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I
_
ME1351 : HEAT AND MASS TRANSFER
_
[ ANNA UNIVERSITY SYLLABUS
hanical Engineering -(R'egulation
For B.E. VI Semester Mec
----
CONTENTS
:201l4f,
l. CO~OU,~TlONt_
Mechanism of Heat Transfer
- Conduction
BaSIC
Concep!) ...
f
'
.
d Radiation - General Differential
equation
0
Heat
Convection an
..
d C I'
.
y mdncal
Con ducii
UCllon - FOllrl'er Law of Conduction - Cartesian an
.
Steady State Heat . Conduction
C oor d·lilates - One Dimensional
Conduction through Plane Wall, Cylinders and Spherical Systems _
Composite Systems - Conduction
with Internal Heat Generation _
Extended Surfaces - Unsteady Heat Conduction - Lumped Analysis _
Use of Heislers Chart.
2. CONVECTION
Basic Concepts - Convective Heat Transfer Coefficients - Boundary
Layer Concept - Types of Convection
- Forced Convection
_
Dimensional Analysis - External Flow - Flow over Plates, Cylinders and
Spheres - Internal Flow - Laminar and Turbulent Flow - Combined
Laminar and Turbulent - Flow over Bank of tubes - Free Convection _
Dimensional Analysis - Flow over vertical plate, Horizontal
plate,
Inclined plate, Cylinders and Spheres.
3. PHASE CHANGE HEAT TRANSFER
AND HEAT
EXCHANGERS
Nusselts theory of condensation
- Pool boiling, flow boiling,
correlations in boiling and condensation. Types of Heat Exchangers _
LMTD Method of Heat Exchanger Analysis -- Effectiveness
_ NTU
method of Heat Exchanger Analysis - Overall Heat Transfer Coefficient _
Fouling Factors.
4. RADIATION
CHAPTER 1: CONDUCTION
~H;at
1.1.1.
1.1.2.
1.1.3.
1.2.
Critical
I? I
1:2:2:
1.3.
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Data
Book
is
1.3.5.
1.3.6.
1.4.
Radius
of Insulation
Critical Radius of Insulation
Solved Problems
Heat Conduction
1.3.1.
1.3.2.
1.3.3.
1.3.4.
Basic Concepts - Diffusion Mass Transfer - Fick's law of diffusion _
Steady State Molecular Diffusion - Convective
Mass Transfer
_
Momentum, Heat and Mass Transfer Analogy _ Convective
Mass
Transfer Correlations.
Transfer
1.1
Modes of Heat Transfer _
I. I
Fourier Law of Conduction
.. .1.2
General Heat Conduction Equation in
Cartesian Co-ordinates
. 1.2
1.1.4. General Heat Conduction Equation in
Cylindrical Co-ordinates
1.9
1.1.5. Conduction of Heat through a Slab or Plane Wall..
.1.14
1.1.6. Conduction of Heat through a Hollow Cylinder
1.16
1.1.7. Conduction of Heat through a Hollow Sphere
1.17
1.1.8. Newton's Law ofCooling........................
.
1.19
1.1.9. Heat Transfer through a Composite Plane Wall
with inside and Outside Convection
1.19
1.1.10. Heat Transfer through Composite Pipes (or) Cylinders
with Inside and Outside Convection
1.22
1.1.11. Solved Problems 0" Slabs
1.25
1.1.12. Soilled University Problems 011 Slabs
1.74
1.1.13. Solved Problems 011 Cylinders
1.111
1.1.14. Solved University Problems 011 Cylinders
1.144
1.1.15. SO/lied Problems 011 Hollow Sphere
1.160
Basic Concepts, Laws of Radiation - Stefan Boltzman Law, Kirchoff
Law - Black Body Radiation - Grey Body Radiation Shape Factor
Algebra - Electrical Analogy - Radiation Shields - Introduction to Gas
Radiation.
5. MASS TRANSFER
Note : . (Use of Standard Heat and Mass
pernulled 117 the University Examination).
Transfer
.
1.167
for a Cylinder
1.167
1.169
1.179
with Heat Generation
Plane Wall with Internal Heat Generation
Cylinder with Internal Heat Generation
Internal Heat Generation - Formulae Used
Solved Problems 011 Plane Willi with Internal
Heat Generation
Solved Problems 011 Cylinder with
Internal Heat Generation
Solved Problems Oil Sphere with
lnternul Heat Generation
1.179
1.183
1.185
1.187
1.196
1.202
T;~~·~·~t~·Fi;~~·::::::::·
..::::·.:·.·.:::·.:::::::::::::::
..::=:':::::::
::~~~
~.i:SI.
1.4.2. Temperature Distribution and Heat
Dissipation in Fin
1.4.3. Application.........
.
.
1.206
1.21
r.:
1.4.'-l.
Fill Ftliciellc)'
·· .
..1.217
1.~.5. Fin rfkcriVt'ness.
1.~.6. Ftll"lllllllicUsed.....
I A. 7. So/I't!d Proh/ellH"
1.4.8. SII/I'd U"itl(!f.5i~I' Prublctn«
1.4.9. Pr()hkllll/Or
Practice ················ ..·
15.
Transient Heat Condul~tioll (or) Unsteady
Conduction
" .. 1.217
1.2IS
.
1.219
1.245
1.263
1.264
1.264
1.266
1.269
tteot AII(I~I'jiJ ........•..........•..•......••....•...•..•.........•.•...••.
Heat Flow in Semi-lnfiutie Solids
1.288
SO/lied Problems - Semi-illfillite
Transient Heat Flow in an Infillite
2.7.
2.8.
... 1.329
1.5.8.
1.332
I.S.9.
1.351
2.9.
1.374
2.1.1.
HEAT TRANSFER
-..-..-..-..-..-..- -..-..-..-..-..-..-
Dimensions
... 2.1
2 1.2. Buckingham 1I Theorem.
. ..
2.1.3. Advantages cf Dimensional Analysis
.
.
2.14. Limitations of Dimensional
Analysis
Dimensionless Numbers and their Physical
2.2.
2.11.
2.2.1. Reynolds Number (Re)
2.2.2. Prandrl Number (Pr)
2.2.3. Nusselt Number (Nu)
2.4
2.4
2.5
2.11.3.
2.12.
~e\Vlonion and Non-Newtollioll
Fluids
2.6
EL;;;~~~:~~::;~p·:~:.::::·:::::.::::~~~:::
..:::::::::::::::::::~~::::::::::::::~:7;
2.3.
L~;~;~·:::············································";'8
2.4.
2 ~ I. Types of Boundary
2.:.~. !iydrodynalllic
Boundary
2.).). r~lenn;]IUoUfldarylayer
i~lIlve~~~:lt~;;:~
2
1.~··~·r:··
y .. ·
'l'~~'"rC' ..···..·..:
-u. Types ofC~nveoc!i
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······· .. ······· .. ····2·9
·····
2·9
,..:::::::::::::::::::::::::::::::::2: 9
onveC!rOfl
011....
··
.
2.9
2.9
2.10
for Combination
Flow...
2.13
of
Formulue Used lor Flow Over Balik of Tubes
Solved Problem
it
Cylinder
-Internal
Flow
Formulae usedfor Flow tit rough
Cylinders (lnternul flow)
S;)lved Problems - Flow through Cylinders
(lnteruat Flow)
Solved University Problems - Internal Flow
Formulae Used/or Free Convection
Solved Problems 011 Free Convection (or)
Natural Convection
2.12.3. Solved University Problems - Free Convection
Problems for Practice
TII'o Murk Questions {lilt! Allswers
CHAPTER III:
3.1.
for
Free Convection
2.12.1.
2.12.2.
2.13.
2.14.
Coefficients
and Spheres
~;~~t~::t~~r~,~~::;
(~~?••.•.••••••••••••••.•••••••••••.•••••••••.•.••••.•.•
;;
2.2.6.
for
Formulae Usedfer Flow Over Cylinders
and Spheres
Solved Problems - Flow Over Cylinders
Flow through
2.11.2.
2.4
Coefricients
Flow over 'lalli, of Tubes
2.11.1.
2.3
Significance
2.9
..2.10
Problems 011 Flat Surfaces - Forced Convection
Solved University Problems 011 Flat Surfaces Forced Convection
How over Cylinders
2.10.1.
2.10.2.
2.2
2.3
.
..
Laminar and Turbulent
.. 2.15
Layer Thickness, Shear Stress and Skin Friction
Coefficient for Turbulent Flow
2.IR
Heat Transfer 1'1'0111 Flat Surfaces - Formulae Used
2.23
2.9.2.
:~~
ocificient
( .3
Boundary
2.9.1.
2.10.
CHAPTER II : CONVECTIVE
2.1.
I)i 111(~ns
iona I A IIa lysis
l leat Transfer
2.8.1.
2.8.2.
1.30R
Solids
Plate
Free (or) Natural Convection
Forced Convection
The Local and Average Heat Transfer
Flat Plate - Laminar Flow
The Local and Average Heat. Transfer
Flat Plate-Turhulcnt
Flow
2.6.1.
1.306
Solved Problems - lufintie Slllitl,·
SO/lied University Problem" - Infinite Solids
1'11'0Mark QlleJtiOlIl
& AII.'"II'en
1.6.
2.5.
2.6.
State
1.5. I. l3ior Number .
.
1.5.:? Lumped Heat Anal)' is
.
1.5.3. Solved Problems -1_llIlIped lleat AII(I~I/JiJ ....•.........•.
'.5.4.
Solllcd University Prohlelll.,·-Llllllped
I. -.S
1.5.6.
1.5.7.
2.4.3.
2.4.4.
( 'onteuts
2.26
2.83
2.115
2.116
2.117
2. 122
2.123
2.124
2.126
2.127
2.129
2.150
2.162
2.162
2.165
2.194
2.217
2.219
PHASE CHANGE HEAT TRANSFER AND HEAT
EXCHANGERS
Boiling and Condensation
~.I
3.1.1.
Introduction
.
).1
.... 3.1
Boiling
.
3.1.2.
Condensation
3.1
3.1.3.
. .
.3.1
Applications
.
3. 1.4.
C.4
Heat and Mass Tram/a
Contents
3.1.5.
3.1.6.
3.1.7.
Boilll1g Heat Transfer Phenomena
Flow Boiling...
········ .. · ·
Boiling Correlations
3.1.8.
3.1.9.
Solved Prohlellls
Solved A11IUIUniversity
3.2
3.4
J.)
3.7
3.23
Problems
.
3.1 10. Condensation.
3.1.11. Modes of Condensation ·
3.1.12. Filmwise Condensation
3.1.13. Dropwise Condensation ..·
3.1.14. Nusselt's Theory for Film Condensation
3.1.15. Correlation for Filmwise Condensation Process
3.29
3.29
3.29
3.30
3.30
3.30
3.1.16.
3.2.
Solved Problems Oil Laminar Flow,
Vertical Surfaces
3.1.17. Solved Problems Oil Laminar Flow,
Horizontal Surfaces
3.1.18. Solved Anna University Problems
3.1.19. Problems for Practice
Heat Exchangers
3.2.1.
3.2.2.
3.2.3.
3.2.4.
3.2.5.
3.2.6.
3.2.7.
3.2.8.
3.2.9.
3.32
3.54
3.61
3.65
3.66
Introduction
Type of Heat Exchangers
Logarithmic Mean Temperature Difference (LMTD)
Assumptions
Logarithmic Mean Temperature Difference for
Parallel Flow
Logarithmic Mean Temperature Difference for
Counter Flow
Fouling Factors
Effectiveness by Using Number of
Transfer Units (NTU)
3.66
3.66
3.73
3.73
3.73
3.77
3.81
Flow Heat £\:cllangers
,
3.2.10. Problems 011 Cross Flow Heal Exchangers (or)
3 2 I Shell and Tube Heal Exchangers
3'2'1~' Solved Anna UI1iversity Problems
3'2'13' Solved Problems Oil NT(! Method
m1
3'2' 14' ; b"1University Solved Problems
"
. ro ems for Practice
3.2.15. Two M k
.
..· ..·..·
·
ur Questions and AI1swers
Introduction
Emission Properties
3.82
4.29.
Electrical Network
by Using Radiosity
3.1 09
3.117
3.J24
3.138
4.30.
4.31.
Radiation of Heat Exchange
Solved Problems
3.145
3.146
..
·
·
·
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·
·
Emissive Power
4.1
Monochromatic
Emissive Power
4.2
Absorption, Reflection and Transmission
4.2
Concept of Black Body
4.3
Planck's Distribution
Law
4.4
Wien's Displacement Law
4.4
Stefan-Boltzmann
Law
·
4.5
Maximum Emissive Power
4.5
Emissivity
4.6
Gray Body
4.6
Kirchoff's Law of Radiation
4.6
Intensity of Radiation
4.6
Lambert's
Cosine Law
4.7
4.16. Formulae Used
4.7
4.17. Solved Problems
4.8
4.18. Solved University Problems
4.25
4.19. Radiation Exchange Between Surfaces
4.31
4.20. Radiation
Exchange Between Two Black Surfaces separated
by a Non-absorbing
Medium
4.31
4.21. Sha pe Factor
4.36
4.22. Shape Factor Algebra
4.36
4.23. Heat Exchange Between Two Non-Black (Gray)
Parallel Planes
4.37
4.24. Heat Exchange Between Two Large Cocnentric Cylinders or
Spheres
4.41
4.25. Radia tion Shield
4.45
4.26. Solved Problems
4.27. Solved Problems 011 Radiation Shield
4.28. Solved University Problems
CHAPTER IV : RADIA nON
4.1.
4.2.
4.3.
4.4.
4.5.
4.6.
4.7.
4.8.
4.9.
4.10.
4.11.
4.12.
4.13.
4.14.
4.15.
3.82
Problems on Parallel Flow and Counter
4.1
4.1
C.5
Analogy for Thermal
and Irradiation
for Three
4.49
4.60
Radiation
Gray Surfaces
4.79
Systems
4.IOO
4. 104
4.105
4.32.
University
Solved Problems
4.33.
Radiation
from Gases and Vapours
4.129
4.34.
M ea n Bea m Length
4.154
4.153
4.35.
Solved
Problems
· 4.155
4.36.
Problems for Practice
4.166
4.37.
Two Mark. Qlte.5tiOl1!iand Al1swers
4.168
~C~.6~~R~ea~t~a~n~d~~~a~s~s~r.~ra~n~sfi~e_r
-
=C-H-AP-T-E-R-V-:~M7.A~S~S~T~RA~NS~F~E~R~-----------------5.1.
5.2.
5.3.
5.4.
5.5.
5.6.
5.7.
5.8.
5.9.
~.I O.
~.II.
5.12.
5.13.
5.14.
5.15.
5.I6.
5.17.
5. J 8.
5.19.
5.20.
5.21.
5.22.
5.23.
5.24.
5.25.
5.26.
5.27.
5.28.
5.29.
5.30.
5.31.
J ntroductlon
·
.
Modes of Mass Transfer
·..·
·
· ·..·
Diffusion Mass Transfer ..·
·..· ··
·
·
·
Molecu~ar ~iffusion
·..·
·
·
·..·..·
·
Eddy Dlffuslon
Convection Mass Transfer
·..·
·
Cocentrations
·· · ·..··
··
·..··..··
Fick'~ Law of Diffusion
·..·..·..·
Steady State Diffusion through a Plane Membrane
So/J'ed Problems Oil Concentrations
Solved Problems Oil Membrane
Solved Univeristy Problems on Membrane
Steady State Equimolar Counter Diffusion
Solved Problems Oil Equimolar Counter Diffusion
Solved University Problems 011 Equimolar
Counter Diffusion
Isothermal Evaporation of Water into Air
Solved Problems on Isothermal Evaporation
of Water into Air
Solved University Problems Oil Isothermal Evaporation
of Water into Air
Convective Mass Transfer
Types of Convective Mass Transfer
Free Convective Mass Transfer
Forced Convective Mass Transfer
Significance of Dimensionless Groups
Formulae Used for Flat Plate Problem.')
Solved Problems on Flat Plate
Anna University Solved Problems 011 Flat Plate
Formulue Used for Internal Flow Problems
Solved Problems on Intemal Flow
University Solved Problems
Problems for Practice
Two Mark Questions and Answers
------
5.1
S.1
S.1
5.2
5.2
5.1
5.2
5.3
5.4
5.6
5.17
5.21
5.23
5.26
cr
Basic Concepts
CF
General Differential
Equation
5.31
5.34
0"
Fourier Law of Conduction
C7
Internal Heat Generation
c:r
Extended Surfaces
c-
Unsteady Heat Conduction
cr
Solved Problems
(7'
Solved University
5.35
5.44
S.54
5.54
5.54
5.S4
5.54
5.56
5.57
5.65
5.68
5.69
5.72
5.75
5.76
ANNA UNIVERSITY SOLVED QUESTION PAPERS ........ S.1 - S.71
DO
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Chapter 1: Conduction
Problems
CHAPTER-I
1.CONDUCTION
1.1 HEAT TRANSFER
Heat transfer can be defined as the transmission
from one region to another region due to temperature
of energy
difference.
1.1.1 Modes of Heat Transfer
* Conduction
* Convection
* Radiation
Conduction
Heat conduction is a mechanism of heat transfer from a region
of high temperature to a region of low temperature within a medium
(solid, liquid or gases) or between different medium in direct physical
contact.
In conduction, energy exchange takes place by the kinematic
motion or direct impact of molecules. Pure conduction is found only
in solids.
Convection
Convection is a process of heat transfer that will occur between
a solid surface and a fluid medium when they are at different
temperatures.
Convection
is possible only in the presence offluid
medium.
Radiation
The heat transfer from one body to another
without any
transmitting medium is known as radiation. It is an electromagnetic
wave phenomenon.
'2
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I.': Heat tnd HII.\.\' Transfer
1.1.2 Fourier
Conduction
tal" or Conduction
Rate of heal conduction is proponional 10 the area mea lIred
iirection of heat OO\V and io the temperature gradient
1.3
O.
1101'111:11 to the
in that direction.
Q
O.
°C.·Ch)
Element volume
\ here
A - Area in 111-
dT _ Temperature gradient in k/m
dr
k - Thermal
nducti iry in W/m"Thermal conducti
to conduct heat.
Fig. 1.1.
it)' is defined a the abilit
fa
ub tan e
[The negative sign indicates that the h at 0 w in a dire ti n
along which there is a decrease in temperature]
1.1.3 General heat conduction
cartesian coordinates
equation
Consider a small rectangular
Net
heat
conducted
into
element
element
fide
The energ balance of this rectangular
from first law of thermodynam ics.
element
all the
coordinate
directions.
Let q x be the heat flux in a direction
q¥
d x be the
heat flux in a direction
The rate f heat' fl '"nine. t th
the face AB 0 i
in
from
f face EF
and
H.
e Iernent .In x diirection
I Q,
dx, d I and
of face ABO
dz
I
through
... (1.2
d: as shown in Fig.I.I.
where
=>
Net heat
conducted into
element from
all the coordinate
directions
l
Heat
generated
element j
t
Heat
st red
=
l
hermal
The rate
heat fl \
the fa e EFGH i
Q +dx
in rhe
elern nt../
...
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k
nducti
btain d
ernperature
1
\\ ithin the
i
1.1
ity, W/mK
gradient
f tJre e Iernent In
. x directi
ut
Q
ax
-k
-d
x
) dx
aT
I
d:
n thr
ugh
/4 Heata~_
..
Subtracting ( 1.2) - (1.3)
Ox - Q(I' + dxl
= -k
Conduction 1.5
~
Net heat conducted
directions
.
aT dydz _I-k ~.QI dydz=
x
l. . ox
AX
~[ :.[
kx :]
+
~
aT Jdx dy dZ]
ax [kx ax
into element from all the coordinate
M
ky :]
+
![
k, :]
] dx dy dz
... (1.7)
er dydz + kx -8of dydz +
= ·-k. -ax
x
Heat Stored in the element
.t
We know that,
=>
Q _Q
.I'
(.I'
+ dx)
or] dx dy dz
= .1_
ax [kx ax
{
... (1.4)
He~t stored}
m the
element
Mass Of} { SpeCifiC}
the
x heat of the
element
element
= {
m x Cp?<
!
Similarly
Q)' .- Q (y + (M =
~
[k
y
:;]
dx
dy dz
{
Rise in }
temperature
of element
aT
at
,.
.•. (1.5)
x
aT
p x dx dy dz x Cp x
at
[v Mass = Density x Volume]
••• (1.6)
Heat stored in }
er
{ the element
= p Cp
dx dy dz
at
... (1.8)
Adding (1.4) + (1.5) + (1.6)
Net heat conducted = a~
[k'l: g: Jdt dy dz +
Heat generated within the element
Heat generated within the element is given by
Q
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=
q dx dy dz
... (1.9)
1.6 Heal ami Mass Tran.~ler
Substituting
(\'1) ~
eqllation
(1.7), (1.8) and (1.9) in equation
I;_ [k\ ~ 1 c [k,. c~] a~ [k: ~ J]
..
,\
+ ....
0'
+
0·
0_
Conduction
{I.I)
Case (i) : No heat sources
In the absence
dx L~l' dz
reduces
of internal
heat generation,
equation
or
..•
(1.10)
to
02r +-+02r 02r
ax2
0,2
az2
P Cp or dx dy dz
+ q dx dy dz
I.?
at
This equation
equation.
oc
at
is known as diffusion
equation
(1.11 )
(or) Fourier's
Case (ii) : Steady state conditions
Considering
the material
is isotropic.
In steady
So,
with time.
k, = ky = k, = k = constant.
reduces
&r &r -~r]
-+-+-k+q=pCax2
0~
c:z2
or
P at
to
So,
state condition,
or
at'
iJ2r
02r
0'2
o:z2
q
+k
does not change
equation
=0
...
(1.10)
(1.12)
(or)
iJ1r
in
+-+-+0,1 &2
q
k·'
p Cp
=--
or
It is a general three dimensional
in cartesian coordinates
or
)
q
V-T + k
at
k
a:
through
the temperature
O. The heat ~onduction
+-+-
Divided by k,
where,
=
at
...
heat conduction
(1.10)
In the absence
becomes :,
is known
of internal
as Poisson's
equation.
heat generation,
equation
is nothing but how fast heat is diffused
changes of temperature
with time.
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equation
...
k
a: =;: Thermal diffusivity
= -- m2/s
.
.
..
pCp
.
Thermal diffusivity
a material during
This equation
= 0
(or)
This equation
is known
as Laplace
equation.
(1.12)
(1.13)
I. 8 Heal and Ma.H' Transfer
Conduction J, 9
Out! (iii): One dimensional steady state "e(lf condllcliOll---
If the temperature varies only in thex direction the e
.
, ,quatloll
(1.10) reduces to
q
o2T
----
1
ax';
z-
0
•.• (1.14)
I.;
In the absence of interns I heat generation,
becomes:
equation ( 1.14)
'"
(1.15)
1.1.4 General
Heat
Conduction
Equation
in Cylindrical
Co-ordinates
The general heat conduction
equation
in cartesian
coordinates derived in the previous section is used for solids with
rectangular boundaries like squares, cubes, slabs etc. But, the
cartesian coordinate system is not applicable for the solids like
cylinders, cones, spheres etc. For cylindrical
solids, a cylindrical
coordinate system is used.
Consider a small cylindrical
as shown in fig.I.2.
element of sides dr, dcj> and dz
Case (iv): Two dimensional steady slate "eat conductio"
If the temperature varies only in the x and y directions,
equation (I. 10) becomes:
...
In the absence of internal heat generation,
redcues to
the
(1.16)
equation
I
(I. 16)
: dr
I
J~_(r,4J,z
tYT
ax2
if1
-j----
oyl
/
/
=0
...
(I. 17)
Elemental
'
volume
/
dz
Case (,~: Unsteady state, one dimensional, without internal
heal generation
:
Q(r+dr)
, oin un~teady state, the temperature
i.e.,
-a,
:t:
O. So, the general conduction
changes with time,
equation (I. J 0) reduces to
Fig.J.2
()
o:
("'"
...
(1.18)
The volume of the element
dv = r d~ drdz .
I Let us assume
that thermal conductivity
and density p are constant.
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k, Specific heat ep
1.10 Hea/I.Jf1J Mass Transfer
The energy balance of this cylindrical
from first law of rhrmodynamiocs.
I
Net heat
conducted into
element from
all the coordinate
directions
Conduction 1.11
element is obtained
Heat entering in the element
(Heat
}
I stored
= ~
in the
Ne! heat conducted into element from
Heat leaving from the element through (~, z) plane in time de.
L element
...
Q r + ell' = Q r + _E_(O
)dr
or
r
(1.19)
(II/ the co-ordinate
Net heat conducted
time de.
= Qr-
directions
-
/
~ de
az
into the element through (~, z) plane in
0,. + dr
= -
-#- (Q ) dr
= -
:,.
Heat entering in the element through (r, ~) plane in time de
Q_ = -k (r d~ dr)
or
r
[-k (rd~.dz). [ :Jde Jdr
Heat lea ing from the element through (r, ~) plane in time de.
Q=
T
d: =
Net heat conducted
time de.
k (dr d~.dz).
o
Q= + oz (Qz) dz
+ aT]
L1/:
or
or
de
into the elem~nt through (r, ~) plane in
= Q= -
Qz + dz
Net heat conducted
through (~, z) plane
. = - -#-- (Q_) dz
az
=
in time dB.
cr de
or
Or=-k(rd~dz)
Heat}-'
generated
+
within the
{
element
(~, z) plane
through
-
[~T
_!_ ~TJ de
a,.1 r cr
k (dr. rd~.dz)
+
I... (1.21)
! [k (rde.dr). [~J de J dz
Heat entering in the element through
Q~ = -k tdr.dz)
-- k. [OZTJ.
0:;2
(dr~rd~.dz) de
(z, r) plane
in time de.
or de
ra~
Heat leaving from the element through (z, r) plane in time de.
Net heat couducred
through (r,~) plane
_
- k
OZT
.
[oz2] (dr.rdq,.dz)de
l
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...
(1.20)
I. 12 Heat and Mass Transfer
Net heat conducted
into the element
rime de.
Conduction 1.1J
(z, r) plane in
t,
Net heat conducted into element from all the co-ordinate
- ,.a~ (Q+) rd$
Q, - Q~ ., d~ =
= _ __L [
ro$
=
through
directions
= k (dr
-k tdr dz). or de] rd¢
ro~
1_
...
k -04>a ['-r D~
-OT] (dr dtkIf dz) de
= k
Total heat generated
... (1.22)
= Net
r&T + .!.._ OT]
de
0,.
heat stored in the element
Substituting
equation
or x de
...
00
C~~n
(1.19)
::::::>
k (dr rd$ dz) de
(dr rd~ dz) de
(1.23). (1.24) and (1.25) in (1.19)
1- &T + L or + .L &T
_ 0,.2
,. a,.
,.2
acp2
+ q (dr rd</>dz) de
1.20, 1.21 and 1.22J
= k (dr rdql dz) de
r- &T + iYT +_}_ oT
p (dr rd$ dz) ep L'T x de
re
oz2
or2
,. or
I a2r J+-;2 ~)~2
Divided
::::::>
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(1.25)
r
+
[Adding equation
is equal to
Increase in internal energy
= p (dr rd$ dz)Cp
Lor2
. .. (1.24)
The increase in internal energy of the element
the net heat stored in the element.
+
k (dr rd~ dz)
by
Heat stored in the element
Net heat conducted into element from all the co-ordinate
directions
k ~; (dr rd$ dz) de
k
within the element is given
Q = ci (dr rd</>dz) de
_r2
WJ
(1.23)
Heat generated within the element
[l_ 8$2
&TJ (dr rd~ dz) de
.
/Net heat conducted _ . ['
cPT .
L~hrOugh (z, r) plane - k 7i
(dl rd$ dz) de
I -&T +&T-/
r 0.2T
+.!.._ aT
r2 ocp2 i3z2 _
or2 r or +--
rd</>dz) de
k
by (dr rd$ dz) de
r &T + _!_,. 8,.OT_ + .L 0$2
&T + &TJ' + .
oz2
q =- p. C, :-
L or2
,.2
I
~!
J-
OZ~
Conduction 1./5
.
From Fourier law of conduction,
Q=-kA
...
It is a ~general
in c~'lindrical
flT
OIl
three dimensional
heat conduction
=
=>
equat' IOn
Q.dr = -k A dT
ae
.
the above equation
Integrating
l_ aT
ex
L
=>
r
a,.
= 0
...
1.1.5 Conduction
T2
Q f dr = - kA
=>
=>
l,. . drdT J 0
...
=
of heat through
o
TI
Q [L - 0] = -k A [T2 - Tj]
Q = kA [T 1 - T 2]
a slab or plane wall
T 2'
~T overall
R
where
~T = T1-
Fig 1.3
...
(1.30)
L
kA
Q =
L --.J
(1.29)
TI- T2
Q
Let us consider a small
elemental area of thickness 'dx'.
...
L
--J d~'1---
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T2
Q [x] =-k A [T]
(1.28)
Consider
a slab of
uniform thermal conductivity
k, thickness
L, with inner
temperature
T I, and outer
temperature
f dT
TI
L
(1.27)
(or)
_!__ _{__
,. dr
1
TI'
o
&T + _!_ aT
tl e limits of 0 to L
T2
L
and no heat generaion ,
( 1.26) becomes:
13,.2
etween
f Q dr = - f k A dT
o
equation
b
and TI to T2·
=>
If the flow is steady, one dimensional
dT
dr
(1.26)
co-ordinates.
+ L aT + .L &T + &T + ~
r r
,.2 13cp2
az2 k
we knoW that,
R=
T2
C. - Thermal resistance of slab.
~~~~~~~~--------~----------/.16 Heal and MLI.\"J Tram/er
-
l.L
6 C' nduction of Heat Through Hollow Cylinder
Conduction 1.17
0
onsidcr a hollow cylinder
ofillllcr radiu rl' outer radius r2,
inner tcrnperatllr~
T I' outer
temperature
T2 and thermal
Q==
Q
cOllducti, it ".
Let II c 11 idcr a small
elemental area of thickness "dz"
From Fourier
law
c nduction, we know that,
Q=-kA
T2
~
of
Q ==
27tLk [T, - T2]
In(;n
...
(1.31)
...
(1.32)
TI- T2
1
('2rl )
--In2nLk
.1Toverall
R
where
dT
-
dr
Fig 1.4
1
R = 2nLk
(r21 Thermal resistance of the hollow cylinder.
Inrtr
Area of a cylinder is 27trL
1.1.7 Conduction of Heat Through Hollow Sphere
A = 27trL
s,
Q = -k27trL
Consider a hollow sphere of
inner radius rl, outer radius r2, inner
.ernperature T I, outer temperature
T2 and thermal conductivity k.
dT
dr
d,.
Q x -
,. = -k27tL dT
from rl to "2 and TI to T2·
Integrating the above equation
~, dr
QJ
T2
r = - k27tL f dT
rl
TI
Let us consider
a small
elemental area of thickness 'dr'.
From
Fourier
law of heat
conduction, we know that
Q = -kA dT
dr
Area of sphere is 4m-2
Q [111'2 <In rJl
=
= - k27tL [T2 -
Q /11 [:~ 1 = 27tLk [T, - T2J
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TI1
Fig 1.5
A = 47tr2
Q = - k 4m2
ciT
e1r
.,.
(1.33)
p'n
eep
•
------------------
Conduction 1.19
1.1.8 Newton's
Law of Cooling
Heat transfer by convection
lntegrating
...
12
r
Q
dr:;;;
:::>
_
,.1
rl
j 47tk dT
'2
(-
. r
'1
1
Q I d~ == - 41tk . dT
:::>
'2
1
12
Q \=1-1 == - 41tk [T]
:::>
r
Q
:::>
:::>
Q
:::>
Q ==
I
'2
h - Heat transfer
co-efficient
in W Im2K
T s - Temperature
of the surface in K
T a: - Temperature
of the fluid in K.
Q=
r1 r2
are subjected
T1-
...
(1.34)
Q
to conduction.
T2
r2 - rl
ilT overall
R
...
(1.35)
Convection
where
_
r2 - '1.
R - 4 k(
1t
Plane Wall with
r2 - '1
41tk[T1-T2]
41tk (rl r2)
:::::>
in m2
Consider a composite wall of thickness L1, L2 and L3 having
thermal conductivity
kl> k2 and k3 respectively. It is assumed that
the interior and exterior surface of the system are subjected to
convection at mean temperatures
T and T b with heat transfer coefficient hQ and hb respectively. Within the composite wall, the slabs
(r2-rl)==
rl r2
Q==
to heat transfer
1.1.9 Heat Transfer Through a Composite
Inside and Outside Convection
== - 41tk[T2 - Td
41tk [T, - T2]
:::::>
A - Area exposed
Tl
r1
lL- l1
'1
(1.36)
where
11
T)
I
is given by Newtons law of cooling
on both sides
.
- Iherrnal resistance of hollow sphere.
A
'1 '2)
Fig 1.6
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~
il
Conduction /.21
1.20 Heal and Mass Transfer
From Newton'S law of cooling, we know that,
Adding both sides of the above eq ua tiIons
Heat transfer by convection at side A is
Q = ha A [Ta - T, J [From equn. (1.36)]
...
(1.37)
...
(1.38)
...
(1.39)
=> Ta - Tb = Q·hA[_1_ + _!j_ + -+
L2
a
k( A
L3 + I
hb A
k2 A
k3 A
_I
+
L2
L3
+-+-+
[ ha A
k( A
k2 A
k3 A
I]
hb A
1
Heat transfer by condl1ction at slab (I) is
Q = k, A [T, - T.,]- -[From equn. (1.29)]
=> Q=
L(
__s_
Heat transfer by conduction at slab (2) is
Q=
k2A[T2-T3]
L2
=> Q =
~Toverall
R
...
(1.42)
where
Similarly at slab (3) is
Q=
k3A[T3-T4J
...
(lAO)
...
(1.41)
Thermal resistance , R = R a + R I + R 2 + R 3+ R b
L3
Heat transfer by convection at side B is
We know that,
We know that,
To-T,
=Qx_1
[From equn. (1.37)]
haA
T,-T2
=Qx~
. k( A
T2 - T~) = Q x~ L2
k2 A
T3 - T4 =Q x ~
[From equn. (1.38)]
[From equn. (1.39)]
k3 A
[From equn. (1.40)]
hb A
[From equn. (1.41)]
R=_l_
UA
Ta-Tb .:::...
=> Q= __
_I_
UA
=> Q = U A [T a - T b ]/
where
'(0"
T4-Tb=Qx_'_
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...
IS t he overall heat transfer co-efficient
(W /m2K).
(1.43)
~I
r
1.24 Heal and Mass Transfer
-~
Q = .1Toverall
R
•"
Conduction I. 25
(1.48)
t,1.1J Solved Problems
where
On Slabs
fZJ Determine the heat transfer through the plane of length
6 111, heigh' 4 m and thickness 0.30 m. The temperature of
inner and outer surfaces are 100 C and 40 C. Thermal
conductivity of wall is 0.55 WlmK.
0
0
Give" :
Inner surface Temperature, T I = 100° C + 273 = 373 K
we know that,
Outer surface Temperature, T 2 = 40° C + 273 = 313 K
I
R= VA
Thickness, L = 0.30 m
Area, A = 6 x 4 = 24 m2
Ta-Tb
~
Q=
~
Q = VA [To - Tb J
Thermal conductivity, k = 0.55 W/mK
_I_
VA
...
(1.49)
where
U = Overall heat transfer co-efficient, W/m2K
Tofilld:
I. Heat transfer (Q)
Solution :
We know that, heat transfer through plane wall is
Q = .1Toverall
R
HMT DOlO book (C P Kothandaraman)
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[From Equn. 110. {I. 30) or
page no. 43 (Sixth editiont]
1.• 6 Heat and Mass Transfer
where
Conduction 1.27
Tofi"d:
Thickness
of insulation (L2)
SOIUlion:
Let the thickness of insulation be L2
We know that,
373-313
0.30
= 2640 watts
Q =
AToverall
R
[From £qun no. 1.42 (or) HMT Data book
page no. 43 & 44 (Sixth edition)]
where
0.55 x 24
AT=Ta-Tb
Q = 2640 watts I
Result:
R =
(or) T)-T3
L
I +_)_+
ha A
k) A
__~
+ __L3 +_ J
k2A
k3A
hb A
Heat transfer, Q = 2640 W
In A wall of 0.6 m thickness having thermal conductivity oJ
::::> Q =
1.1 WlmK. The wall is to he insulated with a material having
an average thermal conductivity of 0.3 WlmK.lnllerandouter
surface temperatures are 1000 C and 10 C respectively. If
heat transfer rate is 1400 Wlm1 calculate the thickness oj
insulation.
Wall
Insulation
I-I-I
Given:
given. So, neglect that terms.
0
0
I
haA+
Heat transfer
Heat transfer per unit area,
0/A = 1400 W/m2
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L)
L2
k) A
k2 A
--+--
ha' hb and thickness L3 are not
[T)-T31
L)
L2
k)
k2
-+-
of
Outer surface Temperature,
T) = 10° C + 273 = 283 K
co-efficients
[T)- T31
Thermal conductivity of wall,
kJ = 1.2 W/mK
Inner surface Temperature,
TJ = 1000° C + 273 = 1273 K
~
L3
I
k2A + k3A + hbA
::::>Q=
Thickness of wall, LJ = 0.6 III
Thermal
conductivity
insulation, k2 = 0.3 W/mK
L)
k)A+
~nl
1273-283
::::> 1400 =
0.6 +
.!:1.
1.2
OJ
[L2 = 0.0621 ~
Result :
Thickness
.'
of I11sulatJOll, Lz
= 00621
.
m
ZFF
1.28 Heal and Mass Transfer
III The wall of cold room is composed of three layer. Til
{I
layer is brick 30 em thick. The middle layer is cork e OilIer
thick, the inside layer is cement J 5 em thick. The temp 20 c",
of the outside air is 25° C and 011 the inside air is _20~~/II'es
film co-efficient for outside air and brick is 55.4 Wlm2/( . ~he
co-efficient for inside air and cement is J 7 Wlm2 K. Fin~ ~i/",
~wro~
~
Conduction
1.29
Tofintl:
Heat flow rate (Q/A)
solution:
. .
b
Heat flow through composite wall IS given y
Take
~Toverall
k for brick = 2.5 WImK
Q ==
k for cork = 0.05 WlmK
R
[From Equn no. 1.42 or
HMT DolO book page No. 43 and 44]
where
k for cement = 0.28 WlmK
Given:
Thickness of brick, L3 = 30 em = 0.3 m
Thickness of cork, L2 = 20 em = 0.2 m
Thickness of cement, L) = IS em = 0.15 m
Inside air temperature.T a = -200 C + 273 = 253 K
Outside air temperature,
Film co-efficient
Film co-efficient
=> Q =
Tb = 2S0 C + 273 = 298 K
for inner side, ha = 17 W/m2K
for outside, hb = 5S.4 W/m2K
=> Q/A
kbrick = k3 = 2.S W/mK
1
L(
L2
L3
-+-+-+-+ha
k(
k2
k3
kcork = k2 = O.OSW/mK
1
hb
253 - 298
=> Q/A ==
kcement = k) = 0.28 W/mK
1 + 0.l5 +_Q1__+..Ql_+_l_
0.28
0.05
2.5
55.4
17
Inside
Cement
Cork
Brick
k(
k2
k)
Outside
I Q/A == -9.S W/m2!
The negative sign indicates that the heat flows from the outside
into the cold room.
Result:
Heat flow rate, Q/A == -9.5 W/m2
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1.30 Heat and Mass Transfer
(!] A wat! of II cold room is composed of three layer; Tile (JUte,
layer is brick 20 em thick, t"e middle layer is Cork 10 c",
thick, the inside layer is cement 5 em thick. The temperature
of the outside air is 25° C and tltat on the inside air is -20 C
TI,efilm co-efficient for outside air and brick is 45.4 WI",2 K
and for inside air 011(1 cement is 17W/m2 K.
0
Find i) Thermal resistance ii) The heat flow rate.
Conduction 1.31
Film co-efficient
for outside air and brick, hb = 45.4 W/m2K
Film co-efficient
for inside air and cement, ha = 17 W/m2K
K) = 3.45 W/mK
K2 = 0.043 W/mK
K( = 0.294 W/mK
Tofind:
Take
I. Heat flow rate
k for brick = 3.45WlmK
2. Thermal resistance of the wall
Ii for cork = 0.043 WlmK
sotutio» :
k for cement = 0.294 WlmK
Heat flow through composite wall is given by
Given :
Q=
In ide
Outside
ement
Cork
Brick
kJ
k2
kJ
~Toverall
R
[From Equn (1.42) (or)
HMT Data book page No.43 &44J
where
~T=T{/-Tb
I
L(
L2
L)
R =--+--+--+--+-ha A
kJ A
k2 A
k) A
I
hb A
=>Q
j
kne
f brick LJ = 20 em = 0.2 m
=> O/A
f ernent LJ = 5 em = 0.05 m
=> QIA
UI
ide air temp ~rature, Tb = 250 C
III ide air len perature, To = -200
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273 = 298 K
273 = 253 K
253 - 298
_1_ + 0.05 + __Q:l_ + 0.2 + _1_
17
0.294 0.043 3.45 45.4
lOlA =-17.081 W/m21
/. 32 Ileal and Mass Transfer
TIle negative sign indicates that the heat flows fro~
into the cold room.
COnd/l(;liml I. JJ
e OUt.
s~
(i;"C'II .'
<h,
<T
(PT2
Inner layer middle laye
For Unit Area
~
I
k,
LI
L2
L)
R =-+-+-+-+ha
kl
k2
k)
(
outer layer
k2
kJ
I
hb
_1_ +0.05 + _QL + _Q1_ +_1_
17
0.294 0.043
3.45
45.4
Thermal conductivity
of inner layer, k I = 8.5 W/mK
Thermal conductivity
of middle layer, k2 = 0.25 W/mK
Thermal conductivity
of outer layer, kJ = 0.08 W/mK
Inner thickness, LI = 2S em = 0.25 III
~ IR = 2.634 KJW I
Midddle layer thickness, L2 = S ern = 0.05111
Rault:
Outer layer thickness,
W/m2
I) Heat flow rate, Q/A
= -17.081
2) Thermal resistance,
R = 2.634 K/W
LJ = 3 em = 0.03 III
Inside wall temperature,
OUI
ide wall temperature,
T t = 6000 C + 273 = 873 K
T4 = 50° C
Tofintl .'
I. Equivalent electrical circuit
(I] A furnace wall is made up 0/ three layers, imide layer MI~~
2. Heat flow per m2
thermal conductivity 8.5 WlmK, the middle layer MIll'
conductivity 0.15 WlmK, the outer layer with cO/l(luClivi~
3. Thermal resistance
0.08 WlmJ(. T,.'lerespective thickness of the inner, n~i{ldlea~
outer layers are 15 em, 5 em , and 3 em respectively- T~
4. Interface temperatures
inside {/~d outside wall temperatures lire 6000 C and ~O·~
respectively. Draw the equivalent electrical circuit f~
conduction 0/ Ileal through the wall and find t"ernl ~
resistance, heat flewlml
and interface temperatures4
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273 = 323 K
1.34 Heal and Mass Transfer
.---=--------------..._
Solution:
Conduction 1.35
electrical circuit for COIl(llICtioll
1. Equivalent
873 - 323
9.25 + 0.05 + ~QL.
8.5
0.25
0.08
~IQ/A
W/I11~
= 909.97
3) Thermal Resistance
2. Heat flow through composite wall is given by
L\ToveraJl
Q=
[From Equn. 110. (1.42) (or)
R
(Neglecting ha' hb terms)
HMT Data book page No. 43 & 44J
where
L\T=Ta-Tb
=T,-T4
For unit area
,
R
=
L,
L2
L3
I
ha A + k, A + k2 A + k3 A + hb A
~ Q =
0.25 -I-
0.05
+ 0.03.
8.5
0.25
0.08
4) Interface temperatures
We know that,
[Convective heat transfer co-efficients ha and hb are not given.
So, neglect rh{lt terms]
= - ...
-.-.--~.-.----
r, - '14
=:)
()
--
-,-{---
=-">
Q =
T,-T2
R,
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T, - 1'2
T:2 - T 3
= -R-,-= --~
=
T 3 -- T4
R3
... (
1)
I: 36 Hear and Mass Transfer
Conduction
1.37
Resllll:
I. Heat flow per m2, Q/A = 909.97 W/m2
klA
2. Thermal Resistance, R = 0.604 K/W
TI -T2
Q/A
909.97 =
T) = 664.23 K
873 - T2
0.25
8.5
I T2 = 846.23 Kj
@ A mild steel tank of wall thickness
20 111111 contains
water fll
100" C. Estimate lite loss of heat per square metre area of lite
tank surface, if lite 11IIIk is exposed 10 OIl atmosphere til
15° C. Thermal conductivity of steel is 50 WlmK, while hem
transfer co-efficient for lite outside wid inside II,£, tank are
10 WI1112K and 28.50 WI",2K respectively. WIItII will he lite
Silllilarly
(1)
T 2 = 846.23 K
3. Interface temperatures,
LI
kl
o = T2-T3
=:>
,
where,
lemperalllN'
R2
Oil lite
outside o.f lite tank wall.
Given :
L2
R2=-k2A
Thickness of steel wall
Q = T2-T3
Inner water temperature
L2
k2A
Q/A
T a = 100° C + 273 = 373 K
Atmospheric
air temperature
T2 - T3
Th= WC+273=288K
~
k2
Thermal
conductivity
steel, k I = 50 W /mK
909.97 = 846.23 - T3
0.05
0.25
I T3 = 664.23 K
Inside
LI = 20111111
= 0.02 In
I
Scanned by CamScanner
of
Inside heat transfer
co2
efficient, ITa = 2850 Whn K
Outside heat transfer coefficient, hb = I 0 W/I11~K
Outside
1.36 Heal and Mass Transfer
where
Conduction
Q = TI - T2
LI
1.37
Reslllt:
I. Heat flow per m2, Q/A = 909.97 W/m2
klA
Q/A
TI - T2
2. Thermal Resistance, R = 0.604 K/W
=-L- I
3. Interface temperatures, T 2 = 846.23 K
~
T 3 = 664.23 K
873 - T2
909.97 = ----=0.25
8.5
I T2 = 846.23 Kj
@] A mild steel tank of wall til ick ness 20 111mcontains
water (It
/ 00° C Estimate tile loss of heat per square metre area of tile
tank surface, if tile tank is exposed to an {Itmo."plwre (It
/50 C. Thermal conductivity of steel i...50 WlmK. while heat
transfer co-efficient for tile out s ide and in ...ide tile ttlnk are
JOWl",] K am/ 2850 Wlm2 K respectively. What will he the
Similarly
Q = T2-T3
(1) ~
temperature
R2
Thickness of steel wall
L, = 20mm = 0.02 m
Q = T2 - T3
k2A
T2 - T3
Ta=100°C+273=373K
To
Atmospheric air temperature
Th= 15°C+273=288K
ha
Thermal conductivity
steel, k I = 50 W ImK
~
k2
909.97 = 846.23 - T3
0.05
0.25
/ T3 = 664.23 K
Outside
Inside
Inner water temperature
L2
Q/A
01 the tank wall.
