fibre glass insulation. The following data are given:
ID of pipe
= 2 cm
Thickness of pipe
= 0.2 cm
Thickness of insulation
= 2 cm
Heat transfer coefficient (inside) = 10 W/m 2 K
Heat transfer coefficient (outside) = 5 W/m 2 K
Conductivity of insulation
= 0.05 W/m K
Conductivity of steel
= 46 WIm K
Inside fluid temperature
= 200°C
Ambient temperature
= 30°C
Also find the net heat loss from the pipe.
[29.8 W]
2.26 A steel pipe (15-cm ID, 10-mm thick) with thermal conductivity of 34.6 W/m°C and provided
with an insulation, 2.5-cm-thick, having a thermal conductivity of 0.025 W/m°C carries steam at
200°C. The inside heat transfer coefficient is 500 W/m2°C. On the outside of the insulation, the
heat transfer coefficient is 300 W/m 2°C. Ambient air is at 20°C. The insulation fails at 190°C.
Will it survive under these conditions?
[199.5°C]
2.27 A steel pipe of 200-mm inner diameter and 2-mm thickness is covered with 20-mm-thick fibre
glass insulation (k = 0.05 W/m 0 C). The inside and outside convection coefficients are 10 W/m 2 °C
and 5 W/m2 °C. Calculate the overall heat-transfer coefficient based on the inside diameter of the
pipe.
[l.608 W/m 20 C]
2.28 A 160-mm diameter pipe carrying saturated steam is covered by a layer of lagging of 40-mm
thickness. (k = 0.8 W/m 0 C) Later an extra layer of lagging, 10-mm thick (k = 1.2 W/m 0 C) is
added. If the surrounding temperature remains constant and the outer heat transfer coefficient for
both cases is 10 W/m2 °C, determine the percentage change in the rate of heat loss due to the extra
lagging layer.
[i.e. a decrease by 0.2%]
2.29 A steel tube of 2.5 cm ID and 2-mm wall thickness discharges a chemical flowing at 300°C. The
tube is insulated with glass wool 5-cm-thick. The inner and outer heat transfer coefficients are 1200
and 12 W/m 2 K, respectively. (a) Calculate the overall heat transfer coefficient based on inner area.
(b) If the ambient temperature is 25°C, determine the heat loss per unit length of pipe. (c) Will it
help to increase the insulation thickness?
kins= 0.1 W/mK, kpipe = 45W/mK
[rC7 < r2 (= 14.5mm)] adding insultation will help reduce the heat loss
Steady-State Heat Conduction-One Dimension
2.127
2.30 A small electric heating application uses 0.183-cm diameter wire with 0.071-cm thick insulation.
The thermal conductivity of insulation is 0.118 W/m K and the surface convection heat transfer
coefficient is 34 W/m2 K. Calculate the critical thickness of insulation and the effect of employing
critical insulation thickness. Assume that the temperature difference between the surface of the wire
and surrounding air remains unchanged.
[16%]
2.31 A 2-mm diameter wire with 8-mm thick layer of insulation (k = 0.15 W/m 0 C) is used in a certain
electric heating application. The insulated surface is exposed to atmosphere with convective heat
transfer coefficient 40 W/m 2 °C. What percentage change in heat transfer rate would occur if the
critical thickness of insulation is used? It may be assumed that the temperature difference between
the surface of the wire and the surrounding air remains unchanged.
[15%]
2.32 Calculate the critical radius of insulation for asbestos with k = 0.17 W/m °C surrounding a pipe
and exposed to room air at 25°C with h = 4.0 W/m2 °C. Calculate the heat loss from 6-cm diameter
pipe at 200°C when covered with critical radius of insulation and without insulation.
[138.64 W]
2.33 An aluminium pipe carries steam at 110 °C. The pipe (k = 185 W/m K) has an inner diameter
of 10 cm and an outer diameter of 12 cm. The pipe is located in a room where the ambient air
temperature is 30°C and the convective heat transfer coefficient is 15 W/m 2 K. Determine the heat
transfer rate per unit length of the pipe.
To reduce the heat loss from the pipe, it is covered with a 5-cm thick layer of insulation (k = 0.2
W/m K). Determine the rate of heat loss per unit length and the percentage reduction in heat loss
by the insulation. Neglect the convective resistance of the steam.
[69.4%]
2.34 A 1-mm diameter wire is maintained at a temperature of 400°C exposed to a convective environment
at 40°C with h = 120 W/m2 K. Calculate the thermal conductivity which will just cause an insulation
thickness of 0.2 mm to produce a critical radius. How much of insulation must be added to reduce
the heat transfer by 75% from that which would be experienced by the bare wire.
[134 mm]
2.35 The thermal conductivity of a certain material varies according to the following relation:
k =a+ bT+ cT2
where a, b and c are constants. Derive an expression for computing heat loss per linear metre for
a hollow cylinder made of this material. Assume that the cylinder ends are perfectly insulated.
Take T1 and T2 as temperatures at inner and outer surfaces which have radii r1 and r2 , respectively.
Q
L
2tr(Ti-T2)[
-=----
lnr2
b
C 2
2]
a+-(Ti+T2)+-(I'i +TiT2+T2)
2
3
1j
2.36 A sphere of 40 cm radius is lagged to a radius of 50 cm, the inner and outer surface temperatures
of the lagging being 230°c and 65°C, respectively. Find the rate of heat leakage, if the thermal
conductivity of the material of the lagging is 5.5 W/m2 K per cm thickness.
[228 W]
2.37 A spherical vessel of 50-cm radius contains a liquefied gas at -183°C. It has two jackets of lagging,
each 10-cm thick. The inner layer has thermal conductivity of 0.05 W/m °C and the outer layer
0.065 W/m °C. Determine the rate of heat leakage into the container if the outermost surface is
at 25°C.
[253 W]
2.38 A hollow metal sphere has the following dimensions:
Inner radius .... 5 cm, Inside fluid temperature ... 330°C
Outer radius .... 5 .2 cm, Outside fluid temperature . . . 24 °C
2.128 ,
2.39
Heat and Mass Transfer
Insulation of 1-cm thickness and having a thermal conductivity of 0.05 W/m °C is fixed on the
outside of the sphere. Assuming that the values of hi and h 0 are both 10 W/m 2 °C, calculate the
total thermal resistance. Also, calculate the values of the overall heat transfer coefficient based on
the inner area of the metal sphere and the heat rate.
[30 W]
Show that the critical radius for insulation on a sphere is r = 2 k/h.
.. /= h2k]
[r ,cntica
0
2.40 A spherical shell, of radii r 1 and r2, is made of a material with thermal conductivity k = k0 P..
Derive an expression for the heat-transfer rate, if the surfaces are held at temperatures T1 and T2 ,
respectively.
2.41
Estimate the rate of evaporation of liquid oxygen from a spherical container 1.8-m inside diameter
covered with asbestos insulation. The temperatures at the inner and outer surfaces of the insulation
are -183 °C and 0°C respectively. The boiling point of liquid oxygen is -183 °C and the latent heat
of evaporation is 212.5 kJ/kg. Thermal conductivity varies linearly from 0.157 W/m Kat 0°C to
0.125 W/m Kat -183°C.
[29.8 W]
1
ANSWER KEY
Multiple Choice Questions
2.1 (d)
2.7 (b)
2.2 (a)
2.8 (c)
2.3 (c)
2.9 (d)
2.4 (b)
2.10 (d)
2.5 (c)
2.11 (c)
2.6 (d)
2.2 F
2.3 T
2.4 F
2.5 F
2.6 F
True/False
2.1 T
Fill in the Blanks
2.1 contact, m 2 K/W.
2.4 Logarithmic, geometric
2 .2 isotropic
2.5 7.0
2.3 Laplace
3
STEADY -STATE,
ONE-DIMENSIONAL HEAT
CONDUCTION WITH
INTERNAL HEAT GENERATION
Concept Review
INTRODUCTION
• 3.1
•
Numerous practical cases exist in which internal heat generation must be accounted for. Situations in
which thermal energy is generated due to conversion from some other energy form include
•
•
•
•
Dissipative processes in current-carrying electrical conductors
Exothermic chemical reactions such as the curing of concrete
Neutron absorption in the fuel element of a nuclear reactor
Absorption of gamma rays in external nuclear reactor components like a biological shield, cladding, etc.
Temperature distribution within a medium involving heat generation and heat flux at a specific
location need to be determined. Heat generation may be uniform or non-uniform (a function of position
or temperature).
• 3.2
3.2.1
PLANE WALL WITH UNIFORM HEAT GENERATION
•
Symmetrical Boundary Conditions
The temperature distribution in a plane wall of
thickness 2L when both exposed surfaces (x = ±L)
are at the same temperature Tw is given by
q(L2 -x2)
T(x) = Tw +--2-k--
r---+- X
! Tmax
(3.1)
where q is the rate of uniform heat generation
per unit volume (W /m 3) and is constant. Clearly,
the maximum temperature occurs at x = 0 (the
plane of symmetry where dT = 0) and is equal to
dx
t:;:;
._, J;
111
Tw
Tw
lll
t:~:\.,
-JVV'-q
x=-L
Fig. 3.1
x=O
x=+L
Temperature distribution in a plane wall
3.2 , Heat and Mass Transfer
The temperature distribution in nondimensional form can be expressed as
T(x)-Tmax
(3.3)
Tw -Tmax
The heat flux at any location x can be found by obtaining dT from Eq. (3.1) along with Fourier' s law.
dx
Note that with heat generation, the heat flux is not independent of x.
Heat flow rates at the two ends are determined to be
Q(- L) = - q AL
and
Q( +L) = + q AL
[Negative sign implies that heat transfer is in negative x direction]
Total heat-dissipation rate,
Bout
=IQ(-L) l+I Q(+L) I= q(2 AL)
This must equal the heat generation rate,
£gen= q(:V-) = q(A.2L)
With Convective Boundary At x = L, the control surface energy balance gives
It follows that
(3.4)
3.2.2
One Surface Insulated
qcond ____...
____... qconv
Adiabatic face
h,
[ :: =oJ
T(O)=Tmax
f f f
,~
Fig. 3.2
r_
L
------I
Temperature distribution in a wall with one adiabatic surface
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3 .3
This is a special case when one surface is adiabatic (perfectly insulated), (x = 0) and the other
maintained at a prescribed temperature, Tw. The temperature distribution is
q(L2 -x2)
T(x) = Tw +--2-k__ ,
(3 .5)
This expression is same as Eq. (3.1) but with a difference. The
wall thickness in this case is L and not 2L.
With Convective Boundary
x=+l
x=-l
Exposed surface temperature,
(3.6)
where L is the plane wall thickness.
3.2.3
Asymmetrical Boundary Conditions
For a plane wall with thickness 2L experiencing uniform heat
generation and surfaces held at temperatures T1 (at x = -L) and T2
(at x = +L), the temperature distribution can be expressed as
q ( L2 - x2) ( T- - T. )
l
T(x)=---+
_2_ _
1 x+-(I;+I;)
2k
2L
2
(3.7)
Plane wall experiencing
uniform heat generation
Fig. 3.3
Location where maximum temperature exists is calculated from
x=~(I;-I;)
q 2L
• 3.3
(3.8)
LONG SOLID CYLINDER WITH UNIFORM HEAT GENERATION
•
Consider a long solid cylinder of radius R and a
surface temperature Tw of thermal conductivity k
in which the constant rate of uniformly distributed
heat generation is q. The temperature distribution
in this radial system is
I/;
h, Too
R
(3.9)
Maximum centreline temperature, Tc (when
r = 0) is
l
(3.10)
Fig. 3.4
long solid cylinder with uniform heat generation
3.4 , Heat and Mass Transfer
so that we can express the temperature variation in the cylinder as a ratio of the temperature excess to
the maximum excess as
T(r)-Tw
R2 -r2
Tmax -Tw
r2
(3.11)
Heat-transfer rate at any radial location is given by Fourier's law as,
.
dT
(-2rq)
Q=-k(2rcrL)-=-k(2rcrL)
-dr
4k
= (rcr2 L)q
At r = R, the surface heat dissipation rate is
(3.12)
IQ= (rcR2 L)q = .Egen =q(volume)I
The centreline is a line of symmetry (r = 0) where the temperature gradient must be zero.
With Convective Boundary From an energy balance,
or
3.3.1
(3.13)
Current-Carrying Conductor
Heat generation rate, Qgen = / 2 R,,
where the electrical resistance, R,, = p ~ where L is the length of the conductor and Ac the cross-sectional
area equal to rcR2 •
4
-R2[ 2]
where i is called the current density. Temperature distribution is T(r) = Tw + q4 k 1- ~
and
T.
max
qR2
pi2R2
i2R2
-T. = - = - - = - w
4k
4k
4kke
where k. is the electrical conductivity (
=}).
(3.14)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
• 3.4
3.4.1
3 .S
HOLLOW CYLINDER WITH UNIFORM HEAT GENERATION
Hollow Cylinder with an Insulated Inner Surface(r = r1, : : = 0)
•
!!!
h,
r_
Insulated - - r - - - . i
surface
Fig. 3.5
Heat generation in a hollow cyclinder with an insultated inner surface
The temperature excess over any point over the outer surface temperature is
qr-2
r2
r
r
4k
1i
1i
1i
l
T(r)-I; = - 2 [( 1- 2 ) +2 ( - ) 2 ln-
(3.15)
Maximum temperature will occur at r = r 1.
Heat dissipation rate through the outer surface must be equal the thermal energy generated from the
energy conservation standpoint. That is,
(3.16)
With Convective Boundary
(3.17)
3.4.2
Hollow Cylinder with an Insulated Outer Surface(r = r2 , : : = 0)
The radial temperature distribution is
q,?:
r
'i
T(r) =I;+- 2/n-+ 4k
'i
r2
( )2- (-1ir )2]
[
(3.18)
Maximum temperature will occur at r = r 2 .
With Convective Boundary
T.=T+
I
=
q(-,:2 -r.2)
2
2hfi
I
(3.19)
3.6 , Heat and Mass Transfer
!!!
Insulated
surface
Fig. 3.6
3.4.3
Heat generation in a hollow cylinder with an insulated outer surface
Hollow Cylinder with Both Surfaces Maintained at Constant Temperatures (r = r 1,
T=T 1, r=r2, T=T2)
Fig. 3. 7
Heat generation in hollow cylinder with both surfaces maintained at constant temperatures
The radial temperature distribution is
T(r)-I;
(3.20)
z; -I;
Maximum temperature will occur at a radius rm(r 1 < rm< r2) given by
1/2
r_m
=
qh2-,n-4k(I;-z;)
2q
zn(~ J
(3.21)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, 7
Maximum temperature at r = rm will be
(3.22)
SOLID SPHERE
• 3.5
•
For a solid sphere of radius R, thermal conductivity k, outer surface temperature Tw and volumetric rate
of uniform heat generation, q , the temperature distribution in the radial direction is
T(r)=T + q(R2 -r2)
(3.23)
6k
w
R
l l l
h,
Fig. 3.8
r..
Heat generation in solid sphere
and the maximum temperature (at r = 0) is
(3.24)
With Convective Boundary
(3.25)
where T= is the adjoining fluid temperature and h is the surface heat-transfer coefficient. Heat dissipated
at the exposed surface (r = R) must equal the heat generated, £gen and equals
.
.
(4 )
Q(r=R)=Egen=q 31CR3
• 3.6
(3.26)
HEAT GENERATION IN NUCLEAR FUEL ELEMENT
Consider a cylindrical fuel rod with internal heat generation varying with position as follows:
where R is the radius of the solid fuel element.
•
3.8 , Heat and Mass Transfer
R
1 1 1
h,
T(O)
r_
~q
I q= ifo {I -(r/R)2} I
Heat generation in nuclear fuel element
Fig. 3. 9
The temperature distribution is given by
(3.27)
where Tmax = Tc (at r = 0) is the centreline temperature. Maximum temperature drop in the fuel rod is
T..
max
-T..
w
3 - R2
=
-~
16 k
(3.28)
Heat-transfer rate at the surface (r = R) is
. q (2rcRL)R
0
(3.29)
Q=--4
With Convective Boundary
Maximum (centre) temperature in the fuel element is
T..
max
• 3.7
R{ hl 4k
3R}
q0 - -+-=T +
-
4
(3.30)
HEAT GENERATION IN NUCLEAR FUEL ELEMENT WITH CLADDING
•
Temperature distribution in the cladding is
(3.31)
Temperature distribution in the fuel element is
Maximum temperature in the fuel rod is
T
max
=T.. + %~ {-3-+_!_/n
w
4
4kF
kc
Cladding (kc)
Re}
Rr,
(3.33)
Fig. 3.10
Heat generation in
nuclear fuel element with
cladding
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, 9
Solved Examples
(A) PLANE WALL WITH EQUAL SURFACE TEMPERATURES
Example 3.1 ~ Derive an expression for the temperature distribution in a plane wall in which there are
uniformly distributed heat sources which vary according to the linear relation
where qw is a constant equal to the heat generation rate per unit volume at the wall temperature Tw· Both
sides of the plate are maintained at Tw and the plate thickness is 2L
Known: Volumetric heat generation within a slab according to a prescribed linear relation.
Find: Temperature distribution in the plane wall, T(x).
Schematic:
I
q=qw[I -i~(T-Tw)]
I
1---L~L-j
Plane of symmetry(x= 0)
Fig. 3.11
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Constant properties.
Analysis: The governing differential equation is
dx 2
k
k
qw ~ qw (T - Tw)
=--+-~--~
k
k
Let 0 = T- TW'· m 2 = ..,Aq W I k
d x2
Hence,
d2 0
dx 2
q
k
- - - m 2 0 =-_:!!._
dx 2
(1)
3.10 ,
Heat and Mass Transfer
The solution will be of the form
(2)
(3)
(4)
The boundary conditions are the following:
dT
At x = 0, -
dx
= 0 (by symmetry)
At x=±L, T= Tw
From Equation (1)
a20 =m20- qw =m2
dx 2
k
[e-~J
km 2
From Equation (4),
a20
-=m2 [ Cemx +C e-mx]
dx2
I
2
Equating the two expressions,
0 - qw =Cemx +C e-mx
km2
I
2
Substituting for 0 from Equation (2),
ao + CI emx + C2 e-mx - km2
qw =CI emx + C2 e-mx
or
~
~
Applying the first boundary condition to Equation (3),
or
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, I I
Applying the second boundary condition to Equation (2),
0(x)lx=L =0(L)= T (x = L)-Tw = 0
or
C=
or
-qw
km2 (eml +e-ml)
The temperature distribution is then given by
0(x)= qw +C[emx+e-mx]
km 2
qw [
coshmx]
= km 2 l- coshmL
T(x)=Tw+( qw )[•- coshmx]
km 2
coshmL
or
(Ans)
Example 3.2 ~ A large 2-kW electric heater (25 cm x 25 cm x 0.1 cm) can be idealised as an infinite plane
wall. The heater element is exposed on both sides to ambient air at 30°C and a convective
heat-transfer coefficient of 85 W/m 2 K. The heater material has a thermal conductivity of 0.55 W/m K. Determine
the maximum steady state temperature in this heater and the location at which it occurs.
Known: A large electric heater approximated as a plane wall experiences internal heat generation. Heat
is dissipated to the surrounding air on both sides.
Find: Maximum heater temperature, TmaxC 0 C); Location at which Tmax occurs, x (m) .
Schematic:
Heater element
k=O.SSW/mK
lll
lll
h=85W/m2K
h=85W/m2K
-J\N'--
---r~=Jo c
0
Qgen:2kW
2L=C).I cm
x=-L
'
'
Fig. 3.12
x=+L
3.12 ,
Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) One dimensional conduction. (3) Constant properties.
(4) Uniform heat-transfer coefficient.
Analysis: Volumetric internal heat generation,
_
Qgen
q=-
¥-
where ¥- is the volume of the heater element.
¥-=0.25mx0.25mx0.001m = 62.5x10-<i m3
q=
2x1Q3 W = 32x106 W/m3
62.5x10-<i m 3
The governing differential equation is
d 2T q
-+-=0
dx 2 k
Integrating twice with respect to x, one gets
dT =- qx +C
dx
k
I
and
The two constants of integration can be evaluated by invoking two boundary conditions (BC).
BCI: By symmetry, at x = 0, i.e., centreline, T= Tmax and the temperature gradient,
dT =O
dx
0 = - q(O) +C1
k
BCJJ:
At X = ±L, %ond = %onv
(from surface energy balance)
or
or
or
where
and
( qL) = h[Tw -T-]
-k -k
!Tw -T- =q LI hi
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, I 3
Temperature distribution is thus given by
q(L2 -x2)
T=T + - - - w
2k
Differentiating with respect to x and equating to zero we have
q
dT
~
-=0=0+-(0-2x)
dx
2k
qx
. .
.
--=0. This is possible only atx=O.
k
location of maximum temperature is at x = 0, i.e., at the centreline.
Maximum temperature, (at x = 0) is
(Ans)
or
1
=30°C+(32xl06 W / m3 ) ( 0 ·05 m)[
O.OSm
+
]
100
100x2x0.55W / mK 85W / m2 K
= 30°C+l95.5°C
= 223,s c
(Ans)
0
(B) PLANE WALL WITH ONE SURFACE INSULATED
Example 3.3 ~ Heat generation occurs at a uniform volumetric rate of 1.5 x IOS W / m3 in a 0.3-m-thick plane
wall (k = 16 W /m 0 C) with one surface maintained at 250°C and the opposite surface effectively
insulated. Determine (a) the steady-state temperature distribution in the wall, (b) the location and value of the maximum
wall temperature, (c) the steady-state heat flux at the isothermal surface, and (d) the average temperature in the wall.
Known: Heat is generated in a wall with one isothermal surface and the other insulated surface under
specified operating conditions.
Find: (a) T(x), (b) Tmax and Xmax• (c) Tavg·
I-
Schematic:
L = 0.3 m
-I
---++--k= l6W/m°C
672°C
Insulated side
250°C
--,J\f\/'--+-
Egen
q= I.Sx I05W/m3
x=O
x=L
Fig. 3.13
3.14 ,
Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) One-dimensional conduction (in the x-direction).
(3) Uniform heat generation. (4) Constant properties.
Analysis: (a) The governing differential equation is
d 2 T =-g_
dx2
k
Integration twice yields
dT =- qx +C
dx
k
I
and
qx2
T(x) = -2k+C1x+ C2
The constants of integration C1 and C2 can be evaluated from the specified boundary conditions, viz.,
T(x = 0) = 250 °C
and
dTI
-o
dx (x=0.3m)
(insulated surface)
Hence
(l.5x10 5 )(0.3) C
0 -__ ----+ I
16
~
Also,
The temperature distribution is thus given by
T(x) = -(l. 5 xl0 6 )x2 + 2812.5x+ 250
2x16
= 250 - 4687.5 x 2 + 2812.5 X
or
I T(x) = 250 + 4687.5 (0.6 x - x 2) I
(Ans) (a)
(b) Maximum temperature will obviously occur where dT!dx = 0, i.e., at the insulated surface (x = 0.3 m).
Therefore,
= 0.3 m
Tmax = 250 + 4687.5 (0.6 X 0.3 - 0.3 2)
xmax
and
= 671.875°C = 672°C
(c) Steady state heat flux at x = 0.3 m is clearly zero while that at x = 0 will be
q(x=O)=-kdTI
dx (x=O)
= -k[O+Ci] = 16x2812.5 = -45x10 3 W /m2
(Ans) (b)
(Ans) (b)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.15
(-ve sign implies that heat flows to the left as x is positive towards the right).
From energy balance,
or
%ut =lq(x=O)l=45xl0 3 W/m2
q L = 1.5x105 W/m3 x0.3m = 45x103 W/m2
and
Hence OK
Thus, the heat flux at the isothermal surface is
q =45 kW/m2
(Ans) (c)
(d) The average wall temperature is determined from
L
L
JT(x) dx= J(250+2812.5-4687.5x2)d.x
0
0
1+ 2812.5 x2 ]L - 4687.5 x33]L
2
= 250x
0
0
= 25 0L+ 2812.5 L2 _ 4687.5 I}
2
T.
avg
3
= 250 + 2812.5 L- 4687.5 L2 = 250 + 2812.Sx0.3
2
3
2
= 531.25°C = 531 °C
4687.Sx0.32
3
(Ans) (d)
Example 3.4 ~ Consider a 0.2-m thick wall (k = 4 WI m 0 C) with volumetric uniform heat generation rate of
2 kW/ m3 • The air inside a chamber at 70°C is heated by this wall with an associated
convection coefficient of 20 W/m2 0 C. To guard against heat losses from the wall to the outside, a strip heater
placed on the outer wall provides the required uniform heat flux. The outside conditions are characterised by
a convection coefficient of 5 W/ m2 °C and the quiescent ambient air at 30°C. (a) Sketch the temperature
distribution in the wall if the heat generated is prevented from going outside the chamber. (b) Calculate the
boundary surface temperatures of the wall under steady operating conditions. (c) What would be the heat flux
supplied by the strip heater? (d) If there is no heat generation within the wall, determine the steady-state outer
wall boundary temperature with the same constant heat flux.
Known: A wall with uniform heat generation and a strip heater with constant heat flux. Convective
boundary surface conditions.
Find: (a) T(x) with outer surface of wall adiabatic. (b) T (x = 0) = T 1 and T(x = L) = T2 • (c) q0 (W/m2)
(d) Ifq=O, T(0)=T1•
3.16 ,
Heat and Mass Transfer
Schematic:
Strip heater ~CD
Outside air
Wall
Inside air
-JV\!'-+- q= 2000W/m3
!!!
!!!
k=4Wlm °C
h0 =5W/m 2°C
h;=20W/m2°C
=,I-=70°c
r
r=,0 =Jo c
0
L=0.2m
Fig. 3.14
Assumptions: ( l) Steady operating conditions exist. (2) Uniform volumetric heat generation.
(3) One-dimensional heat conduction. (4) Constant properties.
Analysis: (a) The temperature distribution, T(x) under the specified conditions is shown below:
T(x) c
0
r oo, i-+- -------------------------------- ------------;-- ---,- - - - roo, i
T=, o-+- ------- --- -- -- ---- ----- - ------- --- ---------- - - -·--- ___ ,______________ - --~ - - - - - - - - - - - - - ~ - - ~ x(m)
x=L
x=O
Fig. 3.15
We note that there is no scope for escape of heat generated in the wall to the outside air because of
strip heater. Hence, at x = 0 (outer boundary of the wall), the surface can be considered adiabatic with
dT = 0 and I; = Tmax. The temperature profile is parabolic.
dx
(b) The governing differential equation is
d 2T
q
dx 2
k
-+-=0
or
d (dT)
dx dx
=-,;q
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.17
Integrating,
dT =-qx +C
dx
k
I
Further integration yields
Boundary conditions
At x=O, dx =0
~
le, =01
IL At x = L, T = J;
~
C2 =J;+--
I.
dT
qL2
2k
q(L2 -x2)
T(x) = - - - + J ;
2k
qL2
T(O)=J; =J; + -
and
2k
At the inner wall, control surface energy balance gives
qcond,in
=qconv,out
-k dTI
=h-(r-2
-T~.,·)
'
dx x=L
-k ( -qL)
k =h-(J:
' 2 -T~.,·)
or
qL
J;=T~.,.+hi
Substituting numerical values,
y; = 700C+ (2000W/m3)(0.2m) = 900 C
2
and
20W/m2 °C
T. = 9ooc+ (2000W/m3)(0.2m)2 = 1000c
1
2x4W/m°C
(Ans) (b)
(Ans) (b)
(c) At the outer wall surface (x = 0) subjected to convective process, we have
Heat flux,
% =h 0 ( 7; - T~,o)
= 5 W/m2°C (100- 30)°C
=350 W/m2
(Ans) (c)
3.18 , Heat and Mass Transfer
(d) Without heat generation: (q = 0)
The equivalent thermal circuit is
qo
- - - qleft
r=,o
T1
j
l/h 0
qright-
T2
Uk
1/h;
Fig. 3.16
350 w /m2 =
or
cz; -
cz; -
30)oc +
70)oc
1/5W/m2 °C (0.2m/4 W/m 0 C)+(l/20W/m2 °C)
or
350=5T1 -150+ 10 T1 -700
or
15T1 = 1200
The steady-state temperature at x = 0 is
(Ans) (d)
(C) PLANE WALL: DIFFERENT END SURFACE TEMPERATURES
Example 3.5
~ A plane wall is a composite of two materials, A and B. The wall of material A has uniform
heat generation q=l.5xl06 W/m 3 , kA = 75 W/m°C, and thickness LA= 50 mm. The wall
material B has no generation with k8 = 150 W / m °C and thickness L8 = 20 mm. The inner surface of material
A is well insulated, while the outer surface of material B is cooled by a water stream with T= = 30°C and
h = 1000 W/m 20 C.
(a) Sketch the temperature distribution that exists in the composite under steady state conditions.
(b) Determine the temperature T0 of the insulated surface and the temperature T2 of the cooled surface.
Known: Plane wall A with internal heat generation has one side insulated and the other makes an interface
with the other wall B without heat generation but exposed to convective environment.
Find: (a) Temperature distribution under steady operating conditions. (b) Insulated surface and cooled
surface temperatures, T0 and T2 •
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, I 9
Schematic:
rx
T1
Water
Insulated
surface(To)
<iA = 1.5 MW/m 3
"ifa=O
®
@
-v\/\f'-+ Egen
kA=75W/m°C
ks= 150
W/m°C
!!!
Cooled
surface (T2)
T"'= 30°C
h= IOOOW/m2°C
I - LA= 50 mm -----+-J-+-- L8 = ---1
20mm
Fig. 3.17
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Constant properties and uniform heat
transfer coefficient. (3) No contact resistance. (4) Inner surface of wall A is adiabatic.
Analysis: Plane Wall A: Energy balance
·O
Jn
(no ya1 inflow)
·
· _ ·O
- Eout
+ £gen lff.1
(st~y state)
Hence,
or
Hence, the steady-state heat flux,
qA =qALA =(l.5xl06 W/m 3 )(0.05m)=75000W/m2
(since dT = 0)
dx
At x = 0 (insulated surface), T=To =Tmax
We note that
where I; = Tw is the interface temperature.
Plane wall B
The temperature T2 of the exposed surface of plane wall B is determined from
At the interface, qA = q8 and since there is no internal heat generation in wall B (i.e., q8 = 0 ), the
heat flux at the exposed surface is same as qA, i.e., 75000 W/m 2 .
Therefore, the temperature of the cooled surface is
r,. =T + qA = 3ooc+ 75000W/m2
2
h
1000 W/m 2 °C
00
10s c
0
(Ans) (b)
3.20 , Heat and Mass Transfer
To find T1 we observe from the following thermal circuit:
- - - qs
~
~
~
••-~VVv~---11•,__-~VVv---e•
Rcond, 8
Rconv
Fig. 3.18
or
Interface temperature,
T.
1
1
0 ·02 m +
=30°C + 75000 w /m 150W/m
]= 115 °C
°C 1000W/m °C
2[
2
Else
m)
I;= 105°C+ ( 75x 103 W/m2 x20xI0- - - - = 115°C
3
150W/m°C
The temperature T0 is then determined to be
-L2
T =T.+~
o
I
2k
A
W/m )(50x 10- m)
c
= l 1 5 °(l.5x10
C + - - - - - - - - - - 1400
6
3
3
2
(Ans) (b)
2 x75W/m 0C
The temperature distribution is qualitatively sketched below:
Parabolic
Tl: 115 ° C 1-----
-------- - -------------------------- ~ c
T2 = I 05 °c 1--------
-------- ---- -- ------------------ ___ j_ __
________
1- ---------,------------------
_____ ,......,._ ___ _______________
I
I
®
T~=30 ° C1
0
----------------------------------------------------------- +
x=LA
Fig. 3.19
-------------------------
+--------- - - - - ;
(Ans) (a)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.21
Comment
•
•
In plane wall A, the temperature profile is parabolic while in plane wall B, the temperature variation is
linear.
Thermal circuit representation is not possible in a slab with heat generation.
•
The convective resistance (~onv =
.!..)
is much greater than the conductive resistance C~ond = Ls ).
h
k
8
Since t:.T = q Rih, the temperature drop across wall B (T, - T2) = 10 °C) is significantly less than the
surface-to-fluid temperature drop (T2 - T= = 75 °C).
Example 3.6
~ A plane composite wall comprising three layers A, B and C has its outer surfaces exposed
to a fluid flow characterized by T= = 40°C and h = 1250 W /m 20 C. Under steady operating
conditions, for which uniform heat generation occurs within the middle layer B at a volumetric rate ifs while
the layers A and C do not experience any heat generation, the interface temperatures are T1 = 275 °C (between
A and B) and T2 = 225 °C (between B and C). The thicknesses of the three walls are LA= 30 mm, Ls= 60 mm,
and Le:= 25 mm. The thermal conductivities of walls A and C are kA = 25 WI m °C and kc= SO WI m 0 C.
Neglecting interfacial contact resistances and assuming one-dimensional conduction, calculate (a) the volumetric
rate of thermal energy generation, ifs, (b) the thermal conductivity of the middle wall ks, and (c) the maximum
temperature in the wall.
Known: A plane wall with heat generation is sandwiched between two other walls without heat generation.
Convection process on exposed surfaces.
Find: (a) Heat generation rate, q8 (W/m 3 ). (b) Thermal conductivity, k8 .
Schematic:
r2 =22s c
T1=275°C
0
®
!!!
©
®
h= 12SOW/m2°C
Too=40°C
h= 12SOW/m2°C
Too=40°C
k8 =?
kc=2S
W/m°C
LA=30mm f.------ L8 =60mm -----+-JLc=2Smmi-
kA=2SW/m °C
--J
lll
Fig. 3.20
Assumptions: ( 1) Steady-state, one-dimensional conduction.(2)
Uniform heat generation. (3) Negligible contact resistances. (4)
r 1=21s c
0
Constant properties.
Analysis: (a) Applying energy balance on plane middle wall B:
®
t,/J - £out + £gen = t,!
(no hJa:ifnflow)
(steat\iate)
Hence,
or
I- 2L=60mm --1
(2L)·q8 =q, +q2(W!m 2 )
~ !"?fs =(q, +q2) 12Lj
x=-L
(A)
Fig. 3.21
x=+L
3.22 , Heat and Mass Transfer
We construct the thermal circuits for walls A and C (without heat generation) to evaluate the heat
fluxes q 1 and q2 •
T..
T1
~
1/h
-
qI
LA/kA
Fig. 3.22a
I;_ -T=
(275-40)°C
q, = (1/ h)+(LA I kA) = [
1
0.03m
]
1250W/m2 °C + 25W/m°C
= 117500W/m2
r..
Lclkc
1/h
Fig. 3.22b
1i - T=
(225 - 40)°C
q2 = 4 +_!_ = [ 0.025m
1
kc h
50W/m°C + 1250W/m2 °C
J
= 142308W/m2
Substituting for q 1 and q2 in equation (A), one gets
_
% + q2 117500+ 142308
qB = ~ =
0.06
=4.33x106W/m 3
(Ans) (a)
(b) The temperature distribution T(x) in the middle wall (with heat generation) with end surfaces at
T1 and T2 is
(Ii-Ti) (Ti +Ii)
2)
q(L2-x
T ( x) = ---+
- - x+ - 2kB
2L
2
dT = q (0-2x)+(1i-1i_)
dx 2kB
2L
Heat flux at x = + L, i.e., q2 can be found from Fourier's law:
q2 =-kB dTI
=-kB[-qL - Ti.-1iJ
dx x=+L
kB
2L
(I;_-7i)kB
=qL+---2L
(B)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.23
(q2 -qL)2L
or Thermal conductivity, k s = - - - - -
(i; -i;)
=
[142 308-(4.33x106 )(0.03)] (0.06)
(275-225)
= 14.9W/m°C
(Ans) (b)
(c) Maximum temperature will occur in plane wall B where
dT =O
dx
'ifsx i;-i;
--+--=0
ks
2L
i.e,
X
=
(½ -J; )(ks)=
(225-275)
14.9
'ifs
0.06
4.33x10
X
2L
6
=-2 .87x I0- 3 m or -2.87 mm
The maximum temperature in the wall is obtained from Equation (B) after substituting x = 0.00287 m.
(4.33x 106 )[ 0.032 -(-0.00287 2 )] (225-275)(-0.00287) (275+225)
T_
=----~------~+--------+
max
2 X 14.9
0.06
2
= 382°C
Example 3. 7
(Ans) (c)
~ A cylindrical rod, IO mm in diameter and 40 mm long with its longitudinal surface well insulated
is made of a semiconductor material (k=2 W/m°C and p=2xl0-5 ohmm). Its ends are
maintained at 100 and 0°C. Determine (a) the midplane temperature, (b) the heat flow rate at each of the two
ends, if the current carried by the rod is 10 A, and (c) the maximum temperature in the rod and its location.
Known: Current flows through a rod with insulated curved surface generating heat and maintaining rod
ends at different temperatures.
Find: (a) T(x=O). (b) Q (x=-L) and Q (x=+L). (c) Tmax• xmax·
Schematic:
/= 12A
k=2W/m°C
Pe=2x 10-snm
(a)
3.24 , Heat and Mass Transfer
Midplane or plane
of symmetry
r 0 = 100°c
Xmax =
-0.0154m
:
;
x=-L
x=O
I-
0.02m - - i - - 0.02m
x=+L
-I
(b)
Fig. 3.23
Assumptions: (1) Steady operating conditions. (2) One-dimensional (axial) conduction. (3) Constant
properties. (4) Uniform heat generation.
Analysis: (a) The axial temperature distribution with midplane (plane of symmetry) as the origin (x = 0)
is given by
T.) (r. + T.)
-q(L2 - x2) ( _2
T._
- _1 x+
T(x)=----+
2k
2L
where
_
·
£gen
q--- - ¥-
2
J2p.(2L/ (~
4 )n2
12 J?..
-(~)D
_2_ _
1
= - - - -1/ - - -
2 (2L)
(~)D
2 (2L)
= _!iPe 12 = .!ix 2xI0- 5 Qmx(lOA)21~1
1t2 D4
1t2
(O.Olm}4
A2Q
= 324.23 x 103 W/m3
Temperature at the midplane,
Tix= O = q L2 +I;+ I;= 324.23x 103 W / m 3 x (0.02 m) 2 + (O+ 100)°C
(
)
2k
2
2 X 2 W / m °C
2
= 82.4°C
(Ans) (a)
(b) The heat flow rates at the two ends can be determined using the Fourier' s rate equation:
Q = _ k ( 1t D 2 ) dT = _ k ( 1t D 2 ) [ - q x + I; - I; ]
x
4
dx
4
k
2L
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.25
At x=-L
Q =( 2 .Y!___)(~xO.Ol2 m2)[(324.23x10 3 W/m3 )(-0.02m)
1
m°C
4
2W/m°C
(0-100)°C]
0.04m
=-0.117 W
(Ans) (b)
At x=+L:
Q =( 2 .Y!___)(~xO.Ol2 m2)[(324.23x103 W/m3 )( +0.02m)
2
m°C
4
2W/m°C
(0-100)°C]
0.04m
(Ans) (b)
=+0.902 W
[-ve sign simply indicates that heat flows at the left face in the negative x direction]
Check:
Energy balance
t
-Eout +£gen=
(no heat inflow)
•
_-
_
_
2
Egen -q¥--/ R. -
/.
(steady state)
/2 p(2L)
102 x2x10-5 x0.04
~ n2
(¾)xo.ol2
( )
4
= 1.019W
£out =IQ, l+Q2 = 0.117 + 0.902 = 1.019 W
Hence OK
(c) Maximum temperature will occur where dT = O
dx
~
i.e,
Location,
X
max
=(I;_-J;)!!_
2L
q
(0-100)°C
2 W/m °C
0.04m
324.23 103 W/m3
xmax = - - - - x - - - - - -
= - 0.0154 m
or
-13.4 mm
(Ans) (c)
And
= (324.23x10 3W/m3 )(0.02 2 -0.01542) m 2 + (0-100)°C ( _ 0_ 0154 ) m+(O + 100) 0 c
max
2x2 W/m°C
0.04m
2
T..
= 101.7°C
(Ans) (c)
3.26 , Heat and Mass Transfer
Example 3.8
~ Find the temperature distribution in a circular rod of length L and diameter D in which the
heat generated per unit time per unit volume increases linearly along the length of the rod
in the form
q=ax
The boundary conditions are x = O, T = T1 and x = L, T = T2• The curved surface of the rod is insulated and
temperature variation in the radial direction may be neglected. Where does the maximum temperature occur?
For what value of 'a' will it occur at the centre of the rod?
Known: Non-uniform heat generation in a circular rod with insulated curved surface and specified end
temperatures.
Find: Location x at which Tmax will occur. Value of constant a if Tmax occurs at the centre of the rod
(x = L/2).
Schematic:
x=O,T=T 1
x=L,T=T2
I
I
ff
I
~ q=ax
L Insulated surface of
~---------------.,--------~
L
x
circular rod
Fig. 3.24
Assumptions: ( 1) Steady state, one-dimensional (x-direction) conduction. (2) Lateral surface is adiabatic.
(3) Constant thermal conductivity.
Analysis: The governing differential equation is
d 2T
q
ax
dx 2
k
k
(1)
subject to the following boundary conditions.
BC/
BC II
x=O
x=L
Integrating equation ( 1),
dT =- ax2 +C
dx
2k
1
Further integration gives
From BC I:
From BC II:
(2)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.27
a/}
]
c, =-L1 [ i;+--Ti
6k
Substituting for C1 and C2 into Equation (2),
ax3 x [ y: +--T.
a/}
] +T.
T=--+1
1
6k L 2 6k
Rearranging, the temperature distribution in given by
T-T. = ~(Y: -T. )+ ax (L2 - x2)
I
L 2
I
6k
or
(3)
.
dT O
For maximum
temperature, - = .
dx
Differentiating Equation (3) w. r. t. x and equating it to zero,
Y: - T. aL2 ax2
0=+-2
__1 + - - L
6k 2k
or
(Ans)
At x=L/2,
~ = {2k [½ -Ti + aL
2
a
2
L
112
]}
6k
Squaring both sides,
L2 = 2k{J;-'fi + aL2 }
4
a
L
6k
2k(J; -Ti) 2kL2
=----+-aL
6k
Cross multiplying,
2k(J; -Ti)
I}
L2
L2
aL
4
3
12
3.28 , Heat and Mass Transfer
Hence
24k (Ji -T2 )
a=-----
(Ans)
L3
Example 3.9
~ A current-carrying laterally insulated copper conductor (k=381 W/m°C, Pe =2xl0~ ohm-cm)
of 10.mm diameter and 1-m length connects two large plates with one face maintained at 80°C
and the other held at 40°C. Develop an expression for temperature profile in terms of nondimensional variables and
determine
(a) the position and value of the maximum temperature if the maximum allowable current is 180 A
(b) the heat flow rate at both ends of the conductor. Indicate the temperature profile
Known: A current-carrying copper conductor with adiabatic lateral surface is connected between two
plates held at different temperatures.
Find: (1) Position and value of maximum temperature. (2) Heat dissipation from the two ends.
Schematic:
x=O, T=T 1=80°C
x=L T=T2 =40°C
Insulated conductor
(k=381 W/m °C, Pe=2x l~Qcm)
t
-""f\_,.__E.
--,, v v ~ gen
D=IOmm
t
/=180A
L= Im
Fig. 3.25
Assumptions: (I) Steady-state, one-dimensional conduction. (2) Constant properties. (3) Uniform
volumetric heat generation.
Analysis: Under steady operating conditions, the energy balance gives
Using Fourier's rate equation,
.
.
dT
Qx = Q,ond,in = - k Ac dx
.
dT d [ -k Ac -dTJ dx
dxdx
dx
.
Qx+dx =Qcond out =-k Ac-+-
'
2
· = -d- [ -k AcdTJ dx=+kAc--dx
d T
E;· 0 -Eout
dx
dx
dx 2
Rate of internal volumetric thermal energy generation,
Q,-l=-EP" i!-Q,.~
I
Hence
-+-
I
~- - - - - - - - - - _ __!
d 2T
k,{-dx+q,{dx =0
dx 2
X
dx
Fig. 3.26
--l
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
d2T q
-+-=0
or
dx 2
3.29
(1)
k
This is a non-homogeneous, second order ordinary differential equation to be solved with the two
boundary conditions at the two ends, specified as follows:
BCI:
T(x= 0) = 'fi
BCII:
T(x=L)= I;_
First integration yields
dT
X
-+q-=C,
dx
k
Second integration gives
x2
T+ q-=C1x+C2
2k
Applying the boundary conditions,
C=I;_-'fi+qL
1
L
2k
or
Thus,
2
I;_-'fi
T(x)= [ - +qL]
- x+T.1 -qx
L
2k
2k
or
Adding and subtracting I; on the left side,
2
T(x)-T. +Y: -Y: =+(Y: -T.)~+ q I} [~-(~) ]
I
2
2
2
I L
2k L
L
or
3.30 ,
Heat and Mass Transfer
Defining nondimensional variables:
e T(x)-7i
i; -Ii
and
X
~=-,
L
the temperature distribution can be expressed as
Defining nondimensional heat generation parameter,
qJJ
H =2k(J; -7i)'
one can write
0=(1-~)+H~{l-~)
or
10={1-~)(l+H~)I
To get the maximum temperature, we differentiate 0 with respect to ~ and equate it to zero.
d0
a~=o=c-1) c1+m)+c1-~) CH)
or
-1-H~ +H-H~ =0
or
H-1=2H~
x
H-1
~=L= 2H
Substituting numerical values,
H=
qJJ
2k(J; -Ii)
=
12 p
where
_ {180 )2 (2x 10-8 )
{1t/4)2 D 4 -
{1t/4)2 {0.01)4
= 1.0505 x10 5 W/m3
H =(l.0505x10 5 )(1) 2 /2(381)(80-40)
= 3.4465
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.31
Position of maximum temperature,
x=(H-1)£=(3.4465-1)(1) =0.3 55 m
2H
2x3.4465
(Ans) (a)
Maximum temperature is determined from
0=(1-~)(1+ H~)=(1-i)( l+H
z)
= (1 - 0.355) (1 + 0.355 X 3.4465)
= 1.434
With
Tmax = J; + 0 ('fi - J;) = 40°C + 1.434 (80 - 40)°C
= 97.4°C
(Ans) (a)
Temperature gradient,
dT d
d0
-=-[0('fi -i;)+i;J = (Ti -i;)dx dx
dx
= ('fi -J;)[-l+H-2xH]
(since L = l m)
= ('fi -J; )[(H - l)-2Hx]
= (80 - 40) [(3.4465 - 1) - 2 X 3.4465 x]
1~
or
= 97.86-275.nxl
Heat dissipated from the 80 °C side is
(1t )
dT
Q.1 =-kA,,-=-k
-D2 -dTI
dx
4
dx x=O
3.32 , Heat and Mass Transfer
= (-381 mV:,C )( ¾x0.I2 m2 ) [97.86 - 273.72 x 0]°C/m
=-292.8 W
(Ans) (b)
(-ve sign simply implies that heat flows in the negative x direction)
Q2 =(-381 W / m C) (¾x0.I2m2 }97.86-273.72]°C/m
0
=+ 532.2 W
(Ans) (b)
Total heat dissipated= I Q1 l+I Q2 l=Eout
= 292.8 + 532.2 = 825 W
Check: Since £out = £gen
£gen
=q(¾D L)=(1.0505x10 W/m ¾x0.I2 m xlm)
2
5
3 )(
2
=825 W
Hence OK
The temperature profile will be as follows:
__
----------:...;;-.;..-_.,.._
80
40
0
0.355
1.0
X (m)
Fig. 3.27
Example 3.1 0 ~ Two large steel plates at temperatures 95 °C and 70 °C are separated by a 2.54-cm diameter
steel rod, 30-cm long. The rod (k = 43 W Im K) is welded to each plate. The space between
the plates is filled with insulation, which also insulates the circumference of the rod. Because of the voltage
difference between the two plates, current flows through the rod, dissipating electrical energy at a rate of 12 W .
Determine the maximum temperature in the rod and the heat-transfer rate at each end. Check your results by
comparing the net heat flow rate at the two ends with the total rate of heat generation.
Known: A steel rod welded to two large steel plates at different temperatures is laterally insulated.
Find: Maximum temperature in the rod, Tmax {°C), heat-flow rates at both ends Q0 and QL {W).
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
Schematic:
Q0 -
3.33
1 k=43W/mK
-*
D=2.54cm
L=Q3m
_ _ _! __ _
Insulated _J
x=O
x=L
T(0)=95°C
T(L)=70°C
Fig. 3.28
Assumptions: ( l) Steady-state conditions. (2) One-dimensional heat conduction in the rod (3) Uniform
heat generation in the rod.
Analysis: The relevant heat conduction equation is
d 2T =-g_
dx 2
k
where the rate of uniform volumertic heat generation is
_
12 W
Qgen
q=--=
"!:...n 2 L "!:...(2.542 x10-4 m2 ) (0.3 m)
4
4
= 78941.0 W/m3
Boundary conditions: T= T1 at x = 0
T= T2 at x = L
Integrating twice successively,
dT =-qx+C
1
dx
k
qx2
T(x) = --+C1x+C2
2k
Applying the boundary condition at x = 0,
Applying the boundary condition at x = L,
C = I;-T; + qL
1
L
2k
Hence, the temperature distribution is
-L) -
'E T.
2
T(x)=T.+ ( _L__!.+L x-:f_
1
L
2k
2k
(1)
3.34 ,
Heat and Mass Transfer
The maximum temperature occurs at x = xm where the first derivative dT = 0:
dx
(2)
or
(3)
The location at which Tmax will occur is
X
m
(I; -I;)k L
=~-~+qL
2
Equation (1) can now be written for x = xm
from Eq. (3)
-
2
=T. + qxm
I
2k
Substituting the values of xm, we have,
= T,
1
+q [(z; -z;) +~]2
k
2k
qL
2
78941.0W/m3 [{(70-95°C)}(43W/mK) 0.3m]2
= 95°c +------'-- -=-------=------'-+-(2){ 43 W/m K) (78941.0W/m3){0.3m)
2
= 10s c
(Ans)
0
The heat flow rate from the rod at x = 0 is
Gi =-kAdTI
dx x=O
= -k ( ~ D2 ) [½~I; + ; ~]
=-{43W/mK) (~x0.0254 m2)x[(70-95)oc + 78941W/m3 x0.3m]
4
0.3m
2x43W/m K
=-4.18 W
(Negative sign merely indicates that the heat flow is in the negative x direction.)
(Ans)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.35
From Equation (2)
dT I
= r; - 'Ii + qL - qL
dx x=L
L
2k
k
T -Ti qL
-2 - - - L
2k
-25°C _ (78941W/m3 )(0.3m)
0.3m
2(43W/mK)
=-358.7 Kim
Heat flow rate from the rod at x = L is
QL = (-43 W/m K)( ¾x 2.542 x 10-4 m2 ) (-358.7K/m)
=+7.82 W
The sum of the heat losses from the rod is
(Ans)
This is the same as the electrical power generated.
Example 3.11
~ An electric current of 34000 A flows along a flat plate 1.25-cm thick and 10-cm wide. The
temperature of one surface of the plate is 80 °C and that of the other is 95 °C. Find the
temperature distribution across the plate, and the value and position of the maximum temperature. Also, calculate
the total amount of heat generated per metre length of plate and the flow of heat from each surface of the plate.
The end effects along the short sides of the plate may be neglected, i.e. it may be assumed that no heat
flows across these surfaces. It may also be assumed that the ohmic heating is generated uniformly across the
section. The resistivity of the steel is 12 x I o-6 ohm cm, and its thermal conductivity is 54 W/m K.
Known: Current flows along a flat plate generating heat with different specified boundary surface
temperatures.
Find: Temperature distribution, T(x), x and Tmax; Heat-generation rate per m length of the plate, Q/L;
Heat flow rates from both end surfaces, Q1 and Q2 .
~ k = 54 W/m K }
i~ . 1 p= 12x 10--oncm
Schematic:
x=0.552cm
r- 2 L = 1.25 cm ---1
x=O
x=L
Fig. 3.29
3.36 ,
Heat and Mass Transfer
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Homogeneous and isotropic material.
(3) Uniform volumetric heat generation.
Analysis: The governing differential equation for this essentially steady-state, one-dimensional problem
with uniform heat generation is
d2T
q
dx 2
k
-+-=0
where
q = heat generated per unit volume
= pi2 where p is the resistivity, and i the current density.
Current density,
I
34000A
i=-=-----=2720 A/cm 2
A 1.25cmx10cm
q = (12 x 10-6 ohm cm) (2720 A/cm2)2 I 1W 2 I
lohm A
= 88.78 W/cm3
= (88.78) (l0-3) (10 6) kW/m 3
= 88780 kW/m3
The differential equation then becomes
d2T + (88780) (10 3 ) = O
dx 2
54
d2T + 1644089 = 0
dx2
or
Integrating this equation successively twice, one obtains
dT
-=-1644089 x+c,
dx
- -1644089
T- - - x2 + C1x+ C2
and
2
where C 1 and C2 are arbitrary integration constants.
The boundary conditions are
T = 95 °C at x = 0
T= 80°C
at x = 0.0125 m
Thus, from Equation (A):
Also, from Equation (A):
80 = - (822044.45) (0.0125) 2 + C1 (0.0125) + C2
or
c. = 10273.56 - 1200
= 9073.56
(A)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.37
Hence, the temperature distribution is given by
!T = -822044.45x2 +9075.56x+95!
(Ans)
This indicates a parabolic temperature profile.
Differentiating the above equation with respect to x and equating it to zero, the position of the maximum
temperature can be identified.
dT
- = -1644089x+9075.56 = 0
dx
or
~
x = 5.52xl0-3 m
(Ans)
x=0.552 cm
The value of the maximum temperature is then found to be
Tmax = -
(822044.45) (3.52 X 10-3) 2 + (9073.56) (3.52 X 10-3 ) + 95
=-25 + 50 + 95
= 120°c
The heat flow from the plate across the surface at 95°C per m length of plate is
. .
)(dT)
Q, = ~-L) =-(-kA -
dx
(Ans)
= kAC1
x=O
= (54) (0.10 X 1) (9073.56) (10-3) kW
=49.0 kW
The heat flow from the plate across the surface at 80 °C per m length of plate is
. .
(Ans)
(dT)
Q2 = Q(+L) = -kA dx
x=0.0125 m
= -(54) (0.10 X 1) (- 20551.11 + 9073.56) (10-3) kW
=-(54) (0.10) (-11473.55) (l0-3) kW
= 61.97 kW
(Ans)
The sum of Q1 and Q2 , i.e., (49 + 61.97) = 110.97 kW must be equal to the total heat generated per
unit length of plate by ohmic heating, which is
Qgen = q¥- =(88780 kW/m3) (lm X 0.0125 m X 0.10 m) = 110.975 kW
(Ans)
(D) LONG CYLINDER
Example 3.12 ~ An electric heater uses nichrome wire of 2-mm diameter and 110 micro-ohm-cm resistivity
as a heating element 25 A current flows through the wire (k = 17.5 W/m 0 C) and the heat
generated is dissipated to 20°C air with a heat-transfer coefficient of 46.5 W/m2 °C. Determine (a) the heattransfer rate per m length, (b) the surface temperature, and (c) the centreline temperature.
3.38 , Heat and Mass Transfer
Known: Heat generation in an electrical heating element under specified conditions.
Find: (a) Heat rate per unit length, Q. (b) Surface temperature, T. (c) Centreline temperature, ~.
L
Schematic:
Air
h=46.SW/m 0 c r==20°c
~
/=2SA
k= 17.SW/m°C
Pe= I I Ox I0-6Qcm
Fig. 3.30
Assumptions: (I) Steady-state, radial conduction. (2) Constant conductivity and uniform convection
coefficient.
Analysis: Electrical resistance, per metre length,
R.=p..!:_ = p___£_ = I !Ox 10-8 Qm x Im
4
re
re x (0.00 I m) 2
r;
= o.35 n
Heat generated per unit volume,
q=Qgen /rcrJL
I 2J?.
25 2A 2 x0.35Q I lW I
= rcrJ L = rcx(I0-3 m)2 x Im IA 2 n
= 69.63 xl06W /m3
and, the heat flow per m length,
Q = /2(R_ / L)
L
= (25)2(0.35) = 218.8 W
(Ans) (a)
Surface temperature,
T =T + qr;, = 20 0 C+ 69.63x!0 6 W / m3xJ0-3 m
s
= 2h
2x46.5W / m2°C
= 768.7°C
(Ans) (b)
Centreline temperature,
69.63x 106 W /m3 x(I0-3 m)2
4xl7.5W / m°C
=768.7°C+--------= 769.7°C
(Ans) (c)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
Example 3.13
3 .3 9
~ A chemical reaction is being carried out at constant pressure in a packed bed between the
coaxial cylinders with radii of 1.2 cm and 1.8 cm. The entire inner wall is at a uniform
temperature of 500 °C and there is almost no heat transfer from this surface. The reaction releases heat at a
uniform rate of 550 kW/m 3 throughout the reactor. The effective thermal conductivity of the packed bed is
0.55 W/m K. Determine the temperature of the outer wall.
Known: Heat is generated in a chemical reactor's packed bed between two coaxial cylinders with insulated
inner surface.
Find: Outer wall temperature, To (0 C).
Schematic
Q
k=0.55W/mK
r;= 1.2cm
(inner surface insulated)
""---¥4'+-- r O = 1.8 cm
T0 = ?•
Fig. 3.31
Assumptions: (1) Steady-state conditions. (2) Constant thermal conductivity. (3) Uniform volumetric heat
generation. (4) One-dimensional (radial) heat conduction.
Analysis: The governing differential equation is
!~(r
r dr
dT)=-g_ or ~(r dT)=- qr
dr
k
dr
dr
k
Integrating with respect to r,
rdT = qr2 +C
1
dr
2k
or
dT
dr
qr C1
r
-=--+-
2k
where C1 is an arbitrary constant.
Boundary condition I
dT
At r= ,;, = 0 (no heat transfer from the inner wall)
dr
0=-q,;+Ci
2k
,;
qr
lq,;2
2k
r 2k
or
dT
dr
-=--+---
3.40 ,
Heat and Mass Transfer
Integrating with respect to r,
T(r)=+ q
2k
[1r lnr- r2 + C
2
]
2
Boundary condition II
At r= ri, T= Ti
,?-]
2
T =q- [ r?- lnr, _..L +C2
2k
I
I
I
Substituting for C1 and C2 in the expression for temperature distribution, we have
q,f[
1] q[
r2]
11(r) = T - - lnr, -- +- i2- lnr-1
2k
I
2 2k I
2
l}
q,f { Zn-+r
= T +--qr2
1
2k
11
2
qr;.2
r
r
4k
'i
'i
4k
l
T(r)=i;+- [ 2ln-- ( - ) 2 +1
or
Temperature of the outer wall is
T, = 5000C = (550x10 3 W /m3 )(0.012m) 2
0
4(0.55W/mK)
= 484.2°c
2
+l]
[ lln(l.8cm)-(l.8cm)
1.2cm
1.2cm
(Ans)
Example 3.14 ~ A concrete column used in bridge construction is cylindrical in shape with a diameter of
I m. The column is sufficiently long so that the temperature variation along its length may
be neglected. Treating the column as solid concrete with an average thermal conductivity of 0.93 W/m K,
determine the temperature at the centre at a time when the outside surface temperature is measured to be
80°C. The heat of hydration of concrete may be taken to be 0.7 W/kg and an average density for concrete of
2300 kg/m 3 may be assumed.
Known: Heat generation in a cylindrical concrete column.
Find: Temperature at the centre, T (r = 0).
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
Schematic:
3 .41
Concrete, TR= 80 ° C
R=O.Sm
K}
k=0.93W/m
p = 2300 kg/m3
T0 : I•
Q/m=0.7W/kg
Fig. 3.32
Assumptions: (1) Steady-state radial conduction. (2) Uniform heat generation. (3) Constant properties.
Analysis: The appropriate differential equation is
_!_~(r
dT) = _g_
r dr
dr
k
(A)
The boundary conditions are the following:
At r=O,
dT =0
dr
(by symmetry)
And at r=R,
Integrating Eq. (A) with respect to r,
dT
qr2
r-=---+C
dr
2k
I
dT
qr C
- = - - + -1
dr
2k r
or
From the first boundary condition, !C1 = O.!
Further integration yields,
[T]r,r. = [ -qr2
-JR
4k
0
qR2
T, -T = - -
or
R
o
4k
The temperature at the centre,
qR2
J;, =TR+4k
or
{(0.7 W/ kg)(2300 kg/ m3 )}(0.5)2 m2
T =80°C+------------0
4(0.93W/ m°C)
= (80 + 108) °C
= 188°C
(Ans)
3.42 , Heat and Mass Transfer
Example 3.1 S ~ The rate of heat generation per unit volume in a long solid cylinder of radius R is given by
q =a+br2
where a and b are constants and r is any radius. The cylinder is undergoing heat transfer with a medium at
a temperature T_ and the surface heat-transfer coefficient is h. Find the steady state temperature distribution
in the solid.
Known: Heat is generated non-uniformly in a long solid cylinder exposed to convection.
Find: Steady-state temperature distribution.
Schematic:
h,T..
Ill
-vVV'-+- q= a + br2
)
Fig. 3.33
Assumptions: (1) Steady-state conditions. (2) One-dimensional (radial) conduction. (3) Constant
properties.
Analysis: The appropriate differential equation is
.!!...(r dT) =_qr
dr dr
k
(a+ br2 )r
k
The boundary conditions are
(I)
_O
( dT)
dr r=o
-k(dT)
= h(T,.=R -T_)
dr r=R
Integrating the differential equation with respect to r, we have
(II)
dT
ar2 br4
r-=----+C
dr
2k
4k
I
le, =OI
From BC I:
dT
dr
ar
br3
2k
4k
Integrating, again,
ar2 br4
T=----+C
4k
I6k
2
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.43
From BC II:
aR bR3 aR2 bR4
C2 =-+-+-+-+T
2h 4h
4k 16k
=
Substituting for C2 in the expression for T, the temperature distribution is given by
ar2 br4 aR bR3 aR2 bR4
T=----+-+-+-+-+T
4k 16k 2h 4h
4k 16k
=
a
aR
b
bR3
T-T =-(R2 -r2 )+-+-(R4 -r 4 ) + = 4k
2h 16k
4h
or
Example 3.16
(Ans)
~ An internally cooled copper conductor of 4-cm OD and 1.5-cm ID carries a current density
of S x I03 A/cm 2. The temperature of the inner surface is maintained at 70 °C, and it may
be assumed that no heat transfer takes place through the insulation surrounding the copper. Determine the
equation for temperature distribution through the copper. Hence find the maximum temperature of the copper,
the radius at which it occurs, and the heat-transfer rate internally. Check that this is equal to the total energy
generation in the conductor. For copper, take k = 380 W/m K and the resistivity p = 2 x I0-8 ohm metre.
Known: Heat generation in a copper conductor under specified operating conditions.
Find: Temperature distribution, T(r); maximum temperature, Tmax {°C); Radius at which Tmax occurs,
r(°C); Heat transfer, Q{W).
Schematic:
r;=0.7Scm
Copper conductor:
k=380W/mK }
p=2x 10-S nm
i=Sx IOlA/cm2
Outer surface insulated
Fig. 3.34
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Uniform volumetric heat generation.
(3) Constant properties.
Analysis: The governing differential equation is
3.44 , Heat and Mass Transfer
The volumetric heat generation,
q = p i2 = (2 x 10-8 ohm m) (5 x 103 x 10 4 A/m 2)2
1~1
A 2 ohm
= 5 X 107 W/m3
The differential equation can also be written as
_!_
_:{_(r dT) = _9_
rdr
dr
k
!{_(r dT) = -qr
or
dr
dr
k
Differentiating with respect to r, one obtains
dT
qr2
r-=--+C
dr
2k
I
dT
qr C1
-=--+dr
2k
r
or
(1)
The boundary conditions are
dT
=0
dr
-
and
at r = 2 cm
[There is no heat transfer due to insulation at this radius]
at r= 0.75 cm
T= 70°C
It follows that the maximum temperature occurs at r = 2 cm where the temperature gradient, dT = 0
dr
Substituting the boundary condition at r = 2 cm in Eq. ( 1), we have
(5x10 7 )(0.02)
C1
0=------+2(380)
0.02
c, =26.316
Integrating Equation (1), we have
qr2
T=--+C1 Zn r+ C2
4k
Substituting the boundary condition at r= 0.75 cm, we get
70 = -
( 5x107 ) ( 0.75x10-2 ) 2
+26.316 ln(0.75xl0-2 )+c2
4 (380 )
= -1.85 - 128.76 + C2
C2 = 200.61
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.45
The equation for temperature distribution is therefore,
(5x10 7 ) r 2
T=-~-~-+26.316 lnr+200.61
4(380)
!T(r) = -32894.7 r 2 +26.316 /n r+ 200.61!
or
(Ans)
where r is in metre.
The temperature profile is obviously parabolic.
The maximum temperature occurs at the outer radius because the outer surface is adiabatic with dT!dr = 0.
Hence, r = 0.02 m.
(Ans)
Hence, the maximum temperature of the copper is
Tmax = -(32894.7) (0.02)2 + 26.316 fn (0.02) + 200.61
= 84.5°c
(Ans)
To calculate the heat-transfer rate internally, it is first necessary to find the temperature gradient at
r= 0.0075 m.
Thus
dT)
26.316
( -d
=-2x32894.7r+(26.316) (1)
- =-2x32894.7x0.0075+-r r=0.0075m
r
0.0075
= 3015.38 °C/m
The heat transfer internally is in the direction of negative radius. Hence, per unit length,
.
~-r)
= - ( -kA dT)
dr
= (380 W/m K) (1t X 1.5 X 10-2 X 1) m2 (+3013.38°C/m)
= 53996 W = 54 kW
(Ans)
Check
The result can be checked as all the heat generated in the conductor must be dissipated internally.
~-r) = qx(Volume perm length)
= q [ ¾(ni
-nnx1]
= (5x 107 W /m3 ) [
= 53996 W
¾( 4 -1.5 )x 1Q-4m x lmJ
2
2
2
or 53.996 kW = 54 kW
Hence OK
3.46 , Heat and Mass Transfer
Example 3.17 ~ Derive an expression for the maximum s1::ady state temperature in the long hollow nuclear
fuel element shown in Fig. 3.35 in which q units are being generated per unit time per unit
volume.
-
h, T~
······················ r0
:··t···· ·
Insulated
Fig. 3.35
Known: Heat generation in a long hollow nuclear fuel rod exposed to convection.
Find: Expression for maximum temperature, Tmax·
Schematic:
h, T~
ttt
Insulated
surface
Long hollow cylindrical nuclear fuel element
Fig. 3.36
Assumptions: (1) One-dimensional conduction. (2) Steady-state conditions. (3) Uniform volumetric heat
generation.
Analysis: The conduction equation for steady state, radial heat flow with heat generation reduces to
(1)
With the boundary conditions:
(2)
and
(3)
From Equation ( 1)
!!_(r dT) =_qr
dr
dr
k
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
dT
qr2
r-=--+c,
dr
2k
3.47
(4)
From Equation (2)
or
Equation (4) then becomes
dT
qr
dr
2k
C1
r
dT
dr
qr
2k
qR/ 1
2k r
-=--+-
or
-=--+--
(5)
qr2 qR,.2
T(r) = --+-lnr+ C
2
4k
2k
(6)
Integrating,
From Equation (3) with Equation (5) and Equation (6),
-k [
--+-- =
qRz
2k
qJ?..2 ]
2kR2
[ qi?} qJ?..2
h - - + - l n Rz
4k
2k
+C
2
-T
=
]
C = T + qRz - qJ?..2 + qR} - qJ?..2 ln R
2
2
= 2h 2h R2 4k
2k
Substitution of constants C1 and C2 in Equation (6) yields
T=T
=
+-
2
qR2 [ 1 R1]
2h
Rf
+-
qR22 [ 1 -r2-]
4k
Rf
+-
qR12 In ( - r )
2k
R2
dT
The maximum temperature occurs where dr = 0, i.e., at r = R 1
T..
max
or
=T
=
+ [1- ]+ [1- ]+
qR2
2h
J?..2
R}
qR}
4k
J?..2
R}
(!i)
qJ?..2 ln
2k
Rz
Tmax =T= +i(R2-R2)(_!_+_l_)+ijR;
/n(R•)
1
2 2
2k hR
2k
R
2
2
(Ans)
3.48 , Heat and Mass Transfer
Example 3.18 ~ (a) A long stainless steel bar of 20 x 20 mm square cross-section is perfectly insulated on
three sides and is maintained at a temperature of 400°c on the remaining side. Determine
the maximum temperature in the bar when it is conducting a current of I000 A. The thermal and electrical
conductivity of stainless steel may be taken as 16 W/m K and 1.5 x I04 (ohm cm)- 1 and the heat flow at the
ends may be neglected.
(b) What is the centre temperature of a stainless steel rod of 20 mm diameter with similar properties to the
bar as in (a) and an outer temperature of 400°c, when conducting 1000 A?
Known: Stainless steel bar of 20-mm square cross section, insulated on three sides, carries current and
thereby generating heat.
Find: (a) Tmax for steel bar. (b) T0 for steel rod of 20-mm diameter under identical conditions of (a).
Schematic:
a=20mm=L
Uninsulated side, Tw=400°C
Insulated
a=20mm
1= IOOOA
{ ke= l.5x I04(flcm)-I
k= 16W/mK
A=a2=(20x IQ-3m)2
=400 x IQ-6m2
x=L
Steel bar
Fig. 3.37(a)
1= IOOOA
{ k= 16W/mK
ke = 1.5 x I04 (n cm)-1
Steel rod
Fig. 3.37(b)
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Uniform heat generation. (3) Constant
properties.
Analysis: (a) Heat generated within a material due to passage of an electric current,
£gen
But
L
R. = P A
= Qgen = I 2 J?.
where p is the resistivity.
[p is the reciprocal of electrical conductivity]
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.49
2
JllkWll lkW I
1000A ) [
1
(
q = 400xl0-6m2
(l.5x10 4 )(10 2) Q- 1m- 1 A2 Q 103 w
kW
=4166.67-3
m
As heat can only flow out of the bar at the uninsulated side, the maximum temperature is at the opposite
side. Let x be measured across the bar from the side at maximum temperature towards the uninsulated side.
Now, we have the governing equation:
Integrating twice successively, we get
dT =- qx +A
dx
k
qx2
T=--+Ax+B
2k
The boundary conditions are the following:
I . Atx= O,
IL At x = 0.02 m,
dT
dx = 0
->.
T= 400°C
~
_,,
IA = 0 I
B = 400+_i_(0.02)2
2k
Substitution of the boundary conditions yields the general equation
T = q [(0.02)2-x2]+400
2k
Since T= Tmax when x = 0, it follows that
1'.
max
= {(4166.67 X 1Q3 W /m3) (0.02m)2}+ 400oC
2(16W/m K)
= 452.1°c
(Ans) (a)
(b) For steady state, radial conduction with uniform heat generation, we have
Integration yields
Further integration gives
dT
qr2
r-=--+A
dr
2k
dT
qr A
-=--+dr
2k r
-2
T=-!l!_+A Zn r+B
4k
3.50 , Heat and Mass Transfer
The boundary conditions are the following:
I . At r = O,
dT
dr = 0
11.Atr=R,
T=TR
!A = 0 !
_._
_,.
(from symmetry)
(outer surface temperature)
qR2
B=TR +-4k
The radial temperature distribution is then given by
T = T. + !L (R 2 - r 2 )
R 4k
where
q=(~J p=(1t:
2
r:. =(1t: y
0
o~~l2
(1.5xl~
= 6.755 x 106 W/m 3
The centre temperature (also the maximum temperature) of the stainless steel rod, TmaxCat r= 0) is
qR2
Tmax =TR+ 4k
= 4000C+ (6.755x 106 W/m 3 ) (O.Ol2m 2 )
4(16W/mK)
= 410.6°c
(Ans) (b)
Example 3.19 ~ A thin hollow tube with 6-mm OD and 4-mm ID carries a current of I000 A. Water at
30°c is circulated inside the tube for cooling the tube. Taking the heat transfer coefficient of
the water side as 35000 W/m 2 K, estimate the surface temperature of the tube if its outer surface is insulated.
The electrical resistivity of the material is 0.1 n mm 2/m, and its thermal conductivity is 18 W/m K.
Known: Heat generation in a thin hollow tube with adiabatic outer surface.
Find: Tube surface temperature, T0 •
Schematic:
r;=2mm
111
Insulated surface (T0 )
Water
h = 3SOOOW/m2K
T==30°C
{
k= 18W/mK
p=O.IX 1o-6nm
Fig. 3.38
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.5 I
Assumptions: (1) Steady-state radial conduction. (2) Uniform heat generation. (3) Constant properties.
Analysis: The governing differential equation is
.!.~(rdT)=-9_
r dr
dr
k
rdT = -qr2 +C
dr
2k
I
dT -qr C1
-=--+dr
2k
r
or
or
qr2
T=--+C1 Zn r+C2
4k
or
As the outer surface is insulated, dT I = 0
dr r=,;,
le,= ,;,21
2qk
Since the inner surface is being cooled by a fluid at a temperature T=, the heat generated inside the
wall would be transferred to the fluid at the inner surface. Thus the heat generated per unit length is
- ( ro2 - 'T
-·') -- -k dT I
(2 TCri )
Q.gen -- qrc
dr r=r;
= h.2 re ri (Ti -T=)
T =
'
or
qr'f
-q(r2 -r.2)
i
o
+T
=
2h,;
q
T =--+-r2 Zn r-+C =
2
'
'
4k 2k O
q
q(r2-r.2)
0
;
2h r.I
q{
+T
=
r!}
C2 = T= +-(r'!-r.2)-r'!- Zn,;-2 hf;
2k
2
0
I
I
O
I
Substituting for C1 and C2 in the general expression for T, the temperature at the outer surface at
r = r 0 is given by
q ro
qro
q ro
q
,; 2 ]
- 2
- 2
- 2 [
2 -r.2)-T =--+-Zn r +T +--(r
2 Zn ,;l - ( -ro )
O
4k 2k
O
= 2hr; O I
4k
or
0
2
2
q (r0 -r.2)+qro
[-2/n,;+(,;)
= = 2h,;
I
4k
I
ro -1+2lnr]
0
T-T
2
l
To = T= + q-(r2
o -r.2)
i
+ q- ro2 [ 2 [n ro + ( 'i ) -1
2hlj
4k
'i
r0
2
3.52 , Heat and Mass Transfer
. h
.
_ Qgen 12Re 12rL
Vo Iumetnc
eat generation rate, q = - - = - - = - ¥
ACL A~L
ii'=(IH)'xp=[
,(./-,;'J xp
- =[
1000
]2 XO I X10-6
q
1t(32 -2 2 )(10-6 )
.
= 4.052 x 108 w/m 3
Substituting T= = 30°C, h = 35000 W/m 2 K, ri = 0.002 m, q = 4.052 x 108 W/m3, r 0 = 0.003 m and
k= 18 W/m K in the above expression, the surface temperature of the tube is
4.052 X 108 (0.0032 -0.002 2 )
T =30+--------0
2 X 35000 X 0.002
+ 4.052xl08 x0.0032
4xl8
[2cn 0.003 +(0.002) -1]
2
0.002
0.003
(Ans)
Example 3.20 ~ A long hollow cylinder (k = 0.5 W/m K) has inner and outer radii as 5 cm and 15 cm
respectively. It generates heat at the rate of I kW/m 3. If the maximum temperature occurs
at a radius of 10 cm and the temperature of the outer surface is 50°C, find
(a) the temperature at the inner surface
(b) the maximum temperature in the cylinder
Known: A hollow cylinder with uniform internal heat generation and specified outer surface temperature
has maximum temperature occurring at the mid-thickness.
Find: (a) Inner surface temperature, 7i (0 C) ; (b) Maximum temperature, Tmax ( 0 C) .
Schematic:
k=O.SW/mK
Fig. 3.39
Assumptions: (I) Steady operating conditions. (2) Uniform volumetric internal heat generation.
(3) Constant thermal conductivity. (4) One-dimensional (radial) heat conduction.
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.53
Analysis: Energy balance
Ein -£out +£gen= .........,,___,
Est
.......__..
No heat enters
thesystem
Steady state
The governing differential equation is
!!_(r dT) =_qr
dr dr
k
or
Integrating with respect to r, we have
dT
qr2
r-=--+C
dr
2k
I
dT
qr C1
-=--+dr
2k r
or
Since maximum temperature occurs at r = 10 cm, i.e., at the mid-thickness, the temperature gradient
at r = O. l m must be zero. Hence, the boundary condition is
dTI
-o
dr r=O.lm
qr? (1000W/m 3 )(0.lm)2
C = O+-- = - - - - - - - ~
1
2k
2x0.5W/mK
= 10
dT
qr IO
-=--+dr
2k r
or
Integrating again with respect to r, we get
-2
T(r)=-~+10 Rn r+C2
4k
The other boundary condition is
At r=r2 =0.l5 m, T=T; =50°C.
Then
q,f
C2 =T;+--10 In 1i
4k
(103 W/m3 )(0.15 m)2
=50°C+~--~-~+10 In (0.15m)
4x0.5W/mK
= 80.22°C
3.54 , Heat and Mass Transfer
The temperature distribution is then given by
qr2
T(r)=--+10 /n r+80.22
4k
(a) Temperature at the inner surface is
I; =T(r=0.05m)
(10 3 W/ m3 )(0.05m)2
----,--~----,--+IO £n (0.05m)+80.22°C
4(0.5W/ mK)
= 49.0°C
(Ans) (a)
(b) Maximum temperature in the cylinder will occur at the mid-thickness i.e. at r =re= 0.1 m.
Tmax =T(r=O. Im}
10 3 XO 12
--· -+10 In 0.1 + 80.22
4x0 .5
= 52.2°C
Example 3.21
(Ans) (b)
~ A resistance wire of 6-mm diameter (~;re= 15 W/m K) is heated by passing current through
it. The heat dissipated per metre length of the wire is I 00 W. The wire is covered by an
insulation (k;nsulation = 0.35 W/m K) of 12-mm outer diameter. The outer surface of the insulation is exposed
to an inert gas at 26°C. The outer surface temperature of the insulation is measured to be 60°C. Determine
(a) the convection coefficient at the outer surface of the insulation, (b) the temperature of the wire at the outer
surface, and (c) the temperature at the centre of the wire.
Known: Heat generation in a wire covered with insulation that is exposed to convective environment.
Find: (a) h; (b) T1; (c) T(O).
Schematic:
Q= IOOW/m
Gas
!!!
k;nsulation = k2 = 0.35 W/m K
Fig. 3.40
Assumptions: (I) Steady operating conditions exist. (2) Constant properties. (3) One-dimensional
conduction. (4) Uniform heat generation.
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.55
Analysis: (a) The heat generated in the round resistance wire is dissipated by convection to the inert gas.
Performing an energy balance
Heat dissipated,
where
=
R
conv
1
h(21triL)
The convection coefficient is
h=
QIL
(21tri) (7; -T=)
lOOW/m
=-----------(21tx0.006 m) (60-26)°C or K
(Ans) (a)
=78W/m 2 K
(b) Radius of the current-carrying wire= Inner radius of the insulating layer, r 1 = 0.003 m
Outer radius of the insulating layer, r2 = 0.006 m.
Conduction resistance due to insulation per metre length is
R
ln(ri/'i)
- -----'-2 1tk, L
cond -
ms
In( 0.006/0.003)
= 21t(0.35 W/mK) (lm) = 0.3l 5 K/W
Hence, the wire surface temperature = Inner temperature of insulation, T1 = T2 + Q Rcond
I;= 60°C+(100W) (0.315 K/W) = 91.5°C
(Ans) (b)
(c) Volumetric heat generation rate,
_ Q £gen
100 W
q=-=--=
¥- 1tf L 1tx0.0032 m 2 xlm
= 3.5368x106 W/m3
Centre temperature is also the maximum temperature of the wire. It follows that
qf
T(O) = Tmax = I; + - 4 kwire
(3.5368 x 106 W/m3 ) (0.003m)2
= 91.5°c + - - - - - - - - - - - 4(15W/m K)
=92°C
(Ans) (c)
3.56 , Heat and Mass Transfer
Comment: The current-carrying conductor (wire), having heat generation, cannot be represented by a
thermal circuit element.
Example 3.22 ~ A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for
which a thorium fuel element (k = 57 W/m K) is encased in graphite (k = 3 W/m K) and
gaseous helium flows through an annular coolant channel as shown in Figure 3.41. Consider conditions for
which the helium temperature is 600 K and the convective heat transfer coefficient at the outer surface of the
graphite is 2000 W/m 2 K. If the rate of volumetric heat generation is I08 W/m 3, find the temperatures at the
inner and outer surfaces of the fuel element.
Coolant channel
with helium gas flow
L
Fig. 3.41
Known: A hollow composite cylinder comprising a layer of thorium (Juel element) encased in graphite
layer exposed to flowing helium gas.
Find: Inside and outside temperatures of fuel element, Ji and r;.
Schematic:
Coolant channel
(helium)
h = 2000W/m2K
T= =600K
- - - + - Graphite
Insulated _ _.....,.
surface
(k2=3W/mK)
~~-------,...--Thorium
(k1=57W/mK
q= IOSW/ml)
Fig. 3.42(a)
Assumptions: ( l) Steady-state, radial conduction. (2) Constant properties, uniform heat generation and
uniform convection coefficient. (3) Negligible thermal contact resistance. (4) The inner surface of the
thorium layer is insulated.
Analysis: Under steady operating conditions,
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.57
heat-transfer rate, Q= qx1t(,f-,;2) L
= (108 W/m3 )[1t(O.Ol l2 -0.0082)m2 xlmJ
(per unit length)
= 17907 W
The inner surface is insulated so that the entire heat that is generated within the thorium fuel element
flows radially outwards and is dissipated by convection to coolant helium flowing through the annular
channel.
dT
Clearly, - = 0 at r =,;
dr
(Boundary condition I)
Also, T= 1i at r = 1i
(Boundary condition II)
Let us construct a thermal circuit:
r..
Rgraphite
Rconvection
Fig. 3.42(b)
The thermal resistances are
ln(tj I 12)
Rgraphite
= 21t"2L
ln(14mm/1 lmm)
21tx3W/mKxlm
= 0.012 79 K/W
R
.
convection
1
h (2 1t tjL)
=----
=------------(2000W/m2 K)(21tx0.014mxlm)
= 0.005 684 K/W
.
With Q=
y: -T
2
~phite
00
,
+ R.:onvection
Temperature of outside surface of the fuel element is
1i = 600 K + (17907 W) (0.01279 + 0.005684) K/W
=930.9 K
Consider the heat generation in the fuel element of conductivity lei-
(Ans)
3.58 ,
Heat and Mass Transfer
The appropriate heat equation is
.!_~(rdT)=- q
r dr
dr
lei
;(r~:)=-t
or
Integrating,
dT
qr2
r-=--+C
dr
2ki
I
Since dTI
=0
dr r=ri
Temperature gradient,
-=--+-=--+-dT
qr
C1
qr
q,f
dr
2ki
r
2ki
2lcir
Integrating again
q,f lnr qr2
T (r ) = - - - --+C2
2ki
4ki
With T(r = 11) = Ji,
C2 -_ qr.J:
4ki
q,f ln11
----'---=-+ Y:2
2ki
The temperature distribution can then be expressed as
The temperature of the fuel element at its inside surface is determined by replacing r with 'i. It follows that
q r.J:
1j
1j
1j
4ki
11
11
11
l
Ti =T(r = 1j)=1i + - [1- (- )2+2 (- )2 /n-
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.59
Substituting values,
T. =
1
93 o. 9
K + (10 8 W/m3 )(0.0llm)2
4x57W/mK
[i-(~) +
2
11
2
2 (~) /n~J
11
II
(Ans)
=938.0 K
(E) SPHERE
•
•
Example 3.23 ~ The average heat production during ripening of oranges is estimated to be 320 W/m 2 . If
the orange is assumed to be a homogeneous sphere with k = 0.14 W/m K, compute (a) the
temperature at the centre of the orange, and (b) the heat flow from the outer surface. Take the diameter of
orange as 8 cm and external surface temperature as I0°C.
Known: Heat generation in a sphere (orange) under specified conditions.
Find: (a) T(x = 0) = TmaxC 0 C); (b) Qout (W)
Schematic:
Orange
R=4cm
Q
-=320W/m2
A
k=0.14W/m K
Fig. 3.43
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Constant properties. (3) Uniform
volumetric heat generation.
Analysis: (a) Temperature at the centre of the orange (a sphere of radius R) is
qR2
Tmax = Tw + 6k
4
rcD 3 re
Volume of a sphere ¥- = -rcR3 = - - = -x0.083 m3
'
3
=
6
6
268 X I0-6 m3
Area of a sphere, A = 4 rc R 2 = re D 2
_
qA
Qgen
q=¥
=
¥ =
6qxrcD 2
re D 3
6q
D
= -6-x320W/m2 = 24000 W/m3
0.08m
3.60 , Heat and Mass Transfer
T_
max
=l0 0 C+(24000)W/m3 (0.04)2m 2
6x0.15W/mK
= 52.67°C
(Ans) (a)
(b) Control volume energy balance gives:
'
-Eout+Egen=
'
(no heat inflow)
(steady state)
or
Heat flow from the outer surface,
= 24 x10 3 W/m3 x268x 10-6 m2
=6.43 W
Example 3.24
(Ans) (b)
~ A solid sphere of radius R is generating heat uniformly at the rate of qW lm2 • The thermal
conductivity of the solid is given by k = k0 (I + aT). Assuming the surrounding temperature
as T~ and the heat-transfer coefficient as h, prove that the steady state radial temperature distribution can be
expressed as
qR2[(')2]{qR
1}
+ -+T~+3h
a
2
I
T(r)=--+
- - I- a
3ak,,
R
Known: Solid sphere with variable thermal conductivity experiences uniform volumetric heat generation.
Find: Steady-state one-dimensional temperature distribution, T(r).
Schematic:
Q
R
Surroundings
h, T~
lll
Solid sphere
k(T)=k 0 [I + a T]
Fig. 3.44
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.61
Assumptions: (1) Steady-state, one-dimensional (radial) conduction. (2) Constant and uniform heat
generation rate.
Analysis: Energy balance in the differential control volume
£gen= Q(r)
or
q(i1tr3)=-k(T)(41tr2) dT
3
dr
or
k(T)dT = - qr dr
3
or
k0 (1+a.T)dT=- qr dr
3
Integrating,
[ r2]
qr2
k T+a.- =--+C
0
2
6
(A)
Control surface energy balance
(at r= R)
(at r= R)
- kdTI
=h[T,.=R -T
dr r=R
00 ]
[-qR]
-k(T) - - = h[T -T]
3k(T)
w
00
Substituting for Tw at r = R in Equation (A), we get
J
a. 2 =---+C
q R2
k0 [ Tw +-T
2 w
6
or
)2]
qR2
qR a.
qR
C=--+k T +-+- T + 6
°
3h 2
3h
[
(
00
00
Substituting for C in Equation (A), we have
)2]
a.T2
qr2 qR2
qR
q R a.
k [r+--]=--+--+k T +-+- T + 0
2
6
6
°
3h 2
3h
[
00
(
00
3.62 , Heat and Mass Transfer
R
R) (
R)
2
q ( 2 - r 2 ) 2 2 ( T +q - + T +q - 2
T 2 +-T=---~x--+a
6
a k0 a = 3 h
= 3h
or
or
where
2
2
qR2- [ 1- ( -r ) ] - {( T +l ) -l } =0
T 2 +2 - T a
3ak0
R
w
a
a2
or
This is a quadratic equation and taking its positive root, we obtain,
or
Example 3.25
2
1
qR2 [ 1- ( -r ) ] + [ -+T=
qR
1]
T(r) = --+
+-
a \ 3ak
0
R
3h
2
Hence proved.
a
~ A solid sphere of I 0-cm radius generates heat according to the law q = 5000 (I+ 2r) [W/m 3]
Obtain the temperature distribution if the solid exchanges heat with surrounding fluid at
I00°C. The thermal conductivity of the material is 0.5 W/m K and the unit thermal conductance is 20 W/m 2 K.
Also find the maximum temperature and the temperature at the surface.
Known: Heat generation in a solid sphere exposed to convective environment.
Find: T(r), Tmax and Tw( 0 C).
Schematic:
Fluid
Solid sphere
!!!
k=0.5W/m K
q= 5000 (I+ 2r)
h=20W/m2°C
T== I00°C
Fig. 3.4S(a)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3 .63
130
127.92
125
u
- r--__
~ K T(r)
120
~
0
i="
115
"
110
~
105
0
4
2
8
6
109.SS
10
r(cm)
Fig. 3.4S(b)
Assumptions: ( l) Steady-state, one-dimensional heat conduction. (2) Constant thermal conductivity.
Analysis: The governing differential equation is
J_~(r2 dT)=-"g_= -5000(1+2r)
r 2 dr
dr
k
k
or
Integrating with respect to r,
r 2 dT = -5000[~+ 2r4J+c
1
dr
k
3
4
(a)
dT = _ 5000 [~+~]+ C1
dr
k
3 2
r2
or
Further integration yields
(b)
Boundary conditions required equal the number of constants of integration.
BC/: At r
dT
=0, ctr=
0
From Equation (a),
(by symmetry)
3.64 ,
Heat and Mass Transfer
Then, the radial temperature gradient is
dT =- 5000
[!'..+ r
dr
3
k
2]
(c)
2
and the temperature distribution is given by
5000
T=---[r 2 +r3]+C
2
6k
(d)
BCII At the outer surface of the sphere at r = R,
qconv
= qcond
h (T,.=R - T.. ) = - kdTI
dr r=R
With equations (c) and (d), we have
or
Substituting for C2 in Equation (d), we have
5000
T(r)=I27.92---(r 2 + r3)
3
(Ans)
Maximum temperature (at r = 0) is
Tmax
= 127.92°C
(Ans)
Surface temperature (r = R = 0 .1 m) is
Tw = 127.92- 5000 (O.F +O.P)
3
= 109.58°C
The temperature distribution is shown in the schematic.
(Ans)
(Ans)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3 .65
(F) NUCLEAR SYSTEMS
•
•
Example 3.26 ~ It is proposed to contain radioactive wastes in a composite spherical shell comprising an inner
layer of lead (k = 34 W/m 0 C) of SO-cm ID and 5-cm thickness followed by an outer layer
of stainless steel (k = IS W/m 0 C), I-cm thick. The rate of uniform volumetric heat generation by the wastes
filled in the container is 0.5 MW /m 3 . For proper and safe disposal of the radioactive material, the shell is to
be submerged in the sea waters which are at IO °C with an associated convection coefficient of 500 W /m2 0 C.
Examine the soundness of the proposal in the light of the fact that the lead has a melting point of 601 K.
Known: A composite spherical system with radioactive wastes in the cavity is enclosed by a layer of
lead and finally covered by stainless steel layer to be submerged in oceanic waters for safe disposal.
Find: Inner surface temperature of lead, T1( 0 C). T1< Melting point of lead for safety.
Schematic:
Q
Lead (k1 = 34 W/m 0 C)
Sea water
T= = I0°C
h=SOOW/m2°C
lll
Stainless steel (k2 = IS W/m 0 C)
r1 =0.2Sm
''
''
I
I
I
''
''
'
Fig. 3.46(a)
r,
T2
R1ead
T3
Rscainless steel
T=
Rconvection
Q
Fig. 3.46(b)
Assumptions: (1) Steady state, one-dimensional conduction. (2) No contact thermal resistance. (3) Constant
properties. (4) Uniform heat-transfer coefficient.
Analysis: Control volume energy balance:
Y-Eout+ £gen=
(no heat inflow)
i
(steady state)
Hence,
or
q(W/ m 3 )¥-(m 3 )
= Q,outcond(W)
3.66 ,
Heat and Mass Transfer
The rate of heat transfer in the radial direction is
Q= q
(~Df)
= (0.5x106 W/m3 {~x0.53 m 3 ) = 32725W
Clearly, referring to the thermal circuit, one can easily write
Q=
J;_ - T=
Rlead
= AToverall
L Rth
+ Rstainlesssteel + Rconvection
1- ( 1 -1)
Rlead = 41tki
1i
1j
1
(
1
1
)
1
1
)
= 41t(34W/m°C) 0.25m - 0.30m
= 1.56 X 10-3 °C/W
Rstainless steel
= 4 lk (_!_ - _!_)
1t2 1i
13
1
(
= 41t(15W/m C) 0.30m - 0.31m
0
= 0.57 X 10-3 °C/W
R
.
convectmn
1
h ( 4mj)
1
500 W /m2 oc (41t XO .3 l2 m2)
= --~= - - - - - - - - - -
= 1.656 X 10-3 °C / W
Hence, the total thermal resistance is
Rtotal =
L Rth = 10-3 (1.560 + 0.570+ 1.656)°C/W
= 3.787 X lQ-3 °C/W
It follows that
1;_ -T= = QRtotal = (32 725W)(3.787 X 10-3 °C/W)
= 123.93°C
Inner surface temperature, J;_ = 10 + 123.93 = 133.93 °C or ""407 K
Since this is well below the maximum allowable temperature (melting point of 601 K), the proposal
for disposal of radioactive wastes seems to be sensible.
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3 .6 7
Example 3.27 ~ The heat generating rate per unit volume, q, at any radius, r, of a solid rod is given by
q=
qo[1-(~J]
where q0 is the heat generation rate at the centre of the rod of radius R. Develop an expression for maximum
temperature at the centre of a rod of 40-cm diameter, when the heat-generating rate at the centre is
24xl06 kJ/h m3 and the temperature at the surface of the rod is 20°C. Assume thermal conductivity of the
material of the rod as 200 kJ/m h 0 C.
Known: Heat is generated at the rate of q0 {1-(r I R) 2 } (W /m 3 ) in a solid rod with specified wall temperature.
Find: Centreline temperature, T (r = 0) = TmaxC 0 C)
Schematic:
p
D=40cm
\..
Fig. 3.47
Assumptions: ( l) Steady-state, one-dimensional heat conduction. (2) Constant thermal conductivity.
Analysis: The governing differential equation is
_!_~(dT) +i=O
r dr dr
k
or
~(rdT)=-qr =-%r[l-(r ! R)2]
dr
dr
k
k
=-
~ {r-;}
Integrating with respect to r, we have
r dT =- q0
dr
k
where C 1 is an arbitrary constant.
{!:.-~}+c
2
4R2
1
(A)
3.68 ,
Heat and Mass Transfer
The symmetry condition at the centreline suggests that the maximum temperature occurs at the
centreline where the slope of temperature with respect to r, i.e., (dT/dr) is zero.
Applying this boundary condition, we get
{Q __4RO_} +C
Ox dT(O) =- q0
dr
k 2
1
2
Therefore
Dividing Equation (A) by r to bring it to a readily integrable form, we have
_.!:__}
dT __ % {!._ _
dr
k 2 4R2
Further integration with respect to r yields
{r2
-% - - r4- } +C
T(r)=2
k
4 16R2
To evaluate the constant of integration C2, we use another boundary condition.
At r= R, T(r) = Tw
Then
Substituting this value of C2 into Equation (B), we get
{r
3q-0 R-2 -q-0 -2 - -r4-}
T(r)=T +
w
16k
k
4 16R2
or
T(r) = T. + l_ qoR2 - qor2
w
16 k
4k
{1-
(r I R'f}
4
This is the expression for temperature distribution in the solid rod of radius R.
Maximum temperature will occur at r = 0.
(B)
3.69
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
Substituting the relevant numerical values, we have
T
=200c + 1._ x _(2_4_x_l_06_k_J_/m_3_h_)_(0_.2_m_)2
max
16
200kJ/mh°C
= 920°c
(Ans)
Example 3.28 ~ A fuel element of a nuclear reactor in the form of a large uranium slab (k = 27.6 W/m 0 C) is
8-mm thick and is covered on both sides by 2-mm thick aluminium cladding (k = 237 W/m 0 C).
The element generates heat at a constant volumetric rate of 4.5 x 108 W/m 3. The two exposed surfaces are
maintained at constant temperature by an external cooling arrangement. The coolant temperature is 95°C and
the heat transfer coefficient is I 5000 W/m 20 C. Determine (a) the temperature of the exposed surfaces, (b) the
temperature at the uranium-aluminium interfaces, (c) the maximum temperature in the composite slab, and
(d) the temperature gradient in the two claddings.
Known: Uniform heat generation in a nuclear fuel element wrapped by aluminium coverings with an
dTI
e~te~nal coo'.ing arr~ngement..
Find. {a) Tc, {b) TL, (c) Tmax• (d) dx cladding
Schematic:
Fuel element (Uranium)
(kF= 27.6 W/m C) ~
0
Plane of symmetry
w
i Tmax
Cladding (Aluminium)
~ (kc=237W/m C)
0
! !!
!! !
Too=95°C
h= 15000W/m2°C
T =95°C
h= 15000W/m2°C
00
~fgen
;
q=4.5x 1ioaw/m3
i
Tc
b = 2 mm ---I
Tc
!
f-- 2 L = 8 mm ---1
/-b=2mm
Fig. 3.48
Assumptions: ( 1) Steady-state, one-dimensional conduction. (2) Constant and uniform heat generation.
(3) Constant properties and uniform heat-transfer coefficient.
Analysis:
(
Heat generated within)= (Heat transferred by conduction)= (Heat dissipated by convection from)
the large uranium slab
through aluminium claddings
the exposed surfaces to the coolant.
By symmetry, half of the heat generated will pass through the left side cover and the other half through
the right side cover.
£gen= q(A2L) = (4.5 x 108 W/m 3 }(8x 10-3 m){A)m 2
E
~ = 3.6 x 106 W/m 2
A
3.70 ,
Heat and Mass Transfer
Half of this, i.e., 1.8 x10 6 W/m 2 will flow on either side of the plane of symmetry.
2
£gen/
Q.,ond
kc (TL -Tc)
- - =- =~-~
A
A
b
= ~nv = h(Tc -T=)
It follows that
~
Exposed surface temperature,
T,
= 95oc+ 1.8xl06 W/m2
c
15000 W/m2 °C
= 21s c
(Ans) (a)
0
Also,
~
Uranium-aluminium interface temperature,
T,
=2150C+ (l.8xI06 W/m2)(2xl0-3 m)
237W/m°C
L
= 2Jo.2°c
(Ans) (b)
For a plane wall of thickness 2L with wall temperatures TL and volumetric heat generation q,
qL2
Tmax
=I;,= TL+2kF
= 230.2oc+ (4.5x10s W/m3)( 4xl0-3 m)2
2x27.6W/m°C
= 360.6°C
(Ans) (c)
Temperature gradient in each side cover,
dTI
=Tc-TL
dx cladding
b
=
(215- 230.2)°C
0.002m
= -7600°C/m
(Ans) (d)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, 7 I
Example 3.29 ~ The shielding wall for a nuclear reactor is exposed to gamma ray radiation, which results in
heat generation according to the following equation:
q =<foe-ax
where q0 and a are constants. Using this relationship for q, determine the temperature distribution in the shield
if the wall is L units thick and its left and right-hand faces are maintained at temperatures T1 and T2 , respectively.
(a) Find the location of maximum temperature. Also, obtain an expression for the maximum temperature in the wall.
(b) If T1 = T2, find the temperature distribution as also the maximum temperature.
Known: Heat generation associated with nuclear reactor shielding.
Find: Temperature distribution within the shield. Location of maximum temperature . Expression for
maximum temperature in the wall.
Schematic:
L
-.JV\!'-+- E. gen
L
Shielding wall of
nuclear reactor
Fig. 3.49
Assumptions: ( l) Steady state prevails. (2) Heat generation is a function of location. (3) One dimensional
conduction. (4) Constant properties.
Analysis: The appropriate differential equation is
or
d2T
q
dx2
k
d 2T
- 'loe-ax
dx 2
k
Integrating, we get
dT
q- e-ax
-=--o--+C1
dx
-ak
(1)
Further integration yields,
(2)
The boundary conditions are
T(O) = T1
T(L) = T2
3.72 , Heat and Mass Transfer
Substituting these boundary conditions in Equation (2), we get
qo
T1 =--+C2
-a2 k
Also,
-
L
-
T = qoe-a +C L+T + ~
2 -a2k
,
, a2k
T2 -T1 q0 ( e-aL -1)
C =--+---1
L
kLa 2
Substituting the values of C 1 and C2 in Equation (2), one obtains
x +Clo- x [ e_aL _1 ] +T +CloT =cioe-ax
- - + (T2 -T1 ) 1
-a2k
L a2k L
a2k
(3)
The temperature distribution therefore is given by
T _T, = (T. _ T, ) ~ + iio [ e-aL _ 1] ~ _ iio [ e-ax _ 1]
or
1
2
1
L
a2 k
L
a2 k
(Ans) (a)
To find the location for maximum temperature in the reactor shield, the temperature gradient dT must
be equated to zero.
dx
Thus, Equation ( 1) can be equated to zero to determine the value of x for Tmax·
dT
-q e-ax T -T
-q
-=0=-0__ +_2_ _1 +-o-(e-aL -l)
dx
-ak
L
kLa2
or
1 -T
2 ) +q-0 -ak(l
-[ak(T
e_ax_
- - -e-aL}]
L
kLa2 q0
q0
or
1 -T
ak(T
-ax= ln [ =-2 ) + - 1 (1- e_aL )]
q0
L
aL
1 [ak(T -T 1( _ )]
1
2 ) +- 1-e aL
x=--/n =- --a
q0
L
aL
or
(Ans) (a)
The maximum temperature can be found by substituting this value of x in Equation (3)
(b) If T1 = T2 , the value of x can be found for maximum temperature, simply by putting T1 equal to T2 ,
in the above equation.
Thus
1 [1-e-aL]
x=--ln
---
a
aL
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3, 73
Substituting in Equation (3 ), we get
T.
max
= T.I
+qo [~(e-al -1)-e-ax +l]
a2k l
_7i +q- -'-----'-1-e-al ){ /n (I-e-al
- - - )} - exp {/n (I-e-al)
- - - + }]
Tmax -
0
[(
a2 k
al
al
al
1
_7; + q [(I-e-al)
(I-e-al) -(I-e-al)]
----/n - - - +1
or
Tmax -
or
0
a2 k
al
Tmax
al
al
= 7i + ~
[ B In B-(B-1)]
2
a k
where
(Ans) (b)
(G) MISCELLANEOUS
•
•
Example 3.30 ~ The annular clearance of 0.3 mm between two cylindrical surfaces is filled with a lubricating
oil (p = 1220 kg/m 3, µ = 0.95 poise and k = 2.42 W/m 0 C). The outer cylinder of 6-cm radius
rotates at 9500 rpm. If both the walls of the cylinders are maintained at 70°C, determine the maximum
temperature in the oil. Heat generation due to viscous dissipation may be taken as q = µ (
!~ J.
Known: Oil between two cylindrical surfaces experiences viscous heat dissipation.
Find: Maximum temperature of the oil.
Schematic:
Moving surface
.-~------,70°c
c=0.3mm _..
(k=
[µ = 0.95 poise
= 0.095 N s/m2]
Moving
Stationary
(b)
Fig. 3.50
(1) Steady-state conditions. (2) Oil is an incompressible fluid with constant properties.
(3) Linear velocity profile in the annular gap. (4) One-dimensional heat conduction.
Analysis: The governing differential equation is
Assumptions:
d 2T
q
dy 2
k
-+-=0
3.74 , Heat and Mass Transfer
The volumetric heat-generation rate is
The annular gap being very small, linear velocity variation can be assumed, i.e., u = 0 at y = 0, and
u = Vat y= c.
.
d"
du Au V - 0 V
ve 1oc1ty gra 1ent-=-= - - = dy By
c-0 c
v2
q=µc2
2rcN
where V = R ro = R x - 60
2 TC
= 0.06x-x9500 = 59.69 mis
60
d 2T
and
Integrating,
µV 2
kc2
-=---
dy2
dT
µ V2
dy
k c2
-=---y+C,
Further integration gives
µ v2 y2
T(y) =----+Cy+C
kC2 2
I
2
Boundary conditions
I. At y = 0, Tw =70°C
IL At y = c, Tw = 70°C
From BCI:
From BCII:
E:ffil
T(y)=-Tw+~~:[1- 2; ]
~ ly=c/21
=?Ooc+ (0.096Ns/m2 )(58.69 m/s 2 ) =S?. 60 C
8x2.4W/m°C
(Ans)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
Example 3.31
3, 7 S
~ An insulating material (k = 0.4 W/m C) slab, 5-cm thick, is sandwiched between and is in
0
physical contact with two parallel electrodes. The exposed electrode surfaces are subjected to
convection characterised by ambient air temperature T_ = 25°C and the convection coefficients h 1 = IO W/m 2 °C
and h2 = 15 W/m 2 0 C. Dielectric heating occurs in the slab at a uniform volumetric rate of 35 kW/m 3• Determine
(a) the electrode surface temperatures, and (b) the position and value of maximum temperature in the slab.
Known: Dielectric heating in an insulating slab under prescribed operating conditions.
Find: (a) Surface temperatures, T1 and T2 . (b) Location and magnitude of maximum temperature, xmax and Tmax·
Schematic:
;
lnsul~ting
mate~ial (k)
Electrode I
Electrode 2
;
;
;
!
:
:
~q
;
;
/- L-j
Fig. 3.51 (a)
Assumptions: (1) Steady operating conditions . (2) One-dimensional conduction. (3) Uniform heat
generation. (4) Constant properties.
Analysis: Consider a differential element of thickness dx. The control volume energy balance gives
£in - £out
+ £gen =
fsi
0
(steady state)
or
Qx -Qx+dx +q(Adx) = 0
or
-! [
Qx] dx + q A dx = 0
or
_.!!_[-k A dT]dx=-q A dx
dx
dx
d 2T
+k-=-q
dx 2
or
The controlling differential equation then is
d 2T
q
-- -dx2
k
We define excess temperature 0 1
-
<2x
.JVV'- q
=7i - T= and 0 =I;_ - T=
2
f-- dx --j
If
Fig.3.51 (b)
-
<2x+dx
3.76 , Heat and Mass Transfer
Hence,
d 2 0 =-:!....
dx2
X
Integrating twice, we get
To evaluate C1 and C2 , we need two boundary conditions. These
are the following:
BCI At x = 0, control surface energy balance gives
Ql(cond) [in the -ve X direction]= Ql(conv)
Ql(conv) -
h,Too
-
Ql(cond)
t t t x=O
Fig. 3.5 I (c)
or
BCII
At x=O, 0=0 1
10, =C2I
Hence,
The temperature distribution is then given by
e
- q x2 h181
0(X ) ---+--+ 1
2k
k
(A)
x=L, 0 = 02
At
- L2 i.9
0 =-L+_',_l L+0
2
2k
k
I
(B)
From energy balance for the system:
r./J -£out +£gen=
(no hlifnflow)
E!
(stea,:tate)
i.e.,
Heat generated within is dissipated to the surrounding air by convection from both exposed surfaces
Hence
(C)
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.77
Substituting for h2 02 from Equation (C) into Equation (B),
1
( q L-h,0
h
]=-
2
qL2 + h,01L +e
2k
k
I
0,[1+ h,L +.!i_]= qL + qL2
k h2
h2
2k
or
Excess temperature 01 (=Ti - T_) is
l
_ [ -1 +LqL
h2 2k
0,=~------~
[ 1+(h 1LI k )+(h 1 I h 2 )]
(35x10 3 W/m3 )(5xl0-2 m)[(l/15)+(0.05/2x0.4) ]m2 °C/W
=--------------------[
!+ (10W/m2 °C x0.05m) +(10W/m2 °C)]
0.4 W/m°C
15W/m2 °C
= 11.s c
0
The electrode surface temperature on the left side is
Ti =T_ +01 = (25+77.5)°C = 102.5°C
(Ans) (a)
02 can be determined from Equation (B).
e = T,. -T = _ (35000)(0.05)2 + (10x77.5)(0.05) + 77 _5
2
2
-
2x0.4
0.4
= 65°C
The electrode surface temperature on the right side is
T2 = (25 + 65)°C = 90°C
(Ans) (a)
The locate the position where maximum temperature will occur, we differentiate Equation (A) with
respect to x and equate the resulting derivative to zero.
d0 =O=- q X + h,_01
dx
k
k
i.e.,
x
~
X
= ~(h,_01 )= h,_01
qk
q
= (10W/m2 0 c)(77.5°C) llOOcml
max
35000W/m3
lm
= 2.214 cm
(from the left side electrode)
The magnitude of maximum temperature would be
(Ans) (b)
3.78 , Heat and Mass Transfer
= 102 _5 _ (35000)(0.02214) 2 + 10x77.5x0.02214
2x0.4
0.4
= 123,95°c = 124°c
(Ans) (b)
Example 3.32 ~ In a spark-ignition engine the heat dissipated through the piston crown is 3500 W . Find the
minimum crown thickness required for an aluminium piston (k = 210 W/m K) if the maximum
temperature of the piston is confined to 300°C and the coolant temperature is 35°C.
Known: Piston crown subjected to heat transferred by gases in an internal combustion engine.
Find: Minimum thickness of piston crown.
Schematic:
Cylinder
Heat flux,
qg
L
Piston crown
Fig. 3.52
Assumptions: (1) Steady-state, one-dimensional conduction. (2) Constant properties. (3) Uniform heat flux.
Analysis: Let the thickness of the piston crown be t, the heat transferred by combustion gases through
convection and radiation to the piston be Qgen and the piston radius be R. Consider a differential elemental
strip dr at a radius r.
.
dT
Q, = -k(2rcrt)dr
and
.
.
d
.
Q,+dr = Q, + -(Q, )dr
dr
Heat given by the gases to the differential element dr
Qgen =(2rcrdr)qgen
Under steady state conditions
Q,+dr = Q, + Qgen
or
or
Q,+dr - Q, = Qgen
dTJ dr=2rcrdrqgen
-d [ -k(2rcrt)dr
dr
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.79
-kt.!!_(r dT) = r q
dr
dr
gen
or
or
r
-d ( rdT)
- +-
or
dr
-O
dr
kt qgen -
dT
r qgen
C,
dT = 0
dr
at
r= 0
Integrating once, we get
-+---=dr 2 kt
r
or
Integrating again, we have
Boundary conditions:
BCI
BCII
T= Tw at r=R
First BC gives
and the second BC yields
C
2
= T_ + R2 qgen
w
4
kt
The temperature distribution in the crown is obtained by substituting the values of C 1 and C2 in
Equation (A). It follows that
T = T.. + (R2 - r2 ) qgen
w
4
kt
The maximum temperature will occur at r = 0, where the temperature gradient is zero.
T_
max
The minimum crown thickness,
= T_ + R2 qgen
w
4kt
3.80 , Heat and Mass Transfer
or
where A = area of the piston crown = 1tR2
Substituting the numerical values given in the problem, we have
3500W
t. = - - - - - - - - - mm
41t(210W/m K)(300-35)°C
= 5 x to-3m or 5 mm
(Ans)
, MULTIPLE CHOICE QUESTIONS
3.1
Which one of the following is the appropriate expression for steady-state, one-dimensional, constant
thermal conductivity heat conduction equation for an infinite slab with internal heat generation?
d 2T
q
dx 2
k
(b) T=O
(a) - + - = 0
d2T - 0
(C) -+q=
dx 2
3.2
The exposed surface (x = 0) of a large slab of thickness L and thermal conductivity k is at T0 . The
volumetric internal thermal energy generation is q and the other surface (x = L) is perfectly insulated.
The maximum temperature in the slab will be equal to
(b) l'o+qL2
6k
-r2
-r2
(c) r, +-q0
2k
3.3
(d) T, +-q0
8k
The steady-state temperature distribution in a composite of two materials A and B is shown below:
To,----®
®
I - 6cm -1-- 3 cm -j
Fig. 3.53
T==40°c
h= I03 W/m 2 °C
3.81
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
The maximum temperature in the composite is
(a) 155°C
(b) 195°c
(c) 240°c
(d) 210°c
3.4 A plane wall of thickness 2L has uniform heat generation rate per unit volume g with the two sides
of the wall held at the same temperature T8 • For temperature- dependent thermal conductivity k = k0
( 1 + aT), the temperature at any distance x, measured from the midplane is
1 [( T +1 ) 2g(L2
-x2)]
(a) T(x)=--+
~
(l
s
ko<l
(l
112
(b) T(x)
(d) T(x)
3.5
gx ]1/2
=--+ [(r +-)2-1
<l
s
1
2
~
ko<l
gx ]1/2
=--+ [(r +-)2-koa
1
<l
1
s
~
In a long solid tube, cooled at the inner radius 'i and insulated at the outer radius 12, heat is generated
uniformly at the rate of q (W/m3 )within the solid. The coolant is at T_ while the inner and outer
surface temperatures are J;_ and r;_ respectively. The convection coefficient h can be expressed as
(a) "if(r.J-f)/212 (r;_-r_)
(b) q(r.J-f)/21j(J;_-T_)
(c) 2 lj ( J;_ - T_) /q (r.} - f)
(d) 212(r;_-T_)/q(r.J-f)
3.6 In a solid sphere of thermal conductivity k and radius r 0 with volumetric heat generation q exposed
to convective environment (h, T_), the temperature at the centre is given by
(d) T_
1]
qr;[l
+2
1/k
3.7 Heat is generated in the cylindrical fuel element of a nuclear reactor according to the relation
"if=%[ l-(r!R)2 J. Assuming steady-state, one-dimensional conduction, (Tmax -Tw) is given by
R2
(a)~
4k
R2
(b) ~
6k
3CJi R_
(c) _o
2
(d) 3%R2
16k
8k
3.8 A round wire (k = 15 W Im 0 C) of 6-mm diameter has uniform electrical heat generation. The
centre and surface wire temperatures are 210°C and 45°C respectively. The rate of volumetric heat
generation is
(a) 3.62x107 W/m3
(b) 5.5x108 W/m3
(c) 2.65x10 4 W/m3
(d) 1.lx109 W/m 3
3.9 A piston crown of thickness t and thermal conductivity k has the outside surface temperature Tw
while the maximum temperature is Tmax· If the crown receives heat Q from the gases, then
(a) t=
(c) t =
Q R2
4k(Tmax -Tw}
Q
41tk(Tmax -Tw)
(b) t=
(d) t =
41tk(Tmax -Tw}
Q.
Q
4k(Tmax -Tw)
3.82 , Heat and Mass Transfer
3.10 Consider steady one-dimensional heat flow in a plate of 20-mm-thickness with a uniform heat
generation of 80 MWlm3. The left and right faces are kept at constant temperatures of 160°C and
120°C respectively. The plate has constant thermal conductivity of 200 WIm K.
(A) The location of maximum temperature within the plate from its left face is
(a) 15 mm
(b) 10 mm
(c) 5 mm
(d) 0 mm
(b) The maximum temperature within the plate in °C is
(a) 160
(b) 165
(c) 200
(d) 250
, TRUE/FALSE
3.1 If a plane wall with uniform heat sources has its two surfaces maintained at different temperatures,
the maximum temperature will always occur at the centreline.
3.2 For a circular rod with insulated lateral surface experiencing steady state, one- dimensional uniform
internal heat generation, the governing differential equation is
.!!_(rdT)+ qr =O
dr
dr
k
3.3 For a cylindrical rod of radius ,;, and of thermal conductivity k with heat generation per unit volume
q and exposed to an ambient at T_ with convection coefficient h, the maximum temperature can
be deduced from
Tmax IT_ = [l + H(2 + B)]
.
.
qr.
.
.
hr.
where H ts the heat generation number = - -0 - and B ts the B1ot number = - 0 •
4hT_
k
3.4 The conduction resistance concept in systems involving heat-generation effects is incorrect.
3.5 Heat is generated in both the fuel rod and the cladding in a nuclear reactor.
, FILL IN THE BLANKS
3.1 In a solid cylinder of radius R and thermal conductivity k with uniform volumetric internal heat
generation q and convection at the surface exposed to an ambient at T_ with convection coefficient h,
hR
Tmax - T_ __ where Bi =k
3.2 If the heat is generated within a material due to the passage of an electric current, the rate of uniform
volumetric heat generation, q{Wlm3 ) in terms of current density i (= !_) and electrical resistivity
A
pis q= --·
3.3 For steady conduction in a sphere of radius R with uniform internal heat generation the dimensionless
temperature can be expressed as
T - T_
To -T_
=
2 + Bi (1- p 2 )
2+Bi
where Bi= __ and p =--
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.83
3.4 The average heat generated during ripening of oranges of 8-cm average diameter with thermal
conductivity of 0.15 W/m °C is 300 W/m2 • The difference in temperatures at the centre and at the
surface is - - ·
, EXERCISES
3.1 Microwave heating is used to heat up a slab of meat that is 25-mm thick. The slab surfaces are
open to surrounding air at a temperature of 35°C and the heat transfer coefficient between the slab
surface and the surroundings is 17 W/m 2 K. The thermal conductivity of the slab may be taken to
be 0.47 W/m K. Find the microwave heating in watts per unit volume so that under steady state
conditions the temperature at the centre of the slab is 100°C.
[7.2txto3 W/m3]
4
3.2 The heat-generation rate in a 10-cm thick plane wall is 4 x 10 W/m3. The convective heat-transfer
coefficient between each face of the wall and the ambient air is 50 W/m2 K. Find (a) the surface
temperature, (b) the maximum temperature in the wall.
Assume the ambient temperature to be 20°C and the thermal conductivity of the wall as 15 W/m K.
[63.33°C]
3.3 Determine the relation for steady-state temperature distribution in a slab O~ x ~ L in which heat is
being generated at a rate of g(x) = g0 x2 (W /m3> while the boundary surface at x = 0 is insulated and
that at x = L is maintained at zero temperature. Also find
3
(a) Temperature at x= 0
0
(b) Total heat loss from the surface at x = L
[g L3A]
3.4 An 80-mm thick plane wall (k= 0.14 W/m 0 C) is insulated on one side while the other is exposed to
environment at 90°C. The rate of heat generation within the wall is 12 x 104W /m 3 . If the convective
heat transfer coefficient between the wall and the environment is 560 W /m2°C, determine the
maximum temperature to which the wall will be subjected.
[2667°C]
3.5 A 90-mm-thick plane wall (k= 0.18 W/m K) is insulated on one side while the other side is
exposed to environment at 80°C. The rate of heat generation within the wall is 1.3 x 105 W/m3. The
convective heat-transfer coefficient between the wall and the environment is 520 W/m2 °C. Develop
an expression for temperature profile and sketch it qualitatively. Determine
(a) the maximum temperature in the wall
(b) the temperature of the wall exposed to convective process
(c) the rate of heat dissipation per unit area
[11.7k W/m2]
3.6 The inside surface of a large flat plate of thickness L at x = 0 is insulated, the outside surface at
x = L is maintained at a uniform temperature T2 , and the heat-generation term is in the form
g(x) = g 0e-'YX(W/m 3)
where g 0 and y are constants and x is measured from the insulated inside surface. Develop
(a) an expression for the temperature distribution in the plate
(b) an expression for the temperature at the insulated surface (i.e., x = 0) of the plate
(c) an expression for the heat flux at the outer surface, x=L
[640k W/m2]
3
3.7 Uniform heat of 4000 kW /m is generated in a 2-cm thick flat plate. The temperature of one surface
of the plate is 160°C and that of the other is 100°C. Find the temperature distribution across the
plate. Also, calculate the heat flow per unit area from the two surfaces and the centre of the plate.
Assume the conductivity of the plate material as 200 W/m °C.
[640k W/m2]
3.84 , Heat and Mass Transfer
3.8 Determine the maximum temperature in a plane wall of 8-cm thickness in which heat is generated
uniformly at a rate of 750 kW lm 3 and the two surfaces of the plate are maintained at 20°C and
100°C respectively. Take the thermal conductivity of the plate material to be 20 Wlm K.
[103.33°C]
3.9 A 0.255-cm diameter stainless steel wire, 0.3-m long, is used as an electric resistance heater, in
a laboratory experiment. The measured voltage drop across the wire is 20 V and the current is
40 A. The wire surface temperature is 315°C. Find the maximum temperature in the wire. Take
k= 17.2 Wlm K.
[327.34°C]
3.10 Consider the rise in temperature in a body muscle fibre induced by heat generated from exercising
in a gym. Modelling the muscle as a cylinder of 20-mm diameter with a volumetric heat generation
at the constant rate of 6 kWlm3, determine the maximum temperature in the muscle if the thermal
conductivity of muscle fibre is 0.4 Wlm °C and the muscle surface temperature is 37°C. [37.4°C]
3.11 A 3.2-mm diameter 30-cm long stainless steel wire has a voltage of 10 V impressed on it. The outer
surface temperature of the wire is maintained at 93°C. Calculate the centre temperature of the wire.
Take the resistivity of the wire as 70 µQ cm and the thermal conductivity as 22.5 Wlm °C.
[138.15°C]
3.12 A power transmission line of copper, 1 cm in diameter, has a resistance of 0.005 Qlm. It carries a
current of 200 A. Determine the centreline and surface temperature of the wire on a breezeless day.
Assume h = 40 W lm 2K
kcopper= 375 Wlm K
Tambient = 30°C
[189.20°C]
3.13 In a nuclear reactor, a fuel rod has annular pellets that are 6-mm OD with 2-mm-diameter cylindrical
voids. The pellets are an 80% uranium oxide and 20% plutonium oxide mixture with a loading of
fissionable material such that the rod power is 50 kW Im. The thermal conductivity of the 80-20
mixture is approximately 2.16 Wlm °C. Determine (a) the rate of volumetric heat generation, and
(b) the maximum temperature excess in the pellet.
[1336.5°C]
3.14 Prove that the temperature distribution in a solid sphere with uniform internal heat generation is
given by
T=
T+ q,;, + "iii [1-(!_)
-
3h
6k
1b
2
]
where T_ is the ambient temperature and his the heat-transfer coefficient. What will be the maximum
. .
iiro iiro
Tm.. = T_ + - + 3h 6k
3.15 Heat is generated at the rate of 900 x 10 5 Wlm 3 in a long uranium rod of 6-cm outer diameter whose
thermal conductivity is 50 WIm K. The rod is jacketed by an annulus in which water is flowing with
an average temperature of 150°C and the unit surface conductance is 60000 Wlm 2 K. Determine
the maximum temperature of the fuel rod.
[577.5°C]
3.16 The shielding for a nuclear reactor can be approximated as a plane wall of thickness L. Heat is
generated per unit volume at a rate q(x) within the shielding according to the relation
temperature and where will 1t occur?
q(x) = "ifoa.e-{J.x
where q0 is the incident radiation flux and a. is the absorption coefficient of the shielding material.
Steady-State, One-Dimensional Heat Conduction with Internal Heat Generation
3.85
(a) Obtain an expression for steady-state temperature distribution if the inner (x = 0) and outer (x = L)
surfaces of the shielding are kept at T1 and T2 respectively.
(b) Determine the location in the shield at which maximum temperature will occur.
x= -lln[k('fi-7;) +
a
%L
1-e-aL]
aL
, ANSWER KEY
Multiple Choice Questions
3.1 (a)
3.6 (c)
3.2 (c)
3.7 (d)
3.3 (c)
3.8 (d)
3.4 (a)
3.9 (c)
3.5 (b)
3.10 (c), (b)
3.2 F
3.3 T
3.4 T
3.5 F
3.2 pi2
3.3
True/False
3.1 F
Fill in the Blanks
3.1
-R
=L[2+Bi]
4h
hR
r
k ' R
3.4 40°c
4
HEAT TRANSFER FROM
EXTENDED SURFACES
Concept Review
INTRODUCTION
• 4.1
•
By far, the most frequent application that involves combined conduction-convection effects is one in
which an extended surface is used specifically to enhance the heat-transfer rate between a solid and a
surrounding fluid. Such an extended surface is called a fin. Extended surfaces are used to increase the
overall heat transfer from a base structure by providing extra area attached to it, as a passive method.
Extended surfaces or fins have wide industrial applications and are provided to increase the rate of
heating or cooling.
Fins are used with all modes of heat transfer (natural convection, forced convection, boiling, and
condensation) and many types of fluids. One must however note that the benefit of increased heat transfer
with added fins should be balanced against increased pressure drop/pumping power that may accompany
the extended surface area, as well as increased cost compared to an unfinned tube or surface.
Important applications include
+ Fins on solid-state devices,
+ Fins on air-cooled cylinders of internal combustion engines, and
+ Fins on electrical transformers.
The optimisation oflengths or sizes for extended surfaces is a complicated procedure. When considering
the use of fins, the choice of the number of fins, spacing, length, thickness, shape, material, and so on,
will depend on both heat-transfer and fluid-flow considerations. Manufacturing, maintenance, material
and operating costs are also a key consideration in fin design.
• 4.2
FINS OF UNIFORM CROSS SECTION
•
Shown in Figure 4.1 is a schematic of the heat-transfer processes that occur in a fin. We assume the
base is hotter than the fluid, although the same processes apply when the fluid is hotter than the base.
Heat conducts from the base into the fin and is removed by convection at the outer surface of the fin.
