MECH 4406
Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng.
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Welcome to the world of Heat Transfer!
 Fundamentally interesting & practically useful area of mechanical engineering
with many, many applications!
• Energy Sector:
– All thermal-cycle energy systems:
○ Nuclear, Biofuel, Fossil Fuel power-plants convert chemical/nuclear
to mechanical energy (which drives a generator) via heat transfer
– Reciprocating engines and gas turbines
– Radiant transfer to solar cells / solar collectors
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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Welcome to the world of Heat Transfer!
 Building Energy Systems
• HVAC - Heating, Ventilation, and Air-Conditioning
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Welcome to the world of Heat Transfer!
 Environment
• Weather
• Global Warming
• Plume rise
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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Welcome to the world of Heat Transfer!
 Material processing
 Cooling of
electronics
 Biology
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Part 0: Introduction & Review
MECH 4406: Heat Transfer
MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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0.1 What is Heat Transfer?
 Heat transfer or “heat” is thermal energy in transit due to a temperature
differences
Q depends in some way
on (𝑻𝒔 – 𝑻 )
𝑇
Q
𝑇
0.1 Relationship to Thermodynamics
 Recall 1st Law: Conservation of Energy from a control volume perspective
𝐸̇ ≡ rate of energy entering CV
𝐸̇
𝐸̇
𝐸̇
𝐸̇
Control
Surface
𝐸̇
𝐸̇
≡ rate of energy leaving CV
𝐸̇
≡ rate of thermal energy generated within CV
≡ rate of change of energy stored within CV
Control
Volume
• 1st law states:
𝐸̇
= 𝐸̇
Transfer / exchange phenomena
1. Mass transfer: Energy carried with mass entering
or leaving control volume
2. Work: Exchange of energy by moving boundaries
of a CV or shaft work
3. Heat: Transfer of energy due to a temperature
difference
+ 𝐸̇ − 𝐸̇
Transfer or
exchange
phenomena
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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0.1 Relationship to Thermodynamics
 Heat transfer deals with the nature and rates of thermal energy exchange
whereas thermodynamics deals with equilibrium states and the consequences of
heat transfer
 Examples:
• Warm drink placed in a cold fridge
– Thermodynamics:
○ Thermal energy will flow from drink to fridge (2nd law)
○ Energy lost by drink will equal energy gained by fridge (1st law)
○ If fridge is held at constant temperature, eventually drink will cool (equilibrium)
– Heat transfer:
○ How long will it take, I’m thirsty!
○ Mechanisms of energy transfer and how to influence them
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0.3 Unsteady vs. Steady Heat Transfer
 In steady heat transfer, the temperature difference inducing the flow of energy
remains constant.
 In unsteady heat transfer, this temperature difference changes in time.
Examples:
 Steady heat transfer:
• Heat transfer through a wall of a house when exterior temperature is constant and a
furnace is maintaining constant interior temperature
 Unsteady heat transfer:
• Hot piece of metal dropped in cold water; warm beer/pop placed in cold fridge;
cooking; etc.
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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0.4 Modes of Heat Transfer
 Three classical modes:
• Conduction
• Convection
• Radiation
 Additional modes
• Boiling / condensation / melting (Phase change): sufficiently different to warrant
separate consideration
• Dufour effects: energy flux due to concentration gradients
• Laser cooling: physics phenomena that can be used to cool atoms near absolute zero
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Three Classical Modes of Heat Transfer
 Conduction: Heat transfer in a solid or a stationary fluid (gas or liquid)
• Physical mechanism: energy transfer due to the random motion of its
constituent atoms, molecules and /or electrons.
• Occurs through molecular collisions, lattice waves, and movement of free
electrons
 Convection: Heat transfer from a fluid to a surface (or vice versa)
 Physical mechanism: energy transfer due to the combined influence of bulk
and random motion for fluid flow over a surface.
 Radiation: Heat transfer via electromagnetic waves
 Physical mechanism: Energy that is emitted by matter due to changes in the
electron configurations of its atoms or molecules and is transported as
electromagnetic waves (or photons).
 In this course most interested in radiative transfer from
one solid to another
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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Conduction – Physical Mechanism
 Conduction in a fluid (gas or liquid) occurs via molecular collisions and molecular
diffusion
•
Energy transfer is in direction of decreasing temperature (from high T to low T)
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Conduction – Physical Mechanism
 Conduction in a solid occurs via:
• “lattice vibrations” or “lattice waves”
• In an electrical conductor
there is also transport
due to translation of
free electrons
Images from : http://cfbt-us.com/wordpress/?p=1110
MECH 4406: Heat Transfer
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© Prof. M. Johnson, Carleton University
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Conduction – Fourier’s Law
 Fourier’s Law: A rate equation governing heat transfer via conduction
𝑞 ″∝
𝑑𝑇
𝑑𝑥
(1-D case)
is the temperature gradient in the 𝑥-direction
• Constant of proportionality, 𝑘, thermal conductivity (note 𝑘 is a material property with units W/mK)
𝑞 =𝑘
𝑑𝑇
𝑑𝑥
(1-D case)
 Notation conventions:
•
•
•
•
𝑞 is rate of heat transfer in Watts, W (J/s)
𝑞 = 𝑞/𝐴 is heat flux or heat transfer per unit area [W/m2]
𝑞′ would be heat transfer per unit length [W/m];
𝑞‴ would be per unit volume [W/m3]
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0.5 Conduction
 Conduction is governed by Fourier’s Law
• General (vector) form of Fourier’s Law:
𝑞 = −𝑘
𝑞 = −𝑘∇𝑇
Temperature
gradient
Heat flux
• In Cartesian coordinates (with constant k):
𝜕𝑇
𝜕𝑇
𝜕𝑇
𝚤̂ +
𝚥̂ +
𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
• In Cylindrical coordinates (with constant k):
𝑞 = −𝑘
Thermal conductivity
𝜕𝑇
1 𝜕𝑇
𝜕𝑇
𝚤̂ +
𝚥̂ +
𝑘
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
• In Spherical coordinates (with constant k):
𝑞 = −𝑘
𝜕𝑇
1 𝜕𝑇
1 𝜕𝑇
𝚤̂ +
𝚥̂ +
𝑘
𝜕𝑥
𝑟 𝜕𝜃
𝑟sin𝜃 𝜕𝜙
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
8
Conduction – 1D Case
 Application to one-dimensional, steady conduction across a plane wall of
constant thermal conductivity:
𝑞 = −𝑘
𝑞 =𝑘
𝑑𝑇
𝑇 −𝑇
= −𝑘
𝑑𝑥
𝐿
𝑇 −𝑇
𝐿
Heat rate [W]: 𝑞 = 𝑞 𝐴
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Example: Heat transfer through a single pane window
kglass=1.4 W/mK
Tin=20°C
(from A.3, p693
Bergman et al.)
Tout=‒20°C
6 mm
x
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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0.6 Convection
 Physical mechanism – Heat transfer where exchange occurs through motion of
fluid. Combination of two processes:
• Diffusion (random molecular motion)
• Advection (Bulk, macroscopic fluid motion) -- dominant effect
𝑇
 Two classes of convection problems:
𝑇
q
1. Forced convection
–
fluid motion is caused by external forces (e.g. fans or pumps)
2. Natural (free) convection
–
Fluid motion is caused by density gradient driven buoyancy forces
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Schlieren Images of Density Gradients Illustrating Natural Convection
https://www.e-education.psu.edu/worldofweather/files/worldofweather/SchlierenPhotograph.jpg
https://www.e-education.psu.edu/course_prog/course_info/meteo101
MECH 4406: Heat Transfer
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© Prof. M. Johnson, Carleton University
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0.6.1 Newton’s Law of Cooling
 Rate of heat transfer per unit area (heat flux) is proportional to the temperature
difference between solid and fluid
𝑞″ ∝ 𝑇 − 𝑇
𝑇 is the temperature of the solid surface
𝑇 is the temperature of the fluid away from the surface
 “Constant” of proportionality, ℎ, is the
convective heat transfer coefficient [W/m2K]
𝑞 =ℎ 𝑇 −𝑇
𝑇
𝑇
q
Newton’s
Law of
Cooling
or
𝑞 = ℎ𝐴 𝑇 − 𝑇
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Convection – Physical Mechanism
 Velocity (momentum) and thermal boundary layers
𝑦
𝑦
𝑢
Velocity
distribution
𝑞
𝑢 𝑦
𝑢 𝑦
𝑇
Thermal boundary layer
Velocity boundary layer
Temperature
distribution
𝑇 𝑦
𝑇
Heated
Surface
𝑇 𝑦
• Convection coefficient, h [W/m2K], depends on characteristics of the boundary
layers
𝑞 = ℎ𝐴 𝑇 − 𝑇
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
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Convection Coefficient, h
 Depends strongly on flow conditions, fluid properties, and geometry
• Must be calculated; not a property!
Typical Range of h values
[W/m2K]
Forced convection
Gases and dry vapours
20-250
Liquids
100-20,000
Liquid metals
2500-30,000
Natural (aka Free) convection
Gases and dry vapours
2-25
Liquids
50-1000
Convection with phase change
Boiling or condensation
2500-100,000
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Convection Coefficient, h
 Convection coefficient varies with position on along an object
h

0°
90°
180°
 In engineering calculations, we are often only concerned with the average
convection coefficient, ℎ , over the immersed object
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
3
Simple Convection Example
 Estimate heat transfer rate per unit
length of a 100-mm diameter, 33°C arm
extending out the window of a car
moving at 20 m/s (72 km/h) if the
ambient air is –20°C. Assume ℎ = 84
W/m2K.
𝑈
𝑇=33°C
𝑇 = ‒20°C
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Ex: Heat Transfer through single pane window (Take 2!)
 Neglecting radiation, and given hin=20 W/m2K and hout=150 W/m2K, what is the rate of heat transfer
(heat flux) through the single pane window (k=1.4 W/mK) from last lecture? What is the indoor surface
temperature?
1
2
3
T
Tin = 20°C
ℎ
=20 W/m2K
1
𝑞 = ℎ 𝐴(𝑇 − 𝑇 , )
2
𝑞 = −𝑘𝐴
3
𝑞=ℎ
ℎ
𝐴 𝑇
,
,
−𝑇
,
=
𝑘𝐴
𝑇
𝑡
,
−𝑇
,
−𝑇
(3 eq., 3 unknowns: 𝑞, 𝑇
, ,𝑇 , )
N.B.: 𝑞 = 𝑞 = 𝑞 by 1st law (no storage or gen.)
Tout = -20°C
window
(t = 6 mm)
Δ𝑇
𝑘𝐴
=−
𝑇
Δ𝑥
𝑡
=150 W/m2K
 𝑞 is same in each equation!
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
4
Ex: Heat Transfer through single pane window (Take 2!)
1
2
3
T
Tin = 20°C
ℎ
=20 W/m2K
1
𝑞 = ℎ 𝐴(𝑇 − 𝑇 , )
2
𝑞 = −𝑘𝐴
3
𝑞=ℎ
Δ𝑇
𝑘𝐴
=−
𝑇
Δ𝑥
𝑡
𝐴 𝑇
,
−𝑇
,
=
𝑘𝐴
𝑇
𝑡
,
−𝑇
,
−𝑇
,
Tout = -20°C
window
(t = 6 mm)
ℎ
=150 W/m2K
Ex: Heat Transfer through single pane window (Take 2!)
1
2
3
T
Tin = 20°C
ℎ
=20 W/m2K
1
𝑞⁄ℎ 𝐴 = 𝑇 − 𝑇
2
𝑞 𝑡⁄𝑘𝐴 = 𝑇
,
−𝑇
3
𝑞⁄ℎ
,
−𝑇
Tout = -20°C
window
(t = 6 mm)
Solve:
𝑞
ℎ
=150 W/m2K
𝐴= 𝑇
,
Rearrange
Sum all 3 equations:
𝑞
𝑞
𝑞
ℎ 𝐴 + (𝑘𝐴/𝑡) + ℎ
𝑞
1
1
1
𝐴 = 𝑞 = 20 + 1.4/0.006 + 150
MECH 4406: Heat Transfer
,
1
1
1
𝐴 ℎ + 𝑘/𝑡 + ℎ
𝐴=𝑇 −𝑇
,
+𝑇
,
−𝑇
,
+𝑇
,
−𝑇
=𝑇 −𝑇
20 − −20 = 656 W
m
Much more sensible answer!
© Prof. M. Johnson, Carleton University
5
0.8 Radiation
 Thermal radiation is emitted by all matter at finite temperature (T > 0 K).
 Radiation is emitted by liquids and gases as well as solids (all matter)
• In this course, we will focus on radiant exchange between solids
𝑞″
𝑇, 𝜖
𝑞″
𝑞″
Physical mechanism
 Attributed to changes in electron configurations of the
constituent atoms or molecules and transmitted as
electromagnetic waves or photons.
 Radiation does not need a medium for transport
(unlike convection and conduction) and is most
efficient in a vacuum.
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Stefan-Boltzmann Law
𝑞″ ∝ 𝑇
• For an ideal “black body”: 𝑞 = 𝜎𝑇
𝜎 ≡ Stefan-Boltzmann constant = 5.67 × 10‒8 W/(m2K4)
𝑞″
• For “grey body” (closer to real):
𝑞 = 𝜀𝜎𝑇
𝜀 ≡ Emissivity, where 0 ≤ 𝜀 ≤ 1
𝑇, 𝜖
𝑞″
𝑞″
• For real body: 𝑞 = 𝜀𝜎𝑇
where 𝜀 is a complicated function of
angle, wavelength, and temperature
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
6
Simple Example: Emitted Radiation
 Calculate thermal radiation emitted by a 1 m2 black body at:
a. 25°C
b. 2500°C
 Extreme difference shows importance of temperature, however radiation will
also be absorbed from other sources
•
We are interested in Net Exchange of Energy.
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Radiation
 Heat transfer at a gas/surface interface involves
radiation emission (E) from the surface and may also
involve the absorption of radiation incident from the
surroundings (irradiation, G), as well as convection
𝐸 = 𝜀𝐸 = 𝜀𝜎𝑇
E = emissive power [W/m2]
𝜀 = surface emissivity (0 ≤ 𝜀 ≤1)
Eb = emissive power of a perfect blackbody
𝜎 = Stefan-Boltzman constant
(5.67x10‒8 W/m2K4)
𝐺
= 𝛼𝐺
Gabs = absorved incident radiation [W/m2]
𝛼 = absorptivity of surface (0 ≤ 𝛼 ≤ 1)
G = incident radiation (irradiation) [W/m2]
 Radiative exchange among multiple interacting surfaces with different
temperatures, properties, and view factors (visibility of one surface to
another can be exceedingly complex
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
7
Special Case: Radiation in an Enclosure
 For special case of a smaller surface,
fully enclosed by isothermal surroundings
at T = Tsurr
• In this case enclosure acts like a black body (all
reflections are contained) so: 𝐺 = 𝜎𝑇
• Net heat transfer by radiation assuming 𝜀 = 𝛼
𝑞
=𝐸−𝐺
,
𝑞
= 𝜀𝜎𝑇 − 𝛼𝜎𝑇
= 𝜀𝜎 𝑇 − 𝑇
,
 This is an extremely useful special case that we can use in many
engineering problems
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Special Case: Radiation in an Enclosure
 In some cases, it may be convenient to “linearize” this equation to something of
the more familiar form that can be written in terms of a linear temperature
difference 𝑇 − 𝑇
:
𝑞
=
𝑞
𝐴
=ℎ 𝑇 −𝑇
where:
ℎ ≡ 𝜀𝜎 𝑇 + 𝑇
𝑇 +𝑇
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
8
Initial Design of a Radiant Heater
 Example: A small radiant heater uses heated metal strips 6 mm wide with a total
length of 3 m and a surface emissivity of 0.85. To what temperature must the
strips be heated to dissipate 1500 W of energy to a room at 20°C?
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Summary of Basic Heat Transfer Processes
Heat Transfer
Mode
Physical mechanism
Rate equation
Coefficient /
Transport
Property
Conduction
Diffusion of energy due to Fourier’s Law:
random molecular motion
𝑞 = −𝑘∇𝑇
Thermal conductivity,
𝑘 [W/m*K]
Convection
Diffusion plus transfer
due to bulk fluid motion
Newton’s Law of Cooling:
Convective heat
transfer coefficient
ℎ [W/m2K]
Energy transfer by
electromagnetic waves
Stefan-Boltzman Law:
𝑞
= 𝜀𝜎𝑇 (gray body)
Radiation
𝑞 =ℎ 𝑇 −𝑇
Special case of transfer between smaller gray
surface and much larger isothermal enclosure
/ surrounding:
𝑞
Emissivity, 𝜀 [-]
Stefan-Boltzmann
constant, 𝜎 =
5.67x10-8 W/m2K4
= 𝜀𝜎 𝑇 − 𝑇
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
9
Multimode Heat Transfer
Problem 1.87(a): What heat transfer processes are involved in energy
transfer through single-and double-pane windows?
Heat transfer schematic:
qconv,2: Convection from outer surface of
qconv,1: Convection from room air to inner
single (or second) pane to ambient air
qcond,2: Conduction through a second pane
surface of first pane
qcond,1: Conduction through first pane
qrad,2: Net radiation exchange between outer
qrad,1: Net radiation exchange between room
surface of single (or second) pane
and surroundings such as the ground
walls and inner surface of first pane
qrad,s: Net radiation exchange between outer
qconv,s: Convection across airspace between
surface of first pane and inner surface
of second pane (across airspace)
panes
qs: Incident solar radiation during day; fraction transmitted
to room is smaller for double pane
Assignment 1
 Posted on cuLearn.
 Problems from Chapter 1 in Bergman et al.
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MECH 4406: Heat Transfer
© Prof. M. Johnson, Carleton University
10
Part I: Conduction
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
1. Fourier’s Law
 Phenomenological equation for heat transfer by conduction
Cartesian Coordinates:
z
T(x,y,z)
𝑞″ = −𝑘∇𝑇 = −𝑘 𝚤̂
y
x
𝑞 = −𝑘
𝜕𝑇
𝜕𝑥
𝜕𝑇
𝜕𝑇
𝜕𝑇
+ 𝚥̂
+𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑞 = −𝑘
𝜕𝑇
𝜕𝑦
𝑞 = −𝑘
𝜕𝑇
𝜕𝑧
(assuming isotropic 𝑘)
• Heat transfer is in direction of decreasing temperature (hence minus sign) and is
perpendicular to isotherms (lines of constant temperature)
• Allows determination of conduction heat flux from temperature distribution
– Need to get temperature field first to apply Fourier’s law
2
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Fourier’s Law
Cylindrical Coordinates:
(assuming isotropic 𝑘)
𝑞″ = −𝑘∇𝑇 = −𝑘 𝑟̂
𝑞 = −𝑘
Spherical Coordinates:
𝜕𝑇
𝜕𝑟
𝜕𝑇
1 𝜕𝑇
𝜕𝑇
+𝜃
+𝑘
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑞 =−
𝑘 𝜕𝑇
𝑟 𝜕𝜃
𝑞 = −𝑘
𝜕𝑇
𝜕𝑧
(assuming isotropic 𝑘)
𝑞″ = −𝑘∇𝑇 = −𝑘 𝑟̂
𝑞 = −𝑘
𝜕𝑇
𝜕𝑟
𝜕𝑇
1 𝜕𝑇
1 𝜕𝑇
+𝜃
+𝜙
𝜕𝑟
𝑟 𝜕𝜃
𝑟 sin 𝜃 𝜕𝜙
𝑞 =−
𝑘 𝜕𝑇
𝑟 𝜕𝜃
𝑞 =−
𝑘 𝜕𝑇
𝑟 sin 𝜃 𝜕𝜙
3
1.1 Thermal Conductivity, k
 Transport property of a material
• Provides indication of rate at which
energy is transferred via diffusion
processes
• Wide variation among materials
 Large 𝑘 means efficient transport of
thermal energy
• 𝑘 depends on structure of material
• In general:
 𝑘
>𝑘
>𝑘
4
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Thermal Conductivity, k
 For solids:
• Transport occurs via:
1. migration of free electrons,
2. lattice vibrational waves (also known as phonons)
• k = ke + kp
○ For pure metals: ke > kp
○ For alloys: ke  kp
○ For non-metals: ke < kp
 For fluids:
• k = k(T,MW)
– As T increases, speed/collisions increase, and k increases
– As MW increases, speed/collisions decrease, and k decreases
 Thermal diffusivity, a
a
k
C p
where  is density
Cp is heat capacity
5
Thermal Conductivity, k
SOLIDS
LIQUIDS
GASES
 For gases especially, 𝒌 can be a strong function of temperature!
6
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
2. How to get Temperature Field? -- Derivation of Heat Diffusion Equation
 Fourier’s Law (𝑞″ = −𝑘∇𝑇) can be used to calculate conduction heat transfer
from temperature gradients
• But need temperature field to get temperature gradients
• Temperature field also important for many other reasons:
–
–
–
–
Structural integrity
Material properties
Deflections / expansions
Compatibility with coatings, adhesives, etc.
7
2. Derivation of Heat Diffusion Equation (HDE)
 Consider a homogeneous medium with no bulk
motion (no advection) and constant density
 Start with 1st Law:
𝑑𝑧
𝐸̇
= 𝐸̇
+ 𝐸̇ − 𝐸̇
(1)
𝐸̇
I.
Rate energy
accumulated
(stored)
II.
Rate
thermal
energy
generated
III.
Net rate of
energy
transfer or
exchange
through
surfaces
𝐸̇
𝑑𝑦
𝑑𝑥
8
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
2. Derivation of Heat Diffusion Equation (HDE)
𝑑𝑧
𝐸̇
𝐸̇
𝑑𝑦
𝑑𝑥
𝐸̇ = 𝐸̇
I
+ 𝐸̇ − 𝐸̇
II
(1)
III
9
2. Derivation of Heat Diffusion Equation (HDE) – Term III
 Consider Term III: Surface transfer term
𝑑𝑧
𝐸̇
𝐸̇
𝑞
𝑞
𝑞
𝑞
 Using Taylor series expansion:
𝑞
𝑞
𝑞
• In absence of motion, there is no change
in mechanical energy and no work done
• Only need to consider thermal transfer
via conduction
𝑑𝑦
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝜕𝑞
=𝑞 +
𝑑𝑦
𝜕𝑦
=𝑞 +
𝑞
𝑑𝑥
𝑞
𝜕𝑞
=𝑞 +
𝑑𝑧
𝜕𝑧
𝐸̇ = 𝐸̇
I
+ 𝑬̇𝒊𝒏 − 𝑬̇𝒐𝒖𝒕 (1)
II
III
10
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
2. Derivation of Heat Diffusion Equation (HDE) – Term III
 Consider Term III: Surface transfer term
𝑞 +
• In absence of motion, there is no change
in mechanical energy and no work done
• Only need to consider thermal transfer
via conduction
𝑞
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝐸̇
𝑞 +
𝑞
 Using Taylor series expansion:
𝑞
𝑞 +
𝑑𝑧
𝐸̇
𝑞
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝜕𝑞
=𝑞 +
𝑑𝑦
𝜕𝑦
=𝑞 +
𝑑𝑥
𝑞
𝑞
𝜕𝑞
=𝑞 +
𝑑𝑧
𝜕𝑧
𝐸̇ = 𝐸̇
I
+ 𝑬̇𝒊𝒏 − 𝑬̇𝒐𝒖𝒕 (1)
II
III
11
2. Derivation of Heat Diffusion Equation (HDE) – Term III
 Term III
𝐸̇ − 𝐸̇
= 𝑞 𝑑𝑦𝑑𝑧 − 𝑞 +
face
area
𝐸̇ − 𝐸̇
=−
𝜕𝑞
𝜕𝑥
𝜕𝑞
𝜕𝑥
𝑞 +
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝑞 +
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝑑𝑥 𝑑𝑦𝑑𝑧
face
area
𝑑𝑧
𝐸̇
𝐸̇
𝑑𝑥𝑑𝑦𝑑𝑧
𝑞 +
𝑞
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
• Similarly
𝐸̇ − 𝐸̇
𝐸̇ − 𝐸̇
=−
=−
𝜕𝑞
𝜕𝑦
𝜕𝑞
𝜕𝑧
𝑞
𝑑𝑥𝑑𝑦𝑑𝑧
𝑑𝑥
𝑞
𝑑𝑥𝑑𝑦𝑑𝑧
𝐸̇ = 𝐸̇
I
+ 𝑬̇𝒊𝒏 − 𝑬̇𝒐𝒖𝒕 (1)
II
III
12
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
2. Derivation of Heat Diffusion Equation (HDE) – Term III
 Term III
𝐸̇ − 𝐸̇
=−
𝐸̇ − 𝐸̇
=−
𝐸̇ − 𝐸̇
=−
∴ 𝐸̇ − 𝐸̇
=−
∴ 𝐸̇ − 𝐸̇
=−
𝜕𝑞
𝜕𝑥
𝑞 +
𝜕𝑞
𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑥
𝜕𝑞
𝜕𝑦
𝜕𝑞
𝜕𝑧
𝜕𝑥
+
𝐸̇
𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑞
𝜕𝑦
𝑞 +
𝐸̇
𝑞 +
𝑞
+
𝜕𝑞
𝜕𝑦
𝜕𝑞
𝜕𝑧
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝑑𝑧
𝑑𝑥𝑑𝑦𝑑𝑧
𝑑𝑥𝑑𝑦𝑑𝑧 −
𝜕𝑞
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝑑𝑥𝑑𝑦𝑑𝑧 −
𝜕𝑞
𝜕𝑧
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
𝑑𝑥𝑑𝑦𝑑𝑧
𝑑𝑥
𝑞
𝑑𝑥𝑑𝑦𝑑𝑧
𝑞
volume
𝐸̇ = 𝐸̇
I
+ 𝑬̇𝒊𝒏 − 𝑬̇𝒐𝒖𝒕 (1)
II
III
13
2. Derivation of Heat Diffusion Equation (HDE) – Term III
 Term III
∴ 𝐸̇ − 𝐸̇
=−
𝜕𝑞
𝜕𝑥
+
𝜕𝑞
𝜕𝑦
+
𝜕𝑞
𝜕𝑧
𝑞 +
volume
and
−𝐸̇
𝐸̇
𝐸̇
𝑞 +
𝑞
𝜕𝑞
𝜕
𝜕𝑇
=−
𝑘
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
𝑞
∴ 𝐸̇
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝑑𝑧
(i.e. energy transfer at surface is via conduction)
𝜕𝑇
𝜕𝑥
𝑞 +
𝑑𝑥𝑑𝑦𝑑𝑧
 Substitute in Fourier’s Law
e.g. 𝑞 = −𝑘
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝑑𝑥
𝑞
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝑑𝑥𝑑𝑦𝑑𝑧
𝐸̇ = 𝐸̇
I
+ 𝑬̇𝒊𝒏 − 𝑬̇𝒐𝒖𝒕 (1)
II
III
14
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
2. Derivation of Heat Diffusion Equation (HDE) – Term II
 Term II (generation term)
𝐸̇
𝑞 +
= 𝑔̇ 𝑑𝑥𝑑𝑦𝑑𝑧
volume
𝑔̇ is rate of thermal energy
generation per unit volume
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝑞 +
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝑑𝑧
𝐸̇
𝐸̇
𝑞 +
𝑞
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
𝑑𝑥
𝑞
𝑞
𝐸̇ = 𝑬̇𝒈𝒆𝒏 + 𝐸̇ − 𝐸̇
I
II
(1)
III
15
2. Derivation of Heat Diffusion Equation (HDE) – Term I
 Term II (generation term)
𝐸̇
𝑞 +
= 𝑔̇ 𝑑𝑥𝑑𝑦𝑑𝑧
𝑔̇ is rate of thermal energy
generation per unit volume
𝜕𝑞
𝑑𝑦
𝜕𝑦
 Term I (storage term)
𝐸̇
𝑞 +
𝑞
• If no phase change, then:
mass
𝜕𝑞
𝑑𝑧
𝜕𝑧
𝑑𝑧
𝐸̇
𝐸̇ = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧𝐶
𝑞 +
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
𝜕𝑇
𝜕𝑡
𝑞
Rate of change of
temperature with time
𝑑𝑥
𝑞
Specific
heat
capacity
𝑬̇𝒔𝒕 = 𝐸̇
I
+ 𝐸̇ − 𝐸̇
II
(1)
III
16
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
2. Derivation of Heat Diffusion Equation (HDE) – Term I
Term I:
𝐸̇ = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧𝐶
Term II: 𝐸̇
𝜕𝑇
𝜕𝑡
𝑞 +
𝜕𝑞
𝑑𝑦
𝜕𝑦
𝑞 +
𝜕𝑞
𝑑𝑧
𝜕𝑧
= 𝑔̇ 𝑑𝑥𝑑𝑦𝑑𝑧
Term III: 𝐸̇ − 𝐸̇
𝑑𝑧
=
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝑘
+
𝑘
+
𝑘
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝐸̇
𝑑𝑥𝑑𝑦𝑑𝑧
𝐸̇
𝑞 +
𝜕𝑞
𝑑𝑥
𝜕𝑥
𝑑𝑦
 Combine Terms I, II, and III:
Volume (𝑑𝑥𝑑𝑦𝑑𝑧) cancels, leaving:
𝜌𝐶
𝜕𝑇
𝜕𝑡
= 𝑔̇ +
𝑑𝑥
𝑞
𝑞
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝑘
+
𝑘
+
𝑘
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝑬̇𝒔𝒕 = 𝐸̇
I
+ 𝐸̇ − 𝐸̇
II
(1)
III
17
Heat Diffusion Equation, HDE
𝜌𝐶
𝜕𝑇
𝜕𝑡
= 𝑔̇ +
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝑘
+
𝑘
+
𝑘
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
II. Rate
I. Rate
thermal
energy
accumulated energy
generated
(stored)
per volume
III. Net rate of energy transfer into
volume by conduction
Cartesian Coordinates
 Appears complex but really just a clever application of conservation of energy (1st Law)
to control volume (𝐸̇ = 𝐸̇
+ 𝐸̇ − 𝐸̇ ) where energy transfer is exclusively via
conduction
• Can use to find temperature distribution in a stationary medium!
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Heat Diffusion Equation, HDE
Cylindrical Coordinates:
𝜌𝐶
𝜕𝑇
𝜕𝑡
1 𝜕
𝜕𝑇
1 𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝑘𝑟
+
𝑘
+
𝑘
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜙
𝜕𝜙
𝜕𝑧
𝜕𝑧
= 𝑔̇ +
II. Rate
I. Rate
thermal
energy
accumulated energy
generated
(stored)
per volume
III. Net rate of energy transfer into
volume by conduction
19
Heat Diffusion Equation, HDE
Spherical Coordinates:
𝜌𝐶
𝜕𝑇
𝜕𝑡
= 𝑔̇ +
1 𝜕
𝜕𝑇
1
𝜕
𝜕𝑇
1
𝜕
𝜕𝑇
𝑘𝑟
+
𝑘
+
𝑘 sin 𝜃
𝑟 𝜕𝑟
𝜕𝑟
𝑟 sin 𝜃 𝜕𝜙
𝜕𝜙
𝑟 sin 𝜃 𝜕𝜃
𝜕𝜃
II. Rate
I. Rate
thermal
energy
accumulated energy
generated
(stored)
per volume
III. Net rate of energy transfer into volume by conduction
20
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Heat Diffusion Equation, HDE
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
 Common assumption of constant and isotropic 𝑘
Cartesian Coordinates
𝜕𝑇
𝜕 𝑇 𝜕 𝑇 𝜕 𝑇
𝜌𝐶
=𝑘
+
+
+ 𝑔̇
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Or:
1 𝜕𝑇
𝑔̇
=∇ 𝑇+
𝛼 𝜕𝑡
𝑘
where
𝛼=
𝑘
= thermal diffusivity
𝜌𝐶
21
Analysis of Conduction Heat Transfer
 Basic methodology:
1. Identify that conduction applies, relevant coordinate systems, and boundary
conditions
2. Specify appropriate form of heat diffusion equation (HDE) and solve to obtain a
general solution
3. Apply appropriate boundary conditions
4. Solve HDE for temperature distribution 𝑇 𝑥, 𝑦, 𝑧, 𝑡
5. Use Fourier’s law to calculate heat flux
22
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
11
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
Special Cases we consider:
1. One Dimensional
a) Space
i) No energy generation:
ii) With energy generation:
𝜕 𝑇
= 0 or
𝜕𝑥
(shown in Cartesian Coordinates)
1 𝜕𝑇 𝜕𝑇
𝑟
=0
𝑟 𝜕𝑟
𝜕𝑟
𝜕 𝑇 𝑔̇
+ =0
𝜕𝑥
𝑘
iii) 1D conduction with varying
: Extended Surfaces / Fins
b) Time (Lumped heat capacity)
23
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
(shown in Cartesian Coordinates)
Special Cases we consider: (continued)
2. Multi-Dimensional
𝜕𝑇
𝜕 𝑇
=𝑘
𝜕𝑡
𝜕𝑥
a) Space-Time (no energy generation)
𝜌𝐶
b) Two-space (no energy generation):
𝜕 𝑇 𝜕 𝑇
+
=0
𝜕𝑥
𝜕𝑦
24
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
12
Example 1: Plane Wall
 For a 1D, plane wall with constant thermal conductivity, what is the temperature
distribution in the wall?
25
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
13
Steady, 1-D Conduction with no energy generation and constant 𝑘
The Plane Wall:
𝑇
Heat diffusion
equation:
Temperature distribution:
𝑇
𝑞
Heat Flux:
x
x=0
Heat Rate:
x=L
Thermal Resistance:
𝑑 𝑇
=0
𝑑𝑥
𝑇 = 𝑇 − (𝑇 − 𝑇 )
𝑥
𝐿
(𝑇 − 𝑇 )
𝐿
𝑘𝐴
𝑞 =
(𝑇 − 𝑇 )
𝐿
𝐿
𝑅,
=
𝑘𝐴
𝑞 =𝑘
26
Steady, 1-D Conduction with no energy generation and constant 𝑘
The Cylindrical Wall:
Heat diffusion 1 𝑑 𝑟 𝑑𝑇 = 0
equation: 𝑟 𝑑𝑟 𝑑𝑟
Temperature distribution: 𝑇 = 𝑇 + 𝑇 − 𝑇
𝑟
𝑇
𝑟
Heat Flux: 𝑞 =
𝑘(𝑇 − 𝑇 )
𝑟 ln 𝑟 ⁄𝑟
Heat Rate: 𝑞 =
2𝜋𝐿𝑘(𝑇 − 𝑇 )
ln 𝑟 ⁄𝑟
𝑇
L
ln 𝑟/𝑟
ln 𝑟 /𝑟
Thermal Resistance: 𝑅
,
=
ln 𝑟 ⁄𝑟
2𝜋𝑘𝐿
27
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Steady, 1-D Conduction with no energy generation and constant 𝑘
The Spherical Wall:
𝑟
1 − 𝑟 ⁄𝑟
Heat diffusion 𝑇 = 𝑇 − 𝑇 − 𝑇
1 − 𝑟 ⁄𝑟
equation:
1 𝑑
𝑑𝑇
𝑟
=0
Temperature distribution:
𝑟 𝑑𝑟
𝑑𝑟
Heat Flux: 𝑞 =
𝑇
𝑇
𝑟
Heat Rate: 𝑞 =
Thermal Resistance: 𝑅
𝑘(𝑇 − 𝑇 )
1⁄ 𝑟 − 1⁄𝑟
𝑟
4𝜋𝑘(𝑇 − 𝑇 )
1⁄ 𝑟 − 1⁄𝑟
,
=
1⁄ 𝑟 − 1⁄ 𝑟
4𝜋𝑘
28
Boundary and Initial Conditions
 To determine the temperature distribution in a medium, we need to solve the heat
diffusion equation (HDE)
• Solving the heat diffusion equation requires appropriate boundary conditions (and initial
conditions if problem is time dependent)
 General from of the HDE is 2nd order in space and 1st order in time
• Requires 6 boundary conditions (2 for each coordinate direction)
• Requires 1 initial condition
29
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Common Boundary Conditions for the HDE
 Several types of boundary conditions are common:
• Specified temperature: 𝑇 𝑥, 𝑦, 𝑧, 𝑡 = 𝑇
• Specified heat flux: 𝑞 = −𝑘
– Finite heat flux: −𝑘
=𝑞
,
(Dirichlet)
(Neumann)
= constant = 𝑞
– Adiabatic (insulated surface): −𝑘
• Convection surface condition: −𝑘
• Radiation condition: −𝑘
,
=0
= ℎ 𝑇 − 𝑇 (0, 𝑡)
= 𝜀𝜎 𝑇 − 𝑇
• Combined convection and radiation
• Interface condition: 𝑇 = 𝑇 ; −𝑘
= −𝑘
30
Example 2: Plane wall of a furnace
 What is the heat flux, 𝑞 , through the wall of a furnace?
• Because of large temperature differences must assume 𝑘 varies linearly with
temperature: 𝑘 = 𝑘 + 𝑘 𝑎𝑇 = 𝑘 1 + 𝑎𝑇
31
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Example 3: Conduction in Steam Pipe with 2 possible boundary conditions
𝑟
𝑇
𝑟
𝑇𝟐
𝑇 ,ℎ
32
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Part I: Conduction
Section 3: Electrical Analog for 1-D, Steady Conduction
with No Thermal Energy Generation
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
A Game to Start!
 Set of Numbers: 1 – 9
 Take turns choosing numbers
• Once a number is chosen, it is gone
 First person to collect 3 numbers that add to 15 wins
• Must be 3 numbers though, i.e. 2 or 4 numbers that add to 15 don’t count
 Prof. vs. students; numbers chosen by poll.
• Can I beat the parallel processing capability of your ~175 young minds??
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
An Alternate Approach to Steady Heat Transfer Problems?
 Consider the simple
plane wall again:
HDE:
𝜌𝐶
𝑘
+
𝑘
Steady
T2
T1
=
0=
+
𝜕
𝜕𝑇
𝑘
𝜕𝑥
𝜕𝑥
No Energy Generation
Assume constant
thermal conductivity, 𝑘
Integrate once:
Integrate again:
x=0
Boundary conditions:
x=L
Solution:
𝑇 𝑥 =
Fourier’s Law: 𝑞 = −𝑘
+ 𝑔̇
1D
k
q
x
𝑘
𝜕 𝑇
=0
𝜕𝑥
=𝐶
𝑇 𝑥 = 𝐶 𝑥+𝐶
@𝑥 = 0, 𝑇 = 𝑇
∴𝐶 =𝑇
@𝑥 = 𝐿, 𝑇 = 𝑇
∴ 𝐶 = 𝑇 − 𝑇 /𝐿
𝑥+𝑇
𝑑𝑇
𝑑𝑥
Heat flux: 𝑞 = 𝑘
𝑇 −𝑇
𝐿
Heat rate: 𝑞 = 𝑘𝐴 𝑇 − 𝑇
(plane wall)
𝐿
5
3.1 Electrical Analogy
 Ohm’s Law:
𝑹𝒕𝒉𝒆𝒓𝒎𝒂𝒍
𝑅
• 𝑉 = 𝐼𝑅 or 𝑅 = 𝑉 ⁄𝐼
V1
V2
𝑻𝟐
𝑻𝟏
• Analogue with 1-D conduction:
𝐼
𝑞
 (V1 – V2)  Electrical Potential
 (𝑇 − 𝑇 )  Thermal Potential
 I  Current or flow of electricity
 𝑞  Heat transfer rate or “flow” of thermal energy
 R = DV / I  Electrical Resistance
 𝑹 = 𝜟𝑻 / 𝒒𝒙  Thermal Resistance
Heat rate: 𝑞 = 𝑘𝐴 𝑇 − 𝑇
(plane wall)
𝐿
Since 𝑅 =
By analogy
(for this case): 𝑅
=
Δ𝑇
𝐿
=
𝑞
𝑘𝐴
6
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
3.1 Thermal Resistance Concept for 1-D Steady Heat Transfer
 N.B.: Only useful/valid for 1-D, steady heat transfer, with no thermal energy generation,
and with constant 𝑘 and ℎ
• 1-D Conduction:
– Fourier’s Law: 𝑞 =
(𝑇 − 𝑇 )
∴𝑅
=
𝐿
𝑘𝐴
(Note this form for Cartesian
coordinates only!)
• Convection:
– Recall Newton’s law of cooling: 𝑞 = ℎ𝐴Δ𝑇
∴𝑅
=
Δ𝑇
1
=
𝑞
ℎ𝐴
• Radiation:
– For special case for gray body radiation without obstruction
𝑞 = 𝜀 𝜎𝐴(𝑇 − 𝑇
𝑞 =ℎ 𝐴 𝑇 −𝑇
)
∴𝑅
where ℎ ≡ 𝜀𝜎 𝑇 + 𝑇
𝑇 +𝑇
=
1
ℎ 𝐴
 So long as assumptions are valid, resistances can be added in series and parallel like circuits
7
Electrical Analog – Thermal Resistances
 That’s neat, but is it useful?
• Consider familiar problem of heat transfer through single pane window
𝑇
𝑘
ℎ
𝑇
ℎ
𝑇
𝑇
𝑞″
𝑇
𝑞″
𝑇
1
ℎ 𝐴
𝑞″
𝑇
𝐿
𝑘𝐴
𝑇
1
ℎ 𝐴
8
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Thermal Resistance Concept
 E.g.: Heat transfer through
a composite wall:
 No longer strictly a 1D problem.
If (𝑘 – 𝑘 ) >> 0 then 2D effects
become non-negligible and problem
must be solve by other means.
Area, A
𝑇
ℎ
1
2
A
𝑘
3
4
C
𝑘
B
𝑘
D
𝑘
5
Area,
AC
E
𝑘
𝑇
ℎ
𝐿
𝑘 𝐴
 If 1D assumptions are still desired,
thermal circuits are typically added
together as are electrical resistors in
series / parallel as appropriate:
𝑇
𝑇
1
ℎ 𝐴
𝑇
𝐿
𝑘 𝐴
𝑇
𝐿
𝑘 𝐴
𝑇
𝐿
𝑘 𝐴
𝑇
𝐿
𝑘 𝐴
𝑇
1
ℎ 𝐴
9
Thermal Resistance Concept
 If 1D assumptions are still desired, thermal circuits are typically added together as are
electrical resistors in series / parallel as appropriate
 Resistors in series:
Replace:
where: 𝑅
with:
𝑅
𝑅
=
𝑅
𝑅
𝑅
𝑅
 Resistors in parallel:
Replace:
𝑅
𝑅
with:
where:
1
1
1
1
=
+
+
+⋯
𝑅
𝑅
𝑅
𝑅
𝑅
○ Special case of two resistors only in parallel: 𝑅
=
10
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Overall Heat Transfer Coefficient
 In composite systems, it is often convenient to work with an overall heat transfer
coefficient defined in an analogous way to Newton’s law of cooling:
A
B
C
𝑘
𝑘
𝑘
𝑞 = 𝑈𝐴Δ𝑇 =
Δ𝑇
𝑅
𝑈𝐴 =
1
𝑈=
𝑅
1
𝑅
𝐴
𝑈 ≡ Overall Heat Transfer Coefficient [W/m2K]
 In the building industry it is practice to use a term called the “R value”, defined as:
• “R-value” =
⁄
=
(inverse of the overall heat transfer coefficient)
– Typically reported in units of [
°
]; sometimes reported as R-value per inch
°
• Metric equivalent is the RSI-value reported in units of of [
] or sometimes as RSI-value per mm
11
Radial Systems
 What about 1D radial systems?
𝑇
• Consider a pipe section
 To apply electrical analog, system must be 1D and we need:
• Temperature distribution
• Heat flux
• Thermal resistance
𝑟
𝑇
𝑟
 HDE in cylindrical coordinates with constant k:
1 𝜕𝑇 𝑔̇ 1 𝜕
𝜕𝑇
1 𝜕 𝑇 𝜕 𝑇
= +
𝑟
+
+
𝛼 𝜕𝑡 𝑘 𝑟 𝜕𝑟 𝜕𝑟
𝑟 𝜕𝜙
𝜕𝑧
If 1D in 𝑟 and no thermal
energy generation:
𝑑
𝑑𝑇
𝑟
=0
𝑑𝑟
𝑑𝑟
(1)
12
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Radial Systems
𝑑
𝑑𝑇
𝑟
=0
𝑑𝑟
𝑑𝑟
(1)
𝑇
𝜕𝑇
=𝐶
𝜕𝑟
Integrate once:
𝑟
Integrate again:
𝑇 = 𝐶 ln 𝑟 + 𝐶
𝜕𝑇 𝐶
=
𝜕𝑟
𝑟
• for 𝑟 >0,
(2)
𝑟
(2a)
𝑇
𝑟
(3, General Solution)
Applying Boundary conditions:
 @𝑟 =𝑟 ,𝑇 =𝑇 :
 @𝑟 =𝑟 ,𝑇 =𝑇 :