Given:
L2
R2=-k2A
where,
Oil tile outside
I
Scanned by CamScanner
of
Inside heat transfer coefficient, "a = 2850 W 1m2 K
Outside heat transfer coefficient, lIb = 10 W/m2K
1\ -----\
...-
I. 38
!...~:!.!~
and Mass Transfer
-------
To .Ii" tI :
..
_--- ----------_
"
i) Ileal loss per square metre area of the tank Surface ( /
"
II)
T ,Hl1\"doutsi c temperature,
, Q 1\)
T2
Conduction 1.39
We know that,
TG -Tl ~
Q =
T1I -T,
=R;--
-R-
Solutio" :
T __
-T,
=> Q = _G
Ra
AToverall
Heal loss,
Q = ---R---
where
[From Equn, I/o. (I.42j~
I-IMT Data book page NO.4J & 44J
1
where, Ra = -, A
1£1
I-H=Ta-Tb
I
L,
L2
R = --+--+--+ha A
k, A
k2 A
LJ +__
I
k3 A
hb A
_ Ta-T,
I/h
=> Q/A -
a
=>
843.66
=
373 - T,
1/2850
(Neglect L2, L3 terms)
=>
IT,
= 372.7
Similarly
=> f)/A
I
LJ
I
-+--+-h{/
kJ
fib
3 73 - 288
'_.) Q/;\
,
--'-
----- --------
:- (V/\
- _. ---.
-I-
2S50
_O_:_Q£ + .L
50
Ll
where, R, = k,A
'()
T,-
--------.-
-, S43.6(j
Willi ~1
-_- .-._------ "_-
1
kJA
=> Q/A
Scanned by CamScanner
T2
L,
KI
_~
T,-T2
__
R,
= T2-Tb_
Rh
1.40 Heat and Mass Transfer
=>
843.66
=
372.7 - T2
0.02
Conduction
50
I T]
= 372.4
Hot gas temperature.
K
I
Result:
I. Heat loss per m2, (Q/A)
o
=
2. Outside surface temperature,
A steam boiler furnace
843.66 \Vlm2
T2 = 372.4 K
is made of fire clay. rite
temnerature imide the boiler furnace
r
susface
to
""(11
slirrouluJi"K
8.2 kWln'; and interior wall sill/ace temperature is 1080"('.
Calculatefor external surface
I. Sur/ace temperature
2. Convective conductance
Given:
Scanned by CamScanner
10 inside
= 25.2
103 W/m]
Convective
transfer
conductance
at interior,
x
"0
of the wall
= J 2.2
= 58
from external
surface
Illleril r wall
of the
W/m2K
W/mK
surface
to surroundinc
10·) WlrnK
.
+ 273
1'1 = 1080°(,
urface temperature,
= 1353
is
K
Tafind :
thermal conductance of the wall is 58 WII1IK, heat flow ~v
extemal
from gases
10
transfer co-efficient at tlte interior surface is 12.2 1¥111I'/(,
radiation from
Heal now b) radiation
wall. Ol{ 1 = 25.2 kW/m2
QI{2 = 8.2 kW/11l2 = 8.2
.., (II,
caSel"
inside surface of the wall is 25.2 k Wlm], cOllveclio"
T b = 50 C + 273 = 323 K
Thermal
I..J I
C
Room air temperature.
Ileal now by radiation
hOI CQJ
is 2100 t, roo-
temperature is 50 'r, heat flow by radiation from
Ta = 2100
i) External
surface
ii) E.\lerll,il
C
temperature,
'1']
nvc rive c nductancc.
hh
Solution :
We kn: \
Ihill
Total
}
hca:
e n Ic r i n ~
the \I all
()
=
Heat
nvc
Inn fer
ti
h~
n (II interior
l le at
radiation
Irat I . fer
h)
at interior
r
,I
"
1.42 Heal (11/(/ Mass
TrLlm!er
('(lnductiun
We kn \\ !haL
o
'1 - T,
.-E-- _ R
T, - T2 _ T2 - '1'"
T-l"_.-!!----R
R
-s:
I
-
1.43
Resull:
I. [.x!<.:rfl<ll .urface temperature,
0
,,/;
"'2 = 703.4 K
(I
2. _xlemal convective
co-efficient,
hh = 77.3 W/m~K.
T, - TZ
~Q==~
@ A wall is constructed
of several layers. The first layer
o/I/Ia.\'OII(/ry brick 2() cm thick (lJ tlt er m al
conductivity fl.66 WlntlC, lite second layer consists of 3 em
thick 1/101'1111' of tltermal conductivity 0.6 WlmK, lite third layer
consists oJ 8 em thick lime MOlle of thermal conductivity
0.58 WlmK and the outer layer consists of 1.2 em thick plaster
0/ thermal conductivity 0.6 WlmK. The heat transfer coefficient Oil the interior 1I11(1 exterior of the wall lire
5.6 WIIII1K and II Wlm2K respectively. Interior room
temperature is 2rc and outside air temperature is -5" C.
consists
J
WJ- T2
RI
LA,I--.V't"" -
1353 - T2
I
-8
3 .644 ==
l''- Re5istance = con d uctancc J
External surface temperature,
T 2 = 703.9 K
Calculate
(I) Overall heat transfer co-efficient
Heal loss due to radiation at exterior
b) Overall thermal resistance
c] The rate of heat transfer
Hear 10>5by
vection at exterior
= T! -I hea entering - Heat
loss by radiation
at exterio
d) The temperature {II the junction between the mortar and
lite limestone
Given :
= 37.644
- 8.2 Y 103
k2
--..'Qc = _9.!
rr
~Tl
---- lib A
II
T ,= 29, 4
!
/ A / IT! - Th) -= 29,441
!
hb / I / [ O~.9 - 323 j :: 29,444
F.\lCrnal
'omccl; e CO-Cf!lCI'CIlt I
__
, 'b::
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)
77.3 W/III-K
I
eDT,
2
I
asouary
Mortar
I . lib
.
Limestone
Plaster'
I
-1 ----'
/I('a1 and Mm.' Transfer
,=
, (1I1;1:-ollary. 1., =
Thickncss
Thcnna I (I IH.lIII..'1ivity ~ I
1 hidlll::-'l
.!-'
~_
ondu
L4 = I.~ em == 0.012111
uvity,
"4
z:
Interior
111;31
Exterior
hctlllrallsfcrco-cfficient,
lntcr:
kl A
L
~
k A
k A
k" A
-
2C)S - 268
O/A
0.08 + 0.012 + .L
0.58
0.6
II
0.20 + 0.03
0.66
0.6
I
5.6
0.6 W/mK
Ileat trallsfer
h" = 5.6 W/m2K
trnnsfcr, co-efficient
V. e knov
hh = II W/m2K
r room ll:rnpcratIJrc, To = 22" C
Tb = -
Ollh';dc air Itl1lpcratIJre
h,A
III
Thi
OJ1dll
_'_'1_ ~ "1
I
"3 == 0.58 W/I11K
f Plasicr.
0 -
"2:;- 0.6 W/IIlK
ue. L, = ~ ern == 0,08
o C
273
per L1l1itarea, Q/A
34.56
W/m2
that.
Q = UA (Til _ T b) [From equation no.I.4J I
Heat transfer,
= 295 K
=
where. U - overall heat transfer co-efficient
- 27" = 26R K.
Tolind:
. ) (h c.:rall
lint transfer
co-efficient,
11) (h'~rjJII !lrcnlltll
reo istancc,
. Ih·allrall·fl:r/lll-.
(()fA)
U
(I<)
34.56
U == _.::.._:_.:.;:_::295 - 268
d) '111(' temperature at the junction
Ilw limes!
between
ihc Mortar
and
Overall heat transfer co-efficient,
U = 1.28 W/m2K
ne. (Tl)
We know (hat,
Solution :
Overall Thermal
.II "11 Il( \\ thrnugll
C)
'\ I )\I'J'
COIJIP(
sire wall
R = --
by
/ h()1JI Equn.
I (I
I
hI A
I/O.
(In) or
I IAtr /)11/0 hook j}(lJ,{(' No. -I3cC-I-Ij
"'1.'
"
is given
resistance
(R)
LI
L2
LJ
L4
--+--+--+--+-kl A
k2 A
k3 A
k4 A
I
hb A
For unit Area
11
H
\111
1.../5
T,,- T"
_.
ThLTm. I 'Olldllclivily,
I..n~ss
Conduction
-,
0.66 \\ IIIIK
tivuy Oflllol1ar.
of limcst
~1l~:\S
Thermal
.~
0 em r-, 0.20 rJ1
f me rtar, L_ = 3 em == 0.0" rn
henna!
. hi
_~ ~
,~._,_c_
__=
I
L,
L)
h(l
k1
"2
R = -+ -+--
LJ
L4
+-+-+kJ
"<j
1
hb
_I + 0.20 + 0.03 + O.OS + 0.012 +_1
56 0.66
0.6
0.58
0.6
II
I;,
L,
I.,
.1
1<
\
/\
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0.78
K!WJ
I 46 Hid
__:_____ eo a17 Mass Tran.~fer
Interface
te mperature between mortar and
------..'
the linrel'IOIIe, T -..........
Interface temperatures relation
. 3
"1-T2
Q _ To--TI
--~=-R-I-=~
"2-T3
_ T3
-"4
-~-==~R
..>
T4-l
Conduction 1.47
5
4
TS-Tb
=>
278_3 - T3
Q=
=~
~
T,,-' TI
Q=
k2A
Ra
278.3 - T3
Q/A
Q
295 - TI
~
A
k2
)/ha
295 - TI
Q/A
34.56
=
11170
278.3 -- T3
0.Q3
0.6
295 - T,
34.56
Temperature
1/5.6
IT, = 288.8 K)
between Mortar and limestone (T]) is 276.5 K
Result:
I) Overall heat transfer co-efficient,
U = 1.28 W/Il12K
R = 0.78 K/W
= 34.56 \\ /m2
2) Overall thermal resistance,
T,-
T2
=> Q=---
R,
3) Heal transfer, Q/A
,1) Temperature
288.8 - T2
Q
L,
k,A
288.8 -, T2
Q/A
L,
",
34,56
__ _3!8.8 - ~~
0.20
0,66
I
IT? = 27S,3E]
I
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between mortar and lilllcstone,(T3)
= 276,5 K
[!J The wall of (/ refrigerators is made lip oftwo mild steel plates
2.5 111111 thick with (I 6 em tltick glass wool ill between the
plates. The interior temperature is' -20" C, while tile outside
of the refrigerator is exposed /0 40" C. Estinuue tile heat flow:
Thermal conductivity of steel alit! glass wool are 23 WllltI(
and 0.015 WlmK respectively.
(Madurai A{/III01'Oj
Ll = L = 2,5 111111 =-= 0.0025
J
L'l = () ern = 0.06 III
III
University /l.'OI'-I.}-I)
I. 48 Heal and Mass Transfer
-
I
Mild steel
Glass wool
Conduction 1.49
Convective
Mild steel
heat transfer coefficient is 1I0t given.
So, neglect ha' lib terms
Ta
k,
k2
I
.
L,
~
• I-
k3
Tb
I
L2
-/-
L3
~QIA.
..,
Ta = -200 C; Tb = 400 C
k, = k3 = 23 W/mK;
Q/A
I
i) Heat flow, (0)
Heat flow through composite slab is given by
R
+ 0.0025
23
Q == -14.99 W/m2!
The '-ve' sign indicated that the heal flows from the outside
into the refrigerator.
Solution:
,1 To vera II
-20 - 40
0.0025 + 0.06
23
.015
k2 = 0.015 W/mK
Tofind :
Q
- -.---.
[From Equn. no. (1.42)]
where
o
Result :
i) Heat flow, Q == -·14.99 W/m2.
IZ!fl The inside temperature of the refrigerator is -1tl" Cand outside
surface tempera/lire
is 30t) C and area is 301111. This
refrigerator consists of 2.2 ntnt of steel at the inner surface,
15 111m plywood at the outer surface and J() em ofglass wool
ill between steel (lilt!plywood. Calculate the heat Ion and the
capacity of the refrigerator
ill tons of refrigeration.
Assume k(sleelj == 20 WlmK. k(p()'",oot!) == 0.05 WlmK.
k(g/as!J'H'oo/)== 0.06 I-Vlm/(.
Given :
Inside
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Temperature,
T, =_IOL'C
f'273
=-=263
· 1.50 Heat and Mass Transfer
Conduction /.51
kI
k2
k)
Steel
Glass wool
Plywood
T4 == 30° C + 273 == 303 K
Outside temperature,
Area, A ==30 m
~~
~ T)
~DT2
(DTI
where
2
Thickness
of steel, L) == 2.2 mm == 0.0022 m
Thickness
of plywood,
Thickness
of glass wool, L2 == 10 cm == 0.10 m
L3==
co-effficients ha and hb are not
15 mm == 0.015 m
Thermal conductive
of steel, k) == 20 W ImK
Thermal conductivity
of plywood,
Thermal conductivity
of glass wool, k2 == 0.06 W/mK
k3 == 0.05 W/mK
Toflnd :
263 - 303
Q
0.002 + 0.10
+ 0.015
20 x 30
0.06 x 30
0.05 x 30
I Q =-610.1
W==-0.610KWI
The -ve sign indicates that the heat flows from the outside
into the refrigerator.
i) Heat loss, Q
2) Capacity of the refrigerator
We know that
Solution:
3.5 kW ==I ton
Hear flow throu?h composite
Q=
(Convective heat transfer
given. So, neglect that terms)
T)-T4
wall is given by
0.610
3.5
ton
==0.174 ton
t1To vera lJ
R
:=)O.610kW==
[From Equn. no.(/.42) ~
HMT Data book page No.43 & 441
Scanned by CamScanner
:=) Capacity
of the refrigerator
==0.174 ton
'{
r 52 HealandMassTrc.!'!!f!!_.-------"-~
Result:
I' id
.iqut surface conductance,
Heat transfer, Q ::::610 W
Capacity of the refrigerator::::
f1D A s'tetlm to liquid
heut exclulIlger
copper Oil lite steam S/{Ies.
I lie
area
of 25
re.\'ISIIV/~V of
is filII
.
Steam
"
, , temperature, Tb = ) ) 00 C·
_,+ 27"., -- .,83
K
Liquid temperature: T(I = 70" C + 273 = 343 K
(I
.Vfller
t>
OJ
deposil 011 lite steam side is {}.0015Kiw. The steun, (I1lti/it
-sCa/!
surftlce C{JIIt/uel'IIIee lire 5400 WI", 2K and 56() W. 14~
reJpeClive~y.Tile Itealeds'lell",
I 53
Steam surface' con d uctance, hh = 5400 W/Ill2K
0,174 ton
11.5 •em
nickel
and• •0.1 en, p / (II'"n
.'"' ;,
I
7'1
.. '
cOllslrllelell ",illi
Conduction
h(/ - 560 W/m2K
Ou C (flllillefllet[ ~",2k
is' al 70° C.
"ql/id
k2(copper)
= 350 W/IllK
k) (Nickel)
= 55 W/IllK
Toflnd :
i) Overall
heat transfer co-efficient, (U)
2) Temperature
drop across the scale deposit. (T, -- T4)
Solutio II :
Calculate
J) Overall steam IO/iquidllelll
transfer co-efficient
2) Tempemllire drop IICrOS,\'lite settle deposit
Heat transfer through composite wall is given by
r
Q = ~Toverall
Take
Front Equn. no, (I. 42) or
lIMT Data hook page No.43 & :J:J}
R
k(copper) = 35(1 W'ImK filii/ k(Nickel)
= 55 WlmK.
where
Given :
Inside
Liquid side
[
Outside
R = _I_+_L_I_ ... L2 ,L)
I
ha A k A . kA kA + _-
"b
T
,
1\2
Steam
side
2
3
A
Ra + R, + R2 + RJ + Rb
T2
Til 11(/
G:~~.:
R,~ value is given,
R" = k~~ = 0.00) 5 K/W
~
Copper
R
I
---t--
lin A
_-'-
__
560 x25.2
Thickness
of Nickel , L I -- 0 ).- em = 0,5
x
10-2
Thickness
of copper
t'
x
10--111
Resistivity
of scale,
'
L'2-- 0 , I cm=O.1
R_l = 0,0015
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K/W
L,
"I
A
+ 0,5 x 10-2 + 0, ) x 10-2
350 x 25,2 + 0,0015
55 x 25,2
+
111
I
5400 x 25,2
~ ~IR
J.~58~x~IO~-_J~K~/W~1
/.5.1
Conduction/.55
fllJ A wall of fllrll(~c~ is made up of 13 em thict: of fire day,of
(I
thermal condllctlv/~" fJ.6 WlmK alU160 em thick of red brick
of conductivity
0.8 . WlmK. Tire inner ~nd outer surface
.
. I.
temperature of wall are 1"000 C and 75 C Determine
0
0
1.
Tile amount of heat loss per square metre of lire
furnace wall.
2.
It is desired 10 reduce lite thickness of lite red brick
layer ill litis furnace to half by filling in the space
between lite two layers by diatomite whose k = 0.11J
+ 0.00015 T. 'Calculate lite thickness of the material.
Give" :
Furnace
Overall heat transfer co-efficient,
U
k,
k2
Fire clay
Red Brick
25 W/m2K
Temperature drop (T~.) - T4 ) across the scale is given by
L1T
Q=-Rsca/e
,
[.:
I·
I:H = T3 - T4]
L1T
25.2 x W = 0.0015
~ltlT=37.8°CI
Result:
Overall heat transfer co-efficient, (U) = 25 W/m2K
Temperature drop across the scale, (T3 - T41 = 37.80 C
L,=13cm=
k, = 0.6 W/mK
L2 = 60 em = 0.6 111
k2 = 0.8 W/mK
T, = 1000° C + 273 = 1273 K
T3 = 75° C + 273 = 348 K
Tofind:
f the furnace wall.
I) Heat loss per square metre 0
.2) Thickness
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0.13m
of the material. whose k=OIII+O.OOOIST
.
__ ._-" - ---
.
Sol"tim, :
CondUCI;OJ1 1.57
2.
1. I teat transfer through composite
wall is given by
Diatomite
...
Q = Il To\'<:.~~~I
R
where
[From EC/III1. no. (1.42)'»,
JlMT Data book page No.43 & 44]
Red brick
13
Furnace
K.
.'
We know that,
TI -T3
~Q
=
·-----;---·i,_ .~+~+-1h~-A+ k, A + "2 A
Neglect unknown terms
~Q
348
= ...-1273·-_. __.._
--
Q
kJ A
=
T, - T.f
R
T I - T2 _ T2 - T3
R2
= -R-,-
= T r T4
••• ( I)
R3
hb A
=> Q =
T,- T2
R)
1273-T2
Q =
L,
k, A
=> Q/A
=
=> 956.8 =
1273 - T2
1273 - T2
I.,
k,
1273 - 1"2
-0.13
0.6
Scanned by CamScanner
,.: A = 1m2 J
---------------------
/. 58 Heal and Mass Trumjer
Conduction J .59
T) - T4
Q=
L3
.956.8 =' 358.8
.. '
,
L2
kJ A
k;
. T) - T4
QIA =
__!2__
kJ
k = 0.111 + 0.00015 T
::::::>k2=0.111 +0.00015T
T4 - Outer surface temperature of red brick'
k) - Thermal conductivity
L) _ Halfofthe
k2 = 0.111 + 0.00015 [T2: T31
of red brick
= 0.111 + 0.00015 [1065.6 +
2
2
thickness of the red brick == 0 6 = O.3m
I k2 = 0.243 WimK I
T) - 348
Q/A ==
0.3
D.8
~
Substitute
~
956.8 =
0.243
~.
L2 = 0.091 m
Thickness'ofthe
T
T2- 1.
R2
Q=
1065.6-706.8
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(
~.
IT
Q=
,
956.8 == 358.8
T) - 348
~ IT) = 706!KJ
706.81
k2 value in Equation (2)
0.3
(I)~
is
Given thermal conductivity for diatomite
where
~
.... '(2)
diatomite, L2 = 0.091 m
Remit:
Heat loss, Q = 956.8 W 1m2
Til ickness of the, d iatOl;nite, L2.=' 0.091 m
),
'"
.
I (10 Heat and Muss
r,'(/l1s[er
-'[ijl A [urnace wull hi made of inside silica hrie~
~
•
•
'J ther
contlllctivill' 1.7 W/mK, 12 em thick and outside m ",~
•
••
~ ~ IV
(Ig'l~f'
brick of thermal conductivity .1••1 ,,'/mK, 22 em thic ",,
temperature Oil the inside of the wall of the silica bk: rh,
. magnesite. bri
92(1UC (111(1olltsult!
nc«t. sur/ace tem'Pe r'rk'~
rmUre'
120" C. Calclliate the heatflow tit rough tit is compos I'te IVaI(~
lf the ('011 tact resistance between the two wall is 0.003Ktlt
find tile temperlllllre of the surfaces at the illter/ace.
~;:::-::--------where
~C~o'~ld1!_''!'£Cli(}~!J_._61
Sf = TI -lJ
R = --I_+~+~+~
ha A
kI A
I
Given:
k2
(DT2
(~TI
Neglect
Magnetic
brick
brick
unknown
1
h A
b
L,
+_J_+
I
k) A
fib A
terms (11(1, hb and L))
TI -T3
=:>Q=------__.:__-
~~T3
Silica
k3 A
I + __LI + __L2
___
h(/ A
kl A k2 A
r----.---r------~
k1
...
k)- A
LI
L)
--+--kl A
k2 A
TI - T3
Q=
Thermal conductivity
1'1 - T3
Thickness of silica, LI = J 2cm = O. J 2 III
Thermal conductivity
of magnesite,
Thickness of magnesite,
~
[where, Rc is contact resistance
between walls]
RI + R2 + Rc
k2 = 5.5 W/mK
= 22 ern = 0.22 m
Inner surface Temperature,
T J = 920" C + 273 = J J 93 K
Outside surface temperature,
T 3 = 120 C + 273 = 393 K
Contact
RI + R2
of silica brick, kl = J.7 W/mK
0
Q=
1193 - 393
resistance between t, .•vo wall, Rc = 0.003 K/W
Tn find:
Temreralurc of thee SUI
surf lace at the Interface,
.
1193 - 393
Q/A =
(T 2)
0.12 + 0.22 + 0.003
1.7
5.5
So/utioll :
HCallraJl!.,fer
.
thr
" ....
oUb" composite
wall is given by
R
II MT 0(1£0 hook page No.
Q :;
7042.9 W /m2
{From £qlll1. no. (J.42)fI)
Scanned by CamScanner
43 c( 4J
I
1.62 Heal and Mass Tra:.:'.:.:ls~ife=-r
~
Conduction 1.63
We know that,
Given:
Brick
Inner
Insulation
Timber
Outer
Cold
Hot
1193 - T2
Q=
Ll
kl A
1193 - T2
Q/A
0.12
Diameter
of the aluminium
Thermal
conductivity
rivet, d = 4 em
= 0.04
m
1.7
of the aluminium
1193 - T 2
7042.9
=
rivet,
kriv<:l
0.12
Area of the surface,
1.7
= 200
W ImK
A = O.I m2
Th ickness of the brick, L I = 12 ern = 0.12 m
IT2 = 695.8 K
I
Thermal
conductivity
Thickness
Result:
Thermal
of the Insulating
conductivity
Thickness
IE] A composite insulatine wall has three layers of material /JeI~
together by 4cm dian:eter aluminium (k :: 200WlmK) riV~
per O.J m2 of surface. The layers of materials consist 0
12 em thick brick (k= 0.90 WlmK) with hot surface 01
of the brick, kl = 0.90 W/mK
Thermal
material,
L2 = I.S em = 0.015 111
of the Insulation,
k2 = 0.170 W/mK
of the timber, L3 = 22 em = 0.22111
conductivity
of the timber, k) = 0.11 W/mK
temperature,
T 1 = 2200 C + 273 = 493 K
Cold surface temperature.
T 4 = ISO C + 273 = 288 K
Hot surface
220°C, 22 em thick timber (k = O.J 10 WlmK) with colt!sl'rf{lC~Tofind:
at 150 C. These two layers are interposed by a third layer0,
.. I .
J 70 W/"'~'
IIlSU ttttng material 1.5 em thick of conductivity O.
CI
I
.
.~r~
a ell ate the percentage of increase ill "eat trans)'
due to rivets.
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Percentage
of increase
in heat transfer rate.
1.6./ Heal Gild Mas ,. Transfer
_ ...
----------
Solution:
(',,"dUCli(J11
I.fij
Lrivel
Heat transfer (without
rivet)
Rrivd
Q
:::
krivCI x Arivet
R
where
0,12 + 0.015 + 0.22
R
17(/ A
20a x 1.25 x 10-.1
[Front fI!IIT d 1/(1 book '0I') I~e /IV.';] &
~T == T I - lt~
Rriv":l == 1.42 K/W
1.1
L..,
L"
kl A
k] A
k,A
We know that
h,
__
"1"1- T-t
L__
_I _ +- _'-
==
Rc:quiv:llenl
=> Q
~--
L
L,
--T--
kl A
k1 A
k, !\
II
Ila !\
.'
Rwall
-
I{riwt
I x ~1.·~_~i.!!IEivct_.
Arca \\ ithour rivet
/ill !\
-I-
R
-_1I{rivel
L .. , (I)
where
=
<)
R==
__ I
",,/\
+ __LI
klA
l.~
+_k!A
I.}
I
f----
f--.--.
kjA
"I,A
{From 1I.\.It dat« hook page no. n/
Neglect
O.(H)
O. I
O. I 70 A 0.1
- ----
~-----
Rivet
idcriug
area
0.1
"b terms
LI
L,
LJ
klA
kzA
kjA
R==--+---+--
+-. _Q·~L_
R == _.Q.JL_ +
0.015
O.QO x 0.1
0.170 x 0.1 0.11 . 0.1
'----------_
ow con
0.11
11(/.
rivet
I{ = 22.2 K/W
~'rrf.l J2
Substituting
- It/t! " (O.04)~
[!\rca~~~~).-~:·.I>S_-·
10-3
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JlI21
R value ill Equn (I)
_1.---R.;qul\'al.:nl
==_L ,(i~'!
22.2
.. 1.~5,( 10-·;)
0.1
r
is 23') C.
fC!"'flt!rllfllre
iusid«
resiSfllll
c! of.5
",I'glcclillf:
thr th ermal
nun mortar joint between marble lI11dbrick.
I ttiefotlowing.
fill
1. Overall
transmitance
•. He II loss through
for the wall
II,e 14'(/11
:. Temperature of lire brick - pine interfa e
T I - T~
-I. Thermal
conductivity
0/ tire mortar.
[Assume ,\lorlar brick interface temperature i 21" C,
Given :
heat loss is reduced
. :e='
, n.age (}
ing thickne s of mortar
v
Q
p~
by J 0%/
• T,
TI
rivet.
I
e
Q,
lnvidc
0
T
l{l"
=
4.13-9.22
15 .13
hb
ha
/ 100
'% ]
., ickncvs of marble, L = 75 mm = J.O 5 m
~
A lurJ(e composite
wall ill made
up (Jf 75 mm tnJlrhltl
thermal conductivity 1.25WlmK, 7fJmm brick 0/ conducfit1
fJ.fi2WlmK,
2fJ5mm pine
of conductivity
O,J2WIK
,'iur/tu:e .onductance
2fJ0(' and
(J.25WlmK. 1he (Julsj
. ..Ie uJ
I,,, 33Wlm2X and m'ilu
I,,, 12Wlm2K. Outside
tempet"
i y of marble, k4 = ',25
Lhickncw
I.} = 70 mm = O,07() m
brick
IUft j
T hcrmal c. nducii
'I hickncss
it)' of brick, k3 = 0,62 W/mK
of pine, 1'2 = 205 rnrn - 0,205 m
Thermal c nductivity of pine k2 = 0,12 W/mK
Thicknc:
s f plaster, LI = 175 mrn = 0,17
'1hcrmal conductivity
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ImK
aI
I 75mm Imide ptaster 0/ conductivity
unit surface conductance
'l hcn (II conducti
m
of plaster, kl = 0,25 W/mK
.e
!_._(i8 .!"I<!~I/ and Mass li'ow/el'
--------_ .__._----...
Outside
surl;lce cOllductance
,
Inside surtilce
Outside
conducrauce
"
hl == 3'J WI I11K
2. -----~
---._---
('OIlC/llc1ioll
h == I'"- U'/
'K
'0
no 111-
ICllIpt:ratlln..:, Tb::: 20" C + 273 :.::293 K
To == 2]0 C + 273
Insidl.:· kll1pl:ralllrc,
---
__
oo
12
0.25
0.12
1.25
0.62
2.69
Q/A == ;'.11 W/m /
__rate,
-.-- -_.
r.lcallransfer
(U)
~-----
2
of brick-pine
interface, (TJ)
I lcat transfer,
Q = IJ x ACTa" Tb)
IFrom cqun, "0. (1.43)1
of the mortar
[Mortar hrick interface lelllperu/ilre i.v] It:
heal/ass
is reduced
1.11 = lJ y. (296 - 293)
by 10%/
SOIIlI;oll :
Heat transfer through composite wall is given by
We know that, Interface temperatures relation
,\Tovl.!rall
R
where
1'4 - 1 ~ _ 15 - ""
Sf:; Ttl ·-lb
1
R = 7;-;;11.- -I-
33
We know that,
-L Thermal conductivity
~=
_
3
2. Heal luss. (Q)
()
••
_L + 0.175 + _Q)J!~~+ O.O.~, + _QJ~~ + ,_!_
2% K
'C)I/\
trunsuuuance,
3. Tempcnuurc
-.
= .5 111111 == 0.005 111
l'vlortar joinllhid:lh.::-;s
To/illll:
I. Overall
==
-------.
I.
= ~-
LI
"'1-"
L2
+ "'2"
LJ
-
-Ti;,--
. , . (,)
L,J_'
+ kJ ,,- + k,J" + lib A
(1)
::::>
Q=
/.,'
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I{"
( 'r,IIIItIl:I!lIn
}) .12
I 'II
,'f !
II. ()
(1.1 ~
'f I
2() •. 1
II. 115
0.12
Il,
c=
~I) -:-
C
I)
.I~
'1',
L"
K\
293.22 K I
TClllpcralUJ'c
T
or brick - pille interface
lical loss is reduced
RI
2() .22 K
I
by 10%
L\
KIA
CmlsitierillJ;
tl,iclme.\·s of tire mortar,
295.9 - T
Q/A
L\
TI
kl
295.9-T2
1.11
\T2
Inside
0.175
0.25
295.12
KI
0
Mortar brick interface temper~tllre is 21 C
1
-'
=>T4=21°C
Q
T4 = 210 + 273
295.12-T3
~
k2/\
££IIIi1t1M'6~'_
Scanned by CamScanner
I. 72 Heal and Mass li'an~k/'
_--_._----.---_.
- ..__ .-.. _---_----_-
Mortar thickness,
1'4 = 5 Illlll = 0.005 III
0.99::
We know that,
T5 -- 293.03
0.075
1.25
IT5:: 293.08 K]
294 - 293.08
Q ::----~
L4
T(,- Tb
(2) => Q =
k4 A
---_._Rb
Q/A =
Q= T6-2~~.
0.92
0.005
k4
1
hhA
Q/A=
T6 - 293
_--
.i.
Thermal conductivity
T6 - 293
0.99
= --.'
3
of the Mortar, k4 = 538 x 10- W/I11K
..-~-
j_
Result :
33
I. Overall transmittance,
293.03 ~"]
2
U = 0.37 W/m K
1. Heat loss, Q = 1.11 W/1112
3. Temperature
ofhric~ - pine interface = 293.12 K
T< - T(J
=:~
L~
~~ ;\
Q/A
15 - 293.03
0.075
1.25
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4. Thermal
3
COlldllctj\,jl~
of the Mortar := 5.38;.10-
\\'/IllK
/. 74 Heal and Mass Trans er
1.1.12 SOLVED UNIVERSITY
PROBLEMS
ON SLAIl
.r
Inner surface temperature
f1] A [urnace wall consists of three layers. Th~
litem thickness is made offire
brick (k= 1.04 Wlt,,~Yer~
intermediate layer of 25 em thickness
is made if ). 1~
o nzlll'o
brick (k = IJ.69 WlmK) followed by (I 5cm/hick co
' ~~
(k = 1.37 WlmK). When the furnace is in continuou'
tire inner surface of tirefurnace
trcrele II'
,\ °perllJ'
~
is at 800 C while t'
I~
concrete surface is at 50"C. Calculate the rate oifl,
~
0
Ire 0llt
per unit area 0/ tire wall, the temperature
.
rellllo~
at the illl""
r., '
•
er lice q
tirefirebrick (lilt/ masonry brick am/the lel1l1Jerlilll
I
Ire 11/ IN
interface 0/ tire masonry brick (111(/ COil crete.
'
Outer surface temperature,
T
800 C
0
Conduction
2
1.75
1+ 73=1073K
T 4 = 50 C + 273 = 323 K
0
Tofilltl :
I) Rate of heat loss per unit area of the wall, (QI A)
2) Temperature
brick, T2
at the interface of the fire brick and masonry
.
3) Temperature
at the interface of the masonry brick and
concrete, T3.
j
Solution :
(i) Heat loss per square metre (QIA)
[Anna Uuiv -June'06j
Heat transfer Q = ~ Toverall
,
Give" :
R
where
Fire
Inner
side
Masonry
brick
brick
( T,
<PT2
Concrete
wall
~DTJ
[From HMT data book
page no.43 & 44 (Sixth editionj]
Outer
side
<DT4
=> Q =
I
L,
LJ
L3
--+--+---+--+ha A
k, A
k2 A
k3 A
I
hb A
[Convective heat transfer co-efficients ha, hb are not given.
So, neglect that terms]
Thickness
of firebrick,
Thermal conductivity
Thid;ness
L) ==. I Ocm == 0, 10m
of
fire brick,
kl == 1,04 W/Il1K
of masonry brick; L2 == 25cm = 0.~5 III
Thermal conductivity of masonry brick, k? = 0,69 W/IllK
Th ickness of '-"0 ncre t e wa II,L.1 = 5cI11 == 0,05- m
Thermal conductivity
of concrete
Scanned by CamScanner
wall, k3 == 1.3 7 W/n1K
Q/A
_~--:---:-
~C~'o_o'lI(jIIlCliO"
I. 7
Similarly
Q= TrTJ
R2
(I)~
where.
LJ
RJ = _-_
-
k2 A
ii) Iff1nftK,~ tepualU,es (T] and Tj)
We
T2 - T3
~ Q=
that,
J rdL-rf see temperat
ures re lat iem
~
Q/A
1515.24
I
= 927.30 - T)
0.25
0.69
ere,
~78.30KJ
'r ,- T2
0"---'
~-
Result :
I. Q/A = 1515.24 W/m2
kl A
'1',- T2
Q/A
2. T2=927.30K
3. T3=378.30K
III All external wall of
1073··· "'2
1515.24 = _-'
.,.0.10
1.04
[li~_~~~.
3·~i.·.Kl
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II house is made up of 10 em commo«
brick (k = O.7 WlmK) followed by II 4 em tayer (II gypsum
plaster (k = 0.48 WI",K). Whu: thickness of Iml.,·(v plIL'ked
insulatlon (k = 0.065 WlmK) slwulll be lidded to reduce
the I,eat loss II"ougl, the wal! by 80%.
[May 200-1 _ AIIIW Univ. Ocl-99 & Oct 20tJ/- ,\4. Uj
1.78 Heat {llId Mass Tran.lfer
Give" :
Conduction
Thi(;kness of brick, LI = 10 em = O.lm
Thermal conductivity
of brick, kl = 0.7 W/IllK
Thickness
L2 = 4 em = 0.04 III
L2 kL) -,.1]
R. = _l_A [-1-'ha ,_.!::Lk,. +-+-+
k
-r
2
of gypsum,
I. 79
. where
Thermal conductivity
of gypsum, k2 = 0.48 W/mK
Thermal
conductivity
of insulation
Brick
Gypsum
,3
Ib
[The term's /7'1', and hb are not eziven . S0, neg Iect t Iiat terms)
R = _l_[.!::L
k- = 0 065 WI IllK
A
'J'
Considering
J:L + _s_ 1
+
k,
k2
k)
two slabs, i.e., neglect L3 term
sr
[''- A = I 1112)
Q =----
_s_+~
k,
k2
kI
k2
6T
100 ",,' - .....__::::...:_~
_QJ_ -I- 0.04
0.7
0.41\
Ilril:~
(iYI ~IIIII
()
W,
()
" I
.(d')
o
I
I
1il [lnd :
Thi ·!.:II('. S 01' msulnrion
to reduce tile
wall hy 80%, (L1)·
r Assume heat
CO) == 100 WI
II 'III luss is red" .cd h)' IilJ% e111l:10 insulllt ion. So, hcnr trnnsfcr
IS
.
transfer
I
leA
t I
. thrOl1uh
oss
10.1
0.7
I ().()il I
OJIS
1'1
(l.()(,.
I
II
'
r~
0.0. ~Ii 111/
.
Result :
Solution :
J [eat
flow rate, Q =
,1T overall
R
'[From I-1M?'data hook page /11)
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Thickness of insulauon
,/1
4J I~
I.J
0.0 XX III
__j
I.SO Heat atld Mass Tratlsfer
-II) A composite
wt,lI consists of 10cIII Ihick lay~
Conduction /.8/
brick, " = fl. 7 WlmK and 3('''' II,ick pla.'iter,1c == 0.5 W/~
ilI.mltlti"g nUlterialof"
= fI.ORWIIIIK is to he tltldedlo (~
the I,et,tlrtlmft!r 111T0III:" II,e wull by 411%. Fintl its th' rr~
.
{Dec- 200-1 .11I1I1I Uuiv
& Dec-lO(lj
Id~
Considering two slabs, i.e., neglect LJ term
:ll/l/u Viii!
Give" :
Thickness of brick, L I = 10 em = 0.1 111
Thermal conductivity
~IOO
of brick, k I = 0.7 WImK
0.7
Thermal conductivity
of plaster, k} = 0.5 W/mK
Thermal conductivity
of insulation, k 3 =: 0.08 W/mK
I
Insulation
k,
k,
[Assume heat
_QJ_ + 0.03
Thickness of plaster, L2 = Jcm = 0.03 m
~:---~ast~
~T
__
= -~
~ I IH 20.28 K I
=
Heat loss is reduced by 40% due to insulation. So, heat transfer
is60 W.
~T
Q = -
I
R
~T
Q=
I
(~+~+~l
1
A
-
----_._._
----_. -_.__ !
f---. L I - --ic- - I. !---..j
kl
1
:::) 0.1
0.7
(_L11(/ + .!:Lkl + .!:L
.!:l_ +_I 1
k2
k
lib
+
0.5
L3
7
0.08
0.08
1
= 20.28
+ 0.03 +..!::L = 0.338
:::) 0.08 =0.135
j
The terms I/{/ and "" are not given. So, neglect that ten11S.
Scanned by CamScanner
0.5
~1
l
R
A
0.7
:::) 60 r _QJ_ + 0.03 +
0.7
0.5
0.08
sr overall
Heat now rate, Q -=
R = J__
k3
_!_ [_QJ_ + 0.03 + _!::L_
to reduce the heat loss through !Ii
Solulio" :
where
k2
20.28
60 =
Ttl find :
Thickness of insulation
wall by 40%, (LJ).
transfer (Q) = 100 WI
0.5
I.S2 Heal 0/1(/ Mass nOl15jer
I L = 0.0108
In
( 'ouduct inn 1.83
I
'If/jiuff
:
i) II
RemIt:
o A sw{ace
10. s per quar
ii) Interface
L) = 0.0 I08 111
Thicknc s of insulation,
;It
is made lip of 3 layers one of fire brick 0
insulating brick and one of red brick. The inner (III 'd Oil)
nIl
HlIIII
surface temperatures are 900 C lind so: C rejfJective~ll.
n
respective co-efficient of thermal conductivity of thelal'~
are 1.2, 0.14 and 0.9 WlmK lind tile thickness of 20 em,8
and L/ em. Assuming close bonding of the layer.\·at
interfaces. Find the heal/on' per square meter and illlerf~
temperatures.
[/11. U OCI-9 J
0
SO/lIlioll
metre, (011\)
(T2 and T )
temperature:
:
(i) /JI'(I1/0.H per square metre (QIA)
/Frolll
Tavera II
I kat Iran fer. Q ==
IIMT data hook
[lag(' 170 . .J3 Gild ./4
R
J
where
1.1
L2
I- I
k2 A
--j--
Given :
lnsulatins
Fire
Inner
(bT ~
<In I
side
(
.I-
k2
kl
,
(
T3
k3
KIA
[Convective
heat
transfer
So" ucglcct
tha: 1 rill J
-....0 =
kIA
f-- LI --tc- L
T, = 900 C + 273 = 1173 K
of fire brick,
Thermal
of
conductivity
knes
Thic~ne
Thicklle
s
01
T4 = 30° C + 273 = 303 K
Thermal conductivity
Thi
k2A
L_
k2 A
it
k, = 1.2 W/rnK
iusulatins o brick '- k? = 0.14 \\/111
of red brick,
LI = 20 CI11 = 0.2
III
brick, L2 = 8 CIll
= 0.08
r re d brick,
L3 = II em
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= 0.11 III
111
LI
I.,
L.
~
~
1173 - J03
k) = 0.9 W/rnK
f Iire brick,
f' III ulatiug
"4
--'1'1'-- ----
~'~
Q/A
L
1
k, A
hb A
-'-~-
co-efliciellts
0
Outer temperature,
Thennal COlldllCti
L2
--
h"A
LI
--j--
Inner temperature,
LI
--j--
brick
brick
side
-~0
Outer
Red brick
0._
0.08
ill
1.2'
0.14
0.9
h(/, hb arc nut given.
/. 84 Heal and Mass rransfer
(ii) Interface teperatures (T) am/ Tj) ~
We know that, Interface temperatures relation
Condur,
I.85
---------------____:_._-;0/1
Q/A
TI - T"
Q
R
1004.457 - T3
----.::....
0.08
0.14
1011.2546=
where
LI
R1=---
kl A
TI - T2
LI
kl A
Result:
(i) Heat loss per square meter (Q/A)
Q/A = 1011.2546 W/m2
(i i) Interface
temperatures
(T 2 and T 3)
TI - T2
Q/A
T2 = 1004.457 K
L,
1;1011.2546=
1l7J-T2
0.2
1.2
IT = I 004.4 57 K I
2
Similarly
T3 = 426.597 K
[I) The wall of a furnace
is made lip of 2.f0 mm fire clay of
thermal conductivity 1.05 WlmK, 120 mm thick of ins Illation
brick of conductivh, O.15 WlmK anti 200 mm thick red brick
of conductivity
0.85 WlmK. The inner and outer surface
temperatllre oj wall are 850 C tlntl 65- C respectively.
Co/cilIate the temper(l/ures at the contact surfaces.
G
[Bharathida an
niversity
L1=2S0mm
5m
ov- 95}
Given:
Thickne
of fire cia
Thermal conducti
Scanned by CamScanner
=0.
ity, kl = 1.05 W/mK
Thi knes of insulation
bri
Thermal
2 = 0.15 W/mK
ndu tivity,
., ~ = 120
m = 0.1
In
---
/. 86 Heal and Mass Transfer
Thickness
of red brick, L3 == 200 nun == 0.2 rn
Thermal conductivity,
__
Conduction 1.87
[Heat transfer co-efficients ha and hb are not given.
So, neglect that terms]
k3 == 0.85 W/mK
Inner surface temperature,
T 1 == 850 + 273 == 1123 K
Outer surface temperature,
T 4 = 65 + 273 == 338 K
Fire
Insulation
clay
brick
• T,
k,
k2
L,
L2
Redbrick
QfA
_s_+ ~
k,
Q/A
kJ
T
slab is gi en by
1123-338
0.25 + 0.12 + 0.2
1.05 0.15 0.85
= 616.46 W/m21
Q/A
T, - T2
Q=--
1~
[From H \17 data book page no. 43 & 4.{
'erall
where
~ Q=
fA
Scanned by CamScanner
k3
= ------
Souaion :
....
+S_
k2
We know that,
ro eft composite
k3 A
TI-T4
=
<D
I
Heat
~
k, A + k2 A
°nj
(~T2
T,-T4
Q =
~
=
RJ
>
1.88 Heal and Mass Transfer
Conduction 1.89
1123 - T2
=> 616.466 =
0.25
1.05
.'
Similarly
wall made up of 7.5 cm offire plate anti O.65 em of
mild steel plate. Inside surface exposed /0 hot glls lit
650" C anti outside air temperature 27" C. The convective
Ilea/transfer co-efficient for inner side is 60 WIm1K. The
cOnl'ective heattmnsfer co-efficient for outer side is 8 Wlm1K.
Calculate the heat lost per square meter area of the furnace
wal! and also find outside sutface temperature.
{M U. April-98]
T2 - T3
Q=~-
(I)=>
@] A furnace
Given:
where
Fire
Mild steel
plate
plate
Outside
Jnside
T(I'
=> Q=
n2
n,
k2
f.c-- L, -----fo--- L2 --l
976.22 - T3
Q/A
L2
Thickness of fire plate, LJ == 7.5 em == 0.075 III
k2
Thickness of mild steel, L2 == 0.65 cm == 0.0065 m
976.22 - T3
=> 616.46
==
T/J'
hb
Ita
k,
=>
( TJ
0.12
0.15
Inside hot gas temperature,
Outside air temperature,
Convective
T a = 650 C + 273 = 923 K
0
0
0
T b == 27 C + 273 == 300 K
heat transfer co-efficient
for
inner side, ha == 60W/m2K
Convective
heat transfer co-efficient
for
outer side, hb= 8 W/m2K.
Result:
Tofi IIlI :
(i) 1'2 == 976.22 K
(ii) T3;: 483.05 K
Scanned by CamScanner
(i) Heal lost per square meter area, (QI A)
(ii) Outside surface temperature,
(T3)
1.90 Heat and Mass Transfer
Solution :
Conducnnn I.91
(i) Heat lost per square meter area, (QIA)
Thermal
conductivity
for fire _ plate (Refract ory clay)
k, = 1.0035 W/IIIK.
{From H.UT data book page no. 9 (I- iflh edition
•
Thermal
.
conductivity
or page
•
.
I)
JlO.
-
r.~
(ii} Olltside surface temperatllre,
We know that, Interface temperatures
retarion
.
Sitt" edlt,
for mild steel plate
k2 = 53.6 W/I11K
... ( I)
[From HMT data book page liD.
where
Heat flow,
Q
Toverall
R
where
TrTb
~
Q=
J
LI
~
--+---_
kJ A
k2 A
fib A
TJ - Tb
QIA
J
fib
~Q
2907.79
[The term LJ is not given. So, neglect that term I
T) - 300
,
8
I
Ta- Tb
~Q
=
T3 = 663.473 K
I
Result :
(i) Heal lost per square meter area, (Q/A)
Q/A
.. Q/A = 2907.79 W/m2
Q/A =
I
QI1\
923 - 300
_, + 0.071+ 0.0065.
60
1.035
53.6
= 2907.79
(i i) Outside
I
t-"8
surface temperature,
(T 3)
.. T] = 663.473 K.
W/m2!
(b!
Scanned by CamScanner
( 'flliilliNil/il
1\'1
-
[]
I.)
"j
I
h(l II
I A t. IJ, .:Iori/.
t I' ''fill ..
Fh
Imllilll(inA
bl'k~
brick
I
Il'hul
I 'J 1
1'1Ilt!