The fin is usually made of a high-thermal-conductivity material to facilitate the flow of heat from the base
to the tip. With heat being convected away from the surface, the temperature of the fin decreases from
base to tip. Often much of the fin is at a temperature not significantly different from the base temperature.
To calculate the heat-transfer rate from a simple fin, we invoke the following assumptions:
+ The heat-transfer coefficient is uniform over the fin surface.
+ Temperature varies along the length of the fin, and does not vary across the fin.
+ The fin's thermal conductivity is constant.
+ The fin shape is constant over the length of the fin.
+ There is no internal heat generation.
4.2
, Heat and Mass Transfer
Temperature, T(x)
Perimeter, P
h, T=
It
Temperature, T(x)
Perimeter, P
Cross-sectional area, Ac
h. T=
\
r:-x-1 ~Ld~
(a)
Fig. 4.1
Geometry and nomenclature of fins of uniform cross-section (a) rectangular fin and (b) circular spine
• Steady operating conditions exist.
• There is no thermal contact resistance between the fin and the base material.
Consider a small element of length dx and write the steady-state energy balance as
{
Net rate of heat gain by
}-{Net rate of heat loss by convection}= 0
conduction in the x direction
i.e.,
through lateral surfaces
( Qcond,x - Qcond,x+dx ) - Qconv
=0
or
With
Qcond x
'
= - kAc dTI and simplifying, we have
dX X
d 2T
hP
---[T(x)-T ]=0
dx 2 kAc
=
(4.1)
where the cross-sectional area Ac, the perimeter P, the heat-transfer coefficient h, and the thermal
conductivity of the fin material k are constant.
This fin equation must be solved for T(x). It can be applied to any fin with a cross-sectional area that
does not vary with x, with constant thermal conductivity, and a heat-transfer coefficient that does not vary
along the fin. Once the temperature distribution is determined, the heat-transfer rate can be calculated.
To solve the fin equation, it is necessary to specify boundary conditions. Typically, the temperature is
known at the base of the fin (i.e., at x = 0). At the tip of the fin (i.e., at x = L), several different boundary
conditions are possible. We can specify the temperature or the heat-transfer coefficient, or the tip might
be treated as an insulated surface. Each of these is an approximation to the real behaviour.
Solutions for fins with four different tip boundary conditions are given in Table 4.1. Note that in
Case 4, the terms preceded by (h!mk) are associated with the convection from the tip. If h is set to zero
(insulated tip) in these terms then the expressions for the convective tip fin Case 4 reduce to the insulated
tip fin expressions Case 2. Likewise, if a fin is very long (mL > -3), then Case I is the limit of Case 2.
Heat Transfer from Extended Surfaces
Table 4.1
r
4.3
Thermal Performance of Uniform Cross-section Fins (m 2 = hP!kAJ
Case
Tip boundary condition
Thermal performance
1
Very long fin (L ~ oo)
T(L) = T=
0(x) = T(x)-T= =exp(-mx)
0b
Tb-T=
Qb = Qfin = mkAC(Tb -T=) = mkAceb
I
'llfin = mL
2
Adiabatic (Insulated) tip
0(x)
T(x)-T=
coshm(L-x)
-dTI
0b
Tb -T=
cosh mL
-0
dx x = L
Qb = Qfin = mkAceb tanh mL = ~hP kAC eb tanh mL
tanhmL
11 fi n = ~
3
Fixed tip temperature
T(L) = TL(known)
( TL -T= Jsinh mx + sinh m(L-x)
0(x)
T(x)-T=
lrb-T=
sinh mL
cosh mL - ( TL - T= J
Tb -T=
'llfin=--m-L-s-inh~m-L-~
l
4
Convection from the tip
0(x)
T(x)-T=
cosh m(L-x)+(h/ mk)sinhm(L-x)
dTI
eb
Tb -T=
cosh mL + (h ! mk)sinh mL
h[T(L)-T=]=-kdx x= L
Q. -Q· _ mkA 0 {sinh mL + (h/ mk)cosh mL}
b- finc b coshmL+(h ! mk)sinhmL
1lr,n
[
sinhmL+(h/ mk)coshmL]( I
mL
= coshmL+(h! mk)sinhmL
j
Figure 4.2 illustrates the schematic representation of the four boundary conditions at the tip of a fin
for the one-dimensional, steady-state heat transfer in a fin of uniform cross section.
4.2.1
Corrected Length
The expressions for the fin with convection from its tip are cumbersome compared to those of the insulated
tip fin. We can obtain a reasonable approximation to the convective tip fin if in the insulated tip fin
expressions we use a corrected length, Le, to account for the additional area of the fin tip.
(4.2)
For the pin fin, Le= L + D/4, and for a rectangular fin, Le= L + t/2, where tis the fin thickness.
4.4
,
Heat and Mass Transfer
lease 21
Tb
T(x)
dTl
dx
=O
=L
T=
~ L
,I
0
lease 41
=h(h-T=)
Tb
T=
i=-;- L
~ L
0
0
!For all cases:
Fig. 4.2
,I
Tlx=o=Tbl
Schematic representation of four boundary conditions at the fin tip
FIN EFFICIENCY AND FIN EFFECTIVENESS
• 4.3
•
The temperature of a fin drops along the fin, and thus the heat transfer from the fin will be less because
of the decreasing temperature difference towards the fin tip. To account for the effect of this decrease in
temperature on heat-transfer, we define fin efficiency as
Actual heat-transfer rate from the fin
rJ fin = Q.fin ,max = Ideal heat-transfer rate from the fin if the
entire fin were at the base temperature
Qfin
(4.3)
where the fin efficiency depends on
• the fin geometry,
• the fin material, and
• the heat-transfer coefficient.
When the fin efficiency is known, the rate of heat transfer from a fin can be found from
(4.4)
The performance of the fins is judged on the basis of the enhancement in heat transfer relative to the
no-fin (bare) case and is expressed in terms of the fin effectiveness E p defined as
Heat Transfer from Extended Surfaces
r
4.5
Heat-transfer rate from
Qfin
Qfin
thefinofbaseareaAb
£fin=-.--=-----=-------Qnofin
hAb(Tb -T=) Heat-transfer rate from
(4.5)
the surface of area Ab
Here, Ab is the cross-sectional area of the fin at the base and Qnofin represents the rate of heat-transfer
from this area if no fins are attached to the surface.
Fin efficiency and fin effectiveness are related to each other by
(4.6)
The ratio of the finned surface to the plane one is called the finning factor.
.
.
Fmmngfactor
= [ Afin + Aunfin
J
(4.7)
Ano fin
FINS OF NON-UNIFORM CROSS-SECTION
• 4.4
4.4.1
•
Circumferential (Annular) Fins
Fin-efficiency relations have been developed for fins of non-uniform (variable) area of cross-section. For
a circumferential or circular fin Figure 4.4 of constant thickness (annular fins) and rectangular profile,
the fin efficiency can be mathematically expressed as
)11 { mr c)- /1 { mr }K 1 { mr c)
2 I mr }K { mr c) + K mr )I mr c)
r/tin = C
1
where
K 1 { mr1
2
0 (
2
1
1
0 (
2
1
1{
(4.8)
2
2r1 / m
C2
= 2
2
r2c - rt
Fin parameter, m = ~2h/ kt
I0 = modified zero-order Bessel function of the
first kind
K 0 = modified zero-order Bessel function of the
second kind
/ 1 = modified first-order Bessel function of the
first kind
K 1 = modified first-order Bessel function of the
second kind
Fin heat-transfer rate,
r2c = r2 + t/2
L=r2 -r 1
Afin= 2n (~c
,_!---.-_r___________._ r2
(4.9)
Fig. 4.3
Annular fin of a rectangular profile
-11)
4.6
, Heat and Mass Transfer
4.4.2
Triangular Fins
For a straight fin with triangular profile shown in Figure 4.4, the fin efficiency is given by
1 / 1 (2mL)
llr,n = ;;i, I0 (2mL)
where
m = ~2h/kt
and
A 60 = 2w~L2 + (t/2)2
(4.10)
The fin heat-transfer rate is expressed as
IQ
fin= r/jin h·2w[L2 + (t/2) 2 ]112
('z;, - T~)I
(4.11)
w
Efficiency of straight fins of rectangular, triangular and
parabolic profiles as well as efficiency of circumferential
fins ofrectangular profile can be obtained both analytically
using the relevant expressions for fin efficiency as well as
graphically.
Fig. 4.4
A straight triangular fin
FIN ARRAYS
• 4.5
4.5.1
:=J
•
Overall Surface Efficiency
For an array of N fins, the total or overall surface efficiency, r/overall (similar to the fin efficiency) is defined as
Qtotal
(4.12)
11 overall = -. Q
max
where Qmax = hAtotal (Tb - T~) is the maximum possible heat-transfer rate from the total surface.
N = number of fins
Anofin=wH
Aunfin = wH-N(wt)
Afin= 2Lw+ wt
""2Lw(one fin)
Atotal = N Afin + Aunfin
Fig. 4.5
A plane wall with a rectangular fin array
In terms of the fin efficiency, an expression for the overall surface efficiency is
)
NAr, 0 (
11 overall = 1 - -A-- 1 - llfin
total
(4.13)
Heat Transfer from Extended Surfaces
r
4. 7
The total heat-transfer rate,
(4.14)
For a fin array of annular fins shown in Figure 4.6 comprising N fins, the total heat-transfer rate is
.
NAfin( l-1Jjin )] 0b
Qtotal
= h Atotal [ 1- -A-total
(4.15)
Atotal = NA fin+ 2nr1 (H - Nt)
where
l'--------,
~----,1t
H ~----'
'---~T
~---------+-r2
An array of annular fins
Fig. 4.6
4.5.2
Effectiveness of a Fin Array
The overall effectiveness for a finned surface or the effectiveness of a fin array is defined as the ratio
of the total heat-transfer rate from the finned surface to the heat-transfer rate from the same surface if
there were no fins.
£ fin,overall
Qtotal,fin
h{Aunfin +'llfinAfin)(J;,-T=)
total,nofin
hAnofin (Tb -T=)
= Q.
(4.16)
Referring to Figure 4.7, if there are N fins and (N- 1) gaps of spacing S between fins, then we have:
Fins
(4.17)
NLTJfin + (N - l)S
Nt+(N-l)S
{4.18)
where
H=(N-1) S+Nt
One must note that the total heat-transfer rate Q101a,
calculated by using both the concepts will be same.
t
---+- -+--
s
L
Total heat-transfer rate,
IQtotal =efinarray(HW)h(Ti,-T=)I
T
1
~
Parent surface
H
Fig. 4. 7
A typical fin array
4.8
, Heat and Mass Transfer
Solved Examples
(A) VERY LONG FIN
•
•
Example 4.1 ~ A very long rod, 25-mm in diameter has one end maintained at I 00°C. The surface of the rod
is exposed to ambient air at 25°C with a convection heat-transfer coefficient of IO W/+m 2 K.
(a) What are the heat losses from the rods constructed of copper and stainless steel? (b) Estimate how long
the rods must be to be considered infinite.
Take k for copper= 398 W/m K, k for stainless steel = 14 W/m K.
Known: A long cylindrical rod is exposed to ambient air.
Find: (a) Heat loss from the rod made of copper and stainless steel. (b) Length of the rod to be considered
an infinitely long fin.
Schematic:
Air
T
Tb= I00°C
Ab =Ac= nD2/4
P=nD
"'--"--~-------+--~£:>_E2Smm
k = 398 W/m K (Copper)
k = 14 W/m K (Stainless steel)
T(x)
~----------~---- X
0
Fig. 4.8
Assumptions: (1) Steady conditions prevail (2) Uniform heat-transfer coefficient. (3) Constant thermal
conductivity. (4) Infinitely long fin. (5) Negligible radiation heat-loss. (6) One dimensional conduction
along the fin.
Analysis: The temperature distribution and the heat-transfer rate, subject to the assumption of an infinitely
long fin, are given by
0
T- T
-=--=-=e-mx
0b J;,-T-
and
~ ( J;,-T- ) =mkAc0b
Q.1 =vhPkA
f~
2
where
m = Jf/2=( k~~~/4
=
:~
Heat Transfer (ram Extended Surfaces
r
4. 9
For copper, the heat-loss is
Qr= (lOW /m2 K)(1tx0.025m)(398W /mK>( ~x0.0252 )m2 (100-25)°C
(Ans) (a)
=29.38 W
For stainless steel,
Qr= (l0)(1tx0.025))(14>( ~x0.0252 )c75)
(Ans) (a)
=5.51 W
Since there is no heat loss from the tip of an infinitely long fin, we can write to a successful
approximation: tanh mL"" 0.99 or mL ~ 2.65. Hence, a rod may be assumed to be infinitely long if
For copper,
m=~k4Dh =
L =
~
4xlOW/m2 K
= 2 .0m-l
398W /mKx0.025m
2 ·65 = 1.32 m
2.om- 1
(Ans) (b)
For stainless steel,
m=
L =
~
4xlOW /m2 K
= 10 _69 m_ 1
14W/mKx0.025m
2 ·65
= 0.248 m
10.69m- 1
(Ans) (b)
Comment: For more accurate prediction of temperature variation, a larger value of mL (4.6 in place of
0
2.65) would be required. With mL = 4.6, __!,_=e-mL =e--4·6 = 0.01, and T(L~) = 25.75°C"" T~.
0b
Example 4.2 ~ Two long rods of the same diameter, one made of brass (k = 85 W/m K) and the other
made of copper (k = 375 W/m K), have one of their ends inserted into a furnace. Both rods
are exposed to the same environment. At a section I0.5 cm away from the furnace, the temperature of the
brass rod is I 20°C. At what distance from the furnace end, the same temperature would be reached in the
copper rodl
Known: Two long rods (fins) of different materials protrude from the furnace with the same end
temperatures.
Find: Distance, L 2 for copper rod.
4.1 0
,
Heat and Mass Transfer
Schematic:
Furnace
L
C Brass (k =85W/m K)
_r-_~I
L1 =O.IOSm
1
Ji - - - .
r 1= 120°c
C Copper (k = 375W/m K)
Furnace
2
l_lf---._ ! -_ _____.
L2 = l
Tb
Fig. 4.9
Assumptions: (1) Steady operating conditions exist. (2) Constant area of cross-section. (3) Very long fin.
Analysis: The temperature distribution along the length of a very long fin is given by
0
0b
where
Ii, -T-
m = ~khAPc =
{IE
-vu
h(trd) _
k(trd2 /4)
As
T(x = L 1) = T(x = L 2) = 120°C, 0 1 = 02
Hence
0, = 0b e-m,L,
and
82 = 0b e-m2L2
With 0 1 = 02 , we have
or
(Ans)
Example 4.3 ~ In an experiment to determine the thermal conductivity of a long solid 2.5-cm-diameter rod,
its base is placed in a furnace with a large portion of it projecting into the room air at 22°C.
After steady-state conditions prevail, the temperatures at two points, IO cm apart, are found to be I I0°C and
85°C, respectively. The convective heat-transfer coefficient between the rod surface and the surrounding air is
28.4 W/m 2 K. Determine the thermal conductivity of the rod material.
Known: Temperatures at two points, a distance apart, on a rod exposed to convective environment.
Find: Thermal conductivity of rod material, k.
Heat Transfer from Extended Surfaces
Schematic:
r
4.1 I
T(x)=85°C
0=2.5cm
lll
Air
Furnace
h = 28.4 W/m2 K
T= =22°C
Fig. 4.10
Assumptions: ( l) Steady-state conditions prevail. (2) The rod is considered very long compared to its
diameter. (3) Uniform heat-transfer coefficient. (4) Let x = 0 at the first temperature point (T (0) = 110°C).
Analysis: The temperature distribution in the long rod is given by
T(x)-T=
T(O)-T=
hnD
k (nD 2 ! 4) =
[}E
VW
Inserting the proper values, we obtain
85-22 = e-m(o.t)
110-22
Solving for m,
e-O.lm
or
-0. lm = ln{0.7159) = -0.3342
Then
or
= 0.7159
m
= 3.342 m- 1
4h
m 2 = - = 11.169 m-2
kD
Thermal conductivity of the rod material,
k = ~ = 4x28.4 W/ m2 K
m2 D
l l.l69m-2 x0.025 m
=406.8 W/m K
(Ans)
Example 4.4 ~ An induction coil is used to heat a spot on a long, 12.5-mm-diameter, 1.5 per cent C carbon
steel (k = 36 W/m 0 C) reinforcing bar for an intricate bending operation. Assuming that
95 per cent of the energy applied to the coil appears in the bar, estimate the required power input to maintain
a spot on the bar at 650°C when the convection coefficient on the bar is 85 W/m 2 °C and the ambient air
temperature is 20°C.
Known: A long circular bar requires a spot on it at midpoint at high temperature for bending operation.
Find: Energy input required to hold a spot on the bar at 650°C.
4.12
,
Heat and Mass Transfer
Schematic:
Tb=6S0°C~pot
rbar (D = 12.5mm)
=================~============~•====IL~=
k = 36W/m°C
2
Ill
h=85W/m °C
T==20°C
Air
Fig. 4.11
Assumptions: ( l) Steady-state conditions. (2) The real physical situation is approximated as two infinite
fins with a base temperature of 650°C. (3) Constant properties and uniform heat-transfer coefficient.
Analysis: Fin parameter,
f'E
hnD
k(nD 2 /4) =
VW
4x85W / m2 oc
= 27.487 m-t
36 W / m°C X 12.5 X lQ-3 m
We note that 95% of energy applied to the coil appears in the bar.
Power input required = 2Qfin(L~=/ 0.95
Qfln = mk,{0b =
mk( ¾)n (Tb -T= )
2
1t
= 27.487 m- 1 x36W/m °Cx-x0 .0125 2 m2 x(650-20)°C
4
=76.5W
Hence, power input to the coil is
2 x 76.5 W/0.95 = 161 W
(Ans)
Example 4.5 ~ A carbon steel (k = 54 W/m K) rod with a cross-section of an equilateral triangle (each side
5 mm) is 8 cm long. It is attached to a plane wall, which is maintained at a temperature of
400°C. The surrounding environment is at 50°C and the convective heat-transfer coefficient is 90 W/m 2 K.
Calculate the heat dissipated by the rod.
Known: A rod of equilateral triangular cross section protrudes from a hot isothermal plane wall.
Find: Heat-dissipation rate, Q(W).
Schematic:
I II
h=90W/m2K
T= =S0°C
r, =400•c j,_,_ _ _ _ _L_=_8_c_m_ _ _ _
Lx
~
x=L=
5mm
/1/1
;
'
ih
/ ; 0=60° !
b -----+<
Fig. 4.12
Assumptions: ( l) Steady operating conditions. (2) One-dimensional conduction. (3) Fin is very long
(mL > 2.65). (4) Constant thermal conductivity. (5) Uniform heat-transfer coefficient.
Heat Transfer (ram Extended Surfaces
r
4.1 l
Analysis: Treating the rod as infinitely long, the heat-transfer rate is
Q=~hPkAC eh
where
eh= (Tb - T~) = (400 - 50) = 350°c
Consider the equilateral triangle (each side 5 mm) .
. e=-h
sm
5
h = 5 sin 60° = 4.33 mm
1
1
Ac =-bh=-(5)(4.33)mm2 = 10.825x1Q-6 m2
2
2
m = ~ hP =
kAc
{90W/m2 K)(5+5+5){10-3 )m
(54W/mK){10.825xl0-6 )m2
48 _06 m-l
mL = (48.06m- 1 C)(0.08m) = 3.845
Now, the heat-dissipation rate from the rod is
Q = ~{90W/m2 K)(0.015 m)(54 W/mK){10.825x10-6)m2 {350°C or K)
=9.8W
(Ans)
Comment: We note that e-mL = 0.021, i.e., approaching zero. Hence, our assumption of infmitely long
rod is valid.
(B) FINS OF FINITE LENGTH
•
•
Example 4.6 ~ Two rods of identical size and shape are both supported between two heat-sources at I00°C
and are surrounded by air at 25°C. One rod is known to have a thermal conductivity of
14.2 W/m K and its midpoint temperature is measured to be 49°C. If the midpoint temperature of the other
rod is measured to be 75°C, what is its thermal conductivity?
Known: Two rods of different materials but same configuration, supported between two ends, at the
same temperature have different midpoint temperatures when exposed to same environmental conditions.
Find: Thermal conductivity of the second rod, k2 , (W/m 0 C).
Schematic:
Air
Air
Ill
Tm 2 =75°C " \
100°c
100°c
I.
2L
100°c
= li-,-h_._r-:25~k 1=14.2W/mK
2s·c
= .~~-h~·-r_~_~_2_s_·_c_½h k2 = l
:i
Tb= 100°c
Fig. 4.13
4.14
,
Heat and Mass Transfer
Assumptions: (1) Steady-state conditions. (2) Constant properties and uniform convection coefficient.
(3) One-dimensional conduction. (4) Adiabatic fin tip.
Analysis: A rod supported between two ends that are maintained at the same temperature and are exposed
to the convective environment can be looked upon as a fin of half the rod length with the midpoint
temperature as the temperature of the adiabatic fin tip.
Fin parameter, m =
C
Dimensionless fin parameter, mL = ~hP L2 = _f_
kAC
Ji
where C is same in both cases.
Midpoint temperature of the rod of length 2L, Tm = Tip temperature of the fin of length L, TL
Temperature distribution is given by
T-T-
Ii, -T0L
or
eh
=
coshm(L-x)
=
coshmL
1
coshmL
0
In the first case, cosh "'1.L = _b
0L I
~
and
TL -T-
Ii, -T-
=
coshm(L-L)
coshmL
0
cosh mL = i
0L
where
0
In the second case, cosh m 2 L = _b
0L 2
::.)= ff1L= }
cosh-'( ::,)=m,L= F,
cosh-• (
and
(A)
(B)
ab= 100-25 = 75°c, eL, = 49-25 = 24°c
and
0L 2 = 75-25 = 50°C
cosh- 1 (0b/0L I ) = cosh- 1 (75/24) = 1.8059
cosh- 1 (0b/0L 2 ) = cosh- 1 (75/50) = 0.9624
(A)/(B) yields
v,;;{k;"
or
= 1.8059 = 1.8764
o.9624
k2 = 3.521 k 1 = 3.521 x 14.2 W/m K = 50 W/m K
:. Thermal conductivity of the other rod is
k2 =50 W/m K
(Ans)
Heat Transfer from Extended Surfaces
r
4.1 5
Example 4.7 ~ A copper fin (k = 387.6 W/m 0 C) of I O-cm 2 cross-section, 15-cm perimeter and 12-cm length,
protrudes from a surface into a convective environment of h = 25 W/m 2 K and temperature
T- = 32°C. The fin temperatures at a distance of 3 cm and 7 cm from the surface are measured to be 250°C
and 200°C respectively.
(a) What is the fin base temperature?
(b) What is the heat lost by the fin?
(c) What is the fin efficiency?
Known: A copper fin of prescribed dimensions is losing heat to a convective environment.
Find: (a) Fin base temperature,Tb {0 C); (b) Heat lost by fin, Q6 n(W); (c) Fin efficiency, 'lltin·
Schematic:
// /
1 / h=25W/m2 K,
T= = 32°C
L= 12cm
x=O
T1 = 250°C, x1 = 3 cm
x=L
Copper fin
(k = 387.6 W/m 0 C)
T2 =200°C, x2=7cm
Ac= 10cm2
P=IScm
Fig. 4.14
Assumptions: (I) Steady state prevails. (2) Fin tip is insulated. (3) One-dimensional conduction.
(4) Uniform cross-section.
Analysis: The temperature distribution along the length of a fin is given by
(I)
To evaluate C 1 and C2, let us first determine the value of fin parameter, m.
m2 = hP = (25W / m2 K)(l5cm) IIOOcml
k Ac (387.6W I m 0 C)(!Ocm2) Im
= 9.675
m=3.ll m- 1 and
mL=(3.ll m- 1) (0.12 m)
= 0.3732
Also tanh mL = 0.3568
At x 1 = 0.03 m, mx 1 = 0.0933 and T1 = 250°C
At X2 = 0.07 m, mx2 = 0.2177 and T2 = 200°c
Substituting these values in Equation (I), one gets
250 - 32 = c,eo.0933 + c 2e-o.0933
200-32 = C,eo.2111 +C2e-o.2m
4.16
,
Heat and Mass Transfer
Simplifying,
1.0978 C1 +0.9109 C2 =218
(2)
1.2432 C1 + 0.8044 C2 = 168
(3)
Multiplying Equation (2) by 0.8831,
0.9694 C1 + 0.8044 C2 = 192.51
1.2432 C1 + 0.8044 C2 = 168
-0.2738 c, + 0
= 24.51
or
C1 =-89.5
and
C2 = [168 - (1.2432) (-89.52)] / 0.8044
or
C2 =347.20
At the base or root of the fin, (x = 0),
eb = Tb - T= = c,e + C2e-0 = c, + C2
0
0b = C1 + C2 = -89.5 + 347.20 = 257.7°C
Fin base temperature, Tb= 257.7 + 32 = 289.7°C
(Ans) (a)
Rate of heat transfer from the fin,
Qfin =mk Ac0b tanhmL
= (3.llm- 1)(387.6 W/m 0 C) (10 X 1Q-4m2)(257.7°C)(0.3568)
= 110.84 W
(Ans) (b)
Fin efficiency,
llr
tanhmL
mL
=95.6%
Example 4.8
0.3568xlOO
0.3732
(Ans) (c)
~ A fin in the form of a ring of 1.25-mm thickness and 25-mm OD and 25-mm long is
employed on an electric device to dissipate heat. The thermal conductivity of the fin material
is 204 WI m C. The heat-transfer coefficient between the effective outer surface of the fin and the I 5°C
surrounding air is 30 W/m 2 C. The heat-dissipation rate is 1.1 W. If the device is also of the same outside
diameter of 25 mm, evaluate the device temperature with and without the fin.
0
0
Known: A electric device is equipped with a hollow fin of the same outer diameter exposed to ambient air.
Find: Electric device surface temperature with and without the fin.
Heat Transfer from Extended Surfaces
Schematic:
r
4.17
Air
T== IS°C
h=30W/m2 K
!!!
k=204 W/m K
~-~-+---------~
J_
t= 1.25 mm
A= ~(D2-D2)
C
4
0
I
T
L=25mm
Electric device
------,
Heat transfer from inner surface
is negligible
ID;=D -2t=(2S-2x 12.S)mm=22.Smm I
0
Fig. 4.15
Assumptions: ( l) Steady operating conditions exist. (2) Constant thermal conductivity. (3) Heat-transfer
from the inner surface of the fin is negligible. (4) The fin tip is adiabatic.
Analysis: Without fin
Heat-transfer rate, O.wo = hAs (Tb - T=)
where As is the area of the base or primary surface (with fin) as well as the area of the electric device
exposed to air (without fin) =~D; =~x0.025 2 m2
4
4
Surface temperature of the exposed area of electric device without fin is given by
T. -T = O.wo
b
=
hA
s
l.lW
I;, =15°C+----------=89.7°C
(30W / m 2 0 C)(~x0.025 2 )m 2
4
With fin
O.w =mk A e; tanhmL =O.wo
where
4(30 W/m2 K) (0.025m)
(204 W / mK)(0.025 2 -0.0225 2 )m 2
= 11.13 m- 1
(Ans)
4.18
,
Heat and Mass Transfer
mL = (11.13 m- 1)(0.025 m) = 0.2782
tanh m L = 0.271
and
Base temperature,
4Q
I;,*=T +
= mk1t(DJ-D?)tanhmL
= 15oc+
4xl.1W
(1 l.13m- 1)(204 W / mK)(1t)(0.025 2 -0.0225 2 )m 2 (0.271)
= (15 + 6.1)°C = 21.1°C
(Ans)
Comment: For the same heat dissipation, the device temperature with fin will be 68.6°C lower than
that without fin.
Example 4. 9
~ An electric motor casing made of cast steel (k = 60 W/m°C) has a diameter of 36 cm and
a length of 40 cm. A number of fins of the same material are provided to dissipate 400 W
of heat to the surrounding air at 30°C with an associated convection coefficient of IO W/m 2 0 C. Each fin, 8-mm
thick with a radial length (height) of 10 mm, runs the entire length of the motor casing which has a surface
temperature of 60°C. Determine (a) the number of fins needed, (b) the temperature at the centre of fin, (c) the
percentage increase in heat-dissipation rate due to addition of fins, and (d) the fin efficiency.
Known: Longitudinal fins of rectangular cross-section are provided on an electric motor casing to enhance
heat dissipation to the ambient air.
Find: (a) Number of fins required, N. (b) Temperature at the centre of the fin, T(x = L/2) (c) Percentage
increase in heat-transfer rate because of fins.
Schematic:
Air
I =0.4m
1
rb=~o · c
(k= 60~/m C)
Electri c motor
casing
W=0.4m
h= IOW/m2°C
r = =30°c
0
r r r
.....r· ··
•
- J A typical fin
I~
L = 10mm
Fig. 4.16
Assumptions: (1) Steady operating conditions prevail. (2) Constant thermal properties. (3) Heat-transfer
coefficient is uniform on finned and unfinned surfaces. (4) One-dimensional heat conduction.
Heat Transfer fram Extended Surfaces
r
4.19
Analysis: The rate of heat dissipation,
Qwithfin =Qunfin +Qfin
(a)
= hAunrinCI'i, -T=)+NmkAJ!i, -T=)tanhmLc
Now,
Aunfin = Area not covered by fins
=A. -Afin
= (re D L) - (N Wt)
Noting that length of the cylindrical motor casing, l is also the width of the rectangular fins, W, we have
Aunfin =(rcx0.36mx0.4m)-(Nx0.4mx0.008m)
= 0.4524 m 2 -(0.0032 N)m2
eh =Th-T= =(60-30)°C=30°C
Fin parameter,
m= ~ hP =
k Ac
h2(W +t) =
kWt
2x10W/m2 °Cx(0.4+0.008m)
60W/m °Cx0.4mx0.008m
= 6.52 m- 1
Ac
Corrected length, Lc =L+p
= O.Olm+( 0.00 32 m 2 ) = 0.014m
2x0.408m
mLc = (6.52 m- 1)(0.014 m) = 0.091
tanh mLc = 0.0907
and
cosh mLc = 1.004
Substituting the appropriate numerical values in Equation (a), one has
400 W = (10 W/m20 C) (0.4524 - 0.0032 N) m 2 (30°C)
+ Nx (6.52 m- 1) (60 W/m 0 C) (0.0032 m 2 ) (30°C) (0.0907)
or
or
400 = (135.72 - 0.96 N) + 3.41 N
2.446 N = 264.28
Hence, the number of fins provided,
N = 264.28 = lOS
2.446
Temperature distribution is of the form
e(x) _ coshm(Lc -x)
eh
coshm4
(Ans) (a)
4.20
,
Heat and Mass Transfer
At x = L/2, the temperature at the centre of the fin is
0(x =L/2)=
0
b
coshmLc
xcosh(m4/2)
30°c
= 1.00 4 xcosh(0.091/2)
= 30 xl.001/1.004 = 29.9°C
T(x = L/2) = 30 + 29.9 = 59.9°C
(Ans) (b)
Heat-transfer rate without fins,
Qwithoutfin =h(1tDL)(Tb -T=)
= (lOW /m20C)(1tx0.36mx0.4m)(30°C)
=135.72W
Percentage increase in heat dissipation due to addition of fins is
Qwi~ - Qwithout X 100
Qwithout
= (400W-135.72W) (lOO) =195 %
135.72W
(Ans) (c)
Fin efficiency,
0.0 907 = 0.997 or 99.7%
0.091
(Ans) (d)
Comment: The excess temperature at the base is 30°C and that at the centre of the fin is 29.9°C. Even
at the fin tip, 0(L) = 30/1.004 = 29.88°C. Thus, we find that throughout the length of the fin there is no
noticeable drop in temperature from the root to the tip. An ideal fin is one in which 0(x = 0) = 0(x = L).
Thus the fins in this case are predictably very efficient (TJr= 99.7%).
Example 4.10 ~ In a gas turbine plant, turbine blades mounted to a rotating disc are exposed to a gas stream
at I 200°C with a combined convection and radiation heat-transfer coefficient between the
blades and gas stream of 200 W/m 20 C. In a proposed blade cooling scheme that involves passing air as a coolant
through the disc, the blade root (base) temperature is maintained at 300°C. The blade profile has a perimeter
of 11 cm and a cross-sectional area of 6 cm 2• The blades are made from lnconel with a thermal conductivity
of 20 W/m 0 C. If the maximum permissible blade temperature is 1050°C due to metallurgical limitations, will a
blade length (height) of 5 cm be satisfactoryl If so, determine the blade tip temperature, and the rate of heat
transfer from each blade to the cooling air. What will be the temperature at the centre of the blade. Account
for the heat-loss from the fin tip by using the 'corrected length' in the insulated fin tip expressions.
Known: A gas turbine blade is subjected to a gas stream. Blade profile and base temperature are known.
Find: Blade tip temperature, TL( 0 C). Heat-transfer rate, Q(W).
Heat Transfer from Extended Surfaces
Schematic:
r
4.21
Blade tip (h = 1)
i
L= 5cm
Gas stream
X
/ / ;h=200W/m2°C
Cooling air
Fig. 4.17
Assumptions: ( l) Steady-state, one-dimensional conduction. (2) Uniform heat-transfer coefficient and
constant properties.
.
~
6~
Analysis: Corrected length, Le = L+ - = 5 cm+-11cm
p
= 42.82 m-1
mLc = (5.545xl0-2 m) (42.82m- 1 )
= 2.374
cosh mLc = 5.4189 tanh mLc = 0.9828
Temperature distribution along the blade length can be expressed as
coshmLc
At x = L, the blade tip temperature is
= 12000C+ (300- l 200)oC XCOSh( 42.82m-l X0.545X 10-2 m)
5.4189
= 1200 - C900) (l .0273) = 1030°c
5.4189
(Ans)
As the blade's maximum temperature (i.e., at the tip) is less than the maximum permissible value of
1050°C, the proposed cooling scheme is satisfactory and the blade height of 5 cm is adequate.
4.22
, Heat and Mass Transfer
The heat transferred from the gases to the blade equals, under steady operating conditions, the rate of
heat-transfer from the blade to the coolant air. Therefore, the fin heat-transfer rate is
Qfin = m k 4, eb tanh mLC
= (42.82 m- 1)(20W /m°C)(6 x 10-4 m2 )(1200- 300)°C(0.9828)
=454.SW
(Ans)
Temperature at the centre of the blade (at x = L/2 = 2.5 cm) is
T(x = L/ 2) = 12000C _ 9000C X cosh( 42.82 m-t X 3.045x 10-2 m)
5.4189
= 871.6°C
Example 4.11
(Ans)
~ An aluminium fin is attached to a transistor which generates heat at the rate of 300 mW.
The fin has a total surface area of 9 cm 2 and is exposed to surrounding air at 27°C. The
contact resistance between the transistor and the fin is 10- 4 m2 °C/W and the contact area is 0.5 cm 2 . Estimate
the temperature of the transistor, assuming the fin is uniform in temperature and the convection coefficient is
11 W/m 2 0 C.
Known: An ideal aluminium fin attached to a transistor loses heat-by convection. Contact resistance is
specified.
Find: Transistor temperature, Ts( C).
0
Schematic:
h= 11 W/m2°C
T==27°C
>----III_____,!
Transistor
TL
Contact resistance
Fin
Fin
T
Ts
-
Tb
T=
Q ~
~
Acontact
h Ar,n
~-------------x
Fig. 4.18
Assumptions: ( 1) Steady-state conditions. (2) Fin is of infinite thermal conductivity.
Analysis: As the fin is uniform in temperature, the fin is ideal with no conduction resistance. It loses
heat to the surrounding air only by convection.
Heat-generated= Heat-dissipated= 300 mW= 0.3 W
As
Q= Ll T,,verall
Rtotal
Heat Transfer from Extended Surfaces
r
4.23
transistor temperature,
= 27oc+0.30W[10-4 m2 oc / W +
0.5xl0-4m 2
l
]
llW / m 2 °Cx9xl0-4 m 2
= 57.9°c
(Ans)
Example 4.12 ~ A semicircular ring (cross-section I cm x I cm and radius IO cm) is fixed to a wall as shown.
Determine the rate at which heat is dissipated to the surroundings from the surface of the
ring if the wall is at a temperature of 200°C and the surrounding air is at 25°C.
Wall
R= 10cm
I cm
Diem
Fig. 4.19
The surface heat-transfer coefficient is 11 W/m 2K and the thermal conductivity of the material of the ring is IOW/m K.
Known: Heat is lost to ambient air from heated wall to which a semi-circular ring is fixed as an extended
surface.
Find: Rate of heat dissipation, Qfin (W) .
Schematic:
///
Wall
0b=Tb-T=
h= 11 W/K
T= =25°C
Insulated end
(centre of the ring)
k= IOW/m K
=200-25
= 11s c
0
X
Square cross section
Fig. 4.20
Assumptions: (l) Steady-state, one-dimensional conduction. (2) Constant properties and uniform
convection coefficient. (3) The centre of the ring (by symmetry) is idealized as an adiabatic fin tip.
4.24
, Heat and Mass Transfer
Analysis: The semicircular ring circumference = ,r R
= 10 1t cm
I
By symmetry, the fin length, L=-x!01tcm
2
= 0.05 1t m
Heat rate, Q = 2 m k Ac0b tanh mL
i.e., twice that for a fin of half the ring's length.
~
Fin parameter, m = V
k Ac
For a square cross-section
.!_= 4a =-4-=400m-t
O.Olm
Ac
a2
m=
l lW/m2 Kx400m- 1 = 20 _976 m-t
IOW / mK
mL = (20.97 m- 1)(0.05 pm)= 3.295
tanh mL = 0.997
Q=2x(20.976m- 1 )(lOW / mK)(O.Olm) 2 (175 C)(0.997:
(Ans)
=7.32 W
Example 4. 13 ~ Consider a transistor, idealised as a long cylinder of I-mm radius and 1.5-cm length. On this,
a sheath of I-mm thickness (k = 0.1 W/m K) is put on it. The rating of transistor is 150 mW,
and if the transistor is to function properly, its surface temperature (at radius I mm) should not exceed 80°C.
The ambient air is at 30°C (h = 15 W/m 2 K). How much heat-can be dissipated into the atmosphere by the
cylindrical surface? The remaining heat-is conducted away by three brass leads (k = 120 W/m K). If circular wire
leads of 0.3-mm diameter are used, what should be the length of these leads exposed to the atmospheric air?
Known: A transistor with a sheath has three brass leads exposed to atmospheric air.
Find: Heat-dissipation rate. Length of leads (fins).
Schematic:
Sheath
Transistor
//;
T==30°C
h= 15W/m2K
L= 1.5cm
Fig. 4.21a
Heat Transfer from Extended Surfaces
4=.ck=
r--
4.25
Atmosphere
Cylindrical
surface
r
120W/~K
D=~mm
.:T
:s
L
Tb=80 °C
/
//r==J0°C
h= ISW/m2K
Wire~ad (fin)
Fig. 4.21b
Assumptions: ( l) Steady-state conditions. (2) One-dimensional (radial) heat-flow. (3) Constant properties.
(4) Constant heat-transfer coefficient. (5) End effects are negligible. (6) Heat-transfer from the tip of the
wire is neglected. (7) Contact resistance is negligible.
Analysis: Heat-transfer in the radial direction from the transistor surface is
.
t:.T
Q=--Rt,overall
where
t:.T= T; - T= = (80 - 30)°C
= 50°C
Overall thermal resistance,
Rt,overall
=Rt,cond + Rt,conv
= -1
-ln,;,+---2nkL
r; h(2rc,;,L)
l
= _l_[ln(,;, l r;) +-1
2n L
k
h,;,
l
[/n(2mm/lmm)
l
]
=2rc(l.5xl0-2 m)
O. lW / mK +(15W/m 2 K)(2xl0-3 m)
=427.22K/W
Therefore, the heat dissipated from the cylindrical surface,
.
50°C
(10 3 mW)
Q(r)= 427.22K /W
lW
= 117 mW
(Ans)
The transistor rating is given as 150 mW.
Hence the remaining heat to be dissipated through three brass wire leads (assumed to be fins with
base temperature equal to the surface temperature of the transistor) is
Qfins
=Q-Q(r)
=(150-ll7)mW
=33mW
4.26
,
Heat and Mass Transfer
Per lead, i.e., for a single fin, the heat-transfer rate is
Qr,n = m k Ac 0b tanh m L
= 33mW l3= 11 mW
Now,
m = ~khAPc =
= ~k4Dh =
h(n D)
k(nD 2 14)
4(15W l m 2 K)
(120 W I m K)(0.3 x I0-3 m)
= 40.825m- 1
It follows that,
1 lx I0-3 w = ( 40.825m- 1}(120W I m K)~(0.3x J0-3 m)2
4
X (80- 30)°C X tanh (40.825 m- 1) L(m)
= 0.0173 tanh(40.825 L)
or
tanh(40.825L) = O.Ol l = 0.6353
0.0173
40.825 L = tanh- 1 0.6353 = 0.75
or
Hence, the length of the leads,
L = (0.75 I 40.825) m = 0.0184 m
= 1.84 cm
(Ans)
Example 4.14 ~ A I-metre long, 6-cm diameter cylinder placed in an atmosphere of 30°C is provided with
IO longitudinal straight fins (thermal conductivity k = I SO W/m K). The fins are 0.80-mm
thick and extend 2.5 cm from the cylinder surface. The heat-transfer coefficient between the cylinder and the
atmospheric air is 25 W/m 2 K. The surface temperature of the cylinder is I 60°C. Calculate
(a) the rate of heat transfer from cylinder to air,
(b) the percentage increase in heat transfer by providing fins in comparison to cylinder without fins,
(c) the temperature at the end of fins, and
(d) the efficiency of fins.
Known: Longitudinal straight fins on a cylinder exposed to convective environment.
Find: (a) Q(W), (b) % increase in Q after providing fins, (k= 150 Wlm K), (c) TL( 0 C), and (d) 'lltin·
Schematic:
IO longitudinal straight fins
Air
rs;;::======:::Jl====:::,.,l_ t=0.8mm
!!!
-----,------------'-.._LL=2.Scm
Tb= 160°C
T==30°C
Length of cylinder,
I= I m = Width of fins, b
h=2SW/m2K
Cylinder
0=6cm
/=Im
Fig. 4.22
Heat Transfer fram Extended Surfaces
r
4.27
Assumptions: (1) Steady operating conditions exist. (2) Uniform area of cross-section of fins. (3) Constant
thermal conductivity and uniform heat-transfer coefficient. (4) No thermal contact resistance.
Analysis: (a) Without fins
Heat-dissipation rate from a bare cylinder with no fin is,
Qwo = Qno fin =h As (T,, -T=) = h(TCD l) (T,, -T=)
= (25W /m2 K)(rcx0.06mxlm)(160-30)°C
=612.6W
With fins
Heat transfer from the finned cylinder to atmospheric air is
Qw =Qunfin + Qfin
Qfin =N mk Ac eh tanhmLC
Ac
where the corrected length, Lc =L+p
Area of cross-section:
A =ht=lmx0.8xl0-3m = 0.8xlQ-3 m2
Perimeter:
P=2 (b+t)=2 (l+0.8x 10-3)=2.0016 m
~
m= ~kA =
Fin parameter,
25W/m2 Kx2.0016m
150W/mKx0.8x10-3 m2
= 20.42 m-1
Lc =0.025m+(0.8x10-3m2 /2.0016 m) = 0.0254m
mLc= (20.42 m- 1) (0.0254 m) = 0.5187
cosh mLc = 1.1376, tanh mLc = 0.4767
Excess temperature at the base of the fins (x = 0), i.e., at the cylinder surface is
Fin heat-transfer rate is
Qfin = (10) (20.42m- 1) (150W /mK) (0.8x10-3m2 ) (130°C) (0.4767)
= 1518.5W
Heat loss from the unfinned surface of the cylinder is
Qunfin =hAunfin eh =h[rcDl-N ht] eh
= (250W /m2 K) [(rcx0.06mxlm)-(10x0.8x10-3 m2 )] (130°C)
= 586.6W
Hence, total heat-dissipation rate is
Qw =1518.5+586.6=2105W
(Ans) (a)
4.28
, Heat and Mass Transfer
(b) Percentage increase in heat-transfer after fins are provided
= Qw :--Qwo X 100 = (2105-612.6) (100) = 243.6%
Qwo
(Ans) (b)
612.6
(c) Temperature at the end of fins is
1
cosh mL
130
1.1376
1ix=L) =T=+---(Tb-T=)=30+--
= 144.7°C
(d) Fin efficiency,
(Ans) (c)
'llfin
= 0.4 767 xlOO =91.9%
(Ans) (d)
0.5187
Example 4.1 5 ~ A long T-shaped beam is attached to a
vertical wall at a temperature T0 and acts
as a fin as shown in Figure 4.31.
Making suitable assumptions, show that the heat-transfer
rate per unit length from the surface of the fin to the
surroundings (T-) is given by
To .... ......... ......b....... . . . ..... ......
h
mkb(T0 - T-) { COS h mLI
sinhmL 1
f--- Li ---j
Fig. 4.23
where
m=i
Known: A tee-shaped beam of specified dimensions protrudes from a vertical isothermal wall, acting as
a fin and is exposed to convective environment.
Find: Heat-transfer rate per unit length, Qfin·
Schematic:
---.j
b
1- -
Q,,n i_l 0
L1
~ Q,
+
T1
b
l1
x=O
(a)
~-
T1
i_
CD
--1
b
I-(b)
Fig. 4.24
T1
i
L1
_J_
®
Heat Transfer fram Extended Surfaces
r
4.29
Assumptions: (1) Steady-state conditions prevail. (2) One-dimensional conduction. (3) Constant thermal
conductivity and uniform heat-transfer coefficient on all finned surfaces. (4) No internal heat generation.
(5) No contact resistance.
Analysis: The T -beam can be looked upon as an extended surface comprising three fins, one of length L 1
and the two others of length L2 each, as shown in the schematic.
Consider a length l of the beam.
We note that II and III are identical. It follows that
Qu =Qm = h(2L.l)(T.. -T_)TJr.II
where
and
.
tanhmL2
Tim = fin efficiency= - - mL2
m=~:~ =~~:~=JR
.
.
tanhmL 2
Qu =Qm=2hL2 1(7'..-T_)--mL2
(1)
Now,
The differential equation is
d2T - hP (T-T )=0
dx 2 kAC
=
where
(2)
The boundary conditions are
At x= 0, T= T0
and
at x=L 1, T= T 1
The general solution of Equation (2) is
T= T_+A cosh mx+B sinh mx
The boundary conditions yield:
and, T 1 = T_ + A cosh mL 1 + B sinh mL 1
IB = [(T.. -T_)-(Ta -T_)coshmL J!sinhmL I
1
1
The temperature distribution is thus given by
T = T_ +(Ta-T_)coshmx+
(1'.. -T_)-(Ta -T_)coshmL1
To determine T 1, we have the following condition.
QI =2Qu
dTI
tanhmL 2
-kbl=4hL2 1(7'..-T_)
dx x=L,
mL2
.
smhmL1
sinhmx
(3)
4.30
, Heat and Mass Transfer
-kbm [ (To-T_)sinhmL 1 +
or
l
(i;-r_)-(To-T_)coshmL 1
.
coshmL1
smhmL1
tanhmL2
= 4h£i (i; - r_)-mL2
.
cosh2 mL 1
-(To-T_)smhmL 1 -(J;-T_)cothmL 1 +(To-T_)-.- s1nhmL1
or
4h
4h kb
=--(i;-r_)tanhmL2 =-x-(i;-r_) tanhmL 2
kbm2
kb 2h
= 2 (Ti - r_)tanhmL 2
Rearranging,
(cosh mL 1 - sinh2 mL1 )
(i;-r_)(2tanhmL2 +cothmL1) = (I;,-T_)---.- - - smhmL1
or
(J;-T_)=
(To-T_)
sinh mL1 ( 2 tanh mL2 + coth mL1 )
(4)
(·: cosh2 mL 1 -sinh2 mL 1 =
1)
. =-kbdTI
Heat-transfer rate per unit length, Qfin
dx x=O
From Equation (3),
dT
[(i;-r_)-(To-T_)coshmL,J
-=O+(To-T_)sinhmx.m+coshmx.m.~---.-----~
dx
smhmL1
and
Substituting for (T1 - T_) from Equation (4), we obtain
-kb dTI
dx x=O
Q.fin
=-kb·
m
[
To-T_
sinh m L1 sinh mL1 (2 tanh mL2 + coth mL 1)
mkb(7'o-T_){
h L
cos m 1
sinh mL 1
(To-T_) coshmL,]
1
}
sinh mL 1 (2 tanh mL 2 + coth mL1 )
QED
Heat Transfer from Extended Surfaces
r
4.31
Example 4.16 ~ A 12.5-mm diameter and 15-cm long iron rod (k = 5 I W/m 0 C) projects from a heat-source
at I 60°C into an environment at I 5°C. If the convection coefficient is 45 W/m 2 °C, determine,
treating the tip of the rod non-adiabatic:
(a) the temperature distribution in the rod
(b) the temperature at the free end
(c) the heat-flowing out of the heat-source
(d) the heat-flowing out at the free end
(e) the effectiveness of the fin
(f) the efficiency of the fin.
Known: An iron rod protruding from a heat source dissipates heat to a convective environment.
Find: (a) T(x) (b) TL (c) Qb (d) QL (e) Er (f) llr
Schematic:
h=45W/m2°C
Air
T== IS°C
///
Iron rod (k=SI W/m C)
0
L=O.ISm
x=L
Heat source (Tb= I 60°C)
Fig. 4.25
Assumptions: ( 1) Steady-state conditions prevail. (2) Constant properties and uniform heat-transfer
coefficient. (3) Heat loss from the fin tip is not negligible. (4) One-dimensional conduction.
Analysis: (a) The temperature distribution in a fin of finite length with non-adiabatic fin end is given by
0
T(x)-T=
cosh m(L- x)+(h/mk)sinh m(L-x)
eb
Tb -T=
coshmL+(h/mk)sinhmL
~
hrcD
[}E
m= Vk Ac= krcD 2 !4 =
VW
where
=[
4x45W/m2oc
]=16.803m-t
51W/m°Cx0.0I25m
eb = Tb - T= = 160 - 15 = 145°c
mL = (16.803 m- 1)(0.15 m) = 2.5205
sinh mL = 6.1772
and
cosh mL = 6.2576
(him k) = (45 W/m 2 0 C) I (16.8 m- 1) (51 W/m 0 C) = 0.05251
0
0b
cosh 16.8(0.15- x)+0.0525sinh 16.8(0.15-x)
cosh2.5205+0.0525lsinh2.5205
4.32
,
Heat and Mass Transfer
or
T(x) = 15 oc + 145 oc x {cosh 16.8(0.15 - x)+ 0.0525 lsinh 16.8(0.15 - x)}
6.582
!T(x) = 15+22.0 [cosh 16.8 (0.15-x)+0.05251 sinh 16.8 (0.15-x)
]!
(Ans) (a)
The temperature profile is sketched graphically below.
160
""""""-,
140
120
100
u 80
~
~
1-
"-.. ~ ffx,
---
60
40
20
r-----
0
0
0.03
0.06
0.09
0.12
0.15
X (m)
Fig. 4.26
(b) Temperature at the free end,
145
T(x = L) =TL= 15+--= 37°C
6.582
(c) Heat-flowing out of heat-source,
Q
=Q
base
where
fin
(Ans) (b)
=MsinhmL+(hl mk)coshmL
cosh mL + ( hi mk) sinh mL
M= mkAc0b
= (l6.8m- 1)(51 W I m 0 C)( ¾x0.0125 2 ) m2 (145°C)
= 19.41 W
il
~base
=(l 9.4 l W) (6. l 772)+(0.05251)(6.2576)
6.582
= (19.41 W) (6.5057/6.582)
= 19.2 W
(Ans) (c)
(d) Heat flowing out at the free end of the rod,
QL =h (1tD 2 14) (TL -T=)
= (45W l m 2 0
=0.12 W
c{
¾x0.0125 2 ) m2 (37-15)°C
(Ans) (d)
Heat Transfer from Extended Surfaces
r
4.33
(e) Fin effectiveness, Er = ~fin
Qnofin
where
Qnofin = h Ac0b = (45W / m2 0
c{
¾x0.0125 2 ) m2 (145°C)
=0.8W
_ 19.2W _ 24
0.8W
Hence
(Ans) (e)
Er-
(f) Fin efficiency,
llr =-1-[sinhmL+;;icoshmL] = -1-
mL coshmL+~sinhmL
mk
65057
2 -5205 6582
= 0.392 or 39.2%
(Ans) (f)
Ac
Comment: Corrected length, Lc =L+-=L+(D/ 4)=0.15m+(0.0125 / 4)m=0.153125 m.
p
145
eb
= 15+
= 37°C. Thus, using the adiabatic
cosh mLc
cosh (16.803 x 0.153125)
fin tip assumption and incorporating corrected length is much simpler- and almost precise- than the
cumbersome analysis based on convective tip condition.
Tip temperature, TL= T= +
Example 4.17 ~ A circular disc-shaped transistor is mounted on an insulated board and dissipates 0.25 W in
steady state operation. A hollow copper tube (k = 400 W/m K) is proposed to be attached
to the transistor as shown to reduce the temperature of the transistor. Estimate the operating temperature of
the transistor with and without the tube if the external surface of the tube is exposed to atmospheric air at
25°C with a convective heat-transfer coefficient of 40 W/m 2 K.
T
1
0.25mm
15mm
Copper tube
Transistor
Fig. 4.27
Known: Dimensions and surface convection conditions of a hollow cylindrical fin mounted vertically on
a disc-shaped transistor.
Find: Transistor temperature with and without the fin .
4.34
, Heat and Mass Transfer
Schematic:
- - - - r O = 7.S mm
= T
t=0.2Smm
Air
Too=25°C
.....---• --r;=7.2Smm
L= 15mm
h=40W/m2K _
Cylindrical fin
1
(k=400W/mK)
Transistor (Q = 0.25 W)
Fig. 4.28
Assumptions: ( l) Steady operating conditions. (2) Constant thermal conductivity and uniform convection
coefficient. (3) Heat transfer from the exposed surface of the transistor and the internal surface of the
tube is neglected. (4) Thermal contact resistance between the transistor and the tube is negligible.
Analysis: Since the thickness of the hollow tube (cylindrical fin), t<< ri, the extended surface can be
treated as a rectangular fin.
Power dissipated by the transistor= Qbase =Qfin = 0.25W
The fin heat-transfer rate is
RP
0b(h/mk)+tanhmL
Q. =m k A
------tin
c l +(him k) tanh mL
,....h-(2_1t_,;,_) fE
F.m parameter, m = - - =
k(21t,;,t)
kAC
=v,:i
40W/m 2 K
,__________
= 20m- 1
400W /mKx0.25x 10-3 m
and mL=(20 m- 1) (0.015 m)=0.3
tanh mL = 0.2913
h
40W/m2 K
=------=0.005
mk 20m- 1 x400W /mK
Substituting the known values, we have
0.25W = (20m- 1)(400W /mK){21tx0.0075mx0.00025 m) x0 x{ 0.00 5 +0. 2913 }
b l+(0.005)(0.2913)
= 0.027886
Excess temperature at the base of the fin (transistor surface) is
0b = 0.25 / 0.027886 = 8.97 K = Tb - T=
The operating temperature of the transistor with fin is
Tb,with = T= + 0b = 25°C + 8.97 K (or 0 C)
=34°C
(Ans)
Heat Transfer from Extended Surfaces
r
4.35
Without fin, the heat-transfer rate or power dissipated,
Q =0.25 W =h(1tr;)(Tb -T-)
It follows that
T: -T
b
=
=__g__=
h(1tr;)
0.25W
(40W / m2 K){1tx0 .0075m) 2
= 35.37 K
The transistor temperature without fin is
Tb.without= 250c
+ 35_37oc
= 60.4°C
(Ans)
The fin is thus a very effective cooling agent that reduces the base temperature by about 26°C.
Example 4.18 ~ Derive an expression for the temperature variation along a pin fin with uniform internal
heat-generation q(W / m3 ) and insulated end. The fin base temperature is Tb and the ambient
temperature is T~· When can the heat-flow into the fin at the base be zero!
Known: Fin with internal heat-source and adiabatic tip.
Find: Temperature distribution along the fin.
Schematic:
///
r=T(x)
5~"'"
I
x=L
(d0/dx)=O
Fig. 4.29
Assumptions: ( l) Steady-state, one-dimensional conduction with uniform volumetric heat-generation.
(2) Constant thermal conductivity and uniform convection coefficient.
Analysis: Energy balance for differential control volume of thickness dx is
'2x - '2x+dx- '2conv + £gen = 0
or
or
or
or
4.36
, Heat and Mass Transfer
With
dT
d0
dx
dx
a2e
q
dx 2
k
and
- - m2 0+-=0
dx 2
where
dx 2 '
hP
m2=-
k4,
The solution to this differential equation is
0(x) = Acosh(mx) + B sinh(x) + (q /km2 )
Boundary conditions:
q
ab at x= 0
BC(l)
0=
BC(2)
d0 =0
dx
at
A=0 - b
ael
x=L
km2
= 0 = Asinh (mL) m+ B cosh (mL)m
dx x=L
~
B = -Atanh (mL) = -( eh - ~ 2}anh (mL)
The required temperature variation along the fin length is
0(x) = T(x)-T= = ( 0b - ~)cosh(mx)+~-(ob - ~ ) tanh(mL)sinhmx
0=
or
q +(eh - q 2 )[cosh(mx)-sinh(mx)tanh(mL)]
km 2
(Ans)
km
Heat-flow into the fin at the base, Qr = 0 if
= O= (eh -~)[msinh(mx)-mcosh(mx)tanh(mL)t
(de)
dx
0
x=O
or
km
0 = ( 0b - ~ 2 )m[O-(l)(tanhmL)]
(Ans)
i.e.,
(C) NUMBER OF FINS ON PLANE WALL
•
•
Example 4.19 ~ Cooling fins (k = 26 W/m 0 C) of rectangular section protrude perpendicular to an isothermal
flat surface at 300°C and project into a fluid at 30°C. There are 12.5 fins per I00 mm and
the fins have a thickness of 3 mm and a length of 30 mm. The convection heat-transfer coefficient for all surfaces
is estimated to be 45 W/m 2 °C. Use the corrected length to compensate for the convective loss of heat-from
the fin tip. Determine
Heat Transfer from Extended Surfaces
r
4.37
(a) the heat-transfer rate per unit area of the primary (base) surface
(b) the temperature at the tip of each fin, and
(c) the fin efficiency.
Known: Dimensions and base temperature of straight rectangular fins. Convective fluid conditions.
Find: (a) Heat-loss per unit base surface area. (b) Temperature of fin tip. (c) Fin efficiency.
Schematic:
Fluid
II I
T~=30°C
h=4SW/m2°C
Tb=300°C - ~ - - - - - ~ - ~ ! 3 m m
L=30mm
--E_T
Rectangular fins
(k = 26 W/m 0 C)
12.S fins per I 00 mm
Fig. 4.30a
Primary (base) surface
Anofin=WH
Aunfin = WH-NWt
Afin
T
l
p=Bmm
= 2 LW+ Wt
(one fin)
100mm per 12.Sfins
or
12Sfins for H= 1000mm
=Im
Fig. 4.30b
Assumptions: ( l) Steady-state conditions prevail. (2) Uniform heat-transfer coefficient. (3) Constant
properties. (4) Negligible radiation effects.
Analysis: Pitch of fins, p = 100 mm / 12.5 = 8 mm
Thickness of fin, t = 3 mm
Space between fins, S = 8 - 3 = 5 mm
Hence, un-finned area per fin, Aunfin =SW= (0.005 m) ( l m)
= 0.005 m2
The temperature distribution along the length of the fin is given by
T(x)-T~
coshm(Lc -x)
coshmLc
At the fin tip,
4.38
, Heat and Mass Transfer
{Ef.
{IT
where m = VkAc =fit=
2x45W/m2 °C
26W/m °Cx0.003m
= 33.968 m- 1
Lc=L+f=30mm+(3~m)=3I.5mm or 0.0315m
mLc = (33.968 m- 1 ) (0.0315 m) = 1.07
cosh mLc = 1.6292
tanh mLc = 0.7895
:. Temperature at each fin tip,
TL = 30°C + (300 - 30)°C I 1.6292
= 195.7°C
(Ans) (b)
Fin efficiency,
tanhmLc
0.7895
TJr =---=---=0.738
mLc
1.07
or 73.8%
(Ans) (c)
Thermal resistance with fins,
1
Rth=-----h(Aunlin +TJrAfin)
If W = H = Im, the base surface area (no fin) is
Anofin = WH = I m X 1 m = 1 m2
Aunfin = WH - NWt
= W (H -Nt) = lm (lm-125finsx0.003 m per fin)
= 0.625 m2
Afin = [2 LW +Wt]N
= [(2x0.03xl)+(lx0.003)] 125
= 7.875 m2
R
th -
I
45W/m2 °C(0.625+0.738x7.875) m2
= 0.00345°C
:. heat-transfer rate per unit prime surface area is
Q = Tb - T= = (300 - 30)°C [ 1k W ]
Rth
=78.2kW
0.00345°C/W 103 W
(Ans) (a)
Heat Transfer from Extended Surfaces
r
4.39
Example 4.20 ~ Stainless steel (k = I 5.2 W/m 0 C) pin fins, 6 mm in diameter and 3-cm long, are installed on
the base surface in contact with 24°C air. The fins are spaced on I cm x I cm centres. The
base surface temperature is 230°C and the mean convective heat-transfer coefficient is 55 W/m 20 C. Calculate
the total rate of heat-transfer per square metre of surface area.
Known: Dimensions and base temperature of stainless steel pin fins. Surrounding air conditions.
Find: Heat-loss rate per square metre.
Schematic:
Air
Ill
h=55W/m2 K
T==24°C
~ - - Stainless steel fins
(k = I 5.2 W/m C)
0
Fig. 4.lla
Assumptions: (1) Steady-state conditions. (2) Constant properties. (3) Uniform heat-transfer coefficient.
Analysis: Unfinned surface:
1t
= l cm 2 --x0.62 cm 2 = 0.717x 10-4 m2
4
l
Thermal resistance, R unfin -- hA- unfin
- - - - - - - - - - = 253.5°C/W
(55W/m2 0 C)(0.717x 10-4 m2 )
Corrected fin length,
L =L+ Ac =L+1tD2 14
C
p
1tD
D
0.6cm
= L+- = 3 cm+--= 3.15 cm
4
{Ef.
4
{lI
4x55W/m2 °C
m=VkAC =fkD= 15.2W/m°Cx0.006m
= 49.115 m- 1
mLc = (49.115 m- 1)(3.15 x 10-2 m) = 1.547
tanh mLc=0.9133
We can use the electrical analogy method to solve this problem.
4.40
, Heat and Mass Transfer
Thermal resistance,
Rrm=-----m k Ac tanh mLc
= [(49.115 m- 1}(15.2 W/m c) ( ¾x0.006 2 m2 ) (0.9133)]
0
= 51.87°C/W
Equivalent thermal resistance,
= RunfinRfin
R
eq
253.5x51.87
253.5+51.87
Runfin+Rfin
=43°C/W
~ Qunfin ~
Fig. 4.31b
Therefore,
Q = Tb -T= = (230-24)°C
Req
43°C/ W
= 4.78 W per I cm 2
=4.78W(l0
cm
lcm
lm
4
2
Example 4.21
2
2
) ( lkW
10 3 W
)= 47 _8 kW/ml
(Ans)
~ Copper-plate fins of rectangular cross section having I-mm thickness and I 0-mm height,
and thermal conductivity 380 W/m °C are attached to a plane wall maintained at a
temperature of 230°C. Fins dissipate heat by convection into ambient air at 30°C with a heat-transfer
coefficient of 40 WI m1 0 C. Fins are spaced at 8 mm on centres. Neglecting heat-loss from the fin tip,
determine
(a) the fin efficiency, (b) the area-weighted (overall) fin efficiency, (c) the net rate of heat-transfer per m1 of
plane wall surface, and (d) the heat-transfer rate from the plane wall if there were no fins attached.
Known: Dimensions of straight rectangular fins attached to a plane wall. Base and ambient temperatures.
Conductivity and convection coefficient.
Find: (a) 'lltin; (b) '11 0 ; (c) Qfin,total; (d) Qnofin ·
Heat Transfer from Extended Surfaces
Schematic:
4.41
Air
N= 125fins perm
!!!
-;f=smm
:_-.c:::....-="""
r
=.t-
T==30°C
h=40W/m2 K
Copper fins
k= 380 W/m °C
Aunfin = HW - NWt (for N fins)
= Wp - Wt (for one fin)
Afin = 2LW + Wt (for one fin)
= (2L+ t)W
= N(2L + t) W for one fin
Anofin = WH = I m x I m = I m2
Atotal = Afin + Aunfin
= [(2L + t) + (p- t)]W
= (2L + p)W (per fin)
Fig. 4.32
Assumptions: (1) Steady operating conditions. (2) One-dimensional formulation. (3) Constant and same
thermal conductivity for fins and wall. (4) Uniform and same heat-transfer coefficient with and without
fins. (5) Radiation effects negligible. (6) Insulated fin tip.
Analysis: (a) Fin parameter,
m=~:;c = h2i:+t)=!J
(since W >>> t)
2x40W/m2oc
=14.5Im-I
380W/m°Cx0.00lm
mL = (14.51 m- 1)(0.0 I m) = 0.1451
tanh mL = 0.144
and
Q
1hP k 4, 0b tanhmL
Fin efficiency, 'llfin =--,---!!.!!- = -V
Qdealfin
h(P L)
= tanhmL
mL
0
b
0.144 =0. 993 or 99 .3%
0.1451
(Ans) (a)
(b) Area-weighted or overall surface efficiency,
Qtotal,fin Qfin + Qunfin
'llo=-.- Qideal fin
Qideal fin
=--------
= I- Afin (I) = I- (2L+t)W (I)
A
'llfin
(2 L+ ) W
'llfin
total
P
{(2X0.01)+0.00 l}
= l------x(l-0.993)
{(2 xO.Ol)+0.008)
= 0.995 or 99.5%
(Ans) (b)
4.42
, Heat and Mass Transfer
(c) Heat-transfer rate per unit plane wall area,
Qtotal,fin = Qfin + Qunfin = llfin N h
Arin 0b + h A,mfin 0b
= Nh0b [llfinArin + 4infin] = ( 40W / m20 C)(230-30)°C X 125 X
[ {0.993 x (2 x 0.0 l + O.OOl)m}+ {lm x (0.008-0.00 lm)}
= (40 x 200) [3.4819 W (l kW/10 3 W)] = 27.85 kW
(Ans) (c)
(d) Heat-transfer rate per m 2 from the plane wall without fins,
Qno fin =hAnofin(Tb -T=)=h(W H)(Tb -T=)
= (40 W/m 2 C) (lm x lm) (230 - 30) °C
= 8000 W or 8 kW
0
(Ans) (d)
(D) BAR WITH TWO ENDS AT SPECIFIED TEMPERATURES
Example 4.22 ~ Two ends of a circular rod of length (2L), perfectly insulated on its lateral surface, are held
at the same temperature (T0) . The left half of the rod has uniform heat-generation at the
rate of (q) W/m 3 while the right half portion has no heat generation. Under steady conditions and assuming
constant thermal conductivity, develop expressions for the temperature distribution, and the location of maximum
temperature.
Known: An insulated rod with its two ends at the same temperature experiences internal heat-generation
in its left half portion.
Find: Expression for temperature variation. Location of maximum temperature in the rod.
Schematic:
®
®
I-
L
©
L
-J
1::~<:~A~=Gl=q;:::::::::::::::::1
Lx
x=O
x=L
T= To
T=h
x=2L
T= To
Fig. 4.33a
Parabolic profile
T(x)
T(x)
······:················································· Tmax
9 qL2 j
32-k-j
To ...............
x=O
..J .. . ...
qL2!4k
1......... ... .....
x=3!4L
Straight line
t. . . .. . ..................................... To
x= 2L
x= L
Fig. 4.33b
c!:
=cons~
Heat Transfer (ram Extended Surfaces
r
4.43
Assumptions: (1) Steady operating conditions. (2) Uniform volumetric internal heat-generation. (3) Constant
thermal conductivity.
Analysis: Left half portion AB: (with heat generation)
The governing differential equation is
d2T q
-+-=0
dx 2 k
d 2 T ___ :f...
dx 2
k
or
Integrating twice with respect to x, we have
dT = -qx +C
dx
k
I
and
The constants of integration C1 and C2 are to be evaluated from the boundary conditions (BC)
BCI Atx=O, T=T0
BCII At x=L, T= TL
It follows that
The temperature distribution in part AB of the circular rod is then given by
qx2 q Lx x
T(x)=--+--+-(T, -T, )+T,
2k
2k L L
o
o
or
.
dT
L.
T emperature gra d1ent,
-at x = 1s
dx
dTI
dx x=L(AB)
=C, _ qL
k
qL TL -To qL
=-+---2k
L
k
or
dTI
dx x=L(AB)
_TL -To - qL
L
2k
4.44
, Heat and Mass Transfer
Right half portion BC: (without heat generation)
The governing differential equation is
d2T
-=0
dx 2
Integrating twice yields
dT -C
dx - 3
and
To obtain C3 and C4 we invoke the following boundary conditions:
At x = L, T = TL and at x = 2L, T = T0
It follows that
TL =C3L+C4
1'o = C3 (2L)+C4
and
Subtracting one from the other,
TL -To =-C3L
=>
lc3 =-(TL ~l'o)I
or
The temperature distribution in part BC of the rod is expressed as
IT(x) = (2TL -To)-z(TL -To)I
Temperature gradient at x = L is
dT I
- C - -( TL -To )
dx x=L(BC) - 3 L
At midway (x = L), the control surface energy balance gives
~lx=L(AB)
or
= ~lx=L(BC)
-k dTI
--k dTI
dx x=L(AB)
dx x=L(BC)
Since k is constant throughout the rod,
dTI
dTI
dx x=L(AB) = dx x=L(BC)
Heat Transfer (ram Extended Surfaces
TL -To
qL
(TL -To)
L
2k
L
r
4.45
-----=----
or
qL2
2(TL -To)= 2k
or
Substituting for Tu the temperature variation in the left-and right-half portion of the rod can be
expressed as
(x = 0
to
x = L)
(Ans)
Also,
(Ans)
(x=L
to
x=2L)
The temperature profile is shown in the schematic.
Maximum temperature will occur only in the left half portion AB because temperature gradient is
constant (not a function of x) in the right half portion BC.
To determine the location for maximum temperature, dTI = 0
dx AB
[3 2 J
=O+ qL
4k L- L2 xlx =O
or
dT
dx
or
3 4x
--L L2
or
lx=¾LI
2
(Ans)
. maximum
.
. ascertame
. d 1"f the second denvattve
. . d 2tT.s negative.
.
That 11r,x) at x = -3 L is
1s
~
4
J
d 2T qL2 [
4
q
dx2 = 4k O- L2 = -k(-ve)
Hence,
4.46
, Heat and Mass Transfer
=To+qL2[~-~J=To+qL2 x~
4 8
4k
4k
8
9 q ll
Tmax
= 1'o +Ilk
(Ans)
Example 4.23 ~ A brass fin (k = 137 W/m 0 C) of I0-mm diameter and 250-mm long is fixed between two walls,
one at I 80°C and the other at 80°C. Heat is being dissipated by the fin to the surrounding
air at 30°C with an associated convection heat-transfer coefficient of 22 W/m 2 0 C. Determine
(a) the temperature of the fin midway between the two walls,
(b) the total heat-loss from the surface of the fin,
(c) the distance from the I 80°C wall where the fin surface temperature would be minimum, and
(d) the minimum fin temperature along its length.
Indicate the temperature variation along the length of this fin.
Known: A brass fin of circular cross-section spans two walls at different temperatures and dissipates
heat-by convection to the ambient air.
Find: (a) Temperature of fin at x = L/2
(b) Total heat-transfer rate from the fin
(c) Location at which the fin has the least temperature
(d) Minimum fin temperature
Schematic:
Air
Ill
T= = 30°C
t
D= 10mm
t
L=0.25m
L____.. X
-
Brassfin(k = 137W/m C)
0
T(O)= 180°C
0(0)=0b= IS0°C
T(L)=80°C
0(L) = 0L = S0°C
Fig. 4.34
Assumptions: ( l) Steady-state one-dimensional conduction. (2) Constant thermal conductivity. (3) Heattransfer coefficient is uniform on all fin surfaces.
Analysis: The temperature distribution along the length of a fin of uniform cross-section is given by
T(x)-T=
[(TL -T=)l(Tb -T=)] sinhmx+sinhm(L-x)
Tb - T=
sinh mL
(a) Temperature at the mid-section of the fin is
T(x=LI 2) = T= + (~b -T=)
smhmL
[(TL -T= )sinh m~+sinh m~J
Tb -T=
2
2
(A)
Heat Transfer fram Extended Surfaces
r
4.47
The fin parameter,
~ (E_
4x22W/m2 °C
m=VkAC =fin= 137W/m°Cx0.0lm
= 8.015 m- 1
mL = (8.015 m- 1) (0.25 m) = 2.0036
and
sinh mL = sinh 2.0036 = 3.6406
sinh mL = sinh 1.0018 = 1.178
2
emL = e2.0036 = 7.416
Substituting the values in Equation (A), one gets
T(x =I)= 3ooc+ (180-30)oc [ (80-30)oc (1.178)+ 1.178]
2
3.6406
(180-30)°C
= 94.7°C
(Ans) (a)
(b) Total heat-dissipation rate,
Q=mk Ac[0b +0d (-co_~_h_m_L_-_1)
smhmL
= (8.0m- 1 ) (137W /m°C) (~x0.012 m2 )(150+50)°Cx (co~h 2 .00 36 -l) = 13.15W (Ans) (b)
4
smh2.0036
(c) Minimum temperature along the fm length can occur only if 0(L) > 0(L) for adiabatic fm.
0(L) for adiabatic (insulated) fin is
150°c
0L = 0b!coshmL=----= 39.7 C
cosh2.0036
0
As 0L(= 50°C) > eL,ins fin• Tmin does exist and its location is given by
(Ans) (c)
(d) Minimum temperature is
(150°C) [ --xsinh(8.015x0.223)+sinh8.015(0.25-0.223)
50°C
3.6406 150°c
Tmin = 30°C+-'-----'-
= 78.9°C
J
(Ans) (d)
4.48
,
Heat and Mass Transfer
The temperature profile along the fin length is indicated below:
180 ~ - ~ - - - - ~ - - - - - - ~ - ~ - ~ - - ~ - ~
+ --
_
I
a_
T
l
78.9
30-+--~------i--~-~------i--~-~-~-~~----<
0
25
so 75 100 125 150 175 200 !225 250
x=L/2
L_.. 223
x(mm)
Fig. 4.35
Example 4.24 ~ Both ends of a 6-mm diameter U-shaped copper rod (k = 398 W/m K) are rigidly fixed to
a vertical wall, the temperature of the wall being 100°C. The developed (total U length)
length of the U-shaped rod is SO-cm and it is exposed to ambient air at 30°C. The combined convection and
radiation heat-transfer coefficient for the rod is 30 W/m 2 K. Determine (a) the temperature at the centre of
the rod, and (b) the rate of heat-transfer from the rod.
Known: A U-shaped rod of uniform circular cross-section with its ends held at the same temperature is
exposed to ambient air.
Find: (a) Temperature at the centre of the rod, Tc( 0 C). (b) Heat-transfer rate, Q(W).
Schematic:
Wall
"'-l
Tb= I00 °C
Air
i - - - - - -- - 4 - - -
-~=======~=====:::'
D=6mm
f f f
h=30W/m2K
T==30°C
h = I 00 °C · 1 - - - - - - - + - Copper rod (U-shaped)
[k = 398 W/m K]
Fig. 4.36
Assumptions: (I) Steady operating conditions. (2) Constant properties and uniform heat-transfer
coefficient. (3) One-dimensional formulation.
Heat Transfer fram Extended Surfaces
r
4.49
Analysis: The problem can be treated as an extended fin with prescribed tip (end) temperature. For such
a case, the temperature distribution is given by
0
T(x)-T- (0L /0b) sinhmx+sinhm(L- x)
=---=~~~-------sinhmL
The ends of the U-shaped fin are both at 100°C, i.e., Tb= 100°C and TL= 100°C. Ambient air
temperature T- = 30°C. Hence, eh= 0L = 100 - 30 = 70°C.
rcD2
For a circular cross-section, AC = -4- and P = rcD
Fin parameter,
m=(::,r =Hf;= 398~/:;:~'.o~6m
and
= 7.09 m- 1
mL = (7.09m- 1) (0.50 m) = 3.545
sinh mL = 17 .295 and cosh mL = 17.324
sinh mL(L - x) = sinh (mL/2) at the centre of the rod= sinh (3.545/2) = 2.8569
Hence,
0(x=Ll2)
(70/70)(2.8569)+ 2.8569
11.295
---=---------
ab
= 0.33
Temperature at the centre of the rod is determined to be
J;, = 1i.x=Ll2) =T- +e(x=L/2)
= T- +0b(0.33) = 30+(70)(0.33)
= 53.1°C
(Ans) (a)
Heat-transfer rate from the rod is
Q = Qb +QL
=
Mb (cosh mL - 0L /0b) + ML (cosh mL-(0b /0L ))
sinhmL
where Mb =mkAceb and ML =mkAceL
Since 0L = ab, 0b/0L =0L/0b = 1 and Mb= ML= M, we have
. 2M(coshmL-1) 2mkAc0b
Q = - - - - - = ---(coshmL-1)
sinh mL
sinh mL
2x7.09m- 1 x398W/mKx~(0.006m) 2 (70°C)(17.324-1)
= - - - - - - - - - ' - 4_ _ _ _ _ _ _ __
17.295
= 10.54W
(Ans) (b)
Comment: By symmetry, we could have analysed the rod as a fin of length L = 25 cm with base
temperature Tb= 100°C and the centre temperature, Tc as the insulated fin tip temperature. The heat-loss
rate from the entire rod would be twice the heat-dissipation rate from an adiabatic tip fin of length 25 cm.
4.50
,
Heat and Mass Transfer
Example 4.25 ~ A bar of square cross-section connects two metallic structures. One structure is maintained
at a temperature of 200°C and the other is maintained at 50°C. The bar is I00 mm long, has
a cross-section of 20 mm x 20 mm and is constructed of mild steel with a thermal conductivity of 60 W/m 0 C.
The surroundings are at a temperature of 20°C and the heat-transfer coefficient between the bar and the
surroundings is 10 W/m2 °C.
Derive an equation for the temperature distribution along the bar and hence calculate the total heat-flow
rate from the bar to the surroundings.
Known: Dimensions and end temperatures of bar (fin) of square cross-section.
Find: Total heat-transfer rate from the bar (fin) to the surroundings.
Schematic:
Surroundings
MS bar: k=60W/m°C
Qcond, i ~
Ill
Qcond, out
h= IOW/m 2 °C
T= =20°C
_____.. Qcond, out
• T(O) = 200°C
A
cL=IOOmm
D 20 mm
20mm
Section A-A
Fig. 4.37
Assumptions: ( 1) Steady-state, one-dimensional conduction along the fin without internal heat-generation.
(2) Constant thermal conductivity and uniform heat-transfer coefficient.
Analysis: Energy balance: Fin
Qcond, in
=Qcond,out + Qconv
where the heat-conduction rates may be calculated from the knowledge of the temperature distribution.
Total heat-transfer rate from the bar to the surroundings,
Qtotal = Qconv = Qcond,in -Qcond,out
The general solution for temperature distribution is
0(x) = C,emx +C2e-mx
where 0=T-T~
The boundary conditions are
T(x = 0) = 200°C, i.e., 0(0) = 200 - 20 = 180°C
T(x = L) = 50°C, i.e., 0(L) = 50 - 20 = 30°C
Hence,
0(0) = 180°C = c,em(O) + C2e-m(O)
or
i.e.,
0(0) = C, +C2
C2 = 0(0)-C, =180-CI
(a)
Heat Transfer fram Extended Surfaces
r
4.5 I
Also,
30 = C1 (emL -e-mL )+I80e-mL
or
30-I80e-mL
C,=--,-----,(emL -e-mL)
(b)
The fin parameter,
P=2 (20+20) mm=80 mm=0.08 m
where
1 m2
A = (20x20)mm2 ( -) = 0.4xI0-3 m2
1Q6 m2
C
Then,
m= 10W/m2 oc x
60W/m °C
0.08m
= 5.7735 m-1
0.4xIQ-3 m2
mL = (5.7735m- 1)(0.lm)
and
= 0.57735
e-mL = 0.5614
emL = 1.7813
Substituting these values in (b ), we get
C2 = 238.24
(from (a))
!0(x)=-58.24emx +238.24e-mx I
(Ans)
C1 = -58.24
and
and the temperature distribution has the form
The conduction heat-transfer rate can be evaluated by Fourier's law:
.
d0
Qcond =-kAc-
dx
d
= - k Ac -[-58.24 emx + 238.24 e-mx]
dx
=-k A [-58.24memx +(238.24) (-m)e-mx]
= mk Ac [58.24emx +238.24e-mx]
4.52
, Heat and Mass Transfer
mk Ac =(5.7735m- 1 ) (60W/m 0 C) (0.4xl0-3 m2 )
Now,
=0.1386
Qcond = 8.07 emx + 33.0 le-mx
Hence, at x = 0,
Qcond in= (8.07)e0 +(33.0l)e-O = 41.08 W
and,
at x=L=O.l m,
Qcond out= (8.07)(1.7813)+(33.01)(0.5614)
= 14.375 + 18.532 = 32.907 W
From the conservation relation,
Qtotal = Qconv = Qcond,in -Qcond,out
= ( 41.080- 31.907) W
=8.17 W
(Ans)
Example 4.26 ~ Two parallel, opposite, isothermal plates maintained at I 50°C and I00°C are 250-mm wide
and 150-mm deep. Fifty rectangular aluminium fins (k = 240 W/m K) and 15-mm long, each
I-mm thick, are joined to opposite plates. The convection conditions correspond to a heat-transfer coefficient
of 100 W/m 2 °C and the ambient air temperature of 25°C. Determine the heat-transfer rates at the two bases.
Derive the expression you use.
Known: A stack of 50 fins between two parallel opposite plates held at different temperatures and
subjected to convection conditions.
Find: Heat-dissipation rates at the two extreme ends.
Schematic:
W=250mm
II
B= 150mm
Ambient
air
Lr--=-=-r==--=~
T
!N=SOfins!
L= 15mm
l_ .___ _ _ _ _. , __ _ _ _ __ y
h= IOOW/m2°C
h=I00 °C
T==25°C
Fig. 4.38
Assumptions: ( 1) Steady-state, one-dimensional conduction. (2) Negligible contact resistance. (3) Constant
properties. (4) Uniform convection coefficient. (5) No internal heat-generation.
Analysis: The general solution for the temperature distribution in fin is
0(x) = T(x)-T= = C1emx +C2 e-mx
and
d0
dT
dx
dx
Heat Transfer fram Extended Surfaces
r
4.53
Boundary conditions
At
x = 0, T = T0
and
00 = 1'o -T=
~
00 = C1 + C2
(1)
At
X
= L, T= TL
and
0L = TL -T=
~
0L = c,emL + C2e-mL
(2)
From equations (1) and (2),
0L = C1emL + (0 0 - C1 )e-mL = C1 ( emL - e-mL) + 0 0 e-mL
C - 0L - 0oe-mL
I - emL -e-mL
or
and
Therefore
0 emx - 0 e-m(L-x) + 0 em(L-x) _ 0 e-mx
0(x) = L
O
O
L
emL -e-mL
=
80 [em(L-x) - e-m(L-x)] + 0L [emx - e-mx]
We note that
eX-e-X
---=sinx
2
Hence,
0 sinhm(L-x)+0L sinhmx
0 ( x0)
=--------,
sinhmL
The fin heat-rate is
.
Q
tin
dT
d0
=-kA - = - k B t cdx
dx
= -k Bt{-m0o coshm(L-x)+ m0L coshmx}
sinh mL
sinh mL
At the base, x = 0:
Q.fi
- kB t { m0o
tanh mL
m0L }
sinh mL
(3)
Q.tinL= k Bl { m 00
'
sinhmL
m0L }
tanhmL
(4)
m,
O-
At the base, x = L
For each fin,
Ac= Bt= 0.15 m X 0.001m = 0.15 X 10-3 m 2
4.54
, Heat and Mass Transfer
P=2 [B+ t]
= 2 [0.150 + 0.001] m = 0.302 m
100Wlm °Cx0.302 m
m=(!!!._]112 = 240Wlm°Cx0.15xl0m
2
Fin parameter,
3
kAC
2
=28.96 m- 1
mL = 28.96 m-1 x 0.015 m = 0.434
sinh mL = 0.448,
tanh mL = 0.409
00 = 150 - 25 = 125°c
eL = 100 - 25 = 15°c
Hence, from equations (3) and (4):
0.409
28.96m-l x75°C]
0.448
[28.96 m- 1 X 125°C
28.96m- 1 x75°C]
Qfm O = (240WIm 0C)(0.15m)(O.OO 1m) X [ 28 ·96 m-l X 1250c
'
= 144.1 W
(from the top plate)
·
Qrm,L = (240WIm 0C)(0.15m)(O.OO 1m) X
0.4 48
0.409
(into the bottom plate)
=99.7 W
Total heat-transfer rates are
Qo = N Qtin,o +{(W -Nt)B}h00
= (50 x 144.1 W) + (0.25 - 50 x 0.001) m x 0.15 m x 100 W/m 2 °C x 125°C
(Ans)
= 7205 + 375 = 7580 W
QL =-NQtin,L + {(W-Nt)B}h0L
= (-50 x 99.7W)+(0.25-50 x0.001) mx 0.15 m xlOO Wlm 2 0c x 75°C
= - 4985 + 225 = - 4760 W
Negative sign indicates that no heat-is dissipated at the bottom plate.
(Ans)
Qtotal = Qo + QL = 7580-4760
=2820W
Example 4.27 ~ Consider a I-mm thick steel panel (k = 60 W/m 0 C) on which copper tubes are soldered
parallel to each other 35-cm apart. Saturated steam at I atm (I00°C) passes through these
tubes and condenses into water at I00°C. Effectively the tube surface temperature is maintained at I00°C. The
atmospheric air is at 25°C with a surface heat-transfer coefficient of 15 W/m 2 °C. Modelling the panel as a flat
fin, determine (a) the heat-loss rate from the panel with equal end temperatures per metre run of the tube,
(b) the fin efficiency (c) the minimum panel temperature, and (d) the mass flow rate of the steam condensed
per day per tube. The latent heat of condensation of steam at I atm is 2256.5 kJ/kg.
Known: Copper tubes parallel to each other are attached to a steel panel and exposed to atmospheric air.
Find: (a) Total heat-loss rate per metre run of the tube (b) Fin efficiency (c) Minimum panel temperature
(d) Steam condensation rate per tube per day.
Heat Transfer from Extended Surfaces
Schematic:
!steam 100°c
! Steam I00°C
r
4.55
! Steam I00°C
Copper tubes
. -2.
.
Qr,n - Qb = Qconv
1::"7 L= 35cm -----I
<!tube = Qr,n = 2Qb
Air
~
r- --+-
h= 15W/m2°C
Adiabatic tip
U2
Too=25°C
L/ 2 ---+j Adiabatic tip
Tb(I00 °C)
Fig. 4.39
Assumptions: (1) Steady operating conditions. (2) One-dimensional conduction. (3) Negligible thermal
contact resistance (4) Uniform heat-transfer coefficient. (5) Constant thermal conductivity.
Analysis: (a) The steel panel acts as an extended surface of a plate fin with one copper tube as its base
at Tb= 100°C and the other copper tube at TL= 100°C. Heat-is conducted from the tubes in a direction
perpendicular to the tubes and through the panel and the heated panel in turn dissipates this heat-to the
atmospheric air by convection.
Then
2 M (cosh mL - l)
sinh mL
The temperature distribution across the panel spanning the length L, i.e., spacing between the two
tubes is indicated in the schematic. By symmetry, the location of minimum temperature will be midway
between the tubes, where the temperature gradient will be zero. The rate of heat-loss from both faces of
the tube is 2Qb which equals the heat-loss from the fin (panel) surface between the two tubes, i.e., Qfin·
4.56
, Heat and Mass Transfer
Method I To calculate ~ we can consider the tip of the fin to be the midway point x = ~ = 17.5 cm
with the fin length L/2 from the base (x = 0), i.e., one side of the tube at Tb= 100°C. Similarly, heat-loss
rate Qb from the other side of the tube also at Tb= 100°C can be determined. It follows that for a finite
fin of length L/2 with insulated end the heat-loss rate is,
Qi, = M tanh(mL/2)
Also,
Qrm (panel)= 2 Qb = Qconv
and
Qtube =2Qb =2Mtanh (mL/2)
In the present case,
ab = aL = T,, -T= = (100 - 25)°C = 75°C
L=0.35 m, k=60 W/m°C, w= 1 m
(per metre tube run)
t= 1 x 10-3 m, h = 15 W/m2 °C.
Fin parameter,
m=
J:;c
where perimeter, P "" 2 w = 2 m, cross-sectional area, Ac = wt = 1mx10-3 m = 10-3 m2
Hence,
m= [
Then,
15W/m2 o Cx2m ] 112 = 2236 m_ 1
60W Im °Cxl0-3 m2
mL/2 = (22.36m- 1)(0.175m) = 3.913
Heat-loss rate from both sides of the tube,
Qtube = Qrm = 2 Qi, = 2 M tanh (mL/2)
where
M=~hPkAC ab
= ~15W/m2 °Cx2mx60W/m°Cx10-3 m2 x75°C
= 100.623 W
Thus,
Qtin = 2 X 100.623 W X 0.9992
=201.1 W
(Ans) (a)
Method II We can consider the panel to be a fin of length L with equal end (tube) temperatures
Tb= TL= 100°C.
Heat-loss rate from the fin with equal end temperatures is given by
Q.
_ 2 M [cosh mL -1]
fin -
sinh mL
Heat Transfer fram Extended Surfaces
r
4.57
With mL = 22.36 x 0.35 = 7.826, the heat loss from the fin with both ends at the same temperature is
Q.
_2x100.623[cosh7.826-1]
sinh7.826
-Q·
panel -
fin -
=201.1 W
Clearly,
(Ans) (a)
Qtuhe = Qpanel = 201.1 W
(b) Maximum heat-dissipation rate from the ideal fin (of infinite thermal conductivity) and of length
L = 0.35 m is given by
Qmax = h A8 (Ii, -T.,.,)
= h(PL)eh
= (15W/m2 °C)(2 mx0.35m)(75°C)
=787.5W
Fin efficiency,
'llrin
Else
= 9rin = 20l.1 W = 0.2553
Qmax
787.5 W
or 25.53%
tanh(mL/2)
'llrin =
mL/2
= tanh 3·913 = 0.2553
3.913
or
25.53%
(Ans) (b)
(c) Minimum panel temperature will occur at x = L/2 where dT = de= 0.
dx dx
Therefore,
e
e
Else
e . = 2 eh sinh (mL/2) = 2 X 75°C X sinh (3.913) = 30C
(L/ 2 ) =
sinh mL
mm
and
eh
750c
min = cosh(mL/2) = cosh3.913
= 30c
sinh 7 .826
Tmin = emin + T.,., = (3 + 25)°C = 28°C
(Ans) (c)
(d) Mass-flow rate of the condensate per day per tube is
m= Qtuhe =
iirg
201.1 or J/s 124x3600 sl
2256.5 X 103 J /kg
1 day
=7.7 kg/day
(Ans) (d)
4.58
, Heat and Mass Transfer
(E) CIRCUMFERENTIAL (ANNULAR) FINS
•
•
Example 4.28 ~ The cylinder of a model aircraft engine is made of an aluminium alloy of thermal conductivity,
k = 186 W/m 0 C. Under typical operating conditions, the outer surface of the 20-mm long
cylinder is maintained at 250°C and is exposed to an ambient at 32°C. The surface convection coefficient in
the slip-stream of the propeller, h = 75 W/m 2 0 C. Eight annular fins of rectangular profile are provided. The fins
are of uniform thickness, t = I mm. The base radius of the fins is 11 mm and the end (tip) radius is 22 mm.
Estimate the increase in heat-dissipation rate by the addition of such cooling fins .
Known: Dimensions and base temperature of annular aluminium alloy fins of rectangular profile. Ambient
conditions.
Find: Increase in heat-dissipation rate due to fins .
Schematic:
l-r2=22mm~
rI=
L=
11mm
11mm
r- ---+-
Ambient
.---+-~-~--~~
!!!
J_
Imm~---~
T~~
h=75W/m2°C
T==32°C
/=20mm
k= l86W/m°C
Fig. 4.40
Assumptions: (I) Steady operating conditions exist. (2) One-dimensional radial conduction in fins.
(3) Uniform convection coefficient and constant properties. (4) Negligible radiation.
Analysis: In the absence of fins, the heat-transfer rate from the cylinder is determined from Newton's
law of cooling to be
Qwithout fin = h Awithout fin (
7i, - T= ) = h (2 1t 'i L )( Ti, - T= )
= (75W/m2 °C)(21tX 0.011 m x0.02 m) (250- 32)°C
=22.6W
The fin parameters needed are
I
Le =L+t/2 = 11 mm+-x I mm= 0.0115m
2
I
'ic = 1i +t /2 = 22 mm +-mm= 0.0225 m
2
Heat Transfer fram Extended Surfaces
'ic
- = 0.0225m/0.0115m = 1.96
1j
AP =Let= (0.0115 m)(lx 10-3 m) = 11.5 x 1Q-6m2
[l
l/2
!} 12
c
_h_
k AP
1/2
= 0.0115312 (
75
)
186xl 1.5x10-6
= 0.231
From Figure 4.40, the fin efficiency is
TJ f ""0.96
Alternatively
m=
[E = 2x75 = 28.398
fki 186x0.001
mr1 = 28.398 x 0.0115 = 0.3266
I 0(mr 1) = 1.0273
K 0(mr 1) = 1.3039
I 1(mr 1) = 0.1656
K 1(mr 1) = 2.8243
mr2c = 28.398 X 0.0225 = 0.639
I 1(mr2c) = 0.3364
21j Im
K (m1j) 1 (mric)- 1 (m1j) K (mric)
1
1
1
T J r = - - x1 ~ - ~
~~~
~~~
~'rc -f /0 (m1j)K1(mric)+ K0 (m1j)I1 (mric)
2x0.0115/28.398 x (0.2843)(0.3364)-(0.1656)(1.2043) = 0 _97
[0.02252 -0.01152] (l.0273)(1.2043)+(1.3039)(0.3364)
Heat-transfer rate from the finned surface is
Qfin =TJ, Qfin,max =TJ, hAf(ann) (Ti, -T=)
= TJ,h[2n(,rc-f] (Ti,-T=)
= 0.97(75W /m20C)[ 21t(0.02252 -0.01152 )m2 ](250-32)°C
= 37.27 W (per fin)
Qfms = N Qtin = 8x37.27 = 298.14 W
Heat-loss rate from unfmned surface is
Qunfin = hAunfin {Ti, - T=)
= h[21t1j (1-Nt)](Ti, -T=)
= (75W /m2 0 c){21t x0.0l lm(0.02-8x0.00I)}m2 x(250-32)°C
=13.56W
Qtotal fin= 298.14 W + 13.56W = 311.7W
r
4.59
4.60
, Heat and Mass Transfer
Increase in heat-dissipation rate as a result of the addition of fins is
Qincrease
=Q,otal, fin - Qwithout fin
= (311.7 - 22.6) W = 289.l W
(Ans)
Comment: Overall effectiveness of fins,
Er= Q,otal,fin / Qwithoutfin = 311.7W /22.6W
= 13.8
The contribution of fins in this case is to augment the heat-dissipation rate by a factor of nearly 14.
This accounts for the widespread use of fins.
Example 4.29 ~ Steam in a heating system flows through 2.5-cm OD tubes whose walls are maintained at
I80°C. Circular aluminium-alloy fins (k = 186 W/m K) of 6-cm OD and a constant thickness
of I mm are attached to the tubes. The space between the fins is 3 mm. The surrounding air is at 25°C and
the surface heat-transfer coefficient is 40 W/m 2 K. Determine the improvement in heat-transfer rate from the
tube per unit length as a result of providing fins, and the overall effectiveness of the finned tube.
Known: Dimensions and base temperatures of circular fins on the tube of a heating system. Surrounding
air conditions.
Find: Increase in the rate of heat-transfer to the surrounding air.
Schematic:
~1=Sc~
Tube wall (Tb= 180 °C, k = 186 W/m K)
__l
~--~f=lmm
t
~--~symm
fs
rrr
_l
h=40W/m2K
T==25°C
Air
Circular fins of constant
thickness attached to a tube
D2 =6cm
----j
f--- r2=3cm - J
Fig. 4.41a
Fig. 4.41b
Assumptions: (I) Steady-state conditions. (2) Radiation effects are negligible. (3) Thermal conductivity
does not vary with temperature. (4) Negligible contact resistance. (5) Heat-transfer coefficient is uniform
over the finned and unfinned surfaces.
Analysis: Heat-loss from the tube (without adding fins)
Heat-transferred to surrounding air by convection from the tube wall per metre tube length is
Qwo
= h(1tD1L){1i,-T=)
= (40 W I m K){1tx0.025mx lm)(l80-25)°C
=487W
Heat Transfer fram Extended Surfaces
Heat-loss from the tube (with addition of fins)
Qtotal,fin =N { Qfin +Qbare(unfin)}
Heat-transfer from the bare (unfinned) surface of the tube is
=h Aunfin {½ - T= )
Qbare
where
Aunfin =1tD1S=1t(0.025m)(0.003m)
= 2.356 X 1Q-4m2
Qbare = (40W /mK){2.356xl0-4 m2 )(180-25)°C
= 1.46 W
Area of the entire finned surface is
where
t
'ic = 1i +- = (30 + 1/2)mm = 0.0305 m
2
Arm= 21t(0.03052 -0.01252 )m2
= 4.863 X 10-3 m2
Maximum fin heat-transfer rate,
Qfm,max
= hAtin (Ti, -T=)
= (40W /mK)(4.863xl0-3 m2 )(180-25)°C
=30.15W
m=
ff, [
=
112
2x40W/m2 K
= 20 _739 m_ 1
]
186W/mKx0.001m
mr1 = (20.739 m-1) (0.0125 m) = 0.2592
mr2c = (20.739 m- 1) (0.0305 m) = 0.6325
2x0.0125m
(20. 739 m- )(0.0305 -0.0125 ) m
=---- - -2 - - - 2- -2
1
= 1.55744
r
4.61
4.62
, Heat and Mass Transfer
= (1. 55744 ) X (3.7571)(0.3326)-(0.1308)(1.2206)
(1.0175) (1.2206)+(1.5275) (0.3326)
=0.97
Hence,
Qfin = Tlr Qfin max= (0.97)(30.15W)
=29.25W
Per metre length, the number of fins,
N=
=
lm
(S+t)mm
1000mm = 250
3mm+lmm
Hence, total heat-transfer rate from the tube with 250 fins and 250 interfin spacings per metre length is
Qw = Qtotal,fin = N (Qunfin + Qfin) = 250 {1.46W + 29.25W}
=7677W
Therefore, the improvement in heat-dissipation rate as a result of adding fins is
Qw -Qwo = 7677 W-487 W
=7190W
(Ans)
The overall effectiveness of the finned tube is
Heat transfer with fins, Qw
Efin' overall = H eat trans1er
c
. h
wit
out fims, Q.wo
= 7677W = lS.S
487W
(Ans)
Example 4.30 ~ An annular (circumferentiaQ aluminium alloy fin (k = 160 W/m K) is mounted on a 25-mm
OD heated tube. The fin has an end radius of 37.5 mm and has a constant thickness of
0.4 mm. The tube wall temperature is 220 °C and the ambient temperature is 15 °C. The average convection
heat-transfer coefficient is 30 W/m 2 K. Calculate (a) the fin efficiency, (b) the rate of heat-dissipation from the
fin, and (c) the temperature at mid-height of the fin.
Known: Dimensions and base temperature of an annular fin. Ambient conditions.
Find: (a) Fin efficiency, (b) Heat-loss rate.
Heat Transfer from Extended Surfaces
Ill
Schematic:
\--
r I -----,
r
4.63
h=30W/m2K
T= = 15 °C
Al-alloy (k = I 60W/m K)
'-----+---~__l_
t=0.4mm
--L---------j-~
T ------~
L=r2-r 1
L, '
2 ---------
= (37.5- 12.5) mm
=25mm
Fig. 4.42
Assumptions: ( l) Steady-state, one-dimensional radial conduction. (2) Constant properties and uniform
convection coefficient. (3) Negligible contact resistance.
Analysis: Length of the fin,
L = End (tip) radius, r 2 - Base radius, r 1
= 37.5-12.5 = 25 mm
Le= L+t/2 = 25+0.2 = 25 .2 mm
fie= 1i +t/2 = 37.5+0.2 = 37.7 mm
m=
~ hP = {2h =
kAc
Vkt
2x30W / m 2 K
= 30 _62 m-t
160W / mKx0.4xl0- 3 m
m'i = (30.62m- 1}(O.Ol25m) = 0.383
mric = (30.62 m- 1)(0.0377m) = 1.154
The fin efficiency is given by
Tlr =
( 2 'i I m) K 1 ( m'i ) / 1 ( m fie )- 11 ( m 'i ) K1 ( m ric)
2
( tfc - 'i ) I 0 ( m 'i ) K1 ( m fie ) + K 0 ( m 'i ) 11 ( m fie )
(2x0.0125m / 30.62m- 1 )
(0.03772 -0.0125 2)m 2
K1 (0.383)/1 (l.l54)-/1(0.383)K1(l.l54)
=--------x------------I 0 (0.383)K1(l.l54)+K0 (0.383)/1 (l.154)
= (0.6454) [(2.3326) (0.6792)-(0.1951) (0.4692)]
(l .0374) (0.4692)+(1. l 584) (0.6792)
= (0.6454) (l .1721) = 0.7565
=75.7 %
Maximum heat-transfer rate,
Qmax
= h[ 21t('22c -'i2) ](I;, -T=)
= (30 W / m 2 K) [ 21t(0.03772 -0.0125 2)m2] (220-15)°C
=48.9W
(Ans) (a)
4.64
,
Heat and Mass Transfer
Rate of heat-dissipation from the fin,
Qfin = llr Qmax = (0.757) (48.9 W) = 37 W
(Ans) (b)
The temperature distribution is of the form
0
0b
I 0 (mr)K1 (m0.c)+K0 (mr)I1 (mric)
/ 0 (m 'i_) K1 (m 'ic)+ K 0 (m'i) / 1 (m 'ic)
Temperature at mid-height of fin (r = r 1+LI 2) = (12 .5 + 12.5) mm= 25 mm, is to be determined.
mr = (30.62 m- 1) (0.025 m) = 0.7655
I 0 (mr) = 1.1526
k0 (mr) = 0.5981
e(r = 25 mm)= (220 _ 15)0 C X (l.1526)(0.4692)+ (0.5981)(0.6792)
(l .0374) (0.4692)+(1.1584) (0.6792)
= 205x0.744
= 152.5°C
Hence, the temperature at the mid-height of the fin is
T (L/2) = 152.5°C + l5°C = 167.5°C
Example 4.31
(Ans) (c)
~ Circular aluminium fins (k = 237W/m K) of outer diameter D2 = 6 cm and constant thickness
t = 2 mm are attached to an aluminium tube of outer diameter, D1 = 3 cm and wall
temperature Tw = I 20°C. The thermal contact resistance between a fin and the tube is estimated to be
0.0002 m2 K/W. Heat-is transferred to the surrounding air at T= = 30°C with a combined heat transfer coefficient
of h = 60 W/m 2 K. Calculate the rate of heat-dissipation from a single fin. What would be the heat transfer in
the absence of the contact resistance.
Known: Circular aluminium fin is installed on a tube. Dimensions, contact resistance and convective
conditions are prescribed.
Find: Heat-dissipation rate with and without contact resistance.
t----------~f--- r2 =3 cm
Schematic:
t-------<------1-- r 1 = 1.5 cm
Tb
t=2mm
,.._ L = 1.5 cm -I -t-
Rc,c=0.0002
m 2 K/W
~
h=60W/m2 K
k=237 W/m)
t t r==Jo c
0
Air
Fig. 4.43
Assumptions: ( l) Steady operating conditions exist. (2) Uniform heat-transfer coefficient. (3) Constant
properties. (4) One-dimensional conduction.
Heat Transfer (ram Extended Surfaces
r
4.65
Analysis: Fin length (height),
1
L = -(D2 - Di)= 1i -lj = 3-1.5 = 1.5 cm or 15 mm
2
t
Le =L+-=15+1=16mm=0.016m
2
Ap =Lt=
16x2xI0-<im2
C
= 3.2 x 10-5 m-2
t
'ic = 1i +-= 30mm+lmm = 31mm
2
1j = 15mm = 0.015m
'ic = 31mm =2.067
'i
15mm
Fin parameter
~ = L c _h_
k AP
(
1/2
]
1/2
60
= (0.016) 312 [
]
237x3.2xI0-5
= 0.18
From the graph for circular fins, with ~ = 0.18 and r2c = 2.067, the fin efficiency is
lj
TJr= 0.96
.
Tw-T
Tw-T=
The rate of heat-transfer, Q =
= = -"--Rt,c + Rrm
Rtotal
Tb -T=
'li,-T=
Tb -T=
Rrm = ~ = TJdlmax = Tlr h Atin (Tb - T= )
=
1
1
=--------------TJr h21t[rfc -1j2 ] 0.96x60W /m2 Kx21t[0.03l2-0.015 2 ]m2
=3.754 K/W
=
R
t,c
0.0002m2 K/W
= 1.0 6 l K/W
21tx0.015mx0.002m
Total resistance= 3.754 + 1.061
=4.815 K/W
Rate of heat-dissipation with contact resistance is
Q = (120-30)°CorK =lS.,W
w
4.815K/W
(Ans)
4.66
, Heat and Mass Transfer
Rate of heat-loss without contact resistance is
Q
WO
= Tw -T= = Tb -T= = (120-30)°CorK
Rrm
Rrm
3.754K/W
=24.0 W
(Ans)
Clearly, thermal contact resistance is detrimental to fin performance and must be minimized to
maximise the heat-transfer rate.
(F) TRIANGULAR FINS
•
•
Example 4.32 ~ To increase the heat dissipation from an air-cooled, flat cylinder wall, it is contemplated to
install fins. Two options are under consideration: Triangular fins and Rectangular fins . For
both types of fins, the pertinent operating conditions and specifications are given below:
Material of the fins ... Aluminium
(k = 268 W/m °C, p = 2707 kg/m 3)
Base temperature . . ...
Tb= 650°C
T~ = 50°C
Environment temperature .. . ... .
Convection coefficient . . .
h = 120 W/m 2 °C
t = 25 mm
Thickness (at the base of triangular fin)
Length of the fin . . .. . . .. . . .
L = I 00 mmz
Compare the heat-loss per unit mass in the two cases.
Known: Dimensions and specifications of rectangular and triangular fins exposed to convective
environment.
Find: Heat-loss per unit mass (W/kg) in the case of (a) Triangular fin , and (b) Rectangular fin .
Schematic:
h = I 20W/m °C
lll
T= =650°C
T
h= 120W/m2°C
T= = SOoC
t=25mm
l_
T
t=25mm
Tb=650°C ·
~-------....L
---- L = I 00 mm ------j
Triangular aluminium fin
Rectangular aluminium fin
Fig. 4.44
Assumptions: ( 1) Steady-state conditions. (2) Constant properties. (3) One-dimensional representation.
(4) Uniform heat-transfer coefficient. (5) Radiation effects negligible. (6) Negligible thermal contact
resistance between wall and fins. (7) No internal heat-generation.
Analysis: Triangular fin
Excess temperature at the base,
. parameter, p = ~2hf
L
Fm
--kt
whe,e
J=~l+C'J' =
1+ (
25mm
2 x 100mm
)2 = 1.0078
Heat Transfer (ram Extended Surfaces
p = [2 X 120W /m2 °C X 1.0078 X 0. lm]112
269W /m°C X 0.025m
= 1.90 m- 112
and
pfl 12 = (1.90 m- 112 ) x (Jfu) m 112 = 0.60
2 pfl 12 = 1.20
Fin efficiency,
_ 1 1, (2 p n12 )
Tlr - p fl/2 Io (2 p fl/2)
=
0 7147
1 / 1(1.20)
~-=
.
=0.8547
0.6 / 0 (1.20) 0.6 X 1.3937
Heat-flow rate from the fin per unit width is
.
k tp0b 1, (2 pfl 12 )
Qr = n12 Io (2 p n12)
(268W /m 0 C) (0.025 m)(l.90 m- 112 )(600°C) 0.7147
=-'------'--'----------'----'-X-0.1112 m 112
1.3937
= 12 386W
Else,
lkW
= 2 X 1.0078 X 120W /m2 °Cx0.lm X 0.8547 X 600°C I-I
1Q3 W
=12.4kW
Mass of fin per unit width,
m=
p( ~t)cw) = (2707kg/m3>(0.lm \0.025m )(1 m)
= 3.38375kg
Heat-dissipation rate per unit mass,
flr = 12386W = 3660 W /kg
m
3.38375kg
Rectangular fin
Fin parameter, m = (.!!.!_]112
kAC
h2(W + t) =
kWt
120W /m 2 °Cx2(1 + 0.025m) = 6_06 m-l
268W/m°Cxlmx0.025m
r
4.6 7
4.68
, Heat and Mass Transfer
Corrected length,
Ac
Wt
L =L+-=L+--P
c
2(W + t)
t =( 0.1+0.025)
=L+ 2
2 - m=0.1125m
mLc = (6.06 m- 1){0.1125 m) = 0.6817
tanh mLc = 0.5926
Heat-flow rate per unit width is
Qf = m k Ac eb tanh mLC
= (6.06 m- 1)(268W/m 0 C)(l m x 0.025 m)(600 °C) (0.5926)
= 14435 W
Mass of rectangular fin per unit width is
m = p (LtW) = (2707 kg/m 3) (0.1 m x 0.025 m x I m)
= 6.7675 kg
Heat-transfer rate per unit mass is
Qr = 14435W = 2133 W /kg
m
6.7675kg
The following table presents these results in proper perspective:
Heat flow
Triangular fin
Rectangular fin
Per unit width (W/m)
12 386
14 435
Per unit mass (W/kg)
3660
2133
Hence, while a rectangular fin promises better heat transfer per metre width, the converse is true in
case of heat-flow per kg mass. Heat loss from a fin of triangular profile is 1.7 times more than a fin of
rectangular profile on per unit mass basis.
Example 4.33 ~ A triangular fin made up of stainless steel (k = 16.3 W/m K, p = 7817 kg/m 3) is attached to
a surface maintained at 450°C. The fin thickness is 6 mm, length 25 mm and is exposed to
a convective environment at 90°C with an associated heat-transfer coefficient of 30 W/m 2 K. Calculate (a) the
fin efficiency, (b) the temperature at the middle of the fin, and (c) the rate of heat loss from the fin per unit
mass of the fin.
Known: A triangular fin of prescribed dimensions exposed to a convective environment.
Find: (a) Temperature at the centre of the fin, T(x = L/2) ( 0 C). (b) Heat-flow rate per unit mass, Q (W /kg).
m
Heat Transfer from Extended Surfaces
Schematic:
Ruid
r
4.69
h=30W/m2K
Too=90°C
Stainless steel
(k= 16.JW/m K, p=7817kg/m3)
T
Tb=450°C
t- 6 mm •-·---·-·-·-·-·-·-·-·--·---·--·--·---·-·-·-·-·-·-·-·-·-·---·-·-·-·-·--·-·-·-·-·-·-·--
J_
x__J
L=25mm
Fig. 4.45
Assumptions: ( l) Steady operating conditions. (2) Constant properties. (3) Uniform convection coefficient.
(4) Negligible radiation effects.
Analysis: For a straight fin with a triangular profile,
[
f=~l+(t/2L) 2 = l+(i:25 )
2ll/2
=l.0072
Fin parameter,
p = ~2hf L = [2 x 30W /m 2 K x 1.0072 x 0.025m]
kt
16.3W /mKx0.006m
Fin efficiency, llr
112
= 3.93
I1 (2 pfl12)
p D12 I0 ( 2 p D12 )
l
I1 (2x3.93x.J0.025)
=-----x~-----(3.93 X-}0.025) I0 (2 X3.93x .J0.025)
= _ l _ I1 (l.243)
0.6214 I0 (l.243)
= - 1- x 0 ·7502 = 0.8465 or 84.7%
0.6214 1.4262
Temperature distribution along a triangular fin is given by
0
eb
T(x)-T
J0 (2pxll2)
T;, - T
I0 (2 p D 12)
00
=--00
At the midpoint, x = L/2 = 0.0125 m
2 p xll 2 = 2 X 3.93X.J0.0125 = 0.8788
/0(0.8788) = 1.2043
0(L/2) = 1.2043 = 0.8444
eb
1.4262
(Ans) (a)
4. 70
,
Heat and Mass Transfer
Temperature at the centre of the fin is
T (x = L/2) = T= + (Tb - T=) (0.8444)
= 90°C + (450 - 90)°C (0.8444) = 394°C
(Ans) (b)
Heat-dissipation rate per unit depth (or width) is
.
pkt0b I 1(2pfl 12)
Qr=
Io (2 pfll2)
Ji
= 3.93x16.3x0.006x(450-90) x 0.7502 = 460 _3 w
,Jo.025
1.4262
Mass of the fin per unit depth (or width),
m = p ~t W = ( 7817 :~ )( 0 ·025 m; 0 ·006 m )um)= 0.5863 kg
Hence, heat-loss per unit mass is
Qr= 460.3W
m
0.5863kg
= 785.l W/kg
(Ans) (c)
(G) ERROR IN TEMPERATURE MEASUREMENT
Example 4.34 ~ A thermometric device is a hollow brass tube with an inside diameter of 0.03 D and an
outside diameter of 0.05 D which extends from the wall to the centreline of a pipe of
diameter D through which air is flowing. The convective heat-transfer coefficient on the outer surface of the
tube is given by Hilpert's correlation:
Nu= O. I 74(Re)0·61 '
The thermal conductivities of air and brass are
k,.;r = 0.028 W/m °C and kbrass = 116 W/m °C
Calculate the thermometric error for air Reynolds numbers of 4000 and 40 000 based on tube OD.
Known: Temperature measurement of air flowing through a pipe by a thermometric device in the form
of a hollow brass tube.
Find: Thermometric error for RE= 4000 and 40000.
Schematic:
L=0.5 D
__ _ _ _J_ _ Air:
kA=0.028 W/m °C
Brass: k8 = 116 W/m °C
Fig. 4.46
Heat Transfer from Extended Surfaces
r
4. 71
Assumptions: ( l) Constant properties. (2) Brass tube as a finite fin with insulated tip.
Analysis: Thermometric error is defined as
T ( x = L) - T=
cosh m ( L - L)
coshmL
=--coshmL
The thermometric device (protruding into the pipe up to its centreline) has length, L = 0.5 D, D being
the pipe diameter.
Fin parameter,
m=
p
where
=
Ac
Perimeter
Cross-sectional area
=
~:;c
1t(0.05D)
1tD2 [0.
052 _ 0 _032 ]
=
125
D
and
k = k8
4
Then
where
k
k
h = Nu - = O. l 741 (Re)06I8
d
d
= 0.174 ~(Re)0.618
0.05D
= 3.48 kA (Re)0.618
D
= 5.59
3.48 X 0.028 (Re)0.618
116
= 0.162 Re0.309
For Re = 4000 and Re = 40 000, the thermometric error is computed and tabulated below:
Re
mL
Error = 1leash mL
4000
2.10
0.241 or 24.1%
40 000
4.28
0.0277 or 2.77%
We note that the error decreases as Re and hence h increases.
(Ans)
4. 72
,
Heat and Mass Transfer
Example 4.3S ~ A steel tube carries steam at a temperature of 320°C. A thermometer pocket of
iron (k = 45 W/m K) of 1.5-cm diameter and I-mm thickness is used to measure the
temperature. The error to be tolerated is 1.5% maximum. Estimate the length of the pocket necessary to measure
the temperature within this error. The diameter of the steel tube is 9.5 cm. Suggest a suitable method for
locating the thermometer pocket. The following data can be assumed: Convection coefficient, h = 80 W/m 2 K;
Tube wall temperature Tw = I 20°C.
Known: A thermometer pocket is used for measuring temperature of steam flowing through a tube.
Find: Length of pocket to ensure that the maximum error does not exceed 1.5%.
Schematic:
Thermometer pocket
Steam
D=9.Scm
h=80W/m 2 K
Ts= T= = 320°C
Fig. 4.47
Assumptions: (1) Steady-state conditions. (2) One-dimensional conduction. (3) Constant thermal
conductivity. (4) Uniform convection coefficient. (5) Adiabatic tip.
Analysis: Assuming the tip to be insulated, the temperature distribution along the length of a fin is given by
0
T-T~
coshm(L-x)
coshmL
At the fin tip (x = L ), we have
or
0
0b
coshmL
TL -T.
Tw - T.
cosh mL
---=---
where
(A)
(Steam temperature)
(Tube wall temperature)
Maximum error in temperature measurement
=.!.2_= T.-TL =l_ TL
100
T.
T.
TL =1-0.015=0.985
r.
Thermometer reading is
TL= (320) (0.985) = 315.2°C
Heat Transfer (ram Extended Surfaces
r
4. 73
Substituting known values in equation (A), we get
315.2-320
120-320
----=
-4.8
1
=---200 coshmL
200
coshmL = -=41.67
4.8
mL = cosh-1 (41.67) = 4.423
or
Now
m=
{hP ~ {I
vii
Vk7td8 vT8
=[
=
=
]1/ =42.16m-1
80W/m2 K
(45W /mK)(0.001 m)
2
42.16 L = 4.423
Length of the thermometer pocket necessary is
L = (4.423)(100 cm/lm)
(42.16m- 1)
= 10.5 cm
(Ans)
Comment: As the length of the pocket is more than the diameter of the steel tube, the thermometer
pocket is to be located in an inclined manner.
, MULTIPLE CHOICE QUESTIONS
Fins are provided to enhance heat-transfer by increasing
(a) the heat-transfer coefficient
(b) the mechanical strength to the equipment
(c) the heat-transfer surface area
(d) the level of turbulence
4.2 For the case of very long fin, the fin efficiency is
4.1
(a)
[K
VhAc
~
(c)~hAf
(b) ~hPkAC
(d) tanh mL!mL
4.3 The fin efficiency is defined as the ratio of the actual heat-transfer from a fin to
(a) the heat-transfer from an infinitely long equivalent fin
(b) the heat-transfer from an identical fin of infinite thermal conductivity
(c) the heat-transfer from an identical fin with an insulated tip
(d) the heat-transfer through the base area of the same fin
4.74
4.4
,
Heat and Mass Transfer
A thin fin of length L has its two ends attached to two parallel walls with uniform temperatures
of T1 and T2 • Heat is lost by convection to ambient air at T=. The heat-rate is given by (symbols
have their usual meanings):
(a) Q =mk Ac (01 +0 2 )tanhmL
(b) Q=mkAc(e, +0 2 )[
c:::;:;
1]
(c) Q=mkAc(01 +0 2 )[coshmL-(1/sinh mL)]
4.5
4.6
(d) Q=mkAc(e, -ei)[(l+sinhmL)]
coshmL
The efficiency of a straight triangular fin of base thickness t (m 2 = 2 h !k t) is defined as
__l_I1 (2mL)
(b)
_ 2 I 2 (2mL)
()
a TJr Tlr 2mL I0 (2mL)
mL / 1 (2mL)
1 I1(2mL)
= tanhmL
(d)
(c) TJr = mL I0 (2mL)
TJr
2mL
The area-weighted fin efficiency for a finned heat-transfer surface comprising unfinned and finned
portions is given by
(d) ~TJ+l-~
(a) ~TJ-1+~
4.7
where ~ = Afin / Atotal and TJ = fin efficiency
The fin effectiveness of a rectangular fin of minimum weight with width b is given by
I
(a) 0.005 (
;!Y
;!Y
I
(b) 0.889
(!: y
I
(c) 0.889 (
4.8
4.9
4.10
(d) none of the above
Two rods, one of length L and the other of length 2L are made of the same material and have the
same diameter. The two ends of the longer rod are maintained at 100°C. One end of the shorter
rod is maintained at 100°C while the other end is insulated. Both the rods are exposed to the
same environment at 40°C. The temperature at the insulated end of the shorter rod is measured
to be 55°C. The temperature at the midpoint of the longer rod would be
(a) 40°C
(b) 50°c
(c) 55°c
(d) 100°c
A plane wall 1 m x 1 m is provided with 100 fins of 2-mm thickness and 2-cm length. The finning
factor is
(a) 4.8
(b) 1.8
(c) 5.0
(d) 22.8
A fin has 5-mm diameter and 100-mm length. The thermal conductivity of fin material is
400 W m- 1 K- 1. One end of the fin is maintained at 130 °C and its remaining surface is exposed
to ambient air at 30°C. If the convective heat-transfer coefficient is 40 W m-2 K- 1, the heat loss
(in W) from the fin is
(a) 0.08
(b) 5.0
(c) 7.0
(d) 7.8
Heat Transfer (ram Extended Surfaces
r
4.75
, TRUE/FALSE
4.1 When mL = 3, a finite length fin, end insulated, will behave as a fin of infinite length.
4.2 Two long slender rods 1 and 2 are made of different materials with thermal conductivities k 1 and
k2. The temperatures at distances x 1 and x2 from the base are same. Then k2 = k 1(xif x 1)2.
4.3 In terms of heat-dissipation relative to the cost of material, a tapered fin is a poor choice compared
to a constant cross-section fin.
4.4 By placing the thermometer well obliquely (inclined to the vertical) in a pipe, more error will be
introduced in temperature measurement.
, FILL IN THE BLANKS
4.1 An ideal fin is one with __ thermal conductivity or __ thermal resistance.
4.2 A composite pin fin with inner material of thermal conductivity k 1 and area of cross section A 1 and
outer material of conductivity k2 with area A 2 has a perimeter P with convection coefficient h. The
fin parameter m is given by
m=--·
4.3 __ fins are attached to the cylinder head of a motor cycle engine while pin fins are used in
computer chips to improve cooling.
4.4 Four fins made up of silver, copper, aluminium and stainless steel are attached to a metal wall
heated to a uniform temperature of 150°C. The fins are exposed to 25°C environment. The fin end
temperature will be maximum in case of __ fin.
4.5
The ratio ( A tin+ A unfin
Ano fin
l
is called the - - ·
, EXERCISES
The end of a very long cylindrical stainless steel rod is attached to a heated wall and its surface is
in contact with a cold fluid.
(a) If the rod diameter were doubled, by what percentage would the rate of heat removal increase?
(b) If the rod were made of aluminium, by what percentage would the heat-transfer rate change
from that of the stainless steel?
ksteel = 16.17 W/m K
kaluminium = 204.7 W/m K
[1166%]
4.2 On an 80-cm long and 10-cm diameter cylinder, 20 longitudinal fins are provided. The cylinder is
placed vertically in an atmosphere of 40°C. The thickness of the fins is 1 mm. The fins protrude
by 5 cm from the cylinder surface. The convection heat-transfer coefficient is 25 W/m 2 K.
Calculate the rate of heat transfer if the surface temperature of the cylinder is 200°C. Also, find
the temperature of the fin at the centre. Thermal conductivity of the fin material is 75 W/m K.
[139.41°C]
4.3 A certain internal combustion engine is air-cooled and has a cylinder constructed of cast-iron
(k = 60 K). The longitudinal fins on the cylinder have a length of 15 mm and thickness of 3 mm.
4.1
4.76
4.4
4.5
4.6
4.7
4.8
4.9
, Heat and Mass Transfer
The convection coefficient is 68 Wlm 2 0 C. The cylinder length is 400 mm. Calculate the heat
loss per fin for a base temperature of 230°C and environment temperature of 38°C .
[162.5 W]
An aluminium pot used for boiling water has 15-cm long handle of circular cross-section (d = 4.5 cm).
The heat-transfer coefficient is 15 Wlm 2 °C. The thermal conductivity of aluminium is 210 W lm°C
and the surrounding air is at 27°C. What would be the temperature at the grip? The temperature of
the boiling water may be taken as 100°C.
If the aluminium handle is replaced by a stainless steel handle (k = 17.4 WIm 0 C) of the same
size, what will be the new grip temperature?
[60.6°C]
A 50-mm-OD copper tube is 1-m long and has 8 longitudinal fins
as shown below. The fins are 3-mm thick and extend 25 mm from
the outside surface of the tube. If the outer surface temperature of
the tube wall is 250°C and the surrounding convective environment
has the heat-transfer coefficient of 15 W lm 2 K and the temperature
80°C, determine the percentage increase in heat-transfer due to fins
as compared to that for an unfinned tube.