𝑇 = 𝐶 ln 𝑟 + 𝐶
𝑇 = 𝐶 ln 𝑟 + 𝐶
Solve 2 eq., 2 unknowns to get final solution
𝑇 𝑟 =𝑇 +
𝑇 −𝑇
ln 𝑟/𝑟
ln 𝑟 /𝑟
NOTE: Temperature distribution is
non-linear!
13
Radial Systems
 Next, find heat flux
𝑇
• From Fourier’s Law:
𝑑𝑇
−𝑘
𝑟
𝑇 −𝑇
Heat flux is: 𝑞 = −𝑘 𝑑𝑟 = 𝑟 𝑙𝑛 𝑟 ⁄𝑟
Heat rate is: 𝑞 = 𝑞 𝐴 =
∴𝑞 =
Heat
Rate
Thus, by analogy:
𝑘2𝜋𝑟𝐿 𝑇 − 𝑇
𝑟
ln 𝑟 ⁄𝑟
𝑘2𝜋𝐿
ln 𝑟 ⁄𝑟
Thermal
Resistance
𝑅
𝑇
𝐴 = 2𝜋𝑟𝐿
𝑇 −𝑇
(Note that the heat flux varies with radius
and heat rate does not – this is logical of
course but important)
Thermal
Potential
=
𝑟
𝑙𝑛 𝑟 ⁄𝑟
𝑘2𝜋𝐿
14
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Radial Systems
 What about thermal resistance for convection on outside?
• As before: 𝑞 = ℎ𝐴𝛥𝑇 so now, 𝑅
=
𝑇
=
𝑟
𝑇
𝑟
 Spherical systems:
• Just like previous example:
1-D spherical HDE:
𝜕
𝜕𝑇
𝑟
=0
𝜕𝑟
𝜕𝑟
Heat rate:
Resistance:
4𝜋𝑘 𝑇 − 𝑇
𝑞 =
1⁄𝑟 − 1⁄𝑟
𝑅
=
1⁄𝑟 − 1⁄𝑟
4𝜋𝑘
15
Example: Critical Insulation Radius
 A laboratory experiment requires heating of samples using an air to water heat
exchanger. (A heat exchanger is simply a mechanical engineering device to efficiently
transfer thermal energy between fluids.) There is a short run of tubing from the
external heater to the chamber containing the heat exchanger. The tubing is stainless
steel, with 20 mm O.D. and a wall thickness of 2 mm. The tubing sits in stagnant lab air
at T=15°C. The water in the tube is at T=55°C.
Stainless steel tubing
kss = 15 W/mK
20 mm O.D.
16 mm I.D.
a) Calculate the rate of heat loss from the tubing
assuming ℎ = 6000 W/m2K and ℎ = 4 W/m2K
b) How does the heat loss change as we wrap
varying thicknesses of insulation around the
tubing?
Chamber for
heating samples
Heater
Heat
exchanger
16
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
Special Cases we consider:
1. One Dimensional
a) Space
i) No energy generation:
ii) With energy generation:
𝜕 𝑇
= 0 or
𝜕𝑥
(shown in Cartesian Coordinates)
1 𝜕𝑇 𝜕𝑇
𝑟
=0
𝑟 𝜕𝑟
𝜕𝑟
𝜕 𝑇 𝑔̇
+ =0
𝜕𝑥
𝑘
iii) 1D conduction with varying
: Extended Surfaces / Fins
b) Time (Lumped heat capacity)
1
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
(shown in Cartesian Coordinates)
Special Cases we consider: (continued)
2. Multi-Dimensional
𝜕𝑇
𝜕 𝑇
=𝑘
𝜕𝑡
𝜕𝑥
a) Space-Time (no energy generation)
𝜌𝐶
b) Two-space (no energy generation):
𝜕 𝑇 𝜕 𝑇
+
=0
𝜕𝑥
𝜕𝑦
2
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Part I: Conduction
Section 4: Conduction with Thermal Energy Generation
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Conduction with Thermal Energy Generation
 Examples:
•
•
•
•
Electrical generation in wire
Nuclear reactions (absorption/deceleration of neutrons)
Microwave heating
Chemical reactions
– Endothermic reactions could be negative generators (glow sticks)
4
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Conduction with Thermal Energy Generation
 Start with HDE (shown here in Cartesian coordinates):
𝜌𝐶
Assume:
• Steady
• 1D in 𝑥
• Constant 𝑘
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
0=𝑘
𝜕 𝜕𝑇
+ 𝑔̇
𝜕𝑥 𝜕𝑥
which
simplifies to:
𝑑 𝑇 𝑔̇
+ =0
𝑑𝑥
𝑘
Integrate once:
𝑑𝑇 𝑔̇
+ 𝑥+𝐶 =0
𝑑𝑥 𝑘
Integrate again to
get general solution:
𝑇 𝑥 =−
𝑔̇
𝑥 +𝐶 𝑥+𝐶
2𝑘
 Note:
• Temperature distribution is no longer linear.
• Resistance concepts previously described no longer apply.
5
Example in a Radial System
 An electrical current of 62.5 Amps passes through a
30-mm diameter Nichrome wire with an electrical
resistance of 8×10−2 ohms/m. The wire is insulated
with 5 mm of soft vulcanized rubber which is
exposed to air at 20°C and h=25 W/m2K.
a) What is the temperature at the center of the wire?
b) What is temperature at the outer surface of the wire?
• Assume: 𝑘wire=10 W/(m∙K), 𝑘insul=0.13 W/(m∙K)
30-mm
diameter
Nichrome wire
𝑘wire=10 W/(m∙K)
𝑇 =20°C
ℎ=25 W/m2K
5 mm insulation
(O.D. = 40 mm)
𝑘ins=0.13 W/(m∙K)
 Solution:
• Break into two problems to solve: (1) solution for wire using HDE with thermal energy generation
(2) solution for insulation with 1D conduction and convection to surroundings.
• Details to be discussed in class.
6
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Critical Insulation Radius
 Consider an insulated cylinder / wire / pipe:
ℎ, 𝑇
Insulation
𝑟
𝑇
𝑇
𝑟
𝑇
𝑇
𝑞=
=
𝛥𝑇
=
𝑅
𝑅
𝑅
𝛥𝑇
+𝑅
Observe:
•𝑅
•𝑅
2𝜋𝐿(𝑇 − 𝑇 )
ln( 𝑟 /𝑟 )
1
+
𝑘
𝑟ℎ
Convection
=
ln 𝑟 ⁄𝑟
2𝜋𝑘𝐿
𝑇
𝑅
=
1
2𝜋𝑟 𝐿ℎ
increases with increasing 𝑟 (more insulation)
decreases with increasing 𝑟 (more area for convection)
18
Critical Insulation Radius
 𝑅
has a minimum as insulation is added
• If Δ𝑇 is held constant, then 𝑞
has a maximum!
19
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Critical Insulation Radius
 Find maximum:
𝑑𝑞
=0=
𝑑𝑟
 Which gives result: 𝑟
1
1
−
𝑘𝑟 ℎ𝑟
−2𝜋𝐿(𝑇 − 𝑇 )
ln( 𝑟/𝑟 ) 1
+
𝑘
𝑟ℎ
=
𝑘
ℎ
 Same procedure for spherical system gives:
𝑟
=
2𝑘
ℎ
20
Summary of Results for 1-D, Steady Conduction with No thermal energy generation
21
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Critical Insulation Radius – Practical Considerations
 Consider typical parameter values:
• Typical insulation:
𝑘=0.03 W/(m∙K)
• Typical convection value (in free air): ℎ=10 W/(m2K)
• 𝑟
=
= 0.003 m or 3 mm
 In most cases, 𝑟
>𝑟
• Generally only relevant for small diameter cases like wires (*but can still be very
important!)
22
3.5 Thermal Contact Resistance
 In composite systems, there may
be appreciable temperature drops
across interfaces between materials
 This effect is known as
Thermal Contact Resistance, Rt,c
Rt ,c 
T A  TB
qx
• Note, text book defines contact resistance on a unit area basis:
Rt,c 
Where:
Rt,c 
T A  TB
q x
TA  TB TA  TB TA  TB A


 Rt ,c A
qx
qx
qx
A
23
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Thermal Pastes to Reduce Contact Resistance between Heat Sinks and Chips
24
3.5 Thermal Contact Resistance
25
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
3.5 Thermal Contact Resistance
26
Review Question
 In a product you are designing, conduction heat transfer occurs through a piece of brass (k=100 W/mK)
and on through a piece of steel (k=50 W/mK) touching the brass. Recognizing that there will be a
thermal contact resistance between the two metals, which of the following is a realistic representation of
the temperature variation through the materials?
(a)
(b)
T
brass
(c)
steel
x
(d)
T
brass
steel
T
x
brass
steel
x
brass
steel
x
T
27
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Part I: Conduction
Section 5: Extended Surfaces & Fins
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Using Fins to Enhance Heat Transfer
 In many engineering problems, we want to enhance heat transfer
 Given 𝑞 = ℎ𝐴 𝑇 − 𝑇 , options include:
• Increase 𝑇 − 𝑇
– Lower 𝑇 (cool surroundings): – can be very expensive and usually impractical
– Raise 𝑇 : may also be impractical (material limits)
• Increase ℎ :
– Switch from natural to forced convection (fan noise / cost / power consumption)
– Change fluid (may not be possible / practical)
• Increase 𝐴 :
– Extended surfaces = Fins!
○ E.g.: car radiators, CPU heat sinks, lawnmower engines, etc.
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Objective of Fins
 To enhance heat transfer between a solid and an adjoining fluid
5
Heat Transfer from Fins
 A variety of fin shapes are possible which influence mathematics of the solution
 Ideally fins will have large thermal conductivities to minimize temperature drop
along fin and maximize 𝑇 − 𝑇 with fluid over entire surface.
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Fins in Cooling Applications
 CPU cooling device (Alpha PAL8045):
(array of pin fins)
(bottom view showing
mount for CPU)
 Heat exchanger fins:
5.1 Analysis of Conduction / Convection from a Fin
General Conduction Analysis
 Assumptions
• Steady state
• 1D in 𝑥
– Assume a well-designed fin so that 𝑘 is high relative to ℎ, which
means temperature variation perpendicular to fin is negligible
• Negligible radiation from surface
– Conservative assumption that will underestimate heat transfer
• No thermal energy generation within fin
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Conduction / Convection from a Fin
1st Law energy conservation
𝑞 =𝑞
+ 𝑑𝑞
(1)
Apply Taylor series expansion for 𝑞
𝑑𝑞
𝑞
=𝑞 +
𝑑𝑥
𝑑𝑥
Sub into (1)
𝑑𝑞
𝑞 =𝑞 +
𝑑𝑥 + 𝑑𝑞
𝑑𝑥
𝑑𝑞
∴
𝑑𝑥 + 𝑑𝑞
=0
𝑑𝑥
Apply Fourier’s Law:
𝑑𝑇
and
𝑞 = −𝑘𝐴
𝑑𝑥
𝑑𝑞
𝑑
𝑑𝑇
= −𝑘
𝐴
𝑑𝑥
𝑑𝑥
𝑑𝑥
Sub (3) into (2):
∴ −𝑘
Which using
chain-rule gives:
𝑑𝐴 𝑑𝑇
𝑑 𝑇
+𝐴
𝑑𝑥 + 𝑑𝑞
𝑑𝑥 𝑑𝑥
𝑑𝑥
=0
(2)
𝑑𝑞
𝑑𝐴 𝑑𝑇
𝑑 𝑇
= −𝑘
+𝐴
𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑥
(3)
(4)
Conduction / Convection from a Fin
1st Law energy conservation
𝑞 =𝑞
+ 𝑑𝑞
(1)
Apply Taylor series expansion for 𝑞
𝑑𝑞
𝑞
=𝑞 +
𝑑𝑥
𝑑𝑥
Sub into (1)
𝑑𝑞
𝑞 =𝑞 +
𝑑𝑥 + 𝑑𝑞
𝑑𝑥
𝑑𝑞
∴
𝑑𝑥 + 𝑑𝑞
=0
𝑑𝑥
Apply Fourier’s Law:
𝑑𝑇
and
𝑞 = −𝑘𝐴
𝑑𝑥
𝑑𝑞
𝑑
𝑑𝑇
= −𝑘
𝐴
𝑑𝑥
𝑑𝑥
𝑑𝑥
Sub (3) into (2):
∴ −𝑘
MECH 4406: Heat Transfer
Which using
chain-rule gives:
𝑑𝐴 𝑑𝑇
𝑑 𝑇
+𝐴
𝑑𝑥 + 𝑑𝑞
𝑑𝑥 𝑑𝑥
𝑑𝑥
=0
(2)
𝑑𝑞
𝑑𝐴 𝑑𝑇
𝑑 𝑇
= −𝑘
+𝐴
𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑥
(3)
(4)
Prof. M. Johnson, Carleton University
4
Conduction / Convection from a Fin
∴ −𝑘
𝑑𝐴 𝑑𝑇
𝑑 𝑇
+𝐴
𝑑𝑥 + 𝑑𝑞
𝑑𝑥 𝑑𝑥
𝑑𝑥
(4)
=0
Sub in Newton’s Law of Cooling:
𝑑𝑞
−𝑘
(5)
= ℎ𝑑𝐴 𝑇 − 𝑇
𝑑𝐴 𝑑𝑇
𝑑 𝑇
+𝐴
𝑑𝑥 + ℎ𝑑𝐴 𝑇 − 𝑇
𝑑𝑥 𝑑𝑥
𝑑𝑥
=0
Rearranging gives:
𝑑 𝑇
1 𝑑𝐴
+
𝑑𝑥
𝐴 𝑑𝑥