I
kl A
,,~ A
II", 1..1 II lid "b nn
Lj
I
IIO(
"I
A
hi) /I
II,lv·,1. So. nogl
tthar rerrns]
6 0
Q/A
0.23 0.115
--+-OL
0.27
kI
872 W/m2!
I--- Ll--~-Thi kness
f fire bri k. L] =
r i kness
f insulating
L
j
ern
= 0.23
Result:
III
Rate of heat lost per square meter, (QI A)
brick, L_ = I!. - ern
= 0.115m
Q/A
Thermal c ndu tiviry of fire brick
e al conductivity
of insulating
perature difference,
brick, k2 = 0.2 \\/rr
[!] TI,e
inner
dimension
of a freezer
cabinates
are
60 em x 60 em. The cabinates wall consists of /HIo 2 mm
6 T = 650 K
thick steel wall (k = 40 WlmK) seperated by a 4 em layer of
Tofind:
fiber
10
= 872 W/m2
k I = 0.72 W/rnK
per square meter, Q/A
SOlUlvm:
glass
insulation
(k = 0.049 WlmK).
D
TI,e inside
C and the outside
temperature
is 10 be maintained at _I5
temperature
on a hot summer day ;J 45° C. Calculate the
maximum amount of heat transfer, assuming a heat transfer
To /erall
R
[From IIMr data book page no.
co-efficient
4J'
of 10 Wlm] K both on inside and outside of tile
cabinate a/JO calculate outer surface temperature
of tile
cabinate.
[M. U. Oct-2002}
Scanned by CamScanner
JUt"
/. 94 Heat and Mass Transfer
Give" "
Conduction 1.95
where
Fiber
Steel
L\T=Ta-Tb
Steel
glass
(DT2
<Drl
kl
<DT3
(
k3
k2
Area, A = 60 ern x 60 ern = 0.36 m2
L
I~
l:J
)
ha A
k A
k2 A
kJ A
hb A
J
J
~ Q =
258 - 318
Q/A
_'--
Thickness of steel, L) = L3 = 2 mill = 0.002 III
Thermal conductivity
,
R = --+--+--+--+-
+ _0.002_ +
) OxO.36
0.04
+ 0.002
0.049xO.36
40x0.36
+_,_
)Ox0.36
of steel, ", = k3 = 40 W/mK
Thickness of fibre glass, L2 = 4 ern = 0.04 III
Thermal conductivity
of fibre glass, k2 = 0.049 W/IllK
/Q
=-21.25W]
[The negative sign indicates that heat flowsfrom outside 10 inside]
Inside Temperature,
To = _150 C + 273 = 258 K
(ii) Outer surface temperature
Outside Temperature,
T b = 450 C + 273 = 318 K
We know that,
Heat transfer co-efficient,
110 = lib == 10 W/1112K.
T -Tb
Q = __:::_[/-"-
R
Tofind :
(i) Maximum amount of heat transfer, (Q)
(ii) Outside surface temperature,
Solutio"
40x0.36
(T4)
"
(i) Maximum (111101111'
O/hNI"Tnm/er
Heal flow, Q ~ .1Tovera"
R
[From /I MT data book page 110.43 & 4t
Scanned by CamScanner
-21.25=
0.36 x 10
(T~
TI - T2 _ Tr
R,
-
T3
R2
T 3 - T4
= T4 - Tb
R3
Rb
.•. (I)
-1. 96 Heal and Mass Transfer
Conduction I. 97
Result:
(i) Maximum amount of heat transfer,Q
:::-2 I .25 W
Tojiml:
(i) Rate 0 f heal loss per m2 of tank surface area (QI A)
T4 = 312.09 K
(ii) Outer surface temperature,
(ii) Tank outside surface temperature (T2)
l!1 A mild steel tank of wall thickness 10 mm Contllins IIIUle'l Soilltio" :
90 C. Calculate tile rate of heat loss per ml Of tank surfll/J.
area when the atmospheric temperature is 15 C. rite tile,.".
conductivity of mild steel is 50 WlmK ami tile hea: trUIU!,
co-efficient for inside ami outside tile tank are 2800 "II WlmlK respectively. Calculate also tile temper(lturt~
tile outside surface of tile tank.
[M U. Apr-2000]
0
Heat loss, Q =
.1Toverall
R
0
where
.1T = T{/- Tb
__ 1_+_S_+~
R - haA
k, A
k2A
+_!1_+_I_
kJA
hbA
[LJ' ~ not given.So, neglect that terms]
Give" :
Inside
k
T(I>
ha
<P 1
~P2
=> Q/A
= _1_+
ha
_s_ +_1
k,
hb
363 - 288
Thickness
of wall, L, = 10 mm = 0.01 m
Inside temperature
Atmospheric
Q/A =
for inside, ha = 2800 W/J112K
Heat transfer co-efficient
for outside, hb == II W/J112K
of mild steel, k = 50 W/mK
0.01
I
50
II
+-+-
[ji:__~19.9W/2]
T b = 15° C + 273 == 288 K
Heat transfer co-efficient
Thermal conductivity
2800
of water, T a = 90° C + 273 == 363 K
temperature,
1
8
__..
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1. 98 Heat and Mass Transfer
We know that,
Conduction 1.99
_ T2-T
- ---..Q_
Q
Rb
==>
". (I)
I T 2 = 362.5 K I
Result:
(i) Heat loss per m2 surface area, Q/A = 8)9.9 W/m2
where, Ra= haA
(ii) Outside surface temperature, T2 = 362.5 K
363 - T)
Q=
QIA
1
I2!l Consil/ering the heating surface of a steam boiler to be plane
haA
wall oftllickness 1.2 em and having k = 50 WlmK. Determine
the rate of heat flow and surface temperatures for tile
363 - T)
following
1
data.
Flue gas temperature
1000°C
Boiling water temperature 200° C
363 - T)
Heat transfer co-efficient on gas side 100 Wlm1 K
_1_
Heat transfer co-efficient on steam side 500 Wlm2K
819.9 = ----!....
2800
[Manonmanium
Sundaranar University April- 97]
Given:
Thickness
LI
(I) ~
where, R, = k A
I
Q=
conductivity,
k, = 50 W/mK
. LI
Flue gas temperature,
T,-T2
___!j_
k,
819.9
Thermal
T)-T2
kl A
Q/A
L, = 1.2 ern = 0.012 m
362.7-T2
0.01
50
Scanned by CamScanner
kl
Tb
Ta
hb
ha
(DT2
( TI
T a = 10000 C + 1273 K
I-
Boiling water temperature,
Ll
..j
Tb = 2000 C + 273 = 473 K
Heat transfer
co-efficient
Heat transfer co-efficient
Thermal
. '
conductIvIty
h
100 W/m2K
on gas side, a =
, 2K
id h =500W/m
on steam 51 e, b
ild steel k = 50 W/mK
of rru
,
1.100 Heal and Mass Transfer
._-----
Tojintl:
(i) Heat transfer
(i i)
- -~---_._.
rate. Q/A
1I1' t'act' teperutures,
COlldl/(:/ioll
( I) ::':> Q;-:
(T, [lllll Tz)
Solution :
lIeat
II uisfer.
\T=T
(I
R=
Q
- l
L,
.1
A
~
k ~.'\
T -T
_!.I_I-
R(I
C.J'
- '1'
__T (I_I
(.)/A
= ,r_I'1 '1'_1
,,,A
(l).~59 ::: -
_1.
100
_-
..
A
1,A
Q = -_-"'--_=---_-
=:>
'I
1_7. -1'1
_ ___:_
[lZ L, values an:' II t uiveu. oo, negle~.t(h,l!(
R=
['.: R =_1 _I
(I' A
(I
II"
}
1
L,
h A
", A
Q==
(I
T1-
I
A
Q=
T2
R,
I
_-~--+--
TI - T_
LI
kl A
QfA =
I
100
I
Q/A
T1 - T2
~
kI
800
=> Q/A =
0.012 +_1
50
500
65,3 59 ==
Interface temperatures
relation
1 - Tb
o = .is.
-R~ =
Scanned by CamScanner
619 - T2
0.012
SO
65.)59 W/m2 1
=>
J. J 0 J
IT2 = 603.3 KJ
1.102 Heal andMass
Transfer
Result:
~
(i) Heat transfer, Q/A = 65,359 W/m2
(i i) Surface
Conduction 1./03
L2 = 10 em = 0.1 m
T2=603.3K
[ll) A composite slab is made ofthree
15em=0.15m
1,,:;::
T 1 = 619K
temperatures,
L) = 12 em = 0.12 m
layers 15 em, 10 c",
12 em thickness respectively. Tirefirst layer is made 0' m ~
'J a'e~
with k = 1.45 WlmK, for 60% of tire area and lire re
S"
material
with k = 2.5 WlmK.
material
with k = 12.5 WlmK for 50% of area and res,;
The second layer is madt;
material with k = 18.5 WlmK. The third layer is madeols;",
material of k = 0.76 WlmK. The composite slab is expOil
on one side to warm at 26 C and cold air at -20· C n
inside heat transfer co-efficient is 15 Wlm2 K. The outsideh,
k'a
= 1.45 W/mK,
k'b==2.5
Ala = .60
W/mK,
Alb=·40
k2a == 12.5 W/mK,
A2a = .50
k2b == 18.5 W/mK,
A2b = .50
k) = 0.76 W/rnK
T a == 260 C + 273 = 299 K
0
Tb == -20
C + 273 = 253 K
ha == 15 W/rn2K
0
transfer co-efficient
is 20 WI",2 K determine
heat flow rt
and interface temperatures.
[MU Nov-~
hb == 20 W/m2K
Tofind :
(i) Heat flow rate, (Q)
(ii) Interface temperatures, (T, , T2, T3 and T4)
Solution:
Heat flow,
A,a = 60%
A2a = 50%
k1a
k2a
(DT,
A,b = 40%
k,b
(DT2
Q==
[From HMT data book. page
.1Toverall
no.43 & 44}
R
A) = 100%
(
where
(
T)
A2b = 50%
k2b
Scanned by CamScanner
k)
-
I
LI
- --+--+-- A a h a A Ik I
L2 +_3_+_
L
I
A2kJ-
A k
3 '3
Abh
b
1.104 Heal and Mass Transfer
COl/duc/ion 1.105
Similarly
•.. (I)
... (3)
where
0.1
12.5 x 0.5 = 0.016 K/W
I Ra
K/W
= 0.066
I
[R20
=
0.016 K/W]
0.1
18.5 x 0.5 = 0.0108 K/W
... (2)
IR2b
R 10 -- k
I
Ria
L
In
I
x A
= 0.1724
---
0.15
1.45 x 0.6
10
o. I 724 KlW
(3) ~
.)
= 0.15 K/W
I
=
0.0064 ~
) xO.76
:\3k3
IR3 = O.)578~
Rb = _1_ ..= _I Ab hb
I x 20
~0.05
Substitute R In and Rib value in (2)
(2) => R 1 =
Kiwi
R~ = ~;_c__QJl_
I
RIb = 0.15 KIW
0.0108
R2 == 0.016 x 0.0108
0.0161-0.0108
I R2
Kiwi
Rib = __ L_;_I_ _
0.15
klb x Alb
2.5 x 0.4
I
==
=
K/\\]
o.
I724 x O.) 5
0.1724 + 0.15
RI =0.08 K/W
I
Scanned by CamScanner
(I) =:>
Q =
---=..:29~9--..::..2~53:----0.066 -1- 0.08 + 0.0064 + 0.15789 + 0.05
1.106 Heat and Mass Transfer
Conduction 1.107
(ii) Interface temperatures (Tl' T2, T3 and T.f)
TrT4
(4):::>
We know that,
Q==~
127.67=
279.532 - T4
0.15789
[T4 = 259.374 K
Result:
T -T]
(4)~
I
(i) Heat now rate, Q = 127.67 W
Q==T
a
(ii) Interface temperatures, (TJ, T2, TJ and T4)
299- T]
T] = 290.57 K
0.066
TJ = 279.532 K
I
T4 = 259.374 K.
IT] == 290.57 K
(4):;" Q== T]-T2
. ~
R]
127:67 == 290.57 - T2
0.08
IT2 == 280.35 K!.
..
i
."(4) ~ Q ==--1:_l
1: T
)
T2 == 280.35 K
127.67 == 299 - T]
0.066
"J
R2
Ii.
127.67 == 280.35 0.0064 .
[!! 279.532 KJ
==
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Afurnace wall consists of steel plate of20 mm thick, thermal
conductivity 16.2 WlmK lined on inside with silica bricks
ISO mm thick with conductivity 2.2 WlmK and on the outside
with magnesia brick 200 mm thick, of conductivity
5.1 WlmK. TIre inside and outside surfaces of the walt are
maintained at 650 C and 150 C respectively. Calculate the
heat loss from the wall per unit area. If the heat loss is
reduced to 2850 Wlm2 by providing an air gap between steel
and silica bricks, find the necessary width of air gap if the
thermal conductivity of air may be taken as 0.030 WlmJ(.
D
D
[Madurai Kamaroj University April 97J
1.J08 Heat and Mass Transfer
Give" :
kI
k2
k3
________
where
----------------------~C~·o~,,~d~uc=Il~·o~n~/~./
6.T= TI-T4
T2
I
LI
kl A
Ll
k2 A
L3
k) A
I
R =--+--+--+--+ha A
Steel
Silica
Magnesia
Steel plate thickness,
L, = 20 mm = 0.02 m
Thermal conductivity
of steel, kl = 16.2 W/mK
Thickness of the silica, L2 = 150 mm = 0.150 m
Thermal conductivity
Thermal conductivity
TI - T4
Neglecting unknown terms (ha and hb)
TI-T4
Q=------LI
~
L3
--+--+-kl A
k2A
k3 A
of silica, k2 = 2.2 W/mK
Thickness of the magnesia,
'-3 = 200 mm = 0.2 III
of magnesia, k3 = 5.1 W/mK
Inner surface temperature,
T I = 6500 C + 273 = 923
Outer surface temperature,
1'4 = 150 C + 273 = 423 K
923 -423
Q = ---~----0.150
0.2
-0,~.0:..::..2_
+--+-16.2xl
2.2xl
Q =
500
0.1086
Thermal conductivity
I =
4602.6 W/m2
of the air gap kair = 0.030 W/mK
Q
Tojind:
(i) Heat loss [without considering
air gap]
Heat loss is reduced to 2850 W1m2 due to air gap. So, the new
thermal resistance is
(ii) Thickness of the air gap
Q=
SOIIlI;oll :
~T
Rnew
Heat transfer through composite
considering air gap]
Sf
R
Scanned by CamScanner
5.lxl
0
Heat loss reduced due to air gap is 2850 W/m2
Q=-
hb A
wall is given by Iwith~
/./10 Heal and Mass Transfer
Conduction 1.111
923 - 423
Rnew
=
1.1.13 Solved Problems On Cylinders
2850
o A Itollow cylinder 5 em inner radius and 10 em outer radius
I
= 0.1754
K/W
Thermal
resistance
of air gap
Rair
has inner surface temperature of 200 C anti outer sur/ace
temperllture of 1000 C. If the thermal conductivity is
70 WlmK,jind heat transfer per unit tength.
0
Rnew
Given:
Inner radius, "1 = 5 ern = 0.05 m
Rnew - R
=
Outer radius, r: = 10 cm = 0.1 m
0.1754 - 0.1086
Inner surface temperature,
T 1 = 200 + 273 = 473 K
Outer surface temperature,
I Rair == 0.066 K/W I
T2=100+273=373K
Thermal conductivity, k = 70 W/mK
We know that,
Ttl find :
Lair
Heat now per unit length
Rair == k air )( A
[.: A = 1m
Lair
0.066 == 0.030)( 1
::::>
I
Heat transfer through hollow cylinder is given by
6Tovcrall
Q= __::.:...::.:..:::oc
R
1
3
Lair == 1.98)( 10- rn _
3
In
I
[r2]
R=--ln2n:Lk
rl
.I
.
) - 4602 W/m2
(i) Heat loss (Wit rout air gap -
.
(ii) Thickness
[From equn. 110.1.32 or
HMT data book page 110.43 & 44J
where
Thickness of the air gap == 1.98 x 10Result:
Solution :
_
of the air gap, Lair - I.
98 x 10-3 rn
::::>
Q =
I
--/11
2n:Lk
[r2- ]
rl
-----...
&.~
\
Scanned by CamScanner
/. 1/2
Heal and Mass Transfer
=>
Q
-----~
2itkL (1', - T2)
/11
Conduction I.J 13
outer temperature,
[;:n
T2 = 27.9° C + 273 = 300.9 K
Heat transfer, Q = 120 W
Toft"d:
Thermal conductivity, k
=> Q/L
SolutiOJl :
Heat transfer through hollow cylinder is given by
2rrx 70(473-373)
=> Q/L = -----___:_
L\ Taverall
IIl[O~O~ 1
Q=--R
I Q/L = 63453.04 W/m = 63.453 kW/m./
[From equn. 110.1.32 or
HMT data book page 110.43 & 44]
where
Result:
R = _I_
Heat transfer per unit length, Q/L = 63.453 kW/m.
211Lk
III Determine thermal conductivity of asbestos powder pllckedu
~
Q
__ I
211Lk
between two concentric copper pipes 25 111m and 36 m
diameter length. The inner pipe housint; has (I heating Coi/I
which 120 HI power is supplied. The average telllpef(/Iu't~
inner (111(1outer pipes are 42.,r C (11/(127. 9° C re.\pectively
111 [r2]
rl
111 [r?
-=- ]
rl
315.4 - 300.9
~ 120 = -------I
111[_0,_0'_8]
211 x I x k
0.0125
Give" ..
Inner diameter, D, = 25 mrn
Inner radius,
/k
r, = 12.5 mm
=0.0125111
I
Result:
T2
Thermal conductivity, k == 0.48 W/mK.
Outer diameter, D2 = 36 mm
Outer radius, r2 = 18 mm = 0.018
III
Inner temperature, T, = 42.4° C + 273
=3J5.4K
9
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0.48 W/IllK
[':L=lmj
J.1J4 Heal and Mass Transfer
Condllclion 1.11 5
III A hotlow cylinder 5 em inner diameter an~'
diamel~r has inner surface temperature of 200
c", Olllrt
surface temperature of 100 C Determine Iteatj1C llnd 0111(1
,
Ow tilr
the eylmder
per metre length. Also determine tit e temper QlIg
.
of the point half wa) between 'he inner and out er Sur I:11/41r
0
0
1 ft!,
Take k = ] WlmK.
473 - 373
o
0
I
2n L x I
::::>
ln [0.05 ]
0.025
[OIL = 906.47 W/m I
Gil'm:
(ii) Temperature
dl = 5 em = 0.05 m
between
Put T2 = T and
'1 = 0.025 m
inner and outer surfaces, (T)
in heat transfer equation
1"2 = r
d = 10 em = 0.1 m
::::>
r = 0.0- m
0
[ -r
J
--In
2rcLk
T 1 = _00t> C = 4 3 K
1
'1
rl
T 2 = 100e C = 373 K
T]-T
::::>
k = I W/mK.
Q/L = ------::::> r=
_/ /I [0.0375]
__ 1
2rc x 1
0.025
Tofind:
(i) Heat flow per meter,
(ii) Temperature
::::> 906.4 7 = -..-:...:.-=----=---
(Q/L)
between
inner and outer
surfaces,
_I_ / [.QJ>J 75 ]
2rc /I 0.025
(T).
IT = 414.5 K I
(i) Heat flow per meter (Q/L)
[From HMT(kll(/~(X
page /10.43 & ~j
~ T overall
R
Result:
(i) Heat flow per meter, Q/L = 906.4 7 W /m
where
(ii) Temperature
between
inner and outer surfaces,
T=414.5
R = --/11
I
2nLk
0.025 + 0.05
2
::::> r = 0.0375 III
473 - T
Solution:
Q=
+ r:
.: r= -2-
l _1_
I" 1
1"]
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K.
1.116 Heal and Mass Transfer
Illller d'
~ of tile pipe is 25 em, wall thickness hi 2 em ti,' 1.lallieler
Condllction
r:tl An insulated steel pipe carr) ing a hot liquid.
.
".
•
'
Temperature
1.1/7
of hot liquid, T a == 100° C + 273
lelliless
insulation IS 5 em, temperature of hot liquid is 10 of
temperature of surrounding
is 20° C, inside heat tr 0 (',
.
/
2K
w.
co-efficient is 730
/m
an d outside
Ileal tr alls/er
.
2
(IIIS/
co-efficient is 12 Wlm K. Calculate tile IIeat loss per er
'"elre
length of the pipe.
0
Ta == 373 K
Temperature
of surrounding,
T b == 20° C + 273
Tb == 293 K
Inside heat transfer co-efficient, ha == 730 WIm2K
Take kstee/:::: 55 WlmK,
killslliatillg
lIIateria/::::
0.22 W/"'K
Outside heat transfer co-efficient, hb == 12 W/m2K
Given,'
ksteel == 55 W/mK
kinsulation == 0.22 W/mK
Tofind,'
Heat loss per metre length
SO/lItiOI1 :
Heat flow through composite cylinder is given by
Q =
~Toverall
R
[From eqllll. no. 1.48 or HMT data book
page no. 43 & 45 (Sixth edition})
where
Inner diameter,
~T=Ta-Tb
d( ::::25 em
Inner radius, 1'( :::: 12.5 ern
R
h =0.125 ml
radius, 1'2 = /'( + thickness
2nL
of wall
0.14Sml
radius, 1'3== r-, + thickness
== 0.145
~3
Ta- Tb
=:> Q =
0.125 + 0.02
[1'2
[h:" +
2rrL
of insulation
0.05
== 0 195 mJ
Scanned by CamScanner
r ' +
harl
III [~~ 1
III [~~ 1
+
k(
k2
,j
+--
hbr3
","78
"7
1.118 Heat and Mass Transfer
373-293
Q-
;:::>--
L
I
I
~
Hot air temperature,
1
m + 111 [~] -------+~
I [.145]
-21t [ 730)(.125 +
11
55
0.22
----
Conduction 1.119
12x.19j
T a == 40° C + 273 == 3 13 K
Inner diameter, d, == 10 cm == O. I m
r, == 5 cm == 0.05 m
Inner radius,
Intermediate radius, r2 == r, + 4 em = 5 + 4 = 9 cm = 0.09 m
Outer radius. rJ = 1'2 + 3 ern = 9 + J== 12 ern = 0.12 m
@/L
==
281.178 W/m]
k,=o.IW/mK
k2 == 0.32 W/mK
Result:
Heat transfer per metre length, Q/L == 281.178 W/m.
ha == 50 W/m2K
hb == 15 W/m2K
III Hot air at 40° C flowing through a steel pipe of 10 f.
Outer temperature of air, Tb == 10 + 273 = 283 K
diameter. The pipe is covered with two layer of ;nsulali"l
material of thicklless
4 em an d 3 em and Ihtu
Tofind:
corresponding
t"ermal
cOlldllctivities
are 0.1 a.1
Heat lost per metre length of steam pipe
0.32 WlmK. The ills ide and outside convective heat tram/e
co-efficient are 50 WlmlK and 15 WlmlK. Tile OUID Solution:
temperature is 10° C. Find the neat toss per meier Itngrl
Heat flow through composite cylinder is given by
of steam pipe.
[From equn. no. 1.48 or
llToveraJl
Q =-.=.:..::.:.=.:.
R
Given:
HMT data book page 110.43 & 4jj
where
llT=
Ta-Tb
R
2.L
I
2nL
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[,,~,
+
f
, I
"
I
J. J lU
Heal and Mass na"'::,jPI'
Q
=>-=
L
Inner air temperature.
T a '"
I T a = 363 K I
Inner diameter
Q/L
24.37 W/m
Heat tran
fer p r unit
f the copper,
d1 '" 5 em
radius,
1'1 '" 2.5 cm
Result,'
len III
II'I '" 0.025
/
= _4.
\ 1m
Thermal
[1] Air at 90° C flows
ill a copper tube or 5 ell, .
.
'J
tnner d,u,"
with thermal con d uctiviry 380 Wlm/( and wirt, O. elflr
wall which is healed from tit ' (/111 ide by water II( 1]0'1
A scale of O. 4 em thick i Iep ositcd 0" lite outer surfa I,
tile tube whose th ermul on du uivitv is I.S2 W/mK. nl.
(111(1water side unit urfa e ontluctance are 220 1I1"r
(In,l3650 W/m? K resp tctivelv.
alc ulate
ndu tivitv.
k, = 80 \\/mK
I the
PI cr, 1'_ '" Inner radiu
wall
r r diu'
L11
1'2
0.025
~_
0.03}3
radiu
, r« '"
=
l. Overall
Harer 10 air tran imittance
I
drop a ross II,e scale deposit.
L11
id icrnp
r lure
thicknes
0.032
= 0.0361nl
f water,
Tb = LO°
of
0.007 III
'2 + thickness
r,
2. Water 10 air h eat ex h ang e
3. Temperature
III I
of .cale
0.004
27
= "9" K
Give" :
\ ter
Thermal
ndu livily
k = 1.82 W/I1lK
n .e
fair, ha = 2 .. 0 W/m-K
n e
f water, h
. urfa
e
nul,
d
urfa
e
ndu
I
= "650 Whn-K
To filld :
vera ll he I Iran
2
.eff
ieru
v arer I air heal Iran f r, Q
) Temperature
Scanned by CamScanner
er
dr p
ihe
ale dep
,
II, (
T
, -
T)
2
1.122 Heat and Mass 1'ransfer
--
Solution:
Heat flow through composite cyl inlier is .
given
T overall
Q
Conduction I. 123
We know that,
by
Heat transfer, Q = U A
R
[Fro", C(i7L1n
I
. 110. J 4
HM'{ (.uta
book pag
. •
e 110.43 & I
where
where
U - overall heat transfer co-efficient
Ta- r,
~T
R = _I
2nL
l-I
harl
A - Area = 21t rJ L
+
In
l:~\
k(
+
6T= Ta-Tb
In [:~ 1
k2
Q
+-L
Q/L
hbr3
~
Ta-T
~ Q=
.L
2nL
T
l
In
h:rl
l:~I
k(
-739.79
U x 21t r3 L x (T a - T b)
=
= U x 21t r3
x (T a - T b)
= U x 2 x It x 0.036 x (363 - 393)
IU = 109.01 W/m K I
2
In (~~ 1
k2
1
h
Overall
+ I
heat transfer co-efficient,
U = 10901 W/m2K.
b'3
Interface temperatures
Q=-=
I II [.036)
JD2
2n:L 220~.02S
1.82
Ta - Tb = Ta - T I
T
R
R
-
T3 - Tb
1
3650~. :
Rb
\ here
'j
ha
hea f1
fr m out ide to inner S·'
1
R)=-
-
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Ra
2nL
...
(1)
1/('(1' (/lid M(I,VV
/ 1)1/
'l'''(lmjl'I'
'f)
'f I
('(111{/," I 1m I 12
~()
o] 12(1 """ 1,,111'( "'"m('ll'( , I"(j"
..
"1m lillie,
111('11" 11,/,11 1111"111,,1 ('''''(/11('111111 H W./nIK ,
/111
(
.
, "/1 "",t!
HI/'" two /11)11"1II/I"tl/IIIIIIIII
('11('" hlllllll)( II /""·,,,,£,U II/
.H 11,,11. Tlu: th rr nut] ('111/(/11('1",11 II/ Ihe /Iffl Inwlll//flll
"",11·,1,,/
ts 1I.1I,f W/",K
11111/ III", (lj th « ,rl'l'/iII" I,
11.// W/",K. till' trmprrutur» 1'./ tI", /111/111' tub» I'llf/II('I' I"
24(r C "lid IIIII'I~/I"('
IIIII,I"'C ,I/Ifjll"e 1'./11,,' IIIfIIllIlll/1l If
MJ" C. (',,/('11/"'(' 1/11' /11,1,1 11/ ""III/lI'f metre h'IIl1lh 1// pipe
(//11/1"(' IlIlcrjIlC(' tempcrutur«
helHiCCII thr twn IlIyef.f of
/'' 1;'11
711 .
r
'1
()
11f-J7' t"A 1'1('/'/ "lpI'
~.
L
11'
ill.I'II/'" irm,
Gille" :
-739.79
-7.6 K
7.6 K
I
J'I'J
I sea I e depo
Temperature across tne
T, - T2 == 7.6 K
Inner diameter,
d I == 120 mm
1', == 60 111111
Result:
I) Overall heat transfer
2) Heat exchange
.
co-efficient,
;;::
Q/L = - 739.79 W/lll
j1
[Negative sign indica I S 1/7(11 h
II
OW
109.0 I \Vln/
'd L
from 0/11 /
Outer diameter,
d2 == 140111111
1'2 == 70 mill
inner side]
J) Temperature
II', == 0.060 III 1
dr par
the
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al dep
it,
[ '2 == 0.070
Ill]
~("fr'beMMMM'."W:r::
/.1]6 Heal and Mass Transfer
radius,
r3
= r2 + thickness of insulation
= 0.070 + 0.055
I
[1"3 = 0.125 In
radius,
Conduction 1.127
----;:ieat
transfer co-efficients ha and lib are not given. So, neglect
th,t rerms
[
r, + III [;~ j + _::j;; IJ
111 r2j
I
~ R=27tL
r4 = rJ + thickness of insulation
k,
k3
k2
= 0.125 + 0.055
[r4 = 0.18 m
I
Thermal conductivity,
TI- T4
z» Q=
[
k, = 55 W/mK
I
k2 = 0.05 W/mK
21tL
k3 = 0.11 W/mK
Inner surface temperature,
Outer surface temperature,
T, = 240 C + 273 :: 513 k
III [;:
1
+
k,
111 [;~ j + In [;;
I]
k3
k2
0
~
T 4 = 60° C + 273 :: 333k
Q
L
513 - 333
I
Tofind :
between
0.060
- [
21t
i) Heat loss per metre length of pipe (Q/L)
ii) Interface
temperature
insulation (TJ)
Ill [ 0.070]
two layer;
~I =
Q/L
55
75.83 W/m
In [ 0.125 ]
+
0.070
0.05
III
[...Q.:..!!_] ]
0.125
+----=0.11
I
We know that,
Sohaion :
Heat flow through composite cylinder is given by
6Toverall
Q = _;::..;..:;.:_=
R
Interface temperatures relation
IFrom ,equn. 1I0lil
Q = ---'---'-
T, - T4
T, - T2
HMT data hook page 1I0.m.
R
R,
where
(I) ~
Q =
T, - T2
where
[ III
I
I"1--21tL
[;:2 ] J
,
> -
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kI
... (I)
I. 128 Heal and Mass Transfer
COlldll(·,irm I. 1]1,1
-----
Q
_I
21tL
[/11
[~Jl
k
Q/L
•
I
TZ .- TJ
_--=---;:..___
[~t;
IJ
2.
512.7-TJ
75.83 =
I"
I
21t I
=> 75.83
=> [_·rJ__
[.QJ.£]j
0.070
0.05
3 7_2_.7_K_-]
Resut«:
I) Heat loss per metre length of pipe, Q/L '" 75.83 WIlli
2) Interface temperature between two layers of insulntion
512.7 K
TJ = 372.7 K.
(I) => Q
III A steel pipe of /70 "'''' inner ,dame/er tllldl90 """ outer
diameter
where
with thermal
conductivity
with two layers of insutation.
[
55 WlmK is covered
rile thickness
IIf 'lie first
layer is 25 mm (k = 0./ WlmK) III1tI the second layer
thickness is 40""". (k = 0./8 WlmK). rile temperature of
the steam and inner surface of tile steam pipe is J20· C
R2= _1-
21tL
allll outer surface
::::>Q
of the insulation
is Sf)" C. Ambient
air
temperature
is 25~ C. rile surface co-efficient
(IIId outside
surfaces
tire }JO
respectively.
Determine
IIII! heat loss per metre letlgll. of
Wlm]K
for inside
alltl 6 Wlm]K
tlte steam pipe and layer of cantuct temperutares t,,"1 atso
calculate
10
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the overall IIellt transfer
co-efficiellt.
I. 130 Heal and Mass Transfer
Given:
Conduction 1.131
Thermal conductivity
Radius,
of first layer, k2 = 0.1 WImK
'4 = r3 + thickness
of insulation of second layer
= 0.12 + 0.040 In
Ir4=0.16ml
Thermal conductivity
air
of second layer, k3 = 0.18 W ImK
Temperature of steam and inner surface of the steam pipe
T a = T I = 3200 C + 273
ITa=TI=593KI
T4 = 80° C + 273
Outer surface of the insulation,
IT4 =353 K
Inner diameter,
d) = 170 mm
Temperature of air, T b = 25° C + 273
'1 = 85 rnrn
~Tb=298KI
I '1 0.085 m I
=
Outer diameter,
d2 = 190 mm
I = 0.095 m I
"2
Radius,
of steel, kl = 55 W/mK
') = '2 + thickness
·
f firslla)~
of insu Iatlon 0
= 0.095 + 0.025 m
I,)
= 0.12 III
Heat transfer co-efficient at inner side, ha = 230 W/m2K
Heat transfer co-efficient at outer side, hb = 6 W/m2K
Tofind:
r2 = 95 mm
Thermal conductivity
I
I
Scanned by CamScanner
i) Heat loss per metre length, Q/L
ii) Contact temperatures,
(T2 and T3)
iii) Overall heat transfer co-efficient, U
Solution:
Heat transfer through composite cylinder is given by
~Toverall
Q=_:.....::..::.=
R
j
I.
i.
I I L' llcut utu! AlII.\',I'
'/hl/I,I/,·,.
We I,IIUW Ihnt.
1kill
()
U'I' A x 6'1'
o
U
1l'lIl1sfcl'.
2nl'4L
x ('I'o-T,)
OIL :: U x 2n 1'4 (To - T b)
[.: A = 2n 1'4 L)
T(/ -T b
368.5
= U x
2 x n x 0.16 (593 - 298)
Q=
Overall heallrilnsfcr
co-cflicicnl, U :: 1.24 W/m2K
/trter/ace temperatllres
21tL
T( - T2
T2 - T3 = T3 - T4
R(
R2
T4 - Tb
"'--
Rb
593 - 298
--------------------------._---+ [I
I
I
21t L
r 230 0.085
x
II 0.095]
[ _0.085 55
+
where
[/11 [.::~ 1]
_I
2nL --k --
_____!--,
111 [~]
[
0.1
r f)/I,
-
1 + [III [W]] + 6 0.16
-~/-=-I
l
x
0.18
-.
f):-
___
'_r ,_-
6~~_~
1
2nL
Scanned by CamScanner
'11._
R3
... (I)
1.134 Heat and Mass Transfer
Conduction 1.135
593 - T2
=> Q/L = ----!:...__-
r/n[~]l
-I
27t
IT2 = 592.9 K
=>
(I) => Q
592.9 _....;.._
-T3 __
=-_.
I [ 0.12 ]
_1_
"o:o9f
[
2X7t
0.1
55
593 - T2
3.21 x 10-4
=> 368.5
368.5
~
[TJ
= 4?5.88
KJ
Result:
I
I) Heat transfer, Q/L = 368.5 W /m
2) Interface temperatures,
T2 - T3
= ----
T2 = 592.9 K
TJ = 455.88 K
R2
3) Overall heat transfer co-efficient, U = 1.24 W/m2K
where
R2=
1
In [;~ 1
.L
[
27tL
k2
J
=> Q =
[2] A steel
['n [~] 1
27tL
k2
diameter
wit" thermal
50 WlmK of 6 mm inner thickness
Steel pipe is covered
carrying
saturated
steam.
material
of 5 em thickness. The thermal conductivity
the insulating
_I
of 20 em outer
pipe
conductivity
with insulating
of
material is 0.09 WlmK. TIre inside film "eat
transfer co-efficient is /100 Wlmz K and outside film heat
transfer co-efficient is 12 Wlm] K. It is fo und til at the heat
loss is more and
it is proposed
to add another layer of
6 em thick insulating material of Slime quality without
=> Q/L =
592.9 - T~
.)
challgi"g
outer conditions.
Delermine
reduction ill heal trails fer.
Givell :
Ctrse (i)
Outer diameter,
d2 = 20 Clll
Outer rad ius,
'2 = 10 cm
("2 == O. 10 III j
Scanned by CamScanner
lire percentuge
of
1.136 Ileal a"d Mass Trm1.~·le"
Conduction 1.137
ToJinti :
Percentage of reduction in heat transfer.
SolUlioll :
ca« (i)
Heat flow through composite cylinder is given by
01 =
Inner radius,
~ToveralJ
R
~T
1'1 == "2 - thickness
o. 0 - 0.006
._I "_:
O_. 094 III I
__
Thermal conductivity
Radius,
I'J
r
o. ] 0 -1- 0.05
I'J
(J_Q_]] ['"
50
of insulating
I 01
material,
=
1.2386T
k2 = 0.09 W/mK
Inside heat transfer
co-efficient,
Outside heat transfer
co-efficient,
2
Ita = 1100 W/f11 K
lib = 12 W/m2K
Case (ii)
6Toverall
011 = _;___
R
Case (ii)
r: + thickness
of insulation (old)
. I'all0n (new)
+ thickness of msu
0.101- 0.05 + 0.06
I
1'3 =
r
1
b 3
_,
I
27txI [ 1100x.094
+ ['11 .094 + __ !_.15]
.I~ +-- I
III
Thermal conductivity
1'3
2
I
.11'
01 =---------------------------
of insulation
I = O. I 5 I
Radius,
I
[~n
In
In [~~
ha'rl +-.!....kl~+-k,:_:.._+-h
of steel, k 1 = 50 W/mK
= r2 + thickness
=
27tL
0.2 111 /
Scanned by CamScanner
27rL
0.09
1
12x.15
1.138 Heal and Mass Transfer
011 =
I
2),[xl
I QII
=
Percentage
r
-d..
1
r
[
~o
+ [I n [.094
In
I
I0 ]
1100x.094
-+~~+
I
0.09
0.772,1T
--
-------------------
~C~'o~nd~u~c~/io~n~/~./~3~9
Given:
I
~
I
of reduction
in heat transfer
Steam pipe diameter,
1.238,1T - 0.772 tlT
x
100
d I '= 15 em
radius,
'1'=
7.5 em
1.238 tlT
0.075
1'1'=
1.238 - 0.772
x 100
Magnesia
1.238
diameter,
d2 = 25 em
= 12.5 em
radius"2
Result :
Percentage
Asbestos
of reduction
in heat transfer,
Q = 37.7 %.
1'2
'=
0.125 m
d3
'=
30 em
radius.v ,
'=
15 em
diameter,
I1fI A J 5 em outer diameter steam pipe is lagged l~~
with maen esi a of til erma I COli ducllfl'
o
0
diam111
0,05 WlmK and further lagged with 3 em
d
,
, 'I 007 W/1fJ
lanllnllleti asbestos of thermal conductiv! y,
(
doPl
Inner temperature
of steam is 20 0 C all "
of sit
temperature is 25° C. Calculate the mass
pI
Thermal
conductivity
of Magnesia,
Thermal
conductivity
of Asbestos,
Inner steam temperature,
0
, e AsS
condensed per hour for 120 m length of pIP'
latent heat of steam is 1900 k l/kg.
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I
1'3=0.'5ml
/0
, meter
dill
ml
Outer temperature,
kl '= 0.05 W/mK
k2 = 0.07 W/mK
T a = 200 C + 273 = 473 K
0
T b = 250 C + 273 = 298 K
Length of the pipe, L = 120 m
Latent heat of the steam, hfg = 1900 kJ/kg
J. J 40 Heat and Mass Transfer
Tofind:
___ ----------------------Mass of the steam condensed
h
per our.
~C~o~n~d~~~·II~-O~,,~J.~J~4J
Q = 10,294 W
~
Solution:
[Heat transfer, Q = 10,294
Heat transfer through composite
.1T
Q =
li
.
cy Inder IS given by
overall
Q = 10.294 kW
~
= 10.294 kJ/s
R
where
w]
= 10.294 )( 3600 kJ/h
Q = 37058.4 kJ/h
Mass of steam condensed per hour
R =
m =
_9__
l'Jg
27tL
37058.4
1900
-~
I
Ta-Tb
Q
[
2nL
)
+ In [~~
kJ
harl
I
111 [;~
+
k2
I
In=
19.5 kg
I
Result:
+_1_
hil3
Heat transfer co-efficients ha and hb are not given.So,neglca
that terms
Mass of the steam condensed per hour = 19.5 kg.
@] A steel pipe lias 18 em inner diameter (k = 70 WlmK)
with 1.4 em wall thickness. A liquid temperature pussillK
through tile tube is 200· C and ambient air temperature is
~ Q =
23· C. Tile inner unit surface conductance of tile liquid is
690 Wlm2 K. Calculate tile heat trailsfer rate antl the over
all heat
transfer
thermocouple
473-298
~ Q ==
111 [~]
2 x 7( x 120
[
0.05
Scanned by CamScanner
__
1+
[:l§l]
0_07
til 170· C.
co-efficient
embedded
for this system
IllIlfway
through
of tl
tire pipe
1.142 Heal and Mass Transfer
Given,'
____
~~~----------------------~C~o~nd~u~c/~
Solution:
Heat transfer at halfway is given by
~T
-
Q=
R
where
Inner diameter,
d 1 = 18 cm
radius,
rl = 9 cm
[_I
R =
=>
21tL
kJ
hdJ
To- T"
Q=
1n['2r 1]
[ I
Thermal conductivity
radius,
I
of steel, k) = 70 W/mK
r2
= rl + wall thickness
r2
= 0.09 + 0.014
r2
= 0.104 m
Liquid temperature,
ha'J +
21tL
kJ
[Put 1'2 = ,]
=>
I
To = 200 C + 273 = 473 K
T b = 230 C + 273 :: 296 K
Inner surface conductance,
ha = 690 W/m K
2
at half way, T h = 1700 C + 273
To-T"
Q=
21tL
0
Ambient air temperature,
Temeperature
[;:1]
+ In
[_I
+ In[:']]
ha,)
kJ
-[~"":"":';;__In
473 - 443
=> Q/L =
1
[-0.097---=-J
I
I
0.09
2 x 1t
690 x 0.09 +
70
Th = 443 K
Tofind:
where
I. Heat transfer at halfway
[Q/L = 10,976 W/m
2. Overall heat transfer co-efficient.
Scanned by CamScanner
r +'2 - 097
- .
m
I' = -
I
2
_!_. 144 Heal and Mass Tramjer
we know
Q=
~
_-------------~C~o~n~du~clion
1.145
Gillen:
Q =
l
10,976 - U x 2 x 1t x 0. 104 x (473-296)
~
lu
I.
= 94.89
W/m2K.1
Ambient air
Result:
I) Heat transfer at halfway,
Q/L = 10,976 W/Ill
Inner diameter of steel, d I '"'5 em = 0.05 m
2) Overall heat transfer
U = 94.89 W/m2K.
co-efficient
Inner radius, rl = 0.025 m
Outer diameter of steel, d2 = 7.6 em = 0.076 m
Outer radius, r2 = 0.038 III
1.1.14 University Sol\'ed Problems
Radius, r) = r2 + thickness of insulation
On Cylinder
= 0.038
f1) A steel tube with Scm I D, 7.6clII OD III1lI II = 15 W/",,(:
covered with (III insuiative covering oflllicllllfsJlc",'
II = 0.2 W/","c. A 1101 Iteas (II JJO°C with I, = 400 U~I'
flows inside II,e lube. Tile outer surface of lire i/lSMJ.
.
0 uri If
ss exposed to cooler air (II JO"C with I, = 6 ",f«'
Calcuate the heat toss from tilt! tube 10 tile (Iirfor [0"
tile tube (IIId the temperature
drops reslillillg fro.:
1llbt
tllerm,,1 resistances of tile IIuI gIll' flow, tile stetl ,
insutatto« layer lind tile outside air.
=
0.058 III
Thermal conductivity
of steel, kl = 15 W/moC
Thermal conductivity
of insulation,
Hot gas temperature,
T a = 330° C + 273 = 603 K
"
-
k2 = 0.2 W/mOC
Heat transfer co-efficient
at inner side. ha = 400 W/m2°C
Ambient air temperature,
T b = 30° C + 273 = 303 K
Heat transfer co-efficient
at outer side, lIb = 60 W/m20C
Length, L = 10 m
[May 2005 . AIII/(ll
Scanned by CamScanner
r)
+ 0.02 m
1.146 Heal and Mass Transfer
Toftnd:
____
------------------------~C~O~~~~"~~~J·IJ
i) Heat loss Q
ii) Temperature drops; (T - T ) (T
Q
,
I,
I -T
and (T _ T ),
J
) (T
2,
Q ::>
-
2 -IJ)
603 303
°
1
[
1
1
[ 038]
2 x 7t X 10 400 x 0.025 + 15 In 0:025
b
Solution:
+_1 In[_0._05_8]+
1
]
0.2
0.038 -60-x-'0;"".0-5-8
Heat flow
IQ = 7451.72 WI
Q = .1Toverall
R
where
[From equn 1.48 or HMTtk
page no.43 & 45 (Sixrtn e~
We know that,
Interface temperatures,
R =_1_
21fL
1
+ -In
k3
{r...i+_1
I .
'3
(I)~Q
hb'4,
=
~ Q =
21fL
TQ-T1
1
1
-x--21fL
[-h'11"1+ -'kJ 111[''J2]+_1k2 In['3]
'2
:3 In [;: h-~4J
(The terms K3 'and z, are not given. So neglect Ihaltenf
hQ'1
7451.72 =
~TQ-TI
= 11.859K
[Temperature drop across hot gas flow, TQ- T 1 = 11.859 KJ
-
Scanned by CamScanner
l. 1-18 Heal and Mass Transfer
~
.
Conduction 1.149
drop across the InsUlation, T2 - T 3 = 250.75K \
~
T 3- Tb
Q =:
(1) ~
= __ T..:....1 _- _T-=..2
I
_I
2itL
Inl!i \
kl
rl J
T 3- Tb
=:
2~L
... R 1=-I [I
2nL
_
451.72
=
Rb
_
(h:rJ
.
,.
kllll( h.
T..!.I
__
--T-=2~---_I In l- 0.038
0.025
2 x it x 10 IS
1 - r,
1)
= -----~~----~ 3
7451.72
2 x ~ x 10 ( 60 x ~.0581
~ T 3 - 1 b = 34.07 K
emperature drop across the outside air, T 3 - T b = 34.07 K
Temperature drop across the steel tube, TI- T2 = 3JIOK'
(I)=> Q = T2-T3
Result:
R2
= ---=--~--2~L
(-~2
In [~~
(ii)
1]
(_I In(!'l\·l
2nL k2
.: R2 = _I
[
7451.72 =
T2 - T3
1
[_I In [ 0.058 ] ]
2 x n x 10 0.2
0.038
Scanned by CamScanner
'2
7451.72 W
Q
(i)
T2 - T3
T a - T ( =: 11.859 K
T(-T2
= 3.310K
T2-T3
=
2S0.7SK
T 3 - T b = 34.07 K
1.150 Heat and Mass Transfer
fJI A steel tube (k= -/3.26 WI",Kj of 5.08 CIII •
'"""ef .
I!J and 7. 62 em outer diameter is cOveredHlil"2
dlQ",.
---Thermal
.