Take kcopper = 385 Wlm K
Assume fin tip to be insulated.
Fig. 4.48
[238.7%]
A saucepan of 20-mm diameter is exposed to 99°C temperature during a cooking process. The
temperature of air in the kitchen is 25°C and the convection coefficient between the handle
surface and the ambient air is 8 W lm 2 K. The thermal conductivity of the handle material is 17
Wlm K. The temperature in the last 10-cm of the handle used for hand grip should not exceed
35°C. Assuming heat-transfer from the fin tip to be negligible, estimate the length of the saucepan
handle.
In order to reduce the thermal resistance at the surface of a vertical plane wall (50 x 50 cm),
100 pin fins (1 cm diameter, 10 cm long) are attached. If the pin fins are made of copper having
a thermal conductivity of 300 Wlm K and the value of the surface heat-transfer coefficient is
15 W lm 2 K, calculate the decrease in the thermal resistance. Also calculate the consequent
increase in the heat-transfer rate from the wall if it is maintained at a temperature of 200°C
and the surroundings are at 30°C. Assume heat transfer from the tip is negligible.
[731.5 W]
Long rectangular fins of steel of 20-mm height and 1-mm thickness are arranged on a surface at a
distance (pitch) of 10 mm each. What is the rate of heat-transfer from the surface if
Surface temperature = 200°C
Ambient temperature = 30°C
Thermal conductivity of steel = 50 WIm K
Heat-transfer coefficient= 10 Wlm 2 K
[8.14 kW per m2 ]
A straight fin of uniform circular cross-section of 2-cm 2 area, 15-cm length and 200 Wlm °C
thermal conductivity has its two ends maintained at 100°C and 50°C respectively. The ambient
air is at 30°C and the convection heat-transfer coefficient is 15 Wlm 2 0 C. Determine the total
heat-loss from the fin to the ambient. What is the minimum temperature in the fin and where
does it occur?
[0.49W]
Heat Transfer from Extended Surfaces
r
4. 77
100
90
80
u 70
0
i=' 60
50
40
30
0
0.05
0.1
0.15
x(m)
Fig. 4.49
4.10
A thin rod of length Lis attached to two parallel walls whose temperatures are T1 and T2 (T1 > T2).
It loses heat-by convection to the ambient air at T=.
Develop an expression for the temperature distribution along the length of the rod and the
heat-transfer rate from the rod. Determine the location of the minimum temperature in the
rod.
Q = mkA(01 +02) ( co_sh mL-1)
smhmL
4.11
4.12
4.13
A thin rod of copper of 1.5-cm diameter and a thermal conductivity of 300 W/m K is thermally
joined to two parallel walls maintained at the same uniform surface temperature of 70°C. Air
flows in the space between the two walls spaced 20 cm apart and provides convective heat-transfer
coefficient of 60 W /m 2 K at the surface of the rod . The ambinet air temperature is 30°C. Determine
(a) the temperature of the rod at its midpoint, and (b) the heat-transfer rate from the rod.
[19.29 W]
Formulate the governing relation for the temperature distribution in a circumferential fin of constant
thickness.
A circumferential fin of rectangular cross-section surrounds a 2.5-cm diameter tube. The length
of the fin is 6.4 mm, and the thickness is 3.2 mm. The fin is constructed of mild steel (k = 43
W Im 0 C). If air blows over the fin so that a heat-transfer coefficient of 95 W/m 2 °C is experienced
and the temperature of the base and air are 260°C and 93°C, respectively, calculate the heat transfer
from the fin.
[25.8 W]
4.78
, Heat and Mass Transfer
4.14
4.15
4.16
Heat-is transferred from the hot water inside the aluminum (k= 237 W/m K) tube, 25-mm outer
diameter and 22-mm inner diameter, to the room air through the outside finned surface. Annular
aluminium fins (1-mm thick and 15-mm long) are spaced 1-cm apart and are press-fit to the tube;
contact resistance between the fins and the tube is about 2.75 x IQ-4 m 2 K/W. To eliminate the
contact resistance, the press-fit fins are replaced with the brazed fins. The water is at 85°C, the
water-side heat-transfer coefficient is 950 W/m2 K, the room temperature is 25°C, and the air-side
convective heat-transfer coefficient is 25 W/m2 K. The total outside area (fins and bare tube) is
0.447 m 2, and overall surface efficiency is 96%. Determine the heat-transfer rate per metre length
(a) with the press-fit fins, and (b) with the brazed fins.
[(a) 553 W, (b) 418 W]
A plane wall is equipped with an array of straight fins of triangular profile (3.8 cm length, 0.32 cm
thickness, spaced 1.5 cm centre-to-centre). The surface temperature is 120°C and the surrounding
air temperature is 25°C. The heat-transfer coefficient is 20 W/m2 °C. The fin material has a thermal
conductivity of 43 W/m°C. Calculate the total or area-weighted fin efficiency.
[0.856 or 85.6 % ]
A mercury thermometer is placed in a well filled with oil is required to measure the temperature of
compressed air flowing in a pipe. The well is 140-mm long and is made of steel (k = 58 W/m K),
1.0-mm thick. The temperature recorded by the well is l00°C while the pipe wall temperature is
50°C. The air film conductance outside the wall may be assumed to be 29 W/m2 K. Estimate the
true temperature of air.
[104.78°C]
1
ANSWER KEY
Multiple Choice Questions
4.1 (c)
4.5 (c)
4.9 (c)
4.2 (c)
4.6 (d)
4.10 (b)
4.3 (b)
4.7 (b)
4.4 (b)
4.8 (c)
4.2 T
4.3 F
4.4 F
True/False
4.1 T
Fill in the Blanks
4.1 infinite, zero
4.3 Annular
4.5 finning factor
4.2 ~hP/[k1A 1 /k2 A 2 ]
4.3 silver
MULTIDIMENSIONAL
HEAT CONDUCTION
5
Concept Review
INTRODUCTION
• 5.1
•
Conduction heat transfer for relatively simple geometries like slab, cylinder and sphere with one-dimensional
approximation has so far been considered. Reasonably accurate and acceptable solutions are usually
obtained in several cases. However, one often encounters more complicated geometries or non-uniform
temperatures along the boundaries. In such situations, multidimensional heat conduction occurs.
There are essentially four methods to solve problems involving heat transfer in more than one dimension.
• Analytical Method
The method involves the solution of the governing differential equation for conduction
in the appropriate coordinate system subject to the initial and boundary conditions. However, rigorous and
complex mathematical treatment except in simple geometries makes this method of limited use.
• Graphical Method
This is a versatile method applicable to many irregularly shaped two-dimensional
geometries of practical interest. Isotherms and heat flow lines are drawn which form groupings of
curvilinear figures. The heat flow across the curvilinear section is given by Fourier's law. Assuming
unit depth of material,
.
M
M
Q = -k(f:.x)(l)f:.T/ f:.y = -kf:.T;,verall = -k(7i -7;)
N
N
(5.1)
If the sketch is drawn as illustrated in Figure 5.1, such that t:.x"' t:.y, the heat flow is proportional to
t:.T across the element since Q is constant within the same heat flow lane . t:.T = t:.ToveraufN and M is the
number of heat flow lanes. The ratio MIN is known as conduction shape factor.
Constant heat flow lines Q1and Q2
)Isotherms
L ... distance
perpendicular to the
plane of the figure
Boundary isotherm T 1
Fig. 5.1
Arrangement of arbitrary isotherms and constant heat-flow lines
5, 2
,
Heat and Mass Trans(er
• Analogical Method
The differential equations associated with heat conduction and those governing
2
2
steady-state distribution of electric potential are analogous. Thus ~ <1> + ~ <1> = 0 where <I> is the electric
ox 2 oy 2
potential. In this method, a model of the given geometry is used with analogous boundary conditions.
Equipotential lines on the model surface are plotted, which correspond to isothermal lines in the
temperature field .
Thus, the temperature distribution can be inferred.
• Numerical Method
In this method, the body is divided into a number of discrete subvolumes.
The centre of each subvolume is called a node and the nodes are connected by 'fictitious conducting
rods'. Energy balance on each node gives a set of algebraic equations which are solved to get the
temperature field by any of the following methods: Relaxation method, finite difference method, finite
element method.
TWO-DIMENSIONAL STEADY-STATE HEAT CONDUCTION
• 5.2
•
Use of Conduction Shape Factors A simple, though approximate, approach to determining the
heat-transfer rate by conduction in two dimensions is by using the conduction shape factors. In 2-D heat
transfer, let the two surfaces of a solid medium are held at temperatures T1 and T2 and k is the thermal
conductivity of the medium. The heat transfer rate with no heat generation within the medium is then
given by
(5 .2)
where S is called conduction shape factor and has the dimension of length. It may be noted that for
liquids and gases where convection is dominant, the above equation is not applicable.
Table 5.1 gives conduction shape factors for some selected two-dimensional systems.
Table 5.1
Conduction Shape Factors
No
System
1.
Isothermal sphere
buried in a semiinfinite medium
2.
Horizontal isothermal
cylinder of length L
buried in a semiinfinite medium
Schematic
Restrictions
z > D /2
·· r··································
··· ·········· ·········· ···· ······
·.
~!!:a
L >> D
L >> D
L » 3 D /2
Shape Factor
2rcD
l-D / 4z
2 rcL
cosh - 1 (2z / D )
2rc L
/n (4z/D)
(Continued)
Multidimensional Heat Conduction
Table 5.1
r
5.3
(Continued)
No
System
3.
Vertical cylinder in a
semi-infinite medium
Schematic
Restrictions
L >>D
Shape Factor
2rcL
ln(4L/D)
·:: ....... ·.·
4.
2rcL
Conduction between
two cylinders of
length L in infinite
medium
i!i~i!i!i!i!i!i!i!i l'i !~!
········i~··•••••••w•••••••••;i••••••••••••
5.
6.
7.
8.
Horizontal circular
cylinder of length L
midway between
parallel planes of
equal length and
infinite width
T2
00
n:rn : : : : : : : : : : : : :il: : : : : : mrr
00
Z> D/2
L >> z
!!i! !1·tMf6tt~l~I!i!i
1
1
~ Q ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ j~ ~ ~ ~ ~ ~ ~ ~ ~
~Ji
00
Circular cylinder of
length L in a square
solid of equal length
w>D
L >>> w
Plane wall
Onedimensional
conduction
Conduction through
the edge of adjoining
walls
2rcL
/n(8z/D)
D > L/5
2rcL
/n(l .08w/ D)
A
L
0.54 D
(Continued)
5.4 , Heat and Mass Transfer
Table 5.1
(Continued)
No
System
9.
Conduction through
corner of three walls
with a temperature
difference of fiT1-2
across the walls
10.
Schematic
Restrictions
Shape Factor
D << and
width of
walls
0.15L
None
2D
L
....... L.
Disk of diameter D
and T 1 on a semiinfinite medium of
thermal conductivity k
and T2
Solved Examples
(A) TWO-DIMENSIONAL STEADY STATE CONDUCTION
Example 5.1 ~ A rectangular plate of 1-m width in the x direction, infinite in the y direction, has a temperature
distribution given by T(x, 0) = I 00 sin nx imposed on the y = 0 edge. Determine the temperature
distribution T(x, y).
Known: Semi-infinite plate with width, W = Im, with the boundary y = 0 subjected to temperature
distribution, given by T(x, 0) = 100 sin 1t x.
Schematic:
T(x, = )=O
T(O, y)=O
T(I, y) =O
~---.-~--.x
0
T(x, O)= 100 sin1tx
Fig. 5.2
Assumptions: ( 1) Steady-state conditions prevail. (2) Uniform thermal conductivity. (3) Let T(O, y) =
T(l, y) = T(x, oo) = 0
Multidimensional Heat Conduction
r
5.5
Analysis: The general solution to the two-dimensional heat conduction equation is
Ir= (A cos AX+ Bsin AX)( Ce-"-Y + De"-Y )I
The boundary conditions are
T(O,y)=O:
A=O
T(l,y) = 0:
A= mt
T(x,oo)=O:
D=O
=
T = L, Cne-n1ty sin mtx
n=l
T(x, 0) = IOOsin1t x
B.C.
100 Sin 1tX = C1 Sin1tx+ l,in21tx+ ...
C1 = 100; C2 = C3 = ... = Cn = 0
Hence, the temperature distribution is given by
!T(x,y) = 10oe-1tY sin1txl
(Ans)
Example 5.2 ~ A two-dimensional fin of thermal conductivity k, thickness 2L and with a base temperature
Tb may be considered to extend to infinity in one direction. The fin experiences a finite
heat-transfer coefficient h and an ambient fluid temperature T~. Show that if the coordinate system is taken at
the centre of the fin thickness, the temperature of the fin, T, at any point is given by
T-T~
~(
- = 2...::..,
Tb - T~
sinA L
)
.
"
exp(-Anx)cosAnY
n=I AnL + Sin An LCOS AnL
Known: Two-dimensional fin exposed to convective environment,
Find: Temperature distribution.
Schematic:
y
Fluid h, T=
///
(
20-Fin
0 t-------....-L-----+-----,- oo
1--------+-L----------- X
Tb
Fig. 5.3
5.6
, Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Constant thermal conductivity.
Analysis: Let 0 = T- T=.
The governing differential equation
a20 a2 0
ax2 ay2
-+-=0
The boundary conditions are the following:
(1) At x = o, e = eh
(2) At x = oo, e = o
(·: T= T=)
ae
(3) Aty=O, ay =0
- +L ,
(4) A ty-_
(symmetry).
ae -
ay - he.
The solution of the preceding equation is
0 = X(x)Y(y)
Substituting this solution into the governing equation, we get
1 d2 X
1 d 2Y
- --=---=')}
2
X
Y dy2
ax
Therefore,
d2X -A2X = 0
dx2
d2Y +A2Y=O
dy2
The general solution to the preceding equations is
X = ce'Ax + De-AX
and
Y = AcosAy+ BsinAy
Therefore,
0 = (Ce'Ax + De-J...x )(AcosAy+ BsinAy)
At x = oo, e = o,
0 = (CeA= + De-A= )(AcosAy+ BsinAy)
= (CeA= +O)(AcosAy+ BsinAy)
C=O
Also,
0 = De'Ax(AcosAy+ BsinAy)
Multidimensional Heat Conduction
r
5. 7
From boundary condition (iii), we get
ae = O= -De-AX AsinAy+ De-AX BA.cos Ay
ay
or
Since D -:t- 0, B = O; the solution thus reduces to
0 = ADe-h cos Ay = a e-h cos Ay
where a= AD, which is another constant. From boundary condition (4), we get
ae = -A,ae-AX sinAy
ay
h0 = A.ae-J...x sin AL
or
h0 = -Aae-J...x sinAL
Subtracting one from the other, we get
0 = 2A.ae-J...x sin AL
Since a -:t- 0, sinAL = 0 = sinn1t
or
'I
11,n
1234
=n1t
n = ' ' ' ,...
L '
Therefore,
0 = aexp(-Anx)cosAnY
There exists a different solution for each integer n and each solution has a separate integration constant
an. Summing these solutions, we get
0=
=
Lan exp(-Anx)cosAnY
n=I
From boundary condition (1), we get
=
ab=
Lan COSAnY
n=I
This is an expression of 0b in terms of the Fourier cosine series, where an are Fourier coefficients:
20b sinA.nL
a=--~~-~-n AnL+sinA.nL COSAnL
Substituting the value of an in the final solution, we get
(Ans)
5.8 , Heat and Mass Transfer
Example 5.3 ~ Consider a two-dimensional rectangular fin of constant thermal conductivity, which has a
thickness L in the y-direction and is semi-infinite in the x-direction. The base temperature of
the fin is Tb and the ambient temperature is T~· The heat-transfer coefficient is large. Develop an expression
for the steady state temperature distribution.
Known: A two-dimensional rectangular fin with finite thickness in y direction and is semi-infinite in the
x-direction.
Find: Temperature distribution.
Schematic:
y
L
~--~-------------+-X
L
Fig. 5.4
Assumptions: ( 1) Steady operating conditions. (2) Constant thermal conductivity.
Analysis: Let 0 = T - T=. The differential equation for this problem is
a20 a20
dx2 + dy2 = 0.
The boundary conditions according to the selected reference frame are
(1) Atx=oo,0=0
(2) At x = o, e= eb
(3) Aty=L,0=0
(4) Aty=0,0=0
The solution of the preceding equation is
0 =X(x) Y(y)
Substituting the solution into the governing differential equation, we get
_!__ d 2 X = _ _!_ d 2 Y = ±A, 2
X
dx 2
Ydy 2
Here the sign ')..,2 must be chosen such that the homogeneous y-direction results in a characteristic
value problem. A positive value of the constant, say (+A.2), will satisfy all the prescribed values of the
boundary conditions. Therefore,
d2X -)..2X=O
dx 2
and
The solutions of these equations are as follows:
X = C1eh + C2e-h
Multidimensional Heat Conduction
r
5. 9
and
Therefore,
From boundary condition (1) at x = oo, 0 = O; i.e.,
Therefore, C1 = 0, and
0 = C2e-h (C3 cosAy+ C4 sinAy)
From boundary condition (4), at y = 0, 0 = O; i.e.,
0 = C3C2e-Ax
Therefore, C3 = 0. The solution then reduces to
0 = C2 e-hc4 sin Ay
0 = De-h sinAy
or
From boundary condition (3), at y = L, 0 = 0. Therefore,
0 = De-h sin AL
Since D -:t- 0, sin AL= 0. This expression is satisfied for A= 0, re/ L, 2rc/ L, ... or, in general,
nTC
An= L' where n = O,l,2,3, ...
Therefore,
0 = Dexp(-Anx )sinAnY
There exists a different solution for each integer n and each solution has a separate integration constant
Dn. The general solution will be the sum of these individual solutions. Therefore,
0=
- exp(-Anx)sinAnY
LDn
n=O
Since An= 0 for n = 0, no contribution is made by the first term. We can now write
-
0 = LDn sin AnY
n=I
From boundary condition (2) at x = 0, 0 = 0b; i.e.,
- sin AnY
ab = LDn
n=I
5, I O ,
Heat and Mass Trans(er
This is an expression of 0b in terms of the Fourier sine series, where Dn are Fourier coefficients. Further,
Therefore, the steady-state temperature distribution is given by
(Ans)
Example 5.4 ~ Compute the temperature distribution in an infinitely long two-dimensional rectangular bar,
shown in Fig. 5.10.
y
T=O
T=O
b
T=O
L
X
T= 100
Fig. 5.5
Known: An infinitely long 2-D rectangular bar.
Find: Temperature distribution.
Schematic:
y
T=O
T=O
b
T=O
L
~-----~--~--x
T= 100
Fig. 5.6
Multidimensional Heat Conduction
r
5.11
Assumptions: Steady state prevails.
Analysis: The differential equation for this problem is
a2r + a2 r =O
dx 2
ay
2
and the boundary conditions are
(l)Atx=L,T=O
(2) At x = 0, T= 0
(3) At y = 0, T= 100
(4) Aty=b, T=O
The product solution of the following form is assumed: T = XY. Substituting the preceding product in
the general equation, we get
Dividing by XY throughout, we get
1 d2T
1 d 2T
---=--=')}
Xdx 2 Ydy2
where ').,,2 is constant. Thus, we get
The solutions to the preceding two equations are as follows:
X = A cos Ax+ BsinAx
T = (A cos Ax+ B sin Ax) (ce-AY + De"'Y)
From boundary condition (2), we get
0 = A( Ce-"'Y + De"'Y)
Therefore, A= 0. From boundary condition (1), and using A= 0, one obtains
0 = B sin AL( Ce-"'Y + De"'Y)
From boundary condition (4), and using A= 0, we obtain
0 = BsinA( Ce-Ah+ DeM)
Since B -:t- 0, Ce-Ah = 0. Therefore,
Ce-Ah = - DeAh or C = - De2Ah
5.12
,
Heat and Mass Transfer
Substituting the value of C in the preceding equation, we get
0 = Bsin'A,L (-De2Ahe-AY + De2Ah)
0 = Bsin'A,x (Ce-Ah +DeAh)
0 = BD sin'A,L[eAY +eA(2b-y)]
0 = K sin'A,L[ eAY -e')..,( 2b-y)
J
O= K sin'A,L[eAY -e-AY eAh ]
e-Ah
0 = K sin'A,L[eA(y-b) _e-A(y-b)] eAh
0 = 2 K sin'A,LeAh sin'A,(y- b)
This holds true if sin 'A, L = 0, or
1t
A=(2n+l)L, where n=0,1,2,3 ...
Therefore,
1tX
[(2n+1)1tb]
1t
T=2Ksin(2n+l)-exp
~~- sinh{2n+l)-(y-b)
L
L
L
L
1t sinh(2n+ 1)-(y-b
1t
T = = Cn sin(2n+ 1)-x
)exp [(2n+ l)1tb]
n=O
L
L
L
or
From boundary condition (3), we get
~
.
1tX
.
1t
[(2n+1)1tb]
100 = "'-Cnsm(2n+l)-x
smh(2n+l)-(-b)exp
~
L
L
L
or
100 =
f A,,
n=O
where
sin(2n+ 1) 1tX
L
.
A,,= en smh(2n+
1) (1th)
L exp [(2n+1)1tb]
L
The preceding equation can be recognised as the Fourier sine series. Therefore,
L
f
L
A,, =3_ 100 sin(2n+1)1tx dx
0
L
Let (2n + 1) [1t x)/L] = t. Therefore,
1t
(2n+I)-dx=dt or dx= L dt
L
(2n+ l)1t
Multidimensional Heat Conduction
r
5.13
Therefore,
200 (2 n+t)1t
L
.
f sm t dt
(2n+l)L1t L1t
A,,=----
0
200
[
] (2n+l)!t
= (2n+l ) 1t -cost 0
200
) [1-cos{2n+1)1t]
2n+ 1 1t
=(
_ 200 [1-(-1)]- 400
- (2n+1)1t
- (2n+1)1t
Therefore,
C =n
400
( 2 n+l)1t
1th
1
cosech {2n+l)Lx e( 2 n+t)
Thus,
400 ~
1
1tx
(b-y)(2n+l)1t
(2n+l)1tb
T=-I,---sin(2n+l)-sinh-'---'--'----'---cosech-'------'-1t n=o{2n+l)
L
L
L
(Ans)
(B) GRAPHICAL METHOD
•
•
Example 5.5 ~ A structural member fabricated from a material with a thermal conductivity of 60 W/m °C
has the cross-section shown in Figure 5.7. The temperature of the end faces are T1 = I00°C
and T2 = 0°C, while the remaining sides are insulated.
Insulated
i.----0.2m
-
-------------------
I
1
0.2m
1-0.lm
Fig. 5.7
(a) Estimate the temperature at the position P.
(b) Using the graphical flux plot method, estimate the conduction shape factor and the heat transfer rate through
the strut per metre length.
Known: A structural member subjected to a temperature difference.
5.14
,
Heat and Mass Transfer
Find: (a) Temperature at the point P; (b) Shape factor, and heat rate per metre length.
Schematic:
._···...i-...
j-O.I m
Symmetry
isotherm
···--v
-J
Fig. S.8
Assumptions: ( 1) Two-dimensional conduction.
Analysis:
(
Symmetry isotherm
I
4
3
2
z0.2
Fig. S.9
(a) Using the flux plot method, we construct a heat flux plot. Note that the line of symmetry which
passes through the point P is an isotherm as shown above.
Temperature at location P is (by symmetry)
T(P) =(I;+ J;)/2 = _!.(100+0)°C =50°C
2
ML
(b) Heat rate, Q = k S ( I; - I;) and shape factor, S = - N
From the plot,
4.2L
S0 = -4- = 1.05 L
(Ans)(a)
Multidimensional Heat Conduction
r
5, I 5
For the full section of the structural member,
M=M0 =4.2
But
N=2N0 =8
Therefore
so
S=-=0.53 L
2
(Ans)(b)
It follows that the heat-transfer rate per unit length is
Q = (60W /m°C){0.53){100-0)°C
=3180 W/m
(Ans)(b)
Example S.6 ~ The crass-section of a furnace wall is shown below. Using the graphical method, evaluate (a)
the shape factor, and (b) the heat-transfer rate per unit length.
k= l.2W/m°C
T
soo c
0
2m
J_
so c
0
3m
T
4m
1
Sm
Fig. S.10
Known: A furnace wall with specified surface temperatures and material thermal conductivity.
Find: (a) Shape factor; (b) Heat rate perm length.
Schematic:
a
b;······B=J·· ·
b
r2 =so·c.____________.
a
Furnace wall crass sectian
a
Symmetrical section
Fig. S.11
Assumptions: ( 1) Two-dimensional, steady-state heat conduction. (2) Constant properties.
5.16
,
Heat and Mass Transfer
Analysis: The heat-transfer rate per unit length is
Q_ = 4~k(T. - J:)
I
I
'
(A)
2
where S is the shape factor for the symmetrical section. Note that due to symmetry the heat flow rate
will be four times that through the symmetrical section.
Adiabatic
-------------- -- __
(
Heat flow lane
M=8.S
Adiabatic
\_,
N= I [
;
2!
3!
s
3
2
Fig. 5.12
Selecting three temperature increments (N = 3), we construct the flux plot as shown above.
We note that the conduction shape factor is
S= Ml
N
Per unit length,
~= M = 8.5 =2.83
I
N
(Ans)(a)
3
Heat-transfer rate, from Equation (A) is
Q = 4(2.83) ( 1.2W /m°C){800-50)°C
I
= 10.2 x 103 W /m or 10.2 kW /m
Example 5.7 ~ Determine the shape factor
and the heat flow per unit
depth of the V-grooved channel shown below,
using a flux plot:
(Ans)(b)
T1 = 300°C
1--1 m--1
f-+---1 m--+-1
T
l ·~ - - - -"
2m
k= I.OW/m°C
4m
Fig. 5.13
Multidimensional Heat Conduction
r
5, I 7
Known: Dimensions and surface thermal conditions of a V-grooved channel.
Find: Flux plot, shape factor, heat flow per unit depth .
Schematic:
r- m ------- m i
I
Im
2
--I
-r
T1 =:300 °C
2m
< L - - - -_ _ _ _ _ _ _ _ _ .
l
4m
Fig. 5.14
Assumptions: (1) Steady operating conditions. (2) Two-dimensional conduction. (3) Constant properties.
Analysis: Because of symmetry about the midplane a-a, only one-half of the cross-section will be
considered here.
The flux plot is constructed with 6 temperature increments (N = 6) and from the plot, we obtain M"" 7.
Shape factor for half section is
S = M l where l is the depth of the object
N
7-1=1.111
6
For the complete system, the conduction shape factor is
S = 2 X 1.7 [ = 2.34[
(Ans)
Heat-flow rate per unit depth is
Q
s
l=kl(r;-z;_)
= ( 1.0W/m°C) (2.34) (300-100)°C
=468 W/m
(Ans)
T 1= 300°C
I
2
,---+----+----l 3
--f----+----1------14
-r-----1-----l----l 5
M=7
2
6
Fig. 5.15
h
Symmetry adiabatic
5.18
,
Heat and Mass Transfer
(C) CONDUCTION SHAPE FACTORS
•
•
Example S.8 ~ Radioactive waste material is stored in a spherical container of 3-m diameter which is buried
in the ground (k = 0.52 W/m 0 C). The distance between the top surface of the tank and the
ground surface is IO m. As a result of radioactive decay process, 1250 W of heat is released at the start of
the storage process. (a) If the soil surface temperature is 36°C on a warm summer day, estimate the outside
surface temperature of the container under steady-state conditions. (b) Show the representative isotherms and
heat flow lines in the soil.
Known: A sphere containing radioactive waste is buried deep underground.
Find: Sphere's outside surface temperature.
Schematic:
Fig. S.16
Fig. S.17
Assumptions: ( l) Steady-state conditions. (2) Constant properties. (3) The earth is a semi-infinite medium.
Multidimensional Heat Conduction
r
5, I 9
Analysis: (a) Conduction shape factor for this geometric configuration is
s
2rcD
l-D /4z
3 m_)- = 20.165 m
- -2-rc_(_
l-3m/4(11.5m)
Heat transfer rate, Q=Sk(I;-I;)
Hence, sphere surface temperature,
= 3 6 ° C + - -1250W
-------(20.165 m)(0.52 W/m 0 C)
= 155.2°C
The representative isotherms and heat-flow lines in the soil are shown in the schematic.
(Ans)(a)
(Ans)(b)
Example S.9 ~ A cylindrical heating element of 5-cm diameter and 1-m length is inserted tighdy into a hole drilled
in the ground. The rate of heat dissipation by the heater is SO W. The average heater surface
temperature is found to be 60°C. Estimate the thermal conductivity of the soil if the soil surface temperature is 25°C.
Known: An electrical cylindrical heater is driven vertically into a hole beneath the earth's surface.
Find: Thermal conductivity of the soil.
Schematic:
Fig. S.18
Assumptions: ( l) Steady operating conditions exist. (2) The earth is a semi-infinite medium with constant
properties. (3) Heater surface is approximated as isothermal.
Analysis: Conduction shape factor,
2rcL
s ln(4L
/ D)
= 1.434m
2rc(lm)
/n(4x lm)
0.05m
5.20 , Heat and Mass Transfer
Heat-transfer rate is
Thermal conductivity of soil is
k=
Q
s(i;-i;)
sow
(l .434 m) (92-25)°C
=0.52 W/m °C
(Ans)
Example 5.1 0 ~ Saturated steam at 7 bar (Tsat= I 65°C, h1g = 2066.3 kJ/kg) is piped vertically into the earth at
a mass flow rate of 250 kg/h as a part of an oil reclamation project. The thermal conductivity
of the earth is 1.4 W/m K and the outside diameter of the pipe is 10 cm. The ground surface temperature is
I S0 C. How deep will the steam travel before it condenses completely?
Known: Vertical isothermal pipe carrying steam buried in the earth.
Find: Length of pipe into the earth, L (m).
-
Schematic:
P= ?bar
T,at= 165 °C
Earth surface
h1g = 2066.3 kJ/kg
Fig. 5.19
Assumptions: ( l) Steady operating conditions exist. (2) The ground temperature is uniform. (3) Pipe wall
conduction resistance and the steam-side convective resistance are negligible, i.e., i;,ipe ='.f.team· (4) The
earth is a semi-infinite medium.
Analysis: Energy balance
Under steady-state conditions, the amount of energy released by the condensing steam= Energy conducted
to the earth.
i.e.,
Qcondensation = Qearth,conduction
mh1g =Sk(I;-I;)
(A)
Multidimensional Heat Conduction
r
5.21
where
S = Conduction shape factor for the pipe and earth
=
2rcL
(L>>D)
ln(4L!D)
[From Table 5.1]
Substituting the appropriate numerical values,
mh1g
S=~--~
from Equation (A)
k(J; -i;)
=
(250/3600kg/s) (2066.3x10 3 J/kg) 1 1 WI
lJ/s = 683.3 m
(l.4W/mK)(165-15)°C or K
Hence
S=
2rcL
( 4 L) = 683.3
Zn -
0.1
or
2rcL = 683.3 Zn (40 L)
or
L = 108.75 Zn (40 L)
Iterative procedure is necessary to determine the depth of pipe into the ground, L.
Let L= 1000 m
RHS= 108.75 Zn (40x1000)= 1152.38
With this value of L = 1152.38 m,
RHS = 108.75 Zn ( 40xl 152.38) = 1167.8
With
L = 1167.8 m,
RHS= 108.75 Zn (40 x 1167.8)= 1169.26 m
With
L = 1169.26 m,
RHS = 1169.4 m
With
L = 1169.4 m,
RHS = 1169.4 ~
Convergence reached.
L= 1169.4 m
Hence, the steam will travel about 1170 m into the ground before it is completely condensed.
(Ans)
Example 5.1 I
~ A pipe of SO-cm-diameter carrying steam at a temperature of I65°C and another pipe of
20-cm diameter conveying water at I 5°C are laid deep below the ground surface with
centre-to-centre distance between the two pipes being 1.5 m. The two pipes run parallel to each other and
each pipe has a length of 50 m. The soil has a thermal conductivity of 0.43 W/m 0 C. If the mean flow velocity
of water passing through the pipe is 1.2 m/min, compute the heat transfer rate and the resulting rise in the
water temperature.
Known: Two pipes carrying steam and water and parallel to each other are buried under the earth's surface.
Find: Heat-transfer rate and increase in water temperature.
5.22
, Heat and Mass Transfer
Schematic:
Fig. 5.20
Assumptions: ( l) Steady operating conditions. (2) Two-dimensional conduction. (3) Pipe wall resistance
is negligible. (4) Constant thermal properties. (5) The earth is an infinite medium.
Analysis: For the specified configuration, the conduction shape factor is
2 1t( 50m)
=70.33m
2 -0.5 2 -0.2 2 )
4x
1.5
cosh- 1 (
2x0.5 x0.2
Heat-flow rate;
Q = Sk (T1 - T2)
= (70.33 m) (0.43 W/m 0 C) (165 - 15)°C
=4536 W
(Ans)
Mass-flow rate of water,
.
1t
m= pAcV=p 4 D? V
= (1000 kg/m3
>( ¾x0.2 m )(~·i mis)= 0.628 kg/s
2
2
Rise in water temperature,
t:.T=_j]_ =
4536J/ s
mCP
(0.628kg / s)( 4180J I kg°C)
1.13°c
(Ans)
Multidimensional Heat Conduction
r
5.23
Example 5.12 ~ A row of one metre long and 25-mm diameter nuclear fuel rods (still radioactive waste) are
buried at a distance of 5 m from the earth's surface where the thermal conductivity of the
soil is 1.2 W/m K. Calculate the rate of heat loss from the fuel rods through the soil to the atmosphere if the
surface temperatures of the rods and the ground are I 50°C and I0°C, respectively. The rods are laid parallel
to each other with a centre-to-centre distance of 20 cm.
Known: A row of equally spaced parallel isothermal cylinders buried in the ground.
Find: Rate of heat transfer.
Schematic:
Fig. 5.21
Assumptions: ( 1) Steady-state conditions. (2) The earth is a semi-infinite medium. (3) Constant soil
properties. (4) Isothermal rods.
Analysis: For the given configuration, the conduction shape factor is
2rcL
s
(per cylinder)
2w
. h2rcz}
In{ - sin
-rcD
w
[L > > D and w > 1.5 D]
Substituting the values,
2rcxlm
S=--=----------~
/n[ 2x0.2m sinh 2rcx5m]
rcx0.025m
0.2m
= 1.607 m
Rate of heat loss from the fuel rods is
Q=S k(I; -I;)
= (1.607 m) (1.2 W/m K) (150 - 10)°C or K
=270 W
(Ans)
5.24 , Heat and Mass Transfer
Example 5.13 ~ A 30-cm OD pipe, 90-m long, with a surface temperature of 95°C carries steam, buried at
a depth of I m below the earth's surface whose temperature is -5°C. The average thermal
conductivity of the soil is 0.7 W/m 0 C. (a) Calculate the heat loss per day. (b) Also estimate the thickness of
85% magnesia insulation (k = 0.062 W/m 0 C) necessary to achieve the same insulation with a total convection
coefficient of 25 W/m 2 °C on the outside of the insulation.
Known: An isothermal pipe carrying steam is buried underground.
Find: (a) Heat loss per day. (b) Equivalent thickness of 85% magnesia insulation.
Schematic:
!!!
85% magnesia insulation
k;ns = 0.062 W/m °C
h=2SW/m2°C
r==-s c
0
Pipe
Fig. 5.22
Fig. 5.23
Assumptions: (1) Steady-state conditions exist. (2) Two-dimensional conduction. (3) Constant thermal
properties and uniform heat-transfer coefficient. (4) The surfaces of pipe and earth are isothermal.
Analysis: (a) Conduction shape factor for the given configuration is given as
2rcL
S=-----cosh-1 (2z ! D)
2 rc( 90m)
(2x Im)
1
=218.79m
cosh- - - 0.3m
Hence, the rate of heat loss from the pipe is
Q=Sk(T.,-I;)
Multidimensional Heat Conduction
r
5, 2 5
= (218.79 m) (0.7 W/m C) (95 - 5)°C
0
= 15 315 W
Heat loss per day
= ( 15315 .:!_) (3600x24s) (lMJ)
s
lday
106 J
= 1323 MJ
(Ans)(a)
(b) To find the equivalent thickness of insulation we equate the same heat loss with that through an
insulated pipe with convective surface.
Q=l5315W= 2 1tL(7i.-T=)
_1_ 1n,;,+ _l_
kins
,;
h,;,
or
or
15 315 W =
21t{90m)(l00oC)
l m°C
r (m)
lm 2 °C }
{ - - - - /n - o- -+ - - - - 0.062 W
0.15m 25Wx,;,(m)
21tx90xl00 _l_/n~+-115315
0.062 0.15 25,;,
or
3.69 = 16.13 /n ( ~ ) + 0.0 4
0.15
,;,
Solving by trial and error:
r0 (m)
RHS
LHS
0.1 8
3.16
3.1 6
0.1 9
4.02
4.02
0.186
3.685
"" 3.69
Hence, r0 = 0.186 m
required thickness of 85% magnesia insulation
=,;, -,; = (0.186 - 0.15) m (1°1°;m)
=3.6 cm
(Ans)(b)
Example 5.14 ~ A pipeline carrying crude oil is laid at a depth of 2 m below the earth's surface. The thermal
conductivity of the soil is 0.43 W/m K. The pipe of 0.6 m outside diameter is equipped with
0.15-m thick cellular glass insulation (k = 0.069 W/m K). The heat loss per metre length of the pipe is 92 W
and the temperature of the heated oil is I 60°C. Estimate the temperature of the surface of the soil.
Known: An insulated oil pipe is buried under the earth's surface.
5.26
, Heat and Mass Transfer
Find: Earth's surface temperature.
Schematic:
Fig. 5.24
Assumptions: ( l) Steady operating conditions. (2) Constant properties. (3) Two-dimensional conduction
through soil and one-dimensional conduction through insulation. (4) Oil convective resistance and pipe
wall conductive resistance are negligible.
Analysis: Conduction shape factor of the specified geometry,
Sw" =cosh-2•t2
1 D:
=
l
2 1t{lm)
=2.893m
cosh-1 (2x2m)
0.9m
Heat-transfer rate,
I:iit -~oil
- - - ln -D_2 +- - 21t kins L
Di Ssoil ksoil
Hence, the soil surface temperature is
= 160°C - 92
w[
l
Zn o. 9 m +
l
]
21tx0.069W/mKxlm 0.6m (2.893)(0.43W/mK
= 160°C - 92 W [0.935 + 0.804] K/W
= 160°C - {92 W X 1.739 K/W)
=0°C
(Ans)
Multidimensional Heat Conduction
r
5, 2 7
Example 5.1 5 ~ A steam pipe is provided with an insulation RUF (rigid urethane foam) as shown in Figure 5.25.
The pipe and insulation are encapsulated in a steel casing of square cross-section. Calculate
the rate of heat loss from the pipe per metre length.
!!!
Steel pipe
(20-mm-OD, 5-mm-thick)
h= 15W/m 2 °C
T==25°C
RUF (k = 0.026 W/m 0 C)
Steel casing
(JO-mm-square) 0.5-mm-thick
Fig. 5.25
Known: An insulated steam pipe carrying steam is exposed to ambient air.
Find: Heat-loss rate per metre length.
Schematic:
Air
!!!
h= 15W/m 2 °C
T==25°C
Fig. 5.26
Assumptions: (I) Steady operating conditions exist. (2) Two-dimensional geometry. (3) Constant
properties and uniform heat transfer coefficient. (4) Pipe wall resistance and steam side convective
resistance are negligible.
Analysis: For a circular isothermal cylinder of length L centred in a square solid of equal length, the
conduction shape factor is given as
s
2rcL
/n (1.08 w / D)
= 13.0 m
2rcxlm
/n{l.08x0.03m}
0.02m
[since L ~ w and w > D]
Energy Balance
i.e., the heat lost by the pipe by conduction through the insulation= The heat dissipated to ambient air
by convection
or
5.28 , Heat and Mass Transfer
where
S .... Conduction shape factor
k ..... Thermal conductivity of insulation
h ..... Convection coefficient
T= ... Ambient air temperature
T1 •••• Pipe's outer surface temperature"' steam temperature
T2 . . . . Steel casing outer surface temperature
Heat-transfer rate,
Q = ~ T,,verall
Rtotal
-
J; - T=
+ Rconv
Rcond
I;-T=
- {)k + h1A}
where
A ... Convective surface area= PL
P ... Perimeter of square casing= 4 x 0.03 m = 0.12 m, and
L . . . Pipe length = I m
Substituting the proper values,
Q
(125-25)°C
{cl3m)(0.0~6W/m 0C) + (15W/m 20 ~)(0.12m 2 )}
(per metre length)
=28.5 W
Comment: As we have neglected steam-side convective resistance, pipe wall resistance and contact
resistance, the calculated heat loss is rather overestimated.
Example 5.16 ~ A 25-mm OD heating rod is eccentrically embedded in a I00-mm OD cylinder as shown
in Figure 5.27. For the conditions shown, (a) determine the heat flow from the heating rod
per unit length. (b) If the same heat loss takes place under insulation without any eccentricity, what will be the
outer diameter of insulation for the same values of temperature.
120°c
1.25cm
k= IOW/mK
Fig. 5.27
Multidimensional Heat Conduction
r
5.29
Known: Hot rod eccentrically embedded in a circular system having a specified thermal conductivity.
Find: (a) Heat loss perm length. (b) OD of insulation for the same heat loss.
Schematic:
k= IOW/mK
01 =2.Scm
D2=10cm
z= 1.25cm
r1 = 1.25cm
Fig. 5.28
Assumptions: ( l) Steady-state conditions exist. (2) Two-dimensional conduction. (3) Constant properties.
(4) Length L > > diametrical dimensions.
Analysis: (a) For this system, the heat loss per unit length is
Conduction shape factor,
2reL
S=--~-----~
cosh-1 (Di2+D}-4z2)
2D1D2
2re(lm)
=--~-------~
cosh- I {
102 +2.52-4(1.25)2)
2xl0x2.5
=4.77 m
Q = (4.77m)(lOW/mK)(l20-20)°C[ lkW J
10 3 W
= 4.77 kW
(Ans) (a)
(b) If concentric insulation loses the same amount of heat per unit length, the conduction shape factor
must be same, i.e., S = 4.77 m.
But for this geometry,
S =2 rel/ /n(D2 / D,)
or
D2 =exp ( 2 rel )
S
D1
5.30 ,
Heat and Mass Transfer
outside diameter of insulation layer without eccentricity will be
D2 =D1ff2rrLIS)
= (2.5 cm)xexp [ 21txl(m)J
4.77m
= 9.33 cm
(Ans) (b)
Comment: If eccentricity is avoided, D 2 is reduced from 10 cm to 9.33 cm. Hence some saving of
insulation material can be expected.
Example 5.17 ~ Liquid nitrogen is stored in a 3-m sphere buried in the earth (k = 0.17 W/m K) with its
centre 4 m below the surface. A 10-cm-thick layer of insulation (k = 0.05 W/m K) covers
the sphere. The nitrogen is at -I 96°C, and the surface of the earth is at I 0°C. Nitrogen vaporizes (and the
vapour is vented to the surface) because of heat transfer from the earth to the tank, thus maintaining a constant
temperature and pressure in the nitrogen. The enthalpy of vaporization of the nitrogen is 200 kJ/kg. Determine
(a) the heat transfer rate to the nitrogen (in W), and (b) the nitrogen evaporation rate (kg/h).
Known: An insulated sphere containing liquid nitrogen is buried underground under specified conditions.
Find: (a) Heat transfer to liquid nitrogen. (b) Liquid nitrogen boil-off rate.
Schematic:
: :fo~~!~tlC?~: : : : : : : : : : : : : : : : :
•:•: •: •:•:<.k;~i:
=;:o:osw,iii:k}:•: •: •:.•:::
. ..................
. .•:::
. .•:.•
.
. .. .... .... .... .... .... .... .... .... .... .... . .
.... .. .... .. ...... ...... ...... ...... ...... ...... ...... ...... ...... .... . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... .. ...... ...... ...... ...... ..
r,
~
--JV\f\r----o--J\f'\f\r-___.,
Q
Fig. 5.29
Assumptions: (l) Steady operating conditions exist. (2) Constant properties. (3) Two-dimensional
conduction in the soil. (4) Convective resistance at the tank inside and the conductive resistance of the
pipe wall are negligible.
Analysis: The rate of heat transferred to liquid nitrogen from the earth, (Qin), W
= ~ate of evaporation ofliquid nitrogen, ri1 (kg/s) x Latent heat of vaporization ofliquid nitrogen, hrg(I/kg)
i.e., Qin =mhfg
Multidimensional Heat Conduction
r
5.31
Total thermal resistance, R101 = Rinsulation + Rsoil
1
Rins = 4 n kinJ
~ ~
] = 2 n ~ins { ~;
~J
l
Rsoil=~
soil
Conduction shape factor, S is
2nD
0
S=
as z >D /2
0
l-D0 / 4z
(.!. __ )_!_
l
l
= 0.0663 K/ W
2n(0.05W/mK) 3 3.2 m
Rinsulation
Rsoil
Sksoil
(l-D0 /4z)
l-(3.2m/4x4m)
- - " - - - ----'------'---=0.234 K/W
2nD0 ksoil 2nx3.2mxO.l7W/mK
Hence, R,otal =Rinsulation +Rsoil = 0.0663 + 0.234 = 0.3 K/W
Heat-transfer rate,
.
[10-(-196) ]°Cor K
Q = - - - - - - = 686 W
0.3K/W
(Ans)(a)
Rate of evaporation of liquid nitrogen,
m=il._=
/trg
686W
llJ/sl3600sl= 12.35 kg/h
200xl03 J/kg l W lh
(Ans)(b)
Example 5.18 ~ A thick-walled concrete duct (k = 0.75 W/m 0 C) of square cross-section is I0-m long. The
outer dimensions of the duct are 200 mm x 200 mm and the thickness of the duct wall is
30 mm . If the inner and outer surfaces of the duct are at I 20°C and 35°C respectively. Calculate the rate of
heat conduction through the walls of the duct.
Known: A square flow passage made of thick-walled concrete has its inside and outside surfaces at
different temperatures.
Find: Rate of heat loss by conduction.
Schematic:
.£..= 20cm = I 43
b
14cm - ·
(> 1.4)
k=0.7SW/m °C
-r T
37.5cm 30cm
1+--b= 14cm--+- _....-··
3-cm
Thick wall
l
Fig. 5.30
5.32
,
Heat and Mass Transfer
Assumptions: (l) Steady operating conditions exist. (2) Constant properties. (3) Two-dimensional
conduction.
Analysis: Conduction shape factor for the above configuration is
S=
=
2rcL
0.93/n(0.948a/b)
2rc(l0m)
=222.77m
0.93/n(0.948x 20cm/14cm)
Rate of heat loss by conduction is
Q = Sk(I; -I;)
= (222.77 m) (0.75 W/m 0 C) (120 - 35)°C
= 14.2 x 103 W or 14.2 kW
(Ans)
Example 5.19 ~ The brick chimney shown in Figure 5.31 is 6-m high. Estimate the total rate of heat loss
through the chimney if the inside surfaces are at a uniform temperature of I 20°C while the
outside surfaces are maintained at 30°C. The thermal conductivity of the brick used is 0.72 W/m 0 C.
80cm
T
10cm
60cm
1~Fig. 5.31
Known: A brick chimney of prescribed geometry with its inside and outside temperatures.
Find: Heat loss rate through the chimney.
Schematic:
--j 10cm f--
T2
Insulated
(Line of symmetry)
@
CD
10cm
T2
\T1
Q)
10cm
r--JOcm
T
l
20cm
Insulated
(Line of symmetry)
Fig. 5.32
Assumptions: ( l) Steady-state conditions. (2) Two-dimensional conduction. (3) Constant properties.
Multidimensional Heat Conduction
r
5.33
Analysis: The conduction shape factor method can be used. By virtue of symmetry, we concentrate on
one of the four corner sections of the chimney. We note that the planes of symmetry across which no
heat is transferred are shown as insulated in the schematic. The three elements that make up the corner
section are also indicated. Conduction shape factors for each of the three elements are
(l) S _ _6 _ 0.3m x 6m _ 18
1- L O.lm
m
(2) S2 = 0.54 D = 0.54 x 6 m = 3.24 m
(3) S3 = A3 = 0.2mx6m = 12 m
L
O.lm
The sum total of the shape factors is
I,s =S1 +S2 +S3
= 18 + 3.24 + 12
= 33.24 m
Heat transfer rate from the corner section is
{?comer
= L, S k ( 7; - T2 )
= 33 .24 m X 0.72 W/m °C X (l20-30)°C( lkW
10 3 W
J
=2.l54kW
Hence, the total rate of heat loss is
Q=4x2.l54kW
=8.62 kW
(Ans)
Example 5.20 ~ A laboratory furnace has outside dimensions of 120 cm by 90 cm by 60 cm and is made of
1S-cm thick brick (k = 0.80 W/m 0 C). Inside and outside surface temperatures are S40°C
and 40°C respectively. Determine the heat transfer rate through the walls of the furnace.
Known: A laboratory furnace with thick firebrick walls loses heat through its walls.
Find: Heat lost through the walls.
T
Schematic:
T2=40°C
ik=0.8W/m °C
.
----kflWcm
1S-cm-thick wall
around the furnace
,-------11 -----.iT
L3 =60cm
.
·
·
)
)
12
~
.:,<,;
'1/0ii
\,'\,
_L
T1 = 540°C
(Temperature inside
furnace)
Fig. 5.33
(Outside surface
temperature)
5.34 , Heat and Mass Transfer
Dimensions: Outside: Li xL 2 xL3 = 120 cm x 90 cm x 60 cm
Inside: 4x /2 x /3
= (Li -2t)x(L 2 -2t)x(L 3 -2t)
= (120 - 2 x 15) cm x (90 - 2 x 15) cm x (60 - 2 x 15) cm
i.e.,
90cmx60cm x30cm
t,
t,
t,
.............................................
Assumptions: (1) Steady-state conditions. (2) Constant properties. (3) Two-dimensional conduction.
Analysis: Conduction shape factor S for the following components needs to be determined:
Plane walls (six), edges (twelve) and comers (eight).
Stotal = Swails + Sedges + Scorners
Plane walls (6)
From Table 5.1
Swails = ~ where t is the wall thickness on all sides and A is the total wall surface area.
t
= 2[(90x60)+(60x30)+(90x30)]
=2 [5400+1800+2700]cm2 (
lm2 )
104 cm2
= l.98m2
Hence,
S
walls
= Awalls = 1.98m2 = 13.2 m
t
0.15m
Edges (12)
Sedges =0.54L
where
L = 4 [90 + 60 + 30] cm= 720 cm= 7.2 m
sedges= (0.54) (7.2m)=3.89m
Corners (8)
scorners =0.15 t
=8x0.15x0.15 m=0.18 m
Therefore,
stotal = 13.2 + 3.89 + 0.18
= 17.27m
Multidimensional Heat Conduction
r
5.35
Heat-loss rate,
Q=Sk(I;-Ti)
=(17.27 m) (0.80 W/m K) (540-40)°CI !kW I
103 W
= 6.91 kW
Example 5.21
(Ans)
~ A steel pipe of 15-cm outside diameter with a wall thickness of 7.5 mm (k = 60.5 W/m C) is
0
buried 1.8 m below ground level. An insulating layer (k = 0.43 W/m 0 C), 5-cm thick, covers the
pipe. The ground thermal conductivity is estimated to be 1.9 W/m 0 C. If the steam passing through the I-km-long
pipe is at I80°C and the ground surface temperature is -I 0°C, calculate the rate of heat loss from the pipe.
Known: An insulated steam pipe is buried underground.
Find: Heat loss from the pipe, Q.
L T,urface = -I 0°C
Schematic:
k,oil = 1.9 W/m °C
z= 1.8 m
Fig. 5.34(a)
r1 = (7.5-0.75) cm
=6.75cm
r2=7.5m
r 3 = (7.5 + 5) cm
Pipe
(ksteel= 60.5 W/m C)
= 12.5cm
0
Insulation
(k;ns= 0.43 W/m 0 C)
Fig. 5.34(b)
Assumptions: (I) Steady-state, one-dimensional conduction without heat generation. (2) Constant
properties. (3) Convective resistance between the steam and the pipe is negligible.
Analysis: The thermal circuit is shown below:
Q
T,team
T,urface
R;nsulation =
Fig. 5.35
Rsoil=-1ksoilS
5.36
, Heat and Mass Transfer
The thermal resistances are:
"T>
i'steel
= ln{ 7 .5/ 6 ·75 ) =0.277xlo-6°C/W
21t X 60.5 X 1Q3
_
~nsulation -
ln{12.5/7.5) _
-4o
2 1tx0.43 xl0 3 -l.89xl0 C/W
1
R.on=~s
soil
where S is the conduction shape factor.
For a circular pipe embedded in a semi-infinite medium with z > r 3,
S=
21tL
ln[(z/13)+~(z/1S-1
J
21tx103
= ---,,=--------;:::======,
1n[{1.8/0.125)+ ~{1.8/0.125) 1J
2
-
= 1870.46
1
Rsoil = - - - - - = 2.814 x 10-4°C/W
1.9 X 1870.46
Total thermal resistance,
Rtotal
= Rsteel + Rinsulation + Rsoil
= 0.277 X lQ-6 + 1.89 X 104 + 2.814 X IQ-4
= 4.7073 X 10-4 °C/W
Rate of heat loss from the pipe is
Q = ¾team - ¾urface
l?iotal
[180-(-10)] 0 c
=------4. 7073 X IQ-4 °C/W
=403.7 W
(Ans)
Example 5.22 ~ A disc-shaped electronic device of 20-mm diameter dissipating 80 W power is mounted to an
aluminium alloy block (k = 186 W/m 0 C) at 30°C. The thermal constant resistance between
the device and the block surface is 0.5 x I o---4 m2 °C/ W. Estimate the temperature of the device finally reached
assuming that all the power dissipated by the device was conducted into the block.
Known: Disc-shaped electronic device mounted flush with an aluminium alloy block with specified
contact resistance.
Find: Temperature of device reached, T1•
Multidimensional Heat Conduction
Schematic:
r
5.37
1~--- Electronic device (D= 20mm, Q=SOW), T 1 =?
I
Rt.c=O.Sx I0-4m2°C/W
~---Aluminium alloy block
(k= 186W/m C)
T2=30°C
0
Fig. 5.36
Assumptions: ( l) Steady-state two-dimensional conduction. (2) The device is at uniform temperature.
(3) The block acts a semi-infinite medium.
Analysis: The thermal circuit for conduction heat transfer between the device and the block is
Ti
T2
~
Rt, C
Rth
Fig. 5.37
1For the given physical system, the conduction shape factor S = 2 D and R,h =-1- = Therefore ,
Sk 2k D
or
l
0.5 X 10- m °C/W
l
= 80W [ ------ +---------~
4
2
(1t/4 ){0.02m)2
[ 2(186 W/m°C){0.02 m)]
= 23.5°C
The temperature the device will reach is
T 1 = (30 + 23.5)°C = 53.5°C
(Ans)
Example 5.23 ~ A radioactive sample is to be stored in a protective box with I0-cm-thick walls having inside
dimensions of 4 x 4 x 12 cm. The radiation emitted by the sample is completely absorbed
at the inner surface of the box, which is made of concrete (k = 1.37 W/m 0 C. If the outside temperature of
the box is 25°C, but the inside temperature is limited to a maximum of 50°C estimate the maximum allowable
rate of radiation from the sample.
Known: Dimensions of the box containing radioactive sample with specified inside and outside surface
temperatures.
Find: Maximum permissible heat loss from the sample.
5.38 ,
Heat and Mass Transfer
Schematic:
Corners (8)
Edges [4 + 8]
...··
24cm - - - - - - - - -
Fig. 5.38
Assumptions: ( l) Steady-state conditions prevail. (2) Two-dimensional conduction exists. (3) Constant
thermal conductivity.
Analysis: Energy conservation: Under steady operating conditions, the net heat flow through the walls,
edges, and corners of the box is equal to the rate of radiation (energy generation) from the sample.
Q = kS t:.T
Using relations for the shape factor:
+ For plane walls (ends)
s wall (end) =
LA1
where L is the thickness of the wall= 10 cm and area,
A1 = 11 x /1 = 4 cm x 4 cm = 16 cm2
16cm2
s wall(end) =--=1.6cm
10cm
+ For plane walls (sides)
A2
s wall (si de) =
L = Li x Lz /L =
4cm 12cm
10cm
Fig. 5.39(a)
=4.8cm
+ For corners of three walls of equal thickness
T2 (outside)
Scorner = 0.15L
= 0.15x 10cm = 1.5 cm
Fig. 5.39(b)
Multidimensional Heat Conduction
+ F or edges of two adjoining walls of equal thickness
Sedge = 0.54 w where w is the length of edge .
5.39
. . ..-~=-~···· -~
LC]
. / .__ I
sedge, = 0.54 Li = 0.54 x 4 cm=2.16 cm
sedge, = 0.5 4 1-i = 0.54 x 12 cm =6.48 cm
r
. .
Ti (ms,de)
. / ;;
L
~
T2 (outside)
Fig. 5.39(c)
We note that there are 4 side walls, 2 end walls, 8 comers, 4 edges with w = /2 = 12 cm and 8 edges
with w = / 1 = 4 cm.
Total shape factor is computed by adding the shape factors for the walls, comers and edges.
S=2 Swan(end) +4Swan(end) +8Scomers +4Sedge, +8Sedge,
l
sixwall sections
eight comres
twelve edges
[
= (2xl.6)+(4x4.8)+ ~ +(4x6.48)+(8x2.16) cm
= 77.6 cm or 0.776 mm
And, the heat flow (and the rate of radiation) is calculated as
Q=kS(I;-T2 )
= (l .37W/m°C){0.776m){50-25)°C
(Ans)
=26.6 W
Example 5.24 ~ Consider a km long network of 15-cm outside diameter steam heated steel pipe
(k = 60.5 W/m 0 C) (wall thickness is 7.5 mm) that is buried 1.75 m below ground level.
A 5-cm-thick layer of insulation (k = 0.43 W/m 0 C) covers the pipe. The ground thermal conductivity is estimated
to be 1.9 W/m 0 C. If the steam is at I80°C and the surface temperature of the ground is -9°C, determine
the heat loss from the pipes
Known: Insulated steam pipes buried below ground level.
Find: Total heat loss from the pipes.
L T,urface=-9 oC
Schematic:
::::::::::::::•i :=:l •.TS ifr ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
s...m
Tsteam= 180°C
•••••••••••••••••••••••• t ••••••••.••············································?r?uhd············•••••
••••••••••~
:•· ;-•·•·•·•·• •·•·•·•·•·•·•·•·•·•••••••• L~•i km ••••·•·•·•·•·•·•·•·•·•·•·•·•·•·•·•·•·• ~•(••••••
f
' / ~' ' ' : ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' '
>
...... . .... . ....................................................
····· ·. ~
•···•··•·····•••-•-----•••••••---------------------------------••••••••••••si~~·;;i~w. ••••••••••••••••••••••••-------~
••••••••••••••••••••••••••••••: •
·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·. ·, ·. ·
Fig. 5.40(a)
5.40
,
Heat and Mass Transfer
D2= 2 r2= 150mm
r1 = 75mm-7.5mm =67.5mm
ksteel= 60.5W/m °C
r3= (75+ 50) mm= 125 mm
k;ns= 0.43 W/m °C
D3=250mm
ksoil= l.9W/m°C
T,team= 180oC
T,urface=-9°C
Rsteel
R;nsulation
Fig. 5.40(b)
Assumptions: ( 1) Steady-state conditions. (2) Heat transfer is one-dimensional. (3) All properties are constant.
(4) There is no internal heat generation. (5) Convective thermal resistance on the steam side is negligible.
Analysis: The heat transfer rate, Q = ~~verall / l?iotal· The overall temperature difference is ~T = Tsteam - Tsurface·
Since the convective heat transfer resistance between the steam and the pipe is assumed negligible,
the total thermal resistance is Rtotal = Rsteel + Rinsulation + Rsoil·
For constant properties, the individual thermal resistances are determined from
ln('i / 'i)
ln(,j / 12)
1
R.teel = 2 k
L Rinsulation = 2 k L Rtotal = ~
1t steel
1t ms
sot!
where S is the conduction shape factor.
The soil thermal resistance requires a conduction shape factor. For a circular tube enclosed in a semiinfinite medium with z > D, one has
S=
21tL
!n[(2z/ D3 + ~(2 z! D3 ) 2 - 1
J
The three resistances are calculated to be
R
steel
R
= ln('i / 'i) =
/n(75/67.5)
=0 27?x10-6 0 C/W
21tksteel L 21t(60.5W/m°C){l0 3 m)
·
ln(l25/75)
= l 89 1Q-4 0 C/W
insulation= 2 1t(0.43 W/moC){l0 3 m)
·
21t(103 m)
S = --;-------'-;::=======~ = 1886.3 m
[( (2)(1.75m))+ ((2)(1.75m))-i]
0.25 m
0.25m
R
1
__
I_
- - - - - - - = 2.79 x I0-4°C/m
soil - k . S 0 C) (1886.3 m)
(l.9W/m
sot!
Therefore, the total resistance is
Rtotal = Rsteel + Rinsulation + Rsoil
= (0.277 X 10-6 + 1.89 X 10-4 + 2.79 X lQ-4) = 4.684 X 1Q-4°C/W
Multidimensional Heat Conduction
r
5.4 I
The total heat transfer rate is
Q= ~Z:,veran = [lg0-(- 9)]°C
403.5xl0 3 W or 403.5 kW
4.684 x 10-4 °C/W
Rtotal
(Ans)
(D) RELAXATION METHOD
•
•
Example 5.25 ~ By using the Relaxation Method, determine the temperatures at the points I, 2, 3 and 4
under steady operating conditions.
3so c
0
I
2
3
4
50°C
150°C
250 °C
Fig. 5.41
Known: A square section with four isothermal surfaces.
Find: Steady-state temperatures at points 1,2,3 and 4.
Schematic:
350 °C
I
2
3
4
50°C
150°C
250 °C
Fig. 5.42
Assumptions: ( l) Two-dimensional heat conduction. (2) Steady operating conditions.
Analysis: The residuals are given by
Ri = T2 + I;, - 4 7; + 50 + 350
R2 =I;+~ -47; +350+150
R3 = I; + ~ - 4 I;,+ 50 + 250
R4 =T2 +T;,-4~ +250+150
5.42
,
Heat and Mass Transfer
or
R, = 400 + I; + I; - 4 7;
R2 = 500 + I; + T4 - 4 I;
R3 = 300 + I; + T4 - 4 7;
R4 = 400+7; +7;-4T4
Let the initial guess values of the specified node temperatures be
z; = 200°c, ½ = 250°C, ½ = l50°C,
~ = 200°c
Unit size operation and operation table:
T,
R,
T2
200°c
Initial values
T3
R2
250°c
R3
150°c
T4
R4
200°c
-4
i:'17; = 1
-4
L'1T2 = 1
-4
i:'17;=1
-4
i:'1T4 = 1
-2
Total with £17; = i:'17; = £17; = £1~ = 1
-2
-2
-2
Sum total of this effect of unit block operation
=-2-2-2-2 =-8
The values of R,, R2 , R3 and R4 with initial estimated temperature values are found to be
R, =400+250+150-4{200)=0
R2 = 500+200+200-4{250)= -100
R3 = 300 + 200 + 200-4(150) = + 100
R4 =400+250+150-4{200)=0
Total : 0
Increment size= 0/(-8) = 0.
Hence, block operation in this case is not necessary. The relaxation table is prepared and presented
below to estimate the final temperatures in steady state when the four residuals ultimately become zero.
Initial values
T,
R,
T2
R2
T3
R3
T4
R4
200
0
250
-100
150
+100
200
0
-25
0
-25
0
Final
200°c
0
225°C
+25
100
-25
0
0
11s 0 c
200°c
The temperatures are
r. = 200°c, i; = 22s c, T = 11s c, T = 200°c
0
3
0
4
(Ans)
Multidimensional Heat Conduction
r
5.43
Example 5.26 ~ Determine the temperatures at interior nodes I, 2, 3 and 4 of a square cross-section
subjected to prescribed surface temperatures using the finite-difference method:
100°c
50°C
I
2
3
4
200°c
300 °C
Fig. 5.43
Known: A square section with specified uniform surface temperatures.
Find: Temperatures at the four prescribed nodes.
Schematic:
y
, 1 0 0°c
T
,-
I
2
3
4
Lly =.:lx
50°C
--
L_
fu<
n
200°C
X
0
300 C
Fig. 5.44
Assumptions: ( l) Two-dimensional conduction. (2) Steady-state conditions.
Analysis: Using the finite-difference method, we follow the Gauss-Seidel iterative procedure carried out
sequentially until acceptable convergence is achieved.
Finite difference equations for each of the four interior nodes can be written in the form:
l
T; = 4L, Tneighbours
Substituting the given values of surface temperatures, we have
l
l
7; =-[T2 +7;+100+50]=37.5+-(7;+7;)
4
4
l
1
4
4
l
l
T2 = -[7; + ~ + 100+200] = 75+-(7; + T4 )
J; =-[7; +~ +300+50]=87.5+-(7; +T4 )
4
4
l
l
4
4
T4 = -[T2 +I;+ 200 + 300] = 125+-(7; + J;)
5.44 , Heat and Mass Transfer
The iteration solution is presented in the following table, starting with initial estimates of T1, T2 , T3 , and T4 .
Iteration #
Tl (OC)
T2 (OC)
T3 (OC)
T4 (OC)
0 (initial)
100
150
175
250
1
118.75
162.50
175
206.25
2
121.88
156.25
168.75
209.38
3
118.75
157 .8 1
170.31
206.25
4
119.53
156.25
168.75
207.03
5
118.75
156.64
168.14
206.25
6
118.95
156.25
168.75
206.45
7
118.75
156.35
168.85
206.25
8
118.80
156.25
168.75
206.30
By now, the convergence is approximately 0.1°C.
(Ans)
, MULTIPLE CHOICE QUESTIONS
5.1 Select the correct answer:
Isotherms are constant temperature lines and they are
(a) parallel to insulated surfaces
(b) parallel to heat flow lines
(c) perpendicular to both insulated surfaces and heat flow lines
(d) bisecting the comers of isothermal boundaries
5.2 Select the incorrect answer:
(a) Diagonals of curvilinear squares in a flux plot bisect each other at 90° but do not bisect the comers.
(b) Flow lines are perpendicular to isothermal boundaries.
(c) Isotherms are perpendicular to insulated boundaries.
(d) Isotherms and flow lines form a network of curvilinear squares.
5.3 If the number of isotherms is M and the number of heat flow lanes is N, the conduction shape factor
S is defined as
(a) MIN
(b) NIM
(c) (M-l)IN
(d) (N-l)M
5.4 Conduction shape factor has SI unit of
(a) m- 1
(b) m2
(c) m
(d) dimensionless
5.5 Using the graphical method to solve a two-dimensional heat conduction problem, if N is the number
of heat flow lanes and M is the number of squares in each flow lane for the entire contiguration, the
conduction shape factor S is
(a) MIN
(b) NIM
(c) MN
(d) MN 2
5.6 The correct steady-state finite-difference heat conduction equation of node 7 of the rectangular solid
is shown in Figure 5.45
1
2
1
(a) 7; =-(z; + 7; +Tio+ I'i1)
(c) 7; =-(J;;+J;+Tg+I;1)
4
Multidimensional Heat Conduction
2
3
r
5.45
4
5
6
7
8
9
10
11
12
Fig. 5.45
, TRUE/FALSE
5.1 The analytical method used to deal with two-dimensional conduction problems yields solutions that
are exact at any point in the conducting material.
5.2 The numerical method of solution gives temperatures anywhere in the solid involving two-dimensional
steady-state conduction.
, FILL IN THE BLANKS
5.1 If the number of heat-flow lanes and the number of temperature increments are 15 and 5 each, then
the shape factor is equal to - - ·
5.2 When two different physical phenomena can be described by the same equation, the two phenomena
are said to be - -·
5.3 In the numerical method of analysis, the differential equation and the boundary conditions are
expressed as a set of __ equations.
, EXERCISES
5.1 Show that for a semi-infinite plate of width L, having the boundary condition for T(x, y)
T (0, y) - T1 = 0
T (L, y) - T1 = 0
T (x, oo) - T1 = 0
T (x, 0) - T1 = (T2 - T1)
the temperature distribution is
4[e-1t/LY
( ) sm-+-e. 1tx I (31t/LYsm--+
) . 31tx ...J
T-T.1 =--
7;.-T;
1t
L
3
For T1 = 0 and T2 = I00°C, plot isotherms of 25, 50, and 75°C.
L
5.46
, Heat and Mass Transfer
y
00
f
T1
T1
L
X
T1-T2
Fig. 5.46
X
L/4
L/2
Tao(C)
p
y
25
0.271
0.415L
50
0.518
0.210
75
0.749
0.092
25
0.199
0.514
50
0.414
0.281
75
0.668
0.128
5.2 A sphere 1-m OD containing a radioactive material is buried such that its uppermost point is 2 m
below the earth's surface. If the outside surface temperature of the sphere is 425°C and the thermal
conductivity of the soil is 0.9 W Im K, determine the heat lost by the sphere. The soil surface
temperature is 25°C.
[2585 W]
, ANSWER KEY
Multiple Choice Questions
5.1 (c)
5.5 (a)
5.2 (a)
5.6 (c)
5.3 (c)
True/False
5.1 T
5.2 F
Fill in the Blanks
5.1 3
5.2 analogous
5.3 algebraic
5.4 (c)
6
TRANSIENT HEAT CONDUCTION
Concept Review
INTRODUCTION
• 6.1
•
There are many situations in the engineering application of heat transfer where steady-state conditions
do not exist and the temperature of a solid varies with time. We need to know what (temperature or
heat-transfer rate) happens when (time) and where (spatial location) in the material in the case of transient
conditions. The central concept in predicting the unsteady-state behaviour of systems is to understand
the interplay or relationship between the internal (conductive) resistance and the external (convective)
resistance to heat flow.
LUMPED CAPACITY FORMULATION
• 6.2
•
In the lumped systems (Figure 6.1) the temperature varies with time but remains uniform throughout the
system at any time. The temperature of a lumped body of arbitrary shape of mass m, volume ¥- , surface
area A 8 , density p, and specific heat CP initially at a uniform temperature r; that is exposed to convection
at time t = 0 in a medium at temperature T= with a heat-transfer coefficient h is expressed as
Heat-transfer coefficient, h
Surrounding fluid temperature, T=
/
T= f(t)
Initial temperature, T;
-
Density, p
Volume, -\l
Specific heat, Cp
\
Fig. 6.1
Surface area A,
for heat transfer
'~
Arrows indicate convective
heat transfer into the system
In a lumped system, the body temperature depends only on time
since internal temperature gradients are neglected
6.2 , Heat and Mass Transfer
T ( t )- T~ = e-at
T;-T~
(6.1 (a))
where
(6.1 (b))
is a positive quantity whose dimension is (timet 1•
We note that Lc is the characteristic length, customarily defined as the ratio of the solid's volume to
surface area, i.e., Lc =¥-I A •. The above relation can also be used to determine the temperature T(t) of
a body at time t or, alternatively, the time t required for the temperature to reach a specified value T(t)
can be evaluated from.
(6.2)
The error involved in the lumped system analysis is negligible (less than 5%) when
(6.3)
where Bi is the Biot number, a dimensionless parameter.
For systems with negligible internal resistance (Bi < 0.1 ), the temperature-time history can also be
expressed as
_T(~t)_-T_~ = exp[-(Bi){Fo)J = exp{-t/'t)
J;-T~
(6.4)
where Fo = at is the Fourier number. It is a common way of nondimensionalizing the time variable. It
1i
can be looked upon as the ratio of the rate of heat conducted through a body to the rate of heat stored at
the prescribed time. 't = p CP 4 is known as the thermal time constant. During one time constant interval,
h
the temperature ratio would change to lie of its initial value.
Once the temperature T(t) at time t is available, the rate of convection heat transfer between the body
and its environment at that time can be determined from Newton's law of cooling as
(6.5)
The total amount of heat transfer between the body and the surrounding medium over the time interval
t
= 0 to t is simply the change in the energy content of the body,
(6.6)
Transient Heat Conduction
r
6.3
The amount of heat transfer reaches its upper limit when the body reaches the surroundings temperature
T=. Therefore, the maximum heat transfer between the body and its surroundings is
(6.7)
FINITE INTERNAL RESISTANCE: ONE-DIMENSIONAL SYSTEMS
• 6.3
•
The evaluation of the Biot number is the initial step in solving any transient heat conduction problem. If
Bi_h(¥-/As) is greater than 0.1 then the lumped mass approximation is no longer valid.
k
When the lumped system analysis is not applicable, the variation of temperature with position as well
as time can be determined using the transient temperature charts (Heisler charts) for a large plane wall,
a long cylinder, a sphere, and a semi-infinite medium. These charts are applicable for one-dimensional
heat transfer in those geometries. Therefore, their use is limited to situations in which the body is initially
at a uniform temperature. All surfaces are subjected to the same thermal conditions, and the body does
not involve any heat generation.
These charts can also be used to determine the total heat transfer from the body up to a specified
time. Transient heat-transfer charts are called Grober charts. The Heisler and Grober charts consider a
wide range of Biot numbers.
We must recognise that all the charts are valid only for Fo > 0.2. The convection coefficient h is
constant and uniform. And, there is no heat generation in the body.
Using a one-term approximation, the solutions of one-dimensional transient heat conduction problems
are expressed analytically as
Plane Wall
~ T(x, O)=T;
,t)-T=
e( x,t )wa ll -_ T(xT-T
=
I
= Aie-1.} Fo cos(A 1x !L) ,
0
h, Toj
L
[Fo > 0.2]
(6.8)
x~, T=
I-- 2L ------J
Long Cylinder
T(x, 0) = T;
e( r,t )cy l -_ T(r,t)-T=
T-T
I
= Ai e- 1../Fo J o ("- I rh:o ) ,
h, T=
)
=
'o
[Fo > 0.2]
r~, T=
(Continued)
(6.9)
6.4 , Heat and Mass Transfer
(Continued )
Sphere
( )
0rt
'
T(r,t) - T=
~ - T=
=~-~-
sph
(6 .10)
h, T~
where the constants A 1 and A. 1 are functions of the Biot number only, and their values are listed in
Table 6.1 against the Biot number for all three geometries.
Table 6. 1 Coefficients used in the one-term approximate solution of transient one-dimensional conduction (Bi = hL/k
for a plane wall of thickness 2L, Bi = h rJk for a long cylinder or a sphere of radius rJ.
Plane Wall
Long Cylinder
Sphere
Bi
11. 1 (rad)
A1
11.1 (rad)
A1
11. 1 (rad)
A1
0.01
0.0998
1.0017
0. 1412
1.0025
0.1730
1.0030
0.02
0.1410
1.0033
0. 1995
1.0050
0.2445
1.0060
0.03
0.1732
1.0049
0.2439
1.0075
0.2989
1.0090
0.04
0.1987
1.0066
0.2814
1.0099
0.3450
1.0120
0.05
0.2217
1.0082
0.3142
1.0 124
0.3852
1.0149
0.06
0.2425
1.0098
0.3438
1.0148
0.4217
1.0179
0.07
0.2615
1.0114
0.3708
0.0 173
0.4550
1.0209
0.08
0.2791
1.0130
0.3960
1.0197
0.4860
1.0239
0.09
0.2956
1.0145
0.4195
1.0222
0.5150
1.0268
0. 10
0.3111
1.0160
0.4417
1.0246
0.5423
1.0298
0.15
0.3779
1.0237
0.5376
1.0365
0.6608
1.0445
0.20
0.4328
1.031 1
0.6170
1.04483
0.7593
1.0592
0.25
0.480 1
1.0382
0.6856
1.0598
0.8448
1.0737
0.30
0.5218
1.0450
0.7465
1.0712
0.9208
1.0880
0.40
0.5932
1.0580
0.8516
1.0932
1.0528
1.1164
0.50
0.6533
0.070 1
0.9408
1.1143
1. 1656
1.1441
0.60
0.7051
1.0814
1.0185
1.1346
1.2644
1.1713
0.70
0.7506
1.0919
1.0873
1.1539
1.3525
1.1978
0.80
0.7910
1.1016
1.1 490
1.1 725
1.4320
1.2236
0.90
0.8274
1.1107
1.2048
1.1902
1.5044
1.2488
(Continued )
Transient Heat Conduction
L6.5
Table 6. 1 (Continued)
Plane Wall
Long Cy linder
Sphere
Bi
11, 1 (rad)
A1
A1 (rad)
A1
11, 1 (rad)
A1
1.00
0.8603
1.1191
1.2558
1.2071
1.5708
1.2732
2.00
1.0769
1.1795
1.5995
1.3384
2 .0288
1.4793
3.00
1.1925
1.2102
1.7887
1.4191
2 .2889
1.6227
4.00
1.2646
1.2287
1.9081
1.4698
2.4556
1.7201
5.00
1.3138
1.2402
1.9898
1.5029
2.5704
1.7870
6.00
1.3496
1.2479
2.0490
1.5253
2 .6537
1.8338
7.00
1.3766
1.2532
2.0937
1.5411
2 .7165
1.8674
8.00
1.3978
1.2532
2.1286
1.5526
2 .7654
1.8921
9.00
1.4149
1.2570
2.1566
1.5611
2 .8044
1.9106
10.0
1.4289
1.2598
2.1795
1.5677
2 .8363
1.9249
20.0
1.4961
1.2699
2.2881
1.5919
2.9857
1.9781
30.0
1.5202
1.2717
2.3261
1.5973
3.0372
1.9898
40.0
1.5325
1.2723
2.3455
1.5993
3.0632
1.9942
50.0
1.5400
1.2727
2.3572
1.6002
3.0788
1.9962
100.0
1.5552
1.2731
2.3809
1.6015
3.11 02
1.9990
00
1.5707
1.2733
2.4050
1.6018
3.1415
2.0000
Note that unlike lumped capacity model, Bi is differently defined here.
Bi= h L where L is the half thickness of plane wall.
k
hr
0
For infinite cylinder and sphere of radius ,;,, Bi=k
The error involved in one-term solutions is less than 2 per cent when Fo > 0.2.
Using the one-term solutions, the fractional heat transfers in different geometries are expressed as
Plan e Wall
(--2:_J _1_ 0
-
Qmax
sinA 1
O,wall
w all
(6.11)
A1
Cylinder
Sphere
(6.12)
= 1_ 30
(--2:_J
Q
max
sph
sin\ - A1 cos A1
O,sph
A3
I
(6.13)
6.6 , Heat and Mass Transfer
The quantities J0 and J 1 are zeroth and first order Bessel functions of the first kind and their values
are listed in Table 6.2.
Table 6.2
The zeroth-and first-order Bessel functions of the first kind
X
Jo (x)
JI (x)
X
Jo (x)
JI (x)
0.0
1.0000
0.0000
1.3
0.6201
0.5220
0.1
0.9975
0.0499
1.4
0.5669
0.5419
0.2
0.9900
0.0995
1.5
0.5118
0.5579
0.3
0.9776
0.1483
1.6
0.4554
0.5699
0.4
0.9604
0.1960
1.7
0.3980
0.5778
0.5
0.9385
0.2423
1.8
0.3400
0.58 15
0.6
0.9120
0.2867
1.9
0.2818
0.58 12
0.7
0.8812
0.3290
2 .0
0.2239
0.5767
0.8
0.8463
0.3688
2.1
0. 1666
0.5683
0.9
0.8075
0.4059
2.2
0. 11 04
0.5560
1.0
0.7652
0.4400
2.3
0.0555
0.5399
1.1
0.7196
0.4709
2.4
0.0025
0.5202
1.2
0.671 1
0.4983
2.6
- 0.0968
- 0.4708
SEMI-INFINITE MEDIUM
• 6.4
6.4.1
•
Infinite Convection Coefficient at Surface
T
(Specified Surface Temperature, T (0, t) = TJ
00
For the special case of h ~
the surface temperature
T8 becomes equal to the fluid temperature T=, and the
above equation reduces to
00 ,
~ 00
T,
l
Ti
T(x,t)-T',,
T(x,t) - T;
- - - = erf [ -x- or --Ti - T,,
2 fcii
T,, - Ti
,'
X
(6.14)
= erfcCJcxr) (r;, = constant)
Fig. 6.2
Schematic diagram and nomenclature for
transient conduction in a semi-infinite solid
z
f
The error function is defined as erf ( z) = exp (-ri 2 ) dr]
0
where rJ is a dummy variable and !erf (0) = O!
and as z ~ oo, erf(z) ~ 1 Also, !erfc(z) =l-erf(z)!
where the quantity erfc (z) is the complementary error function.
(6.15)
Transient Heat Conduction
L6.7
Table 6.3 gives the values of complementary error function, erfc (z).
Table 6.3
Gaussian complementary error function
z
erfc(z)
z
erfc(z)
z
erfc(z)
z
erfc(z)
z
erfc(z)
0.00
1.0000
0.52
0.4621
1.04
0.1414
1.56
0.0274
2.08
0.00322
0.02
0.9774
0.54
0.4451
1.06
0.1339
1.58
0.0255
2.10
0.00298
0.04
0.9549
0.56
0.4284
1.08
0.1267
1.60
0.0237
2.12
0.00272
0.06
0.9324
0.58
0.4121
1.10
0.1198
1.62
0.0220
2.14
0.00247
0.08
0.9099
0.60
0.3961
1.12
0.1132
1.64
0.0204
2.16
0.00225
0.10
0.8875
0.62
0.3806
1.14
0.1069
1.66
0.0189
2.18
0.00205
0.12
0.8652
0.64
0.3654
1.16
0.1009
1.68
0.0175
2.22
0.00169
0.14
0.8431
0.66
0.3506
1.18
0.0952
1.70
0.0162
2.26
0.00139
0.16
0.8210
0.68
0.3362
1.20
0.0897
1.72
0.0150
2.30
0.00114
0.18
0.7991
0.70
0.3222
1.22
0.0845
1.74
0.0139
2.34
0.00094
0.20
0.7773
0.72
0.3086
1.24
0.0795
1.76
0.0128
2.38
0.00076
0.22
0.7557
0.74
0.2953
1.26
0.0748
1.78
0.0118
2.42
0.00060
0.24
0.7343
0.76
0.2825
1.28
0.0703
1.80
0.0109
2.46
0.00050
0.26
0.7131
0.78
0.2700
1.30
0.0660
1.82
0.0101
2.50
0.00041
0.28
0.6921
0.80
0.2579
1.32
0.0619
1.84
0.0093
2.55
0.00035
0.30
0.6714
0.82
0.2462
1.34
0.0581
1.86
0.0085
2.60
0.00024
0.32
0.6509
0.84
0.2349
1.36
0.0544
1.88
0.0078
2.65
0.00018
0.34
0.6306
0.86
0.2239
1.38
0.0510
1.90
0.0072
2.70
0.00013
0.36
0.6107
0.88
0.2133
1.40
0.0477
1.92
0.0066
2.75
0.00010
0.38
0.5910
0.90
0.2031
1.42
0.0446
1.94
0.0061
2.80
0.00008
0.40
0.5716
0.92
0.1932
1.44
0.0417
1.96
0.0056
2.85
0.00006
0.42
0.5525
0.94
0.1837
1.46
0.0389
1.98
0.0051
2.90
0.00004
0.44
0.5338
0.96
0.1746
1.48
0.0363
2.00
0.0047
2.95
0.00003
0.46
0.5153
0.98
0.1658
1.50
0.0339
2.02
0.0043
3.00
0.00002
0.48
0.4973
1.00
0.1573
1.52
0.0316
2.04
0.0039
3.20
0.00001
0.50
0.4795
1.02
0.1492
1.54
0.0294
2.06
0.0036
3.40
0.00000
6.4.2
Specified Surface Heat Flux [ q. =-k
::l.,
= conrtant]
q,[ A' ( x' )
T(x,t)-T = I
k
--exp - - - -xerfc [-X-)]
1t
4(lf
2,Jat
(6.16)
6.8 , Heat and Mass Transfer
Surface heat flux, q8 (t) =
6.4.3
k(T. -i;)
r:;;-;;:;
vrca.t
(6.17)
Finite Convection Coefficient at Surface
-k{
:t
=[q,(t}=h[T_ -T(O,t)]}
The analytical solution for one-dimensional transient heat conduction in a semi-infinite solid subjected
to convection is given by
2
~T(~x,t~)-~i; = erfc(-x-)- exp ( hx + _h
a.t) [erfc (-x-+ _h.Jai_a.t )]
T,,. - J;
2.Jai
k
k2
2.Jai
k
• 6.5
(6.18)
MULTIDIMENSIONAL SYSTEMS: PRODUCT SOLUTIONS
•
Transient problems for which two-dimensional and even three-dimensional effects are important are
frequently encountered.
Using the principle of superposition called the product solution, the transient temperature charts can
also be used to find solutions for the two-dimensional transient heat-conduction problems involving in
geometries such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or plate, and even
three-dimensional problems associated with geometries such as a rectangular prism or a semi-infinite
rectangular bar. Of course, all surfaces of the solid should be subjected to convection to the same
fluid at temperature T,,., with the same convection heat transfer coefficient h, and the body should not
involve any no heat generation. The solution in such multidimensional geometries (Figure 6.3) can be
expressed as the product of the solutions for the one-dimensional geometries whose intersection is the
multidimensional geometry.
For example, the product of the one-dimensional infinite slab solution and the infinite cylinder solution
can describe the transient heat transfer in a short cylinder:
0{r,x,t~hort
= 0{x,t}infinite x0{r,t}infinite
cylinder
slab
cylinder
(6.19)
0( x,y,t) longrectangularbar = 0wall ( x,t)x 0wall (y,t)
(6.20)
+ For a rectangular bar,
+ For a semi-infinite plate,
0(x,y,t)semi-infplate =0wall (x,t)X0semiinf (y,t)
(6.21)
+ For a semi-infinite cylinder,
0(r,x,t)semi-inf cyl =0 infcyl (r,t) x0semi-inf (x,t)
(6.22)
Transient Heat Conduction
r
6.9
• For a rectangular parallelepiped,
le{x, y, z, t) = ew.11 ex, t)xew.11 (y, t)xew.11 cz, t)I
(6.23)
• For a semi-infinite rectangular bar,
e(x,y, z, t)=0wanCx, t)x0wanCY, t)x0semiinf.(z, t)
x
(6.24)
I Two dimensional
X
Finite cylinder
Semi-infinite cylinder
y
y
~----+---x
Finite body
Semi-infinite body
Three dimensional
z
Semi-infinite bar
Parallelepiped
Fig. 6.3
• 6.6
y
Multidimensional systems
HEAT TRANSFER IN MULTI-DIMENSIONAL SYSTEMS
•
The total heat transfer to or from a multidimensional geometry can also be determined by using the onedimensional values. The transient heat transfer for a two-dimensional geometry formed by the intersection
of two one-dimensional geometries 1 and 2 is
(6.25)
6.10 ,
Heat and Mass Transfer
Transient heat transfer for a three-dimensional body formed by the intersection of three one-dimensional
bodies I, 2 and 3 is given by
lQ:.1.,.m ~ lQ:J +[ Q:J HQ:Jl
+ ( Q:ax
l[ Jl[ 1l
I -( Q:ax
(6.26)
I -( Q:ax
PERIODIC VARIATION OF SURFACE TEMPERATURE
• 6.7
•
Consider the surface of a semi-infinite solid whose temperature varies in a periodic fashion (Figure 6.4).
The variation of surface temperature T0 in excess of the mean surface temperature Tmin• i.e., 00 =( To -Tmin)
can be expressed as
0o =0o,max COS rot
where 0o,max = ½,max - Tmin
To.max= Maximum surface temperature ro = 21tfwithfbeing the frequency of temperature oscillations.
The excess temperature at any depth x may be written as
l
j
l
j
le= 0o,max exp -Jrofia. x cos rot - x Jrofia, I
(6.27)
Temperature, T
>+-----
Periodic time - - - - + <
Mean temperature
// Temperature variation at
··~ distance x from the surface
Time for one complete-----.j
surface temperature variation
t=O ~ - - - - - - - - - - - - - - - - - - - - - - + - T i m e, t
Fig. 6.4
• Time lag:
Periodic temperature variation with time
IM=~ I
(6.28)
x
2cxro
TQI
• The velocity of the temperature wave:
.
2
+ Heat-flow rate: Q = r;:::;:; kA 00 max
..;ro ex
·
Iv= ~I
(6.29)
(6.30)
( 6.31)
Transient Heat Conduction
r
6.11
Solved Examples
(A) TRANSIENT TEMPERATURE VARIATION
Example 6.1 ~ At a certain time, the temperature distribution in a long cylindrical fire tube, 30-cm inner
radius and SO-cm outer radius, is given by
T=800+ 1000 r-5000 r2
where T is in °C and r in m. The thermal conductivity and thermal diffusivity of the tube material are 58 W/m K
and 0.004 m2/h, respectively
Find
(a) the rate of heat flow at inside and outside surfaces per unit length
(b) the rate of heat storage per unit length
(c) the rate of change of temperature at inner and outer surfaces
Known: Temperature distribution in a long hollow tube of given dimensions.
Find: (a) Qi and Q,, [W/m]; (b) Est (W/m); (c) (dT!dt)°C/h at ri = 0.3 m, and r0 = 0.5 m.
Schematic:
....···;;.::·····
.......······· \
'
\ .. -··
....···
.)•::·
.... -·······=-::.,
···...
..
.
\ ...
·- .......;.....--··/
..... -··
............................./>·,,::::::::....~~ 30cm }
/v
~
............-
{
k=S8W/mK
a=0.004 m2h
,__~
~~=SOcm
L
'<
Fig. 6.5
Assumptions: ( 1) Constant properties. (2) One-dimensional (radial) heat conduction. (3) No internal heat
generation.
Analysis: The temperature profile is given by
T(r) = 800 + 1000 r - 5000 r 2
Differentiating with respect to r,
dT
-=1000-lOOOOr
dr
The temperature gradient or slope is not constant and depends on the radius. The larger the radius,
the lesser the slope.
6.12 ,
Heat and Mass Transfer
(a) Rate of heat flow at inside surface, per mere length.
.
dTI
Qi= -k(21t,;_ L) dr r=r;
= -(58W/m K)(21tx0.3mxlm)[ 1000-(10000 x0.3)°C/m]
=+218.655 x 103 W or 218.655 kW
(Ans) (a)
(in the outward direction, being positive)
Similarly, the rate of heat flow at the outside surface per metre length is
Q =-(58W/m K)(21tx0.5mxlm) [1000 - 10000 x 0.5]°C/m
0
= 728.85 x 103 W or 728.85 kW
(Ans) (a)
(in the outward direction, being positive)
(b) Control volume energy balance,
Ein - Eout +
~
=Est
(no he t generation)
or
Rate of change of thermal energy storage per unit length is
Est = (218.655 - 728.850) kW
=-510.195 kW
(Ans) (b)
(negative sign implies that there will be a decrease in internal energy of the system).
(c) The general heat conduction equation in cylindrical (polar) coordinates for one-dimensional heat
conduction without internal heat generation is
!i_[rar] - _!_ ar
rar
ar
(l
at
Rate of change of temperature,
ar
a ( ar)
a
a
- =a- - r - =--[1000
r-10 000 r2 ]
at
rar ar
rar
a
= -;:-[1000-20000
r] = <l [1000
-r--20000
J
At the inner surface [ri = 0.3 m],
(ar)
at
= 0.004 m2/h [ 1000 -20000]
0.3
r=lj_
= -66.67°C/h
(Ans) (c)
Transient Heat Conduction
r
6.13
At the outer surface, [r0 = 0.5 m] ,
= 0.004[lOOO -20000]
( <JT)
dt
0.5
r= 'o
=-72°C/h
(Ans) (c)
The negative sign indicates that temperature at both inner and outer surfaces will decrease with the
lapse of time. This is to be expected as the rate of heat storage is negative.
(B) LUMPED PARAMETER MODEL
•
•
Example 6.2 ~ In the lumped parameter analysis, it is customary and convenient to consider h a constant. For
natural convection heat transfer, if the convection coefficient is given by h =CST113 , where C
is a constant and .6.T=T -T-, invoke the lumped capacitance approximation and obtain an expression for the
time required for the .6. T to reduce to SO% of its initial value. Compare it with the estimated time based on
the constant value of h if .6.T = SS.S°C at t = 0.
Known: Lumped capacity formulation with convection coefficient as a function of temperature difference
between surface and fluid.
Find: Expression for time required if t:.T is reduced by half. Comparison with constant h case.
Schematic:
Qconv= hA(T - T=)
/ . ···;~;~;~·;j;·············..... /- ····1·····-... _c::~~-~-~~I volume
\..\..__ r(O)=T;
~E,t
·.
"···...
T(t)=T
··- .. ___
/
I h=C(T-T=)l/31
..··
Fig. 6.6
Assumption: ( 1) The system is spatially isothermal. (2) Constant properties.
Analysis: Applying an energy balance to the control volume,
to- tg/=
Est or -EOIII =Es,
£ out+
-CA (T-T=)413 = dd (p¥-CPT)
t
or
Separating the variables and integrating,
dT
dt
or
T
dT
f(T-T~t3
=-[_Q_J (r-r=t3
p¥-CP
l
1
- [ -CA
- - dt
p¥- CP [
or
6, I 4 ,
Heat and Mass Transfer
t--3pCP¥-[(
- ~ T-T )-1/3 - (T-T )-1/3]
CA
=
I
=
or
(Ans)
With
(given)
and 7; -T= = 55.5°C, the time required is
3pC ¥t =
p
[ 2113 -
C A(55 .5y;3
pC ¥-
1] - 0.2044 _ P _
-
(Ans)
CA
With li = constant, the estimated time is
t=P~pVln(J;-T=)= 0.6931 pCPV
hA
T-T=
hA
= (pCP¥-)(0.6931)
C(55.5)113A
[In 2 = 0.6931]
0.1817 pCP¥CA
(Ans)
With increasing t, the heat transfer decreases due to a decrease in li as well as a decrease in (T-T
00) ,
Example 6.3 ~ A large aluminium plate [p = 2707 kg/m 3, CP = 0.896 kJ/kg °C, and k = 204 W/m 0 C] of thickness
L = 0.10 m, initially at a uniform temperature T; = 200°C, is cooled by exposing it to an
air stream at T_ = 40°C. Determine the time required to cool the aluminium plate from 250 to 75°C if the
heat-transfer coefficient between the air stream and the surface is h = 80 W/m 2 0 C.
Known: Transient cooling of a plate exposed to air stream.
Find: Time required, t.
Schematic:
Air
Aluminium plate}
~ - - - - - - - - - + - ~ k = 204 W/m °C
Initially
T==40°C
h =80W/m2°C -----=--
T;= 250 °C
T(t) = 75 °C
p = 2707 kg/m3
Cp = 896 J/kg °C
L=O.IOm
Fig. 6.7
Assumptions: (1) Lumped system analysis is applicable. (2) Constant properties. (3) Uniform surface
heat-transfer coefficient.
hL
Analysis: To simplify analysis, if Biot number, Bi (= __
c where Le= ¥-I A8 ) is less than 0.1, we can
k
claim that internal temperature gradients can be neglected so that, T= f(t) and the temperature at any
point in time is uniform throughout the plate.
Transient Heat Conduction
r
6.15
Characteristic length,
¥- AL L
L =- = - = - = 0.05 m
c
A8
2A
2
(considering both su,faces of the plate)
Bi= (80W / m2oC)(0.05m) =0.0196 (< 0.1)
204W / m°C
Temperature of the plate as a function of time is then given by
T(t)-T~ e-bt
T;-T~
where
~
ln( T;-T~ )=bt
T(t)-T~
b=~=-h-=
80W / m2°C
pCP¥- pCpLc (2707kg/ m3 x896J / kg°Cx0.05m)
= 6.597 X I0-3 s-I
T-T
250-40
I
~ =---=6
T(t)-T~
75-40
And
Therefore, the time required is
I
t=-ln 6
b
ls
x ln 6 = 2716 s or 45 min 16 s
6.597xI0-3
(Ans)
Example 6.4 ~ A stainless steel ingot (cylindrica~ 10 cm in diameter and 30-cm long, passes through a heat
treatment furnace which is 6-m long. The initial ingot temperature is 92°C, and it must reach
827°C, in preparation for working. The furnace gas is at I 267°C and the combined convective and radiant
heat-transfer coefficient is I OS W/m 2 K. In order that the required conditions be satisfied what must be the
maximum speed with which the ingot moves through the furnace.
Take: k=23 W/m Kand a=0.44 xl0-5 m2 /s
Known: A cylindrical steel ingot undergoes heat treatment in a furnace .
Find: Speed of ingot passing through the furnace .
Schematic:
! 01F.-· )
D= IOcm
Stainless steel:
k=23W/m K
a= 0.44 x I o-s m2/s
Gas
h=IOSW/m2K
Furnace
T = 1267°C
~
.
S<eel ,goo,
=
T(t)=827°C
( ) T, 92 C \
-V=?
··.
r- L= 30 cm -I
------------d=6m------------
Fig. 6.8
Assumptions: (I) The ingot is spatially isothermal (Negligible internal temperature gradients). (2) Constant
properties and uniform heat transfer coefficient.
6.16 ,
Heat and Mass Transfer
Analysis: The Biot number for lumped capacity model is
hLC
Bi=-k
~D2 L
DL
L = ¥- = _ _,_4_ __
C
A
1tD2
2(D+2L)
s
2--+1tDL
where
4
10cmx 30cm
= 2.143 cm= 0.02143 m
2(10cm+2x30cm)
Bi= (105 W/m2 K) (0.02143 m)/23 W/m K
= 0.0978 (< 0.1)
Lumped capacitance analysis is appropriate.
Temperature response is given by
Zn (J;-T=)
=(Bi)(at)
T-T=
L~
or
:. time required,
L~
l nJ;-T=
t=--X
--aBi
T-T=
=
(0.02143m)2
xln (92-1267)°C
5
2
(0.44 X 10- m /s)(0.0978)
(827-1267)°C
= 1048 s
Speed of ingot,
V = Distance through the furnace length, d (m)
Time,t(s)
= ~(3600s) = 20 _6 m/h
1048s
lh
(Ans)
Example 6.5 ~ A right circular cone-shaped body made of lead with a base diameter of 12 cm and a height
of 8 cm is subjected to a cold-air environment at 5°C. The initial uniform temperature of the
body is 43°C and it takes 25 min for the body to reach 25°C. Determine the convective heat-transfer coefficient.
Properties of lead: k = 35.3 W/m K and ex= 24.1 x Io---6 m2/s.
Known: A right circular cone is exposed to convective cooling process under the given conditions.
Find: Heat-transfer coefficient, h [W!m 2 K].
Transient Heat Conduction
Schematic:
T
1
T;=43°C, T=25°C
t=25min
H=Scm
Lead - - [k=35W/m K
a=23.43x lfr-6m2/s]
·········
r
6.17
Ambient
air
T ~=S°C
h=?
' ···
············· ··················· ······•······························
·:::::..
1 + - - - - D = 12cm - - - - - - + J
Fig. 6.9
Assumptions: (1) Lumped heat capacity model is valid. (2) Constant properties and heat transfer
coefficient.
hL
Analysis: Biot number, Bi =-c
k
where
RH
=
6cmx8cm
= -----,-,,,-----
3.JR 2 + H 2
3 [62 + 82 ]112 cm
1.6cm
Bi= h(l.6x 10-2 m)
35.3W/mK
As h is not known, the validity of lumped system approximation will be confirmed later.
Fourier number,
Fo = ~: = ( 24.1 10-6 m2 s}(25 x 60 s )/(1.6 x 10-2 m}2 = 141.21
C
The transient temperature distribution is given by
T(t)-T~
e-(Bi)(Fo)
T;-T~
or
(Bi) (Fo)=ln
T;-T~ =In ( 43 - 5)°C =0.642
T(t)-T~
{25-5)°C
Bi= 0 ·642 = 4.545 X 10-3
141.21
(«0.1)
Hence, the lumped capacitance analysis is appropriate.
The heat-transfer coefficient is
h = Bi·k = (4.545 x l0- 3 ) (35.3W/m 0 C)
Le
= 10.0 W /m 2 °C
l.6x10-2 m
(Ans)
6, I 8 ,
Heat and Mass Transfer
Example 6.6 ~ A triangular fin is 25-mm thick and 15-cm long at the base and is 5-cm wide (perpendicular to the
base surface).The fin is made of aluminium (k = 204 W/m °C, p = 2707 kg/m 3, CP = 0.896 kJ/kg 0 C)
and has an average heat-transfer coefficient along the sides and ends of 90 W/m 2 °C. There is no convective heat
transfer at the base. The fin is initially at a temperature of 225°C and is suddenly exposed to ambient air. It takes
6.2 min for the fin to attain 42°C. Determine the ambient air temperature, justifying any specific assumption made.
Known: A triangular aluminium fin is suddenly exposed to convective environment.
Find: Ambient air temperature, T=( 0 C).
Schematic:
Aluminium fin
r = 2707 kg/m3
cP = 896 J/kg 0 c
k = 204W/m 0 c
Initially
T i= 225°C
Assumptions: (I) Internal temperature gradients are neglected. (2) No convective heat loss from the base
of the fin. (3) Constant properties and uniform heat transfer coefficient.
Analysis: Let us first determine the Biot number for lumped system approximation.
Bi = h Lc where Lc is the characteristic dimension defined as Lc = ¥k
Volume,¥-=
(1 ab )L= I(2.5 cm) (5 cm) (15 cm)= 93.75 cm
2
2
Heat-transfer surface area, As =
where
/4
3
(½ab) x 2 + (cL) x 2
c=
J(~Y
+b2 =~1.25 2 +5 2 =5 .154cm
J
As= [½(2.5 x 5)x 2+(5.154 x 15)x 2 cm 2
= 167.12 cm 2
Lc =93.75cc/167.12cm 2 = 0.561 cm
Hence,
Bi= hLc = (90W/m20 C){0.56lxl0-2 m)
k
204W/m°C
= 24.75 xJ0-4
(« 0.1)
Hence, the assumption of lumped system model is justified.
The transient temperature response for this case is given by
T-T
--=-=exp (-Bi Fo)
T;-T=
Transient Heat Conduction
r
6.19
Fourier number or nondimensional time,
Fo =at=_!_.!_
pCP L~
L~
=
204W/m°C
x 6.2x60s = 994
2707kg/m 3 X 896J/kg°C 0.00561 2 m2
Bi Fo = (24.75 x!0-4) (994) = 2.46
T-T- =e-2·46 = 0.0854
T.-Tor
0.0854 J':.-0.0854 T= =T-T=
0.9146 T= = 42 - (0.0854 X 225) = 22.785
or
:. ambient air temperature,
(Ans)
T= = 22.785/0.9146 = 24.9°C
Example 6.7 ~ An egg with a mean diameter of 40 mm and initially at 20°C is placed in a boiling water pan for
4 minutes and found to be boiled to the consumer's taste. For how long should a similar egg for
the same consumer be boiled when taken from a refrigerator at 5°C? Take the following properties for the egg:
k = 10 W/m °C
p = 1200 kg/m 3
c = 2 kJ/kg °C
h = 100 W/m 2 °C
Known: An egg is boiled to the consumer's taste from an initial temperature.
Find: Time required, t (s).
Schematic:
t=4 min when
T;=20°C
t=? when
T;=S°C
_- - - -_- -_-_-_-_-_-_-_- ~e--------C+----
---=-=- -~
-=:=:=:~ ~
= ·-=-=-=-= --=:=:=:=:=~-
Boiling water
(T~ = I00°C)
- D=4Qmm
Fig. 6.11
Assumptions: (I) Constant properties and uniform heat-transfer coefficient. (2) Lumped capacity model
is valid.
Analysis: Biot number,
hL
Bi = - c
k
.
where
(100W/m2 oc{ 40x~0-3m)
B1 = - - - - - ~ - - ~ = 0.0667
10W/m°C
(< 0.1)
6.20 , Heat and Mass Transfer
Therefore, lumped parameter analysis is justified and internal temperature gradients are neglected.
Time required,
t= pcLc In~
h
e
0i =J; - T= = 5 - I 00
where
0 = T -T= = T- I 00
Tis the temperature to which the egg should be boiled to the consumer's taste.
t= 4 min= 240 s when T, = 20°C
To find the temperature T after 240 s, we use the following expression
[-hAt]
T(t)-T= =exp - T;-T=
pc¥-
or
= IOOoC + (20 - IOO)oC exp [-
100 W/m2 oc x 240 s x 6
]
1200 kg/m3 X 2 X 103 J/kg°C X 0.04 m
= 82.15°C
When Ti= 5°C,
Time required,
and
T(t) = 82.15°C,
t = 1200kg/m3 x 2000J/kg°C x (0.04/6) m In{
100 W/m2 °C
(5-100)°C }
(82.15-100)°C
= 267.5 s or 4 min 27.5 s
Example 6.8
(Ans)
~ A solar energy storage bin contains 1786 rocks that can be idealized as 75-mm diameter
spheres. Thermophysical properties of the rock are k = 3.81 W/m °C, CP = 837 J/kg °C and
p = 2560 kg/m 3. How long would it take to raise the temperature of rocks to 94°C using 95°C air from a solar
collector? How much energy is stored in the bin under these conditions if the bin is initially at 23°C and the
average heat-transfer coefficient is 29 W/m 2 °C?
Known: A rock-bed solar energy storage system has rocks heated by air from solar collector.
Find: Time required, t (s). Energy stored, f:.£81 (J).
Schematic:
Air
h=29W/m2°C
r==95°c
Rock
(0=75mm)
k=3.81W/m °C
Cp= 837 J/kg °C
p = 2560 kg/m 3
Fig. 6.12
Transient Heat Conduction
r
6.21
Assumptions: (1) Lumped parameter analysis is applicable. (2) Constant properties and uniform
heat-transfer coefficient.
The characteristic length of a sphere is D/6.
Analysis: The Biot number is
Bi= hLc = (29W/m2 0 C)(0.075/6m)
k
3.81W/m°C
(< 0.1)
= 0.095
Hence, the lumped capacitance method is reasonable.
The time constant is
p ¥-CP p CpLc p CPD
't=--=--=-hA.
h
6h
't
(2560 kg/m3 )(837 J/kg 0 C) (0.075/ 6m)
29W/m2 °C
= 923.6 s
From the transient temperature response and solving for t, we have
t ='tln(T;-T
00
T-T
00
)
= 923.6 s /n[ 23 - 95 ]
94-95
= 3950 s or 1.1 h
(Ans)
From first law for a closed system:
Q-W=AE
energy stored in the rock = Heat entering the system
Work interaction = 0
or
p¥-Cp(T-T;)=Qin
rcD3
AES! =Qin =pCP-6-(r-i;)
= (2560 kg/m3) (0.837 kJ/kg 0 C) ( ~x0.075m
J
X (94 - 23) °C
=33.6 kJ
The storage bin contains 1786 rocks.
Hence, the total energy stored in the bin is
AEst,total = (1786) (33.6 kJ)
= 60 x 103 kJ or 60 MJ
(Ans)
6, 2 2 ,
Heat and Mass Transfer
Example 6. 9
~ An aluminium sphere weighing 5.5 kg, initially at 300°C is immersed in a fluid at I0°C where
the film heat-transfer coefficient is 60 W/m 2 0 C. Calculate
(a) the time required for the sphere to reach 20°C
(b) the temperature of the sphere after 2.2 minutes
(c) the instantaneous rate of heat flow after 1.5 minutes
(d) the heat transferred between third and fourth minute
For aluminium: density= 2700 kg/m 3, specific heat= 0.9 kJ/kg K and thermal conductivity= 200 W/m K.
Known: An aluminium sphere is immersed in a fluid under specified conditions.
Find: (a) Time required, t (s) for T(t) = 20°C
(b) Temperature, T(t) after 2.2 min
(c) Instantaneous heat-transfer rate, Q(kW) after 1.5 min
(d) Heat transferred, Q [between 3<d and 4th min] (kJ)
Schematic:
Fluid
T(t)=2Q °C oh=60W/m 2 °C
Tj = 300 °C
T= = I0 °C
Aluminium:
p = 2700 kg/m 3
Cp= 0.9 kJ/kg K
k=200W/mK
Sphere (m = 5.5 kg)
Fig. 6.13
Assumptions: (1) Constant properties. (2) Lumped capacity model is valid so that T= f(t). (3) Uniform
heat transfer coefficient.
Analysis: (a) Biot number,
Bi=
hLC
k
h(¥-/ A) =!!._(rcD 3/6)= hD
k
k rcD 2
6k
To find diameter D, we note that
¥-= rcD3 = m =
5.5kg
= 2.037 xI0-3 m3
6
p 2700kg / m3
(6¥-)l/J =0.1573m
and
D=-;-
Hence,
Bi= (60W/m2oC)(0.1573m) =0.007 86
6(200W/m K)
As Bi <<< 0.1, the lumped parameter analysis is justified.
Time required,
t=pCpLc /n~=pCPD zn( T;-T=)
h
0
6h
T-Too
= (2700kg/m3)(0.9 x 103J/kg K)(0.1573m) x Zn (300- 10)°C
6(60W/m 2 K)
(20-10)°C
= 3575 s or 59.6 min
(Ans) (a)
Transient Heat Conduction
r
6.23
(b) Temperature of the sphere after 2.2 minutes is
T (t) = T= + (J; -T=)exp[-__!:!__]
pCpLc
= 10°C + (300 - 10)°C exp [
60W/m2 Kx{2.2x60s)x6
]
2700kg/m3 x900J/kgKx0.1573m
=266°C
(Ans) (b)
(c) Instantaneous heat-transfer rate at t = 1.5 x 60 s is
Q=h A (T-T=)=h A (J;-T=) e-hAtlpCP¥-
=h(1tD2)(J;-T=)exp[- p~~)
=(60W/m2 K)(1tx0.15732 m2) (300-10)°C
6x60W/m2 Kxl.5x60s
]
xexp [ - - - - - - - - - - - - 2700 kg/m3 x900 J/kgKx0.1573 m
= 1242.6 W
(Ans) (c)
(d) Heat transferred between third and fourth minute= Q (240 s) - Q (180 s)
After 4 minutes
n.
>q240s) =p¥-Cpe.I exp[1-e-hAl/p¥CPJ
= mCp(J; -T=)exp[l-exp(-ht /pCPLc)]
= (5.5 kg) (0.9 kJ/kg K) (300 x 10) K
x exp [ 1-exp {
}]
60W/m2 Kx60x4s
2700kg/m3 x900J/kgKx0.1573/6m
= 290.4 kJ
After 3 minutes
~isosJ = (5.5 kg) (0.9 kJ/kg K) (290 K)
x exp
[l
{
60W/m2 Kx60x3s
}]
-exp - 2700kg/m3 x900J /kgK x0.1573/6m
=223.85 kJ
(h_4min] = 290.4 - 223.85
= 66.55 kJ
(Ans) (d)
6.24 , Heat and Mass Transfer
Example 6.10 ~ During a manufacturing process, two brass plates are to be joined face to face at a bonding
temperature of 500 K. The dimensions of the plates are 0.6 cm by 5 cm by IO cm and their
initial temperature is 300 K. For bonding the plates have to be kept in an oven and the bonding time should
not exceed 5 min. The mean convection heat transfer coefficient is 80 W/m 2 0 C. Determine the minimum
temperature of the oven if the bonded faces are in complete thermal contact. Properties of brass to be used are
p = 8530 kg/m 3
K= 137 W/m °C
cp = 395 J/kg c
0
Known: Two rectangular brass plates are bonded together by keeping them in an oven for a specified time.
Find: Oven temperature, T= (K).
Schematic:
1
l
T(t)=500K
t=5min
10cm
Brass plates
p = 8530 kg/m3
Cp=395J/kg °C
Oven
h=80W/m2°C
T= =1.
<,c,<'<'
I···
,1 /
0.6 0.6
cm cm
Fig. 6.14
Assumptions: ( 1) The plates' temperature is space-wise uniform. (2) Constant properties. (3) Negligible
radiation effects.
Analysis: Let us first evaluate the Biot number to find if the lumped capacitance method can be applied
in this case.
hL
Bi= __c where Le
k
¥-
A,.
Characteristic length,
{(2 x 0.6) x 5 x 10}cm3
[(2 x 10 x 5)+(2 x 1.2 x 10)+(2 x 1.2 x 5)] cm2
= 0.44 cm
Hence,
or
0.0044 m
Bi (86W/m 2 0 C) (0.0044m)
137W/m °C
= 0.00257
Transient Heat Conduction
r
6.25
As Bi << 0.1, the lumped capacity method is applicable. Then
T-T= =e-(Bi)(Fo) = A(say)
T;-T=
T - T= =AT; - AT=
or
Therefore, the oven temperature is
T-AT;
1-A
Now, Fourier number,
Fo = _!_.!__=
(137W/m 0 C){5 x 60s)
pCP L~ (8530kg/m3 )(395J/kg 0 C)(0.0044m) 2
= 627.2
A= exp (-Bi Fo)
= exp [- (0.00257) (627.2)] = 0.2
It follows that the oven temperature is
T = 500K-(0.2)(300)K = 550 K
=
Example 6.11
(Ans)
1-0.2
~ A thin shell made of aluminium (k = 237 W/m °C, a= 97.1 x I0- 6 m2/s) of 8-mm diameter and
0.4-mm thickness falls off a conveyor vertically down to the ground. The shell temperature
is initially at 95°C throughout. The uniform convection coefficient is 80 W/m 2 °C and the ambient air is 25°C
as the shell gets cooled during the IS m distance covered by it. Estimate the temperature finally reached when
the shell eventually hits the ground.
Known: A thin steel shell drops off to the ground and gets cooled by ambient air in the process.
Find: Final temperature of the shell, T(t) (0 C).
Schematic:
D = 8 mm
o=0.4mm
Initially
Distance,
Air
s= I Sm
T;=95 °C
Mild steel :
g
h=80W/m2°C
Ill r==2s c
0
Ground
Fig. 6.IS
k = 237W/m °C,
a=97.I x I0-6m2/s
6.26 , Heat and Mass Transfer
Assumptions: (1) Lumped system formulation is acceptable. (2) Constant thermal properties and heat
transfer coefficient.
Analysis: Biot number,
¥- <> 1.e.,
.
. h"1ckness = O.2 mm
Bi= -hLC where -=-,
sem1-t
k
A 2
Bi= (80 W/m2 °C) (0.2 X 10-3 m)
237W/m°C
(«< 0.1)
= 67.5x1Q-6
Hence, the lumped capacity model is appropriate.
The transient temperature response is
where
As Le= 0.2 mm,
RC=
1 1
237W/m°C
(0.2x10-3 m)
(97.lx10-6m2 /s) 80W/m2 °C
= 6.1 s
For vertical motion of shell falling under gravity from rest, we have
1
s= 0 +-gt2
2
where g is gravitational acceleration= 9.81
.
time, t =
m/s 2
2x15m
{Ii"
= 1.75
"Jg=
9.81m/s
I----
S
2
:. temperature of the shell after it hits the ground,
T(t) = T= +(I; -T=)e-tlR.C,
= 25°C + (95 - 25)°C e-l.?Ss/ 6 ,ls
= 77.5°C
(Ans)
Transient Heat Conduction
r
6.27
Example 6.12 ~ Consider a container of fluid with a uniform initial temperature T; exposed to an environment
at T_(T_ < T;) as shown in Figure 6.24. Draw the thermal circuit and determine the
temperature-time history of the fluid in this two-lump heat-capacity system.
Container wall
Fluid
CD
Fig. 6.16
The thermal capacity of the container fluid is Pi'\f'Cp, and that of the container walls is Pf"'2C"'. Show that the
final solution for T1(t), in dimensionless form, can be expressed as
7j ( t) - T'f; - r_
"'2 e"'1' - ~ emit
"'2 - 111i
m2 - ""i
Known: A fluid is contained in a composite system exposed to surrounding medium as shown in the
schematic.
Find: Temperature-time history of fluid. Thermal circuit. Temperature variation in non-dimensional form.
Schematic:
Container wall
Initially
T;
Surrounding
medium T=
Fluid
Fig. 6.17
Assumptions: ( 1) Lumped capacity formulation is applicable so that internal thermal gradients are
negligible. (2) Constant properties and uniform convection coefficients.
Analysis: Writing the energy balance equations for lump 1 (fluid) and lump 2 (container), we have
Lump 1 (fluid): Thermal energy leaving each system equals rate of decrease of stored (internal) thermal
energy
(2)
6.28 , Heat and Mass Transfer
These two linear differential equations for T 1(t) and Ti(t) have to be solved simultaneously to determine
the temperature-time history of the container fluid.
The two initial conditions are
T 1(t= 0) = Ti(t= 0) = 1j
Rearranging equations (1) and (2), one gets
dJ; _
d 1j + ~ (1j - ½) =0
dt P1-Jte'p,
(3)
h 1A 1
h 2A 2
---(1j-J;)+--(J;-T )=0
P2~CP2
P2~CP2
(4)
00
dt
h1A1 /p 1-Jte'Pt =!Ji, h1A1 /p 2 ~CP, =~
Let
and
hi.Ai.lP2~CP2 =h:i
Equation (3) can be rewritten as
or
(5)
Differentiating with respect to t,
dJ;
dt
l d 21j
h, dt2
d1j
dt
-=-+--
(6)
From Equation (4)
dJ;
dt
=~ 1j + b3T
00 - (
~ + b3 ) J;
(7)
Equating (6) and (7), and substituting for T2 from (5),
_!_ d 21j + d1j =b T. +b T -(1, +b
bI dt2
dt
2 I
3 00
"2
3
)(T. +_!_b d1j)
dt
I
I
or
d 21j d1j
- + - ( b1 +b2 +b3 )+b11j(b2 +b3 -~)= b1b3T
dt2
dt
or
d 21j
d1j
dt2 +(b,+~+h:J) dt +b,h:J1j =h,h:JToo
00
{ D 2 + D(b 1 + ~ + h:J)+b 1h:J} 1j(t)=b 1h:J T
00
where
d
D=dt
and
d2
D2=dt 2
Transient Heat Conduction
r
6.29
The general solution of the above equation involving only T1 is
!Ti(t)=Too +C,e"'t1 +C2e"'21 I
(8)
where m 1 and m2 are given by
and
The arbitrary constants C1 and C2 can be evaluated from the two initial conditions.
At t = 0, T 1(t) = Ti = constant
At t=O dJ;(t)=O
' dt
It follows that
(9)
and
or
(10)
From Equation (9),
From Equation (10),
Hence,
or
and
After substituting the values of C1 and C2 in Equation (8), the temperature-time history can be expressed
in the final form as
6.30 , Heat and Mass Transfer
In dimensionless form, one can write
Hence proved.
The thermal circuit is shown below:
1
Fig. 6.18
where the thermal resistances are defined as
= -1
-R
R
t, -
h,A,
h2A2
t,
and, the thermal capacitances are defined as
ct =p,-JtCP ct =P2-JLiCP
I
I
2
2
From the electrical analogy shown above, when the switch S is closed, the two thermal capacitances
Ct I and Ct 2 are charged to the potential Ti . At time t = 0, the switch is opened and the capacitances
discharge through the two thermal resistances Ri I and Ri2 .
Example 6.13 ~ A plane wall, I 0-mm thick, is fabricated from plain carbon steel [p = 7850 kg/m 3,
CP = 0.43 kJ/kg °C, k = 60 W/m 0 C) with an initial uniform temperature of I 5°C. One side
of the wall is exposed to furnace gases at I000°C with a convection coefficient of I00 W/m 2 °C, while the
other surface is in contact with water at 30°C with a heat-transfer coefficient of 25 W/m 2 0 C. How long will
it take for the wall temperature to reach 720°C! Derive the expression you use.
Known: A plane wall is exposed to two different convective environments on the two sides.
Find: Time required t (s) to attain a specified temperature, T(t).
Schematic:
Gases
r-
L= 10mm
~
Initially
l l l
h 1= IOOW/m2°C
T;=T(O)
= l5°C
T(t)=720°C
T= I = 1000°C
h2 = 25 W/m2 °c
T=2=30°C
lll
Water
Fig. 6.19
Transient Heat Conduction
r
6.31
Assumptions: (1) Lumped capacity model is valid. (Wall is spatially isothermal). (2) Constant properties
and uniform heat transfer coefficients on both sides of the wall.
Analysis: The Biot number,
. h4
.
AL L
Bz = where Le = half-thickness = - = - = 5 mm
k
2A
2
As there are two different convection coefficients h 1 and h2 with different fluid temperatures T~ I and
T~,
let us take the larger value of h to find it Bi < 0.1 so that lumped parameter analysis is applicable.
2
With h 1 = 100 W/m2 °C,
Bi= (100W/m2 0 C) (0.005m)
60W/m°C
= 8.33 X lQ-3
(« 0.1)
:. the lumped capacity model is valid.
Applying the energy balance:
1- tJ
Eout +
=Est
.
dT
Qconv,out =-mep -d t
or
or
h1A[T(t)-T~J +h2 A[T(t)-T~J =-p¥-CP : :
( h, + h2 )r(t)-( h,r~, + hi r~,) =
or
-p:
cp ::
¥- AL
-=-=L= 10 mm
But
A
Let
A
h1 + h2 = a
And,
h1T~I +h2 T~2 = b
dT
a T(t)-b=-pLC p dt
or
a [r(t)-(~)]=-pLCP : :
1
or
pLC T(t)
i;
0
or
dT
Jdt---a-P J -T(-t)---(b-/a-)
pLC
t=---P ln[T(t)-(b/a)fi;Ctl
a
pLCP
T, -(bla)
t=--/n-'-1- - - - - ' - - I
a
T(t)-(b/a)
,
6.32 , Heat and Mass Transfer
From the problem statement,
a=h 1 + h2 = 100 + 25 = 125 W/m 2 °C
and
=(100W / m20C) (l000°C)+(25W / m20C) (30°C)
= 100750 W/m 2
bl a= 100750W/ m2 =S0 60C
125W/ m2 °C
time needed,
t = (7850kg/m3 ) (O.Olm)(430J/kg°C) £ { (15-806)°C }
125W/m 2 °C
n (720-806)°C
= 599.2 s = 10 min
(Ans)
Example 6.14 ~ A copper-constantan thermocouple junction which may be approximated as sphere of 3-mm
diameter is used to measure the temperature of an air stream flowing with a velocity of
3 mis. The initial temperature of the junction and air are at a temperature of 25°C. The air temperature
suddenly changes to and is maintained at 200°C.
(a) Calculate the time required for the thermocouple to indicate a temperature of I S0°C. Also, find the thermal
time constant and the temperature indicated by the thermocouple at that instant.
(b) Discuss the suitability of this thermocouple to measure unsteady state temperature of a fluid when the
temperature variation in the fluid requires a time period of 3 s.
Assume:
h = I SO W/m 2 °C
cP = 377 J/kg c
0
p = 8685 kg/m 3
k = 29 W/m 0 c
Known: A thermocouple junction is used to measure temperature of an air stream.
Find: (a) Time required to reach 150°C in air stream at 200°C, and thermal time constant. (b) Suitability
of thermocouple for time period of 3 s.
Schematic:
Leads
Air
T==200°C
h= ISOW/m2°c-
D=3mm
k=29W/m°C
p = 8685 kg/m3
cP = 377 Jtkg c
0
Thermocouple
junction
Fig. 6.20
Transient Heat Conduction
r
6.ll
Assumptions: (1) Lumped capacity formulation is appropriate. (2) Conduction heat loss through leads is
negligible. (3) Radiation effects are ignored. (4) Constant properties.
Analysis: Characteristic length for a sphere of diameter D is
Biot number,
Bi= hLc = hD = (150 W/m2 0 C)(3 x 10-3 m)
k
6k
(6)(29W/m 0 C)
=2.6x 10-3, i.e., less than 0.1.
Energy balance
Ein-
i
i
+
no heat loss
=Est
no heat generation
.
dT
. =mCp -dt
Qconv,m
or
_ p C ¥--d(_T_-_T-_)
p
dt
-
- ~ d t d(T-T-)
pCP¥T-T-
or
Integrating between appropriate limits,
:. time required,
-pCPD (T-T )
t = - - - I n ---6h
I;-T= _ (8685kg/m3 )(3 X lQ-3 m) /n (150-200)°C
(6)(150W/m20C)
(25-200)°C
= 13.67 s
(Ans)
6.34 , Heat and Mass Transfer
Thermal time constant,
pCP¥- pCPD
't=--=--
hA.
6h
0
= (8685kg/m )(377 J/kg C) (3 X 10- m)
3
3
-'-----='----'---'-------=--'--'------'-
(6) (150W/m20C)
= 10.9 s
Now,
(Ans) (a)
_T_(t_)_-_T__ exp[--h_A_._t ]=e-tlt = e-1
I;-T_
pCP¥-
(ift ='t)
Temperature indicated by the thermometer at one time constant, t = 't is
T(t) = T_ +[ ( I;-T_) ·e-1]
= 200°c + [(25 - 200)°C (0.36788)]
= 135.6°C
(Ans) (a)
As the time period required for unsteady state temperature variation is only 3 s as against 10.9 s which
is the thermal time constant, the suitability of this thermocouple for accurate temperature measurement
is questionable.