𝑑𝑇
1 ℎ 𝑑𝐴
−
𝑑𝑥
𝐴 𝑘 𝑑𝑥
𝑇−𝑇
=0
(6)
In principle, this can be solved for general cross-sectional shape
with appropriate boundary conditions, but it can get ugly!
Fins of Constant Cross-section
 Start from Equation (6) and simplify:
𝑑 𝑇
1 𝑑𝐴
+
𝑑𝑥
𝐴 𝑑𝑥
𝑑𝑇
1 ℎ 𝑑𝐴
−
𝑑𝑥
𝐴 𝑘 𝑑𝑥
𝑇−𝑇
=0
(6)
𝑃 (perimeter)
If 𝐴 is constant:
= 0,
𝐴 = 𝑃 𝑥 , and
= 𝑃, where 𝑃 is fin perimeter
Thus (6) simplifies to:
d 2T hP
T  T   0

dx 2 kAc
MECH 4406: Heat Transfer
(7)
Prof. M. Johnson, Carleton University
5
Fins of Constant Cross-section
 Continue from (7):
d 2T hP
T  T   0

dx 2 kAc
(7)
Define: 𝜃 𝑥 ≡ 𝑇 𝑥 − 𝑇 such that
=
and sub into (7) and rearrange to standard form:
𝑑 𝜃
−𝑚 𝜃 =0
𝑑𝑥
where:
ℎ𝑃
𝑘𝐴
𝑚=
(8)
• Eq. (8) is a linear, homogeneous, 2nd order ordinary
differential equation with solution:
𝜃 𝑥 =𝐶 𝑒
ℎ𝑃
𝑘𝐴
𝑚=
+𝐶 𝑒
(9)
• Need two boundary conditions to solve for 𝐶 and 𝐶
• 1st boundary condition:
• Assume temperature difference at the base of
the fin, 𝜃 , is known; i.e., 𝑇 − 𝑇 is known
2nd Boundary Condition: Fin Tip Conditions
Case A: Convective Tip
Tb
dT
k
dx
T(x)
Case B: Adiabatic Tip
 hTL  T 
Tb
x L
dT
dx
T(x)
T
L
𝑞
T
L
= ℎ(𝑇 − 𝑇 )
x
𝑞
=0
x
Case C: Known T @ Tip
Tb
0
x L
T(x)
T
x L
 TL
Tb
Case D: Infinite Tip
T x  T
T(x)
T
L
x
MECH 4406: Heat Transfer
T

𝑇=𝑇
x
Prof. M. Johnson, Carleton University
6
Uniform Cross-section Fin Solutions – Case A
Case A: Convective Tip
Tb
T(x)
k
dT
dx
𝜃 𝑥 =𝐶 𝑒
 hTL  T 
x L
L
𝑞
@𝑥=0, 𝜃 = 𝜃
1st b.c.:
T
(9)
+𝐶 𝑒
∴𝜃 =𝐶 +𝐶
𝑑𝜃
2nd b.c. - 1st Law balance at tip: −𝑘
𝑑𝑥
= ℎ(𝑇 − 𝑇 )
Sub (11) into (9):
𝑘𝑚 𝐶 𝑒
−𝐶 𝑒
x
=ℎ 𝐶 𝑒
= ℎ𝜃
+𝐶 𝑒
(10)
(11)
(12)
Solve (10) and (12) for 2 unknowns (𝐶 and 𝐶 ) to get solution:
Temperature
Distribution in Fin:
𝜃
𝑇−𝑇
cosh 𝑚 𝐿 − 𝑥 + ℎ⁄𝑚𝑘 sinh 𝑚 𝐿 − 𝑥
=
=
𝜃
𝑇 −𝑇
cosh 𝑚𝐿 + ℎ⁄𝑚𝑘 sinh 𝑚𝐿
Heat Transfer
Rate through Fin:
𝑞
= 𝑞 = −𝑘𝐴
𝑑𝜃
𝑑𝑥
=
ℎ𝑃𝑘𝐴 𝜃
where:
𝑚=
ℎ𝑃
𝑘𝐴
sinh 𝑚𝐿 + ℎ⁄𝑚𝑘 cosh 𝑚𝐿
cosh 𝑚𝐿 + ℎ⁄𝑚𝑘 sinh 𝑚𝐿
15
Hyperbolic sines and cosines
 Recall:
sinh 𝑥 =
1
𝑒 −𝑒
2
𝑑
𝑑𝑢
sinh 𝑢 = cosh 𝑢
𝑑𝑥
𝑑𝑥
cosh 𝑥 =
1
𝑒 +𝑒
2
𝑑
𝑑𝑢
cosh 𝑢 = sinh 𝑢
𝑑𝑥
𝑑𝑥
tanh 𝑥 =
𝑒 −𝑒
𝑒 +𝑒
=
sinh 𝑥
cosh 𝑥
𝑑
1
tanh 𝑢 =
𝑑𝑥
cosh 𝑢
𝑑𝑢
𝑑𝑥
16
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Uniform Cross-section Fin Solutions – Case B
Case B: Adiabatic Tip
Tb
dT
dx
T(x)
𝜃 𝑥 =𝐶 𝑒
1st b.c. - @𝑥=0, 𝜃 = 𝜃
x L
T
L
𝑞
(9)
+𝐶 𝑒
0
𝑑𝜃
𝑑𝑥
2nd b.c. - Adiabatic tip so:
=0
(10)
∴𝜃 =𝐶 +𝐶
(11)
=0
Sub (11) into (9) and divide by 𝑚:
x
𝐶𝑒
−𝐶 𝑒
(12)
=0
Solve (10) and (12) for 2 unknowns (𝐶 and 𝐶 ) to get solution:
Temperature
Distribution in Fin:
𝜃
𝑇−𝑇
cosh 𝑚 𝐿 − 𝑥
=
=
𝜃
𝑇 −𝑇
cosh 𝑚𝐿
Heat Transfer
Rate through Fin:
𝑞
= 𝑞 = −𝑘𝐴
=
where:
ℎ𝑃
𝑘𝐴
𝑚=
ℎ𝑃𝑘𝐴 𝜃 tanh 𝑚𝐿
17
Uniform Cross-section Fin Solutions – Case C
Case C: Known T @ Tip
Tb
T(x)
T
x L
𝜃 𝑥 =𝐶 𝑒
 TL
(9)
+𝐶 𝑒
1st b.c. - @𝑥=0, 𝜃 = 𝜃
∴𝜃 =𝐶 +𝐶
2nd b.c. - @𝑥=𝐿, 𝜃 = 𝜃
∴𝜃 =𝐶 𝑒
(10)
T
L
+𝐶 𝑒
(11)
𝑇=𝑇
x
Solve (10) and (11) for 2 unknowns (𝐶 and 𝐶 ) to get solution:
Temperature
Distribution in Fin:
𝜃
𝑇−𝑇
𝜃 ⁄𝜃 sinh 𝑚𝑥 + sinh 𝑚 𝐿 − 𝑥
=
=
𝜃
𝑇 −𝑇
sinh 𝑚𝐿
Heat Transfer
Rate through Fin:
𝑞
= 𝑞 = −𝑘𝐴
𝑑𝜃
𝑑𝑥
=
ℎ𝑃𝑘𝐴 𝜃
where:
𝑚=
ℎ𝑃
𝑘𝐴
cosh 𝑚𝐿 − 𝜃 ⁄𝜃
sinh 𝑚𝐿
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Uniform Cross-section Fin Solutions – Case D
Case D: Infinite Tip
Tb
T
T(x)
x
 T
T
Can find answer from Case C solution and taking limits:
𝜃
𝑇−𝑇
𝜃 ⁄𝜃 sinh 𝑚𝑥 + sinh 𝑚 𝐿 − 𝑥
Case C
Solution: 𝜃 = 𝑇 − 𝑇 =
sinh 𝑚𝐿
=

x
𝜃 ⁄𝜃
Heat Transfer
Rate through Fin:
𝜃
𝑇−𝑇
=
=𝑒
𝜃
𝑇 −𝑇
𝑞
=
𝑒
−𝑒
+
𝑒
As 𝐿 → ∞, 𝜃 → 0
𝜃
𝑒
−𝑒
=
𝜃
𝑒
Temperature
Distribution in Fin:
0
=
𝑒
𝑒
−
𝑒
−𝑒
𝑒
𝑒
−𝑒
0
=
𝑒
𝑒
𝑒
=𝑒
where:
𝑚=
−
𝑒
−
𝑒
𝑒
𝑒
𝑒
0
𝑒
−
𝑒
0
ℎ𝑃
𝑘𝐴
ℎ𝑃𝑘𝐴 𝜃
19
𝑻(𝒙) and 𝒒𝒇𝒊𝒏 Solutions for Fins of Constant Cross-sectional Area
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Fin Example 1
 A long 5 mm diameter pin fin has a base temperature of 100°C and protrudes
into air at 25°C with h=100 W/m2K. What is 𝑞 if material is:
a) Copper
b) SS 316
c) Al2024
 How long would each of the above fins need to be before they could accurately
be considered infinite?
21
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Engineering Simplification: “Corrected” Fin Length, Lc
 A convective tip is the most common / most accurate fin tip boundary condition, yet it
is also the most complex
 A useful engineering approximation is to instead use the simpler adiabatic tip solution
with a “corrected” fin length, Lc
• “Move” convection that would have occurred on tip to sides of fin (i.e. increase fin
length accordingly) and assume new tip is adiabatic
Lc
L
𝐿 = 𝐿 + 𝑡 ⁄2
t/2
t
Active tip
with
convection
Adiabatic tip
𝑑𝜃
𝑑𝑥
=0
Engineering Simplification: “Corrected” Fin Length, Lc
 Same concept for a round pin fin (or any other fin)
Note for the annular find this is a
rough engineering approximation
𝜋𝐷Δ𝐿 = 𝜋𝐷 ⁄4
∴ 𝐿 = 𝐿 + 𝐷 ⁄4
• Generally, errors using the corrected length approximation
are negligible if (ht/k) or (hD/2k) ≤ 0.0625
23
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
5.2 Fin Performance
 Consider Electrical Analog concepts: 𝑅 = Δ𝑇 ⁄𝑞
• Compare heat transfer with and without a fin
Tb
𝑞
Tb
qno fin
Ab
𝑅
=
𝑇 −𝑇
𝑞
𝑅
=
𝑇 −𝑇
𝑞
• Paradox: we use fins to increase the area for heat exchange, but the fin itself is an
added thermal resistance!
– Is it worth it? Is it a good fin? Can we improve the design?
24
Fin Effectiveness, 𝜺𝒇𝒊𝒏
 Fin Effectiveness, 𝜀
“Is a fin worth the
effort”?
𝑞
Tb
𝑅
𝜀
=
R fin
Tb
=
T
Δ𝑇 𝑇 − 𝑇
𝜃
=
=
𝑞
𝑞
𝑞
;𝑞
=
𝜃
𝑅
Fin Heat Transfer Rate
Heat transfer rate without fin
𝑞
𝜀
=
𝜀
𝑞
=
ℎ𝐴 𝜃
𝑞
=
𝑞
ℎ𝐴 𝑇 − 𝑇
MECH 4406: Heat Transfer
RCONV
Tb
Tb
qno fin
Ab
∴𝜀
Rb  RCONV 
=
𝑞
𝑞
T
1
; 𝑞
hAb
=
𝜃
𝑅
=
Δ𝑇 𝜃
=
= ℎ𝐴 𝜃
𝑅
𝑅
1
ℎ𝐴 𝜃
=
𝑅
𝑅
Prof. M. Johnson, Carleton University
2
Fin Effectiveness, 𝜺𝒇𝒊𝒏
 For any rational design, 𝜀
should be as large as possible
𝜀
• Use of a fin is rarely justified if 𝜀
=
<2
∴𝜀
=
 Think about what this means practically:
• For 𝜀
≥2, then 𝑅
Fin Heat Transfer Rate
Heat transfer rate without fin
𝑞
𝑞
=
𝜃
𝑅
1
ℎ𝐴 𝜃
=
𝑅
𝑅
< 𝑅 /2
– i.e. resistance with fin added is notably less than original resistance from the surface due to
convection
○ True for small ℎ, since
is large
○ True for high 𝑞
 Still need an additional performance measure
• Fin might be effective but is it efficient / well designed?
26
5.2.2 Fin Efficiency, 𝜼𝒇𝒊𝒏
 Fin efficiency compares actual fin performance
to “ideal” fin performance
𝑞
𝑞
𝜂 =
=
𝑞 ,
ℎ𝐴 𝜃
Tb
T(x)ideal
T(x)actual
 Ideal fin (i.e. max theoretical heat transfer)
• Maximizes Δ𝑇 (maximizes 𝜃)
• Would occur if 𝜃 = 𝜃 all along fin
 Also recall that 𝑅
𝑅
=
=
1
ℎ𝐴
𝜂
MECH 4406: Heat Transfer
Tb
so:
Very useful result to be able to
express 𝑅 in terms of 𝜂
Prof. M. Johnson, Carleton University
3
Fin Efficiency as a Design Tool
 Data for more complex fins can be usefully tabulated or plotted in terms of fin
efficiency
Fin Data in Terms of Efficiency
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Design Implications for Fin Efficiency
 Considering results shown on previous figures:
• If 𝐿 ↑ then 𝜂
↓ (𝑇
decreases with 𝐿)
• If 𝑘 ↑ then 𝜂
↑ (𝑇
increases with 𝑘)
↑ if shape is refined (e.g. rectangle to parabola)
• 𝜂
30
Fin Example 1b
 Consider again fins from previous example
• What are the efficiencies and effectivenesses for each fin?
a)
Assume the fins are different lengths as required to treat them as “infinite”
○ I.E.: 𝐿
= 18.7 cm; 𝐿
= 3.5 cm; 𝐿
b) Assume fins are all 20 cm long (𝐿
= 12.6 cm;
= 20 cm)
31
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
5.3 Other Performance Considerations for Fins
 Thermal contact resistance
• Especially important if fins are added to base later in manufacturing as a separate
piece
• Common concern for fins added to CPUs for example
Tb

Rt,c
Rfin
T∞
Rb
R fin  Rt ,c
Thermal Pastes to Reduce Contact Resistance between Heat Sinks and Chips
33
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
5.4 Annular Fins
 Starting from general fin equation:
𝑑 𝑇
1 𝑑𝐴
+
𝑑𝑟
𝐴 𝑑𝑟
𝑑𝑇
1 ℎ 𝑑𝐴
−
𝑑𝑟
𝐴 𝑘 𝑑𝑟
For annular fin:
𝐴 = 2𝜋𝑟𝑡 ,
𝑇−𝑇
=0
𝑑𝐴
= 2𝜋𝑡
𝑑𝑟
𝑑𝐴
𝐴 = 2𝜋 𝑟 − 𝑟 ,
= 4𝜋𝑟
𝑑𝑟
Substituting and simplifying:
𝑑 𝑇
1
𝑑𝑇
1 ℎ 2
+
2𝜋𝑡
−
4𝜋𝑟
𝑑𝑟
2𝜋𝑟𝑡
𝑑𝑟
2𝜋𝑟𝑡 𝑘
𝑑 𝑇 1 𝑑𝑇 2ℎ
+
−
𝑇−𝑇
𝑑𝑟
𝑟 𝑑𝑟 𝑘𝑡
𝑇−𝑇
=0
=0
5.4 Annular Fins
𝑑 𝑇 1 𝑑𝑇 2ℎ
+
−
𝑇−𝑇
𝑑𝑟
𝑟 𝑑𝑟 𝑘𝑡
Substituting
𝜃 =𝑇−𝑇
and
=0
𝑑𝜃 𝑑𝑇
=
𝑑𝑟 𝑑𝑟
And rearranging to standard form:
𝑑 𝜃 1 𝑑𝜃
+
−𝑚 𝜃 =0
𝑑𝑟
𝑟 𝑑𝑟
where: 𝑚 =
2ℎ
𝑘𝑡
Modified Bessel Equation
of zero order!
From applied math can get solution:
𝜃 𝑟 = 𝐶 𝐼 𝑚𝑟 + 𝐶 𝐾 𝑚𝑟
MECH 4406: Heat Transfer
where 𝐼 and 𝐾 are modified, zero-order,
Bessel functions of first and second kind.
Prof. M. Johnson, Carleton University
7
Bessel & Modified Bessel Functions
 Bessel function of First kind of order n:

J n x   x n 
m0
2
 1m x 2 m
m!n  m !
2 mn
 Bessel function of Second kind of order zero:
m 1
2
 x
   1 hm
Y0  x    J 0  x  ln      2 m
2
 
 2
 m 1 2 m!

x 2m 

  Euler constant
h  1
1
1
 ... 
2
m
 Modified Bessel Function of First Kind, order :

x 2 m 
I  x    2 m 
m!m    1
m0 2
 Modified Bessel Function of Second Kind, order :
K  x  

2 sin 
I  x   I  x 
Bessel & Modified Bessel Functions
 Appendix B.4 and B.5 in Bergman et al.
• Also calculate directly in Excel, e.g. BESSELJ(x,n); BESSELI(x,n)!
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Annular Fin Solution
 Consider Case B boundary conditions only:
•𝜃 𝑟
= 𝜃 and adiabatic tip, 𝑑𝜃 ⁄𝑑𝑟|
𝜃
𝐼 𝑚𝑟 𝐾 𝑚𝑟
=
𝜃
𝐼 𝑚𝑟 𝐾 𝑚𝑟
= 0, gives following solution for temperature distribution:
+ 𝐾 𝑚𝑟 𝐼 𝑚𝑟
+ 𝐾 𝑚𝑟 𝐼 𝑚𝑟
where 𝐼 𝑚𝑟 =
𝑑 𝐼 𝑚𝑟
𝑑 𝑚𝑟
and 𝐾 𝑚𝑟 =
𝑑 𝐾 𝑚𝑟
𝑑 𝑚𝑟
are modified 1st-order Bessel functions of 1st and 2nd Kinds
• Applying Fourier’s Law to calculate heat transfer rate from annular fin:
𝑞 = −𝑘𝐴
,
𝑑𝑇
𝑑𝑟
𝑞 = 2𝜋𝑘𝑟 𝑡𝜃 𝑚
= −𝑘 2𝜋𝑟 𝑡
𝑑𝜃
𝑑𝑟
𝐾 𝑚𝑟 𝐼 𝑚𝑟 − 𝐼 𝑚𝑟 𝐾 𝑚𝑟
𝐾 𝑚𝑟 𝐼 𝑚𝑟 + 𝐼 𝑚𝑟 𝐾 𝑚𝑟
• And using fin efficiency and resistance are simply (like previous results) :
𝜂 =
𝑞
ℎ2𝜋 𝑟 − 𝑟 𝜃
=
𝑞
; and since: 𝑅
ℎ𝐴 𝜃
,
=
𝜃
𝑞
𝑅
,
=
1
ℎ𝐴 𝜂
Annular Fin Solution
 Bessel function solution is useful in design calculations and can easily & accurately be implemented in
Excel, Matlab, etc.,
 For hand calculations (depending on your calculator), it may be easier to use charts/tables written in
terms of fin efficiency (see examples in text)
• Especially true for other, more complex fin shapes
etc.
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
5.5 Fin Arrays
 In most engineering situations, multiple fins are employed…
5.5 Fin Arrays
 In most engineering situations, multiple fins are employed…
• Simple to extend what we’ve already covered:
For Fin
 fin 
qf
q f ,ideal
qf 
 fin 

For Array
qf
hA f  b
b
Rf
q with fin
q without fin
MECH 4406: Heat Transfer
 array 
qtot
qtot ,ideal
q fins  Nq f 
 array 