.s C", lIljt~,
insulation (k ;:;:
0.208 WlmK) the Inside
SUI':I'.
l
. lelllne,
JQceolt~tl~
receivers heat fro", a 1101gas at the
r
Q/llfe!
with heat transfer co-efficient of 28 WI",2g
Hot gas temperature, T a = 3 16° C + 273 = 589 K
Ambient air temperature, Tb = 30° C + 273 = 303 K
3J6'(
.
outer surface exposed to the amh,ent
air Q/30.C. lJI~ilt
. ~
"'Ilk
,."<t
transfer co-efficient of 17 WI",]/(. Calcula/e"
eallollA _
3m length of the tube.
[Madras University
Given:
OCI ~J9~
Conduction 1.151
conductivity of insulation, k2 = 0.208 W/mK
Heat transfer co-efficient at inner side, ha = 28 W/m2K
Heat transfer co-efficient at outer side, hb = 17 W/m2K
Length, L ==3 m
Tojind:
i) Heat loss,
Q
Solution,'
Heat flow, Q ==
where
.1Toverall
R
[From HMT data book page no,43 & 45}
R==-
I
21tL
Steel tube thermal conductivity,
Inner diameter
k, = 43.26 W/mK
of steel, d, = 5.08 cm == 0.0508 m
Inner radius, r, == 0.0254 m
Outer diameter of steel, d2 == 7.62 cm == 0.0762 m
Outer radius, r2 == 0.0381 m
Radius, r3 ::: r2 + thickness
of insulation
Radius, r3 == 0.0381 + 0.025 m
'3 ::: 0.0631
m
Scanned by CamScanner
~Q=
_1_ + _, In
27tL [ harl
kJ
[;2] + tin [;n
['4]+_1]
k;"I In'3
ht/'4
J
2
I. /52 Heal and Mass Transfer
[The terms k.3 and r4 are not' given. So ---
Conduction 1.153
--, neglecllh
Q =
T
~~
a-Tb
_I [_I _I [r2]rj +I-In[.!i]l -....._
Give" :
~I
+
hell
21tL
k) In
589
Q
= 1129.42
<,
1
+ ';',
h~!
r2
303
I
[ 28 x 0.0254
+ _I _--------[Om '
+-I-In[
0.0631]
I
0.0381 +~006 . JI
0.208
IQ
2
43.26
In ~I
Inside temperature,
W
Result:
Heat loss, Q = 1129.42 W
Inner diameter,
dl = 80 mrn = 0.080 m
Inner radius,
rl
= 0.040 m
Wall thickness,
== 5.5 mm
Radius,
r2 = rl + thickness
= 0.040
Urll
/1) A hot steam pipe having
(Ill inside !)'urfi,,:e lempera/
2500 C has (In inside diameter of 80 111mand a wall 'hick~
of 5.5 mm. II is covered with a 90 mm layer ofinsu/~
having thermal conductivity. 010.5 WlnrK followed
V
. '"ofb)'1
mm layer of insulation having thermal conductlv/~ .D
WI,
.r t ula/l '
mK. Tlte outside surface temperature o,,ns
lODe
A ",etlltl
. Calculate heat loss per metre lellgtll. JSU
I
conductivity of the pipe as 47 Wlm K.
M d.
[ a ras University Apr 2002, Baralh~yar
Scanned by CamScanner
U . /,silY Apr
/1/ve
.
0
T 1 = 250 C + 273 = 523 K
of wall
+ 5.5. x 10-3 m
I "2 = 0.0455 111 I
Radius,
r3 = r2 + thickness
= 0.0455
of insulation (I)
+ 90 x 10-3 m
r3=0.1355111
Radius, r4 = r3 + thickness of insulation (II)
= 0.1355 + 40 x 10"4 = 0.1755
111
3
m
1.154 Heal and Mass Transfer
Thermal conductivity
of pipe k == 47
Thermal conductivity
of Insulation (I)
Thermal conductivity
.
Outs Ide temperature,
of insulation (II)
'
I
W/rnK
_
, k2 - 0.5 W
_
IrnK
' k3 - 0.25 W
T 4 == 20°C + 273
IrnK
In
_1_
[~I
+
21tL
== 293 K
Tofind :
Heat transfer per metre length.
~ 0 == _---;~:__:_:----=5~2=-3 =-- !:229~3
Solution:
_I_
Heat flow through composite cylinder is given by
21tL
~Toverall
Q = __;:::....:...:.:..:..:..:
R
l
/n[O.0455]
0.040
47
+
_
j
In[O.1355j
,n[0.1755
0.0455 +
0.1355
1
0.5
0.25
[From HMTdalabool
[lage no,43 & 4J}
where
~ lOlL = 448.8 W/m I
Result:
Heat transfer, Q/L = 448.8 W/m.
o
=> Q ==
[;n
21tL
of the pipe.
In
In [~]
[ h~rl + -k-=-I--=- + k2
[;4 ]
In
I
_::.--3_+_
+ kJ
hbr4
Heat transfer coefficients
ho' hb are not given. So, n~
hat terms.
Scanned by CamScanner
A thick walled tube of stainless steet lt: = 77.85 kJlllr m·CJ
25 mm ID alll150 111m OD is covered with a 25 mm layer of
ashesto~'lk = 0.88 k.l/hr m"Cj. If the inside walltenrperc,ture
of the pipe is maintained at 550" C and the outside of the
insulator at 45" C. Calculate the I,eatloss per meter lengt"
[Madras University April 1995. EEEl
Given:
Inner diameter of steel, dl == 25 nun
Inner radius, rl == 12.5 rnm => 0.0125 m
Outer diameter, d2 == 50 rnrn
----
__ ------------------
_:C~·o:n~d~uc~tl~·o~n~/.
solution:
Heat now through
m
yl inder :
ite
To crall
T:
i en b\
{From Hi fT
R
I
R=-
o
2 L
r:e no n
l~
0.0
rn
rn=
f-1
I
haT I
er
on e ti e heat tran
-effi len
ha and h are n t
o ne le t tha term .
m
T(I- T
(it un
inless
el
=>
Ihr rn° . = _._5
3
1=
J/~
=
o
rn
me
.021
::::)
l"'[:~\ In l:~\1
_I
2nL
KI
k2
_
VIm 'C
Ta- T
=>
IL
Similar!
~II err al C nducti
it
0
as
.st s.
2 = 0.88
kJ/hr 1'(
= 0.24
Wlm:i_
2
·1
(I
l'n
_1-
2n
= 5500
1
21t
To find:
I. Heat loss per metre
Scanned by CamScanner
[Q/L
length
l:~\ In l:~\1
kl
k2
550 - 45
IL
l' [~lll/n[~11
0.012
0.025
2 \ .625
0.244
n
= \ 103.9 W/m
rJ
I
ive
1.158 Heal and Mass Transfer
Result:
(i)Q/L=
----------------------~==~~
Conduction1.J 59
II03.9W/m.
Solution:
Heat flow through composite cylinder is given by
ill A steam pipe of 12 em outer diameter is at 197.
lagged to a radius
of 10 em with asbestos
conductivity of 1WlmK. The temperature
is 25" C and heat transfer co-efficient
!iT overall
R
Q=
C.llt
'J Ihtl"'.
Of
[From HMT data book
0" SUr
'J
page no. 43 & 45]
rO"ft~i.
outside;s 12 IJI ..,
"/fIIll
Calculate the heat loss per meter length of tl'e pipe.
[Madras University, OCll9r,
l
R=_1 11 +
21tL
haTl
Given:
=> Q =
--1
21tL
[
I
+
harl
Neglecting unknown terms
Ta-Tb
d)=12cm
=> 0 =
r) = 6 em => 0.06 m
r2 = 10 em => 0.1 m
k) = 1 W/mK
470
- 298
.:..:....::..---=c::...=:..----
=> OIL =
Ta = 1970 C + 273 = 470 K
_1
21t
T b = 250 C + 273 = 298 K
[In [~]
I
+
1
12xO.1
1
hb = 12 W/m2K
[OIL
Tofind:
I. Heat loss per metre length
Scanned by CamScanner
=
804.01 W/m \
Result:
Heat loss per meter length = 804.01 W/m.
1.158 Heat and Mass Transfer
Result:
(i) Q/L = 1103.9 W/m.
solution:
Heat flow through composite cylinder is given by
IIJ A steam pipe of 12 em outer diameter is at 197.
lagged to a ra d·IUS 0if 10· em witli ashestos 01' hc'/'i
'J t trill
conductivity of 1 WlmK. The temperature 01" sun
•
D
--
Conduction 1.159
'J
Q=
~Toverall
R
{From HMT data boole
page no.43 & 45]
oundi
is 25 C and heat transfer co-efficient outside is 12U'/~~
Calculate the heat loss per meter length of the pipe.
[Madras University, OCI199il
R = _I
Given:
27tL
[]
harl
+ In [~]
k,
+ In
=> Q =
Neglecting unknown terms
Ta-Tb
d,=12cm
T)
=> Q =
= 6 em => 0.06 m
T2=IOcm=>0.lm
k, = 1 W/mK
T a = 1970 C + 273 = 470 K
T b = 250 C + 273 = 298 K
470 298
:::) Q/L = --_'!~~~--:I
27t
[/n[~]I + 12 0.) 1
1
x
hb = 12 W/m2K
[Q/L = 804.0) W/m)
Tofind:
I. Heat loss per metre length
Scanned by CamScanner
[;n +
k2
Result·
Heat Joss per meter length = 804.01 W/m.
_I ]
htl"3
1.160 Heat and Mass Transfer
Co"duction /./6/
1.1.15 Solved Problems on Hollow sPher~
-----Jo;ide
0
It; covered
10
illsicle (:;
Ill,
rite
IIl)lt,.
temperatures (Ire 500" C and 50" C respecli"el),.
the rate of heat flow through this sphere.
5 00" C + 2 7J - 77 J K
outside temperature, T b = 50 C + 2'73 == J23 K
II] A hollow sphere (k = ~5 WlmK~of 120 m~
and 350 mm outer diameter
ins ulatlon (k=JO WlmK).
temperature, T a
C:;'1;4,
clllt"
roftnd:
Heat loss, (Q).
Solul;Oll :
Heat loss through hollow sphere is given by
Given:
6Toverall
Q=
[From HAfT data book page /JO.
R
43 (~45(.5ixlh cdiuon]
where
R =-
I
4n
=>0 = ---------------
_I [_IhJf + _I [.l.._..!_]+ _Ik [.l_ fJI J+_Ilib"; ]
4n
kl'l
'2
Z'1
Heat transfer co-efficients ha and hb are not given.So, neglect that
terms.
Thermal conductivity of sphere, k I = 65 W/mK
=>Q=
Inner diameter of sphere, d I = 120 rnrn
Radius,
_I [_I [ITj- r;1]+ k;-1 [Ir:;-"'3I] J
4n
rl = 60 mm = 0.060 m
kl
Outer diameter of sphere, d2 = 350 mm
773 - 323
•. [*[O~ - o:7shk[o:75 -oiss]]
Radius, r2 = 175 mm = 0.175 m
Radius, rj = r2 + thickness of insulation
~ [§ = 28361 W I
rj = 0.175 + 0.010
Refllll:
Ir3=0.185ml
Thermal conductivity of insulation,
Heal transfer, Q = 28361 W
k2 = 10 W/rnK
12
Scanned by CamScanner
_j
}./62 Heal and Mass Transfer
o A ho/low sphere 1.2 Inner diameter
m
.~__
alld
Conduction 1.163
R == .t,
[-!_47r
11,./ + _Ikl'l [J_ --,:;I] +-h I]
7
diameter is having a thermal ClJnlfUClivil 1. /Jr, o~
The inner surface temperature is 70 K an/ Of I "'/IIf(
lemperatllre is 300 K. Determine,
Oilier slill~
2
a
(i) heat transfer rate
(ii) Temperature
fII it fffdillS
b'2
T)- T2
0=--of 650 "'III.
-4~-.
[h~-:-12-+ -k-II [,.11 - ;2 ]+~]_(1
Given:
11b'2
..
[The terms ha and' hb are not given. So, neglect that terms]
d) == 1.2 m
~) -. T2
T
... (I) .
r) == 0.6 m
•• 1,
d2 == ).7 m
70 - 300
r2 == 0.85 m
k) =J W/mK
T) == 70 K
~ [0
T2 == 300 K
r = 650 mm = 0.65 m
= - 5896.1iW]
[The negative sign indicates that heat flows from outside to inside]
To/inti :
(i) Heat transfer
rate, (Q)
(ii) Temperature
at a radius
(ii) Temperature at a radius = r = O.6~m
Put 1'2 = T and '2 = ,. in equation ( J )
of 650 111111
Solution:
(I) :::> Q = -------
(i) Heal transfer rate, (Q)
Heat transfer, Q =
.
L\T
._ 5896. J 4 == __
overall
R
.
where
4~
[Front IiMT data book page na
4.1&/';
47r
[T == J30.1ii]
Scanned by CamScanner
TJ -T
[* [*-f-]]
____:_7..::__0~T
__
[+ I
.
1
0 6 - 0 ~5
lJ
...
I. /64 Heal alld .V{ClS.I.!!_U:_'_ls-=-;fi_e,_" -----
_
....----h
ReJlt/t:
Q = - "896.1
(i) Heat trallster
rate,
(ii) Telllperallin.:
at a radiu
W
\I
re
sr=T,-Tb
f 650111111
T = I 0.15 K
ill A hoI/ow Jphere hus inside surface
I.:!J.
(11111 the»
olltJl~/e
.Htrj(l~e
.r
tempeT(UIITe
OJ
I{
= -;
J (/0'
[t-
I
t'f JO'[
temperatllre
/,J k =111WI",K. (a/CII/flfe
(I) hcat tost by CQllt/II"
C.
_
"
((lUll fIJI
inside diameter oj.\ CIII and outside dlflmeler of 15 CIII("
.
if
. r.
iii
'ieat/m'l. bvJ COIltI uctton, / equal/oil/or.'
(I I' I{I ill wul]
"
ere« .
equut to sphere area.
r \ ladras "iverSI/Y,lprlr
*
=T,-T:
[*- /- -1-1
IIbr.-
T, - T_
_I
I-_I-
l
h 1'1
-I
k,
r,'J'
and\1 II b aree n01'given.
lh(·t'·rITI·;'
,
III
'.
_1-2
hbJ'~
.o. ueplcct
thai terms)
"
:;-:, (
Give« :
T, = J 000 C f- 27 J = -7
K
T 2 = 300 C + 27 J = 30" K
k, = 18 W/IllK
1
4•
d, = 5 em "" 0.0:5 III
1', = 0.025
'--
III
d2 = 15 CIII = 0.1 - III
b~[
0 ~_5 - 0 ~J:
iJ
(ii) 1Jt'{/( lost ( If the 111'('(1 is (!lJIIIIIIO
f2 = 0.07 - III
Tofiud :
.. O. 7. - 0.0 S
(i) Heal 10SI, Q
)::; )
(ii) Heat lost (Ifthe
area i equal
I III plain
,\all area
So! 111;011 :
(i) Heut tost (Q)
[Frolll /1 \ tt
Heat 11m.
() = _L\_'_'o_e_ra_1I
R
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.1
III
II
,l!
[L:.:0.5m
lJ
l
the pluin Willi area] Q/
!}!!!_~·~====I~-··---~
l\~2~(rf
/. /66 Heat (mel MaH Transfer
~ Critical Radius of !""ularion 1.167
+ r~)
CAL RADIUS OF INSULATION
eRIT.I
.
~
We know that,
Addition
of insulating
material on a surface does not reduce.
f heat transfer rate always. In fact under certain
mount 0
.
.
the a
it actually increases the heat loss up to certain thickness
.
stances I
'.
.
circum on
.
The radius of insulation for which the heat transfer IS
. ulall .
....
.
of InS
. ca lied critical radius of insulation and .the corresponding
. um IS
.
maxim
.
lied critical thickness. If the thickness IS further
'ckness Ie; ca
.
rill
d he heat loss Will be reduced.
increase ,t
~T
R
TI - T2
L
kA
1
L
1
I
Critical
Radius
= rc
Critical
thickness
= rc - rl
1
I
1
1
1
1
573 - 303
0.05
IS x 21t(0.0252 + 0.0752)
Q
1
1
1
1
10,= 3S17.03W I
1
r
Result:
i.-c-:::::
.I{i) Heat lost, 0 = 2290.22
1
1
=,~_.J_,
_,rc'-----J
W
(ii) Heat lost (If the area is equal to the plain wall area),
0, = 3S17.03 W.
Fig 1.8
1.2.1Critical Radius of Insulation For A Cylinder
Consider a cylinder having thermal conductivity
ro inner and outer radii of insulation.
Heat transfer,
Q
k. Let
r, and
Tj-Ta;:
In (~)
21tkL
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[From equII.no.(I.31)}
P'IZWfiJlfwttd6'1f{WiUMM
('1'11/('(1/
/«(Ir/i",v
of Imllllll/rm
I 1(i(J
() ..,
,
I'tl )
III ( "1
I1J A II dl!t'Irit'fI/
I
I
Here Ao
C(I-(:fllclt'"1
Th« ttiermat CfJlltllI('I"'i~v
,
I '
/
T; - Toc
Q=
(~l/ {J", "'''Illit 1I1It/ / III",
llill"'I!It!I'
I" uir nt lJ"C 11", convectio» heat ITf1I1.~fer
heIHIt!t'II lilt' wire ,.."r{fln' flllt! uir ts /J If!1m) K.
;1
7tkL
wlrr
d/JJlpfllt,,"]fI(JW
I
I
I
11,£,crtticul
"111111,\'
of ;I1,wlllllm,
oftlu: wire
II,ic:kllt!!i," of insulatinn,
lenlpt!Ttllllre
---
I" (;~ J
----+--Znkl,
2nrOLh
uf wire is (J. J81 WI",/(. (;11/('1111"1'
tll1Il tI/.WI dt~/ermille the
~fit is insutoted 1(1 II,e crttical
( June 2006 - Anna Univ]
Give" :
Length of the win', L = 10 rnm
To find the critical radius of insulation,
respect to 1"0 and equate it to zero.
difTereniiale~l
Diameter
of the wire, d = 1 mm
Radius of the wire,
dQ
0- (T; - T,,oJ [2n~Lro
---
In
2rrkL
- 2n~Lr[1
(;0, J+ 2nhiLro
J
2nkLro
_
I
Heat transfer,
Q = 200 W'
Surrounding
temperature,
Thermal
T b = 2SoC + 273 = 298 K
conductivity
between
of wire, k = 0,S82 W/mK
=0
2
2nhLrO
ro~+~rc
O,S x 10-3 m
Convection
heal transfer co-efficient
surface and air, hb = 15 Wm2K.
since (T; - To:)"# 0
~
r = O.S mm =
air
Q, hb T b
(wire
jd-Olr""--------:-·,:::<·o--....,.'_:]
I
1 ~ __
_L
.:
1'0 filld :
I. Critical
radius of insulation,
2, Temperature
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of wire, >10
r('
the wire
II
U
1.1 O nea
I and ~{(]5SrrollSfer
ri(ieal
.
Radius
of bmdatio1/
1.1 I
of 6 IfInI diameter
with 1 mm thic« insulatiolt
co-efficient
rio
"I' the ill!iliialing surface and air is lJ 'Jim) K find
bt'tl.·t!<c
.
'.'
.
'
critical Ihlckness
0/ insulation 'IIId also find the
t/'(
. I I
,/'.
/{Iut! 0/ change tn tne neat transfer rate if the critical
piTC el, • lJ .
Tt,dius .s used.
.
Sol.ti"" :
r,] A ""re
.
.
t:J. ':::::(1.11 WI",K). 1/II,e COIII'eCIH't'ht'allransft'r
We I..lhl\\ rhat.
('ritical rldius of insulati
k
ll.' =-II
:; 0.58_
IS
Gil·tn:
I 'c = 0.0388 ~
Heal transfer through an insulated
dl==6J1Hll
'1 == 3 mrn = 0.003
wire when critical ra
is IL~ is siven by
.
m
air
'2 =='1 + 2 == 3 + 2 = 5 mill
= 0.005 m
Q==
k==O.IIW/mK
In (~~)
___ +_12itL
kl
hb ==25 W/m2K
111/.
Tofind:
I. Critical
200 = ------------I
2it x 10
200 :;
2. % of change
Ta - 298
[In [g.~~:]
0.582
thickness
in heat transfer
Solution :
I. Critical
+
J
J 5(0.0388)
k
radius,
r
. I'
T(/ - 298
c
= _Q.JJ_
c =
25
[From equn. no. (I.50)}
h
J
= 4.4 x 10-3 m
4.4 x 10-3 m
I
0.146
Crilicallhickncss,
~ era:: 327.28 K J
Ie='e -'1
= 4.4
x
10-3 - 0.003
= 1.4 )( 10-3 m
Relult:
Ie"
.
rrtrca]
.
radlllS of insulation,
Crilicallhickllcss,
rc = O.0388mm
2. Tcmpcralu re o ,. llcwire,T
I
a=327.28K(or)
e
54 .i>:
2
D
...
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Ie = 1.4 x 10-3
m (or) 1.4 mill
po
1.172 Ileal and Mass TranJ/er
2. Heat transfer
Q _
I -
through an inslliated ~~:-~
'''"els'
, given by
(T _ T )
a
b
~j
[In [~l + __
__I
27rL
k,
12.57 - 12.64 x '00
=
J
1rdQ1Q
Page
2
12.64
~
II04J~1
27rL (T a - Tb)
0.55 %
Result:
I. Critical
= -In--;(--=O-=.0--=-0-5:--)
---
0.003
thickness,
'e = 1.4 x 10-3
2. Percentage
of increase
radius = 0.55 %.
25 x 0.005
in heat transfer by using critical
II] A wire of 7 mm diameter is covered witlt
2nL (T a - Tb)
Q, =
Critical Radu
if I.
.
~~~o~n~s~ul~m~/o~I1~I.~/7~3
[FronIHA'
hi r
----_+
O. 1)
~,
.--~
material (k = I W/mK). rite wire temperature
12.64
Heat flow through an insulated wire when erilicalradi
used is given by
(III
insulating
(l1U1 ambient
temperature
are BOt}C and 15° C. If the inside convective
Ileattrtlllsfer
co-efficient
tllickness
is B.2 Wlm2K, find tile minimum
of insutation and also find tile percentage of
increase ill the heat dissipation.
Given:
[In
__I
[;~~ ]
2nL.
+
kJ
1
]
d,
= 7 I11Ill
/', = 3.5 111111 = 3.5 x
hbrc
10-3 III
k = I W/Il1K
T a = 80° C + 273 = 353 K
=
T b = 15° C + 273 = 288 K
3
111[4.4 x 10- )
_.:__0_._0_03_-
+
0.11
Q2=
I ---10-1
25 x 4.4 x
ha = 8.2 W/m2K
.
Tofind :
2rrL (T a - Tb)
I. Minimum
12.572
.
)
.
fl
bv uSing
... J e-rcentage of increase ill heat 0" .
thickness
2. % of increase
of insulation
in the heat dissipation.
I
critical
r~dills
=
Q2 - QJ
x
100
OJ
= ••• ,
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Solution:
I.
ritical radius. "c =
Critical Radiu5 ot I . I .
. 'J nsu at/on 1175
r-------_~
------
k
II
89.74 W/m
Percentage
Irc = 0.121901
of increase
Q2 - QJ
\
----'-
01
in heat dissipation
100
x
Ic=rc-rl
Criti al thickness,
= O. 12 19 -
-,-8-,-9_.7_4 _-_;1...:..1_;.7-=-2 x 100
11.72
3.5 x 10- 3
'c=0.11801
\ Minimum Insulation
thickness,
2. Heat loss without
insulation
665.69 %
tc - 0.118 m
2nL (T a - T b)
2nL(353
Result:
) . Minimum
insulat ion thickness, 'c = O. J J 8 III
2. Percentage
of increase in heat dissipation
III A steam pipe J 0 em inner diameter J J em outer diameter is
covered with an insuluting substance (k = J WlmK). The
steam temperature
and the ambient temperatures are
200" C and 20" C respectively. If tire convective heat transfer
co-efficient
between the insulating surface alii/ air is
2
8
K. Find the critical radius oJ ;I/\'II/al;OI1 (lilt! the
heal 10.'11per metre of pipe for the vulue of r(~ And II/SO
- 2RR)
1
Wim
8.2 x 3. 5 x I 0-:1
find the outer surfcc temperature.
\ Ql/L = 11.72 W/m \
Given :
I kat los. with insulation
d, = 10cm=0.1111
== 0.05
III
d == 11
III
/'1
1'2 == 0.0))
k.l
I
.
= 0.11
III
ITI.
kl == I \\' lllK
- - ~~ == ~
=. "
-_.-:--
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= 665.69 %.
\\'
r - -
-
I. J 76 Heat and Mass Transfer
Cri/ical Radius oj Insulation t.tr;
I
R=-
21tL
[ --+-In
I klI [1'-'I2] + k;If']
3
In '2
hal'l
I" '
I
I,
)
t
Q =
I [I --+ -In
I ['-'I2] +-InI f'3]
21tL
kl
k2 '2
--
hal'l
Tofind :
(i) Critical radius of insulation, "c
+ _Ik3 In[''3
4]+ _Ih '4 ]
b
[The terms ha, k2 and k3 are not given. So, neglect that terms]
(ii) Heat lost per meter at "c
(iii) Outer surface temperature,
TJ
I
1
I
r
Ta-Tb
Solution:
1. Critical radius of insulation (rcJ
k
rc =_
Put I' 2 = I' c and 1'1 = 1'_, in the above equation.
h
r = I
c
8
Irc = 0.125 III
I
2. Heat lost per meter at 'rc'
.
~=----~4~73~2~93~----~1
[From HMT do/a book page no:. 43&Jj)
L
21t
We know that,
H ea t tl ow, Q -_ £1ToveraJl
R
[ I I (0.125)+
-I n 0.055
I
8xO.125
[1-~
621 W1m I
where
13
__j
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Heat Conduction With Heat (,eneralion 1.i79
1.178 /leal (/lid Moss 71'aIl5fer
(iii) Ollter SIIr/ace temperatllre (T3)
1.3
We kllo\\ that,
t-IEATCONDUCTION WITH HEAT GENERATION
.
.
In many practical cases, there
rn Typical examples are
Ie syste .
11
E lectrJc"1 COl s
).. Resistance
). Nuclear
QfL
= 91.S K
,,11251
I
reactor
•1
III
in the fuel bed of boiler furnaces.
Consider a slab of thickness
shown in fig. 1.10.
fin
Heat 1 t per meter,
utcr
offuel
1.3.1 Plane wall with internal heat generation
Result:
riti alradius
within
In electric coi I and resistance heater, heat is generated due
10 electric current
flowing in the fire. In nuclear fuel element,
heal is generated by nuclear fission.
T - 293
621
[s
generation
heater
4. Combustion
2~
IS a heat
ulati
n rc=O.125m
Con ider a mall elemental
L, thermal conductivity
area of thickness dr .
IL = 621 W/m
urfa e temperature,
T
= 391.8 K
Qg
~
()
L
Fit! I. If}
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k, as
I 180 Heal an
•
dMmSff~a~n~sift_er
.
_
Heal Conduction
1 w of conductLOn, we know that
From Founer sa.
.
dT
Q =-kA Heat transfer at r,
x
dx
. " (1.5 I)
•
Heat con
I
ducted out at x + dx
:::>T+
-
k
' .. (1.52)
saJ11e
tWO
• " (1.53)
We know that,
Sides.
T
boundary
ee -
q A dr
conditions;
•
(1.56)
CI = 0
k
APply T = T w' x =
L
2"
+ (L)2
q
k 2 + C2
= 0
2
d T +_!_ dx = 0
dx2
k
::::)
'"
_!_ i.. x2 + C2
2
,
dx2
C2
.•
ApP IY
::::)T w = -
::::)
Cjx+
The temperature on the two faces of the slab (Tw) is the
because it loses the same amount of heat by convection on
(1.56):::>
Qx + Qg = Qx+dx
2
kA d T +
2
r:
Heat generated within dx
,
x2
--=
2
dT _ kA d T dx
Qx+dx = -kA dr
dx2
Qg= q A dr
~q'
With Heal Generation /.18/
...
(1.54)
Integrating above equation
Substituting
(1.54) ::::)
J
+J qk dx = f 0
C t and C2 value in equation (1.56)
•
ddx2T2
I q
•
.,
qL
2
T = - 2 k .r- + 0 + Tw + 8k
••• (1.55)
::::)
q
--.
T = T w + -8k
_"
Integrating
(1.55)::::) J~
+: Jx=JC
t
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2
2
(L - 4x )
... (1.57)
J.l8
2 Heaton
d Mass Transfer
Heat Conductiun
-~
ature T max (at the centre) is b
'
temper
0 ta'
The Olaxunulll
, (1,5 7),
Int~
,
:: 0 in Equation
by putllng x
~\'ith
internal heat generation
Z CyJlll,
'
1.3.
'der a cylinder of radius r and thermal condUclivit
k
'I
I' d
Y ,
ConSI
, genera ted (Qg) 111 t re cy III er due to passag'e of an e Iectnc'
~eat IS
rrenl,
( 1,58)
with Heat Generation 1./83
ell
m Fourier's
Fw .
,
law of conductIOn,
qr
d2T
_+= 0
r dr2
k
we know that
'
...
(1.60)
'"
(1.61)
Heat flow rate
i :
Q:: _qA
Integrating
L
2
,
Heat transfer by convectIOn
r
Q :: h A (Til' - Too)
=> Q::
1- q AL :: h A
.L q AL :: hAT
II' -
hAT
11'::
Til'::
i:I+I~=Jo
I dr2
k
dT
q r2 _ C,
_
+--dr
k 2
(Til' - T ex,)
2
.
dT
qr
-+dr
2k
h A Too
,.
h A Too + 2 q AL
Integrating
dT + j q r J
j Tr
2k
qL
Too +?Ji'
=
Surface or Wall temperature
C I /n r + C2
:::> T::
'"
qr2
- --
(1.59)
4k
Apply boundary
(1.61):::> Tw=-
+ C I In r + C2
conditions
~I'J
-+C,
4k
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~I
-
[PUI T :: T w ' r = roJ
dM~s~~a~n~if_e,
_
J.}84 Heal an
Heal Cnnduclion
wilh Heat Generalion
1.185
. ,( I 63) and ( 1.64)
Equat rng
.
hx2nroL(Tw-TaJ)
2L':=
APply CI an
1('0
d C value in Equation (1.61)
2
q
- h x 2 x (Till - Too)
'0 q
. 2
qr
::> T '" -
_+0+
4k
T'" T II'
(V6
T +
HI
4k
>
'0 q := 2h TIII-
2h Too
21t T 11':= '0 q + 2h T co
+3- [rg - ,2J
4k
'oq
-T 00 +-2h
Till -
At centre
r :: 0,
::>
T'" TllIax
q
- T +Tmax - II' 4k
temperature,
(1.65)
...
(1.66)
Similarly,
For sp here , temperature
Tmax
T
II'
+ qro
4k
at the centre
q'r2
0_
T C = T III + __
6k
'" (1.6.'
We know that,
Heat generated
1.3.3 Internal Heat Generation
2
...
?
[roJ
• 2
Maximum Temperature,
T 111-- T00 + roq
211
•
Q = 11'0 Lq
.,'
- Formulae used
(1.6))
Forplane wall "
Heat transfer due to convection
Q = h x 211ro L (Till - TO')
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, " (1.641
I. Surface temperature,
TII'=
1.Maximum temperature,
T rna'
+ qL
T
00
2h
ciL2
T 1\1 +-8k
"
tk\l{ (Ilk' M(l~S
Ikt/t Conduction with Heat Gene rutton
. \ 1./87
TroflSjer
Solved Problenls
1).4,
r; Fluid temperature,
K
o
_ Thickness, m
k - Themal conductivity,
Generation
I An tltctric c;lrrellt is. passed
~ _ Ilt':It generation, W 1m3
h _ Heat transfer co-efficient,
Plane Wall with Internal Heat
011
.
2
W Im K
W/mK.
l'
through
a plane wall of
W"'C" generates ''£'(It at tile rate of
t/,;ckneS J5f} mm
50,000 Win/. tt« convective "eat transfer coefficient
between w(llI and ambient air is 65 WI",] K. ambient air
It",peratttre is 28°C (1/1(1the thermal ('ontlll('tivity of tile wal!
",alerial is 22 WlmK.
Calculate:
J, Sur/(Iu temperature
2. Maxim"m temperature in tile w(,11
Q
I. Heal generation,
q = V
2. Jlaximu", temperature
qr2
Tmax = Tw+ 4k
.
•
Heatgeneration, q
=
50,000 W 1m
Convective heat transfer coefficient.
J. Surface temperature
T w =T
Given:
Thickness, L = 150 mm = 0.150 m
(X)
rq
+2h
where
V - Volume - 1t r2 L
r - radius - n.
For sphere
I. Temperature at the centre
Ambient air temperature.
Thermal conductivity,
T
3
h = 65 W/mlK
= 28°C + 273 = 301 K
I
k = 22 W/mK
Tofind:
I. Surface temperature
2. Maximum temperature
in the wall
Solution:
Weknow that.
Surface temperature
Tw = T" +
301 +
lilt,
1~
[FI'IIIII Fil'I. I/O. 1.5YJ
50 000 )( 0.150
2)( 65
358.6 K
I
&z_i1.4fufJHl
Scanned by CamScanner
as
H lS:J ~Tr~a~11J~if4.::t!r:__
!~~?:.~~~
c -_
1.188_
Heal Conduction
_
with Heal G
.
eneratlon 1.1119
J
• llperalure
"""" un urn It I
q. L-,
i 1
-
, Ina(
r From £~/I"" n I .j8j
.1'It' +-8k
.. 3 8.6 +
.
0
50,000 x (0.150)2
8)( 22
.,,,,It .'
I $UI f~ c 1~'II)p\'rllllrC Til' - )' lUi K
Tnl(\\
-~"
~'" = 402.6
['I '" rk 'rril' """,'III is 1'4"·WC/IAN'''J:A u ptunr "'(III I'flh/t .•
I.:J'
\
•
""1'lJ
: ("'!!Ii fMI!,J 1.'(1 ",," "'10/(, "'/tJeA 1;,\' ".\','cI II' h"1I1 " JI"id III
.1"" trlll'flllll'If
.l!4fol""rl
v v] tlll'rl,rt(
InM
NilI' i,\ 0.( ,
/I,. I,' ",,I. Thr'lfIiII
.1(L.6
I
)03
T",;:o T
t
qL
+211
lIS x I05 )\ 0.025
-It
1'1'1'-C' !.'kll' t'J ,",';"'11/'" (h,· 11''''1,,'rlllllr,' '~llhi" rl,lIt
I ~ I 11111
K
i.\· .'.( ... ",1\, ('"iI"111I11' tb« Arill
lit
.' iser.
\\ '\ \\
8k
1f,·1.1) K
Surfllce ICl1Ipcl'lllure.
,til' C '1)r
.
T max -- T w + - qL2
. lun lemperalure,
MaX,111
_ T + 65 x 105 x (0.025)2
423 w
8 x 25
423= Tw+20.31
T Itl"~ '" 364.9 K
-: M~,"nll1ltllrllllwrf\IUrt',
51'11;011 :
I. oW Ihal,
XI
,7W~Ill~KJ
{l 120m
1].411electric current is pm;,H'" II",IIIgl, a co"'pol'ilf! "",11n",de
"I' of '''-0 layers. First layer is steel of 10 em thickness m,d
ftrond laycr;s bran of 8 em thickness. The outer surface
'tmptralllre of steel ami brass are maintained at 120~C and
_-3 =·L . K.
rtlJi"l::
litJ! rnn-fer
65"Crespectively Assuming tluu the contaa between 01'17 slab
is perfect and lite heat generation
is 1,65,000 Wlmj•
Dtltr"'ine
CO-efficient, (h)
I, Heal flux througn lite Oilier surface of brass slab
2, Inter/ace temperature.
-.
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\
Man' 7ram/er
1190 Heaton d . "
,
.. ~-- ~ -==
,
fi aee! ls ~5 WIttiK. K/or brass is Hfllt', ~
Ttlkt k or,
IIPr'(
'"
(I)
Given:
'l1sfer through
tccl,
Hca t tra
Sf
T
R
T)-T2
[': R= l]
kA
LI
k)A
Let interface
temperature
T2 is greater than TI, So,
T2-T)
Thickness of steel, L) = 10 ern = 0.10
...
(2)
...
(3)
111
Thickness of brass, L2 = 8 CI11 = 0.08 m
Heat transfer
Surface temperature of steel, T) = 120 C + 273 = 393 K
through
brass
is given by
0
~T
Outersurface temperature of brass, T) = 65° C + 273 = 338K
R
Heatgeneration, qg = 1,65,000 W/m3
T2 - T)
L2
k2A
k) =45 W/mK
k2 = 80 W/mK
To filld:
Total heat transfer
1) Heat flux. thr
.
~~~
S0111;011:
I
ou
gh
[Adding
the surface of the brass slab, q?
ce temperature T
'
,.
Q=
-
L2
k2A
Let
q -H
1
eat flux. th
q _H
rough the surface
2
eal flux th
rOugh the surface
Scanned by CamScanner
T - T)
2
+---
of the steel slab.
of the brass slab.
(2) + (3)]
with He
,=
C
enenu ion 1.193
45
.J;. =
~
QJ
L
T - .>38
m equation
I ,
0.08
80
He.a generation,
q
1,65,00
1,65,000 - 3 .020
Heat nux throu
= T (4'4.5
- 1O()(J1 - 1,78,6]6.3
=T2(144.'41-[5,16,6361
- 3]8000
1,.1i .I)(J(J+
'
"
16636
= T2 [1454.51
h he
surfa e f the brass slab
Rault:
(i)
::::>
1,3 ,980 \ 1m2
q2
T
q2 = I 30,980
(ii) T2 = 468.
1m2
K.
T2=46~.6 K
lJ A plane wall 10 em thick generates heat at the rate of
4 x 104 WlmJ when an electric current is passed through it.
Theconvective heat transfer co-efficient be/ween eachface
of tile wall and the ambient uir is 50 WIlli]K. Determine
(a) the surface temperature
(b) tlte maximum
air tempera/lire on the wall. AJ .ume
lite ambient air temperature to be 20°C and tile thermal
conductivity of the Willi material to be 15 W/",K.
[ 10 lr ts
Scanned by CamScanner
/17
iv r 'tv :-Ipnl 8)
1.194 Heaton
T
dM~sva.~n~~~er
__ ------------_
Heat Conduction
with Heat G
_:.:..:.-__ ------_.:_.:..._.:.~~~elnerat
te wo/l of Im thick is poured with
~~
~nN~n
or concrete generates 150 W/m3 h
.
e
~
Jr(ltlO1l 'J
eat. If hoth th
hY"
or the wall are maintained
at 350 C.
e
.faCes 'J.
. Find th
sur)"
m temperllture
In tIre wall.
e
(t1 ,4 C
Gi,'t" :
L ::::10 em == 0.10 m
Thickness,
.
ion J. J 95
'_4xI04W/m3
He.atgeneration, q . heat transfer co-effie ient, h = SO W 1m2 K
Convective
.
. temperature Too = 20 C + 273 = 293 K
Ambient air
'
•
",~I"'U
[Madras Univenity'
' Apri '19 9]
0
Thennal c.onduetivity, k = 15 W/mK.
1
Thickness, L == ~
G'plt" . :
Heat generation,
Tofind:
I. Surface temperature
q == 1SO W 1m3
0
T w = 35 C + 273
Surface temperature,
r., == 308 K
_. Maximum temperature in the wall.
qL
SMwW":
Surfac.e temperature,
T w = To:: + -
Tofind:
Maximum temperature
2h
{From equn no(1.59 ]
= 293 +
4 x 104 x 0.10
Solution:
Maximum temperature
2 x 50
Tmax
Maximum temperature,
in the wall
qL2
==
qL2
Tw+8k
Thermal conductivity
T max- - T w + -
of concrete,
8k
k = 1,279 W/mK
{From HMT data book.
(From equn no. (1.58)]
Page No.18 (Sixth editiont]
= 333 + '4 x 104 x (0.10)2
8 x IS
~ max = 336.3
Result:
Surface temperature T _
,
Maximum t
emperature
w -
K.I
l
T max = 322.6 K. \
333 K
.
T
, max = 336.3 K.
Scanned by CamScanner
1SOx (1)2
T max = 308 + -----~--8 x 1279 x 10-3
ReSUlt:
Ma .
X1mum temperature,
Tmax
= 322.6
K.
1.19 fle(ll o/ld Moss 7i'ol7sfer
--
" problems on Cylinder with internal
1••3 ;}
rn A copper wire of 40 mm diameter ~carries 250
l!.J
n
A--J
Heat Conduction
heat ge IICtatio
= 1.56
J 75 WlmK, calculale
"e i,
I. Hea/lransfer co-efficienl
ambient air.
2. Maximumlemperalllre
between
156 W/rn.
Heat generated,
q=-
. Q
q = _
lind
I
Tw = 2500
273=523K
ler,
o
'w
2
[From £'1111'1 no.t / (2)1
4 Y 175
52J.()7 K.
KJ
52io7
h
IV~ kunw
II'UIIIUII
•
q r:
-1-4k
523 "'" 124140 / (O.()2U)2
Lrrnux
Sy I utlun ;
Wlm.1)
- 10° . + 27 - 2RJ K
TQfind.'
'
_"
max -
k - 175 WltnK
.
(I::: 124140
temperature
'1'
Resistance, R .. 0.25 x 10-4 (2 ern/ length
T
(0.020)2 x I
We know that,
Maximum
2) MaxilllUllltclIlpl;;/,uturc
1'2 /. I.
1I %
_.!..;J 5~6,--_
1I x
in tile wire.
urrent, I = 250 A.
I) IIcul IrllIl.,'cr
C"
.
o
we ,'I"
ucrcnt,
1.197
wire .S.lIr/ace
Radiu , r'" 20 mm = 0.020 m
'1hermal conductivity,
-
:::----..::__-
V
Diameter, d '" 40 mrn '" 0.040 rn
Ambient air temperature, T
.
allon
We kllow that,
Gillen .'
urfacc temperature,
Heat C·lenl!f
102 Wlm
Y
==
A (Illd l
resiSlanceofO.25 x Iv·
cmllellglilsurfacetem
III.I·u
.
.
250'
ell'
,
bi
'Per(lfll
cO'{1perWife IS
anc 'If ant U!I1t air t empert
'tOIJ
10. C. If lite Ihermal cOlldllctivity of tile c opper illite
(I
IV'
.h
Wit
max.
thill- ,
SIl/'li,CC
l'Ollll)CI'l'Illll'U,
'1'
'"
T
I-
12R
(2 0)2 x (0.2
I'll
II
/ F/'(JIII
. -3
_1(1 I
/':1/1111 1/11. (/
U.(U) ~ 121114~
It
1.62 W/I,; II I
Scanned by CamScanner
1/5)!
With fI .
Heal Conduction
_--------.:.::~~e'(I(l11
Gen eru/lnn
/./98 Heal and M(J$sTransfer
'null :
I. Heattransferco-efficient,
h = 5.17 W/m2K.
2. Maximum temperature,
T max = 523.07 K.
IJ kW /teater. TIre copper surface
.
temperature
tDJJbielll air te"'Peratwe is 11"C, outside
is 11 kWI",1X. Tlrenrud conductivily
13 x 103
q =
I
7t x r2 x
llJ A cop~r wire of 1m long is used as a heating e/e~",
.
is /j
III.
u~.
Surface co-elfi ~
and resist an
copper lITe 15 WlmK and 0.11 n respectively.
~
Ca/culale
follJlwillg
Surface temperature,
•
Tw = T
co
+ ~
2h
{Fro", Equn.
,,0./.65J
1. DilDlldU of copper wire
2. /lJlJe of CIUTUII flow.
295 + r x
1573
G;reJr :
4138
r2
Length, L = 1m
2"1.1)(103
Heal transfer, Q = 1HW = 13 x loJ W
4138
1278
r x 2200
Surface temperature, T", = 1300" C + 273 = 1573 K
Ambeint air temperature,
Too= 220 C + 273 = 295 K
1278
1.88
=
r
Outside surface co-efficient,
Ir
or
Heal transfer co-efficeiet,
ml
1.47xIQ-3
Id 2.94 x 10- I
h = 1.1 kW/m2K
3
=
= 1.1 x loJ W/m2K
ThennaJ conductivity, k = 15 WImK
Weknow that,
Q
Resistance, R = 011 O.
Toruul:
13 x 103
1) Diameter of copper wire, d
::)
2) Rate of current flow , I
Scanned by CamScanner
1.19fJ
!
1
= 12R
= )2 x 0.21
= 248
A
I
m
2 (J
/
I () ) m.
.-----
,
. -
: ([~
f JI
I
II is pcmtd through a "ainle"
MK, Jmm ill diamet
t
f.
The resiui
II
"ttl ..
if] jiO em and 'he I n;(lh ofth.
. uid iu J /(1'
;;;iJh heal/ramIer
()
'41ht
IJ.O,)') (2
== J2p.
::; (20(1r / ((UfJ"Ij
~i,
>-e/li.","
(
uku« the eentre I mp fa[ure 'iflhe
[Modr
k
==
enlly, Ap J (,_~
t Lm
J9~
()
/L
q = ----,
)960
-=-,-
(J/IQ-"r / I
_.
.,.
(
..l~
'f C - ~-:; = )< 3
"'!
.- m1 (
q alue in Equn
/ loj
I)
(
4 y 19
...
(J)
{From Equn no (/ ,.,."
Area
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(
tre temperature
of wire, T
= 399.' K
Heat Conduction
with Heat Generation 1.203
1.202 Heat and Moss Transfer
250
4/3 1tr3
1.3.6Solved problems on Sphere with
Internal heat generation
= 250 x 4 x 1t x (0.050)2
q
rIl A sphere of J 00 mm diameter having thermal conduelill'
G
0.18 WlmK. The outer surface temperatu Fe IS. BOCllyOI7
250 WI",z of e"ergy is released due to I,eat sou rce. CalclI Qlld
I. Heat generated
4/3 1t x (0.050)3
= 15.0~0 W/m31
/tllt
2. Temperature at the centre of the sphere.
Temperature at the centre of the sphere
Give" :
qr2
Diameter of sphere, d = 100mm
- T +Tc w
6k
r = 50 mm = 0.050 m
[From Equn no.(/.66))
15000 x (0.050)2
Thermal conductivity, k = 0.18 WlinK
= 281 +----'---'-
6 x 0.18
Surface temperature, T w = 80C + 273 = 281 K
Energy released, Q = 250 W/m2.
~c
Tofmd:
= 315.7K
]
I
1. Heat generated,
r-
q.
I
2. Temperature at the centre of the sphere
2.Centertemperature, T c = 3 15.7 K
SOIUlwlI :
Heat generated,
q = _g_
V
q/A
q/A '"
Q/A
V
250
4/) nr3
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[.: Q/A
.
3
I. Heat generated, q = 15,000 W/m
= 250 W/m2]
T
1.204 Heal and Mass Transfer
Fins 1.205
~~~-----------------t.4FINS
~
It is possible to increase the heat transfer rate~
c. Th
t:
Y Inere .
the surface of heat transler.
e surraces used for in creasln. asln,e
transfer are called extended surfaces or fins.
g heal
1.4.1 Types of fins
Some common types of fin configuration
are shown' In fiIg.!.I!.