(Ans) (b)
Example 6.1 S ~ A copper wire, 0.8-mm diameter and 50-mm long, is placed in an air stream whose
temperature increases with time according tor_= (10 + 5 t)°C where tis the time in seconds.
If the initial uniform temperature of the wire is I0°C, determine the transient temperature response of the wire.
The convection heat-transfer coefficient between the air and the wire is 55 W/m 2 0 C. Properties of copper:
k = 401 W/m °C
p = 8933 kg/m 2
CP = 0.385 kJ/kg °C
Known: A copper wire is exposed to air stream whose temperature rises with time.
Find: Temperature-time history.
Schematic:
Air
r_= (10+ St) ·c
h=40W/m2°C
L Copper wire
Initially
T;= 10°c
~---------~'Tl D=0.8mm
,__ _ _ L=SOmm
Copper: p=8933kg/m3, Cp=385J/kg°C and k=401 W/m°C.
Fig. 6.21
Assumptions: (1) Internal thermal resistance of copper wire is neglected. (2) Constant wire properties
and uniform convection coefficient.
Analysis: As T_ is also a function of time, one must develop the expression for transient temperature
response of the wire by going back to basics.
Energy balance: Rate of increase of stored energy, Est
= Rate of heat transfer by convection from air to wire, Qin
Transient Heat Conduction
r
6.35
dT
mCP dt=hA. (r= -T)
dT
p ¥-CP dt=h A8 [10 + 5 t - T]
or
dT = ~ (10+5t-T)
dt p¥-CP
or
~=b
p¥-CP
Let
dT
-+bT =b (10+5 t)
dt
Let
(1)
(2)
Differentiating with respect to t,
dT
-=C -Cbe-bt
dt
I
2
Then,
dT
-+bT=C
-C2 be-bt +bC0 +bCt+bC
e-bt
I
I
2
dt
(3)
Comparing equations (1) and (3)
C1 +bC0 +bCif = b (10 + 5 t)
C1 +bC0 =10b
and
10 b + bC1t = 10 b + 5 b t
or
1c, =51
and
~ ic =10-¾I
C0 =¾(lOb-5)
0
Substituting for C0 and C 1 in Equation (2),
T(t) = 10-~+5t+C e-bt
b
Initial condition: At t = 0, T = Ti = 10°C
The temperature-time history can then be expressed as
Ir=
10 + 5t+ ¾(e-b 1 - 1)1
2
6.36 , Heat and Mass Transfer
b =~=
where
h(rcDL)
pCP ( ~
p¥-CP
DL)
2
4h
4x55W/m2 °C
p CPD
(8933 kg/m 3 ){385 J/kg°C)( 0.0008 m)
= 0.08 s- 1
5
T = l O+ 5 t + - [ e-o.ost -1]
0.08
Ir= 10 +st+ 62.s{e-O.os, -1}1
or
The temperature- time plot is shown below:
250
230
T (t)
210
190
170
150
fJ
130
i='
110
/
90
70
/
50
30
10
0
----r
/
V
/
/
/
/
\/V
/
/
/
/
10
20
30
40
50
60
t (s)
Fig. 6.22
Example 6.16 ~ A potato ( ¥ = 3.1 x I Q--4 ml, As= 0.025 m2), initially at a uniform temperature of 20°C, is
placed in a microwave oven. The oven supplies 300 W of heat to the potato for 7.5 minutes.
The temperature of the air in the oven is 200°C, and the convection heat transfer coefficient is 4 W/m 2 K.
Determine (a) the temperature of the potato after 7.5 minutes of heating, and (b) the rate of heat transfer
from the air to the potato in the oven at that time.
The thermophysical properties of the potato are
p = 1055 kg/ ml
Derive the formula you use.
k = 0.498 W/m K
CP = 3.64 kJ / kg K
Transient Heat Conduction
r
6.37
Known: A potato is heated in a microwave oven for 7 .5 minutes.
Find: (a) Temperature of potato T (°C) after t= 7.5 min. (b) Rate of heat exchange between air and
potato, Q(W).
Schematic:
Air
T;=20 °C
h=4W/m2K
T =200°C
--....J"\..f\-- Qgen = 300 W
00
A5 =0.025m2
k = 0.498 W/m K
p = I 055 kg/m 3
CP = 3640 J/kg K
T(t) = 1
-\l=3.lxlQ-4m3
t=7.5min
Fig. 6.23
Assumptions: ( l) Constant properties of potato. (2) Uniform heat transfer coefficient. (3) Lumped capacity
model is applicable.
Analysis: Energy balance:
.
dT
Qconv +q As =mCP dt
or
dT
h As(Too-T)+q As=pfLCPdt
or
dT + ~ ( T -T) = ~
dt pfLCP
pfLCP
or
OO
T-Too =0,
Let
Also, let
and
where the characteristic length, Le = ¥As
dT
d0
dt
dt
7;-Too=0;
(A)
6.38 , Heat and Mass Transfer
Substituting these values in Equation (A), one has
~
~
Eliminating the non-homogeneity by introducing the transformation
0' =0-~
a
d 9' = d9
dt dt
and
0=~+0'
a
(b )
d0'
-+a
-+0' =b
dt
a
d0'
-+b+a0' -b = 0
dt
or
Separating the variables and integrating it, one gets
9' d0'
t
e;
o
J-,
=-a Jdt
8
/n 0' -/n0; =-at
or
or
/n (
f:)
=-at
!0' =0; exp(-at) I
or
0 - (b/a) = (0i -b/a)exp(-at)
or
0 = 0i exp(-at)+- {1 - exp (-at)}
a
b
where 0i =I;-T= = 20 - 200 = -180°C
hA
a= hlp4c = - -8p
p¥-CP
(4 W/m2 K) (0.025m2
>( !~)
= - - - - - - - - 3- - - - 3- - (1055kg/m3)(3.lx lQ-4 m )(3.64x 10 J/kgK)
=3000 Kor °C
Transient Heat Conduction
r
6.39
e-at =exp[-(84x I0-6 s- 1) (7 .5 x 60s)]
= 0.9629
0=T(t)-T= = (-180°C) (0.9629) + (3000°C) (1 - 0.9629)
=- 62°C
Temperature after 7.5 min is
T= 200 - 62 = 138°C
(Ans) (a)
Rate of heat transfer at that time from air to potato is
Q=h As (T= -T(t)) =(4 W/m 2 K)(0.025m 2 )(200 - 138)°C
(Ans) (b)
=6.2 W
Example 6.17 ~ Toy cars made of steel sheet (Cp = 0.46 kJ/kg 0 C) with a mass of 0.12 kg and a total surface
area of 0.06 m2 are immersed in a molten plastic bath maintained at a temperature of 200°C.
The plastic layer on the cars has an average thickness of 0.3 mm, a thermal conductivity of 0.2 W/m °C and
sets at I2S 0 C. The ambient air temperature is I S°C. Neglecting internal temperature gradients, estimate the
time to be allowed for the plastic coat to set after the cars are withdrawn from the both. The free convection
heat transfer coefficient between the car and the surrounding air is given by the following approximate relation:
1/4
h =1.3 ( d T )
0.1
(W/m2oc)
Known: Toy cars removed from a molten both of plastic are allowed to cool in air.
Find: Time required for plastic coat to set on the car.
Schematic:
Quiescent air
Plastic coated toy car
(m =0.12kg, A =0.06m2)
Plastic coating
o=0.3mm
k=0.2W/m°C
T(t=O) = T;=200°C
0
0
.
T(t)
QIA - - --J""'~---~""
otkp
1th
T(t=t)=T= 125°C
--QIA
Fig. 6.24
Assumptions: ( 1) Lumped heat capacity analysis is valid. (2) Mass of plastic car and radiation effects
are neglected.
Analysis: Energy balance:
6.40 , Heat and Mass Transfer
Heat flux,
where total thermal resistance, ~otal comprises conduction resistance offered by plastic coat and convection
resistance. Conduction resistance due to steel sheet is neglected.
8
coating
= -Thickness
- - - -of~plastic
--~kp Plastic's thermal conductivity
0.3 x 10-3 m
0.2W/m°C
= 1.5xl0-3 m2 °C/W
This can be neglected compared to l/h.
It follows that U"" h.
Also since T,,. is constant, T(t)-T,,. =0=AT
dT(t) = d0
d0
dt
and
d0
90.25
mCP-=-hA0=-l.3
025 A0
dt
(0.0·
= -2.312 A 0'-25
Separating the variables and integrating,
9
I
Afdt
J0'-d025 =- 2.312
mC
9;
P
0
where
0 = T-T,,. = 125-15 = l 10°C
and
Therefore,
[
9-1.25+1 ]" 0 = 2.312 X 0.06
-----xt
0.12 X 460
-o.2 5 185
Time required is then determined to be
t = -4 X O.12 X 460 X [ 110-0.25 - 185-0.25]
-2.312 X 0.06
= 60 s or 1 min
(Ans)
Example 6.18 ~ A bronze {(75% copper (Cu), 25% tin (Sn)} bearing blank consists of half of a right circular
cylinder as shown in Figure 6.39. The blank is removed from a heat treatment furnace at
480°C and is cooled in 25°C air to 200°C (in a vertical position) for a forging operation prior to finish machining.
How long does it take to cool the sleeve if the heat transfer coefficient in the vertical position is 70 W/m 2 Kl
Assume: k = 31W/m K
p = 8670 kg/m 3
CP = 0.343 kJ/kg K
Transient Heat Conduction
r
6.41
Fig. 6.25
Known: A bronze blank comprising half of right circular cylinder is cooled in a vertical position.
Find: Time required.
Schematic:
I
r0 = 2 (7.S) = 3.75cm
{
r;=r0 -t=2.Scm
1
L=IScm
Initially
T(t=O)
T;=480°C
l
. /itf
Air
T~ =2S °C
h=70W/m2K
,.
/1
"'
Bronze sleeve
T(t)=200°C
Fig. 6.26
Assumptions: (I) Lumped capacitance formulation is applicable (i.e., the blank is space-wise isothermal).
(2) Constant properties. (3) Uniform heat transfer coefficient.
Analysis: In the vertical position the surface area for heat transfer comprises the inner-and outer-half
cylindrical surfaces, the edge areas and the top half-washer-like surface. Heat transfer to the air from the
bottom surface can be neglected.
In order to ensure that lumped capacitance method is applicable, the Biot number (Bi) should be less
than 0.1. To calculate the Biot number, one must know the characteristic length, Lc.
hL
¥Bi = __c where Lc =
k
As
Area (inside cylindrical surface)= 1t,;L = (7t) (2 .5 cm) (15 cm)= 37.5 7t cm 2
6.42 , Heat and Mass Transfer
Area (outside cylindrical surface)= rer;,L = (re) (3 .75 cm) (15 cm)= 56.25 re cm 2
Half-washer area = ½re[ (3.75}2-(2.5}2] cm 2 = 3.9 re cm 2
Edge area= 2(1.25 cm) (15 cm)= 37.5 cm 2
Total surface area, As= 37.5 + re [37.5 + 56.25 + 3.9] = 344.28 cm 2
Volume= base area x height
:. ¥-=(3.9re)cm 2 (15 cm)= 184 cm 3
¥184cm3
Characteristic length= Le=-=----= 0.5347 cm
As 344.28cm 2
Bi= hLc = (70W/m2 K)(0.5347xI0- 2 m) =O.Ol 2
k
31W/mK
The Biot number is less than 0.1. Hence, internal temperature gradients can be neglected and the
lumped capacitance model is valid.
Thus,
(200-25)°C
[
(70W/m2 K)(s)
]
3
2
(480-25)°C exp (8670kg/m )(343J/kgK)(0.5347xI0- m)
or
or
0.3846 = exp [-0.0044 t]
or
ln 0.3846 = -0.0044 t
Time required to cool the sleeve,
t= -0.9555/ (-0.0044 s- 1) = 217 s or 3.62 min
(Ans)
Example 6.19 ~ A steel tube of 20-cm length with internal and external diameters of IO and 12 cm is
quenched from S00°C to 30°C in a large reservoir of water at 10°C. Below 100°C, the
heat transfer coefficient is I .S kW/m 2 K. Above I 00°C, the average value of heat transfer coefficient may be
taken as O.S k W/m2 K. The density of steel is 7800 kg/m 3 and the specific heat is 0.47 kJ/kg K. Neglecting
internal thermal resistance of the steel tube, determine the quenching time.
Known: A steel tube of given dimensions is quenched in water. The convection coefficients are different
in the two stages of the cooling process.
Find: Quenching time, t (s).
Schematic:
Water
r.. = 10°c
Steel:
p = 7800 kg/m3
Cp= 0.47 kJ/kg K
(J
f.---
L=20cm
Fig. 6.27
Transient Heat Conduction
r
6.43
Assumptions: (1) Temperature of the tube is spatially uniform. (2) Constant properties. (3) Negligible
radiation effects.
Analysis: Internal thermal resistance of the tube is neglected. Hence, the tube surface temperature will be
same at every location at any instant, i.e., T = f(t). Lumped capacity model is valid.
h
. .
h ,
Volume of tube,¥. .ft
S1gm
1cant or c aractenst1c 1engt , -'t = - - - - - - - Surface area of tube, A.
2 ~ D2 JL = ~(D 2 -D2 )L
¥- = [~
4D
0-4
1
40
I
= ~(122 -10 2 )(20) = 691.15 cm3
4
A = Curved surface area + End surface area
1t
= 1t(D +D;)L+ 2 (D +D;)(D -D;)
0
0
0
1
= 1t(D +D;)[L+ 2(D -D;)]
0
0
= 1t (12 + 10) [20 + (12 - 10)/2] = 1451.42 cm2
4 = ¥- = 691.15cm3 = 0.4762 cm
4,
1451.42cm2
Cooling process during quenching of the tube in a pool of water takes place in two stages due to two
different convection coefficients.
I. For cooling from 500°C to 100°C
T(t,)-T~
'Ii., -T~
e-"iAtifp¥-CP
= exp[-_.!!t._]
pCP4
__!!_n_ =- In (T(t,)-T~) where 'Ji.,= T(t=O)
or
pCpLc
7i.-T~
or
Substituting the appropriate numerical values in the above expression, one gets
t = (7800kg/m3)(0.47xl03J/kgK)(0.4762xl0-2 m) 11Wlx/n{500-10°C}
1
0.5x103 W/m2 K
lJ/s
100-10°C
=59.17s
6.44 , Heat and Mass Transfer
II. For cooling from 100°C to 30°C
CpLc
( J; - T~
t =p- - In -~'- -
Time required,
hi
2
Initial temperature now,
J
T(t2)-T~
z; =T(t1) = 100°c as T(t= O) = 100°c
2
t _ (7800)(0.47x!0 3)(0.4762x 10-2 )
2 1.5 x I 03
/n (100-10)
30 - I 0
= 17.50 s
Hence, total quenching time required
t= t1 +t2 =59.17+ 17.50=76.67s-a=77s
(Ans)
Comment: Temperature averaged heat transfer coefficient,
h = h/:.1;_ + h/1T2
t:.T
0.5 X (500-100) + 1.5 X (100-30)
500-30
= 0.6489 kW /m 2 K
hL
Biot number, Bi=-:= (0.6489x!03)(0.4762x!0-2 )/41 =0.0754 (< 0.1)
[Assuming ksteel = 41 W!m K]
Hence, our assumption of negligible internal thermal resistance is justified.
Example 6.20 ~ An electronic device, which generates 60 W of heat, is mounted on an aluminium finned
heat sink having a mass of 0.32 kg and specific heat of 935 J/kg 0 C. Under steady state
conditions, it attains a temperature of I I0°C in ambient air at 30°C. If the device is initially in equilibrium with
the environment, calculate its temperature 7 minutes after the power is switched on.
Derive the expression you use. The device and the heat sink may be assumed to be nearly isothermal.
Known: An electronic device mounted on a heat sink is energized after being initially at ambient air
temperature.
Find: Temperature of the device after 7 min.
Schematic:
Ambient
air
T(=)= I l0°C
T;=T~=30°C
m=0.32kg
Cp = 935 J/kg °C
T(t= 7 min)=?
Electronic device
mounted on a heat sink
Fig. 6.28
Transient Heat Conduction
r
6.45
Assumptions: ( 1) Internal temperature gradients can be neglected because the device is spatially isothermal,
i.e., T= T(t). (2) The device is primarily aluminium. (3) Initially the device is in equilibrium with ambient air.
Analysis: Control volume energy balance:
.
.
dT
dt
-Q,,ut conv + Egen =m CP -
or
'
dT
.
or
Egen =mCPdt+hA 8
[
T(t)-T=
]
(1)
Under steady-state conditions, the energy storage term ( mCP : ) will vanish as
dT =O
dt
Then,
=h,{[T{oo)-T=]
£gen
(2)
Substituting for £gen from Equation (2) in Equation (1), we have
or
dT
hA.[T(t)-T= -T(oo)+ T=] = -m CP dt
or
mCP dT
T(t)-T(oo)=--hA8 dt
or
dT
_ h,{dt
T(t)-T(oo)
mCP
Integrating between the limits: t= 0, T(t) = T;. = T= and t= t, T(t) = T(t), we get
T(t)
dT
J T(t)-
Too
or
[;~Jt
dt
hA 8 t
In T(t)-T(oo)
T= -T(oo)
mCP
From Equation (2), h/4 =£gen ![T(oo)-T=]
T(t)-T(oo)
T= -T(oo)
[
exp -
£gent
mCP[T(oo)-T=]
Hence, the transient temperature response is given by
]
6.46 , Heat and Mass Transfer
Substituting the appropriate numerical values, we get
T(7 min)= l 10°C + (30 - l 10)°C ex
P
[
60W x(7x 60s)
]
(0.32kg)(935J /kg 0 C)x(l 10-30)°C
= 82.1°C
Example 6.21
(Ans)
~ Pulverised coal pellets, approximated as spheres of I-mm diameter, are preheated by passing
them through a cylindrical tube maintained at I 200°C before being injected into the furnace.
The pellets suspended in air flow move with a speed of 3.5 mis. The initial and final temperatures of the coal
pellet are 30°C and 700°C. The dominant mode of heat transfer is radiation and the pellet is very small compared
to surface area of the tube. Determine the length of the tube required for this preheating. State clearly the
assumptions made. The following thermophysical properties of coal can be used:
p = 1350 kg/ml
CP = 1.26 kJ/kg °C
k = 0.26 W/m 0 C.
Known: Coal pellets are preheated as they move through a tubular furnace subjected to radiative heating
process.
Find: Length of tube required, L (m) to heat pellets to 700°C.
L Cylindrical tube
Schematic:
Airflow
-
-
1-+-----
V= 3.5 m/s
L
-----+-I
lnitially
T;=S0°C
Tsur= I 200°C
E;n = Qrad
Cool pellet
D= Imm
T(t) = 700°C
Fig. 6.29
Assumptions: ( 1) Lumped heat capacity model is appropriate. (2) Coal pellets have an emissivity of 1.
(3) Only radiation heat exchange is present.
Analysis: Energy balance
1-Eout +~=Est
.
dT
or
Qrad =p¥-CPdt
or
cr Ae(T.4 -T4) = p¥-C dT
sur
p dt
Transient Heat Conduction
r
6.47
1t
Withe= 1, ¥-=-D 3 and A= 1tD2 we have
6
'
¥-
1tD3 !6
D
-=--=A
1tD2
6
Separating the variables and integrating, one gets
p¥-Cp T(t)
dT
CJ
4
t
-AEJ - Jat
T.sur -T
T;
4
o
pC DT(t)
or
dT
J-60
T. -T
t-
p
4
7;
sur
4
A standard form of integral is:
dx
1
a+x
1
x
J- = -ln--+-tana -x
4a
a-x 2a
a
1-
4
=
4
3
3
=
Noting that x T, a J'.ur, T;_ = 30°C or 303.15 K,
T(t) = 700°C or 973.15 K, T.ur = 1200°C
T(t)
dT
1
T. + T IT(t)
J - - -4 = - - I n ~
T;
J'.t - T
4 J'.~
T.ur - T T;
T IT(t)
1
+--tan- 1 2 J'.~
T.ur T;
=-+[{In T.ur ~T(t) -In T.ur ~T;_ }+2{tan-1 T(t) _tan- 1
4 T.ur
T.ur
T.ur
T ( t)
1
= 4(1473.15K)3
J;_
T.ur
l}]
T.ur
[{z (1473.15+973.15)K I (1473.15+303.15)K}
n (1473.15-973.15)K
n (1473.15-303.15)K
973.15K
303.15K
+ 2 { tan- I - - tan- 1 - -}]
1473.15 K
1473.15 K
°
= (78.2xl0- 12 K-3 ) [{1.5877-0.4175}+2{0.5838-0.2030}] = 1.51x10-1 K- 3
Also,
pCPD
60
(1350kg/m3 ) (1260J/kg 0 C) (O.OOlm)l lWI
6x5.67xl0-8 W /m2 K 4
lJ/s
= 5x109 s K3
°
Time taken, t= (5x109 s K 3) (l.51x10- 1 K-3)
= 0.755 s
Length of furnace, L =Vt= (3.5 mis) (0.755 s)
=2.64 m
(Ans)
6.48 , Heat and Mass Transfer
(C) PLANE WALL
•
•
Example 6.22 ~ A steel pipeline of 1-m diameter and 40-mm wall thickness is effectively insulated on its
exterior surface. Hot oil at 60°C is pumped through the pipe with a convective heat transfer
coefficient at its inner surface of 500 W/m 2 0 C. Before the flow commences, the pipe walls are at a uniform
initial temperature of - 20 °C. Determine, after a lapse of 8 min: (a) the temperature of the outer surface of
the pipe wrapped by insulation, (b) the heat flux from the oil to the pipe, and (c) the energy transferred from
the oil to the pipe per metre pipe length. The properties of the pipe material are
p = 7832 kg/m 3
cp = 434 J/kg c
0
k = 63.9 W/m 0 C.
Known: The wall of a pipe is suddenly subjected to convection on its inner surface while its outer surface
is insulated.
Find: (a) Temperature of adiabatic outer surface of the pipe, T (x = 0, t = 8 min) {0 C), (b) Heat-transfer
rate per unit area q (W!m 2) at 8 min, (c) Energy transferred, Q (Jim) after 8 min.
Schematic:
x=O
T(O, t)
x=L
Initially
T;=-20°C
T(L, t)
Oil
L=40mm
lll
Adiabatic
outer surface
(insulated)
Inner surface
LPipewall
Fig. 6.30
Assumptions: (I) One-dimensional conduction in x. (2) Exterior pipe surface is insulated. (3) The pipe
wall is approximated as a plane wall because thickness is much less than the diameter. (4) Constant
thermal properties of the pipe material.
Analysis: For a plane wall of thickness 2L the axis of symmetry is at x = 0 (centreline) and the characteristic
length is L, the half thickness. In this case, one surface is insulated (dT!dx = 0) and is thus equivalent to
a case of plane wall of 80-mm thickness with midplane being the insulated surface. Thus,
Biot number,
Bi= hL = (500W !m20C) (0.04m)
k
63 .9W/m °C
=0.313
Transient Heat Conduction
r
6.49
Thermal diffusivity, a.=k/pCP
=
Fourier number,
63.9W/moc
=l.88x10-5m2/s
(7832kg/m3 ) ( 434J/kg 0 C)
Fo= a.t = (l.88x10-5 m2/s)(8x60s)
(0.04m) 2
£2
=5.64
As Bi> 0.1, the wall is not spatially isothermal and the lumped capacity model is invalid. Also since
Fo > 0.2, approximate analytical series solution is appropriate. Heisler and Grober charts can also be
employed.
(a) Temperature at the outer adiabatic surface is like the centreline temperature of a plane wall of
80 mm thickness. Therefore,
'I
- 7;,
- T=
- 1 exp ( -11,12 Fo )
0o,wall --A
J;-T=
For Bi= 0.313, from Table, 6.2:
A 1 = 1.0467 and \
= 0.5311 rad
80 ,wall = 1.0467 exp {-(0.5311)2 (5.64)}
= 0.2133
Hence,
J;, =T= +0.2133(7;-T=)
= 60 + 0.2133 (-20 - 60)
= 42.94°C
(Ans) (a)
(b) Heat flux, i.e., heat transfer rate per unit area from the oil to the inner surface of the wall is given by
q (x=L, t=8min)=h [T--T(x=40mm, t=480 s)]
To find,
T (L, t), we use the following expression
0 ( x,t
)I 11 -_T(x,t)-T= -_ A 1e-J.., F cos ('I 1 ~)
2
wa
J;-T=
0(L,t) = A 1 exp(-l? Fo)cosAi
0
11,
[·:
180°
= 0.2133 cos (0.5311 rad) x - 1t rad
= 0.1839
T (L, t) = T= + 0.1839 (J;-T=)
= 60 + 0.1839 (-20 - 60)
= 45.29°C
L
x=L]
6.50 , Heat and Mass Transfer
Heat flux from oil to the wall at t = 8 min is
q = (500 W/m 2 0 C) (60 - 45.29)°C
=7355 W/m 2
(Ans) (b)
(c) Energy transferred to pipe wall from oil after 8 min duration can be found from the following equation:
sin A-1
-- 1- 0o, wall--
( -Q- )
Qmax wall
A-1
= l-(0_ 2133 ) sin(0.531 lrad)
A-1
= 0.7966
Qmax =Q =mCP (T=
-7;)
= p-JLCP ( T= - J;)
Per metre pipe length, the volume, ¥- = rcD L
Q = 0.7966 p(rcDL)Cp(T= -7;)
= (0.7966) (7832 kg/m 3) (rcx lmx0.04m)x 434 J/kg C) {60 - (-20)°C
= 27.22 x 106 J or 27.22 MJ
0
Example 6.23
(Ans) (c)
~ Consider the nose of a missile re-entering the earth's relatively dense atmosphere with a very
high velocity and creating huge heating effect. The nose section can be idealized as a 6-mm
thick stainless steel plate with one surface adiabatic and the other exposed to convective environment. The
uniform surface heat transfer coefficient is estimated to be 3390 W/m 2 K and the initial uniform temperature is
50°C. (a) Neglecting radiation effects, determine the maximum allowable time for heat dissipation if the effective
air temperature in the vicinity of the nose section is estimated to be 2100°C. Metallurgical considerations limit
the metal surface temperature to 1100°C. Also compute the inside surface temperature under the specified
conditions. The following thermophysical properties of steel may be used:
p = 7900 kg/m 3,
CP = 611 J/kg K,
k=25.4 W/m K
Known: The nose section of a re-entry missile modelled as a plane wall with one surface adiabatic is
suddenly exposed to surrounding atmosphere.
Find: (a) Time required, t (s). (b) Inside adiabatic (insulated) surface temperature, T(x = 0) or To( 0 C).
Schematic:
\-l=6mm
n
----1
Ambient
air
Initially
T;=S0°C
Adiabatic surface
(x=O), T0 = 1
111
T==2100°C
h=3390W/K
T(x=l)= I l00°C
X
Fig. 6.31
Transient Heat Conduction
r
6.5 I
Assumptions: (1) Nose section approximated as a plane wall or infinite plate with inside surface adiabatic
(insulated) (x = 0). (2) Constant thermal properties and uniform heat transfer coefficient.
Analysis: The lumped capacitance analysis Biot number is
hLC
Bi=-k
where
¥- AL L
. .
.
Le=-=-=- (considering both sides of the plate)= 3 mm
A. 2A 2
Bi= (3390W/m2 K)(3x 10-3 m)
25.4W/mK
=0.40
As this is greater than 0.1, the lumped capacity model, in which T= T (t), neglecting internal thermal
resistance, is not appropriate.
Hence, in this case, T = T (x, t) one can use transient temperature charts or one-term approximate
analytical solution.
Analytical method
For a plane wall with one surface insulated, the thickness is L = 6 mm which is also the characteristic
dimension in determining the Biot number,
Bi= hL = (3390W/m2 K)(0.006m)
k
25.4W!mK
=0.80
From Table 6.2, at Bi= 0.80 for a plane wall,
A1 = 0.7910 rad and A 1 = 1.1016
Fourier number, Fo = at
L2
where
a=_!__=
pCP
lS.4W/mK
=5.26x10-<im 2 /s
(7900kg/m3 )(611J/kgK)
Fo= (5.26x10-<im2 /s)(t,s) =0. 146 t
(0.006m) 2
We note that
- T(x,t)-T=
0wall ( x,t) ---J;-T=
At x = L i.e., at the exposed surface, the maximum permissible temperature is 1100°C.
_12p,
'I
0wan(L,t) =T(L,t)-T=
- - - - = A 1e '"I 0 cos11,1
J;-T=
6.52 , Heat and Mass Transfer
Substituting the appropriate values, we have
1100-2100
50-2100
- - - - = 1.1016 exp (-0.79!2 x 0.146t) x cos(0.791rad)
or
exp (-0.79l2 x0.146t) = 0.6298
or
-0.09135 = Zn 0.6298
:. time required,
_.
5 06 s
t_--0.4623
----0.09135
(Ans) (a)
The inside (adiabatic) surface temperature,
T0 , i.e., T (x = 0, t = 5.06 s) can be found as follows:
T, -T
0o,wall =~=A
e-At Fo
T-T
I
I
[at x = 0]
-
= 1.1016 expl-(0.791}2 (0.146)(5.06) J
= 0.6936
Hence,
To= T_ +0.6936(1;-T_)
= 2100°c + (0.6936) (50 - 2150)°C
= 678°C
(Ans) (b)
Transient temperature charts
Slightly less accurate but perhaps more convenient method is to use the transient conduction charts.
Jix=~-T-1
'J; T_
x=L
l
110~-2100
50 2100
= 0.4878
With -X = 1 and Bi= 0.8, 1ix-L)
- - T_
""0.7, from the chart,
L
To-T_ =L
=
¼ T_ = ( Jix=~ - T_ )/( Jix=L~ - T_)
J; T_
'I; T_
To T_
= 0.4878/0.7 = 0.6969
Inside surface temperature,
To =T_ + 0.6969 (J; - T_)
= 2100°c + 0.6969 (50 - 2100)°C
= 671°C
(Ans) (b)
Transient Heat Conduction
r
6.53
From the chart, with Bi= 0.8 and 1'o -T= = 0.6969, we get
T;-T=
time needed,
at
Fo= 0.73=L2
t= (0.73)(0.006m) 2 = S.O s
5.26 x 10-6 m2 Is
(Ans) (a)
Example 6.24 ~ A large steel plate, 75-cm thick [k = 15 W/m °C, a= 3.96 x IQ-6 m2/s] is initially at a uniform
temperature of 50°C. Suddenly both of its surfaces are raised to and maintained at 450°C.
Determine
(a) the temperature at a 25-cm plane from the surfaces one hour after the sudden change in surface temperature
(b) the instantaneous heat-transfer rate across the above plane per m2 at I h
(c) the total heat flow across the above plane in I h
(d) the temperature at the midplane of the plate after one hour
Known: A large plate is subjected to sudden temperature change of the surface under specified conditions.
Find: (a) T(x=0.25 m, t= l h) (0 C), (b) Q / A(t= l h) [kW/m 2], (c) Q/A [M = l h] (kJ/m 2), (d) Tc, i.e.,
T (x = 0.375 m, t = l h) (0 C).
Schematic:
i----L=75cm----
T5 =T==450°C
x- _J
r . = T= =450°C
25cm1
h ~ 00
Time, t= I h
~-+--
Initially
T;= S0°C
Steel plate
(k= ISW/m°C,
a =3.96x IQ-6m2/s)
Fig. 6.32
Assumptions: (1) One-dimensional transient conduction. (2) Large plane wall subjected to sudden change
in surface temperature to environment temperature (I'. =T= ), i.e., Bi= hL/k = oo ash ~oo. (3) Constant
thermal properties.
Analysis: Lumped capacitance Biot number,
hLC
Bi= --:=ooo
k
Clearly, the lumped capacity model is ruled out.
(»> 0.1)
6.54 , Heat and Mass Transfer
Now, to use the transient temperature charts,
Fo > 0.2
Fo = at where Le is half-thickness
In this case,
1i
= (3.96x10--<:i m 2 /s) (3600s) = O.l
(0.375m) 2
As Fo < 0.2, we cannot use the charts either.
Let us then go back to basics for analytical solution. It is noteworthy that in the expressions that follow
L is the total plate thickness and not half thickness, i.e., L = 0.75 m.
(a) To determine the temperature at any distance x measured from the surface (not from the midplane),
we have
T-T = T-7'.
00
(ash~ oo)
T; - Too T; - T.
l
~
2
(n1tx)
1
•
= -4 £.. -exp
- (n1t)
at } sm
-7t n=l,3,5 n
L
L
From the problem statement,
x 25cm 1
-=--=-
L
75cm
3
Also,
J
(",X)' <xt=n' [ ( 0_7~m x3.96x!Q-' m' /sx3600s]
= 0.25 n 2
T-7'.
4[1-e-0· 25 sm-+-e. 1t 1 2 ·25 sm1t+-e--<:i·
.
1
. 51t]
25 sm-=T;-T. 1t 1
3 3
5
3
= ~[(o.7788) (0.866)+G}o.1054){o) +¼(0.00193){-0.866)]
= 0.8583
or
T (x = 25 cm)= T. +0.8583 (T;-T.)
= 450 + 0.8583 (50 - 45)
= 106.7°C
(Ans) (a)
Transient Heat Conduction
r
6.55
(b) Instantaneous heat transfer rate per m2 at t = 1 h is
. 4k A ( T. -i;) £..
~ exp- (n7t)
Q=-atcos-2
L
L
n=l,3,5
= 4kA(1'.•
L
-T.)[
'
n1tx
L
1t
3
57t]
3
e-0· 25 cos-+e-2·25 cos1t+e-6· 25 cos-
= 4 (l 5 Wlm OC)(lm)2 ( 45 0- 50)oc [(0.7788){0.5)+(0.1054)(-1)+(0.00193){0.5)]
0.75m
= 9.12x103 W
= 9.12 kW (per m1)
(Ans) (b)
(c) Total heat flow across the plate after one hour is
(kAL)(
Q =42 - - T.-7;
7t
a
) "-'
~ 21 { 1-exp [ - (n1t)
n1tx}
at cos-2
n=l,3,5
n
)
L
L
= _i_[(15W/m°C)(lm2 )(0.75m)( 450-50)°C]
1t2
3.96xl0-6 m2 /s
x [(1-0.7788){0.5)+(¼)(1-0.1054){-1)+ 215 (1-0.00l93)x(o.s)J
= 14.35 X 106 J
= 14.35 MJ (per m1)
(Ans) (c)
(d) Midplane or centreline temperature is
~
~
2
• (n7t)
1
(n1t)
-expat sm
1t n=l,3,5 n
L
2
-T.= -4 "-'
-7; - T.
= 0.9474
Hence, the midplane temperature of the plate after one hour is
Tc= 450 + 0.9474 (50 - 450)
=71°C
(Ans) (d)
6.56 , Heat and Mass Transfer
(D) LONG CYLINDER
•
•
Example 6.2S ~ A long cylinder of 20-cm radius of a material with k = 170 W/m K and a= 9.05 x Io-7 m2/s
is initially at a uniform temperature of 6500C. The cylinder is quenched in a medium at 75°C
for heat treatment with h = 1700 W/m 2 K. Find the time at which the process should terminate to ensure that
the temperature attained at a depth of 20 mm from its surface is 2500C. What will be the temperature on
the axis of the cylinder at this time? Calculate the amount of energy transferred from the cylinder per metre
length during this period.
Known: A long cylinder is quenched in a heat treatment process.
Find: Time t so that T(r/r0 = 0.9, t) = 250°C. Centreline temperature, T0 . Energy transfer, Q(t).
Schematic:
2cm
r= 18cm, r/r0 =0.9
r=1o=20cm-~-----------.,,,._~
Medium
r
r-o ---·-·- ------------·--·-·------·---·---·-·---·----·-------- ---- ___ j
T==75°C
h= 1700 W/m2 K
Long cylinder
k= I 70W/m K, a=9.05 x I0--7 m2/s
Fig. 6.33
Assumptions: (1) One-dimensional (radial) conduction. (2) Constant properties. (3) Transient-temperature
charts can be used (Fo > 0.2).
Analysis: We note that
T(r/r0 = 0.9) = 250°C,
1'; = 650°C
T= = 75°C,
Hence,
T-T= = 250- 75 = 0 _30435
T;-T=
650-75
=A 1 exp(-Af Fo)J0 (A1r/,;,)
(A)
For this case,
. h~ (1700W/ m2 K){0.2m)
Bz = - = - - - - - - - - = 2.0
k
170 W/m°C
Bi= 2.0, A- 1 = 1.5995
At
and
A 1 = 1.3384
J 0 (A- 1 rj,;,) = J 0 {1 .5995 X 0.9) = J 0 {1.43955) = 0.5451
Hence, from (A), we have
1'o -T=
_
( 2
)
00 -_ - - A 1 exp -A. 1 Fo
T,-T=
= 0.30435 I 0.5451 = 0.5583
(B)
Transient Heat Conduction
r
6.57
Hence,
-In ( 0.5583/1 .3384)
Fo = - - - - - ~ = 0.34175
1.59952
at
r.2
0
Time required,
(0.34175)(0.2m)2
t=------
9.05x10-1 m 2 /s
= 15105 s = 4.2 h
(Ans)
Also, the centreline temperature is obtained from (B):
1a = T~ +(0.5583) (J;-T~)
= 75+(0.5583)(650-75)
(Ans)
=396°C
per unit length, energy transfer,
where
= 1- 2 x0.5 699 x0.5583
1.5995
= 0.602
Q = (0.602{~){1t,;,2) (i:-r~)
= (0.602)(
170 W/mK )(1tx0.22 m2 )x(650-75)K
9.05 x 10-7 m2 /s
= (0.602) (1.358 X 1010 J
= 8.2 x 109 J or 8.2 GJ
Alternatively
We note that for Bi=2.0 and rl,;, =0.9, from the chart,
T(rh;,) = 250-75 =0. 54
J;, -T~
1a - 75
6.58 , Heat and Mass Transfer
:. centreline temperature,
T = 75 +(250-75)
0
0.54
(Ans)
=399°C
T;,-T=
T;-T=
399-75
650-75
= 0.5636
<J,t
Fo=-=0.35
from the chart,
r2
0
(0.35)(0.2m)2
t = - - - - - = 15470 s
9.05 x 10-7 m2 Is
time required,
(Ans)
=4.3 h
From the chart,
__g_= 0.60
Qmax
Q = 0.60 X 1.358 X 10 10 J
= 8.15x 109 J
(Ans)
Example 6.26 ~ Two long cylinders of 10-cm diameter, one of copper and the other of asbestos, are placed
in a furnace. The initial temperature of cylinders is 30°C and the temperature in the furnace
is I000°C. (a) How long should the two cylinders be kept in the furnace to reach the centreline temperature
of 428°C? (b) Also, determine the temperature at a distance of I cm from the surface after this time period
in the two cases. The combined convection and radiation heat transfer coefficient is 100 W/m 2 0 C. Use the
following properties:
k (Wlm 0 C)
a (mm 2/s) __,
_______
Copper
385
117
Asbestos
0.1 I05
0.288
Known: Two long cylinders made of copper and asbestos are heated in a furnace under identical conditions
in a convective environment.
Find: (a) Time, t (s) to reach a specified centreline temperature, T0 for both cylinders. (b) Temperature,
T( 0 C) at a radius of 4 cm, (I cm from the surface).
Transient Heat Conduction
Schematic:
Copper cylinder
6.59
I cm
T;= 30°C
T(O, t) = 428°C
D=IOcm
Furnace
r
T== I000°C
h= IOOW/m2°C
Surface
Asbestos cylinder
T;= 30°C
T(O, t) = 428°C
D=IOcm
Fig. 6.34
Assumptions: ( l) One-dimensional heat conduction. (2) Constant thermal properties and heat transfer
coefficient. (3) Fourier number, Fo > 0.2 so that one-term approximate solutions are applicable.
Analysis: Copper cylinder
Bi= hr,,= (IOOW/m 20C)(0.05m)
k
385W/m°C
Biot number,
= 0.013
(« 0.1)
At Bi= 0.013, from Table 6.2:
A,= 1.0022
A. 1 = 0.1121 rad
and
The dimensionless form of the temperature distribution for an infinite cylinder is
[Fo>0.2]
Centreline temperature, T (0, t) = 428°C
At
r= 0, J 0 (0) = l
T(O,t)-T=
T;-T=
-'A. F
A 1 e 2, 0
428-1000
30-1000
- - - = 1.0022 exp (-O.ll2J2xFo)
or
(0.5887 / 1.0022) = exp (0.112 J2 Fo)
or
(J,f
Fo=42.2 = -
or
,._2
0
time required,
42.2 X (0.05 m) 2
t=-----ll7xl0-6m2/s
= 901.7 s "' 15 min
(Ans) (a)
6.60 , Heat and Mass Transfer
At 1-cm from the cylinder surface, the radius, r = 5 cm -1 cm = 4 cm.
Then
T(r=4cm,t=901.7s)-T
_12Fo
('I
)
- - - - - - - - - = A 1 e ''1 J0 11,1 rl,;,
I; -Too
00
Now
A1rlr0 = (0.1121) (4 cm/ 5 cm)
= 0.08968
J0 (0.08968) = 0.9978
T(r=4cm)-Too = 1.0022 e(--0.11212x42.2) x 0.9978
I;-Too
= 0.5884
:. temperature at a radius of 4 cm,
T= 1000 + (30 - 1000) (0.5884)
= 429°c
(Ans) (b)
Asbestos cylinder
. h,;, (100W/m2 0 C) (0.05m)
Bz=-=-------k
0.1105W/m°C
= 45.25
From Table 6.2, for Bi= 45.25
A 1 = 1.5998
and
A1 =2.3516
From Table 6.2,
4cm] = J0 {1.8813)
J 0 (A1 rl,;,) = J 0 [ 2.3516x-5cm
= 0.2927
Dimensionless excess temperature,
0
o,cyl
or
J428-1000)oc = 0 5897
(39-1000 ) 0 c
.
0.5897 = A 1 exp(-Af Fo)
= 1.5998 exp {-2.35162 Fo)
or
ln (0.5897 I 1.5998) = - (2.3516) 2 (Fo)
at
Fo = 0.1805 = ,._2
0
Transient Heat Conduction
time required,
r
6.61
t= (0.1805) (0.05m) 2 /0.288x10-<i m2 /s
= 1566.6 s ""26.1 min
At 1 cm from the surface, i.e., at r = 4 cm, the temperature can be determined as follows:
0(r t)I
'
cyl
=T(r,t)-T=
T-T
1
A e-~FoJ (')..,
I
o
00
I
(Ans) (a)
!_)
,;,
= 1.5998 exp[-2.35162 x0.1805]x0.2927
= 0.1726
Hence, temperature at a radius r = 4 cm is
T= 1000 + (30 - 1000) (0.1726)
= 832.6°C
(Ans) (b)
Comment: The thermal conductivity of asbestos is much less than that of copper. This is reflected in
almost uniform temperature (internal temperature gradients negligible) in case of copper cylinder while
in asbestos cylinder the temperature varies from 428°C at r = 0 to 832.6°C at r = 4 cm. The time required
to reach the same centreline temperature is understandably more in case of asbestos cylinder.
(E) SPHERE
•
•
Example 6.27 ~ The glass beads (k = 0.7 W/m K, CP = 0.75 kJ/kg K and p = 2800 kg/m 3) of 0.6-mm-diameter
initially at 520°C throughout are sprayed into 20°C air and allowed them to harden as they
fall to the ground with a constant downward velocity of 4.7 m/s as a part of the manufacturing process. The
centre temperature of the beads should not exceed 40°C. The average convection heat transfer coefficient
is 234 W/m 2 K. Calculate the surface temperature of the beads when the centre reaches 40°C, and (b) the
distance through which the glass beads should fall from rest. Is the lumped capacity model validl
Known: Glass beads are heated and sprayed in cold air for hardening as they hit the ground.
Find: Surface temperature of the beads, T (r0 , t) (0 C).
Schematic:
G l a s s ~ 0 _3 mm
1,;ti••-1 0
s= V.t= l
i
" " . . .1 U
Air
T;=s20°c
g=9.81 m/s2
V=4.7m/s
~ T ( r0 ,t)=l
r(o, t) =40°c
Fig. 6.35
lll
r==20°c
h=234W/m2K
6.62 , Heat and Mass Transfer
Assumptions: (1) Glass beads are approximated as spherical objects. (2) One-dimensional conduction in
radial direction. (3) Constant velocity, properties and heat transfer coefficient. (4) Fo > 0.2 so that one-term
approximation is justified.
Analysis: Lumped system formulation
L _ ¥- _ 4/31t,;;1
C A. - 41t,;,2
Characteristic length,
,;,
3
Biot number,
. h4, h,;, (234 W/m2 K)(0.3x10-3 m)
Bz= - = - = - - - - - - - - - k
3k
3x0.7W/mK
(< 0.1)
= 0.0334
Hence, the lumped capacity analysis is acceptable.
Surface temperature, T(r0 , t) = T(O, t) = 40°C
For greater accuracy, the one-dimensional model is preferred. For a sphere, in this case:
Bi= h,;, = (234 W/m2 K)(0.0003m) = O.lO
k
0.7W/m K
Dimensionless centre temperature,
e (O t)=T(O,t)-T- = (40-20)°C =0.0 4
sph
'
T;, -T-
(520-20)°C
Thermal diffusivity,
k
0.7W/mK
P CP
(2800 kg/m3 ) (750 J/kg K)
a=--=--------= 3.33x10-7 m2 /s
From Table 6.2, with Bi= 0.10, for a sphere:
A1 = 0.5423 rad
and
A 1 = 1.0298
sin A1 = sin (0.5423 rad)= 0.5161
Transient temperature response is given by
0sph(r,t) =0sph(O,t) sin(A1r/,;,)/(A1r/,;,)
To find the surface temperature, we have
T(,;,,t)-T-
i:. - T-
T(O,t)-T-
sinA.
- - - - x - -1
I;, - TA1
= (0.04) (0.5161/0.5423) = 0.038
T(,;,,t)=T- +0.038(T;,-T-)
= 20°C + (0.038) (520 - 20)°C
=39°C
(Ans) (a)
Transient Heat Conduction
r
6.63
r,2
Time elapsed, t = Fo·..E_
We note that
a
0sph (0 , t)· -- A ,e-A./ Fo
l
{0sph(O,t)}
0.04
Fo=--ln
~ - - = - -l- - / n
--
or
-Af
0.5423 2
A1
1.0298
= l 1.05
(> 0.2)
t = (l 1.05){0.3 x 10-3 m) 2 I 3.33 x 10-7 m 2 Is
= 2.98 s
The distance through which the beads should fall from rest is
s = Vt= (4.7 mi s) (2.98 s)
= 14.0 m
(Ans) (b)
Example 6.28 ~ A solid steel [ k = 22.4 W/m °C, p = 7689 kg/m 3, CP = 0.46 kJ/kg °C ] spherical ball bearing
of 25-mm OD leaves the furnace at a uniform temperature of 600°C. It is quenched in
oil that is maintained at 40°C and the heat transfer coefficient between the bearing's surface and the oil is
1.5 kW/m 2 0 C. Determine: (a) the centre temperature, and (b) the temperature 1.25 mm from the surface
after the bearing has been in oil for 30 s. (c) Calculate the heat lost by the bearing during the first 30 seconds.
Known: A spherical steel ball bearing is quenched in oil and subjected to a convection process.
Find: (a) Centre temperature, T(O, 30 s) [0 C]. (b) Temperature at 1.25 mm from surface [r= (12 .5-1.25) =
11.25 mm], T[0 .01125 m, 30s] (0 C). (c) Heat (energy) lost during first 30 s, Q [kJ].
Schematic:
surface
Fig. 6.36a
Ball bearing (D =22.4 mm, =600°C)
Oil
T= =40°C
h= ISOOW/m2 °C
Fig. 6.36b
6.64 , Heat and Mass Transfer
Assumptions: (1) One-dimensional transient heat conduction in radial direction [T= f(r, t)]. (2) Properties
and convection coefficient are constant. (3) Fo > 0.2 so that transient temperature charts are applicable.
Analysis: Let us first determine if the Biot number, Bi is less than O.1 so that the lumped capacity model
is valid and T= f(t) with negligible internal temperature gradients. The characteristic length, Lc for a
spherical geometry is
Bi= hLc = (1500W/m2 0 C) (0.00417m) = 0279
k
22.4W/m°C
As Bi > 0.1, the negligible internal thermal resistance concept is not applicable. Hence, Heisler and
Grober charts will be used.
(a) For the purpose of using the charts: Lc =,;,
. h,;, (1500W/m2 0 C) (12.5x10-3 m)
Bz=-=-'-------'---'----k
22.4 W/m°C
= 0.837
k
22.4W/m°C
(7689kg/m3 ) (460 J/kg 0 C)
a=--=--------P cp
= 6.33x10-6 m2 /s
Fourier number,
Fo =at= (6.33xl0-6 m2 /s) (30s) = 1. 215
,;,2
(0.0125m) 2
(> 0.2)
For Bi= 0.837 and Fo = 1.215, from the transient temperature (from the chart) we get
To-T- z0.09
I;-Tcentre temperature after 30 seconds is
To =T- +0.09(7;-T-)
= 40°C + (0.09) (600 - 40)°C
= 90.4°c
(b) Temperature at r = 11.25 mm after 30 s can be found as follows:
From the chart, with Bi= 0.837
and
~= l 1. 25 = 0.9
r;, 12.5
(Ans) (a)
Transient Heat Conduction
r
6.65
Hence,
'li_rtr.=0.9) = T= +(J;-T=)[(0.09) (0.74)]
= 40°C + 0.0666 (600 - 40)°C
= 77.3°C
(c) Energy balance
Ior
Eout +
(Ans) (b)
t
=Est
Q,ut = - Est = - m CP (½nal - J;nitial)
or
For maximum heat transfer, the final temperature of the sphere will approach the temperature of the
surrounding fluid, i.e., T=. Then
We can determine (Q/Qmax) or (Q!QJ from the Grober chart for Bi= 0.837 and Bi2Fo = (0.837) 2
(1.215) = 0.851. From the chart, we read
Q/Qmax = 0.925
Mass of the spherical ball,
rcD3
6
m=p-F=p--
= (7689 kg /m3 )( ~x0.0253 m3 ) =0.0629 kg
heat lost by the bearing during first 30 s is
Q=0.925Qmax = (0.925) X (0.0629 kg) X (600 -40)°C (0.46 kJ/kg 0 C)
= 15.0 kJ
(Ans) (c)
Comment: We note that the centre temperature (90.4°C) is about 13°C higher than that at 1.25 mm
from the surface. Also, in the first half-a-minute, 92.5% of the heat is ultimately lost. Note also that the
characteristic length for using the charts is different from that used for the lumped parameter analysis.
As (Bi)iumpedmodel > 0.1, the bearing is not spatially isothermal and the temperature is a function of not
only time but also radius.
Example 6.29 ~ An 8-mm diameter sphere [ k = 20 W/m °C and a.= 0.024 m2/h ] is initially at a uniform
temperature of 400°C. It is subjected to two-step cooling process. The sphere is first cooled
in air at 25°C with a convection coefficient of 15 W/m 2 °C until a temperature of 335°C is reached. Subsequently,
the sphere is cooled in a well-stirred water bath held at 25°C with a heat-transfer coefficient of 6 kW/m 2 °C.
Calculate (a) the time required to accomplish the cooling process in both steps if the centre temperature of
the sphere finally reaches 50°C, (b) the surface temperature of the sphere at the end of the cooling process.
6.66 ,
Heat and Mass Transfer
Known: A sphere is cooled in two stages first in air and then in water under prescribed conditions.
Find: (a) Time required in two steps, t1 and tn. (s) (b) Surface temperature after cooling, 'f. (0 C).
Schematic:
Water
Air
Sphere
(D=Smm)
k=20W/m °C
a= 0.024 m2/h
!!!
h= ISW/m2°C
T==25°C
!!!
h=6kW/m2°C
T==25°C
t,=?
t11 = 1
I Step I I
I Step II I
Fig. 6.37
Assumption: ( l) Constant properties and uniform heat transfer coefficient in both stages of cooling.
Analysis: Cooling in air
Lumped capacitance Biot number is
. hLc
h(¥-I A) h(rcD 3 /6) hD
Bz=-=---=----=k
k
(rcD 2 )k
6k
= (15W/m2 0 C) (0.008m) = O.OOOl
6(20W/m 0 C)
(< 0.1)
Hence, internal temperature gradients are negligible.
The temperature- time history is given by
T-T
_ _=_
= e-(Bi)(Fo)
T;-T=
335-25
- - - =exp-[O.OOlxFo]
400-25
or
0.8267 = e-I0- 3 Fo
or
(/n 0.8267) (-10 3) =Fo
or
Fo = 190.35 = ext / L~
where
LC =D/6
Hence, time required for air cooling is
Fo(D /6) 2
a
(190.35)(8xI0-3 /6) 2 m2 =SO.S s
(0.024/3600) m2 Is
(Ans) (a)
Cooling in water
Bi= hD
6k
(6000W!m2 0 C)(0.008m) = 0.4
6(20W/m 0 C)
As Bi > 0.1, the lumped parameter analysis is not justified. However, one-term approximation can be
used provided Fo > 0.2 (to be verified later).
Transient Heat Conduction
r
6.6 7
The Biot number is now defined as
Bi= h,;, = (6000W/m2 °C)(0.004m) = 1. 2
k
20W/m°C
From Table 6.2, with Bi= 1.2,
A1 = 1.6624 rad
and
A1 = 1.3144
Dimensionless temperature difference at the centre of the sphere is
_To-T=_ A exp (-A.2)
1
1 Fo
Ii_ -T=
0o,sph - - - -
50- 25 = 1.3144 e-(1.6624)2 Fo
335-25
or
or
Zn [
25
] = -(1.6624)2 Fo
310xl.3144
Fo = at=
,;,2
-2.791 = 1.0l
-(1.6624) 2
(> 0.2)
Time required for water cooling is
_For;_
tn
(l.Ol)(0.004m) 2
_
24
(0.024/3600)m2 /s] - • s
-a- [
(Ans) (a)
Total cooling time is
t= t 1 +tn = 50.8 + 2.4 = 53.2 s
To find the surface temperature, we note that
-r -- I
,;,
and
T,.-T=
sinA.1
- - , -. ,,ie_12F,
"1 ° - Ii_-~
~
= 1.3144 exp [-(1.6624)2 (1.01)] sin(l. 6624 rad )
1.6624
= 0.0483
Hence, the surface temperature of the sphere is
T,. =T= +0.0483(1i_-T=)
= 25 + 0.0483 (335 - 25)
=40°C
Alternatively
Using the transient temperature charts:
T, -T
50-25
With Bi= 1.2 and ~ = - - - = 0.081, we read from the chart.
Ii_ - T= 335 - 25
(Ans) (b)
6.68 , Heat and Mass Transfer
Fo = 1.0 (same as obtained analytically)
t = 2.4 s
To find T8 , with r/r0 = l and Bi= 1.2, from the chart, we read
7'.-T= =0.58
To-T=
¾-T= = ¾-T= x J;, -T= = {0.58){0.081)
T; - T=
1'o - T=
T; - T=
= 0.047
Hence, the surface temperature of the sphere at the end of cooling is
Ts= 25 + (0.047) (310)
=39.6°C
(Ans) (b)
Comment: For step II, the initial temperature must be uniform and Fo > 0.2 . As in step I, the lumped
capacity model was valid, the final temperature of 335°C was uniform and became the initial uniform
temperature for the second stage. For Step II, Fo was assumed to be greater than 0.2 and ultimately it
turned out to be 1.0 l (> 0.2). Hence, the assumption was justified.
(F) SHORT CYLINDER/SEMI-INFINITE CYLINDER
Example 6.30 ~ A man is found dead at 7 o'clock in the morning in a room whose temperature is I 7°C. The
surface temperature on his waist is measured to be 24°C and the heat transfer coefficient
is estimated to be 9 W/m 2 K. Considering the body as a 28-cm diameter and 1.8-m long cylinder, estimate the
time of death of that man. Assume that the man was healthy when he died with an initial uniform temperature
of 37°C. Take the properties of the body to be k = 0.62 W/m K and a= 0.00054 m2/h.
Known: A man presumably healthy before death is found dead in a room at 7 a.m.
Find: The time of death.
Schematic:
Room
air
111
Tc=I
+
H= I.Sm
T == 17°C
h=9W/K
L=OJ_.9m
l
.......
·......
lniti~lly
T;=31°C
T(r= r0 = 0.14m,
x=O,t)=24°C
X
O
,
Human body modelled
...... as a short cylinder
___..:. k=0.62W/m K
a= 0.00054 m2/h
~-~·:··o:~a·~·:=..i
Fig. 6.38
Assumptions: (1) The man is approximated as a finite cylinder. (2) Two-dimensional transient conduction
in axial and radial directions. (3) Constant properties and heat transfer coefficient.
Transient Heat Conduction
r
6.69
Analysis: Lumped capacity formulation
Let us first find the Biot number for lumped system formulation.
Bi = h 4 where Le is the characteristic length defined as ¥- .
k
A.
For a finite cylinder of radius,
r 0 = 0.14 m and height, H = 2, L = 1.8 m
¥rcr;,2 H
,;,H
L --------=--c - A. - 2rcr;,H +2rcr;
2(r;, +H)
= (0.14m)(l.8m) = 0 _065 m
2(0.14+1.S)m
Bi= (9W/m2 K)(0.065m) =0. 943
0.62W/mK
As Bi> 0.1, the lumped capacity analysis is not appropriate. However, as a guesstimate we can still
find the time, neglecting the internal temperature gradients.
Initial temperature, Ti= T(t = 0) = 37°C
Temperature, T(t) = 24°C
Room air temperature, T= = l 7°C
T(t)-T= = exp (-Bi) (Fo) = exp (-t/'t)
J;-T=
or
Time elapsed after the man's death is
J;-T=
t = -Lek ln --'--ha. T(t)-T=
=
(0.065 m)(0.62 WIm K) Zn (37 -17)°C
(9W/m2 K)(0.00054m2 /h) (24-17)°C
=8.7h
The man might have died at around 10:20 in the night.
Two-dimensional solution
For a more reliable and reasonable estimate of the time of person's death, we model the body as a finite
cylinder viewed as an intersection of a long cylinder and a large plane wall.
6.70 , Heat and Mass Transfer
The transient temperature distribution is expressed as a product solution.
{ T(r,x~t)-T=} ={T(r,~-T=} x{T(x,~-T=}
I; T=
20
T;_ T=
cyl
T;_ T=
wall
The temperature measured at the surface of the body's waist (x = 0, r = ,;, ) is 24 °C. The dimensionless
temperature is
0(r=,;,,x=O,t)lshort =0(,;,,t)lcyl x0(o,t)lwall
cylinder
24-17
0(,;,,0,t)=-- = 0.35
37-17
Now
As time, tis unknown, the product solution cannot be obtained in a straightforward manner. Furthermore,
using the Heisler charts or one-term approximation is subject to the condition that Fo > 0.2 for a fair
degree of accuracy. In the absence of information about time, t, Fo cannot be determined. Hence, the
trial and error solution is called for.
Trial I: t = 8 h [t = 8. 7 h from the lumped system approximation]
Long cylinder: Fourier number or dimensionless time,
F;,*=a.t = (0.00054m2 /h)(8h) =0_ 22
(0.14m) 2
,;,2
As F;,* > 0.2, one-term approximation is valid.
Bi*= h,;, = (9W/m2 K)(0.14m) = 2 _03
k
0.62W/mK
From Table 6.2, for an infinite cylinder, with Bi* = 2.03
A; = 1.6056 rad
and
A;= 1.3410
0(,;,,t)=A; exp(-A; Fo*)J0 (A;}
= 1.3410 exp {-1.60562 x0.22)J 0 {1.6056rad)
From Table 6.3,
J 0 (1.6056 rad)= 0.4522
Hence,
0(,;,,t=8h)lcyl = 1.3410 exp (-0.5682) (0.4522)
= 0.344
Large plane wall t = 8 h
. hL (9W!m2 K)(0.9m)
Bz=-=------k
0.62W/mK
= 13.06
Fo= a.t = (0.00054m 2 /h)(8h) = 0 _0053
L2
(0.9m) 2
Transient Heat Conduction
r
6. 71
As Fo < 0.2, one-term approximation is not appropriate. Transient-temperature charts cannot be used
either. An exact solution is of the form
0 ( x, t)lwall =~A,, exp (-A;,Fo) COS ( An
Z)
At the centre, x = 0 and cos ( An x O) = l
The coefficient An is
A = __4_s_in_A--'n=----
n
(a)
2An +sin2An
where An satisfy the equation:
An tan An= Bi
The first four roots of the above equation, viz., A1,A2 ,A3 and A4 for the specified value of Bi= 13.06 are
obtained from Table 6.4. The values of coefficients A 1, A 2 , A 3 and /4i are then calculated from equation (a).
The coefficients An and A are tabulated below:
IBi= 13.061
11.4553
1.2656
4.3776
--0.403
7.3288
0.223
10.314
-0.148
0(0 ' t)I wall = A I e-'AiFo + A2 e-"-iFo + A3 e-AjFo + '-'4
A e-"-iFo
= (1.2656 exp 1.45532 x 0.0053) + (-0.403) exp ( 4.3776 2 x0.0053)
+(0.223) exp (7.32882 x 0.0053)+(-0.148) exp (10.3142 x0.0053)
= 0.971
0(,;,,0,t)l2D = (0.344) (0.971)
= 0.334
As this is less than 0.35, let us reduce Fo, i.e., time t.
Trial II: Long cylinder: t = 7.4 h and
Fo* = (0.00054m 2 /h)(7.4h) = 0 _204
(0.14m) 2
(> 0.2)
one-term approximation is adequate.
0(r,,,t)lcyl = 1.341 exp (1.6056 2 X 0.204) X (0.4522)
=0.359
Plane wall
at
L2
(0.00054m 2 /h)(7.4h)
(0.9m) 2
Fo=-=-'------'----
= 0.00493
6. 72 ,
Heat and Mass Transfer
0(0,t)lwall = 1.2656 exp (-1.4553 2 x0.00493) + {-0.403)exp (-4.3776 2 x0.00493)
+{0.223) exp (-7.32882 x0.00493)+{-0.148) exp (-10.314 2 x0.00493)
=0.969
0(r,, =0.14 m, x=O, t=7.4 h)
= (0.359) (0.969) = 0.348
As this is almost equal to 0.35, no more iteration is warranted.
t = 1.4 h
or
7 h 24 min
Comment: The man might have died in the night about 7.4 h earlier, i.e., at around 11.30 p.m. Lumped mass
technique indicated the man's death 8.7 h earlier, i.e., at about J0.20 p.m. For a rough estimate, judging by the
relatively simple and less time-consuming lumped capacitance analysis, the result was not wide off the mark!
Example 6.31
~ A short brass cylinder of 8-cm diameter and 15-cm height is initially at a uniform temperature
of 400 K. The cylinder is now placed in atmospheric air at 300 K, where heat transfer takes
place by convection with a heat transfer coefficient of 65 W/m 2 K. Determine (a) the centre temperature of the
cylinder, (b) the centre temperature of the top surface of the cylinder, (c) the temperature at the mid-height
of the side, and (d) the total heat transfer from the cylinder, 15 min after the start of the cooling.
Known: A short cylinder is allowed to cool in atmospheric air under specified conditions.
Find: (a) T(r=O, x=O, t= 15 min) (K), (b) T(r=O, x=L=7.5 cm, t= 15 min) (K), (c) T(r=r0 , x=O,
t= 15 min) (K), (d) Q (kJ).
Schematic:
Top surface [x = L]
T
T
L
X
H= 15cm
'o
O...____.._--+-.c..+
Short brass
cylinder
t= 15min
Initially
T;=4POK
:
Air
!!!
T ==300K
h=65W/m2 K
L
. . . · · · · · · I · · · · · · · · ·. J_
r-
D=Bcm
~
Fig. 6.39
Assumptions: ( 1) The transient heat conduction is two-dimensional and the temperature varies both in
the radial and axial directions, i.e., T= f(r, x, t). (2) Properties and heat transfer co-efficient are constant.
(3) Fo > 0.2 so that approximate analytical solution is justified.
Properties: At average temperature, (400 + 300) K/2 = 350 K
Brass: k= 116 W/m K, CP = 377.5 J/kg K, p = 8530 kg/m3,
ex= k/p
CP = 3.6 x 10-s m2/s.
Analysis: The short (finite) cylinder can physically be formed by the intersection of an infinite cylinder
of radius r 0 = 0.04 m and an infinite plane wall of thickness 2 L = 0.15 mas shown in the schematic. The
Transient Heat Conduction
r
6. 73
solution for a two-dimensional geometry of a short cylinder is the product of one-dimensional solution
for plane wall and cylinder. Thus
0(r,x,t) lsho_rt =0(x,t)I lane wall x0(r,t)linfinite
cylinder
P
cylinder
Plane wall: Half-thickness,
L = 0.15 m/2 = 0.075 m
Biot number,
Bi= hL = (65W!m2 K)(0.075m) = 0 _042
k
116W/mK
Fourier number,
Fo= at= (3.6xl0- 5 m2 /s) (15x60s) = 5 _76
L2
(0.075 m) 2
From Table 6.2: For Bi= 0.042
A1 = 0.2033 rad
and
A1 = 1.0069
At x = 0 (centre of the cylinder):
0(x,t)lx=O = A1 exp(-Af Fo)
= 1.0069 exp (-0.2033 2 x 5.76)
0(0, t) = 0.7935
Infinite cylinder
,;, =0.04 m
Bi= h,;, = (65W/m2 K)(0.04m) = 0 _0224
k
116W/mK
Fo = at= (3.6 x 10-5 m2 /s) (900 s) = 2025
,;,2
(0.04m) 2
From Table 6.2, for Bi= 0.0224:
A1 = 0.2102 rad
and
A 1 = 1.0056
At r = 0 (centreline of cylinder):
0 (r, t)lr=O = A 1 exp (-A? Fo)
= 1.0056 exp[-0.21022 x20.25]
0(0, t) = 0.411
(a) Hence, the centre temperature of the cylinder is given by
0(r,x,t)lcentre = (0.7935) (0.411) = 0.326
6.74 , Heat and Mass Transfer
T (0, 0, t) = T= + 0.326 ('I; -T=)
= 300 K + 0.326 (400 - 300) K
=332.6 K
(Ans) (a)
(b) To evaluate the centre temperature at the circular top surface of the cylinder, r = 0 but x = L.
Plane wall: 0(x,t)lx=L =A 1 exp(-Af Fo) cosA.1
= (0.7935) (cos 0.2033 rad)
= 0.777
Infinite cylinder
0(0,t)lcyl = 0.411
0(0, L, t) = 0(L, t) x 0(0, t)
= (0.777) (0.411) = 0.3194
T (0, L, t) = T= +('l;-T-)(0.3194)
= 300 K+ (0.3194) (400- 300) K
= 331.94 K
(Ans) (b)
(c) At the mid height of the cylinder side,
0(r=,;,,x=O,t =t) = 0(,;,,t)I cy1X0(0,t)lwall
0(0, t) = 0.7935
0(,;,,t)=0(r,t) Ir=,;,
= A1 exp(-Af Fo) J 0 ( \ )
where
J 0 (\)=J0 (0.2102 rad)= 0.9887
0(,;,,t) =(0.411) (0.9887)=0.4064
Therefore, applying the product solution, we get
0(,;,,0,t) = (0.7935) (0.4064) = 0.3225
and
T(,;,,O,t) = T= +0.3225(1;-T=)
= 300 K + 0.3225 (400 - 300) K
=332.25 K
(d) Mass of the short cylinder,
m=p¥- = p1tr; H = (8530 kg/m3 )[ 1t (0.04 m)2 (0.15 m)]
= 6.43 kg
(Ans) (c)
Transient Heat Conduction
r
6.75
Maximum heat that can be transferred from the cylinder,
Qmax =mCp(Ii.-T-)
= (6.43 kg) (377.5 J/kg K) (400 - 300) K = 242.8 x 10 3 J
= 242.8 kJ
Now, we find the dimensionless heat transfer ratios for both geometries.
Plane wall
( _JL_) -(_JL_)
Qmax
I
Qmax
-1-0
0,wall sin\
wall
Ai
= l-(0. 7935 ) sin(0.2033 rad)
0.2033
= 0.212
Infinite cylinder
=(_JL_)
( _JL_)
Q
Q
max
With
2
=1-20 o,cyl J, A
(\)
max cyl
I
J 1(A 1) =J1 (0.2102) = 0.1045
( _JL_) = 1 - 2 (0.411) (0.1045)/0.2102
Qmax 2
= 0.5914
:. the heat transfer ratio for the two-dimensional short cylinder is
(Q:1 =(Q:J +(Q:..J, HQ:Jl
= 0.212+(0.5914) (1-0.212)
= 0.678
Hence,
the total heat transfer from the cylinder during the first 15 minutes of cooling is
Q = 0.678 Qmax
= (0.678) (242.8 kJ)
= 164.63 kJ
(Ans) (d)
Comment: This problem could have also been solved by using Heisler and Grober charts. However, for
values of Bi < 0.1, it is preferable to go in for one-term analytical solution for greater accuracy.
6, 7 6 , Heat and Mass Transfer
Example 6.32 ~ Short plastic cylinders 6-cm long and 3-cm in diameter are uniformly heated in an oven as
preparation for a pressing operation. Initially, the cylinders are at 25°C. The manufacturing
requires that no portion of the plastic be below 200°C. The oven temperature is 250°C, and the heat-transfer
coefficient on the cylinders is 8 W/m 2 K. The plastic properties are: k = 0.30 W/m K, and pCP = I040 kJ/m 3 K.
The supplier of the oven states that the plastic cylinders should be in the oven for 15 minutes. Determine (a) the
minimum temperature in the cylinder (in °C) after 15 min (b) the maximum temperature in the cylinder (in °C) after
15 min, and (c) the rate at which heat must be added to the oven if I00 cylinders per minute are heated (in W).
Known: Short plastic cylinders are heated in the oven.
Find: (a) Minimum temperature of cylinder, (b) Maximum temperature of the cylinder, and (c) Heat
transfer rate for 100 cylinders.
Schematic:
T
Oven
111
2L=0.06m
Tmax=
T(ro, L, t= 15min=?)
lniti~lly
T;= ~5 °C
Tmin= T (0, O, t = 15 min)=?
1
h=8W/m2K
T==250°C
....JV\J'-
Plastic cylinder
(pCP = I040 kJ/ml K,
k=O.JOW/m K)
Fig. 6.40
Assumptions: (I) Two-dimensional conduction in x-and r-directions. (2) Constant properties. (3) There is no
internal heat generation. (4) The heat transfer coefficient is uniform over the surface. (5) Radiation is ignored.
Analysis: (a) The characteristic length for lumped capacity formulation is Lc = ¥-IA. The volume is ¥- = 2rcD 2
L/4 = 2rc (0.03 m) 2 (0.03 m)/4 = 4.241 x 10-5 m3, and the surface area is A = 2rcDL + 2rcD2L/4 = 2rc (0.03 m)
4 24lxI0-5
(0.03 m) + 2rc (0.03 m)2/4 = 0.00707 m2, which gives a characteristic length of ·
= 0.006 m.
Therefore
0.00707
'
hLc
(8W/ m 2 K}(0.006m)
k
(0.3W/ mK)
Bi=-=-'------'--'----'-=0.16 (> 0.1)
Thus, the lumped capacity model is not exactly valid. A two-dimensional transient conduction analysis
would be more appropriate.
A short cylinder is created from the intersection of an infinite plane wall and an infinite cylinder:
[T(x,t)-T=]
[T(r,t)-T=]
[T(r,x,t)-T=]
T; - T=
short
= T; - T= plane X T; - T= infinite
cylinder
wall
cylinder
Transient Heat Conduction
r
6. 77
Thus, we need to evaluate the dimensionless temperatures for the plane wall and the infinite cylinder at
both the centreline and at the edge of the end of the cylinder. For the plane wall, the characteristic length is
L=0.03 m
Fourier number,
Fo = a.t = _!!._ =
(0.3W/mK)(900s)
= 02885
2
2
L
pCPL
(1040x10 3 J/m3 K)(0.03m}2
As Fo is greater than 0.2, we can go ahead with the one-term approximation.
Biot number,
. hL (8W/m2 K)(0.03m)
Bz=-=-------=0.8
k
0.3Wm/K
At this Biot number from Table 6.2, we obtain A. 1 = 0.7910 and A 1 = 1.1016. The non-dimensional
temperature at the centreline of the infinite plane wall (x = 0) is given by
'1 2
00,plane wall -[T(0,900s)-T=]
- -----'---~-_ A, exp (-11,,
Fo )
I; -T=
plane
wall
= 1.1016 exp (0.7910 2 x0.2885) = 0.9197
For the infinite cylinder, its Biot and Fourier numbers are
. h,;, (8W/m2 K)(0.015m)
Bz = - = - ' - - - - ~ - - ~ = 0.4
k
0.3W/mK
Fo = a.t = _!_!__ =
(0.3W/mK)(900s)
= 1.1 54
2
2
3
3
,;,
p CP ,;,
(1040 x 10 J/m K)(0.015 m}2
At this Biot number from Table 6-2 we obtain A 1 = 1.0932 and A. 1 = 0.8516. The non-dimensional
temperature at the centreline of the infinite cylinder (r = 0. t = 900s) is expressed as:
00,infinite = [T{0, 900s )-T=]
cylinder
I; - T=
infi,nite
cylinder
J
= A 1 exp(-Af Fo) = 1.0932expl-(0.8516}2 (1.154) = 0.4734
We note that
(
T(O,O,t)-T=)
= 0wall (O,t)x0cyl (O,t)
I; -T=
short
cylinder
Hence,
Tmin -_ T (0,0,900s )- T= + ( I; _ T= )[T(0,900s)-T=]
-----'----'--x [T(0,900s)-T=]
---'---~I; - T=
plane
I; - T=
infinite
wall
cylinder
= 250°C + (25°C - 250°C) (0.9197) (0.4734) = 152°C
(Ans) (a)
6.78 , Heat and Mass Transfer
(b) The same approach is used to find the temperature at the edge of the end of the cylinder. We
determine the surface temperature of the plane wall (x = L = 0.03 m) and that of the infinite cylinder
(r = r 0 = 0.015 m). For the plane wall,
[
T(0.03m,900s)-T=]
[
(
)]
(
)
1j-T=
plane= A 1 exp -A,rFo cos \x/L
wall
= 80,wall xcos\
= (0.9197)(cos 0.7910)
= 0.9197 X 0.703 = 0.6467
For the infinite cylinder,
T(0.015m,900s)-T=]
- - - - - - - . . -[
- A1 exp (-A21 Fo )] J 0 ( A1r/ro )-8
[0 ,cy1 xJ0 ( \ )
1j-T=
mfimte
cylinder
= (0.4734){J0 (0.8516)} = 0.4734x0.8263
= 0.3912
Finally, for the edge temperature,
Tmax = T,(0.03 m, 0.15 m, 900s)
_
-T
+ (T _ T )[T(0.03s)-T=]
~-~-=
1
=
1j - T=
plane
x [T(0.015s)-T=]
1j - T=
wall
infinite
cylinder
= 250°C+(25°C-250°C) (0.6467) (0.3912)
= 193°c
(Ans) (b)
(c) For the heat transfer into one of the short cylinders, we use the following expression.
[LL =[Ll,+[LlHLll
along with the appropriate expressions for an infinite plane wall and an infinite cylinder.
For the heat transfer into the plane wall,
= 1-0 plane sin\ = 1-(0.9197) sin(0. 79 lO) = 0.173
(_JL_)
Qmax plane
Ai
0.7910
0
'wall
wall
For the heat transfer into the infinite cylinder,
=1-20 O,mfimte
· · J,(\) =1-2(04734) 03879
(_JL_)
l 8516 =0569
·
Qmax infinite
cylinder
'I
cylinder
11,l
•
•
Transient Heat Conduction
r
6. 79
The maximum possible heat transfer into one cylinder is
Qmax = mCP (T=
-J;) = p¥-CP (T= -J;)
=(1040xl0 3 J/m 3 K) (4.24lxl0- 5 m3 )(250-25)°C
= 9924.3 J
= (9924.3 J) [0.173 + 0.569 (l - 0.173)] = 6385 J
(Ans) (c)
For the heat-transfer rate to process 100 cylinders per minute:
Q = (6385 }/cylinder) (100 cylinders/ min) (l min/ 60 s)
= 10642 W = 10.64 kW
Example 6.33
(Ans) (c)
~ A long lnvar (36% nickel steel) rod [k = 10.7 W/m K and a= 0.0103 m2/h) 10-cm diameter
has one end machined flat perpendicular to the axis. The rod is initially at 320°C and is set
to cool in 20°C air such that the heat transfer coefficient on the surface is 43 W/m 2 K. What is the temperature
at a point 2 cm below the cylindrical surface and 4 cm in from the flat end after 20 min?
Known: A long rod with one flat end (semi-infinite cylinder) is allowed to cool in convective atmosphere.
Find: Temperature, T(r = 3 cm, x = 4 cm, t = 20 min) [0 C] .
f-- D= 10cm --1
Schematic:
Air
r==20°c
h=43W/m2 K
lni~ially
T;=~20°c
0(r, X, t) = 0cyl(r, t) X0semi-inf (X, t)
T(r, x, t) =?
X
lnvar
k= 10.7W/mK
a=0.0103 m2/h
~--_._,_....._.r
r0 = 5cm
Semi-infinite cylinder
Fig. 6.41
Assumptions: (l) Two-dimensional transient conduction in radial and axial directions. (2) Constant
properties and uniform heat-transfer coefficient. (3) Fo > 0.2 so that one-term approximation is valid.
Analysis: Solution is obtained as a product solution using the one-dimensional long cylinder and
semi-infinite solid solutions.
T(r,x,t)-T=
- - - - - = ecyl(r,t)x0semi-inf (x,t)
J; -T=
solid
6.80 , Heat and Mass Transfer
Long cylinder
Location at 2 cm below the cylindrical surface, i.e., at a radius, r = (5 cm - 2 cm)= 3 cm.
Time, t = 20 min= (20/60) h or 1/3 h.
h,;,
Bi=-
The Biot number,
k
= (43W/m2 K)(0.05m) = 0 _20
10.7W/mK
With Bi= 0.20, for an infinite cylinder, from the chart,
A.1 = 0.6170 rad
and A 1 = 1.0483
r
3cm
r = 3 cm, - = - - = 0.6
,;, 5cm
At
Hence,
at
(0.0103m2 /h)(l/3h)
(0.05m) 2
Fo=-=-------
where
,;,2
= 1.373
Also
J 0 ( \ rl,;,) = J 0 (0.6x0.617) = J 0 (0.37) = 0.9655
Hence,
ecyl(r, t) = 1.0483 exp (-0.6170 2 xl.373)x 0.9655
=0.60
Semi-infinite solid
Location at 4 cm in from the flat end surface, i.e.,
x=0.04 m
T(x,t)-T=
esemi-inf(x = 0.04 m, t = 1/3h) = - - - -
T. -T-
(hx h at)[ erfc [- x- +h.Jai)]
k k
2.Jai -k-
x)
[2.Jai
2
= 1- erfc - - +exp - + 2
Now,
_x_=
2.Jai
0.04m
= 03413
2
2~(0.0103m /h)(ll3h)
h.[cii 43W/m2 K~(0.0103m2 /h)(ll3h)
- - = - - - ~ - - - - - - = 0.2355
k
10.7W/mK
hx = (43W/m K) (0.04m) = 0 _1607
2
k
10.7W/mK
Transient Heat Conduction
r
6.81
h [hfcii) =0.05544
2
2
- -a.t
= -k2
k
erfc = [ 2
Jn,)
= erfc (0.3413) = 0.6293
2 a.t)
hx+hexp= ( - =exp(0.1607+0.05544)=1.2413
k
erfc[
~+
2"a.t
k2
hJai)=
erfc {0.3413+0.2355} = erfc (0.5768) = 0.4147
k
Substituting these values, one gets
esemi-inf(x,t) = 1- 0.6293 + (1.2413) (0.4147)
=0.8855
Therefore,
0(r,x,t) = (0.60) (0.8855) = 0.5313
Temperature at the specified position is
T(r, x, t) = T= +0.5313(~ -T=)
= 20°C + (0.5313) (320 - 20)°C
= 179.4°C
(Ans)
Comment: Fourier number, Fo = 1.373 (> 0.2). Hence, the approximate analytical solution is reasonably
accurate. Alternatively, with less effort we can get the answer by making use of the appropriate charts.
From the chart. With Bi= 0.20 and Fo = 1.373
T(O,t)-T= = 0.625
~-T=
From Figure 6.70, with Bi= 0.2 and rlr,, = 0.6
T,
(r/r0 )
-T
= = 0. 96
~-T=
ecyl (r, t) = (0.625) (0.96) = 0.6
From the chart, with hx = 0.1607 and
~ = 0.3413
k
2"a.t
esemi-inf(x,t) = 0.88
0(r, x, t) = (0.6) (0.88) = 0.528
T(r, x, t) = 20 + (0.528) (320 - 20)
= 178.4°C
(Ans)
6.82 , Heat and Mass Transfer
(G) SEMI-INFINITE MEDIUM
•
•
Example 6.34 ~ Two large bodies I and 2 with thermal conductivities and thermal diffusivities k 1 and k2,
and a: 1 and CXi are initially at temperatures T1 and T2 at their plane surfaces. If the plane
surfaces are placed in contact with each other, show that under equilibrium conditions the contact surface
temperature, T5, is given by
Jci;) +( k2T2 t .ja;)
(/<ii Ja;) +(k2 t .ja;)
( /<iTj/
f.
A large mass of steel (k=43 W/m °C and a= l.17x 10-5 m2/s) at l00°C with one plane face, is dropped into
water (k = 0.589 W/m Kand a= 1.41 x I0-7 m2/s) at I 5°C. Determine: (a) the temperature at the surface of contact.
(b) How much time will elapse before the temperature at a location 2.0 m inside the surface will reach 95°C.
Known: Two large bodies at different temperatures are brought into physical contact with each other.
Find: (a) Contact surface temperature, T.. (0 C), (b) Time required, t (days) at a depth of 2 m below the
steel surface to reach 95°C.
Schematic:
Steel
x=O, t=O
Ts
(90.S?°C)
q1
CD
T2
s·q
(1
Lx
Fig. 6.42
T2 <T;, <J;
k (W/m °C}
a m2/s
Steel
43.0
1.1 7 X I0- 5
Water
0.589
1.41 X I0- 7
Assumptions: ( 1) Constant properties. (2) Contact surface remains at constant temperature Ts (no contact
resistance). (3) Both bodies are large, (A 1 andA 2 are high), treated as semi-infinite media.
Transient Heat Conduction
r
6.83
Analysis: Surface energy balance
Heat lost by body 1 (metal)= Heat gained by body 2 (water)
q1I
= Q ="1(1;-r.)
x=O
q2
Al
I -Q-_k2(T.-Ti)
r:;;:-;::-;
x=O A2
Now,
Jre<X/
'11C(J,2f
=q2(x=O)
k1Cz;-T.) "2(T.-Ti)
ql(x=O)
Jrca t
1
Jrca t
2
or
or
or
Contact surface temperature in equilibrium condition is
T_ =
s
(/ci1i /.[ci; )+~7; / ..[ci;)
( lei 1.[ci;) + (lei 1..[ci;)
Hence proved
Substituting the appropriate numerical values in the above expression, the equilibrium surface temperature is
(43 W/m 0 C) (100°C)
(0.589W/m 0 C)(l5°C)
J1.11 x 10-5 m 2 /s
Jl.4lx 10-7 m2 /s
-,=====~-+-,=====~-
T,=
=
(
43W/m°C
( 0.589W/m°C ]]
J1.17 xI0- 5 m2 /s + Jl.4l xI0- 7 m2 /s
(l.257lxl0 6 ) +(0.02353x 106 )
12571 + 1568.58
= 1.2806 X 1Q6 = 90.57oC
14139.58
(Ans) (a)
The mass of steel initially is 100°C and its surface temperature is suddenly changed to 90.57°C. For
a depth of 2 m such that the temperature reached is 95°C, the time required can be estimated form the
following expression.
T(x,t)-T.
c ()
- - - =er1c
z
r. -r.
where
X
z=--
J4at
6.84 ,
Heat and Mass Transfer
95-100
erfc (z) = - - - - = 0.53
90.57-100
or
z=0.444= _x_
~4cu
From Table 6.3,
Time needed for T(x = 2 m, t) = 95°C is
(x / z) 2
(2 .0m / 0.444) 2 I lday I
t = -4a = 4(1.17xI0-5 m2 /s) 24x3600 s
= 5.0 days
Example 6.3S
(Ans) (b)
~ A large mass of a material (thermal conductivity, k = 395 W/m °C, Thermal diffusivity,
a= 11.25 x I0-5 m1/s) is initially at a uniform temperature of I00°C. Suddenly its surface
temperature is dropped to and held at 0°C. Determine (a) the temperature 1.8 cm below the surface after
2 s, (b) the time needed to decrease the temperature to 52°C at a point 3 cm below the surface, (c) the time
required for the temperature gradient at the surface to reach 4.7 Klem, (d) the depth below the surface at
which the rate of cooling is maximum after 100 s and the maximum cooling rate (K/h), (e) the maximum heat
removed per m1 area during I00 s, and (f) the maximum heat transfer rate per m1 after I00 s.
Known: A large mass of material initially at 100°C is suddenly cooled to and maintained at 0°C.
cJT
Find: (a) T (x = 1.8 cm, t= 2 s) (0 C); (b) t (T= 52°C, x = 3 cm); (c) t (s) for dx = 4.7 Klem; (d) x (cm)
(cJT)
dt
.
for maximum cooing after 100 s, -
(K/h); (e) Q (J/m2) ; (t) Q(W).
max
Schematic:
T (0, t) = T5 = T= = 0°C
T Initially
X
x= 1.8cm
T;= I00°C
T (x= 1.8cm, t= 2s) =?
L
Semi-infinite medium
[a= 11.25 x IQ-Sm2/s]
Fig. 6.43
Assumptions: ( 1) The large mass is approximated as a semi-infinite solid subjected to sudden change in
its surface temperature (h~
x
00 ) .
(2) Constant material properties.
l 8xI0-2 m
Analysis: (a) z =--=-==·======
2
Jcii 2~1 l.25x 10-s m /s x 2s
2
=0.6
From Table 6.3, erfc(z) = erfc (0.6) = 0.3961
For a semi-infinite medium,
T(x,t)-1';
_
I;, T;
= erfc
[
x )
C:: = 0.3961
2vat
Transient Heat Conduction
r
6.85
T(x=l.8cm, t=2 s)='Z;+0.3961 ('f.-7;)
or
= 100°C + 0.3961 (0 - 100)°C
(Ans) (a)
=60.4°C
T(x,t)-1; 52-100
(b) - - ~ = - - - = 0.48 = erfc (z)
T. -1;
0-100
From Table 6.3,
z = 0.50 = x/2.Jai
Hence, time needed is
(x/0.5) 2
(0.03m/0.5) 2
t=---=------4a
4xll.25x10-5 m2 /s
(Ans) (b)
=8.0 s
(c) We note that the instantaneous heat-flow rate,
Q=-kA(aT)
= kA(l;-7'.)
ax x=O
~TC<ll
The temperature gradient at the surface (x = 0) is
1;
( ar)
a
X
-r.
= ~TC<ll
x=O
Therefore, time required
I )
2
1
(100-0)°Cor K
- rc(ll.25x10- 5 m2 /s)
(100cm)
4.7K/cm - lm
= 128 s
(Ans) (c)
(d) Rate of change of temperature with time, i.e., rate of cooling is given by
ar -(T-T.)x e-x,!4at
at
Fa x - - l
2
s
t312
Differentiating with respect to x, and setting the resulting derivative to zero, we have
j_(ar)=
ax at
(7;-T.) [e-x2/4at+x·e-x 2!4at(-~)J=o
2.fia-1312
4at
6.86 , Heat and Mass Transfer
[1-~J
or
2a.t
=0
x = ~2a.t
or
= ~2(11.25xl0- 5 m2/s) (100s) llOlO~ml
(Ans) (d)
=15cm
maximum rate of cooling,
(ar)
at
= -(100-0)°Cx0.15 m x exp-{0.15 2 m2/4xl 1.25x10-5 m2/sxlOOs}
max
2 ~(J. X 11.25 X lQ-5 m2/s
( 100 S) 312
= -0.242°C per s or -871 K/h
(Ans) (d)
(-ve sign implies cooling)
(e)
The maximum heat removed during 100 s per unit area is
Q= 2 k A(T,. - I;)
~Tea.It
=
2(395W/m 0 C)(lm2)(0-100)°C
~rcxl 1.25x10-5m2/s/100s
= -42.0 x 10 6 J or -42.0 MJ
(Ans) (e)
(-ve sign indicates heat lost by the body)
(f)
The maximum heat transfer rate per unit area after a period of 100 s is
.
kA(I;-T,.)exp(-x2 !4a.t)
Q=---~~-~TC(J.t
(395W/m°C) (lm2) (100-0)°C exp-(
0 · 152 m2
)
4xl 1.25xl0-5 m2 /sxlOOs
~rcxl 1.25xl0-5 m2/sxlOO s
=-127.44 x10 3 W or -127.44 kW
(Ans) (f)
(-ve sign indicates heat loss).
Example 6.36 ~ (a) A very strong blizzard suddenly reduces the surface temperature of the earth in an open
area to -25°C and stays that way for a 36 h period of time. If the ground was initially at
I 5°C, estimate how far the freezing temperature would penetrate, neglecting any latent heat effects due to the
soil moisture, using a value of thermal diffusivity of 0.83 mm 2/s.
(b) If the weather front moves in, drops the temperature quickly to -25°C, then becomes calm instead of
blizzard so that a heat transfer coefficient of 28 W/m 2 K exists at the surface. The thermal conductivity of the
soil is 0.43 W/m K. Estimate the depth of penetration of freezing temperature under these conditions in 6 h.
Transient Heat Conduction
Known: Temperature imposed at the earth's surface initially at 15°C.
Find: (a) Depth of penetration of freezing temperature, x (m) for t = 36 h, h ~
(b) Depth of penetration, x (m) fort= 6 h, h = 28 W/m2 K.
Schematic:
Semi-infinite solid
r
6.87
00
=i
t
Soil
k=0.43W/mK
a=0.83x lo-6m2/s
x=?
_+_
T(x, 36h)=0 °C
Initially
T;= IS °C
Fig. 6.44
Assumptions: ( 1) One-dimensional heat conduction. (2) Soil is treated as a semi-infinite medium with
constant surface temperature ( h ~
due to blizzard. (3) Latent heat effects due to soil moisture are
negligible. (4) Thermal properties of soil are constant.
Analysis: (a) Constant surface temperature
00 ) ,
hfoj
As - - =
k
00
(
since h ~
00 ) ,
the transient temperature response can be expressed as
T(x,t)-1';
c (
----=er1c
-x-)
7'. -1';
2Jai
(0-15)°C = erfc(
x(m)
(-25-15)°C
2 ~0.83x 10-0 m2 /sx 36x3600 s
or
or
l
0.375 = erfc (1.5245 x)
From Table 6.4:
For erfc (z) = 0.375, we read
(after linear interpolation)
Z = 0.627 = 1.5245 X
0.627
depth, x = - - = 0.41 m
1.5245
(b) Convection at the exposed surface
Now
t= 6, h = 6x3600s, h = 28 W/m2 Kand k = 0.43 W/m K.
hfoj 28W/m2 K , - - - - - - - - - - - = - - - - x ,J0.83xl0-6 m2 /sx6x3600s = 8.72
k
0.43W/mK
T(x,t)-T= = [0-(-25)]°C = 25 =0. 625
T;-T=
[15-(-25)]°C 40
From the chart, read
X
C" = 0.56
2 "\/CXt
(Ans) (a)
6.88 , Heat and Mass Transfer
Depth of penetration,
x [T= 0°C, t= 6 h] = 2 x0.56x0.134
= 0.15 m
(Ans) (b)
Example 6.37 ~ A thick aluminium block at 20°C is subjected to constant surface heat flux of 1000 W/m 2•
Determine the temperature of the block at the surface and at a depth of 0.5 m from
the surface after 20 minutes. If the block were made of concrete instead of aluminium, what would be the
corresponding temperatures under identical conditions. Comment on the results obtained.
Material
Thermal conductivity (Wlm 1 0 C)
Thermal diffusivity(m 1 /s)
Aluminium
237
9.71 x 10- s
Concrete
1.06
5.19 x Io-7
Known: A thick slab is subjected to constant surface heat flux.
Find: T (0, 20 min), T (0.5 m, 20 min) for aluminium block and concrete block.
Schematic:
T (x, t)
46.46°C
2 1. 6oc __f Aluminium
T;= 20°C ,____-_-_-_-_-_-_-_-_-.:...c-=-=-.::.--=-c...:-=-. ~8.3
Block
Initially
T;=20°C
0
O.Sm
Distance from surface, x(m)
(a)
(b)
Fig. 6.45
Assumptions: ( 1) The block is sufficiently thick to be considered a semi-infinite solid. (2) Constant
properties. (3) All heat flux is absorbed by the block.
Analysis: For a semi-infinite solid exposed to specified surface heat flux,
l/2]
4at
x
x
- - exp (- - - )-xerfc (- - )
[~
k
1t
4at
4at
%
2
T(x,t)-T = -
'
The surface temperature, T(O, t) can be expressed as
T.. =T+ % )4at
S
I
k
1t
2
Transient Heat Conduction
r
6.89
For aluminium block
T.
s,Al
= 200 C+ 1000W/m2 4x9.71x105 m/sx(20x60)s
23 7 W/m2 °C
1t
= 20 + ( 4.22 X 0.385) =21.6°C
(Ans)
For concrete block
T,. concrete = I; + -% ~4a.t
'
k
1t
=20+ 1000 f4x5.19xlQ-7x20x60 =20+(943.4x0.0282)
1.06
1t
v
= 46.6°C
(Ans)
At a depth of x = 0.5 m after t = 20 min, the temperature is found to be the following:
For aluminium block
x2
4a.t
0.52
4x9.71x10-5 x20x60
=0.5364
and
( _±_)
4a.t
112
= 0.7324
exp(-_±_)= exp(-0.5364) = 0.585
4a.t
2 )
erfc ( ~
4a.t
TAI
112
=erfc(0.7324)=0.300
(x = 0.5m) = 20 + 4.22[(0.385 x 0.585)-(0.5 x 0.3)]
=20.3°C
For concrete block
0.5 2
= - - - - - - - = 100.35
4a.t 4x5.19x10-7 x20x60
x2
exp[-_±_]= exp(-100.35) ss 0
4a.t
( _±_)
4a.t
2 )
erfc ( ~
4a.t
112
112
= .J100.35 = 10.02
=0
(Ans)
6.90 , Heat and Mass Transfer
T.:oncrete (x = 0.5m)= 20+943.4 [{0.0282) {0)-{0.5){0)]
(Ans)
=20°C
Comment: Thermal diffusivity ex of aluminium is about 50 times that of concrete. The larger the value
of ex, the faster is the heat spreading process within the medium . The heat flux supplied to the concrete
accumulates near the surface due to low thermal conductivity and diffusivity of concrete. This causes the
surface temperature to rise to high value. In the case of metals like aluminium with higher conductivity
and diffusivity, the temperature rise, at the surface will be much less At x = 0.5 m, for aluminium the
temperature reduces to 20.3°C and for concrete it falls to almost the initial temperature 20°C. This shows
that the heat penetrates faster and further in aluminium compared to concrete.
(H) MULTI-DIMENSIONAL CONDUCTION
•
•
Example 6.38 ~ In the heat treatment of a steel billet [k = 43 .3 W/m °C, a= 1.471 x 10- 5 m2/s] in the form
of a rectangular parallelepiped, measuring 0.61 x 0.61 x 1.22 m, it is required that the billet
be heated to just about 370°C. To do this, the billet, initially at a uniform temperature of 35°C, is placed in
an oven which provides an atmosphere of inert gas at 400°C with a convective heat transfer coefficient of
568 W/m 2 0 C. How long must the billet remain in the oven for its centre temperature to reach 370°C.
Known: A steel billet in the shape of a rectangular parallelepiped is required to be heated to a specified
temperature.
Find: Time required, t (h).
Schematic:
Inert gas
-f
2 L1 =0.61 m
Initially
T;= 35°C
T(O, 0, t) = 370°C
f----------i----t_=
1
----{
k=43 .3W/m°C
a= 1.471 x I0-5m2/s
l ._____ _ _____,
111
T==400°C
h = 568 W/m2 °C
Fig. 6.46
Assumptions: ( 1) Three-dimensional transient heat conduction. (2) Constant properties and convection
coefficient. (3) Fo > 0.2 so that approximate one-term analytical solution is applicable.
Analysis: We note that
00 lrectangular =01,wan (O,t) x02,wan(O,t)x 0 3,wan (O,t)
parallelop1ped,
As
0 = T(0,0,0,t)-T= = 02 (0 t)·0 (0 t)
0
T-T
I
,
3
,
I
=
Transient Heat Conduction
r
6.91
Bi= hLi_ = (568W/ m2 °C)(0.305m) = 4 _0 = Bi
2
I
k
43 .3W/m°C
Fo =at= (1.471x 10-s m2 /s) (t(s)) = 158.13 x 10-6 t Fo
2
I
q
(0.305m) 2
Bi =hL 3 = (568W/m2 °C){0.61m) = 8 _0
3
k
43 .3Wlm°C
Fo =at= (1.471x 10-s m2 /s) (t(s)) = 39.532 x lQ-6 t
3
L1
(0.61 m) 2
From Table 6.2, we obtain the values of A. 1 and A 1 for each value of Bi.
Wall #
Bi
}.,1 (rad)
1, 2
4.0
1.2646
3
8.0
1.397 8
Fo
1.f Fo
1.2287
158. 13 X lQ- 6 t
25 2.88 X lQ- 6 t
1.2570
39.53 X lQ- 6 t
77.24 X lQ- 6 t
0o = ef (0, t). 03 (0, t)
=[A I e -200x10-• -¼alll
i1 [A e-77.24 x J0-6' l
I
Jwall 3
1
Taking log on both sides,
/n = 0o = 2 x [zn Al -(252.88 x 10-6 t )] walll ,2 +[In Al -(77.24 x I0-6 t)] wall 3
or
370-400)
/n ( - - = 2 [!nl.2287-(252.88xl0-6 t)] + [!nl.2570-(77.24xl0-6 t)]
35-400
or
30
583 x I0-6 t = 2 /n 1.2287 + /n 1.2570 - /n 365
= 3.1393
time required,
3.1393x 106
583
t= - - - - s = 5385 s"' 1 h 30 min
(Ans)
Example 6.39 ~ A cylindrical granite block (k = 2.5 W/m K and a= 1.15 mm 2/s) of 50-mm diameter as well
as height is to be compared for transient thermal response with a cubical granite block
with each side 50-mm long. Both blocks are initially at 25°C throughout and both are exposed to hot gases at
525°C in a furnace on all of their surfaces with a heat transfer coefficient of 50 W/m 2 K. Estimate the centre
temperature of each geometry after IO, 20 and 60 min.
Known: A cube and short cylinder of the same material are heated up from the same initial temperature
in the same convective atmosphere for the same duration.
6.92 , Heat and Mass Transfer
Find: T(r = O,x = O,t)lfinite
and T(O, 0, 0, t)I cub e for t = 10, 20, and 60 min.
cylinder
Schematic:
Short cylinder
(L = r0 = 0.025 m)
T
- - - _j
Hot gases
L
2L
X
t
L
_l
Cube (L=0.025m)
Initially
T;= 25°C
!!!
Initially
T;= 25°C
h=SOW/m2K
T =525°C
00
2l
2L
Granite:
k=2.5W/mK,
a= I.ISx lo-6m2/s
Fig. 6.47
Assumptions: (1) Two-dimensional conduction in radial and axial directions of the cylindrical block, and
three-dimensional conduction in the case of cubical block (2) Constant thermal properties and uniform heat
transfer coefficient on all surfaces of the object. (3) Fo > 0.2 so that one-term analytical solutions are valid.
Analysis:
Cylindrical block: A cylinder with a specified length (or height) is a finite or short cylinder and should
be analyzed as a two-dimensional geometry. The transient temperature distribution is given by
i.e.,
0(r,x,t)l2D,cyl =0cyl = (r,t)x0wan(x,t)
At the centre of the cylindrical block, r=x=O
0o,short = 0cyl (0, t) X 0wall (0, t)
cylinder
To find the one-dimensional solution of infinite cylinder and infinite plane wall, we use the one-term
approximation analytical method.
For an infinite cylinder
Bi= h,;, = (50W/m2 K)(0.025m) = 0 _5
k
2.5W/mK
For Bi= 0.5, the coefficients A. 1 and A 1 are obtained from Table 6.2:
A- 1 = 0.9408 rad and A 1 = 1.1143
With
Fo= at
r;
(l.15xl0-6m 2 /s)(t,min) 160s I
(0.025m) 2
lmin
= 0.1104 t
where t is in min, we have
Transient Heat Conduction
r
6. 93
= 1.1143 exp l-(0.9408}2(0.1104t)j
= 1.1143 exp (-0.0977 t)
with t in min
For an infinite plane wall
Bi= hL = (50W/mK) (0.025m) = 0 _5
k
2.5W/mK
From Table 6.2, for Bi= 0.5
A1 = 0.6533 rad
and
A 1 = 1.0701
at
Fo = - = 0.1104 t because L = r 0
Also,
£2
0wan(O,t) = A 1 exp(-Af Fo)
= 1.0701 exp [-(0.6533}2{0.1104t)]
= 1.0701 exp (-0.04712 t)
with t in min
Hence, the product solution gives
0(0,0,t)l 20 cyl = 1.1143 exp (-0.0977 t) X 1.0701 exp (-0.04712 t)
or
100,shortcyl = 1.1924 exp (- 0.1448 t) withtinminl
and
7;,,shortcyl = 80,shortcyl (7; -T=)+T=
(a)
where T= = 525°C and
7; - T= = 25 - 525 = -500°C
(b)
To,short cylinder = 525 - 50000,short cylinder
Cubical block
The transient temperature response can be expressed as
( T(x,y,z)-T=)
T-T
1
=
=(T(x,t)-T=)
(T(y,t)-T=)
T-T
1
=
T-T
1
=
cube
walll
(T(z,t)-T=)
wall2
where x= y=z=L = 0.025 m
At the centre of the cubical block:
8 0 lcube =0(0,0,0,t) =[0wan (O,t)]3
T-T
1
=
wall3
6.94 ,
Heat and Mass Transfer
To find the one-dimensional solution using one-term approximation method,
hL
where A 1 and A- 1 for plane wall at Bi= -=0.5 are same.
k
[Note that= L = 0.025 m as in the earlier case]
at
Fo=-=0.1104 t
L2
Also, for the same reason,
ewa11(0,t) = 1.070lexp(-0.04712 t)
Finally dimensionless centre temperature of the cube is
10o,cube ={1.070lexp(-0.04712t)}3with tin mini
(c)
~.cube = 525 -
(d)
and
where
500 0o, cube
The following table succinctly presents the salient results when
i; = 25°c and T= = 525°C
T= 10 min
T = 20 min
T=60min
e o,short cy l Eq. (a)
0.2803
0.0659
0.0002
eo,c ube Eq. (C)
0.298
0.0725
0.000254
384.87
492.05
524.9
376.0
488.75
524.87
To,short cyl Eq.
To,cube Eq. (d)
(b)
(Ans)
Comment: As time passed increases, the centre temperatures of the two blocks increase as expected.
At t= 60 min= 1 h, the temperature~ =T=. Also, the short cylinder (cylindrical block) geometry yields
higher centre temperatures than the cubical block but as the time progressed increases, the difference
between the two decreases and at t = 60 min, the two values are almost the same and both are nearly
equal to the temperature of the hot gases.
Example 6.40 ~ A large billet of steel initially at 275°C is placed in a large radiant furnace where the surface
temperature is maintained at I 250°C. Assuming the billet infinite in extent, determine the
temperature at point P shown in the accompanying figure after 30 min have elapsed. The average properties of
steel are k = 28.2 W/m K, p = 7360 kg/m 3, and CP = 500 J/kg K.
Transient Heat Conduction
r
6.95
.
p
_
Fig. 6.48
Known: A large billet extending to infinity in x, y and z directions is placed in a large radiant furnace.
Find: Temperature at point P (x = 5 cm, y = 25 cm and z = 0)
Schematic:
Initially
T;= 275°C
\
p
Surface temperature
Corner of billet
T, = 1250°C
Large radiant furnace
Fig. 6.49
Assumptions: (1) Conduction occurs in x-and-y-directions. (2) Uniform and constant thermophysical
properties. (3) Uniform temperature in the z-direction.
Analysis: We use the product of one-dimensional solutions to obtain the temperature for two-dimensional
conduction in a semi-infinite solid.
The temperature distribution as a function of one spatial coordinate and time with a sudden change in
surface temperature is given by the following :
Constant surface temperature:
T (0, t) = T8
6. 96 ,
Heat and Mass Transfer
ex,semiinf =
and
ey,semiinf =
0 = 0x semi inf X 0 y semi inf =
·
T(x,t)-7'.
(
x )
c::
2vat
T(y,t)-7'.
(
= erf
y )
c::
2vat
T(x,y,t)-7'.
(
Ii. - T.
= erf
i; T.
I;-T.
= erf
x )
c:: erf ( yc:: )
2vat
2vat
We note that
a = _k_ =
p CP
28.2 WIm K
= 7 .663 x lQ-<i m2 /s
(7360kg/m3 ) (500J/kg K}
t= 30 min= 1800 s
x = 0.05 m, y = 0.25 m = 5 x,
Then
=
X
2Jai
0.05m
= 02129
2~(7.663x10-<i m2 /s)(1800s)
y
5x
2 vat
2 vat
c:: =
c:: = (5)(0.2129) = 1.0643
From the Gaussian error function table (by interpolation):
erf (0.2129) = 1 - erfc (0.2129) = 0.2366
erf (1.0643) = 1 - erfc (1.0643) = 0.8677
Substituting these values,
e= T(5 cm,25 cm, 1800 s)-1250 =(0.2 366 x 0 _8677 )=0.20 54
275-1250
It follows that
Temperature at the point P,
T = 1250-(0.2054)(975) = 1050°C
(Ans)
(P) PERIODIC VARIATION OF SURFACE TEMPERATURE
Example 6.41
~ On cold days, at a certain location, the temperature of the earth's surface varies sinusoidally
between -20°C and 20°C within a period of 24 hours. Determine the amplitude of the
temperature variation at a depth of 40 cm from the earth's surface. Also find the time lag of the temperature
wave at a depth of 40 cm. What is the temperature at that depth 5 hours after the surface temperature reaches
a minimum value? Determine the heat flowing into the surface per unit area every half cycle. For earth, take
k = 0.63 W/m °C, CP = 1.88 kJ/kg °C and p = 1600 kg/m 3•
Transient Heat Conduction
r
6.97
Known: In winter, the earth' s surface varies sinusoidally between -20°C and 20°C during a 24 hour period.
Find: (a) Amplitude at depth x = 0.4 m, (b) Time lag lit, (c) Temperature at x = 0.4 m after 5 h from
when it was minimum, (d) Heat flux per half cycle.
Schematic:
Temperature
excess, 0
Surface temperature
Time, t
0=01'------"'-+-+----''<-+-----1'--',-------.
Temperature variation at
distance x from the surface
Fig. 6.50
Assumptions: (I) Periodic heat flow in semi-infinite solid. (2) Sinusoidal oscillations in temperature.
(3) Constant properties.
Analysis: We note that,
ro=2rc/
I
where f = frequency of temperature wave = 24
I
ro=2rcx- = 0.2618 rad/h
24
The mean temperature of the earth's surface is
T
= Tmax +Tmin = 20+{-20) = 00C
o, mean
2
2
and
0o,amp = Tmax -Tmean = 20 - 0 = 20°C
Thermal diffusivity,
k
0.63W/m°C
a=--=----------pCP (1600kg / m3 )(1880J / kg 0C)
= 20.94 xI0-8 m2 /s = 7.54 x!0-4 m2 /h
(a) Amplitude at depth x = 0.40 m is
ex.max =0o,max exp[-x~ro /2a
where
J
00 = r;,,max - r;,,mean = 20 - 0 = 20°C
~=
0.2618rad/h
=
_
v2a 2x7.54xJ0-4 m /h 13 18
m-I
2
and
x
~ = (0.4 m) (13.18 m- = 5.27
v2a
1)
ex.max = (20°C) exp (-5.27) = 0.103°C
(Ans) (a)
6. 98 ,
Heat and Mass Transfer
(b) Time lag, At
~x2 /2roa. =
(0.4m) 2
2x0.26I8rad/hx7.54x10-4 m2 /h
= 20.13 h
(Ans) (b)
(c) Time for minimum surface temperature is obtained by putting,
.
31t 71t
smrot=- 2 ' 2
Taking the first value,
t= ~ =
2ro
3 1t
= 18 h
2x0.2618rad/h
Five hours later,
t=18+5=23h
Temperature at x = 0.40 m and t = 23 h is
0(x,t) = 0o,max =exp[-x~] Sin( ffit- ~ -
X)
0 = 20 °Cxexp(-5.27)sin[{0.2618x23)-5.27]
= 0.01°c
i.e., T - Tmean = 0.07°C
i. T= 0 + 0.07 = 0.07°C
(d) Heat flow per m2 per half cycle is
(Ans) (c)
Q
2
-= ,,::-;;;k0oamp
A vroa.
,
2 (0_63 W/m 0 C)(20 0 C)[1J/s] [3600s] ( lkJ)
lW
lh
103 J
=---,=============--2
~0.2618 rad/hx7.54x10-4 m /h
= 6457 kJ per m2
(Ans) (d)
Example 6.42 ~ A cold front moves in at a certain location for three weeks and causes a temperature drop
of 20°C at the surface. The average thermal diffusivity of the soil is 0.15 x IQ-6 m2/s. Estimate
the temperature change at a distance of 0.80 m below the ground surface and the time lag for this case. Also
find the burial depth you would recommend to avoid freezing while laying water mains if the specified base
temperature is 5°C.
Known: A prolonged cold wave causes a temperature reduction on the earth's surface with periodic
oscillation.
Find: 0x,amp (x = 0.8 m) and At (x = 0.8 m), Depth x(m) for 0x,amp = 5°C.
Transient Heat Conduction
Schematic:
r
6.99
0
Le (x. t)
Soil:
a =0. 15 x lo-6m2/s
Three weeks duration
Temperature drop= 20 °C
Fig. 6.51
Assumptions: (I) Periodic change in surface temperature is sinusoidal. (2) Constant soil properties.
(3) The earth is a semi-infinite medium.
Analysis: Period of oscillation, tP = 3 weeks = 3 x 7 x 24 h
Amplitude at the surface, 0o,amp = 20/2 = 10°C
At a depth x = 0.80 m, the amplitude of temperature variation is
ex.amp= 00,amp exp[-x~ro /2a,
J
21t
21t
ro=21tf== - - - rad/h
where
tP
and
3x7x24
a=0.15xI0-6 x3600m 2 /h
21t
~
= - - - - - - - - - - - = vl 1.543 = 3.398
2
3x 7 x24 x 2 x 0.15 x 10-6 x 3600 m / h
0x ,amp =100c X e-0.8m x 3.398 m- '
= 0.66°C
(Ans)
Time lag,
lit=-x-=
· ~2 roa,
0.8m
~{(2x21t /( 3x 7x 24)} rad / h x (0.15x 10-6 x 3600 m2 / h)
=218 h
or
or
(Ans)
_5_ = J_x,Joo!2a.- x,Joo!2a. J
0.66
In ( 5- )=(x -x)
o.66
l
2
~
v2a
6.100 ,
Heat and Mass Transfer
2.025
= 0.596 m
3.398m- 1
Hence, the burial depth for laying water mains is
x 2 = 0.80 m - 0.596 m = 0.204 m
(Ans)
Example 6.43 ~ The temperature below the ground is affected by daily variations of temperature above it.
The surface temperature of the earth at a certain location follows a sine wave form varying
between 25°C and 45°C during 24 h. Calculate (a) the amplitude of temperature variation at a depth of 25 cm,
(b) the depth at which the temperature amplitude is just equal to 5% of the amplitude at the surface, (c) the
time lag at a depth of 25 cm, (d) the temperature at a depth of 25 cm, 6 hours after the surface temperature
reaches a minimum value, (e) the heat flowing in and out of the earth per unit area every half cycle. (f) Plot
the variation of surface and sub-surface temperatures as a function of time.
The following thermophysical properties for the earth's soil may be assumed:
p=2050kg/m 3 k=0.52 W/m K and
Cp=l.84kJ/kgK.
Known: The earth ' s surface temperature oscillates periodically in a diurnal cycle in sinusoidal fashion .
Find: (a) 0o,amp ( 0 C) at x = 25 cm, (b) /it (h) at x = 25 cm, (c) T (x = 25 cm) 6 h after T= Tmin•
(d) Q/A(J / m2 )
Schematic:
0( C)
0
Surface (x = 0)
10
Fig. 6.52
Assumptions: (l) The earth is a semi-infinite solid with periodic sinusoidal vanat10n in surface
temperature. (2) Constant properties. (3) There is no internal heat source below the earth's surface.
Analysis: The amplitude of surface temperature variation is
I
45-25
0
-(Tmax -Tmin)=--- = 10 C = 0o max
2
2
Mean temperature,
I
2
= -{45+25) = 35°c
'
Transient Heat Conduction
The transient temperature response is given by
0(x,t)max
0(0,t)max
e-x
V2 a
% ·
l l
~
~
Slll O)[-
dep~ent
attenuallon
r
6.10 I
(A)
-· X
depth-dependent
phase lag
The period of oscillation is TP = 24 h and frequency, f = - 1- = _!_ h- 1
TP 24
Circular frequency of oscillation,
1 1t
ffi= 21tf =21tX-=-rad/h
24 12
The thermal diffusivity,
k
0.52W/mK
I1J/sll 3600sl
<l= pCP = (2050kg/m3 ) (l.84x10 3 J/kgK) lW ~
= 4.96x10-4 m2 /h
~=
,_ _1t_/1_2_ra_d_/_h__ = 16.24 m-1
2x4.96x10-4 m2 /h
The amplitude of temperature variation at a depth, x = 0.25 m is
ex.max =0o.max exp[-x v2cx
~]
= 10°C exp [-0.25mx16.24m-1]
= 0.1125°c
(Ans) (a)
Depth at which the temperature amplitude equals 5% of the amplitude at the surface is given by
0x(5%),max
e-x(5%)~
00,max
or
0.05 = exp [-x (5%) x 16.24]
or
X
(5%) = -ln(0.0 5) = 0.1845 m
16.24
= 18.45 cm
(Ans) (b)
Time lag at x = 25 cm is
At=-x-=---.====0=.2=5=m====
~2roa
= 15.5 h
1trad
2x--x4.96x10-4 m2 /h
12h
(Ans) (c)
6.102 , Heat and Mass Transfer
Temperature at x = 25 cm, 6 h after the surface temperature has attained a minimum value is given by
_0_(x_=_2_5_cm_,t_)
0(0,t),max
=e-Rf·x xsin[rot- ~_ro__ x]
v2a.
. [( 12
1t h
rad x t ) - ( 16.24 m- 1 x 0.25 m} ]
= e-(0.25xl6.24) x sm
Time for minimum surface temperature to occur for the first time is when rot= 3 1t .
2
t= ~ =31tx12 = 18 h
2ro
2X1t
Six hours later, t = 18 + 6 = 24 h
0 (x = 0.25 m, t = 24 h)
= 10°C xexp[-4.06sin{(~x24)-4.06}]
= o.1J1°c
T= (0.25 m, 24 h) = Tmean +0
= 35 + 0.137 = 35.14°C
(Ans) (d)
Heat flow in half cycle, per unit area is
Q
2 keo,max
A= [<Mi.
2x0.52W/mKxl0°CO*) ( ~ )
=
~{1t/12radlh}(4.96xl0-4 m2/h}
=3.28 MJ/m2
(Ans) (e)
Equation (A) can be rewritten as
=
0
x,max sin[ ro(t-At)]
where At is the time lag
00,max
=0.01725 sin [ 1~{t-15.5}]
(B)
Transient Heat Conduction
r
6.103
0(x,t) = 0.1725 sin [15 (t- 15.5°)]
0(0,t) = 10 sin (15 t0 ) (at the surface)
and
(C)
The following table gives the values of temperature excess (amplitude), 0 at the surface (x = 0) and
at a depth x = 25 cm from the surface as a function of time for a wave period of 24 h, calculate from
equations (B) and (C):
Tim e, t (h)
0(x, t)
0(0, t)
Tim e, t (h)
0(x, t)
0(0, t)
0
+0.13 7
0
18
+0.105
-10
3
0.0225
7.07 1
21
+0.171
- 7.071
6
-0.105
10
24
+0.13 7
0
9
-0.171
7.07 1
27
+0 .0225
+7.071
12
-0 .137
0
30
-0.105
15
-0 .0225
-7.071
33
-0.171
+7.071
36
-0.137
0
The results of these surface and sub-surface temperatures are plotted and indicated in the schematic.
(Ans) (f)
, MULTIPLE CHOICE QUESTIONS
6.1
A large concrete slab 1-m-thick has one dimensional temperature distribution:
T = 4-10x+20x2 + 10x3
where T is the temperature and x is the distance from one face towards the other face of the wall. If
the slab material has a thermal diffusivity of 2 x 10 - 3 m2 /h, what is the rate of change of temperature
at the other face of the wall?
(a) 0.1°C/h
(b) 0.2°C/h
(c) 0.3°C/h
(d) 0.4°C/h
6.2
Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different
temperature is valid when
(a) Biot number < 0.1
(b) Biot number > 0.1
(c) Fourier number < 0.1
(d) Fourier number > 0.1
6.3
A plate of surface area As and volume ¥- has one boundary surface subjected to a constant and
uniform heat flux q. If the thermal capacity and density are Cp and p, initial temperature is r;, the
temperature after time t is T(t) and the surrounding fluid temperature is T=, the time, assuming
lumped capacity model, is given by
~i
[a= __!f.i__,
b=
p¥-CP
p¥-CP
6.104 ,
Heat and Mass Transfer
(a) t = _ _!._ln [T(t)-T= -(a/b)l
b
T;-T= -(a/ b)
(b) t=-_!._ln [T(t)-T=-(b/ a)l
a
T;-T= -(b/ a)
(c) t= .!..1n [T(t)-T= +(a/b)l
(d)
b
t=_!_[ T;-T=+(b/ a)
a T(t)-T= +(b/ a)
T;-T= +(a/ b)
l
6.4 The temperature-time relation for a spatially isothermal body can be expressed as
6.5
T(t)-T=
(a) - - - = exp (-Bi Fo)
T;-T=
(b) T(t)-T= = exp (Bi Fo)
T;-T=
T(t)-T=
(c) - - exp ( -Bi)
T;-T=
Fo
T(t)-T=
(d) - - exp ( -Fo)
-.
T;-T=
Bz
Transient temperature profiles in respect of semi-infinite geometry along with the statement are
given as under. Match the profile with the appropriate statement using the codes given below (usual
symbols are used).
Profile
Statement
T
A.
1. Constant surface heat flux
Increasing time
Ts
-------------=
X
x=O
T
B.
2. Sudden jump in surface temperature (Ti< T5 )
T;
Ts
00
x=O
X
T
C.
3. Finite convection at the surface (h < 00)
Ts
T;
x=O
x
Transient Heat Conduction
T
D.
r
6.1 OS
4. Constant surface temperature (Ts < T;)
T=
00
x=O
Codes:
(a)
(b)
(c)
(d)
A
3
4
4
X
B
1
3
C
D
2
4
3
2
2
3
4
2
6.6 Match List I with List II according to the codes given below:
List I
Codes:
(a)
(b)
(c)
(d)
List II
A. Rectangular parallelepiped
l. T=f (t)
B. Short cylinder
2. One-dimensional transient
C. Sphere
3. Two-dimensional transient
D. Spatially isothermal object
4. Three-dimensional transient
A
4
1
2
3
B
3
3
C
2
4
3
4
D
1
2
4
2
6.7 Match List I with List II and select the appropriate answers using the codes given below.
List I
(Parameter)
List II
(Definition)
A. Transient conduction in semi-infinite medium
l. A 1 exp (-Af Fo)
2. pCP D/6h
B. One-term approximation
C. Time constant of thermocouple
3. at/L~
D. Fourier number
4. x / 2.Jcii
Codes:
A
(a)
4
(b)
(c)
(d)
1
3
4
B
1
2
4
C
2
3
2
3
D
3
4
1
2
6.8 Penetration depth d can be expressed in terms of thermal diffusivity, a and penetration time, tP as
r:::;
r:::;
3 .6
1. 8
(a) d=l.8v"''
(b) d=3.6vat
(c) d= r:::;
(d) d=-.-2
vat
(at)
6.106 , Heat and Mass Transfer
6.9
A small copper ball of 5-mm diameter at 500 K is dropped into an oil both whose temperature is
300 K. The thermal conductivity of copper is 400 WIm K, its density 9000 kg/m3 and its specific heat
385 J/kg K. If the heat transfer coefficient is 250 Wlm 2 Kand lumped parameter analysis is assumed
to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,
(a) 8.7
(b) 13.9
(c) 17.3
(d) 27.7
6.10
T~e dimensionless heat-loss ratio for a short (finite-length) cylinder with (QI Qma,Jpiate = 0.25 and
(QI QmruJcylinder = 0.56 is
(a) 0.14
(b) 0.67
(c) 0.81
(d) 0.98
, TRUE/FALSE
6.1
6.2
6.3
For low values of h and high values of k, large temperature differences occur between the inner and
outer regions of a large solid.
Heating or cooling of a road surface can be analysed using the semi-infinite slab model.
The time required by a temperature-measuring instrument after which the temperature difference is
reduced to 63.2 % of the initial temperature difference is called its sensitivity.
~
6.4 In periodic unsteady state conduction, the time lag is defined as At =
x
where ro = 21t f, x is
2 <l ro
depth,! the frequency and <l the thermal diffusivity.
6.5 Transient heat conduction in the semi-infinite cylinder is one-dimensional.
, FILL IN THE BLANKS
6.1 The solution for a multi-dimensional geometry is the _ _ of the solutions of the one-dimensional
geometries whose intersection is the multi-dimensional body.
6.2 Biot number, Bi for a sphere with h = 50 W /m2 K, k = 50 W /m Kand D = 40 cm equals _ _ for
use in transient temperature charts and is equal to _ _ for lumped capacity formulation.
6.3 A thermocouple junction approximated as a I-mm-diameter sphere has k = 35WIm K, a = 12.5 x
10-6 m 2ls, and h = 212 W lm2 K. The thermal time constant is - - ·
6.4 Semi-infinite solid is characterised by a _ _ slab with _ _ thermal diffusivity and _ _ exposure
time.
6.5 Heisler charts are applicable for plane wall (2 L thick), long cylinder of radius r 0 and sphere of
=
radius r0 , for Bi ( \L / h';) greater than _
and Fo (
=~/ ~;) greater than - ·
, EXERCISES
6.1 The temperature distribution in degree centigrade through a wall 0.7-m thick at some time is given by
T = 900 - 1350 x 2
where x metres is the distance measured through the wall. The wall material has the following
properties:
Transient Heat Conduction
6.2
6.3
6.4
6.5
r
6.107
Density .... 2740kg/m3
Specific heat .... 1.34 J/kg K
Thermal conductivity .... 0.796 W/m K
Determine
(a) The rate of change of temperature at the midplane of the wall
(b) The net heat gain or loss in the wall per m2 of wall area per hour
Prove the formulae used for calculation.
[2.71 MJ/hm2]
An aluminium alloy plate of 400-mm x 400-mm x 4-mm size at 200°C is suddenly quenched into
liquid oxygen at -183°C. Starting from fundamentals or deriving the necessary expression, determine
the time required for the plate to reach a temperature of - 70°C.
[l.055 s]
Assume: h = 2x10 4 kJ/m2 h °C, CP = 0.8 kJ/kg °C and p = 3000kg/m3
A stainless steel rod of I-cm outer diameter originally at a temperature of 320°C is suddenly
immersed in a liquid at 120°C for which the convective heat-transfer coefficient is 100 W/m2 K.
Determine the time required for the rod to reach a temperature of 200 °C. (For steel p = 7800 kg/
m3, c = 460 J/kg K, k = 40 W/m K).
[82.19 s]
A sphere of 200-mm diameter made of cast iron, initially at a uniform temperature of 400°C is
quenched in oil. The oil bath temperature is 40°C. If the temperature of the sphere is 100°C after
5 minutes, find the heat transfer coefficient on the surface of the sphere. Take CP = 0.32 kJ/kg K
and p = 7000kg/m3 • Neglect internal thermal resistance.