qt
hAt b
b
R fins
q with array
q without array
Prof. M. Johnson, Carleton University
1
5.5 Fin Arrays
Fin Array (top)
 Treat array like a simple 1-D thermal resistance problem
 Separate area of fins and area of “unfins”
𝑇
𝑅
=
1
𝑅𝑢𝑛𝑓𝑖𝑛𝑠
ℎ𝐴
𝑅
𝑅𝑓𝑖𝑛𝑠
𝑇𝑏
𝑅𝑏𝑎𝑠𝑒
𝑅𝑡𝑐
𝜀
=
𝜂
=
𝑇𝑐
𝑇𝑏
𝑇𝑐
Fin Array (side)
𝑅
=𝑅
=𝑅
+𝑅 + 𝑅
∴𝑞
=
=
𝜃
𝑞
=
𝜃
𝑁𝑞
𝑞 with array
𝑞 without array
𝑞
𝑞
=
,
𝑞
ℎ𝐴
𝜃
+𝑅
𝛥𝑇
𝑅
Fin Array Example
 An Aluminum (𝑘=238 W/mK) array of 8 by 8
square pin fins cools a silicon ship with
𝑇 =75°C; ℎ =250 W/m2K; and 𝑇 = 25°C. The
square pin fins are 0.5 mm wide and 8 mm long
and are spaced 1 mm apart.
0.5 mm
1.0 mm
a) What is the rate of heat transfer from the
chip?
b) What is overall 𝜀 and 𝜂 for the fin array?
8mm
(Neglect radiation and thermal contact resistance)
2mm
𝑇𝑏
𝑇𝑐
43
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
Special Cases we consider:
1. One Dimensional
a) Space
i) No energy generation:
ii) With energy generation:
𝜕 𝑇
= 0 or
𝜕𝑥
(shown in Cartesian Coordinates)
1 𝜕𝑇 𝜕𝑇
𝑟
=0
𝑟 𝜕𝑟
𝜕𝑟
𝜕 𝑇 𝑔̇
+ =0
𝜕𝑥
𝑘
iii) 1D conduction with varying
: Extended Surfaces / Fins
b) Time (Lumped heat capacity)
1
Road Map for Conduction Lectures
HDE:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
(shown in Cartesian Coordinates)
Special Cases we consider: (continued)
2. Multi-Dimensional
𝜕𝑇
𝜕 𝑇
=𝑘
𝜕𝑡
𝜕𝑥
a) Space-Time (no energy generation)
𝜌𝐶
b) Two-space (no energy generation):
𝜕 𝑇 𝜕 𝑇
+
=0
𝜕𝑥
𝜕𝑦
2
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Part I: Conduction
Section 6: Transient Conduction
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
The Biot Number
 Consider heat transfer in a plane wall:
𝑞=
𝑇
Δ𝑇
𝑇 −𝑇
𝑇 −𝑇
𝑇 −𝑇
=
=
=
𝐿
1
𝐿
1
𝑅
+
𝑘𝐴 ℎ𝐴
𝑘𝐴
ℎ𝐴
𝑇
𝑇
𝑘
𝐵𝑖 =
𝑇 −𝑇
𝑇 −𝑇
=
𝑞𝑅
𝑞𝑅
=
𝐿⁄𝑘𝐴
ℎ𝐿
=
1⁄ℎ𝐴
𝑘
ℎ
𝑇
𝑇
𝐿
𝑘𝐴
𝑇
1
ℎ𝐴
MECH 4406: Heat Transfer
 Biot number, 𝐵𝑖
• 𝑩𝒊  Ratio of conductive resistance to convective resistance
(named for Jean-Baptiste Biot, ~1850)
• Can also think of the Biot number as a ratio of
temperature drop within the solid wall to
temperature drop between the surface and the fluid
Prof. M. Johnson, Carleton University
2
The Biot Number
Biot number rules of thumb:
 If 𝐵𝑖 < 0.1:
𝑘
• Rconduction << Rconvection
• Forget about conduction and
concentrate on convection!
ℎ
𝑇
𝐵𝑖 << 1
𝐵𝑖 ≈ 1
𝑇
𝑇
𝑇
 If 𝐵𝑖 > 20:
𝑇
• Rconvection << Rconduction
• Forget about convective heat transfer and concentrate
on conduction aspects.
• Set wall temperatures to fluid temperatures.
𝐵𝑖 >> 1
 If 0.1 < 𝐵𝑖 < 20:
• Both convective and conductive heat transfer must be considered
 The Biot number is a powerful engineering criterion for simplifying problems
An Aside – Recall Heat Transfer from “1D” Fins
 In reality, heat transfer is in more than one dimension
𝜕𝑇
𝜕𝑥
• 𝑞⃑ is in both in 𝑟 and 𝑥, or both in 𝑥 and 𝑦
𝑟
 Calculate Biot numbers:
• For pin fin:
𝑥
𝐿
𝐵𝑖 =
• For rectangular fin:
𝑅
𝐵𝑖 =
𝐵𝑖 =
𝜕𝑇
𝜕𝑟
𝐵𝑖 =
𝜕𝑇
𝜕𝑥
 For large 𝑘, small 𝑡 or 𝑅, and small ℎ:
𝑦
• Temperature gradients in fin in 𝑦 and 𝑟 directions are
negligible
• 𝑇 = 𝑇(𝑥) so that fins can be analyzed as a 1D problem!
(Phew!)
𝑦
𝑥
𝐿
𝜕𝑇
𝜕𝑦
𝑡
𝑥
6
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Transient Conduction
 So far, we have only considered heat conduction at steady state
 But many, if not most, problems in heat transfer involve changing or transient
conditions.
 Consider again the Temperature distribution in a plane wall now allowed to
change with time
7
Transient Conduction and the Biot Number
𝑩𝒊  1
𝑩𝒊 >> 1
𝑇(𝑥, 0)= 𝑇
𝑇(𝑥, 0)= 𝑇
h
𝑇
𝑇
-L
Temperature
Distribution
within solid:
L
𝒙
𝑇 = 𝑇(𝑥, 𝑡)
𝑩𝒊 << 1
𝑇(𝑥, 0)= 𝑇
h
h
𝑇
𝑇
-L
L
𝒙
𝑇 = 𝑇(𝑥, 𝑡)
𝑇
𝑇
-L
L
𝒙
𝑇 ≈ 𝑇(𝑡)
 If 𝐵𝑖 <<1 (usually 𝐵𝑖 <0.1), then 𝑇 = 𝑇(𝑡) only within solid!!
• Can treat solid as a single “lump” of material –
“Lumped Heat Capacitance Model”
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
6.2 Lumped Heat Capacitance Approach
 If 𝐵𝑖 <<1 (usually 𝐵𝑖 <0.1) then can ignore temperature variation in solid and write 1st
law for “lump” of material:
1st Law: 𝐸̇ = 𝐸̇
𝒒𝒄𝒐𝒏𝒗
𝑻 = 𝑻(𝒕) only
(no spatial variation)
+ 𝐸̇ − 𝐸̇
𝐸̇ = −𝐸̇
𝑇 ,ℎ
𝜌𝑉𝑐
@t=0, 𝑇 = 𝑇
𝑑𝑇
= −ℎ𝐴 𝑇 − 𝑇
𝑑𝑡
𝜌𝑉𝑐 𝑑𝜃
= −𝜃
ℎ𝐴 𝑑𝑡
Solution:
(assuming h is
independent of time)
𝜌𝑉𝑐
ℎ𝐴
𝑑𝜃
=−
𝜃
𝜌𝑉𝑐 𝜃
ln = 𝑡
ℎ𝐴
𝜃
𝑑𝑡
(where 𝜃 ≡ 𝑇 − 𝑇 ,
, and 𝑉 is volume)
𝜃
𝑇−𝑇
ℎ𝐴
=
= exp −
𝑡
𝜃
𝑇 −𝑇
𝜌𝑉𝑐
(1)
Lumped Heat Capacitance Approach
 Solution to simple transient conduction using basic lumped heat-capacitance model
(𝐵𝑖 <<1):
𝑻 = 𝑻(𝒕) only
(no spatial variation)
@t=0, 𝑇 = 𝑇
𝒒𝒄𝒐𝒏𝒗
𝑇 ,ℎ
𝜃
𝑇−𝑇
ℎ𝐴
−𝑡
=
= exp −
𝑡 = exp
𝜃
𝑇 −𝑇
𝜌𝑉𝑐
𝜏
where tt is
thermal time
constant:
𝜏 =
𝜌𝑉𝑐
1
=
ℎ𝐴
ℎ𝐴
𝑅
𝜌𝑉𝑐
𝐶
𝜏 =𝑅 𝐶
• Equation is analogous to that of an RC circuit!
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Lumped Heat Capacitance Approach – Non-Dimensional Form
𝜃
𝑇−𝑇
ℎ𝐴
=
= exp −
𝑡
𝜃
𝑇 −𝑇
𝜌𝑉𝑐
𝐵𝑖 <<1
(1)
𝑻 = 𝑻(𝒕) only
𝒒𝒄𝒐𝒏𝒗
(no spatial
variation)
𝑇 ,ℎ
 Can rewrite in non-dimensional form by defining:
ℎ𝐴 𝑡 ℎ𝑡 𝐴
=
𝜌𝑉𝑐
𝜌𝑐 𝑉
𝜃
= 𝜃∗
𝜃
=
∴
ℎ𝑡 𝑘
𝜌𝐿∗ 𝑐 𝑘
=
ℎ𝑡
𝜌𝐿∗ 𝑐
𝐵𝑖
where: 𝐿∗ =
ℎ𝐿∗ 𝑘 𝑡
𝐿∗
=
∗
𝑘 𝜌𝑐 𝐿∗
𝐿
ℎ𝐴 𝑡 ℎ𝐿∗ 𝛼𝑡
=
𝜌𝑉𝑐
𝑘 𝐿∗
𝐵𝑖 =
but
= characteristic length
= 𝛼 = thermal diffusivity
ℎ𝐿∗
(Biot Number)
𝑘
𝛼𝑡
𝐿∗
 Thus, in non-dimensional form, (1) becomes:
𝐹𝑜
@t=0, 𝑇 = 𝑇
𝐹𝑜 =
(Fourier Number)
(dimensionless time)
𝜃 ∗ = exp −𝐵𝑖 ⋅ 𝐹𝑜
Lumped Heat Capacitance Approach
 Lumped capacitance solution:
𝜃
𝑇−𝑇
ℎ𝐴
=
= exp −
𝑡
𝜃
𝑇 −𝑇
𝜌𝑉𝑐
in non-dimensional form becomes:
𝜃 ∗ = exp −𝐵𝑖 ⋅ 𝐹𝑜
𝐵𝑖 =
ℎ𝐿∗
(Biot Number)
𝑘
𝛼𝑡
𝐹𝑜 = ∗
𝐿
MECH 4406: Heat Transfer
Note: that in this form, 𝐵𝑖 and 𝐹𝑜
are defined using a characteristic
length, 𝐿∗ , defined as 𝐿∗ = .
 For a cylinder, 𝐿∗ =
 For a sphere, 𝐿∗ =
=
=
=
=
(Fourier Number)
Prof. M. Johnson, Carleton University
6
Lumped Heat Capacitance - Example
 In a metal forming process copper billets (0.5m by 0.1 m by 0.1 m) are annealed
in air at 𝑇 =25°C with a ℎ=100 W/m2K. The following properties for copper are
applicable: 𝜌 = 9000 kg/m3; 𝑘 = 385 W/mK; 𝐶 = 3831 J/kgK.
a) What is the half-life of the temperature decay?
b) If 𝑇 =1500°C, how much energy is transferred from the billet during this initial
period?
13
6.3 Generalized Lumped Capacitance
 Lumped capacitance model with other boundary conditions (not just convection)
𝑞
𝑻 = 𝑻(𝒕) only
(no spatial variation)
𝑬̇𝒈𝒆𝒏
@t=0, 𝑇 = 𝑇
𝒒𝒄𝒐𝒏𝒗
𝑇 ,ℎ
𝑞𝑟𝑎𝑑
1st Law at any instant in time:
𝑞 𝐴 + 𝐸̇
− ℎ 𝑇−𝑇
+ 𝜀𝜎 𝑇 − 𝑇
𝐴 = 𝜌𝐶𝑉
𝑑𝑇
𝑑𝑡
• Non-linear, 1st order, non-homogeneous ODE. Must be solved numerically.
• Two special cases are interesting analytically.
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Generalized Lumped Capacitance
 Note that at least 2 Biot numbers may be necessary to determine applicability of
𝑞
Lumped Capacitance Model:
ℎ𝐿
≲ 0.1 (as before)
𝑘
ℎ 𝐿
=
≲ 0.1
𝑘
• For radiation boundary condition: 𝐵𝑖
𝒒𝒄𝒐𝒏𝒗
𝑻 = 𝑻(𝒕) only
• For convection boundary condition: 𝐵𝑖 =
(no spatial
variation)
𝑇 ,ℎ
𝑬̇𝒈𝒆𝒏
𝑞𝑟𝑎𝑑
@t=0, 𝑇 = 𝑇
 If both convection are radiation are relevant at the same time, then to use lumped
capacitance approach:
𝐵𝑖
– where where 𝑅
=
,
𝑅
≲ 0.1
𝑅
is equivalent thermal resistance for convection and radiation in parallel
Generalized Lumped Capacitance:
2 special cases
1. No generation, convection insignificant relative to radiation, no applied heat flux:
−𝜀𝜎𝐴 𝑇 − 𝑇
= 𝜌𝐶𝑉
• Solve by separation of variables and integration:
𝑡=
𝜌𝑉𝑐
4𝜀𝜎𝐴 𝑇
ln
𝑇
𝑇
+𝑇
𝑇
− ln
−𝑇
𝑇
𝑑𝑇
𝑑𝑡
+𝑇
+ 2 tan
−𝑇
𝑇
𝑇
− tan
𝑇
𝑇
2. No radiation; just generation, convection, and applied heat flux:
𝑞 𝐴 + 𝐸̇
• Let 𝑎 =
ℎ𝐴
and
𝜌𝑐𝑉
𝑏=
− ℎ𝐴 𝑇 − 𝑇
𝑑𝑇
𝑑𝑡
𝑞 𝐴 + 𝐸̇
𝑏
, so equation becomes 𝜃 = 𝜃 − , and solve:
𝜌𝑉𝑐
𝑎
𝑇−𝑇
𝑏 ⁄𝑎
= exp( − 𝑎𝑡) +
𝑇 −𝑇
𝑇 −𝑇
MECH 4406: Heat Transfer
= 𝜌𝐶𝑉
1 − exp( − 𝑎𝑡)
Prof. M. Johnson, Carleton University
8
6.4
Alternate Interpretation of Fourier Number
 Consider case of solid in which both heat conduction and energy storage are taking place
• Conduction:
𝑞 = −𝑘𝐴
∴ 𝑞 ≈ 𝑘𝐿
• Energy storage:
𝑑𝑇
𝑑𝑥
For characteristic length, 𝐿:
𝑑𝑇 Δ𝑇
≈
𝑑𝑥
𝐿
and 𝐴 ≈ 𝐿
𝑑𝑇 Δ𝑇
≈
𝑑𝑡
𝑡
and 𝑉 ≈ 𝐿
Δ𝑇
=≈ 𝑘𝐿Δ𝑇
𝐿
𝐸̇ = 𝜌𝐶𝑉
∴ 𝐸̇ = 𝜌𝐶𝐿
𝑑𝑇
𝑑𝑡
For characteristic length, 𝐿:
Δ𝑇
𝑡
 ∴Fourier number can also be interpreted as a measure of the relative effectiveness of a solid in
conducting and storing thermal energy:
𝑘𝑡
𝛼𝑡
𝑞
𝑘𝐿Δ𝑇
=
≡ 𝐹𝑜
=
≈
̇
𝜌𝐶𝐿
⁄
𝐿
𝜌𝐶𝐿 Δ𝑇 𝑡
𝐸
17
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Part I: Conduction
Section 6.5: Transient Conduction with
Spatial Temperature Variation
(subtitle: What if 𝑩𝒊 > 0.1?)
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
1D Transient Heat Conduction
 If Bi > 0.1 then lumped heat capacitance no longer applies and we must consider spatial
variations in the solid as well
𝑩𝒊  1
 Consider case temperature varies in 1D in space and in time:
T(x,0) = Ti
• For 1D in 𝑥 with constant 𝑘, HDE reduces to:
𝜌𝐶
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
=
𝑘
+
𝑘
+
𝑘
+ 𝑔̇
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
h
T
𝜌𝐶 𝜕𝑇
𝜕 𝜕𝑇
=
𝑘 𝜕𝑡 𝜕𝑥 𝜕𝑥
T
-L
1 𝜕𝑇 𝜕 𝑇
=
𝛼 𝜕𝑡 𝜕𝑥
L
x
• Solution requires 2 boundary conditions plus 1 initial condition
19
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Transient Conduction: 1D space + time
 Consider a plane wall with convection
𝑇
𝑇 ,ℎ
• Relevant form of HDE:
𝑇
1 𝜕𝑇 𝜕 𝑇
=
𝛼 𝜕𝑡 𝜕𝑥
𝑇 ,ℎ
• Boundary conditions:
𝑇 𝑥, 0 = 𝑇
𝑇
@ 𝑥= 𝐿: −𝑘
= ℎ 𝑇(𝐿, 𝑡) − 𝑇
𝑑𝑇
=0
𝑑𝑥
@ 𝑥=0:
𝑇
𝑑𝑇
𝑑𝑥
(convection at surface)
(symmetry about centerline)
• Initial condition:
𝑇 𝑥, 0 = 𝑇
−𝐿
+𝐿
• Problem is complex if no simplification:
x
𝑇 = 𝑇 𝑥, 𝑡, 𝑘, 𝑇 , 𝑇 , 𝐿, 𝛼, ℎ !!
Non-Dimensionalize Variables
 Need to first simplify math by non-dimensionalizing the variables in the problem
• Define non-dimensional variables:
𝜃
𝑇−𝑇
=
𝜃
𝑇 −𝑇
𝜃∗ =
𝑥 ∗ = 𝑥/𝐿
𝑡∗ =
𝛼𝑡
= 𝐹𝑜
𝐿
• Now recast HDE using these variables:
𝜕 𝑇 1 𝜕𝑇
=
𝜕𝑥
𝛼 𝜕𝑡
𝜕 𝜃∗
𝜕𝑥 ∗
=
𝜕𝜃 ∗
𝜕 𝐹𝑜
and update boundary & initial conditions:
@ x=0,
=0
@ x=𝐿, −𝑘
= ℎ 𝑇(𝐿, 𝑡) − 𝑇
@ t=0, T(x,0) = Ti
𝜃 ∗ = 𝑓 𝑥 ∗ , 𝐹𝑜, 𝐵𝑖
MECH 4406: Heat Transfer
∗
@ 𝑥 ∗ =0,
∗
@ 𝑥 ∗ =1,
∗
∗
∗
∗
=0
= −𝐵𝑖𝜃 ∗ 1, 𝑡 ∗
@ 𝑡 ∗ =0, 𝜃 ∗ 𝑥, 0 = 1
(3 independent variables instead of 8 !!)
Prof. M. Johnson, Carleton University
2
Transient Conduction: 1D space + time
 Problem is much simplified (now 3 variables instead of 8!):
𝜕 𝜃∗
𝜕𝑥 ∗
𝜕𝜃 ∗
=
𝜕 𝐹𝑜
@ 𝑥 ∗ =0,
@ 𝑥 ∗ =1,
∗
∗
∗
∗
∗
∗
=0
= −𝐵𝑖𝜃 ∗ 1, 𝑡 ∗
@ 𝑡 ∗ =0, 𝜃 ∗ 𝑥, 0 = 1
 Non-dimensionalized problem has an exact, infinite series solution:
𝜃∗ =
where: 𝐶 =
𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
4 sin 𝜁
2𝜁 + sin 2𝜁
and: 𝜁 tan 𝜁 = 𝐵𝑖
Transient Conduction: 1D space + time
Solution:
(Transient conduction
in plane wall with
convection)
𝜃∗ =
𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
where: 𝐶 =
4 sin 𝜁
2𝜁 + sin 2𝜁
and: 𝜁 tan 𝜁 = 𝐵𝑖
 Good news is that if Fourier number, Fo > 0.2, we can get an accurate solution using
only the 1st term of the infinite series
• For Fo > 0.2:
𝜃 ∗ = 𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
where: 𝐶 =
4 sin 𝜁
2𝜁 + sin 2𝜁
and: 𝜁 tan 𝜁 = 𝐵𝑖
and 𝐶 and 𝜁 are tabulated in Table 5.1 to quickly look these
values up as a function of Biot number (note footnote to table!)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
• Note Table footnote states that in this table:
• 𝐵𝑖 is defined using 𝐿∗ = 𝑟 as per Fig. 5.6
(NOT 𝐿∗ =
= or which we typically use to calculate Bi)
• Thus, when using this table 𝑩𝒊 and 𝑭𝒐 must be defined
using 𝑳∗ = 𝒓𝒐 when solving for a cylinder or sphere
24
Solution Interpretation: Transient Conduction: 1D space + time
 For 𝐹𝑜 > 0.2:
Transient conduction
𝜃 𝑇(𝑥, 𝑡) − 𝑇
𝜃∗ = =
= 𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
in plane wall with
𝜃
𝑇 −𝑇
convection:
Midplane
temperature, 𝜃 ∗,
(1)
where: 𝐶 =
and: 𝜁 tan 𝜁 = 𝐵𝑖
 Look at solution more closely:
• At midplane: 𝑥 ∗ = 0, ∴ cos 𝜁 𝑥 ∗ = 1
• ∴ Midplane temperature: 𝜃 ∗, ≡
 Rewrite (1) above as:
𝜃 ∗ = 𝜃 ∗, cos 𝜁 𝑥 ∗
MECH 4406: Heat Transfer
,
= 𝐶 exp −𝜁 𝐹𝑜
 Time dependence of 𝑇 at any location in wall is
same as at the midplane !
 Time variation is independent of spatial variation!
Prof. M. Johnson, Carleton University
4
Transient Conduction: 1D space + time
 Solution Procedure:
• Calculate Biot number
(verify Lumped Capacitance not valid!)
• If possible verify Fo>0.2
(otherwise assume and check later)
• Calculate 𝜁
(or get value of 𝜁 from Table 5.1, listed as function of 𝐵𝑖)
• Calculate 𝐶
(or use Table 5.1, listed as a function of 𝐵𝑖)
• Calculate 𝜃 ∗ 𝑥 ∗ , 𝑡 ∗ which gives 𝑇 𝑥, 𝑡
• Calculate 𝑄 𝑡 as necessary
Transient Conduction: 1D space + time
 How to calculate amount of heat transferred, 𝑄?
• 1st Law over any arbitrary time interval: 𝐸 − 𝐸
= Δ𝐸
• For wall, 𝑄 = Δ𝐸 = 𝐸 𝑡 − 𝐸 0 :
∴𝑄 =− 𝐸 𝑡 −𝐸 0
= − 𝜌𝑐 𝑇 𝑥, 𝑡 − 𝑇 𝑑𝑉̶
• Defining: 𝑄 = 𝜌𝑐𝑉̶ 𝑇 − 𝑇
• Then:
𝑄
=
𝑄
− 𝑇 𝑥, 𝑡 − 𝑇 𝑑𝑉̶ 1
= ̶
𝑇 −𝑇
𝑉̶
𝑉
• Finally, if Fo > 0.2:
MECH 4406: Heat Transfer
1 − 𝜃 ∗ 𝑑𝑉̶
𝑄
sin 𝜁 ∗
= 1−
𝜃,
𝑄
𝜁
Prof. M. Johnson, Carleton University
5
Transient Heat Conduction Example: Annealing Steel
At a process plant you are trying to anneal some 100mm thick steel plate to make it less
brittle by heating it to a minimum temperature of 550°C before letting it slowly cool. You
plan to first heat the steel in a gas-fired furnace where the products of combustion have a
temperature of 800°C and a convection coefficient of 250 W/m2K. Assume the steel is
initially at 200°C. Assume radiation is negligible and check later.
a) How long should the steel be left in the furnace?
b) What will be the surface temperature of the steel plate at this time?
c) How much energy is transferred per unit are of the plate during this time?
Assume the following properties for the steel plate:
𝜌
=7830 kg/m3, 𝑘
=48 W/mK, 𝐶
= 550 J/kgK
28
What about a half-insulated plane wall?
• For plane wall with convection on both sides:
𝑇 ,ℎ
𝑇 ,ℎ
– Consider solution from −𝐿 to +𝐿
– Wall thickness is 𝟐𝑳
𝜃∗ =
𝑇 𝑥,Insulated
0 =𝑇
Surface
𝑑𝑇
=0
𝑑𝑥
𝑇
𝑇
−𝐿
𝜃 𝑇(𝑥, 𝑡) − 𝑇
=
= 𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
𝜃
𝑇 −𝑇
(Fo>0.2)
• For insulated wall (convection on one side only):
– Same solution applies BUT:
– Consider solution from 0 to +𝐿 only
– Wall thickness is 𝑳
+𝐿
𝑥
0
29
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
What about a half-insulated plane wall?
𝑇 𝑥, 0 = 𝑇
• For plane wall with convection on both sides:
– Consider solution from −𝐿 to +𝐿
– Wall thickness is 𝟐𝑳
𝑑𝑇
=0
𝑑𝑥
𝑇 ,ℎ
𝑇 ,ℎ
𝑇
𝜃∗ =
𝑇
−𝐿
𝜃 𝑇(𝑥, 𝑡) − 𝑇
=
= 𝐶 exp −𝜁 𝐹𝑜 cos 𝜁 𝑥 ∗
𝜃
𝑇 −𝑇
(Fo>0.2)
+𝐿
𝑥
• For insulated wall (convection on one side only):
𝑇 𝑥, 0 = 𝑇
– Same solution applies BUT:
– Consider solution from 0 to +𝐿 only
– Wall thickness is 𝑳
𝑑𝑇
=0
𝑑𝑥
Insulated
𝑇 , ℎSurface
𝑇 ,ℎ
𝑇
𝑇
−𝐿
+𝐿
𝑥
0
30
Transient Conduction: 1D space + time
 Solution for Cylindrical Geometries:
• Infinite cylinder with 𝐹𝑜 > 0.2:
Temperature
Distribution:
𝜃∗ =
𝑇−𝑇
= 𝐶 exp −𝜁 𝐹𝑜 𝐽 𝜁 𝑟 ∗
𝑇 −𝑇
𝜃∗ =
𝑇 , −𝑇
𝑟
, 𝑟∗ =
𝑇 −𝑇
𝑟
𝜽∗𝟎
(center temperature)
Heat
Transfer:
𝑄
2𝜃 ∗
=1−
𝐽 𝜁
𝑄
𝜁
where:
• 𝐽 and 𝐽 are zero & first order Bessel functions of the first kind
•𝐶 =
•𝜁
MECH 4406: Heat Transfer
(or just get from 𝐶 from Table)
= 𝐵𝑖 (or just get from 𝜁 from Table)
Prof. M. Johnson, Carleton University
7
Transient Conduction: 1D space + time
 Solution for Spherical Geometries:
• Sphere with Fo > 0.2:
Temperature
Distribution:
𝜃 ∗ = 𝐶 exp −𝜁 𝐹𝑜
1
sin 𝜁 𝑟 ∗
𝜁 𝑟∗
𝜃∗ =
𝑇 , −𝑇
𝑟
, 𝑟∗ =
𝑇 −𝑇
𝑟
𝜽∗𝟎
(center temperature)
Heat
Transfer:
𝑄
3𝜃 ∗
=1−
sin 𝜁
𝑄
𝜁
− 𝜁 cos 𝜁
where:
•𝐶 =
(or just get from 𝐶 from Table)
• 1 − 𝜁 cot 𝜁 = 𝐵𝑖 (or just get from 𝜁 from Table)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Part I: Conduction
Section 6.9: Semi-Infinite Solid Solution
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Semi-Infinite Solid
 Another interesting (& surprisingly useful) geometry for which analytical solutions can
be obtained
• Useful idealization for many engineering problems.
– e.g.: Have you ever wondered why if you touch something metal vs. something wood, the metal
feels colder?
– e.g.: Buried pipe…
– e.g.: Heating pavement…
∞
 From HDE: 𝜌𝐶
=
𝑘
 1D in 𝑥, with no generation:
+
𝑘
+
𝑘
+ 𝑔̇
1 𝜕𝑇 𝜕 𝑇
=
𝛼 𝜕𝑡 𝜕𝑥
• Initial Condition: 𝑇 𝑥, 0 = 𝑇
• Internal boundary condition: 𝑇 𝑥 → ∞, 𝑡 = 𝑇
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Semi-Infinite Solid Solution
Note: Even for non-infinite cases, if t is small
enough, semi-infinite solid
approximation can still be useful
Semi-Infinite Solid – Case 1 – Constant Surface Temperature
HDE:
1 𝜕𝑇 𝜕 𝑇
=
𝛼 𝜕𝑡 𝜕𝑥
(1)
• Initial Condition: 𝑇 𝑥, 0 = 𝑇
• Boundary conditions: 𝑇 0, 𝑡 = 𝑇 and 𝑇 𝑥 → ∞, 𝑡 = 𝑇
Can solve using a similarity variable, choose: 𝜂 ≡
𝑥
4𝛼𝑡
⁄
First rewrite differential terms of (1) using this similarity variable:
𝜕𝑇 𝜕𝑇 𝜕𝜂 𝜕𝑇
1
=
=
𝜕𝑥 𝜕𝜂 𝜕𝑥 𝜕𝜂 4𝛼𝑡
⁄
𝜕 𝑇
𝑑 𝜕𝑇 𝜕𝜂
𝑑 𝜕𝑇
1
=
=
𝜕𝑥
𝑑𝜂 𝜕𝑥 𝜕𝑥 𝑑𝜂 𝜕𝜂 4𝛼𝑡
𝜕𝑇 𝜕𝑇 𝜕𝜂
−𝑥
=
=
𝜕𝑡 𝜕𝜂 𝜕𝑡 2𝑡 4𝛼𝑡
⁄
⁄
𝜕𝜂
1
=
𝜕𝑥
4𝛼𝑡
⁄
𝜕 𝑇
1
𝜕𝜂 4𝛼𝑡
⁄
=
1 𝜕 𝑇
4𝛼𝑡 𝜕𝜂
𝜕𝑇
𝜕𝜂
Sub. rewritten differential terms into (1):
1
−𝑥
𝛼 2𝑡 4𝛼𝑡
⁄
𝜕𝑇
1 𝜕 𝑇
=
𝜕𝜂 4𝛼𝑡 𝜕𝜂
𝜕 𝑇
−2𝑥 𝜕𝑇
=
𝜕𝜂
4𝛼𝑡 ⁄ 𝜕𝜂
𝑑 𝑇
𝑑𝑇
(2)
= −2𝜂
𝑑𝜂
𝑑𝜂
Have Transformed 2nd
Order PDE into
Linear, 2nd Order ODE!
36
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Semi-Infinite Solid – Case 1 – Constant Surface Temperature
𝑑 𝑇
𝑑𝑇
(2)
= −2𝜂
𝑑𝜂
𝑑𝜂
HDE is now:
where: 𝜂 ≡
𝑥
4𝛼𝑡
⁄
For this to work, need to also transform boundary & initial conditions:
• 1st B.C.: @𝑥 = 0, 𝑇 = 𝑇 or 𝑇 0, 𝑡 = 𝑇
∴ @𝜂 = 0, 𝑇 = 𝑇
@𝑥 = 0, 𝜂 = 0
• 2nd B.C.: @𝑥 → ∞, 𝑇 → 𝑇
@𝑥 → ∞, 𝜂 → ∞
• Initial condition: @𝑡 → 0, 𝑇 → 𝑇
@𝑡 → 0, 𝜂 → ∞
∴ Both 2nd B.C. and I.C. are equivalent, as 𝜂 → ∞, 𝑇 → 𝑇
 Thus, the similarity variable works, and we can solve Eq. (2) by separation of variables using the updated
boundary conditions @𝑥 = 0, 𝜂 = 0 and as 𝜂 → ∞, 𝑇 → 𝑇
37
Semi-Infinite Solid – Case 1 – Constant Surface Temperature
Solving ODE:
𝑑 𝑇
𝑑𝑇
(2)
= −2𝜂
𝑑𝜂
𝑑𝜂
Separate variables:
𝑑 𝑑𝑇 ⁄𝑑𝜂
= −2𝜂𝑑𝜂
𝑑𝑇 ⁄𝑑𝜂
Integrate once:
ln 𝑑𝑇 ⁄𝑑𝜂 = −𝜂 + 𝐶
Rearrange and redefine constant:
Integrate again:
𝑇=𝐶
𝑑𝑇 ⁄𝑑𝜂 = 𝐶 exp −𝜂
exp −𝑢
𝑑𝑢 + 𝐶
Apply 1st B.C., @𝜂 = 0, 𝑇 = 𝑇 :
∴𝐶 =𝑇
Apply 2nd B.C., 𝑇 𝜂 → ∞ = 𝑇 :
∴𝑇 =𝐶
exp −𝜂
𝑑𝜂 + 𝑇
38
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Semi-Infinite Solid – Case 1 – Constant Surface Temperature
Solution so far:
𝑇=𝐶
exp −𝑢
𝑑𝑢 + 𝑇
(3)
𝑇 =𝐶
exp −𝜂
𝑑𝜂 + 𝑇
𝜂≡
𝑥
4𝛼𝑡
⁄
Recognize the Guassian Error Function:
erf 𝑥 =
2
𝜋
erf 𝑥 =
2
𝑥
𝑥
𝑥
𝑥−
+
−
+. . .
3 ⋅ 1! 5 ⋅ 2! 7 ⋅ 3!
𝜋
=
2
𝜋
exp −𝑢 𝑑𝑢
−1 𝑥
2𝑛 + 1 ⋅ 𝑛!
erfc 𝑥 ≡ complementary error function
erf ∞ = 1
erfc 𝑥 = 1 − erf 𝑥
39
Tabulated erf 𝒙
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Semi-Infinite Solid – Case 1 – Constant Surface Temperature
Solution so far:
𝑇=𝐶
exp −𝑢
𝑑𝑢 + 𝑇
(3)
∴𝑇 =𝐶
2
𝜋
Recognize the Guassian Error Function: erf 𝑥 =
∴𝑇=𝐶
𝜋
erf 𝜂 + 𝑇
2
exp −𝜂
𝜂≡
𝑑𝜂 + 𝑇
𝑥
4𝛼𝑡
⁄
exp −𝑢 𝑑𝑢
∴𝑇 =𝐶
1
𝜋
erf ∞ + 𝑇
2
∴𝐶 =
2 𝑇 −𝑇
𝜋
∴ 𝑇 = 𝑇 − 𝑇 erf 𝜂 + 𝑇
𝑞 = −𝑘
∴
𝑇−𝑇
= erf
𝑇 −𝑇
𝜕𝑇
= −𝑘 𝑇 − 𝑇
𝑑 erf 𝜂 𝑑𝜂
Use Fourier’s
𝜕𝑥
𝑑𝜂 𝑑𝑥
Law to get heat
𝑞 = −𝑘 𝑇 − 𝑇 2⁄ 𝜋 exp −𝜂
4𝛼𝑡
transfer rate:
𝑥
4𝛼𝑡
𝑞 =
41
⁄
𝑘 𝑇 −𝑇
𝜋𝛼𝑡
Transient Heat Conduction Solutions for a Semi-Infinite Solid
Case 1:
Constant Surface Temperature
𝑇 0, 𝑡 = 𝑇
∴
Case 2:
Constant Surface Heat Flux
𝑞 = 𝑞 = constant
𝑇 𝑥, 𝑡 − 𝑇 =
Case 3:
Convection at Surface
−𝑘
= ℎ 𝑇 − 𝑇 0, 𝑡
𝑇 𝑥, 𝑡 − 𝑇
𝑥
ℎ𝑥 ℎ 𝛼𝑡
= erfc
− exp
+
𝑇 −𝑇
𝑘
𝑘
2 𝛼𝑡
MECH 4406: Heat Transfer
𝑇 𝑥, 𝑡 − 𝑇
= erf
𝑇 −𝑇
𝑥
𝑞 𝑡 =
4𝛼𝑡
2𝑞 𝛼𝑡 ⁄𝜋
𝑘
⁄
exp
𝑘 𝑇 −𝑇
𝜋𝛼𝑡
−𝑥
𝑞
𝑥
− erfc
4𝛼𝑡
𝑘
2 𝛼𝑡
erfc
𝑞 =𝑞
𝑥
2 𝛼𝑡
+
ℎ 𝛼𝑡
𝑘
Prof. M. Johnson, Carleton University
5
Semi-Infinite Solid Example 1: Buried Pipe
 How long does it take for a buried water main to reach freezing temperature?
• Assume pipe depth=1.5m (4.9ft), 𝛼
= 0.25 × 10 m2/s, 𝑘=2.6 W/mK,
𝑇 = 8°C. Assume winter conditions with ℎ=10 W/m2K and 𝑇 = −20°C.
43
Semi-Infinite Solid Example 2: Two Solids in Contact
 Why does metal feel colder than wood?
44
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Example: Two solids placed in contact
 1st law surface energy balance gives:
𝑞
,
=𝑞
,
 If negligible contact resistance, single
temperature, 𝑇 at interface
• 𝑇 remains constant
 Using Case 1 solution:
𝑞
,
=
−𝑘 𝑇 − 𝑇
𝜋𝛼 𝑡 /
∴𝑇 =
𝑘
,
=
𝑘
𝛼 𝑇 , +𝑘
𝑘 𝛼 +𝑘
MECH 4406: Heat Transfer
𝑇 −𝑇 ,
𝜋𝛼 𝑡 /
𝛼 𝑇
𝛼
=𝑞
,
,
Prof. M. Johnson, Carleton University
7
Part I: Conduction
Section 7: Steady-State Conduction in
More than One Dimension
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Canada Research Professor
Director of Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Multi-dimensional Heat Conduction
 Although many engineering problems may be approximated to be onedimensional, in general heat conduction is a 2 or 3D phenomenon
 Approaches to solving 2D problems:
• Analytical
– Used for special cases only
– Not terribly useful
• Shape factors
– Used for special cases only
– Some very useful
• Numerical
– Most everything else
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
1. Analytical Solutions to 2D Steady Conduction
 HDE in cartesian coordinates:
• For 2D, steady-state:
1 T  2T  2T  2T