(iii) Splines
(i) Uniform ,traigl,t fin
(iI) Tap
ered straight fin
(iv) Annular fill
z
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,
I.206 Heal and Mass Transfer
Fins 1.207
M P;IIjlns
Fig 1.11
Commonly there are three types of fin
1. Infinitely long fin
2. Short fin (end is insulated)
3. Short fin (end is not insulated)
1.4.1 Temperature distribution and heat dissipation
in fin
Fig. 1.12 (a) and (b) shows the straight fin or longitudinal tin
of rectangular section and circular section respect ively, One end of
the fin is enclosed in a heating chamber and the other end is exposed
to atmospheric air.
Heat.iS transferred across the rcciangulllr fin and circular r~
by conduct~on. From the surface of the fin, heat is transferred 10 air
by COIIVCl:llon. Let Us consider a sm II I
I I
of'tll,'ckncss
". IS at a distance of x rro ., tha Cb cmcll It arelt
dx, which
II In
elise.
¥AA&CiU&CiMiDNiRM
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I.l~~::...=-Heat artd Mass Transfer
!
I
I
_
---------_
Fins 1.209
state
conditions,
heat
balance
e~th
stca dY
or at
A.
follows.
enl IS as
(1,111
conducted out ofth
n dueted into the element = Heat
tCO
.
.
e
!-lea
onvected
to
the.
surroundlllg
air.
-/-heat
c
nl
l(I1le
Q
(
Qr:::
Q.r
...
dx
+
conv
•.• (1.67)
?
]I~
pi -c"
where,
"'C
t::
II
c..
-dx -kA (ddx
- T) dx
( dT)
2
'<x+d\
=-kA
Qeonv
= hA (T - T cx:)
f'l
.!::
2
f':l
OIl
::::
"t:l
::::
::::I
......
0
::: h(P dx) (T - Tcx:)
::::I
rJl
Substituting Qx, Qr.,.dt and Qeonv values in equation (1.67)
~
'-
~
dT = - kA (dT'
- J - kA cilT.J dx
(167) ~ - kA -
...;
.
·tc·
dr
dx
.1.\-
~
+ h(P dx) (T - T.x:)
~
kA d1T = liP (T - T cx:)
dx2
~kA ( T-T.)
d1T _ ~
(tt2
d2T
dx2
L
---=. ::=
... ~"_'_"""---
Scanned by CamScanner
15
kA
.r:
(T _ Toc) = 0
1,210 Heat and Ma.I',I' f'r(III.~I(!r
where,
JIl2
Flns 1.211
lIP
kA
c _
,
bsll'tllting
At X -'-' cr., T = T a. in equation
SlI
C Ie -mrs:. + C 2em".
(T C/. - T a: ) =
(PT
-
dx2
- JIl2
I 70
(l. )
e =0
..• (1,68)
C2c"'OC = 0
[':0 "'1' '
Equation (1,68) shows that the temperature is f ~f,,1
x and III, It is a second order, linear differential equatio a ,UnCtion
Of
solution is,
n, ts Beneral
c",ct. ~ 0, So, [C2 = 01
substituting.
The temperature distribution
Upon the following lin conditions.
... (1.69)
and heat dissipation d
().71)::=:>Th-oc-
C 2 =Ovalucinequation(1.71)
T
ePend!
::=:> rTb-Toc=
Case (i): Infinitely Im,gjin
If a fin is infinitely long, the temperature
that of the surrounding fluid,
at its end is e
qUalto
, '
Suhstitutlllg
- CI + 0
ell
e I lind C2 value in equation (1.70)
(J.70)::=:> T - T a: = (T h - T)a: e -liLT + 0
At x = 0; T = Tb and At x == ac; T:: T",
Tb - Base temperature of fin
From equation (1.69), We know that
o == C I e-nlt + C2emx
,
Temperature
T - Toc == Cle-nu + C enlt
2
SUDstituting
distribution
T-Toe
IIIX
offin, Tb _ Toe = e
... (I.70)
[.: e == T - T«l
At X== 0; T = Tb
Where,
Tb - Base temperature, K
(l.70) ==>
T a. -
Surrounding temperature, K
T - Intermediate temperature, K
...
(1.71)
.\'- Distance, ",
-
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...
(1.72)
v:
I
If
I.
11/-
/I
I
II
~
~.'\
,,- heat transfer co-efficient,
p- Perimeter,
W/1112K
III
k -1 hcrmal onductiviry, W/I11K
r\ - Area.
...
III
11/
\
Q z: /hP -A (T
the fin is obtained by intoeg .
.
rating the
over the entire fin surface.
II through
hea
II
We kn
- 'r ,
//'" " I
w that,
Heat I
t by
Qeo,1\' == hA (T - T"..)
nvection,
Q == hP dx (T - TCf_)
(4jt(ilJ: Fill with insulated end 'Short jill)
Thefin has a finite length and the tip of fin is insulated.
if.
Q == fliP (T - T ) dx
At r= L' dT == o·
'dx
'
o
Q==
Atx = 0; T == Tb
From equation (1.70),
we know
that,
T-1'
---"'e
Tb -T 'F..
- 1'-T "(1',(_)
(T - TrfJ == C I e-II/X + C2ellIX
II/X
dT
dx == C,e-lIIx
r.: e:" J
x (-111) + C2efllx
ApplVilJ~ II Ii.
.
te trst bOLIndary
hI> (T - T,.) J C 11/\ dr
o
:::>
ciT
condition, i.e., at x = L. --=0
0 == C .-1111.",
,e . x -1/1 + C°2\.·
nllli.
hp I'll' r. ) _ I
.
r
l'
1/1.\
III
J
x 111
•
X
III
.r
()
//JC,e-IIIL
e,e-
lIl).
=:
=:
C
I/I(',(;'IIIL
2e
ml
.
. .. (1.74)
Scanned by CamScanner
dy
d "",,r,,,,ufe~r
.",.....-
-
_
/.214 Heal an.
70) we know that,
,
-mx + C emf
2
(T - T a. )==C,e
Fins 1.2/5
equation ( I.
From
Applying the Seeon
(T b -
d boundary condition,
. ting C, and C2 value in equation (1.70)
bsutu
su
i.e., aLt::: 0, 1'::::
(Tb - Tcr.)
(Tb - T a: )
1+ e-2mL x e-mx +
---.:_
"elllX
1+ e+2111L
(T - T ex:) =
Ib
T ) == CleO + C2eO
a.
(Tb - 'I'a. ) '" C, + C2
[
--+_
2mL + C2
Tb - Ta. -- C2 e
(T - T ex:)
-
e-nu
entX]
I + e-2mL
1+ e2111L
[e-mx
+_elllY]
l+e-2mL
(Tb-Tex:)
l+e2l11L
Tb-Tex:
... (1.75)
~ '" [e2mL +1]
. . C2 value in equation ( 1.74)
Subsu,utmg
Tb-Tocl
C '" -,
I
e2mL+1
Multiplying
the numerator
and denominator
e-nu
x __
emL
1+ e·-2mL
emL
x e2nrL
+-----
e-mL
x __
e-mL
1+ elmL
x -e-2111L
e-m(x-L)
enrL + e-IIIL
Tb- r,
e-m(L-x)
em(L-x)
e2mL x e-2nrL + e-2nrL
(,=I + e-2mL
em(.\-L)
+ ------elllL + e-IIIL
e-mL + emL
C, =-------------
.
by emL and e-mL
+
... (1.76)
em(L-x) + e-m(L-x)
em!.. + e-nrL
Scanned by CamScanner
e-IIIL + emL
... (1.77)
1.2/6 Heal. and Mass Transfer
In terms ofhypcrbolic
T - Tif.
function it can be w rltten
.
a
.>:
~
,cos h 11/ (L-x)
cos h In L
traJ1sferred
Q
Heat
fi
. sulaled In,
Temperature distribution of fin with insulated end
JhPkA
('1'
JhPkA
(T b -. T,,) Ian h (/ilL)
b - T,,) Ian h{nrL)
-
fOf 111
.. (1.79)
cos h 111 (L-x)
•.• (1.78)
cos h 111 L
cos II 111 (L-x)
=> T - T IX == (T b - T ,,) ----'--__:,_
cos h m L
=> dT
2. Cooling of motor cycle engine.
= (T b _ T »( _ m sin II In (L-x)
~
c~hmL
1
\\e know that,
Heat transferred. Q = -kA
pJicalions
(,4.3'\P
.'
.
ain appllcallon
of fins are
rhe J1l
I. cooling of electronic
component
dT
3. cooling
of small capacity
4. Cooling
of Iran formers
5. Cooling
of radiators
= -kA (T b - T aJ x _ /11)( sin hili (L-x)
cos hili L
etc,
The efficiency
f a fin i defined a the ratio of actual heal
transferredfin (0 the maximum po ible heal Iran ferred by the tin
llr,n=
Q = kA m (Tb _ T,,) x sin h 111 (L-x)
cos h 111 L
At x = 0,
For insulated
l1fin =
/ilL
1.405 Fin effectiveness
J kA (T b - T cJ tan h (/ilL)
'=kAxlhP
III =
max
end
.
[.:
Ofin
-0
Ian II (mL)
sinh(mL)
cos h (/ilL)
'= kA III (Tb - T,,:) tan II (mL)
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and refrigerator:
IA.4 fin efficiency
dr
Q=kAm(Tb-T"Jx
compres: or
Rf]
It IS defined
.'
~
as the ratio of heat trans er
ilhout fin
Fin effectivenes
, E == Q\\iiholll fin
II
ilh lin 10 heal transfer
1.218
Ht!(/I and Mass Transfer
For insulated
Fins 1.219
end
E=
Fin effectiveness,
tan h (mL)
J
~d
stl
.'
Ilire tlistrlblillOIl
fell,pera
iI)
T _ T rIJ _ cos hill [L-x]
1.4.6 Formulae used
~
[Refer HMT data book page
(I.
LONG
d
l·
kP
1. INFINITELY
[ _1_ < 301
is insulated]
Olrr
FIN (OR) LONG FIN
Temperature distribution
110. 49(SiXth
.
edllj~
';
b)Hea
-
cosh
[mL)
I trails/erred
Q == (hpkA) '12 (T b - Too) tan h (mL).
~4 7 solved Problems
1..
I1J Find tire heat loss J,.om a rO.d of ~ mm in diameter IIntl
where
T b - Base temperature,
T
K
- Surrounding
temperature,
K
T - Intermediate
temperature,
K
x - Distance.
infinitely long when Its base IS maintained at 140" C. TI,e
conductivity of tire material is 150 WlmK andth« lreattransfer
co-efficient on tire surface of the rod i.\·3()~ WIlli] K. TI,e
temperature oJ tire air surrounding the rod is 15" C.
Given:
Fin diameter, d = 3 mm = 3 x 10-3 m
11/
Base temperature,
lIP
Surrounding
kA
T b = 1400 C + 273 = 413 K
Temperature,
Thermal conductivity,
I .ieni, \\' m-K
T,I'. = I So
273::_ nil K
k = 150 W/mK
Hea trans er o-efficien
h = 3(J<J
'K
To find :
" -
·IllUJII :
-:;.
.
.2~
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Fins 1.221
-------------~
1.220 Heat and Mass Transfer
. Tetnperature. T = 20°C + 273 = 293 K
olltldIII g
511rr
\' == 20 em == 0.2 III
llce
T == 60~ C + 273 = 333 K
Oist:l •..
tCInperature.
lcdl!ltC
Illterll
. 'tv k == 200 W/mK
IcOlldueUvi s :
'fhenn:l
== 2L d2
Area
4
IA
6 m2]
== 7.06 x 10-
P - Perimeter
nd
==
1t x 3 x
IP
fofi"d :
I 0-3 III
9.42 x 10-3 m
I
~ \Q = 6.838 Watts
tran!)ler
solutiO" :
Apply A, P, T i» T co' hand k value in equation (I)
10-3 x 150X7.06~
(I) ~Q=(413-288)J300x9.42;(
co-effie ient, Jr
J.
!-tea t
I
ono tin tempe
For l :::>
T - Too
--:::-
rature distribution
[From HMT data book
page 110 ~9J
== e-lIIx
Tb- Too
~==e-/II>.O.2
423 - 293
Result:
0.307
Heat loss, Q == 6.838 Watts.
III (0.307)
\1l A long rod 5 em diameter its base is connected to afurnacl
wall"t 150· C, while the end is projecting into the room a/
20· C. The temperatllre of the rod at distane of 20 cm apart
from its base is 60 C. Tile conductivity of tile material is
200 WlmK.Determine convective heat transfer co-efficienGive" :
== e
_III
x 0.2
== -III x 0.2
-\. \ 8
== -/11)(
0.2
G'' ---:=-S -.9-,-,rj
G
~rr dara book
We know that,
fhP
/II
:= rk~
where
Fumace
1500C 60"C
20cm
e
20D
I
Diameter of the rod, d ~ 5
Cm:: 5 )( 10-2
Base temperature, T b ~ \ 500
Scanned by CamScanner
m
C + 273 =: 423 K
A-
Area
4
[From f/ .
page /lo.~9J
1.224 Heal and Mass Transfer
P _ Perimeter:::
xd = 1( x 0.050
_
1(
A - Area = 4
Ii"": st dissipate d , Q
~o.m;J
for
d2
. 90 mrn, it is treated
solll ti~'': he length of the ro d IS
h
Sil1ce tid
as s ort fin.
d is insLi ate .
He
~t11e ell
::: _l!_ (0 050)2
4
fA :::
(1) :::>
.
Q::o (hPkA)
t transfer,
ASS
y:.
,WI
3 m2/
:2
2.55
=
6.50
= -k-X:=';"~.9-6~x:..!..'~0~-3-
(T b-1 exl) tan h (mL)
[From HMT data book page No49}
1.96 x 10-
30 x 0.157
2
Q
::0
k > 1.96 x 10-3
(90 x 0.69 x 55 x 5 x 1O-3)Y2 (673 - 323)
tan h ( III x 0.09)
30 x 0.157
[k :::369.7 W/mK I
==
Resull:
Thermal conductivity
90 x 0.69
55 x 5 x 10-3
of the rod, k = 369.7 W/mK.
~::o
[!) A carbon steel (k :::55W/mK) 90 mm long rod wil/I croSJ
~
seclional area 5 x Nr] m2 and permiler 0.69 m is allac/ltd
10 a pintle wall which is mailltailled at (Itemperatllre of 400'(.
The surmunding environment is (II 50" C and heat trallS!fr
co-efficient is 90W/1112 K. Calclllale lite Ileat dissipated hy tilt
Q
15.02 Ill-~
::: (90
x 0.69
x
55 x 5 x IO-3)Y2 (673 - 323)
tan h (15.02 x 0.09)
[Q::: 1264.8 Watts]
rod.
Given :
Thermal conductivity,
luult:
Heat transfer,
k = 55 W/mK
Length, L = 90 mm = 0.09 III
I
5 em] area, J 50 mm
~ A stainless steel blade oj 80 mill ,Ollg, b
of the bIad e IS.
perimeter and the temperatur« at tlte use
Til heat
750'C. Tile blade is exposed to hot gas (It J 000 C. te is
• if)
Area, A = 5 x 10-3 1112
Perimeter, P = 0.69 III
Base temperature.
Q = 1264.8 Watts.
D
Tb == 400 C + 273 == 673 K
0
Too
SO" + 273 == 323 K
Surrounding
temperature,
Heal transfer
co-efficient, h == 90 W/m2K.
0=
I
transfer co-efficient
16
Scanned by CamScanner
.face and the gas
betweell tlte blaM SIlT)'
/.226 Heal and Mass Transfer
<.
500 Wlm2KfIIlfltllermal c~.
the heat flow lit tile root of the blalle. A . "'/(.lJpl_ ""
,'f.~"",
e(~t
e l,~(', t
firom the tip of tile blat/e.
500 x 0.150
"0 "'i~
10
30 x 5 x 10-4
11
Given:
Length, L:= 80 mm := 0.080
~
Q
III
:= 5 x 10--4 012
Area" A := 5 cm2
(I) -;?
::: 1.06 x (-250)
[tan It ( 70.7 x 0.080)]
::: _ 265 [tan It (5.65)J
Perimeter, P = 150 mill := 0.150 III
0
Th = 750 C + 273
Base temperature,
Hot gas temperature,
::: _ 265 x 0.999
= 1023 K
Too = 1000 + 273 = 1273 K
Heat transfer co-efficient,
0
~
h = 500 W/m2K.
Btsu1tHeat
: transfer rate, Q = - 2649. W .
Thermal conductivity, k = 30 W/mK.
Tofind:
A" alll",i"illlll
Heat flow
@]
alloy fin of .'i mm thick anti 40 mm 100'g
/fIules from a H'{II/. The base temperature is 420° C ami
pro
"mbient air temperature IS. 250C • 11,e heat transfer coefficient
between aluminiuIII rod (Inti environment is 2.'i Wlm2K.
Calclliate tile heat loss from tile fin of material taking its
thermal comtuctivity as 200 WlmK.
T
Solution:
No heat loss from
the tip of the blade i , .. " tip isS IIlSU
i IHIed.
.
Length
of
the
blade
IS 80 mrn, so, short fin. This is sho t fi
•
•
r 111 end
Insulated type problem.
Q
Heat transferred [Short tin, end insulated]
Q = (hPkA)~
Givtn:
Thickness, t = 5 mm = 0.005m
(Tb- Too) tan h (mL)
[HMT data book page N049J
= [500 x 0,150 x 30.x 5 x
1O-4]~ (1023 - 1273)
= 1.06 x (-250) tan h ( m x 0.080)
In
Base temperature,
T b = 420° C + 273 = 693 K
...
(I.)
Heat transfer co-efficient,
Thermal conductivity,
fhP
tkA
;
, I
.--'
'
..
\
Scanned by CamScanner
Too = 25°C + 273 = 298 K
Ambient temperature,
tan II ( rn x 0.080)
where
Length, L = 40 mm = 0.040 m
[HMT data book flag" No ~9J I Tofind:
Heat loss, Q
h = 25 W/m2K.
k = 200 W/mK.
~''4'el'ri'
I
,,7X
f
d M ,,11'clIIs~~er
Ile~
~
S(}/II,loil :
"
III' 1"llUlh of the tin is 40 nun, it is Ire'll"d
SInce
c
0
<"
tI ~ hUll I'
III
AsslIme cnd is inslIloted.
. ,sf;'rred I Shun till, end insulated I
lleat t rlIl, "
'
.
Q
(hPkA)i/2 (Tb- T .) tan h [ml.]
... (I)
llad .\' tlr(' nuulc
M"e ' e
".
V
IS
.
C~ , b,d, 11/11/
'5 It' neot- 1he ('TO,U !ICU/mlll/i" 'I' III , '.
I. e
I 'S H
'J tile /, III IU/" ,
(II" e
.,rimeter 0/ each blude 1\ 7 em. TI,e "(1\'
• If
, .",1, pc
/ . H( ", "
..
'empera''''e
1.5C , 1 {Jve r thc him e I." If) C. Temperll""" ' (, , ttu:
, fIllll Of
10111"'11 ~ " J2 50° C, th ermul elllll/llcilvily
1/ hi
J
'....
ij tu«
IIIle is
/ blade U
1/1£
J( (lIId hell' tr(llu/a
co-effictem ls /1 n WI 2
~ ;I /1"
19
of ,\(111,,/.\, \(' -t
(II
(I UAT duta I ook p
I
(IRe No,49/
where
2 x Length
P _ Perimeter
Will',
.'
m 1(.
]] ,,,,,,,e
. the /,Clf:ht of 'he bltule IICS:/{'('Iill" .. the " eat f1ow
tC
(lfi to the eml
of the blade.
oe
(Approximately)
fro'" ,Ire g ,
::: 2 x 0.040
Give":
1~leattransfer
[p ::: 0.08 I
III
erOS
.\ _ Area
::: t x L
.
perllne
::: 0.005 x 0.040
IA ::: 2x 10-4m
ctional
area of the blade
ter
7 ern := 7 . 10-2 III
se
p:=
,
rr
Gas temperature,
1
[HMT data book page No.49)
800
1071 K
:=
J
Tb := 12 0°
273 = 1523 K
r
Thermal condu
ti it
k= 22 W/mK.
Heattransfer
o-e
ient. II:::
1
m? = 4. / 10-4 m2
A::: 4.5
273:::
0
if.)
Rool temperature,
2
where
m
Q- 8 W
-
2
110 \ Im K,
Tojiml:
25 x 0.08
200 x 2 x 10-4
Im
Substitute
(I) ~
Q
Heighl of the bl d
So/ulioll :
Negle ling h at n w Ir 111 the end fa e of the bl de [ livellj,
7.07 m-I!
m, h, P, k, A, T b» Too values in Equation,
(I).
= (25 x 0.08 x 200 x 2 x 10-4]Y2 (693 - 298)
tan h ( 7.07 x 0.040)
.--\
Q-= -30-.7-7W-1
Scanned by CamScanner
L
.ihis is hort fin, end in ulatc
I) pc
pr \ lein.
Heal Iran ferr d I h rt fin. end in. ulaied]
Q
=
hPk
'12 Th- Tu
I 11 h
trnL!
Scanned by CamScanner
I Mass Transfer
I. 232 Heal ane /YII o./'
alllte middle of tl'e fin
ern/ure
Ji ferrtP
-:; 1.)2 in Equation (I)
where
_(hP
!11
_
JkA'
p
Pllt"
I
-;::::.>
Perimeter:::: 2 x L (Approx)
-r _ Tet)
'
cos h m [L - Ll2]
_
.s--e->
Ib-Too
cosh(mL)
(I)
2 x 0.050
[p
cos h 26.9 [0.050 - ~250]
Tet)
__.!--:::- ==
'f"_
O.G]
-;::::.>
A - Area:::: Length x thickness
IA
::::0.050 x 0.007
Ib - Too
1- 295
_
~-
3.5 x 10-4m21
;:::J
393 - 295
1-295
;:::J
j
1m::::
(2)
~
140xO.1
T - Too
T b _ Too
I
T -295
393 - 295
=:>
~
T-295
2.049
== 0.6025
393 - 295
Temperature at the middle of the fin
[ Tx= Ll2
cos h (26.9 x 0.050)
iii) Total !lcal dissipated
i
T - Too
Tb-Too
}.234
[T == 354.04 K I
55 x 3.5 x 10-4
26.96 m-I
cos h [26.9 x (0.050)]
---
I
:::: [140
---
x 0.1 x 55 x 3.5 x
10-4]lh
x (393-295)
x tan II (26.9 x 0.050)
2.05
I T = 342.8 K I
[From H MT data book page no. 49J
Q == (hPkA)Y2 (Tb- Too) tan h (mL)
2.05
=47.8
== 354.04 K
I Q = 44.4
I
I
WJ
Result:
1. Temperature at the end of the fin, T1'= L == 342.8 K
Temperature at the end oCthe fin, Tr _ L = 342.8 K
--
2. Temperature at the middle of the fin, T x = LJ2 :::: 354.04 K
3. Total heat dissipated,
.
/
"
Scanned by CamScanner
Q == 44.4 W
1.234 Heal and Mass Trans er
Ial A rectangular ailiminium
fins of 0.5 """ s
~
I
qll(l'e
101'" are attac"ed Oil a ptane pltlte IV/iicll r.,
(lI/rll'
lJ
.'
Il ",.
< '"
80· C. Slirrolillflmg atr temperatllre is 22° C C(lil/IOilltl/ '"
IlIImher 0/'fins required to generate 35>< J 0'-3 ;.v (llclllCitIIr
rr 0lli
t'4
k = 165 WII1IK and II = 10 WI",] K. Ass"""
'efll. 1'; e
e
l
Q*e
from the tip 0/ tilt!fin.
leQll
"0
all
Givell :
Fin dimensions = 0.5 mm square, 12 mm long
.
Fins I'
tin
IS 12 mm and there is n h
·~J5
ofpe
.'
0 eai ]
ellgtlt
Oss frOllllhe
' this is short fin end 1Il1isiated ty
l- {ill. So,
pe problelll
f tile
_ [short fin, end insulated]
.
liP 0
rallsfer
!-Ieat t
[From HMT data boo
k page flo -I9}
J(A)Y2 (Tb- Too) tan h (mL)
.
P
I
- (I
~Q - [10><2>< (0-3>< 165x2.SxI0-7]~)«(3
::;:
h(l11xI2xI0-3)
53-295)
x tan
~ Q ::;: 0.0526 tan h (m x 12 x 10-3)
...
(I)
So, Fin thickness (t) = 0.5 mm = 0.5 x 10-3 rn
Fin breadth, (b) = 0.5 mm = 0.5 x 10-3 m
Fin length, (L)= 12 mm = 12 x 10-3 m
Base temperature, Tb = 800 + 273 = 353 K
Surrounding temperature,
(0)<2><10-3
165 x 2.5 x 10-7
Too ::: 220 C + 273 :::295 K
Heat generation, Q = 35 x 10-3 W
Thermal conductivity, k::: 165 W/mK.
Heat transfer co-efficient, II::: 10 W/m2K.
~
(I) ~ Q
== 0.0526
E
Tojind:
tan h (22 x 12 x 10-3)
0.0135 W per fin
I
Number of tins required.
We know that,
Solutioll :
Fin area, A::: b x I ::: 0.5 x 10-3 x 0.5 x 10--3
Heat generated
Number of fins required
Heat transfer per fill
/A::: 2.5 x 'O-7m2/
Perimeter
p:::
(for rectangular tin)
35 x 10-3 W
0.0135
2 (b + t)
2.57 == 3
I Number of fins
::: 2 [0.5 x IO-J + 0.5 x 10-3)
tr.p:-e-fJ:-·
rn-e-te-r,-p-:::-2-x-, 0---3-m
J
=3
Imlit:
Number of tins required is 3.
Scanned by CamScanner
I
I
36 Heal and Mass Transfer
/.
2
-~
@Tenthin brassfins (k -100 WlmK), O.7S mil, Iflick
~
axial/II on a lm long lind 60 mm diameter
. Q'epi
J
"
eng,,,
Qre4
which is surrounded by 27 C. rhe.fins are e.
e cYIi""
"Xle"d
"e,
from the cylinder surface and the heat transfer
ed I.S c
between cylinder and atmospheric air is 1S Win 2 CO-emC!!;'"
I /( C
~I
the rate of heat transfer ami the lemperalur
. (I/t'14IQ(
e lit Iii
e
fins when the cylinder surface is at 160" C.
Ie e"d OJ
Fins
Thickness. t:: 0.75 x 1Q-3 m
Length, L = 1.5 x 10-2 m
(MU April 2
OOOJ
Given:
We knOW that,
Number of fins = 10
Thermal conductivity, k = 100 W/mK
Heat transferred, Q1 = (hPkA)112 (T, - TI>:»)tan h (mL,)
(From HMT data book page no. 49)
Thickness of the fin, t = 0.75 mm = 0.75 x 10-3 m
Length of engine cylinder, Ley = I m
Diameter of the cylinder, d = 60 rnrn = 0.060 m
where
2 x Length of the cylinder
p _ Perimeter
Atmosphere temperature, Too = 27 C + 273 = 300 K
0
Length of the fin, Lf= 1.5 em = 1.5 x 10-2 III
Heat transfer co-efficient, h = IS W /1112 K.
Cylinder surface temperature
= 2 x 1
Ip = 2ml
A= Area = Length of the cylinder x Thickness
or
Base temperature, Tb = 1600 C + 273 = 433 K
Tofind:
I. Rate of heat transfer Q
2. Temperature at the end of the fin
Solution:
Length of the fin is I 5
. .
.
the fin end is .
. em. So, this IS short fin. Assumms Ihal
Insulated.
Scanned by CamScanner
... (I)
=
1 x 0.75 x 10-3 m
IA = 0.75 x 10- m I
3
2
1H
m
A
=
15 x 2
100 x 0.75 x 10-3
[m = 20 m-1
123H Ib!OI an~_~~~~'\'?~:!~I:r(j·~_
~;__..,_,._
..(.;..:- (hPkA)Yl (Th- Toc) Illn II (nIl;
(I) ~..'> <',
.
jl
• 115 x 2 x 100 x n,75 ' 10 ,1 J ~
1111111(20)< 1.5 x 10.2)
. (4 3 3()())
-e-
z= s
)!
0, :: 1.5)(
133 x 0,29
0, ::: 58,1 W
Heat transferred per fin
=
Heallransferred for 10 fins
[QI::: 581 W
58,1 W
=
0.95
58, I x 10::::: 581 \V
l
...
:=
(2)
0,95
Heat transfer from un finned surface due to c onVection'
Q2 = h A
Tb - Too
_---
IS
1'00+
T
t x Lfi (1'b - 1'(1)
r
(1)
095
sr
:::II x (n d Ley - lOx
1'b-
:=
433 - 300
300+--0.95
[,.' Area of unfinned surface = Area of cylinder - A rea offinl
=15x[(nxO,060xl)-(IOx075
.
x
I 0-3 x
1.5 x 10-2)] 1433-30°1
I 02 = 375.8 Wi·
Result :
.. (3)
I Heat transfer,
Q = 956,8 W
2. Temperature
at the end of the fin, T = 440 K,
So, Total heat transfer, Q = QI + Q2
[illA circum/erential
ITotal heat transfer, Q = 956.8 W I
rectangular fins 0/ 140 mm wide, ami J
mnl thick are/Wed Oil a 200 mm diameter tube. The fin base
temperatllre is 17()" C ami the ambient temperature is 25°C.
We know that,
Estimate/in efficiency
Temperature dis
. [short fin, end insulated]
istnibuuon
T{lke
Q = 581 + 375.8
T T
-
coshm{Lf-x]
if)
Tb - T
=rr_
I
Thermal
(//1(/
heat
conductivity,
Heat transfer
/Oo5S per fin.
k = 220 WlmK.
co-efficiem. h := 140 Wlm2X.
__
__:.__..
cos h (mL )
f
(From IIMT do/a hook pag« /10.49J
~
Scanned by CamScanner
fir
seZW'l1
\
___ -----:----_ .._1.24 Heal and Mass Trans er
0
________
Give":'d
L::: 140 mrn == 0.140 m
WI e,
. k ss t::: 5 mill == 0.005 III
Thlc «ie ,
.
ter d == 200 mm ~ r == 100 111111
== 0 I III
Dlame ,
.
Fin base temperature,
T b = 170
0
~
./
1.24/
I [rze - r II
/'":::
"Ill :::: 0.005
[(0.242» - _. (0. 100)1
lO"'lm21
~>(
[~--.--_j
C + 273 == 443 K
0
Ambient temperature,
Too = 25 C + 273 == 298 K
Thermal conductivity,
k = 220 W ImK.
Heat transfer co-efficient,
40
h = 140 W/m2K.
Tofind:
30
fill
dliciellCY
I. Fin efficiency, T1
2. Heat loss, Q
20
11
10
Solution:
A rectangular fin is long and wide. So, heat loss is calculated
by using fin efficiency curves.
Corrected length, Lc
1.5
o
{From HMT data book page no.50 (Sixth edition)]
2
2.5
3
f_h \0.;
Lc l kAm
1.5
t
= L + 12
= 0.140 + 0.005
2
ILc
From the graph, we know that,
[ flAlT claw book page 110.50]
=0.14251111
'1 + Lc
X . = LeI.)
ans
0.100 + 0.1425
0.2425 m I
-
21t [I'2C2 - I}]
21t [(0.2425)2
0.30650
Scanned by CamScanner
1112.1
- (0.100)2]
05
.
-[ " 1
-kA
,r 1.5
- (0.14_:»
, [~= 1.60 I
III
140
1°·5
220 x 7.125 x 10-"
l
neaI and Mass Transfer
U
/.242
r2(
Curve~
-
r,
==
.
0.2425 = 2.425
0 I
.
'fe'"
61 Oia(J1
Curve value is 2.425
,~
f
rOil] Dr..
t:>'Gph
Fin efficiency, 1') == 28 %
Heat transfer, Q == 1') As h [Tb - TC()]
[From HMT data bookp
=>
Q = 0.28 x 0.30650
d:::: 1.2 ern == 1.2 x 10-2 m
Length, L == 6 cm = 6 x 10-2 rn
. ht or
HelS
ductivity, k = 25 W/mK.
(J1al con
filer
. temperature, T co = 600 C + 273 = 333 K
dmg
Olm
.
_
Sorr
f r co-etliclent, h = 4:> W/m2K.
at trans e
He
ature, T b == 1000 C + 273 = 373 K
temper
Base
X axis. value is 1.60
By using these values, we can find fin efficiency
eter,
10
ft", : Fi efficiency, 1') fin
1
age n050j
10
.
2. Tempera
x 140 x [443 - 2981
,J
3. nea
ture at the edge of the rod, T, =L
t dissipation, Q
4. fin effectiveness,
IQ= 1742.99 Wi
E
So/Illion:
Result:
. .m 'ency (For insulated end)
l. Fin efficiency, TJ = 28 %
J. Fm ell'C'
2. Heat loss, Q = 1742.99 W
TJfin ==
1m A stainless steel cylindrical rod fin 0/1.2 em diameter
tan h mL
and
where
m=
6cm height with thermal conductivity 0/25 WlmK is exposed
to surrounding with a temperature 0/60° C The heattrensfe:
co-efficient is 45 Wlm2K and the temperature at the haseo!
the fin is 1000 C Determine
... (I)
mL
rtf
[From HMT data book page no.49]
hP
kA
P> Perimeter
== nd == 0.0376
m
2
A - Area == TC/4 d == 1.13 x 10-4 m2
I. Fin efficiency
2. Temperature at the edge 0/ the rod.
3. Heat dissipation
4. Fin effectiveness.
Assume fin end is insulated
Scanned by CamScanner
~
m ==
j kAhP
~ [m = 24.4 m-II
45 x 0.0376
25x1.13xl0-4
I
/
I. ~"d
Mass n'ollsjer
tan 11(24.4 x 6 x I o-~~
<I)
=:}
24.4 x 6 x 10-1
thin =
'llin = 0.61 (or) C~!..ill== 61
~
o/j
J) Tenlperalure llllile edge of the rod,
tall h (24.4 x 6 x J 0-2)
::=
Temperature distribution [short tin, end inslilated]
T - TX)
-.-t:s--~ 3
J. J
x ] o-~
J --25 x 0.0376
cos" III [L - xJ
= -------
cos" (/ilL)
cos II 11/ [L - L 1
cos" (IIIL)
,tSIl/I:J. flO• e fJicienc)" 11 = 6) %
2. Temperature at the edge of rod, T, = L == 350.5 K
T - 333
373 - 333
cos" (24.4 x 6 x 10-2)
T - 33J
373 - 33J
0.439
" J~eatdissipation,
Q = 2.48 W
4. Fin effectiveness,
E = ] 2.2
J.
___--:S
Soh'ed lJni~'~rsity"ProbJems
-----------
~-.
IT = 350.5 K I
o alliminilllll rod
All
Temperature at the edge of the rod, Tx = L = 350.5 K.
J) Hetll tlissiplIlitJll[shor:jill, em/ lnstdated]
Q = ("PkA)~ (T s: Tx» tall" (mL)
{ From NUT data book page no.49]
~ = [45 x 0.0376 x 25 x 1.13 x
10-" 1!tS x (373 - 333)
(Ii = 204 WlmK) 2ent in diameler and 2fkm
/ollg prolflllies fro",
a wall which is mainlained al 300·C
TIlt end of the rod is insulated and the sutface o/Ihe rod is
txposed 10 air III
Tile heat transfer co-efficient between
'hi rod stir/ace (II1dair is JOWl",] K. Calculale I/It /Ielillosl
~I' the rod and the temperature
of the rod at a dislance 0/
.woe
JOcm/rollltlll? WII/I.
[Anna Univ- June 2006J
x tan It (24.4 x 6 x 10--2)
19_:r-Q
-=_-~i@-8 W
Givm:
Thermal conductivity
of aluminium
Diameter, d = 2cIlI "'-,0.02 III
Scanned by CamScanner
rod, k = 204 W/mK
r
. N~ t! tlr 111,1 HI,I,I ., 'il/I. er
"
. II· I -
0 'II' -
\.
I\:\~
I Hil c\·nll"\'.
~I"\\
IlIhlhl 1\.11111''-'1,,1111
"I'~II1I;",~t~ I \.'"
11
o.zo "'
I,
_J:'I'
II'
l\)1\" \'
1'1\ -
I '
I
'"
~ I 1\
I-
II) \\1," "
"
1.1 I '" I
()_~IO.()'O('
x-'/Ol
1111
,..1 .I, Il'al h'~1h~ Ih{' hid, l.,.)
.
I\\d
11111 111:-1"11\.'\
1'1' 10\.'111 I I)
.
[<> ,... H),t)7 W
1,11"IO"IYi'('1I
1(1)
I
,VI) ~1I0\V III It.
t II, I - 0,"0 m
Isho!'1 till, 'lid in ..lllIedl
di~ll'ihlllinll
. 'flilif'
'i'I)III
10
I
_IIIIJ,(),III),I)
1
1111 II \\1111
S.Ilt4fil'" :
I hllllt'l 1'1'1' Ih\1 t II, II - 0,0 I III
I t'''~lh l,l'lh(
, 41
1'", - Ill" \ \ I i 1- 10 I "
dl \.'1"111,
\ 'I '11'1' \I'~'"11 \ 1\l'ill
"/,,
P' ,
)0
'()'hill/(L-
d
')1
cos II (",1 )
So, Ihis is short lin.
[From /I M1' III I book p(/~e 110. ·19 (Slxtlt t.:d:'/OII) 1
We knov that,
Heat transferred
[short fill, end insulated]
Q = (hPkA)Y: (Tb- Too) tan It (rnl.)
I)UIX
... (I)
~
[From HMT data book page no. 49 (Sixth editioll)}
= 1OcIII = O. 10m
h [3.13(0.20
T - 303
c
573 - 303
- 0.10)]
h(3.13xO.20)
where,
T - 303
A - Area
::)
= 2!_ d2
4
_ 7t
- 4 (0.020)2
IA = 3.14 x 10-4 rn2 I
P - Perimeter = 7td = 7t x 0.02
.,
[p = 0.0628 m]
~~-J
~
Scanned by CamScanner
0.8727
573 - 303
=>
IT = 538.63 K I
Illu!! :
I. Heat lost by the rod,
Q = 30.07 W
2. Temperature of the rod at a di tance of l Ocrn from the
wall == 538.63 K
-
Fm
1249
'''I; -
Fin . ThickneM.
4"''\'
t • 0.76 mm
Length, L· 1.27 em
WC kllow that,
Diameter of the cylinder, d - 5 rn » 0.05
Armo pheric temperature
1
= 45°
; 27
NUI1Ib<'r of fins = 10
Thermal conductivity
Heal transferred, 01 ::: (hPi<A )1/2 (T b- T (J"J) tan h (mLp
ITt
... (I)
3 ....318 K
",here,
of fin, k = 120 W/nlK
Thickness of the fin, I c: 0.76 111111 = 0.76 )( 10')
Length (heiglu) of the fin L/= 1.27 em = 1.27
Ileal transfer co-efficient. II = 17 W/m2K
p _ Perimctcr:::
::: 2 x \
III
10-2 III
ylinder surface It'lllp<'ralure
2 x Length of the cylinder
\p ::: 2m I
A - Area ::: Length of the cylinder x thickness
or
Ba e temperaiure
T h:: 1SO C + 273 == 4 3 K
Tofind:
::: IxO.76x\0-3
\ A ::: 0.76 x \ 0-3 m2
I. Rate of hear transfer. Q
-.
ernperaru--
at the end of the fin.
III
==
j kA
hP
Solution:
1
e'
ng h 0 he fin is 1._ em S
hi
. o. r IS IS short fin. Assumins
IS insulate
.
-
\ 20 x 0.76 x 10-3
10 en
m == \ 9.30 O1-1J
L.
Scanned by CamScanner
d Mass Transfer
1250 Heal an
we nee d temperature
Q, ::: (hPkA)Y2 (Tb- Too) tan h (ml )
.
f
(1) ~
::: [17 x 2 x 120 x 0.76 x 10-3]~)(
xtanh(19.30xI.27
(423
x 10-2)
::7
Tb - T ~
::7
..:.---:::Tb- r~
::: 1.76 x 105 x 0.240
Heat transferred per fin = 44.3 W
_
cos h(19.30
r-31S
x
423 - 3 IS
10
W
T-3IS
I Q, =443 wi
105
:=>
Q2 =hA~T
~:::
x 1.27 x 10-2)
_
1.030
= 0.970
••. (2)
Heat transfer from un finned surface due to convection is
= h (nd
:::
.-:-----:- -
Heat transferred for 10 fins = 44.3
=443
cos h (mlf)
T - r~
[Q, ::: 44.3 WJ
' put x == L
::: cos h [111 (l - L)]
~._
-318)
at the end of fin. So
,
4 19.94 K
I
Result:
Ley - lOx
t x Lf) x (T b -
I. Heat transfer,
To:»
[.: Area of unfinned surface = Area of cyl inder - Area of fin]
Q = 723.21
W
2. Temperature at the end of the fin, T = 419.94 K
= 17 x [(n x 0.05 x I) -
(10 x 0.76 x 10-3 x 1.27 x 10-2)] x (423 -318)
IQ2 = 280.21
Wi
2.5em diameter tube to dissipate the heat. rile tube surface
temperature is 170° C ambient temperature is 20" C.
So, Total heat transfer, Q = Q, + Q2
= 443 + 2S0.21
[Q = 723.21 wi
We know that,
T-T
Tb-T
co
Calculate tile heat loss per fin.
K == 200 WlmoC for aluminium.
Take I, = 130 WI",2 C and
[Madras University Ocl-99, OCI-2001}
G;Vfn:
Temperature distribution [short fin end insulated]
co
o Aluminiumfins 1.5 em wide and 10 mm thick are placed on a
==
~s h [m (Lf- x)]
cos h (mLj)
-,\,.
Scanned by CamScanner
'
Wide of the fin, b == 1.5 ern == 1.5 x 10-2 m
Th'
Ickness, t == 10 rnrn == 10 x 10-3 m.
~mt e er of the tube, d == 2.5 cm ::: 2.5 x IO'--m
' .
d Mass Transfer
1.252 Ileal (111
~c
temperature,
Ambient temperatureT
Heattral
o
r, ~J7~0.~: + 2:3:::: 443)( <,
00 -
isfer. co-efficient
Them1al conductivity
----~
------FillS I 2'
::: 130)( 0.05 x 200 x 1.5 x 1()-4]~-;---::':~
[xtanh(14.7xI.5XI0-2)
(443-293)
II)~~]
20 l. -I- 273 ::::293 )(
h = ) 30 W 11112 "C
~
k = 200 W/moC
I(
,1 ('lesttra
llsferred by fin, Q = 14.3 W
I·
Tofind:
. hi reclangular jill has II lengll, 0/ J5 mm I~. k
I
d "
.
lie nes«
TI,e tl,ernla con uctlVlty IS 55 IV/m"e T'l
•
11
I ",,1"
"
• I tiejill IS
t:J il.
10 II cOllvectloll envlronmellt at 20" C
~ set! ,VI",] " e. Ca Icu Iate the I,elll loss 1'.0 b(IIId
t-"pO O"/j
J"a~
SO
h$ll
.F/50"e.
Heat loss
..A;"
A sll" "
Solution:
Assume fin end is insulated; So, this is short fin end'
Insulal
type problem.
ed
r,,'UreoJ
Heat transfer [short fin, end insulated]
'" (I)
~:
where
A - Area
=
P - Perimeter
data book I)
age 1I0.49J
Lellgd1,
[Madras Umversity Apr-1002j
L == 35 nun = 0.035 m
s t == 1.4 mm = 0.0014m
Thicknes ,
1.5 x 10-2 x 10 x 10-3
conductivity, k = 55 W/mo C
Thertna I
0
Fluid temperature, Too = 20 C + 273 = 293 K
1.5 x 10--4 m2
Base temperature,
Breadth
x thickness
2[(1.5
I
x 10-2)+(IOx
IO-3)}
h = 500 W/m2 OC
r,p.d:
Heatloss, [Q1
;Wlllioll :
= JhP
~ Since the length of the tin is 35 mrn, it is treated as short fin.
kA
'~sumeend is insulated.
130 x 0.05
200 x 1.5 x 10--4
14.7 m-I
Scanned by CamScanner
T b = 1500 C + 273 = 423 K
Heattransfer co-efficient,
I P = 0.05 m
m
I
2 (b+t]
=
.
It"rpe
Q = (hPkA)Y2 (Tb- Too) tan h (ml.)
[FromllMT
I
I
Heattransferred [short tin, end insulated]
Q = (hPkA)Y2 (T s: T~) tan h (mL)
..• ([)
[ From H MT dol a book page 110.49J
Fins 1.255
_ 2 5 COl ::; 0.025 rn
d d- .
{tlte ro '
::; 0.16 m
eler 0
J..,:;:::: 16 clTl
/.254 Heal and Mass Transfer
where
P _ Perimeter
~
A-Area
[A
==
==
2 x Length (Approx'
11a e
2 x 0.035
ll t ly)
==
0.07 m
I
==
Length
x thickness
==
0.035 x 0.0014
==
4.9 x 10-5 m2
_:
6j,~Oi~(1l {the rod,
ase te
..",81 COo
I
I"
1Ifi~d:
bY the rod.
Heallost
1IIl11ioll:
.aJ11eter
of
500 x 0.07
==
113.9
m- I
1
Substituting h, p, k, A, T b: Too, m, L values in equation (I)
[500xO.07x55x4.9x
L
_...QJ2_ == 6.4 < 30
d - 0.025
. .. h rt tin. Assume end is insulated.
So, thiS 155 0
I
We know that
Heattransferred [short fin, end insulated]
10-5]~x(423-293)
... (\)
Q == (hPkA)I/2 (Tb- Too) tan h (mL)
x tan h (113.9 x 0.035)
I Q 39.8 w·1
the fin, d == 0.025 rn
DI
- 0 16 rn
Length of the fin, L - .
55 x 4.9 x 10-5
(I) ~ Q==
T:;::::2
J11perattlre, b
::; 160 C + 273 ::; 289 K
erattlre, Too
2
e 1)11d'og
tel11P
a-:'
t h::; 15 W 1m K.
I
f r co_eulclen ,
surro
h at trailS e
vective e
. k == 190 W ImK.
Con
dtlctiVlty,
I;en
rite 1
m
[m
600 C + 273 == 533 K
gthO
where
==
P-Perimeter=
Result:
Heat loss, Q == 39.8 W
I P = 0.0785 m\
[I] An aluminium rod 2.5 em in diameter and 16 em IO/lg
protrudes from a wall which is maintained at 260"C The rod
is exposed to an environment at 16"C, Tile convective heat
transfer co-efficient is 15 Wlm1K. Calculate tile Ileal/Osl by
tile rod. Take k = 190 WimK.
-:
[Madurai Komara) University Apri/-97}
Scanned by CamScanner
xd = n x 0.025 = 0.0785
A-Area
=..!_
4
d2
= ~ (0.025)2
~ = 4.9 x 10-4 m2 \
m
r
"6 u t 011(/ Mass transfer
"--I. ]) ne«
III =
R
Fins 1.257
______
---'-',
(Jf
ferred. ()
pi'
ItrailS
~e8
=
{15- x 0.07~
1190 4.9
x
[m
x
de length is 8 ern, it is treated as short fin A
he bla
. Ssume
SiOCe I d
.
su ~te . s: red 1 short fin, end msulated]
JIS iO
1/
'Q~:
10-4
f
= 3.5 1l1~
He
8
t trlll1sler
PkA)~(Tb-TCXl)tanh(mL)
... (1)
Q::; (h .