[446 W/m2 K]
An aluminium sphere weighing 5.5kg and initially at a temperature of 290°C is suddenly immersed
in a fluid at 15°C. The convective heat transfer coefficient is 58 W/m2 K. Estimate the time required
to cool the aluminium to 95°C using the lumped capacity method of analysis. Take
p= 2700kg/m3
k = 205 W/m K
C = 900 J/kg K
[22.5 min]
6.6 A solid copper sphere of 10-cm diameter [p=8954kg/m3 , CP = 383 J/kg K, k = 386 W/m K]
initially at a uniform temperature I;. = 250°C, is suddenly immersed in a well-stirred fluid which is
maintained at a uniform temperature T- = 50°C. The heat transfer coefficient between the sphere
and the fluid is h = 200 W/m2 K. Determine the temperature of the copper block at t = 5min after
the immersion.
[120°C]
6.7 A thermocouple of diameter 0.71mm (Cp = 420 J/kg K; p = 8600kg/m3) is used to measure the
temperature of a gas stream for which the convective heat-transfer coefficient is 600 W/m2 K.
Estimate the time constant of the thermocouple. What is the time period after which an acceptable
reading of temperature can be recorded?
[2.85 s]
6.8 A household iron has a surface area of 500 cm 2 and made of stainless steel (k = 17 .5 WIm K) of
1.4 kg mass and 500 W capacity. The heat transfer coefficient on the surface of the iron is 17 W/m2 K,
when the surrounding air temperature is 30°C. Determine the time required for the iron to reach
l 10°C after being turned on if originally its temperature is 30°C. Take
Density, p (iron) = 7820 kg/m3
Specific heat, c (iron) = 460 J/kg K
Prove the formula used.
[l min 51s]
6.9 A hollow commercial bronze cylinder (50-mm-OD and 10-mm-ID) is 50-mm long. Initially at a
temperature of 25°C, the cylinder (p = 8800kg/m3, CP = 0.42 kJ/kg °C, k = 52 W/m 0 C) is placed
6.108 ,
6.10
Heat and Mass Transfer
in boiling water at 100°C. It takes 250 s for the cylinder to attain a temperature of 75°C. Determine
the heat-transfer coefficient. Justify any assumption made.
[0.0159]
A carburettor body casting has a mass of 1.2 kg and comes out of the injection moulder at a
temperature of 343°C. The mould operator picks the castings up with tongs and hangs them from
a conveyor for transport to an inspection station. The surface area of the casting can be estimated
to be 0.075 m 2 and the heat transfer coefficient 66 W/m3°C. If the air temperature around the
conveyor is 18°C, find how long the casting should be cooled in air before it can be safely handled
without gloves (assuming most persons can tolerate 48°C for a short period of time). Assume for
the casting:
p = 2790kg/m3
6.11
CP = 0.883 kJ/kg °C
k = 168 W/m °C.
[8.5 min]
During the manufacture of large plastic sheets, 10-cm thick, the sheets are brought to a uniform
temperature of 175°C and then allowed to cool to a surface temperature of 90°C in air at 40°C.
The average convective heat transfer coefficient is 11.5 W/m2 °C. Estimate (a) the duration of
cooling, and (b) the centre temperature of the sheets when the surface temperature reached is 90°C.
Thermophysical properties of plastic are
k = 0.23 W/m 0 c, p = 1300kg/m3
and
CP = 1.67 kJ/kg 0 c.
[(a) 94.5 min (b) 158.4°C]
A cylindrical rod of50-mm diameter (p = 8055kg/m3, Cp = 559 J/kg K, k= 20.0 W/m K) undergoes
heat treatment by drawing it through a 6.5-m long oven in which air is maintained at 800°C. The
rod enters at 40°C and attains a centreline temperature of 660°C before leaving. The combined
convection and radiation heat transfer coefficient is 160 W/m 2 K. Estimate the speed (m/min) at
which the rod must be passed through the oven.
[0.6 m/min]
6.13 Oranges of 6-cm diameter (k = 0.45 W/m K, and <l = 1.3 xl0-7 m2 /s) are initially at a uniform
temperature of 26°C. They are to be cooled by refrigerated air at -4°C. The average heat-transfer
coefficient between the oranges and the air is 30 W/m2 K. Determine how long it will take for
the centre temperature of the oranges to drop to 8°C.
[2215 s "" 37 min]
6.14 An apple of 10-cm diameter, when taken out from a refrigerator, has a temperature of 5°C.
Determine the temperature at the centre of the apple after one hour if the ambient air temperature
is 20°C and the heat transfer coefficient is 5 W/m 2 °C. The following properties of apple may be
used:
6.12
k = 0.418 W/m °C
6.15
6.16
p = 840kg/m3 CP = 3.81 kJ/kg °C
[7oC]
A short aluminium cylinder has a diameter of 5 cm and a length of 10 cm. Initially, it is at a uniform
temperature of 200°C. It is plunged into a quenching bath at 60°C with a convection heat transfer
coefficient of 530 W/m2 K. Determine the temperature on the centreline of the cylinder, 2.5 cm from
one end, after one minute has elapsed. Take, for aluminium, k = 215 W/m Kand <l = 8.42 x 10-5 m2 /s.
[102.3°C]
A carbon steel cylinder [k= 60.5 W/m °C, a= 17.7 x10-<im2 /s], 50mm in diameter and 200-mm
long is suspended in an agitated hot oil bath. Initially the cylinder is at a temperature of 15°C. The
mean convection heat transfer coefficient is 125 W/m 2 °C. Estimate the minimum bath temperature
required to raise the centre temperature of the cylinder to l 10°C in 15 min.
[116°C]
Transient Heat Conduction
6.17
6.18
6.19
6.20
6.22
6.23
6.24
6.109
Estimate the centre temperature of a copper cylinder of 0.5-m diameter and 0.5-m length, two
minutes after the surface temperature suddenly changes from an initial 30 °C to 500°C. When will
the temperature at the centre attain 420°C?]
Properties of copper: k = 383 W/m °C, p = 8933kg/m3, CP = 0.41 kJ/kg °C.
[80 s or 30 min]
A glass cylinder [k = 1.4 W/m K, p = 2500kg/m3, CP = 0.75 kJ/kg K] of 20-cm diameter and 40-cm
length is initially at a uniform temperature of 25°C before being placed in an oven for annealing process.
The air in the oven is at 465°C and the heat transfer coefficient is 14 W/m2 K. How long should the
glass cylinder be kept so that every part of the cylinder attains a temperature of at least 425°C?
[5.48 h]
A short metal cylinder, 5-cm in diameter and 10 cm long is initially at a uniform temperature of
800°C. It is quenched in oil which remains at 30 °C. Using transient conduction charts, determine
the temperature after 3 minutes at a point whose coordinates are x = 2 cm, r = I cm with respect
to the centre of the cylinder.
Data: h = 500 W!m 2 K, k = 50 W/m K, a=l.5xI0- 5 m 2 /s
[130.7°C]
A large thick layer of ice is initially at a uniform temperature of -20°C. If the surface temperature
is suddenly raised to -4°C, determine the time required for the temperature at a depth of 15 mm
to reach -12°C. The properties of ice are
p = 920kg/m3
6.21
r
k = 1.88 W/m °C
CP = 2.04 kJ/kg °C
[244 s]
A fire test is to be conducted on a large mass of concrete initially at a uniform temperature of 25°C.
The surface temperature is instantaneously raised to 525°C. Considering the concrete to be a semiinfinite solid, estimate the time required for the temperature at a depth of 30cm to reach 160°C.
For concrete: k = 1.4 W/m K, p = 2300kg/m3, CP = 0.88 kJ/kg K.
[5.5 h]
A large steel slab, initially at a uniform temperature of 400°C, suddenly has its surface temperature
lowered to 20°C. (a) Calculate the time required for the temperature to reach 125°C at a depth of
2.8cm. (b) Also, determine the total heat removed from the slab per square metre during this time.
The thermal diffusivity and the thermal conductivity of the steel are 11.6 xlO-<i m 2/s and 43 W/m K,
respectively.
[(a) 4.5min, (b) 89 MJ]
A thick slab ofbakelite [k= 1.4 W/m °C, <l=7.35xI0-7 m2 /s] has a uniform temperature of27°C.
An electric heater is placed in good contact with the slab's surface. Calculate the temperature of
the slab (a) at the surface, and (b) at a depth of 30-mm, 10 minutes after the heater starts providing
constant surface heat flux of 2500 W/m2 •
[(a) 69.3°C, (b) 35.65°C]
An axial flow compressor has a blade made of titanium alloy (k = 25 W/m °C, p = 4500kg/m3, and
CP = 0.52 kJ/kg 0 C) which is initially at 60°C. The effective thickness, from pressure to suction side, is
10 mm, and in a gas stream at 600°C the blade experiences a heat-transfer coefficient of 500 W/m2 °C.
(a) Use the semi-infinite solid approximation to estimate the midplane temperature of the compressor
blade after 1, 5, 10, 20, 50 and 100 seconds.
(b) Compute the temperatures using analytical method.
(c) Compare them with those obtained by the lumped capacity analysis.
[(a) 66.78°C, (b) 82.57°C (c) 73.54°C]
Comment on your results.
6.110 ,
6.25
6.26
6.27
6.28
6.29
6.30
6.31
Heat and Mass Transfer
The ground at a particular location has an initial uniform temperature of 10°C. The surrounding air
temperature is raised over a short period of time to 35°C. The convection heat transfer coefficient
is 20 W/m 2 K. Estimate the temperature of the soil 3 cm below the earth's surface after one hour.
[13.5°C]
Properties of soil: k = 0.52 W/m Kand a = 1.38 xl0-7 m2 /s.
A thick steel slab [k = 24 W/m °C, a = 0.024 m 2/h] is to be surface hardened by raising the
temperature near the surface, then quenching. The slab is exposed to a gridwork of torches whose
combustion temperature is 2200°C and the result is a heat transfer coefficient at the surface of
1200 W /m2 °C. The slab is initially at a temperature of 25°C throughout. What is the surface
[162.7°C]
temperature and the temperature at 4-cm-depth 1 min after exposure to the torches.
A thick bronze slab [k = 26 W/m K, a= 8.6 x10-6 m2 /s] is initially at a uniform temperature
of 250°C. Suddenly the surface is exposed to a coolant maintained at 25°C. The convective
heat transfer coefficient between the surface and the cooling fluid is 150 W/m2 K. Calculate the
[210°C]
temperature 5 cm from the surface 10 min after the exposure.
A rectangular iron bar (5 cm x 4cm) [k = 55 W/m K, <l = 0.055 m 2/h] is initially at a uniform
temperature of 250°C. Suddenly the surfaces of the bar are cooled in ambient air at 30°C with a
convection heat-transfer coefficient of 530 W/m 2 K. Determine the centre temperature of the bar,
[42.4°C]
4 min after the cooling starts.
A rectangular brick (8 cm x 10 cm x 20 cm) is at a uniform temperature of 40°C. It is put into
an environment at a temperature of 400°C. Find the temperature at the centre of the brick after
150 minutes have elapsed.
h = 19 W/m2 K
Data:
k= 0.38 W/m K
<l = 0.0012 m 2/h
[49°C]
A long steel bar has a rectangular cross-section (50mm x 30mm) and is initially at a uniform
temperature of 20°C. It is placed in a furnace at a temperature of 1000°C. Estimate the temperature
at the centre of the bar after 15 minutes.
[987°C]
Assume: h = 100 W/m2 °C, k = 30 W/m °C, and <l = 0.05 m 2/h.
To compute the thermal diffusivity of the soil on ground, temperatures are recorded and it is found
that the surface reaches a mean temperature at 1900 h and the soil at a depth of 12 cm from the
[0.003 m2/h]
surface attains the mean temperature at 2200 h. Estimate the thermal diffusivity.
, ANSWER KEY
Multiple Choice Questions
6.1 (b)
6.6 (a)
6.2 (a)
6.7 (a)
6.3 (a)
6.8 (b)
6.4 (a)
6.9 (c)
6.5 (c)
6.10 (b)
6.2 T
6.3 T
6.4 F
6.5 F
True/False
6.1 F
Fill in the Blanks
6.1 product
6.4 thick, low, short
6.2 0.2 m, 0.067 m
6.5 0.1, 0.2
6.3 2.2 s
7
BASICS OF CONVECTION
HEAT TRANSFER
Concept Review
INTRODUCTION
• 7.1
•
The concept of convection embraces the process of heat transfer through the motion of a fluid (liquid or gas).
Convection heat transfer involves bulk fluid transport and is always accompanied by heat transfer by conduction.
Convection is the mode of heat transfer that occurs due to fluid motion. Convective heat transfer may
be subdivided according to a number of categories:
(a) The physical cause of motion: free or natural convection, forced convection, mixed convection
(b) The flow configuration: internal or external flow
(c) The presence of turbulence: laminar, transitory or turbulent flow
(d) The nature of flow: single-phase, boiling, condensation, melting, solidifying
HEAT-TRANSFER COEFFICIENT
• 7.2
•
Convective heat transfer is expressed by Newton's law of cooling as
(7.1)
where h is the convection or film heat-transfer coefficient, Ts is the surface (wall) temperature, and T=
is the free-stream fluid temperature. The convection heat transfer coefficient can also be expressed as
h=
-kflmd. (<JT/<Jv)
-' surface
(7 .2)
7'.-T=
where {dT I dy)surface is the temperature gradient in
the fluid at the fluid-surface interface as shown in
Figure 7.1.
The convection coefficient, in general, varies
along the flow direction. The average convective heat
transfer coefficient h for a surface is calculated by
integrating the local convection coefficients hx over
the entire surface length L, as follows .
(
(7.3)
Fig. 7. I
Fluid temperature gradient at the surface
7.2 , Heat and Mass Transfer
The Nusselt number, which is the dimensionless heat-transfer coefficient, is defined as
(7.4)
where k is the thermal conductivity of the fluid and Lc is the characteristic length.
• 7.3 EXTERNAL FLOWS: CONCEPT OF VELOCITY AND BOUNDARY LAYER
•
The boundary layer for flow past a solid surface is a thin layer in which viscous effects are important.
Velocity varies from zero at the wall (no-slip condition) to the free-stream velocity at the edge of the
boundary layer. The boundary-layer thickness,
is defined (arbitrarily) as the distance from the surface
to where the boundary-layer velocity, u reaches 99% of the free-stream velocity, u~. Outside the boundary
layer, viscous effects are negligibly small and are ignored, and the flow can be considered inviscid.
The criterion for ascertaining the flow condition is the Reynolds number. For external flow, the
dimensionless Reynolds number is expressed as
o,
Re = Inertia forces = u~ Lc = p u~Lc
Viscous forces
v
µ
(7.5)
The highly ordered fluid motion characterised by smooth streamlines is known as laminar. The highly
irregular and disordered fluid motion that typically occurs at high velocities is characterized by velocity
fluctuations and fluid mixing is called turbulent. The random and rapid fluctuations of groups of fluid
particles, called eddies, provide an additional mechanism for momentum and heat transfer.
For a flat plate, the characteristic length is the distance xc, from the leading edge. The Reynolds number
at which transition occurs from laminar to turbulent flow is called the critical Reynolds number. For flow
over a flat plate, its value is generally assumed to be
!Rec,= u~xc, !v = 5 X 10 5 1
(7.6)
Both laminar and turbulent flow regions are shown schematically in Figure 7.2 .
Fluid
'--- Flat plate (T,)
-----I
Fig. 7 .2
Laminar and turbulent regions of the boundary layer during Pow over a Pat plate
The friction or drag force per unit area is called shear stress, and the shear stress at the wall surface
is expressed as
't
w
dUI
=µd y y=O
or
(7.7)
Basics of Convection Heat Transfer
r
7.l
whereµ is the dynamic viscosity, u= is the upstream velocity, and Cris the dimensionless friction coefficient. The
property v =µ/ p is the kinematic viscosity. The friction (drag) force over the entire surface is determined from
pu;,
Fo =Cr/42
(7.8)
Drag force on a body is the sum of the pressure and shear forces acting parallel to the flow velocity.
The local skin friction coefficient is defined as
'tw
(7.9)
Cr,x =-1--
-pu2
2 =
For laminar flow over an infinitesimally thin flat plate aligned with the velocity field,
_ 0.664
f,x-..JRe:
C
(7.10)
where the characteristic length used in the Reynolds number is defined as the distance from the leading edge.
The drag coefficient is defined as
C D -
1
Fo
(7.11)
-pAu2
2
=
where A is the projected frontal area for blunt objects or the projected plan form area for flat plates and
airfoils.
For laminar flow over an infinitesimally thin flat plate aligned with the velocity field, we can obtain
the drag coefficient by integrating the local skin friction coefficient over the flat length:
(7.12)
Similar expressions for turbulent flow on a flat plate have been obtained from experimental data:
C _ 0.074
D - ReJl5
for 10 5 < ReL < 107
(7.13)
L
C _
o-
• 7.4
0.455
( log ReL )2.584
for 10 7 < Rei, < 109
(7.14)
THERMAL BOUNDARY LAYER
•
If a fluid at a free stream temperature T= flows over a solid surface at a temperature T8 , there exists a
thin region close to the surface in which the temperature variation in the direction normal to the surface is
significant is the thermal boundary layer. The thickness of the thermal boundary layer 8th at any location
along the surface is the distance from the surface at which the temperature difference (T.-T) equals 99%
of (T8 - T=). The relative thickness of the velocity and the thermal boundary layers is best illustrated by
the dimensionless Prandtl number, defined as
7.4 , Heat and Mass Transfer
Pr= Molecular momentum diffusivity = ~ = CPµ = CP vp
Molecularthermal diffusivity
a
k
k
• 7.5
(7.15)
ANALOGY BETWEEN MOMENTUM AND HEAT TRANSFER
•
The similarities between a fluid boundary layer and a thermal boundary layer indicated that a similarity
exists between the transfer of momentum and heat. The Reynolds analogy relates the convection coefficient
to the friction coefficient for fluids with Pr= 1, and is expressed as
or
~
~
(7.16)
where
h
_ Nu
St -----pCpu= RePr
(7.17)
is known as the Stanton number. The analogy is extended to other Prandtl numbers by the modified
Reynolds analogy or Chilton-Colburn analogy, expressed as
Re.
Cf,x _2x = N11--x Pr 113
(7.18)
or
(7.19)
These analogies are valid for laminar flows over a flat plate and are also applicable approximately for
turbulent flow over a surface of any shape, even in the presence of pressure gradients.
INTERNAL FLOW
• 7.6
•
For steady, incompressible, constant property fluid flowing through a horizontal circular pipe of radius R,
the fully developed velocity profile whose form does not change with axial coordinate x is given by
(Laminar flow)
u
max
I dP
= - - - R2
4µ dx
(7.20)
(7.21)
Basics of Convection Heat Transfer
and
u(r) = Umax [ ~ ] =
2[1-(~J]
V
r
7 .5
(7.22)
where dP is the pressure gradient, and umax and V are the centreline and average velocities, respectively.
dx
T5 = constant
Fully developed laminar flow
Fig. 7.3
Fully developed turbulent flow
Fluid in a tube with constant surface temperature under fully developed laminar and turbulent flow conditions
Pressure loss per unit length,
t:.P
32µV
L
D2
(7.23)
.
pVD
For lammar flow (Re 0 =-- less than 2300),
µ
11P =-12_8µ_Q_L_,
rcD 4
(7.24)
64
Darcy friction factor, f =- - (irrespective of pipe surface roughness)
Re 0
Shear stress at the wall,
l'tw =-~~I
(7.25)
The head loss in a pipe is given in terms of friction factor as
h =f~v2
f
D 2g
Pumping power, Wpump =[ f
("ii)~] (~)
(7.26)
(7.27)
smooth tubes,
For turbulent flow (Re 0 > 4000), the friction factor is obtained from the following equations over the
prescribed range of the Reynolds number.
If =0.3164(Re0
If= 0.184 (Re0
f1 4000<Re <10
14
51
(7.28)
fl/ 5 30000 < Re < 106
(7.29)
0
0
1
7.6 , Heat and Mass Transfer
If= [1.82 log Re0 -1.64] - 2 10000 < Re0 10 7 1
(7.30)
For pipes with rough surfaces, the Moody chart that presents the friction factor fas a function of Re
and relative roughness EID over a wide range is often used.
The following explicit relation to determine f for rough tubes is recommended:
1
[6.9 ({E lD))I.II]
-=-1.8log -+ - Re
3.7
.fl
• 7.7
(7.31)
DIMENSIONAL ANALYSIS
•
In engineering, physical quantities are often represented by their dimensions. The four fundamental
dimensions are Mass [M], Length [L], Time [t] and Temperature [T]. The method of dimensional analysis
is based on the Principle of Dimensional Homogeneity. It is an important tool when complex mathematical
analysis is impractical and qualitative, quick answers are desired. Moreover, by reducing the number of
variables requiring experimental investigation, it helps in formulating the empirical correlations.
By carefully identifying the variables governing a process, one can use Rayleigh or Buckingham-II
method grouping the quantities that form dimensionless parameters, or 1t-terms.
Solved Examples
(A) BOUNDARY LAYER CONCEPT. LAMINAR AND TURBULENT FLOW
Example 7. I ~ Calculate the thickness of the boundary layer on a flat plate at 160 km/h, 15-cm from the
leading edge for an air temperature of I 5°C and I atm. Also calculate the drag rce due to
skin friction (pressure drag is negligible) if the air is in contact with both sides of the plate and the plate is
60-cm wide. At I 5°C, for air at I atm:
p = 1.225 kg/m 3, v = 1.470 x I0-5 m2/s.
Known: Atmospheric air flows over a flat plate with a prescribed velocity.
Find: Boundary layer thickness, (mm); Drag force, F 0
o
Schematic:
Boundary layer
Air
P= I atm
u== 160km/h
T= = IS°C
~
~
~""''"'
[_
x= IScm _ _____,
L-.x
Fig. 7.4
As= 2 X W
=3 (0.15m) (0.6m)
=0.18m2
Basics of Convection Heat Transfer
r
7.7
Assumptions: (1) Steady-state conditions. (2) Air is an ideal gas. (3) Recr = 5 x 10 5•
Analysis: The free stream velocity,
u_ =160 km ( 103 m) ( ~ ) = 44.44m/s
h
1km
3600s
Let us calculate the local Reynolds number, Rex to determine whether the boundary layer is laminar
or turbulent.
u_x
Re=--
Therefore
V
x
(44.44 rn/s)(0.15 m)
= 1.470x10-5 m2 /s
= 4.535x10 5
This is a value just slightly less than Recr = 5 x 10 5 so that we are pretty close to the transition point
but still laminar. Accordingly,
<>(x)= 5.0x = 5.0(0.15m) (10 3 mm)
~Rex
~4.535x10 5
lm
= 1.114 mm
(Ans)
Average skin friction coefficient or drag coefficient,
Cf.L or C _ 1.328 _
1.328
o - ~ReL - ~4.535xl0 5
= 0.001972
(dimensionless)
Average surface shear stress for the plate is
---c
1 2
'tw- f.L-pu_
' 2
Drag force for both sides of the plate,
3 x0.18m2 x44.44 2 m2 /s 2 I lN I
= 0.001972x-xl.225kg/m
1
2
lkgm/s2
=0.4295 N
(Ans)
7 .8 , Heat and Mass Transfer
Example 7.2 ~ Air at standard conditions is flowing over a flat plate which is 1-m long and 0.3-m wide. The
flow is uniform at the leading edge of the plate. The velocity profile in the boundary layer is
assumed to be
t = 2( f)-( fJ.
The free-stream velocity is U = 30 mis. Assume that the flow conditions
are independent of z. Using the control volume a b c d, shown by dashed lines, calculate the mass flow rate
across the surface ab. [Density of air may be taken as 1.23 kg/m 3]. Refer to Fig. 7.5.
u
a
CV
d
C
Fig. 7.5
Known: Air flow over a flat plate.
Find: Mass flow rate across surface ab, mab·
Schematic:
b
a
Air
U=30m/s
p = 1.23 kg/ml
===
y=o= 3.8 x IO-lm
u=U
u(y)=U
~(tJ-(t]J
Flat plate, L = I m, w = 0.3 m
Fig. 7.6
Assumptions: (I) Steady operating conditions. (2) Constant properties.
Analysis: Mass flow rate of air across ab= (Mass flow rate across ad) - (Mass flow rate across be).
)}y
m,, =p[w 6]U-pwU J{2(f)-(f
P [8-{~
= wU
y2 _ _!__ y3
82
82 3
}o]
0
I
= -xl.23 kg/ m3 x0.3mx30 m/sx3.8xI0-3 m
3
= 0.014 kg/s
(Ans)
Basics of Convection Heat Transfer
r
7. 9
Example 7.3 ~ The temperature profile at a particular x location in a thermal boundary layer is given by
T(y) =A - By+ Cy2, where A, B and Care constants. Obtain an expression for the corresponding
local heat-transfer coefficient.
Known: Temperature distribution at a location in the thermal boundary layer.
Find: Local heat-transfer coefficient, hx.
Schematic:
T(y) =A-By+ Cy2
X
- -------+-,j-----
Tw= A
------I
Fig. 7.7
Analysis: At any distance x from the leading edge, the local heat flux may be obtained by applying
Fourier's law to the fluid at y =0, since at the surface, there is no fluid motion, and heat transfer is only
by conduction. That is
From Newton's law of cooling, q= hy_ (Tw -T=)
Equating the two, we have
The wall temperature gradient is 'lJTI
.
dy y=O
The temperature profile is T(y) = A- By+ Cy 2
Differentiating with respect to y,
'lJT
dy
-=0-B+2Cy
Hence,
(dT)
dy
=-B
y=O
It follows that
(Ans)
7. I O , Heat and Mass Transfer
Example 7.4 ~ A barge with a rectangular surface (25-m long x 10-m wide) is travelling down a river with
a mean velocity of 2.S km/h . A laminar boundary layer exists up to a Reynolds number of
S x I 05 beyond which abrupt transition occurs to turbulent boundary layer. Compute (a) the maximum distance
from the leading edge up to which the laminar boundary layer persists, and the maximum boundary layer
thickness at that point, (b) the total drag force on the flat bottom surface of the barge, and (c) the power
required to push the bottom surface through water at the specified velocity. For water: Density= 999.1 kg/m 3
and Viscosity= 1.138 x I 0-3 kg/m s
Known: A barge with flat bottom surface moves in a river with a given velocity.
Find: (a) Lc.(mm); <>c,(mm); {b) F 0 (N); (c) (W).
Schematic:
Water
Bottom surface of barge
u==2.Skm/h
A5 = LW = 25 m x IO m
=2SOm 2
/
L=25m
Fig. 7.8
Assumptions: (1) The flow is steady and incompressible. (2) Rec,= 5 x 10 5 .
Analysis: Maximum distance up to which laminar boundary layer exists is
L _Rec,µ_
er -
5xl0 5 xl.138xI0-3 kg/ ms
pu= - 999.1 kg / m3 x(2.5x1000 / 3600)m/s
= 0.82 m or 820 mm
(Ans) (a)
Maximum laminar boundary layer thickness,
0 = 5Lc, = 5x820mm
er
~Rec,
~5 X 105
=5.8 mm
(Ans) (a)
Reynolds number at the end of the bottom surface with L = 25 m is
(999.1 kg/m3 )( 2 ·5 m/s)(25 m)
3.6
1.138xI0-3 kg/ms
= 1.524 X107
The average friction coefficient over the entire plate is
Cf,L =
O.ReLO~: - 1742
0.074(1.524X 10 )-0.
ReL
= 0.0026
=
7
2-
1742(1.524 X10 7
r
1
Basics of Convection Heat Transfer
r
7. I I
Drag force on the bottom surface of the barge is
-1
Fo =Cr L -p,{u:,
, 2
1 999. lkg/ m3 x250 m2 x (2.5 / 3.6) 2 m2 /s2 [
= 0.0026 x-x
2
IN
]
lkgm /s2
= 156.2 N
(Ans) (b)
Power required to overcome drag is
= (156.2 N) (2.5/3.6 mi s)
= 108.47 W
(Ans) (c)
Example 7.S ~ A small thermocouple probe is placed in a laminar boundary layer near a flat plate at SS°C over
which the water at 35°C and 0.20 mis flows. The probe is used to establish the temperature
profile which is found to be well represented by
At a location where the thermal boundary layer thickness 6, is 21 mm, determine (a) the heat flux from plate
to water, and (b) the heat-transfer coefficient.
Properties of water at 45°C : k = 0.637 W/m°C
Known: Temperature profile for water flow over a flat plate.
Find: (a) q (W/m 2) ; (b) h (W/m 2 0 C).
Schematic:
Water
T= =35°C
y
u= =0.2m/s
Flat plate
><--........._- - - - - - - - ~ ------- ~ (T5 = SS 0 C)
Fig. 7.9
Assumptions: (1) Laminar, incompressible flow. (2) Constant plate surface temperature.
Analysis: For external flow heat transfer the fluid properties are evaluated at the film temperature,
1
1
T. = -(T + T ) = -(55 + 35)°C = 45°C.
f
2
s
=
2
(a) Heat flux, q = Q = -kfluid cJTI
A
dy y=O
(W/m2 )
where T represents the temperature distribution in the fluid and is the temperature gradient at the
surface.
7. I 2 ,
Heat and Mass Transfer
The temperature profile is given by
:; ly=O = [ 2~t - 2 ~f X 3 y 2 l=O
3
Since
ae (ar/ay)
=-'------'-
It follows that
Substituting the values, the heat flux is
q=-(0.637 W/m 0 C)(35-55)°Cx
1. 5
0.021 m
(Ans) (a)
=910 W/m2
(b) Heat-transfer coefficient,
h=-q-
T.-T=
910W/m2
(55-35) °C
=----
= 45.5 W/m2 ·c
(Ans) (b)
Basics of Convection Heat Transfer
r
7.13
(B) INTERNAL FLOW: FLOW THROUGH CLOSED CONDUITS
Example 7.6 ~ Water at 30°C (p = 995.6 kg/m 3, v = 0.782 x I~ m2/s) flows with a velocity of I .S m/s through
a tube. The cross-section of the tube is shown in Figure 7.10. Determine the pressure drop
per unit length.
f- L= I cm --j
Fig. 7.10
Known: Water flows through a tube of prescribed cross section.
Find: Pressure drop per metre length, t:.PIL
Schematic:
Tube cross-section
T
D= 1cm
f-L=lcm-j
1
Fig. 7.11
Assumptions: (1) Inner surface of the tube is smooth. (2) Water is incompressible.
Analysis: Equivalent diameter,
2
4x{ (LxL)+~
8 D x2} 4x(l2 +~xl2)cm2
4xArea
4
D =
= --=,,.----~--=- = - - - - ' - - - - • Perimeter
{
rcD}
2cm+(rcxlcm)
2L+2x 2
4+rc
2+rc
= --cm = 1.389 cm
Reynolds number,
Re= VD. = l.5m /sx0.01389m = 26 643
v
0.782xl0-6 m2/s
7.14 ,
Heat and Mass Transfer
Fanning friction factor,
IF = 0.046 =
Re0· 2
0.046 = 0.006
{26 443)02
Pressure drop per metre length,
I 11~1
~p = 4fF pV 2 = 4x0.006 x 995.6kg/m3 x(l.5) 2 m2 /s2
IN
L
De 2
0.01389m
2
lkgm/s2 IN/m2
= 1933 Palm
(Ans)
Example 7.7 ~ Helium gas (Viscosity= 2.321 x I Q-4 poise) flows in a smooth stainless steel tube I 0-mm-lD
and 0.3-m-long heated electrically to an average temperature of 860 K. The gas temperature
at the start of the heated part is 280 K and at the end of the heated part is 460 K at a mass flow rate of
12.8 kg/h. The pressure at the start of the heating section is 8.4 bar. Determine: (a) the Reynolds number,
(b) the friction factor, and (c) the pressure drop due to wall shear over the 0.3-m-length.
Known: Helium gas is heated while flowing through an electrically heated tube at a specified flow rate.
Find: (a) Reynolds number, Re0 , (b) Friction factor,/ (c) Pressure drop, t:.P (Pa).
Schematic:
Helium
-
m = I 2.8 kg/h
D= 10mm
-Out
T;=280K
Fig. 7.12
Assumptions: (I) Steady state conditions. (2) Helium is an ideal gas.
Analysis: (a) Reynolds number,
VD
Re0 = v
4m
--
1tDµ
Properties at the mean temperature of helium,
I
I
2
2
Tm =-(T; +T,,)=-(280+460)K = 370K:
µ = 2.321 x I0- 5 kg/m s
[I poise= 0.1 kg / ms]
Density at 8.4 bar and 280 K is
8.4x!0 2 kPa
I lkJ I
p= RT= 2.077kJ/kg Kx280K lkPam 3
P
= 1.444 kg/m 3
Basics of Convection Heat Transfer
r
7. I 5
4 X (12.8/3600) kg/s
1t{O.Ol m )(2.32 Ix I0- 5 kg / ms)
= 19 500
( > 2300)
~
Turbulent flow
(Ans) (a)
(b) Friction factor, for 3000 ::; Re0 ::; 5 x I 06 is
/= [ 0.79/n{Re 0 ) - 1.64]-o.2 = [0.79 /n (19 500) - 1.64]--0·2
= 0.0263
(Ans) (b)
(c) Pressure drop,
t:.P = f pV2L
2D
Mean flow velocity of helium is
m
4m
p,{
p1tD2
V=-=--
4 X 12.8 / 3600 kg/s
=---------'---(l.444kg/m 3 ) (1txO .Ol2 m 2 )
= 31.34 m/s
Pressure loss,
t:.P = 0 _0263 x (1.444 kg/m3 )(31.34 m/s) 2 (0.3 m) 11 N s2 I
2x0.0lm
= 560.5 N/m 2
lkgm
or 560.5 kPa
(Ans) (c)
Example 7.8 ~ Water at 20°C flows through a 19-mm-lD tube at a flow rate of 1.5 Umin. Determine (a) the
flow regime; (b) the entrance length; (c) the pressure drop over a distance of 3000 m, and
(d) the corresponding pumping power necessary to maintain the flow. (e) Calculate the pressure drop and the
pumping power if the flow rate is increased to 15 Umin.
Properties of water at 20°C are
p = 998.2 kg/ml,
µ = I 004.2 x I 0-6 kg/m s
Known: Dimensions of tube and water flow rate .
Find: (a) _Laminar/Turbulent flow; (b) Entry length, L.; (c) Pressure drop, t:.P; (d) Pumping power, WP.
r:v
O)U_
t:.P and WP for ten times the flow rate.
Schematic:
Water
LTube
D=0019m
-¥ = 1.5 x I O-l ml/min
T=20°C
L= 3000m
Fig. 7.13
)-
7.16 ,
Heat and Mass Transfer
Assumptions: (1) Steady operating conditions prevail. (2) The inner surface of the tube is smooth. (3) The
velocity profile is fully established. (4) The fluid properties are constant. (5) Constant pressure gradient.
Analysis: (a) Reynolds number,
Re= pVD
µ
where V is the average flow velocity.
Volume flow rate,
¥-=Cross sectional area, Ac x Average velocity, V
Hence,
V= ¥- = 4¥- = 4x(l.5x10-3/60)m3/s
rcD 2
4,
rcx(0.019m}2
= 0.0882m/s
Then,
(998.2kg/m3) (0.0882 m/s)(0.019 m)
Re=---------~--~
1004.2 X lQ-6 kg/m S
= 1665.3
As Re is less than 2300, the flow is laminar.
(Ans) (a)
(b) Entrance length,
Le= 0.0575 Re D
= (0.0575) (1665.3) (0.019 m) = 1.82 m
(Ans) (b)
(c) Pressure drop,
where f is the Fanning friction factor.
For laminar flow,
f = ~ = 16
- -=0.00961
Re 1665.3
Therefore
M = (4x0.00961x3000m) (.!..x 9982 kg x0. 08822 m2 )
0.019 m
2
m3
s2
=23.6x103 kg 11Pa 11 lN I
ms 2 1 N/m2 1kg m/s 2
= 23.6 x 103 Pa
or 23.6 kPa
(Ans) (c)
Basics of Convection Heat Transfer
r
7.17
(d) Pumping power required,
WP= (Pressure drop,AP) (Volume flow rate, -Ji:.)
/ I
= (23.6 x 103 Pa )(1.5 x 10-3 /60 ms3 ) I1Pa1 W
m3 s
(Ans) (d)
=0.59 W
(e) With flow rate increased ten-fold, i.e., ¥- = I5LI min,
( 15 x 10-3 /60)m2/s
V=-'---~---'----
0.882 mis
1t/4x(0.019 m}2
pV D 998.2kg/m3 x0.882m/sx0.019 m
Re=--=------------µ
1004.2x1Q-6 kg/ms
= 16653
As
Re > 2300, the flow is turbulent.
f = 0.079{Re
t>·25 = 0.079{16653r 25
0·
= 6.954 X IQ-3
Hence,
AP= (4x6.954xl0-3 x3000m)(.!..x 9982 kg x0. 8822 m2 )
0.019 m
= 1.7x 106 Pa
and,
or
2
s2
m3
1.7 MPa
(Ans) (e)
WP= AP x ¥- = (1.7x106 Pa)
(1
5
:~o-
3
~
3
)
(Ans) (e)
=426 W
Comment: The power input increases by a factor of more than 700 when the flow rate is increased by
a factor of 10.
Example 7.9
~ Air flows through a 25-mm diameter tube with a mean velocity of 30 m/s. The tube wall
temperature is 280°C and the air temperature increases from 20°C to 260°C. Using the
simple Reynolds analogy, calculate the length of the tube required and the pumping power. For turbulent flow
in a tube, take f=0.046/Re& 2 •
The properties may be taken at the mean film temperature from the tables. These are
k = 38.45 x I0-3 W/m°C
cp = 1.0268 kJ/kg°C
p=0.7306 kg/m 3
µ=26.17x I~ kg/ms
7.18 ,
Heat and Mass Transfer
Known: Air flows through a tube and gets heated in the process.
Find: Tube length and pumping power using Reynolds analogy.
Schematic:
D = 25 mm
~
V=30m/s
L=/
Fig. 7.14
Assumptions: ( 1) Steady state conditions. (2) Fully developed flow through the tube.
Analysis: Reynolds number,
pV D (0.7306kg/m 3 )(30m/s)(0.025m)
Re0 = - - = - - - - - - - - - - - µ
26.17xI0-6kg/ms
= 20 938
(> 2300)
The flow is turbulent.
Fanning friction factor for turbulent flow is
f = 0.046 Rei{1· 2
= 0.046 (20 938)-0· 2 = 0.0063
Using Reynolds analogy,
St=f
2
Nu
hD
µ
k
h
where Stanton number, S t = - - = - x - - x - - = - RePr
k
pV D CPµ pVCP
Heat-transfer coefficient,
h=pVCP x 1
2
= (0.7306kg/m3 )(30m/s)(l026.8 J/kgK) x0.0063
-2
= 70.77 W /m 2 K
For constant wall temperature condition,
Q= mCP ( Tbe -TbJ= h ,4s fl. Tim
where
m=p /4, V=p~D 2 V
4
= (0.7306 kg/m 3 ) ( ¾x0.025 2 ) m2 (30 m/s) = 0.01076 kg/s
Basics of Convection Heat Transfer
r
7.19
Heat rate, Q = (0.01076 kg/s) (1026.8 J/kg K) (260 - 20)°C or K
=2651.4 W
Log mean temperature difference,
_ (260-20)°CorK
l ( 280-20)
n 280-260
=93.57K
Hence, surface area of the tube,
A,. = re D L = ___g_
hAT;m
2651.4 W
(70.77W/m 2 K)(93.57K)
=-------= 0.4m2
Hence, the length of the tube required is
L=--1._= 0.4m2
rcD rc(0.025m)
(Ans)
=5.1 m
Pumping power required is
Wp=mgJt
hr_,=4fLY!_
D 2g
where
= (4x0.0063x5.lm)( 302 m 2/s2 )
0.025m
2x9.81m/s2
= 235.5m
Therefore
.
I l wl
1WP= (0.01076kg/s)(9.81N/kg) (235.5m) -11- lNm lJ/s
=24.86 W
(Ans)
7 .20 , Heat and Mass Transfer
Comment: Reynolds analogy is appropriate for Pr"' I. In this case,
CPµ
Pr =k
=
1026.8x26.17x!0-6
38.45 X 10-3
- 0.699"' 1.
Hence OK
Example 7.10 ~ Air is heated by passing it through a copper tube (25-mm-lD) which is maintained at 300°C.
The air enters at 20°C and leaves at 280°C at an average velocity of 30 mis. Using the
Reynolds analogy, determine (a) the tube length, and (b) the pumping power required.
Properties of air:
At T1 = 225°C:
At Tbm = I S0°C:
k = 0.0394 W/m K
p = 0.7087 kg/m 3
V = 37.73 X I 0--6 m2/s
Cp= 1015 J/kg °C
Pr=0.696
Known: Air is heated while passing through a tube held at constant temperature.
Find: (a) Tube length, L (m); (b) Pumping power, WP {W).
Schematic:
Copper tube
(Tw=300°C)
D=25mm
~80°C
- - _ . V= 30m/s
Air
L =?
Fig. 7.15
Assumptions: (I) Steady operating conditions. (2) Fully developed flow. (3) Constant tube wall
temperature.
Analysis: Mean film temperature is
.!.[
20 + 280 +300] °C = _!_{150+300) °C = 225°C
Tr= Tbm +Tw =
2
2
2
2
Reynolds number,
Re =VD= 30m/ sx0.025 m
0
v
37.73x 10-6 m 2 / s
= 19 878
~
Turbulent flow
Darcy friction factor
f = 0.3164{Re 0
rli
4 = 0.3164 (19878)-o. 25
(Re< 2 x 104)
= 0.0266
We have from the Reynolds analogy,
Stanton number, St= f/8
or
St=___!!:!_= 0.0 266 = 0.00 333
Re Pr
8
Basics of Convection Heat Transfer
r
7.21
Nusselt number
Nu =St (Re Pr)= (0.00333){19878)(0.696)
= 46.08
Convective heat transfer coefficient
k
(46.08) (0.0394 W/m °C)
h=Nu-=-------D
0.025 m
= 72.6 W /m 2 °C
Mass flow rate of air,
rh = p 4, V = (0.7087 kg/m3 ) ( ~x 0.025 2 m 2 ) (30 mis)
= 0.010 44kg/s
From energy balance:
(
Heat transferred by convection)= (Rate of thermal energy)
from tube wall to air
storage of air
i.e.,
Then
Q = ( 0.01044kg/s)( 1015 J/kg °C){280-20)°C
=2754W
Also
Q=h,4.AT,m =2754W
Now,
(300- 20 )-(300- 280)
l ( 300-20)
n 300-280
= 98.52°C
Hence,
Q
2754 W
/4 = hAT,m = 72.6 W/m2 °Cx98.52 °C
= 0.385m2
tube length required,
L = _1__ = 0.385m2
rcD rc(0.025m)
=4.9m
(Ans) (a)
7 .22 , Heat and Mass Transfer
t:.P
f LV 2
Pressure drop = H = = --pg
2gD
{0.0266){4.9 m)(30 m/s)2
--,---~---,--~-~- = 239.3 m
2(9.8lm/ s2 ){0.025 m)
.
(t:.P)
Pumping power WP = mg pg
= (0.01044 kg/s )(9.81 N/kg)(239 .3 m)= 24.5 W
Example 7.11
(Ans) (b)
~ Glycerine at 25°C flows in a pipe (IS-cm ID) with an average flow velocity of 3.6 mis.
Ascertain if the flow is laminar or turbulent. Compute points on the velocity profile from
the pipe wall to the centre of the pipe at increments of IS mm. Plot the data for local velocity against radius
and show the average velocity on the plot. Calculate the radius at which the local velocity equals the average
velocity for laminar flow.
Properties of glycerine at 25°C:
p = 1261 kg/m 3
v = 840 x 10-6 m2/s
Known: Glycerine flows with a prescribed velocity through a pipe.
Find: Nature of flow; Velocity profile; Radius at which u(r) = V
·lllllllt.
Schematic:
Pipe (D = 0.1 Sm)
1f1ycerinp
V= 3.6m/s
T=25°C
90 - - - - - - - - - - - - - - - - - - - - - ~
75 --t---=-- - - - - - j - - - - + - - - + - - - + - - - - - - - - , - - - - + - - - + - - -1
r=R=
60 4--------,-------t-::--,,,.-..+ - - - + - - - - - - - - l - - - + - - - + - - -I
!~ . ..... ....
t-f----+---+-+---=--~ - - - + - - - + - - - -1
30
.-.
E
E
._.
IS
0
... -IS
-30
-45
53
-60
lr=\/Q5~
R
; I ;v·1~: .
i
I ····
5
1
-75
-90
I
Velocity profile
Fig. 7.16
Basics of Convection Heat Transfer
r
7.23
Assumptions: (1) Fully developed flow. (2) Constant properties.
Analysis: Reynolds number,
V= D (3.6m/s) (0.15 m)
Re0 = - - = - - - - - - v
840 x 10-6 m2 / s
= 642.86
As this value is less than 2300, the flow is laminar.
Now
D
R=-=0.075 m or 75 mm
2
r
At r = R (at the pipe wall), - = l and u = 0.
R
From the velocity distribution for laminar flow through a pipe
where Vis the mean (average) flow velocity. This is in keeping with the observation that the velocity
of a fluid at a solid boundary is equal to the velocity of that boundary, i.e., zero. (No slip condition)
At r=60 mm:
u=2(3.6 m/s)[,-U~::J]=259 m/s
Similarly, the following values are also computed at different radii at increments of 15 mm to plot
the velocity profile.
r (mm)
U (mis)
± 75
± 60
± 45
± 30
± 15
±0
0 (pipe wall)
2.59
4.61
6.05
6.91
7.20 (centre of the pipe)
For the condition when local and average flow velocities are same, we have
With
v[,-(il']
f =2[i-(il']=
u=2
1
7.24 , Heat and Mass Transfer
Solving for r, we get
!...=Jo5 =0.707
R
Also average velocity, V = 3.6 mis, the maximum velocity (at r = 0) = 2 V = 7 .2 mis. The velocity
profile is shown in the schematic.
r=0 .707x75 mm=53 mm
(Ans)
Example 7 .12 ~ It was found during a test in which water flowed with a velocity of 2.44 mis through a
tube (2.54-cm ID and 6.08-m long) that the head lost due to friction was 1.22 m of water.
Estimate the surface heat-transfer coefficient based on Reynolds analogy.
Take p =998 kg/m 3 and CP =4.187 kJ/kg K
Known : Water flows through a tube. Frictional head loss is given.
Find: Surface heat-transfer coefficient, h [Wlm 2 K] using Reynolds analogy.
Schematic:
Water
~=254cm
Frictional head loss,
hr= 1.22 mof water
U)
_~___-------___
v_=_2_.4_4_m_/_
s--~)
1---- L=6.08m
Fig. 7.17
Assumptions: (1) Steady operating conditions. (2) Smooth tube surface.
Analysis: For fully developed tube flow, Reynolds analogy can be stated as
f
Nu
Stanton number, St = - = - 8 RePr
Reynolds number,
Prandtl number,
CPµ
pVD
Re0 = - -
Pr=-k
µ
hD
Nusselt number, Nu0 = k
St= hD x_.f:!:.._x_k_=_h_
k
pVD
h = p cpv
u 18)
where f is the Darcy-Weisbach friction factor.
Head lost due to friction,
CP µ
pCPV
Basics of Convection Heat Transfer
r
7.25
f = 11f (2gD)
LV 2
=
(l.22m)(2x9.8lm/s 2 x2 .54x10-2 m)
6.08 m x (2.44 m/s) 2
= 0.0168
Hence, the surface heat-transfer coefficient is
f
h = 8pVCP
= ( 0 ·0; 68 ) (998kg / m 3 )(2.44 m / s)( 4187 J / kg K)
= 21406 W/m 2 K
=21.4 kW/m 2 K
(Ans)
Example 7.13 ~ The internal cross-section of a pipe is an equilateral triangle of 25-mm side. The pressure
drop per metre along the pipe is 0.2 x 10-3 bar at a gas velocity through the pipe of 20 m/s.
The mean constant pressure specific heat of the gas is 1.1 kJ/kg K. Estimate the average heat-transfer coefficient
between the gas and the pipe wall.
Known: Gas flows through a pipe of triangular cross-section and suffers a pressure drop.
Find: Mean heat-transfer coefficient.
Schematic:
b
Equilateral triangle with
each side b =25 mm
b
b
Fig. 7.18
Assumptions: ( l) For gases with Pr"" 1, the Reynolds analogy is applicable .
Analysis: For noncircular ducts, the equivalent diameter,
4x - bh
Perimeter
2
3b
D = 4 xcross-secttonalarea=
•
l
.
,---r;;
=3_h=3_~b2-(b/ 2)2 =3..x:!..l x b
3
3
b
25
=-=-mm= 14.43 mm
J3 J3
According to Reynolds analogy for turbulent flow through a pipe:
Stanton number, St = f
8
where f is the Darcy ' s friction factor.
3
2
7.26 , Heat and Mass Transfer
Nu
hD
µ
k
h
St=--=--X--X--=-Re Pr
k
p V D CP µ p CP V
Also, the pressure drop.
f=2t:.P(D)
pV2 L
or
Therefore,
h
1 2{t:.P)
D
--=-X--X-pCP V 8
L
pV 2
Average heat-transfer coefficient is then determined from
h = pCP V (t:.P)__!!__
4
L pV2
h = CP (t:.P) D
4
L V
or
Replacing D with D. for a noncircular tube,
h = ( 1.1 X 103 J/ kg K) [0.2 X lQ- 3 X 10 5 N ] 14.43 X lQ-3 m
4
lm
m2
20 m/s
s:4W/m 2 K
(Ans)
Example 7.14 ~ A fluid is flowing through a 10-cm-inner diameter pipe in which velocity is uniform over the
cross section of the pipe but the temperature varies linearly from 90°C at the pipewall to
0°C at the centreline. Determine the heat transfer coefficient based on the mean fluid temperature if the heat
flow rate from the wall is 9280 W/m 2.
Known: Flow of fluid through a pipe with uniform velocity and temperature varying linearly with radius.
Find: Heat-transfer coefficient.
Schematic:
r=+R
(Flu_
id)
---- r=O --r=-R
l
E}~~::~:
T(r=R)=
=9280W/m2
qw = _Q
A
r,~
Velocity profile
1
Temperature profile
Fig. 7.19
'.m
Basics o( Convection Heat Transfer
Assumptions: (1) Uniform velocity. (2) Linear temperature profile.
Analysis: Heat-transfer rate, Q = hA.
-Tbm)
r
7.27
(3) Fully developed flow.
(T.
Hence, heat-transfer coefficient is
h =QI A. = 9280 W/m2
T.-Ti, (90-Tb)°C
To find the bulk fluid temperature, we define it as
R
J uT dA
'T'
1b -
J uT 2trrdr
__
O_ _ __
A
_
-
JAu dA
R
J u2trrdr
0
R
or
Tb= J uT rdr
/RJ urdr
0
0
But u = V = const
Hence
R
/RJ rdr
0
0
Tb= JT(r)rdr
Linear temperature variation means:
T(r=r)-T(r=O)
r-0
= R-0
T(r=R)-(r=O)
T(r)-0
or
r. -0
=
r
R
r
90
T(r) = -T.. = - r = 1800 r
R s 0.05
~
Then
R
Ti, = J 1800 r2 dr
/RJ rd r
0
0
= 1800 ~
3
/R /2 = 1200 R
2
= 1200x0.05= 60°C
Hence,
h = 9280 W
/m2 = 309.33 W/m2 °C
{90-60)°C
Example 7. IS
(Ans)
~ Calculate the pressure drop which occurs when water at 30°C flows with an average velocity
of 0.2 m/s through a 500-m long cast iron (Cl) pipe having an inside diameter of 15 cm. Take
roughness, £, for Cl pipe = 2.6 x Io-4 m. Justify any assumptions made. Also find the required pumping power.
Known: Flow velocity through a specified CI pipe.
Find: Pressure drop.
7.28 , Heat and Mass Transfer
Schematic:
Water
d5D=015mm
C Cl pipe)=0.26mm)
T=30°C
V=0.2m/s
i------
L=SOOm
Fig. 7.20
Assumptions: (1) Steady, incompressible flow. (2) Fully developed flow . (3) The pipe involves no bends,
valves or connectors. (4) The piping section involves no work devices such as turbine or a pump.
Analysis: For water at 30°C, p = 996 kg/m3, µ = 0.798 x 10-3 kg/ms
Reynolds number,
pV D (996kg/ m3) (0.2m/s)(0.15m)
Re=--=-'-----'--'-----'-'-----'µ
0.798xl0-3 kg/ms
~
= 37443.6
c.
c.
Re = 37443 .6 and
· · 1actor,
Fnct10n
1or
Turbulent flow
cjD -- 0.000 26 m -- 1.73 x l0-3,
C,
0.15m
can be obtained from
_l ""-1. 8 [6.9 +(e/ D)1.,,l
fl
Re
3.7
= -1.8 lo [ 6.9 +(0.00173)1."] = 6.144
g 37443.6
3.7
2
1-] = 0.0265
f =[6.144
The pressure drop is
= 0.0265( 500 m
0.15m
)(.!.2 x 996 kg/m3 x 0.2 ms
= 1760 Pa or
1.76 kPa
(Ans)
= ~(0.15 m}2 (0.2 m / s)(l760 Pa)= 6.2W
(Ans)
2
2
)
2
Pumping power,
4
Basics o( Convection Heat Transfer
r
7.29
Example 7 .16 ~ Atmospheric air at 25°C flows through a 5-mm inside diameter horizontal circular tube.
Assuming the flow to be fully developed, determine the mass flow rate of the air if the
pressure drop per metre tube length is 250 Palm. If the tube were vertical, would the air flow rate change?
Known: Air flows through a horizontal tube with a specified pressure drop per unit length.
Find: Mass flow rate of air. Effect of change of orientation.
Schematic:
Air
dr=OOOSm
r==2s c -m
)
0
"; = 250 P,/m
P= I atm
L
Fig. 7.21
Assumptions: (1) Incompressible, fully developed flow.
Analysis: For incompressible, fully developed flow of a fluid passing through a horizontal tube, the
pressure drop is given by.
llP= 128µQL
rcD 4
128µ(rh/p) 128th V
=---~=
L
rcD 4
rcD 4
l!J'
or
At
1 atm, 25 °C:
v = 15.62 x lQ-6 m2 / s
p = 1.184 kg/m3
Mass-flow rate of air is
.
rcD 4 (llP IL)
128v
m=----rc(0.005m) 4 (250Pa/m)l 1 N 11 kg ml
= 128x15.62x1Q-6 m2 /s 1 Pa m2 1 Ns 2
= 2.455 x 10-4 kg/s
or
0.884 kg/h
(Ans)
In the case of vertical orientation,
llP *
L
128 µQ
rcD4
where llP* is the piezometric pressure difference = llP + pg L, L being the elevation difference or height
(which is zero in the horizontal orientation)
If
pg<< llP, the effect of the orientation of the tube will be immaterial.
L
pg=(l.184 kg/m3 ) (9.81 N/kg)=ll.615 N/m3
or
Palm
[:.1 Pa= 1 N/m2]
7.30 , Heat and Mass Transfer
t:.P = 250 Pa lm
L
and
Since
pg<< t:.P, the orientation of the tube will not matter.
L
(Ans)
m
V=---p (1tD 2 I 4)
Comment: Average air velocity,
=
4 x 2.455 x 10-4 = 10.56 mi s
1.184 X 1t X 0.005 2
Acoustic (sonic) velocity,
a= ~k RT= .Ji.4 X 287 X 298.15
= 346mls
V 10.56
Mach number, M = - = - - = 0.0305
a
346
Since M << 0.3, the assumption of incompressible (p =const) flow is valid.
(C) DIMENSIONAL ANALYSIS
•
•
Example 7.17 ~ Determine if the following quantities are dimensionless or not:
(b) _P_
p gL,
(a) pVCP~
h
(d) g ~~~Tp2
µ2
The units and dimensions of each of the quantities are given below:
Quantity/Parameter
Units
Dimensions
+ Characteristic length, L
+ Length, x
+ Density, p
+ Velocity, V
+ Gravitational acceleration, g
+ Temperature difference, ~T
+ Dynamic viscosity, µ
+ Coefficient of volumetric thermal expansion, ~
+ Pressure, P
... m
[L]
... m
[L]
... kg/m 3
[ML-3]
... mis
[Lr 1]
... m/s 2
[Lr2]
... °C or K
[T-1]
0
... kg/ms
[ML- 1 r-- 1]
... K-1
[T-1]
... N/m2 or kg/ms2
[ML-I r--2]
(Continued)
Basics of Convection Heat Transfer
r
7 .3 I
(Continued)
Quantity/Parameter
Units
Dimensions
• Specific heat, CP
...
J/kg K or m 2/s 2 K
[L2 r-2 T-11
• Convective heat-transfer coefficient, h
kg m 2
... W/m 2 K o r - - -
[Mr-3 rl]
s3 m2 K
(a)
= M 0 Vt0T 0 = V
(not dimensionless)
(Ans) (a)
(b)
~
= [M0 U't0 ]
(c)
xL
_c
(dimensionless)
(Ans) (b)
= [U ](U ](M- 1L+ 1t+ 1](L-2 t2T1]
µCP
= [M- 1Lt3 T]
~
(not dimensionless)
(Ans) (c)
= [M0 U't0T6]
~
(dimensionless)
(Ans) (d)
(d)
Example 7 .18 ~ Show by dimensional analysis that for forced convection heat transfer
Nu= <!>(Re,
P,,~)
Cl
when frictional heating in the fluid cannot be neglected. Discuss physical significance of each term.
Known: Forced convection including frictional heating effects in the fluid .
Find: To prove that
Analysis:
• The dimensionless numbers Nu, Re and Pr are defined as
hL
k'
pU L
µ '
CPµ
k
Nu=- Re=-- Pr=--
The fourth non-dimensional groups is U 2/CP T, known as Eckert number, Ee.
7 .3 2 , Heat and Mass Transfer
• The relevant parameters with their units and dimensions in the problem statement are tabulated below:
S. No.
Parameter
Units
Dimensions
1.
Characteristic L
dimension
Sy mbol
M
[L]
2.
Density
p
kg/m 3
[ML-3]
3.
Velocity
u
mis
[Lt-I]
4.
Temperature
difference
Tor !',.T (to be
precise)
K
[T]
5.
Viscosity
µ
kg/ms
[ML- 1 r 1]
6.
Specific heat
cp
m2
J
- - or - kg K
s2 K
[L2 r 2T-I]
7.
Thermal
conductivity
K
W
Nm /s
kg m
--or--or-mK
mK
s3 K
[MLr- 3 T- 11
8.
Heat transfer
coefficient
H
w
kg
--or-2
m K
s3 K
[Mr3 T- 11
• Total number of parameters (or variables) is n = 8
• The main dimensions in the case will be m = 4, namely, [M], [L], [t] and [T]
• Obviously, from Buckingham-1t theorem, the number of1t-groups or dimensionless parameters will be
p=n-m=8-4=4
• The non-dimensional relationship will be of the form
1t1 = <p [1t2,1t3,1t4]
t We now select the appropriate repeating variables (equal to the number of principal dimensions, i.e., 4).
RI
L
R3
p
R2
µ
R4
K
• The non-repeating variables are:
NRI
h
NR 2
u
NR 3
cp
NR 4
T
• The four 1t-groups will be
1t1 = NI?i (Ri)° (R2t (R3t (R4/
= h(Lt (µt (Pt (k/
1t2 = NR2 (Ri)° (R2t (R3t (R4/
= U(Lt (µt (Pt (k/
(dependent variable)
Basics o( Convection Heat Transfer
1t3
= NR3 (J?i f (lli
t ( f (~ t
R3
= CP (L)° (µt (pf (k/
1t4
= N~ (J?if (R2t (R3f (~t
= T(L)° (µt (pf (k/
[Note that values of exponents a, b, c, and d in each of the 4 cases will be different.]
+ Writing the dimensions for 1t 1 and equating the powers on both sides:
M: 0 = l+b+c+d ~
le= O!
0=a-b-3c+d ~
!a= I!
!h=OI
L:
t:0=-3-b-3d
~
T:0=-1-d
or
1t1
=h(L)1(µ) 0 (p)0 (k)1
!1t1
= hL/k = Nu!
Equating powers on both sides for
M: 0 = O+b+c+d
~
L: 0=l+a-b-3c+d ~
t: 0=-l-b-3d
~
T:0=0-d
~
le= 11
la= I!
!h=-11
!d=OI
1t2
or
=U(L)1{µf1(p)1(k)o = pUL
µ
r
7.ll
7.34 ,
Heat and Mass Transfer
Equating powers on both sides for
~
!c = O!
L:0=2+a-b-3c+d ~
la=O!
t:0=-2-b-3d
lb= 11
M: 0 = O+b+c+d
~
T:0=-1-d
!d=-11
1t3
= CP ( L )0 (µ )1 (p )0 ( k r'
l1t3 = CPµ/k =
Prl
Equating powers on both sides for
M:O=O+b+c+d
lc=21
L:0=0+a-b-3c+d
~
la=2!
t :0=0-b-3d
~
lh=3I
T :0=1-d
!d=ll
1t4
= T(L)2 (µr3 (p )2 (k)1
1t4
=
TL2p2 k
µ3
The fourth dimensionless number is U 2 which is nothing but 1t~ as can be seen as follows:
CPT
1t31t4
,
1t~
p2u2L2
k
µ3
u2
1t =--=---x--x---=
4
1t31t4
µ2
CPµ TL2p 2 k CPT
i.e.,
Hence proved.
or
Physical Significance of Dimensionless Parameters
Nu = h:) can be interpreted as the dimensionless temperature gradient at the
surface. [Nu= hAT /k] where AT is the difference between the wall surface and fluid temperatures.
AT/L
+ Nusselt number (
Basics of Convection Heat Transfer
r
7.35
This non-dimensional combination (hL/k) measures the severity of the temperature gradient at the
wall.
+ Reynolds number ( Re = p
:L)
can be looked upon as the ratio of the inertia and viscous forces.
J.
2
As Re is increased, the inertia forces become dominant and small
( Re= VL = V // L2 = F:nertia
V
VV L
Fviscous
disturbances in the fluid may be amplified to cause the transition, from laminar to turbulent flow.
For small values of Re, the viscous forces dominate the inertial forces and the viscous forces may
be negligible for value of Re.
+ Prandtl number ( Pr = C;µ J from laminar to turbulent flow can be viewed as the ratio of the
momentum and thermal diffusivities. It represents the relative importance of momentum and energy
transport by the diffusion process. Gases with Pr - 0.7 have the two diffusivities of the same order.
Liquid metals have very small values of Pr.
+ Eckert number [ Ee = ~~) is significant as it denotes the kinetic energy of the flow ( ½p V 2 )
relative to the boundary layer enthalpy difference {p CP t:.T). Eckert number assumes significance
when at high speed, frictional heating becomes considerable. Rearranging, denoting that the heat
diffuses much more readily than does the momentum.
u 2 u 2/ cp dynamic temperature due to fluid motion
Ee= - - = - - - = - ' - - - - - " - - - - - - - - - - - CPt:.T
t:.T
temperature difference
If Ee is small, the viscous energy generation effects due to fluid motion can be neglected
compared to the temperature differences involved in the heat transfer process.
Example 7 .19
~ Atmospheric air at 25°C flows through a 25-mm-lD, 2-m-long smooth circular tube. The
Reynolds number is 16 000. Determine the average air velocity and the mass flow rate.
The pressure drop is measured to be I 30 Pa. If the fluid flowing through the tube is now changed to water at
25°C without changing the Reynolds number, estimate the pressure drop. The heat transfer coefficient with air
is estimated to be 47 W/m 20 C. The Prandtl number of the fluid is believed to influence the Nusselt number
by a factor Pr0-36. Estimate the heat-transfer coefficient with water in place of air.
The following properties can be used:
k(Wlm
C)
Pr
p(kg/m 3)
v(m 2/s)
Air
( I atm, 25°C)
1.184
I 5.62 x I0-6
0.02551
0.7296
Saturated water
(25°C)
997.0
8.94 X 10-7
0.607
6.14
Fluid
0
Known: Air at a specified temperature flows through a tube of prescribed dimensions. Reynolds number,
pressure drop and heat transfer coefficient are given.
Find: m and V for air. t:.Pwater• fiwater·
7.36 , Heat and Mass Transfer
Schematic:
Noc p,o.36
Reair = Rewater
= 16000
C1
AP.,,= 130~
h~;~;~m' •c
)f
25mm
L=2.0m
Fig. 7.22
Assumptions: (1) Steady operating conditions. (2) Fully developed internal flow. (3) Smooth circular
tube. (4) Similarity laws are applicable.
Analysis: Fluid: Air
pVD VD
Re0 =--=-=16000
Reynolds number,
µ
Mean air-flow velocity,
V
Reo Vair
D
~ir
16000x15.62x10-<i m 2 ls
= - - - - - - - - - = 1O.Om 1s
0.025m
(Ans)
Mass flow rate of air passing through the tube,
m=p~D2 V
4
= 1.184 kg x~x{0.025m)2 xlOmls
m3
4
= 5.81 xl0-3 kgls
Nusselt number,
or
20.9 kg/h
W
°C
(Ans)
1 (m°C)
hD
Nu0 =-=47--x0.025mx---k
m2
0.02551 W
=46.0
Fluid: Water
Since LID and Re0 are same in both cases, geometric similarity dynamic similarity are satisfied. To
estimate the pressure drop with water, we equate the Euler number for air and water.
(EuL = {Eu)water
i.e.,
AJ'.iir
APwater
Pair ~Tr
Pwater V,;ater
Mean water-flow velocity,
16000 x8.94 x 10-7 m2 Is
0.025m
=---------
= 0.572 mis
Basics of Convection Heat Transfer
r
7.37
Hence,
M_
water
=~ pair. X Pwater X ( vwater
J2
Ji'.
P air
= JSS.O Pa
air
(Ans)
= 130Pax 997 x(0.572)2
1.184
10.0
As Nu0 ex Pr1· 36
NuD(water)
= NuD (air ) X
[
0.36
Prcwater)
J
Prcair)
0 36
= 46.0 x [ ~ ] .
0.7296
= 99.0
Heat-transfer coefficient with water is
k
h=Nu0 D
= 99.0x 0. 60?W/m°C = 2405W /m 2 °C
0.025m
Comment: The heat-transfer coefficient increases from 47 to 2405 W/m 2 °C when the flowing fluid
is changed from air to water an increase of more than 50 times!
Example 7.20 ~ Consider the flow of toluene at an average temperature of 30°C through a tube of 12.5-mm
diameter at a mean velocity of 0.8 mis. The heat transfer coefficient for toluene is to be
determined.
It is desired to get the information from experiments carried out with water flowing through a 20-mm
diameter tube. (a) Specify the conditions of the experiment, that is the temperature and velocity of the water.
(b) If the heat transfer coefficient for water found experimentally is 3.6 kW/m 2°C, calculate the value for toluene
under the specified conditions.
The properties of toluene at 30°C are
µ = 0.508 x 103 kg/m s
p = 890.7 kg/m 3
k=0.1486 W/m °C
Cp= 1.717 kJ/kg °C
The properties of saturated water are
T(°C)
20
25
30
35
p(kg/m 3)
998.3
997. 1
995.7
994.1
Cp(kJ/kg C)
4.182
4.180
4.180
4.179
µ x I0 3(kg/m s)
1.003
0.8908
0.7978
0.7196
k(W/m C)
0.5996
0.6076
0.6150
0.6221
0
0
Known: Toluene flows in a tube under specified conditions. Experimental data for water is used to obtain
the desired information.
Find: (a) Temperature, Tbm(0 C) and velocity, V(m/s) of water; (b) Convection coefficient, h (W/m2 0 C) for toluene.
7.38 , Heat and Mass Transfer
Schematic:
r,.=:~;'"' ~ O
. .)_v_=o_
. .am_/s_ _o=__._+_mm_(_,_,~jFig. 7.23(a)
Water
~o
V=?
Fig. 7.23(b)
Assumptions: (1) Steady operating conditions exist. (2) Fully developed flow. (3) Smooth surface of tubes.
Analysis: The Reynolds number and the Prandtl number must be the same for the experiment with water
as for toluene.
For toluene:
Reo =pV D = (890.7kg/m 3 )(0.8m/s)(l2.5xl0-3 m) = 17 533
µ
0.508xl0-3 kg/m s
Pr= CPµ = (l.717xl0 3 J/kg°C) (0.508xl0-3 kg/ms)= 5•87
k
0.1486W/m°C
The Prandtl number is a property of the fluid and depends on temperature. The Reynolds number
depends upon the fluid velocity V as well as the properties p and µ. Hence, the correct temperature for
the experiment must first be specified so that the experiment can be carried out at the right Pr.
Re can then be adjusted by an appropriate selection of the velocity V.
{Pr)water ={Pr)toluene = 5.87
As Pr= f (T), let us first evaluate Pr at different temperatures from the given table for water.
T (°CJ
Pr= Crµ
20
25
30
35
6.99
6.13
5.42
4.83
k
Clearly, for (Pr)water = 5.87, the correct temperature lies between 25 and 30°C. By linear interpolation:
Twater = Tbm = 25 + {30 - 25) (15.87 - 6.13) / {5.42 - 6.13)
= 26.8°C
(Ans) (a)
Mean velocity of water can now be determined from the Reynolds number.
(Reo)water =(Reo)toluene = 17 533
=(
p:
D laterat26.8°C
Basics of Convection Heat Transfer
r
7.39
By linear interpolation, at 26.8°C for water,
p = 996.6 kg/m 3
µ = 0.8568 x 10-3 kg/m s
k = 0.6103 W/m °C
V= Re0 µ = (l7533)(0.8568xl0- 3 kg/ms) =0. 754 m/s
pD
(996.6kg/m3 )(20xl0-3 m)
Hence,
(Ans) (a)
The Nusselt number for the water experiment is found to be
hD
Nu0 = k
The value of h found from the correct experiment is 3.6 kW/m2 °C
(3.6x 10 3 W/m 2 0C)(20x 10-3 m) =
Nu0 = - - - - - - - - - - - -
118
0.6103 W/ m0C
Hence, for toluene:
(Nuo)toluene =(Nuo)water = 118
The heat-transfer coefficient for toluene is given by
(
h= Nu0 -
k)
D toluene
(l 18)(0.1486W/m 0 C)
12.5xl0-3 m
= 1400 W/m 2 °C or 1.4 kW/m 2 °C
=
Example 7.21
(Ans) (b)
~ The dimensionless heat transfer coefficient for laminar flow through a circular pipe is given by
( o)"
Nu=l.86 Pel
3
where
Pe
Peclet number
Inside tube diameter (m)
Tube length (m)
L
V
Velocity of fluid (m/s)
Thermal diffusivity (m 2/s)
a
If the velocity of flow of liquid ammonia through a 0.3-cm-bore tube, 0.7-m-long is 0.08 mis, calculate the
velocity of water in the same tube to give the same value of h, the local heat transfer coefficient, for the data
given below:
D
Liquid
p(kg!m 3)
CP (kjlkg K)
k (Wlm K)
Ammonia
480
5.65
0.381
Water
963
4.21
0.68
Known: Laminar flow of liquid ammonia in a circular pipe.
Find: Velocity of water for the same local convection coefficient.
7.40 , Heat and Mass Transfer
Schematic:
w
0
~0=~3,m--
Ammonia
V=0.08m/s
)
)
i - - - - - L = 0.7 m - - - - - i
Water
---V=l
Fig. 7.24
Assumptions: (1) Steady operating conditions. (2) Laminar flow.
Analysis: Peclet number, Pe= Re Pr
VD
v
VD
=-X-=V
CX.
CX.
We note that
V=0.08m/s
D=0.3x10-2 m and
For ammonia:
L=0.7 m.
p = 480 kg/m3
CP = 5.65 kJ/kg K
k = 0.381 W/m K
Thermal diffusivity,
0381 W/m K
= 1.405 x 10-7 m2 /s
ex.=..!__=
p CP 480 kg/m3 X 5.65 X 103 J/kg K
Pe= VD = 0.08 mis X 0.3 X 10-2 m = 170835
ex.
1.405 x 10-7 m2 /s
( D)
Nu= 1.86 PeL
113
2 ]113= 3.612
[
= 1.86 1708.35 X 0.3 X lQ- m
0.7m
Local heat-transfer coefficient for ammonia is
h=.!_ Nu= 0.3 8 lW/mKx3.612
D
0.3xl0-2 m
= 458.7W /m2 K
For water,
p = 963 kg/m3
CP = 4.21 kJ/kg K
k=0.68W/m K
Basics of Convection Heat Transfer
r
7.41
k
and
CX.=--
p cp
0.68W/m K
963kg/m3 x 4.2lx103 J/kg K
=-----~---= 1.678 x 10-7 m 2 Is
For similarity, to give the same value of h for flow of water through the same tube, we have
k
h = 458.7 = NuD
[
=~xl.86 PeD
D
L
]1/3 = 0 ·68 W/m Kxl.86xPe113 x ( 0.00 3 m )1/3
0.003m
Pell3 = 458.7 = 6.7
68.48
or
0.7m
VD
Pe=300.5 = ex.
Hence, the velocity of water is
Pe ex. 300.5 X 1.678 X 10-7 m 2 Is
D
0.003 m
=0.0168 mis
V=--=-------(Ans)
Alternatively Let subscript 1 refer to ammonia and subscript 2 to water.
Nu=hD
With
k '
Since hi = I\, we have
(for the same tube D1 = D2 and L, = L2)
or
or
or
or
Vi = (0.08m/s)(l.678x 10-7 /1.405 x 10-7 ) (0.381/0.68)3 = 0.0168 m/s
(Ans)
7.42 , Heat and Mass Transfer
Example 7.22 ~ The average Reynolds number of air passing in turbulent flow over a 1.5-m-long flat plate
is 2.4 x I06 • Under these conditions the average Nusselt number was found to be equal to
4150. Determine the average heat-transfer coefficient for an oil having a thermal conductivity of 0.1316 W/m °C,
a specific heat of 2.194 kJ/kg °C and a viscosity of 13.2 kg/m h at the same Reynolds number in flow over
the same plate.
Known: Forced convection heat transfer for air flowing over a flat plate.
Find: Average heat-transfer coefficient for oil, with the same Reynolds number as for air.
Schematic:
-- ...-----------, -Air
Oil
Re 1=2.4x 106
Nu1 =4150
I-
L=I.Sm
I - L=I.Sm
----+j
Fig. 7.25
Assumptions: (1) Steady operating conditions.
----+j
(2) Constant Properties. (3) Flow is turbulent over the
entire plate.
Analysis: For turbulent flow over a flat plate,
Nu= 0.037 (Re)415 (Prf 3
Let the subscript 1 denote the air flow conditions and the subscript 2 refer to the oil flow over the
same plate.
By the principle of similarity,
0.037 (Re2
t P,jf
5
3
0.037 (Rei.t 5 P·r/1 3
or
Hence, the average heat-transfer coefficient for oil is
With air
Nu1 =4150, Rei_ =2.4xI06
Basics of Convection Heat Transfer
Prt3 =
=
r
7.43
Nui
0.037 (Rt;)°' 8
4150
= 0.8824
0.037( 2.4xl06 )°" 8
With oil
CPµ
Pr2 = - k
p,,v 3 = [(2.194 X 10 3 J/kg°C) {13.2 / 3600)kg/m
2
s]
113
0.1316 W/m°C
= 3.9393
h2 = 0.1316 W/m°Cx 4150 x[3.9393]
1.5 m
0.8824
Hence,
= 1625 W/m 2 °C
(Ans)
Example 7.23 ~ Show that for pure conduction from a spherical surface into a fluid infinite in extent, the
Nusselt number approaches the value 2.
Known: Heat is transferred from a spherical surface into a fluid by pure conduction.
Find: Nu= 2.
Schematic:
Fluid, T..,
Sphere
of diameter
D=2r 1
T2~T..
---------1---.. r2 ~oo
k = knuid
Fig. 7.26
Assumptions: The spherical surface is approximated as a hollow sphere with outer radius approaching
infinity.
Analysis: Heat transfer by conduction from a hollow sphere with 1i ~ oo is
. 41tk(I;-T..,)
Q = [_!_ _ _!_]
1i
1i
(
)
41tklj I;-T..,
(since T2 ~ T..,)
(a)
7.44 ,
Heat and Mass Transfer
This is also equal to heat transfer by convection and is given by
Q = h ( 41t r/ )(T. - T= ) = 41t h r/ (I; - T= )
(b)
Equating (a) and (b),
hD
k=hlj=-
2
where k is the thermal conductivity of the medium (fluid).
It follows that
hD =l
k
hD
But = Nusselt number, Nu for a sphere with diameter D as the characteristic dimension.
k
Hence proved.
Example 7.24 ~ It is required to estimate the heat transfer from a cylinder SO-mm-diameter and of surface
temperature 140°C when kept in a cross flow of air at a velocity of 4 mis and temperature
20°C. For this purpose, scale-model experiments are performed using a I/5 scale-model with the same surface
and air temperature but different velocities. The following results are obtained from the experiments on the
model:
Velocity of air (m/s)
2.0
5.0
I0.0
20.0
Heat transfer coefficient fl/V/m 2 K)
39.5
71.2
I06.5
165.3
Calculate the heat-transfer coefficient and the rate of heat transfer per metre length of the actual cylinder.
Known: Results of scale-model experiments in the field of crossed flow forced convection heat transfer
over a cylinder.
Find: Heat-transfer coefficient and heat-transfer rate per m length of the actual cylinder (prototype) .
Schematic:
Prototype
Fig. 7.27
Assumptions: Surface and air temperatures are same under actual and experimental conditions.
Analysis: In forced convection (external flow) heat transfer, the Reynolds number for the model and the
prototype must be equal. Then
Basics of Convection Heat Transfer
r
7.45
Since Ts and T~ are same for both model and prototype, the film temperature Tr = _!_(I'. - T~) will
2
also be same. The air property like kinematic viscosity v will be same for both model and prototype.
Thus, we have
Since vm = vp, the model air velocity is determined to be
J
Vm = VP ( ; : = 4m/sx50mm/10mm = 20 m/s
From the given table (experimental results), the heat-transfer coefficient corresponding to a free stream
velocity of 20 m/s is 165.3 W/m 2 K.
h = 165.3 W/m 2 K
(Ans)
Heat-transfer rate from the actual cylinder per metre length is
Q = hA,; (I'. -T~) = h(1tD L)('.f. -T~)
= (165.3 W/m2 K) (1tx 0.05 m x 1m){140-20)°C
= 3115.83 W
(Ans)
Example 7.25 ~ Using a laboratory-scale model in water, the natural convection heat transfer from the casing
of a jet engine of 2.0-m-diameter and 3.0-m-length is to be investigated. The steady-state
operating temperature of the full-size engine casing is I 20°C when the outside air temperature is I8°C. The
model is to be heated to 60°C and immersed in water at 20°C.
(a) At what temperature should the thermophysical properties for air and water be evaluated?
(b) What should be the dimensions of the model?
(c) During a test on the model it is found that 17.2 kW are required to balance the losses by free convection
from the cylindrical surface. What would be the heat loss from the full size engine?
Known: Model testing of a cylinder in water.
Find: (a) R,eference for air and water ; (b) D2, L2; (C) QI .
Schematic:
Still air
T~1 = l8°C
Li= 3.0m _ _ __,
Prototype
Model
Fig. 7.28
7.46 , Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Constant fluid properties. (3) Radiation effects are
not considered.
Analysis: (a) Properties of fluid for free convection in a horizontal cylinder are evaluated at the film
1
temperature defined as Tr = -( 'T.urface + Tfluid)
2
Let suffix 1 denote actual casing (in air) and suffix 2 the laboratory model (in water). Then
1r, = T., +T-, = 120+18 =69oc
2
I
1r, = 'f.2 + T-2 = 60 + 20 = 400c
and
2
2
(Ans) (a)
2
2
(Ans) (a)
(b) The characteristic dimension for horizontal cylinder is its diameter, D.
D1 = 2 m
and L 1 = 3 m
To calculate D 2 and L 2 , the size of the model cylinder, we should equate the dimensionless number,
Ra0 for the model and prototype.
That is
Rao,1 = Rao,2
g ~1 Df ('f., - T-, )P'i
g ~2 Di ('f.2- T-2 ) Pfi
=---~--~-
v22
Vf
or
We note that
T.S1 -Tool =120-18=102°C
and
T82 -T002 = 60-20 = 40°C
The fluid properties at 1r,I = 69 °C (by linear interpolation) are obtained from the property table for
dry air at 1 atm:
1
1
1
v1 = 1.985x10-5 m2 /s 1
~ =
(K) = {69+273.15)K = 342.15 K
4i
"1 = 0.02874 W/m °C
p'i = 0.718
Porperties of saturated water at 1 atm and 1r,2 = 40°C are:
v2 = µ 2 /Pi = (0.653x10-3 /992.1) = 6.582x10-7 m 2 /s
k2 =0.631W/m°C
Pfi = 4.32
~2
= 0.377 X 10-3 K- 1
Basics of Convection Heat Transfer
r
7.47
Substituting the relevant values,
D = 2 om[(6.582x10-7 )
2
·
1.985x10-5
2
(
1
)(0.718)(102)]
342.15x0.377x10-3
4.32
40
113
=0.307 m
(Ans) (b)
Scale ratio,
Dm = D2 = 0.307 m = 0 _1534
DP
D1
2.0 m
Hence, L 2 =L1 x0.1534=3.0x0.1534 m
=0.46 m
(Ans) (b)
(c) Heat-transfer rate,
Dimensionless number,
Nu=hD
k
~
k
h = NuD
Hence, Q = Nu~(1tDL) (T,.-T=)
D
Q = Nu(1tkL) (T,. -T=)
or
Q1 QP Ntti (1t ki 4) (T,. T=J
-=-=--------Q2 Qm Nu2 (1t ki Li) (T,. T=, )
1 -
2 -
For similarity, Ntti = Nu2• Hence,
= 17 _2 kWx 0.02874 x-3-x 102
0.631
0.46 40
= 13.0 kW
(Ans) (c)
Example 7.26 ~ A small body of thermal conductivity k, thermal diffusivity a, and characteristic length Le
is initially at temperature T;. It is exposed to a convective environment at temperature r_
characterized by the convection coefficient h. The temperature of the body after time t is T(t).
Making use of mass (M), length (L), time (t) and temperature (7) as the four primary dimensions, perform a
dimensional analysis of the problem. Obtain the non-dimensional parameters that govern the problem.
Known: Unsteady-state heat transfer involving a small body subjected to convection.
Find: Non-dimensional groups.
7.48 , Heat and Mass Transfer
Schematic:
Body [k , a] with Le
111
Initially
T;= T(t= 0)
Finally T(t)
h,
r..
Fig. 7.29
Assumptions: (1) Constant properties. (2) Being a small body, spatial temperature gradients are neglected.
(3) Uniform convection coefficient.
Analysis: Let us first list the physical quantities, with their symbols and dimensions which are likely to
influence the temperature-time history of the body as given in the following table:
S.No.
Physical quantity
Symbol
Dimensions
1.
Excess temperature of the body above the
environment temperature at any time t.
(T- T~) =0
[T]
2.
Initial excess temperature of the body
above the environment temperature
T; - T~ = 0;
[T]
S.No.
Physical quantity
Symbol
Dimensions
3.
Characteristic length
LC
[L]
4.
Time
T
[t]
5.
Thermal diffusivity
k
[L 2 r 1]
(Xee--
P cp
•
•
•
6.
Convection coefficient
H
[Mr3 T- 1]
7.
Thermal conductivity
K
[MLr3 T- 1]
Number of physical quantities, n = 7
Number of primary dimensions, m = 4
Then, according to the Buckingham 1t-theorem, we get
Number of dimensionless pi-groups= n - m = 7 - 4 = 3
The dimensionless relationship will be of the form
1n1 = 1(n2, n3)1
Essentially, the objective is to determine the body temperature (or excess temperature, 0) after time t.
Thus 0(t) is a dependent variable. Excluding this, the four repeating variables are selected as
R. = 0;, R2 = Le , R3 = a
and
R4 = k
The non-repeating variables are identified as NR 1 = 0, NR 2 = h,
and
NR 3 = t.
Basics of Convection Heat Transfer
Then,
With NR 1 = 0
II1 = 0[0Ja [Lc)b [a]c [k]d
r
[M 0 L0 t0 T0 ] = [T][Tt [Lr [L2t- 1 [MLt-3 T- 1
or
t
Equating powers on both sides for:
=>
!d=OI
L:O=b+/+/ =>
!h=OI
t:0=-cAd
=>
1c=o1
T:O=l+a-/
=>
la=-1!
M:O=d
rr, = 0[0J-1 =>
Irr,= 010i I
II2 = h[0J a [LcP [a ]c [k ]d
f
[M 0 L0 t0 T0 ] = [Mr3 T- 1 ] [T]a[L]h [L2r 1 [MLr3 T- 1
Equating powers on both sides for:
M:O=l+d
=>
Ib = l I
t:0=-3-c-3d =>
!c=O!
T:0=-l+a-d
la=O!
L : 0 = b + 2c + d
=>
rr2 = h(4)1 (kt'
=>
lrr2 = h; I
With NR 3 =t
I13 = t[0j]a[Lc]b[aY[k]d
[M 0 LO t 0 T°] = [t][Tt [Lt [L2r 1
Equating powers on both sides for
M:O=d
=>
!d=OI
L:O=b+2c+/ =>
!h =-2c!
t:0=1-c--J
=>
le= 11
T:O=a-l
=>
la=O!
=>
!h=-21
f [MLr T- t
3
1
t
r
7.49
7.50 , Heat and Mass Transfer
(J,f]
0/0 = f[h LC
'
k 'L~
Since
We recognise that
h4 = Bi
-k
(Biot number)
and
at/2
Lc =Fo (Fourier number)
dimensionless temperature,
!0 10; = /(Bi,Fo)!
(Ans)
Example 7.27 ~ Consider a tube of 25-mm-diameter and surface temperature of 50°C losing heat by natural
convection to still air at I 5°C. In order to estimate the heat loss, model tests with a wire
heated electrically to 270°C are to be carried out in compressed air at I 5°C. The wire is to be 2.5 mm in
diameter. Determine the pressure that would be required, assuming that Nu = cp (Ra) with properties evaluated
at the film temperature.
Known: A tube loses heat to ambient air by free convection. Model testing with an electrically heated
wire dissipating heat to compressed air also by free convection is undertaken.
Find: Pressure of compressed air used for model testing, P(bar).
Schematic:
Q
D=25mm
Tube, T5 = S0°C
Natural Convection heat transfer
Fig. 7.lO(a)
Model testing
Fig. 7.JO(b)
Basics of Convection Heat Transfer
r
7.5 I
Assumptions: (1) Radiation effects are negligible. (2) Constant properties. (3) Air is an ideal gas.
Analysis: From dimensional analysis, for natural convection:
Nu =<l>(Ra)
Hence,
(Nu)tube =(Nu)wire
( Rao )tube = ( Rao )wire
Tube : D = 0.025 m , Film temperature, Tr =.!.(T. + T=)
2
= (50 + 15)/2 = 32.5°C
or
305.65 K
The relevant properties at the film temperature of 32.5°C for air at 1 atm are:
v = 16.315 xio-6 m2/s
~
1
idealgas - 4(K)
1
(305.65K)
Pr=0.7275
Rayleigh number, Ra0 =
g~D3 (1'.-T )Pr
• =
y2
= (9.81m/s2)(0.025m)3 (50-15)°C(0.7275) = 47 972 .4
(305.65 K)(16.315x10-{j)2
Wire D = 0.0025 m,
1
Tr =-(T. + T=) = (270 + 15)/ 2
2
= 142.5°C
or
415.65 K
Properties of air at 1 atm and 142.5°C are
V = 27.74 X10-{i m2 /s
1
1
~idealgas = 4 (K) = ( 415.65 K)
Pr=0.7038
3
( Ra ) . = 47 972.4 = g~D (T. -T=)Pr
0
wire
y2
Unlike incompressible fluids like liquids, gases are compressible, i.e., their density depends both
on pressure as well as temperature. Hence, v =.!: for compressed air will be significantly different
p
from that for 1 atm. All other properties like Cp, µ, k and Pr are practically insensitive to changes in
pressure. They vary only with temperature. Therefore, let us find v from the expression for Rayleigh
number for the wire.
7.52 , Heat and Mass Transfer
=[(9.8 lm/s 2 )(0.0025 m)3 (270-15)°C (0.7038)]112
(415.65K) (47 972.4)
= 1.175xl0-6 m2 /s
But
Y@latm =27.74x1Q-6m2 /s
We note that vis inversely proportional to p [v = µ / p] and p for an ideal gas is directly proportional
to pressure,
P= [p = :T J.
Hence, for air, v 2 = Pi
v1 P2
V[P,142.5°C]
latm
V[latm,142.5°CJ
P(atm)
Hence, the pressure of compressed air is
27.74x10-6
P =- - - x 1atm = 23 .62 atm
1.175x10-6
As 1 atm = 1.013 25 bar,
P = 23.62 X 1.013 25
= 23.93 bar= 24.0 bar
(Ans)
, MULTIPLE CHOICE QUESTIONS
7.1 The hydrodynamic boundary layer thickness is defined as the distance from the surface where the
(a) velocity equals the local external velocity
(b) velocity equals the approach velocity
(c) momentum equals 99% of the momentum of the free stream
(d) velocity equals 99% of the local external velocity.
7.2 Nusselt number is defined as hL/k, where k represents
(a) thermal conductivity of the solid surface
(b) thermal conductivity of solid or fluid whichever is higher
(c) thermal conductivity of the fluid flowing past the surface
(d) thermal conductivity of solid or fluid whichever is lower
7.3 If the number of fundamental dimensions equals m then the repeating variables shall be equal to
(a) m and none of the repeating variables shall represent the dependent variable.
(b) (m + 1) and one of the repeating variables shall represent the dependent variable.
Basics o( Convection Heat Transfer
r
7.53
(c) (m + 1) and none of the repeating varables shall represent the dependent variable.
(d) m and one of the repeating variables shall represent the dependent variable
7.4 Reynolds analogy states that
(a) St= Cr
(b) St= Cr
4
2
(c) St= .jc;
(d) St= 2 Cr
where St is the Stanton number and Crx is the skin friction coefficient
7.5 Water at 12°C (v = 1.225 mm2ls) flows with a mean velocity of 0.2 mis through a 10-mm-diameter
smooth pipe. Considering fully developed flow, the Fanning friction factor and centreline velocity
will be
(a) 0.0392, 0.4 mis
(b) 0.15, 1.5 mis
(c) 0.072, 1 mis
(d) 0.0098, 0.4 mis
7.6 Given below are the ratios of the displacement-and momentum-thickness to the boundary layer
thickness on the left and the boundary layer profile on the right.
o*lo; 0/o
Velocity profile
1. 0.125; 0.097
A. u/u- = 2{y/'6)-{y/'6)2
2. 0.375; 0.139
B. u/u- = (y/'6)'17
3. 0.333; 0.133
C. u/u- = sin{1ty/2'6)
4. 0.363; 0.137
D. u/u- = 3(y/8)/2-(y/'6)3 /2
Mark the correct answer:
(a) lD,
2C,
3A,
4B
(b) lC,
2B,
3D,
4A
(c) lA,
2B,
3C,
4D
(d) lB,
2D,
3A,
4C
7.7 If the turbulent boundary layer on a flat plate is assumed to start at the leading edge, the average
friction coefficient Cr over a length L is given by
(a)
0.005
(R~)°"s
(b)
(c) 0.09(ReL)°" 2
0.072
(R~)°"2
(d) none of the above
where ReL is the Reynolds number
7.8 Match List I (Variable in Laminar Boundary Layer Flow over a Flat Plate Set Parallel to the
Stream) with List II (Related Expression with usual notations) and select the correct answer using
the codes given below the lists:
List I
A. Boundary layer thickness
B. Average skin friction coefficient
List II
1.1.12y~
7 .54 ,
Heat and Mass Transfer
~
v--;z
C. Shear stress at the boundary
3.5
D. Displacement thickness
4. 0.664
~ V
u=x
5. 1.328/ ~u= L/v
A
B
D
Codes:
C
4
2
(a)
3
5
2
4
l
(b)
3
2
l
(c)
3
5
4
l
2
(d)
5
7.9 For flow of fluid over a heated plate, the following fluid properties are known: viscosity= 0.001 Pa.s;
specific heat at constant pressure = 1 kJ/kg K; thermal conductivity = 1 WIm K.
The hydrodynamic boundary layer thickness at a specified location on the plate is l mm. The
thermal boundary layer thickness at the same location is
(a) 0.001 mm
(b) 0.01 mm
(c) l mm
(d) 1000 mm
7.10 Match the following
P : Compressible flow
Q : Free surface flow
R : Boundary layer flow
S : Pipe flow
T : Heat convection
U : Reynolds number
V : Nusselt number
W: Weber number
X : Froude number
Y : Mach number
Z : Skin friction coefficient
(a) P-U; Q-X; R-V; S-Z; T-W
(b) P-W; Q-X; R-Z; S-U; T-V
(c) P-Y; Q-W; R-Z; S-U; T-X
(d) P-Y; Q-W; R-Z; S-U; T-V
, TRUE/FALSE
7.1 Laminar sub-layer is a region near the flat surface in a turbulent zone where velocity variation is
nearly linear.
7.2 The Darcy friction factor for fully developed turbulent flow in a smooth tube for Re0 > 2 x l 04 is
f= 0.316 Re0114
7.3 The friction factor for flow through a pipe of constant diameter is inversely proportional to Reynolds
number.
7.4 Kinematic similarity implies geometric similarity and vice versa.
7.5 Navier-Stokes equations are valid for a two-dimensional compressible, variable viscosity flow.
Basics o( Convection Heat Transfer
r
7.55
, FILL IN THE BLANKS
For the velocity distribution u/u- = 2TJ-2TJ3 +TJ4. the ratios 8* /8 and 0/8, where TJ = y/8, 8* is the
displacement thickness and 0 the momentum thickness, are _ _ and _ _ respectively.
7.2 Benzene at a temperature of 27°C (µ = 59 x 10-5 kg/m s) flows in a tube with 5-cm ID at a flow
rate of 1.5 kg/min. The flow is _ _ _ _, with Re 0 =__.
7.3 For saturated liquid propane at 30°C, the thermal diffusivity is 240 x lQ-6 m2 /h and the kinematic
viscosity is 190 x 10-9 m 2/s. The Prandtl number is _ _ .
7.4 For laminar flow over a 1-m long flat plate (ReL = 2.45 x 10 5) with linear velocity profile, the velocity
boundary layer thickness at the trailing edge is _ _ .
7.5 The maximum velocity in a fully developed turbulent flow through a circular tube is _ _ times
the average velocity.
7.1
, EXERCISES
7.1
Calculate the drop in pressure when liquid refrigerant R-134a at 45°C flows through a 0.5-mmdiameter capillary tube 50-cm-long with a velocity of 0.2 mis. Assume that the flow is fully
developed. Take p = 1125 kg/m 3 and µ = 1.554 ~ 10--4 kg/m s. How does the pressure drop change
if the velocity of flow is doubled?
[3978 pa]
7.2 Water at 30°C flows through a 25-cm diameter concrete pipe having a roughness (E) of 1 mm.
Determine the Darcy friction factor for fully developed flow if the mass flow rate is 2 5 kg/s.
Properties of water at 30°C:
p = 996.0 kg/m3,
µ = 0.798 x 10-3 kg/ms
[0.0291]
7.3
The torque due to the frictional resistance of the oil film between a rotating shaft and its bearing
is found to be dependent on the force F normal to the shaft, the speed of rotation N of the shaft,
the dynamic viscosity µ of the oil, and the shaft diameter D. Establish a correlation between the
variables by using dimensional analysis.
[T=µND 3 f[F/µ ND2]]
, ANSWER KEY
Multiple Choice Questions
7.1 (d)
7.6 (d)
7.2 (c)
7.7 (b)
7.3 (a)
7.8 (c)
7.4 (b)
7.9 (c)
7.5 (d)
7.10 (d)
7.2 F
7.3 T
7.4 F
7.5 T
True/False
7.1 T
Fill in the Blanks
7.1 0.30, 0.117
7.4 7 mm
7.2 laminar, 1079
7.5 1.25
7.3 2.85
FORCED CONVECTION:
EXTERNAL FLOW
8
Concept Review
INTRODUCTION
• 8.1
•
Forced convection heat transfer involving surfaces- flat or curved- subjected to external fluid flow are
characterised by the absence of buoyancy force and by the freely growing boundary layers surrounded
by a free-stream flow region in which there are velocity or temperature gradients.
Flow is driven directly from the action of a fan or a pump, or due to relative motion such as between
a moving surface in a stationary fluid .
Empirical correlations for different idealised geometries, configurations and orientations are presented
in this chapter. These have wide applications in the design and analysis of thermal systems.
Convective heat transfer is related to temperature difference by
(8.1)
Convection correlations for the heat-transfer coefficient, h, are written in terms of the nondimensional
Nusselt and Prandtl numbers, defined as Nu = hL Pr= CPµ, where L is a characteristic length for the
k
k
geometry. The Prandtl number is a fluid property that varies with temperature.
PARALLEL FLOW OVER FLAT PLATE
• 8.2
•
Fluid
Flat plate
Ts
L
Fig. 8.1
• Characteristic length: Linear dimension along the flow direction.
• Properties of the fluid at the mean film temperature,
Tr =..!._(T.+T=)
2
• For forced convection over an isothermal flat plate, the local heat-transfer coefficient is
8.2
, Heat and Mass Transfer
Rex < 5 X 10 5
h X
Nu= _ kx_ = 0 .332 Re X112 Prli 3
Pr> 0.6
(8 .2)
isothermal plate
Re,._> 5 X 10 5
Nu= h xx = 0 0296 Re 415 Pr 113
k
.
X
0.6 < Pr < 60
(8.3)
isothermal plate
with properties evaluated at the film temperature Tr = (I;, + T~) I 2 .
• For forced convection over an isothermal flat plate with a laminar boundary layer, the average
heat-transfer coefficient is
ReL <5xl0 5
hL
NuL = - = 0.664 Re L2 Prli 3
k
Pr>0.6
(8.4)
isothermal plate
• For a plate with both laminar and turbulent boundary layers,
5 x 10 5 < ReL ~ 10 8
hL
)
NuL =-=(0.037Ref5 -871 Pr 113
k
0.6 < Pr< 60
(8 .5)
isothermal plate
• For a plate with only a turbulent boundary layer,
hL
N uL =-=0.037Ref 5Prll3
k
0.6 <Pr< 60
(8 .6)
isothermal plate
The critical Reynolds number for transition from laminar to turbulent flow is 5 x 10 5.
• For forced convection over a flat plate with constant heat flux, the local heat-transfer coefficient is
Rex < 5 X 10 5
112
113
NuX = hxx
k = 0 .453 Re X Pr
Pr< 60
(8 .7)
constant heat flux plate
Rex > 5 X 10 5
hx
415 113
NuX = _ x_
k = 0 .0308 ReX Pr
0.6 <Pr< 60
constant heat flux plate
(8 .8)
Forced Convection: External Flow
r
8.3
To find the average heat-transfer coefficients for a flat plate with constant heat flux, use the correlations
for the average heat-transfer coefficient with constant surface temperature given above.
• For Re Pr > 100 and especially for liquid metals (very low Pr) and heavy oils (very high Pr),
Churchill and Ozoe correlation can be used for laminar flow over an isothermal flat plate:
Nu.~H~rr
0.3387 Re~12 Prll 3
(8.9)
Flat Plate with Unheated Starting Length As shown in Figure 8.2, the velocity boundary layer
development begins at x = 0, while the thermal boundary layer development starts at x = x0. There is
therefore no heat transfer for O < x < x 0 .
-----8
Fluid
r.
x=x0
Fig. 8.2
L
Flat plate in parallel flow with unheated starting length
• For an unheated starting length with laminar flow over a flat plate where the velocity boundary
layer begins at x = 0 and the thermal boundary layer at x = x0.
= 0 ·332 Rex Prll
112
Nu
x
3
(8.10)
[1-(xo /x)3/4f3
• For an unheated starting length in turbulent flow with a thermal boundary layer that begins at x = x0 ,
= 0.0296 Re:_15 Pr113
Nu
x
(8.11)
[1-(xo / x)911of9
Average convection heat-transfer coefficients are the following:
(8.12)
Turbulent h = hx-L
-
l-(x lxt'
[1.25{l-(x
I L)
0
0
0
}]
(8.13)
8.4
, Heat and Mass Transfer
• High-speed Flow
When the free-stream velocity is very high, the effects of viscous heat dissipation within the boundary
layer, can no longer be neglected. Considering the adiabatic case, the wall (surface) temperature may be
significantly higher than the free-stream temperature.
Adiabatic wall temperature, Tad can be related to the stagnation temperature, T0 by means of the
recovery factor, r, defined as
¾d - T=
r =- - where r =Pr*112
(laminar) r =Pr*113 (turbulent)
(8.14)
To-T=
The empirical correlations are
!Nu; = 0.332 Re;u 2 Pr *1/ 3 1
(laminar, Re; < 5 x 10 5 )
!Nu; = 0.0296 Re; 415 Pr* 113 ! (turbulent, 5 x 10 5 < Rex < 10 7)
Nu*= 0.185 Re; Pr*'"
x
(log Re; )2.584
(8.15)
(8.16)
(8.17)
(10 7 <Re;< 109 )
where properties are evaluated at the reference temperature T* which is given by
(8.18)
Stagnation temperature,
where M= is free stream mach number
Heat-transfer rate,
IQ= hA(T. -T,.d )I
(8.20)
• 8.3
FLOW ACROSS A CYLINDER
•
For the cylinder in cross flow as shown in Figure 8.3, the
correlations suggested are given as follows:
• Hilpert's Correlation
INuo =C Ref', Prll 3 I
(8.21)
0
• Characteristic dimension Diameter, D
• Fluid properties at Tr =(T. + T=) I 2
• Constant surface temperature
Fig. 8.3
Circular cylinder in cross flow
Forced Convection: External Flow
8.5
Values of Constants C and n in Eq. 8. I4
Table 8.1
• Fond's Correlation
r
Range of ReD
C
n
40 to 4000
4000 to 40 000
40 000 to 400 000
0.683
0. 193
0.027
0.466
0.6 18
0.805
For 10- 1 < Re < 105, for liquids in cross-flow over cylinders:
INuo =(0.35+0.56Re& )Pr0· I
52
(8.22)
3
• Whitaker's Correlation
(8.23)
for 40 < Re < 105, 0.65 < Pr < 300, and 0.25 < µ= /µs < 5.2 All properties are evaluated at T= except
that µs is at Ts.
• Churchill and Bernstein Correlation
N
{
5/8 )
Re 0
0.62Re01/2
1
. +[1+(0.4 / Pr)2/3r
+C82000)
O3
UD=
415
(8.24)
[Re0 up to 107 ]
[Re0 Pr>0 .2,
For the range 20 000 < Re0 < 400 000, replace
with [1 + (
Reo
282000
)
112
]
• Zhukausakas Correlation
1Nu =CRef)Pr,n(Pr l P,-.)°"
0
25
(8.25)
1
[0.7 < Pr< 500 ; I < Re0 < 10 6]
All properties are evaluated at T= except Prs, which is evaluated at Ts. Values of C and m are listed
in Table 8.3. If Pr~ IO, n = 0.37; if Pr> IO, n = 0.36.
Table 8.2
Values of Constants C and m in Eq. 8. I8
ReD
C
m
1 to 40
40 to 1000
10 3 - 2 X 10 5
2 X 105 -1 0 6
0.75
0.5 1
0.26
0.076
0.4
0.5
0.6
0.7
8.6
,
Heat and Mass Transfer
FLOW OVER SPHERES
• 8.4
•
• Whitaker's Correlation
For a sphere in cross-flow, the average Nusselt number is
Fluid
Sphere
Q
(8.26)
0.7 l< Pr <380, 3.5< Re0 < 7.6xl0 4 , 1.0 <( t J< 3.2
Fig. 8.4
Ts
Flow over a sphere
Fluid properties are evaluated at the free stream temperature T= and µs is evaluated at the surface
temperature, Ts.
FLOW ACROSS BANKS OF TUBES
• 8.5
•
Fluid
Fluid
T=
V
OT
ST
0
0
0
0
0 ...L
0
0
0
i-SL-J
(a)
Fig. 8.S
T=
V
---
0
0
OT
ST
0
o...L
0 '<'D
s ~
0
0
0
Lsl3
(b)
Tube arrangements in a bank: (a) In-line (b) Staggered
• Aligned (in-line) Arrangement In a bank of tubes, the tubes are arranged either in line or staggered
in the direction of fluid velocity Vas shown in Figure 8.5.
(8.27)
• Staggered Arrangement
(8.28)
Forced Convection: External Flow
r
8. 7
The latest correlation, for 0.7 < Pr< 500 and O < Re0 < 2 x 106, the average Nusselt number is
-
hD
Nuo =-=C Re{!.
k
D,max
Pr0 -36
( h )l/4
(8.29)
-
P,;,
T+T
where all fluid properties are evaluated at Tmean = -•--• except Prs which is at Ts.
2
The constants C and n for heat transfer in tube banks of 16 rows or more are given below.
Table 8.3
Constants C and n in Eq. 8.29
Geometry
In-line
Staggered
Rev, max
C
n
0- 100
0.9
0.4
100- 1000
0.52
0.5
1000-2 X 10 5
0.27
0.63
2 X 105 - 2 X 106
0.033
0.8
0- 500
1.04
0.4
500- 1000
0.71
0.5
1000- 2 X 10 5
0.35(ST I SL) 02
0.6
2 X 105 - 2 X 106
0.03(ST I SL) 02
0.8
For tube banks with less than 16 rows, the correction factor F given below should be used .
Nuo =F Nuo
Table 8.4
In-line
Staggered
Correction Factor F to be used in Nuo.N =F Nuo for NL < I 6 and Re0 > I 000
1
2
3
4
5
7
10
13
0.70
0.64
0.80
0.76
0.86
0.84
0.90
0.89
0.93
0.93
0.96
0.96
0.98
0.98
0.99
0.99
pV2
Pressure drop, liP=NLfX~
2
(8.30)
where f is the friction factor and X is the correction factor plotted in Figure 8.5(a) and (b ). X = l for both
square and equilateral triangle arrangements.
Power required, p =m11P Ip
where m=p V(NTSTL)
(8.31)
8.8
,
Heat and Mass Transfer
Solved Examples
(A) FLAT PLATE IN PARALLEL FLOW: LAMINAR AND TURBULENT
Example 8.1 ~ Air at 25°C blows over a 35°C flat surface with a sharp leading edge with a free-stream
velocity of 1.5 mis. Calculate (a) velocity boundary layer thickness 0.5m from the leading edge.
and (b) the overall drag coefficient for this surface if its total length is 0.5 m. (c) Compare the average shear
stress with the local shear stress at the trailing edge. (d) At what point on the surface do they become equal?
Properties of air at 30°C are
p = 1.164 kg/m 3 v = 16.08 x Io--6 m2/s
Known: Air flows over a flat surface with a prescribed velocity.
Find: (a) 8; {b) (\, (c) Tw and (d) x at which tw=tw .
Schematic:
Air
u== l.5m/s
T= =25°C
x=L=0.5m
------==.r----T, = 35°C
Fig. 8.6
Assumptions: ( 1) Steady state conditions. (2) Air is an ideal gas. (3) Constant properties. (4) Recr = 5 x 10 5.
Analysis: Reynolds number,
( < Recr = 5 x 105 )
Re =u=L=(I.5m!s)(O.Sm)=4.664xl04
L
v
==}
Laminarflow
16.08xJ0-6m2 / s
Velocity boundary layer thickness is
= 4.91 L = {4.91) (0.5m) = 0 _0114 m
0
(x=L)
(4 .664xl0 4 }11 2
~ReL
=11.4mm
(Ans) (a)
The overall drag coefficient,
c _1.328 _
f,L -
1.328
~ReL - .J4.664 X 104
= 6.15 X 10-3
(Ans) (b)
and, the average shear stress
'tw = ½ p
u;, Cf,L = ½(1.164 kg/m 3) [ {1.5)2 m 2 /s2 ]( 6.15 X J0-3)
= 8.05 x 10-3 kg/m s2
or
N/m 2
(Ans) (c)
Forced Convection: External Flow
r
8. 9
At the trailing edge, i.e., at x = L the ratio of local to average shear stress is,
(Ans)(c)
Now
- -
'tw ( X ) - 'tw
h
W en
0.664 _ 1.328
1
vx
~
-
v0.5m
Hence, the value of distance from the leading edge, x where the local and average shear stresses are equal is
0.664~0.5 m ] 2
= 0,125 m
1.328
X = [--~--
~ = 0.125 m = 0.25 or
i.e.,
L
(Ans)(d)
1
4
0.5 m
o
Comments: The boundary layer analysis holds good only if 8/ x << 1. In this example, - =
X
0.0114m
0.5m
= 0.0228 (« l).
It may be noted that the shear stress which is infinite to begin with (at x = 0) drops to the average
value( 'tw), one-fourth of the way from the leading edge, and in the rest of the three-fourth of the plate,
it reduces only to one half of the average shear stress.
Example 8.2 ~ Atmospheric air at 25°C and 20 mis flows parallel to and over a flat plate. Calculate (a) the
boundary layer thickness, and (b) the wall shear stress at distances of I mm, 10 mm and
I 00 mm from the leading edge (c) If a second plate is placed parallel to and separated by 4 mm from the first
plate, estimate the distance from the leading edge at which the two boundary layers will merge with each other.
Air ( I atm, 25 °C): p = 1.184 kg/m 3, v = 15.62 x I 0--6 m2/s.
Known: Temperature and velocity of air in parallel flow over a flat plate.
Find: (a) o(x) and 'tw(x) at x= l, 10 and 100 mm. (b) xmerger for boundary layers to merge if the distance
between parallel plates L = 4 mm.
Schematic:
Leading edge
Air
P= latm - - T~ =250C - - u~ = 20 mis - - -
Lx
Xmerger
x=O
Fig. 8.7
Assumptions: (1) Steady-state conditions. (2) Laminar flow. (3) Boundary layer approximations hold good.
Analysis: Reynolds number for the maximum specified distance from the leading edge, x = 100 mm= 0.1 mis
u x
Re=~
x
v
20m/sx0.lm
15.62xl0-6 m2 / s
=l.28xl0 5
(<5xl05 )
8.1 0
,
Heat and Mass Transfer
The flow is thus laminar.
The exact solution gives the boundary layer thickness as
112
=> <>(x) = 5.ox[~J
~=_2:2..._
X
~
U00 X
2
() =5 x112 ( u~ )'1
or
112
= 5 [15.62 xl0-6 m2 /s] ( x 112 )
20m/s
= 4.4187 xl0-3 xx112 (m)
1<>=4.4187v'x(mm)I
X
J mm
JO mm
JOO mm
6(x) (mm)
0.140
0.442
1.397
(Ans) (a)
Surface (wall) shear stress,
l
1
2 = 0.664Re.::- 112 x-pu2
'tW =Cf,x ·-puoo
X
2 oo
2
= 0.332 pu~ x( u;
r
12
X .x- 112
k
(
=0.332xl.184~x 20 m
3
m
s
=0.139.x- 112 (N/m2)
)3/2 x~15.62x10- m2 Is x[ x(m)J-112
6
With x in mm, 'tw = 0.139x(l0-3 t 112 xx- 112
l'tw = 4.394[x(mm)f1 12 (N/m2)1
X
J mm
JO mm
100mm
4.394
1.39
0.4394
(Ans) (b)
We note that as the distance from the leading edge increases, the velocity boundary layer thickness
also increases but the wall shear stress decreases.
1
Boundary layer merger will take place at a distance xmerger when <> = H = 2 mm.
2
8 = 4.4187 112 (mm)
Since
xll2
merger
Hence,
=<>(mm)= ---3.:2_ = 0 4526
4.4187 4.4187
·
xmerger = 0.205 m
or
205 mm
(Ans) (c)
Forced Convection: External Flow
r
8.1 I
Example 8.3 ~ Atmospheric air at 30°C is flowing over a flat plate with an approach velocity of 4 mis. Determine
(a) the distance from the leading edge of the plate where the air velocity is 3.96 mis at a vertical
distance of I cm from the plate surface, and (b) the vertical distance from the plate surface at the same distance
from the leading edge where the temperature will be 30.3 °C. The uniform plate surface temperature is 60°C,
Use the following properties of air at 30°C:
p = 1.1646 kglm3
µ = 1.86 x I 0-5 Nslm2,
cp = 1.007 kJlkg c
k = 0.0265 W/m 0 C.
0
Known: Air flows over a flat plate with a prescribed velocity.
Find: (a) x (cm); (b) y (cm) under specified operating conditions.
Schematic:
Air
P=l,m,
~Edg,of<h•=I
u =4mls - - ~
boundarylayer
= 30° C
Edge of velocity
T:
y
07{x)
Lx ----
o(x)
boundary layer
Flat plate (Ts=60°C)
X
Fig. 8.8
Assumptions: (1) The edge effects are negligible. (2) The critical Reynolds number is 5 x 10 5 . (3) The local
atmospheric pressure is l atm.
Analysis: (a) At a distance x from the leading edge, y(x) = l cm.
u(x) = 3.96m I s= 0 _99
u~
4m /s
u!u~ = 0.99, y = 8 = l cm.
At
Thickness of the velocity boundary layer, o(x) = 1 cm
Using the Blasius' exact solution:
o(x)
5.0
-x- ""
5.0
-JRe: = Jp u~x I µ
or
82
25µ
The length from the leading edge, x for 8 = l cm is
x= (O.Olm}2(1.1646 kg/ m3 )(4.0 m/s)I
25 x 1.86 x 10-5 N s/ m2
lN
I
l kg m/ s2
=1.0 m
(b) The ratio
('f.-T(y)) = 60-30.3 = 29.7 =0_ 99
('f. - T~)
60 - 30
30
(Ans) (a)
8.12
,
Heat and Mass Transfer
The thermal boundary layer thickness BT is typically defined as the value of y for which the ratio
[ (T. -T)/(T. -T=)] is 0.99. At x = l m, 8 = 1 cm, and y = BT under these conditions.
We note that
1
<> T /<> "" Pr113
y(x = lm) =BT= B(x = lm)Prll3
= lcmxPr- 113
where
CPµ 1007 J/kg°Cxl.86xl0-5 Ns/m2
Pr= - - = - - - - - - - - - - - k
0.0265 W Im °C
= 0.7068
y =BT= lcmx(0.7068t 113
= 1.12 cm
(Ans) (b)
Example 8.4 ~ Mercury at 80°C flows with a free-stream velocity of 0.1 mis over a flat plate maintained at
I 20°C. Calculate the average heat transfer coefficient over the length of the plate where the
flow is laminar. Assume that transition from laminar to turbulent flow occurs at Recr = 5 x I 05•
Properties of mercury at I00°C are
k = 9.46706 Wlm °C
v = 9.326 x I0-8 m2ls
Pr= 0.018
Known: Mercury flows over a heated plate with a prescribed velocity.
Find: Average heat transfer coefficient, h(W/m2 °C).
Schematic:
Mercury
u==O.I mis
r==ao·c
c=
Flat plate
,...__ _ _ _ _ _ _ _ _ __.I (T.= I 20°C)
Xcr - - - - . . i
Fig. 8.9
Assumptions: (1) Steady-state conditions. (2) Constant properties. (3) Rec, = 5 x 10 5 •
Analysis: Critical Reynolds number,
Re = u= xc, = 5 X l 05
er
V
Hence, the length of the plate over which the flow is laminar is
V Rec,
9.326 X 10-8 m2 /s x 5 x 10 5
xcr = - - =
u=
=0.4663 m
Peclet number, Pex = Rex Pr
0.1 m/s
Forced Convection: External Flow
r
8. 1l
For liquid metals with small Prandtl number, for laminar flow, the relevant correlation is
!Nu,, = 0.565 Pe!/ 2 I
At the critical point, (x = xcr)
Local heat-transfer coefficient,
hx =Nu,,~= 0.565 {9000f2 X 9.46706 W/m oc
X
0.4663 m
= 1088 W/m2 °C
Average heat-transfer coefficient,
(Ans)
Example 8.5 ~ A 750-W heater is constructed of a glass plate, 50 cm by 50 cm with an electrically conducting
film which produces a constant heat flux. The plate is placed in an air stream at atmospheric
pressure and 30°C with a velocity of 6 m/s. Determine (a) the average temperature difference along the plate,
and (b) the temperature difference at the trailing edge.
Known: Glass plate with constant surface heat flux loses heat to air stream.
Find: (a) Average temperature difference along the plate, (T.-T=)(°C); (b)Temperature difference at the
trailing edge, (
r. -T=) I
x=L ( 0
Schematic:
c).
Air
P = I atm
r==30°c
u==6m/s
~ Glass plate (W=0.5m)
L _ (constant heat flux)
~,:::===============~====::I
,____ L=0.5cm _ _ _..., Q. = 750W
Fig. 8.10
Assumptions: ( 1) Steady-state conditions. (2) Constant heat-flux conditions. (3) Constant properties.
(4) Recr = 5x10 5 •
Analysis: (a) Reynolds number at x = L = 0.5 m is Re= u= L/v
Properties are to be evaluated at the film temperature, 4 = .!_ (T. + T=). Since T. is not known, we take
2
the properties at the free-stream air temperature of 30°C. At 30°C:
v= 16.08 x lQ-6 m2/s
k= 0.02588 W/m °C
Re= ( 6 m/s)(0.5m) =1.866x10 5
16.08x10-6 m2/s
(<5xl0 5 )
Pr= 0.7282
~
Laminarflow.
Local Nusselt number,
Nu,, =0.453 Re}!. 2 Pr113 = \x
k
(Pr~0.6, q.=constant, Re<5xI0 5 ).
8.14
,
Heat and Mass Transfer
(A)
Also
and
Average temperature difference,
L
L
-T )dx= qs J x dx
(T.s. -T )= _!_J(T..
L
s
L kM
00
00
0
U,,.
0
L
=q J
X
8
dx
L 0 kx0.453 ( u: x )112 Pr113
L
q. Jx 112 dx
=
o
Lk(0.453) (u.. /v}11 2 Pr113
=
q.
( I}t2 /3 I 2)
k L(0.453)(u.. /v}11 2 Pr11 3
q.L/k
=---~-----,-,-,--1.5x(0.453)(u.. L/v}11 2 Pr113
)T..-T
( s
..
-
QL/k A.
(B)
0.6795 Rei 2 Pr113
k--q. = 1.5 XO .453 Rel! 2 Pr113 X - (T. - T.. )
L
Also
(C)
Substituting appropriate values in equation (B), we have
(
T.
_
)- 750 Wx0.5 m/(0.02588 W/m°Cx0.5mx0.5m)
T.. -
0.6795x(l.866x10 5 }11 2 (0.7282}113
=220°c
Film temperature,
4 = T. -T.. +T.. +T.. = 220+30+30 = 1400C
2
2
At 140°C:
k = 0.03374 W/m °C
v= 27.45x10-<i m2 /s
Pr= 0.7041
Forced Convection: External Flow
Re=
L
(
_
r
8.15
{0.5 ){ 6 m/ s) =1.093xl0 5
27.45xl0-6m2 / s
)- (750 X 0.5) / (0.03374 X 0.5 X 0.5)
I;, T= - 0.6795x{l.093xl0 5
f {0.7041{
2
3
= 222.5 °C
(Ans)(a)
(b) At the end of the plate (x = L = 0.5 m), the temperature difference is obtained as follows:
(from Equation A)
or
(from Equation C)
= 1.5 X 222.5 = 333.8°C
(Ans)(b)
~ Air at velocity of 3 mis and at 20°C flows over a flat plate along its length. The length, width
Example 8.6
and thickness of the plate are 100 cm, 50 cm and 2 cm, respectively. The top surface of the
plate is maintained at I00°C. Calculate the heat lost by the plate and the temperature of the bottom surface
of the plate under steady state conditions. The thermal conductivity of the plate may be taken as 23 W/m K.
Take properties of air as
p = 1.06 kg/m 3,
v = 18.97 x I0-6m2/s
k=0.02894 W/m K,
Pr=0.696
Known: A flat plate with specified top surface temperature and physical dimensions is subjected to
convective process.
Find: Heat loss from the plate, Q(W), and plate's bottom surface temperature, J;{ 0 C).
Schematic:
Q
T1=I00°C
Air
soc0
j
T==20°C
u==3m/s
kp=23W/mK
T2 (bottom surface)
Q
T=
•
T1
Wv
Rconv (top)
•
T=
T2
Wv
Rcond
•
Q
Wv
Rconv (bottom)
•
Q
Fig. 8.11
8.16
,
Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Transition Reynolds number,
Recr = 5 X 10 5.
Analysis: Heat lost by the plate,
Qtotal =2hA.(1i-T=)
A8 =1.0mx0.5m=0.5m2
where
r, = 100°c, r= = 20°c
Qtotal =2x0.5m2 x(100-20)°Cxh (W/m2 K)
= 80h(W)
To evaluate the convection coefficient, h, we need to ascertain if the flow is laminar or turbulent. For
this purpose, let us evaluate the Reynolds number.
u=L
3m/sxl.Om
ReL = - - = - - - - - v
18.97x10-<i m 2 /s
= 1.58x105
(< 5x105 )
The flow is therefore laminar.
The appropriate correlation is
NuL =0.664 ( ReL )
1/2
(Pr)113
= 0.664(1.58x10 5 )11 2 (0.696)'1 3 = 234.0 = hLI k
The average heat-transfer coefficient is
h = !!._ NuL
L
0.02894 WIm K X 234 _0
1.0m
= 6.772 W /m2 K
Rate of heat transfer is
Qtotal =80x6.772 = 541.8 W
Referring to the thermal circuit,
·
=
Rcond
I; - T=
'li - T=
=
Y:2-L
T.1-T
Q=
+ Rconv(bottom)
Rconv(bottom)
Rcond
+ Rconv(bottom)
-
Rconv(bottom)
1/ h A
=----(
½Jkp A)+ hlA
= _1_=[(6.772W/m2 Kx0.02m)+i]-I = 0 _994
h½ +l
23W/mK
kp
(Ans)
Forced Convection: External Flow
r
8.17
Hence, the temperature of the bottom surface of the plate is
Ti =T= +0.994(7; -T=)
= 20°C + 0.994 X (100 - 20)°C
= 99.53°c
(Ans)
Comment: Had the plate been thin, the bottom surface temperature would also have been 100°C.
Example 8.7
~ Air at a velocity of 3 mis and at 20°C flows over a flat plate along its length. The length,
width and thickness of the plate are I00 cm, SO cm and 2 cm respectively. The top surface of
the plate is maintained at I00°C. Calculate the heat lost by the plate and the temperature of the bottom surface
of the plate for the steady state conditions. The thermal conductivity of the plate may be taken as 23 W/m K.
Properties of air at 60 °C:
p = 1.06 kg/m 3
v = 18.97 x 10- 6 m2/s
k=0.0290 W/m K
Pr=0.696
Empirical correlations:
N"i_ = 0.664 Re~2 Pr' 13 for laminar flow
NuL = 0.037 Re~15 Pr' 13 for turbulent flow
Known: Plate temperature and dimensions. Cold air flow velocity and free stream temperature.
Find: Heat loss from plate, Q(W). Plate bottom surface temperature.
Schematic:
Air
Q
Flat plate (top surface)
T,, top= I 00°C
T==20°cu==3m/s== \
~----i>------~
A=LW= I xO.S=O.Sm2
W=O.Sm
~ >-----L~=_Im_ _ ___, 1 t=0.02 m
r,, bottom= 1
T (bottom surface)
Fig. 8.12a
Assumptions: (I) Steady-state conditions exist. (2) Air is an ideal gas. (3) Constant fluid properties.
(4) Critical Reynolds number, Rec,= 5x10 5.
Analysis: To determine whether the air flow over the plate is laminar or turbulent, let us first find the
Reynolds number. The characteristic length is the length along the flow direction, that is, L = I m. Then
u=L
ReL = - v
(3m/s)(lm)
18.97x10-6 m2 /s
= 1.581x105
Since ReL < 5 x 10 5, the fluid flow is laminar.
The appropriate correlation to be used is
NuL = 0.664 Ref 2 Pr113
= 0.664 (l.581x I0 5 )ll 2 (0.696)1i 3 = 234
8.18
,
Heat and Mass Transfer
hL
NuL=k '
With
Average convective heat-transfer coefficient,
h=!:...NuL = 0.0290W/mK x234
L
l.Om
= 6.786 W/m2 K
Rate of heat loss from the top surface of the plate is
Q = hA (I'.,top -T=)
=(6.786W/m2 K)(0.5m 2 )(100-20)°C
or
K
= 271.45 W
(Ans)
Under steady state conditions, the thermal circuit is
R
r., bottom
t
kA
-
cond -
r., top= I00°C
••--~VVv,----e•
-
.
Q= 271.45 W
Fig. 8. 12b
.
· 11T I'.,bottom -T.,top
.