 t x 2 y 2 z 2
 2T  2T

0
x 2 y 2
(Laplace Equation)
• Solve using separation of variables:
– Let 𝑇 𝑥, 𝑦 = 𝑋 𝑥 𝑌 𝑦
– Get 2 equations for “separation constant”
(eigenvalue / eigenfunction problem)
– Apply boundary conditions to solve for coefficients in eigenfunctions
– Get some sort of infinite series solution
– Mostly an exercise in applied math!
2. Shape Factors for 2D (& 3D) Problems
 Heat flux is a vector quantity
whose direction is normal to
isotherms (lines of constant
temperature)
 “Heat flow lines” indicate
direction of heat flux
(normal to isotherms)
𝑞⃑ = 𝑞 𝚤̂ + 𝑞 𝚥̂
𝑞
𝑞
𝑦
𝑥
isotherm
𝑇
𝑇
𝑇 >𝑇 >𝑇
𝑇
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
2. Shape Factors for 2D (& 3D) Problems
 Shape factors provide a method of calculating heat transfer in multidimensional
problems by:
𝑞
,
= 𝑆𝑘Δ𝑇 = 𝑆𝑘 𝑇 − 𝑇
• 𝑆 is the “shape factor” (See Table 4.1)
• 𝑞 , indicates heat transfer from
boundary(ies) at 𝑇 to boundary(ies) at 𝑇
• Restrictions:
– Problem must be described in terms of 2 temperatures
– Uniform material properties
– Must be of a particular common geometry
Tabulated
Shape
Factors
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Tabulated
Shape
Factors
Tabulated
Shape
Factors
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Tabulated
Shape
Factors
Shape Factors Example
 Calculate heat transfer from a box where inside
wall temperature is T1, outside wall temperature is T2,
wall thickness is L, and inside dimension is 5L.
Solve as:
5L
a) 1D using inside area
b) 1D using outside area
c) 3D using shape factors
L
 Solution:
a)
T
2 T
 65 L  k
L
L
q  150 LkT
q  6 Ai k
MECH 4406: Heat Transfer
b)
T
2 T
 67 L  k
L
L
q  294 LkT
q  6 Ao k
Prof. M. Johnson, Carleton University
5
Shape Factors Example
c) 3D using shape factors
 Solution:
q  6q faces  12qedges  8qcorners
5L
= 150𝐿𝑘Δ𝑇 + 12𝑆 𝑘Δ𝑇 + 8𝑆 𝑘Δ𝑇
= 150𝐿𝑘Δ𝑇 + 12 0.54 5𝐿 𝑘Δ𝑇 + 8 0.15 𝐿𝑘Δ𝑇
𝑞 = 183.2𝐿𝑘Δ𝑇
L
From Table 4.1:
Sedge = 𝟎. 𝟓𝟒 𝟓𝑳 (case 8)
Scorner = 𝟎. 𝟏𝟓 𝑳 (case 9)
 Compare Results:
(c) 3D using shape factors: 𝑞 = 183.2𝐿𝑘Δ𝑇
(a) 1D using inside area:
𝑞 = 150𝐿𝑘Δ𝑇
(18% low)
(b) 1D using outside area:
𝑞 = 294𝐿𝑘Δ𝑇
(60% high)
Thermal Resistance with Shape Factors
 Can extend thermal resistance concept for use with shape factors in 2D steady
heat transfer problems
• If:
𝑞
• Then:
𝑅
,
= 𝑆𝑘 𝑇 − 𝑇
(
)
=
= 𝑆𝑘Δ𝑇
1
𝑆𝑘
12
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Part II: Convection
9. Intro / Convection in External Flows
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Scientific Director, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Introduction to Convection
 So far, while concentrating on the details of conduction, we have only considered
convection as a boundary condition.
 What do we know about convection?
𝑞
• Mode of heat transfer driven by bulk fluid motion
(transfer from solid surface to fluid)
• Governed by Newton’s law of Cooling: 𝑞 = ℎ𝐴 𝑇 − 𝑇
T
Ts
 What about ℎ? Show me the “ℎ”!
• No tables for ℎ as ℎ = ℎ(𝜌, 𝑢, 𝑇, 𝑘, geometry, surface roughness, …)
 How do we calculate / determine values of ℎ?
• Theoretical (limited cases only)
• Experimental / Semi-Empirical (based insight from limited theoretical cases)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Flow over an arbitrary surface
 Boundary Layer (BL) basics:
• More than one boundary layer
– Velocity (Momentum) BL
– Thermal BL
– Concentration BL (if mass transfer is relevant)
T
δt
u
δ
n T
s
• Velocity B.L. thickness (δ):
typically defined where u=0.99 u
• Thermal B.L. thickness, δt:
typically defined where T=0.99T
Flow over an arbitrary surface – Velocity Boundary Layer
 Velocity boundary layer:
• Zone where viscous effects dominate
– Outside the velocity BL there are limited effects
of viscosity, i.e. Bernoulli equation applies.
– Shear stress at the surface, 𝜏 , is a manifestation of
the velocity boundary layer:
𝜏 =𝜇
u
δ
n
𝜕𝑢
𝜕𝑦
– Relates to the local skin friction coefficient, 𝐶 :
𝜏
𝐶 =
𝜌𝑢
• Boundary layer typically starts as laminar toward the leading edge of a surface and
transitions to turbulence further along the surface
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Flow over an arbitrary surface – Thermal Boundary Layer
 1st Law Energy Balance for
zero-velocity fluid on surface:
𝑞
,
=𝑞
𝑞
∴ −𝑘
= −𝑘
,
𝑞
T
,
δ
δt
n T
s
=ℎ 𝑇 −𝑇
,
𝜕𝑇
𝜕𝑛
=ℎ 𝑇 −𝑇
𝜕𝑇
𝜕𝑛
𝑇 −𝑇
(n is coordinate direction normal to the surface/wall)
This defines the local
convective heat transfer
coefficient
−𝑘
∴ℎ=
𝜕𝑇
𝜕𝑛
u
Convective heat transfer coefficient
∴ To find local ℎ, we need the temperature gradient at surface:
𝜕𝑇
𝜕𝑛
𝑇 −𝑇
−𝑘
∴ℎ=
• Requires solutions to governing equations within the boundary layer
 To calculate total convective heat transfer over an entire surface, integrate local ℎ :
𝑞=
ℎ 𝑇 −𝑇
𝑑𝐴 = 𝑇 − 𝑇
ℎ𝑑𝐴
 Can define average convective heat transfer coefficient: ℎ =
∴𝑞 = 𝑇 −𝑇
∫ ℎ𝑑𝐴
ℎ𝑑𝐴 = ℎ𝐴 𝑇 − 𝑇
∴ 𝑞 = ℎ𝐴 𝑇 − 𝑇
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Convection – Finding 𝒉
 Bottom line:
• To find ℎ, we need temperature gradient,
• To determine
, we will need to solve governing equations within the
boundary layer
 Analysis of convection heat transfer is as much a fluids problem (velocity field) as
it is a heat transfer problem (temperature field)
• Although most problems are complex, the basic physics are the same as that of the
simplest case of the heated flat plate
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Part II: Convection
10. Governing Equations for Boundary Layer Flows /
Flow over a Heated Flat Plate & Other EXTERNAL flows
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Scientific Director, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
10. Flow Over a Heated Flat Plate
 Reynolds Number for flat plate: Re =
• Re < 5 × 105: Laminar
• Re > 3 × 106: Fully Turbulent
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Governing Equations: 1. Mass
 Conservation of Mass (continuity):
v 
• At steady state (shown here in 2D):
𝑚̇
=
𝑚̇
𝜌𝑢𝑑𝑦 + 𝜌𝑣𝑑𝑥 = 𝜌𝑢 +
0=

v dy
y
u 
u
𝜕
𝜕
𝜌𝑢 𝑑𝑥 𝑑𝑦 + 𝜌𝑣 +
𝜌𝑣 𝑑𝑦 𝑑𝑥
𝜕𝑥
𝜕𝑦
dx
dy

u dx
x
v
𝜕
𝜕
𝜌𝑢 𝑑𝑥𝑑𝑦 +
𝜌𝑣 𝑑𝑦𝑑𝑥
𝜕𝑥
𝜕𝑦
divide through by 𝑑𝑦𝑑𝑥
𝜕 𝜌𝑢
𝜕 𝜌𝑣
+
=0
𝜕𝑥
𝜕𝑦
If incompressible
(𝜌=constant) then:
𝜕𝑢 𝜕𝑣
+
=0
𝜕𝑥 𝜕𝑦
Gov. Eqs.: 2. Momentum
 Newton’s 2nd Law: Sum of all forces on a CV must equal net rate momentum leaves the
CV (outflow – inflow)
 Two kinds of forces may act on fluid:
• Surface forces (proportional to area)
– Due to fluid static pressure as well as viscous stresses (normal + shear)
• Body forces (proportional to volume)
– eg. Gravity, Magnetism and/or electric fields, centrifugal forces
 Surface Forces:
• Summing all forces in diagram & cancelling
terms gives net forces in x and y:
 
p  yx 
Fs , x   xx 

dxdy

x
x y 

 
p  xy 
Fs , y   yy 

 dxdy
y x 
 y
MECH 4406: Heat Transfer
 yy 
p
p
 xx

 yy dy
y
p
dy
y
 yx 
dy
 xy
dx
 yx

 yx dy
y

 xy   xy dx
x

 xx   xx dx
x
p
p
dx
x
p
 yy
Prof. M. Johnson, Carleton University
2
Gov. Eqs.: 2. Momentum
 Momentum Fluxes (assuming steady flow):
(only x-direction shown)
• Summing & simplifying terms for x and y:
𝜌𝑣 𝑢 +
𝜕 𝜌𝑢 𝑢
𝜕 𝜌𝑣 𝑢
𝑑𝑥𝑑𝑦 +
𝑑𝑦𝑑𝑥
𝜕𝑥
𝜕𝑦
𝜕
𝜌𝑣 𝑢 𝑑𝑥
𝜕𝑦
dy
𝜌𝑢 𝑢
𝜌𝑢 𝑢 +
 Forces & Momentum together:
𝜕
𝜌𝑢 𝑢 𝑑𝑦
𝜕𝑥
dx
𝜕 𝜌𝑢 𝑢
𝜕 𝜌𝑣 𝑢
𝜕𝜎
𝜕𝑝 𝜕𝜏
+
=
−
+
+𝑋
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜌𝑣 𝑢
where 𝑋 represents the sum of all body forces in the 𝑥 direction
If  is constant, expanding derivative and substituting in continuity:
𝜌 𝑢
𝜕𝑢
𝜕𝑢
𝜕
+𝑣
=
𝜎
𝜕𝑥
𝜕𝑦
𝜕𝑥
−𝑝 +
𝜕𝜏
+𝑋
𝜕𝑦
Can be shown that for a Newtonian fluid, stresses depend on velocity gradients:
𝜎
= 2𝜇
𝜕𝑢 2 𝜕𝑢 𝜕𝑣
− 𝜇
+
𝜕𝑥 3
𝜕𝑥 𝜕𝑦
𝜎
= 2𝜇
𝜕𝑣 2 𝜕𝑢 𝜕𝑣
− 𝜇
+
𝜕𝑥 3 𝜕𝑥 𝜕𝑦
𝜏
=𝜏
=𝜇
𝜕𝑢 𝜕𝑣
+
𝜕𝑦 𝜕𝑥
Gov. Eqs.: 2. Momentum
 Momentum Equation:
• Shown for 2D steady flow of an incompressible fluid with constant viscosity:
𝜌 𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑝
𝜕 𝑢 𝜕 𝑢
+𝑣
=−
+𝜇
+
+𝑋
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜌 𝑢
𝜕𝑣
𝜕𝑣
𝜕𝑝
𝜕 𝑣 𝜕 𝑣
+𝑣
=−
+𝜇
+
+𝑌
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑥
𝜕𝑦
where 𝑋 and 𝑌 represent the sum of any body forces in the 𝑥 and 𝑦 directions
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Gov. Eqs.: 3. Energy
 Consider all relevant physical processes acting on a control volume of viscous
fluid:
•
•
•
•
Thermal internal energy (e)
Kinetic energy (V2/2 where V2=u2+v2+w2)
Surface forces (e.g. pressure)
Body forces (e.g. gravity)
 Transport of energy:
• Energy may be advected or transported with the bulk motion of the fluid
• Energy may also be transported by molecular processes (conduction, diffusion)
• Energy may be transferred through work interactions of body and surface forces
Gov. Eqs.: 3. Energy
1. Transport by advection of internal thermal and kinetic energy:
𝐸̇
− 𝐸̇
,
≡ 𝜌𝑢 𝑒 +
,
=−
𝑉
2
𝑑𝑦 − 𝜌𝑢 𝑒 +
𝜕
𝑉
𝜌𝑢 𝑒 +
𝜕𝑥
2
𝑉
2
+
𝜕
𝑉
𝜌𝑢 𝑒 +
𝜕𝑥
2
𝑑𝑥𝑑𝑦
𝑑𝑥 𝑑𝑦
W
E cond, y dy
E adv, y  dy
dy
2. Transport by conduction:
𝐸̇
,
− 𝐸̇
,
≡− 𝑘
=−
𝜕𝑇
𝜕𝑇 𝜕
𝜕𝑇
𝑑𝑦 − −𝑘
−
𝑘
𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑥 𝜕𝑥
𝜕𝑥
𝜕
𝜕𝑇
𝑘
𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑥
3. Work of body and surface forces
𝑊̇
,
E conduction, x
E gen
E advection, x
E cond, x  dx
E adv, x dx
dx
E conduction, y E advection, y
y
x
𝜕
𝜕
= 𝑋𝑢 𝑑𝑥𝑑𝑦 +
𝜎 − 𝑝 𝑢 𝑑𝑥𝑑𝑦 +
𝜏 𝑢 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑥
work done by work done by pressure and viscous forces
body force
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Gov. Eqs.: 3. Energy
 Put all three parts together along with analogous forms in 𝑦-direction, simplify
by dividing out 𝑥 and 𝑦 momentum equations and after considerable algebra
and assuming incompressible fluid and constant 𝑘:
 Thermal Energy Equation for incompressible flow:
𝜌𝑐
𝑢
𝜕𝑇
𝜕𝑇
𝜕 𝑇 𝜕 𝑇
+𝜈
=𝑘
+
+ 𝜇Φ + 𝑔̇
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
where: 𝜇Φ = 𝜇
𝜕𝑢 𝜕𝑣
+
𝜕𝑦 𝜕𝑥
+2
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
(Viscous dissipation term)
Gov. Eq. Summary: How to calculate h?
Shown in simplified form for 2D, incompressible flow:
 Mass:
𝜕𝑢 𝜕𝑣
+
=0
𝜕𝑥 𝜕𝑦
 Momentum:
x-direction:
𝜌 𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑝
𝜕 𝑢 𝜕 𝑢
+𝑣
=−
+𝜇
+
+𝑋
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑥
𝜕𝑦
y-direction:
𝜌 𝑢
𝜕𝑣
𝜕𝑣
𝜕𝑝
𝜕 𝑣 𝜕 𝑣
+𝑣
=−
+𝜇
+
+𝑌
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑥
𝜕𝑦
 Energy:
𝜌𝑐
𝑢
𝜕𝑇
𝜕𝑇
𝜕 𝑇 𝜕 𝑇
+𝜈
=𝑘
+
+ 𝜇Φ + 𝑔̇
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
where:
Solve simultaneously to get
MECH 4406: Heat Transfer
𝜇Φ = 𝜇
𝜕𝑢 𝜕𝑣
+
𝜕𝑦 𝜕𝑥
+2
𝜕𝑢
𝜕𝑥
and hence, ℎ = − 𝑘
+
𝜕𝑣
𝜕𝑦
𝑇 −𝑇
Prof. M. Johnson, Carleton University
5
How to calculate h?
 General solution to governing equations too difficult to solve analytically
 Consider simplest possible problem to see what we might learn
• Laminar flow
over a heated flat plate
T
u
δt
δ
y
Ts
x
 Simplify governing equations using assumptions appropriate for boundary layer
flow in this case
• Can now get a useful analytic solution that helps us approach more complex
problems
Laminar Boundary Layer Solution for a Heated Flat Plate
Need to calculate:
 Trying to solve: ℎ =
 Laminar Boundary Layer on Flat Plate:
𝑇
δt
Re =
Boundary conditions:
@ 𝑦=0, 𝑢=0 (no slip)
@ 𝑦=0, 𝑣=0 (solid boundary)
@ 𝑦=∞, 𝑢=𝑢
@ 𝑥=0, 𝑢=𝑢
𝑢
𝑢 𝑥
𝜐
δ
𝑦
𝑇
𝑥
Re < 5 x 105: Laminar
19
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Boundary Layer Approximations
 Key simplifications that enable a solution to the flow in the laminar Boundary Layer
 Assume:
T
u
• Steady flow
δt
• Incompressible
• Boundary layer approximations valid:
𝑢 >> 𝑣
𝜕𝑝
≈0
𝜕𝑦
𝜕𝑇
𝜕𝑇
>>
𝜕𝑦
𝜕𝑥
𝜕𝑢
𝜕𝑢 𝜕𝑣 𝜕𝑣
>>
, ,
𝜕𝑦
𝜕𝑥 𝜕𝑦 𝜕𝑥
𝜕𝑝
=0
𝜕𝑥
𝜇Φ = 0
Re 
U x

δ
y
Ts
x
Re < 5 x 105: Laminar
𝜌 𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑝
𝜕 𝑢 𝜕 𝑢
+𝑣
=−
+𝜇
+
+𝑋
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜌𝑐
𝑢
𝜕𝑇
𝜕𝑇
𝜕 𝑇 𝜕 𝑇
+𝜈
=𝑘
+
+ 𝜇Φ + 𝑔̇
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
20
Simplified Gov. Eq. for Laminar Boundary Layer
“Boundary Layer Equations”:
 Mass:
𝜕𝑢 𝜕𝑣
+
=0
𝜕𝑥 𝜕𝑦
 Momentum:
𝜌 𝑢
 Energy:
𝑢
𝜕𝑢
𝜕𝑢
𝜕 𝑢
+𝑣
=𝜇
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑇
𝜕𝑇
𝜕 𝑇
+𝑣
=𝛼
𝜕𝑥
𝜕𝑦
𝜕𝑦
 Much simpler problem that can be solved analytically with a series expansion solution
and/or with a bit of numerical integration
• Solved by Blasius (1908) – Graduate student of Prandtl
– See Section 7.2.1 in text
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Blasius Solution to Laminar B.L. (1908)
 Similarity solution -- mathematically reduce PDE to ODE
 Physically represents fact that velocity profile as a function of y/ is similar for all values of x.
𝑈
𝑈
𝑢 𝑥, 𝑦
𝑢 𝑥, 𝑦
𝑦
𝛿
𝑦
𝛿
• If
=
, then
=
or 𝑢 = 𝑢 (𝑈 is constant)
 Formally, this permits a similarity parameter to be used:
•
= 𝑓 𝜂 where 𝜂 = 𝑦
f 
(similarity parameter)
Blasius Solution to Laminar B.L. (1908)
df
0
d
 Substitute similarity parameter 𝜂 = 𝑦
into Boundary Layer Equations:
• Original PDE Boundary layer equations:
𝜕𝑢 𝜕𝑣
+
=0
𝜕𝑥 𝜕𝑦
and
• Become a simpler ODE:
𝑑 𝑓 1 𝑑 𝑓
+ 𝑓
=0
𝑑𝜂
2 𝑑𝜂
𝑢
𝜕𝑢
𝜕𝑢
𝜕 𝑢
+𝑣
=𝜐
𝜕𝑥
𝜕𝑦
𝜕𝑦
with transformed boundary conditions:
@  = 0, 𝑓 =
@  = ,
𝑑𝑓
=0
𝑑𝜂
𝑑𝑓
=1
𝑑𝜂
 Can now be solved with a simple numerical technique (i.e. Runge-Kutta)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Laminar Boundary Layer Solution for a Heated Flat Plate
 Blasius (1908) solution:
• 𝛿=
 x
5𝑥

Re
x
• Local shear stress at wall, 𝜏 :
𝜏 =𝜇
𝑑𝑢
𝑑𝑦
𝜌𝜇𝑢
𝑥
= 0.332𝑢
• Local friction coefficient, 𝐶 , :
𝐶
,
≡
= 0.664 Re
⁄
Laminar Boundary Layer Solution for a Heated Flat Plate
 Prandtl Number, Pr :
• Pr ≡ =
Momentum Diffusivity (a.k.a. kinimatic viscosity)
Thermal diffusivity
• Ratio of diffusivity of momentum and thermal energy
• Pr is proportional to 𝛿 ⁄𝛿
 For laminar boundary layer over a flat plate (assuming Pr > 0.6):
≈ Pr
MECH 4406: Heat Transfer
/
Fluid
Pr
t 
Air
Water
Engine oil
Liquid metal
0.7
5
3000
0.005
1.10
0.57
0.07
5.7
Prof. M. Johnson, Carleton University
9
Laminar Boundary Layer Solution for a Heated Flat Plate
 Solving simplified B.L. equations:
𝜕𝑇
𝜕𝑦
= 0.332 Pr
⁄
𝑇 −𝑇
𝑢
𝑣𝑥
 Thus, for laminar flow over a flat plate:
• ℎ=
⁄
• 𝛿 ⁄𝛿 ≈ Pr
|
/
= 0.332𝑘 Pr
/
/
(solution assuming fixed
plate temperature)
𝛿 = 5𝑥⁄ Re
• Or...
Laminar Boundary Layer Solution for a Heated Flat Plate
 For flat plate (laminar flow) with fixed plate temperature boundary condition:
Nu ≡
ℎ𝑥
= 0.332Pr
𝑘
/
Re
⁄
(valid for Pr >0.6)
 For flat plate (laminar flow) with fixed uniform heat flux boundary condition:
Nu ≡
ℎ𝑥
= 0.453Pr
𝑘
/
Re
/
(valid for Pr >0.6)
 Nu  Nusselt Number, Nu ≡
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Laminar Boundary Layer Solution for a Heated Flat Plate
 How to calculate average h over plate?
∫ ℎ𝑑𝑥
ℎ=
∫ 𝑑𝑥
=
=
𝑘0.332Pr
𝐿
1
𝐿
/
ℎ𝑑𝑥 =
𝑢
𝜐
1
𝐿
⁄
Nu
1
𝑥
/
𝑘
1
𝑑𝑥 =
𝑥
𝐿
𝑑𝑥 = 2
𝑘0.332 Pr
𝑘0.332Pr
𝐿
/
/
𝑢 𝐿
𝜐
𝑢
𝜐𝑥
/
/
𝑘
= 2 𝑁𝑢
𝐿
𝑑𝑥
ℎ = 2ℎ
∴ Nu =
ℎ𝐿 2ℎ 𝐿
=
= 2Nu
𝑘
𝑘
= 2Nu
• Valid for laminar boundary layers
What T to use? Film Temperature
 When
≫ 1 or
≪ 1, what temperature should we use to evaluate fluid properties??
(It matters!!)
 By convention, use “Film Temperature” or “Mean Film Temperature”: 𝑇
• In problems where 𝑇
can’t be evaluated from the start (i.e. if 𝑇 or 𝑇 not specified), must iterate!
1.
Guess 𝑇
2.
3.
4.
Evaluate property data;
Solve problem;
Calculate new 𝑇
and compare with previous guess;
5.
Repeat until 𝑇
MECH 4406: Heat Transfer
=
;
and solution data converges!
Prof. M. Johnson, Carleton University
11
Example: Laminar flow over a heated flat plate
Consider a horizontal flat plate held at Ts=400K, in an air stream at T=300K moving past at 1 m/s
 Find:
a)
δ at x=0.1 m
T, u
b) δt at x=0.1 m
c)
air
Nu and h at x=0.1 m

t
d) Average h from 0 < x ≤ 0.1 m
e)
Drag force on 0.1 m length of plate
(Will solve separately later on)
30
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
12
Example: Laminar flow over a heated flat plate
Consider a horizontal flat plate held at Ts=400K, in an air stream at T=300K moving past at 1 m/s
 Find:
a)
δ at x=0.1 m
T, u
b) δt at x=0.1 m
c)
air
Nu and h at x=0.1 m
d
dt
d) Average h from 0 < x ≤ 0.1 m
e)
Drag force on 0.1 m length of plate
(Will solve separately later on)
32
Solutions for More Complex Geometries
Can gain insight into more complex problems by non-dimensionalizing governing eqns.
 Non-dimensionalize variables:
𝑥∗ =
𝑥
𝐿
𝑦∗ =
𝑦
𝐿
𝑢∗ =
𝑢
𝑉
𝑣∗ =
𝑣
𝑉
𝑝∗ =
𝑝
𝜌𝑉
𝑇∗ =
𝑇−𝑇
𝑇 −𝑇
 Non-dimensionalize momentum equation:
𝑢∗
𝜕𝑢∗
𝜕𝑢 ∗
𝑑𝑝∗
1
∗
+
𝑣
=
−
+
𝜕𝑥 ∗
𝜕𝑦 ∗
𝑑𝑥 ∗ Re
𝜕 𝑢∗
𝜕𝑦 ∗
• Solution will have the form: 𝑢∗ = 𝑓 𝑥 ∗ , 𝑦 ∗ , Re ,
∗
∗
Great simplification as we do not
have to separately consider effects
like m, cp, k, , etc.
– Note 𝑑𝑝 ∗ ⁄𝑑𝑥 ∗ represents influence of surface geometry since 𝑝 ∗ 𝑥 ∗ is a function of surface geometry and
may be obtained independently by considering flow conditions in the free stream.
– So for a given geometry: 𝒖∗ = 𝒇 𝒙∗ , 𝒚∗ , 𝑹𝒆𝑳
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Non-dimensionalized Momentum Equation (cont.)
 Shear stress / friction coefficient solution also has non-dimensionalized form:
𝜏 =𝜇
𝐶 =
𝜕𝑢 ∗
𝜕𝑦 ∗
=
∗
𝜇𝑈
𝜕𝑢∗
𝐿
𝜕𝑦 ∗
𝜏
2 𝜕𝑢∗
=
𝜌𝑈 /2 𝑅𝑒 𝜕𝑦 ∗
∗
𝜕𝑢∗
𝜕𝑦 ∗
∗
= 𝑓 𝑥 ∗ , 𝑅𝑒 ,
∗
𝑑𝑝 ∗
𝑑𝑥 ∗
• So for a specific geometery:
𝐶 =
2
𝑓 𝑥 ∗ , 𝑅𝑒
𝑅𝑒
34
Non-Dimensionalized Energy Equation
 Non-dimensionalize variables:
𝑥∗ =
𝑥
𝐿
𝑦∗ =
𝑦
𝐿
𝑢∗ =
𝑢
𝑉
𝑣∗ =
𝑣
𝑉
𝑝∗ =
𝑝
𝜌𝑉
𝑇∗ =
𝑇−𝑇
𝑇 −𝑇
 Non-dimensionalize energy equation:
𝑢∗
𝜕𝑇 ∗
𝜕𝑇 ∗
1 𝜕 𝑇∗
+ 𝑣∗ ∗ =
∗
𝜕𝑥
𝜕𝑦
PrRe 𝜕𝑦 ∗
• Solution will have the form: 𝑇 ∗ = 𝑓 𝑥 ∗ , 𝑦 ∗ , Re , Pr,
∗
∗
Great simplification as we do not
have to separately consider effects
like m, cp, k, , etc.
– Note 𝑑𝑝 ∗ ⁄𝑑𝑥 ∗ represents influence of surface geometry since 𝑝 ∗ 𝑥 ∗ is a function of surface geometry and
may be obtained independently by considering flow conditions in the free stream.
– So for a given geometry: 𝑻∗ = 𝒇 𝒙∗ , 𝒚∗ , 𝑹𝒆𝑳 , 𝑷𝒓
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
10.5 So what is the Nusselt number really?
 Recall for boundary layer at surface:
𝑞
𝜕𝑇
−𝑘
𝜕𝑦
=𝑞
−𝑘
=ℎ 𝑇 −𝑇
∴ℎ=
𝜕𝑇
𝜕𝑦
𝑇 −𝑇
 In terms of non-dimensional variables:
𝑇∗ =
𝑇−𝑇
𝑇 −𝑇
−𝑘
∴ℎ=