(I) ~
Q = (15 x 0.0785 x 190 x 4.9 x 10-4 ]~ x (533
x tan"
(3.5 x 0.16)
[From HMT data book page n0.49]
- 289)
where
IQ = 41.03 w.j
m
==
lesult :
Heat lost by the rod, Q = 41.03 W
1*
kA
465xO.12
==
I!l A turbine blade 8 em 10llg made of Sillilliess Sleel
32x 4.75 x 10-4
[m == 60.5 m- I
I
(It = 32 Wln,K) has a cross sectionat
area of if. 75 em} tIIltla
perimeter of 12 em: Tilt! bast! temperature of the billtleis
6IHr C Find the qualllily of I,eal given to blade if in the blade
is exposed to Ilot gtlses 850"C. Take heat transfer co-efficient
to be 465 WlmJK.
I) ::>
(
Q == [465 x 0 12 x 32 x 4.75 x 10-4]Y2 x (873 -1123)
x tan h (60.5 x 0.08)
[ Q == -230.2 ~
Given:
length of the blade, L = 8 cm = 0.08111
Thermal conductivity, k = 32 W/I11K.
[-ve sign indicates that heat flows from gas to turbine blades]
ltsuJl:
Area, A = 4.75 cm2 = 4.75 x 10-4 1112.
Heat transferred, Q = -230.2 W.
Perimeter, P = 12em = 0.12 111
Base temperature, Tb = 6000 C + 273 = 873 K
Hot gas temperature, T'Xl= 8500 C + 273 = 1123 K
H
eat transfer co"efficient,"
.
= 465 W/1112K.
-l
S 6 em diameter
in the form of a eyI'inaer
I
.
.
. 20 Iong!'tudinalfins 3. mm
I and 1.2 m long. It IS provided
With
J
.
.r.
Of the eyltnuer.
Ifllck whicl, protrude 50 mm from tire sur,aee 'J
bi t
. 800 C The am len
The temperatllre at the base of the fin IS
•
'jJ
'11
Scanned by CamScanner
•
'.!J Aheatmgunit made
J.2j
i
Heal and Mass Transfer
'n"rnlllres is 2.~"C. Ttiefio» heat 'ral1\·r.
rem,."'.
''Jerco_
l e el,lillller and JIIIS 10 'he :H"rolilldillg
. :lJici(!11 -,
"
.
(Ur IS J
1ft
C leulnlt the rate of heat transfer from 'heli
0 If;/
(/
11111e(/
",1,.
_IrrOllll(lillO.
Take k = 90 '~/'
1 'ImA.
1v(1//, ~.
'
I
V"
0/
"
[ Man0l11110nilllll SlIl1daranar
V
o.
•
(lfl ell
()~I
jl
f t Iie fin is 50 mrn. Assume end is insul a tiS
en.
I'flgth
d IIlSU
. nor1
11lJerSily No
if)
tan IT (mLf)
'"
(J)
~ '9(J
Gil'en:
l}
(From HMr datn book page n0.49]
Oiameter of the cylinder, d = 6 ern = 0.06 rn
\vhere
Ley = J.2 III
Length of the cylinder,
.
J11e
p- Pen
Number of fins = 20
ter === 2 x Length of the cylinder
===
Thickness of fin (t) = 3 mm = 0.003 m
Length of fin, Lf = 50 mrn = 0.050 m
Base temperature,
===
A - Area
)<
x
J11
~hP
=
J kA
Rate of heat transfer
j
Sotution :
I
III
L--
~~
(I) ~
~-r-r-r-~
=_
IOx2.4
90x 3.6 x 10-3
~~f2J
0, = [10 x 2.4 x 90 x 3.6 x 10-3)12 x (353-298)
x tan h (8.6 x 0.050)
01 = 62.16 W.
Thickness,
t = 0.003 J11
Length, L = 0.050 m
{-
Scanned by CamScanner
offin
[A = -3.6 10-3 m21
k = 90 W/mK.
~~~
x thickness
1.2 x 0.003
h = J 0 W Im2K
Toftnd:
d=~jomrt_--
J.2
Length of the cylinder
T IX = 25° C + 273 = 298 K
Film heat transfer co-efficient,
Thermal conductivity,
2
[!_2_.4~
T b = 80° C + 273 = 353 K
Ambient temperature,
. -i-
IS
.
Heat trans
1/.
Q, ::::= (hPkAV2 (1 b- T
VI .
'
lshort fin, end IIlsulated]
ferred
1,5
Ip
0 thi
lated Iype problem.
Heat transferred
per fin = 62. I6 W
1.260 Heal and Mass Trans er
Fins 1.261
Number of fins = 20
So, Total heat transferred,
0, :; 62.16 x 20
:
6jl'tJI 1.alTlcter.
10, :; 1243.28~
Heat transfer from untinned surface due to Cony
0
.
eCllon'
t -x: 3 rnrn :; 0.003 m
. kneSS,
le
In
L === 20 111111 :; 0.0 2 0 111
IS
1,,"l1oth,
02 = h A t1.T
c;:>
k= 45 W/mK
. .
co-efficient, II = 100 W/m2K
etlan
nve
Co
_
erature. 1 b:; 120 C + 27J = J9J K
P
Base tem
d'10 temperature, T C1J = 35° C + 273 = 308 K
0
[ .. Area of un tinned surface= Area of cylinder - A
. .
reaoff!n)
]0 [x x 0.06 x 1.2 - 20 x 0.003 x 0.050J f3'~
,J,)
-
298)
[o_ = 122. 5 W 1
Surroun
r,fif/d:
I'
Q=Q,
I conductivity.
Ine(1J18
= h x [11 d Ley - 20 x t x Lji (T, - TyJ
So,
Total hear rraasf e t,
d === 50 111 :; 0.050 rn
I. Heat
+Q2
~.."
II::>
n 0\ rate per fin, Q
Fio efficienc
Q = _43.28 - 122. ~
..,
,
, Fin effecti ene s. E
).
Q = 1366 If.!
I
)4IJlliDfI :
20 rnrn, it i treated as hart fin.
; e the fin len
Inlllt :
urne
~: '. ilD lated,
Heal rans err
!l A dlcum/uential
rtC/ungulor profile fin on a pipe of
5fJmmlluJutli!uneler nJ mm 'ltick uno 20 mm Mn;:. Thermal
fJntlucti.,il i. 45 WlmK. Convection
co-efficient
is
100 W/",1X. Bote temperature i, /2(1" • and urrounding
fllr lemperu/ule I. JJ" C Determine
I. /legl jll'if41 rut.eper [in
0:;
hP
[sbo
fin, en i ula ed]
)Y2(Tb-T'f)Lanh(mL)
[FfI'Im HMT d. /0
P _ Pcrimc -r
-
[P -
11 /
0.050
(J, J 5 7
mJ
2. FIn ejJkkn(:,
2
. f'lli tj/et:II'Iene.u.
I M(lnriMlflnllHrI /:/H1d(Jf(~nar I/nM;rKlly, /I/{JV ()6/
(
/,/r I
Scanned by CamScanner
..!.. ((),(J5(J P
.. '
Q(j.
I)
page!1O ,')/
)$·'':
s For Practice
~
[t.
J.QO x 0.157
-1 - ,---.:...:.__
1.96 x 10-3
'_3.3 m-
......
1_11_1
O.157
J
. . 1{llet
. 0
(f//II
1/ d/·I
.
x tan" (13.3 ' 0020)
,.-__
IQ
~J
_5.9
Fin efliciell~Y,
2)(
.
'l =
(39'
.J -
I (If II Inllg .' rod
ell~ 'J
1 Otll!
bv pl{lcillg
x 10-) 1y,
J(8)
'1=
Ii (13.3
1llL.
-
Fill effectivelless
tan II IIIL
----_
..
jf~
Etl'ectivcnss, E == 1.56J
Re.m/I:
tan h (13.3 x 0.020)
/-'00
I em dml1leler
b
if ",lIilllfll'
'llel
=:
x
1.96 x 10~J
45 x 0.157
jill (k = 200 WlmK) 3 mm thick 11/1(/7.5 em long
prolrudesfrom II 11',,/1 tu JOO" C The ambient temperature is
50·C wit" " = 10 WIlli] K. Compute heat loss from the jill per
IlIIil dept" of the material. AI!;() calculate its efficiency 11IIt!
effectivelless.
/ AilS Q = 359 Wlm T] = O.9J7, E = 4UI
I, All"llIlIIinilim
-
i. A one meter long, 5 CI1I diameter cylinder placet! ill {I atmosphere
I. Heat flow ' Q -).
- 2' 9 W
w--(
2. Fin efticicncy n - 9
,
'" - 7,7 %
3. FIll dfectiv
cness, E ::: 1.56
~/
Scanned by CamScanner
,
tJ/ 20 ""
ami J ""~' buse ~/lic/(//e.H. Tlter",a/
CIIJI(IIIc1iv,Iy ts 45 WlmK. Convection
coefficielll
is
100!VIm]/(, base temperature is 120vC. SlirrOll/lllillfJ fluu!
((",pertltllre is J5" C. Determine also jill effectiveness. Use lite
eI/nrl
/AII,\' I. Rectangular jill Heat flow, Q == 285 W
Fill effectiveness, E == 11.6
2. Triallgu/ar jill Heat flow, Q = 268 W
Fill effectiveness, E == /(UI
'ltin = 97.7 ~
E==
•
, Oder",;ne lite Iteal flow for-ti} rectangula» fins (ii) Triallgular
x 0,020)
13.3 x 0.020
.'
of
[Ans k = 45 WI",KI
tall" ",L
.I, fill
tall
(II
'C
I
'J'
KI
(It
It 1/1 a furnace.
The rod is 1!\/7o"e(11
.' • (I"C •
....
.
...
() utr. (It
,!(/
'(11 (I c:ollveUIOII co-efflclelll
of 15 WI,II1K .,..,
"I' C wt
.
. t ne
.1/'
Illlfe
melHllred
III II dlSIIlIH:e
0
78 6 ""
"perl
,
:J"
•
n Will'
l de Delermille
(t
lite thermal eOllllliClivilV Of lite "'(II
. I
141.5 •
. '.I
ertat.
/JII'
'
I'.
39.1 nuujrom tue Iurnace end. D'I
.
'.
•
e trlll"'e the
er{l/llre oj the rod.
{Ami T == 773
. "IIIIICe
45 x , 96
()
Jj
I'I! lell'P
=> Q = [I 00
-
1ertl
r,.ob
.-J.9
.fIlII" rOil I em diameter
1IfI,Villa a Ilte----'
"
d II] II
""(1 COlltlU'I'( IVtty.
I/ttll
II' i\·,1i(lced 1/1 a furnace,
Tlte rod is '
V W/IIII\ '.
exposed 10 .
/'H' 0 er its !I'lIr/act! (l1Il1 the COlivcction
'{','fi' tur (/(
Ilj • C
I'
, •
.
co-e tctetu .
/6(1 . I (II 1.5 I,Jljm'" R.. TIle lemperullire
is relit/liS 2651) IS
11/40· C is provided
with 12 IOllgillll/illa/llrfliglll
[ins
(k:::75.6 WIIIlK). The fins art! 0.8 111111 thick and protrude
~.5('111/rolll the cylinder surface. T"e "ellllrall!Jjer co-eflil.ielll
IS 23.25 WI",] K. Catcutatc
lite rate ofhea! IramIe, illlle mr/(Ice
(ell/pert/tllre is 1JO" C
1;1/1.)' Q = /170 JIll
Transtent
. Heal ( ,
1.264 Heal and Mass Transfer
-ondUClion 1.265
}.5 TRANSIENT HEAT CONDUCTION
UNSTEADY S!ATE CONDUCTION
0
( Q.)
where
hennal
k- T
If the temperature of a body does 110t vary .-~
.
"
With ti
.
to be IIIa steady state. But If there IS an abrupt cl
.rne, It is .
.
.
lange In .
sal~
temperature, It attains a steady state after SOlne p'
lis s~r[
.
.
.
ertod. D . aCt
period the temperature vanes with time and the b dv : lJrlng tho
.
0 )'Issa'd
II
an unsteady or transient state.
I to be in
Transient heat conduction Occur in Cooling of
automobile engines, boiler tubes, heating and coolina f Ie enginel
•
•
t> 0 metal b'l
'
rocket nozzles, electric Irons etc.
I lets,
Transient heat conduction can be divided
flow and non periodic heat flow.
in to
h _. Heat transfer
Charac
I
varies on a r
I
.
L - Thickness
A
::J__ _ A x L
Lc~
~ \
A --:;;:-
of the slab
egu ar basIs.
Examples.' Cylinder of an IC engine,
For cylinder:
Surface of earth during a period of 24 ho urs.
Characteristic
V
Lc = -A =
length,
(ii) Non periodic heat flow:
In non p~no. diIC h eat flow, the temperature
~
at any point within
with time.
Example .. Heat' mg a f an
an i
.
ingot 111 a furnace,
~
cool ing of bars.
The ratio of int
I
.
convettion'
erna conductIOn resistance
reSIstance is k
nown as Biot number.
Biot. umbe r -- _Intemal con ducri
.
uctlOn resistance
Surface co
.
nvectlOn resistance
8.I ---.::.
- h Lc
k
Scanned by CamScanner
where
R - Radius of cylinder
1.5.1 Biot Number
to the surface
For sphere:
Characteristic
tI
ength
Volume
- V
Surface Area -
_
Lc -
.
length,
n
where
In periodic heat flow, the temperature
the system vanes non-linearly
W/m2K
length or Significa
. . Ien gth , L c -terlstlc
for slab :
..
Characteristic
WlmK
co-efficient,
Characteristic
Lc -
'.
penodlc heal
(i) Periodic "eat flow
..
conductivity,
lenzth
Lc = V
0'
A
n:R2 L
2n:RL
hA (T - T ) = p
. 'ble
Ne~"!'
V ciT
cit
Internal
where
.
Resistance
R - Radius ot the sphere.
For c"be:
Characteristic
length, Lc
T
V
A
T:::: To at t = 0
LJ
·
Lumped/,etd
fIg 1.13
6L1
.
· 've heat \
convect I
bodv
fronl t h e
'
\Lc~ ~]
.
5
::::
-
-::::>
1.5.2 Lumped beat Analysis [Negligible
internal
In a Newtonian heating or cooling
is assumed as
resistance
process
is considered to be uniform at a given time.
called lumped parameter analysis.
is known as
the temperature
Such an analysis is
p X Cp x V
dt
-hA
respon
determined by relating its rate of change
conve ti e exchange at the surface.
energy with
-hA
~
-
~
/11 t -T
dT
x \ cit
T - TL
b und
t + C\
ry c ndiu
At t =
e of the body can be
of internal
x dt
Integrating
r\ppl
Let us consider a solid whose initial temperature
is To and
it is placed suddenly in ambient air or any liquid at a constant
\.&0)
n·
0
III t o-T \==
· lib titutin
Scanned by CamScanner
energy
pCp V
T-T~
resistancel
resistance
negligible in comparison with its surface
Newtonian heating or cooling process.
T~. The transient
.
dT
dT
temperature
ot
-::::>
L - Thickness of the cube.
The process ill which the internal
Rate of change
internal
-hA (1' - T F)
where
ctlp"cily system.
jj
C\
. .. (\ ,80)
Transient Heat Conduction 1.269
_ Lumped
Heat Analysis
bleJ1lS
I,ed fro
1.268 Heal and Mass Transfer
I SO
~ In [T - Too]
-
-
-hA
p Cp V
t +I
11 [To -
-------T <xlJ
~
y
2 alum'
'"illm slab of 6 mm thick is at 400· C
,
,
50 c"',
d /ellly immersed In water, So Its surface
;0 ~
illS SU (
"
~li',/1" gJld
ed to 50 C Determine tIre time required
IV '/1# /
's lower
~I
Ilre'
DC
~~(r"t
reacl. 120 '
I ~~ lab 10
I for/h{ S
fer co-efficiellt, I. = 100 Wlm1 K
D
=> In [~ ,-_ T.;:.
Mal ,ra"s
I ra~{
,
_ SO x SO em:! = SO x SO x 10-4 m2
:
. OS-
~ Di,(enSlo L'" 6 mm
= 6 x 10- 3 m
. knesS,
,hie
T = 4000 C + 273 = 673 K
.' temperature,
0
Initial
T = 50 C + 273 = 323 K
. I temperature,
co
na
fl
.
mperature T = 120 C + 273 = 393 K
mediate te
'
Inter
f
o-efficient, h = 100 W Im2K
Beat trans er c
, , , (1.81)
0
I
where
0
To - Initial temperature of the sol id, K
T
- Intermediate temperature
of the solid, K
Too - Surface temperature
of the solid (or) F'
temperature of the solid, K
Inal
- Heat transfer co-efficient, W/m2K
A - Surface area of body, ml.
lIioll :
-
Specific heat, Cp = 896 J/kg K
I. In lumped parameter
than 0.1.
i.e.,
T
are
Density, p = 2707 kg/rn '
Note
2. To
{From HMT data book page no. I}
Propertiesof aluminium
Speci fie heat of the body, J/kg K.
- Time, s.
Cp -
t
0
I
I
Density of the body, kg/rn-'
V - Volume of the body, m3
p
'fiwd:
. (t) required to reach 120 C
Time
h
Bj <0.1
-Initial
~
system,
hLc
-
temperature,
- Intermediate
k
Biot number value is less
Thermal conductivity,
k = 204.2 W/mK.
irllab,
< 0.1
Characteristic length
K
.
(or)
L
Significant length, Lc = 2
temperature, K
Too - Surface temperature or Final temperature,
~'~-=:m-=~
Scanned by CamScanner
__'~_L-
K
.J_
1.270 Heal and Mass. Transfer
where
H
_------71:.:.r~alnsielll
_______
eat CondUCTion 127
=ue:
\,/~cre
L - Thickness of slab
K
10 - Initial temperature,
L = 6 x 10-3
c
2
.,.. _ Final temperature
1<1)
1 - Intermediate
I
I'
Ii
,
K
h _ Heat transfer co-eflici('nt
We know that ,
C - Specific
.
hL
B lot number , B.I = __k c
p
,
W /1ll2K
heat, J/kg K
Lc - Characteristic
p _ Density
K
temperature,
length, III
kg/rn!
100 x 3 x 10-3
t - Time,
204.2
Bi = 1.46 x 10-3 < 0.1
S
393 - 323
~iot number value is less than 0.1. So thi "
analysis type problem.
' IS IS lumped heal
(Il :::>
For lumped parameter system,
-1,609
(I)
[From HMT data book page no, 5 i (.'iixlh editioni]
-100 XI]
673 - 323
In (0,2)
'"
[
= e 3 x IO-J )( 896 x 2707
-100 x 1
3 )( IO-J x 896)( 2707
-IOOxl
3 x 10-3 x 896 x 2707
G" .~ 117.1 s I
Nfsu1t :
Time requ ired for the slab 10 reach 120· C is "7, I s,
We know that,
Characteristics
T - T~
(I) ~
"C
length, I.e = :i._
A
[Lc:~p,p"'1
To - T",
Scanned by CamScanner
o A copper rod of oilier diameter 10 mm illilia/~I' a
(I'
umperature of 31J(}UC is sllll(lell~11immerseti ill a water
allOO"C. Determine the lime reqllired/or Ihe rOt/IOreach
2 J O"c. Tali e COli l'eCI;W! Ire(II ITlIIIS/tr co-e//iciol' is
9J WlmlK.
.'.
Transient Heat C
onduc(
value is less than 0.1. So th"
Ion 1.273
Inbcr
'
IS IS lu
,
'011111 blerll.
Illpedheat
l
I'
e pro
. :; I)'P d parameter system,
,~~l~"C(l1and Masj' Transfer
()'/,·.,,, .Dlnmctcr of the rod, D = 20 mm - 0.020
Radius of the rod, R = 0.0 10m
Initial temperature,
Final temperature,
e
In
I ,.,1
To = 380° C + 27 _
3 - 65" K
T = 100° C + 273 ==373,)K
x
t]
... (J)
~~e
T ==210° C +
fo - T
273==483
h ==95 W Im21(
[-hA'
pCp V
IX)
Intermediate temperature,
~.'
Heat transfer co-efficient,
C
/'fo( 1lI(llP
[F rOI1/ H MT data b ookpa
0)
ge nO.5 J
K
n'llntl:
Time (t) required to reach 210° C
Solutlon :
[From H,\fT data book
Properties of copper are
page "o.l)
Density, p = 8954 kg/rn!
Specific heat, Cp = 383 J/kg K
Thermal conductivity, k ::: 386 \\ ImK.
-95 x t
In (0.392)
For Cylinder.
Characteristic length, Lc :::
R
2
5 x 10-3 x 383 x 8954
[1
=169.03s
::: 0.010
/JjMlt:
Time required
2
We know that,
.
hLc
B lot number B,'::: ,
k
to reach 210° C is 169.03 s.
0
] A 5 em thick copper
slab is at 200 C initially and it is
suddenly immersed
in water. So its surface lemperature is
lowered to 90° C. In one test run, the initial temperature is
::: 95 x 5 x 10-3
386
I
Bj ::: 1.23 x 10-3 < 0.1
Scanned by CamScanner
I
decreased by 40° C and the time taken is 6 minutes.
Determine the heat transfer
capacity method
of analysis.
co-efficient by using lumped
't
/-:
. kl1(1'
I'.L .•:'''"~
.
If Iii II
I
il :lllenrper:llure.
\I~
III
lensllCS
-11
T; :::90
0
.
I
I L - V
engt 1. c - A
.'
(hMae
T ::: _000
erarure.
Transtem Heal C . I
______ --------------~~~O~I~Ir.~II~C/~I~OI~/~/
== e [ P x Cp x Lc x t
1 _T
1
.:.------:;To-' fcc
~3 - 363
Ti ie, I = 6 min = 360 s
4
473 - 363
[
-11 x 360
]
8954x 383x 0.025
~
Tofind:
Heallransfer co-efficient.
h
/11 (0.636)
{From IiMr
So/ution:
== e
-h x 360
== -----.::....:...::.._-
8954x 383x 0.025
d,
ata book
Page no.•}
Pr perties of copper are
107.77 W/m2K
~
Density p = 8954 kg/m-'
Specific heat, Cp = 383 J/kg K
Thermal conductivity,
For
,
JtS,,/1 :
k = 386 W/IllK.
Heat transfer
h == 107.2 W/m2K.
co-efficient,
Slab.
Characteristic
length, Lc ==
L
B.A solid copper cy/:"der
2
For lumped
==e
0.05
2
Lc
==
Lc
= 0.02)
again measured
== 0.025
III
I
(IS
J
O
of tile cylinder is
C. Determine
unit surface
conductance by using lumped heat analysis method.
Givtll :
Diameter of cylinder,
parameter system.
-hA
[ pCp V
... (I)
{From f-f "'"T data book page no 5ij
Scanned by CamScanner
diameter is initially at a
water:After J mill utes tile temperature
where L - Thickness of slab
I
of _7 ""
temperature of 25 C and it IS suddenly dropped into ice
D == 7 em == 0.07 m
Radius of cylinder,
R == 0.035 111
Initial temperature,
To == 25 C + 273 = 298 K
Cylinder
ill to ice water. So, final tempcralure
0
is dropped
is 0 C, i.e., Too == 0 C + 273 == 273 K
0
0
6 Heal and Mass Trans er
__
---...:7.:_:ra:::n_:::s,_,i"ent
lieal •
Conduct'
[-h
Time, t.,
min - 180
/
774-273
::.:-:----;::;:
nit surface conductance,
=el~l<180\
h
So/"t;o" :
[From flMT d
Pr pcrties of copper
ala
book
p 10
.,c 1)0 ,
are
,]
[h
Specific heat, Cp = 383 J/kg K
unit surface conductance,
'ylinder.
haracteristic
Lc =
length,
R
0.035
=O.OI75m
=e
L
L
MasSof the sphere,
m = 7 kg
sy rem,
Initial temperature,
To = 320° C + 273 == 593 K
t]
Final temperature,
Too = 25° C + 273 == 298 K
In
T-T
'
•
Girt" :
I Lc = 0.0 I 75 1
F r lumped parameter
h == 1073.21 W/m2K.
aillm;nium sphere weighing 7 kg
d'"
] A"
an Initially
at a
. ",peralure of 320 C 1.'1 sUc/tieniy immers d .
'.
Ie.
e In CI lIqUId at
25'C. Tile cOllve~/lve heal, transfer co-efficient is 50 W/ml K
Valermine the time required for the sphere t0 reuc II 100'C'
0
2
-2
1073.2IW!m2~
=
Ili"ll:
k = 386 W/mK.
Thermal conductivity,
-h x 180
383 x 0.0175 x 89S4
/11 (0.04)
Den ity, p = 8954 kg/m!
[-C-_I~......;.~-x-p x
P
, .. (I)
To - T
[From I-IMT dat a book page n0.57j
Intermediate temperature,
T = 100° C + 273 == 373 K
Heat transfer co-efficient,
h = 50 W Im2K
Tofind:
We know,
Time requ ired for the sphere to reach 100° C
Characteristics
Solulion :
length, Lc = V
[From HMT data book page no.l]
A
(I)
I
'
1.277
298 - 273
Tofi"d:
For
IOn
~
T-Too
Properties of aluminium
JCp:l~c' "1
Density, p = 2707 kg/m3
p
Specific heat, Cp = 896 J/kg K
To-Too
Thermal conductivity,
J
Scanned by CamScanner
are
k = 204.2 W/mK.
I
11{f/Jf I till
fl71.!
" • ~.",
f
110
1/1111'/1'1
Itl~'
'"'/
IIDif';';
1M ~ 1/ I'
/1.lmlr
(
/ ,I
/'
;t
/
,
, ,,
II'
II
'J
tlly'!
III ~ "1 {
'Z'1,/
"
4
/
"'h
. 'J II - '''/,
1 Jd
~/
'
-'
(),(JX5 rn
373 - 2<)~
593 - 2c)~
In (0,25 ) -
For Sph('fe.
I'
h' t: Cti" ic leng h, Lc
-- - SO ----
1 /
It
3
W)61' ()'(J2~31' 27()7
c
-50
~961' 0,0283/
2'1(J7
/
/
=1881.33
O,DRS
---
ltlU/t :
3
Lc
'~I (;
Time required
= (J,(J2IO m 1
to reach
hat,
Bioi
hLc
umber. . 13·I =-k
made
of copper of dtumeter 6 mm
of 2(r C These spheres are
annealed in a annealing
unneuling furnace
50 /. 0.0283
for the sphere
204.2
13i = 6.92 /. I 0- 3
ana Iysi
furnace.
The temperature of the
L\'450" C. Calculate the time required
10 reach
the temperature
heat transfer co-efficient
It = J 0 W/ml K.
of 320 C Take
0
CiVet! :
O. j
/3iot number value is less than 0.1. S ,this
type problem.
s
initially at 1I temperature
~ Thousand sphere
cnow
IO()O • is 1881,33
is lumped heat
umber of sphere
= 1000
Diameter of the sphere,
D = 6 mrn = 0.006 rn
___(
Scanned by CamScanner
1iIIl4"" 'L...
1+,
,u~ . \1,' .11lt' c , R = l. OJ 11\
P
. 1 11,11'
It
UI, I \.:,'I :r:\1 I e.
Inl nll('dinle
II '.
Y
T ==
lellli ernture.
Ir:\H~fa'
.n'·
r = 4' 0"
")
-
-effic ient. h ==..,
,
J;:::-
W 11ll- K
'I(
, red l)fHllmeter
ime rec uired
10 reach
r'" _0° c
the leillperature
S{lIUlilln:
(Fr m H.I(T d I book
crt ies
f
en ·il~·.
-r
I-I
_:----;=-
==e
-hA
[C «v « P
P
... (I )
(From HAn dala b k
page no.5il
, 1'1,0'" that
\\ ~ I'
c,aI racteristics
Lc == "!_
length,
A
pper are
== S
- 4 ,,::/111'
pC'il'heal.
(
The mal
/-'0
S
It
ndu livir).
'8
,,==
W/mK.
-30 x t
593 - 723
293 - 723
.
R
::::>
--
.
j
111 (0.302)
-30 x t
383x 0.001
\1
136.87 s I
x 8954
== 0.001
1m
Rt'Hllt:
Time requ ired to reach the temperature
is 136.87 s
~o " 11.00 I
Jso
().I I
Scanned by CamScanner
',
of 3200 C
·'1~1.. ~
",I,
• t. '"
y. tern,
f r Il.Iln
00
:
..
I'"
,ber value i
10-
Ti) fiff"
.. ,.,
"problem.
. I) II;,
51
1
\
-
,
>h....
/.lBi lieal an d~ Mass Transfer
Transien; He II
_
, d ;cal stuinless steel i"l:0t 170 mm il',.
r:7I A ('ylm r
~
Lc
_
== 0,0425
OlldllClioll
1.2 J
I~
( '(11"1.'1
I "" pll,ue:i t",ollg" II "CIII tremmelllji',r
e'r'''rl
'50 em I} t.I
,
11(,(,(, IV '
10 "" le"gt", T"e tC'mpcrm"'e of '''C'ji",
/"C/,
;,~5(1("
'"
'It'ce • "
I(Idc.
I,ft h
('''III
"
1,\
Tile ;",'1111 1111:(1(Icmperfll",c
I'
' ,I ,mlim" 11111 CI)l/I'C'C:/'I'eleflllrall,\/er
"",.
I~II'IIII} K, C.,lelll,,'e
'
I ,,""'(',~ IImJIIJ:II
"'1/"
""
1/1<1'"111
,o'c{fi
C
~("113'
C'. 'fl
"
==
tlu: "'"xi",,,,,,
a is fI, 46 x I (h~ III}/ ,\',
\),;,,"cl~'r
",h,I'"'~, I,,'III\,
,
~ll.'d I\ld, R
I 'lI).tlh, I
511em
1\11II;IIe"-' kll, Ih
l)" 0 1\\
IHIllal klllll\,f:III1' '. \' II
I-'i II ,11 1,'11 I ,1:11111
:111 ~'" 1,1
O.\IS:i III
,r",
for
I
\:11'1"" ;11111 -, r
"
N\ \ " 1...'
~ == c
. T
1"1"\. "\. ,• "I,',
1 J 1 I"~ ,
I:':P. l ~::~v '1
-
:i 1\1
~O \\) 1\\1\\
Il)
I
_;
I { '~
K
r r.
~\ - r,\,
,I~ \\ HlI\.
.... ~
-
t ,l :-;,
Scanned by CamScanner
[Fran, J /.1 IT 1,1/I I 01) • I .~,J III), '"J
,I
'
\:i \\' '"'1\.
\. I
0, I
' I til' 11\ I,
ue'r value is less than 0,1. So, this is lumped heal
H'O 'problem .
."IYP~
~~,~
I'
II\~'.
cd purruuetcr s S em,
"I' Ih\' steel '\lil, l) .. I 0 111111.. 0, I 0 'II
L\,
s, = 0.099
(I"tl
(l'i,·",. :
105 x 0.0425
45
I I'
"",111('111"'1'" III '\'(fllllle,\'~',"/I'cI is, 4 ,~ .JlI "'I( (lAr
,h"fllUI I t lillll~iI'i'"
\,'
,
hI
'r
'\'J1ced '\I' ," (Ir",
',II
, "'/'i,
11",./ ""'''Ce 1110 fllI(lill 800"
e/,
C'. l' ,
H,
IIIf!
is 12()"
I
~ ,
, II - $,
13i - Hi )1 HlIl\1\ :1'
.. ,( I)
Transient Heat Conduction 1.285
T == 20° C + 273 = 293 K
rature, .
tc(11pe
ature, T = I SO° C + 273 = 423 K
'f)1II
te!11per
fl
d'ate
.
r!11c I
nsfer co-effiCIent, h = 110 W/m2K
1~le
. e heat tra
ccttV
3
cof)V
"" 7850 kg/m
ef)sitY, P
.. ty ex. == 0.044 m2/hr
O
diffuSIVI ,
fhertllal
[; == 1.22 x 10-5 m2/s
r:f)
1.284 Heal and Mass Transfer
~
-0.099
In (0.423)
~
,--I
Ingot speed
=
x 0.46 x 10-5 ~
(0.0425)2
x t
3_4_1_2_.5_3_s--,'
1__
I
Furnace Length
Time
C == 474 J/kg k
.fic heat, p
CI
Spe
d ctivity, k = 43 W/mK
Icon u
erJlla
'fh
5
3412.53
I ingot speed = 1.465 x 10-3 m/s·1
(0
Result:
sphere to reach 150° C
filld: • e requIre. d for the
.
s heat transfer at I SO° C
tantaneou
2. Ins II eat transferred up to 1 SO° C
3. Tota 1
I. TIJll
Ingot speed = 1.465 x 10-3 m/s.
[!) A mild steel sphere of 15 mill diameter is planned to he
cooled by an air flow at 20° C. The convective Iteattransfer
co-efficient is 110 W/m2K. Calculate the following
1. Time required to cool the spit ere from 700 to 150 C
0
0
Solution:
For Sphere,
Characteristic
length, Lc =
R
3
I;
2. Instantaneous
heat transfer rate at 150 C
0
7.5 x10-3
3
3. Total energy transferred up to 150 C.
0
p = 7850 kg/m3
Takefor mild steel
Cp = 474 J/kg K
a = 0.044 ",2/11
We know that,
hLc
k=43 W/",K
Biot number, Bi = -k-
Given:
Diameter of the sphere,
Radius of the sphere,
Initial temperature,
D = 15mm = 0.015
III
~
43
R = 7.5 x 10-3111
To = 7000 C + 273 = 973 K
Scanned by CamScanner
9 x 10-3 < 0.1
Bi = 6. 3
r
TranSient Heat . d
.
em UCllon 1.287
ana),
q
.
ber value is less than 0.1. So, this is I
Blot num
. Ulllpcd h
. woe problem.
I:~I
I ),..For lumped parameter
T - T a)
::::
::;;110 x 7.06 x 10-4 t423 - 2931
.~
q:::: 10.09 W
system,
-hA
x
e[ Cp x V x p
ow upto J 50° C
I heat fl
t}
~fola
.
... (I)
qt
~p'i-
c..> V [T - T 01 [From HMT dala bo k
P
0
",here
(From HMT data book
Volume,
page nO.57]
4
3 .
::: 31tR
V
We know that,
Characteristics
4
V
= 3 x 1t x (7.5 x 10-3)3
Lc = A
length,
V
[-IIOxt
423 - 293
x10-3x 7850
= el474X2.s
pageno.57j
=1.76X10--f>m3.J
==' qt::: 7850 x 474 x 1.76 x 10--f>[423 - 9731
1
Total heat transfer,
qt = -3616 j
973 - 293
{The negative
=>
In(0.191)
-110 Xl
474x2.s
to reach
Result:
1500 C is 139.9 s.
0
{From HMT data book
page 110.57 J
where
A - A rea ::::4 1t R 2 ::::4 x 1t x (7.5
l
A ::::7.06 x \ 0-4 1Ta2
Scanned by CamScanner
1. Time required
for the sphere to reach 1500 C is \39.9 s
2. Instantaneous
heat flow at \ 500 C is \ 0.09 W
3. Tota\ heat flow up to 1500 C is -36 \ 6 J.
2, lnstantuneous Ileal flow at 150 C
q ::::hA [T - T (1)]
coming out ofthe
Iphere 1
x 10-3x 7850
It = 139.9 s \
Time required
sign showSlhaH-R~~s
x \ 0- 3)2
Transient Heat C
onduClion 1.289
hLc
ul1lber s, =
8Ix8·~H~ea~l~a~1I~d~A~la~s_s_Tl_ra_n~sfi~e~r-------___
/.2,, '4 Soh:ed Unh'ersity problems - Lumped H eat All alYsis
8
I5
=
ill An alu",inium plate (k = 160 w/mrc, p == 279
C == 0,88 KJ/kg°C) o/thickness L = 3cm and
0 kgllrrJ
P
,('225"C '
dd
at a Ii'
temperature oJ
IS su
enly immt?rsed at'
'd
'
' d
t""e t
a well stirred f1UI mamtame
at a constant te
"'"(}ill
T tD == 25"C. Take h = 320 Wlm2oC. Deter mine
,tnperatllr
th
e
I
[Dec-2005-Ann
Given:
Q
Thermal conductivity
of aluminium
8
e problem.
lysiSt)'P
for lumper parameter
1 _ TtD _ e
vx p x
..:---:::..,(I)
10 - Tao
[From HMT data book page no . 57 !,IS'/XI h edlllon)]
..
, k = 160 W /moc
= 0.88 x 103 J/kgoC
Specific heat, Cp == 0.88 KJ/kgoC
We knOW that,
Characteristics
L == 3 em == 0.03 m
Initial temperature,
To == 225° C + 273 = 498 K
Final temperature,
Too = 25° C + 273 = 298 K
Intermediate temperature,
T ,;" 50° : + 273 = 323 K
Heat transfer co-efficient?
h = 320 W Im20C
length, Lc = y_
A
(1) =='
r
323 - 298
Tofind:
-320 Xl
= e l 0.88 x 103 x 0.015 x 2790
498 - 298
Time (t) required to reach 50° C.
-320
Solul;on:
In (0.125)
We know that, For slab,
Characteristic
system,
l~:~I]
Univ]
Density, p == 2790 kg/m-'
Thickness,
Illber value is less than 0.1. So ' thiIS IS
. Iumped heat
IlU
iot
I'"
320xO.01S
160
== 0.03 < 0.1\
~
"'fo,,,,
required/or the centre of the plate to reach 50"C e tilrre
k
1'ot "
length, Lc =
L
2
-2.079
2
0.00868
t
s
I
ltsull '
'
\
lIme
I
-I
Scanned by CamScanner
= -
x 103 x 0.015 x 2790
= 239.26
0.03
0.015m\
0.88
x t
r
required
to reach
50° C is 239.26 s.
]
~
~
NVH&/i.~
'.-'
v,
I
Transient Heat Cunduction /.29/
:}
(
1) ...fr i,lu",ilflu",
~lIbc 6 e",
~ - ....
__
011 (I
s;dr is or; ,;
'(5 0 O· C . I 11,\'
. Hulth'"''
Ii'HlPI"Mlllrt'
()J
,
• ,
i
~'/ "lilly
.
(I,
b"'"l!r\"
"
..____''.
c ((
1I, "
!
Q
I'
10' C for M'lflClt h U' 1]0 "r",] /(. 1:'.\·li"'flle u ''1'';'1
uquire'(/ ' for Ih~ cub« to reach (I "''''[la'''"r',
,r te "''PIt
e ''.1 250'
For (J/u"'illi~'"
p
] 7(10 1.1:/",3, C" ::: 900 J/~. ~.
Ilf
, • ]04 "I"'~'
. nunl b er value is less than 0.1. So, this is lumped heat
Slol
lOCI - 00] "'I. J
I.sis type
f ubc. L ;:; 6 em ::: 0.06 m
Thickness
Initial temperature,
To::: SOO
Final temperature,
T
Intermediate
temperature,
Heat tran fer co-efficient,
pe ifie heal.
-T-T.Xl
T::: _'0
I
t]
CpxVxP
, .. (I)
[From HMT data book page no.57]
We knoW that,
V
Characteristics
length,
Lc = A
k = _04 W/mK
[ Cp~hLcx P x t]
To filld:
Time required
for the cube at reach _ -0°
SOllll;Oll :
(I)~
TO-T<Xl
=e
Xl]
-120
=e [ 900xO.01x2700
523-283
For Cube.
Cbaracteri
x
-hA
[
==e
system,
To - T <Xl
It::: 120 W/rn~K
p = 900 J/kg k
onductivity,
.d parameter
For IUlllpe
::: 283 K
Den it). p = _700 kg/rn '
problem.
~~,)
- "::: 773 K
= 10
204
B, == 5.88 x 10-3 < 0.1
II ~,
GiI,tll :
Thermal
120xQ.01
==
773 - 283
tic length
Lc =
L
-120
6 -
In (0.489)
0.06
6
0.01
III
I
G
X
t
900xO.0 I X 2700
== 144.86 ~
Rtsult:
ach 250 C is 144.86 s.
.
r the cube to re
.
Time reqUired fo
0
Scanned by CamScanner
1.292 Heat and Mass Trans er
rn A copper plate 2mm thiclc is heated up to 400
Transient Heat Conduction 1.293
0
quenched into water at 300 C. Find the time r
. C Q"d
eqUlred
the plate to reach the temperature of 50 C H,
fo,
• eOI Ira"
co-efficient is 100 WlmZ K. DenSity OF
sfe,
~ wekn
0
J
cOP ,
[Oct '97 M U. Apr'97 Bharmhiyar u'n'
Given:
k
Blo
.
8800 kglm3. Specific heat of copper = 0.36 kJ/kg K.Pe
Plate dimensions = 30 x 30 em
hL c
. I number, Bj -
IS
=
.
l\lerSlfy}
IOOxO.OOI
386
s, = 2.59 ~ 10-4 < 0.1
Thickness of plate, L = 2 mm = 0.002 m
. num ber value is less than 0.1. So, this is lumped heat
Blot
Initial temperature, To = 400 C + 273 = 673 K
0
. type problem.
lJIalyslS
for lumped parameter system,
Final temperature, T eo = 300 C + 273 = 303 K
Intermediate temperature, T = 50° C + 273 = 323 K
Heat transfer co-efficient, h = 100 W Im2K
T-Too
Density, p = 8800 kg/rn!
[~::vx xt]
p
=e
... (1)
[From HMT data book page no. 57]
Specific heat, Cp = 360 l/kg k
Plate dimensions
We know that,
= 30 x 30 ern
V
Characteristics length, Lc = A
Tofind:
Time required for the plate to reach 50° C.
Solution:
{From HMT data book page no.l]
I
(I)
=>
Thermal conductivity of the copper, k = 386 W/mK
For Slab.
323 - 303
=e
-100
x
[ 360xO.00 Ix 8800
673 - 303
Characteristic length, Lc = _1._
2
=>
In (0.0545)
-100
x
= [ 360xO.001 x 8800
t]
t]
=~
2
I Lc = 0.001 ~
=>
Result:
h 50 C is 92.43 s,
late to reac
Time required for the p
0
Scanned by CamScanner
r
I. ]Y-I
s:
_
1I1'(/I_III!d_M,:.:(I~.~.I~·
l_i_·(/_I1.....
f
1.295
Transient Heat COlldliC/ioll
1.11 II 12 ('III diameter
lOll/: bur illililllly
(II It
.
l!J
'.
II "1/ ,
temperature 0/40" C I.~"I(fcet! III II mediu", (II 65(JoC0Iv.'"
II convective
co-efficient
0/22
WI",} K. Deler"'i",
I' .. '1/,
e ne .II" Ie
required for lite center 10 rca cit 255"C. For lite "'(11
11,e' bar k = 20 WI",K, Dell!Wy
. == 580 /( m • e"cllo. /
fll 3
heal = 1050
'
llkg K.
e
','Pee'/ie-
:= 22 x 0.03
20
Bj:=
[OCI '98 MUj
Given :
Diamctcr of bar, D = 12 em == 0.12
III
Initial temperature,
To == 40 C + 273 =: 313 K
Final temperature,
T
Intermediate
. t number value is less than 0.1. So, this is lumped heat
,
problem.
I)sis type
I)l
d parameter system,
I
Radius of bar, R == 6 CI11 == 0.06111
< 0.1
0.033
BIO
XI]
0
rs)
== 6500 C + 273
temperature,
923 K
T == 255 C + 273 == 528 K
0
Heat transfer co-efficient,
Thermal conductivity,
=:
h == 22 W/1Il2K
k == 20 W/IllK
Density, p = 580 kg/m3
ForIUl11pe
-
T-Too
[~h~vx
=e
p
P
•••
To - Too
(1)
[From HMT data book page no 57}
I Weknow thaI.
Characteristics
V
Lc == A
length,
Specific heat, Cp == 1050 J/kg k
'I' _ T",
To [ln« :
·,I)~
---'
~
5211 - 923)
III [ 313 - 923
Time required (t).
SO/Iltioll
,
[ Cp~hLcx p x
To - 100
:
For Cylinder.
Characteribtic
length. Lc ==
R
2
0.06
2
=:
J
2
105~ 2x O.~3t x 580]
360.11 s
ult:
Time required
I
. 3£'0 II S
for the cube to rcac h 255 C IS
0
I' _
Scanned by CamScanner
t]
=e
u
.0
.
-_£\--
Transient Heat Conduction 1.297
1.296 Heat and Mass Trans er
rn
A steel ball (specific "eat = 0.46 kJlk /(
..
35 "'ImK)
16/
•
g . alld IIr
conductivity =
having J em dia",
trlflo,
eleralld'
at a uniform temperature of 450· C is "
II,ilill/I]
SUudelll
in a control environment in whie" II'e tem Y PlaCed
maintained at 100" C. Calculate II,e li",e,e
~erQIllrt i.
qU"ed/o
ball to attained a temperature of 150.C.
r Ille
Take h = 10 Wlm1K
10)( 8.3 x 10-3
35
::
l
1M. V. April-lOOO,
2001, 2002, Barathiar Uni A .
. Pfl/98]
Given:
Specific heat, Cp = 0.46 kJ/kg
Thermal conductivity,
K = 460 J/kg K
value is less than 0.1. So, this is lumped heat
Siot num b er
. type problem.
aoa1ysiS
ed parameter system,
For I urn P
k = 35 W/mK
Diameter of the sphere,
D ee 5 ern = 0.05 m
Radius of the sphere, R ::s 0.025 In
To - 450" C + 273 - 723 K
Final temperature,
TIlJ - 1000
• -I-
273 - 373 K
tempcrarurc,
T
1500
Heat transfer co-efficient,
h
10 W/m2K
To flnd:
Time: required for the ball to reach
T- T
-
so
To - T
Initial temperature,
Intermediate
Bi:: 2.38 x 10-3 < 0.1
+ 273 • 423 K
J C;.h:.
p x ,]
•••
(
1)
I From H MT data boo. page no. 37J
(f)
We know that,
V
Characteristics
length, Lc"
A
(I) -
To - T""
150 C
0
-10
( 460 ~ 8.33 x JO-J x 7833
_l<
I From "MT
Solution:
data book. pO~Jeno. II
423 - 373
De:nflit)' of steel is 7833 kg/rn)
723 - 373
I
-10
p - 7X 1-'-k-gl-n-13-'~
I.e -
R
J
(),025
3
-_
Ci; - 8.33 )( 10-3 Ill]
Scanned by CamScanner
_Mt
In ...;4.;:..23~-_3_73• ----8-3-)-x-,0-3)(
723-373
460)(·
For .'pllue,
Ch,.ractc:riKtic length,
-e
t]
~
Rtf"lt :
.Time required
78)3
- 5840'B
o. C is 5840.545.
.
• ball to reach 15
for the
n~II'\~'1I1 /"'/11 ('III11/III'/jOll
j.,j ,i" {,I"",III/"",
I.:.t.I
_\",,,"",
"
111",\',\'
l.f 46:
,
,",,1/
,! nl{J
11/111111,
IM'I'i'f{""rtl 'if 2911 ( 1,\' ,\,,,,ItJ,'''~I'/""",'r,\'"" I"~ .I '" II
, J" C 11,1", lu'{"
"'1-,:1111'/,'''' IN """"A' 1':"~',/'1",
, 1:,\'1'
lilt'
,(,,,"If'~" ", ('11111tlu (""111",1,
'1""1"
""
I"
9r.