With Q=- = - - - - - - , the bottom surface temperature 1s calculated to be
R,h
t I kp1a1e A
.
t
T.s, bottom =T.s, top +Q·-k
A
plate
= l00 0 C+ (271.45W)(0.02m)
(23W/mK)(0.5m2 )
= l00.4°C
Example 8.8
(Ans)
~ A kitchen in a restaurant has a large, flat burner plate for frying. Since a great deal of heat
rises from the plate, the cook decides to let a small fan blow over the burner, which is
1.2 m long and located 1.5 m down a level smooth counter from the fan. The total length is 2.7 m. The uniform
plate temperature is I 22°C and the free-stream temperature and velocity are 32°C and 2.1 m/s respectively.
What will be the heat lost per m width by the burner plate?
Properties of air at I atm and 77 °C are
k = 0.03 W/m °C
v = 20.92 x I06 m2/s Pr= 0.700
Known: Forced convection over a flat plate with unheated starting length.
Find: Heat-transfer rate per unit width.
Schematic:
Ak
~ d y a = ; , bo,ad,cy layec
T= = 32 °C u= =2.I mis==:
Thermal boundary layer
T5 = 15°C
m -f-(L-x0 )= 1.2m1
L=2.7m
X
Fig. 8. 13
Forced Convection: External Flow
r
8. I 9
Assumptions: (1) Uniform surface temperature for x > x 0 ; (2) Air is an ideal gas. (3) Recr = 5x10 5
Analysis: Reynolds number at the trailing edge of the plate is
2.lm/sx2.7m
20.92x1Q-6 m2 /s
u L
v
00
ReL = - - = - - - - - -
=2.71x10 5
~
(<5x10 5 )
Laminar flow
Local Nusselt number for laminar flow is given by
Local heat-transfer coefficient is
3/4]-1/3
[
= 0.332 ~ 1-( ~)
Reli2 Prli3
With x=L=2.7 m, x0 =unheated starting length= 1.5 m, Re=2.71x10 5 and Pr=0.700,
we have
[
O75Jli3x(2.71x105 )°" 5 x(0.7f 3
hx=L = 0.332X 0.0 3 W/moC X l-(l. 5 m).
2.7m
2.7 m
= 2.4105 W/m2 °C
Average convection coefficient,
t
_ 2[1-(x0 / L 4 ]
h = ~____,---,--~h
1-(xo/L)
x=L
75 ]
2[1-(1.5/2.7)°"
=-=------=x2.405 W/m2 °C = 3.858 W/m2 °C
1-(1.5/2.7)
Heat-transfer rate per unit width of the burner plate is
Q = 1z[(L-x )xl] (T. -T
0
00 )
= (3.858 W/m2 0 c) [1.2 mxlm] {122-32)°C
=416.7 W
(Ans)
8.20
,
Heat and Mass Transfer
Example 8.9
~ Air at 20°C and at a preasure of I bar is flowing over a flat plate at a velocity of 3 mis.
If the plate is 280 mm wide and at 56°C, calculate the following quantities at x = 280 mm.
(a) Boundary layer thickness
(b) Thickness of thermal boundary layer
(c) Local convective heat-transfer coefficient
(d) Average heat-transfer coefficient
(e) Rate of heat transfer by convection
(f) Local friction coefficient
(g) Average friction coefficient
(h) Shear stress due to friction
(i) Total drag force on the plate
(j) Local mass flow rate through the boundary.
Properties of air at the bulk mean temperature of 38 °C are
Density
= 1.1374 kg/m 3
Thermal conductivity = 0.02732 W/m K
Specific heat
= 1.005 kJ/kg K
Kinematic viscosity = 16.768 x I0-6 m2/s
Prandtl number
= 0.7
Known: Flow of air over a flat plate with a given velocity.
Find: (a) 8 (mm); (b) 8T (mm); (c) hx (W/m 2 K} ; (d) h (W/m 2 K}; (e)
(h) 'tw (N/m2 ); (i) F 0 (N); (j) m(kg/s).
Schematic:
Q(W); (f) Cr,x
(g) Cr,x;
Thermal boundary layer
Hydrodynamic (Velocity)
boundary layer
Air
==
T =20°cu:=3m/s
W=0.28m
T,= 56 °C
A5 =xW
=0.28 2 m2
x=0.28m
Fig. 8.14
Assumptions: (1) Air is an ideal gas. (2) Isothermal surface. (3) Constant air properties. (4) Steady state
conditions. (5) Recr = 5 x 105 ,
Analysis: Note: In the problem statement, bulk mean temperature should read mean film temperature,
Tr = (T;, + T )/2 . Bulk mean temperature is used in the context of internal flow (i;,m = _!_( Tbi +I;,.)).
00
Reynolds number
2
Re =u x= (3m/s)(0.28m) =S .Olxl0 4
x
v
16.768xI0-6 m2 /s
00
Since Rex < 5 x 10 5 , the flow is laminar.
(a) (Velocity) boundary layer thickness,
O= 4.91 x = 4.91x280 mm
~Rex
= 6.14 mm
v'5 .0lx10 4
(Ans) (a)
Forced Convection: External Flow
r
8.21
(b) Thermal boundary layer thickness,
BT= 8Pr 113 ""6.14mmx(0.7fl/3
=6.92 mm
(Ans) (b)
(c) Local convective heat-transfer coefficient,
= 0.332(Rex)l/ 2 (Prf ~
X
= 0.332(5.01 x 1Q4
r2
(0.7yi3 0.02732 W/m K
0.28m
(Ans) (c)
=6.44 W/m2 K
(d) Average heat-transfer coefficient,
X
h = Jhx dx = 2hx= 12.88W/m2 K
(Ans) (d)
0
(e) Rate of heat transfer,
Q = h A.('.f. -T.. )
= (12.88W/m2 K) (0.282 m 2 )(56-20) K
(Ans) (e)
=36.35 W
(f) Local friction coefficient,
Cr,x = 0.664 ( Rex )
-1/2
= 0.664 (5.01 X 10 4 f1 12
= 2.967 X 10-3
(Ans) (f)
(g) Average friction coefficient,
-
(
Cr,x = 1.328 Rex
)-1/2
= 2 Cr,x
= 5.933 X 10-3
(Ans) (g)
(h) (Average) shear stress,
---c
2
'tw - fx .!. pu..
' 2
=(5.933x10-3 )(.!.xl.1374 kg)(3m/s}21 IN I
2
m3
1kg rn/s2
= 0.0304 N/m2
(Ans) (h)
8.22
, Heat and Mass Transfer
(i) Total drag force on the plate,
N
A 8 = 0.0304- X 0.282 m 2
m2
= 2.38 X 10-3 N
(Ans) (i)
(We must use ~w and not local 'tw)
G) Local mass flow rate through the boundary,
m= ~pu= (<>(x)-0)
8
= ~xl.1374 kg x3 m x6.14x10-3 m
8
m3
s
= 1.31x 10-3 kg/ s
(Ans) (j)
Example 8.1 0 ~ Air at atmospheric pressure and temperature of 20°C flows over a plate (80 cm x 40 cm)
at a velocity of 2 m/s. The plate is maintained at I00°C. If the flow is along the 80 cm side,
calculate the heat transfer rate from (a) the first half of the plate, (b) the full plate, and (c) the next half of the plate.
The properties of air at the mean film temperature of 60°C are
k=28.74x 10--3 W/m °C
V= 19.21 x I~ W/m 2
Pr=0.7024.
Known: Atmospheric air flows past a flat plate with a prescribed velocity.
Find: (a) Qfirsthalf• (b) Qtotal (c) Qnexthalf•
Schematic:
Air
u==2m/s T =2o•c -
=
r- First half-I- Next half
-
Flat plate (W=40cm
r.= 100°q
L=80cm
Fig. 8.15
Assumptions: (1) Air is an ideal gas. (2) Constant properties. (3) Radiation effects are not considered.
(4) Uniform plate surface temperature. (5) Critical Reynolds number is 5 x 10 5•
Analysis: (a) Considering the first half of the plate, i.e., L = 0.40 m, the Reynolds number is
R
_ u=L _ 2m/sx0.40m
v - 19.21x lQ-<i m 2 /s
E?r, -
= 41645
~
Laminar flow
(< 5 x 105 )
Average Nusselt number,
-NuL = -hL = 0.664(41645)112 (0.7024)113
k
= 120.45
Forced Convection: External Flow
r
8.23
Average heat-transfer coefficient is
h = NuL !5._ = 120.45x28.74x10-3 W/m°C
0.40m
L
= 8.655 W/m2 °C
Heat-transfer rate from the first half of the plate,
QI= hA ('I'. -T=)
= (8.655W/m2 °C) (0.4mx0.4m) {100-20)°C
= 110.8 W
(Ans) (a)
(b) Now, considering the entire plate, i.e., L = 0.80 m:
Reynolds number,
u=L
v
2m/sx0.80m
19.21xlQ-6 m2 /s
ReL = - = - - - - - -
= 83 290
As this is less than 5 x 10 5, the flow is laminar.
= 0.664(83290) 0 ·5 (0.7024)113 = 170.34
or
-
-
k
h=NULL
=
(170.34)(28.74 X 10-3 W/m 0 C)
0.80m
= 6.12 W/m2 °C
:. Heat lost from the whole plate,
Q = (6.12 W/m2 °C)(0.80 mx0.40m){100-20)°C
= 156.7 W
(Ans) (b)
(c) Hence, the heat-transfer rate from the next half of the plate,
QII = Q-QI
= (156.7 -110.8) W
=45.9 W
(Ans) (c)
8.24
, Heat and Mass Transfer
Example 8.11
~ A residential building has its outside wall insulated with 15-cm thick insulation (k = 0.045 W/m K).
The air inside the building is at 25°C and the free convection heat transfer coefficient between
the inside surface of the wall and air is estimated to be 6 W/m 2 K. The outside surface of the wall is exposed
to wind blowing at 6 mis parallel to the wall 15 m wide and 8 m high. Estimate the rate of heat loss from the
building. Properties of air at O °C are
K = 0.02364 W/m K v = 13.38 x I0-6 m2/s
Pr= 0.7362
Known: Heat is lost through an insulated wall of a building exposed to air by natural convection inside
and forced convection outside.
Find: Heat loss, Q(W).
Schematic:
Outside surface
Inside surface
Still air
r~, =25°c
I
Air,
h 1=6W/m2K
Wall
(k=0.045W/m K)
IL=
0.15m
Fig. 8.16
Assumptions: ( 1) Steady-state conditions. (2) The side of the building is approximated as a large flat
plate. (3) One-dimensional conduction. (4) Critical Re= 5 x 10 5.
Analysis: The thermal network for the given configuration is shown below:
-
. r~,
r.
r.
1
2
r~
Q --~\/\1\~----~,,l\n,~----~,""~--e
R1 =(l/h1A)
R2= (UkA)
Fig. 8.17
Area,A = WH= 15 m x 8 m= 120 m2
The heat loss rate is given by
• Convective resistance (inside)
l?i = -
1
h 1A
1
= - - - - - - = 1.389 x 10-3 K/W
6W / m2 K x 120 m2
• Conduction resistance through the insulation
R2 =~=
kA
O.lSm
=27.78xl0-3 K/W
0.045W / mK x 120 m 2
2
Forced Convection: External Flow
r
8.25
Let us now calculate h2 in order to find R 3 • For flow parallel to the wall surface, L = W = 15 m.
The wind velocity, V = 6 mis. For forced convection heat transfer, the properties of air at the film
temperature, Tr =
½( r., + T-
2 )
must be evaluated. However, since T,,2 is unknown, the air properties
at T-2 = 0°C can be used to estimate h2 •
Reynolds number,
VL
ReL=-
v
=
(6 m/s)(l 5 m) = 6.726 x lQ-6
13.38x10-6 m2 /s
(5x 105 <Re< 108 )
The appropriate correlation is
NuL = L0.037(Red0 -8 -87Ij(Pr)113
= h2L
k
where k is the thermal conductivity of air.
Therefore,
k
h 2 = -x[0.037(Red0·8 -871](Pr) 113
L
= 0.0 2364 W/m K [o.037(6.726x106 }°" 8 -871](0.7362)'13
15 m
=14.0 W/m2 K
+ Convective resistance (outside)
R __I__
1
3 - h A - (14.0W/m2 K}{I20m2 )
2
= 0.594x 10-3 K/W
Total thermal resistance L Rth = Ri + R2 + R3
= {1.389+27.78+0.594){10-3 )K/W = 0.02976 K/W
Heat-loss rate,
. T"°I -T-,
(25-0)°C = 840 W
Q = LRth = 0.02976K/W
(Ans)
Example 8.12 ~ An air cooled motorcycle engine has fins which may be approximated as individual flat
plates of length L = 0.2 m. Disturbances in the free stream cause the transition to occur at
Rex.trans= 2 x IOS. Determine, for a speed of 140 km/h, the average heat transfer coefficient from the fin surface
permitting separate laminar and turbulent regions. Compare this with the result obtained by considering turbulent
flow right from the leading edge. Use the following properties of air:
k = 0.0263 W/m °C v = 15.89 mm 2 /s
Pr= 0.707
8.26
, Heat and Mass Transfer
Known: Air flow over fins idealised as flat plates under specified conditions.
Find: Average heat-transfer coefficient considering mixed flow conditions and considering turbulent flow
over the entire plate.
Schematic:
··· ...h,;,;.;;.,ii.;
u== 140km/h -
X
c_r L=0.2m - - - - - - - i
i-.1
, _ _ _ _ _ _
Fig. 8.18
Assumptions: ( 1) Isothermal flat plate. (2) Constant properties.
Analysis: Mixed flow conditions:
f
l { xc,
haverage
=L
hx,/am dx
0
112
=!:_Nu
=!:_·O 332(u=x)
h
k
=-Nu
k 0296 ( u=x )
=-·O
x.turb
\verage =
x,lam
X
X
x.turb
X
X
V
•
V
•
f
L
hx.turb dx
Pr113 = 0 332 k Pr113
.
415
fzZ:,xv--;-
Pr113 = 0 0296 k Pr113 ( u=
·
V
112
)0.8 -x-
I
115
L1 { 0.332 k Pr113 ( u; )0.5 xc,[ x- 112 dx + 0.0296 k Pr113 ( u; )0.8 L x- 115 dx )
= kP:"
r
{om("; 2x/?
+00296( ";
r¾[,Lxi))
140
3.6
At u= =140km/h-m/s,
Re =u=L = 140 mx
L
)
Xcr
h
x,lam
+
v
3.6 s
0.2m
= 4 _9 xl0 5
15.89x10-6m2 /s
Average heat transfer coefficient is then
Forced Convection: External Flow
= 0.0 263 WI m°C x(0.707)'1 3 {o.664~2x10 5 +0.037[( 4.9x10 5
0.2m
r
8.27
t -(2 x10 }4 J}
5
5
15
=(0.117) [(297 + 674)]
= 114 Wlm 2 °C
(Ans)
Pure turbulent flow
k- k
haverage =-Nu=-0.037
ReL415 Pr113
L
L
= 0.0 263 W lm°C x0.037x(4.9xI0 5
0.2m
t x0.707
5
113
= 155 WI m2 °C
(Ans)
Rex trans XV
Comment: Critical length, xcr =--'---
u(2 x10 5 )(15.89 xlO-<i) m 2 Is
(140 I 3.6) mis
=---------=0.082m
The fin (flat plate) length is L = 0.2 m. Since xcr = O.OS 2 m = 0.41, transition occurs at 41% of the
L
0.2m
fin length. Hence, both laminar and turbulent sections must be taken into account for calculating the
average heat transfer coefficient.
Example 8.13 ~ Carbon dioxide gas at 1.2 bar, free-stream velocity of 4 m/s and temperature of 130 °C flows over
a 2-m-long flat plate maintained at a uniform temperature of 84°C by a coolant flow on the side
opposite to the gas flow. Transition to turbulence occurs at 0.88 m from the leading edge. Determine (a) the average
heat transfer coefficient over the entire plate length, and (b) the rate of heat transfer to the plate per unit width.
Properties of CO 2 gas at I atm and the film temperature of I07 °C are
k = 0.02275 W/m °C
Pr = 0.737
µ = 181 x I0--7 N s/m 2.
Known: CO 2 gas flow over a relatively cold plate.
Find: (a) h and (b) Q.
Schematic:
P= 1.2 bar
r-= 130°c
u-=4m/s
- - - - - - L = 2 m _ _ _ _ __,
Fig. 8.19
Assumptions: (1) CO 2 is an ideal gas. (2) Uniform surface temperature. (3) Constant properties.
8.28
, Heat and Mass Transfer
. of CO 2 at 4 = 1( T,, + T ) = (130+84) = 107 C
Analysis: Density
2
2
0
00
or
380.15 K is
PM
1.2 x 102 kPa x 44 kg/kmol
p =-=-- = - - - - - - - - - " - - - - = 1.671 kg/m3
RT 8.3143 kJ/kmol Kx380.15 K
Transition Reynolds number,
puoo xcrit
1.671kg/m3 x4m/sx0.88m
Re . = - - ~ = - - - - - - - - - cnt
µ
181xl0-7 Ns/m2
= 3.25x 105
Then
ReL = Recrit ( ~ )
xcrtt
= (3.25 X 10 5 )(2 / 0.88) = 7.386 X 10 5
Average heat-transfer coefficient is
J
h- = Nuk
- = -k[ 0.037 Re2° 8 - A Pr113
L
where
L
A= 0.037 Re~~~ - 0.664 Re~~~
= 0.037(3.25x 105 ) 0·8 -0.664(3.25 x 105 )°' 5
= 571.4
i:Jm
h = 0.0 221
J
°C [ 0.037(7.386 X 105 }°' 8 -571.4 (0.737}'1 3
= 12.95 W/m2 °C
(Ans) (a)
Heat-transfer rate per metre width from the gas
Q = h(LW) (T -1',,)
00
= (12.95 W/m2 °C)(2mx lm)(130-84)°C
= 1190 W
or
1.19 kW
(Ans) (b)
Example 8.14 ~ The crankcase of an automobile is 80-cm long, 30-cm wide and I0-cm deep. Find the rate of
heat flow from the crankcase to the atmosphere when the automobile is moving at 90 km/h.
Assume that the vibration of the engine and the chassis induce the transition from the laminar to turbulent
boundary layer over the entire surface. Also, find the drag force on the total surface of the crankcase.
Forced Convection: External Flow
r
8.29
Take the same average convective heat transfer coefficient and average shear stress for front and rear surfaces
as for the bottom and the sides. Crankcase surface temperature is 75°C and atmospheric air temperature is
S0 C. At 40°C for properties of air, take
p = 1.128 kg/m 3
CP = I 005 J/kg K
µ= 19.1 x 10-6 kg/ms
k = 0.0276 W/m K
Pr=0.699
Known: The crank case of a speeding automobile with prescribed dimensions and temperature loses heat
to atmosphere at a specified temperature.
Find: Heat-transfer rate, Q(W) and total drag force, F0 (N ).
Schematic:
Air
u= =90km/h
T= =S°C
--
L=0.8m
------..,
Fig. 8.20
Assumptions: ( l) Steady operating conditions exist. (2) Constant properties. (3) Turbulent boundary layer
over the entire surface.
l 03m ) ( -l h-) =25 mis
. velocity,
.
km ( Analysis: Atr
u= =90h 1km
3600s
Heat-flow rate from the crankcase to atmosphere is
where
As = total surface area
=2 [LW+LH+HW]
= 2 [(0.8 X 0.3) + (0.8 X 0.1) + (0 .1 X 0.3)] = 0.7 m 2
Ts - T= = 75 - 5 = 70°C or
K
Properties at film temperature, Tr = ( T. + T=) I 2
= {75 + 5) 12 = 40°c
(given)
Reynolds number,
Reo =pu=L = 1.1 28 kg/m 3 x (25)m/sx0.8m
µ
19. l X ]0-6 kg/m S
= 1.18] X 106
(turbulent)
8.30
, Heat and Mass Transfer
Assuming turbulent flow right from the leading edge,
hL
NuL = - = 0.037 Ref,15 Pr113
k
= 0.037 (1.181x106 )°" 8 (0.699)1" 3 = 2367
Hence, the average convection coefficient is
h=!!....NuL = 0.0276W/mK x2367
L
0.8m
= 81.65 W/m2 K
The heat-transfer rate,
Q = (81.65 W/m2 K) (0.7 m2 )(70K)
= 4000 W
(Ans)
or 4 kW
Drag force,
-1
F0 =CfL -p A 8 u~
' 2
where
Cn =
'
O.o:: =0.074(1.181x10 r
6
0·2
ReL·
= 0.004516
1
kg
m2
F0 =0.004516x-xl.128-x0.7m2 x(252 ) 3
2
m
s2
= 1.114 kg m/s2
Example 8.1 S
(Ans)
or N
~ A refrigerated truck is travelling on the highway at the speed of 80 km/h in a desert area
where the temperature is 70°C. The body of the truck is considered to be a rectangular box
3 m wide, 2 m high and 4.5 m long. The heat transfer from the front and back end of the box is neglected.
Assume no separation of the air stream from the surface and consider the boundary layer as turbulent over
the entire surface. The heat transfer coefficient for the sides may be assumed to be same as top and bottom.
The temperature at the surface is uniformly at 10°C. For every heat loss, one ton capacity of the refrigerating
unit is needed. Calculate: (a) heat loss from the four surfaces, (b) the tonnage of the refrigerating unit and
(c) the power required to overcome resistance acting on the four sides.
Properties of air at 40°C:
k=0.0276 W/m °C
V= 16.96x lo--6 m2/s
Pr=0.699
Known: A refrigerated truck travelling in a desert is considered to be rectangular box with negligible
heat transfer from the front and rear ends.
Find: (a) Heat loss from four surfaces, Q(W); (b) Tonnage, TR; (c) Power to overcome resistance, W(W).
Forced Convection: External Flow
r
8.31
Schematic:
2m
Fig. 8.21
Assumptions: (1) Steady-state conditions. (2) Heat transfer from front and back ends is neglected.
(3) Turbulent boundary layer over the entire surface. (6) Uniform surface temperature. (5) Air is an ideal gas.
Analysis: Free-stream velocity of air,
u = (80)(1000) = ~m is
=
3600
3.6
Reynolds number,
Re = u=L (80/3.6) m/s(4.5 m) = 58 _962 x 105
L
V
16.96 X I0-6 m 2 /s
Average Nusselt number,
-
hL
NuL = -
k
= 0.037(Red 415 (Pr)
113
(for turbulent flow over the entire surface)
= {0.037)(58.962 X 105}°" 8 {0.699{ 3 = 8567
Average convection heat-transfer coefficient,
h = NuL ! = 8567 x 0 ·0276 W/moc = 52.54 W/m2 °C
4.5m
L
Total surface area of four sides,
A 8 =2(3 X 4.5)+ 2(2 X 4.5) = 2(3+ 2)4.5 = 45 m2
heat-loss rate,
(2 = hAS (T. -T=) = (52 .54 W/m 2 °C)(45 m 2 ){70-10)°C
= 142x 103 W
or 142 kW
We note that I ton of refrigeration= 211 kJ/min.
Tonnage of the refrigerating unit
=
142kW
= 40.4TR
(211 / 60) kW/TR
(Ans) (a)
(Ans) (b)
8.32
, Heat and Mass Transfer
Average friction coefficient over the entire surface,
Cr= 0.074(ReL )-0· 2
= 0.074(58.962 X 105
t.2
= 0.0519
Density of air,
p =PI RT= 101.3 kPa/(0.28 kJ/kg K) (40+273.15) K = 1.127 kg/m3
Drag force,
Power required to overcome frictional resistance,
.
1
W = F0 V = Cr-PA. V 3
2
1
3 )x(45m2 ) (80m)
= (0.0519)x-(1.127kg/m
3.6s
2
= 14.44 x 103 W
Example 8.16
3
or
14.44 kW
(Ans) (c)
~ As a means of supplying freshwater to arid regions of the world, it has been advocated
that icebergs be towed from polar regions. Icebergs that are considered to be best
suited for towing are those that are relatively broad and flat. Consider an iceberg that is I km long by 0.5
km wide and of depth D = 0.25 km. It is proposed that this iceberg be towed at I kmph in the direction
of its length for 6000 km through water whose average temperature (over the trip) is I0°C. As a first
approximation, the interaction of the iceberg with its surroundings may be assumed to be dominated by
conditions at the bottom (I km x 0.5 km) surface. The latent heat of fusion of ice is 334 kJ/kg and density
of ice is 920 kg/ml.
(a) What is the average recession (melting) rate dD at the bottom surfacel
dt
(b) What is the power required to move the iceberg at the designated speedl
(c) If towing costs amount to Rs. 250/kWh of power requirement, what is the minimum cost of freshwater at
the destinationl
The following properties of water at 5°C may be used.
µ = 1.519 x IO-l kg/ms
p = 999.9 kg/ml
Pr= 11.2
k = 0.571 W/m K
Known: Icebergs towed from polar regions to supply fresh water to arid regions.
Find: (a) Average melting rate, dD!dt, (b) Power required to move the iceberg, p, (c) Minimum cost of
supplying water.
Forced Convection: External Flow
r
8.33
Schematic:
Water
T5 =0°C
T== I0°C
Ice
V= I km/h
T
1
~---~
1----
D=250m
L= 1000m
Fig. 8.22
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Forced convection heat transfer
is dominated by conditions at the bottom surface with As = WL.
Analysis: Free-stream velocity= Towing speed of iceberg,
V = l km/h = lOOO m
3600 s
Kinematic viscosity,
µ
V = -
p
1.519 x 10-3 kg/ ms
999.9kg/m3
= -------"-'---
= 1.5192 x 10-6 m2 /s
Reynolds number,
Re = V L = 1000 X 1000 X 106 = 1. 8285 X l0 8
L
V
3600
1.5192
:. boundary layer is likely to be entirely turbulent.
-
08
N~ =0.037 Prll 3 Re2 8 =0.037xll.2ll 3 x(l.8285xl0 8 ) · =336991
Average convection coefficient
h = NuL ! = 336991 x 0. 57 l W/ m K = 192.42 W /m2 K
1000m
L
Energy balance: Energy transferred to ice for melting = Heat transfer by convection from water to ice.
£melting
= !2conv
dD
-xp
xAsxhsr=h- As ( T=-T. ) =192.42xAsx(l0-0)=1924.2 As
dt
,ce
Average recession rate,
1924 ·2
dD =
= 6.262 x 10-6 mis
dt
920 X 3.34 X 10 5
= 0.0225 m/h
(Ans) (a)
8.34
, Heat and Mass Transfer
Average friction coefficient,
-
Cr,L =0.074 Re'r!1· 2 =0.074x(l.8285x10 8 )
-02
·
= 1.6475 X 10-3
Drag force,
1
F0 = Cr x-xpAs V 2
2
2
1 999.9 X 1000 X 500 X ( 1000)
= 31776 N
= 1.6475 X 10-3 X-X
2
3600
Power required,
_
_
1000 m _
_
p - FD V - 31776N x - - - - 8.83x 103 W - 8.83 kW
3600 s
(Ans)(b)
. .
p x Towing time x Towing rate
Mm1mum cost='---,---------.,....-( D- d D x Towing time) As
dt
8.83 kw( 6000 kmJ ( R s ~ J
1 km / h
1 kWh
= ~---~---~~--~-6
( 250-0 .0225 m x 000kmJ(5xl0 5 m2)
h
lkm / h
= Rs 0.23 per m3 of water.
(Ans) (c)
Example 8.17 ~ Oil at 25°C is forced over a 400 cm 2 square plate at a velocity of 1.18 mis. The plate is
heated to a uniform temperature of 55°C. Calculate the heat loss by the plate. Given:
v = 0.00024 m2/s
p =876 kg/m 3
Cp = 1.97 kJ/kg 0 C.
k=0.144 W/m °C
Known: Oil flows over a heated square plate with a prescribed velocity.
Find: Heat loss by the plate.
Schematic:
Square plate
T,= SS°C
V= l.18m/s
Oil
r~=2s c
0
A,= bL= L2=400cm2
L=20cm
f--- L = 20 cm -----j
Fig. 8.23
Forced Convection: External Flow
r
8.35
Assumptions: (1) Steady operating conditions. (2) Constant oil properties. (3) Critical Reynolds number
is 5x10 5 •
Analysis: Reynolds number,
R
_ V L _ 1.18m/sx0.20 m
v - 0.000 24 m2/s
E?r, -
= 983 .3 3 (< 5 x 105 )
~
Laminar flow
Prandtl number,
CPµ cP vp
Pr=--=--
k
k
(1.97x103 J /kg 0 c) (0.00024 m2/s) (86 kg/m3)
= ~ - - - - ~ ~ - - - ~ ~ - ~ = 2876.2
0.144 W/m°C
For high Prandtl number fluids, and laminar flow on an isothermal flat plate, the Churchill---Ozoe
correlation is applicable.
H0~68rr
0.3387 Re~2 Pr113
Nu,=
=
0.3387(983.33}'1 2 {2876.2)'1 3
1/4
213
[ 1+ (0.0468) ]
2876.2
\L
=151=k
Average heat-transfer coefficient,
-
k
h = 2 hx=L = 2 Nu.,_=L L
= 2 x151x0.144 W/m°C = 217.44 W/m2 oc
0.2m
Rate of heat loss from the plate is
Q=hA,(T.-T-)
= (217.44 W /m2
00
)
(400 x10--4m2 )(55-25)°C = 261 W
(Ans)
Example 8.18 ~ Air at 30°C and I atm pressure flows over a flat plate at a velocity of 2 m/s. Calculate the
boundary layer thickness at distances of 20 cm and 40 cm from the leading edge of the
plate. (a) Compute also the mass flow which enters the boundary layer between x = 20 cm and x = 40 cm per
metre depth of the plate.
The plate is heated and maintained at a temperature of 90°C over its entire length. (b) Compute the heat
transferred in the first 40-cm of the plate per metre depth.
(c) Determine the drag force exerted on the first 40-cm of the plate using the analogy between fluid friction
and heat transfer.
8.36
, Heat and Mass Transfer
Properties of air at 60°C are
CP = I .007 kJ/kg K
p = 1.059 kg/ml
k = 0.02808 W /m K
v = 18.96xlo-6 m2/s
Pr= 0.7207
Properties of air at 30 °C are:
p= 1.164 kg/ml
V
= 16.08 X I0--6 m2/s
Known: Air flows over a heated flat plate with a specified velocity.
Find: (a) 8 (mm) atx= 0.2 m and atx= 0.4 m; A rh(kg/s) betweenx= 0.2 m andx= 0.4 m, (b) QIL (W/m).
Schematic:
Air
r~=30°c
u~=2m/s
-- -------------,,
~"::,
...
~
_
L__ (T5 =90°C)
Fx=0.2m-i
l----x=0.4m----
Fig. 8.24
Assumptions: ( 1) Steady state conditions. (2) Constant properties. (3) Air is an ideal gas. (4) Recr = 5 x 10 5•
Analysis: (a) Reynolds number:
At x=0.20 m
Re = _u~_x = (2m/s)(0.20m) = 24875.6
v
x
16.08x10-6 m2 /s
At x=0.40 m
Re = (2m/s)(0.40m) = 49751.2
x
16.08x10-6 m2 /s
:. the flow on the flat plate is therefore laminar since Rex < 5 x 105 •
Boundary layer thickness
<> = 4.64x
~
and
<>(x=o. 2 m)= 4.64x0.2m/J24875.6 = 5.88x10-3 m
or 5.88 mm
(Ans)
<> (x=0.4 m)= 4.64 X 0.4 m/ .J49751.2 = 8.32 X 10-3 m
or 8.32 mm
(Ans)
Mass flow per m depth of the plate between x = 0.2 m and x = 0.4 m in the boundary layer is given by
Ii
i(r)-.!.(X.)
=
m= Jpudy where ..!:!:._
u~
2 8
0
2 8
3
Forced Convection: External Flow
r
8.37
Thus
5
=-pu
<>
8
00
:. mass flow between x = 0.2 m and x = 0.4 m is
Am-.
- .
- ~x=0.4m) ~x=0.2m)
= ~( 1.164 kg/m3 )( 2 mis) [ (8.32- 5.88)(10-3 ) m J
8
= 3.55 X 10-3 kg/s
(Ans) (a)
(b) Reynolds number,
ReL = _u.._L = (2m/s)(0.40m) = 42194
v
18.96 x 10-6 m2 /s
[The flow is laminar]
Average Nusselt number,
r,;::1/3
NuL = 0.664-vReL (Pr)
= 0.664 (42194f 2 (0.7202f 3 = 122.26
:. average heat transfer coefficient,
-
122.26 (0.02808W/mK)
0.40 m
/
h = - - - - ' - - - - - - - ' - - = 8.583 W m2 K
Heat transferred per metre depth is
Q=hA.(J'.-T.. )
= ( 8.583W/m2 K) ( 0.40mx lm) (90-30)K
(Ans) (b)
=206 W
(c) The analogy between fluid friction and heat transfer for laminar flow on a flat plate is expressed as
-St Pr2t3 = _f_
c
2
where Cr is the average sin friction coefficient and St is the average Stanton number given by
Nu/Re Pr.
8.38
, Heat and Mass Transfer
Nu
hL
v
k
Re Pr
k
u= L
CP v p
-
St=--=-X--X--
h
8.583 W /m2 K
= 4.024 X IQ-3
1.059 kg/m3 x2m/sx1007J/kg K
cf2 = (4.024x10-3 ) (0.7202}213
= 3.233x10-3
The average drag force, F 0 = Average shear stress at the wall, 't x Surface area, A.
1F0 = 2Cf p
u;, As
= (3.233x 10-3 ) (1.059 kg/m3 )(2m/s)2 (0.40 m x lm)I
= 5.48 x 10-3 N
lN / I
1kg m s2
or 5.48 mN
(Ans)(c)
Example 8.19 ~ Atmospheric air at 25°C and 40 m/s blows over an array of IO silicon chips, each of I0-mm
length on a side and insulated on one surface. The same power is dissipated in each chip,
maintaining a uniform surface heat flux over the whole cooled surface. Calculate
(a) the maximum allowable power per chip if the chip temperature should not exceed 85°C, and
(b) the maximum power dissipation if turbulence promoter is used at the leading edge.
(c) If the chip array is placed normal to air flow, would it be preferablel
Properties of air at the film temperature of 55°C are
k=0.02772 W/m° C
V= 18.47x I0-6 m2/s
Pr=0.7215
Known: Air flow conditions. Dimensions of array of 10 silicon chips. Maximum permissible chip temperature.
Find: Maximum power dissipated (a) without and (b) with turbulence promoter in place. (c) Desirability
to orient the chip array normal to air flow.
Schematic:
Air
r =2s c 0
~ Ist chip
=
u==40mls_
···C
I
1--f- IOmm
I I I I I I
I
f: IOth chip
1w.sss·q
L·x ·-----------------------------[ Insulated ·----------------------r---l L = O. I m
surface
x=0.095m
Fig. 8.25
Assumptions: (1) Steady operating conditions. (2) Recr = 5x10 5 • (3) Uniform heat flux. (4) Negligible
radiation.
Analysis: (a) Reynolds number,
R
~
= u=L =
v
40m/sx0.lm = 2 _166 xios
18.47xI0-6 m2/s
Forced Convection: External Flow
r
8.39
Therefore, the flow is laminar over all the ten chips without the turbulence promoter.
For laminar flow, ~ ax-0· 5 • Minimum hx is thus associated with the last chip. Local convection
coefficient at x = 0.095 m may be approximated as the average heat transfer coefficient for the last chip.
-
k
h10= h,;=95mm
= 0.453- Re}/.2 Pr 113
X
Re = u..,L = 40 m/s x 0.095 m = 2 _057 x 105
x
v
18.47x1Q-6 m2 /s
hio = 0.453x 0·02772 W /m °C x (2.057 x 105}'1 2 x (0.7215)'1 3
0.095 m
= 53.774 W /m2 °C
Heat dissipation rate,
Q=hA.(J'.-T..,)
= (53.774 W/m2 c)(O.Olm}2 (85-25)°C
0
(Ans) (a)
=0.323 W
(b) For turbulent flow:
hio= 0.0308~ Re415 Pr 113
X
= 0.0308 x 0 ·02772 W/m oc x (2.057 x 105}41 5 x (0.7215}'13
0.095m
= 143.53W/m2 °C
Heat rate,
Q = h 10 A (J'. -T..,) = (143.53W/m2 °C)(0.01)2 m2 (85-25)°C
8
(Ans)(b)
=0.861 W
(c) As h1 > h10 , more heat will be dissipated per chip if the array were placed normal to air flow. The
same heat would be dissipated from each chip.
(Ans)(c)
Example 8.20 ~ In a wind tunnel, a thin, smooth model aerofoil is to undergo a test for lift and drag. The
model is heated to 45°C uniform surface temperature when the free-stream air temperature
is 25°C. The model chord length is 1.5 m, and the undisturbed air velocity is 20 mis. Determine (a) the local
convection coefficient 0.9 m from the leading edge, (b) the average convection coefficient for the entire chord
length assuming Recr = 5 x I05 • (c) the rate of heat transfer to the air per unit aerofoil width, (d) the average
convection coefficient and the heat transfer rate per unit width if due to surface roughening the transition from
laminar to turbulent flow is assumed to occur at Recr = 2.5 x IOS, and (e) the average convection coefficient and
the heat-transfer rate per unit width if a 'tripping wire' triggers transition to turbulence almost immediately
from the leading edge.
Assume that the flat-plate approximation of the model aerofoil is valid and account for both sides of the
model for evaluating the heat-transfer rate.
The following thermophysical properties of atmospheric air at 35 °C may be used:
k = 0.0269 W/m °C
V
= 1.67 X I0--5 m2/s
Pr= 0.706
8.40
, Heat and Mass Transfer
Known: A model aerofoil idealised as a flat plate of specified length and temperature. Air temperature
and velocity.
Find: (a) hx=0. 9 m; (b) ~ and (c) Q for Recr = 5x105 ;
(d) I;., and Q for Recr = 2.5 x 10 5 ; (e) ~ and Q for turbulent flow over the entire surface.
Schematic:
Air
C
L Model aerofoil as flat plate (both sides)
Leading edge
P=latm
-
r..,=2s c 0
?__________.,' A5 =2LW=2x 1.Smx Im
~
~ m -----+j
u..,=20m/s -
7
= 3.0 m2
1-L=I.Sm----i•I
Fig. 8.26
Assumptions: (1) Air is an ideal gas. (2) Uniform surface temperature. (3) The model aerofoil is modelled
as a flat plate. (4) Heat is lost from both sides of the plate. (5) Radiation effects are negligible.
Properties: Air properties need to be evaluated at the film temperature,
Tr = (112) (T. + T..,) = (45 + 25) / 2 = 35° C (Given in the problem statement)
(a) Reynolds number at the end of the plate is
u.., L (20 m/s) (1.5 m)
ReL = - - = - - - - - v
1.67 x 10-5 m2 /s
= 1.796x106
At a distance x = 0.9 m from the leading edge,
Re = u.., x = 20 x0.9 = 1.07 Sxl06
x
V
1.67 X 10-5
Local heat-transfer coefficient is determined from
N~ = 0.0296 Re:15 Pr113 = hx_ x
k
Hence,
hx_ = Nux
! = 0.0296 !(Rex)°"8 (Pr)'13
X
X
= 0.0296x 0 ·0269 W/m °C (1.078x106 }°" 8 (0.706}'1 3
0.9m
= 52.8 W/m 2 °C
(b) Average heat-transfer coefficient (with Recr =5x10 5 ) is calculated from
i. L
NuL = _''L_ = ( 0.037 Ref 15 - A) Pr113
-
k
(Ans) (a)
Forced Convection: External Flow
r
8.41
A= [ 0.037 Regis - 0.664 Reg/ ]
where
= {0.037 (5x10 5 )°" 8 -0.664(5x10s)°" 5 }
= 871.3
Hence,
~ = NuL
z[
0.037 Re2° 8 -871.3] Pr113
= 0.0269 W/m °C [ 0.037 (1.796x106 }°" 8 -871.3J(0.706}'13
1.5m
= 45.65 W /m2 °C
(c) Heat-transfer rate per metre width (considering both sides) is
Q=hLAs(T.-Too)
= (45.65 W/m2 0 c) (3.0m2 ) {45-25)°C = 2740W
(Ans) (c)
=2.74 kW
(d) With Recr = 2.5 X 105,
A= 0.037 (2.5 x 105 )°" 8 - 0.664 (2.5 x 105 )°" 5
= 438.1
It follows that
h = ![ 0.037 (Red 0-8 - 438.1] Pr113
L
= 0 ·0269 [ 0.037 (1.796x106 )°" 8 -438.1] (0.706}'13
1.5
= 52.56 W /m2 °C
Q = h As (T. - T
00 )
= (52.56) (3){45-25)
= 52.56 x 60 W = 3154 W
or
3.15 kW
(e) For turbulent flow right from the leading edge, A= 0 and
h = ![0.037 Re2" 8 Pr113 J
L
= 0·0269 x[ 0.037 (1.796 X 106 )°" 8 (0.706f3 J
1.5
= 59.56W /m2 °C
(Ans) (d)
8.42
,
Heat and Mass Transfer
Q = h A. (T. -T=)= 60 h (W)
and
= 60x 59.56
=3574 W
=3.57 kW
Example 8.21
(Ans) (e)
~ The top surface of a heated compartment consists of very smooth (A) and highly roughened
(B) portions, and the surface is placed in an atmospheric air stream. In the interest of
minimizing total convection heat transfer from the surface, which orientation, (a) or (b) is preferred? If
Tw = I00°C, T- = 20°C, and u- = 20 mis, what is the convection heat transfer from the entire surface for this
orientation?
Im
j- O.Sm -1
T
O.Sm
J_
A
or
B
B
or
A
----------"- T.
uoo, Too
-<
-1
-I
B
A
(a)
B
A
Fig. 8.27
The following properties of air at I atm, 60°C may be used:
k=0.02875 W/m K
v= 19.0Sx 10-6 m2/s
Pr=0.701
Known: Surface characteristics of a flat plate exposed to an air stream.
Find: Orientation that minimizes convective heat transfer from the surface.
Assumptions: ( 1) Steady state conditions exist. (2) Constant properties. (3) Surface B is rough enough
to trip the boundary layer when in the upstream position in configuration (b ).
Schematic:
Air
T= =20°C
u==20m/s
Smooth
Configuration (a)
Fig. 8.28a
Roughened
Forced Convection: External Flow
Air
--
T==20°C
u== 20m/s
Smooth
r
8.43
Roughened
.r. . . . . rT
:©:::::::: 0.5m
0
··.~.·:·:···~
Tw= I00 °C
/ - 0.5 m - - - - . , - - 0.5 m
---i
Configuration (b)
Fig. 8.28b
Analysis: Configuration (b) results in a turbulent boundary layer over the whole surface. Hence, for
minimising convection heat transfer, it is better to have A first, i.e., configuration (a).
(Ans)
Configuration (a)
Reynolds number,
ReL =-u=_L_=
v
20m/sxlm
= l.05xl06
19.05xI0-6 m 2 /s
Critical Reynolds number, Recr = 5 x 10 5 = u= xcr
V
Position of point of transition to turbulent flow is
Recrv
xcr = - - =
(5x10 5 )(19.05xI0-6 m 2 /s)
20m/s
=0.48m
transition to a turbulent boundary layer occurs on block A, just before the rough surface B.
NuL = h L = 0.037x0.701[ ( 1.05 x 106 )°" 8 -871] = 1383
k
h =1383x 0.0 2875 W/mK = 39.76 W/m 2 K
I.Om
Rate of heat transfer,
Q=39.76W/m 2 Kx(100-20)Kx lmx0.5m = 1590 W
Configuration (b)
If B, the highly roughened surface, is first, the boundary layer will be turbulent everywhere.
Then
NuL =0.037X0.701 [ ( 1.05 X106}° 8 ] = 2156
li =!!_Nu= 0.02875W/mK x 2156
L
=62.0 W/m 2 K
Im
(Ans)
8.44
, Heat and Mass Transfer
Rate of heat transfer,
Q=h(WL)(Tw -T=)
= (62 W/m2 K) (0.5mxl m) (100-20) K
(Ans)
=2480 W
Example 8.22 ~ Atmospheric air at a free-stream velocity of 12 m/s and 30°C flows parallel to a 3-m long
flat plate maintained at a temperature of I 50°C. Calculate the heat-transfer coefficient at
the trailing edge of the plate based on the following analogies: (a) Colburn analogy, (b) Prandtl analogy, and
(c) Von K a'rm a' n analogy. The properties of air at the film temperature of 90°C are
k = 30.24 x 10-3 W/m °C
Pr= 0.7132
v = 22.01 x Io---6 m2/s
Known: Forced convection heat transfer for flow of air over a flat plate.
Find: Heat-transfer coefficient, h [W/m 2 0 C] based on (a) Colburn analogy, (b) Prandtl analogy, and
(c) Von Karman analogy.
Schematic:
Air
P= I atm
Flat plate
[_
u== 12m/s
r==30°c
r.= 1so c
0
\~--------------~
~1-------L=3m------~
t==x
Fig. 8.29
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Properties are at the film
temperature, Tr =.!..(T. +T=). (4) Transition Reynolds number, Recr =5xl0 5
2
Analysis: Reynolds number at the trailing edge of the plate,
u=L
v
(12m/s)(3m)
22.0lxI0-6 m2 /s
6
ReL = - = - - - - - - = 1.636x10
Since the Reynolds number is greater than the critical Reynolds number Recr, the flow is turbulent.
(a) Colburn analogy: StPr213 =Cr,x / 2
or
N14. Pr213 =_!_(0.0592Re;z 115 ) =0.0296Re;z°·2
2
RexPr
The Nusselt number for turbulent flow
hx
Nux =-=0.0296Re2_· 8 Pr113
[
k
heat-transfer coefficient is
k
\ =N14. -
X
]
0.6 ~Pr~60
5
7
5xl0 ~Rex ~10
Forced Convection: External Flow
r
8.45
At the trailing edge,
I;_, =!x0.0296(ReL)°" 8 (Pr)'1 3
L
= 3o.2 4 xl0-3 W/m °C x0.0296(1.636x 106 )°" 8 (0.7132}'1 3
3m
= 24.93 W/m2 °C
(Ans) (a)
(b) Prandtl analogy
St=
= Nu.,_
Cf/2
1+5 (
f
C ) 112
(Pr-1)
Re Pr
x
0.0296 Ret 8 Pr
Nux-L =-----;:==========---1+5~0.0296R~o.2 x(Pr-1)
=
h
x=L
0.0296(1.636x106 ) 0 -8 (0.7132)
= 2098
1+ 5 ~0.0296 (1.636 X 106 ) 0 -2 (0.7132-1)
= ! ..,
= 0.03024 W/m °C 2098
L ,vux=L
3m
X
= 21.15 W/m2 °C
(c) Von Ktirmtin analogy
St=
Cf/2
1+5~Cr12 {(Pr-1)+ zn( 1+¾(Pr-1))}
Nu.,_=L =St Rex=L Pr
0.0296 Re:~sL Pr
=------~-------~
1+5~0.0296xRe;?;l, [(Pr-l)+zn{ l+¾(Pr-1)}]
0.0296(1.636 x10 6 }°" 8 (0.7132)
=--~~~~~~~~~~~~~~~~---------~
1+5~0.0296x(l.636x 106 )-o. 2 [(0.7132-1)+ zn{1 + ¾(0.7132-1)}]
=2231.5
k
hx=L =LN'¾=L
= 0.03024W/m°Cx 223 1. 5
3m
(Ans) (b)
8.46
, Heat and Mass Transfer
= 22.49 W /m1 °C
(Ans) (c)
The mean value of the local heat transfer coefficient at the trailing edge (x = L) is
!(24.93+2 l.15+22.49)
3
Example 8.23
=23.0 W /m1 C.
0
~ Atmospheric air at 300 K flows parallel to a flat plate. The velocity of air far away from the
plate is 16 m/s. Determine, at 0.6 m from the leading edge of the plate, the thickness of
(a) the velocity boundary layer, (b) the laminar sublayer, and (c) the buffer zone. At this location, calculate the eddy
diffusivity at (d) 0.3 mm from the surface, (e) 3 mm from the surface, and (f) the edge of the velocity boundary layer.
Known: Temperature and velocity of atmospheric air flowing past the surface of a flat plate.
Find: At L = 0.6: (a) 8, (b)Y1amsublayer and (c) Ybufferzone·
Eddy diffusivity, £ at: (d) x = 0.3 mm, (e) x = 3 mm, and (t) x = 8.
Schematic:
Edge of velocity
boundary layer
Air
u== 16m/s
Flat plate
T== 300K
------L=0.6m _ _ _ __,
Fig. 8.30a
u
u
159_5_m_m_~
Buffer zone
0.655mm
Laminar sub layer 0.1 I mm
Velocity profile in turbulent boundary layer on a flat plate
Fig. 8.30b
Assumptions: ( 1) Isothermal flat plate. (2) Universal velocity distribution. (3) Constant air properties.
Properties: At 300 K for atmospheric air:
p=l.1614kg/m3,
v=l5.89x10-6m 2 /s
Forced Convection: External Flow
r
8.47
Analysis: (a) Reynolds number,
u L
16 m/sx0.6m
Re=-=-----v
15.89x1Q-6 m2 /s
00
= 6. 04 x 105 ( > 5 x 105) ~
Turbulent flow
The velocity boundary layer thickness is given by
~ = 0.381 Re-115
X
8 at x=L=0.6 mis
<> = (0.381)(0.6 m) (6.04 X 105)-115
= 0.01595 m
or
15.95 mm
(Ans) (a)
(b) The characteristic shear velocity or friction velocity is
C 1
u* =ti=
2
f 2:uoo
= fiu~
where Cr is the skin friction coefficient given by
Cr = 0.0592 Re-;!l·2
= 0.0592(6.04x 105)-o. 2 = 0.00413
u*=
And,
0.00413x162 m2 /s2 = 0 _727 mis
+- ~ -
2
(m)x
Y -y v -y
0.727m/s
15.89x1Q-6 m2 /s
= 4.577 X 104 y
Thus
(c) Limit of laminar sub layer: y+= 5
y = 5/(4.577x104 ) = 0.1 lxl0-3 m
or 0.11 mm
(Ans) (b)
or 0.655 mm
(Ans) (c)
(d) Limit of buffer zone: y+= 30
y = 30/( 4.577 x 104 ) = 0.655 x 10-3 m
At y = 0.3 x 10-3 m from the plate surface,
y+=(4.577x10 4 ){0.3x10-3 m)=13.73
(5 < y+< 30)
This location is in the buffer zone. Hence, the eddy diffusivity is
J
6
2
=15.89xl0-6 [ -13.73
5 --1 =27.74xl0- m /s
8.48
, Heat and Mass Transfer
(e) At y = 3 mm or 3 x 10-3 m,
y+= 137.3
(y+ > 30)
This location clearly lies in the fully turbulent zone, and we have
~=y+ -1= 137 ·3 -1=53.92
V
2.5
2.5
£ = {53 .92 )(15.89 x 10-6 m2 /
(t)
s) = 8.57 x I0-4 m1 / s
(Ans) (e)
At y= 8= 0.01595 m,
y+ =730
and
~= y+ -1= 730 -1=291
V
2.5
2.5
£ = (291)(15 .89x 10-6 m 2 / s) = 4.624xI0-3 m1 /s
(Ans) (t)
(B) CYLINDER IN CROSS FLOW
•
•
Example 8.24 ~ An electric resistance wire heater 0.1-mm diameter is placed normal to an air flow. The rate
of heat dissipation per unit length is 17.9 W and the wire surface temperature is maintained
at 40°C. The surrounding air temperature is 20°C. Calculate the velocity of air flow. The following correlation
may be used (below Re 0 = 4000):
H::rr
0.62 Re¥} Pr 13
N-, =0.3+
Properties of air at mean film temperature of 30°C are
k=26.52x 10-3 W/m K Pr=0.707
V= 16.2 mm 2/s
Known: An electric wire heater dissipates heat during cross flow of air over it under forced convection
conditions.
Find: Air speed, V (mi s) .
Schematic:
Electric resistance wire (T,=40°C)
Air
D=0.0001 m
Fig. 8.31
Assumptions: (1) Constant wire surface temperature. (2) Re < 4000. (3) Uniform heat transfer coefficient.
Analysis: This is a case of forced convection heat transfer during flow over a cylindrical surface. Heat
dissipation rate,
Forced Convection: External Flow
r
8.49
Hence, the average heat-transfer coefficient,
-
Q!L
h =-----"---
1tD(T. -T
00 )
17.9 W Im
(1txO.OOOl m) (40-20)°C
=----------
= 2848.9 W /m2 K
Average Nusselt number,
h D (2848.9 W/m2 K) (0.0001 m)
Nu = - - = ~ - - - - - ~ - - - ~
D
k
(0.02652 W/mK)
= 10.74
Using the given correlation, we find Re0
114
Reo = [ (~ -0.3) { 1+(¥r)2/3}
J
0.62Pr113
2/3 1/4 ]
=[
2
{10.74-0.3){1+(~) }
0.707
0.62 (0.107)113
= 463.64
Then, air velocity,
V
V=Re0 D
(463.64) ( 16.2x10-6 m2 /s)
=---------0.0001 m
= 75.1 mis
(Ans)
Comment: This method can be used to measure local fluid velocities with fairly good accuracy. The
accuracy can still be improved if the device is calibrated. Such a device is called hot wire anemometer.
Example 8.25
~ An electrical transmission line of 12-mm OD carries 200 A and has a resistance of
3 x lo-" ohm per metre length. Determine the surface temperature when the wind blows
across the line at 36 km/h and the ambient air temperature is I5°C. Properties of air at I 5°C are
k=0.02476 W/m K
v=l.47xl0-5 m2 /s
Pr=0.7323
Use the correlation:
N!Jo =CR~ Prn where n = 0.37, C = 0.26 and m = 0.6 for I03 < Rei, < 2 x I05 and n = 0.37, C = 0.51 and m = 0.5
for 40 < Re 0 < I000.
Known: Cross flow of air at prescribed temperature and velocity over an electric transmission line of
given specifications.
Find: Surface temperature, T,. ( C).
0
8.50
, Heat and Mass Transfer
Schematic:
Air
D= 12mm
/=200A
R,,= 3 X lo-4n/m
u== IOm/s
r== 1s 0 c
Fig. 8.32
Assumptions: (1) Steady state prevails. (2) Constant properties.
3
36km(10
m)( - lh
. velocity,
. u= =
Analysis: Air
-- - - -) = 10 mis
h
1km 3600s
Reynolds number,
R
_ u= D _ 10 m/sx12x10-3 m
~-
v
-
1.47x10-5 m2 /s
= 8163
For this value of Re0 , C = 0.26 and m = 0.6, n = 0.37
:. Nusselt number,
:. average heat transfer coefficient is
k
h=-x0.26xRe& 6 xP,.0.37
D
= 0 ·02476 W/mK x0.26x8163°- 6 x0.7323° 37
0.012m
= 106.3 W/m2 °C
Rate of heat dissipation = Rate of heat generation
= I2 R., =(200}2 (3x10-4) = 12 W/m
or
Surface temperature,
T =T +(QI L)
s
=
hrcD
= 15oc+
=18°C
12W/m
(106.3W/m2 °C)(rcx0.012m)
(Ans)
Forced Convection: External Flow
r
8.5 I
Example 8.26 ~ Air at 2 atm and 300 K flows across a circular cylinder of 50-mm-OD with a velocity of
16 m/s. The cylinder is maintained at a temperature of 350 K. Determine (a) the drag force
and (b) the rate of heat transfer per metre length of the cylinder.
Properties of atmospheric air at the film temperature of 325 K are
k=0.0282 W/m °C,
Pr=703,
v=l.841xl0-5 m2 /s
Known: Conditions associated with cross flow of air, over a cylinder.
Find: (a) Drag force, F 0 , and (b) Heat-transfer rate, Q.
Schematic:
T5 = 350K
Air
P=2atm
T==300K
V=l6m/s
D=O.OSm
Fig. 8.33
Assumptions: (I) Steady operating conditions. (2) Air is an ideal gas. (2) Radiation effects are not
considered.
Analysis: The Reynolds number
Reo = _V_D = (16 m/s){0.05m) = 86909
v
{1.84 Ix 10-s m2 /s/2)
1
1
1
[Note that v a - a - Hence, at 2 atm, v = -v@tatml
p
p
2
For 104 < Re0 < 2 x 105, Drag coefficient, C0 = 1.1
Drag force per unit length is
1
2I2
Fo -- C0 .-pAV
2
where A is the projected area of the cylinder normal to the flow.
With A = D L = (0.05 m) (I m) = 0.05 m2 and
= __!___ =
p
RT
2 x 101.325kPa
= 2 _173 k /m3
(0.287kJ / kgK) (325K)
g
m2)
1
Hence, F0 ={1.1) ( -x2.173
kg/m3 x0.05m 2 xl6 2 -
2
= 15.3 N
s2
(Ans) (a)
8.52
, Heat and Mass Transfer
For the range 20000 < Re0 < 400 000, the Churchill-Bernstein relation is
0 62 Re0 li 2 Prli 3 {
112}
Nuo = 0.3+ ·
l/ 4 I+{Re0 /282000)
[1+ (0.4/Pr)213]
( 0.62 X 869091/ 2 X 0.703113
= 0.3+
)l 1+ {86909 / 282000 ) J
015
025
[1+{0.4/0.703}2'3 ] .
=222
Average convection coefficient,
-
-
k
h =NuoD
= 222 x 0.0 282 W/m°C = 125.25W/m2 °C
0.05
Heat-transfer rate per unit length is
Q = h (1tDL )(T. -T=) = (125.25~){1t
x 0.05 m x lm){350-300)k
m2oc
(Ans) (b)
=983.7W
Example 8.27 ~ A tube, 2.5-cm OD, is losing heat at a rate of 1800 W/m 2 due to convection alone to a
stream of air flowing across it. If the surface temperature is 80 °C and the air temperature is
20°C, determine the velocity of air, using the Hilpert's empirical correlation. Properties at the film temperature
of 50°C for atmospheric air are
k=0.0283xl0- 3 W/m K
V=l7.95xlo-6m2 /s
Pr=0.698
Known: Operating conditions associated with cross flow of air across a tube.
Find: Velocity of air.
Schematic:
Fig. 8.34
Assumptions: (I) Steady operating conditions. (2) Air is an ideal gas. (3) Uniform tube surface temperature.
Analysis: The convection heat transfer coefficient is determined from
h = Qj As = 1800 W/m 2
T.-T= {80-20)K
= 30W/m 2 K
Forced Convection: External Flow
r
8.53
Working with Hilpert's correlation, assuming the range of Re0 as 40 to 4000 (since the free-stream
air velocity is unknown), we have
-
hD
Nuo = -- = 0.683 Re& 466 Prll 3
k
Nuo = (30 W/m 2 K)(0.025 m) = 26 _5
0.0283 x 10-3 W/m K
Therefore
Re& 466 = {26.5/0.683){0.698fll 3
= 43.74
Reynolds number,
Re0 = {43.74}11°·466 = 3320
VD
V
Free-stream air velocity is
V= Re0 v = 3320xl7.95xI0-6 m2/s
D
0.025 m
= 2.38 mis
(Ans)
Comment: Re 0 being in the assumed range, the correlation used was correct.
Example 8.28 ~ A tube of 20-mm-OD maintained at a surface temperature of 90°C with atmospheric air at
30°C moving in cross flow over it. What is the velocity of air if the steady-state heat flux
from the tube surface is 1.7 kW/m 2 ?
The following properties of air at I atm and 60°C may be used:
k = 28.74xl0- 3 W/m°C
v = 19.2xlo-6m2 /s
Pr=0.702
Use the Hilpert's empirical correlation:
-
hD
Nuo = -
4 - 40
40 - 4000
4000 - 40000
k
= CReP} Prll 3
C
m
0.911
0.683
0.193
0.385
0.466
0.618
Known: A tube losing heat with air moving in cross flow over its surface.
Find: Air velocity.
8.54
, Heat and Mass Transfer
Schematic:
Air
T~=30°C
V=?
~Q
- = 1700W/m2
As
Fig. 8.35
Assumptions: (1) Steady operating conditions. (2) Uniform tube surface temperature.
Analysis: Nusselt number,
hD
Nuo = - where
k
-
(Q/As)D
Nuo=----
k(J'. -T~)
(1.7x 103 W/m2 )(20 x 10-3 m)
(28.74xl0-3 W/m °C) {90-30)°C
= - , - - - - - - ~ - - - ~ - = 19.717
From Hilpert's correlation, assuming Re0 lies between 40 and 4000,
INuo = 0.683 Re&
466 Prll 3
I
Hence,
Re0.466 = NuoPr-l/3
D
= 19.717 x(0.702f1/3
0.683
= 32.48
0.683
Reynolds number,
Reo = {32.48{0.466
= 1753
Since Re0 (= 1753) lies between 40 and 4000, the constants C and mused were appropriate.
VD, th e air
. ve 1ocLty
. .LS
. h Re =W Lt
0
V
V = Re0 v = 1753x 19.2 x 10-6 m2 / s
D
20xl0-3 m
= 1.68 mis
(Ans)
Forced Convection: External Flow
r
8.55
Example 8.29 ~ A copper pin fin, (k = 400 W/m 0 C) 4 mm-diameter and 12-mm-long, is brazed to the surface
of a silicon chip. Atmospheric air at 25°C and free-stream velocity of 20 mis is in cross flow
over the fin. The chip surface temperature is 75°C. Determine (a) the average surface heat transfer coefficient
for the surface of the pin, and (b) the pin heat-transfer rate.
Properties of air at 50°C are
k=0.028 W/m °C
V=l8.2xlo--6m2 /s
Pr=0.704
Known: Dimensions of a pin fin. Air velocity and temperature base temperature.
Find: h(W!m 2 °C); (b) Q(W).
Schematic:
Air
u==20m/s,
r==2s c
0
Silicon chip surface
Copper Pin fin
rb=1s c
0
Fig. 8.36
Assumptions: (1) Steady-state conditions. (2) Constant fluid properties. (3) Negligible contact resistance.
(4) Radiation effects are neglected.
u D (20m/s) (0.004 m)
Analysis: (a) Re = - = - = -'-----'---''-------'- = 4395.6
D
V
18.2 X 10-6 m 2 /s
Using the Churchill- Bernstein equation:
Nuo-0.3+
0.62Reg2pyll3
[l+(0.4/ Prf3]
-[
l/ 4 -1+
(
Reo
)5/8]4/5
282000
3
518 415
0.62{4395.6)1/ 2 {0.704)1/
=0
- = [ l + ( 4395.6 ) ]
. 3+-~-~~-~
~
[1+{0.4/704f3]
282000
= 34.28
Average convection coefficient
h = Nuo .!.... = 34.28x 0.028 W/m oc
D
0.004 m
=240W/m 2 °C
(b) For the fin with tip convection,
m=
$f
r
2
= ( :;
= 24.5 m- 1
4
= [ 400 ;~;00 ;:~~;~ m
(Ans) (a)
T
2
8.56
, Heat and Mass Transfer
Corrected length,
Lc = L+(D/4) = 0.012 m +{0.004 I 4) = 0.013 m
tanh mLc = tanh (24.5 m- 1) (0.013 m) = 0.308
Pin heat-transfer rate,
Q = mk Ac eb tanh mLC
rcD 2
= mk--(Tb -T=)tanh mLc
4
= (24.5 m- 1 ) (400 W /m 0
c)(
¾x 0.0042 m 2 ) (75-25)°Cx 0.308
= 1.90W
(Ans)
Example 8.30 ~ A metallic bar of 25-mm diameter is cooled by air at 30°C, cross flowing past the bar with
a velocity of 2.5 mis. If the surface temperature of the bar is not to exceed 85°C and the
resistivity of the metal is 0.015 x lo--6 ohm-m 3 per metre, calculate (a) the heat transfer coefficient from the
surface to the air and (b) the permissible current intensity for the bus bar.
Known: 02erating conditions of a heated metallic bar in cross flow of air.
Find: (a) h; (b) I.
Schematic:
Air
Fig. 8.37
Assumptions: (I) Steady-state conditions. (2) Constant properties. (3) Constant surface temperature.
Properties: Atmospheric air (at T= = 30°C):
k=0.0265 Wlm °C,
v=l6.2xI0-6 m 2 ls,
Pr=0.707
Pt;,@ss•c = 0.698
Analysis: (a) For cross flow of air, the Zhukauskas correlation will be used:
Pr
Nuo =CRegPr0 ( - )
1/ 4
P,;,
Reynolds number,
u=D 2.5 mis x 0.025 m
Re0 = - - = - - - - - - v
16.2 x I0-6 m2 Is
= 3858
Forced Convection: External Flow
For
103 <Re0 <2x10 5 ,
C=0.26,
m=0.6,
n=0.37
r
8.57
(for Pr < 10).
It follows that
0 25
Nu = 0.26(3858)°" 6 (0.707)°" 37 ( 0 ·707 ) .
0.698
D
= 32.54
The average convection heat transfer coefficient is determined to be
h = Nuo !:_ = 32.54x 0.0265 W/moc
D
0.025 m
= 34.5 W /m2 °C
(Ans) (a)
(b) From energy balance
£
-E,out+Egen=
no heat inflow
£
steady state
Hence,
or
Heat transfer by convection per unit length is
Q~nv = h(1tD){T. -T=)
= (34.5 W/m2 °C) ( 1t X 0.025 m) (85-30)°C
=149.0W/m
This must equal the heat generated in the bus bar.
Electrical resistance, R. = Pe L
Ac
With
1t
Ac=-D2 , one gets
4
0.015x10-6 ohm m3
R. = - - - - - -m- L
~(0.025 m)2
4
= 3.056 x 10-5 ohm/m
12 R. = Qconv = 149.0 W/m
L
or
L
149.0 W/m
2
1 = 3.056 x 10-5 ohm/m
IA ohm I
2
1W
8.58
, Heat and Mass Transfer
Permissible current density,
I= 149 X 105 = 2208 A
(Ans) (b)
3.056
Example 8.31
~ Water at 20°C flows normal to the axis of a circular tube with a velocity of 1.5 mis. The
diameter of the tube is 25 mm. Calculate the average heat transfer coefficient if the tube surface
is maintained at a uniform temperature of 80 °C. Also estimate the heat transfer rate per unit length of the tube.
20+80
Properties of water at 7i = - - - = 500C are
2
Specific heat, CP = 4.1813 kJ/kg K
Kinematic viscosity, v = 0.568 x I 0-6 m2/s
Thermal conductivity, k = 0.6395 W/m K
Prandtl number, Pr= 3.68
Density, p = 990 kg/m 3
Dynamic viscosity of water
µw at 80°C = 3.5456 X 1Q-4 kg/m s
µ at 20°C = 1.006 X 10- 3 kg/m s
Use the correlation:
1/4
Nu= (0.4 Re0-5 + 0.06 Re2tl) p,0.4 ( ~ )
Known: Operating conditions for a heated circular tube in cross flow with water.
Find: Heat-transfer rate per unit tube length, Q I L.
Schematic:
u= = I .Sm/s
Fig. 8.38
Assumptions: (1) Steady-state conditions exist. (2) Uniform tube surface temperature (3) Constant fluid
properties.
Analysis: Working with the given correlation,
1/ 4
Nu= [ 0.4 Re0- 5 + 0.06 Re213] p,.0.4 ( ~ )
Accordingly, with diameter D as the characteristic length,
Re= u= D = (1.5 mis) (0.025 m)
0.568xl0-6 m 2 /s
v
= 66021
Forced Convection: External Flow
r
8.59
Substituting the appropriate numerical values,
-
(
Nu= h D =[0.4(66021)0- 5 + 0.06(66021) 213 ](3.68)0.4 1.00 6 x I0-3
k
3.5456xl0-4
)l/4
= 438.84
Heat-transfer convection coefficient is
h = 438.84 x 0 ·6395 W/m K = 11225.5 W / m2 K
(Ans)
0.025 m
Also,
Heat transfer rate per unit length of the tube,
Q = h As (T. - T= ) = h (1t D L) (T. - T= )
=(11225.5W/m2 K) (1tx0.025 mxlm)(80-20)°C
= 52.9 x 103 W
or
K
or
52.9 kW
(Ans)
(C) SPHERES
•
•
Example 8.32 ~ The decorative plastic film on a copper sphere of I0-mm diameter is cured in an oven at
75°C. After removal from the oven, the sphere is exposed to an air stream at IO m/s and
23°C. Estimate the time taken to cool the sphere to 35°C. Use the following correlation:
1/4
Nu= 2+[ 0.4Re112 +0.06Re 2'l]
(
p,.0.4 : :
)
Properties:
Copper: p = 8933 kg/m 3 k = 400 W/m K
CP = 380
J/ kg K
Air (at T~ = 23°C):
µ = 18.16 X 10-6 Ns/m2
V = I 5.36 X 10-6 m2/s
Pr= 0.709
k=0.0258 W/m K,
µ,@J,c = 19.78 x 10~ Ns/m2
Known: A sphere is cooled in an air stream under unsteady state conditions.
Find: Time required for cooling, t (s) .
Schematic:
r. (t= o) = T;=7s c}
r. (t= t) = r= 3s c
0
Air
0
T==23°C
u== IOm/s
Fig. 8.39
Assumptions: (1) Thermal resistance and capacitance of the plastic film are neglected. (2) Radiation
effects are ignored. (3) The sphere surface temperature is a function only of time.
8.60
, Heat and Mass Transfer
Analysis: Reynolds number,
pu.,,D
u.,,D
µ
V
R(b=--=-
= (10m/s)(l0xl0-3m) =6510.4
15.36xl0-6m2/s
Nusselt number,
Nu =2+[0.4(6510.4)112 +0.06(6510.4)213](0.709)0.4 x { 18 · 16 xlO -6 Ns/m2 }"
D
19.78xl0-6 Ns/m2
4
= 47.38 = hD
k
The convection coefficient, h is
h=!!_Nu = (0.0258W/mK) ( 4738 )
D D
O.Olm
= 122.24 W/m2 K
hLC
Biot number, B i = ksph
where ksph = 400 W/m K,
.. l
h ¥- rcD3 / 6
4 = charactenstlc
engt = - = - - -
A.
rcD2
D
6
Bi= hD = (122.24 W/m2 K)(O.Olm)
6ksph
6x400W/mK
(« 0.1)
= 0.51xl0-'
Hence, the lumped parameter analysis is valid.
Using the lumped capacity model,
_T_(_t)_-T_.,,_ =~=exp [-( Bi)(Fo)]
T(t = 0)-T.,, 0i
= exp[
;;P;]
ht
e
p cp4
ei
= e-httpc.r,
--=-ln-
or
:. time required for cooling process to be completed is
pCpLc 0i pCPD (J;-T.,,)
t=--ln-=--ln
-h
0
6h
T-T.,,
f
= (8933kg/m3)(380J/kgK)(O.Olm) xln (75-23)° C}
6(122.24 W/m2 K)
(35-23)° C
l
= 67 .9 s
or 1 min 7.9 s
(Ans)
Forced Convection: External Flow
hL
.
hL
r
8.61
..
Comment: The Nusselt number, (Nu=-) and the B1ot number (Bi =-c) appear to be similar but
k
k
note that k in Nu is the thermal conductivity of the fluid while k in Bi is that of the solid.
Example 8.33 ~ A spherical thermocouple bead of I-mm-diameter (p = 7800 kg/m 3, CP = 0.4 kJ/kg 0 C) is
required to respond to 99.5% change of the surrounding air temperature in 10 ms. What is
the minimum air velocity at which this will occur? Use the following correlation and properties of air:
Nu= 2 + (0.4Re112 + 0.06Re213 )P,o.4
k = 0.0259 W/m K
v = I 5.06 x I~ m2/s
Pr= 0.703
Known: A spherical thermocouple bead subjected to forced convection heat transfer conditions.
Find: Minimum air velocity, V(m/s).
Schematic:
Spherical thermocouple bead
D = 0.001 m, p = 7800 kg/m3,
Air
CP = 400 J/kg K
h, T~
V=?
T(t)- T ~ = 0.995 (T(O)- T ~)
t = IO milliseconds= 0.01 s
Fig. 8.40
Assumptions: (1) Steady-state conditions. (2) Constant properties. (3) Lumped capacity model is valid.
Analysis: Assuming lumped capacity analysis to be valid, we have
tl
T(t)-T~
[ -hA
---=0.995=exp
- 5T.-T~
p¥-CP
. . d.
.
¥- re D3 I 6 D
. h h
w here L c 1s t e c aractensttc 1mens1on = - = - - - =
As
rcD 2
6
6ht
In 0.995 = - - p CPD
It follows that the average convection coefficient is
pCP D(-ln0.995)
h=------6t
=
(7800 kg / m 3 )(400J / kg K) (0.001 m)(+5.01254)
6x0.01 s
= 260.65 W/m 2 K
Nusselt number,
hD
Nu= -
k
260.65W/m 2 Kx0.001m
=- - - - - - - - -
= 10.064
0.0259W/mK
8.62
, Heat and Mass Transfer
Since
Nu = 2 + [ 0 .4 Re112 + 0 .06 Re213] p,.0.4
10.064 = 2+ [0.4 Re112 + 0.06Re213 ](0.703) 0 A
or
or
l0.0 64 - 2 = [ 0.4 Re112 + 0.06 Re213 ]
0.8685
or
9.588 = 0.4 Re112 + 0.06 Re213
VD
VxO.OOlm
Re=-=-----=66.4V
v
15.06x10-<im2 ls
where
9.588 = 0.4 (66.4 V) 112+ 0.06(66.4 V) 213
9.588 = 3.26 v 112 + o.984 v 213
The free-stream velocity, V can be found by trial and error.
Let
V= 4 mis.
RHS = 3.26( 4)112 + 0.984( 4) 213 = 9
(< 9.588)
Let V = 4.5 mis
RHS = 3.26 ( 4.5)112 + 0.984( 4.5)213 = 9.598"" 9.588
Hence, the air velocity is
(Ans)
V= 4.5 mis
Example 8.34 ~ A spherical thermocouple junction of 2-mm-diameter and emissivity of 0.7 is used I atm
to measure the temperature of a hot gas at 450°C, I atm which is at 3.5 mis through a
large duct whose surface temperature is 250°C. (a) What temperature would be sensed and indicated by the
junction? (b) If the pressure of the gas is doubled, what would be the error in temperature measurement? Gas
properties may be taken as those for air.
Correlation:
0.25
Nu = 2 + { 0.4
v'Re + 0.06 Re213 } p,.OA ( ~ )
Properties of air at I atm, 450°C:
p = 0.488 kg/m 3
Pr= 0.6965
k = 0.05298 W/m K
µ = 34.15 x Io---6 kg/ms
µw = 27.6 XI o---6 kg/ms
Known: Diameter and emissivity of thermocouple junction. Gas and duct temperatures. Flow velocity.
Find: Measured gas temperature. Error in temperature measurement if the pressure is doubled.
Forced Convection: External Flow
Schematic:
r
8.63
Large duct (Tw=250 °C)
fr--
Gas
~ •
Tg=450°C
Thermocouple junction
Tr. i:: =0.7, D=2mm
V= 3.5 mis
Fig. 8.41
Assumptions: (1) Equilibrium operating conditions. (2) Diffuse-gray surface of the junction. (3) Gas
properties are approximated as those of air. (4) Junction is very small compared to the large duct.
Analysis: Under equilibrium conditions:
Heat transfer by forced convection from the gas (g) to the thermocouple junction (t) = Net radiant heat
exchange between the thermocouple junction (t) and the duct wall (w)
= Qradiation (t-.w)
%onv = qrad
{W /m 2)
h ( ~ - i; ) = e cr ( I;4 - r,: )
i.e.,
12:onvection(g -H)
or
or
{W)
Measured gas temperature,
T.=Tg
t
E<J (T.4 -T4)
t
h
w
and, the error in temperature measurement= Tg - 1'r_
To evaluate the convection coefficient, h, we use the given correlation for flow over a sphere.
l/4
Nusph = hD =2+[ 0.4Rell 2 +0.06Re213 ] p,.0.4 ( ....!:.._ )
µw
k
Reynolds number,
pV D
(0.488 kg / m3 ) (3.5m / s) (0.002 m)
Re=--=-'-----"---'----'---'-----'µ
34.15x10-6kg/ ms
=100
Nusselt number,
Nu
sph
= 2 +[ 0.4(100)11 2 +0.06(100) 213 ](0.6965)0 .4 ( 34 .1 5 x l0-6 )l/
27.6x 10-6
=6.83
Convection coefficient,
h = Nu
!:_ = 6 _83 x 0.05298 W/m K
sph D
= 181 W/m2 K
0.002 m
4
8.64
, Heat and Mass Transfer
With
~ =(450+273.15) K=723.15K,
Tw =(250+273.15) K=523.15 K,
cr = 5.67 x 10-8 W/m 2 K 4
£ = 0.7,
_
J; - 723.15
h = 181 W/m 2 K, we have
and
-{0.7 X 5.67 X lQ-8 (J; 4 -523.15 4 )}
181
= 723.15-0.02193[T1 /100]4 +16.425
or
Ii;= 739.58-0.02193 [T1 /100]4 I
By trial and error
Trial
T, (assumed)
T, (ca lculated)
420°C or 693.15 K
689 K
3
689 K
690. 17 K
3
690 .1 7 K
689.8 K
4
689.8 K
689.9 K or 416.7 °C
!Ti = 416.7°c1
Hence,
(Ans) (a)
Error in temperature measurement
= ~-i; = 450-416.7 = 33.3°C
(b) If P = 2 atm, the properties like µ, µw, k and Pr will not change but the density, p will.
Since
p oc P,
_
2atm_
P2 atm - Pt atm X
latm
-
2 X 0.488 kg / m
3
= 0.976 kg / m3
Re= 0.976x3.5x0.002 = 200
34.15xl0-6
Then
Nu= 2 +[ 0.4.Jioo + 0.06(200) 213
J(0 .6965)
04 [ 34 .1 5 x 1Q-G
27.6xl0-6
]o.25 = 11.04
h = 11.04 x 0.0 5298 = 292.5 W/m2 K
0.002
We note that
T.=Tt
g
ecr (T.4 -T4)
t
w
h
4
= 723 _15 _(0.7 X 5.67)(l) +(0.7 X 5.67 X 5 _23154 )
292.5
100
292.5
Forced Convection: External Flow
J; = 733 .31-0.01357(l)
or
r
8.65
4
100
Let
1'r_ = 700 K
RHS = 700.7 K
With
1'r_ = 700.7 K
RHS = 700 .6 K"' 427.5°C
!T1 = 427.5°CI
Error in temperature measurement
= ~ -i; = 450-427.5 = 22.5°C
(Ans) (b)
Comment: The effect of increasing the pressure two fold is to increase the value of h, resulting in
reduced error in temperature measurement (from 33.3°C to 22.5°C).
Example 8.3S ~ Lead shots are manufactured by dropping molten lead (p = IO 600 kg/ml) from a tower
through cool air into a pool of water. The latent heat of fusion of lead is 24.5 kJ/kg. Each
pellet, a sphere of 2-mm-diameter, is cooled while passing through 10°C air. Assume that the shot falls at its
terminal velocity. Calculate the minimum necessary height of the drop tower. The lead must be solidified from
its molten state at 327°C to a solid state before it hits the pool of water at the bottom of the tower.
Use the following correlation and the properties of air:
Nusph
= 2 + [ 0.4 Re112 + 0.06 Re2'l] p,o.4 (µ~/µ. )"4
p = 1.247 kg/ml
µ = 17.7xlo-6kg/m s
v = 14.16 x lo-6 m2 /s
Pr= 0.705
µs at 327° C = 30.6 x I0-6 kg/m s
Known: Lead pellets are dropped through cool air from a height to convert from molten (liquid) to solid
state.
Find: Height of tower.
Schematic:
Fo
t
·----··-
Air
...
T,= 327°C
(melting point)
r.. = 10°c
________.. Qconv
· · · · · -. . ... .. . . . . . . · · · L
!
V
Fig. 8.42
Lead pellet (0 = 2mm)
p = I0600 kg/m3, h,1= 24.5 kJ/kg
8.66
, Heat and Mass Transfer
Assumptions: (1) Constant air properties. (2) Pellet temperature and density remain constant at the melting
point throughout the process. (3) Radiation effects are ignored.
Analysis: Energy balance on the lead shot gives:
-t-E,out +
or
or
tJ
= Est
-!?.:onv ·t = -m \r
-h A8 ( T,, - T=) t = -p ¥- hsf
(-ve sign because the lead pellet is getting solidified)
where t is the time required to completely solidify the pellet.
The terminal velocity, V, with which the pellet is falling = (Distance traversed, i.e., the height of the
drop tower, H!Time interval, t)
H
Hence, t=V
It follows that
H = _V.c....p_D_h--=•-=-f- I
6h(T,,-T=)
or
To obtain the terminal velocity, we apply the force balance:
Gravitational force, Fg = Drag force, F 0
or
2
mg -C
- O A projected .!_
2 Pair V
or
)
(rcD ) 1
2
P ( -rcD
6 - g = Co - 4 - 2Pair V
or
V= [ !PDg ]
6 Co Pair
3
2
1/2
= [.ix 10600 kg/ m3 x O.002 m x 9 .81 m/ s2 ] 112
3
C0 xl.247kg/m3
Iv= 14.91/ (C 112 I
0 )
The drag coefficient, C0 is dependent on the Reynolds number, Re 0 .
VD V(m/s)x0.002m
Re0 = - - = - - - - - - v
14.16x10-6 m 2/s
1Re = 141.243 vi
0
To find C0 one requires Re0 which in tum depends on V, and V is related to C0 . Hence, a trial and
error procedure is necessary.
Forced Convection: External Flow
r
8.6 7
Assuming C0 = 0.4, V = 23.6 mis and Re0 = 3330. From the graph showing C0 against Re0 for a
smooth sphere, we find C0 = 0.4 for Re0 = 3330.
Thus, finally we have
Re0 = 3330,
C0 = 0.4
and
For a sphere, using the given correlation:
Nuo =2+[ 0.4Re'r?+0.06Rel{ 3 ]
p,.cu(~:
V = 23.6 mis
r
4
= 2 + [ 0.4(3330}'12 + 0.06(3330 )213 ] (0.705)°"4 x[ 17 ·7 x IO-<i kg/ms ]
114
30.6x10-<i kg/ms
= 29.64
Average heat-transfer coefficient is
h = Nuo !_ = 29.64x 0.0251W/moc
D
0.002m
=372 W/m2 °C
Hence, the height of the tower is
H = [23.6 mis x 10600 kg/m3 x 0.002 m x 24.5 x 103 J/kg]
6x372 W/m2 °Cx(327-10)°C
= 17.3m
(Ans)
Example 8.36 ~ Atmospheric air at 25°C and a free-stream velocity of 45 km/h flows across a 10-cm-diameter
(a) circular cylinder of 1-m-length, and (b) a sphere. The surfaces of both are maintained at
I40°C. Determine the drag force and the heat transfer rate in each case. The following Whitaker's correlations
may be used:
Circular Cylinder
~=
~=
0.25
Nu = [ 0.4 Re0·5 + 0.06 Re213] p,.OA
Sphere
(
)
Nu = 2 + [ 0.4 Re0·5 + 0.06 Re213] p,.OA (
0.25
)
Properties of air
µ{140°C) = 235.Sxl0-7 Ns/m2
µ = 183.6xl0-7 N s/m2
v = I 5.71xlo-6m2/s
k=0.0261 W/m K
Pr= 0.71
Known: Conditions associated with air flow across a cylinder and a sphere of specified diameter and
surface temperature.
8.68
, Heat and Mass Transfer
Find: F 0 and Q for (a) cylinder, and (b) sphere.
Schematic:
Air
Air
O=O.IOm
u~ = 45 km/h _ ____
T~=25°C
(a) Circular cylinder
(b) Sphere
Fig. 8.43
Assumptions: (!) Steady operating conditions. (2) Uniform surface temperature. (3) Air is an ideal gas
at atmospheric pressure.
Analysis: (a) Circular Cylinder
45 m
u~ = 45 km/h= - - = 12.5 mis
3.6 s
Reynolds number,
Re= u~D = 12.5m/sx0.10 m
v
15.71xl0-6m 2 / s
= 79567
Nucyl = [0.4Re05 +0.06 Re213 ] Pr0.4 (µ~/µw )025
= [0.4(79567)
05 + 0.06(79567f J(0.71 )°'
3
4 ( 183 ·6 X I0-7
235.5 X 10-7
)0.
25
hD
= 183.4 = k
Average convection coefficient is
h = Nu
~ = 183 .4x 0.0261 W/m K
0.10 m
cyl D
=47.87 W/m 2 K
Heat-transfer rate is
Q = h (1t D L) ( I'. - T~)
= (47.87W /m 2 K) (1txO.lO m x lm)(l40-25)°C
= 1730W
or
K
(Ans)
Forced Convection: External Flow
r
8.69
Drag force,
1 A u_2
Fo -C
- o-P
2
where A = frontal area exposed to flow
= DL = 0.10 mxlm = 0.1 m2
The drag coefficient is a function of Reynolds number for a smooth cylinder. For Re= 79567
i.e., 104 <Re< 2 x 105 , C0 (cylinder) "" 1.1.
Air density,
p
= __!__ =
101.3 kPa
= 1.184 k /m3
RT 0.287 kJ/kg Kx298.15 K
g
Hence,
F0 = (u).!.(1.184 kg/m3)(0.lm2 ){12.5 m/s)2
2
= 10.2N
[l N= 1 kg m/s2 ]
(Ans) (a)
(b) Sphere
Nusph
= 2 + Nucyl
= 2 + 183.4 = 185.4
h = 185.4 x 0.0 261 W/m K = 48.39 W/m2 K
O.lm
and, the rate of heat transfer is
Q = h (1tD2 ) (T. -T_)
= (48.39 W/m2 K}(1txO.l2 m2 )(140-25) K
=175W
(Ans) (b)
For a smooth sphere, with Re= 79567 from the graph for drag coefficient as a function of Reynolds
number, we have
1t
Frontal area, A= 1tD2 /4 =-x0.l2 m2
4
:. drag force,
1 A u_2
Fo -C
- D -p
2
= (0.5)(½)(1.184 kg/m3) ( ~x0.0lm2 ) (12.5 m/s)2
=0.363 N
(Ans) (b)
8.70
, Heat and Mass Transfer
(D) TUBE BANKS
•
•
Example 8.37 ~ Pressurized liquid water at 40°C flows, without change of phase, across a 45-cm-wide and
staggered bank of tubes (SrlD=S...ID = 2) of 2.5-cm-outside diameter. The number of tubes
in the transverse row and the longitudinal row is 6 each. For each 30-cm in height of the tube bank, water is
supplied in a 15-cm-inside diameter pipe flowing with a velocity of I m/s. Calculate (a) the temperature of the
water leaving the tube bank and (b) the heat transfer rate per m height of tube bank if the tube wall temperature
is maintained at I 20°C by passing exhaust gases through the tubes. Use the following properties of water:
k = 0.635 W/m K
v = 0.626xlo-6m2 /s
CP = 4.179 kJ/kg K
p = 990.7 kg/m 3
Pr= 1.43
P; = 991. I kg/m3
Pr=4.08
Known: Staggered tube arrangement with flow conditions in a bank of tubes.
Find: Outlet temperature of water, T.(°C).
Schematic:
Water
T
l
V= I m/s
T== T;=40°C
H::::o.Jn,
Tubes:
N=NLxNT
=6 X 6= 36
= 120°c
Fig. 8.44
Assumptions: (1) Steady operating conditions. (2) Tube surface temperature is uniform .
Analysis: We note that
ST +D = 5+2.5 = 3.75 cm
2
2
S0 =~(S¥ ! 4)+Se, ='12.5 2 +5 2 =5.59cm
As S0 >(ST+ D) I 2, the minimum flow area will be on the transverse plane.
Aflow,min =[H-NTD]L
= [30cm - 6x2.5cm] 45cm = 675 cm 2
From equation of continuity:
Vmax At1ow, min
=Ac.pipe Vpipe
maximum velocity of water flowing across the tubes is
( 1t I 4) DJipe x vpipe
Aflow, min
Forced Convection: External Flow
r
8. 71
l
-x152 cm2
= [ 1t
4
{lm/s)=0.2618m/s
675cm2
R
_ Vmax D _ (0.2618m/s)(0.025m)
fn,max V
0.626x1Q-6 m2/s
=1.046x10 4
Since NL=6 (<16), the correction factor, F=0.945
Ren.max< 2 X 105 and
As
> 103
0.2
(
Nun =0.945x0.35 ::
)
(Rfn,max)0 -6 P,..0- 36 (Pr I Pr,J114
02
0 25
= 0.945x0.35(~) · (1.046xl0 4 )°" 6 (4.08) 0 -36 ( 4 ·08 ) .
5
1.43
= 184
Hence,
Ji =Nun jj = 184x 0.635W/mK
0.025m
= 4675W/m2 K
Mass flow rate,
= (991.1 kg/m3 { ~x0.152 m2}1m/s) = 17.51 kg/s
Total heat-transfer surface area of the tubes,
A. =N1tDL=36x1tx0.025mx0.45m
= 1.272m2
Outlet temperature of water,
A•]
Te =Ts -(Ts -T)exp
[ -h1
• C
m P
= 120oC-(120-40)oCexp[- 4675W/m2 Kxl.272m2]
17.51kg/sx4179J/kgK
= 46.2°C
(Ans) (a)
Heat-transfer rate per metre height of the tube bank is
Q=mcp (r. - i; )
=(17.51kg/s) (4.179kJ/kgK) (46.2-40)°C or K
= 453.7 kW
(Ans) (b)
8. 72
, Heat and Mass Transfer
Example 8.38 ~ Air at atmospheric pressure and 20°C flows across a bank of tubes 12 rows high and 6 rows
deep at an undisturbed velocity of 10 mis measured upstream of the tubes. The surfaces
of the tubes are maintained at I00°C. The diameter of the tubes is 26-mm, and the tubes are arranged in a
staggered fashion. The centres of the tubes form corners of an equilateral triangle of sides 45 mm.
(a) Determine the total heat transfer rate if the tubes are 4-m long.
(b) Calculate the heat transfer rate if the tube are arranged in an in-line fashion with the same row spacing.
Compare it to the staggered arrangement.
(c) Compare the pressure drops across the tube bundle for the two orientations.
Known: Surface temperature and geometry of a tube bank. Velocity and temperature of air in cross flow.
Find: Heat transfer rate, Q and pressure drop, t:.P for (a) staggered, and (b) aligned arrangement.
Schematic:
Air
/4 D=26mm, T = I00°C, L=4m
5
-·-·-\jj--·---·--·--·--·---·-·---·---·--·-0 ....._ SL ____.I
V= IOm/s
T= =20°C
I
;
'
0
J
:
!
;
i
P= I atm
__ __! __ _
(a) Staggered
NL =6, NT= 12
N=NLNT=6 x 12=72
0
Air
V= IOm/s
T= =20 °C
P= I atm
--
0
OT
ST=4Smm
0
0
o_L
0
0
0
I- SL -I SL =39mm
(b) Aligned
Fig. 8.45
Assumptions: (I) Steady-state conditions. (2) Negligible radiation effects. (3) Uniform tube surface
temperature.
Properties: Air (1 atm. 25°C)
p = 1.184 kg/m 3
v= 15.62 x I0-6 m 2 /s
k= 0.02551 W/m °C
cp= 1.001 kJ/kg c
P@2o•c = 1.204 kg / m3
Pr@ioo•c =0.7111
0
Pr= 0.7296
Forced Convection: External Flow
Analysis: (a) Staggered Arrangement
For equilateral triangle, ST = S0 = 45 mm
From the tube pitches
s5=st+(s;J
!
SL = ~ ~ 4 52 - 4
2
===39mm
Also,
ST + D = 45 + 26 = 25 _5 mm
Since
- 2 - < So,
2
2
ST+D
Maximum velocity is
S
45mm
vmax =-T-V=-----xlOm/s
ST -D
(45-26)mm
= 23.68 mis
Reynolds number
Vmax D
v
-===---
Ri!o,max
23.68m/sx0.026m
=-----15.62x10-<i m2 /s
= 39 423
With (ST/SL)= 45/39 = 1.154 (< 2), it follows that
C=0.35(~:rs =0.35xl.154°- 2 =0.36
and
m=0.6
Hence, from Zhukauskas correlation,
Nun =0.36Rei~ax Pr0 36 (Pr!P,;)°" 25 xF
where Fis the correction factor= 0.945 for NL= 6
0 25
Nuo = 0.945x0.36x(39423) 0-6 (0.7296)0- 36 ( 0 ·7296 ) .
0.7111
= 174.86
The average convection coefficient is
h =Nuo ~=174.86x 0.02551W/moc
D
= 171.56 W/m 2 °C
Total heat transfer rate,
0.026m
r
8. 73
8.74
, Heat and Mass Transfer
The air exit temperature is obtained from
T.=T.-(T.-~)exp[-
hNrcDL
Pi VNTSrLCP
= l00 0 C-{l00- 20) 0 C exp[-
l
171.56W/m2 °Cx72xrcx0.026m
]
1.204kg/m3 x10m/sx12x0.045mx1007 J/kg°C
= 31.4°c
Log mean temperature difference,
AI'im
er. -~)-CT.-T.)
Zn [(T.-~)l(T.-T.) ]
= c100-20)-(100-31.4) = s 74 .10 c
L
fn(80/68.6)
Q = (171.56W/m20 C)(72xrcx0.026mx4m)(74.1°C)
(Ans) (a)
=299 kW
(b) Aligned arrangement
With
ST= 45 mm,= 39 mm (as in the previous case)
D = 26 mm and V= 10 mis,
Sr
45mm
Vmax =--V=----xlOm/s
Sr-D
(45-26)mm
= 23.68 mis
VmaxD 23.68m/sx0.026m
Reo m a x = - - = - - - - - ,
v
15.62x10-6 m2 /s
= 39 423
-
025
0 ·63
Nuo =FxO · 27 ReD,max
Pr0· 36 (Pr/Pr..).
s
0 25
= 0.945 x0.27x(39 423)0 -63 x(0.7296) 0 -36 ( 0 ·7296 ) .
0.7111
= 180.14
Average heat-transfer coefficient,
h =Nuo ~=180.14x 0.02551W/moc
D
= 176.75 W/m 2 °C
0.026m
Forced Convection: External Flow
Exit air temperature,
T. =T. -(T. -i;)exp[
h N rcD
Pi V NT ST Cp
r
8.75
l
]
176.75W/m2 °Cx72xrcx0.026m
=100°C-(100-20)°Cexp [ - - - - - - - - - - - - - - 1.204kg/m3 x10m/sx12x0.045mx1007 J/kg°C
= 31.74°c
Log mean temperature difference,
AT. _ (100-20)-(100-31.74)
Im Zn (80 / 68.26)
= 73.97°C
Heat-transfer rate,
Q=h (NrcDL)AT;m
= (176.75 W/m 20 C)(72xrcx0.026mx4m)(73.97°C)
(Ans) (b)
=307.6 kW
(c) Pressure drop across the tube bank is determined from
For staggered orientation
With
S
45mm
26mm
_!_ = - - = 1.73 and Re0
D
= 39 423 ""4x 104,
From the graph, the friction coefficient,
f = 0.28
Also, for equilateral triangle tube arrangement, X = 1.
Pressure drop,
3 _x-'-(2_3_.6_8_m_/--'s)_
2
A p = 6 x O28 x 1x _1._18_4_k_cg_/m_
2
= 558 Pa
For in-line (aligned) orientation
With
SL 39mm
-=--=1.5, andRe0 =39423
D
26mm
(Ans) (c)
8.76
, Heat and Mass Transfer
Friction coefficient, f = 0 .3
( S -D)/(S -D)=(ST/D)-1=1.73-1=1.46
T
T
(SL/ D)-1
1.5-1
'
With
Correction factor,
X=0.8
Pressure drop,
AP= NL f
x(pV;ax)
= 6x0.3x0.8[1.1 34 ~g/m3 x(23.68m/s) 2 ]
= 478 Pa
(Ans) (c)
, MULTIPLE CHOICE QUESTIONS
For a circular cylinder in cross flow, if the fluid is a highly viscous oil (Pr > 500), the most
appropriate correlation is
(a) Zhukauskas
(b) Churchill-Bernstein
(c) Hilpert
(d) none of these
8.2 In high speed flows, the heat transfer rates are calculated by the relation:
8.1
(b) hA(T._d -To)
(a) hA(Tw -T=)
(c) hA(Tw -To)
(d) hA(Tw -T..d)
8.3 The largest local convection heat transfer coefficient for laminar flow of a fluid along a flat surface
with the same values of Re and Pr is when the surface has
(a) constant heat flux
(b) constant temperature
(c) an unheated section
(d) none of these
8.4 The value of constant C in the empirical correlation Nu = C Re112 Pr113 for constant heat flux on the
surface of the wall for laminar flow is
(a) 0.0296
(b) 0.3387
(c) 0.332
(d) 0.453
8.5 For air flow over a flat plate, velocity (V) and boundary layer thickness (<>) can be expressed
respectively, as
(a)
~ = i2:'.._.!. (2:'..)\
U=
2 <>
2
<>
8 = 4-:;;.
v Rex
If the free-stream velocity is 2 mis, and air has kinematic viscosity of 1.5 x 10-5 m2 /s and density
of 1.23 kg/m 3, the wall shear stress at x = 1 m, is
(a) 2.36x10 2 N/m2
(b) 43.6xl0-3 N/m 2
(c) 4.36x10-3 N/m2
(d) 2.18x10-3 N/m2
8.6 Air at 300 K flows over a flat plate held at 400 K. At a certain location, the experimental data on
temperature distribution are plotted below. The value of k at the film temperature is 0.03 W/m K.
The value of convection heat-transfer coefficient h (W /m 2 K) at this location is
Forced Convection: External Flow
(a) 25.0
(b) 20.0
(c) 15.0
(d) 10.0
8.7 Water flows at l 5°C past a flat plate. At a
location where the surface temperature is 45°C
and the thickness of the thermal boundary layer
is 15 mm, the local convection coefficient for
the temperature profile:
(1:'....)- 0.5 (1:'....)
T -Tw = 1.5
T= -Tw
Or
Air
2.0mm
ar I
1.5mm
dy
r
8. 77
t-.r
y=O = !ly
1.0mm
3
300K
Or
400K
Fig. 8.46
with kwater = 0.615 W/m K is
(a) 82.0 W/m 2 K
(b) 126.5 W/m 2 K
(c) 61.5 W/m 2 K
(d) 9.65 W/m 2 K
8.8 Atmospheric air at 300 K (k = 0.0263 WIm 0 C) flows over a 25-cm-diameter spherical object
with a mean surface temperature of 500 K. The Reynolds and Prandtl numbers are 5.5 x 104 and
0.707 respectively. The viscosities at 300 Kand 500 Kare 18.46x10-6 and 27.0lx10-6 N s/m 2
respectively. The appropriate correlation is
Nu= 2 + [ 0.4 Re0 -5 + 0.06 Re0 -667 ] p,.0.4 (µ= /µw )°' 25
The average convection heat-transfer coefficient is
(a) 6.9 W/m 2 °C
(b) 15.3 W/m 2 °C
(c) 25.7 W/m 2 °C
8.9
7~
For a fluid flowing past a flat plate ( 3 x I 0 < ReL < I 0
5
(d) 32.8 W/m 2 °C
the average Nusselt number is calculated as
(a) (0.037 Re0 -8 -527) Pr113
(b) (0.037 Re0· 8 -871) Pr113
(c) (0.037 Re0 -8 -702) Pr113
(d) (0.037 Re0 -8 -347) Pr113
, TRUE/FALSE
8.1
8.2
For turbulent fluid flow over a flat plate, the average Nusselt number is twice the local Nusselt number.
The integrated form of the energy equation for forced convection over a flat plate is
~
fu(T-T=)dy =a(E.-)
dx
dy
O
y=O
Flow over the tubes in a tube bank can be analysed by considering flow across a single tube and
multiplying the results by the number of tubes.
8.4 For cylinders in cross flow, the local free stream velocity equals the approach velocity of the flow.
8.5 The recovery factor in high-speed gas flow is given by r = (¾ct -T= )/( To -T=).
8.3
, FILL IN THE BLANKS
8.1
For forced flow external to surfaces, the laminar boundary layer persists to a local Reynolds number
of approximately - - ·
8.78
, Heat and Mass Transfer
The local heat-transfer coefficient for turbulent flows past a flat plate varies according to the relation
h = A .x-0·2 where A is a constant of appropriate dimensions and x is the distance measured from the
leading edge. The average Nusselt number is __ times the local Nusselt number.
8.3 In the case of flow normal to tube banks, there are two general types of orientation - in-line and
staggered geometry. For minimum pressure drop, one would use __ arrangement, and for maximum
heat transfer __ arrangement.
8.4 The separation point, given in terms of the angle 0 from the front stagnation point for a cylinder
in cross flow is about __ for laminar flow and about __ for turbulent flow.
8.2
, EXERCISES
8.1
Determine the thickness of hydrodynamic boundary layer, 8 and the thermal boundary layer thickness
BT at the trailing edge of a thin flat plate at a surface temperature of 15°C in the following two cases:
(a) Atmospheric air at 35°C flows parallel to the flat plate with a free-stream velocity of 10 mis
and length of the plate is such that the flow is just critical (Rec,= 5 x 10 5) at its trailing edge.
(b) Water at 35°C flows parallel to the same flat plate such that the flow is just critical at its trailing edge.
Thermophysical properties for air and water at the mean film temperature,
1
T,r = -(T.s + T= ) = 25 °C are
2
Air:
v = 15.62x10-6 m2 ls,
Pr= 0.7296
Water:
v = 0.894x10--{j m2 ls,
Pr= 6.14
[(a) 5.52 mm, (b) 3.0 mm]
Determine the drag force on a plate 2-m square for the flow of air at 1 atm and 350 K with a free-stream
velocity of 4 mis. Use the approximate expression for average drag coefficient, Cr,x = 1.37 Re~ 112 •
Take p = 0.998kg/m3 and µ = 20.75 x10-6 kglm s
[70.5 x 103 N]
8.3 Air at a temperature of 30°C flows past a flat plate at a velocity of 1.8 mis. The flat surface has a
sharp leading edge and its total length equals 750mm. Calculate
(a) The average skin friction coefficient,
(b) The average shear stress, and
(c) The ratio of the average shear stress to the shear stress at the trailing edge.
Properties of air at 30°C are
p = 1.165kglm3
µ = 6.717 x 10-2 kglhm
v = 16 x 10--{j m 2ls.
[2.0]
8.4 In a certain process, castor oil at 30°C flows past a flat plate. The velocity of oil is 0.08 mis. The
length of the plate is 5 m. The plate is heated uniformly and maintained at 90°C. Calculate the
following:
(a) Hydrodynamic and thermal boundary layer thickness at the trailing edge of plate
(b) Total drag force per unit width on one side of the plate
Use the following correlation:
8.2
Take properties as
p = 956.8 kglm3
k = 0.213 Wlm K
a= 7.2 x 10-s m 2 Is
v = 0.65 x lQ-4 m2 Is
(c) Heat loss per unit width from one side of plate
[6434 W]
Forced Convection: External Flow
8.5
r
8. 79
For each of the following flowing fluids (values ofproperties are tabulated) over a surface, calculate
for a pressure of l atm, a free stream velocity of l m/s, a temperature of 25°C, a characteristic
length of l m and an average convection heat transfer coefficient of 80 W/m 2°C, the Nusselt number,
Reynolds number, Prandtl number, and Colburn j-factor.
Property table (P = l atm, T = 25°C):
k (Wlm °C)
Fluid
a. (111 2 Is)
V (111 2 Is)
Air
0.02551
15.62
Water
Glycerin
0.607
8.94 X 10-7
1.457 X 10-7
0.286
7.878 X lQ-4
15.62 X lQ-6
Mercury
8.51533
1.133 X lQ-7
4.514 X lQ-6
X 10-6
21.41 X lQ-6
8.6 Assuming an average fluid temperature of 25°C and l atm pressure, sketch the vanation of
hydrodynamic and thermal boundary layer thicknesses (6 and 6.J with distance from the leading
edge of a flat plate for the laminar parallel flow (Rex < 5 x 10 5 ) of each of the following fluids:
(a) Air (b) Water (c) Glycerin (d) Mercury
Fluid
Air
Water
Glycerin
Mercury
Pr
0.7296
6.14
8392
0.0251
8.94 X 10-7
7.878 X lQ-4
1.133 X lQ- 7
v (m 2/s)
15.62
X 10-6
Calculate the values of 6 and 61h at a distances of 50 mm from the leading edge for the four fluids
flowing with a velocity of l m/s.
Fluid
Rex
o(mm)
8,h(mm)
Air
3201
4.42
4.91
Water
5.593 X 10 4
1.06
0.53
Glycerin
63.5
31.4
1.54
Mercury
4.413 XlQ 5
0.38
1.29
8.7 A rectangular plate is 120-cm long in the direction of flow and 200-cm wide. The plate is maintained
at 80°C when placed in nitrogen that has a velocity of 2.5 m/s and a temperature of 0°C. Determine
(a) the average heat transfer coefficient, and (b) the total heat transfer from the plate. The properties
of nitrogen at 40°C are
CP = 1.04 kJ/kg K
p = 1.142 kg/m 3
v = 15.63 x lQ-6 m 2/s
and
k = 0.0262 W/m K
[1087 W]
8.8 Air at l 5°C and 1 atm flows with a velocity of 8 m/s past a flat plate which is maintained at a uniform
temperature of l 15°C. (a) Calculate the thickness of the boundary layer 60cm from the leading edge,
and (b) estimate the position of the point of transition to turbulent flow. (c) Also calculate the total
drag force on one side of the plate, per metre width of plate, over the first 60 cm. (d) Calculate the
heat-transfer coefficient at 60 cm from the leading edge of the plate. (e) Hence calculate the heat
transfer rate from one side of the plate to the air, per metre width of the plate over the first 60 cm.
8.80
, Heat and Mass Transfer
Following properties of air at the mean film temperature 65 °C may be used:
p = 1.044kglm3
v = 19.73 x 10-6 m2ls
Pr= 0.702
k = 0.0291 Wlm K
[847 W]
8.9 Air at atmospheric pressure and 40°C flows with a velocity of u= = 5 mis over a 2 m long flat
plate whose surface is kept at a uniform temperature of 120°C. Determine the average heat transfer
coefficient over the 2 m length of the plate. Also, find the rate of heat transfer between the plate
and the air per metre width of the plate.
Properties of air at 1 atm and 80°C:
v =2.107x10-5 m2 ls
8.10
k= 0.03025 Wlm K
Pr= 0.6965
[981.28 W]
Air at 20°C and atmosphere pressure is flowing over a flat plate which is 50cm-long and 20cm wide
and is maintained at 100°C. Find the heat lost from the plate if the air is flowing parallel to 50-cm side
with 2 mis. What will be the percentage increase in heat transfer if the flow of air is parallel to 20-cm
side?
Properties of air at the mean film temperature of 60 °C are
v = 18.97 x 10---Q m2ls
µ = 20.1 x 10---Q N slm2
CP = 1.005 kJ/kg K
k= 0.029 Wlm K
The relevant correlations are
NuL = 0.664(Re)112 (Pr) 113 for laminar flow
NuL = 0.0366(Re)D· 8 (Pr) 113 for turbulent flow
8.11
Helium at 20°C flows at 10 mis over a 2-m-long horizontal flat plate which is maintained at 80°C.
(a) Find the value of the local heat transfer coefficient at a distance of 30 cm from the leading
edge. (b) Also determine the rate of heat loss from the entire plate per unit width, and (c) the drag
force exerted on the plate.
Properties of air at the film temperature of 50°C are as follows:
k= 0.1602 Wlm K
8.12
8.14
v = 1.389 x 10-4 m2ls
Pr= 0.68
[0.0528 N]
Consider a flat plate, 1-m wide and 3-m long, maintained at 85°C. Atmospheric air at 5°C flows
past it at 10 mis. Assuming critical Reynolds number of 4 x 105, calculate (a) the location of the
maximum heat loss point along the plate, and (b) the total heat loss from one side of the plate.
Neglect heat transfer from first 5-cm length of the plate.
Properties of air at 45 °C are
k = 0.02699 W/m °C
8.13
[58%]
v = 17.5x10-6 m2 /s
Pr=0.7241
[5606 W]
Nitrogen gas at 30°C temperature and 83.4 kPa pressure flows parallel to and over one side of a
flat plate (0.5-m wide and 3-m long) with a velocity of 14.5 mis. Calculate (a) the drag force, and
(b) the rate of heat transfer to the air if the plate is maintained at a temperature of 70°C.
Properties of atmospheric air at the mean film temperature of 50°C:
v= 17.74x 10-6 m2ls
p = 1.0564 kglm 3
k= 0.02746 Wlm K
Pr= 0.7114
[1.58 kW]
Water flows over a flat plate measuring 1 m x lm with a velocity of 2 mis. The plate is at a uniform
temperature of 90°C and the water temperature is 10°C. Estimate the length of a plate over which the
flow is laminar and the rate of heat transfer from the entire plate. The properties of water at 50°C are
Forced Convection: External Flow
8.15
8.16
8.17
p = 988.1 kglm 3
v = 0.556 x 10-6 m 2ls
Pr= 3.54
k= 0.648 Wlm K.
r
8.81
[445 kW]
The wing of an aircraft has a polished chromium skin. At a 3000-m altitude, it receives 650 WI m2
by solar radiation. Assuming that the interior surface of the wing's skin is well insulated and the
wing has a chord of 6-m length, estimate the equilibrium temperature of the wing at a flight speed
of 540 kph
[-1°C].
Water flows at a free-stream velocity of 3 mis over a smooth plate which is 2-m long. The bulk
water temperature is 20°C and the plate surface temperature is uniform at 80 °C. Determine (a)
the heat-transfer coefficient at the plate, (b) the drag force, and (c) the rate of heat loss per unit
width.
[926 kW]
A flat plate 1-m-wide and 1-m-long is placed in a wind tunnel. The temperature and velocity of
free stream air are 10°C and 80 mis respectively. The flow over the whole length of the plate is
made turbulent with the help of a turbulising grid placed upstream of the plate. Determine the
thickness of the boundary layer at the trailing edge of the plate. Also calculate the mean value of
the heat-transfer coefficient from the surface of the plate.
The following properties of air at l 0°C may be used:
k= 0.02439 Wlm K
Pr= 0.7336
v= 14.26 x10-6m 2 ls
[204 Wlm 2K]
8.18 A flat plate of 30-cm x 30-cm at 90°C is placed in an air stream of 60 mis. The temperature of air
is 0°C. If the turbulent flow exists over the entire plate, calculate the heat transfer coefficient and
the rate of heat transfer from the upper surface of the plate. Take the following properties of air:
k = 2.8 X 10-2 Wlm K
V = 17.45 X 10-6 m 2ls
Pr= 0.698
[1605 W]
8.19 Air at 1 atm and 25°C flows with a velocity of 10 mis in parallel flow over an isothermal flat
plate, 1 m long. (a) Sketch the variation of the local heat transfer coefficient, h,. (x) with distance
from the leading edge, x along the plate corresponding to the critical Reynolds number ~f 5 x 10 5
(b) Also calculate the average convection coefficient for the entire plate of length L, ft, for the
flow conditions specified above. The following properties of air at 1 atm and 25°C may be used:
k= 25.51 x 10-3 Wlm °C
v = 15.62 x lQ-6 m 2ls
Pr= 0.7296
[17.51 W/m 20C]
8.20 Air flows over a heated flat plate at a velocity of 50 mis. The local skin friction coefficient at a
point on the plate is 0.004. Estimate the local heat transfer coefficient at this point. The following
property data for air are given:
Density= 0.88 kglm 3
Viscosity = 2.286 x 10-s kglm s
Specific heat= 1.010 kJ/kg K
Thermal conductivity= 0.0335 Wlm K
Use the following relation: St Pr213 = Crf2
[113.9 W/m 2K]
8.21
The leading edge of an aircraft wing travelling at a speed of Mach 1 has a de-icer strip that is 6 cm
in length from the stagnation point (along the top and along the bottom) and extends the length
of the wing. How much energy is required at 10.5km altitude to keep the surface of the de-icer
strip at 340 K when the free-stream temperature is 220 K and the pressure is 24.5 kPa.
[1353 W]
8.82
8.22
, Heat and Mass Transfer
Air at 30°C flows over a long heated cylinder (5-cm diameter) whose surface is maintained at
70°C. The direction of the air flow is at right angles to the axis of the cylinder. Using Hilpert's
correlations, calculate the average heat transfer coefficient if the velocity is (a) 2 mis, (b) 20 mis.
Properties of air at 50°C are
k= 0.02735 W/m °C
Pr= 0.7228
v = 1.798x10-5 m2 / s
[(a) 19.55 W/m 2°C (b) 87.55 W/m 20C]
8.23
A long 80-mm-diameter steam pipe whose outer surface temperature is 90°C passes through some
open area which is not protected against the winds. Calculate the rate of heat transfer from the
pipe per metre length when the air is at l atm pressure and 10°C and the wind is blowing across
the pipe at a speed of 60 kmph.
Properties of air at 50°C are
v = 18.2 x 10-6 m2 /s
k= 0.028 W/m K
Pr=0.7038
[1394 W]
8.24 A lightly clad jogger can be modelled as a cylinder 0.35 m in diameter and 1.8 m tall with a
surface temperature of 37°C. For jogging at 8km/h on a day with atmospheric air at 23°C, estimate
the rate of heat loss by convection using the following correlation and air properties:
0.62Rell2 Pr113 [ ( Re
Nu=0.3+------ 1+
[l+(0.4!Pr)2t3f 4
282000
)112]
Properties of air at 30°C are
k= 26.75 xl0-3 W/m °C
8.25
v = 16x10-6 m2 /s
Pr= 0.701
[320 W]
Engine oil at 20°C flows with a velocity of 10 mis across a 25-mm-diameter tube which is held
at a uniform surface temperature of 100°C. Calculate the mean heat transfer coefficient and the
heat transfer rate between the tube surface and the oil per unit tube length.
The following properties of engine oil at the film temperature of 60°C may be used:
k=0.140 W/m °C
Pr= 1075
v=85.6x10-6m2 /s
[12.62 kW per metre length]
8.26
Water at 20°C flows normal to the axis of a circular tube with a velocity of 1.5 mis. The diameter
of the tube is 25 mm. Calculate the average heat transfer coefficient if the tube surface is maintained
at a uniform temperature of 80°C. Also, estimate the heat transfer rate per unit length of the tube.
Use the following correlation:
Nu= (0.4Re 0 -5 +0.06Re213 ) PrOA (~:
r
4
Properties of water at the free-stream temperature of 20°C:
k = 0.598 W /mK
p = 998.0 kg/ m3
µ = 1.002 X lQ-3 kg/ m S
CP = 4.182 kJ /kg K
µw@ 80°C = 0.355 X 10-3 kg/ m S
8.27
[45950 W""46 kW]
Water at 20 °C flows at right angles across a heated pipe 5-cm OD with a velocity of l m/s.
Calculate the average surface temperature of the pipe if a uniform heat flux of 75 kW /m 2 is supplied
at the outer surface.
[36.1°C]
Forced Convection: External Flow
8.28
8.30
8.31
8.32
8.83
Steam at 1 atm and 100°C is flowing across a 50-mm-OD pipe at a velocity of 7 mis. Estimate the
heat transfer rate per metre length if the pipe is maintained at 200°C. Use the Churchill-Bernstein
correlation.
Properties of steam at 150 °C are
k= 0.02861 Wlm °C
8.29
r
v = 28.06 x 10-6 m2 Is
Pr=0.9712
[617 W]
Engine oil at 300 K flows over a heated cylinder, 10 cm in diameter, with a velocity of 1 mis.
Calculate the heat transfer rate if the cylinder surface is maintained at 460 K.
Properties of air at the mean film temperature of 380 K are
k= 0.136 Wlm K
Pr= 233
p = 836.0 kglm 3
v = 16.9 x lQ-<i m 2ls
[21.43 kW]
Develop an expression for the ratio of the rate of heat transfer to water at 37°C from a thin flat
strip of width rrD and length L at zero angle of attack and a tube of the same length and diameter
D in cross flow with its axis normal to the air stream in the Reynolds number range between
40 000 and 400 000. Assume both surfaces to be at 92°C.
[15.656 Re--0.3os1
Air at 30°C flows over a long heated cylinder (5-cm-dia) whose surface is maintained at 70°C. The
direction of air flow is right angles to the axis of the cylinder. Calculate the average heat transfer
coefficient if the velocity is 2 mis. What will be the heat loss rate per unit length.
Properties of air at 50°C are
k= 0.0283 Wlm K
v = 17.95x10-<i m2 ls
Pr= 0.698
[137.85 W]
A spherical storage tank of 5-m-diameter contains LNG (liquefied natural gas). The temperature of
its outer surface is 10°C and the ambient air temperature is 20°C. The surface emissivity is 0.4.
Calculate the heat transfer rate to the gas when the wind velocity is 8 mis.
For flow over spheres:
Nu0 = 2+[ 0.4Re'1j1 +0.06Reif3]P,.0.4 ( ~:
r
4
Take the following properties of air:
µ= 18.17 X IQ-6 kglm S
µs = 17.69 X lQ-<i kg/m S
p = 1.2043 kglm3
CP = 1.0061 kJ/kg K
k = 25.64xI0-3 W/m K
8.33
[8090 W or 8.09 kW]
Copper spheres of IO-mm-diameter are heated to 85°C and then subjected to an air stream at
25°C with a free-stream velocity of 10 mis. Estimate the time required to cool the spheres to a
temperature of 35°C.
Properties:
Copper:
p = 8933kg/m3
CP = 0.387kJ/kg °C
Air: At T~ = 25°C:
k=0.02551 W/m°C
v = 15.62 x lQ-<i m2 /s
µ = 18.49 X IQ-6 kg/m S
Pr= 0.7296
At~= 60°C:
µw = 20.08 X I0-6 kg/m s
[85.2 s]
8.84
, Heat and Mass Transfer
8.34 Water at 25°C is heated by passing it through a tube bank in staggered tube configuration. The
outside diameter of the tubes is 25-mm and their average surface temperature is 100°C. The freestream water velocity before entering the bank of tubes is 0.3 mis. The transverse and longitudinal
pitches are 50-mm and 37.5 mm, respectively. Determine (a) the average convection coefficient,
(b) the number of tube rows in the direction of flow required to raise the temperature of water by
50°C, and (c) the pressure drop across the tube bank.
Properties of water at the bulk mean water temperature of 50°C are
p = 988. l kg/m3
cp = 4.181 kJ/kg c
k= 0.644 W/m °C
µ = 0.547x10-3 kg/m s
Pr= 3.55
Pr@1oo•c = 1.75
0
P@ 2 Soc = 997 kg/m3
[(a) 8215 W/m 2°C (b) 106.5::::107 (c) 5.71xl03 N/m 2 or 5.71xPa]
8.35
A cross-flow steam condenser comprises a square, aligned array of 400 tubes, 2-m-long, with
tube outside diameter of 2.0 cm. The longitudinal and transverse spacings are 6.0 cm and 3.0 cm,
respectively. Saturated steam at 120°C (hrg = 2202.6 kJ/kg) condensing inside the tubes maintains
the tube outer surface temperature at 120°C. The upstream velocity and temperature of the air at
1 atm entering the array are 4 mis and 30°C, respectively. Calculate (a) the total heat transfer
rate for the array, (b) the rate of condensation of steam, (c) the pressure drop across the array
and (d) the fan power requirement. Use the Zhukauskas correlation and the following properties
of air:
P@ 3o•c = 1.164 kg/m3
Pr@ 12o•c = 0.7073
p=l.059kg/m3
cp = 1.001 kJ/kg c
k= 0.02808 W/m °C
Pr= 0.7202,
0
v = 18.96 x 10-{j m2 /s
[(a) 348 kW (b) 0.158 kg/s (c) 686 Pa (d) 3620 W or 3.62 kW]
, ANSWER KEY
Multiple Choice Questions
8.1 (b)
8.5 (c)
8.9 (a)
8.2 (d)
8.6 (b)
8.3 (a)
8.7 (c)
8.4 (d)
8.8 (b)
8.2 F
8.3 F
8.4 F
8.2 1.25
8.3 in-line, staggered
8.4 80°, 140°
True/False
8.1 F
8.5 T
Fill in the Blanks
8.1 5xl0 5
FORCED CONVECTION:
INTERNAL FLOW
9
Concept Review
INTRODUCTION
• 9.1
•
In many heating and cooling applications, a fluid (liquid or gas) is forced to flow by external means
(pump or fan) through a closed conduit (pipe or duct) and is completely confined by the inner surfaces
of the passage. The analysis of such internal flow systems requires determination of friction factor and
convective heat-transfer coefficient so that the pressure drop and the rate of heat-transfer can be evaluated.
This, in tum, enables us to calculate the required pumping power and the tube length.
PARALLEL FLOW BETWEEN TWO PLATES
• 9.2
Figure 9.1 shows schematic representation of parallel
flow between two plates. One of the plates is moving
and the other is stationary. This is known as Couette
flow and it occurs, for example, in the narrow gap
between a lubricated bearing and its journal.
For Couette flow, the plate at y = 0 is stationary
and the plate at y = L is moving with a velocity U.
The temperature distribution is given by the following
expression:
T = T,
o
C
To 7----_-_-_-_-_-_-_-_-_-_-_-_-_-_----1----_-_-_-_ I
C
Stationary plate
Fig. 9.1
+µu2
[2:'..-(2:'L..)2]+ (r,L -T,o)2:'L.
2k L
•
Moving plate
Couette flow
(9.1)
AVERAGE VELOCITY AND TEMPERATURE
• 9.3
The average fluid velocity in a circular pipe of radius R for incompressible flow can be expressed as
.
•
2 R
V = ...!!!_=-Ju(r)rdr
p4, R2 o
(9.2)
where m is the mass flow rate, p is the fluid density, u(r) is the velocity profile and Ac is the crosssectional area.
The average or bulk temperature of the fluid with constant density and specific heat flowing in a
circular pipe of radius R can be expressed as
9.2 , Heat and Mass Transfer
2
R
Ti,= V R2 Ju(r)T(r)r dr
(9.3)
0
The bulk mean fluid temperature at which the fluid properties are usually evaluated in internal flows
is the arithmetic mean of the bulk (average) temperatures of the fluid at inlet and exit. That is
(9.4)
LAMINAR AND TURBULENT FLOW
• 9.4
Flow inside a tube or duct can be laminar or turbulent depending on the flow conditions.
The Reynolds number for internal flow is defined as
IRe=7=~1
•
(9.5)
where Vis the mean flow velocity, D is the pipe diameter and v(= µ/p) is the kinematic viscosity of the fluid.
For a circular tube,
m
V=----,
p(1t/4)D2
and
__________
4m _ 4m _ 4¥-
Re -
1tDµ
1tDvp
1tDv
(9.6)
where ¥- is the volume flow rate of the fluid.
In the case of non-circular pipes or ducts, the Reynolds number, friction factor and the Nusselt's
number are evaluated based on the hydraulic diameter, Dh which is defined as
(9.7)
where Ac and P are the cross-sectional area and perimeter respectively.
The flow is generally said to be laminar if Re< 2300, fully turbulent for Re> 10 000, and transitional
in between.
• 9.5
THE ENTRANCE REGION
•
The region of the flow from the pipe inlet to the point at which the boundary layer merges at the
centreline is called the hydrodynamic entrance region. Beyond this region, the velocity profile is fully
developed and remains unchanged. The length of the region in which the velocity profile is developing
is the hydrodynamic entry length, Lh.
In laminar flow, the velocity profile is parabolic but flatter in the case of turbulent flow.
The length from the pipe entrance required to achieve a fully developed temperature profile is called
the thermal entrance length, Lr The region beyond this is called the thermally fully developed region. The
dimensionless temperature profile defined as (Tw - T)/(Tw - Tb) [where Tw is the tube wall temperature]
as well as the velocity profile remain unchanged in a fully developed flow.
Forced Convection: Internal Flow
r
9 ,3
Hydrodynamic entry length, Lh for laminar flow is
(9.8)
Thermal entry length, L 1 for laminar flow is
14,Iam"' 0.05RePrD "'Pr Lh,lam I
(9.9)
The entry length in turbulent flow is nearly independent of Reynolds number and as a first approximation.
14,turb "'4.,turb "'10
DI
(9.10)
The Nusselt numbers and thus the convection coefficients are much higher in the entrance length and
reach a constant value at a distance L"' 10 D.
CONSTANT HEAT FLUX AND CONSTANT WALL
TEMPERATURE CONDITIONS
• 9.6
•
The rate of heat transfer to or from a fluid flowing through a tube is given by
(9 .11)
The two boundary conditions commonly encountered in engineering practice are the following:
9.6.1
Constant Surface Heat Flux (q 5 = const.)
T
L
0
·
·
·
.