∴ 𝑁𝑢 ≡
and
𝜕𝑇
𝜕𝑦
−𝑘 𝑇 − 𝑇
=
𝑇 −𝑇
ℎ𝐿 𝜕𝑇 ∗
=
𝑘
𝜕𝑦 ∗
𝜕𝑇 ∗
𝐿
𝜕𝑇
=
∗
𝜕𝑦
𝑇 − 𝑇 𝜕𝑦
𝜕𝑇 ∗
𝜕𝑦 ∗
∗
𝐿 𝑇 −𝑇
=+
𝜕𝑇
𝑇 − 𝑇 𝜕𝑇 ∗
=
𝜕𝑦
𝐿
𝜕𝑦 ∗
and
𝑘 𝜕𝑇 ∗
𝐿 𝜕𝑦 ∗
∗
Nusselt number is non-dimensional
temperature gradient at the surface!!
∗
Non-dimensionalized Solutions
 Thus for a prescribed geometry:
𝑁𝑢 = 𝑓 𝑥 ∗ , Re , Pr
 Average Nu for surface by integrating over body becomes independent of spatial
variable so:
𝑁𝑢 = 𝑓 Re , Pr
 HUGE IMPLICATIONS FOR ENGINEERS!
37
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Boundary Layer Analogies
 Recognize that non-dimensionalized governing equations for momentum and
energy have equivalent mathematical forms: they are analogous
 In case where
∗
∗
= 0:
𝑢∗
𝜕𝑢∗
𝜕𝑢∗
1
+ 𝑣∗ ∗ =
∗
𝜕𝑥
𝜕𝑦
Re
𝜕 𝑢∗
𝜕𝑦 ∗
𝑢∗
𝜕𝑇 ∗
𝜕𝑇 ∗
1 𝜕 𝑇∗
+ 𝑣∗ ∗ =
∗
𝜕𝑥
𝜕𝑦
PrRe 𝜕𝑦 ∗
Equations have the
same mathematical
form!
 Can use this observation to directly link non-dimensional solutions:
38
Boundary Layer Analogies
• Compare:
• Define:
𝐶,
𝜏
≡
2
𝜌𝑢
= 0.332Re
Nu
ℎ
=
PrRe
𝜌𝑐𝑢
∴ StPr
/
=
𝐶,
2
⁄
Nu ≡
ℎ𝑥
= 0.332Re
𝑘
≡ St (Stanton number) = 0.332Re
⁄
⁄
Pr
/
/
Pr
(Chilton-Colburn Analogy)
• Relates fluid friction and heat transfer for laminar flow over a flat plate
– Can use a heat transfer problem to solve a drag problem (or vice versa)!
○ E.g. determine ℎ just by measuring drag in isothermal conditions!
∗
– Although developed for laminar flow over a flat plate where ∗ = 0, experimentally shown to be
applicable to turbulent flow over a surface even in the presence of a
pressure gradient
39
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Example: Laminar flow over a heated flat plate
Consider a horizontal flat plate held at Ts=400K, in an air stream at T=300K moving past at 1 m/s
 Find:
a)
δ at x=0.1 m
T, u
b) δt at x=0.1 m
c)
air
Nu and h at x=0.1 m
d
dt
d) Average h from 0 < x ≤ 0.1 m
e) Drag force on 0.1 m length of plate
40
Effects of Turbulence
P
P
t
• Turbulence increases
temperature gradient at wall
•  Turbulence increases h and
thus increases rate of heat
transfer
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Heated Flat Plate Solutions
 Laminar Flow Only:
(valid for Pr  0.6)
Nu x  0.332 Re 1x 2 Pr 1 3
Nu x 
0.3387 Re1x 2 Pr 1 3
1  0.0468 Pr  
23 14
(valid for Peclet Number  100
where Peclet Number  Pe  RexPr)
 Fully turbulent boundary layer:
d  0.37 x Re x1 5
d  dt
Nu x  St Re x Pr  0.0296 Re 4x 5 Pr 1 3
(valid for 0.6  Pr  60)
Heated Flat Plate Solutions
 Mixed laminar and turbulent boundary layer
ℎ =
1
𝐿
ℎ =
𝑘
𝑢
0.332
𝐿
𝜐
ℎ
𝑑𝑥 +
𝑁𝑢 = 0.664 Re
ℎ
⁄
⁄
(assuming an abrupt
BL transition at x = xc)
𝑑𝑥
⁄
𝑑𝑥
𝑢
+ 0.0296
𝜐
𝑥 ⁄
+ 0.037 Re
⁄
− Re
⁄
Pr
• If transition to turbulent BL is at: Re
𝑁𝑢 = 0.037 Re
MECH 4406: Heat Transfer
⁄
− 871 Pr
⁄
𝑑𝑥
Pr
𝑥 ⁄
⁄
⁄
= 5 × 10
valid for:
• 0.6  Pr  60;
• 5 x 105 < ReL  108
Prof. M. Johnson, Carleton University
6
Empirical Methods for Determining 𝒉
 Recognize from theory that in general: 𝑁𝑢 ≈ 𝐶 Re
Pr
 Carefully design an experiment to measure: 𝑞, 𝐴, 𝑇 , 𝑇
• Measure / Calculate: ℎ =
• Vary experiment over a wide range of Re and Pr
– Vary 𝑢 , fluid, scale/size, 𝑇 etc. as necessary to produce a range of Re and Pr
 Correlate data where you expect: Nu = 𝑓 Re, Pr
Empirical Methods for Determining h (Correlations for external flows)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Empirical
Methods for
Determining h
(Correlations for
external flows)
46
Methodology for a Convection Calculation
1. Clarify flow geometry (if not obvious, can you approximate?)
2. Specify appropriate reference temperature (usually Tfilm) and calculate if possible or
guess if necessary
3. Evaluate relevant property data.
4. Calculate Reynolds number.
5. Decide whether local or average h (convection coefficient) is required.
6. Select appropriate correlation based on above.
7. Solve problem – if guesses were required, you may have to iterate and even choose a
different range of correlation!
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
External Convection – Practice Problem
Heat Loss from and Insulated Steam Pipe
A 10-m long, 100-mm diameter steam line wrapped in 50-mm of insulation
(k=0.035 W/mK) runs between two buildings. If the temperature in the steam line
is 110°C, the ambient temperature is −20°C, and the crosswind speed is 10 m/s,
estimate the rate of heat loss from the line.
Hint: As always, document & justify/interpret any engineering assumptions you
need to make during your solution
48
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Part II: Convection
11. Convection in Internal Flows
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Head, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
11.1 Forced Convection in Internal Pipe Flow
Boundary layers
Re =
u(r,x)
u
d
r
x
ro
𝐷
𝜐
=
𝜌𝑢
𝐷
𝜇
u(r only)
For pipe flow:
• Re <2300, laminar
• Re >2300, turbulent
d
Hydrodynamic
entry region
0.05DReD
𝑢
Fully
developed
region
• Laminar flow Hydrodynamic entry length:
𝑥
,
𝐷
≈ 0.05Re
 Differences between external and internal flows:
• External Flow: boundary layer develops without outside constraint
• Internal Flow: confinement eventually restricts development of boundary layer
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Velocity Field in Laminar Pipe Flow
 Consider only fully developed region
of flow: 𝑢(𝑟, 𝑥) → 𝑢(𝑟)
u
𝑢(𝑟, 𝑥)
d
r
x
ro
u(r only)
d
 What is 𝑢(𝑟), i.e. what is the velocity field?
• Consider a ring-shaped fluid element
r
• For laminar, fully-developed flow of an incompressible,
x
constant property fluid in a tube:
𝜕𝑢
𝑣=0
=0
𝜕𝑥
• Since 𝑢 is constant in 𝑥 (along pipe), net momentum change in 𝑥-direction is zero
Velocity Field in Laminar Pipe Flow
 Write force balance for
ring element of fluid:
r 
p
r
x
Σ𝐹 :
𝜏 2𝜋𝑟𝑑𝑥 − 𝜏 +
𝜕𝜏
𝑑𝑟
𝜕𝑟
d r
dr
dr
dr
dx
r
2𝜋 𝑟 + 𝑑𝑟 𝑑𝑥 + 𝑝 2𝜋𝑟𝑑𝑟 − 𝑝 +
p
dp
dx
dx
r
𝑑𝑝
𝑑𝑥
𝑑𝑥
2𝜋𝑟𝑑𝑟 = 0
• Ignoring higher order terms (i.e. dx*dr…), this reduces to:
−
𝑑
𝑟𝜏
𝑑𝑟
=𝑟
𝑑𝑝
𝑑𝑥
• For Newtonian fluid, shear stress 𝜏 = 𝜇 𝜕𝑢⁄𝜕𝑟 so:
𝜇 𝑑
𝑑𝑢
𝑑𝑝
𝑟
=
𝑟 𝑑𝑟 𝑑𝑟
𝑑𝑥
MECH 4406: Heat Transfer
Solve via integration
Prof. M. Johnson, Carleton University
2
Velocity Field in Laminar Pipe Flow
𝜇 𝑑
𝑑𝑢
𝑑𝑝
𝑟
=
𝑟 𝑑𝑟 𝑑𝑟
𝑑𝑥
𝑑
𝑑𝑢
1 𝑑𝑝
𝑟
=
𝑟
𝑑𝑟 𝑑𝑟
𝜇 𝑑𝑥
Recognize that axial pressure gradient,
, is independent of 𝑟 and integrate:
𝑑𝑢 1 𝑑𝑝 𝑟
𝑟
=
+𝐶
𝑑𝑟 𝜇 𝑑𝑥 2
Integrate again:
Boundary conditions are:
𝑢 𝑟 = 0 (no slip)
1 𝑑𝑝 𝑟
𝑢 𝑟 =
+ 𝐶 ln 𝑟 + 𝐶
𝜇 𝑑𝑥 4
= 0 (axisymmetric about centerline)
Subbing in boundary conditions:
𝑢 𝑟 =
1 𝑑𝑝
𝑟
4𝜇 𝑑𝑥
1−
Velocity profile for laminar
pipe flow is parabolic
𝑟
𝑟
5
Velocity Field in Laminar Pipe Flow
 Laminar pipe flow parabolic velocity profile:
 Mean velocity in pipe, 𝑢
𝑚̇ = 𝜌𝑢
𝐴 =𝜌
=
∴𝑢
=−
𝜌𝜋𝑟
𝑟 𝑑𝑝
8𝜇 𝑑𝑥
1 𝑑𝑝
𝑟
4𝜇 𝑑𝑥
1−
𝑟
𝑟
:
𝑢 𝑟 𝑑𝐴 = 𝜌
𝜌2𝜋 ∫ 𝑢 𝑟 𝑟𝑑𝑟
∴𝑢
𝑢 𝑟 =
=
2
𝑟
𝑢 𝑟
𝑢
𝑢 𝑟 2𝜋𝑟𝑑𝑟
𝑢 𝑟 𝑟𝑑𝑟 =
= 2 1−
2
𝑟
1 𝑑𝑝
𝑟
4𝜇 𝑑𝑥
1−
𝑟
𝑟
𝑟𝑑𝑟
𝑟
𝑟
 Next need to solve for Temperature field using the energy equation...
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Convection in Internal Flows
 Newton’s law of cooling in a pipe:
q   h( Ts


surface
temperature
T?

r
)
x
T(r,x)
need a representative
temperature of fluid
 A convenient temperature to use is the mean temperature, Tm, also known as
the mixing cup temperature
• Tm is the imaginary uniform flow temperature that would give the equivalent
thermal energy content of the actual varying temperature fluid
• Can think of Tm as temperature of the fluid you would get if you dumped it into an
insulated (adiabatic) container (cup) and mixed it thoroughly
Mixing Cup Temperature
 Internal energy
of actual flow:
𝐸̇ =
 Internal energy for
same mass of flow at 𝑇 :
𝐸̇ = 𝑚̇𝑐𝑇
 Can then calculate:
𝑇 =
𝜌𝑢𝑐𝑇𝑑𝐴
∫ 𝜌𝑢𝑐𝑇𝑑𝐴
𝑐𝑚̇
 For incompressible flow
2
∴𝑇 =
in a circular tube with
𝑢 𝑟
constant heat capacity, 𝑐:
MECH 4406: Heat Transfer
𝜌 = fluid density
𝑢 = velocity (which may vary over area A)
𝑐 = specific heat capacity of fluid
𝑇 = temperature (which also may vary over A)
𝑢𝑇𝑟𝑑𝑟
Note the fundamental difference
between internal and external flow
For external flow:
• 𝑇 is generally constant
• In steady problems, 𝛥𝑇 = 𝑇 − 𝑇 is
also constant
For internal flow:
• 𝑇 is generally not constant even in
steady problems
Prof. M. Johnson, Carleton University
4
Temperature Field in Laminar Pipe Flow
Thermal boundary
layers
T(r,x)
r
Surface Condition
qs”=const.
Ts>T(r,0)
dt
x
Laminar flow
thermal entry length
ro
T(r,x)
T(r,x)
dt
𝑥
,
𝐷
Thermal
entry region
Shape of
profile
changing
Fully developed
region
≈ 0.05 𝑅𝑒 𝑃𝑟
Compare to hydrodynamic
entry length
𝑥 ,
≈ 0.05𝑅𝑒
𝐷
Profile continues to
change although shape
remains constant
 Different Temperature profiles depending on boundary condition
 Temperatures in fully developed temperature profile continue to change, although
shape of profile remains constant
Need Temperature Field to get 𝒉
 As we did with the flat plate in external flow, to calculate ℎ we start by applying
an energy balance at the wall
• At internal surface of pipe where velocity is zero:
𝑞
∴ −𝑘
=𝑞
,
𝜕𝑇
𝜕𝑟
=ℎ 𝑇 −𝑇
−𝑘
∴ℎ=
,
𝜕𝑇
𝜕𝑟
𝑇 −𝑇
𝑟
Need temperature
gradient at pipe wall to
solve for 𝒉
10
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Temperature Field in Fully Developed Laminar Pipe Flow
 Energy equation for
ring-shaped CV:
𝐸̇ − 𝐸̇
= 𝐸̇ = 𝑚̇𝐶 Δ𝑇
𝑞 +
𝜕𝑞
𝑑𝑟
𝜕𝑟
 Since no velocity in 𝑟-direction, 𝑞 is via conduction only
 Assuming conduction in x-direction is negligible,
no thermal generation, and no viscous dissipation then:
𝑞 − 𝑞 +
𝜕𝑞
𝑑𝑟 = 𝑑𝑚̇ 𝐶
𝜕𝑟
𝑇+
and subbing: 𝑑𝑚̇ = 𝜌𝑢2𝜋𝑟𝑑𝑟
 We get:
1 𝜕 𝜕𝑇 𝑢 𝜕𝑇
𝑟
=
𝑟 𝜕𝑟 𝜕𝑟 𝛼 𝜕𝑥
and
where
𝑇+
𝑞
𝜕𝑇
𝑑𝑥 − 𝑇
𝜕𝑥
𝜕𝑇
𝑑𝑥
𝜕𝑥
𝑇
𝑞 = −𝑘 2𝜋𝑟𝑑𝑥
𝜕𝑇
𝜕𝑟
𝑑𝑥
=𝛼
Need to solve with appropriate boundary conditions
Boundary Conditions for Internal Flow
Constant wall Temperature
(Ts = constant)
Ts
T
DT
Tm
x
Constant heat flux at wall
( q s= constant)
Ts
T
DT
Tm
x
fully developed
region
 Note: unlike with convection in external flow,
in internal flow DT can vary even when flow is steady!
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
11.2 Temperature Field in Laminar Pipe Flow
 Find solution for case where 𝑞 = constant
• Boundary conditions to get started are:
𝑘
𝜕𝑇
𝜕𝑟
and @𝑟=0, 𝑇=𝑇 (centerline Temp.)
= 𝑞 = constant
 Solution:
𝑇
𝑢 𝑟
𝑢
1. Substitute velocity field:
𝑟
=2 1−
𝑟
into energy equation:
1 𝜕 𝜕𝑇 𝑢 𝜕𝑇
𝑟
=
𝑟 𝜕𝑟 𝜕𝑟 𝛼 𝜕𝑥
to get:
1 𝜕
𝜕𝑇
2𝑢
𝑟
=
𝑟 𝜕𝑟
𝜕𝑟
𝛼
𝑇
,
,
𝑞
𝑑𝑇
𝑑𝑥
1−
𝑟
𝑟
Temperature Field in Laminar Pipe Flow
1 𝜕
𝜕𝑇
2𝑢
𝑟
=
𝑟 𝜕𝑟
𝜕𝑟
𝛼
𝑑𝑇
𝑑𝑥
1−
𝑟
𝑟
2. Integrate once (noting that 𝑇 (𝑥) varies linearly with 𝑥 and
𝜕𝑇 2𝑢
=
𝜕𝑟
𝛼
𝑑𝑇
𝑑𝑥
𝑟
𝑟
−
2 4𝑟
+𝐶
and since
is constant:
= 0 at 𝑟 = 0, 𝐶 = 0
3. Integrate again:
𝑇=
2𝑢
𝛼
𝑑𝑇
𝑑𝑥
𝑟
𝑟
−
4 16𝑟
+𝐶
𝑇=
2𝑢
𝛼
𝑑𝑇
𝑑𝑥
𝑟
𝑟
−
4 16𝑟
+𝑇
MECH 4406: Heat Transfer
and since 𝑇 = 𝑇 at 𝑟 = 0, 𝐶 = 𝑇
Prof. M. Johnson, Carleton University
7
Temperature Field in Laminar Pipe Flow
4. Need to write 𝑇 in terms of 𝑇
• Substitute expressions for 𝑇(𝑥, 𝑟) and 𝑢 𝑟 into definition of 𝑇 :
𝑇 =
2
𝑢 𝑟
𝑢𝑇𝑟𝑑𝑟 =
2
𝑢 𝑟
1 𝑑𝑝
𝑟
4𝜇 𝑑𝑥
1−
𝑟
𝑟
2𝑢
𝛼
𝑑𝑇
𝑑𝑥
𝑟
𝑟
−
4 16𝑟
+ 𝑇 𝑟𝑑𝑟
After a bit of manipulation:
𝑇 =𝑇 +
7 𝑢 𝑟 𝑑𝑇
48 𝛼 𝑑𝑥
5. After considerable algebra, can solve for h based on temperature gradient from
solved temperature profile:
𝑘
ℎ=
𝜕𝑇
𝜕𝑟
𝑇 −𝑇
=
48 𝑘
11 𝐷
Forced Convection in Laminar Pipe Flow
 Solution is then:
𝑁𝑢 ≡
ℎ𝐷 48
=
= 4.36
𝑘
11
q s  constant 
 Repeating calculation with constant temperature boundary condition gives:
𝑁𝑢 ≡
ℎ𝐷
= 3.66
𝑘
Ts
 constant 
 Caution: results are for laminar, fully-developed pipe flow only!
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
11.3 Heat Transfer in Entry Regions in Laminar Flow
𝑥
 Laminar thermal entry length is:
,
≈ 0.05 Re Pr
𝐷
• Semi-empirical rule of thumb is that fully developed velocity and thermal boundary layers occur
when:
Gz
=
𝑥 ⁄𝐷
≈ 0.05
Re Pr
where Gz  Graetz Number)
– If ReD = 2300 (point of transition to turbulence):
○ x/D = 80.5 for air
○ x/D = 500 for water
• Laminar entry lengths can be very long a may even be entire pipe system!!
T(r,x)
r
dt
x
ro
dt
Gr-1=(x/D) / (Re Pr) = 0.05
Heat Transfer in Entry Regions in Laminar Flow
 Combined Laminar Entry Length (Both velocity and thermal boundary layers developing)
 Re Pr 
Nu D  1.86 D 
 L/D 
1/ 3
 

 s



0.14
valid for: Ts = constant
0.48 < Pr < 16700
0.0044 < (/s) <9.75
 Thermal Entry Length (or unheated starting length)
Nu D  3.66 
0.0668D L  Re D Pr
1  0.04D L  Re D Pr 
2/3
0.05
valid for: Ts = constant
q"s = const.
4.36
Nu
Ts = const.
fully dev.
3.66
Gr-1=(x/D) / (Re Pr)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
11.4 Empirical Correlations for Turbulent Internal Flows:
 Fully Developed Flow in Circular Tubes (generally accurate within 25%):
• Dittus-Boelter Equation (1930)
Nu = 0.023Re
valid for:
⁄
where:
Pr
𝑛 = 0.4 for heating (𝑇 > 𝑇 )
𝑛 = 0.3 for cooling (𝑇 < 𝑇 )
0.7  Pr  160
ReD  10000
L/D  10
𝑇 – 𝑇 is moderate
• Sieder Tate (1936) [used for larger temperature variations]
Nu = 0.027Re
⁄
Pr
⁄
𝜇
𝜇
.
valid for: 0.7  Pr  16700 ReD  10000
L/D  10
Large variations in 𝑇
𝜇 at surface temp (𝑇 ); other properties at 𝑇
Empirical Correlations for Turbulent Internal Flows:
 Transitionally turbulent flows:
𝑁𝑢 =
𝑓⁄8 (Re − 1000) Pr
1.07 + 12.7 𝑓⁄8 ⁄ Pr ⁄ − 1
f is friction factor from Moody diagram
 Turbulent entry lengths:
• Many different correlations exist but since in general 10  (x/d)fd  60, it is more
often reasonable to neglect the starting length and assume entirely fully developed
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Table 8.4
Convection
Correlations
for flow in
circular tubes
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
11
Heat Transfer Enhancement in Internal Flow
Empirical Correlations for Non-Circular Tubes (e.g. Ducts, annular flows)
 For turbulent flows:
• Use “hydraulic diameter”, 𝐷 , with same circular pipe correlations:
𝐷 ≡
4𝐴
𝑃
where Ac is flow cross-sectional area
and P is wetted perimeter
• In other words, use standard circular tube correlations except that Re is calculated
using “hydraulic diameter”, 𝐷
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Empirical Correlations for Non-Circular Tubes (e.g. Ducts, annular flows)
 For laminar flows:
• Use geometry specific
correlations
• Table 8.1 shows
correlations for noncircular tubes for two
different boundary
conditions
f Re
One Important (Common) Case – Concentric Tube Annulus
 Use hydraulic diameter:
𝜋
4
𝐷 −𝐷
4𝐴
𝐷 ≡
= 4
𝑃
𝜋𝐷 + 𝜋𝐷
=
𝐷 +𝐷 𝐷 −𝐷
𝐷 +𝐷
∴ 𝐷 =𝐷 −𝐷
25
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Empirical Correlations for Non-Circular Tubes (e.g. Ducts, annular flows)
 For laminar flow in a tube annulus:
• Correlations depend on boundary
conditions on inner/outer surface of
annulus
11.5 Energy Balance in Internal Flows
𝑑𝑞
𝑇
𝑇
,
Tm+dTm
𝑚̇ Tm
pv+d(pv)
pv
,
x
𝑞
0
inlet, i
= 𝑞 𝑃𝑑𝑥 = ℎ𝑃𝑑𝑥 𝑇 − 𝑇
dx
L
outlet, o
 Make following assumptions:
•
•
•
•
•
•
Steady- flow,
No conduction in fluid in x-direction
negligible PE and KE,
no shaft work done,
no energy generation
Only pv work and thermal energy changes
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Energy Balance in Internal Flows
 1st law for CV:
𝑑𝑞
= 𝑚̇ 𝑒
−𝑒
𝑑𝑞
𝑑 𝑐 𝑇 + 𝑝𝑣
= 𝑚̇ 𝑐 𝑇 + 𝑝𝑣 + 𝑚̇
𝑑𝑥 − 𝑚̇ 𝑐 𝑇 + 𝑝𝑣
𝑑𝑥
𝑑𝑞
= 𝑚̇𝑑 𝑐 𝑇 + 𝑝𝑣
𝑇
𝑇
,
,
𝑞
𝑑𝑞
= 𝑞 𝑃𝑑𝑥 = ℎ𝑃𝑑𝑥 𝑇 − 𝑇
Tm+dTm
𝑚̇ Tm
pv+d(pv)
pv
x
dx
L
outlet, o
0
inlet, i
Energy Balance in Internal Flows
 1st law for CV:
𝑑𝑞
= 𝑚̇𝑑 𝑐 𝑇 + 𝑝𝑣
𝑇
• For ideal gas: 𝑝𝑣 = 𝑅𝑇 and 𝑐 = 𝑐 + 𝑅 and 𝑐  constant
∴ 𝑑𝑞
= 𝑚̇𝑑 𝑐 𝑇 + 𝑅𝑇
∴ 𝑑𝑞
= 𝑚̇𝑐 𝑑𝑇
= 𝑚̇𝑐 𝑑𝑇
𝑑𝑞
(Same result!)
= 𝑞 𝑃𝑑𝑥 = ℎ𝑃𝑑𝑥 𝑇 − 𝑇
Tm+dTm
𝑚̇ Tm
 Finally:
∴ 𝑑𝑞
,
𝑞
• For incompressible liquid: 𝑐 = 𝑐 and 𝑑 𝑝𝑣 ≪ 𝑑 𝑐 𝑇
∴ 𝑑𝑞
𝑇
• Rearranging:
𝑑𝑇
𝑞 𝑃
𝑃
∴
=
=
ℎ 𝑇 −𝑇
𝑑𝑥
𝑚̇𝑐
𝑚̇𝑐
x
(1)
pv+d(pv)
pv
= 𝑚̇𝑐 𝑑𝑇 = 𝑞 𝑃𝑑𝑥 = ℎ𝑃𝑑𝑥 𝑇 − 𝑇
MECH 4406: Heat Transfer
,
0
inlet, i
dx
L
outlet, o
Prof. M. Johnson, Carleton University
4
Energy Balance in Internal Flows
 Case I: Solve Eq. (1) for Constant Heat Flux boundary condition
𝑞 𝑃
is constant and Eq. (1) can be integrated directly to give:
𝑚̇𝑐
𝑇
,
=𝑇
,
+
𝑞 𝑃
𝑥
𝑚̇𝑐
(Case I: 𝑞 is constant)
𝑇
T
T
𝑇
𝑥
fully developed
region
Energy Balance in Internal Flows
 Case II: Solve Eq. (1) for Constant Surface Temp. boundary condition
𝑇
T
T
𝑇
Temperature different T varies along pipe
due to heat transfer!!
𝑥
• Define local temperature difference: Δ𝑇 = 𝑇 − 𝑇
𝑑 Δ𝑇 = 𝑑 𝑇 − 𝑇
= −𝑑𝑇
• Rewrite Eq. (1) in terms of ΔT:
∴
𝑑𝑇
𝑑 Δ𝑇
𝑞 𝑃
𝑃
=−
=
=
ℎ Δ𝑇
𝑑𝑥
𝑑𝑥
𝑚̇𝑐
𝑚̇ 𝑐
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Energy Balance in Internal Flows
• Separate variables to integrate:
−
𝑑 Δ𝑇
ℎ𝑃𝑑𝑥
=−
Δ𝑇
𝑚̇𝑐
𝑑 Δ𝑇
𝑃
=
ℎ Δ𝑇
𝑑𝑥
𝑚̇𝑐
• Integrate from inlet, 𝑖 to outlet, 𝑜 :
𝑑 Δ𝑇
𝑃
=−
Δ𝑇
𝑚̇𝑐
ln Δ 𝑇
ln
=−
ℎ𝑑𝑥 = −
𝑃𝐿ℎ
𝐴 ℎ
=−
𝑚̇𝑐
𝑚̇ 𝑐
𝑃𝐿
𝑚̇𝑐
1
ℎ𝑑𝑥
𝐿
where: 𝐴 is interior surface area
Δ𝑇 = Δ𝑇
Δ𝑇 = Δ𝑇
Δ𝑇
𝑃ℎ𝑥
=−
Δ𝑇
𝑚̇ 𝑐
Δ𝑇
𝑇 −𝑇
𝑃ℎ𝑥
=
= exp −
Δ𝑇
𝑇 −𝑇
𝑚̇𝑐
=𝑇 −𝑇
=𝑇 −𝑇
(Case II: Ts is constant !!)
(2)
32
Energy Balance in Internal Flows
 Case II: Now consider overall energy balance for length of pipe, 𝑥 :
𝑑𝑞
= 𝑑𝑞 = 𝑚̇𝑐 𝑑𝑇
Rearrange (2):
ln
Δ𝑇
𝑃ℎ𝑥
𝑃ℎ𝑥
=−
=−
𝑇
Δ𝑇
𝑚̇𝑐
𝑞
Sub in (3):
ln
Δ𝑇
ℎ𝐴
=
Δ𝑇
𝑞
ln
Δ𝑇
ℎ𝐴
=
Δ𝑇 − Δ𝑇
Δ𝑇
𝑞
Integrate over length of pipe:
𝑞 = 𝑚̇ 𝑐 𝑇
,
−𝑇
,
(3)
∴ 𝑞 = ℎ𝐴
Δ𝑇 − Δ𝑇
Δ𝑇
ln
Δ𝑇
∴ 𝑞 = ℎ𝐴 Δ𝑇
MECH 4406: Heat Transfer
= ℎ𝐴 Δ𝑇
Δ𝑇
=
Δ𝑇 − Δ𝑇
Δ𝑇
ln
Δ𝑇
𝑇 −𝑇
,
,
−𝑇
− 𝑇 −𝑇
,
,
(Case II: Ts is constant !!)
is the “log-mean
temperature difference”
Prof. M. Johnson, Carleton University
6
Energy Balance in Internal Flows
 Case IIb: where 𝑇 is constant rather than 𝑇 (Generally more relevant)
• Use Case II solution but replace 𝑇 with 𝑇 and
replace ℎ with overall heat transfer coefficient, 𝑈:
Δ𝑇
𝑇 −𝑇
𝑈𝐴
=
= exp −
Δ𝑇
𝑇 −𝑇
𝑚̇𝑐
∴ 𝑞 = ℎ𝐴 Δ𝑇
=
= exp −
Δ𝑇
𝑅
1
𝑚̇𝑐𝑅
Δ𝑇
𝑅
=
=
1
𝑈𝐴
Δ𝑇 − Δ𝑇
Δ𝑇
ln
Δ𝑇
Note: This is one of the most generally useful results of the whole course!
Pipe Flow Example
Water enters a 20 m long, 0.1m diameter pipe at 𝑇 , = 300 K . If the surface
temperature of the pipe is held constant at 𝑇 = 360 K, and the mass flow rate of
water through the pipe is 0.02 kg/s:
a) What is the mean temperature of the water at the exit?
b) What is the rate of heat transfer to the water?
35
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Example restated: What if 𝑻 is constant rather than 𝑻𝒔 ?
Water at a mass flow rate of 0.02 kg/s and 𝑇 , =300K enters a 20 m long, 0.1m
inner diameter pipe. If the ambient temperature of the air surrounding the pipe is
𝑇 =380K, the convection coefficient around the outside of the pipe is
ℎ =40 W/m2K, the thermal conductivity of the pipe wall material is
𝑘
=15 W/mK, and the outer diameter of the pipe is 𝐷 =0.12 m:
a) What is the mean temperature of the water at the exit?
b) What is the rate of heat transfer to the water?
36
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Part II: Convection
12. Heat Exchangers
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Head, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Introduction to Heat Exchangers
 What is a heat exchanger?
• A piece of equipment specifically designed to effectively transfer thermal energy
between two fluids at different temperatures
• In most cases the fluids need to be kept separate (fluid mixing is not an option) and
heat transfer occurs across a solid wall
 Examples:
•
•
•
•
•
•
Car radiator
Heating / cooling coils in different applications including labs and HVAC systems
Power production (condenser units)
Waste heat recovery
Chemical processes
Your nose
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Heat Exchangers
Human Nasal Passage
Car
Radiator
Home Furnace (“Clamshell Heat Exchanger)
Tube-in-tube
3
Heat Exchangers
Shell and tube
Tesla Model 3
4
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Basic Types of Heat Exchangers
 Tube in tube (concentric tube)
• Parallel flow
• Counter-flow
 Crossflow
• Mixed
• Un-mixed
 Shell and tube
 Compact
Types of Heat Exchangers
 Tube-in-tube (a.k.a “double pipe”):
Parallel flow:
Th
T
Tc
x
• Temperature difference is initially large, but decays rapidly with increasing x,
asymptotically approaching zero.
• Outlet temperature of cold fluid never exceeds that of the hot fluid.
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Types of Heat Exchangers
 Tube-in-tube (a.k.a “double pipe”):
Counterflow:
Th
T
Tc
x
• Temperature difference remains large throughout
• It is possible for the outlet temperature of the cold fluid to exceed that of the hot
fluid!
Types of Heat Exchangers
 Crossflow:
Un-Mixed: (Fins prevent mixing of crossflow fluid)
x
y
u
Cross
flow
T(x,y)
Tube
flow
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Types of Heat Exchangers
 Crossflow:
Mixed: (crossflow fluid mixes freely with itself – no fins)
x
Cross
flow
y
u
T(x)
Tube
flow
Types of Heat Exchangers
 Shell and Tube:
• 1 shell pass & 2 tube passes
MECH 4406: Heat Transfer
• 2 shell passes & 4 tube passes
Prof. M. Johnson, Carleton University
5
Types of Heat Exchangers
 Compact:
Two Approaches to Heat Exchanger Analysis
1. Log-mean temperature difference
• Use in problems where all temperatures are known or easily calculated at the outset
2. Effectiveness-NTU
• Use in problems where geometry is known at the outset
 Approaches are complementary.
• In problems where one might require an iterative solution, the other will enable a
direct solution
 Both approaches needed in design and analysis
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
1. Log-Mean Temperature Difference Approach
 Start with energy balance:
ℎ , , 𝑇 , , 𝑚̇
𝑞
ℎ , , 𝑇 , , 𝑚̇
𝑞 = 𝑚̇ ℎ
,
−ℎ
ℎ
ℎ
𝑞
,
= 𝑚̇ ℎ
,
−ℎ
,
,
,𝑇
,
,𝑇,
,
Assuming:
• Negligible heat transfer with surroundings
• Negligible potential and kinetic energy changes
• Steady flow
where: 𝑚̇ is mass flow rate of fluid
ℎ is enthalpy
subscripts h=hot, c=cold, i=inlet, o=outlet
• If no phase change and constant specific heats:
𝑞 = 𝑚̇ 𝑐
,
𝑇
,
−𝑇
,
= 𝑚̇ 𝑐
,
𝑇, −𝑇,
 Locally 𝚫𝑻 = 𝑻𝒉 − 𝑻𝒄 but 𝜟𝑻 will vary with 𝒙
 We need an appropriate average 𝚫𝑻∗ to use with Newton’s law of cooling: 𝒒 = 𝑼𝑨𝚫𝑻∗
1. Log-Mean Temperature Difference Approach
 Locally Δ𝑇 = 𝑇 − 𝑇 but 𝛥𝑇 will vary with 𝑥
• We need an appropriate average 𝛥𝑇 ∗ to use with Newton’s law of cooling: 𝑞 = 𝑈𝐴𝛥𝑇 ∗
 Consider Parallel Flow Tube-in-tube Heat Exchanger
𝑇
,
𝑇
𝑑𝑇
Δ𝑇
Δ𝑇
𝑑𝑞
𝑇
,
Δ𝑇
Δ𝑇+𝑑 Δ𝑇
𝑇,
𝑑𝑇
𝑇,
1
MECH 4406: Heat Transfer
2
𝑥
Assuming:
• Negligible heat transfer with
surroundings
• Negligible potential and kinetic
energy changes
• Steady flow
• No axial conduction
• No phase change and constant
specific heats
• Overall heat transfers coefficient is
constant
Prof. M. Johnson, Carleton University
7
1. Log-Mean Temperature Difference Approach
 Energy balance:
• For hot fluid:
𝑇
𝑑𝑞 = −𝑚̇ 𝑐
• For cold fluid:
𝑑𝑞 = 𝑚̇ 𝑐
,
𝑑𝑇 = −𝐶̇ 𝑑𝑇
𝑑𝑇 = 𝐶̇ 𝑑𝑇
𝑪̇𝒉 = 𝒎̇𝒉 𝒄𝒑,𝒉
,
𝑑𝑇
Δ𝑇
𝑪̇𝒄 = 𝒎̇𝒄 𝒄𝒑,𝒄
Δ𝑇
Δ𝑇+𝑑 Δ𝑇
𝑑𝑞
𝑑𝑇
• Rate of change of Δ𝑇 along 𝑑𝑥 is:
𝑑 Δ𝑇 = 𝑑 𝑇 − 𝑇 = 𝑑𝑇 − 𝑑𝑇
𝑇
,
Δ𝑇
𝑇,
𝑇,
2
1
• Combining equations: 𝑑 Δ𝑇 = −
̇
−
= −𝑈
𝑥
̇
• Since locally, 𝑑𝑞 = 𝑈Δ𝑇𝑑𝐴, subbing this in for 𝑑𝑞: 𝑑 Δ𝑇 = 𝑈Δ𝑇𝑑𝐴
• Integrating from 1 to 2: ∫
,
𝑇
̇
gives: lnΔ𝑇| = −𝑈𝐴
+
+
̇
∫ 𝑑𝐴
̇
+
̇
̇
(Eq. 1)
̇
1. Log-Mean Temperature Difference Approach
• From overall energy balance: 𝐶̇ =
,
• Subbing into (1) from previous slide: ln
,
and 𝐶̇ =
= −𝑈𝐴
,
,
,
,
,
+
,
 Finally, we can calculate heat transfer in heat exchanger as:
∴ 𝑞 = 𝑈𝐴
For parallel flow
heat exchanger:
MECH 4406: Heat Transfer
Δ𝑇 ≡ 𝑇
Δ𝑇 ≡ 𝑇
Δ𝑇 − Δ𝑇
= 𝑈𝐴Δ𝑇
Δ𝑇
ln
Δ𝑇
,
,
−𝑇,
−𝑇,
For counterflow Δ𝑇 ≡ 𝑇
heat exchanger: Δ𝑇 ≡ 𝑇
,
,
−𝑇,
−𝑇,
Prof. M. Johnson, Carleton University
8
Three Special Cases for Counterflow Heat Exchangers:
1. If 𝐶̇ = 𝐶̇ :
• Δ𝑇 = Δ𝑇 = Δ𝑇
2. If 𝐶̇ ≫ 𝐶̇ or 𝐶̇ → ∞
3. If 𝐶̇ ≫ 𝐶̇ or 𝐶̇ → ∞
• Temperature of hot fluid
then remains approximately
constant
• Approximately what occurs
if the hot fluid is a
condensing vapour
• Temperature of cold fluid
remains approximately
constant
• Approximately what occurs
in an evaporator or boiler
HEAT EXCHANGER EXAMPLE USING LMTD APPROACH
For a dynamometer test of an SAE racing engine, a counter flow tube in tube heat
exchanger will be used to cool the engine oil with water. The diameter of the thin-walled
inner tube is 14mm and the inner diameter of the outer tube is 30mm.
If your design conditions are:





𝑇
, = 𝑇 , = 10°C;
𝑇 , = 𝑇 , = 110°C;
𝑇 , = 𝑇 , = 75°C
Flow rate of water through the inner pipe is 0.1 kg/s
Flow rate of oil through the outer annulus is 0.15 kg/s
How long must the tube be? What will be the outlet temperature of the water?
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
2. Effectiveness-NTU Approach
 Define Effectiveness, 𝜀, of a heat exchanger:
𝜀=
actual heat transfer
𝑞
=
maximum possible heat transfer 𝑞
• Maximum possible heat transfer, 𝑞
exchanger
, could be achieved in an infinitely long counterflow heat
19
Effectiveness / Max. Possible Heat Transfer
 Maximum possible heat transfer:
• Hottest temperature in system: 𝑇
(inlet 𝑇 of hot fluid)
,
• Coldest temperature in system: 𝑇 , (inlet 𝑇 of cold fluid)
 Max. energy transfer is smaller of two options:
• Either raise 𝑇 of cold fluid from 𝑇 , to 𝑇
• Or, lower 𝑇 of hot fluid from 𝑇
• Thus: 𝑞
= 𝑪̇𝒎𝒊𝒏 𝑇
,
,
,
so that: 𝑞
to 𝑇 , so that: 𝑞
= 𝑚̇ 𝑐
= 𝑚̇ 𝑐
,
,
𝑇
𝑇
,
,
= 𝐶̇ 𝑇
−𝑇,
−𝑇,
= 𝐶̇ 𝑇
,
,
−𝑇,
−𝑇,
−𝑇,
(Note that if we tried to use 𝐶̇
instead, 1st law tells as that the fluid with 𝐶̇
undergo a temperature change beyond the limits of (𝑇
,
would have to
 𝑇 , ) which is obviously not possible)
 Thus, effectiveness of a heat exchanger is:
𝜀=
MECH 4406: Heat Transfer
𝑞
𝑞
=
𝐶̇ 𝑇 , − 𝑇 ,
𝐶̇
𝑇 , −𝑇,
=
𝐶̇ 𝑇 , − 𝑇 ,
𝐶̇
𝑇 , −𝑇,
Prof. M. Johnson, Carleton University
1
2. Effectiveness – NTU Method
 If effectiveness and inlet temperature are known, q can be directly calculated!
= 𝜀𝐶̇
𝑞 = 𝜀𝑞
𝑇
−𝑇 ,
,
 As shown on next slides:
𝜀=𝜀
𝐶̇
𝐶̇
(𝜀 is a function of two non-dimensional parameters)
, 𝑁𝑇𝑈
𝑁𝑇𝑈 ≡
𝑈𝐴
𝐶̇
(NTU  Number of transfer units)
• Simple algebraic expressions for e and NTU can be derived as in class and are commonly
given in tables/charts as on following slides
Derivation of e / NTU Expression for Parallel Flow Concentric Tube Heat Exchanger
𝑞 = 𝑈𝐴Δ𝑇
= 𝑈𝐴
Rearrange:
𝑇
,
−𝑇, − 𝑇 , −𝑇,
𝑇 , −𝑇,
ln
𝑇 , −𝑇,
ln
𝑇 , −𝑇,
𝑇 , −𝑇,
=
𝑈𝐴
𝑇
𝑞
,
ln
𝑇 , −𝑇,
𝑇 , −𝑇,
= 𝑈𝐴
𝑇
,
where for parallel flow:
−𝑇,
− 𝑇
−𝑇
𝑞
−
,
Sub. in simple energy balance relations: 𝑞 = 𝐶̇ 𝑇
ln
𝑇 , −𝑇,
𝑇 , −𝑇,
Assume for now 𝐶̇
= −𝑈𝐴
= 𝐶̇ , 𝐶̇
,
Δ𝑇 ≡ 𝑇
,
−𝑇,
Δ𝑇 ≡ 𝑇
,
−𝑇,
−𝑇,
,
𝑇, −𝑇 ,
𝑞
−𝑇
,
and 𝑞 = 𝐶̇ 𝑇 , − 𝑇 ,
1
1
+
̇
𝐶
𝐶̇
= 𝐶̇ :
ln
𝑇 , −𝑇,
𝑇 , −𝑇,
=−
𝑈𝐴
𝐶̇
1+
̇
𝐶
𝐶̇
𝑁𝑇𝑈
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Derivation of e / NTU Expression for Parallel Flow Concentric Tube Heat Exchanger
Seeking expression of form: 𝜀 = 𝜀
Currently have: ln
𝑇 , −𝑇,
𝑇 , −𝑇,
𝐶̇
𝐶̇
, 𝑁𝑇𝑈
= −𝑁𝑇𝑈 1 +
𝐶̇
𝐶̇
Just need to rewrite left hand side temperature term:
𝑇 , −𝑇,
𝑇 , −𝑇,
=
𝑇
,
−𝑇
𝑇
But with 𝑪̇𝒎𝒊𝒏 = 𝑪̇𝒉 , 𝑪̇𝒎𝒂𝒙 = 𝑪̇𝒄 :
+𝑇 , −𝑇,
−𝑇,
,
,
𝜀=
= −𝜺
𝐶̇ 𝑇 , − 𝑇 ,
𝐶̇
𝑇 , −𝑇,
=
𝑇 , −𝑇 ,
𝑇 , −𝑇 ,
With further algebraic manipulation:
= −𝜀 +
𝑇 , −𝑇,
𝑇
= −𝜀 +
𝑇 , −𝑇,
= −𝜀 + 1 −
𝐶̇
𝑇
𝐶̇
,
−𝑇
−𝑇 , +𝑇 , −𝑇,
𝑇 , −𝑇,
,
,
𝑇
,
= −𝜀 + 1 +
1
𝐶̇
= −𝜀 + 1 −
−𝑇,
𝐶̇
𝑇, −𝑇,
𝑇 , −𝑇,
𝜀 = 1−𝜀 1+
= −𝜀 + 1 −
𝐶̇
𝐶̇
𝑞⁄𝐶̇
𝑇 , −𝑇,
Which has the
form we want!
=𝜺
Derivation of e / NTU Expression for Parallel Flow Concentric Tube Heat Exchanger
Putting it all back together:
ln 1 − 𝜀 1 +
𝐶̇
𝐶̇
= −𝑁𝑇𝑈 1 +
𝐶̇
𝐶̇
Solving for 𝜺:
1 − exp −𝑁𝑇𝑈 1 +
𝜀=
𝐶̇
𝐶̇
𝐶̇
1+ ̇
𝐶
 Had we instead assumed 𝐶̇
Equation we need to solve 𝜺
𝑪̇𝒎𝒊𝒏
, 𝑵𝑻𝑼
𝑪̇𝒎𝒂𝒙
for a
parallel flow concentric tube heat exchanger:
= 𝐶̇ , 𝐶̇
= 𝐶̇ , we would get same result
 Additional expressions for other types of heat exchangers are shown on the next slides
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Heat Exchanger Effectiveness Formulae
Parallel flow
Counterflow
Heat Exchanger NTU formulae
Parallel flow
Counterflow
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Plots of Effectiveness / NTU Relations
 Parallel and
Counterflow heat
exchangers
Plots of Effectiveness / NTU Relations
 Shell & Tube type
heat exchangers
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Plots of Effectiveness / NTU Relations
 Cross-flow heat
exchangers
A Note on the Overall Heat Transfer Coefficient
 Recall: 𝑞 = 𝑈𝐴Δ𝑇
𝑈𝐴 =
1
𝑅
or
𝑅
=
1
𝑈𝐴
 For a plane wall:
𝑅 = 𝑡 ⁄𝑘𝐴
𝑅 = 1⁄ ℎ 𝐴
𝑅 = 1⁄ ℎ 𝐴
1
1 1
𝑡 1
=
+ +
𝑈𝐴 𝐴 ℎ
𝑘 ℎ
so that: 𝑈 =
1
1
𝑡 1
+ +
ℎ
𝑘 ℎ
 For a tubular or spherical surface, 𝐴  𝐴 :
𝑅 = 1⁄ ℎ 𝐴
1
1
1
1
ln 𝑟 ⁄𝑟
1
=
=
=
+
+
𝑈 𝐴
𝑈𝐴
𝑈𝐴 ℎ 𝐴
2𝜋𝑘𝐿
ℎ𝐴
𝑅 = 1⁄ ℎ 𝐴
MECH 4406: Heat Transfer
𝑅=
ln 𝑟 ⁄𝑟
2𝜋𝑘𝐿
Prof. M. Johnson, Carleton University
6
A Note on the Overall Heat Transfer Coefficient
 For a tubular or spherical surface, 𝐴  𝐴 :
1
1
ln 𝑟 ⁄𝑟
1
1
1
=
+
+
=
=
𝑈𝐴 ℎ 𝐴
2𝜋𝑘𝐿
ℎ𝐴
𝑈 𝐴
𝑈𝐴
∴𝑈 =
1
𝐴
1 𝐴 ln 𝑟 ⁄𝑟
+
+
ℎ
2𝜋𝑘𝐿
ℎ𝐴
𝑈  𝑈 !!
1
∴𝑈 =
𝐴
𝐴 ln 𝑟 ⁄𝑟
1
+
+
ℎ 𝐴
2𝜋𝑘𝐿
ℎ
 However 𝑼𝒐 𝑨𝒐 = 𝑼𝒊 𝑨𝒊 = 𝑼𝑨 so it doesn’t matter how 𝑈
is defined as long as it’s consistent with how 𝐴 is defined!
𝑞 = 𝑈𝐴 Δ𝑇
=𝑈 𝐴
,
Δ𝑇
= 𝑈 𝐴 , Δ𝑇
31
Heat Exchanger Example #2 – Solution Using Effectiveness-NTU Approach
 Revisit previous example on dyno-testing a racing engine. Assume we have a
20m long concentric tube coil available for use as a counterflow heat exchanger.
If we use this, what will be the outlet temperature of the engine oil? (As before
𝑇
, = 10°C; and 𝑇 , = 110°C; through the inner pipe; is 0.1 kg/s through
the outer annulus; )
 Note: although we now know the length of the heat exchanger, we no longer
know the relevant temperatures and can’t directly evaluate Δ𝑇 . LMTD
method of heat exchanger analysis would require iteration. Use 𝜀-NTU method
instead.
32
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Nusselt Number Correlations for LAMINAR flow in an Annulus
Table 8.4
Convection
Correlations
for flow in
circular tubes
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Part II: Convection
13. Free / Natural Convection
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Head, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
Natural Convection
 Heat transfer between a surface and a fluid in which fluid motion is proportional
to density gradients.
• Fluid motion caused by body forces (gravity/buoyancy) acting on the fluid itself
rather than external pumps/fans/motion
 Fluid velocities associated with natural convection are low so in general:
ℎnatural convection < ℎforced convection
 Despite low ℎnat.conv, advantages in engineering applications include:
• Low cost (no $ for pumping fluid), no noise, no pollution
 Typical applications:
• Electric baseboard heaters, refrigeration coils, electronic equipment, power
transmission lines, humans/animals
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Natural Convection
 Base case for study – heated vertical flat plate
Since temperature gradients drive the fluid motion,
there is no distinction between 𝜹 and 𝜹𝒕 .
Turbulent
, T
Ts
𝑅𝑎 ~ 109
d
As before h is determined by solving
governing equations to calculate
temperature gradient at surface
𝑞
𝑥
−𝑘
Laminar
,
𝜕𝑇
𝜕𝑦
𝑦
=𝑞
=ℎ 𝑇 −𝑇
−𝑘
ℎ=
,
𝜕𝑇
𝜕𝑦
𝑇 −𝑇
Governing Equations for Laminar Boundary Layers with Buoyancy
 Previous, simplified momentum equation now also has a body force
• Body force per unit volume is −𝜌𝑔, where 𝑔 is acceleration due to gravity
𝑢
𝜕𝑢
𝜕𝑢
1 𝜕𝑝
𝜕 𝑢
+𝑣
=−
+𝜈
−𝑔
𝜕𝑥
𝜕𝑦
𝜌 𝜕𝑥
𝜕𝑦
• Outside boundary layer:
𝜕𝑝
= −𝜌 𝑔
𝜕𝑥
• Subbing into above:
𝑢
𝜕𝑢
𝜕𝑢 −𝜌 𝑔
𝜕 𝑢
+𝑣
=
+𝜈
−𝑔
𝜕𝑥
𝜕𝑦
𝜌
𝜕𝑦
𝑢
𝜕𝑢
𝜕𝑢
Δ𝜌
𝜕 𝑢
+𝑣
=𝑔
+𝜈
𝜕𝑥
𝜕𝑦
𝜌
𝜕𝑦
where
Δ𝜌 = 𝜌 − 𝜌
4
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Governing Equations for Laminar Boundary Layers with Buoyancy
 Previous, simplified momentum equation now also has a body force
𝑢
𝜕𝑢
𝜕𝑢
Δ𝜌
𝜕 𝑢
+𝑣
=𝑔
+𝜈
𝜕𝑥
𝜕𝑦
𝜌
𝜕𝑦
where
Δ𝜌 = 𝜌 − 𝜌
Buoyancy
force per
unit mass
• If density variation is only due to temperature, then
density term relates to volume expansion coefficient, 𝛽:
𝛽=−
1 𝜕𝜌
𝜌 𝜕𝑇
𝛽≈−
1 Δ𝜌
1𝜌 −𝜌
=−
𝜌 Δ𝑇
𝜌𝑇 −𝑇
𝜌 −𝜌
≈ 𝜌𝛽 𝑇 − 𝑇
𝜌
5
Governing Equations for Laminar Boundary Layers with Buoyancy
 Momentum equation with Boussinesq approximation:
𝑢
𝜕𝑢
𝜕𝑢
𝜕 𝑢
+𝑣
= 𝑔𝜌𝛽 𝑇 − 𝑇 + 𝜈
𝜕𝑥
𝜕𝑦
𝜕𝑦
• For ideal gas, 𝜌 = 𝑃 ⁄𝑅𝑇, so:
𝛽=−
1 𝜕𝜌
𝜌 𝜕𝑇
=
1
𝑅𝑇 𝑃
=
𝑇
𝑃 𝑅𝑇
• For non-ideal gas and compressible liquids 𝛽 is obtained
from thermodynamic property tables (Appendix A)
6
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Non-Dimensionalized Gov. Eqs. with Boussinesq Approximation
𝑥∗ =
 Energy Eq.:
𝑥
𝐿
𝑦∗ =
𝑢∗
 Momentum Eq.:
𝑦
𝐿
𝑇∗ =
𝑇−𝑇
𝑇 −𝑇
𝑢∗ =
𝑢
𝑢
𝑣∗ =
𝜕𝑇 ∗
𝜕𝑇 ∗
1 𝜕 𝑇∗
+ 𝑣∗ ∗ =
∗
𝜕𝑥
𝜕𝑦
PrRe 𝜕𝑦 ∗
𝑢∗
𝑣
𝑢
𝑢 is some reference velocity
(Same as before)
𝜕𝑢∗
𝜕𝑢∗
𝑔𝛽 𝑇 − 𝑇
+ 𝑣∗ ∗ = −
∗
𝜕𝑥
𝜕𝑦
𝑢
𝐿
𝑇∗ +
1
Re
𝜕 𝑢∗
𝜕𝑦 ∗
• Choose reference velocity to simplify form of equations: 𝑢 = 𝑔𝛽 𝑇 − 𝑇 𝐿
• Re =
is then Re =
 Define, Grashof number, Gr =
(𝐆𝐫 ≡ ratio of buoyancy
and viscous forces)
Non-Dimensionalized Gov. Eqs. with Boussinesq Approximation
 Momentum Eq.:
 Energy Eq.:
𝑢∗
𝜕𝑢∗
𝜕𝑢∗
1
∗
+
𝑣
= −𝑇 ∗ +
𝜕𝑥 ∗
𝜕𝑦 ∗
Gr
𝑢∗
𝜕𝑇 ∗
𝜕𝑇 ∗
1
𝜕 𝑇∗
+ 𝑣∗ ∗ =
∗
𝜕𝑥
𝜕𝑦
Pr Gr 𝜕𝑦 ∗
𝜕 𝑢∗
𝜕𝑦 ∗
Grashof number
Gr =
 Expect solutions of form:
𝑁𝑢 = 𝑓 𝑥 ∗ , 𝐺𝑟, 𝑃𝑟
𝑔𝛽 𝑇 − 𝑇 𝐿
𝜈
(ratio of buoyancy
and viscous forces)
or
𝑁𝑢 = 𝑓 𝐺𝑟, 𝑃𝑟
8
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Forced vs. Natural Convection
 Forced Convection
 Natural Convection
u
, T
Important physics:
• Ratio of inertial vs.
viscous forces
• Re  Reynolds Number
Important physics:
• Ratio of Buoyancy vs.
viscous forces
• Gr  Grashof Number
• Nu ≡
• Nu ≡
= 𝑓 Re, Pr
= 𝑓 Gr, Pr
Natural Convection
 Grashof Number:
Gr =
𝑔𝛽 𝑇 − 𝑇 𝐿
𝑔𝛽 Δ𝑇 𝐿
=
𝜐
𝜐
Turbulent
 Rayleigh Number: Ra ≡ GrPr =
𝑔𝛽 Δ𝑇 𝐿
𝜐
𝜐
𝑔𝛽 Δ𝑇 𝐿
=
𝛼
𝜐𝛼
• Transition to turbulence for vertical heated flat plate is a function of 𝑅𝑎 :
𝑅𝑎 =
𝑔𝛽 Δ𝑇 𝑥
≈ 10
𝜈𝛼
, T
Ts
𝑅𝑎 ~ 109
d
Laminar
𝑥
𝑦
 In general for engineering calculations:
𝑁𝑢 =
ℎ𝐿
= 𝐶Ra
𝑘
MECH 4406: Heat Transfer
𝑅𝑎 = 𝐺𝑟 Pr =
𝑔𝛽 Δ𝑇 𝐿
𝜈𝛼
Prof. M. Johnson, Carleton University
5
Nusselt Number Correlations for Natural Convection
Nusselt Number Correlations for Natural Convection
Define
𝑨𝒔
𝑳=
𝑷
(Eq. 9.29)
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Nusselt Number Correlations for Natural Convection
 Long horizontal cylinder:
 Sphere:
Valid for 𝑃𝑟 ≳ 0.7 and 𝑅𝑎 ≲ 10
13
Nusselt Number Correlations for Natural Convection in Cavities
Horizontal Cavities:
• If 𝑤 ⁄𝐿 and 𝐻 ⁄𝐿 ≫ 1, and 𝑅𝑎 <1708, 𝑁𝑢 =1
• If 3 × 10 ≲ 𝑅𝑎 ≲ 7 × 10 :
14
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Nusselt Number Correlations for Natural Convection in Cavities
Vertical Cavities:
H
15
L
Nusselt Number Correlations for Natural Convection in Cavities
Inclined Cavities:
 Relevant to flat plate solar collectors
 Heat transfer depends on the magnitude of
relative to a critical angle 𝜏, whose value
depends on 𝐻 ⁄𝐿 (See Table 9.4 in textbook)
 Heat transfer also depends on the magnitude of
𝑅𝑎 relative to a critical Rayleigh number of
𝑅𝑎 ,
= 1708/ cos 𝜏
• See correlations 9.54-9.57 in textbook
16
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Natural Convection Example
What is the rate of heat loss through the top
surface of a polystyrene cooler sitting in air at
𝑇 = 25°C?
1m
0.5 m
Assume for the cooler:
• Wall thickness, 𝑡=0.02 m; 𝑘=0.027 W/mK.
• Inner surface temperature, 𝑇 , =2°C
Neglect radiation and check later!
17
Mixed Forced and Natural Convection
 Natural convection:
 Forced convection:
• Flow is set by a balance of momentum (𝑀)
and viscous forces (𝑉)
• Flow is set by a balance of buoyancy (𝐵)
and viscous forces (𝑉)
• Reynolds number, Re =
• Grashof number, Gr =
 Relative importance of natural and forced convection is determined by the ratio of 𝐵 to 𝑀
𝑀𝐵
Gr
𝐵
= 𝑉 =
Re
𝑀
𝑀
𝑉
• If
≪ 1, only consider forced convection
• If
≫ 1, only consider natural convection
• If
≈ 1, both are relevant
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Mixed Forced and Natural Convection
 If both natural and forced convection are relevant, consider whether the
combined forces assist or oppose each other:
Transverse
Assisting
Opposed
𝑢 forced
𝑢forced
𝑢forced
𝑁𝑢 = 𝑁𝑢
+ 𝑁𝑢
𝑁𝑢 = 𝑁𝑢
− 𝑁𝑢
• For rough engineering estimates, 𝑛 ≈ 3
19
Prof. M. Johnson, Carleton University
MECH 4406: Heat Transfer
10
Part III: Radiation
14. Introduction to Thermal Radiation
MECH 4406: Heat Transfer
Prof. Matthew Johnson, Ph.D., P.Eng
Head, Energy & Emissions Research Lab. (EERL)
Mechanical & Aerospace Engineering
Carleton, University
Ottawa, ON Canada
What is Radiation?
 Electro-magnetic Radiation: Energy exchange by electromagnetic waves
• All matter at finite temperature, emits radiation.
• Emission related to energy release due to oscillation or transition of electrons
– James Maxwell (1864) – accelerated charges give rise to electromagnetic waves
– Simple Model: Bohr’s atom:
○ Electron is continually accelerating
and therefore according to Maxwell
econtinually emitting electromagnetic waves.
• Oscillations are maintained by internal energy of matter
– Temperature related process so amount of energy emitted scales with temperature
• No medium required to transport energy
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
What is Thermal Radiation?
 For opaque materials (most solids and some liquids), radiation measured at
surface originates from <1mm below surface
• Adjacent molecules strongly absorb emission from interior molecules
• Thus radiation from solids is a surface phenomena
 For gases, some liquids, and semi-transparent solids, emission of radiant energy
is a volumetric phenomenon.
• More complex – beyond scope of this course!
 Thermal Radiation: Radiation in band of electromagnetic spectrum that
interacts with a material and affects temperature and internal energy.
Transport of Energy by Radiation
 Dual theory (both are ~correct!):
• Propagation of Electromagnetic waves:
– 𝑐 = 𝜆𝜈 where: 𝜆 ≡ wavelength [mm]
𝜈 ≡ frequency [1/s]
𝑐 ≡ speed of light [2.998x108 m/s in a vacuum]
• Photons (quanta)
– Energy of a photon:
e  hv 
hc

h  Planck’s const. = 6.626 × 10–34 Js
– Note: small 𝜆 = high energy; large 𝜆 = low energy
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Electromagnetic Spectrum
Given:
𝒆 = 𝒉𝒗 =
𝒉𝒄
𝝀
small , high n
= large energy
large , low n
= low energy
Visible light:
400-700 nm
5
Electromagnetic Emission from a Surface
 Two important characteristics:
1. Spectral Distribution
Energy in
angular
band
Energy in
wavelength
band

If you measure the energy
emitted by a surface in a series
of small wavelength bands, it is
usually not evenly distributed
over all wavelengths
MECH 4406: Heat Transfer
2. Directional Distribution

If you measure energy emitted
by a surface it varies with
direction (angle)
Prof. M. Johnson, Carleton University
3
Black Body Radiation
 A black body is an idealized surface with properties chosen to greatly simplify
radiant heat transfer analysis
 Blackbody surfaces are “references” or “standards” for which other “real”
surfaces may be compared
 Blackbodies:
•
•
•
•
Absorb all incident radiation at all wavelengths (no reflections)
Emit maximum possible energy at any given temperature
Emit uniformly on all directions
Emission is still a function of temperature and wavelength
 A close approximation to a blackbody is a small opening into an irregular cavity
Blackbody Radiation
E
E,black body at Ts
E,real body at Ts
E,gray body at Ts

MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Blackbody Radiation
 Spectral Emissive Power of a Blackbody
• Given by Planck’s Distribution
 Emitted radiation varies continually
with wavelength
 At any wavelength, magnitude of
emission increases with temperature
 Shift to left (to shorter wavelengths) as
𝑇 is increased
 Significant portion of energy from sun is
in visible region (black body at 5800K)
Blackbody Spectral Emissive Power
 Spectral Emissive Power of a Blackbody
• Given by Planck’s Distribution:
E  ,b  , T  
C1
 expC 2 T   1
5
where:
𝐶 and 𝐶 are first and second
radiation constants:
𝐶 ≡ 2𝜋ℎ𝑐 = 3.742 × 10 [W ⋅ 𝜇m /m ]
𝐶 ≡ ℎ𝑐 /𝑘 = 1.439 × 10 𝜇m ⋅ K
ℎ  Planck’s const. = 6.626 × 10–34 Js
𝑘  universal Boltzmann constant,
1.3805 × 10–23 [J/K]
𝑐  2.998 × 108 [m/s]
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Blackbody Radiation
 Wien’s Displacment Law:
• To find wavelength of maximum emissive
power:
E  ,b