For 111"",1111",,, iliA,' P 17m) *1:/111" C
9
'c.
J
'f'''',~/'''
"'",,!
'I'
A 10J 11'/",1:,
3V _ 3
1
lUI J/k
2,()J)
4;- - :::; 0,0786,11
I: A',
,,; -
10 I
I
/tI/.{/, 0('(-99, 11(/I'(/fltiu/, U,'
1/, Nov
GII'i'II "
Jphi,re
I
Mass, III ;:: 5,5 kg
;:: .!
'.,
,
!'riSIIC length, Lc
, C~Brnct
luitial tcmpcrature,
To:: 290 C + 17 J :: 563 I(
Final temperature,
T :: 15° C + 273 :: 288 I(
Intermediate
96/
3
0
temperature.
Heat transfer co-efficient.
Thermal conductivity.
0,0786
=--
3
r:: 95" C + 273 == 3681\
r Lc = 0.0262 m
" :: 58 W Im21(
k :: 205 W /I11K
, , knoll' that,
I
IL
III
1 C
Biotnumber, Bj = -k-
Density. p = 2700 kg/m3
58 x 0.0262
Specific heat, (p = 900 J/kg K
205
Tofind,'
Time required to cool the aluminium
Bj=7.41 x 10-3<0.1
at 95° C
. nunl ber value is less than 0.1. So, this js lumped heat
BiOt
~I)sis type problem.
Solution:
Density, p
mass
volume
For lumped parameter
= _!!!_
V
t
T-T'IJ
V
:: • .!!!_
= 2.037 )( 10-3 Ill]
Scanned by CamScanner
[C(l~I~xp
x]
_=c
p
To - T
5.5
2700
IV
system,
I
J)
/
...
(1)
From HMT data boo]:page 1/0.57/
V
Charactcrist ics length, Lc :: A
/,2')1)
Transient Heat Conduction 1.301
= 55.55 lIs m DC
1.300 Heat and Mass Transfer
[k
-h
T - T r;r,
(I)
~
To-
~
Thenna
= e [ 900 x ~.~~62 x 2700)(
[.: lIs = WI
.
:;::7865 kg/m!
DenSIty, p
1 d'ffusivity,
ex = 0.06 m2/hr
p x t
Tr;r,
368 - 288
563 - 288
~
]~
= e [ Cpx Lex
= 55.55 W/mDC I
I
0.06
3600
=
t]
m2/s
= 1.66 x 10-5 m2/s
368 - 288]
-58
In [ 563 _ 288 = 900 x 0.0262 x 2700)( t
I t = 1355.36 s I
Specl'ficI heat ,
c, = 0.45 kl/kg DC
= 450 l/kg
Heat transfer co-efficient,
DC
h = 140 kllhr m2 DC
= 140 x 103 l/hr m2 DC
Result:
Time required to cool the aluminium to 95°C is 1355.6s
= 140 x 1031/36005
ill Alloy steel ball of 1.5 em diameter heated to 700' Calld
= 38.8 lIs m2 DC
h = 38.8 WI m2 DC [.: lis = W]
quenched in (I bath at 100" C. The material properties of
tk« ball are thermat conductivity, k = 200 kJ/m hr' C,
Density,
p = 7865 kg/m}, Thermal
dijjusivity
a= 0.06 ml/lt, Specific heat, Cp = 0.45 kJlkg"C, Conveai«
heattransjer co-efficient, It = 140 kJII" ml• C. Determint
the temperature of lite ball after 10 seconds.
Given:
(Manonnranillm Sundaranar
University Nov- 96)
Diameter of the steel ball, D = 1.5 cm = 0.015 m
Radius of the ball, R = 7.5 x 10-3 m
Time, t = 105
Tofind:
Temperature
of the ball after 105.
Solution:
For Sphere,
Characteristic
length, Lc ==
R
3
Initial temperature, TO = 700 C + 273 = 973 K
7.5 x 10-3
Final temperature, Tctj'= 100 C + 273 = 373 K
3
0
0
Thermal conductivity k - 200 kJ
•
-
1m hr" C
'=
200 x 103 Jim hr" C
'=
200 )( 103 JIm x 3600 soC
Scanned by CamScanner
m2 DC
1.302 Heal and M,
_
W
ass rransf'e .
e know that,
I
Bior number,
Transient
~cI
B, == ~
38.8 . .., 5).5 -
B·I --
anal
balf to cool
I . 746
-hA
.
== e ( CpXV . p
~
To- Tco
Charact
400" C.
.
SIS
IUn
lped heal
Gille" :
Diameter of the ball.
0 = 12 rum = 0.012 m
Radius of the ball, R = 0.006 m
t]
-
We know that,
(0
10-3
x
Blot numb
0.1
.
er value is I
ysis type probl em.
ess than 0 . I . S o.lhi
F
-or lum ped pa
( rameler S\·_,stem
T-T
/11
... (I)
HMT d uo book page no 5ij
[froll/
0
Initial temperature,
TO = 800 C + 273 = 10
Final temperature,
T
Thermal conductivity,
:=
1000 C + 273 := 3 r K
k = 205 kJ/m hr K
205x
1000J
3600 s mK
. .
ensne, length, Lc == :y_
A
T-T<Xl
T
== e
0- T<Xl
(I)=>
[c
56.94W/mK
:hL
p
X
C
P
[. .:
".
Density, p == 7860 kg/m3
Specific heat, Cp = 0,45 kJ/kg K
=>
T - 373
[
-38.8
973 _ 373 == e 450 x 2.5-x-I-0--3-x 7865 x 10]
= 450
Heat transfer
=>
T-373
973 - 373
J/kg K
co-efficient,
h = 150 kJ/hr
"
J
= ~lltH
0.957
11I~
3600
S 1\\- "
= 41.()6 W/ \\\~"
To find:
Result:
T· ITIpcralurc
or 11Cbali
I
(i) Temperature
nner 10 s is 947.2 K.
Scanned by CamScanner
/
of ball after I 0 S~'(
(ii) Time for ball to cool to 400"
'.
:/,
:
MI"" I
C- " .
(i) Temperatllrl!' 4 hall afer I (J second atul tu, 11
~I
.
///11
~ ;l11OY " ,/,ed ill a but]) at 100" C. Tire "'fller/all"'II"
!) '. if,'il
1.1
I '" h{lIl arc k :: 205 Ii.///II I" K, P
7f!f,/,
/I(. _1 Ie('145 kJ/Ii~ K. Ir :: 150 KJ/ ltr 1112 I<. IJK'
k
==
11i.:01 ( 'fit/dill
---------------~
htlll of 12 nun diameter h e atrt!
l
1"£
I~
v
___ .Iss(1_'~/.'i.:...:ie.:...:".:.....;{
/{e
J. 304 Heal and Mass Transfer
(I~
(a"d
Lief ion
-41.667
SoiuJion:
Case (i) Temperature of ball af/~~ J 0 sec
o.oo~ , 7860
==e
For Sphere,
Characteristic
~
length, Lc == _.!i_
~3
I
,
== ;,__
0_.0_06_
_ 3
[LC
==
0.002 m
-:)
.
'J Ttn1t
fior hal/to
T-T?J
To - T
41.667
x
==---:----
0.002
673 - 73
1073-373
56.94
Bj == 1.46 x 10-3 < 0.1
1
I
... (I)
{From HMT data book page no.57]
We know that,
Characteristics
lenzth
o
I
L
'c
= VA
•.. (2)
Scanned by CamScanner
t
'If
1
-41.667
0.002 7860
I
==e
- 373 ] ==
1(7) - J 73
450
system,
=e [ CpXV x p x t
[ 450
:::> 111 (673
Biot number value is less than 0.1. So this is lumped heal
analysis type problem.
-hA
40lr C
I
hLc
Biot number 'IB· ==- k
T-Ta;
coo/to
T:= 4000 C + 273 ==673 K
(pi III
We know that,
For Jumped parameter
1031.95 K.'
== 143.849
-41.667
0.002
7M60
I
A
sJ
llsII1, :
of ball after 10 sec. t: 1032.951\
1-13.8-19.
(ii) Time for ball to cool 10 -IOO~C. t
(i)Tcmperaturc
I.J05
Transient Heat Conductio" /.307
are
conditions
1.306 Heat and Mass Tram; "er
1.5.5 Heat Flow in Semi-Infinite
fhe
Solids
A solid which extends itself infinitely in all dir •.
. k.nown as In
. fuute so I'dI. If' an infinite
.
splice IS
solid' .. cellon'
' S Of
..
IS spill '
middle by II plane, each half IS known as semi infi .
IIIlite
nile Solid
In II semi infinite solid, lit any instant oftitnc tl
' '
, lere IS 81
II point where the effect of heating (or COoling")"
.
Ways
.
.
.
"
,\I One f'
boundaries IS not felt at all. AI this POIIII the IClllll' .
0 lIs
cralnre re .
unchanged.
ilia Ins
OllOd3r)'
b
- T·
I. 'f(x. 0) -
,
2. T(O, t) ::: To for t
0
u= Ti for t
0
). T(Cl,
I tical solution
for thi case is given by
'n: :::
r :~
o
Y
!b'
err \ 2
Tj - T~o:!--
(l
t
...
( I.R2)
_
. ,rf indicates "error function of' and the definition
f ction is gencrally available
in mathematical texis.
~(rror U~~IatiOnof error values are available in data hooks.
where e
where
Initial temperalure
r 0 - Surface temperalure
~'j -
1111= 0
usually til '
(,(_ Thermal
I - Time,
diffusivity,
m2/s
S
x - Distance. m
Tj -Initial
temperature,
K
To- Surface temperature
T, - lmermediaic
temperature,
l. In semi infinite
solid, heat transfer co-effici~1l1 or biot
x
FiR I.U Semi II,/illite P/flle
Consider a semi infinite bod
. .
the +VC x directio
'rl
. y and It extends to infinity in
n.
Ie entIre bod
. . ..
temperature
'I'... 1 .
Y IS Initially
at uniform
I Inc ud Illg the
su rface
. _
.
tempcrature at r == 0 .
d
at " - O. The surface
.
IS su denly raised to T
2
dT
equation
number value
i.e.,
I
dT
is 00.
"
or
is
d\,2 == a dt
Bi
2. Tj -Initial
T
0-
S
temperature,
urfacc temperature
Tx - Intermediate
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K
Nolt
o·
The governing
(or) final temperature, K
K
(or)
tempcrature
fl'11al temperature. K
K
/;'(I!,siC11I 1/c'lIl Conduct
~--1.308 Ileal and Mass trails er
1.5.6 Solved problems - Semi Infinite Solids
h :=
ill A /t"ge concrete high WI~1'
illili{/Ih fit 1,-::------flllperfit
70 C mill stream water is direcled
U
fill lilt! Ilirrl.
the surface temperature is sliddellly
low
Determine lilt! lime required
" em from the surface.
55" C
10 reach
~ 0J
Ure of
e , '''flY \.
·0 I"u/
,
eret/ to 40.
IS W,
6.I,a Iu C
C
IlIlld.
I
f'Pt" of
Given :
So.• ihis is semi infinite
lfinlic s('I'id,
I
..
For
5C;:I~IT0
:=
"':--T
Initial temperature. Tj =70° C + 273 = 343 K
solid type p
T·- 0
elf [..,j.\' _]
- (l I
I agl'
(From IIMT
l
110.58(Sl.r/ll
e
I
Final temperature
or
Tr - To
== ('11 (Z)
_:.--
Surface temperature, To = 40° C + '273 = 313 K
I
Tj- To
Tx = 55° C + 273 = 328 K
Intermediate temperature,
Sailitioll :
{From HI/7'd
.
0((1
l
b 001(I. page
(Si.l'lh
Properties of concrete are
edition)/
1.1:..
prllhklllh~';)llJ"1
r:
crf (Z)
0.5. corresponding
0.48
Z
.. ,.
2300~O" ..
• Il~"-I co-cltlcl~'Ill"
::::.
is 111)1~i\~'11. S,•.
::::.
II - "I).
.r
=
2...;at
= 0.4') ...., f) (, 1I1~1,
0.48
.r
2jOi
0.48
2
d
Scanned by CamScanner
I
Z is O.-lS
I From nut data
1"'Ke
\I.e know that,
==
IlIlh;,
0.5
k = 1.2790 W/IIIK
1.2790
1.lh, ll;)~
err (Z
110./8
Thermal diffusivity,
J
er] (Z)
0.5
Time (t) required to reach 55° C
i
Z = 2/0 I
343 - 3 J3
Ttl filltl :
Ii
.r
328 - 3 13
Depth. .r = 4 em = 0.04 III
Thermal conductivity,
\\ here
,/O.59(Sixlh
edil/(
.
,..
,...
a: ~.: .:.._:_~.;.
.. :...
1.312 /feat alld Mass Transfer
Transient Heal Conduclioll 1
111 A large wall 2 em tltick II(U~
is SII'"e"'Perlll lire 30' ,
illitial(" and the W(/1/ temperfltllre
.
maintained
at 4()O(}.C. FWd
"lle,,/), r'
(
(1lSerl "~d
I. TIll' t(''''peratllre at (I deptlt of 0 8
.5IIT/(lel' of the wall (ifter lOs. . . ell, firo",'hi
2. Instantaneous
Item flow rare tl"olloll
per m-, per hour.
e ,
1l
:
.
So "lioIn thiS
. pro blem heat transrerC co-e ff"icrenr h IS. 1I0tgiven.
'
it as cf). t.i.e ., h ~
00.
lake I
We know that,
1
"'nt SlIr'
JQCt
h
Take a = O.OOH /112/It,.,k = 6 W/m0C.
Bj value
CI)
[B; ==
=>
[Apr'97 MUj
Given:
Thickness,
hLC
k
Biot numbcr, B;
. C/). So , this
IS
CI)
I
is semi infinite solid type problem
J.. = 2 em == 0.02 m
Cns£' (i)
Tj == 30° C + 273 == 303 K
Initial temperature,
To == 400
Thermal diftusivity,
(.( == 0.008 1112/h
= 2.22
Thermal conductivity,
For semi infintie solid,
C + 273 == 673 K
Surface temperature,
0
x
T.,. - To
'rj- To
c=
where,
Z
Tofind :
I. Temperaun.,
ofthcwall<lfter
2. Instamaneon,
per hour.
I
::.:
I lt
= O.OOS m, t
=>
Z:=
== 3600 s
IZ
the surface
(I. .x) at a depth
0 f 0 .8 em
from
I
lOs.
I that stir ~ace ,
heat flow rate (qx) throug 1
Scanned by CamScanner
6
0.008
!?2? x 10 6xlO
2'1':'" == 0.S4iJ
di g erf (Z) ;S 0.7706
111
.
ni61
I eli (Z) := O. ~
.
2
= 10 s, a := 2 .22 x 10- 1II /s.
Z == 0.84S, correspOI1
=::>
. .. (I)
== 2Jc11
Putr
Time, I == 10 s
Case (iij
lata book page 110. 58
(
x
Depth,.r == O.~ em = 0.8 x JO-·2 III
m
jJHT
7)
('1/ (Z)
T, - To:"
Tj- To
Case (i)
Time.
t]
[From
10-6 m2/s
k == 6 W/m°C.
= 0.008
[-2/atX
a
crf
[Re/e
rHA
IT dala boo k page no. 59]
~ ..'
LV
r
1.314 II eat and Mass T
ransfer
Tx - To
(1) =>
T, - To
:: 0.7706
Tx - 673
Transient Heat Conduction 13 J5
sltlnlllneous
heat flow rate at a depth ·1"300
2. I n
0./
mmand
on surface after 7 hours.
3. Tottll heal energy after 7 hours.
303 - 673 ::: 0.7706
Take k == 0.75 WlmK,
Tx - 673
::: 0.7706
-370
::: 387.85 ~
Case (ii)
a = 0.002 m2/hr.
Gillen:
Initial temperature,
T. = 25 C + 273 = 298 K
final temperature,
To = 700 C + 273 = 973 K
0
0
Depth, x = 300 mm = 0.300 m
Instantaneous
heat flow
Time, t = 7 hr = 25,200 s
[-x
-_.:..___:..::_ e 4a. t
a = 0.002 m2/hr
Thermal diffusivity,
2
k[T() _ Tj]
]
= 5.5 x
JWrt
{From HMTd
ala ilatA
page no. 58(Sixlh ed"
Thermal
conductivity,
10-7 m2/s
k = 0.75W/mK.
ItU)ftIJ'
t = 3600 s ( iven)
Tofincl :
(
6 (673 _ 303
-"'Ix =
)
In /2.22 / IO-I'i;; 3600
I qx=13982.37
/c
[-(0.008)2,
4x2.22xIO-liYJ6fXJ
.
temperature,
2. Instantaneous
heat flow, q.x
3. Total heat enrgy, q,
Solution:
W/m2.!
In this problem
Result :
take it as 00. i.e.,
Intermediate
temperature,
Heat flux, qx = 13982.37
Tx = 387.85 K
B.lot number
temperature of 25' C a~d
ftJ
u C nd re/tlal
wa II temperature is suddenly raised to 70O {t
constant there after. Calculate the following
l
(I
jro/tllk
ill plane (It (I depth of 300 nlfn
surface after 7 hours.
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heat transfer co-efficient
h ~
h is not given. So
00.
We know that,
W/m2.
!iJ A very thick wall initially at
1. Temperature
r,
I. Intermediate
'I
hLc
B· = k
h = 00
B
.
. . fi't
olid type problem.
B j value is 00. So, this IS semi 111 uu e s
J.316
____
Heat__:~.:.::~~.
and Mas' :_'.!_J!:
s 'T!_IiilI11S,er
.c
I. For semi infintie
SOlid~
Transient Heat Cd'
on UCllOIl 1.317
~_
T, - T 0
T;- To
-------~
[~l
~ erf
::=
[From HMT I
? qx
(.Ola
=>
= erf.
book
Page n~~
.
. . . ( I)
(Z)
where
0.75 (973 - 298)
Ii x 5.55 x 10-7 x 25,200
2
W/m .
----...J
::= 483.36
q.r.__
]. Total heat energy
x
z
[From HMT data book page 110..17/
q,.::= 2k [TO - T j]
2J(Xt
J 7t~
2 x 0.75 (973 - 298) x
2/S.SSx
Z
=
25,200
7t x 5.55 x 10-7
q = 121.72
10-7 x 25,200
r
r------
1.2iJ
x
106 J/m2
Result :
Z = 1.27, corresponding
=>
J
0.3
Z
-(03)2
[ 4x555 I' -7 ]
X e
. x 0 x2),200
I. Temperature
at a depth of 300 mm, Tl = 346.9 K
2. Instantaneous
heat flow, qx = 483.36 W/m2
err (Z) is 0.92751
I elf (Z) = 0.927SI I
3. Total heat energy,
q, = 121.72 x 106 J/m2
[Refer /-I MT data book page 110j9j
(I)
=>
T,. - TO
~ if large cast iron (It 750 C is taken out from afurnace and
0
= 0.92751
its one of its surface
Tj-TO
(1/ 45° C. Calculate
T,. - 973
=>
=>
tile following
0
I. The time required to reach tile tempertltllre 350 C lit tt
= 0.92751
depth of 45 nun from tile surface
2. ill.'itallllllleoll.'i
heat flow rate at a ,Ieptll of 4.' 111111
and
298 - 973
I Tx
is suddenly lowere{llIIul maintained
=
346.9
K
l
Olt .\. IIr
l'ace after 3() minutes.
3. n,ta/II(,'al e~,er{:y after 2 I" for ingot.
2. 1nstantaneous heat flow
Take a == (J.(M m2//".
[;~~
]
e
[From
Scanned by CamScanner
k = 48..l w/",K.
Gh'ell :
HMT
d
book !,ng
ala
C/1041]
1
Illiti a I klllpl.:ratlirc
r ""'7S0~ C + 273 =' 10s»' K
S
' I
318K
'llrfa.:.
l: klllpl.:raturc,
1'0 == 45° C + 273- -
~
Intertnediats
temperature,
Tx == 350° C +- 2
.r = 4- mm =: 0.045 m
Depth.
j
I
1.3/8 Heal and Mass Transfer
Transient Heat Conduction
:~432~
I?
a = 0.06 m2fhr
Thermal diffusivity,
Thermal
Tofind :
er. I
'. The time required
to reach
2. Instantaneous
heat flow
surface after 30 minutes.
~
of 4Ure
_ J Sf}' C
;;:;0.432. c rresponding
at a depth
mill an '-'
Z is 0.41
0.41
[Z
the temperat
\\e k.now that
Z
after 2 hr.
In this problem heat transfer
ta .e il as -r.,. i.e., h ~ 'ZI.
co-efficient
0.045
0.41
h is nOI gi en.s,:
We know that.
Bi
0.432
==
Z
k == 48.5 W/mK.
conductivity,
3. Total heat energy
Solution ;
er/(l)
.
erf
:::: 1.66 ~ 10-5
1.319
")
:::- 0.41f
hLc
number B· = __
,
f
k
~
h = a»
I
1_8_I_A_2_s_1
L_t
Time required
OJ value i r/). $0, this i semi infinite
solid type problem.
1.lnstunlflneous
t
reach
350°C
is 181.42
s.
It eat flow
I. For semi lnflntle solid,
T -To
., j -T()
=erf
[2yCt.t
~]
I From
- elf (7)
_Q_2 J -l.!!
1023 - 18
0
err (Z)
Scanned by CamScanner
II MT dar a book PQ}?,c110 58J
where
.x
Z = 2fo1
[From HUT data book page no. 58}
I:: 30 minute
I:: 1800
,
(Given)
.-.,ILI;"
Transient Heat COlldut:tion 1.311
~
Gilt"· .. ,tc:tnperature
T i = 6000 C 2 3::: 8 3 K
a
\olU
e tc:tnperature,
TO =50 C.1. - 1"' ::: ' _ ' K
sur f ac
Thermal diffusi\,ity,
sign show , that heat I
f
osr rom th e .In!!
3. Total heat ellergr
~o 1
a
=
["egati\e
.
[From
H.\fT d
a/a book
3600
.
Or e tJl;:.
La
I
ito.
== 2 Y 48.5018
IT'
.
.
X
it /
1.66;
10--'
\
Toji/ld :
\. Temperature
I
\
\qT == -803.5
I
106 J/m2 \
/
1.11 x IO~
\~
51
2 hr = 7,200
l Ime IS given,
=
k = 1.2 W/mK.
Thermal conductivity,
- 1023)
m-) Ilr
--0.004
\
,
0.004 m- hr
(T x) at a depth of 3 em after 6 minutes.
2. How much time (t) required,
the temperature at J de
of 3 em will reach to 350 C.
0
[Negative
1'1)
how
\
that heat lost from the ingot}
3. Cumulative
Result :
heat (qT) at a depth of 3 till within first
hour.
I. Time
required
2. Instantaneous
to rca h the temperature
350 C is 1&1.1:1: Solution:
0
n w, qx = -I ()1)72SA W/m2
Ilcat
3. Total heat cncr 'y, 'IT
- -XO".5
/
In this problem hcat transfer
take it as
i.e., h ~ 00.
10(' J/m2
surfuce temperature
ICIIII'et'lIlIIrc
of' Mill' C ilndin
h .'illddc"ly IlIlI'act/
105(1"(. CII/(wl~1
(I
n is not given
CY).
We know that,
@) A IlIrlo:C .,111" ;1I;1;lIlIy III
co-efficieut
.
\3 lot nllmber,
u.,
B· = I
k
h =
CIJ
tile! /i,llol\';"1o:
I. T"IIIf1"rtIll1rC
2. II,,,,'
.
""h'III111ll'
11/11 tlc'[lIII
.'
or 3
required.
nil I(fit'r
(, ",;IIIIIt'S.
3 011 will rc aclt '0 3 -' 0" C.
3.
.,
01
'.'
,'''111''/1111''''
.
.
,
Tuk«
= 0.004
Scanned by CamScanner
III)
(~,.3 <'III
.
,~I~I
"
1/,,· 11"1111(1/1'
[l1c",' til II tI"I'0l
i~
, IIIre! /II II ilt'p'
III,' tc'/IIpcNl
~H'IIIIITf"'~
lu 1111"'"
.
"1 Iwe III,IIT.
11·"1",, .f"s
IIr. I. == 1.1,,-,,,,1\·
,
\
B·I vai
. 00. S 0, this., IS seuu .' III 1-III itcI.: solid type pro
a lie IS
oV
~~~-
/.322 Heal and Mass Trails er
Case (i)
Depth, x = 3 em = 0,03 III
Time, t = 6 minutes
= 360 s
For semi infintie
solid,
Tt - To
Tj-TO
=erf
T.x _ To
T. _ To
=>
C'
Transient Heat Conduction 1.323
jet;;)
- 0 03 m
=== 3 ern -,
0
Depth, x 'ate temperature,
T x::: 350 C + 273 = 623 K
d
loterrne t
Forsemi infintie solid,
0:::
erf r· !c.1
Tj-To
l2...;ut
[21at]
[From HMT d ata book
= erf (Z)
."
page no 58]
(I)
Tx- To
== erf
(Z)
Tj-To
where
where
x
Z
Z
0.Q3
Z
2-/1.11 x 10=6
r:
LZ
==
623 - 323
873 _ 323
x 360
==
erf (Z)
0,545 = erf (Z)
O,75J
Z == 0 ' 75 ,correspond'
2#
==
erf (Z) == 0.545
II1g erf (Z) is 0,71116
elj(Z)
erf (Z) == 0,71116
[Refer HMTd ala bo k
== 0,71116
0
== 0,545, corresponding
Iz
= 0,53]
(Refer HMT data book page no. 59]
pageno.59}
We know that,
==>
x
~
873 - 323 == 0,71116
Z = 2-:;at
x
0,53
:=
2-:;at
0,53
:=
2jl.ll
0,03
Scanned by CamScanner
Z is 0,53
x 10:::()x t
pa
. I .,325
__
7;'ans I.e 11]' Heat COl1dUCI1O!1__
I
/
Mass transfer
'!ea!_
llII!i__
:._:;:__---
/.m
(0.5))'
(2)'
---I
)'
~.
I. " .
-
}-le
I 06W 1m
qo
llt
C
0.25
::=
_ 0 10 III
- 30 em - ..
Distance, .
.
:= (l00 s
\0
lllinutes
Tillle, t :.::
11,'IUll
Depth,
x =-= J 1.:11\ == 0.03
lime,
I ==
1:1)1'semi infinite
?73 = 29R K
1-I = 2S
. , temp erClture
Om! (iii)
1
.
Oux, -'0
@'
(0.03)'
1.11" \0-('" t
_ 0 25 MW/m2
n
\.
11\
I hr == 3600 s
-
solid.
Total hcnt energy
md:
.
after 10 minute ..
T)
r.fi 1 Surfocc rem pcrat '" e (0
.
2.TcmpcJ'(\tlll
['I r
lN~'gativc
= -424
x
1.11
IQ-6
Heat fill',
-1'06l/~
Si!!n shl)ws
that heal
nil em from Ihe surface
.
50"";011 :
-~
It
.
..c (T .\ .) (\t a cit. ranee
.. , (I)
<to'
),',
n I
I
iv e III) 5!()
/1, rrun 11M'! data how, J1 ,I';
/}
.
I
k
fJ(/"
(' I/O.
/11 A'IT rial I
lost from Ihe slabl
R,·.\'IIII:
)I)t)
I. l"
2.
"=
7I-U I\.
Propcn ic: of aluminium
I == 721.6 S
3. 4 T == -42.4 x 106 J/m(
III A ".,1 Infl,,;t, 'lab of 01" ml" I"m I, «spose« to 0""",,,
hetll Jlllx til nie Sllr/ace
1)
'
Thcrmal
diffusi"ity.
Thermal
cOlldllCtlvlt).
=.::>
025
'.
2
Xt1. ! R )' 10 II III /~
(I.
••
I
...
k
'''-
7.04.2
W/mK.
.
204.2 (Ttl - 29~) _.
-~- R4.1.. n
«oo
;-='10 (;.
10(' :-
0/ O. 25 \1Wlm2. Illitialtemperatilri
u/lhe slab is 25° C. ClIICIllllle tlte :lllr/llce telllpeNltllre il/ttr
10 milllltes and alsu /illt/tlte
temperatllre
3(J em from lite SlIr/ace after 10 milllll .
es
til if tlistallcl!/if
(ii) For sellli infintie
'1 \. - T 0
_
-:--1- '-T
.
i-
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~olid
0
-r \ ... x _. 1
l'1.
/')(11
.
/1'/,0111
Ii IllW
I
I-IM7 dat a
100
J)II
e.
58)
FTTtt
1.326
Heal and Mass Transfer
Transient Heat CondUction 1.327
T.\.- To
~
== erf
T'-T
I
(Z)
0
... (2)
I
e s Ia b initially at a temperaturo of J 20° C alld itS
peralllre is suddenly lowered to 0° C. Calculate
, .Iace tent
'11
where
i
Z
==_
.
x
2.jOi
Alatg
slit}'
•
foJlowtng
the The time reqllired for tire temperatllre gradient at the
J.
2)84.18
[Z
== 0.667
.f.
e to reacll 6°C/cm
Sllt,oc.
. .
2. The {Iepth (~t which tire rate of cooling IS m(Iximum
after two nun ute.
Take thermal diffusivity, a = 0.612 m1/II.
0.30
Z
! ,
x 10-6 x 600
]
Given:
Z == 0.667, corresponding
elf (Z) is 0.65663
[From HMT dala book
,---___
T, = 120° C + 273 ::::393 K
Final temperature,
TO = 0° C + 273 ::::273 K
page nu.59)
Thermal diffusivity,
[ell (Z) = 0.65663]
(2)
Initial temperature,
a
=
----m
0.612
3600
"('.f -To
CI
T'-l
I
0
0.612 m2/h
0.65663
1.7xIO-4m2/s
=
T{ - 785.61'(
298 - 785.68
2/h
To find:
= 0.65663
I, The time required
for the temperature
surface to reach 6°C/cm
gradient at the
2. The depth at which the rate of cooling is maximum after
two minute.
I Tx ::; 465.4SK]
Sfllul/OII :
Tcmperaturc
Resutt ,
at 30 em is 465.45 K
at a distance
.In this problem
take It a
.
S C1.l. l.e.,
I. Surface
temperature,
2. Temperature
TO = 785.68 K
at a distance
of 30 em, T, = 465.45 K
h ~
heat transfer
co-efficient
II is not given. So
tJ).
We know that ,
".
Btot number
B. .
'I
h
hL
k
c
= 00
~
\ Bj = 00 \
13·
.
'. . I'd t .pe problem.
,V.lue IS 00. So, this is semi infi,"" sou )
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A
1.32H
------__
",,'
neat and Moss 7'.
lan·Ve,.
Case (i)
---.--
~
I
I,
Transient Heal 'onduction I.J2~
We know Ih(l1.
Temper(llllrc
=>
---:>
!!radicnl
<iT
<ix
_6nc.
<iT
<ix
(inC
.-~.:.--
cll
<ix
- .MIOne
IS
-
~oo
dT
~ -fo1lt
dx
273 - 393
em
= 600
0== 74.89 s I
IO-2m
---
Case (ii)
III
1==2 min
We know
= 120 s
that,
For maximum
Heat transfer.
:::::::>
x=
:::::::>
:::::::>
qo
dT
/2
x 1.7 x -10-4 x 120'
Result:
ciT
k ,._
cix
required for the temperature
6 "Czcrn is 74.89 s.
J. The time
[Heal flux,q J
o
A
2. The depth at which
two min lite is 0.20
~
dx
k
~=
600 nC/m
k
rate
I x = 0.20 m I
ciT
k -.
cix
-
cooling
x=~
Q == kA ciT
dx
Q
A
I
1.5.7Transient
A solid
600 °C/I11
K
gradient
the rate of cooling
to reach
is maximum
after
111.
Heat Flow in an Infinite Plate
which
extends
space is known
as infinite
Consider
an infinite
Shown in fig. I. 15, which
Tj• It is slIcicienly
exposed
itself
infinitely
in all directions
flat plate of uniform
is initially
thickness
2L as
at a uniform temperatllre
of
to a large mass of fluid having
a
1cnlperfltllr" Tv. Thi. temperature
is aS~lJIned to be constant
IhrOllnl
. '1" ie p I a te. is extended
. r 1 out the process
of cooling or heating.
10 InC'
.
Illlty In the y and z directions.
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of
solid.
,
/.330
----
Heal and MaS\" Tor "f,
__ ..:.:._:::.
~1~/ans.ler
Transient Heal Conduction 1.331
~
r.: r,
==f[~' ht,~~)
tion we know that conduction resistance is
this equa
,
fro~
The temperature
history becomes a function of
liulble.
neg hLc] Fourier number [a-2t and the dimensionless
L
I :: OU JTIber [ - k ,
(.!-)which indicates
the location of point within the
I-I
I
0
"
I
.!f3rneter
L
e temperature
is to be obtained.
The dimensionless
dale Vi h er
~rameter
is replaced
by [ ~ lin case of cylinders and
l~ 1
-x
I
.pberes.
Heisler has prepared
Fig. 1.15 Infinite plate
The heat transfer C
f .
plate and the fluid
O-e ficlent between the
center of the I
.on both sides is assumed t b surface of the
p ate IS selected as tl
..
0 e constant. The
ie orgln.
The governing different'
I
.
d2T
ra equatIon is
==i_ dT
dx2
a dx
The boundary
,
2.Atx==0,
3.Atx==±L,kA
bo
The solution
Ulldary conditio
of the
problems.
These charts have been
onstructed in non-dimensional
parameters.
The charts are
: 9Jitablefor problems with a finite surface and internal resistance.
I for suchcase the biot number lies between 0 and 100.
These heiler charts were further extended
!rober.
and improved by
NOle:
fj
dT
cJx
==0
For infinite solids
Take
'
ciT
T i-Initial
dx ==hA [TO - T ifJ
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solutions
steady state conduction
0 -
of the ab
.
.
. .'
ove dIfferentIal
n IS gIven by
for graphical
The heiler and grober charts are used to solve the problems
ofsUddenim'rnersion of plate, cylinder
.
.'
or sphere 1I1toa
fluid,
co di .
.n Itlons are
I. At t ==0 1" _-
charts
temperature
T C7;J - F'mal temperature
.'
equation
with these
- K.
- K.
To - Center line temperature - K.
It. T x - Intermediate
temperature
- K
11~lfinite solids, biot number value is in between 0.1 and 100
r.e 0 I
., . < B, < 100.
Transient
Heat COlldUClioll
I.JJJ
l
2
0.05
I
Given:
We knoW that,
lhic.kncss,
Initial
BiotllUl\lber.
L == 5 ern ::::0.05 rn
I
Finaltcmperature
T
.r = 10 rnrn == 0.0 I 0 III
Time
1 minute
t v-
Heat transfer
1800
==673K
::::900 C + 273::: 363 K
Distance,
Biotllumber
== 60 s
0.025 III \
hlc
I3j = k
T == 4000 C + 273
temperature
L.
i in between O. I and 100.
value
i.e .. 0.\ < I3j
0.025
204.2
0, thi
\ 00.
i infinite solid type prubleill.
h = 1800 W/m2K
co-efficient,
To Ii IItI :
Case (i)
1. Mid plane temperature
2. Intermediate
(TO) lifter I min
( T .r ) at a
temperature
distance
ofO.OIOm
X axis ----)fourier
Sotution :
rt-
-
Propel! ies of aluminium
l diffusivity,
/
I
at
from the mid plane.
Thr-rma
I
A
1'0111
IJ '1'( doto hun' I
7/1
iHC
(1
'l hernia I conduct ivity, k
~-= ~4. J R
== 204.2\\
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/mK.
'(lg~ 110 11
iter
r ;n'il/;te fI £)11'. rej
"pera(llre fio 'J'
J
{Tv calculate III/ p ane et
..
f!e;\'/er "hart
n
.
6- (S'/x III e(//t/O ) HAfT data book p .Ige no.o:
.,
lIumber
==
L;
~
==
10-6 x 60
(0.025)2
. nt Heat Conduction J .335
Trans re
1.334 Heal and Mass Trans er
.'
hLc
Curve
#
k
\ 800 x 0.025
204.2
X axis value is 8.08, curve value is 0.22 F
can find corresponding Y axis value is O. \ 9 lFr' rom that, ~t
om graph\.
(II)
distance
0
f 0 0 \ m from mid p\ane
.
ra\Ure at a
1et1\\)e
HM
T data book page
no 66 (Sixth
.
J
[Refer
charI
,~eisler
_ hLc::
.
Blot number, B, - k
Y. at-is ~
x _ _Q_.O\ :: OA
Cut'le ~ ::
0.015
0.11
Lc -
. 0 4 from that, 'Weca
. 0 11 curve va\ue is . .
. ova\ue \S . ,
'j. aY.\S .
'{ axis va\ue is O.9S.
lInG corresponding
Q\
o
II
QO
0:
o
II
at
--
_
-
8.08
Lc2
Y ax is = TO - Too
~=
\r.
= O.\ 9
T-T
1
0.11
00
TO-363
=0.19
673 - 363
TO = 421.9 K
\
Mid plane temperature or
Center line temeprature, To = 42 \.9 K J
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edition).
\x-363
41\.9 - 363
~
=O.9S
Trans I'ent Heal Conduction 1.33
/.336
Heal and Mass Transfer
Temperature
at a distance
of 0.0 I 0 fr
420.72 K.
om the mid
plane'
IS
.....cificheat,
5y-
fIIe
To = 421.9 K
2. Temperature
at a distance
T, = 420.72 K.
(l
c::::
896 J/kg K
·tfuSIY1t)'.
11'ef11lal
dl
Result,'
I. Mid plane temperature,
of 0.0 10m fro
I
In tIe rn'd
I
.....,alcoo
p
'.
ductivlty.
k:::: 204.2W/mK.
ll"
_L
plane
Slob.
eharac
If
.'
length,
tenSue
Lc::::
0.120
2
120 "'''' thlck is illitiall
600"
C
I'
.
temperature oJ .
. tIS slIddenly
itnmer!)"edin IIYIi "' .II
"0"('
.
I'
.
I.'
qUId
til 1~
,re'HI ttng III (I teat transfer c()-e//iiciefl'
0/
I.
== 0.06 ffiJ
~~c:....-__
---
.1'
1400 WlmlK. Calculate the [ollowing
We knoW that,
I. Temperature (It tile center line after I millute.
2. Temperature (It tile surface
3. Total thermal energy removed per unit area
L ::;:120 mill == 0.120
C
273 = 873 K
final temperature,
1'0 = 1200 C
273 = 393 K
[Bi = 0.41
iJ
.'
. olid type problem.
0.\ < Bi < \ 00, So this IS infinIte 5
.. 'illiteplate, refer
ture/or In}'
[To calculate mid plane temperah tJ
'H
ar
i AfT data book page 110.65 H e IS. Ire
e
CUt (i)
at the center
2. Temperature
at the surface
3. Total thermal
energy remover I pe r unit area
Solution "
Density,
J(
h = 1400 W/Il12K.
co-efficient,
Toflnd :
I. Temperature
Properties
a, == k
\400
O.Q§
204.2
~
T, = 600
0
h Le
Biot number,
III
Initial temperature,
Heat transfer
2
== -
III A slab of aluminium
Give" :
Thickness,
10-6 m2/s
:::: 84.18
.'
.
t"
11111.: a ter
I' III inute
I
at
X axis -+ Fourier number::;
"AIK)I.~
..
book page ,,0 / J
[From HAn data
of aluminium
are
L;
0-6 )I. 60
:::~)2
l. t= \ minute=
I
60 51
p = 2707 kg/rn-'
~
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1.33
Transient Heat Conduction 1.339
Hear and Mass Transfer
I "J
I..tIll
tUre at t
V" re(1lpera
i.e., x = Lc
he surface
\ x = 0.06 m
urvc -...
I -:;
.c HMT data book page nO.66 - Heisler chart]
[ Rejer
:::; 0.411
hLc
hL
Cune
-...
k = 0.411
rresponding
Y axis
.
Biot number,
curve ~
x == Lc -
X a,XIS ~
X axis value is 1.403, Curve
can find
alue is 0 4
alue is 0.62.' II. From that, lI.e
.
1
B· = -
k
I
~=
= 0.411
I
0.06
. 0 411
X a,XIS va ue IS.
,
curve value is I.From that, we can
•
:) rresponding Y axis value
IS
0.85.
hLc
-k-= 0.411
1-
Too = 0.85 ~
~~
- Too
1.403
Y axis=
TO - T
---
= 0.62
hLc= 0.411
k
To - 393
= 0.62
873 - 393
TO = 690.6 K
I Centre line temeprature, To = 690.6 K
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I
Y axis = T x -T
To-Tee)
ee)
= 0.85
Tx - 393 = 0.85
690.6 - 393
T, = 645.96 K
Tronsie,,' Heo: CQMuction 1.341
1.340 Heal and Mass Transfer
C4S~ (Iii)
Total thermal enercv
. e.J re moved
or
Total heat energy removed.
[Refer HMT data b
0.8
ook Po
_ h2 at
----..:....
X axis
ge 110.6 }
0.6
k2
Q
- (1400)2
-~I(M>£,
x 84 18
/"
04
~ .
(240.2)2~
\ X axis
Curve ~
==
0.171J
hLc
-k-
\0
\()2
1400 x 0.06
204.2
I
Curve
_g_ = 0.24
(J):::::>
hLc = 0,411 \
-k-
i--s
Qo
:::::>
Q
X axis value is 0.171, curve value is 0.411. From that we
can find corresponding Y axis value is 0.24.
'
Y·axis = - 0 = 024
0
.
We know that,
x Q
o
\Q
Rts .. " :
00 = pCp LITj
-
T
[Refer HMT data bool
page "0.6J)
= 2707 x 896 x 0.120 x [873 - 3931
~~7::
0.24
= 0.24 x \39.7 x lO~
... (I)
0
=
106 J/1I12
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I. Temperature at the centre line, TO = 690.6 K
2.Temperature at the surface, T, = 645.96 K
3. Total heat energy removed, Q = 33.52 x l()6 J/m2,.
1.342 Heat and Mass Transfer
I1J A long steel CYlinder 15
.
temperature of 350 C. It is Icm lI'am
. eler 01';0'
p
ocea
Ill(
a/"l1/t
Calculate tl,e fOllowing
0 Q fur"ac
Y 01 Q
•
.
1 ,.,..,.
. lime reqUiredfor tile ax'
e Qt 950'
Transient Heat Conduction 1.343
~ we~(10
hL c
. number, Bi - k
r.
610t
150 x 0.0375
20
IS temperatu
2. C·orrespondtng temperatur
re 10,.elle!, 820'
.
e at a radius 01"
C
t'me.
'J 6 e", III
Take a = 6.11 x 10-6 m2/s, k = 20 W.
0.1<Bj < 100 , So this is infinite solid type problem.
'/tng, /1 :::: J 50 WI",1!.
Given:
Diameter,D=
[ Bi == 0.281251
:::>
IhQl
15cm==0.15m
Radius, R ==7.5 ern ==0.075 m
I cast (i)
Initial temperature,
Tj ==35° C + 273 == 308 K
I
Final temperature,
T
ex)
==950° C + 273 == 1223 K
a == 6.1 I x 10-6 m2/s
AXIS.
.
temp erature (or) centre__Ime temperature
I
To==8200C+273=1093K
I
To ==1093 K
Time (t) ==?
k == 20 W/mK
[Refer HMT dolo book page no.68(Sixth edition))
h == 150 W/m2K
Tofind:
Curve
I.Time required
2. Corresponding
for the axis temperature to reach 820°e.
temperature
x 0.075
= 0.5625
20
Yaxis
To-Too
T, - Too
For Cylinder,
Characteristic
K
ISO
at a ra d IUS
. 0 f 6 ern at that time.
Solution:
hR
length, Lc == R
2
0.075
2
0.0375 m ]
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1093 - 1223
308 - 1223
= 0.142
. 0 5625. From that, we
Y axis value is 0.142, curv e value
. 2IS5 .
can fiInd Corresponding X axis. value IS
.
'*li!J/l::
L
"
'-:;-'
1.344 Heal and Mas. Transfer
Transient Heat Conduction I.~
X ,axis
0. t
,. ., -- -""'25
R2
.
==
. 0 8 X axis value is 0.5625. From that,
.
. 08on diIn g Y axis alue IS . ).
value
CurVe
corres P
(LIl d
15
Tr - T
.;J)J
yaX
is ==-
= 0.85
To-Ta.,
(J. t
-=
R2
2.5
T, - T if.
_:.----
0.85 th-----~
-:=
To - T,FJ
-R2
U t
6. J J
x J ()-6 x t
(0.075)2
= 2.5
hR == 0.56
k
= 2.5
/t = 2301.55/
== 0.85
Case (ii)
Intermediate radius,
Tr - 1223
r = 6 em = 0.06 m
1093-1223
[r = 0.06 mj
== 0.85
_8200C+273=10931<
[ .: To-
{Refer HMT data book page no. 69(.S Ixth edilion)]
Curve
r
R
=
0.06
Result:
0.8
·dt-230
I. Time require
, -).5s
0.075
.
2. Intermediate
X axis
=
hR ,
k
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tern perature, T r
:=:
1112.5 K.
1.346 Heat and Mass Transfer
o
Transient Heat Conduction 1.347
A sphere of 30 mm dia
meter is .
temperature of 450°C It is t
,nitlally III
·
p
aced
in·
Q IIJ,'
center lme temperature rea h
llir at 2'
c es 3500
~oC
transfer co-efficient /, = 15 W/nr2
C "'ith tire IIIrtiJ ~
immersed in a water at 220C . K. After thlll fir locoJ ~
wlIh he
e $ph
of 5500 W/m1 K until the cent
. attrQlISJer Cn.... tlti
if,
~ for
. .
. Characteristic
I Lc
Calculate the following:
3. Surface temperature after cool'
,
To = 4500 C + 273 == 723 K
Initial temperature,
= 30 mm == 0.030 m
Final temperature, T = 220 C + 273 == 295 K
0
.
Intermediate
temperature, T -- 350 C + 273 == 623 K
OC)
Radius, R = 15 mm = 0.015 m
Density, p = 3100 kg/m'
Specific heat, Cp = 1005 J/kg K
For lumped parameter system,
-hA
[ CpxV x p x
= 22 W/mK,
t]
.. , (I)
==e
Thermal diffusivity, ex. = 6.6 x 10-6 m2/s.
V
Characteristics
Solution:
Case (i): Cooling in air
We know that,
(I)
::)
T-TOC)
where,
s, = T
h
= 15 WIm2K
length, Lc == A
[Refer HMT data book
page no. 57]
.[Cp:hLc' p 'I]
==e
_xt]
hLc
Biot number,
10-3 <0.1.
analysistype problem.
For lumped heat analysis (Cooling in air)
Given:
Diameter of sphere, D
= 3.4 x
Biot number value is less than 0.). So. this is lumped heat
mg III lVtlJer
a= 6,6 x 10-6 m1/s.
I
15 x 5 x 10-3
22
B·I
Take p = 3100 kg/mJ, C p = 1005 Jlkg, k = 22 WI"",
0.015
3
= 5 x 10-3 m
B·I
1. Time required/or cool,'n'o' ,
Gina"
2. Time required for coolin»G in Water
Thermal conductivity, k
3
=
ter /llre t
v-tf!~
elllperOll4re r~
from 350°C to 60°C
= R
length Lc
623 - 295
723 - 295
15
[
==e lo05x 5 x 10-3 x 3100
fi
'
;
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.r~I.;'.