0
qs = constant
Fluid
i i i i i
·············1 ··
-
-Tbe
. . . . . .J .
t _ _ _ _.,
,___ _ _ _ _t L _
Fig. 9.2 Axial temperature variation for heat transfer in a circular tube under constant wall heat flux condition
9 .4 , Heat and Mass Transfer
This surface condition is approximated in applications such as electrical resistance heating, nuclear
heating or radiant heating. Tube wall heat flux can be expressed as
(9.12)
where hx is the local heat transfer coefficient, Ts the tube wall (surface) temperature and Tb the bulk
temperature of the fluid at the specific location. When hx = h = const, the surface temperature Ts, must
change because during heating or cooling the fluid temperature in internal flow must change. Hence, if
qs = const., it follows that (Ts - Tb)= constant.
The heat transfer rate can then be expressed as
(9 .13)
The bulk temperature of the fluid at the outlet is
qSAS
T..be =T..b.1 +-• C
m
(9.14)
P
where As is the surface area equal to perimeter, P which is constant, times the tube length which increases
as the fluid moves along. Hence, with qs P m and CP constant, the fluid temperature will increase linearly
in the flow direction.
In the fully developed region, the surface temperature also will increase linearly in the flow direction
because for qs = const, (Ts - Tb) has to be constant during the fluid flow.
9.6.2
Constant Tube Surface Temperature (T5 = const.)
T
r.
T
1
L
0
Fluid
0
-.)
~_-_-__-_ --------------------- r
__ ____________________________ L
L--------
Fig. 9.3
Axial temperature variation for heat transfer in a circular tube under constant surface temperature condition
Forced Convection: Internal Flow
r
9.5
This condition is realised when the surface is subjected to phase change like evaporation or condensation
so that the surface is maintained at a constant temperature. The heat-transfer rate is
(9.15)
where h is the average heat transfer coefficient, A 8 is the surface area equal to rcDL for a circular pipe
of length L and AT1m is the logarithmic mean temperature difference.
With
(9.16)
the fluid outlet temperature is
(9.17)
LAMINAR FLOW IN A TUBE
• 9.7
•
The parabolic velocity profile in a fully developed laminar flow through a circular tube of radius
R (Re 0 < 2300) is given by
(9.18)
where V is the mean fluid velocity.
The maximum velocity occurs at the centreline (r = 0) and equals
lumax = 2v1
(9.19)
The pumping power requirement is determined from
(9.20)
where ¥- is the volumetric flow rate and AP is the pressure drop along the tube length. Pressure drop
in a pipe flow is given by
(9 .21)
where f is the Darcy friction factor.
~
~
(Re< 2300)
(9.22)
+ Fully Developed Laminar Flow Through A Circular Tube
!Constant heat flux: Nu= 4.364!
(9.23)
!Constant surfacetemperature: Nu= 3.658!
(9.24)
9.6 , Heat and Mass Transfer
+ Developing Laminar Flow in the Entrance Region
Constant Surface Temperature
average Nusselt number,
For hydrodynamically developed and thermally developing flow, the
_ 3 66
Nuo -
·
+
0.0668(D/ L)Re0 Pr
(9.25)
213
1+0.04[(D/ L)Re0 Pr]
For both hydrodynamically and thermally developing flow with large difference between the surface
and fluid temperatures,
Nu = 1.86 ( -Re0- Pr)
D
LID
113
0 14
( ....!:!:_ ) ·
(9.26)
µw
where
+ µw is evaluated at the tube wall temperature while all other fluid properties are at the bulk mean
temperature, Tbm =(Thi+ The)/2.
+ 0.48<Pr< 16 700
+ 0.0044 < µ/µw < 9.75
113
0 14
+ (RePr) (....!:!:_) - > 2
LID
µw
Constant Heat Flux
Local Nusselt number,
Re0 Pr
Nu0 = 1.302 ( - -)
LID
1/3
-0.5
[0.00005<L*< 0.0015]
(9.27)
where (LID) I (Re0 Pr) = L • is known as the dimensionless axial distance.
!Nu0 = 4.364+8.68 (1000 L*)--0· 506 exp(-41L*)!
for
L* > 0.0015
(9.28)
+ Colburn analogy can also be applied to evaluate the convective heat-transfer coefficient in laminar
flow through tubes by the relation
St Pr213 = j78, where f = 64/Re
• 9.8
TURBULENT FLOW HEAT TRANSFER IN CIRCULAR TUBES
•
For fully developed flow for a smooth circular pipe (or duct) the Dittus-Boelter equation can be used
subject to the specified restrictions.
hD
Nu0 = - = 0.023(Re0 ) 415 (Pr}ll
k
+ The thermophysical fluid properties must be evaluated at the bulk mean temperature, Tbm·
+ 0.7 <Pr< 100
+ The exponent n = 0.4 for the fluid getting heated, and n = 0.3 for the fluid getting cooled
(9.29)
Forced Convection: Internal Flow
r
9. 7
+ LID> 60
+ The correlation is valid for both cases namely, T = const and q = const.
+ Reynolds number, Re0 < 1.2 x 10 5 and> 2300.
8
9.8.1
8
Gnielinski Correlation
It is the more recent and recommended one which is valid for both smooth and rough tubes, for both
constant wall temperature and constant heat flux conditions and for a wide range of 0.5 <Pr< 2000 and
3000 < Re0 < 5 x 10 6 • Fluid properties are at the bulk mean temperature. It is expressed as
=_(_f_I8_)_(R~e=0 =-_10_0_0_)P_r_
Nu
D
1+12.7.Jr78(Pr213 -1)
(9.30)
where the Darcy friction factor, f = (0.79 Zn Re0 - 1.64)-2 for smooth tubes.
The value off for rough surfaces can be read from the Moody diagram.
9.8.2
Sieder-Tate Equation
For 0.7 ~Pr~ 17 600 and Re~ 10 000, this is used when the variation in fluid properties is large due to
a large temperature difference. All properties except µw which is evaluated at the tube wall temperature
are found at the bulk mean temperature. The correlation is
(9 .31)
LIQUID-METAL HEAT TRANSFER
• 9.9
•
For fluids like liquid metals which have low Prandtl numbers (0.02 to 0.003), the following expressions
for fully developed, turbulent flow are recommended:
Constant Wall Temperature
!Nu= 4.8+ 0.0156 Re0- 85 p,.0. 93 1
(9.32)
!Nu= 6.3+0.0167 Re0 -85 p,.0. 93 I
(9.33)
Constant Heat Flux
The above correlations are valid for 10 4 <Re< 10 6 and for 0.004 <Pr< 0.1. Fluid properties are to
1
be evaluated at the bulk mean temperature, ¾m = -( Thi +The).
2
The other correlations used for fully developed turbulent flow of liquid metals are the following:
Constant Wall Temperature
[Pe= Re Pr> JOO]
!Nu=5.0+0.025 (RePr) 0 ·8 !
(9.34)
Constant Heat Flux
!Nu= 4.82 +0.0185(RePr) 0 -827 !
3.6x103 < Re< 9.05x10 5
and
102 <Pe<104
(9.35)
9 .8 , Heat and Mass Transfer
Solved Examples
(A) COUETTE FLOW: FLOW BETWEEN PARALLEL PLATES
Example 9.1 ~ A heavy lubricating oil (k = 0.145 W/m K, µ = 0.8 N s/m 2) flows in a journal bearing can
be idealised as parallel flow between two large isothermal plates, both maintained at 20°C.
The top plate moves at a constant velocity of IO m/s while the bottom plate is stationary. The oil gap
is 2 mm. Calculate (a) the maximum oil temperature, (b) the shear stress in the oil, (c) the heat-transfer
rate to each plate per unit area, and (d) the mechanical power wasted by the viscous dissipation in the
oil per unit area.
Known: Oil properties, journal and bearing temperatures. Journal speed. Parallel flow of oil between
two plates.
Find: (a) Maximum oil temperature (b) Shear stress in the fluid, and (c) Heat flux fom the oil to the
plates (d) Power dissipated per unit area.
Schematic:
y
/
Moving plate
.. .L__ _______+-_ _____, ______________________ _1 - - - V= IOm/s
L
mmmmm~--===iilttttimmmimmmm
=:=:=:::=:::=:::=::
_::\::::: u(y) :=: :=:=: :=:=:=:=:=:=:=:=:=:=:=:=:=:=
--------------------------------------------------------------------------------------------------------------------------------
-- ------ - ------ ---------------- ------------ --- --- --- ---- ------------------------------------------------------------------------------------------------------------------------------------
····1- ---- -- ---::::::::::::: ---- ----- ----- -- ---- --,
Q.
~ Stationary plate
Fig. 9.4
Assumptions: (1) Steady-state conditions. (2) Oil is incompressible with constant properties. (3) Clearance
is very small so that Couette flow approximation is valid. (4) Body forces like gravity are negligible.
Analysis: Continuity equation
du+dv=O
dX dy
du =O
dX
since u = fly) and v = 0 as there is parallel flow between two plates (no normal component of velocity).
Velocity profile is uniform since x-component of the velocity is constant in the flow direction.
Momentum equation
O ~ =µ
du -d
du ,-v
p[u ~
- y ::::}
2
X
du C
.
.
Integratmg
twice, - = 1
dy
y
dy 2
d
d2u
-=0
dy2
Forced Convection: Internal Flow
Boundary conditions:
u(y=O)=O
and
u (y = 0) = 0 + C2
u(y=L)= V
~
C2 = 0
and
The plates are isothermal and there is no change in the x-direction. Hence, T= T(y).
Energy equation
d 2T
or
µ V2
-=--dy2
k £2
Integrating twice,
2
dT =-.!::(V) y+C
3
dy
k L
Boundary conditions
T(y = 0) = T0
and
T(y = L) = T0
(v)
2
T, = -µ- o
2k L
·L2 +C L+T,
3
o
µV2 [y y2]
T(y)=T, + - - - o 2k L L2
r
9. 9
9.10 ,
Heat and Mass Transfer
Maximum temperature will occur when the temperature gradient, d T = 0.
dy
Hence, differentiating the expression for T(y) with respect to y and equating the resulting derivative
to zero, we have
dT = O= O+ µV [_!_- 2y]
2
dy
2k
L
~
y=L/2
£2
µV 2 [L/2
L2/4]
µV 2
T..max = T,@y=L/2 =T,0 +-- =T,0 +-2k -L- - £2
8k
1~1
= 200C+ 0.8Ns/m2 x10 2 m 2/s 2
8x0.145W /mK
lNm/s
(Ans) (a)
=89°C
Shear stress,
_ dul _
V _ 0.8Ns lOm/s
'to-µ- µ----X--dy y=O
L
m2
0.002m
= 4000 N/m2 or 4 kN / m2
(Ans) (b)
Heat transfer rates per unit area at the plates are
QI
A y=O
= -k
dTI
dy y=O
=-k µV2 (l-0)=-µV2
2kL
2L
= -(0.8 N s/ m 2)(10 m/s) 2 / (2x0.002m)
=-20000W/m2
QI
or
-20kW/m2
dTI
µV2
µV2
2
=-k=-k-(l-2)=+-=+20kW/m
A y=L
dy y=L
2kL
2L
(Ans) (c)
(Ans) (c)
Thus, the heat fluxes at the two plates are equal in magnitude but opposite in sign.
Power wasted by viscous dissipation in oil, p = F V = 't AV
Per unit area,
kN
f,J='t V=4-xlm2 xl0m/s
m2
=40 kW
(Ans) (d)
Forced Convection: Internal Flow
r
9. I I
(B) CIRCULAR TUBES: LAMINAR FLOW
•
•
Example 9.2 ~ Experimental data were obtained during the performance studies of convection heat transfer
for a fluid flowing in a 20-mm-diameter tube. The data were curve fitted to get the following
expressions for the velocity and temperature profiles at a particular axial location.
u(r)=2.4ll-(100r)2J
(mis)
T(r) = 70 - 213(0.1875 + 6.25 x I 06 r4 - 2.5 x I Ol r2]( 0 C)
and
where the radius r is in metre.
Determine (a) the mean velocity, and (b) the bulk (or mixing up) temperature of the fluid at this axial position.
Known: Velocity and temperature profiles at a specific location for a fluid flowing through a circular tube.
Find: (a) Mean fluid velocity. (b) Bulk (or mean) fluid temperature, Tb {0 C).
Schematic:
r0 = 1.0mm or 0.01 m
- F l u i d flow
u(r) and T(r) specified
Fig. 9.S
Assumptions: (1) Incompressible fluid flow {p = const).
Analysis: The velocity distribution (radial) is given by
u(r) = 2.4ll-(100 r}2 J{mis)
Mass flow rate of the fluid flowing through the circular tube of radius r0 = 0.01 m is given by
,;,
,;,
f
m= pu(r)Ac (r) = p Ju(r)(21trdr)
0
0
,.
= 21tp J{ 2.4 r-2.4r3 x 104 dr}
0
= 21tp [2.4 r2 /2 - 2.4 X 104 r4 /4 ]~o=O.Olm
= 21tp[l.2 X 0.012 -0.6 X 104 X 0.014 ]
= 21tp x 0.6x0 .0l2 = 377 x 1Q-6p kg/s
Mean (average) velocity,
m
m
V=u - = - av pAc
p1tr;
[3 77 X 10-6 p ]{kg/s)
= ------~-2 = 1.20 m/s
p (kg/m3 )x1tx(O.Olm)
(Ans) (a)
9 .12 ,
Heat and Mass Transfer
(b) Temperature variation (radial) is given by
T(r) = 70-213[ 0.1875 + 6.25 X 10-6 r4 -2.5 X 10 3 r3]
{°C)
Bulk (or mean) temperature is determined to be
~
Ya
fpu(r)Cp T(r)d Ac fu(r)T(r) 2rcrdr
~=_o_______ =_o_______
(p A V)cp
C
377 X 10-6
2 1C
,;,
= - - - [2.4- 2.4 x 10 4 r 2 ] [ 70- 213{0.1875 + 6.25 x 106 r4 -2.5 x 10 3 r 2 }] rdr
3.77 X 10-4 0
f
21C X 10 4 ,;,
= ---f[72 .15+556.5 x 10 3 r 2 -1.5975 x 1010 r 4 + 3.195 x 10 13 r6] rd r
3.77 0
21C X 10 4
0.01
= ---(72.15 r 2 /2+556.5 X 10 3 r4 /4-1.5975 X lOIO r6 /6+ 3.195 X 1013 r8/8]
3.77
0
21C X 10 4
=---x27.35625xl0-4 =45.6°C
3.77
(Ans)(b)
Example 9.3 ~ Oil at S0°C enters a duct of 1-m length, where heat is being removed at a constant heat flux
rate of 3000 W/m 2. The duct has an equilateral triangle cross-section with 3-cm sides. Determine
the wall temperature at the exit of the duct. The mean velocity of oil is 0.1 mis. The properties of oil are
Density (p) = 850 kg/m 3
Specific heat (Cp) = 2.3 kJ/kg °C
Dynamic viscosity (µ) = 0.0064 N s/m 2
Thermal conductivity (k) = 0.372 W/m °C
Known: Oil flowing through an equilateral triangular duct is cooled under constant wall heat flux conditions.
Find: Wall temperature at exit (0 C).
Schematic:
qw= 3000W/m 2
0
Oil
Inlet - ----+Tb;=S00C
! ! !
Equivalent diameter
De=4Ac/P
! !
_____. V = 0.1 mis
i
L= Im
Fig. 9.6(a)
- - ---- Exit
Forced Convection: Internal Flow
r
9, I 3
Triangular duct: P= 3 a= 9 cm
I
A= - ah
C
2
I
a
h
= - (3 cm)(2.598 cm)
2
a
= 3.897cm2
60°
h =a sin 60° = a'flcm = 2.598 cm
2
a=3cm
Fig. 9.6(b)
Assumptions: ( l) Steady operating conditions exist. (2) Constant properties. (3) Constant wall heat flux.
(4) Developing, laminar flow.
Analysis: Heat transfer rate during cooling of oil under constant heat flux condition is
Q=qwAs =mCP (Tbi -Tbe)=h As (Tb -Tw)
'-v-----'
constant
Surface area for heat transfer,
As= 3(aL) = 3x0.03mxlm = 0.09m 2
Q = (3000W/m2)(0.09m2) = 270W
Mass flow rate of oil is
m=pAc v
= ( 850 kg/m 3 )( 3 .897 x 10-4 m2 )( 0.1 mis)
= 0.0331 kg/s
Fall in temperature of oil over 1-m length is
_g__
T. -T. _
270W
bi
be - mCP - (0.033lkg/s) (2300J/kg 0 C)
= 3.55°C
And, exit oil temperature, Tbe = 50 - 3.55 = 46.45°C
Heat flux,
qw = const = h(Tb -Tw)
or
3000 W/m 2 = h(x=L) [Tbe -Twe]
wall temperature at the exit of the duct is
T =T. _ __k_=4645oC_3000W/m2
we
be h
·
h
(x=L)
(x=L)
(A)
9.14 ,
Heat and Mass Transfer
To find the local heat-transfer coefficient at the duct exit (x = L), let us first ascertain the flow conditions
Reynolds number,
pVD0
Re=--
µ
where D 0 is the equivalent or hydraulic diameter given by
D = 4 ~= 4 x3.897x10-4m2
e
p
9x10-2 m
= 0.01732 m
D =_!!__= 3 cm =~3cm
Else
e
.J3 .J3
= 0.01732 m
Re (850kg/m 3 )(0.lm/s)(0.01732m) I IN I
(0.0064Ns/m2 )
lkgm/s 2
( « 2300)
= 230
The flow is therefore laminar.
To ascertain whether the oil flow is developed or still developing, let us calculate the parameter
Re Pr(D0 IL)
Prandtl number,
Pr= Cpµ = (2300 J/kg °C)(0.0064 N s/m 2 ) 1__!_!_111 w I
k
0.372W/m°C
lNm lJ/s
= 39.57
Re Pr(D0 IL)= (230){39.57)(0.01732/1)
= 157.6
For fully developed flow,
(!:_)
D
~ _}_ Re Pr~ _}_(230 ){39.57)
fullydeveloped
192
192
~ 331.8
1m
As - 1- =
= 57. 74 which is less than _}___ Re Pr, the flow is still developing.
D0 0.01732m
192
The value of hx can be found from the following equation:
Nu= 1.30[ Re Pr( D0 IL )]' 13
= 1.30 [157 .6] 113 = 7.0
Forced Convection: Internal Flow
But
r
9. I 5
't_x=L)De
k
Nu(x=L)=
local heat-transfer coefficient at the duct exit is
h. _ = 7 x O.3 72 W I m °C = 150.35 W / m1 °C
''lx-L)
0.01732m
Substituting this value of h(FL) in Equation (A), one gets
= 46 .450
T
we
c-(
3000W/m2 )
150.35W/m2 °C
(Ans)
=26.5°C
Alternatively
For
qw = const,
and
L* > 0.0015
Nu= 4.364+8.68(1000 L*)--0.506 e-41L'
1
£._LID_
------
With
RePr
157.6
= 0.00634
Local Nusselt number,
Nu= 4.364+ 8.68(1000 X 0.00634)--0.506e-41x0.00634
= 6.992
h = Nu -
k
D0
0.372
= 6.992 x - - -
0.01732
= 150.18 W /m1 °C
and
T
we
= 46.45 -( 3000 ) = 26.5 °C
150.18
(Ans)
=26.5°C
Example 9.4 ~ Lubricating oil at a temperature of 60°C enters a I-cm-diameter tube with a velocity of
0.3 mis. The tube surface is maintained at 28°C. Assuming that the oil has the following
average properties, calculate for the tube length of 10 m, (a) the outlet temperature of the oil, (b) the heattransfer rate, (c) the log mean temperature difference, and (d) the arithmetic mean temperature difference.
p = 865 kg/m 3
k = 0.14 W/m K
cP = 1.78 kJ/kg c
0
v = 9 x Io-6 m2/s
9 .16 ,
Heat and Mass Transfer
Known: Oil flows through a tube under specified conditions.
Find: {a) The {0 C), (b) Q(W) , (c) 11T1m and (d) t:.Tam·
Schematic:
Tube
~----~-------~--..:r.=2s c
0
Oil
0
D= I cm
- V = 0.3m/s
-Tbe=?
L=IO
Fig. 9.7(a)
60°C
t
Tbe
tir e
surface
2s·c ~ - - - - - Tube
--- - - - - ~ ' - , 2s c
1
0
Fig. 9.7(b)
Assumptions: (1) Steady-state conditions prevail. (2) Constant tube wall temperature. (3) Constant
properties. (4) Thermally developing, laminar flow.
Analysis: Reynolds number
Re =VD= 0.3m/sx0.0lm
0
v
9xJ0-6 m 2 /s
= 333.3
Prandtl number,
CP vp
1780J/kgKx9xI0-6 m 2 /sx865kg/m 3
k
0.14 W/mK
Pr=--=-----''-----------'--
= 98.98
Hydrodynamic entry length,
Lh = 0.0575 Re0 D
= 0.0575 X333.3X0.01 m
=0.192 m
Thermal entry length,
4 = 0.03347 Re0 Pr D
= 0.03347 X333.3 X98.98 X0.0 J m
=llm
Forced Convection: Internal Flow
r
9. I 7
For 4. << L 1, the velocity profile may be assumed to be already developed and only the temperature
profile is developing.
Accordingly to evaluate the convection coefficient, we use the Hausen's correlation,
-
Nu0 = J(Re0 , Pr,D/ L)= 3.66+
0.0668 (DI L)Re0 Pr
l+(0.04)[(D/ L)Re0 Pr]
213
hD
= k
Substituting proper values,
Nuo = 3_66 +
0.0668(0.01/10)(333.3) (98.98)
1+ (0.04) [(0.01/ 10) (333.3) (98.98)
f
3
= 3.66 + 1.56 = 5.22
Average heat-transfer coefficient,
h =!.._ Nu = 0.14 W/m Kx5.22
D
O
O.Olm
=73.1 W/m2 K
Mass flow rate of oil flowing through the tube,
m= p¾ D 2 V=(865kg/m3 { ¾x0.0l2 )m2 (0.3m/s)
= 0.02038 kg/s
From energy balance:
or
Exit oil temperature,
¾e = Tw + (¾i - Tw) exp (- h ~ D L
mCP
l
= 280C+( 60 _ 28) oc exp (-73.1 W/m2 KxO.OlmxlOm) = 450 C
0.02038 kg/s X 1780 J/kg K
(Ans) (a)
9 .18 ,
Heat and Mass Transfer
Heat-transfer rate,
Q = (0.02038 kg / S )(l .78 X 103 J/kg C)( 60-45)°C
0
=544 W
(Ans) (b)
Log mean temperature difference,
~i:-~r.
~T; = ----'-----'-ln(~T; / ~T.)
m
where
~I; = Tb; - Tw = (60 - 28)°C = 32°C
and
~T. = Tbe -Tw = (45-28)°C = 17°C
~T,
=
32-17 =23.7oc
/n(32 / 17)
1
32+ 17
Im
(Ans) (c)
Arithmetic mean temperature difference,
~¾m = 2(~T; + ~¾) = - 2 (Ans) (d)
= 24.5°C
Comment:
Check: Q=h(rcDL)~Tim =(73.1 W/m2 K) (rcx0.0lmx10m)(23.7°C)=544W
Example 9.S ~ A small air-cooled condenser is to be designed. The air passes through a number of small
circular ducts which have essentially a uniform surface temperature. The ducts are 5 mm in
diameter and 40-mm long. The air enters with a mean flow velocity of 4.75 mis at 24°C and the wall temperature
is 52°C. Calculate the average heat transfer coefficient and estimate the exit air temperature.
The following properties of air at 27°C may be used:
V
= 15.89 X 10-6 m2/s
CP = I .007 kJ/kg K
Pr= 0.707
Jlw@Sl°C
k = 0.0263 W/m K
= 19.64 X I 0-6 kg/ms
Known: Air flows through a circular duct under constant tube surface temperature condition.
Find: Convection coefficient, h (W/m 2 K); Exit air temperature, Tbe( 0 C).
Schematic:
Circular duct (D = 5 mm)
Air I
Tb;=24°C
T, = 52°C (constant)
-----+-0
-----+- V=4.8m/s
\.
~-------------~
L=40mm
Fig. 9.8(a)
Forced Convection: Internal Flow
r
9, I 9
52°C
T
(T5 = con st)
,H;
Jo c
0
1
24°C
Fig. 9.8(b)
Assumptions: ( l) Steady operating conditions exist. (2) Air is an ideal gas. (3) Constant wall temperature.
Analysis: To ensure whether the air flow through the duct is laminar or turbulent, let us first find the
Reynolds number.
Re= VD= (4.75m/s)(0.005m) = 1495
v
15.89 xl0-6 m 2 Is
( < 2300)
Hence, the flow is laminar.
For a circular tube, Lh,e = 0.0575 Re0 D and Lt,e = 0.04305 Re0 Pr D
Hydrodynamic entry length, Lh,e = 0.0575 (1495) (0.005 m) = 0.43 m
Thermal entry length, Lt,e = 0.04305 (1495) (0.707) (0.005 m) = 0.2275 m
The duct length is 0.04 m which is much less than both Lh,e and Lt,e both the velocity and temperature
profiles are developing simultaneously. The Sieder- Tate correlation which is valid for the combined entry
length case, is therefore the most appropriate.
Then
Nu= l.86{Gz}11 3 (µ/µw }°'1 4 = hD
k
=l.86(132{3(l8.46xl0-6)0.14 =9.39
19.64 X lQ-6
:. average heat-transfer coefficient,
D
Graetz number, Gz = Pe - = Re Pr (D/L)
L
= (1495) (0.707) (5 mm I 40 mm)= 132
h=.!5.._Nu= 0.0263 W/m K x 9 _39
D
0.005 m
=49.4 W /m 2 K
Heat-transfer rate,
.
p
For an ideal gas, p = -
RT
(Ans)
9.20 ,
Heat and Mass Transfer
101.325 kPa
0.287 kJ/kg Kx(27+273.15)K
= 1.1762 kg/m3
Mass flow rate of air,
m=p~D2 V=(l.1762 kg/m3 { ~x0.0052 m2 )c4.75m/s)
= 1.097 X IQ-4 kg/s
Log mean temperature difference,
AJim =(A~ -AT.)/!n(A~ I AT.)
= (52-24)-(52-30) = 22 90c
28
.
/n-
22
Q = (49.4 W/m2 K)(1tx0.005m x 0.04m)(24.9°C)
=0.773W
Exit air temperature,
Q
'Eb e ='Eb', +-• C
m P
0.773 W
= 24°c + - - - - - - - - - - - = 31°c
(l.097x10-4 kg/s)(1007 J/kgK)
As the fluid properties are evaluated at ¾m which requires Ti,. but Ti,. was not given with the calculated
value of Tbe=31°C ¾m =( 24 ; 31 )°c=27.5°C
Properties used were at Tbm = 27°C which is very close to the value obtained (27.5°C). Hence, no
more refinement in the result is required.
Example 9.6
~ Dry sand is to be heated by passing it steadily downwards through a pipe heated by steam
at 1.25 bar condensing on the outside. The inside wall of the pipe is held at the saturation
temperature of the steam and the thermal resistance between the pipe wall and the sand is negligible. The
sand is assumed to flow through a mixer so that it is fed to the pipe at a rate of 30 litres per hour at a
temperature of 41°C. The pipe has 30-mm ID and is 5-m long. For slug flow with constant wall temperature,
the Nusselt number is 5.78.
Estimate the temperature of the mixed sand leaving the heater, using the following properties of sand:
k=0.35 W/m K
CP = 1.0 kJ/kg K
p = 1600 kg/m 3
Known: Dry sand flows through a pipe under slug flow conditions.
Find: Temperature of sand leaving the heater, The (0C).
Forced Convection: Internal Flow
r
9.21
Schematic:
Pipe
(D=30mm, L=Sm)
Sand
·--...__
.. ······
Tbe
Fig. 9.9
Assumptions: (1) Steady slug flow. (2) Thermal resistance between pipe wall and sand is negligible.
(3) The inside wall temperature is same as the saturation temperature of steam condensing on the outside.
Analysis: For slug flow with uniform velocity profile,
INu = 5.78 =TI
h = (5 ·78 )(0. 35 ) =67.43W / m2 K
0.03
Mass rate of flow of sand,
m= (30!) ( 1m33 ) (1600~)
= 48kg/h
3
h
10 1
m
For flow through a closed conduit at constant surface temperature, only the bulk temperature increases
along the conduit. On a differential element at a distance x from the pipe inlet, the energy balance is
or
or
9.22 , Heat and Mass Transfer
As pipe wall thermal resistance is neglected, Tw = Tsat @ 1. 25 bar= 106°C
Substituting appropriate numerical values, we have
(67.43)(1tx0.03x5.0) = /n[ 106-41]
(48/3600)(1000)
106-Ti,.
2.383 =
or
or
[
or
61 ]
/n[ 106-Ti,.
61 ] = e2.383 = 10.84
106-Ti,.
T.. = 106 - ( ~ ) = (106 - 6)°C = 100°C
be
10.84
(Ans)
:. temperature of mixed sand at the pipe exit= 100°C
Example 9.7
~ In a solar water heating system, water at a temperature of 27°C enters a 20-mm internal
diameter tube at a Reynolds number of 1200. The tube is heated by concentrated sun rays
such that constant tube surface heat flux condition exists. If the bulk temperature and the centreline temperature
at the tube exit are 67°C and 46°C, respectively, determine (a) the constant heat flux supplied to the tube,
(b) the length of the tube, and (c) whether boiling occurs in any part of the tube.
Properties: Saturated water [I atm, Tbm = i(Tb; + Tbe) = (27 + 67)/2 = 47°C]
p =989 kg/m 3
CP = 4.18 kJ/kg K
µ = 0.577x I0-3 kg/ms
Pr= 3.77
k = 0.640 W/m K
Known: Laminar flow of water through a tube supplied with constant wall heat flux. Operating data are
provided.
Find: (a) qw (Yv/m 2), (b) L (m), and (c) Tw,ma,/ 0 C).
Schematic:
Water
x=O, Tb;=27°C
x=L, Tbe=67°C
Exit
Inlet
Fig. 9.10
Forced Convection: Internal Flow
r
9.23
Assumptions: (1) Fully developed flow. (2) Steady operating conditions. (3) Constant properties.
(4) Uniform heat flux at the tube wall.
Analysis: The bulk temperature at the outlet is given by
The constant wall heat flux is then determined from
24k(Ti,.-To)
qw =-7-
=
R
24(0.64W/mK)(67-46)°C orK
O.Olm
(Ans) (a)
=4608 W/m2
Reynolds number, Re= 1200
(< 2300)
Re=pVD
µ
~
Laminar flow
Average water velocity,
Reµ
V=pD
V = (1200) (0.577 x 10-3 kg/ms) = 0 _035 mis
(989kg/m3 )(0.02 m)
Mass flow rate,
m=p!!.n2 v
4
= (989 kg/m3 )( ~x0.02 2 m2 )<o.035m/s) = 0.01088kg/s
Heat-transfer rate,
The tube length required is
(0.01088kg/s)(4180K/kgK)(67-27)°C or K
= -'--------"----'--'-----"---'-'------( 4608W /m2 ) (1t x0.02 m)
=6.28 m
Maximum wall temperature will occur at x = L.
(Ans) (b)
9.24 ,
Heat and Mass Transfer
3 qwR
Tw max =To+--'
4
= 460
k
c+i (4608W/m2 )(0.0lm)
4
0.64W/mK
= 100°c
(Ans) (c)
Boiling point of water at 1 atm = 100°C. Hence, there is a possibility of boiling occurring only at the
outlet of the tube.
Aliter: Thermal diffusivity,
pCP
0.640W/mK
1.548xI0-7 m2 /s
(989kg/m3 )(4180J/kgK)
dT
-
ATi,
--
k
<l=--=---------
dx
L
2 qwa
---
VRk
length of the tube is
L
(I'iie -Ti,i)VRk
2qw<l
(67-27)°C(0.035 mis) (O.Olm) (0.64 W/mK)
2 (4608W/m2 )(1.548x 10-7 m2 /s)
=6.28m
Comment: For fully developed laminar flow, the thermal entry length,
4,th =0.05 Re0 Pr0
= 4.52 m
= {0.05){1200)(3.77)(0.02)
(<L = 6.28 m)
Hence, the fully developed flow condition is satisfied.
Example 9.8
~ Air is to be used to cool a solid material undergoing a heat-generation reaction. Holes of
I0-mm diameter are drilled through the material. The thickness of the plate is 8 cm and
the thermal boundary condition at the surface of the holes is considered to be a uniform heat flux. The air
enters the holes with a uniform velocity profile at 1.5 m/s, and a temperature of 20°C. Estimate the rate at
which heat is dissipated by the air per hole if the maximum allowable temperature of the material is 200°C.
The following properties of air at 310 K and I atm may be used:
k= 0.027 W/m K
p = 1.1389 kg/m 3
cp = 1.0064 kJ/kg K
v = I 6.70 x Io--6 m2/s
Known: Air is used to cool a solid material experiencing heat generation. Constant wall heat flux
conditions are prescribed.
Find: Rate of heat removal by air, Q[W].
Forced Convection: Internal Flow
Schematic:
r
9.25
,.__ _ _ L=Scm - - -----+-1
, - - - -i--r------~L._____,Solid material
Air
Tb;=20oC _
t
Hole
___,___ D= 10mm
V= I.Sm/s
+i--- Tse =200°C
q. = con st
(Maximum surface
temperature)
Fig. 9.11 (a)
T5 , max= 200° C
=Tse
Tbe = 53.37 °C
Fig. 9.1 l(b)
Assumptions: (1) Steady state conditions. (2) Air is an ideal gas . (3) Fully developed velocity and
temperature profiles. (4) Uniform surface heat flux.
Analysis: As air outlet temperature is not specified, it is necessary to initially evaluate the thermophysical
properties at an estimated average bulk temperature, say 310 K.
Reynolds number,
pVD
VD
µ
V
Re=--=-
= (l.5m/s)(O.Olm) = 898
16.70x J0-6 m 2 Is
Hence, the flow is laminar.
For laminar flow under constant heat flux condition,
hD
48
Nu=-=k
11
which is independent of either Reynolds or Prandtl number.
(< 2300)
9.26 ,
Heat and Mass Transfer
Convection coefficient,
h = 48
~ = (48)(0.027W/mK) = l 1. 78 W /m2 K
11 D
11
O.Olm
Therefore,
q. = Jl_ = h [ T. - Ti,] = const
Heat flux,
A.
or
(1)
Mass flow rate of air is
m=p AcV=p~D2V = (1.1389 kg/m3 )(~x0.0I2 m2 )(1.5m/s)
4
4
= 0.134 X lQ-3 kg/s
Heat removal rate,
Q=mCp(li,. -Ti,J=hA.(T.. -Ti,.)
Q mCP
q, - - - - - ( ' E -'E )
or
s -
As - rcDL be
bi
= (0.134x10-3 kg/s)(1006.4J/kgK) (Ti, _ 20 )
rc(O.Olm)(0.08m)
•
lqs =53.73(Tbe -20)1
or
(2)
Equating (1) and (2):
53.73 li,. -(53.73x 20) = 11.78 X 200 -11.78Tbe
or
65.5 Ti,. = 3430.6
or
i;,. = 52.31°c
And, the bulk mean temperature is
Tbm = (52.37 + 20) I 2 = 36.18°C or 309.33 K
=310 K as assumed earlier. Hence, a second iteration is not in order.
From Equation (1) or Equation (2),
Heat flux, q8 = 1739 W/m 2
and
Heat-dissipation rate,
Q=q.(rcDL)= (1739 W/m2 )(rcx0.0lmx0.08m)
=4.37 W
(Ans)
Forced Convection: Internal Flow
Example 9.9
r
9.27
~ Water at l0°C is to be heated to 40°C in a thin-walled tube of 2-cm ID at a mass flow
rate of 0.01 kg/s. The outer surface of the tube is subjected to a uniform and constant wall
heat flux of 15 kW/m 2• Neglecting any entrance effects, determine the
(a) Reynolds number
(b) heat-transfer coefficient
(c) length of the pipe needed
(d) inner tube surface temperature at the exit
(e) friction factor
(f) pressure drop in the pipe
(g) pumping power required if the pump efficiency is 50%
Known: Water is heated in a tube under constant wall heat flux conditions.
Find: (1) Re; (b) h; (c) L; (d) Ts,e; (e)f; (f) M; (g) p.
Schematic:
~ate~
L. Tbe=40°C
Tb;= 10°c-Or,
••
_r---:0~0~~0~2~m~-------,• •
- - - - - - - L------+i
Fig. 9.12
Assumptions: (1) Steady operating conditions. (2) Entrance effects are neglected. (3) Uniform and constant
heat flux at the pipe surface.
Properties: Water (Tiim = (10+ 40)/2= 25°C):
cp =4.18 kJ/kg°C
µ=0.891xl0- 3 kg/ms
p =997kg/m3
k=0.607 W/m °C
Analysis: From energy balance:
Q = mCP (I'iie - Ti,i) = qw (1t D L) = h (1t D L) (T. -Ti,)
Heat rate is
Q = (0.01 kg/s) (4180 J/kg°C) ( 40-10)°C
=1254 W
Reynolds number,
Re=pVD = 4m
µ
1tDµ
4x0.0lkg/s
1tx 0.02 m x 0.891 x 10-3 kg/ms
=714.5
(< 2300)
(Ans) (a)
Hence, the flow is laminar and fully developed because entrance effects are neglected.
9.28 ,
Heat and Mass Transfer
For laminar flow (qw = const):
hD
48
Nu=-=k
11
Heat-transfer coefficient,
h= 48X~= 48X0.607W/m°C
11
11
D
0.02m
= 132.4 W /m2 °C
(Ans) (b)
Length of the pipe required,
Q
L=
=
qw (1tD)
1254W
15000W/m2 X1tX0.02 m
= 1.33 m
(Ans) (c)
Inner tube surface temperature at the exit is
T.e = ¾e +
q;
= 40 0 C+ 15000W/m2
132.4 W/m2°C
= 153.3°c
(Ans) (d)
Friction factor,
f = 64 =~=0.0896
Re 714.5
(Ans) (e)
Pressure drop,
where velocity,
V=
m
=
p(1tD 2/4)
4x0.01 kg/s
2 =0.032m/s
997kg/m3 x1tx(0.02m)
llP = 0.0896( 1.33m)(997 kg x 0.0322 m2/s2)
0.02m
m3
2
= 3.03 Pa
(·: 1 pa= 1 kg/ms2 )
(Ans) (t)
Pumping power,
mM
f,J=-
PTJ
(O.Olkg/s) (3.03 Pa)
= -'-------'---'-----'(997 kg/m3 )(0.50)
=6.lxl0-5 W
3
Jpam
(·:
-8- = I W )
(Ans) (g)
Forced Convection: Internal Flow
Example 9. 10
r
9.29
~ A constant heat flux of 8.5 kW/m 2 is applied on the outside wall of an insulated pipe ( 12- mm
ID and 5.5-m long) under the insulation. Water flowing through the pipe is heated from 20°C.
The maximum allowable inside wall temperature is 90°C. Determine the minimum velocity of water through
the pipe. Are entrance effects negligible?
Known: Water flows through a pipe subjected to constant wall heat flux .
Find: Velocity of water, V (mis).
Schematic:
Wa~r
(D
+ -- - - - -
qw = 8.5 kW/m2
Twe = 90oC = Tw,max
+ + + + +
(j ~r~
L= 5.5 m - - - - - - - - i
Fig. 9.13
Assumptions: (I) The flow is laminar, steady and fully developed. (2) Constant water properties. (3)
Constant wall heat flux.
Analysis: From energy balance:
or
Solving for velocity,
(A)
To evaluate The• we note that
Tw - Tb
=qwh =constant for fully developed flow.
ITbe = Twe -(qw/h)I
(B)
To calculate h, for laminar flow (qw = const):
1Nu=¥=~=4.3641
Fluid properties need to be evaluated at the bulk mean temperature,
I
Tbm = (7;,; +The).
2
Since The is unknown, as a first approximation, let
I
I
7;,m =-(Ti,;+ Twe) = -{20 + 90) = 55°C
2
2
(C)
9.30 ,
Heat and Mass Transfer
Thermal conductivity of water at 55°C, k = 0.649 W Im °C
Then, from Eq. (C):
h = 48 .!5._ = 48 X 0.649 W/m°C
11 D
11
0.012 m
=236 W/m2°C
Also, from Eq. (B):
T.. -r_ -(
be -
we
qw
/h)-9ooc-(8.5x103W/m2)
236 W/m2oC
=54°C
1
Bulk mean temperature, ¾m = -(20 + 54) = 3 7°C
2
Properties of water at 37°C are:
k= 0.626 W/m °C
p = 993.2 kg/m 3
cp = 4.178 kJ/kg 0 c
µ = 0.693 X lQ-3 kg/m S
Pr=4.63
[Since the value of k above is not much different from the assumed value, we accept Tbm = 37°C].
From Eq. (A), one gets
Minimum mean water flow velocity is
4x8500 W/m 2 x5.5m
V=------------------993.2 kg/m3 x0.012 mx4178 J/kg °C x(54-20)°C
=0.11 m/s
(Ans)
Reynolds number,
pV D
(993.2 kg/m3)(0.llm!s)(0.012m)
Re=--=-----------µ
0.693 X lQ-3 kg/m S
=1892
Since Re < 2300, our assumption of laminar flow is justified.
Entrance effects are negligible if L > Lh,lam and> Lt,Iam·
Hydrodynamic entrance length, Lh,lam "" 0.05 Re D
""0.05x1892x0.012 m"" 1.14 m
Thermal entrance length, 4,Iam "" Pr Lh,lam
""1.14 m x 4.63"" 5.27 m
Since the pipe length of 5.5 m is greater than both Lh and L1, entrance effects can be ignored and the
flow can be considered fully developed as assumed.
(Ans)
Forced Convection: Internal Flow
Example 9.11
r
9.31
~ One kg/h of dry air is to be heated by passing it through an electrically heated tube. The
tube has I-cm ID and the heating section is 50-cm long. An unheated section of tubing
precedes the heated section so that the flow enters the heated section with a fully developed velocity profile.
Determine (a) the heat flux, and (b) the maximum temperature of the air leaving the heating section under the
design constraint that the maximum temperature of the tube wall cannot exceed 200°C. The temperature of
the air entering the unit is 20°C.
Known: Dry air gets heated as it passes through an electrically heated tube.
Find: (a) qw (b) The (0 C).
Schematic:
D=0.01 m
Air
- r h = I kg/h
rh;=2o·c-
L=O.Sm
Fig. 9.14
Assumptions: (1) Steady operating conditions. (2) Constant wall heat flux. (3) Fully developed velocity
profile. (4) Laminar flow.
Analysis: As the outlet bulk temperature is to be found, we have to assume it as an initial guess to
calculate the average bulk mean temperature.
Let
The= 100°C
T,_h = I'iii + Ti,. = 20 + 100 = 6ooc
m
2
2
Properties of air at 60 °C:
k=28.74x 10-3 W/m K
v = 19.21 x 10-6 m2/s
CP = 1.008 kJ/kg K
p = 1.0597 kg/m 3
Pr=0.7024
Reynolds number,
4m
4(1/3600kg/s)
Re0 = - - - = - - - - - - - - - - - - - - - 1tDpv 1t(O.Olm)(l.0597 kg/m3 )(19.21x10-6 m2 /s)
=1737
(<2300)
Hence the flow is laminar. The velocity profile is fully developed.
Thermal entrance length,
Lth = 0.04305 Re Pr D = 0.04305 X 1737 X 0.7024 X 0.01 = 0.525 m
Since L < Lth, the flow is thermally developing.
Now
L* = LID = 0.5m/0.0Im
RePr
= 0.041
(1737) (0.7024)
9.32 ,
Heat and Mass Transfer
For L* > 0.0015, local Nusselt number,
Nu= hD = 4.364+8.68(1000£*) exp (-41L*)
k
= 4.364+8.68(1000x0.041) exp (-41x0.041)
= 4.61
Local heat-transfer coefficient,
h=Nu kl D
= (4.61 )(0.02874 WIm K) = 1325 W /m 2 K
O.Olm
The wall temperature will be maximum at the end of the heating section i.e. at x = L = 0.50 m.
qw = h(Tw -Ti,)lx=L
= 13 .25 (200 -Ti,) IL=O.Sm
lqw =13.25 {200-Ti,e)I
(1)
But
Therefore,
=
qw1t(O.Olm)(0.05m)
+ 20 0 C
3
(1/3600 kgls)(l.008 X 10 J/kg K)
ITiie = 0.056 qw +201
From equations (1) and (2), we have
¾e = 0.056 [13.25(200-Ti,e)]+20
or
or
The= 0.742 (200 - The)+ 20
1.742 The= 168.4
The = 96.68°C
With this value as the second guess, we have
¾m = 20 + 96.68 = 58.34cc
2
(2)
Forced Convection: Internal Flow
r
9.ll
Properties of air at 58.34 °C:
v = 19.04 x 10--{j m2/s
k=28.62 x IQ-3 W/m K
CP = 1.008 kJ/kg K
p = 1.065 kg/m 3
Re=
4(113600)
Pr=0.7026
= 1744
1t X 0.0lx 1.065 X 19.04 X 10--{j
Now
L* = LID =
Re Pr
0.5/0.0l
= 0.0408
(1744) (0.7026)
Nu =4.613
h= 4.613(0.02862W/mK) =l 3 .2 W/m2 K
O.Olm
Equations (1) and (2) will now become
qw=l3.2 (200-Tbe)
and
~
ITii.=200-{qw/13.2)1
!Tiie = 0.056 qw + 201
It follows that
qw [o.056+-1- ] = 180
13.2
(Ans) (a)
:. heat flux, qw = 1366 W /m2
And
Ti,.= (0.056)(1366)+20
=96.5°C
Since this value is only slightly different from the previous value,
Tbe=96.5°C
(Ans) (b)
Example 9.12 ~ Water is heated by flowing through a horizontal tube of I 0-mm inner diameter maintained
at constant surface temperature of 90°C. Water can flow using either (a) a small power
pump so that the mean flow velocity is 0.1 mis or (b) a larger pump with the average flow velocity of 0.6 m/s.
Water enters the tube at 40°C and is to be heated by 30 K. Determine the length of the tube needed to
effect this rise in water temperature.
9.34 , Heat and Mass Transfer
Properties of water (Tbm = SS C):
0
k = 0.649 W/m K
Pr= 3.25
µ = 0.504 x I0-- 3 kg/m s
Cp=4.183 kJ/kg K
p (at Tb;= 40 K) = 992.1 kg/m 3
Known: Water flowing through a horizontal tube at constant wall temperature with specified wall
tern perature.
Find: Length of the tube when (a) V= 0.1 mis; (b) V= 0.6 mis.
Schematic:
Tube (T5 =90°C)
[
_
Water
Tb;=40°C-~
______
0_=_2:~·s_c_m_ _ _-_ _ _
v_~~
-
Tbe=70°C
L
Fig. 9.15
Assumptions: (1) Fully developed flow . (2) Steady operating conditions. (3) Constant tube wall
tern perature.
Analysis: (a) Flow velocity, V = 0.1 mis:
Mass flow rate of water entering the tube,
m=p~D2V
4
= 992.1 (:~ )x¾x(o.01)2 (m)2 xo.i(
7)
= 7.792 X lQ-3 kg/s
4m
4x7.792x 10-3 (kg/s)
Re0 = - - = - - - - - - - - - - - 1tDµ 1tx0.0l(m)x0.504xl0-3 (kg/ms)
= 1968.5
(< 2300)
[laminar flow regime]
For constant tube wall temperature:
Nusselt number,
hD
Nu0 =-=3.66
k
Convection coefficient,
h= 3 _66 x 0.649 (W/mK)
O.Ol(m)
= 237.4 W/m2 K
Forced Convection: Internal Flow
r
9.35
Tube length,
mcp ( r.w - 1:b.1 )
L=--ln
hrcD
Tw -The
= 7.792x 10-3 (kg/s)x4183(J/kgK) x /n {90-40)K
237.4(W/m2 K) xrcxO.Ol(m)
(90-70)K
=4.0 m
(Ans)
(b) Flow velocity, V= 0.6 mis:
Mass flow rate of water,
m=p~D2V
4
= 992.l(kg/m3 )x~x(0.01)2 (m)2 x0.6(m/s)
4
= 0.04675 kg/s
Reynolds number,
4m
Re0 = - rcDµ
4x0.04675(kg/s)
rcxO.Ol(m)x0.504xl0-3 (kg/ms)
=----------= 11810.7
(>2300)
Using Gnielinski correlation:
Friction factor,
f =(0.79 /n Re-l.64f2
=(0.79 /n 11810.7-1.64}2
=0.03
Nusselt number,
(f !8)(Re0 -1000 )Pr
Nu0 = - ~ ~ - - - ~ - 1+ 12.7 ( f I 8 )'12 ( Pr213 - 1 )
(0.03/8 )( 11810.7 -1000 )( 3.25)
=~-~----~~~
1+12.7 (0.03/8}'1 2 ( 3.25213 -1)
= 68.28
Convection coefficient,
k
h=Nu0 D
_ 68.28x0.649 W/m K
0.01 m
=4432 W/m2 K
[turbulent flow regime]
9.36 , Heat and Mass Transfer
Length of the tube,
mcp ( rw - r.b.'
L=--ln
h1tD
Tw -Tbe
J
= 0.04675 (kg/s)x4183(J/kgK) Zn (90-40)K
4432.0(W/m2 K)1txO .Ol(m)
(90-70)K
= 1.287 m
Check: .£= 1.2 87 m = 129
D O.Olm
( > 10)
(Ans)
OK.
Comment: The tube length required is reduced to almost one-third for the same temperature rise of water
when the flow velocity is increased six-fold.
Example 9. 13 ~ Water at a bulk mean temperature of 80°C and an average velocity of 0.15 mis flows inside
a 25-mm inside diameter, thin-walled copper tube. Atmospheric air at 20°C and with a freestream velocity of IO m/s flows across the tube. Estimate (a) the overall heat transfer coefficient, and (b) the
rate of heat transfer per unit length of the tube. For air flow, use the following correlation:
Nu= 0.3 +[ 0.62Re112 Pr'13 x{l+(0.4/Pr)213
r" x[1+ (Re /282 ooo)5'
4]
8
r
5
Known: Water flowing through a tube is subjected to a cross flow of air under specified operating conditions.
Find: Overall heat-transfer coefficient, U (W/m 2 K); Heat transfer rate per metre length, Q I L (W/m).
Schematic:
Fig. 9.16
Assumptions: ( 1) Steady-state conditions. (2) Constant properties. (3) Tube wall resistance is negligible.
(4) No fouling.
Properties: Water (1 atm, Tbm = 80°C):
p=971.8 kg/m3
k=0 .670 W/m °C
µ = 0.355 x 10-3 kg/ms
Pr= 2.22
.
80+20
Au (1 atm, T1 "" - - = 50°C ):
2
v = 1.798 x 10-5 m2/s
k= 0.02735 W/m °C,
Pr=0.7228
Forced Convection: Internal Flow
r
9.37
[As tube surface temperature is not provided, the closest approximation is T8 "" Tbml
Analysis: (a) Overall heat-transfer coefficient,
U=[~+~r
To evaluate inside heat-transfer coefficient, hi, let us first determine the Reynolds number to find
whether the flow of water through the tube is laminar or turbulent.
pV D (971.8kg/m3)(0.15m/s)(25xl0-3m)
Re=--=------------µ
0.355 X lQ-3 kg/ m S
= 10 265.5
( > 2300)
The flow is clearly turbulent.
Using the Dittus-Boelter correlation with n = 0.3 as water is being cooled.
h-D
Nu= - 1- = 0.023(Re)0-8 (Pr) 03
k
I\= ( 0 ·67 Wlm°C)(0.023)(10265.5)0·8 (2.22)03
0.025m
= 1267 .28 W /m2 °C
To determine outside heat-transfer coefficient, h 0 , we use the specified correlation for cross flow of
air over a cylinder.
3
m/s)(25x 10- m) _ 13 904
Re0 -_ u=D -_ -(10
'----'--------'- v
1. 798 x 10-5 m 2/s
Nu= 0.3+ 0.62(13 904)112 (0.7228)113 [i+( 13904 ) 518 ]
[1 + (0.4 / 0.7228)2/3
t
4
415
282 000
= 64.91 = hD
k
ho= 0.02735W/moc {64.91) = 71.0 W/m2oc
0.025m
U =[
l
+_!_] = 67.25 W/m2°C
1267.28 71
(Ans) (a)
The heat loss rate per m length of the tube is
Q
L = U A (J;n -7;,ut)
=U(1tD)(Tbm -T=)
= (67 .25 W /m2 0 C)( 1t X 0.025 m)(80 - 20)°C or K
=317 W/m
(Ans) (b)
9.38 , Heat and Mass Transfer
(C) CIRCULAR TUBES: TURBULENT FLOW
Example 9.14 ~ A solar concentrator focuses sunlight on a bank of molybdenum alloy tubes which are
16- mm ID, and 360 mm long. A pyrometer scan indicates an average surface temperature
of I 650°C for a tube and the air enters the tube at 60°C with a mass-flow rate of 0.035 kg/s. Estimate the
exit air temperature.
Known: Air gets heated as it passes through a tube whose wall is subjected to focused sunlight.
Find: Exit air temperature, Tbe( 0 C).
Schematic:
Fig. 9.17(a)
T5 = I 6S0°C
Air
CD
m= 0.035 kg/s
L=0.36m
Tb;=60°C
Tbe
Fig. 9.17(b)
Assumptions: (1) Steady operating conditions exist. (2) Air is an ideal gas. (3) Constant properties. (4) Kinetic
and potential energy effects are negligible. (5) The tube has smooth surface. (6) Fully developed flow.
Analysis: Energy Balance
Heat transferred to flowing air by convection from the heated tube wall
= Rate of enthalpy rise of air
h(rcDiL)(Ts -Tb)= mCp(Tbe -TbJ
Predicting outlet temperature of air, Tbe requires knowledge of the convection coefficient, h from
the Nusselt number (Nu= hD/k) so that the convective heat transfer can be matched by the change
in the enthalpy of air. Properties of air are to be evaluated at the bulk mean fluid temperature,
1
Tbm = Tb = -( Tbi + Tbe) but Tbe is unknown. Hence, we estimate Tb, then check out energy balance and
2
re-evaluate, if need be, till a close correspondence is reached.
Forced Convection: Internal Flow
r
9.39
Trial I: Let Ti, = 200°c
The relevant properties of air are
CP = 1.023 kJ/kg K
k= 0.03779 W / m K
µ = 2.577 X IQ-5 kg/m S
Pr= 0.6974
T,, = 1650°C
At
µw = 6.061 kg/ms
As the difference between average temperatures of surface and fluid is very large, we prefer to use
Petukhov equation for better accuracy, provided the flow is turbulent.
Re = 4m =
4x0.035kg/s
D
1tDµ 1txO.Ol6 mx2.577x10-5 kg/ms
= 108 080 (> 2300)
For Re0 > 10 000,
and
~
Turbulent flow
0.5 ~Pr~ 2000,
Nu
(f I 8) Re Pr
(....!:!:.._)n
1.07+12.7(f /8) 0·5 (Pr 21 Ll) µw
where friction factor,!= (0.790 ln Re - l.64)-2 and n = 0.11 because Tw >Tb.
[for 10 4 <Re<I06 ]
f= (0.790 ln 108 080 - 1.64)-2 = 0.0177
Nusselt number,
Nu_ hDi _
k
(0.0177 /8)(108080)(0.6974)
(2.577) 0· 11
1.07+12.7(0.0177 /8) 0 -5 (0.6974213 -1) 6.061
= 161.06
h = (161.06) (0.03779 W / m K) I 0.016
= 380.41 W/m2 K
The inside surface area of the tube is
A.= 1tDiL = 1t(0.016m)(0.36 m) = 18.lxIQ-3 m2
Q = hA.(f. -Tb)= (380.41 W/m2 K)(18.lx10-3 m2 )(1650-200)K
= 9.98 X 103 W = 9.98 kJ/s
This equals mCP ( 'Ii,. - Thi )
:. outlet air temperature,
_g__-6ooc
9.98kJ/s
T.b -T.
- b"+
+--------•
' mCP
(0.035kg/s)(l.023kJ/kgK)
= 338.8°C
9 .40 ,
Heat and Mass Transfer
:. Bulk mean air temperature
Tb =.!_(60+338.8)=199.4°C ::::200°C
2
As the calculated value is almost equal to the one assumed, no more trial is necessary.
Tbe = 338.8°C
(Ans)
Comment: Hydrodynamic and thermal entry lengths for turbulent flow inside a circular tube are
4. ""4h ""10 x 16 mm, i.e., 160 since the tube length, L = 360 mm is greater than Lh"" L1h, the assumption
of fully developed flow is valid.
Example 9.15 ~ A pipeline heater, 3-m long and 40-mm diameter, is used to heat a supply of carbon dioxide
(CO 2) from 280 K to 520 K. The walls of the heater are heated to a constant uniform
temperature of 600 K. Calculate (a) the mean flow velocity of CO 2, and (b) the heat transfer rate, using the
following correlation and the thermophysical properties of carbon dioxide.
For turbulent, developed pipe flow:
Nu0 = 0.03955 Reij 4 p,.m
Properties of CO 2 at 400 K are:
p = 1.3257 kg/m 3
k = 0.0243 WI m K
CP = 942 J/kg K
µ = 1.90 x I 0-5 kg/m s
Pr= 0.737
(c) What will be the velocity and mass rate of flow of CO 2 if the heater diameter is reduced to 25 mm, all
other parameters remaining unalteredr
Known: CO 2 is heated while flowing in a pipeline heater maintained at a constant surface temperature.
Find: (a) Average velocity, V(m/s); (b) Heat-transfer rate, Q(W); (c) Effect of reduction in diameter on
velocity and mass-flow rate.
f"" Pipeline heater
Schematic:
L Tw=600K
CO 2
Tb;=280 K - - 0
D=r.04m
) - - Tbe=520K
,__ _ _ _ L=3m - - - - ~
Fig. 9.18(a)
Tbe
(520K)
Tb;
(280K) ~ - - - - - - ~
Fig. 9.18(b)
Assumptions: ( 1) Steady operating conditions exist. (2) Fully developed, turbulent flow. (3) Constant properties.
Forced Convection: Internal Flow
r
9.41
Analysis: Heat-transfer rate,
or
T - T. · )
k
VD
=-X1tDLx0.03955 ( _P_ )
Tw -Tbe
D
µ
3/4
or
mCP ln (
or
p~D2 VCP ln [ r_w - bi = 1tkLx0.03955 _pVD)
_
4
Tw-Tbe
µ
bi
w
T.]
(
(Pr)
314
113
(Pr) 113
V 114 xl.3257x~x(0.04) 2 x942 x zn{ 600 - 280 }
or
4
600-520
3/4
=1tx0.0243x3x0.03955x [ 13257 x0.0 4 ]
1.90x10-5
x(0.737)113
v114 = 1.4441
Mean flow velocity, V = (1.4441) 4 = 4.35 m / s
(Ans) (a)
Mass flow rate, m=p~D2 V=(l.3257kg/m3{~·x0.042 m2 ) (4.35 mis)
=0.00725 kg/s
Heat transfer rate, Q = mCP (The - TbJ
=0.00725kg/s x942 J /kg Kx(520-280) K
= 1638 W
(Ans) (b)
Check: Reynolds number,
R
fn
=pVD =(1.3267x0.04x4.35)
µ
1.90 X 10-5
= 12150
(> 2300)
Hence, the flow is indeed turbulent.
A close look at the above expressions reveals that if the pipe diameter is reduced from D 1 to D 2, then
the velocity and mass flow rate, Vi and mi are given by
Therefore, for D2 = 25 mm,
V2 = 4.35 mis (25/40t 5 = 45.61 m/s
(Ans) (c)
9.42 , Heat and Mass Transfer
11½ =0.00725 kg/s {25 / 40f3 = 0.0297 kg/s
and
(Ans) (c)
These observations are valid only for the prescribed empirical correlation.
Example 9.16 ~ Water at 20°C is to be heated by passing it through the tube. The surface of the tube is
maintained at 90°C. The diameter of the tube is 1.3 cm while its length is 9 m. Find the
mass flow rate so that the exit temperature of water will be 60°C. The properties of water are:
cP = 4.174 kJ/kg K
p = 895 kg/m 3
k = 0.64 W/m K
v = 0.62 x 10- 6 m2/s
~ = 4.25 x I0- 3 K- 1
Use the correlation: Nu = 0.023 Reo.s p,').l
Known: Water is heated as it flows through a tube under constant wall temperature conditions.
Find: Mass flow rate of water, m(kg/s).
r---
Schematic:
X
- - dx -j
r
T;=20°C- D=0.04m -
T
-
T+dT
---Te=60°C
.....----- L = 9 m
c.v
Fig. 9.19
Assumptions: (1) Steady state conditions. (2) Constant properties. (3) Changes in kinetic and potential
energy are negligible.
Analysis: Reynolds number for flow through a tube is
4m
4m
4m
Re= --Re= - - - = - - - - - - - - - 1tDµ
1tDvp 1tx0.04x0.62x10-6x995
=51.6xl03m
Prandtl number,
CPµ CPvp
Pr=--=--
k
k
= 4.174xl0 3 x0.62x10-6 x995 = 4 _023
0.64
Nusselt number,
Nu= 0.023(Re)0-8 (Pr}°-4
= 0.023 (51.6 X 103 m.)0· 8 ( 4.023)0.4
=236.4 m. 0·8
[Note: The index of Pr should be 0.4 not 0.3 as mentioned in the problem statement since water is
getting heated].
Forced Convection: Internal Flow
r
9.43
Applying energy balance on the control volume (C.V) shown in the schematic, we have (Rate of heat
transfer to the water in the CV) = (Rate of increase of enthalpy of water in CV)
h(rcDdx)(Tw -T) = mCPdT
or
or
hrcDdx
dT
mCP
Tw -T
Integrating between limits: x = 0, T= r; and x = L, T= r., we get
h re.D
( Tw - 1'; )
-L
= [ -ln(Tw -T) ]r;
T = In - -
or
mCP
Tw -T.
,
J
= /n[ 90 - 20 = 0.8473
90-60
hD
Nu=k ,
With
h = Nu.!5_ = 236.4 m. 0 -8 x 0 ·64 = 3782.4 m. 0 -8
D
0.04
(3782.4)(m. 0 ·8 )(rcx0.04x9) = 0 _8473
ril.x4174
m. 0 _2 = (3782.4)x(rcx0.04 x9)
or
(4174)(0.8473)
1. 2 1
Hence, the mass flow rate is
m=(l.21)110 -2 = 2.59 kg /s
(Ans)
Example 9.17 ~ Hot engine oil at I 400C is to be cooled to I OOOC, by passing it through a circular tube
of 2-cm ID. The wall of the tube is maintained at 40°c. If the flow rate of oil is 0.8 kg/s,
determine the length of the tube.
Properties of engine oil:
T(0 C)
p (kg I m3)
CP (k)lkg K)
v x / 04 (m 2 /s)
k (Wlm K)
Pr
40
87 6.05
I .9674
2.4
0.144
287
100
840.01
2.219
0.203
0.137
276
120
828.96
2.307
0.124
0.133
175
140
816.94
2.350
0.080
0.132
116
Known: Inlet and desired outlet temperature of oil flowing through a tube with uniform wall temperature.
Find: Required tube length.
9 .44 ,
Heat and Mass Transfer
Schematic:
Engine
oil
- \40oc
Tbi(11::o.s\(g/S
Fig. 9.20
Assumptions: (1) Steady operating conditions exist. (2) Constant fluid properties. (3) Fully developed
flow (L >> D). (4) Constant tube wall temperature. (5) Negligible potential energy, kinetic energy and
flow work effects.
Analysis: Bulk mean temperature of oil,
r.b = 140 + 100 = 1200c
2
m
Reynolds number for a circular tube is
ReD=pVD = 4m = ~
µ
=
1tDµ
1tDpv
4 X 0.8 kg/s
= 4955
3
4
2
1t{0.02 m)(828.96 kg/m )(0.124 x 10- m /s)
~
Turbulent flow
Prandtl number, Pr= 175
For flows characterised by large property variations as in this case and since Pr> 100, the DittusBoelter equation cannot be used but the Seider- Tate equation is applicable.
Nusselt number,
0.14
Nu 0 = 0.027 Re& 8 Pr113
(
~
)
= 0 _027 ( 4955 )°" 8 ( 175 )11 3 [828.96 kg/ m3 x 0.124 x 10-4 m2 / s]0 · 14
876.05 kg/ m3 x 2.40 x 10-4 m2 / s
=89.44 = hD
k
Average convection coefficient,
h={89.44)~ = (89.44)(0.l 33 W / mK) = 595W/m 2 K
D
0.02m
To determine the tube length, for constant surface temperature cooling, we have from energy balance:
mCP (Tbi -Tbe) = h(1tDL)t:i.lim
where
t:i.r;m = [(rw -Tbi )-(Tw -Tbe
)]Vzn( TwTw =TbeTbi)
Forced Convection: Internal Flow
r
9.45
After rearranging, we have,
595W/ m2 Kx1tx0.02mxL{m) =ln(40-140)
0.8kg/ sx2307 J/ kgK
40-100
Tube length required,
L=
0.8 x 2307 In 100
60
1tX0.02x595
=25.2 m
(Ans)
L 25.2
Comment: - = - - = 1260 >> 50 and hence, the assumption of fully developed conditions throughout
D 0.02
is justified.
Example 9.18 ~ An air compressor used in a large automobile body shop is located in an inside equipment
room. Fresh air at 6°C and 0.97 bar is conveyed to the compressor from outside through a
30-cm diameter 16-m long circular duct. The duct runs along the ceiling of the facility, where the temperature
is 35°C. The volumetric flow rate of air entering the duct is 27 m 3/min. Calculate (a) the temperature of the
air when it reaches the compressor (0 C), and (b) the heat transfer rate (W).
Known: Air is heated in a circular duct under specified conditions and is being supplied to an air compressor.
Find: (a) Air temperature leaving the duct and entering the compressor, Tb.( 0 C); (b) Heat transfer rate, Q(W).
Schematic:
[
-
Air
Circular duct
Tw=35oC
. P;=0.97 bar -
~--D__,_;_J_o_c_m_ _ _ _ _ _ _ _ _ ___,~- T o air compressor
.\L,=27m3/min
x=O
Tb=Tb;=6°C
x=L= 16m
Tb=Tbe=?
Inlet
Exit
Fig. 9.21(a)
T
__-,.1-
T
tire
Tbe=?
1
Tb;
(6°C)~--------~
Fig. 9.21(b)
Assumptions: (1) Flow is fully developed. (2) Duct wall is isothermal (at constant temperature). (3) Steady
operating conditions. (4) Duct's inner surface is smooth.
9.46 , Heat and Mass Transfer
Analysis: Heat-transfer rate,
Now,
LMTD=
AJ;-AJ'.
AT
l n -1
AJ'.
where
and
Substituting for LMTD and noting that AJ; -AT. = 'Ii,. -Thi, Equation (A) can now be expressed as
.
(The-Thi)
AT
mCp('li,. -Thi)=h(1tDL)
l n -1
AJ'.
or
AT
1
mCP =h(1tDL)/lnAJ'.
or
-Zn AJ'. = h1tDL
AJ;
mCP
or
Ar. = A y; exp [ -( h 1t D LI mCP)
J
AJ; =35-6=29°C
Mass flow rate of air
m= ·*= }l ¥-=
Pi
RJ;
97kPax27/60m3 /s
0.287kJ/kgKx279.15 K
= 0.5448 kg/s
Surface area,
1tD L=1t(0.3 m){16 m)=4.8 1tm2 =15.08 m2
AT = 29ex [- 15.08h
•
p 0.5448CP
or
l
l¾e = 35-29 exp [ -27.68 h/CP
JI
Properties of air are to be evaluated at the bulk mean temperature, 'liim =.!_ ('li,i +'Ii,.). But The is
2
unknown. Hence, the solution calls for a trial and error procedure.
The duct is pretty long (L = 16 m) and Tw = 35°C with Thi= 6°C. First guess therefore can be
1
The=24°C, so that, Thm =-(6+24)=15°C
2
Forced Convection: Internal Flow
r
9.47
Trial I: The relevant properties at 15°C for air are:
k= 0.02476 W/m °C
Pr= 0.7323
µ = 1.802 x 10-5 kg/m s
cp = 1001 J/kg 0 c
Reynolds number,
pV D
4m
4x0.5448kg/s
Re0 = - - = - - = - - - - - - - - - " - - - µ
1tDµ 1t(0.3 m)(l.802xl0- 5 kg/ms)
= 1.283 x 105
(> 2300)
~
Turbulent
Since !:.... = 16 m = 53.3 » 10, the flow is fully developed.
D 0.3m
Using the Dittus-Boelter correlation:
hD
Nu0 = - = 0.023(Re0 ) 0·8 (Pr)D- 4
k
h = 0.023!.-(Reo)0 ·8 (Pr)°" 4
D
or
= 0.023x 0.0 2476 W/m°C x(l.283x10 5 )D- 8 x(0.7323) 0 .4
0.3 m
= 20.46W /m2 K
The= 35-29 exp [ -27.68 x20.46/1007]
=18.5°C
The(estimated) = 24°C
but
The(computed) = 18.5°C
Hence, another trial is necessary with The (computed) of first trial as The (estimated) for the second trial.
Trial II: Let Thm = 185 + 6 = 12.25°C"" 12°C
2
Properties of air at 12°C are:
k = 0.02456 W /m °C
cp = 1006.45 J/kg 0 c
µ = 1.7888 x 10-5 kg /m s
Pr= 0.733
4m
4x0.5448 kg/s
Re 0 = - - = - - - - - - - - - - n D µ 1tx0.3 mxl.7888x10- 5 kg/ms
= 129.26x103
9 .48 ,
Heat and Mass Transfer
h = 0.023x 0.0 2456 W/m°C x(129.26x10 3 ) 0 ·8 x(0.733) 0 .4
0.3 m
= 20.42 W/m 2 °C
Tbe = 35 - 29 exp [- 27.68 X 20.42 / 1006.45]
= 18.46°C"' 18.5°C
=
As Tbe(computect) Tbe(estimated) = 18.5°C, this is the final answer. Hence, the temperature of air exiting
the duct and entering the compressor= 18.5°C
(Ans) (a)
Heat-transfer rate,
Q=mCP(Tbe -Tbi)
={0.5448 kg /s) (1006.45 J / kg 0 C) (18.5-6)°C
= 6855 W
(Ans) (b)
Example 9.19 ~ In a gas-cooled nuclear reactor, the 20-mm diameter, 0.8-m long coolant tubes are used.
Helium as a coolant gas enters such a tube at 427°C with a mass flow rate of 0.5 kg/min
and leaves at 827°C. (a) Calculate the constant tube wall temperature and the rate of heat removal. (b) If helium
is replaced with air as a coolant, find its flow rate and exit temperature for the same rate of heat transfer and
tube surface temperature.
Properties at the bulk mean temperature of 627°C, I atm are
Fluid
I-----Helium
Air
Pr
k (Wlm 0 C)
µ(kglm s)
Cp(kJ!kg 0 C)
---------------1
5.193
0.330
41.4 x Io-6
0.654
0.062
39.81 x Io- 6
1.121
0.720
Known: Helium is heated through a tube under constant wall temperature conditions.
Find: (a) Tube surface temperature, T.( 0 C). Heat removal rate, Q(W). (b) Mass flow rate, m(kg / min)
and outlet temperature, Tbe( 0 C) for air as a coolant in place of helium.
m
C r. = r
D = 20 mm
Schematic:
Helium
Tb,=427°C
m= 0.5 kg/min
)
-
-
Tbe=827°C
.
:""'_-_-_-_-_-_-_-_-_-_ _L_=_O-.S-m-~~~~=======-
Fig. 9.22
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Constant tube surface temperature.
(4) Fully developed flow.
Analysis: (a) To ascertain if the flow is laminar or turbulent, Reynolds number needs to be determined.
_pV D _ 4m _
4(0.5/60)kg/s
Re0 - - - - - - - - - - - - - - - - - µ
1tDµ 1t(0.02m)(41.4x10-6 kg/ms)
=12814
The flow is turbulent.
Forced Convection: Internal Flow
r
9.49
( > 10). Pr is 0.654.
LID=0.8 m I 0.02 m=40
Using the Dittus-Boelter correlation:
Nu 0 =0.023 = (Re0 ) 0 -8 (Pr)D.4
=0.023(12814) 0 -8 (0.654)0.4
=37.5
Average heat transfer coefficient, h = !.._ Nu0
D
= 0.3 30W/m °C x37.5 = 619 W /m 2 °C
0.02m
For flow through a tube with constant wall surface temperature (T. = const), we have
hrcDL =In T,. -Ti,i
mCP
-Ti,.
r.
hrcDL]
r. -Ti,. = er. -Ti,i) exp [ - -.
-m Cp
or
L)]
T,. [ 1- exp [ - hrcD
mcp
or
[ L]
= ¾e - ¾i exp - hrcD
mcp
Now,
hrcDL
mCP
(619W/m 20C)(rcx0.02mx0.8m) = 0 _7188
(0.5/60)kg/sx5193J/kg 0 C)
Hence,
Tube wall surface temperature,
r. = [827°C - 427°C exp (-0.7188 )] / 1 -exp (0.7188)
= 1201°c
(Ans) (a)
Rate of heat removal,
Q=mCp(Ti,. -Ti,J=(0.5/60 kg/s)(5193 J/kg C)(827 - 427)°C
0
= 17 310 W or 17.31 kW
(Ans) (a)
(b) For the same values of cooling rate, Q and surface temperature, T., with air as a coolant, the
pertinent relations are:
Q = mcp CI'ii. -Ti,J
T,. -Ti,. = exp[- h~DL]
T,.-Ti,i
mCP
9 .SO , Heat and Mass Transfer
4m
Re0 = - -
1tDµ
and
hD
Nu 0 = - 0.023(Re0 ) 0 -8 (Pr) 04
k
We have to evaluate both m and The for air using the properties of air at 900 K. An iterative procedure
is necessary.
Step I: Assume some reasonable value of The (0 C)
Step 2: From energy balance, calculate the flow rate min terms of The• Thi remaining same.
=17310W=m(kg/s) (1121 J/kg°C) (Tb.-427)°C
15.4415
m=---
(1)
i;,. -427
Step 3: Using this value of m with assumed The• find Re0 .
4m
1t(0.02m)(39.81x10-6 kg/ms)
16x105 m
Step 4: With this value of Re0 , evaluate h from Dittus- Boelter equation.
k
h = -x0.023(Re0 ) 0 -8 (Pr) 04
D
0.062W/m°C
= -----x0.023(16x 10 5 }° 8 (0.72}° 4 xm0 .s
0.02m
(2)
Step 5: Now we compute ½e using these values of m and h from the following relation.
h1tDL]
I;,.=~ -(~-7;,i)exp [ - .--
mCP
= 1207°C-(1207-427)°C ex [- 1tX0.02 mxO_g
p
1121J/kg°C
(.!!._)]
m
(3)
1rb. =1201 - 18oexp [-44.84x10-6(h/m)ll
Step 6: Compare this calculated The with assumed The" Repeat the procedure, till the two values of The
(assumed and computed) nearly match. The results are tabulated below.
Tb.(OC)
(Assumed)
m(kgls)
Equation (1)
h(W/m 2 °C)
Equation (2)
Tb.(OC)
Equation (3)
650
6.924 X 10- 2
678.3
704.3
2
710
5.456 X 10- 2
560.6
715
3
715
5.362 X 10-2
552.8
715.7
4
716
5.343 X 10r2
551.27
715.9
Trial
Forced Convection: Internal Flow
r
9.5 I
In the fourth trial convergence is reached. Hence, with air as the coolant, m=(5.343xl0-2 kg/s)
(60 s/1 min)= 3.2 kg/min
Tbe = 716°C
And
(Ans) (b)
Comment: For achieving the same heat removal rate with air, the flow rate required will be 6.4 times
greater than that with helium .
Example 9.20 ~ Water is heated by flowing through a horizontal tube of I-cm diameter maintained at constant
surface temperature of 380 K. Water can flow using either (a) a small power pump so
that the mean flow velocity is 0.08 mis or (b) a larger pump with the average flow velocity of 0.5 mis. Water
enters the tube at 310 K and is to be heated by 40 K. Determine the length of the tube needed to effect this
rise in water temperature.
The following properties of water may be used (Tbm = 330 K):
k = 0.650 W/m K
µ = 0.489 x I0-3 Pas
Pr= 3.15
cP = 4.184 kJ/kg K
p(atTb;=310K)=993 kg/m 3
Known: Water flowing through a horizontal tube at constant wall temperature with specified wall
temperature.
Find: Length of the tube when (a) V = 0.08 mis; (b) V = 0.5 mis.
Schematic:
Tube, Tw=380K, D= I cm
Water
Tb;=310K --"'<---+-
V=0.08m/s (a)
V= 0.5 mis (b)
L--------.i
Fig. 9.23
Assumptions: (1) Fully developed flow. (2) Steady operating conditions. (3) Smooth inner surface of the
duct. (4) Constant tube wall temperature.
Analysis: (a) Average water velocity, V = 0.08 mis :
Mass flow rate of water entering the tube,
m=p~D 2 V
4
=993 (kg/m3 )x~x(o.01)2 (m)2 x0.08(m/s)
4
=6.239x 10-3 kg/s
Reynolds number,
4m
4x6.239xl0-3 (kg/s)
Re0 = - - = - - - - - - - - - - - - 1tD µ 1t x O.Ol(m)x 0.489 x 10-3 (kg/ms)
=1624.5
(<2300)
[laminar flow regime]
9.52 ,
Heat and Mass Transfer
For constant tube wall temperature, Nusselt number is
hD
Nu0 =-=3.66
k
Convection coefficient is
h=3.66x
0.65 (W/mK)
( )
0.01 m
=237.9 W / m 2 K
Tube length,
(r_ -'Eb.)
mCP
1
L=--ln
w
h TC D
Tw - ¾e
= 6.239xI0-3 (kg/s)x4184 (J/kgK) x/n(380-310) (K)
237.9 (W/m 2 K)xTCx0.01 (m)
380-50 (K)
(Ans) (a)
=2.96m
(b) Average water velocity, V= 0.5 mis
Mass flow rate of water,
m=p~D2 V
4
=993( kg/m3 ) x~ x (o.01)2(m}2x 0.5(m!s)=0.039kg/s
4
Reynolds number,
4m
4x0.039 (kg/s)
Re0 = - - = - - - - - - ~ ~ - - , - - - TC D µ TCX0.0l(m)x0.489xI0-3 (kg/ms)
=10153.4
(> 2300)
[turbulent flow regime]
Using Dittus-Boelter correlation, the convection coefficient is
k ( Re )0.8 ( Pr )0.4
h = Nu 0 - k = 0.0230
D
= 0.023x
D
0.65 (W/mK)
08
04
( ) x(I0153.4) · (3.15) · = 3795.4W /m2 K
0.01 m
Length of the tube,
(r_ -'Eb.)
mCP
1
L=--ln
w
h TC D
Tw - ¾e
=
0.039(kg/s)x4184 (J/kgK) x/n{(380-3IO)K}
3795.4 (W/m 2 K)xTCx0.01 (m)
(380-350)K
=1.16m
(Ans) (b)
Forced Convection: Internal Flow
r
9.53
Check:
!:_= 1.1 6 m =116 (> 10) OK
D
Example 9.21
0.01 m
~ Atmospheric air at I6°C enters a 12.5-mm ID circular tube held at a constant wall temperature
of I00°C with a mean velocity of 30 mis. Using the Dittus-Boelter correlation, calculate the
exit air temperature and the pressure drop in cm of water for a tube length of (a) IO cm, and (b) I00 cm.
Properties of air:
___
T( C)
,__
p(kg/m3)
Cp(Jlkg K)
k(Wlm K)
v(m 2/s)
Pr
20
1.204
1007
0.02514
15.16 X 10-6
0.7309
30
I . I 64
I007
0.02588
16.08 X Io- 6
0. 7282
40
I . I 27
I007
0.02662
17.02 X Io- 6
0.7255
50
1.092
I007
0.02735
17.98 X 10-6
0.7228
0
Known: Air is heated while passing through a tube at constant surface temperature with a specified velocity.
Find: Exit temperature of air, Tbe( 0 C) and pressure drop, M ( cm H 2 0) for (a) L = 0.1 m and (b) L = 1 m.
;;;::D=O.Ol25m
Sch~adc,
Tb;=l6°c-
ll1_
[ : T,= 1oo•c
--V=30m/s
)-rbe=?
i------- L ------.i
T
Tube surface, T,= I00 °C
Tbe=75. l°C
(b)
Tbe=25.3°C ·
(a)
Tb;= l6°C
~----~--------~~L
0
0.1 m
I.Om
Fig. 9.24
Assumptions: (I) Steady operating conditions. (2) Air is an ideal gas . (3) Constant tube surface
temperature. (4) Smooth tube.
Analysis: From energy balance:
where
or
-/n (-7'._-_Tb_
e ) = _-h_1t_D_L
r. -1i,;
mCP
9.54 ,
or
Heat and Mass Transfer
Exit air temperature,
(A)
Properties are to be evaluated at the bulk mean temperature ¾m = .!_ (¾i +Ti,.) . Since The is unknown,
2
a trial-and-error procedure is adopted.
(a) Length L = 0.1 m: Let Ti,.= 24°C so that
¾m = 16+24 = 200c.
2
Mass flow rate of air, m= p Ac V = p ~ D 2V
4
= {1.204 kg/m3 ) ( ~x 0.0125 2 ) m2 (30 m/s) = 4.4326 x 10-3 kg/s
Reynolds number,
Re -_VD __ (30m/s)(O.OI25m) __ 24736
v
15.16x1Q-6 m2 /s
~
bl fl
1ur u ent ow
7',
Using Dittus-Boelter equation:
Nu= hD = 0.023{Re)°" 8 (Pr)°" 4
k
Convection coefficient,
h = 0.023x 0·02514 W/mK x(24 736)°" 8 (0.7309)°" 4
0.0125 m
= 133.47W/m2 K
hrcDL
{133.47W/m2 K){rcx0.0125mx0.lm)
mCP
{4.4326xI0-3 kg/s){1007 J/kg K)
- - = ~ , - - - - ~ - - , - - , - - - - - - - - , - - = 0.1174
From Equation (A), the exit air temperature is
¾e = 100 °C -(100 - l6) 0c exp (-0.1174) = 25.3°C
16+ 25.3
Check: ¾m = - - - = 20.65°C "" 20°C as assumed.
2
No more trial is therefore necessary.
Pressure drop,
(Ans) (a)
Forced Convection: Internal Flow
where friction factor, f = [1.82 log10 Re -1.64
r
9.55
r2
= [1.82 log 24736-1.64f2 = 0.0248
:. for L = 0.1 m, the pressure drop is determined from
3 ){30m/s)2 1
AP=0.0248 ( -O.lm
- -){1.204kg/m
----- - - - lN
- -I
0.0125 m
2
lkgm/s 2
=107.5 N/m2 I lcm H20 I
98.1 N/m 2
(Ans) (a)
=1.096 cm H2 0
(b) Length, L = 1.0 m
To assume a reasonable value of The• we note that
hrcDL =a-L.a(b) =4._b) =~=lO
mCP
a(a) 4._a) 0.1 m
a(b) =0.1174x10"" 1.174
½e ""100 -(100 -16)exp(-1.174) =74°C
1
1
2
2
½m = -(Tiii + ¾e) = -(16+ 74) = 45°C
From the given table, by interpolation, the properties of air at 45°C are
p = 1.109 kg/m 3 ,
CP = 1007 J/kg K,
v = 17.5 x 10-<:i m2/s
k = 0.02699 W/m K,
and Pr= 0.7241
Following the same procedure as in (a), we have
m= (1.109)( ~x0.01252}3o) = 4.083x10-3 kg/s
Re= 30xO.Ol 25 =21429
17.5 X 10-6
( > 2300)
h = 0 ·02699 x0.023(21429)°" 8 (0.7241)°" 4 = 127.27 W/m2 K
0.0125
a= hrcDL = 127.27xrcx0.0125xl.O = 1. 216
mCP
4.083x10-3 x1007
:. exit air temperature,
¾e = 100-(100-16)exp[-1.216] = 75.1°C
(Ans)
9.56 ,
Heat and Mass Transfer
Check: ¾m = 16 + 75 · 1 = 45.5°C"" 45°C
2
No more trial is therefore necessary,
f =(1.821og21429-1.64r2 =0.0257
Pressure drop,
1.0
302
AP= 0.0257x--xl.109x- = 1026N/m2
0.0125
2
=10.46 cm H2 0
(Ans) (b)
Example 9.22 ~ Air at atmospheric pressure and I00°C enters a 4-cm-diameter, 2-m-long tube with a velocity
of 9 mis. A I kW electric heater is wound on the outer surface of the tube. Find (a) the
mass flow rate of air; (b) the exit temperature of the air, and (c) the wall temperature at the outlet assuming
uniform heat generation.
Known: Air flows through a tube and electrically heated under uniform surface heat flux conditions.
Find: (a) Air flow rate; (b) Air exit temperature; (c) Tube wall temperature at exit.
Schematic:
Air
P= I atm
V=9mls
Tb;= 100°c
r•-0.. . . . . .____ _
o=t,m~-}-r~
t
or 373.ISK
t
.-------L=2m------~
Fig. 9.25
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Uniform heat flux. (4) Fully
developed flow. (5) Air is an ideal gas.
Analysis: Uniform and constant surface heat flux,
.
E
q = JL = ~ =
lOOO W
= 3979 W/m 2
w
A. rcDL n(0.04m)(2m)
From energy balance:
h A. (T. - Ti, )exit = mCP (Ti,. - ¾i ) = qw A.
where the mass flow rate of air is
m=p;n1« AcV
p TC
=---D2 V
RTi,i 4
=(
101.325 kPa
)(~xo.042 m2) (9 m/s)
0.287 kJ/kg Kx373.15 K 4
=0.0107 kg/ s
(Ans) (a)
Forced Convection: Internal Flow
r
9.57
Since qw = h (T.. - Ti,.), the wall temperature at the outlet is
(A)
We note that
Air outlet temperature is
¾e = ¾i +(Q/mcp)
With Cp(at ¾i =100°C)=1009 J/kg°C,
r.. = 1000c +
1ooo w
(0.0107 kg/s) (1009 J/kg°C)
be
= 192.6°C
1
Bulk mean temperature, 'I'iim = -( ¾i + ¾e)
2
=.!.(loo+ 192.6) = 146.3°C
2
We recalculate The with CP(at ¾m =146.3°C)=1014 J/kg°C.
r.. = 1000c +
be
1000 w
(0.0107 kg/s) (1014 J/kg°C)
=100+92.17 = 192.17°C"" 192.2°C
Now 'I'iim = (100+ 192.2)/2 = 146.1 °C
and CP at 146.1 °c"" 1014 J/kg 0c.
Hence, finally
The
= 192.2°C
(Ans) (b)
To determine r•• , from Eq. (A), we need to determine the convection heat transfer coefficient, h. To
ascertain if the flow is turbulent, we find the Reynolds number. Properties of air are to be evaluated at
146.1 °C. These are obtained by interpolation from the property table:
k = 0.03416 W /m °C
µ = 2.368 X 10-5 kg/m S
Pr=0.7033
pVD
4m
4(0.0107 kg/s)
- µ - 1tDµ - 1t(0.04m) (2.368x10-5 kg/ms)
Re---------,---~---~--~
=14383
(> 2300)
9 .58 ,
Heat and Mass Transfer
Hence, the flow is turbulent.
(fully developed flow).
L/D= 2m/0.04 m = 50
Using Dittus- Boelter equation:
Nu= hD = 0.023(Re)°" 8 (Pr}°- 4
k
[n=0.4 since air is gettingheated]
= 0.023 (14383)°" 8 (0.7033)°" 4
hD
=42.35=k
Convection coefficient,
h = Nu.!5_ = 42 _35 (0.03416 W / m°C)
D
0.04 m
= 36.17 W /m 2 °C
It follows from Eq. (A):
T. = 192 _20
se
c+(
3979 W/ m 2 )
36.17W/ m 2 °C
=302.2°C
(Ans) (c)
Example 9.23 ~ Air at 2 bar pressure and 200°C temperature gets heated as it flows through 2.5-cm-diameter
tube with a velocity of IO m/s. A constant heat flux condition is maintained at the wall and
the wall temperature is 20°C above the air temperature all along the length of the tube. Calculate:
(a) Heat transfer per unit length of tube
(b) Increase in bulk temperature over a 3-m length of tube
Use the following correlation and air properties:
Nu = 0.023 Re0-8 pl).4
µ = 2.57 x 10- 5 N s/m 2
k = 0.0385 W/m K
Known: Air gets heated in a tube under constant heat flux conditions.
Find: (a) QI L (W / m); (b) ~Ti, {°C)for L = 3 m.
Schematic:
Air
P=2bar
Tb;=200°C
V= IOm/s
-------L=3m------~
Fig. 9.26
Assumptions: (1) Steady flow conditions exist. (2) Uniform and constant wall heat flux. (3) Air is an
ideal gas. (4) Fully developed flow.
Forced Convection: Internal Flow
Analysis: Heat-transfer rate for constant surface heat flux (qw = const) condition is given by
Q=qwAsh
(Tw-Ti,)
(rcDL)
'---..r---J
constant throughout
Also,
where ATb is the increase in the bulk temperature of air.
Reynolds number,
Re= pV D
µ
where
P
2x10 2 kPa
I 1 kJ I
p =RT= (0.287 kJ/kg K) {200+273.15)K lkPa m 3
=1.4728 kg/m3
Then
Re= (1.4728 kg/m3 ) (10 mis )(2.5 x 10-2 m) 11 N s2 I
2.57 x 10-5 N s/m2
1kg m
(> 10 000)
=14 327
~
Turbulent
Prandtl number
Pr= CPµ = (1025J/kg K)(2.57x10- 5 Ns/m2 )11WI 1kg I
k
0.0385 W/m K
lJ/s 1N s2
= 0.684
Nusselt number,
Nu= h D =0.023(Re)°" 8 (Pr)°" 4
k
= 0.023(14327)°" 8 (0.684)°" 4
= 41.75
Convection heat-transfer coefficient,
k
41.75(0.0385 W/mK)
h =Nu-=--~---~
D
0.025 m
= 64.3W/m2 K
Heat-transfer per metre tube length is
r
9.59
9.60 ,
Heat and Mass Transfer
= (64.3 W/m 2 K) {1tx0.025 m)(20°C or K)
=101.0W/m
(Ans)
For 3-m-long tube,
Q ={101.0W/m){3m) = 303 W
m=mass-flow rate of air = pAcV
where
1t
=p-D 2 V
4
= ( 1.4728 kg/m 3 ) (
~ x 0.0252 m2 ){10 mis)
= 7.23x 10-3 kg/s
It follows that
303 W = (7.23x10-3 kg!s) (1025 J/kg K)(ATi,)K
Increase in bulk temperature of air over L = 3 m is
AT, =
h
303
(7.23x10-3 )(1025)
=40.9°C
or
K
(Ans) (b)
Comment: Note the use of proper area of the tube in calculations. To find the mass flow rate, one
Ac(=~
must use the cross sectional area
n 2 ) while for finding the heat transfer rate, surface area
4
A.(= 1tDL) should be used.
Also note that air properties like k, CP and µ are essentially independent of pressure and depend
primarily on the temperature of the fluid. However, the density of air depends strongly on both
temperature and pressure. The tabulated properties of air are invariably for P = I atm. In the
present case, the pressure of air is 2 bar and hence p must be evaluated from the ideal-gas equation
(P = p RT).
Example 9.24 ~ Hot air enters at 100°C an uninsulated sheet metal duct of 20-cm-diameter with a mass flow
rate of 4.5 kg/min and after a distance of 5 m cools to 80°C. The heat transfer coefficient
between the outer surface of the duct and the ambient air at 5°C is estimated to be 6 W/m 2 0 C.
(a) Determine the rate of heat loss from the duct over the 5-m length.
(b) Calculate the heat flux and the duct wall temperature at x = 5 m.
The following properties of air may be used:
CP = I.009 kJ/Kg °C
k = 0.02953 W/m °C
µ = 20. 96 x I o---6 kg/ms
Pr= 0.7154
Forced Convection: Internal Flow
r
9.61
Known: Hot air flows through a duct.
Find: (a) Q(W); (b) qw(L){W/m 2 ); Tw(L)( 0 c).
Schematic:
Cold ambient
air
Hot air
r==s·c
h0 =6W/m2"C
r
L-
Duct (D =0.2 m)
'0.~.____________·_J
.(.
Tb;= I00°C -
l_...., Tbe=80°C
m=4.Skg/min
~-;
i-----L=Sm-------+1
Fig. 9.27(a)
Assumptions: (1) Steady operating conditions exist. (2) Uniform heat-transfer coefficient at the duct's
outer surface. (3) Duct wall thermal resistance is negligible. (4) Constant properties.
Analysis: (a) From energy balance, the heat loss from the duct is determined to be
Q = mCP (¾i - The)
= ( 4 ·5 kg)(1009_{_)(100-80) c
0
60 s
kg°C
=1513.5W
(Ans) (a)
(b) Thermal circuit (at x = L = 5 m) is shown below:
-
Tbe
Tw(L)
T=
qw(L) ••--~VVv,----tet----VVv---e•
h;(x=L)
h0
Fig. 9.27(b)
Heat flux at x = L = 5 m is given by
To evaluate inside heat-transfer coefficient, we first calculate the Reynolds number.
Re_ p VD _ p D (
m )- 4 m
- - µ - - µ prcD 2 /4 - rcDµ
4(4.5/60 kg/s)
rc(0.2m){20.96x10-<i kg/ms)
=----'------'----
=22 780
(>2300)
9.62 ,
Heat and Mass Transfer
The flow of air is turbulent and the Dittus-Boelter correlation can be used since ( L/ D) = ~ = 25 ( > 10).
0.2m
As air is being cooled, with n = 0.3, we have
Nu =0.023(Re)°" 8 (0.7154)°" 3
=63.7
Hence,
,.
_ N ~ _ 63 7 0.02953 W/m °C
uD . X
0.2m
'\(x=L) -
= 9.405 W /m 2 °C
Heat flux,
{80-5)°C
qw (L)= [-l-+_!_Jm2oc
9.405 6 W
= 274.74 W /m2
(Ans) (b)
Tube wall temperature at x = L = 5 m is determined from the thermal circuit representation.
or
=Soc+ 274.74 W/m2
6W/m2 °C
= 50.8°C
(Ans) (b)
Comment: This case is neither related to constant wall temperature nor constant wall heat flux condition.
For qw =const, Q is 1513.5 W. This is because as x increases, both hi and (Ti,(x)-T=) decrease. This
is reflected in the reduced value of qw towards the pipe exit.
Example 9.25 ~ Air at I atm and 260°C flows at 12 m/s through a thin-walled metal tube of 25-mm diameter
with the surroundings at I 5°C. The combined convection and radiation heat transfer
coefficient between the exposed tube surface and the surrounding air is 18.7 W/m 2 K.
(a) Determine the heat loss from the tube per metre length. (b) Calculate the percentage decrease in the
heat loss if an insulation of thermal conductivity 0.173 W/m K and 25-mm thickness were added to the tube.
The combined convection and radiation coefficient for the outer side of the insulated tube is 11.4 W/m 2 K.
Thermophysical properties of air at 260°C:
k = 0.0425 W/m K
Pr= 0.680
v = 42.15 x Io---6 m2/s
0
8
Correlation for Internal Turbulent Flow: Nu 0 = 0.023(Re0 ) - (Pr)0.4
Forced Convection: Internal Flow
r
9.63
Known: A tube carrying hot air at a given velocity loses heat to the surroundings with and without insulation.
Find: (a) Rate of heat transfer from the bare tube (b) Rate of heat transfer from the insulated tube.
Schematic:
lll
h0 = 18.7W/m2 K
r =o =l5°C
Air
-------+-
V= 12 m/s
D;=0.025m
Tb;= 260°C = T= I
h;=?
Thin-walled tube
Fig. 9.28(a)
r;= 1.25 x I0-2m
iii
r0 = 1.25+ 25
h0 = I I .4W/m2 K \
=3.75cm
T=0 = l5°C
k = 0.173 W/m K
Fig. 9.28(b)
Assumptions: (I) Steady-state conditions. (2) Fully developed internal flow. (3) Negligible temperature
drop across the tube wall. (4) Constant properties and uniform heat transfer coefficients.
Analysis: (a) Rate of heat loss (without insulation):
where
= 260-15 = 245°C or K
A= rcDL = rcx0 .025x 1
= 7 .854 x 10-2 m2 per m length
Overall heat-transfer coefficient,
Here h0 = 18.7 W/m2 K, h; needs to be evaluated.
VD (12 mis) (0.025 m)
Re =-=--~~-~=7117.44
0
v
(42.15x10-6 m2 /s)
9.64 , Heat and Mass Transfer
The flow is thus turbulent.
Using the correlation:
Nu0 =0.023 (R~)°" 8 (Pr)°" 4
=0.023 {7117.44)°" 8 (0.680)°" 4 = 23.8
k
17;. =Nu0 x -
Di
= 23 _8 x 0.0425 W/mK
0.025 m
=40.46 W/m2 K
l
l
U= [ - - + -
40.46
]-I
18.7
=12.79W/m 2 K
Qwo = (12.75 W/m2 K}(7.854x10-2 m2 ){245 K} perm length
=246.lW/m
(Ans) (a)
(b) Heat loss rate (with insulation):
where
1
,;
,;,
,; 1 ]
17;.
k
,;
,;,h0
-1
ui = -+-ln-+-[
[
= _1_+ 0.0125 ln 3.75 + 0.0125 x-140.46 0.173 1.25 0.0375 11.4
=7.5 W/m2 K
-1
J
[Note that h0 in this case is 11.4 W!m 2 K]
Heat-transfer rate (with insulation),
Qw = Ui (1t Di L) Ai:iveran
= (7.5 W/m2 K} (0.07854 m2 ){245 K} perm length
= 144.3W /m
Percentage decrease in heat loss
= Qwo_ - Qw X 100
Qwo
= (246.1-144.3)x lOO= 4 1.4 %
246.1
(Ans) (b)
Forced Convection: Internal Flow
r
9.65
Example 9.26 ~ A thick stainless steel pipe (D; = 20 mm and D0 = 40 mm) (k = 15 W/m 0 C) is heated by
passing an electrical current through it which provides a uniform volumetric heat generation
rate of I MW/m 3 of pipe material. The outer surface of the pipe is well insulated, while water flows through
the inside at the rate of 0.1 kg/s.
(a) If the water inlet temperature is 20°C and the desired oudet temperature is 40°C, what is the required
length of the pipe?
(b) What is the temperature of the inner surface of the pipe at the outlet end?
(c) Sketch the axial variation of the temperature along the inner surface of the pipe.
(d) What is the pressure drop in the pipe?
(e) What is the location and value of maximum pipe temperature?
Known: A pipe with outer surface adiabatic (insulated) experiences uniform heat generation, Flow rate,
inner and outer diameters, inlet and exit temperatures of water flowing through the pipe.
Find: (a) Pipe length required to attain the desired exit temperature of water. (b) Pipe inner surface
temperature at the outlet end. (c) Axial variation of pipe inner surface temperature, (d) Pressure drop in
the pipe. (e) Location and value of maximum pipe temperature.
Schematic:
Insulated surface =i
I
i~
Qcond
r-
t
Tw, max(L)=Ts(r0 )
~'.(~):a~J;;;ainless steel)
I------- L - - - - -I
Fig. 9.29
Assumptions: ( l) Steady operating conditions exist. (2) Outer surface of pipe is insulated. (3) Uniform
volumetric heat generation. (4) Constant properties. (5) Fully developed flow conditions. (6) Onedimensional radial heat conduction in pipe wall. (7) Pipe surface is smooth.
l
Properties: Water: At bulk mean temperature, 'Ii:,m = -(20+40)=30°C:
2
p = 996kg/m3
CP = 4178 J/kg °C
µ = 0.798x 10-3 kg / ms
Pr= 5.42
k=0.615 W/m c
0
Analysis: (a) Considering the control volume about the inner pipe and performing an energy balance, one has
Qin = Est = mCP (Ii:,. - TbJ
The rate of uniform internal heat generation, £gen= Qin because the pipe's outer surface is effectively
insulated.
Egen --¥---!!_(D
- q - q 4 o2 -Di2 )L- m· Cp (1:be -7:)
bi ·
9.66 , Heat and Mass Transfer
From which
Length of the pipe required,
L = 4mCp(Ti,. -Tbi) = 4(0.lkg/s) (4178J/kg 0C) (40-20)°C
q1t(Di-D?)
(10 6 W/m3 ) 1t (0.042 -0.02 2 ) m 2
(Ans) (a)
=8.87 m
(b) At the outlet end, if h is the local convection coefficient and T. (L) is the inner pipe surface
temperature.
Qin = Qconv = h(7tDiL) [T,,(L)-Ti, 0 ] = q ~(D'i - D?)L
It follows that
(A)
To determine the convection coefficient, h, one must first evaluate the Reynolds number,
pVD
4m
4(0.1 kg/s)
Re= - -1 = - - = -----'-----''---'---- = 7978
µ
1tDiµ 1t(0.02m) (0.798x10-3 kg/ms)
~
Turbulent flow.
As LI Di =8.87 m/0.02 m=443 (>10), the temperature profile is fully developed.
Using Dittus-Boelter correlation,
Nu =0.023(Re)0-8 (Pr)0.4
[n = 0.4 as the water is getting heated up]
hD
Nu = -1 = 0.023(7978) 0·8 (5.42) 0 .4
k
=59.8
h=..!5.._Nu= 0.615W/mocx59.8
Di
0.02m
= 1839 W/m2 °C
Substituting in Equation (A),
T. (L) = T. (r,) = 400C+ 106 W/m2 (0.04 2 -0.02 2 )m 2
1
•
•
4x1839W/m 2 °Cx0.02m
=48.2°C
(c) Heat flux at the pipe surface,
Q
£gen
·
q8 = - - = - - = constant as Egen is uniform.
7tDiL 7tDiL
(Ans) (b)
Forced Convection: Internal Flow
r
9 .6 7
For constant pipe surface heat flux condition,
h and ('I'. - Tb) = constant for fully developed flow.
Entrance effects are confined within a tube length of 10 diameters
i.e., thermal entry length, L e,th = 10 x 0.02 m=0.2 m
The axial variation of inner surface temperature is sketched below:
T(°C)
T,(L) = 48.2°C
'
hermal
entranc~ Fully developed region
~r=e=io~n~'- - - - - - - - - - ~ - - - - - x(m)
Le,th = 0.2m
x= L
Fig. 9.30
The physical situation corresponds to uniform wall heat flux and Tb increases linearly with x. In the
fully developed region, Ts too increases linearly with x.
(Ans) (c)
(d) Pressure drop in the pipe is
L pV 2
t:.P=f D-2-
where
V=_!?!_= (0.1 kg/s / 996kg/m3 )
(1t I 4)(0.02 m) 2
p Ac
and
f = [0.79/n Re-1.64]-2
0 _32 m/s
= [0.79 /n 7978-1.641- 2 = 0.0336
t:.P = (0. 0336 ) ( 8.87 m) ( 996 kg/m3 x 0.32 2 m2 /s2 ) I l N I
0.02m
2
lkgm/s2
(Ans) (d)
= 760N /m 2 or Pa
(e) The maximum wall temperature clearly exists at the insulated outer surface at the exit (x = L), i.e.,
Tw ,max
= 'f.(,;,)
With heat generation in the pipe wall in which there is radial one-dimensional conduction, the
governing differential equation is
_!__:{_(r dT) = -q
r dr
dr
k
9.68 , Heat and Mass Transfer
Separating variables and integrating with respect to r, one gets
dT
q r2
r-=---+C
dr
2k
I
Boundary condition I
-o
dTI
dr r=r;,
qr
dT
dr
C1
r
-=--+2k
Integrating again, the temperature distribution is
-qr2
T(r)=--+C1 Zn r+C2
4k
Boundary condition 2
T(r= ,;_) = T8
(qr2) Zn ri +C
-qr.21
T(s) = - + - 0
4k
2
2k
-q,r qr;,2
C2 = - - - - Zn r,+T..
Hence,
4k
2k
I
S
The temperature distribution is then determined to be
-2
q 2
2
qr;,
r
T(r)=--(r -r, )+-Zn-+T..
4k
11
S
q(,;,2-,r) q,;,2
"c,
I
2k
At the insulated pipe surface (at r = r0 ),
Tw max= T(r = ,;,) = - - - - - + - - l n - + T.(r=r.)
'
4k
2k
11
•
At the outlet end of the pipe, the inner surface temperature is
T.(L) = T.(,;.) = 48.2°C
Maximum pipe temperature that occurs at the exit end (at x = L) is
T..
= 1Q6 W/m3 [ (0.02m)2-(0.0lm)2] + 106 W/m3(0.02m)2 Zn 0.02m + 4820C
4(15W/m 0C)
w,max
=52.4°C
2 x 15W/m°C
O.Olm
(Ans) (e)
Forced Convection: Internal Flow
r
9.69
Example 9.27 ~ Water at 60°C flows at the mass flow rate of 10 kg/s through a pipe with a 75-mm inside
diameter. Calculate the heat-transfer coefficient using (a) the Colburn analogy, (b) the Prandtl
analogy and (c) the von Ka'rma'n analogy.
Properties of water at 60°C:
p = 983.3 kg/m 3
µ = 0.467 X I0-3 kg/m s
k = 0.654 W/m°C
Pr= 2.99
Known: Flow conditions of water flowing through a pipe.
Find: Heat transfer coefficient using (a) Colburn analogy, (b) Prandtl analogy, and (c) von Ka'rma'n analogy.
Schematic:
Water
::::, O
__)__
o=_o~},_s_m_ _ _ _ _ _) Fig. 9.31
Assumptions: (1) Steady operating conditions exist. (2) Constant properties. (3) Smooth pipe. (4) Fully
developed flow.
Analysis: Reynolds number,
pVD
4m
Re0 = - - = - -
µ
1tDµ
4(10kg/s)
1t{0.075m)(0.467x10-3 kg/ms)
=------..C.....-'----=363523
(> 2300)
The flow is turbulent.
(a) Colburn analogy
Nusselt number,
Nu0 = f Re0 Pr113
8
f=0.184(Re0 rlis
where friction factor,
[3x 10 4 <Re0 < 10 6]
1 = o.1s4{363523r0·2 = 0.0142
Nu0 = 0·0142 {363523){2.99f3
8
=930.2
:. heat-transfer coefficient, h = !!.._ Nu0
D
= 0.654 W/m °C x 9302
0.075m
= 8112 W /m2 °C
(Ans) (a)
9.70 , Heat and Mass Transfer
(b) Prandtl analogy
(f /8)Re0 Pr
Nu0 = - - = = - = ' - - I+ 5~f /8 (Pr- I)
_ (0.0142/8)(363523){2.99)
-1+5~(0.0142/8)(2.99-1)
=1359.3
k
h=-Nu0
D
= 0.654 W/m °C X 1359.3
0.075m
=11853W /m1 °C
(Ans) (b)
(c) Von Karman analogy
(f /8)Re0 Pr
Nu0 = - - - - - - - ------
1+ 5~(! /8) [(Pr-1)+ zn{l +¾(Pr-I)}]
(0.0142 I 8) (363523) (2.99)
=--===------------
1+512J~~T(2.99- l)+ln{1 + ¾(2.99-1)
}J
=1187.83
k
h=-Nu0
D
= 0.654 W/m oc xl 187.83
0.075m
=10358 W /m1 °C
(Ans) (c)
(C) NON-CIRCULAR TUBES: TURBULENT FLOW
Example 9.28 ~ Hot air at atmospheric pressure and 80°C enters an 8-m long uninsulated square duct of
cross-section 0.2 x 0.2 m that passes through the attic of a house at a rate of 0.15 m3/s.
The duct is observed to be nearly isothermal at 60°C. Determine the exit temperature of the air and the rate
of heat loss from the duct to the attic space.
Known: Heat loss from an uninsulated square duct as hot air gets cooled while passing through it.
Find: Air temperature at duct exit, The ( 0 C). Heat-loss rate, Q(W).
Forced Convection: Internal Flow
r
9, 7 I
Schematic:
Hot air
I atm, Tb;=80°C---;;;;;;;;~---¥=0.1Sm3/s
Fig. 9.32(a)
Tb;
(80°C)
T
Tbe
L'lT;
1
T
L'lTe
Tw=60°C
_1_
Constant wall temperature
Fig. 9.32(b)
Assumptions: ( l) Steady-state conditions. (2) The inner surfaces of the duct are smooth. (3) Air is an
ideal gas. (4) Fully developed flow. (5) Constant tube wall temperature.
1
Analysis: Properties of air need to be evaluated at the bulk mean temperature, Tbm =-(Thi+ The).
2
However, the exit air temperature, which is likely to be less than inlet air temperature of 80°C (because
the duct wall temperature is 60°C which is less than Tb) is not known.
Assuming Tbm = 350 K:
k=0 .030 W/m K
Pr= 0.700
CP = 1.009 kJ/kg K
V = 20.92 X lQ-6 m 2 /s
p = 0.9950 kg/m 3
4A
For a non-circular duct, the characteristic dimension is equivalent diameter, De = __c
For square cross section of each side, a= 0.2 m,
p
Area of cross section, Ac= 0.2 m x 0.2 m = 0.04 m 2
Perimeter, P = 4 a= 4 x 0.2 m = 0.8 m
9.72 , Heat and Mass Transfer
n. = 4x0.04m2 = 0.2 m
0.8m
VD.
Re=--
Reynolds number,
v
where V is the mean air velocity.
With
V= ¥- =O.l 5 m3/s=375m/s
•
'
Ac
0.04m2
(3.75 m/s) (0.2m)
Re=---~
=35851
20.92 x 10-6 m2 /s
Since Re> 10000, the flow is turbulent.
Entry length, 4, ""4 =10 D. =lOx0.2 m=2m which is much less than the duct length of 8 m.
Hence, the flow can be considered fully developed. The Prandtl number, Pr= 0.7 which is between 0.6
and 160. As the air is being cooled, the exponent in the Dittus-Boelter correlation is 0.3. The Nusselt
number can be determined from
Nu= hD. = 0.023 (Re)°" 8 (Pr)°" 3
k
= 0.023(35851)°" 8 (0.700)°" 3
= 90.96
The convection heat-transfer coefficient is
h = Nu _k = ( 90 _96 )~(o_.0_30_W_/_m_K~)
n.
0.2m
= 13.64 W/m 2 K
For constant wall temperature case,
or
mCP ln T:b'' -T_w = h (PL)
The -Tw
or
Forced Convection: Internal Flow
r
9, 73
The mass flow rate of air is
m=pAcV=p-¥-=(0.9950 kg/m 3 }(0.15 m3 /s)
= 0.14925 kg / s
Substituting the appropriate values,
o
T. = 60 C + 80 - 60 C ex
0
be
{
)
[
P
-
13.64 W/ m 2 K}(0 .8m)(8m)l
( 0.14925 kg/ s) (1009J/ kgK)
(
= 11.2°c
(Ans)
The rate of heat loss is
Q = mCP (Ti,; - Tbe )
= (0.14925 kg/s) (1009J/kg K){80-7l.2)°C
= 1325 W
(Ans)
Comment: The bulk mean temperature with the calculated value of Tbe is z;,m = .!_{80 + 71 .2) = 75.6°C or
2
348.75 K which is quite close to 350 K, the value of Tbm assumed. Therefore no more iteration is necessary.
Example 9.29 ~ Air at 0.1 S kg/s and 24°C enters a rectangular duct which is S-m long and 75 mm by I SO mm
on a side and has relative roughness of I0- 3. The duct wall temperature is maintained constant
at I 3S0 C. Determine (a) the temperature of the air at the exit, and (b) the heat transfer rate from the duct
to the air flowing inside the duct. Use the Gnielinski correlation.
Known: Air is heated while passing through a rectangular duct with specified surface temperature and length.
Find: (a) Tbe ( 0 C), and (b) Q(W).
Schematic:
Duct (Tw = I JS C)
0
£ID.= I 0-3
L=Sm
Fig. 9.33
Assumptions: (1) Steady operating conditions. (2) Constant properties. (3) Fully developed flow .
(4) Negligible kinetic and potential energy changes and flow work effects.
Properties: At an assumed value of Tbm = 50°C, the properties of air are
Pr= 0.7228
µ = 1.963 x 10-5 kg/ms
CP = 1.007 kJ/kg K
k = 0.02735 W/mK
9.74 , Heat and Mass Transfer
Analysis: Equivalent diameter,
4Ac
4ab
D -----• - P - 2(a+b)
=
4[0.075x0.150]
2[0.075+0.150]
=
4x0.01125 m2
0.45 m
= O.lm
Reynolds number,
m
pVD.
Re=-- where V=µ
pAC
m n.
(0.15 kg/s)(0.1 m)
Re=--=-----'-------'--'------'--Ac µ
(0.01125 m2 )(1.963xI0-5 kg/ms)
(> 2300)
= 67 923
=> Turbulent flow
Friction factor, f for relative roughness (e/D.) = 10-3 is given by Haaland's explicit relation:
_1 = -1. 8 log[6.9 +(E/D)l.lll
fl
Re
3.7
(IQ-3
)I.Ill
6.9
=-1.8 log--+
[
67923
3.7
= 6.616
f = {1/6.616 )2 = 0.0228
Using the Gnielinski correlation:
Nusselt number,
Nu=
{f/8) (Re-lOOO)Pr
1+ 12.7 {f /8)11 2 (Pr213 - 1)
{0.0228/8)(67923-1000){0.7228)
=- - - - ' - - - - - - - - 1+ 12.7 {0.0228/8}11 2 (0.7228213 - 1)
=159.15
hD.
Nu=-=159.15
k
The heat-transfer coefficient,
h = (159.15)(0.02735 W/mK)
0.1 m
= 43.53 W /m 2 K
Forced Convection: Internal Flow
r
9.75
The rate of heat transfer from the isothermal duct to the flowing air is
Q = mCP (The -TbJ= h(P L)A1tm
(Tw -TbJ-(Tw -Tbe)
where
(Tbe -Ti,J
m(;:=;:) m(;:=;:)
dT,. =
=
Hence,
or
or
The exit air temperature is
Substituting the appropriate values,
Tb =135oC-{l35-24 )°c exp[- 43.53 W/m2Kx0.45 mx5 mJ
•
0.15 kg/s X 1007 J/kg K
(Ans) (a)
=77°C
The rate of heat transfer is
Q = rhCP (Ti,. -TbJ
~:1
=(0.15kg/s) (1007J/kgK) (77 - 24) 0 cl 110
(Ans) (b)
=8.0 kW
Comment: Q can also be computed by first finding A7tm which is
A7tm=
(Tb.-Ti,J
zn(Tw-Tbi)
Tw -T,b e
=
77-24
zn(l35-24)
135-77
=8l.65oc
9, 7 6 , Heat and Mass Transfer
Q = h(P L)Li'I;m
= (43 .53 W/m 2 K)( 0.45 m X 5 m )( 8l.65°C) I 110:: I
=8.0kW
Note that Tbm = 24 +7? = 50.5°C which agrees well with the value used (50°C) for evaluating
2
thermophysical properties of air.
Example 9.30 ~ In the initial design study of steam power plant, calculations are carried out to investigate
the effect of different methods of heating the boiler feed water on the overall capital cost
and efficiency of the plant. One method involves making use of the geometry of the furnace to incorporate
a large coiled duct around the perimeter, and pumping the feed water through the duct before it enters the
steam raising part of the furnace. The feed water flows at a rate of 5.0 kg/s through a rectangular duct of
8 cm x 4 cm cross-section. The walls of the duct are maintained at I 70°C throughout. The feed water enters
the duct at 20°C and is heated by I 30°C. Determine the heat transfer coefficient and the length of the duct
required, based on the following correlations:
(a) Dittus-Boelter
(c) Petukhov
(b) Sieder-Tate
(d) Gnielinski
The following properties of water may be used:
At Tw = 170°C:
µ = 0.160 x I0- 3 kg/ms
Tbm = 85°C:
µ = 0.333 x I0- 3 kg/ms;
k = 0.673 W/m K
CP = 4.201 kJ/kg K
Known: Flow of water through a rectangular duct.
Find: Heat-transfer coefficient, h (W!m 2 K) and Duct length, L (m).
Schematic:
Tb;=20°C
T = I 70°C
Lr~e= 150°C.
------------------------------------------------------------,
: ------------------------------------------------------------;
L
------------------------------------------------------------'
I '??' m=:s~ls Ii
a=Bcm
ml
----------- -- ------- --- -------- -- -------- --- ------- ----- -- - ,
,.__ _ _ _ _ L _ _ _ _ _ _.,
•········•••••··········•··••••············••..•...·.·.·.·••..••········••. b = 4 cm
Fig. 9.34(a)
Fig. 9.34(b)
-------------------------------------------------------------:
Assumptions: (1) Steady operating conditions. (2) Fully developed flow. (3) Smooth surface of the duct.
(4) Constant duct surface temperature.
Analysis: (a) Dittus-Boelter correlation
Hydraulic or equivalent diameter,
4Ac
D=-
P
e
2ab
a+b
= 2 (8)( 4 ) cm= 0.0533 m
8+4
Re=
pV/, P~, [p: l
=
= :~
Forced Convection: Internal Flow
r
9. 77
:. the flow is turbulent.
CP µ
(4.201x10 3 )(0.333xl0-3 )
Pr=-=~--~~--~
k
0.673
=2.08
Nu= 0.023(Re)°" 8 (Pr)°" 4
(as water is being heated, n = 0.4)
= 0.023(250250}°" 8 (2.08)°" 4
= 642.2
:. heat-transfer coefficient,
h =Nu!_= {642.2)(0.673)
n.
o.0533
=8104 W/m 2 K
(Ans) (a)
Heat-transfer rate,
Q = mCP (The -TbJ= (5){ 4.201){150-20)
=2730.65 kW
Also, Q = h(P L)AT,m
where
P = 2(a+b) = 2(0.08+0.04) = 0.24 m
AT-AT. {170-20)-(170-150)
AT,m = l (i A'Ji. )e =
l ( 170 - 20 )
n -AT
n 170-150
0
= 64.52°C
or
K
Hence, the required duct length,
L=
Q
=
2730.65x103 W
h(P)AT,m (8104 W/m2 K){0.24 m)(64.52 K)
= 21.76 m
(Ans) (a)
(b) Sieder-Tate correlation
0.14
Nu= 0.027 Re41 5 Pr113
(
~)
= 0.027 (250250 )°" 8 (2.08f 3 ( 0.3 33 X IQ-J )O.l
0.160x10-3
= 795.5
4
9.78 , Heat and Mass Transfer
h =Nu!_= {795.5)(0.673)
n.
o.0533
= 10 039 W/m2 K
L=
And,
(Ans) (b)
Q
=
2730.65 X 103
h(P)AJ;m (10039)(0.24)(64.52)
= 17.57 m
(Ans) (b)
(c) Petukhov correlation
(f !2)RePr
( µ
Nu - - - ~ - - - - = = , - - - - - 1.07 + 12.7.J f I 2 (Pr213 - 1) µw
)O.ll
wheref=Fanning friction factor= (1.58lnRe-3.28r2
f = {1.58ln250250-3.28r2 = 3.736x10-3
Nu=
(1.868 x 10-3)(250250){2.08)
(0.333x 10-3 ) 0· 11
1.07 + 12.7(1.869 X lQ-3f 2 (2.08213 -1) 0.160 X 10-3
= 756.935
h = (756.935)(0.673) = 9552 W /ml K
(Ans) (c)
0.0533
And,
L=
2730.65x103
= 18_46 m
(9552) (0.24) (64.52)
(Ans) (c)
(d) Gnielinski correlation
(f !8)(Re-lOOO)Pr
( µ
)o.ll
Nu= 1+12.7.Jf !8(Pr213 -1) µw
where f = Darcy friction factor= (0.79 lnRe- l.64r2
f = (0.79xln250250-1.64r2
= 0.01495
And,
Nu= (0.01495/8)(250250-1000){2.08) ( 0.333 x lQ-3
1+ {12.7){0.0495/8f2 (2.082/3 -1) 0.160 X 10-3
)O.ll
= 780.3
h = (780.3)(0.673)
0.0533
=9847 W/m2 K
(Ans) (d)
Forced Convection: Internal Flow
L=
And
r
9, 79
2730.65 X 10 3
{9847){0.24 ){64.52)
= 17.9 m
(Ans) (d)
Comment: In cases (b), (c) and (d), the variation in viscosity of water over a wide range of temperature
has been accounted for. Of these, the Gnielinski correlation is the most reliable.
Example 9.31
~ Water at a temperature of 50°C enters a 1-m-long square tube with cross section 15mm
by 15mm. The water mass flow rate is 13.5 kg/min, and the tube is maintained at a constant
surface temperature of 90°C. Determine (a) the water outlet temperature, (b) the heat transfer rate from the
tube to the water, (c) the pressure drop, and (d) the pump power requirement, assuming 65% pump efficiency.
Properties of water:
T ( C)
p(kg/m 3)
1-----325
987.2
0
Cp(k)lkg 0 C)
k(W/m 0 C)
µ(kg/ms)
Pr
0--6
3.42
4.182
0.645
528 x I
330
984.3
4.184
0.650
489 x I0--6
3.15
335
982.3
4.186
0.656
453 x I0--6
2.88
Use the Gnielinski correlation:
Nu=-(~f/_8~)(-=Re=-~IO_OO~)_Pr_
1+12.7 Jrls(Pr2i3 _l)
where f=(0.79/nRe-l.64r2
Known: Water flow rate through a constant wall temperature square tube.
Find: (a) Tbe (°C); (b) Q{W) ; (c) t:.P (Pa); (d) p (W).
Schematic:
Water
1 r.=90°c
N
J.
\
Tb;=50°c------\j a= 15mm
m= 13.5 kg/min
'\
~--------------------'
L= Im
Fig. 9.35
Assumptions: ( l) Steady operating conditions exit. (2) Constant tube wall temperature . (3) Fully
developed flow.
Analysis: Energy balance:
IQ = mcp (Tbe-1i,;) = h /4 ~ Tim I
where
_ ~T, - ~T. _ (I'. - Tb;)- (I'. - Tbe)
~Ti---------------~
m ln(~T, /~T.) /n [(T.-Tb;)/(T.-7i,.)]
(1)
9.80 ,
Heat and Mass Transfer
Hence
(where A. =PL)
or
or
The water outlet temperature is determined from
(2)
To find the exit temperature of water, one requires the value of h which in tum requires Nusselt
number, Nu. To evaluate Nu, one would need the thermophysical properties of water at the bulk mean
temperature, ¾m [ =
½(¾i - The)Jfor use in the specified correlation.
Since The is unknown, a trial solution is unavoidable.
Trial I: With ¾i = 50 °C, T. = 90 °C, and L = I m, one can assume Tbm = 325 K (or 52°C) to begin with.
Mass flow rate,
Average water velocity,
V = m/pa2 =
(13.5/60)kg/s
(987.2 kg/m3 ) (0.015m)2
V= lOOO =1.013 mis
987.2
Equivalent diameter,
4a2
D0 =4(4,/P)= 4 a =a=I5mm
or
0.015 m
Reynolds number,
pV n.
(987.2 kg/m3 ) (1.013m/s)(0.015m)
µ
528x 10-<i kg/ms
Re=--=------------= 28409
=> Turbulent flow
Forced Convection: Internal Flow
Friction factor,
f = (0.79 In Re - 1.64t2
= (0.79 /n 28409-1.64r2 = 0.02396
Nusselt number,
Nu=
(f /8)(Re-I000)Pr
1+12.7
Pr213 -1)
Ji/8(
(0.02396/8)(28409-1000){3.42)
=------;:====----1+ 12.7 ~(0.02396/8)( 3.42213 -1)
= 149.1
Then, the heat-transfer coefficient is calculated to be
h =!_Nu= 0.645 W/m oC x149.1
n.
0.015 m
= 6411.3 W/m2 °C
Exit water temperature can be calculated from Equation (2).
0
(
)
[
The= 90 C- 90-50 °C exp -
h(W/m2 °C)(4x0.015mxlm)l
(13.5/60)kg/sxCP (J/kg c)
0
ITiie = 90-40exp[-o.2667(h/Cp)JI
or
The = 90 - 40 exp [-0.2667 X 6411.3/ 4182]
= 63.4°C
1(
50+63.4
¾i + The ) = - - = 56.7 C
2
2
0
Tbm (calculated)= -
-::/- Tbm (assumed) (= 52 °C)
Trial 2 : With Tbm = 56.7°C ,.,330 K, p = 984.3kg/m3
V = lOOO = 1.016 m/s
984.3
Re= 984.3xl.016x0.015 = 30 675
489 X 10-<i
f = (0.79 /n 30675-1.64r2 = 0.0235
r
9.81
9.82 ,
Heat and Mass Transfer
(0.0235/8)(30675-1000){3.15)
=------;====----1+ 12.7 ~( 0.0235/8)(3.15213 -1)
= 153.4
h= 153.4x 0·650 = 6647W/m2 °C
0.015
Hence
¾e =90-40exp[-0.2667x6647/4184]
= 63.8°C
Then
Tbm =_!.(50+63.8)=56.9°C
2
Since the calculated value of Tbm = 56.9°C"" 56.7°C (assumed), no more trial is necessary.