𝜆
0
𝑇 = 2897.8 𝜇𝑚 𝐾
 Stefan-Boltzmann Law
• Integrate Spectral Emissive Power to find
total emissive power:
𝐸 =
𝐸
,
∴ 𝐸 = 𝜎𝑇
𝐶
𝑑𝜆 = 𝜎𝑇
𝜆 exp 𝐶 ⁄𝜆𝑇 − 1
𝑑𝜆 =
𝜎 = 5.670 × 10
W/m K
Band Emission, F, of a Black Body
 Amount of radiant energy from a blackbody over a finite wavelength band
𝐹
𝐹
MECH 4406: Heat Transfer
→
→
=
∫ 𝐸 , 𝑑𝜆
𝜎𝑇
=𝐹
→
−𝐹
→
Prof. M. Johnson, Carleton University
6
Band Emission, 𝐹, of a Black Body
Use Table 12.2 for simplified (tabulated) calculations:
Simple Radiation Band Pass Example
A window on the international space station (ISS) receives energy at a rate of
100 W from the sun. If the glass transmits visible light with a spectral
transmissivity, 𝜏 , of 0.9 while blocking light at all other wavelengths, what is the
net amount of energy transmitted through the window?
14
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Environmental Radiation
Spectral distribution of incoming
short wavelength solar radiation
Spectral distribution of outgoing
long wavelength environmental radiation
Environmental Radiation
Energy balance on the atmosphere for moderate temperature and cloudy condition
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
8
Energy Balance for Earth
 Energy inputs to the earth:
• Sun (flux of electromagnetic radiation)
• Tides (consequence of earth / moon gravitational
system & viscous dissipation)
• Stellar / cosmic radiation (negligible)
• Decay of radioactive isotopes within earth
(geothermal)
• Heat release from consumption of nuclear and fossil
fuels
– Although small portion of solar energy goes to
produce fossil fuels, this is entirely negligible
compared to usage rates
 For constant earth, heat radiated must equal sum of
heat inputs
17
Energy Balance for Earth – Inputs
 Incoming Solar Flux:
𝑞
= 1 − 𝑟 𝜋𝑅 𝑆
• Solar constant, 𝑆 ≈ 1361 W/m2
– Energy flux from sun just outside earth’s atmosphere
– Spread over multiple wavelengths
– Varies very slightly during the year (elliptical orbit) and with sun activity
• Mean Radius of Earth, 𝑅 = 6.378 × 10 m
• Average reflectivity, 𝑟 ≈ 0.3 (biggest contributor is cloud cover)
𝑞
= 1 − 0.3 𝜋 6.378 × 10
1361 = 1.22 × 10
W
Or averaged over entire surface area of earth (𝐴 = 4𝜋𝑅 = 5.11 × 10
𝑞
=
1 − 𝑟 𝜋𝑅 𝑆
=
4𝜋𝑅
m )
1 − 𝑟 𝑆 = 238.2 W/m
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
9
Energy Balance for Earth – Inputs
 Energy Input for Tidal Dissipation
• 𝑞
≈ 2.4 × 10 W (2.4 Terawatts) (Munk & McDonald, 1960)
• 0.002% of energy absorbed from sun
 Geothermal Energy Input
• 𝑞
≈ 32.5 × 10
W (32.5 Terawatts) (VanHerzen, 1967)
• Includes average heat flow from earth’s interior via conduction (~99%) plus volcanoes and hot
springs (~1%)
• 0.03% of energy absorbed from sun
 Anthropogenic Input from Consumption
• Currently ~15 Terawatts (15 × 10 W)
• 0.012% of energy absorbed from sun
 ∴Solar energy dominates. All other inputs can be treated as perturbations
19
Energy Balance for Earth – Outputs
 Outgoing flux from Earth
𝑞
= 4𝜋𝑅 𝜀𝜎𝑇
• 𝑇 is mean surface temperature of earth ≈286 K (13°C)
• 𝜀 is average emissivity at earth’s surface
 For equilibrium:
𝑞
=𝑞
1 − 𝑟 𝜋𝑅 𝑆 = 4𝜋𝑅 𝜀𝜎𝑇
1 − 𝑟 𝑆 = 4𝜀𝜎𝑇
• If 𝑇 ≈ 286 K, 𝜀 ≈ 0.628
20
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Simple Perturbation Analysis to Earth’s Energy Balance
 Consider perturbation, 𝑃 [W/m2], to earth’s energy balance:
1 − 𝑟 𝑆 + 𝑃 = 𝜀𝜎𝑇
Define 𝛿 as ratio of 𝑃 to solar input: 𝛿 =
𝜀𝜎𝑇 =
1−𝑟 𝑆+𝑃 =
1−𝑟 𝑆 1+𝛿
Resolving for 𝑇 :
𝑇 =
1−𝑟 𝑆
4𝜀𝜎
⁄
1+𝛿
⁄
∴𝑇 =𝑇
1+𝛿
⁄
≅𝑇
1+ 𝛿
Right hand side can be represented with Binomial Series:
Earth’s nominal
temperature
1+𝑥
= 1 + 𝑛𝑥 +
𝑛 𝑛−1
𝑛 𝑛−1 𝑛−2
𝑥 +
𝑥 +. . .
2!
3!
If 𝛿 is small, only need 1st term
∴ Δ𝑇 = 𝑇 − 𝑇
= 𝛿𝑇
21
Simple Perturbation Analysis to Earth’s Energy Balance
 Consider perturbation, 𝑃 [W/m2], to earth’s energy balance:
∴ Δ𝑇 = 𝑇 − 𝑇
𝛿=
= 𝛿𝑇
Tides
Geothermal
0.0047 W/m2
0.0636 W/m2
Human Energy
Consumption
0.029 W/m2
𝛿
0.0000197
0.000267
0.000123
Δ𝑇
0.0014 K
0.019 K
0.0088 K
𝑞
 On their own these direct effects are insignificant relative to solar energy input
22
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
11
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
 Energy Input:
𝑞
= 1 − 𝑟 𝜋𝑅 𝑆
 Energy Output:
𝑞
= 4𝜋𝑅 𝜀𝜎𝑇
 Steady state average temperature
𝑞
=𝑞
1 − 𝑟 𝜋𝑅 𝑆 = 4𝜋𝑅 𝜀𝜎𝑇
1 − 𝑟 𝑆 = 4𝜀𝜎𝑇
𝑇 =
1−𝑟 𝑆
4𝜀𝜎
⁄
23
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
1−𝑟 𝑆
4𝜀𝜎
Steady state Earth Temperature:
𝑇 =
Sensitivity of 𝑇 to 𝑟 and 𝜀: 𝑑𝑇 =
𝜕𝑇
𝜕𝑇
𝑑𝑟 +
𝑑𝜀
𝜕𝑟
𝜕𝜀
𝜕𝑇
1 1−𝑟 𝑆
=
𝜕𝑟
4
4𝜀𝜎
=−
𝜕𝑇
𝜕𝑟
∴ 𝑑𝑇
For small changes in 𝑟 and 𝜀:
=−
1
4 1−𝑟
=−
⁄
−𝑆
4𝜀𝜎
1−𝑟 𝑆
4𝜀𝜎
×
1−𝑟
1−𝑟
⁄
𝜕𝑇
1 1−𝑟 𝑆
=
𝜕𝜀
4
4𝜀𝜎
⁄
=−
𝑇
4 1−𝑟
𝜕𝑇
𝜕𝜀
1 1−𝑟 𝑆
4
4𝜀𝜎
𝑇
=−
4𝜀
⁄
1 − 𝑟 𝑆 −1
4𝜎
𝜀
⁄
1
𝜀
𝑇
𝑇
𝑑𝑟 + −
𝑑𝜀
4 1−𝑟
4𝜀
Δ𝑇 = −
𝑇
𝑇
Δ𝑟 −
Δ𝜀
4 1−𝑟
4𝜀
24
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
12
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
 Sensitivity of 𝑇 to small changes in 𝑟 and 𝜀:
Δ𝑇 = −
𝑇
𝑇
Δ𝑟 −
Δ𝜀
4 1−𝑟
4𝜀
 Consider perturbation required for 1°C change in earth’s temperature
• For change in atmospheric reflectance only:
4 1−𝑟
4 1 − 0.3
Δ𝑟 ≅ −
Δ𝑇 =
= −0.00979
𝑇
286
• For change in emissivity only:
4𝜀
4 0.628
Δ𝜀 ≅ −
Δ𝑇 =
= −0.00878
𝑇
286
∴
Δ𝑟 −0.00979
=
= 3.2%
𝑟
0.3
∴
Δ𝜀 0.00878
=
= 1.4%
𝜀
0.624
 Both 𝜺 and 𝒓 are strongly influence by pollution
 This is the essence of the greenhouse effect
Small
perturbations
indeed!
Similar results no
matter what values
of 𝜀 and 𝑟 are
originally assumed
25
Changes to Reflectivity and Emissivity
https://climate.nasa.gov/evidence/
© U.N. Arctic Monitoring and Assessment Programme
Average Sept. Minimum Arctic Sea Ice Extent (loss of 13.1%/decade)
26
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
13
Environmental Radiation
Spectral distribution of incoming
short wavelength solar radiation
Spectral distribution of outgoing
long wavelength environmental radiation
Environmental Radiation
Energy balance on the atmosphere for moderate temperature and cloudy condition
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
1
Energy Balance for Earth
 Energy inputs to the earth:
• Sun (flux of electromagnetic radiation)
• Tides (consequence of earth / moon gravitational
system & viscous dissipation)
• Stellar / cosmic radiation (negligible)
• Decay of radioactive isotopes within earth
(geothermal)
• Heat release from consumption of nuclear and fossil
fuels
– Although small portion of solar energy goes to
produce fossil fuels, this is entirely negligible
compared to usage rates
 For constant earth, heat radiated must equal sum of
heat inputs
17
Energy Balance for Earth – Inputs
 Incoming Solar Flux:
𝑞
= 1 − 𝑟 𝜋𝑅 𝑆
• Solar constant, 𝑆 ≈ 1361 W/m2
– Energy flux from sun just outside earth’s atmosphere
– Spread over multiple wavelengths
– Varies very slightly during the year (elliptical orbit) and with sun activity
• Mean Radius of Earth, 𝑅 = 6.378 × 10 m
• Average reflectivity, 𝑟 ≈ 0.3 (biggest contributor is cloud cover)
𝑞
= 1 − 0.3 𝜋 6.378 × 10
1361 = 1.22 × 10
W
Or averaged over entire surface area of earth (𝐴 = 4𝜋𝑅 = 5.11 × 10
𝑞
=
1 − 𝑟 𝜋𝑅 𝑆
=
4𝜋𝑅
m )
1 − 𝑟 𝑆 = 238.2 W/m
18
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Energy Balance for Earth – Inputs
 Energy Input for Tidal Dissipation
• 𝑞
≈ 2.4 × 10 W (2.4 Terawatts) (Munk & McDonald, 1960)
• 0.002% of energy absorbed from sun
 Geothermal Energy Input
• 𝑞
≈ 32.5 × 10
W (32.5 Terawatts) (VanHerzen, 1967)
• Includes average heat flow from earth’s interior via conduction (~99%) plus volcanoes and hot
springs (~1%)
• 0.03% of energy absorbed from sun
 Anthropogenic Input from Consumption
• Currently ~15 Terawatts (15 × 10 W)
• 0.012% of energy absorbed from sun
 ∴Solar energy dominates. All other inputs can be treated as perturbations
19
Energy Balance for Earth – Outputs
 Outgoing flux from Earth
𝑞
= 4𝜋𝑅 𝜀𝜎𝑇
• 𝑇 is mean surface temperature of earth ≈286 K (13°C)
• 𝜀 is average emissivity at earth’s surface
 For equilibrium:
𝑞
=𝑞
1 − 𝑟 𝜋𝑅 𝑆 = 4𝜋𝑅 𝜀𝜎𝑇
1 − 𝑟 𝑆 = 4𝜀𝜎𝑇
• If 𝑇 ≈ 286 K, 𝜀 ≈ 0.628
20
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Simple Perturbation Analysis to Earth’s Energy Balance
 Consider perturbation, 𝑃 [W/m2], to earth’s energy balance:
1 − 𝑟 𝑆 + 𝑃 = 𝜀𝜎𝑇
Define 𝛿 as ratio of 𝑃 to solar input: 𝛿 =
𝜀𝜎𝑇 =
1−𝑟 𝑆+𝑃 =
1−𝑟 𝑆 1+𝛿
Resolving for 𝑇 :
𝑇 =
1−𝑟 𝑆
4𝜀𝜎
⁄
1+𝛿
⁄
∴𝑇 =𝑇
1+𝛿
⁄
≅𝑇
1+ 𝛿
Right hand side can be represented with Binomial Series:
Earth’s nominal
temperature
1+𝑥
= 1 + 𝑛𝑥 +
𝑛 𝑛−1
𝑛 𝑛−1 𝑛−2
𝑥 +
𝑥 +. . .
2!
3!
If 𝛿 is small, only need 1st term
∴ Δ𝑇 = 𝑇 − 𝑇
= 𝛿𝑇
21
Simple Perturbation Analysis to Earth’s Energy Balance
 Consider perturbation, 𝑃 [W/m2], to earth’s energy balance:
∴ Δ𝑇 = 𝑇 − 𝑇
𝛿=
= 𝛿𝑇
Tides
Geothermal
0.0047 W/m2
0.0636 W/m2
Human Energy
Consumption
0.029 W/m2
𝛿
0.0000197
0.000267
0.000123
Δ𝑇
0.0014 K
0.019 K
0.0088 K
𝑞
 On their own these direct effects are insignificant relative to solar energy input
22
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
 Energy Input:
𝑞
= 1 − 𝑟 𝜋𝑅 𝑆
 Energy Output:
𝑞
= 4𝜋𝑅 𝜀𝜎𝑇
 Steady state average temperature
𝑞
=𝑞
1 − 𝑟 𝜋𝑅 𝑆 = 4𝜋𝑅 𝜀𝜎𝑇
1 − 𝑟 𝑆 = 4𝜀𝜎𝑇
𝑇 =
1−𝑟 𝑆
4𝜀𝜎
⁄
23
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
1−𝑟 𝑆
4𝜀𝜎
Steady state Earth Temperature:
𝑇 =
Sensitivity of 𝑇 to 𝑟 and 𝜀: 𝑑𝑇 =
𝜕𝑇
𝜕𝑇
𝑑𝑟 +
𝑑𝜀
𝜕𝑟
𝜕𝜀
𝜕𝑇
1 1−𝑟 𝑆
=
𝜕𝑟
4
4𝜀𝜎
=−
𝜕𝑇
𝜕𝑟
∴ 𝑑𝑇
For small changes in 𝑟 and 𝜀:
=−
1
4 1−𝑟
=−
⁄
−𝑆
4𝜀𝜎
1−𝑟 𝑆
4𝜀𝜎
×
1−𝑟
1−𝑟
⁄
𝜕𝑇
1 1−𝑟 𝑆
=
𝜕𝜀
4
4𝜀𝜎
⁄
=−
𝑇
4 1−𝑟
𝜕𝑇
𝜕𝜀
1 1−𝑟 𝑆
4
4𝜀𝜎
𝑇
=−
4𝜀
⁄
1 − 𝑟 𝑆 −1
4𝜎
𝜀
⁄
1
𝜀
𝑇
𝑇
𝑑𝑟 + −
𝑑𝜀
4 1−𝑟
4𝜀
Δ𝑇 = −
𝑇
𝑇
Δ𝑟 −
Δ𝜀
4 1−𝑟
4𝜀
24
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Sensitivity of Earth’s Energy Balance to Changes in 𝜺 and 𝑟
 Sensitivity of 𝑇 to small changes in 𝑟 and 𝜀:
Δ𝑇 = −
𝑇
𝑇
Δ𝑟 −
Δ𝜀
4 1−𝑟
4𝜀
 Consider perturbation required for 1°C change in earth’s temperature
• For change in atmospheric reflectance only:
4 1−𝑟
4 1 − 0.3
Δ𝑟 ≅ −
Δ𝑇 =
= −0.00979
𝑇
286
• For change in emissivity only:
4𝜀
4 0.628
Δ𝜀 ≅ −
Δ𝑇 =
= −0.00878
𝑇
286
∴
Δ𝑟 −0.00979
=
= 3.2%
𝑟
0.3
∴
Δ𝜀 0.00878
=
= 1.4%
𝜀
0.624
 Both 𝜺 and 𝒓 are strongly influence by pollution
 This is the essence of the greenhouse effect
Small
perturbations
indeed!
Similar results no
matter what values
of 𝜀 and 𝑟 are
originally assumed
25
Changes to Reflectivity and Emissivity
https://climate.nasa.gov/evidence/
© U.N. Arctic Monitoring and Assessment Programme
Average Sept. Minimum Arctic Sea Ice Extent (loss of 13.1%/decade)
26
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
Radiative Emission – Intensity of Radiation
 Spectral Emissive Power [W/m2/mm]:
• Radiation emitted per unit surface area at a specific l
⁄
𝐸 𝜆 =
𝐼
where 𝐼
,
,
𝜆, 𝜃, 𝜙 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙
is rate of emitted radiant energy at wavelength 𝜆
in the (𝜃, 𝜙) direction
 Total, hemispherical emissive power [W/m2]
• Integrated over all wavelengths
⁄
𝐸=
𝐸 𝜆 𝑑𝜆 =
𝐼
,
𝜆, 𝜃, 𝜙 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙 𝑑𝜆
Radiation Exchange Definitions
Radiosity, J
Irradiation, G
emitted
Total Energy Reflected
= 𝜌 ≡ Reflectivity
Total Irradiation
reflected
Total Energy Absorbed
= 𝛼 ≡ Absorbtivity
Total Irradiation
surface
Absorption, 𝛼
Transmission, t
Total Energy Transmitted
= 𝜏 ≡ Transmissivity
Total Irradiation
𝜌+𝛼+𝜏 =1
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
7
Radiation Exchange Definitions
 Irradiation, 𝑮: Radiant energy incident on a surface
⁄
𝐺 𝜆 =
𝐼
,
𝜆, 𝜃, 𝜙 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙
where 𝐼
• Total irradiation: 𝐺 =
(Directionally and wavelength dependent)
is rate of incident radiant energy at 𝜆
,
(over all wavelengths and directions)
𝐺 𝜆 𝑑𝜆
 Radiosity, 𝑱: Sum of energy emitted and reflected from a surface
• Includes all radiant energy leaving surface
⁄
𝐽 𝜆 =
𝐽=
𝐼
,
𝜆, 𝜃, 𝜙 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙
𝐽 𝜆 𝑑𝜆
Modelling “Real” Surfaces: Emissivity
 Emission from a real surface is a function of both 𝜆 and 𝑇 (as for a blackbody)
but also of direction
 Simplest model for a real surface is to “correct” black body emission with an
“emissivity”:
⁄
∫ ∫ 𝐼 , 𝜆, 𝜃, 𝜙, 𝑇 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙
𝐸 𝜆, 𝑇
=
Spectral Hemispherical Emissivity: 𝜀 𝜆, 𝑇 ≡
⁄
𝐸 , 𝜆, 𝑇
∫ ∫ 𝐼 , 𝜆, 𝑇 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜙
Total Hemispherical Emissivity:
𝜀 𝑇 ≡
∴𝐸
MECH 4406: Heat Transfer
∫ 𝜀 𝜆, 𝑇 𝐸 , 𝜆, 𝑇 𝑑𝜆
𝐸 𝑇
=
𝐸 𝑇
𝐸 𝑇
= 𝜀𝜎𝑇
Prof. M. Johnson, Carleton University
8
Blackbody Radiation
El,black body at Ts
El
El,real body at Ts
El,gray body at Ts
l
View Factors
 View factor, 𝐹  fraction of radiation leaving surface 𝑖 that strikes surface 𝑗
𝐹 =
MECH 4406: Heat Transfer
1
𝐴
cos 𝜃 cos 𝜃
𝑑𝐴 𝑑𝐴
𝜋𝑅
Prof. M. Johnson, Carleton University
9
View factors for
2-D Geometries
View factors for
3-D Geometries
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
10
Charts of View Factor Equations
Charts of View Factor Equations
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
11
View Factor Relations
 Reciprocity:
• 𝐴 𝐹 =𝐴 𝐹
but note that 𝐹 ≠ 𝐹
 Summation Rule for an Enclosure:
𝐹 =1
 Superposition applies:
𝐹 =
𝐹
View Factor Relations
 For planar or convex surfaces:
• 𝐹 =0
 For concave surfaces:
• 𝐹 0
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
12
Example: Calculating View Factors
 Find 𝐹
and 𝐹
for each of the following geometries:
(a) Small sphere of area 𝑨𝟏
under a concentric
hemisphere of area
𝑨𝟐 = 𝟑𝑨𝟏
(b) Disk sitting centered
beneath a hemisphere
(c) Long, square rod
centered in a long
cylinder. Also find 𝑭𝟐𝟐
39
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
13
Blackbody Radiation Exchange
 Consider transfer between two black bodies:
𝑞→ = 𝐴𝐸
𝐹
,
Fraction of Energy
leaving 𝑖 that
strikes 𝑗
Total Energy
leaving surface 𝑖
• No need to consider reflected energy since all is absorbed by black bodies!
• Similarly: 𝑞
→
= 𝐴𝐸
,
𝐹
Blackbody Radiation Exchange
 Since 𝑞 → = 𝐴 𝐸
surfaces is:
,
𝐹 and 𝑞
𝑞 = 𝐴𝐸
,
→
= 𝐴𝐸
𝐹 − 𝐴𝐸
∴𝑞 =𝐴𝐹
𝐸
,
,
𝐹 , net energy exchange between
and since: 𝐴 𝐹 = 𝐴 𝐹
𝐹
−𝐸
,
,
=𝐴 𝐹 𝜎 𝑇 −𝑇
 Net exchange from surface 𝑖 in multi-surface system:
𝑞 =
MECH 4406: Heat Transfer
𝐴 𝐹 𝜎 𝑇 −𝑇
Prof. M. Johnson, Carleton University
1
Example: Problem 13.14 Drying Oven
 A drying oven consisting of a long semi-circular duct of D = 1.5 m dries material
covering the base of the oven. If the oven walls are maintained at 1200 K, what
is the drying rate per unit length (kg/s*m) if a water coated layer of material is
maintained at 350 K during the drying process? Blackbody behaviour may be
assumed for the water surface and the oven walls.
42
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
 In general, radiation may leave a surface due to emission and reflection, and
when reaching a second surface may be partly absorbed and reflected again
 Can solve this more realistic case with some key assumptions
• Each surface is isothermal
• Each surface is opaque (energy is either absorbed or reflected – no transmission
through)
• No directional dependencies
– Emission, absorption, and reflection are diffuse
• No wavelength dependencies
– Can show that this means 𝜀 = 𝛼
43
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
2
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
 Heat transfer, 𝑞 , from a grey surface:
𝑞
𝑞
− =𝐸 −𝐽
𝜀𝐴
𝐴
𝑞 =𝐴 𝐸 −𝛼 𝐺
𝑞 =𝐴 𝐽 −𝐺
But 𝐸 = 𝜀 𝐸 and 𝛼 = 𝜀
But radiosity 𝐽 = 𝐸 + 𝜌 𝐺
𝑞 =
𝑞 =𝜀 𝐴 𝐸 −𝐺
0 (opaque)
and 𝜌 + 𝛼 + 𝜏 = 1
But 𝐺 = 𝐽 −
and ∴ 𝛼 = 1 − 𝜌
𝑞 =𝜀𝐴
𝐸 −𝐽 +
𝐸 −𝐽
1−𝜀
𝐴𝜀
=
𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍
𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝑞
𝐴
44
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
 Net heat transfer, 𝑞 , from a grey surface:
𝑞 =
𝐸 −𝐽
1−𝜀
𝐴𝜀
=
𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍
𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆
(1)
𝑞
𝐸
1−𝜀
𝐴𝜀
𝐽
• If emissive power a surface would have if it were a black body (𝐸 ) exceeds the radiosity
(𝐽 ), then there is net heat transfer from the surface
• If 𝐽 exceeds 𝐸 , then there is net heat transfer to the surface
• In limit where 𝐴 becomes very large,
→ 0, just as it would if 𝜀 → 1
– Thus, a single large surface surrounded by smaller surfaces acts like a blackbody
45
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
3
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
 For a gray surface in an enclosure:
1.
2. Net heat transfer
from the surface:
Irradiation to the surface is the sum of all
radiosities from other surfaces in the enclosure:
Using reciprocity:
𝐴𝐺 =
𝐹 𝐴𝐽
𝐴𝐺 =
𝐴𝐹 𝐽
𝑞 =𝐴 𝐽 −𝐺
Subbing in (2):
(𝐴 𝐹 = 𝐹 𝐴 )
Which simplifies to:
𝑞 =𝐴
𝐹 𝐽 (2)
𝐺 =
𝐽 −
Rewrite noting ∑
𝑞 =𝐴
𝐹 𝐽
𝐹 = 1:
𝐹 𝐽 −
𝐹 𝐽
∴Net heat transfer from the surface:
𝑞 =
𝐴𝐹
𝐽 −𝐽
=
𝑞
(3)
46
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
• This suggests that the net radiative heat transfer from
surface 𝑖, 𝑞 , is sum of components related to radiative
exchange with other surfaces
• Each component can be represented by a network
element:
 Rearranging (3), net heat transfer from
gray surface in an enclosure is:
𝐽 −𝐽
𝑞 =
=
𝑞
𝐴𝐹
 (𝐽 −𝐽 ) is the driving potential

 Subbing in (1):
𝑞 =
𝑞 =
𝐴𝐹
is the space or geometrical resistance
𝐸 −𝐽
1−𝜀
𝐴𝜀
𝐸 −𝐽
=
1−𝜀 ⁄ 𝐴 𝜀
𝐽 −𝐽
𝐴𝐹
(4)
• Can solve using a “Radiation Network”
47
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
4
Radiation Exchange between Opaque, Diffuse, Grey Surfaces
 Instead of networks, can also use “Direct Approach” to solve (4):
𝑞 =
𝐽 −𝐽
𝐸 −𝐽
=
1−𝜀 ⁄ 𝐴 𝜀
(4)
𝐴𝐹
 For every surface where 𝑇 is known, write (4a):
𝐸 −𝐽
=
1−𝜀 ⁄ 𝐴 𝜀
 For every surface where 𝑞 is known, write (4b):
𝑞 =
𝐽 −𝐽
𝐴𝐹
𝐽 −𝐽
𝐴𝐹
 Gives a system of 𝑁 linear algebraic equations that can be solve for 𝐽 , 𝐽 , 𝐽 , … 𝐽
 Then solve for 𝑞 or 𝑇 using:
𝑞 =
𝐸 −𝐽
1−𝜀 ⁄ 𝐴 𝜀
48
Radiant Exchange in Grey, Two Surface Enclosures
 Simplest case is
for a two-surface
enclosure
• Because there are only two surfaces:
𝑞 = −𝑞 = 𝑞
 Writing radiation network:
• Since 𝐸 = 𝜎𝑇
∴ 𝑞 = −𝑞 = 𝑞
=
𝜎 𝑇 −𝑇
1−𝜀
1
1−𝜀
+
+
𝐴 𝜀
𝐴 𝐹
𝐴 𝜀
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
5
Radiation Shields
• High reflectivity (low 𝛼 = 𝜀) surface(s) inserted between
two surfaces to reduce radiation exchange
𝑞
1−𝜀
𝐴 𝜀
1
𝐴 𝐹
1−𝜀 ,
𝐴 𝜀,
1−𝜀 ,
𝐴 𝜀,
1
𝐴 𝐹
1−𝜀
𝐴 𝜀
𝑞
𝜀
∴ 𝑞 = −𝑞 = 𝑞
=
𝜎 𝑇 −𝑇
1−𝜀
1−𝜀
1−𝜀
1
+
+
+
𝐴 𝜀
𝐴 𝐹
𝐴 𝜀
𝐴 𝜀
+
1
1−𝜀
+
𝐴 𝐹
𝐴 𝜀
𝑞
,
𝜀
,
𝐴 ,𝑇
Example: Problem 13.33
 A flat-bottomed hole 8-mm in diameter is bored 24-mm deep into a diffuse, grey
material having an emissivity of 0.8 and a uniform temperature of 1000 K.
a) What is the radiative power leaving the opening of the cavity?
b) What is the effective emissivity, 𝜀, of the cavity (defined as the ratio of the radiant
power leaving the cavity to that of a blackbody with the same area as the cavity
opening at the same temperature as the cavity)
51
MECH 4406: Heat Transfer
Prof. M. Johnson, Carleton University
6
MECH 4406
Major Topics
0. INTRODUCTION TO HEAT TRANSFER AND HEAT TRANSFER MODES
Book
Chapters
Approx. #
of Lectures
1
2
I. CONDUCTION
Introduction; thermal resistance concept
2
3
3,4
4
5
3
Introduction to convection; boundary layers; convection coefficients
6
1
Convection in external flows; flat plate boundary layer
7
2
Convection in internal flows; Nusselt number relationships
8
2
HEAT EXCHANGERS: Log-mean temperature difference (LMTD) method
11
1
HEAT EXCHANGERS: NTU-Effectiveness method
11
1
Free / natural convection
9
1
Steady state conduction, 1D & 2D; Fins
Transient conduction; Lumped capacitance, semi-infinite solids
II. CONVECTION
III. RADIATION
Introduction to radiation, environmental radiation
Viewfactors, grey bodies, grey body exchange
MECH 4406: Heat Transfer
12 , 13
2
13
2
Prof. M. Johnson, Carleton University
7