(Ans) (a)
:. water outlet temperature, The= 63.8°C
Heat-transfer rate from the tube to water can be calculated from Equation (1).
Hence,
Q = mCP (¾e - ThJ
= ( 135 kg)(4.184~)(63.8-50)°C
60 s
kg C
= 13.0 kW
(Ans) (b)
Else
AT.
The -Ti,i
63.8-50
= 32.6cc
tm = zn(R) = zn(9~0--6~~8)
I/::1
Q = h( PL )Alim= ( 6647W/m2 °C) ( 4x0.015mx lm )( 32.6°C) 0
= 13.0 kW
Pressure drop,
AP=
t(iJ(½pV
2)
I
=0.0235( 1.0m )(.!.x984.3 kg xl.0162 m2
lN
0.015m 2
m3
s2 lkgm/s2
= 796 N/m2
or
Pa
I)
(Ans) (c)
Forced Convection: Internal Flow
r
9.83
Pumping power requirement,
fO = f:.P X (
m/p)/rJ
lW I
984.3kg/ m3 1Pa m3 / s
= 796 Pa )13.5/ 60)kg/ sl
0.65
=0.28 W
(Ans) (d)
Ts= 90°C ,___ _ _ _....-___________ 90°C
Tube surface
- - - - , Tbe=63.8°C
-----L=lm----Inlet
Exit
Fig. 9.36
I
I.Om
0.015m
Comment: Since L D. = - - - = 66.7 which is greater than 60, the assumption of fully developed
flow is justified. For Gnielinski correlation, 0.5 ~Pr~ 2000 and 3000 <Re< 5 x I 0 6 • These conditions
are also satisfied.
Example 9.32 ~ Air initially at 30°C, I atm, enters a triangular duct of 2-cm side and 2.5-m-long at the rate
of 12 kg/h.An electric heating coil of 150 W is wound uniformly over the length. (a) Calculate
the heat-transfer coefficient and the temperature of the wall at the exit. (b) If the triangular duct is replaced
by a square duct of 2-cm side, what will happen to the exit wall temperature?
Known: Air is heated under constant wall heat flux conditions while passing through a triangular duct.
Find: (a) Heat transfer coefficient; Exit wall temperature. (b) Exit wall temperature if the duct has square
cross-section.
Schematic:
_ Qelec
qw- T
Qelect = I 50 W
m= 12kg/h
Air
Tbi::: 300c £..J, :.=======~ a = 2cm
(a) Triangular duct
Fig. 9.37(a)
9 .84 ,
Heat and Mass Transfer
Duct (Tw= 70°C)
Air
15 cm
Tbe m= 12kg /h
Qelect= 150W
=qwAs
I atm, Tb;=85°C
¥=360m3/h
L= 11 m
(b) Square duct
Fig. 9.37(b)
\af ouc.t
,r\al"iu e ouc.t
sa.uar
T (°C)
T(°C)
95.4
i 6°C
r.-Tr
87.4
14.8
30
Inlet
Exit
Fig. 9.37(c)
Assumptions: ( l) Steady operating conditions exist. (2) Constant properties. (3) Flow is fully developed.
(4) Inner surface of the duct is smooth. (5) Constant wall heat flux.
Analysis: (a) Triangular duct
Equivalent (or hydraulic) diameter for a non-circular duct is given by
A
D =4-c
e
p
Cross-sectional area,
Ac =.!..(base){height)
2
l
.
l
.Ji
2
2
=-axa sm 60° =-(2cm) 2 x -
2
= 1.732 cm 2
Wetted perimeter, P = 3 a= 3 x 2 cm= 6 cm
Forced Convection: Internal Flow
r
9.85
D =4x L 732 cm2 = 1.1547 cm
e
6cm
Density of air at 1 atm, 30°C,
P
101.325kPa
I lkJ I
p =RT= (0.287kJ/kgK) (30+273.15)K lkPam3
= 1.1646 kg/m3
Equation of continuity:
Mean velocity of air,
m
(12/3600)kg/s
(l.1646kg/ m3 )(1.732x10-4 m2 )
V=--=----------p Ac
= 16.525 m/s
Rate of heat transferred to air, Q = Electrical coil rating= 150 W
Exit air temperature,
Q
'Eb e = 'Eb', + • C
m P
150W
=30°C + - - - - - - - - - - - = 74.8°C
(12/3600)kg/sx1005J /kg°C
Bulk mean temperature,
¾m = .!.(Tbi +Ti,.)= .!.(3o+74.8)°C
2
2
= 52.4°C
Properties of air at Tbm = 52.4 °C are
k= 0.0282 W/m
0
c
Pr= 0.703
v = 18.46 x lQ-<i m 2/s
To find the heat-transfer coefficient, let us first determine whether the flow is laminar or turbulent.
Reynolds number,
VD.
Ren=-'
V
=
(16.525m/s) (l.1547x10-2 m)
18.46x10-6 m2 / s
= 10 337 (> 2300)
~
Turbulent fl.ow
9.86 ,
Heat and Mass Transfer
L
D0
=
2 ·5 m
=216.5
1.1547x 10-2 m
(>60)
:. the flow is fully developed.
Using the Dittus-Boelter equation with n = 0.4 as the air is getting heated:
Nu = 0.023 (Re) 0-8 (Pr)0.4
= 0.023 (10337) 0-8 (0.703)0.4
hD0
=32.5=k
:. heat-transfer coefficient,
h =!__Nu= 0.0282W/m oC x32.5
D0
1.1547x10-2 m
= 79.4 W/m2 °C
(Ans) (a)
Q Q
150W
q =-=-=
w
A. 3aL 3x0.02mx2.5m
Heat flux,
= 1000 W/m2
Now,
Exit wall (surface) temperature,
Twe = ¾e + {qw I h)
=74.8oc
+(
1000W/m2 )=87.4oc
79.4W /m2 °C
(b) Square duct
4Ac 4xa2
De -P- - --a
- -4a
Equivalent diameter,
=0.02 m
Mean flow velocity,
V=_!±!__=
pAc
Heat rate,
(12/3600)kg/s
= 7 _16 rn/s
3
2
1.1646kg/m x(0.02m)
(Ans) (b)
Forced Convection: Internal Flow
r
9.87
¾e = ¾i +(Q!mCP)
= 30°c + (
150W
)
(12/3600)kg/s xl005J /kg°C
= 74.8°C
Tbm = (Ti,i +Ti,.)12 = 52.4°C
Air properties remain unchanged.
Re= VD.= (7.16m/s)(0.02m)
v
18.46x1Q-6 m2 /s
= 7753
(> 2300)
k
k
n.
n.
~
Turbulent flow
Heat-transfer coefficient,
h =-Nu= -x0.023(Re)0-8 (Pr)0.4
= 0.0282 W / m°C X 0.023(7753)0-8 (0.703)0.4
0.02m
= 36.4 W /m2 °C
But Q = h(PL)(Twe -Tb.)
exit wall temperature,
T
we
= r,_ +
Q
be
h(4aL)
= 74.8oC+
150W
(36.4W/m2 C) (4x0.02mx2.5m)
0
(Ans) (b)
=95.4°C
Comment: Square cross section of the duct gives higher surface area and hence lower value of heat flux.
This yields lower convection coefficient and hence higher surface temperature at the outlet.
(D) LIQUID METAL HEAT TRANSFER
•
•
Example 9.33 ~ Liquid bismuth flows at a rate of 4.5 kg/s through a 5-cm-diameter stainless steel tube. The
bismuth enters at 41 5°C and is heated to 440°C as it passes through the tube. If the tube
wall is at a temperature 20°C higher than the bismuth bulk temperature by maintaining constant heat flux along
the tube, calculate the length of the tube required to effect the heat transfer.
Use the following properties at 427.5°C:
µ = 1.34 X 10-3 kg/m s
CP = 0.149 kJ / kg K
k= 15.6 W/m K
Pr= 0.013
Use the following correlation (for I04 < Re < I0 6) for qw = con st:
Nu= 6.3 + 0.0167 Reo.ss p,o.93
9 .88 ,
Heat and Mass Transfer
Known: Liquid bismuth is heated in a tube subjected to constant wall heat flux .
Find: Tube length, L (m).
Schematic:
J/= T,b,
a
Liquid
bismuth
41s c 0
(D=O.OSm)
m=4.Skg/s
t t t
t t
\.....
- - 440°c
t t
qw=const, (Tw-Tb)=20°C
Fig. 9.38
Assumptions: (I) Steady operating conditions. (2) Constant properties. (3) Constant heat flux at the tube
surface.
Analysis: Heat-transfer rate is determined from
Q = mCP~T = (4.5 kg/s){l49 J / kg K) (440-415) K
= 16762.5 W
For a circular cross section,
Reynolds number,
4m
Re = - 1tDµ
4(4.5kg/s)
= -------''-----1t (0.05 m)(l.34 x 10-3 kg/ ms)
= 85 516
Substituting appropriate values in the given correlation,
Nu= hD = 6.3+0.0167 (85516) 0 -85 (0.013) 0 -93 = 10.88
k
Hence, the heat-transfer coefficient,
h =Nu~= 10.88 x 15.6W Im K
D
0.05m
= 3395 W / m2 k
For constant wall heat flux condition,
The length of the tube required is
L=
Q
h(1tD)(Tw -7;,)
= 1.57 m
16762.5W
(3395W / m2 K)(1tx0.05m)(20) K
=-----------(Ans)
Forced Convection: Internal Flow
r
9.89
Example 9.34 ~ Liquid sodium at 350°C enters a 50-mm ID pipe with a flow rate of 5 kg/s and is heated
to 450°C. The pipe wall is maintained constant at 500°C. What length of the pipe would
be required to achieve thisl
Properties of liquid sodium at 400°C are
k = 71.2 W Im K
p = 853 kg/ m3
cp = 1296.5 J/ kg K
v = 3.184 x I0--7 m2 /s
Pr= 0.004978
Known: Liquid sodium flow through a circular pipe under constant wall temperature conditions.
Find: Pipe length, L.
ill
Schematic:
Liq~id
Sodium
-
C
Tw=500°C
-
D=0.05m
~------------(~)~r..,=•so·c
rb;=350°c m=5kg/s
-
- .
------L------.
Fig. 9.39
(1) Steady operating conditions. (2) Constant fluid properties. (3) Constant pipe wall
temperature. (4) Fully developed velocity and temperature profiles.
Analysis: For flow through a tube under constant surface temperature conditions:
Assumptions:
mcp
L=--ln
hrcD
( r_w - r,_b.' )
Tw -Ti,.
k
where
h=Nu0 D
k
= - [ 4.8+ 0.0156 Re& 85 Pr0- 93
D
J [10 < Re < 10
4
0
6
and
0.004 <Pr< O.l]
Properties at the bulk mean temperature of (350 + 450)/2 = 400°C are known.
Then, for a circular cross section,
4m
4x5 kg/s
Re0 = - - - = - - - - - - - - - - - - - - rcDvp rc(0.05m)(3.184x10-7 m2 /s)(853 kg/m3 )
= 4.688x10 5
Hence,
(< 106 )
h = 7 1. 2 W /mK x[4.8+0.0156(4.688x10 5
0.05m
}°' 85 (0.004978}°" 93 ]
=17433W/m2 K
Length of pipe,
(5 kg/s)(1296.5 J/kg K)
(500-350)
L = -'-----'--'--------'---In - - - = 2.60 m
(17433 W /m2 K)(rcx0.05 m)
500-450
(Ans)
9.90 ,
Heat and Mass Transfer
, MULTIPLE CHOICE QUESTIONS
When all conditions are identical for fluids flowing through a pipe with heat transfer, the velocity
profiles will be identical for
(a) liquid heating and liquid cooling
(b) gas heating and gas cooling
(c) gas heating and liquid cooling
(d) cooling and heating of any fluid
9.2 Consider a hydrodynamically fully developed flow of cold air through a heated pipe of radius r0 • The
velocity and temperature distributions in the radial direction are given by u(r) and T(r) respectively.
If um is the mean velocity at any section of the pipe, then the bulk mean temperature at that section
is given by
9.1
,;,
(a)
,;,
Ju(r)T(r )r2 dr
(c)
0
9.3
Ju(r)T(r) dr/2rc1
0
2
(d) -
,;,
2
Ju(r)T(r) dr
umro o
Match four correct pairs between List I and List II.
List I
List II
A. Nusselt number
1. VD/a.
B. Reynolds number
2. hip CP V
C. Prandtl number
3. at!L2
D. Peclet number
4. hD/k
5. CPvp/k
6. pVD/µ
Codes:
A
B
C
D
(a)
1
2
4
3
(b)
2
6
5
3
(c)
2
4
5
(d)
4
6
5
9.4 For laminar flows in circular pipes, the hydrodynamic entry length Lh is correlated by
(a) 0.0575 D Ren
(b) 0.023 D Ren
(c) 0.04305 D Ren Pr
(d) 0.037 D Ren Pr
9.5 Oil flows through a 20-cm diameter, 200-m-long pipeline with a velocity of 20 cm/s. With specific
gravity 0.89, Prandtl number 260 and viscosity 0.02 kg/m s, the
(a) hydrodynamically developed but thermally developing laminar flow
(b) fully developed laminar flow
(c) hydrodynamically and thermally developing flow
(d) fully developed turbulent flow.
9.6 Nusselt number for a pipe flow equation heat-transfer coefficient is given by the Nun= 4.36. Which
one of the following combinations of conditions do exactly apply for use of this equation?
(a) Laminar flow and constant wall temperature
(b) Turbulent flow and constant wall heat flux
Forced Convection: Internal Flow
r
9. 9 I
(c) Turbulent flow and constant wall temperature
(d) Laminar flow and constant wall heat flux
9.7 When the temperature differences between the fluid and wall are large, viscosity effects must be considered
)n
. .
.
Nu ( variable properties) ( µb
and the effects of property vanat10ns may be estimated from
µw where
Nu(constant properties)
n is
(b) 2/3
(d) 0.14
(a) 0.5
(c) 0.8
9.8 The Dittus-Boelter equation for Nusselt number in fully developed turbulent flow in smooth tubes
for cooling of the fluid is given by
(a) 0.023Re0· 8 p,.0.4
(b) 0.023Re0 -8 Pr0· 2
(c) 0.023Re0· 8 p,.0. 3
(d) 0.023Re0· 2 p,.0.4
where Re is the Reynolds number and Pr, the Prandtl number.
9.9 For fully developed turbulent flow of water through a pipe with constant wall temperature, the
correct temperature difference in the expression for convective heat transfer is
(a) arithmetic mean temperature difference
(b) difference between the inlet and exit water bulk temperature
(c) log mean temperature difference
(d) difference between the wall temperature and the bulk mean water temperature
, TRUE/FALSE
9.1
9.2
9.3
9.4
9.5
The Dittus-Boelter equation for developed turbulent pipe flow is not applicable for constant wall
heat flux condition.
The Graetz number is defined as Gz = (LID) Re Pr
The value of the Nusselt number is always greater than unity.
Local Nusselt number in the thermal entrance region for laminar flow in a circular pipe with fully
developed velocity profile is always greater for constant wall temperature case than for constant
wall heat flux condition.
Unlike laminar flow, the heat-transfer coefficient and the friction factor in turbulent flow depend
strongly on the surface roughness.
, FILL IN THE BLANKS
9.1 A conduit with a triangular cross-section with all three sides of length 5.2-cm carries water at 20°C.
The hydraulic diameter, Dh is - - ·
9.2 For liquid sodium at 100 °C (k = 86.2 W/mK) flowing through a circular tube of 50-mm diameter
with parabolic temperature profile and almost flat velocity profile, the heat transfer coefficient is
_kW/m2 K.
The heat transfer coefficient for laminar flow of saturated ammonia at 35°C (k= 0.4579 W/mK)
through a 25-mm ID tube under constant wall temperature condition is __ W/mK.
9.4 An NTU more than __ implies that irrespective of the inlet temperature the fluid flowing through
a tube will attain the tube wall temperature at the outlet.
9.3
9, 92 ,
Heat and Mass Transfer
, EXERCISES
9.1 An incompressible fluid is flowing between two parallel plates, each of infinite width, and separated
by a small vertical distance L. The upper plate, maintained at a uniform temperature of TL, is moving
with a velocity U. The lower plate, maintained at a uniform temperature of T0 , is stationary. Assuming
the flow to be steady, obtain the velocity distribution and temperature distribution between the plates.
A heavy lubricating oil(µ= 0.8 N s/m 2 and k = 0.15 W/mK) flows in the clearance space between
a shaft and its bearing. The shaft and the bearing are kept at 30°C and l0°C respectively and the
clearance between them is 2 mm. Determine the maximum temperature rise and heat flux to the
plates of shaft and bearing for a velocity of U = 6 mis.
[35°C]
9.2 The velocity and temperature variation in a pipe of 20-cm diameter can be approximated by the
following relations
u = 185 r (0.85 - r) (m/min)
T= 95 (l - 2r) {0 C)
where r is the distance measured in radial direction from the pipe surface. The axial velocity of
flow is u and Tis the temperature.
If the temperature of the pipe wall is maintained at 120°C, calculate the heat transfer coefficient
based on the bulk mean temperature of the fluid for a heat loss of 500 W per metre length of the
pipe. Assume the flow in the pipe to be laminar.
Take v = 21.l x 10-6 m 2/s
[20.3 W/m 20 C]
9.3 Lubricating oil flows through a 2.5-cm inner diameter, 9-m long copper tube at a velocity of 2.3 mis.
The oil enters the tube at 25°C and the tube is heated by steam condensing on its external surface
at l atm with a heat-transfer coefficient of 11 kW/m 2 0 C. Determine (a) the hydrodynamic and
thermal entry lengths, and (c) exit temperature of the oil and the rate of heat transfer to the oil. The
accompanying table lists the thermophysical oil properties at various temperatures:
Temperature (0 C)
p (kg/m 3)
cp (kJ/kg
0
C)
17
27
37
47
890.0
884.1
877.9
871.8
1.868
1.909
1.951
1.993
µ x 106 (m 2/s)
1122
549.7
288.2
161.7
k (W/m 0 C)
Pr
0.145
12 900
0.145
6400
0.145
3400
0.143
1965
µ x 104 (kg/ms)
9990
4860
2530
1410
µ@,oooc = 0.01725 kg/m s
[2090 W]
9.4 Engine oil at 60°C with a flow rate of 12 kg/min enters a 2-cm ID tube which is maintained at
120°C. Assuming hydrodynamically developed flow, determine the length of the tube required to
heat the oil to 80°C.
The properties of engine oil at 70°C:
p = 858.03 kg/m 3
Pr= 770
Cp=2.089 kJ/kg K
k=0.139 W/m K
v = 0.607 x lQ-4 m2/s
[63.6 m]
Forced Convection: Internal Flow
r
9.93
9.5 Water at a uniform temperature of 25°C flows through a long tube (I-cm-diameter) with a velocity
of O.15 mis. After a fully developed velocity profile has been attained, the wall temperature is raised
to a constant value of 80°C. Obtain the value of the heat transfer coefficient after (a) a distance
of 7 .5 cm, and (b) the temperature profile is also fully developed.
Use the following properties of water at 25°C:
v = 0.8933 x 10-6m2/s k= 0.6076 W/m K
Pr= 6.14
[222.2 W/m 2 K]
9.6 Water at 20°C with a flow rate of 0.015 kg/s enters a 2.5-cm ID tube which is maintained at a
uniform temperature of 90°C. Assuming hydrodynamically and thermally fully developed flow,
determine the heat transfer coefficient and the tube length required to heat the water to 70°C.
Water properties at 20°C:
p = 1000.5 kg/m3 ; CP = 4181.8 J/kg K; V = 1.006 X 10-4 m2/s
Properties at 45°C are
k= 0.638 W/ m K;
V = 0.613 X 10-4m2/s.
p = 992.3 kg/m 3;
CP = 4180 J/kg K;
The average Nusselt number for the tube,
Nu= 3.657
[10.72 m]
9.7 Hot air flows through a 20-mm x 20-mm square duct and is cooled along the length at a constant
rate. At a certain section, the inner tube wall temperature is measured to be 150°C. At another
location, 1-m downstream, the wall temperature reduces to 100°C. The mass flow rate of air is
3 x 10-4 kg/s. Calculate (a) the constant wall heat flux, qw(W/m2), and (b) the average air temperature
at the exiting section, Tbe( 0 C).
Properties of air:
p = 0.8148 kg/m3
µ = 2.42 x 1o-5 kg/ms
CP = 1.016 kJ/kgK
k = 0.03511 W/mK
[180°C]
9.8 Water at the rate of 0.5 kg/s is forced through a smooth 25-mm ID tube of 15 m length. The inlet
9.9
9.10
9.11
9.12
water temperature is 283 K and the tube wall is at a constant temperature of 313 K. What is the
exit water temperature ?
Data: Average properties of water
Cp=4180 J/kg K, µ=0.8x 10-3 Pas, k=0.57 W/m K
[310.28 K]
In a heat exchanger air flows through a long 8-cm ID (inside diameter) tube at a bulk velocity of
3 mis and is heated by hot gases on the outside of the tube. The temperature of the tube wall is
120°C. The air enters at 40°C and leaves at 60°C. Find the heat transfer coefficient for air, using
the Dittus-Boelter equation. The thermophysical properties of air at 50°C are
v= 17.95x lQ-6 m2/s
Pr=0.698
k=0.0283 W/m K
[14.0 W/m 2 K]
The radiator of a motor car is constructed of finned tubes. The tubes are 0.6-cm ID and 60-m
long. Cooling water flows through them with a velocity of 1 m/s and 600C temperature. Calculate
the heat-transfer coefficient on the tube wall. Assume that the greater part of the flow is fully
developed. The Gnielinski correlation and the following properties of water may be used:
µ = 0.467 x 10-3 kg/ms
k= 0.654 W/m K
p = 983.3 kg/m 3
CP = 4.185 kJ/kg K
[7644 W/m 2 K]
A 3-m long, 25-mm diameter tube is held at 100°C by steam jacketing. Water flows through the
tube at the rate of 180 kg/hat 20°C. Calculate the rate of heat transfer from the tube to the water.
For water the properties at 45°C:
CP = 4.180 kJ/kg K
k= 0.637 W/m K
µ = 0.596 x 10-3 kg/m s
[10 kW]
p = 990.1 kg/m 3
Within a condenser shell, water flows through one hundred thin-walled circular tubes
(diameter= 22.5 mm, and 5-m length) which have been arranged in parallel. The mass flow rate of
9.94 , Heat and Mass Transfer
water is 65 kgls and its inlet and outlet temperatures are known to be 12°C and 28°C, respectively.
Predict the average convection coefficient associated with water flow. Take the thermodynamic
properties of water at 20°C.
[5975 Wlm 2 0 C]
9.13 Air with an average velocity of 10 mis at 300 K enters a copper tube of 11.2-mm-diameter and
2.5-m-length. The tube wall is maintained at 373 K by condensing steam at one atmosphere. Using
LMTD method, determine the temperature of air at the outlet of the tube. Average properties of air are:
k=0.02624 Wlm K
Cp= 1.005 kJ/kg K
p = 1.174 kglm 3
v = 1.568 x 10-5 m 2ls
Pr= 0.1
Nu=3.66+ 0.0668(DIL)RePr
l+0.04(DI L)RePr
9.14
9.15
9.16
9.17
9.18
9.19
Re<2300
Nu = 0.023 Rf!J.8 p,'J.4
Re>2300
where D and L are diameter and length of tube respectively. Assume heat transfer coefficient to
be constant and neglect conduction thermal resistance of copper.
[372 k]
For pasteurization of milk, a 20-mm-diameter tube with its wall maintained at a constant temperature
of l15°C is used. The milk enters the tube with a flow rate of 1.5 kgls at a temperature of 35°C.
Determine the required tube length if for the sterilization of milk, its temperature is increased by
40°C. The thermophysical properties of milk at the bulk mean temperature are
k = 0.6 W Im °C
CP = 3.85 kJ/kg °C
p = 1030 kglm 3
v = 2.058 x lQ-<i m 2ls
[6.15 m]
Water flows at the rate of 2 kgls through a 40-mm-diameter tube, 4-m-long. If the water enters
the tube at 25°C and the surface temperature is maintained at 90°C, what is the exit temperature
of the water? What is the rate of heat transfer to the water?
Properties of water at 35°C are
CP = 4.178 kJ/kg K
µ = 0.72 x 10-3 kglm s
k= 0.623 Wlm K
p = 994 kglm 3
Pr= 4.83
[167 kW]
Water at 24°C flows at 0.8 mis in a smooth, 1.5-cm-ID, tube which is maintained at 30°C. The system
is perfectly clean and quiet and the flow remains laminar until a noisy air compressor is turned on. The
flow then suddenly turns turbulent. Compute the value of turbulent and laminar heat transfer coefficients.
Use the following properties of water:
k= 0.613 Wlm K
p = 997 kglm 3
Pr= 5.83
µ = 0.855 x 10-3 kglm s
[149.5 Wlm 2K]
A circular tube of 150-mm diameter and 1.5-m length carries air at 20°C flowing with a mean
velocity of 15 mis. The tube's surface is maintained at a constant temperature of 160°C. Compute
the convection coefficient and the heat-transfer rate from the tube surface to the air flowing inside
it, using the Reynolds analogy.
The following properties of air may be used:
v = 22.14 mm2ls
k= 30.84 x 10-3 WI m K
Pr= 0.10
[3.17 kW]
The main trunk duct ofan air conditioning system is rectangular in cross section (400 mm x 800 mm)
and has air at atmospheric pressure and at 20°C flowing with a velocity of 7 mis. Estimate the
heat leakage per metre length per unit temperature difference. Relevant physical properties are:
v = 15.06 x lQ-<i m 2ls
a= 7.71 x 10-2 m 2ls
k= 0.0259 Wlm K
Use the correlation:
Nu0 h =0.023 Re 0415h p,'J.4
[48.14 Wlm K]
A horizontal rectangular tube, 3 cm x 5 cm, made of galvanised iron (surface roughness = 0.15
mm) carries water at a rate of 2 kgls. Determine the length required to heat the water from 30°C
to 50°C if the wall temperature is maintained at 90°C. What will be the pressure drop?
Forced Convection: Internal Flow
9.20
9.21
9.22
9.23
9.24
r
9. 9 5
The properties of water at 40°C are
p = 992.1 kg/m 3
k= 0.631 W/m °C
CP = 4.179 kJ/kg °C
µ = 0.653 x 10-3 kg/m s
Pr= 4.32
[l.68 kPa]
Air at 2 bar and 55°C enters a square duct (0.2 m x 0.2 m) with a velocity of 1.5 mis. The 8-m
long duct is maintained at a constant surface temperature of 10°C. Determine (a) the heat-transfer
coefficient, (b) the heat-transfer rate, and (c) the outlet temperature of the air. Dittus-Boelter
correlation may be used for turbulent flow.
[2.58 kW]
Consider an uninsulated square duct of cross-section 15-cm x 15-cm, and 11-m long. Hot air at
1 atm, 85°C enters the duct with a volumetric flow rate of 360 m3/h. The duct is observed to be
nearly isothermal at 70°C. Determine (a) the exit air temperature, and (b) the rate of heat loss from
the air as it flows through the duct.
[994 W]
Calculations are carried out to study the effect of different methods of heating the boiler feed
water on the overall capital cost and efficiency of the plant. One method involves making use of
the geometry of the furnace to incorporate a large coiled duct around the perimeter, and pumping
the feedwater through this duct before it enters the steam raising part of the furnace. The duct is
of rectangular cross-section measuring 80 mm x 40 mm and the wall temperature is nearly 170°C
throughout. The feedwater flows at a flow rate of 300 kg /min, enters at a temperature of 20°C
and is heated to 150°C. Estimate the required length of the duct.
[19.4 m]
A 5-m-long rectangular tube, 3 cm x 5 cm, carries water at a rate of 1.6 kg/s. The tube is equipped
with an electric resistance heater which provides uniform heating throughout. (a) Determine the
power rating of the heater required to heat the water from 25°C to 55°C. (b) Also find the inner
surface temperature of the tube at the outlet. (c) What will be the pressure drop? For turbulent
flow, the Dittus-Boelter equation may be used.
The properties of interest are
p=992.3 kg/m 3
k=0.6316 W/m K
Cp=4.179 kJ/kg K
µ = 656.6 x lQ-<i kg/m s
Pr= 4.34
[108.5°C]
Liquid mercury flows through a 5-cm ID stainless steel tube which is 5-m long at the rate of
0.5 kg/s. Calculate the heat transfer coefficient if the mercury enters at a temperature of 20°C and
leaves at a temperature of 80°C. The wall is maintained at a constant temperature of 100°C.
Use the correlation:
Nu0 = 5 + 0.025(Pe)0 -8
Properties of mercury at the bulk mean temperature of 50°C are
p = 13473 kg/m 3
CP = 138.6 J/kg K
k= 8.836 32 W/m K µ = 1.423 x 10-3 kg/m s
[1495 W/m 2 K]
, ANSWER KEY
Multiple Choice Questions
9.1 (c)
9.6 (d)
9.2 (d)
9.7 (d)
9.3 (d)
9.8 (c)
9.4 (a)
9.9 (c)
9.5 (a)
9.2 F
9.3 T
9.4 F
9.5 T
9.2 13.8
9.3 67
9.4 5
True/False
9.1 F
Fill in the Blanks
9.1 3.0cm
10
NATURAL (FREE) CONVECTION
Concept Review
INTRODUCTION
• I 0.1
•
Free or natural convection is the mode of heat transfer that is created by the buoyancy force which arises
as a result of the motion of the temperature difference between the surface and its surrounding fluid.
In natural convection heat transfer fluid motion occurs by natural means (without any external agency).
Like all viscous flows, free convection flows can be laminar or turbulent. The flow regime in natural
convection is governed by a dimensionless number called the Grashof number, which represents the ratio
(Buoyancy force) I (Inertia Force) (Viscous force) 2 acting on the fluid and is expressed as
(10.l)
where Le is the characteristic length, which is the height L for a vertical plate or vertical cylinder, and
the diameter D for a horizontal cylinder or a sphere. The correlations for the Nusselt number Nu= hLJk
in natural convection are expressed in terms of the Rayleigh number defined as
(10.2)
~ is the coefficient of volumetric thermal expansion of the fluid which represents the variation of its
density with temperature at constant pressure. For an ideal gas, it is expressed as ~ = 1/T, where Tis the
absolute reference temperature in K.
VERTICAL FLAT PLATE
• 10.2
•
Consider a heated vertical plate of height L maintained at a constant surface temperature T8 transferring
heat by free convection to the surrounding fluid at a temperature T The velocity and temperature profiles
are as shown in Figure 10.1.
00 •
+ Exact Solution Local Nusselt number at a distance x from the bottom edge of the vertical isothermal
plate is given by
INux = 0.48Gr,V Pr {0.861 + Prf I
4
112
114
(10.3)
Average Nusselt number for a plate of height L is expressed as
INuL = 0.64G,e Pr {0.861 + Pr(' I
4
112
4
(10.4)
10.2,
Heat and Mass Transfer
Note that
(10.5)
It follows that
t---'------------~ T=
(10.6)
Surface
and
Velocity profile
( 10.7)
• Integral Method Solution The velocity and thermal
boundary layer thicknesses are assumed to be of equal
thickness.
The velocity profile is assumed to be
u=O
u=O
X
Ly
0
Fig. I 0.1 Temperature and velocity distributions
near a heated vertical plate in free convection
(10.8)
where u0 is some arbitrary function of x and has the units of velocity.
Velocity boundary layer thickness is
(10.9)
Local Nusselt number,
!Nux = 0.508G,;1 14 Pr112 (0.952+Pr)- 114 !
(10.10)
Average Nusselt number,
hL
NuL = - = 0.679 Gr{f 4 Pr112 (0.952 + Prtll 4
k
• McAdam's Correlation
(10.11)
For fluids whose Prandtl number is close to unity, e.g., for air and other gases:
INuL=0.59Ra~ 4 1 (10 4 <RaL <10 9 )
(10.12)
INuL = 0.10 Ra~ I (10 < RaL < 10
(10.13)
3
9
12 )
All fluid properties are evaluated at the film temperature Tr = T. + T= .
2
• Churchill-Chu Correlation
-
[
9/ 16]-4/ 9
NuL =0.68+0.67Ra~ 4 1+(0.492/Pr)
(10.14)
Natural (Free) Convection
NuL = {0.825 + 0.387
Raf 6 [ 1+ (0.492 / Pr)9116
r
8127
r
(>Ra't_,)
r
I0.3
(10.15)
1
All fluid properties are to be evaluated at the film temperature Tr=-( T. + T=)
2
• Constant Heat Flux: ( q, = const)
hX
(
•
)1/5
Nux = k=0.60 Gr.,_ pr
[ 10 5 <Gr.,_*< 10 11 ]
laminar range
(10.16)
where
(10.17)
Average convection coefficient,
- 5
h =4\=L
INux =0.17(G,;,* Prf 4 1 [2x10 13 <G,;,* Pr<10 16 ]
turbulent region
(10.18)
All fluid properties are evaluated at the local temperature.
Average convection coefficient, h = hx = constant
• Vertical Cylinders The general criterion is that the correlations for a vertical flat plate are applicable
to vertical cylinders as well, provided
D
35
->--
(10.19)
L -Gr{ 4
• Inclined Plates Figure 10.2 shows the sign convection for angle of inclination of flat surfaces under
natural convection.
Hot surface facing
upwards
e
Fig. I 0.2
Hot surface facing
downwards
Orientation of the flat hot surface in terms of negative and positive angles of inclination from the vertical
10.4,
Heat and Mass Transfer
For downward-facing hot plate,
!Nu=0 .56(RaL cos0)lf 4 I (10 5 <RaLcos0 < 1011 )
(10.20)
For upward-facing heated surface,
Nu=0.145
l(
)
J
(
GrLPr l/3 -(G,;,Pr) l/3 +0.56 G,;, Prcos0 )1 / 4
(forGrLPr<10 11 ,GrL>G,;,
and
(10.21)
-15°<0<-75°)
All physical properties are evaluated at the effective temperature,
(10.22)
with ~ evaluated at T= + 0.25 (T. -T=)
The value of critical Grashof number, Grc depends on the angle of inclination 0 (taken negative), as
listed in the following table:
e (deg)
- 15
5 X 10 9
- 30
1Q9
- 60
1Q8
- 75
1Q6
• Rectangular Blocks/Short Vertical Solid Cylinders
(10.23)
!Nu=0.55Ra~ 4 1 (I0 4 <RaL <10 9 )
The characteristic dimension Lc is given by
(10.24)
where 4I is the longer of the two horizontal dimensions and Ly is the vertical dimension.
HORIZONTAL PLATES
• 10.3
•
Figure 10.2 shows the flow pattern from the surface of the horizontal plate under different conditions in
free convection. The characteristic length, Lc is defined as
L _ A. __
P_la_te_s_u_rf_a_ce_a_r_ea_
c - P Perimeter
(10.25)
• (a) Upper Surface of Heated Plate/(b) Lower Surface of Cooled Plate
!NuL = 0 .54Ra~ 4 I (10 4 <RaL <10 7 )
(10.26)
Natural (Free) Convection
r
IO.S
(10.27)
//////, 1////////////////////////////////////////////,
///////////////////////////////////, 1////////////,
Fluid T=
Plate, T5 > T=
(a) Upper surface of a heated
horizontal plate
(b) Lower surface of a cold
horizontal plate
(c) Lower surface of a heated
horizontal plate
(d) Upper surface of a cold
horizontal plate
Fig. I 0.3
Natural convection flow patterns for a horizontal plate under different conditions
(c) Lower Surface of Heated Plate I (d) UPf>er Surface of Cooled Plate
INuL = 0.27 Ra~ I (10 <RaL < IOlO)
4
(10.28)
5
HORIZONTAL CYLINDERS
• 10.4
•
The flow pattern on the outside of a long horizontal heated cylinder at a surface temperature of T8 is
shown in Figure 10.3.
For an isothermal cylinder,
INuo
=T=C
(10.29)
Ra8I
where
RaD
10- 10 to 10-2
C
n
0.675
0.058
10 2
1.02
0.148
10-2 to 104
0.850
0.188
104 to 10 7
0.480
0.250
10 7 to 10 12
0.125
0.333
10- 2 to
Horizontal cylinder, T5
Natural convection flow over
a heated horizontal cylinder
Fig. I 0.4
10.6,
Heat and Mass Transfer
• Churchill-Chu Correlations
For constant wall temperature case for all values of the Prandtl number:
Nuo = 0.36+
O.Sl 8Rags
[l+(0.559 / Pr)9116]419
{
Nuo = 0.60+
[t0-6 < Ra0 < 109 ]
2
6
0.387 Rali
}
0
[l+(0.559 / Pr)9116]8127
(10.30)
[10 9 <Ra0 <10 12 ]
(10.31)
SPHERES
• 10.5
•
• Yuge's Correlation
1Nu=¥=2+0.43Rag4 1 for l<Ra0 <10 5
!Nu=2+0 .50Rah 4 !
(10.32)
and Prz l
for 3xl0 5 <Ra0 <8xl08
and
10::; Nu::; 90
(10.33)
• Churchill Correlation
Nuo =2+A / B
where
(10.34)
A =0.58 9Rag 4
B = [l + (0.469 / Pr)9116 ]419
[Pr :2'. 0.7, Ra0 < 10 11 ]
SIMPLIFIED CORRELATIONS FOR AIR
• 10.6
•
Surface
Laminar
(10 4 <Ra< 109)
Turbulent
(Ra> 109)
Vertical plate or cylinder
h = 1.42 (f'..T I L) 114
h = 1.31 (f'..T) 113
Horizontal cylinder
h = 1.32 (f'..T I D) 114
h = 1.24 (f'..T) 113
Horizontal plate: Heated plate facing
upwards or cooled plate facing downwards
h = 1.32 (f'..T I L) 114
h = 1.52 (f'..T) 113
Heated plate facing downwards or cooled
plate facing upwards
h=0.59 (
• 10.7
• Horizontal Rectangular Cavity
f',.:
1/ 4
)
ENCLOSURES
•
If the hotter plate is at the top, there are no convection currents and
heat transfer is downwards by conduction (Nu= 1). When the hotter plate is at the bottom, significant
convection currents set in for RaL > 1708, and the heat transfer rate increases.
Natural (Free) Convection
r
10.7
For large aspect ratio (H/L), as a first approximation,
INuL =0.069 Rat p,.o.o I (3x!0 <RaL <7x!0
3
5
74
(10.35)
9)
3
1708]+
Nu=l+l.44 [ 1 - - + [Rali
_ L__ l ]+
RaL
18
(10.36)
The notation [ ]+ indicates that if the quantity in the bracket is negative, it should be set equal to zero.
All fluid properties are at f =(7; + 7;) I 2
• Inclined Rectangular Cavities
-
For HIL ;;=: 12, and tilt angles 0 up to 70°,
1708
1708(sinl.80)l ·6
cos0))
l+ [((RaL
]+
-~--1
l/3
NuL =1+1.44 [ 1 - - - -] + [ 1 - - ~ - - ~ RaL cos0
RaL cos0
5830
T1
T2
I
H
TL
J_
H
f.--- L ~
(b)
(a)
Fig. I 0.5
(10.37)
1
(c)
Free convection in rectangular cavities: (a) Horizontal (b) Inclined (c) Vertical
For horizontal cavities, 0 = 0°. All properties at mean air temperature ( 7; + 7;) I 2.
• Vertical Rectangular Cavities
from
Nu=0.18 (
For vertical horizontal enclosures, the Nusselt number can be determined
p
r RaL
0.2+Pr
)0.29
l<H I L<2
{ any Prandtl number
}
(10.38)
RaL Pr I (0 .2 +Pr) > I03
Pr
Nu=0.22 ( ---RaL
0.2+Pr
)o. s(H)-u
2
L
4
2<H IL<10
}
{ any Prandtl number
(10.39)
RaL < 1010
t For 10 < (H/L) < 40, 104 < RaL < 10 7 :
INuL =0.42Rat Pr04
012 (H I L)-0· 3 1
(10.40)
10.8,
Heat and Mass Transfer
t For l < (H/L) < 40, 106 <RaL <10 9 :
!NuL = 0.046 RaL"' I
(10.41)
cz;
All fluid properties are to be evaluated at
+ Ti)/ 2 .
• For 5 < (H/L) < 110, 10 2 < RaL < 10 7 , choose the largest of the following:
!NuL = 0.060 Raf 3 I
[
-
NuL = l+
{i
(10.42)
0.104Ra0 ·293
L
+ (6310 / RaL )1.3 6 }
3]ID
(10.43)
R aL 0.272
NuL = 0.242 [ - -]
(HI L)
-
(10.44)
CONCENTRIC CYLINDERS
• 10.8
For concentric horizontal cylinders, as illustrated in Figure 10.6, the
rate of heat transfer through the annular space between the cylinders
by natural convection per unit length is
•
Fluid
(10.45)
where
k = 0.386 (
~
k
and
pr
0.86l+Pr
)l/ (Ra* )
4
1/4
(10.46)
cyl
- - - - D o _ _ __,
r
[ln(D0 ID;) RaL
•
Ra 1 =~--~~cy
(10.47)
I} (D:-315 + D-315) 5
C
I
0
Fig. I 0.6 Natural convection in a
horizontal cylindrical annulus
CONCENTRIC SPHERES
• 10.9
•
For a spherical enclosure, the rate of heat transfer through the space between the spheres by natural
convection is expressed as
(10.48)
where
k
(
~=074
k
·
pr
0.86l+Pr
)l/ ( Ra*sph)
4
1/4
(10.49)
Natural (Free) Convection
r
I0.9
(10.50)
and
(10.51)
The quantity kNu is called the effective thermal conductivity of the enclosure, since a fluid in an
enclosure behaves like a quiescent fluid whose thermal conductivity is kNu as a result of convection
currents. The fluid properties are evaluated at the average temperature of ('I';_+ T0 )!2.
• I 0.10
TURBINE ROTORS / ROTATING CYLINDERS / DISKS/ SPHERES
•
+ Cooling of Turbine Blades
Nu=0.0246 [
Pr.117 GrL ]0.4
l+0.495Pr213
(Gr>1012 )
(10.52)
(10.53)
where
+ Rotating Horizontal Cylinders Peripheral speed Reynolds number,
Re00 =1t D 2 ro / v
For Re00 > 8000 in air,
(10.54)
• I 0.11
COMBINED FREE AND FORCED CONVECTION
In mixed forced and natural convection, the Richardson number is defined as
•
(10.55)
The range in which mixed forced and natural convective effects are important is
0.1 <Ri < 10
(10.56)
The Nusselt number for mixed convection can be approximated by a relation of the form
(10.57)
where the Nusselt numbers for pure forced and pure natural convection are found from known correlations.
The plus sign is used when forced flow is aiding or perpendicular to the natural flow and the minus
sign is used when forced flow opposes natural flow. For vertical geometries, n = 3, while for horizontal
geometries, n = 3. For cylinders or spheres, use n = 4.
I0.10, Heat and Mass Transfer
Solved Examples
(A) EXACT ANALYSIS, INTEGRAL SOLUTION
•
Example I 0.1
•
~ A vertical plate 30-cm-high is maintained at a temperature of I7°C in quiescent air at a
temperature of 37°C. Using the results of the integral method, find (a) the thickness of
thermal boundary layer, and (b) the local heat-transfer coefficient at the bottom edge of the plate.
Properties of air at 27°C are
k = 0.0263 W/m K v = 15.89 x I0-6 m2/s
Pr= 0.707
Known: A cold vertical plate is exposed to warm stagnant air.
Find: (a) Thermal boundary layer thickness, OT; (b) Local heat-transfer coefficient at the bottom edge, hx.
_,,,,.--~<---,._,,.-- -,,~-- -,. . . . '-
Schematic:
} Stagnant ;ur, -J
r
T==37"C. ':)
_./
!-.- ~~ ./
X
g
L=0.3m
s= 17°C
x=L' ~ - - 8"" 8T
Fig. 10.7
Assumptions: ( l) Air is an ideal gas at l atm. (2) The vertical plate is isothermal. (3) OT"' 8. (4) The
flow is laminar, steady and incompressible. (5) Negligible viscous dissipation.
Analysis: Properties are to be evaluated at the mean film temperature,
l
l
7'r = 2(T. +T=) = 2{17+37) = 27°C
A.
1-',dealgas
=-I-=
I
7'r{K) {27+273.15)K
=000333K-l
·
The boundary layer development is given by
~ = 3.93 (0.952+Prf 4 G'"x.-l/4 Pr-112
X
Natural (Free) Convection
r
I 0.1 I
where
x=L=0.30 m,
At
G,;_ =
(9.81m/s2 ) (0.00333 K- 1) (0.3 m}3 (37-17) K
2
(15.89x1Q-6 m2 /s)
= 6.99x107
Therefore
~ = 3.93 (0.952 + 0.707)0.2 5 ( 6.99 X 107 )-0.25 (0.707t· 5 = 0.058
X
Velocity boundary layer thickness at the bottom edge of the plate is
8 (x = 0.3 m) =(0.3 m) (0.058) = 0.0174 m = 1.74 cm
Since BT"" 8 for fluids with near unity Prandtl number like air,
(Ans) (a)
BT= 1.74 cm
The local Nusselt number,
2x
N"x_ = -
8
By definition,
hx X
Nu=-x
k
Hence,
k
k 2x 2 k
\ = -; N"x_ = -; ·
=
r; T
local heat-transfer coefficient at the bottom edge of the plate,
\ = 2x0.0263W/m K = 3 _02 W/ml K
0.0174 m
(Ans) (b)
Example I 0.2 ~ A 0.6-m high and 3-m wide wall is maintained at 80°C in an atmosphere of 24°C. Neglecting
end effects and radiation, determine the rate of heat loss by free convection. Use the following
correlation for a vertical flat surface:
Nux = 0.509(PrY 12 (Pr+0.952)- 114 (GrxY' 4
where all properties are at the film temperature, Tf = 52°C:
k = 0.0282 W/m°C v = 18.42 mm 2/s
Known: A heated wall is exposed to still air.
Find: Heat loss rate by free convection, Q[W].
Pr= 0.7035
I
0.12,
Heat and Mass Transfer
Schematic:
i - - - - - - - W=3 m - - - - - - . i
Quiescent air
r.. =24°c
x=O
x=L
Fig. 10.8
Assumptions: ( l) Steady operating conditions exist. (2) Air is an ideal gas. (3) Constant properties.
(4) Radiation and end effects are neglected.
.
80+24
Analysis: Film temperature, Tr =(T. +T.. ) I 2 = - - - = 52°C or
2
325.15 K
(based on T.. as mentioned in the problem statement)
~idealgas = 1/(325.15 K)
Local Grashof number,
g~x 3 (T-T.. )
Gr.X =
s
y2
where x = L = 0.6 m
(9.8lm/s 2) [ l/(325.15K)] (0.6m) 3 (80-24)°C
(18.42 x 10-6 m2/s)2
= - - - ~ ~ - - - ~ - - - - - - = 1.076 X 109
Hence, local Nusselt number,
Nux =0.509
(0.7035) 112 (0.7035 +0.952tll 4 (l.076 X 109 )1/ 4 = 68.16
local heat-transfer coefficient, h (x = L) =
z
Nu(x=L)
= 0.0282W/m oc {68.16)= 3.2 W/m2 oc
0.6m
Average heat-transfer coefficient,
-
4
4
h =3 \x=L) = 3 x3.2 = 4.27 W/m 2°C
Therefore,
Rate of heat loss,
Q=h(LW)(7'.-T.. )
= ( 4.27 W/m 20C) (0.6 m x 3 m) (80 - 24)°C = 430.4 W
(Ans)
Example I 0.3 ~ Two vertical plates, 150-mm high, at 90°C are placed parallel to each other in a tank of
water at 24°C. Determine the minimum spacing which will prevent interference of the free
convection boundary layers.
Use the following thermophysical properties of water at 57°C:
= 504 X 10-6 K- 1
v = 4.97 x 10-7 m2/s
~
Pr= 3.15
k = 0.650 W/m K
Natural (Free) Convection
r
I 0.13
Known: Two parallel vertical plates placed in water.
Find: Minimum spacing between the plates to avoid interference of free convection boundary layers.
Schematic:
- - - 8 _ _...,.,___ 8
T
H= 150mm
Vertical plate
(T.=90°C)
t----+----',,o----+--- Boundary layers
1
L
Fig. 10.9
Assumptions: (I) Negligible radiation heat transfer. (2) Water in the tank is quiescent.
Analysis: Minimum spacing required between the two plates = L = 2 8
where o ... boundary layer thickness at the trailing edge of the plate
Mean film temperature,
I;-= T,; +T= = 9o+ 24 = 57°C{= 330.15 K)
2
2
Local Grashof number with x = H (at the trailing edge),
Gr,.= g~x3 (T,; -T= I v 2 )
9.81 m/s 2 X504 X IQ-6 K- 1 X0.153 m3 X{90-24)°C
= - - - - - - - - - - - - - - - - = 4.46 X109
(4.97 x 10-7 m2/s)2
Gr,,_ Pr= ( 4.46 X109 ){3.15) = 14.04 X109
As Gr Pr> 10 9, the flow is turbulent.
Boundary layer thickness for turbulent flow,
0.565x[1+0.494Pr213]
o(x)-------G1·l p,,s11s
O(H) = 0.565 X 0.15 m [l + 0.494 X3.152/3]
(4.46xIQ 9 )°" 1 X(3.15) 8 /l 5
= 0.01027 m or 10.27 mm
minimum spacing required, L = 2 o= 20.54 mm
(Ans)
1
Comment: Note carefully that ~ for water at 57°C (330.15 K) is 504 x 10-6 K- 1 and not - , i.e.,
1
--=3.03xI0-3K- 1, because ~=III;- only for ideal gases.
330.15
Tr
10.14, Heat and Mass Transfer
Example I 0.4 ~ A hot plate of 40-cm height and 70-cm width at I 50°C is exposed to the air at 30°C.
Calculate the following using the integral (approximate) solution:
(a) Maximum velocity at 30cm from the leading edge of the plate
(b) Mean velocity at 30cm from the leading edge of the plate
(c) Boundary layer thickness at 30cm from the leading edge of the plate
(d) Local heat-transfer coefficient at 30cm from the leading edge of the plate
(e) Average heat transfer coefficient over the entire surface of the plate
(f) Total mass flow through the boundary
(g) Total heat loss from both sides of the plate
(h) Rise in temperature of the air passing through the boundary
Known: A hot vertical plate is exposed to quiescent air.
Find: (a) umax
(b) uav
(c) 0 (x)
(d) h (x) (at x = 0.30 m).
(e) h1
Schematic:
(h) ~¾ir·
(g) Q
(t) mtot
Trailing edge
~W-=-0-.7-0-m----~(x = L = 0.40 m)
A5 =2 WL
= 2 (0.70 m) (0.40 m)
Thermal boundary layer
=0.56m 2
Stationary fluid
r~= Jo c
0
X, U
Leading edge
Lo
Fig. 10.10
Assumptions: ( 1) Steady operating conditions. (2) Air is an ideal gas. (3) Local atmospheric pressure
is 1 atm.
1
Properties: At film temperature, Tr = -( 150 + 30) = 90°C :
2
k = 0.03024 W /m °C
Pr= 0.7132
p = 0.9718kg/m3
v=22.0lxI0-6 m2 /s
~=
cp = 1.oos kJ /kg c
l
= 0.00275 K- 1
{90+273.15) K
0
Analysis: (a) Grashof number (at x = 0.30 m),
Gr. = g ~ x3 ( I'. - T=)
X
y2
(9.81m/ s2 ) (0.00275 K- 1 ) (0.30m)3 (150-30)°C
= -'------'--'------'----'---'---'----(22.0lxl0-6 m2 / s)2
3
0 40) =4.277x108
and Grdat x=L=0.40 m)=l.804xl0 8 x ( -·0.30
1.804 X 108
Natural (Free) Convection
r
I 0.15
Maximum velocity based on integral solution is
1/2
u,. =5.17 [ g~(r.-r.,.,) ]
Pr+0.952
where
,;:;::
xlf2
1
V
=5.17vv'x x-x
x
1/2
(Pr+ 0.952)
1
=5.17.J'l.804x108 x 22 ·0lxl0-6 x
0.30
(0.7132+0.952)
= 3.06 mis
Hence
4
Umax
(Ans) (a)
=-X3.06=0.453 m/s
27
(b) Mean velocity, uav (at x = 0.30 m) = 27 umax
48
=(!D (0.453)=0.255 m/s
(Ans) (b)
(c) Boundary layer thickness, <>(x) (at x = 0.30 m) is
<>(x) = 3.93 x [0.952+ Pr]l/4 = 3.93 x 0.30 mx [
Gf"x Pr2
= 0.0137 m
or
0.952+ 0.7132 ]114
1.804x108 x0.7132 2
(Ans) (c)
13.7mm
(d) Local Nusselt number,
Nu,.= 0.508 Pr112 (0.952+Pr)
-1/4
2x ~ X
=8
k
G~1 4 = -
~ = 2k = 2x0.03024 W/m°C
Hence
8
0.0137 m
= 4.415 W / m2 °C
(e) Average heat-transfer coefficient,
-
-
k
hi_,= NULL
NuL = 0.677 G,t14 Pr112 (0.952+Prf 114
= 0.677( 4.277 X108 )' 14 (0.7132)112 (0.952+ 0.7132t 114
= 72.4
(Ans) (d)
I
0.16,
Heat and Mass Transfer
~ = 72 .4 X 0.03024 W/m °C
0.40 m
= 5.472 W / m2 °C
(Ans) (e)
(f) Total mass flow through the boundary is
m = 1.7 vw
Gr.
]
L
2
Pr (Pr+0.952)
[
p
tot
1/4
= 1.7x0.9718kg/m3 x22.0lxl0-6 m2 /sx0.70m
[
4.277x108
(
)
0.7132 2 0.7132+0.952
= 3.815 kg/s
IM
J
(Ans) (f)
(g) Total heat loss from both sides of the plate is
Q= ~ (2 WL )( 'f. - T-)
=(5.472 W/m2 °C) (2x0.4mx0.7m)(150-30)°C
= 367.7 W
(Ans) (g)
(h) Rise in temperature of air passing through the boundary is calculated from
AT.. =
air
367.7W
3.815 X 10-3 kg/s X 1008 J/kg°C
= 95.6°C
Example I 0.5
(Ans) (h)
~ Compare the boundary layer thickness o and the displacement thickness o* on a vertical
flat plate in (a) air, and (b) water at surface temperatures of 93°C and fluid temperatures
of 70°C at a location 0.3 m from the bottom edge of the plate.
Properties: Air ( I atm, 82°C)
k = 0.03038 W /m °C
V=21.47 xlo--6m2 /s
k = 0.671 W/m °C
v = 353.3 xl0-9 m2 /s
Pr= 2.14
~
Pr= 0.699
Water (82°C):
= 652.3 x I o---6 K- 1
Known: Natural convection from the heated vertical surface to the ambient fluid: (a) air and (b) water.
Find: 8 and 8* for (a) air, and (b) water.
Natural (Free) Convection
Schematic:
r
I 0.17
r.=93°c
r.=93°c
Turbulent
Turbulent
Quiescent air
Still water
r .. =11°c
r.. =11°c
Tx=0.3m
Laminar
X
X
!x=O
Fig. I 0.11
Laminar
!x=O
-~----~~
------~~
(a)
(b)
Boundary layer on a vertical flat plate with free convection on the surface in (a) air, and
(b) water as the surrounding medium
Assumptions: ( l) The fluid is a stagnant medium. (2) Constant properties
Analysis: The characteristic dimension in the case of a vertical hot plate is the height, x above the bottom
edge of the plate.
Hence, x = 0.3 m.
l
Mean film temperature, Tr = -( T. + T.. ) = {93 + 71 )/2 = 82°C
2
,
A
l
l
l
(a) Au: 1-'idealgas = 7'r = {82+273.15)K = 355.15 K
Grashof number
(9.8lm/ s2 ) [ 1/(355.15 K)] (0.3 m)3 ( 93-71) K
=----~----~~---,------(21.47 x 10-6 m2 / s)2
= 35.6 X 106
Rayleigh number,
Ra,,_ = Grx Pr= 35.6 X 106 X 0.699 = 2.488 X 10 7
As this value is less than 109, the boundary layer is laminar.
The laminar boundary layer thickness is given by
<>iam
= 3.93{0.952 + Pr{ 4 Pr-ll 2 Gr-ll 4
X
At x = 0.3 m from the bottom edge of the heated plate,
<>1am
= 3.93 X 0.3 m X [0.952 + 0.699f 4 (0.699rl/ 2 (35.6 X 106 r' 14
= 0.0207 m
or 20.7 mm
(Ans) (a)
10.18, Heat and Mass Transfer
Displacement thickness,
<>~am = 0.562 <>iam = 0.562 X 20.7 mm
= 11.63 mm
(Ans) (a)
(b) Water: Grashof number,
Gr. = g ~ x3 ( I'. - T=)
X
y2
_ (9.81 m/ s2) (652.3xl0-6K- 1) (0.3 m)3 {93-71) K
(353 .3x l0-9 m2/s)2
= 30.452 X 109
Rayleigh number,
Ra,,_ = Gr.,_ Pr= 30.452 x 109 x 2.14
= 6.5167 X 1010
This value being greater than 109, the boundary layer is turbulent.
The turbulent boundary layer thickness is given by
<>turb
= 0.565 ( 1+ 0.494 Pr213 f' 0
Pr- 8115
Gr-lllO
X
At x = 0.3 m from the bottom edge of the heated plate,
<>turb
= 0.565 X 0.3 m X [l + 0.494 + 2.142/3 f- 1 (2.14)-8115 X (30.452 X 109 )-0.I
= 0.0107 m or 10.73 mm
(Ans) (b)
Displacement thickness,
= {0.272) {10.73 mm)
=2.92 mm
(Ans) (b)
Example I 0.6 ~ A vertical plate maintained at 90°C is placed in still atmospheric air at 40°C. The plate is
0.80-m long and 1-m wide. (a) Determine the height at which the boundary layer will turn
turbulent if the onset of turbulence takes place at Ra= 109 . (b) If the fluid is water instead of air, what will be
the maximum height for laminar convective flow! (c) Using the integral method, find the values of boundary
layer thickness, and local heat transfer coefficient at 0.2, 0.4, 0.6 and 0.8 m for air. (d) Also determine the heat
loss from both sides of the entire plate to ambient air.
Properties at 65°C are tabulated below.
Fluid
k (W/m 0 C)
v(m 2/s)
Pr
Air
29.lxl0-3
19.71 X lQ- 6
0.7017
2.957 X lQ- 3
Water
658.5 X 10-3
440.8 X lQ- 9
2.74
554.7 X lQ- 6
Natural (Free) Convection
r
I 0.19
Known: An isothermal heated vertical plate exposed to quiescent fluid.
Find: Xtrans for air and water; o(x); hx and Q for air.
Schematic:
Turbulent
Vertical plate
(L=O.Bm , W= Im)
Quiescent
air/water
RaL = 109
Ts=90°C
T
1
Laminar
X
y
Fig. 10.12
Assumptions: ( 1) Steady operating conditions. (2) Quiescent air/water. (3) Constant properties.
Analysis: Rayleigh number,
g~~ ('f.-T=)Pr
RaL = ----'----'--v2
where Lc is the characteristic length which is height in the case of a vertical surface.
Fluid: Air With Lc = x, one has
( 9.81 m/s2) ( 2.957 X 10-3 K- 1) (x )3 m 3 ( 90-40)°C( 0.7017)
Gr=--------------------( 19.71x 10-6 m 2/s)2
= 3.734 X 109 x 3
Ra=GrPr=3.734x109 x 3 x0.7017
=2.62x10 9 xx3
r r
3
xtrans = ( 2.6::rl09
3
= (2.6~~ 109
(Ans) (a)
=0.725 m
Water
(9.81)(554.7x10-6 ){x}3{90-40)
Gr=~~--'----~---,-----= 1.4 x 1012 x 3
( 440.8 X 10-9)2
Ra= Gr Pr= 1.4 x 1012 x 2.74 x 3
= 3.837 X 1012 x 3
109
(
Xtrans = 3.837 X 1012
)l/J
=0.064 m
In case of water, turbulence regime starts at a much shorter height.
(Ans) (b)
10.20, Heat and Mass Transfer
Calculations of Boundary Layer Thickness
Using the approximate integral method:
Fluid: Air Boundary layer thickness,
<>1am
()
= (3.93)[Pr+Pr0/52]114
x
Gr/ 4
[upto xtrans = 0. 725 m]
= 393 [0.7017+0.952]114
X
lam
·
0.70172
[3.734x109 xx3
f
4
= 0.0215 x 114
At
x=0.2 m,
8 = 0.0215 (0.2)114 = 0.0144 m
or
14.4 mm
= 0.4 m,
8 = 0.0215 (0.4)114 = 0.0171 m
or
17.1 mm
X
8 = 0.0215 (0.6)114 = 0.0189 m or 18.9 mm
x=0.6 m,
At x = 0.8 m, the boundary layer is turbulent,
Hence,
<>turb = 0.565 x (1 + 0.494 Pr213
)'1' 0 Prst1s Gr,z-1110
= 0.565x [1 + 0.494 X 0.7017213 ]1110 (0.7017t8115 X (3.734 X 109 x3 rll!O
= 0.07784 X x 0-7
At
x=0.8 m,
8 = <>turh = 0.07784 x (0.8) 0·7 = 0.0666 m or 66.6 mm
Calculations of Local Heat Transfer Coefficient
Fluid: Air (For Ra< 109, i.e., up to x = 0. 725 m)
Local Nusselt number,
_
Nux(lam) -
Pr
Pr+0.952
114
0.508 [ - - - - ]
114 _ hy_x
R~ - k
Local free convection heat-transfer coefficient,
k[
hy_ = 0 ·508 ;
Pr
] 114 [g~x3 (T. -T.,.,)Prl
-P-r+-0-.9-5-2
v2
= 0.508x0.0291 [
X
0.7017
]
0.7017+0.952
1/4
= 2.70 (xt· 25
At
x=0.2 m,
hy_ =2.70 (0.2)--0· 25
= 4.04 W / m2 °C
114
(2 _62 xl09 x3}'14
Natural (Free) Convection
r
I 0.21
h,. = 2.70 (0.4 )-0·25
x=0.4 m,
= 3.40 w / m2 °C
h,. = 2.70 (0.6 )-0·25
x=0.6 m,
= 3.07 W / m2 °C
At
x = 0.8 m, the boundary layer is turbulent.
Ra > 109
For
or
for x > 0.725 m,
Nux =0.0295 Gr;f 5 PrJll 5 [1+0.494 Pr213r215= h,,_x
k
Local heat-transfer coefficient,
h,,_ =~x0.0295 G,2 4 Pr7115 [1+0.494 Pr213r 04
X
= 0 ·0291 X 0.0295 X(3.734 X 109 x 3 }°" 4 (0.70 l 7f 15 X [l + 0.494X0.7017 213 ]-0.4
X
= 4.30 x 0 -2
At
x=0.8 m,
h,,_ =4.30 {0.8)°" 2 =4.11 W /m2 °C
The values of o(x) and hx are tabulated below for air:
H eight, x (m)
8 (x), mm
hx (Wlm 2 °C}
0.2
14.4
4.04
0.4
17.1
3.40
0.6
18.9
3.07
0.8
66 .6
4.11
(Ans) (c)
Heat loss from both sides of the vertical plate is
Q = hL (2 L W) ( 7'. - T~)
where
hL
IL
=-f
h,,_ dx
Lo
(xt.2
I
L
1
)L
= - x 4.30 x0 -2 dx = - x 4.30 L
L
1.2
0
O
f
= 4.30 x ro.2 = 4.30 x 0.8 o.2 = 3.427 W/m2 oc
1.2
1.2
10.22, Heat and Mass Transfer
Hence,
Q = (3.427 W/m 2 °C) (2 X 0.8m X lm) {90-40)°C
=274W
(Ans) (d)
(B) VERTICAL PLATE/TUBE
•
•
~ The fre~onvection correlations for a vertical plate are:
Example I 0. 7
Nu= 0.59(RaL)1 ' 4
(104 <Ra< 109 )
Nut. = 0. I 0( RaL yt 3
(109 <Ra< 10' 3 )
For air at I atm and a film temperature of I 50°C:
k = 0.0357 W/m °C
v = 28.95 x 10-6 m2/s
Pr=0.683
Deduce the simplified equations for the average convection coefficient in terms of i1.T (°C or K) and L (m).
Known: Vertical plate exposed to free convection air.
Find: Simplified expressions for hL.
Schematic:
T
Still air, T
00
L
l r,=
I 50°C, I atm
i1..T=Ts-Too
Fig. 10.13
Assumptions: ( 1) Air is an ideal gas (~ = 1/Tr{K )) . (2) Constant properties. (3) Quiescent air.
Analysis: Both correlations are of the form:
-
hLL
NuL = C(Radn = K
where
g~Dt:,,.T-Pr
RaL=-"-'----y2
=
(9.81 m/ s 2 )[1/ {150+273.15) K]D t:,,.T{0.683)
2
(28.95xl0-6 m 2 / s)
hi,= 0.0 357 C(l.89xl0 7 t:,,.T.Dr
L
For laminar boundary layer conditions (RaL < 109 ):
C = 0.59,
n = 1/4
=l.89xl0 7 t:,,.T-D
Natural (Free) Convection
~ =0.0357x 0.59x(1.89xI0 f
7
1/iL = 1.39(~T/L)1' 4
For
C=0.10
(
~:r
I 0.23
4
(Ans)
1
and n = 1/3
~ = 0.0357 X 0.10 X ( 1.89 X 107
Hence,
4
r
t3 (~T)
1/iL = 0.95{~T)113 I
(Ans)
Comment: It is noteworthy that the average convection coefficient is independent of the characteristic
length in case of turbulent boundary layer condition.
Example I 0.8
~ Consider a design for a nuclear reactor using natural convection heating of liquid bismuth.
The reactor is to be constructed of parallel plates 1.8-m tall and 1.2-m wide, in which
heat is generated uniformly. Estimate the maximum possible heat-dissipation rate from each plate if the surface
temperature of the plate is not to exceed 875°C and the lowest possible bismuth temperature is 325°C.
Properties of liquid bismuth at 600°C are
~=0.124x 10--3 K- 1
Pr=0.01022
k = 15.60 W/m°C
v = 1.048 x 10--7 m2/s
Known: Natural convection heating of molten bismuth in a nuclear reactor comprising vertical parallel plates.
Find: Maximum heat dissipation rate, Q(W).
Schematic:
f
L= I.Sm
l
Fig. 10.14
Assumptions: ( 1) Steady-state conditions exist. (2) Constant properties.
Analysis: The characteristic length for vertical plate in free convection heat transfer is the plate height.
L=H= 1.8 m
Rayleigh number,
RaL
g~ L3 ( 'f. - T= ) Pr
= GrL P r = - - ~
--~y2
(9.8lm/s2 )(0.124x 10-3 K- 1)(1.8m) 3 [(875-325)°C orK](0.01022)
( 1.048 x 10-7 m2 /s) 2
= - - - - - - - - - - - - ~ - - - - - - ~ - - - = 3.63 X 1012
10.24, Heat and Mass Transfer
Since RaL > 109 , the boundary layer is turbulent. The appropriate accurate correlation is
- { 0 825
NuL-
.
l/6
0 .38 7 RaL
+[1+(0.492 / Pr)9it6t27
}2
= 634.7 = hL
k
Heat-transfer coefficient is
h=!:.._NuL = 15.6W/m oc = {634.7)
L
1.8m
= 5500 W/m2 °C
Rate of heat dissipation from both sides of each plate is
Q = h(2 WH)(T,. -T=)
= (5500 W/m20C)(2 x 1.2 m x 1.8 m)(875 -325) °C
= 13.07 x 106 W
or 13.07 MW
(Ans)
Comment: Note that we are in turbulent flow. Therefore h does not vary up the plate. The entire plate
will operate at the limit of 875°C.
Example 10.9
~ A hot plate of 15 cm 2 area, maintained at a temperature of 200°C, is exposed to still air
at 30°C. When the smaller side of the plate is held vertical the convective heat transfer
rate is 14% higher than that when the bigger side of the plate is held vertical. Determine the dimensions of
the plate. Neglect the internal temperature gradients of the plate. Also, calculate the heat transfer rate in both
cases. Use the correlation
The properties of air are given in the table below:
T (OC)
CP (kjlkg K)
µ (kg/ms)
30
1.165
1.005
18.6 x I o- 6
0.0267
115
o.91 o
1.009
22.65 x I o- 6
0.0331
200
0.746
1.026
26.0 x I o- 6
0.0393
p (kg!m 3)
-----'---------
k (Wlm K)
---------1
Known: Hot vertical plate of given area and constant temperature exposed to quiescent air with smaller
side vertical and then larger side vertical.
Find: Dimensions of the plate. Heat-transfer rate in the two cases.
Natural (Free) Convection
r
I0.2S
Schematic:
Steel air
T==30°C
(11)
(1)
Q 1= l.14Q 11
Fig. 10.15
Assumptions: ( l) Steady operating conditions. (2) Air is an ideal gas. (3) Radiation effects are negligible.
Analysis: The properties of air for both cases are to be evaluated at the film temperature,
Lr= _!_(T.. + T ) = (200+ 30) = 1150c
2 s =
2
From the given table:
Pr= CPµ = (l.009xl0 3 J/kgK)(22.65xl0-6 kg/ms)ll Wl=0_ 6904
k
0.0331 W/mK
l J/s
Characteristic length, L = height.
v = .!: = (22.65/0.910)(10-6 ) = 24.89 x 10-6 m2 /s
where
and
p
__
l__
A
1-'idealgas -
l
_ _l_(K-1)
Tr(K) - (ll5+273.15)K - 388.15
Gr.= (9.81 m/s2 )[l/388.15 K)]L3(m 3 )(200-30)°C or K
(24.89 x 10-6 m2 /s) 2
L
= 6.935 X 109 /)
10.26, Heat and Mass Transfer
GrLPr = (6.935 x 109 D)(0.6904)
and
= 4.788x109 D
The convective heat transfer rate is
Q1 = h1A.(7'. -T=)
This is 14% more than that in the second case.
Q11 = h11A.(T. -T=)
qi = .!1_ = 1.14
Qu
hu
-
hL
NuL = -
Since
k
= 0.59(GrLP,)'l4
h = 0.59!x[ 4.788x109 D]'t4]
L
= 0.59x0.0331x(4.787x109 )114 xL-114
= 5.137 L- 114
With smaller side vertical, L = b,
h1 = 5.137 b- 114
With larger side vertical, L = a
h11 = 5.137 a- 114
h = 1.14 = 5· 137 b-114 = ( ~ ) 114
Therefore =1
' h11
5.137 a- 114
b
~ = (1.14) 4 = 1.689 ~ a= 1.689b
or
b
Since, area A 8 =ab= 15 cm2
1.689b2 = 15 ~
b = ~15/1.689 = 2.98 cm
a= 1.689 x 2.987 = 5.03cm
and
:. dimensions of the plate are ax b = 5.03 cm x 2.98 cm
In the first case,
(Ans)
h1 = 5.137 (b)- 114 = 5.137 (2.98 X IQ-2m)--0.25
= 12.36 W/m2 K
Heat-transfer rate,
Q1 = h1A.(7'. -T=)
= (12.36 W/m2 K)(15 x 10-4 m2)(200-30)°C
or
K = 3.15 W
(Ans)
Natural (Free) Convection
r
I 0.27
In the second case,
hn = 5.137(5.03 x l0-2 m)-0·25 = 10.84 W/m2 K
Heat transfer rate,
Qu=(l0.84 W/m2 K)(l5xl0-4 m2 )(200-30)°C
or
K
=2.76W
Check: Q 1 / Qn = 3.15 I 2.76 = 1.14
Also,
h1 I hn = 12.36/10.84 = 1.14
Example I 0.10 ~ An aluminium alloy plate, heated to a uniform temperature of 227°C, is allowed to cool,
while suspended vertically in a room. The ambient air and surroundings are at 27°C. The
plate is 0.3-m-square with a thickness of 15 mm and an emissivity 0.25.
(a) Develop an expression for the time rate of change of the plate temperature, assuming the temperature to
be uniform at any time.
(b) Determine the initial rate of cooling when the plate temperature is 227°C.
Take: Aluminium alloy:
k= 186 W/m °C,
p = 2770kg/m 3,
cp = o.983 kJtkg c.
0
I
Properties of air at 1i = -(227 + 27) = I 27°C:
2
k = 0.0338 W/m °C,
v = 26.41 x I o-6 m2/s,
Pr=0.690
Known: An isothermal aluminium alloy plate is vertically suspended in a room losing heat by both natural
convection and radiation.
Find: (a) Expression for time rate of change of the plate (b) Initial rate of cooling.
Schematic:
Quiescent air
T= =27°C
Tsur= T=
=27°C or 300.15K
Vertical plate (E = 0.25)
Area, As= 2 bL
T; = 227°C or 500.15 K
Fig. 10.16
Assumptions: (1) Uniform plate temperature. (2) Quiescent room air. (3) Surroundings are large compared
to the plate.
Analysis: (a) Applying the energy balance, we have
0
0
¢-£out +¢n = Est
10.28, Heat and Mass Transfer
Clearly,
.
or
Qconv
.
dT
+ Qrad = -mCP -
dt
dT
=-p¥-CPdt
h (2 bL) (T. -T-)+ e(2 bL)o (T.4 -T.t) = -p (b Lt)CP dT
or
dt
where T8 is the uniform plate temperature at any instant.
Then, the initial rate of cooling is
- dT = -2-[ii(T-T )+eo(T4 -T4 )]
dt p tCp
• •
sur
(Ans) (a)
(b) To determine (-dT!dt), one must find the value of h for which let us first calculate the Rayleigh
number. Characteristic length for the vertical plate is its height, L = 0.3 m.
Rayleigh number,
1
where
~ideal gas =
Hence,
RaL =
1
Tr = 400 K
(9.81 m/s2 )[ 1/( 400 K)](0.3m}3 {227-27)K(0.690)
2
= 1.31x108
(26.41xI0-6 m2 /s)
Since RaL < 109, the appropriate Churchill-Chu correlation is
_NuL = 0.68 + 0.67 Raf 4 [ 1+ (0
492)9116
~
]-4/
9
=0.68+0.67(1.31x108 ) 1/ 4 [ 1+(0.492/0.690}91 16 J-4/9 =55.52
Average convection coefficient is
_
_
k
(55.52)(0.0338W/m °C)
/
2
h = NuL - = - - ~ - - - - ~ = 6.255 W m °C
L
0.3 m
Substituting the proper values in the expression derived above, one gets
2
dT
--=-------------x
dt (2770 kg/m3 )(0.015 m) (983 J/kg°C)
[ 6.255 W /m2 °C{227 - 27)K + (0.25)(5.67 x IQ-8 W /m2 K3 ) {500.154 -300.15 4 } K 4 ]
=0.099°C
or K / s
(Ans) (b)
Natural (Free) Convection
Comment: The assumption of uniform plate temperature is valid if
-
where
-
-
htotal = hconv + h,ad and
r
I 0.29
B{ = htoti Lc) < 0.1
Lc = ¥-A. = 2bLt
bL= 2t = 0.0075 m
hrad = E cr ( I'. + I'.ur )( I'.2 + I'.~r )
= 0.25 X 5.67 X 10-8 {500.15+300.15) (500.152 +300.152 )
= 3.86 W /m 2 °C
htotal = 6.255 + 3.86 = 10.115 W/m 2 °C
Bi= 10.ll5x0.0075 = 4 _08 xl0-4
(« 0.1)
186
This validates our assumption.
Example I 0.11
~ An isothermal vertical plate 25-cm-long and 12-cm-wide is maintained at 90°C in still air
at 30°C. Find the total heat transferred by free convection from both sides of the plate if
the air pressure is (a) I atm, (b) 0.1 atm, and (c) IO atm.
Properties of dry air at atmospheric pressure and 60°C are
k = 0.0288 W/m °C
V = 0.068 m2/h
g~/va: = 57.7 x I06 (ml K)- 1
Pr=0.700
Known: A heated vertical plate is exposed to dry still air.
Find: Heat loss by natural convection in air at (a) l atm, (b) 0.1 atm, and (c) 10 atm.
Schematic:
Still air
r-=Jo c
0
P= I atm / 0.1 atm / IOatm
T
L=0.25m
1
Fig. 10.17
Assumptions: (l) Steady operating conditions. (2) Air is an ideal gas. (3) Radiation heat transfer is neglected.
Analysis: (a) P = 1 atm
Rayleigh number Ra =
'
L
g ~ D (T - T )
s
V<J,
= =
g ~ D (T - T ) Pr
s
V2
=
10.30, Heat and Mass Transfer
where
1
~ideal gas = T,
and
L is the plate height.
f
.
1
90+30
Film temperature, 4 = -(T. + T_) = - - - = 60°C
2
2
or
333.15 K
Substituting the given property values at the film temperature of 60°C and 1 atm,
1-) X (0.25) 3 m3 X (90-30)°C
RaL = 57.7 X 106 ( m3 K
or
K
(<109 )
=5.41x10 7
The appropriate correlation is
u
JV~
=0 68
•
1/4
06
. 7 RaL
+
4N
[1+(0.492/ Pr) 9116 ]
=
hL
(for 10-1 < RaL < 10 9 )
k
Hence, the free convection heat-transfer coefficient is
_0.0288W/m°C[o68
0.67(5.41xl0 7 )°" 25
h------ · +
4/9
0.25m
{1+(0.492/0.7)9 116 }
l
= 5.15 W/m 2 °C
Heat loss from both sides of the plate by natural convection is
Q=hA.(T.-T_)
=(5.15 W/m2 °C) (2x0.25 mx0.12 m)(90-30)°C
= (5.15x3.6) W =18.54 W
(Ans) (a)
(b) P = 0.1 atm
Since
l
RT
~
Vo.I atm = V1 atm X 0.1 atm
v=.!:: and
-=p
p'
p
V2 = V1 X
Pi
Pi
1
p
(for an ideal gas)
1 atm
_R
RaL -_ g~L\T. - T_)Pr
10 _2
2
aL I atm X
(viatm/0.1)
= 5.41 X 10 5
Then
0.0288 [0.68+0.67(5.41x105 )°" 25 ]
h =--x
= 1.68 W/m2 C
419
9116
o.2 5
{1+(0.492/0.7) }
Hence,
Q = (1.68 X 3.6) W = 6.05 W
(Ans) (b)
Natural (Free) Convection
r
I 0.31
(c) P= 10 atm
1 atm
Since
VlO atm
= Vl atm X -10 atm
RaL IOatm =(5.41x10 7 )(10 2 )=5.41x109
The correlation for RaL > 109 is
-[o
NuL -
825
.
2
0.387 Raf! 6
hL
]
+ { 1+ (0.492 / Pr)9l l6 }8/ 27 = k
Then
00288 [
+ 0.387 x (5.41x109) ] 2
h = -·- - x 0825
·
8121
= 23.9 W/m2 °C
0.25
{1+ (0.492 / 0.7)9/16}
Q = (23 .9) (3.6) W = 86.04 W
and
(Ans) (c)
Example I 0.12 ~ A 4-mm-thick steel wall of dimensions 1.5-m height and 0.6-m width (k = 52 W/m K) is
provided with insulation of 10-cm thickness(k =0.15 W/m K). The interior surface of the
steel wall is measured to be 75°C. The surrounding air is at 25°C. The exposed outer surface of the insulated
wall experiences natural convection. The recommended correlations are
Nlli_ = 0.59(R~f
N11i.. = O.IO(R~)113
109 <R~<I0' 2
Calculate (a) the heat-transfer coefficient, (b) the rate of heat transfer across the composite wall, and (c) the
insulation's outer surface temperature.
Known: A two-layer composite wall has its exterior surface subjected to free convection.
Find: (a) Heat-transfer coefficient, h(W/m 2 0 C), (b) Rate of heat transfer, Q[W] , and (c) Outer surface
temperature T0 ( 0 C).
Schematic:
Quiescent air
T =25°C
00
L= I.Sm
T=75 °C
kstee1=k1 = 52 W/m K
CD
I
k;nsulation= k2 = 0.15 W/m K
~--~~
L 1=0.4cm -I
f+-- L2= 10cm ----+j
Fig. 10.18a
W=0.6m
10.32, Heat and Mass Transfer
T;=75°C
--I 0.4 1 -
IO cm
cm
Fig. 10.18b
Assumptions: (l) Steady operating conditions (2) Constant properties. (3) Negligible radiant heat
exchange. (4) One-dimensional conduction. (5) Air is an ideal gas at 1 atm.
Analysis: The equivalent thermal circuit of the two-layered wall is shown below:
T;=75°C
-
T0 =!
T==25°C
Q ---Vvv--~---Vvv--~---Vvv~--L1
L2
I
R1 = k1A
R2= k2A
R3= hA
Fig. 10.19
The rate of heat transfer through the composite wall is
I;, - T=
'I'; - I;,
Q = 1/ h A = [(L 1/ki A)+(£i /~ A)]
.
Convective heat flux, %onv = h ( I;, - T=)
and, conductive heat flux, %and= ('I';
-1;,y[ ~/ + ~]
%ond = %onv = q =
Q/A·
To determine h, we need properties of air at the film temperature, Tr = I;,+ T= . But T0 is unknown.
2
Hence a trial and error procedure is necessary.
As an initial guess, let I;, = 40°C
1
so that Tr = -( 40 + 25) = 32.5°C
2
Properties of air at 30°C are
k = 0.02607W /m °C
v = 0.6315 x 10-5 m2/s
Pr=0.7275
Natural (Free) Convection
Rayleigh number,
RaL = 1700 =
g A I} (T. -T. )
'"'
00
s
y2
Pr
where L is the characteristic length equal to height of a vertical surface = 1.5 m
1
~idealgas =
1
4 = {32.5+273.15)K
= 0.00327 K- 1
(9.81 m/s 2)[ (0.00327K- 1)] (1.5 m)3 {40-25)K(0.7275)
=-'-------'--------------( 1.6315 x 10-5 m2/s}2
= 4.44 X 109
The relevant empirical correlation is
hL
(
)t/3
NuL = k = 0.10 RaL
h =0.lOx 0.02607W/m2 oc x[4.44xl09]'13
1.5
2
= 2.857 W /m °C
Then,
%onv =
h (I;, - T.. ) = 2.857W /m2 °C ( 40-25)°C
= 42.85 W/m2
Now,
i;-i;,
%o
nd
{75-40)°C
= (Li/lci)+(Li/kz) = ( 0.004 m )+(
0.10m
)
52 W/m°C
0.15 W/m°C
= (1.5 m;c)(40°C) = 52.5W/m2
Clearly, T0 must be increased.
Let T0 = 45°C
Then,
1
35°c
4 = -(45+25)=
2
At 35°C
k = 0.02625W/m °C,
v = 1.655 x 10-5 m2/s,
Pr= 0.7268
r
I 0.33
10.34, Heat and Mass Transfer
(9.81)[ 1/(35 + 273.15)] (1.5)3 ( 45- 25)(0.7268)
RaL = - - ~ - - - - ~ - - - - - - -
(1.655 X 10-5 }2
= 5.7x109
h = 0.10 X 0·02625 X (5.7 X 109 }'13
1.5
=3.13W/m2 °C
qconv ={3.13) {45-25) = 62.5
and
%ond ={1.5){75-45) = 45.0
W/m2
W /m2
qconv•W/ml
Difference
52.5
45.0
42.85
62.5
9.65
-17.5
0 - 9.65
T.0 - 40
----=--17.5-9.65 45-40
~
T"' °C
40
45
By linear interpolation, we have
I;, =41.8°C
Steady state heat flux is
q= (L,/ki +£i/ki)
=
(75- 41.8)°C
20c
[(0.004/52)+ (0.1/0.15)]~
w
= 49.8W/m2
Heat-transfer coefficient is
h=-q-= 49.8W/m2
T;,-T~
{41.8-25)°C
= 2.96 W/m2 °C
(Ans) (a)
Rate of heat transfer across the composite wall is
Q = q(W L) = ( 49.8; ){o.6mxl.5m)
=44.8 W
(Ans) (b)
And, the outer surface temperature of insulation is
T0 = 41.8°C
(Ans) (c)
Natural (Free) Convection
r
I0.3S
Interface temperature,
T* = T _ L1 = 75 _ (49.8 X 0.004)
' q~
52
= 75 - 0.004 = 74.996°C
There is hardly any temperature drop in the steel wall. In the insulation wall temperature falls by
33.2°C due to conduction resistance, and convective resistance causes another temperature drop by 16.8°C
as shown in the temperature distribution.
Example I 0.13 ~ A vertical metallic plate, 3.5-m high and 2-m wide, is exposed to the sun and receives a net
solar energy flux of 600 W/m 2• The plate is insulated on the back side and painted black
so that all the incident solar radiation is absorbed by the plate, which in turn dissipates heat to the surrounding
air at 30°C. Determine the equilibrium average temperature of the plate.
Known: A plate subjected to constant heat flux loses heat by free convection to the surrounding air.
Find: Mean plate surface temperature, Ts (0 C).
Schematic:
Ts =1.
L=3.5m
Quiescent air
r~=Jo c
0
Insulated backside
Incident solar radiation flux, q 5 = 600W/m2
Fig. 10.20
Assumptions: ( 1) Constant surface heat flux. (2) Negligible radiation heat loss. (3) Constant thermophysical
properties of air.
Analysis: This is a case of constant heat flux at the vertical surface. All properties of the fluid (air) are
to be evaluated at the film temperature Tr. However, for determining Tr, one requires surface temperature
Ts, which is unknown.
To begin with, let us assume an approximate value of 10 W/m2 °C (for free convection problems), so
that surface-to-fluid temperature difference (Ts - T=) or t:.T is estimated to be
t:.T = qs = 600 W / m2 = 60oc
h 10W/ m2 °C
Then
I
I
'fr = 2('f. + T=)= 2('f. -T~ )+ T~
t:.T
60
=T +-=30+-= 60°C
~
2
2
10.36, Heat and Mass Transfer
Properties of air at 60°C are
v = 18.96 x lQ-6 m2/s
k = 0.02808W /m2 °C
Pr= 0.7202
1
A.
= - - - - - = 0 003 K- 1
1-'tdealgas
{60+273.15)K
·
With x = 3.5 m, the modified Grashof number,
G,:* = G,: N11 = g~x3 AT hx
X
X
..X
y2
k
(9.81m/s2) (0.003 K- 1)(3.5m}4 (600W/m2)
(0.02808W/m0C) (18.96xI0-6 m2/s)2
=---------~-~----
= 2.627 X 1014
Gr;Pr = 2.627 X 1014 X 0.7202
= 1.892 X 1Q14 (> 1Q14)
This value indicates fully developed turbulent flow. For the turbulent region, the local heat-transfer
coefficient hx can be determined from
(q8 = const)
\ = (0.17/0.02808W/m oc) [ 1.892 X 1014] 1/4
3.5
= 5.06 W/m 2 °C
In the turbulent heat transfer,
"' = k
h,;c - (Grx*)114 - ( x 4)114 - x
1vu,.
i.e., hx does not vary with x and we may take this as the average value.
h = 5.06 W /m2 °C. This is less than 10 W /m2 °C assumed earlier. Recalculating AT, one finds
AT= qs =
h
600W /m2 = 118.6oC
5.06W/m2 °C
New film temperature would be
T,f = T
~
+AT= 30 + 118.6 = 8930C
2
2
Natural (Free) Convection
At 89.3°C, the properties of air are:
k = 0.0302W /m °C
v = 21.94 x lQ-<i m2 /s
~=
Pr= 0.7134
l
= 0.00276 K- 1
{89.3+273.15)K
Gr_*= (9.81){0.00276){3.5)4 (600)
Then
(0.0302) (21.94 X 1Q-<i)2
x
= 1.677 X 10 14
Gr;Pr = 1.196 x 1014
and hx is calculated from
\ = ~(0.17) (Gr; Prf 4
X
= (0.02808)(0.17){1.196 X 1014 )'14
3.5
= 4.51 W /m2 °C
New temperature difference is
AT = (T. - T )
= average
s
= qs = 600 = 133 °C
h
4.5 l
The average surface temperature is
T. = 133 + 30 = 163°C
AT
133
New film temperature, Tr = T= + - = 30 + - = 96.5 °C
2
2
At 96.5°C,
k = 0.0307W /m °C
v = 22.96x 10-6 m2 /s
1
(96.5 + 273.15) K
~ = - - - - - = 0.0027 K- 1
Then
•
(9.81)(0.0027){3.5)4 (600)(0.712)
G~Pr=~~~-~~~~~~
(0.0307) (22.69 X 10-<i}2
= l.074xl014
Hence
\ = ( 0 ·0307 )(0.17){1.074 X 1014 }'1 4
3.5
= 4.8 W/m2 °C
Pr= 0.712
r
I 0.37
10.38, Heat and Mass Transfer
Now
t:.T = qs = 600 = 1250c
h
4.8
The average surface temperature is
r. = 125+30 = 1ss c
(Ans)
0
As hx = 4.8 W /m 2 °C is not much different from previous hx = 4.51 W /m 2 °C, no further iteration is
necessary.
Example I 0. 14 ~ Heat is dissipated at the rate of 4.5 W from the front surface of a vertical circuit board,
0.15-m high and 0.15-m wide. The back surface of the board is insulated while the front
surface is exposed to still air at 25°C.
(a) Determine the maximum surface temperature of the circuit board, assuming uniform heat flux.
(b) If uniform surface temperature condition is assumed, what would be the steady-state temperature of the
isothermal board? Sketch the temperature distribution in the two cases.
Properties of air:
T (0 C)
k (Wlm 0 C)
V (m 2 /s)
Pr
t-------------40
0.02662
17.02 x Io-6
0.7255
45
0.02699
17.50 x Io-6
0.7241
50
0.02735
17.98 x Io-6
0.7228
Known: Vertical circuit board dissipates heat to quiescent air.
Find: (a) Tmax ( 0 c)I
; (b) I;. (0 c).
qs=const
Schematic:
------Circuit board----~
L = 0.15 m (A, = 0.15 m x 0.15 m)
=0.0225m 2
U2
q =200W/m~
s
Still air
r_=25°C
--
X
_J
L=
T
T,
0.15m
1
-U2
Tu2
---Q=4.5W
X
_J
Uniform heat flux, q 5
Uniform surface
temperature, T5
(a)
(b)
Fig. 10.21
Assumptions: ( l) Board subjected to uniform surface heat flux or uniform surface temperature, Ts.
(2) Quiescent air at l atm is an ideal gas.
Analysis: (a) For constant heat flux condition,
qs
=Q=
As
4 ·5 W =200W/m 2
O. 0225 m2
Natural (Free) Convection
r
I 0.39
We note that for this case
!qs =hATu 2 ! where
!ATu2 =Tu2 -T_!
The maximum temperature on the circuit board will occur at x = L.
Since A~= 1.15(x I Lf 5 ATu 2 , it follows that
!Tmax =TL= T_ + 1.15ATu2 I
The average heat-transfer coefficient, h is evaluated from a vertical isothermal plate correlation but
based upon the temperature difference ATu 2 • Obviously a trial solution is necessary.
Trial # I: Let Tu 2 = 65°C so that A Tu 2 = 65 - 25 = 40°C or K
1
1
2
2
Tr= -(Tu2 +r_) = -(65+25) = 45°C
and
~ideal gas = 1/( 45 + 273.15) K = 1/318.15 K
The Rayleigh number is
RaL = g~/} ATu2 Pr/v 2
(9.81m/s 2 ) [ 1/(318.15 K}](0.15m}3 ( 40 K}(0.7241)
=----~---~-------(17.5x10-<i m2 /s)2
= 9.842 x 106
The appropriate correlation is
NuL = 0.68 +
~
( < 109 )
Laminar flow.
0.67Rat
4; 9
[ 1+ (0.492/ Pr)91' 6 ]
= 0.68+
0.67(9.842 X 106 }'1 4
4/9
[ 1+ (0.492/0.7241)91' 6 ]
= 29.55
h = NuL }!_ = 29.55 x 0 ·02699 W /m °C = 5.32 W /m2 °C
L
0.15 m
Hence, the calculated heat flux is
q. =hATu2 =(5.32 W/m2 °C){40°C)=212.8 W/m2
Since this value is greater than 200 W/m2, we need to reduce Tu2 •
Trial # 2: Assume Tu 2 = 63°C so that ATu2 = 63- 25 = 38°C
and
By interpolation, at 44°C
k = 0.02692 W /m2 °C
Pr=0.7244
10.40, Heat and Mass Transfer
~ = 1/{44+273.15)K
v=17.40xl0-6 m2 /s
Ra = (9.81)[1 / 317.15](0.15)3(38)(0.7244)
(17.4 X 10-6)2
L
=9.5x106
N = O 68
U
·
+
0.67(9.5 X 10)1/ 4
4 /9
[ 1+ (0.492 / 0.7244 )9116
J
=29.3
h=29.3x 0 ·02692
0.15
5.26W/m2 °C
Therefore
qs = {5.26){38) = 199.88 W/m 2
( s:200 W /m 2 )
No more trial is necessary.
Hence, Tu 2 = 63 °C
and
Tmax =TL =25+(1.15){38)=68.7°C
(Ans) (a)
(b) For the uniform surface temperature case, the procedure for estimating the average heat transfer
coefficient is the same.
Hence,
T;, = Tu 2 = 63°C
(Ans) (b)
L'lh =43.7°C
.!.... =I
L
f;;
I
= const Case (a)
0.5
L'lTu2=38°C
Case (b)
Fig. 10.22
The temperature distributions for both cases (a) and (b) are quite different and are shown in the
sketch. For case (a), i.e., constant heat flux condition, t:.1'y_ - x while for case (b ), i.e., constant surface
temperature condition, t:.T = I;, - T= = (Tu 2 - T=) from x = 0 to x = L. We recognize that in both cases,
Q=4.5 Wand h=5 .26 W/m 2 °C.
Natural (Free) Convection
r
I 0.41
Example I 0.15 ~ In order to reduce the heat loss per unit time per unit area from a 20-cm thick vertical
furnace wall made of fire brick, the outer surface is coated with aluminium paint. The inner
surface of the wall is at a temperature of 1275 K and the temperature of the outside air and surrounding
surfaces is 295 K. Calculate the reduction if
(i) the thermal conductivity of fire brick is given by the expression
k = [ 0.87 + 0.00058 (T - 373)] W/m K where Tis in K
(ii) the natural convection heat transfer coefficient from the outer surface is given by h =1.34 ( d T)ll 3 W /m2 K
where dT is the temperature difference between the outer surface and the ambient air in K,
(iii) the emissivity of the outer surface without any paint is 0.95, and
(iv) the emissivity of aluminium paint is 0.2.
For simplicity, take the temperature distribution in the wall to be linear and use the value of k at the mean
wall temperature.
Known: A vertical furnace wall has its outer surface coated with paint exposed to convective and radiative
environment.
Find: Reduction in heat flux after the outer surface is coated with low emissivity paint.
Schematic:
Surroundings
CD
1 11
Firebrick
k(T)
Air
-<2cond
K
!
Q.conv
Qrad
(T2
1-L=0.2m-J
' e = 0.95 (without paint)
e =0.2 (without paint)
Fig. 10.23
Assumptions: ( 1) Steady-state conditions. (2) Thermal conductivity of furnace wall material linearly varies
with temperature . (3) Ambient air and surroundings are at the same temperature.
Analysis: (a) When £ = 0.95
The surface energy balance gives:
%ond
=% onv + qrad
or
k (T. - J:)
m 1
-T.4
)
L 2 = Ji (J:2 _ T= ) + cr £ (J:4
2
sur
or
[0.87 + 0.00058 (Tm - 373)] ( T. -J:)
y
10.42, Heat and Mass Transfer
or
[0.87 + 0.00058 (Tm - 373)] W/m K (1275 - T2) K/0.2 m
= { 1.34(7; - 295f 3 W / m2 K }(7; -295)K
+(5.67xl0-8 W / m2 K 4 )(0.95)[7;4 -295 4 ]K 4
or
5 (1275 -7;) [o.81+0.00058{
(1
27 ~+7; )-373}]
LHS
= 1.34 = 1.34 (T,-295)"' +5.3865[
c~
0 J-2.95']
(a)
RHS
7; will have to be found by trial and error.
Hence, 7; will lie between 498 and 500 K.
LHS
RHS
LHS-RHS
4537.0
4503.78
+ 33.22
498.8
4533.3
4533.3
0
500
4527.59
4578.36
- 50.77
498
By linear interpolation, we can find T2 = 498.8 K and LHS = RHS = q = 4533.3 W/m 2 [from
Equation (a)] corresponding to (LHS - RHS) = 0 as mentioned in the accompanying table.
(b) When £ = 0.2
%ond
=%onv + qrad
5(1275-7;)[ 0.87+0.00058{
(1 21~+ ½)-373}]
= l.34(T,-295)"' +(5.67x0.2) [
c~
0 r-2.95']
Trial I
Let 7; = 600 K
LHS = 4041.25 W
RHS = 4134.9 W
LHS < RHS
LHS = 4051.25 W
RHS = 4091.35 W
LHS < RHS
LHS = 4061.25
RHS = 4048.0 W
LHS > RHS
Trial II
Let 7; = 598 K
Trial III
7; = 596 K
(b)
Natural (Free) Convection
r
I 0.43
Hence, T2 will lie between 596 K and 598 K. For difference (LHS - RHS) = 0, T2 = 596.5 K and
q = 4058.8 W/m 2 [from Equation (b)] by linear interpolation as shown in the accompanying table.
596
LHS
RHS
LHS -RHS
406 1.25
4048.0
+ 13 .25
596.5
4058.8
4058.8
0
598
4051.25
409 1.35
- 40.1
Thus, there is a reduction in heat flux of the order of (4533.3 - 4058.8)
=474.5 W/m 2
(Ans)
Example I 0.16 ~ The dimensions of bricks made in a factory are 7.5 cm (height) by 22.5 cm by 15 cm. The
temperature of the brick leaving the kiln is 350°C and the brick is exposed to cooling air
at 35°C. Determine the instantaneous rate of cooling as the brick leaves the kiln.
Following properties of air may be used:
k = 0.0384 W /m K,
v = 34.39 xi~ m2 /s,
Pr= 0.685
Known: Bricks from a kiln are exposed to quiescent cooling air.
Find: The rate of cooling of brick, Q[W].
Schematic:
Quiescent air
r-=35°c
22.5cm
7.5cm
Brick (T5 = 350 °C)
Fig. 10.24
Assumptions: (1) Brick surface temperature is uniform. (2) Air is an ideal gas.
Analysis: Characteristic length,
= LHLv
L
c
where
LH+Lv
LH = 0.225 m, Lv = 0.075 m
L = 0.225 X 0.075 = 0 _05625 m
C
0.225 + 0.075
10.44, Heat and Mass Transfer
Nusselt number,
Nu= 0.55(Raf 4
[ I 0 4 < Ra< 109 ]
g ~ _q ('.f. - T=) Pr
Ra= Gr P r = - - ~ - - ~ -
where
y2
I
I
~ideal gas = - {-) = ( 350 35 )
= 2.1475 X 10-3 K-I
7'r K
+
+273.15
2
9.8lm/s2 x 2.1475 x 10-3 K-I x 0.05625 3 m 3 x {350-35)K x 0.685
Ra=---------------------(34.39 x lQ-6 ) 2 m 4 / s 2
= 6.84 X 10 5
Nu= hLc = 0.55(6.84 x 105
k
f = 15.8
4
Heat-transfer coefficient,
h =Nu.!._= 15 _8 x 0.0384 W/ mK
Le
0.05625 m
= 10.8 W/m 2 K
Surface area of the brick is
As= 2 (7.5 X 22 .5)+ 2 (22 .5 X 15)+ 2 (15 X 7.5)
= 1237 .5 cm 2
or
0.12375 m 2
Instantaneous cooling rate is
Q = h As ( '.f. - T=)
=(10.8W/m 2 K)(0.12375 m 2 ){350-35)°C
=421 W
Schematic:
(Ans)
Solar radiation
q. = 300 W /m 2
Quiescent air
r==2s c
0
qconv
,----T.=?
Plate ( 1 m x I m)
Insulated bottom
Fig. 10.25
Assumptions: (I) Steady-state conditions. (2) Free-convection heat transfer only from the top surface.
(3) Radiation heat transfer negligible.
Natural (Free) Convection
r
I0.4S
Analysis: Under steady-state conditions
£in= £out
or
or
Thus the free convection heat transfer from the plate to the air in steady state,
Qconv = (300 W/m2 ) (Im X Im}= 300 W
or
As h depends on Grashof number which in turn requires I;,, calculating I;, is not so easy. One has
to follow the trial and error procedure.
Initial guess: Let I;, = 70°C
Mean film temperature,
I;,+ T= I (70+27 )
Tr =-2-=2
= 48.5°C or 321.65 K
Properties of air at 48.5°C:
k=2.795 xI0-2 W/m K
Pr= 0.704
v = 18.07 xI0-6 m2/s
~ = 1/321.65
K- 1
Characteristic length,
L= A
p
(for a horizontal plate)
= lxJm2 =0.25 m
4xlm
Example I 0.17 ~ The roof of a house is inclined at an angle of 25° with the horizontal. The temperature of
the roof surface is estimated to be 0°C. The surrounding air is at -20°C. Determine the
rate of heat loss from the roof surface which is 1-m square.
Following properties of air may be used:
k = 0.0233 W/m K
Pr= 0.718
V=ll.lxlO---om2 /s
Known: An inclined roof of a house is exposed to ambient air losing heat in the process.
Find: Rate of heat loss, Q[W].
Schematic:
Quiescent air
T= =-20°C
Fig. 10.26
10.46, Heat and Mass Transfer
Assumptions: (1) Steady operating conditions. (2) Air is an ideal gas and quiescent.
Analysis: Film temperature,
'4 = 0+(-20) =-l0°C = 263.15 K
2
1
~ideal gas=-= 1/263.15 K
Tr
The appropriate correlation is
NuL =0.56(RaL cos0)114
where 0 is the inclination with the vertical.
Rayleigh number,
RaL =GrL Pr=
=
g~(r. -r. )n
•
y2
=
Pr
(9.81m/s 2 )(1/263.15) K- 1 (0+20)°C (L)3 m3 (0.718)
(11.lxl0-6 )2 m2 /s 2
= 4.345 X 109 I}
RaL cos0=4.345x109 xcos65°xU
= 1.836 X 109 I}
For 1 m square area, L = 1 m
Average Nusselt number,
hLL
NuL = - = 0.56 (1.836 X 109 X 1)0 -25 = 115.92
k
Average heat-transfer coefficient,
hr,= (115.92)(0.0233W/mK)
lm
=2.70 W/m2 K
Hence,
Heat flux, q = hdT,, -T=)
= (2.7 W/m 2 K) [O - (-20)]°C or K
= 54 W/m2
Rate of heat loss,
Q=qA=(54W /m2 )(1mxlm) =54 W
(Ans)
Natural (Free) Convection
r
I 0.47
Example I 0.18 ~ A 0.5-m by 0.5-m square horizontal plate with bottom surface insulated is subjected to a
constant wall heat flux of 15 kW/m 2 and dissipates heat by convection to water at 20°C.
Estimate the average top surface temperature of the plate.
Properties of water:
T ( C)
k (Wlm C)
v X I 0 (m /s)
,_______
--------0
0
6
2
Pr
f3 x I 04 (K- 1)
20
0.599
I .006
7.02
1.82
30
0.618
0.805
5.42
3.21
40
0.634
0.659
4.31
3.87
50
0.648
0.556
3.54
4.49
Use the following correlations:
Nu = 0. 13 Ra 113
(Ra < 2 x 108)
Nu=0. 16 Ra 113
(2x 108 <Ra< 10 11 )
(for the top surface of a horizontal heated plate with constant heat flux)
Characteristic length, Le = A/P.
Properties of water to be evaluated at
1
T.=T.-0.25 (T.-T- ) except~ which should be evaluated at (7.+r-).
2
Known: A horizontal heated square plate with constant surface heat flux and bottom surface insulated
loses heat to water.
Find: Average surface temperature of the plate.
Schematic:
0.5m
Fig. 10.27
Assumptions: (1) Steady operating conditions. (2) Constant wall heat flux . (3) Water is quiescent.
(4) Constant properties.
Analysis: Heat flux, qs = h (I;. - T=)
Average surface temperature of the plate,
T. = T + qs
s
= h
Since h is unknown, we can estimate it by first assuming a suitable value of Ts and using an appropriate
correlation.
10.48, Heat and Mass Transfer
Let
r. =40°C.
Then
T., = T,, -0.25(T,, -T=)
= 40 - 0.25( 40- 20) = 35°c
Properties of water at 35°C (from the given table by interpolation):
k=0.626 W/m °C
v=0.732x10-<:i m 2 /s
1
fiat-(T,,+T=)=30°C
2
is
Pr=4.865
3.21xl0-4 K- 1
.h
A/ 0.5mx0.5m . 5 m,
Hence, wit Lc = s P = - - - - = 012
4x0.5 m
g~_q (T. -T=)Pr
Ra=--~-~v2
(9.81m/s2 ) (3.21xl0-4 K- 1 ) (0.125 m)3 (40-20)Kx(4.865)
=~--~~---~------,------(0.732xl0-6 m2 !s}2
= 1.117x109
The relevant correlation is
Nu= 0.16(Ra)'13
= 0.16(1.117 X 109
f = 166
3
h =Nu.!_= 166x 0.626W/moc
Lc
0.125 m
Then
= 831.35 W/m2 °C
T. = T. + q. = 200C+
s
=
h
15000 W/m2
831.35 W/m2 °C
=38°C
Since the assumed value was 40°C, let us try T. = 38.8°C, so that
1
T., = 38.8- 4(38.8- 20) = 34.1 °C
7r = 38.8+20 =29.4oc
and
2
By interpolation:
k= 0.6246 W/m °C
v = 0.745x10-<:i m 2 /s
Pr=4.965
~ = 3.13xl0-4 K- 1
Natural (Free) Convection
r
I 0.49
9.81 X 3.13 X 10-4 X 0.125 3 x( 38.8-20) X 4.965
Ra=----------------( 0.745 X 10-6 )2
=1.0lx10 9
Nu= 0.16(1.0lx 109 yi3 = 160.5
h = 160.5 X 0 ·6246 = 802 W/m 2 °C
and
0.125
T.. = 20+ I 5000 = 38.7°C ("" 38.8°C)
s
802
average surface temperature of the plate is
I;,= 38.7°C
(Ans)
Example I 0.19 ~ Consider a 0.6 x 0.6 m thin square plate in a room at 30°C. One side of the plate is
maintained at a temperature of 74°C, while the other side is insulated. Determine the
rate of heat transfer from the plate by natural convection if the plate is (a) vertical, (b)
horizontal with hot surface facing up, and (c) horizontal with hot surface facing down.
Properties of air at the film temperature of 52°C are
k = 0.028 W/m °C
v = 18.4 x I0--6 m2/s
a= 26.2 x I0--6 m2/s
Known: Heat dissipation from a hot square plate by free convection
Find: Surface heat loss from the plate for three geometries: (a) vertical, (b) horizontal: hot surface up,
(c) horizontal: hot surface down.
Schematic:
T,=749)
(a)
\
/
F
,C~· = 74°c
6666666666 6
,,,,,,,,n,,,,,,,,h
V\.-lr=74
°c
s
(c)
(b)
Vertical plate
~
Hot surface up
(Horizontal plate)
Hot surface down
(Horizontal plate)
Fig. 10.28
Assumptions: ( 1) Steady operating conditions exist. (2) Air is an ideal gas. (3) Radiation effects are not
considered.
Analysis: (a) Vertical: The characteristic length in this case is the height of the plate, which is L = 0.6 m.
Coefficient of volumetric thermal expansion, ~ideal gas = -1() where film temperature, Tr = .!. (I;, + T=) =
Tr K
_!_(74 + 30) = 52°C
2
or
325.15 K
2
10.50, Heat and Mass Transfer
Prandtl number,
Pr=~= 18.4 x lQ-6 m2 /s = 0.702
a 26.2 x 10-6 m2/s
The Rayleigh number is RaL =
g~D(J:-T)
s
=
V<l
(9.81 m/s 2)[1/(325.15K)] (L,m}3 (74-30)°C
(18.4x10-6 m 2/s) (26.2x1Q-6 m 2/s)
= -------~----,-~------,-= 2.7537x109 D
With
L=0.6 m,
RaL = 5.948 X 108
Using the Churchill-Chu correlation for a vertical plate (RaL < 109):
.., = O 68
,vu
.
+
= 0.68+
0.67Rat
[1+(0.492/ pr)91' 6 ]
4/9
0.67(5.948 X 108 }'14
4/9
[1 + (0.492/0.702)91' 6 ]
=80.89 = hL
k
Hence, the heat-transfer coefficient,
h=
! Nu= 0.028 W /moC X 80.89
L
0.6m
= 3.775 W /m2 °C
Surface heat loss from the vertical plate is
Q=hA_(T,.-T=)
= (3.775W /m2 0 c)(0.6mx0.6m)(74-30)°C
(Ans) (a)
=59.8W
(b) Horizontal: Hot su,face up
The characteristic length L for the horizontal plate is given by
L =A.IP= 0.6mx0.6m = 0.15 m
4x0.6m
The Rayleigh number is
RaL =2.7537Dx109 =2.7537x0.15 3 x10 9
= 9.3x 106
Natural (Free) Convection
r
I 0.5 I
The appropriate correlation for this case is
Nu= 0.54(Raf 4 = hL
k
= 0.54(9.3x106 }'1 4
=29.82
Hence, the convection coefficient is
h =!Nu= 0.028 W/moc x29.82
L
0.15 m
= 5.566 W/m2 °C
Heat-transfer rate,
Q=h A. (T,.-T=)
= (5.566 W/m2 °C)(0.6 X 0.6)m2 {74-30)°C
(Ans) (b)
=88.2W
(c) Horizontal: Hot su,face down
For horizontal plate, L = A./P = 0.15 m
RaL = 9.3x 106
The relevant correlation is
Nu=O.I5(Ra)
114
hL
=k
= 0.15(9.3x106 )'14
=8.2835
The convection coefficient is
h=! Nu= 0.028W/mocx8.2835
L
0.15 m
= 1.546 W /m2 °C
surface heat loss is
Q = h A. (T,. -T=) = (1.546 W/m 20C)(0.36 m2 ) (74-30)°C
(Ans) (c)
=24.5W
Example I0.20 ~ A cubical furnace of 1-m-side rests on a concrete floor. If the top and the sides are at
90°C and the ambient air is at 24°C, determine the rate at which heat is dissipated to the
surroundings. Neglect the heat loss due to radiation and conduction through the base of the furnace.
The following properties of air at the film temperature of 57 °C may be used:
k=0.0285 W/m°C
V=l8.9xl0-6 m2/s
Pr=0.703
10.52, Heat and Mass Transfer
Known: Heat is lost from the top and sides of a cubical furnace to the surroundings by free convection.
Find: Rate of heat loss, Q{W) .
Schematic:
Cubical furnace
T5 =90°C
Still air
T =24°C
00
Im
Im
Fig. 10.29
Assumptions: ( 1) Steady operating conditions. (2) Quiescent air. (3) Constant properties. (4) Air is an
ideal gas. (5) The local atmospheric pressure is 1 atm. (6) Heat transfer by radiation and by conduction
through the base is negligible.
Analysis: Film temperature, Tr =.!.(I'.+ T
00 )
2
= (90 + 24 )/2 = 57°C
Coefficient of volumetric thermal expansion,
A.
l-'1dealgas
1
1
= - = - - - - - = 3 03xI0-3 K- 1
Tr {57+273.15)K .
Total heat dissipation rate from the furnace to the surrounding air by natural convection is
The Rayleigh number is
(9.81m/ s2 )(3.03 x 10-3 K-l )(U ,m3 ){90-24 )°C(0.703)
=-------------------(18.9 x lQ-6 m2 / s)2
= 3.861 L3 X 109
where L is the characteristic length.
• Vertical surface Characteristic length,
L = height, 1 m
RaL = 3.861 X 109
Natural (Free) Convection
r
IO.SJ
Using the appropriate Churchill-Chu correlation:
l
l
2
Nu = 0.825 +
0.387 RaL1/6 s 27 }
[1+{0.492/ pr)9116 ] I
= 0 _825 +
0.387(3.861x109 ) 8127
1/6 }2
[ 1+ {0.492/0.703)9116 ]
=186.74
k
lz..ertical = lz.ide = Nu L
= 186 _74 X 0.0285 W /m °C
lm
= 5.32 W /m2 °C
+ Top horizontal su,face (hot su,face facing up)
Characteristic length,
L=A.=lmxlm
P
4m
= 0.25 m
RaL =3.861x(0.25)3 x109
= 6.033x107
107 < RaL < 1010 ,
Since
Nu= 0.15 (RaL)'13
= 0.15(6.033x10 7
f = 58.83
3
k
14.op = hhorizontal = Nu
L
= 58 _83 X 0.0285 W /m °C
0.25m
= 6.7W/m2 °C
Noting that /4op = A.ide = 1m x 1m = 1m 2 , the heat-transfer rate is
Q = [(6.7W /m2 °Cxlm2 )+(4x5.32W/m2 °Cxlm2 )] {90-24)°C
= 1847W
(Ans)
10.54, Heat and Mass Transfer
Example 10.21
~ A transformer is cooled by immersing in an oil bath housed in a cylindrical tank 1.25-m
long and of 0.75 m in diameter. Neglecting the energy transfer from the bottom of the
tank and assuming laminar boundary layer, estimate the surface temperature of the tank if the electrical energy
dissipation is 1.4 kW. The entire loss of energy is assumed to be by free convection to the ambient air at 25°C.
The following simplified relations for air may be used:
VerticaI cyIin d ricaI su rface:
h --1.42(dLT)"4
HorizontaI top su rface:
h --1.32(dLT)"4
where dT"' (T.- r_) 0 c, h is the convection coefficient in W/m 2 °C and Lis the characteristic length in m.
Known: Natural convection heat transfer from the lateral and top surfaces of a cylindrical tank to
surrounding air.
Find: Surface temperature of tank, Ts (0 C).
Schematic:
Top surface (D=0.75m)
Quiescent air
T==25°C
T
Cylindrical surface (T5)
L= 1.25 m
<2cyl
<2elect
Bottom surface (insulated)
Fig. 10.30
Assumptions: (I) Steady-state conditions. (2) Heat loss from the bottom is neglected. (3) Laminar
boundary layer. (4) Only free convection heat transfer is considered.
Analysis: From energy balance:
Qelect
= Qcyl + Qtop
1400 W = hcyl 4,yl f:.T + htop Aiop f:.T
or
= l.42(t:.T}11 4 (rcDL)L-ll 4 f:.T+ l.32(t:.T}11 4 ( ¾D 2 }n/4(14 t:.T
Note that the characteristic length for a vertical cylinder is its height L = 1.25 m and that for a horizontal
A. (rc/ 4)D2
plate, L=-=---=D/4=(0.75/4)m
P
rcD
Hence
1400 = t:.T 514 [ (1 .42)(nDL314 )(~ D 714 4 ll 4 )
J
= f:.T 514 [ {(1 .42)(.1l' X 0.75 X 1.25 314 )}+ {J.32 X .1l' X 0.75 714 X 4-314 }
J
Natural (Free) Convection
r
I 0.55
1400 = 4.8415~r1.2s
or
~T = I; -25 = (1400/4.8415)08
= 93.1°C
Surface temperature of the tank is
I;= 25+93.l = 118.1°C
(Ans)
Example I 0.22 ~ An electrically heated square plate 75cm by 75cm has one surface thermally insulated with
the other surface dissipating heat by natural convection into atmospheric air at 30°C. The
mean surface temperature of the plate is S0°C. The plate is subjected to uniform surface heat flux and is inclined at
an angle of 50° from the vertical. Determine the heat-transfer rate from the plate for (a) the heated surface facing
up and (b) the heated surface facing down. For a square plate, take the side length as the characteristic length.
Critical Grashof number Grc
e (deg)
Grc
-30
109
-60
108
T (K)
K (Wlm K)
v x / 0 6 (m 2 /s)
Pr
300
0.0263
15.89
0.707
350
0.0300
20.92
0.700
Known: An inclined square plate with one surface insulated loses heat to air under constant heat flux
conditions.
Find: Heat loss rate for (a) the heated surface facing up, and (b) the heated surface facing down.
Schematic:
Hot surface facing
upwards
Quiescent air
r~= 3o•c
Hot surface facing
downwards
Square plate(l=W=0.7Sm
(a)
r.= so·q
(b)
Fig. 10.31
Assumptions: (1) Steady-state conditions. (2) Air is an ideal gas. (3) Constant properties. (4) Uniform
heat flux.
Analysis: For constant wall heat flux (q8 = const), the physical properties are to be evaluated at
T. = J;-0.25(7;-T~)
=50-0.25(50-30)=45°C
or
318.15K
10.56, Heat and Mass Transfer
By linear interpolation (from the given table):
k = 0.0263 + (0.0300 - 0.0263) (318.15 - 300)/(350 - 300)
= 0.02764 W/m K
v = lQ-<i [ 15.89 +(20.92 -15.89)(18.15 / 50)] = 17.7x 10-<i m 2 /s
Pr= 0.707 + 0.363 (0.7 - 0.707) = 0.7045
Coefficient of volumetric thermal expansion, ~ is, however, to be evaluated at
T~ + 0.25(T. - T~)= 30+0.25(50-30)
= 35°C or 308.15 K
1
~ideal gas=
1
'fr {K) = 308 _15 K = 3.245x10-3 K- 1
(a) Heated su,face facing up: (8 negative)
The correlation for free convection on an inclined plate with q8 = const is
hL
[ Raf 3 -Rar 3 ] +0.56 ( Raccos0 )114
Nu=-=0.145
k
for
RaL <1011 , GrL>G,;,
and
-15° <0< -75°
Critical Grashof number for the angle of inclination 0 = -50° is
G,;, = 109 +(10 8 -109 ){-50-(-30)}/{-60-(-30)}
= 4x108
Rae = G,;,Pr = ( 4 X 108 )(0.7045)= 2.818x 108
RaL = GrL Pr where the characteristic length L is taken as the length of a side for a square. Hence,
L=0.75m.
Then
(9.81m/s 2 ) (3.245 x 10-3 K- 1 )(0.75 m}3 (50-30)K(0.7045)
=~--~~----~--~-----(17.7x10-<i m 2 /s)2
= 6.04x108
J
(
hL
[( 6.04x108 )113 - (4x108 )113 +0.56 4x108 xcos(-50) )114
Nu=k=0.145
= 86.65
Natural (Free) Convection
r
I 0.57
convection coefficient,
h =Nu!!.._= 86 _65 x 0.02764 W/mK
L
0.75m
= 3.193 W/m2 K
Heat-dissipation rate,
Q=hA.(T.-T~)
= (3.193 W/m2 K) (0.75 mx0.75 m) (50-30)K
=35.9W
(Ans) (a)
(b) Heated su,face facing down: (8 positive)
For an inclined plate (q 8 = const) for this case:
!Nu= 0.56 (RaL cos0)114 ! for+ 0 < 88°, 10 5 < RaL < 10 11
Substituting numerical values,
hL
~
Nu= - = 0.56(6.04x108 cos 50°) = 78.61
k
h = 78 _61 X 0.02764 W/m K
0.75 m
= 2.897W/m2 K
Heat-loss rate,
Q = h As (T,. - T~ )
= (2.897 W/m2 K)(0.75 2 m2 )(20 K)
(Ans)
=32.6W
Example I0.23 ~ (a) A power amplifier is mounted vertically in air at I 30°C. The case is made of anodized
aluminium (emissivity 0.8) with a surface area of about 3800mm 2 and a height of 40mm.
Assuming a uniform case temperature of 24°C, estimate the power dissipated from the unit.
(b) If the above unit is mounted horizontally with the hot surface facing upwards, estimate the rate of heat transfer.
Properties of air at Tf = 77°C:
k = 0.0297 W/m K
Pr= 0.706
V = 20.6 XI 0--6 m2/s
~
= 2.86xl0-3 K- 1
Known: Power amplifier, approximated as a flat plate, of given dimensions dissipating heat by convection
and radiation in quiescent air.
Find: (a) Power dissipated (vertical orientation), (b) Heat-transfer rate (horizontal orientation, hot surface up).
10.58, Heat and Mass Transfer
Schematic:
Quiescent air
T-=24°C
Lr =24°c
sur
T
A=3800mm2
l
H=40mm
(a) Vertical orientation
Fig. 10.32a
Lr =24°c
sur
Quiescent air
T-=24°C
----L=H
t=·:
rs = 1Jo c
0
(b) Horizontal orientation
Fig. 10.32b
Assumptions: (1) Power amplifier represented as a flat plate. (2) Air is quiescent. (3) The unit is relatively
much small compared to the enclosure.
Analysis: (a) Vertical orientation
Power dissipated by the unit,
Q = Qconv + Qrad
Now
g~~(T.-T-)Pr
RaL = - - - - - - y2
9.81 x 2.86 x 10-3 x{l30-24)x 0.706 x L3
=--------------(20.6 X lQ-6 ) 2
=4.94xl09 D
For vertical orientation, the characteristic length is height.
i.e.,
L=H=0.04 m
RaL = 4.94 x 109 x (0.04) 3 = 3.16 x 10 5
Natural (Free) Convection
,:;:-- = 068
,vuL
·
0.67Raf4
+
4/9
= 068
·
+
6
[i+(o;:2r' J
r
I 0.59
0.67(3.16x10 5 )'14 = 1287
4/9
·
[i+(~:~~~r,6J
- k
0.0297
h = NuL x- = 12.87x-- = 9.55 W/m2 K
L
0.04
Qconv = 9.55 X 3800 X 10-<:i X {130- 24) = 3.85 W
Qrad = 5.67 X 10-8 X 3800 X 10-<:i X 0.8 X [ 403.154 - 297.15 4 ]
(Ans) (a)
=3.21 W
Q = 3.85+3.21 = 7.06 W
:. power dissipated,
(b) Horizontal orientation
Characteristic dimension,
4 = A.
p
A.= 3800 mm2
L=40mm
W=~= 3800 =95mm
L
40
P=2 (L+ W)=2 (40+95)=270mm
3800
Le= - - = 14.1 mm or 0.0141 m
270
Hence,
RaL = 4.94 X 109 X (0.0141)3 = 1.377 X 104
For horizontal plate with hot surface up:
NuL = 0.54(RaJ 14 = 0.54( 1.377x104
f = 5.85
4
- k
0.0297
h = NuL x- = 5.85x-- = 12.32 W/m2 K
L
0.0141
Qconv
= h A.(T. -T=)
= 12.32 X 3800 X lQ-<:i X (130- 24) = 4.96 W
(same as before)
Total heat-transfer rate,
Q = 4.96+3.21 =8.17W
(Ans) (b)
Comment: The natural convection heat-transfer accounts for about (3.85/7.06) i.e. 54.5% of the total
power dissipated by the power amplifier, when the unit is mounted vertically.
This contribution is (4.96/8.17) i.e. 60.7% of the total power dissipation when the unit is mounted
horizontally. Power dissipated is greater in the horizontal position.
10.60, Heat and Mass Transfer
(D) HORIZONTAL CYLINDER
•
•
Example I 0.24 ~ An athlete may be approximated as a cylinder of 0.3-m diameter and 2.0-m height. The
average body temperature of the athlete is 35°C and the ambient air temperature is 25°C.
Determine the heat loss from his body under the following conditions:
(a) The athlete lies on the ground for rest. (b) The athlete runs a 400-m distance in SO seconds.
Thermophysical properties of air at the mean film temperature of 30°C:
k = 0.0265 W/m K
Pr= 0.707
v =16.2xlQ-6 m2 /s
~ = 0.0033 K- 1
Use the following correlations:
Free convection:
O.Sl8Rag4
Nuo = 0.36 +
[1+(0.559/ Pr) 9116
T'
9
Forced convection:
Nuo
=0.30+
0.62Re0112 P,m
[1 + (0.4 / Pr)2'l
1'4
[ 1+ ( __
Reo_ )" 2 ]
282000
Known: An athlete considered to be a cylinder of prescribed dimensions subjected to specified convection
conditions.
Find: Body heat loss for given temperature conditions for (a) free convection, and (b) forced convection.
Schematic:
I
L=2.0m
(j)
)
Lr
Quiescent air
T-=25°C
s =35°C
D=0.3m
(a) Horizontal cylinder approximating the athlete at rest (free convection)
Fig. 10.33a
Air
T-=25°C
V=Bmls
--
I
L=2.0m
T,= 35 °C
(b) Flow across a vertical cylinder (forced convection)
Fig. 10.33b
Natural (Free) Convection
r
I 0.61
Assumptions: (1) Steady-state conditions. (2) Athlete is approximated by cylindrical form having uniform
surface temperature.