.1,
,.
Transient Heat Conduction 1.349
1.348 Heat and Mass Transfer
In [623-295]_
_
-~
)( 10-3 '(100)(
It;: 276.3 sJ
Time required
_-
-15
723-295
t
k
~=3.75
"'"
22
for cooling in ..
air IS 276.3 s.
Case (ii) Cooling in water
333 - 295
:::623 - 295
B·lot num b er B· == --hLc
,
I
k
where h - Heat transfer
= 5500 W/m2K
co-efficient
I
For sphere
Characteristic
length, Lc =
R
3 75 Y axis value is 0.1158. From that, we
Curve value IS . ,
.
8
d·
X axis value IS 0.4 .
:In find correspon
109
•
Xaxis=
3
== 0.1158
~
R2
== 0.48
0.015
3
I
=
Lc
5 x 10-3 m
_E!_ == 3.75
I
To- T
__
00_
k
= 0.1158
r, - Too
5500 x 5 x 10-3
B·I
22
a. t == 0.48
I Bi = 1.25 ]
R2
I·d
0.1 < Bi < 100. So, this is infinite so I ty
pe problem.
E:...L == 0.48
R2
for infinite solids (Cooling
•
•
Initial temperature,
I
•
Final temperature,
in water)
0
T, = 350 C +
0
Too = 22 C +
0
Centre line temperature,
To == 60
273 ==623 K
273==295
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(0.015)2~
333 J(
C + 273 ==
[Refer HMT data
6.6 x 10-6 x t == 0.48
K
bOOK pO
geno)1
t
== 16.36 s
. 1636 S
]
1. in water
The time required for CO mg
IS·
~.
' ..... '
.~~
"....':
7
/ I~II IfNI/lind M'I"e 'l"il'"
.
n", II)
,
, ''''?,t'f'
~rt-'J/"
(
r ~
::().J
;:;
To - Trl..
~
333 - 295
,
~
T, - 295
J,(J ~
(j (1' I
1'1 It I) fi1IJ(",M
X lf~I'"
<.
I
"'1; ~lJit;,-"
rrr =
'
.,,·,.r-kJ
k
2. Time required for cooling in water, t = 16.36 s
- ,,7
Curv« v~llI(; ,I} I. X axii. o,'},
,
after cooling in water is 306.4 K.
L Time required for cooling in air, t = 276.3 s
22
villo
I
In'" :
. ')(J(J / C).{j I
i1,d~
3()6.4K
surface temperature
III'
:: 0,3
J. Surface temperature after cooling in water, T, = 306.4K.
value ; ~ ),75,
com: pon
M d'
mg Y
1.5.9Solved University
Problems
on Infinite Solid!
I1]A slab of aluminium
L:
R
hR
T3.75
I:
I
10cm thick is originally at II
temperature of SOODe. It is suddenly immersed in a liquid
at /{)ODC resulting in a heat transfer co-efficient of
1200 Wlm2 K. Determine the temperature at the centreline
and the surface 1 minute after the immersion. Also
calculate tire total thermal energy removed per unit arell
of tireslab during this period. Tireproperties of aluminium
for the given conditions are
a = 8.4 x 10-5 m]ls
k = 215 WlmK
p « 2700 kglm3
C = 0.9 kllkg K
[ May 2005. Anna un.iv.]
~
~.l~
Scanned by CamScanner
I.>.,
1:_1~_~ Iltffl/ tmtl A111.\·.~
-
---
(ti"ctn .'
~
7h"""re,.
----~..::::::.--....c----="=>-_
Transient Heat Conduction 1.353
llli '''"eSS,
L
10 ell)
0.10
111
Initial
• T
hllllP~1'II11I1'
273 .. 373
h::::: 1100
c\)- 'I)'icirlll,
Prop,.rtico!tof "'''",ini"",
"'('
p - .700
kg/Ill
Density,
.73 ... 77
T - 100"
FillaltcmpCl'lItlll'("
I kilt lntllslcl'.
~.
(11
~
Thermal
condul:ti,
uv, k ::: .. I" W/mK
Specific
heat,
== 0.9
I
.
Fourier number
X aXIs
-+
r:l_!axis. -+ Fourier number
Curve
-. Temperature
at the surface
3. Total thermal
energ
entre line after 1 minute.
removed
per unit area.
I
We know that
length
for slab,
c
--+ =
=2.0161
hLc
k
1200 x 0.05
215
Solution:
Characteristic
L 2
=
[.: t = 1 minute = 60 s]
. 103 J/kg~ K .
at t.he
i!.!._
8.4 x 10-5 x 60
(0.05)2
k.l/kg K
I. Temperature
Carve -> "~"
0.2791
. 0279. From that, we
. 2. 016 , cu rvevaluels.
X axis value IS
. 064
Y axis value IS . .
can find corresponding.
L
0.10
L =_=c
2
2
ILc = 0.05 m I
hLc = 0.279
hLc
s, =T
Biot number,
~
I
k
1200 x 0.05
215
Bj = 0.279/
at
. In
. between 0 . ) and 100,
Biot number value IS
blefll.
i.e., 0.1 < B, < ) 00, So, this IS., infinite. sorid type pro
lJ
•
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I
1
I
I
I,
I(
I
, i) :::::
8.4 . 10-5 m2/s.
1lJjilld:
L'
1\
Whn2K
Thermal ditTusivit
p " O.
leu late IIIid plane temperature for infinite plate. refer
rTo ca
bOOkpage I10 .6S (Sixth edition) - Heisler chart]
·t'fdllIS
~~
00"
j
I
I
(
== 2.016
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\).119
~2
Tran .
_-=-:=--=-:-7_~:..::.:~sl~en'!!_t!}_H.~e(Jalend'
Temperature at the surface T _
0
llello" 1.357
.
, x - 598.28 K
3. Total thermal
energy removed per urut. area
Q = 33.04 x 106 J/m2
'
Q
Qo = 0.34
r1I A large iron plate of lOcm tl,iclcnessad'
.
I.:J
n ongma/lyat 800-C
is suddenly exposed to an environment at O.C
•
,1"1"...
",'here'''e
convectIOn co-efficient IS 50 Wlm1K . Calcula'e 'he
ttmperature at a depth of -Icm from one OJ'he
.£
facts
100 seconds after the plate is exposed to t'he tnv"OIf~IIL
.
We know that ,
How. muclr
.£ th
l
.. energy has been lost per unit area oJ
e pille
durtng this time.
[June 2006 - Anna. Univ]
[Refer HMT data bookp
age no.63 (Sixth td"1I101lj!
= 2700 x 0.9 x 103 x 0.10[773-373)
Qo = 97.2 x 106 J/m2.
GiI'en:
Initial temperature,
Ti = 8000 C + 273 = 1073 K
Final temperature,
Too = 0 C + 273 = 273 K
Convective
From graph, we know that,
Distance,
Q
Qo = 0.34
0
heat transfer co-efficient,
h = 50 W/m2K
x = 4 em = 0.04 m
time, t = 100 s
Io flnd :
I. Temperature
Q = 0.34 'Qo
= 0.34
of iron plate, L = 10 em = 0.10 m
Thickness
(Tr) at a depth of 0.04m from one end of
the plate.
6
97.2
10
2. Total thermal energy lost per unit area, Q
Q = "3.04 ' 106 Jim·
Solution:
Properties of iron are
l raj
Q=33.0-l
Thermal
1
10 Jlnt
Th erma I diff
.'
I· USIVlty,
• ~.Sldl:
I. Temperature
at the centre
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line
conductivity,
To::: 629 K
k = 72.7 W/mK
2
- 20 .34 x 10-6 m /s.
(l -
1. 358 Heal and Mass Transfer
Density,
p = 7897 kglmJ
Specific
heat, Cp = 452 J/Kg K.
Transient Heat Conduction 1.35
______
.> curve::::
For Slab
Characteristic
L
length,
L
2
c " -:::
0
.10
-....::.
hLc
k
50 x 0.05
72.7
2
§i"ve
:::: 0.0343/
~
We know that,
Biot number,
8· == _hLc
can
k
I
is 0 183 curve value is 0.0343. From that, we
X xis va Iue I . ,
a
ding
Y axis value is 0.92 [From graph}.
find correspon
50 x 0.05
72.7
:LC = 0.0343
TO- Teo = 0.92
Ti - Teo
[Note:
Biot number value is less than 0.1. So, thisislumped
heat analysis i.e., Neglecting internal resistance.
Bur
we have to find temperature at a depthof 4cm from
one end. So, we can go for Heisler Chart}
Y axis =
TO - 273
To calculate mid plane temperature, refer HMT data bookpage
X axis
== 0.92
Ti - Teo
Case (i)
no.65(Sixth
To - Teo
= 0.92
1073-373
edition).
.
~ Fourier number
=>
at
. plane temperature or Centre I·me temperature, To
Mid
Lc
Temperature at a depth ofO. 04 m from mid plane.
(0.05)2
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0.813J
1009 K
Case (ii)
20.34 x /0-6 x 122
I X axis ~ Fourier number
TO == 1009 K
= -2
I,
[Refer
HMT
H""'A
66 (Sixth edition)
data book page no. -
1.360
Heal and U
ass Transfer
X axis ~ B'
lot numb
Curve
::::
t- ",~
c
can
er, B.",
I
1
h Lc
Transient Heal Conti .
_-------.....::..=.:::.:..:-:::::..~UC~/lIloon 1.36/
~
k:::
0.0343
2
~Fourier
number = h 11 t
X 8,,15
0.05'" 0.8
k2
. X axis value is 0 03
find corresponding
4~, curve Val
y
aXIS value'
.
Ue IS 0.8 F
IS 0.90
. tOrn th
.
= (50)2 x (20.34 x 10-6)x 100
(72.7)2
I
at, "'e
X axis = 0.962 x 10-3 ]
x
L"'0.8
c
hLc
I
Curve
=
@urve
-k-
50 x 0.05
72.7
= 0.0343/
X axis value is 0.962 x 10-3, curve value is 0.0343. From
that, wecan find corresponding Y axis value is 0.02.
hLc
k = 0.0343
0.6
Tx - 273
1009 - 273
= 0.90
r, = 935.4 K
Temperature
at a depth of 0.04m from one endoftheplate,
T, = 935.4 K
Q
Qo
0.4
0.2
10-5
10-4
10-3
10-2
1 xlOl
10-1
h2a t
Case (iii)
k2
Total thermal
energy lost per unit area, Q
67 (.'Six/h edition)}
[Refer HMT data book page no.
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Y8)(15·=_Q_=002
Q
.
o
... (I)
Transient Heal Conduction 1.363
_
1.362 Heat and Mass Transfer
We know that.
..
edition)]
Io
Q
= 7897 x 452 x 0.10 [1073 - 273]
=
. iry k::: 42.S W/mK
0.285 x 109 J/m21
,henna
go 0.02
\ conduCtlVI
,
~fj"d:
. e temperature,
I. center lin
=
2. Temper
Q = 0.02 x Qo
= 0.02 x 0.285 x
109
To
ature inside the plate
~"';o":
fll Platt:
106J/m2
Q=5.7x
25 min :::0.0125 m
80
.inutes::: \
s
3 Ill
fit1le, t :::;
fficient, h = 285 W/m2K
sfer
co-e
O
tlest tra
. iry a.::: 0.043 m2/hr
I ditTuSIVI ,
fhef(lllJ
= \. \ 9 x \ 0-5m2/s.
Vista
[Refer HMT data book page no.63 (Sixth
(lCe,·
r - \.
Characteristic
leogt
hL
c
Result:
2. Q = 5.7 x 106 J/m2
Weknow that,
111 A large steel plate 5 em thlck is initially at a uniform
temperature of 400· C. It is suddenly exposed on hoth sides
to a surrounding at 60·C wit" convective "eat transfer
co-efficient of 285 Wlm}K. Calculate the centre line
temperature and the temperature inside the plate 1.25 em
from tile mid plane after 3 minutes.
Takek for steel = 42.5 WlmK, a for steel = 0.043 m1/hr.
[Nov'96
Given:
Thickness, L = 5 em = 0.05 m
I ..
n.llIal temperature,
Flnalte
mperature,
L
= Q;Qi
2
2
= 0.025 m]
I. Tx = 935.4 K
;,
~
~
=
\ 2S em from the mid plane.
.
r, = 400 C + 273 = 673 K
0
TaJ = 600 C + 273 = 333 K
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MUj
Biot number,
~
hLc
_ _]8S .::-Q.:021.
a, = k ~i::
42.5
0.167~
id
O.\ < s, < \ 00, So, t
hi . infinite soh ty
pe problem.
IS IS
.
temepratul
rature or Mid plane
line tempe....ook page no. 651
{Tocalculate centre
.
HMT data vvv
for infinite plate, refer
Case (i)
Transient Heat Conduction 1.365
J.364 Heat and Mass Transfer
~
X axis ~ Fourier number
Ti - Tet)
= at
Lc2
=
~:::0.64
673 - 333
1. 19 x I 0-5 x I 80
1--------
=0.64
(0.025)2
-
\ X axis ~ Fourier number = 3.42 ,
~
550.6 K
Curve
= hLc
k
~~
=
I
Curve
285 x 0.025
42.5
.
Temperature (T;t) at a distance of 0.0125 m from mid plane
= 0.167
{Refer HMT data book page no.66}
hLc
u., = 0.167 \_
= -k-
X axis ~ Biot number,
X axis value is 3.42, curve value' .
can find corresponding' Y r..
.
IS 0.167. From that, we
axis value IS 0.64
x
e urve ~
::: --Lc
a, :::k::: 0.167
0.0125
0.025
==
0.5
X axis value is 0.167, curve value is 0.5. From that, w
== 0.64
TO-T "',
T
= 0.64
tan find corresponding
hLc
-k-= 0.167
T 1- T
(I,;:::
Y axis value is 0.97.
0.97 ~~...,_:>....--~
O.S
To-Tu:
0.6
i - T."
h~
k
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0.\67
Transient Heat Conduction 1.367
1.366 Heal and Mass Transfer
:::::1 hour == 3600 s
fme t
Y axis ==
T, - Ta>
P == 998 kg/m3
.
2
fer co-efficient, h == 6 W 1m K
tieat trans
h at C == 4180 J/kg K
specific e , p
al conductivity, k = 0.6 W/mK
Therm
k _
0.6
al
diffusivity,
ex
==
-p
C
998
x 4180
Ther m
p
I
To-Too
Tx-Too
= 0.97
To-Too
T, - 333
----'''----
550.6 - 333
'.
penslty,
== 0.97
= 0.97
(ex == 1.43 x 10-7m2/s.\
Temperature at a distance of 1.25 em from the mid plane is 544 K.
Result:
ToFind:
Center line temperature (To)
1. Centre line temperature, To == 550.6 K
2. Intermediate temperature, T x == 544 K
Solution
For
III A 10 em diameter apple approximately spherical in shape is
taken from a 20° C environment
length, Lc ==
and placed ill a refrigerator
is 5° C and average
where temperature
Sphere.
Characteristic
==
heat transfer
Thermal conductivity
= 0.6 WlmK.
We know that,
hLc
Biot number, Bi ==
[Apr'98
0.05
3
Gc ~ 0.016 mJ
coefficient is 6 Wlml K. Calculate the temperature at the centre
of the apple after a period of 1 hour. The physical properties
of apple are density = 998 kglm3. Specific heat == 4180 J/kg K,
R
3
M.UJ
Given:
k
:::
~
0.6
Diameter of sphere, 0 = 10 em = 0.10 m
Radius of sphere, R = 5 em = 0.05 m
Initial temperature, T, = 20° C + 273 == 293 K
F'
I
.
ma temperature, Too= 5° C + 273 = 278 K
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~
~
~.
0.1 < Bi < 100. So,
lid type problem.
. . . finite so
this IS In
1. 368 Heal a"d Mass Transfer
I"fi"ile Solids
d
--------b [To calculate centre I'me temp
ala ook page no.71 (S'IX th editionj]
"
erature
for sphere ' reler
c.
Ii
Xaxis
MT
= ~
R2
=
1.43 x 10-7 x 3600
,tsll/f: Center line t~l1ll;pr:ltLln.:. r
(0.05)2
I X axis = 0.20 I
Curve
hR
k
I
0.5
] A long steel cy! i,,"« 1_ cm ,/illmtltr and illi,ially ., 20' C
0
i> plac,,1 in 0 [ur nil<' at 820· C wi,h h = 14 IfI m ' K.
\ C.lc.
,I" ume "if"ir_" fur ,I.. «<is " .. p"II'."
I
lote
800· .,t I 0 CIIlc "IIIte ,I.. corr<spo",lillRttn",,,a,."
.,. r.di. of 5.4 <III ,II ,IIU' time- Physic,,1 proper,i" of
'0
Z
stee! CITe k :: 21 WI",K, a=
0.5 I
X axis. value is 0 20
find corresponding
Y '. ,curve. value is 0.5. From that w
axis value IS 0.86.
' e can
~ Y axis = To - Too
Ti-Too
~90.9K
".eI,
6 x 0.05
0.6
.Curve
c,
= 0.86
[0(.'('99
GiI'tn:
Diameter
Radiu
f c lin lcr. \) :: 1- em :: O.\ - til
f phere,
R -
Final temperature.
Heat transfer
or
Axi temperature
cm " 0.06 111
273:: 293 K
T, ==
T = 8,0· C
273 = 109- K
2K
_etli~i';llt. II - 140 W/ln
} .,
= 800· C
21)
o
Centre line temperature
4
Intermediate
radiu , r:: SA ~t1I::\ 0-6
0.05mIls.
III
1
Thermal dillusi it)" u. ;::: 6.1
Th
. I·:: 2 \ W /In K
ermal conducti It),,"
~.
~
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M. Vj
II
Initialtemperatur·.
hR _
k - 0.5
6.11 )( Itrb m /!.
= ,01' K
Transient Heat Conduction 1.371
1.370 Heat and Mass Transfer
To find:
_____
I. Time (t) required for the axis temperat
ure to reach 800
0
2. Corresponding temperature
(T ) at a
r
Solution:
d'
ra IUS of5 . 4 ern
C.
1073 - 1093
.
::: 293 - 1093
i or Cylinder,
_ 0.06
Characteristic length, Lc ::: R
-2
2
curve V a
0.03 m 1
lue is 0.4 Y axis 0.025. From that, we can find
'.
X axis value IS 5.
l.,rr~~I[)()n(Jlm~
We know that,
Biot number B.::: hLc
k
'I
~
::: 140 x 0.03
21
Ir--B-:::-0-.2-1
j
at
0.1 < Bj < 100. So, this is infinite solid type problem.
':ase (i)
-
R2 -
= ~
~ X axis
Axis temperature
}
or
To ::: 8000 C
Centre line temperature
R2
=
5
5
5 x (0.06)2
t =
TO::: 8000 C + 273 = 1073 K
It
Time (t)?
=
(6.11 x 10--6)
2945.9
sJ
Case (ii)
[Refer HMT data book
page
Curve ::: hR
no. 68(Sixth edition)]
Intermediate
k
:::
140 x 0.06
21
radius, r = 5.4 em'
::::0 054 m
[Refer HMT dala book p g
::: 0.4
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Curve
_L
= ~::::0.9
R
0.06
. h dition)]
a e no.69(szxt e
I. J 72 Heal alld Mass Transjer
Trans,
LJ
lent neat Conduction
X axis
hR
K
==
, ~ems
140 x 0.06
for practice
~uminium
cylinder 5 em in diameter and initially ot
/. 200 C is suddenly
exposed to a convection environment
'
at
11
== 0.4
21
I. 373
70llC and I, = 525 Wlm2 K, Determine the temperature at a
radiUS 0/1.25 em and the heat lost per unit length 1minutes
nfter the cylinder is exposed to environment.
Curve value
. is 0.9 ,.~X axis va I lie IS. 0 4
find corresponding Y axis val
. 0
.. FrollJ that
lie IS .84.
' We can
Take P = 2700 kglm3, C = 0.9 kJlkg K, k = 215 WlmK
a= 8.4 x 10 52'
m Is.
R
0.9
I Oct' 2002 M U]
]. A slab of rubber of thickness 40 em, initially at a uniform
temperature of 300 e. It is exposed to air at 30" C, the
convection co-efficient being 240 kJII" m1 "C. Assuming that
il is a large slab, find the mid plane temperature after 15
11
hR
Y axis == Tr -T <r.l
TO-T(1J
minutes.
== 04.
- k
{Manonmanium
== 0.84
J.
uniform temperature of 200 C is suddenly immersed in an
oil both at 20°e. The convection heat transfer co-efficient
between the fluid and the surface is 500 Wlml "C How long
will it take for the centre plane to cool to 100"C
{ Madurai Kamara} University Apr'97]
11
== 0.84
Tr-1093
1073 - 1093
(a = 1.25 x 10-5 m1ls, p = 7833 kglmJ,
Cp = 465 Jlkg ° C, k = 43 Wlmll C) thickness 5 em initially at
Tr -T (1J
TO-T(1J
A steel plate
Szmdaranar University Nov'96]
== 0.84
4
~1076.2KJ
R~sull :
.'
initially at a uniform
, A metallic sphere of radiUS 10 mm IS
I'
't .
I d b"jirsl coo »s I tn
temperature oif400° C. It is "eallrea e
'I '
ntralltmperalure
air (/, = 10 WI",2 K) at 20° C unll ~/S ce
both 01 20" C
hed m awaltr
reaches 335° e. It is then quenc
if Ihe sphere cools
re
with h = 6000 WI",1K until the cenl 0 uired/orcooling
uJ Ihe time req
from 335°C to 50° C. Co",p e
. oJ ro.nertieso/sphert
.
fi L wing phYSIC p r
'" air anti water for the 0 ,0
J
IT"
. nne required for I
.
2945.9 s.
t re aXIS temperature
to reach 800°C is
2. Temperature (T
r) at a radius of 5.4 em is 1076.2 K.
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A
~
I
74 Heat and Mass Transfer
1. 3
~onduction,
In c
motion or ~irect impact of molecules. Pure conduction is found
only in sohds.
p= 3000 kglnr3,
C
= 1000 Jlkg K
Define Convection.
k=20 WlmK
4. convection is a process of heat transfer that will occur between
solid surface and a fluid medium when they are at different
a
.
.
temperatures. Convection IS possible only in the presence of
a= 6.6 xlfr6 m2ls.
[ Bharathiyar
UniversityA
pr'97]
A 15 em thick plate initially at 20 C is suclden/u Pt'
U Into a
furnace at 1100 C. The values of thermal condu CtilV,ty
.
J
0
5.
Conduction 1.375
energy exchange takes place by the k'memauc.
I
fluid medium.
:J
0
I
and diffusion co-efficient
of plate are k = 30 WlmK
I
an
d
a= 0.042 mllhr. The average heat transfer co-efficient;s 350
WI",2 K. Find the temperature
after 5 minutes of heating.
at tire surface and at the centre
[Nov'97 Manonmanium Sundaranar University]
Define Radiation.
5. The heat transfer from one body to another without any
transmitting
medium is known as radiation. It is an
electromagnetic wave phenomenon.
6. Sttrte Fourier's law of conduction.
[Apr'97, Oct' 98 Madrasllniv ,
May'04, May'05 , June'06 Anna Univ]
The rate of heat conduction is proportional to the area measured
normal to the direction of heat flow and to the temperature
1.6Two mark Questions and Answers
gradient in that direction.
1. Define heat transfer.
Qa-A
Heat transfer can be defined as the transmission of energy from
one region to another due to temperature difference.
2.
Q= -kA dT
dx
What are the modes of heat transfer?
1. Conduction
2. Convection
3. Radiation
3.
dT
~
where, A - Area in m2
dT _ Temperature gradient, KIm
d.x
What is conduction.
Heat conduction is"amechan ism of heat transfer from a region
of h~gh temperature to a region of low temperature within a
medlUm(solid
I'IqUiid or gases) or different medium 10
. direct
" .
I
phYSicalcontact.
I
Scanned by CamScanner
k - Thermal conductivity, W/mK
7
.
'97 M U. Oct' 99 M lJ.'
.,
bility of a substanceto
. defined as the a
. 'ty
Define Thermal conductivi .
.'
Thermal conductivtb'
conduct heat.
IS
[Apr
Conduction 1.377
1.376 Heat and Mass Tra17~r('_,.__
n
s. Write down the three dimell!i;0IU1/~u
where
cIon e
in Cartesian co-ordinate system,
Sf:::::
qUatioll
heat conduction
I
L _ Thermal resistance of slab
R:::::V:
[May'05 & June'06 A
The general three dimensional
,
,
cartesian co-ordinate IS
T - T2
nna Univ.)
e
.
qUatlon in
hi kness of slab
J..,- T ic
l( - Thenna
I conductivity of slab
2
a2T + a2T + a T + ._~ = .L a~
ax2
ay2
az2
k
A - Area
01
a
\
,
where
q. - Heat generator - W /m2
a - Thermal diffusivity
9.
lotion for conduction of heat througlr a
"'"'te down tire eql
, h 1I0wcylinder.
o
6.T overall
Heat transfer, Q =
R
J2 "rl
- 1112/s
eqllation
where
6.T=T\-T2
[May'05 & June'06 Anna Univ.]
1
R=_'n
27tLk
Write down the three dimensional
heat conduction
in cylindrical co-ordinate system.
The general three dimensional
cylindrical co-ordinate is
heat conduction
equation in
~2)
_Thermal resistance of slab
r
I
L _ Length of cylinder
k _ Thermal conductivity
2
.s. = j_a aTae
crT + .L aT + .L ;/T + a T +
or2
r ar
r2 a$2
8z2
k
10. List down the three types of boundary conditions.
1. Prescribed temperature
[Dec-2005. Anna Univ]
rOuter
2 rl -
radius
Inner radius
.
13. Write down equatIOn fi
spl,ere
2. Prescribed heat flux
3. Convection boundary conditions
11. Write dow" the equation for conduction of heat through a slab
or plane wall
Heat transfer, Q = ~ T owrall
R
~c
Scanned by CamScanner
t through /IO/low
or condllction of I,ea
Heat transfer, Q :::
~ T overall
~
R
Conduction 1.379
1.378 Heal and Mass Transfer
..,n tIle eq
I
til! ile t 0
I~"r'pes of cylinder.
U. State Newtons law of cooling or convection la;;-----'__
[ April'98 M Uj
r
Heat transfer by convection is given by Newtons I
aWofcool'
Q = hA (Ts - Too)
uation for heat transfer throug" composite
109
where
A - Area exposed to heat transfer in m2
B
h - heat transfer coefficient in W/m2K
Convection
A
Ts - Temperature of the surface in K
hb
Too- Temperature of the fluid in K
15. Writedown tire equation for heat transfer tllrouo/I
a compOSlIe
.
to
plane wall.
[ April'97 M u.}
Tb
d) T2
(DTI
L\T overall
Heat transfer, Q::::
R
(D
(i)T)
where
LI
L2
,
L\T::::Ta-Tb
I
R=-
21tL
Heat transfer Q _ 6T overnll
R
61-'1' u- 'l b
--.s_+
,
hbA
A - Area
11(/
-
Ileut transfer' coe.f'Tiicienr
"
'
,
at Inner
diameter
hI; - Ileat transfer"
" coe ffici
ic lent at outer side
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2.) ___l.2In r
I
+ ---
lr,
~3 )
.-l..-+-+
k
hart
hbr3
k
2
1
.
•
uation
I steady state conductIOneq
lOna, .
17. Write down one t/imens
without internal/leat generatIOn.
where
R - -L+
L2
L3
I A
,--+--+
'.
kl A
k2A
k3A
L - Thickness of slab
In r
)
ciT ::::0
QX2
'
•
I conduction equation
wo dinrenslOna
18. Write dow" steady stllte, t
without heat generlltion.
1.380 Heal and Mass Transfer
/9. Write down the general equation for one dim .'
state heat transfer in slab or plane wall witbout ell.~/O"(l1
l
SIf.'(fdy
leut gener .
llt1o"
a2T + a2T + a2T = .L aT
ax2
ayl
az2
<X:
al
20. Define overall heat transfer co-efficient.
.
The overall heat transfer by combmed
in terms of an overall conductance
efficient 'U'.
[ April'97 MU
modes is Usually
.J
expressed
or overall heat trail l'
sler Co-
Conduction U81
~1
AddlUO of insulating material on a surface does not reduce
ount of heat transfer rate always. Infact under certain
t~eam s tances it actualiy increases the heat loss upto certain
Ircum
.
c. nesS of insulation. The radius of insulation for which the
thick
"
" I ra d'IUS0 f iI11SUIa tiion,
sfer is maximum
IS ca IIe d cntica
heat tran
".
.
.
thickness IS called critical thickness.
and t Iie c orresponding
.Ii
U~M
fins or Extended surfaces.
.
eat
surface 0 f heat transfer. The surfaces used for increasing
.
k
C
nown as
transler
ar e called extended surfaces or sometimes
Heat transfer, Q = UA dT.
21. Write down the general equation for one dimensional steady
state heat transfer in slab witll heat generation.
[Oc('99 MUj
22. What is critical radius of insulation or criticaltllickness.
[May'04 & Dec'04 Anna Univ - Nov'96 ,Oc('97 MUj
fins.
24. Stale the applications of fins.
The main application
= rc
Critical
radius
Critical
thickness
= rc - rl
offir.s are
I. Cooling
of electronic
2. Cooling
of motor cycle engines.
components
3. Coo[ingoftransformers
4. Cooling
Q
.
.
ible to increase the heat transfer rate by mcreasmg
[t IS possi
. hthe
of small capac Iity compressors.
25. Define Fin efficiency.
[Dec'04 & Dec'05 - Anna Uni!']
[ Nov'96, Oct'97 M U]
e ratio of actual heat
f ' define d as th
f
The efficiency of a III IS
ibl h at transferred by the III
. m POSSI e e
transferred to the maxunu
Qfin
Tlfin = Qmax
26. Defille Fin effectiveness.
~~_r_I
r,,c
J
'th fin to that
, of heat transfer WI
F·III et of,'ectivenes SI'S the ratto
without fin
Scanned by CamScanner
Dec'05 - Alllla Univ]
[Dec'04 &
[Nov '96. Ap,'2001 M Uj
(
Conduction 1.383
J,382 Heat and Mass Transfer
Qwithout fi~
17. What is meant by steady state heat condu CIOn?
tt
If the temperature ofa body does not vary Wit. I1tlln
. .
to be in a steady state and that type of cond uctlon
. IS
.e,kIt is said
steady state heat conduction.
n°Wn as
18. What is meant by Transient heat conduction
conduction?
or unsteady state
lfthe . temperature of a body varies with time ' it' IS salid to be in
transient state. and that type of conduction is kno wn as transient
' a
heat conduction or unsteady state conduction.
29. What;s Periodic heat flow.
In periodic heat flow, the temperature
varies on a regular ba SIS,'
Example:
1.Cyl inder of an Ie engine.
2. Surface of earth during a period of 24 hours,
30. What is non periodic heat flow?
In non periodic heat flow, the temperature
the system varies non linearly with time.
. meant by Lumped heat analysis?
{Oct'98 M VJ
.
I'
heatmg or coo mg process the temperature
111 a n~out the solid is considered to be uniform at a given time.
throUg analysis is called Lumped heat capacity analysis.
_ such an
. meant by Semi-infinite solids?
'~ {Oct'99 MVj
ot
IS . I'nfinite solid, at any instant of time, there is always a
\ JJ. Jfh semi
i
111 ~
h re the effect of heating or cooling at one of its
lint
w
I
pO
. e is not felt at all. At this point the temperature remains
boundanesd In semi infinite solids,
.' the blot number value 'ISCXl.
unchange .
IS wtonian
J1, '.JI/,{lt
"
,Fin effectiveness ~ Qwith fin ~
at any point within
34.
WI,at is meant by infinite solid?
id hi ch extends itself infinitely in all directions of space
A soh W I
is known as infinite solid.
, fi't
ol'lds the biot number value is in between 0.1 and
In m iru e s
,
'00.
0.1 < B; < 100.
35. Define Biot number.
,
'
0 the
It is defined as -the ratio of internal conductive resistance t
surface convecti~ resistance,
Internal conductive resistanc.:_
Bj == Surface convective resistance
Examples:
I, Heating of an ingot in a furnace.
2, Cooling of bars.
31. What is meant .b~ Newtonian
II.T·
I,eatin~ f or cooling process. ?
The process
in w hiIC h the internal
.
'~ r sistance IS
. assume d as
,
neg
igfble
in
comp'
"
.
as
N'
anson with Its surface resistance ISknown
"
ewtonlan
heatin g. or cooling
. process.
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hLc
Bj== k
.
.
.'
., Bioi Number".
36. What IS the sIgnificance oJ
,
U APr'2002 M. V]
[ NoV 96 IYI, ,
,
. Semi infinite
. '.
d heat analySIS,
..
.
d to find Lumpe
.~
B lot number ISuse ."
solids and Infinite solids
1.(
l
1.384 Heat and Mass Tram,fer
If
\
n, < 0.1 ~ Lumped heat Rnalysis _________
B. = r:IJ ~
I
Semi infinite solids
0.1 < B; < 100 ~
Infinite solids.
37. Explai/l the significance of Fourier number,
.,
"
It is defined as the ratio of characteristic
temperature wave penetration
Fourier Number
( Apr'2002 Mu)
body dime
.
nSl01l to
depth in time.
Characteristic
body dimension
Temperature
wave penetration
=
depth in time.
It signifies the degree of penetration
«r BastC Concepts
cr Dimensional
Analysis
cr
Boundary Layer Concept
«r
Forced Convectlon
«r
.
d Turbulent Flow
Lammar an
of heating or cooling effect
ofa solid.
38. What are the factors affecting the thermal conductivity?
1. Moisture
[Apr'9 7 M. u.]
2. Density of material
«r Internal Flow
3. Pressure
r::?
4. Temperature
it problems
39. Explain the significance of thermal diffusivity. [Oc/'98 M u.j
The physical significance of thermal diffusivity is that it tells us
how fast heat is propagated or it diffuses through a material
during changes of temperature
40. What are Heisler charts?
with time.
[Oc/'99 M u.j
In Heisler chart, the solutions for temperature distributions and
~eat flows in plane walls, long cylinders and spheres with tinite
Internal
.
.
. and sure"
lace resistance
are presented. Heisler
c Iiar ts are
nothing but a a na Iytica
. I solutions
.
.In the torm
.
of graphs.
/
I
" ,t.
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Free Convectlon
cr Solved Problems
5. Structure of material.
~
Chapter 2: Convection
C?
Solved Unlversl y
------
-
Ctl~PTER - II
2. fONVECTIVE
HEAT TRANSFER
~~~===========================
2.1.DIMENSIONAL ANALYSIS
Dimensional analysis is a mathematical, method which makes
of the study of the dimensions for solving
several engineering0
~.e
•
prohlems. This method can be applied. to all types of fluid
resistances, heat flow problems and many other problems in fluid
mechanics and thermodynamics.
2.1.1.Dimensions
In dimensional analysis, the various physical quantities used in
fluid phenomenon can be expressed in terms of fundamental
quantities. These fundamental quantities are mass (M), length (L),
lime (T), and temperature (0).
The dimensions of commonly used quantities in heat transfer
analysis is listed in Table 2.1 with reference to MLeT where
M
Mass,
-
L -
Length,
0
-
Temperature,
T
-
Time.
Velocity V
=
For example,
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L :: LT-I
Distance - - T
Time
~
/
] 2
Heal and Mass 7i'C/I/!Jjer
. ,
r--j---------------- .
-__
No.
Length
2,
Area
1
Velocity
1"',,, I pr
In
uch
h is kn "'0
"hi
"
~cOreI1'l'
rem t te t1 foil ws.
l3IJCkil1gharn 1t the
..If I~cr< 3re n v3fiabk
in a dimensi nail' h moe
,
lntalll In funds mental
dimensi I n inen
h he
<'loai n and I the-c
Quantity
I.
applie.
61'ffef(Oua o'~ hire I. required
TUb/~~
S.
learl
,,,i,bt
Or<.1 ran ed inl
n
dimen ink"
le
dio
i IIle icrrn arc ailed 1t term ..
>
term . The
'"
1cl1
Acceleration
1.1. . ,.d._nlage.
3
Mass
I.
Density
It exprc
varia
2.
fJ
01 Dimension_I An.lysls
the fun ti nal relation
se
ionic s term'.
\
n:tical
hip between
'olutiun
the
in a implified
It en
W
T
r 11'1 the experilllental
3.
data or direct solution of
P
[ern.
\I
ne 'eric'
Q
4.
The re
of tests call be applied to a large
problems with the help 01
Q
dimen
II
k
W/I11K
(L
11111s
J/kg K
i nal analysis,
2.1.4. Limitations of Dimensional AnalysiS
The omple,e inlo""alion
is not provided by dimensional
onsh
\. analy is. It onl indical's Ihal Ihere is some rel,II
,P
ani
between the paral~leters.
al
No inf r nation i given aboul Ihe in"rn n,,,,,h
m ,f
2.
the ph~ -Jl:al phenornenon.
2.1.2. Buckingham 1t Th eorem
A more general sit
be profitably
.
.
employed uation
is one'in wllie. I1. dimensional
analysis may
.
In which
tl 'ere IS
. no governing
Scanned by CamScanner
3.
DimcII~o!,al
analysi
electidn
f ariables.
doe
not give allY c1llC re~ Irding Ihe
~
24
Ileal and Mass Transfer
2.2.
DIMENSIONLESS NUMBERS AND TH
SIGNIFICANCE
EIR PI-t'rSIC
_------
I
Pr andt'
~l
2.2.1. Reynolds Number (Re)
(1"
ellcc
~Lon\lecli\le
num ber
provides
tivencss of the momentum
a
measure
and energ t
Heal T.
ranifer
of th
e relative
Y ran sport by d·ffi .
I USlon.
It i defined a
2.2.3. Nusselt Number (Nu)
It is defined as the ratio of the heat flow b
.
.
Y convectIOn process
gradient to the h
fl
nde,. an urut temperature
Ll.
.
eat ow rate by
onductlon under an umt temperature gradient th
h
.
c
roug a stationary
thickness of L metre.
qconv
Nusselt Number (Nu)
qcond
Velocity
L
\I
==
Length,
rn/s ,
where
m,
2.2.2. Prandtl Number (Pr)
0
h
f h
t e rnomentum
diffusivity
k A t1T
=T
L
L
1]
Nu
... (2.3)
Length, 111,
k
-
The Nusselt number is a convenient measure of the convective
heat transfer coefficient. For a given value of the Nusselt number,
the convective heat transfer coefficient is directly proportional to
thermal conductivity of the fluid and inversely proportional to the
significant length.
to the thermal
MOlllentum diffusivity
Thermal diffusiviry
as the ratio of product of inertia force and
buoyancy force to the square of viscous force.
It is defined
E~~;J
v
... (2.2)
_
Kinematic viscosity,
m2/s,
Therrnal ciiffusivity,
m2/s.
(l
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Gr
f
Thermal conductivity, W/mK.
2.2.4. Grashof Number (Gr)
Pr ==
I
Heat transfer coefficient W/m2K,
-
L
H
p
Kinematic viscosity, 1112/s.
Reyn Ids
be
.
num r. is therefore
'
m nltude of the i
.
, a measure
of relatlv!.'
e merna force to til
.
n "'.
e VISCOUSforce Occurring in the
It i the ratio
difiusi it)'.
I
... (2, I)
/
u -
"here
h
=~
=
Inertia force x Buoyancy force
(viscous forceY
e.. U2 L2 x e P g iH LJ
(J.' U L)2
/
Convective Heal Trail fer
Heal and Mass Transfer
J.6
'~Ow
1. L.a
:J
g x p x L3 x L1T
\12
~;r
where
1I111lnar
1
the
'" (2.4)
Coefficient
L
Length,
v -
Kinematic
visco it)', 1112/s
Temperature
diffe renee K.
-
.
(lmetlllles
.'
[luid move
floW is
I
IJ
p -
~T
.
of expa nSIIIII,
.
K-I
. of tOw.
~f
a smooth
and
joIl0\\5 .. in in
an order!
13)
m,
er rel1la
al
called
III
lavers
. .
COlltll11101lS
tream line now. In this
and each
.
.
fluid panicles
..
path. The fluid panicles
sequence
2i
without
mixing
in each
with each
other.
Grashof Number has a role'
played by Reynolds numb
. f III free convection
I
er III creed convection.
2.2. 8 .
illlilar to thai
Turbulent Flow
I
In addition
~.
Ire uentl
the lam.inar
t
b erved
:~rbul~nt n w. Th
irreg~t1ar flow
This type of flow
path of any individual
h \\
irregular. Fiu.2.1
type of fl~w, ::I distinct
In nature.
the ill iantaneou
L
called
panicle
is zig-zag
and
velocity
in laminar
and
turbulent 11 w.
2.2.5. Stanton Number (St)
It is the ratio of N
I
nu b
usse t NUlllber
III er and Prandtl number.
Turbulent flow
to th
e product
<II
:J
of Reynold
0>-
gE
_ 0
c'Q)
>
II>
Nu
St
Laminar flow
E
~
x
Jl Ce
k
Time
Fig. 1./.
eJU:
Jl
St
2.2.6. Ne
==
x
2.3. BOUNDARY LAYER CONCEPT
Jl Ce
k
Th
e concept
II
ie staning pint
__}1__
pUC
P
...
(2.5)
Vilonion and N
on-Newto .
he fluids wl .
mon Fluids
the N
uch obe h
ewtonioll fll .
y t e Newton's I
.
io flu: lids and thos
.
aw of viscosity are called
n ulds .
e which do not ohey are called nonT
.ed by Prandtl forms
layer as propOs
.
.
.
f h equations of mOllon
for the simplification
0 t e
f a boundary
and energy.
.
. 1
along a stationary
towS
. When a real fluid i.e., VI COli fluid,
.
.
. 'ontact Wit. h th e
sohd boundary
a layer of fluid which comes III C
.
wa
bo
fl id which callnot slip a )
undary surface. Thu the la er of III
'.
d d laver
th b
dation ThiS retar e .
e ollndar),
urface and undergoe
retar
.
f h fluid. So,
Iu h
. d' ent layer 0 t e
n
er
cause
retardati
II for the a .lac
.
. .....
of the
s
.
edlale VIUlllt)
Illall I egi n is developed
ill
the 1111111
I
r":
Scanned by CamScanner
Convective
_______
boundary
surface
in which
the
velocity
increases rapidly from zero at boundary
velocity of main stream.
of the~o
.
Wing
urface and "
fiUid
( PJ)roaches the
/
HydrodynamiC
:.3.2.
Boundary
boundary .' layer
111 drodynamiC
hy
99% of free stream velocity.
[hall
In this concept, the flow fie Id over
b d . .
regions .
a 0 y IS divided
rfJ% of free stream temperature.
int o~
1. A hi'thin region near the body calle e d t Ite bounda
I
were me velocity and temperature
grad'
ry ayer,
.
.
lents are large
.2. The region outside the boundary la yer w Ilere velo . .
temperature gradients are very nearl
city and
stream values.
Ileal y equal to their free
The thickne s of the boundary la -er ha .
distance from the surface at whi I I I)
S been def.lIled as the
rcn tne ocal velocit
reaches 99% of the extern' I I . .
I Y Or temperature
.
a ve ocrry or temperature.
a solid surface
of the fl UIid IS
. less
velocity
Thermal Boundary layer
2,3..3
111thermal boundary
layer, temperature
2.4. CONVECTION
convection is a process
2.9
Layer
.
The layer adjacent to the boundary.
IS k nown
layer. Boundary layer is formed wheneve r tl iere IS
. rei' as bounda ry
between the boundary and the fluid.
at,ve motion
Heal Transfer
'J'
of the fluid is less than
of heat transfer that will occur between
and a fluid
medium
when they are at different
temperatures.
2.4.1. N'.!wton's Law of Convection
Heat transfer
from the moving fluid to solid surface is given by
theeljuation,
Free stream
velocity
U""
This equat ion is referred
where
h
A
Til'
Trailing
edge
_
_
Too _
to as Newton's
law of cooling,
Local heat transfer coefficient
in W Im2K,
2
Surface area in m ,
Surface (or) Wall temperature
Temperature
in K,
of the fluid in K.
Fig.1.1. BoUII dary layer 011fit/t plate
2.4.2. Types of Convection
2.3.1. Types of B oundary Layer
l. Hydrody namic. boundary
I. Free convection,
(or)
Velocity • bo un d ary layer
2.
Themlal boundary
2. Forced convection.
Iayer
layer.
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2.4.3. Free (or) Natural convection
.
to change in denSity
If the fluid motion is produced
ue d f heat transfer is
res I .
di t the 1110 e 0
:u tltlg from temperature
gra len s,
said to be free or natural convection.
d
2./0
Convective Heal Transfer
Heal and Mass Transfer
2.1 J
I'
\
2.4.4. Forced Convection
If the fluid motion is ar1ificialh
.
.' created b
external force like a blower or fan , tl iat type of YI Illeans II " illl
known as forced convection.
leat transfer' IS
Q
THE LOCAL AND AVERAGE HEAT T
COEFFICIENTS FOR FLAT PLATE RANSFER
- LAMINAR FLO
At the surface of the flat plate heat fl
W
,
ow may be wrl'tt
_ Q.
ell as
q - A = hx(lll'-Ta-.J
A
h, == 0.332 x ~ x (Re)05 (Pr)OJ33
1c(8T)
8y
'" (2.6)
y =0
:::J
l
Local heat transfer
coefficient, h x J
We know that,
(~n,.-0
rL': Re = Ur-; 1J
Till - r, x ~
vx
T", - T so x ~
x ( 8e )
8n
'1
1)
=
°
We know that,
Local N usselt ~
number, Nu, J
x 0.332 (Pr)O)]]
h:o; x
k
0.332
8e
== 0.332 (Pr)O.133
[ .: (8 )
11 11 = 0
T", - T~ x ~
-\I -; x ~ x 0.332 x (Pr)OJ33
T '" - TO')x -~
y
_ •. ,ubstituting
'\\
0
=
(~T)
UY
k
J-
.. , (2.8)
J
T".-Ta',)
x
J
x X III x x-' x 0.332 (Pr)O)]]
(T", - T~) x ~
(,aT)
oy
x;k (R e)05 (Pr)0333 x_
x 0.332 x PrO)]]
x rill
_ ~
x
'\1 -;
in equation
x 0.332 (Pr)OJ33
o
k
oS (pdJ33 dx
1.L SO'".-' 32 x -x (Re)
L
o
(2.6)
y=O
Scanned by CamScanner
'
... (2.7)
0.332 x ~ (Re)05 (Pr)om
I
I
J
2.12
Heat and Mass Transfer
Convective Heat Transfer
L
.!_
L
fo""".s s : x .~k (UX)O.5
-;o
'~A"erageN~~elt}
= 0.664 (Re )05 (Pr )0333
•
number, ~_1I....t._J
(Pr)O m dx
ll3
... (2.10)·
.
--.!_ LX 03"2
. ..) x k x ( ~U
)0.5 x (Pr)03J3
x
m eq
uation (2.7) and (2.9), we know that,
h = 2 h .r
11.
F rO
°
tHE LOCAL AND AVERAGE HEAT TRANSFER
:2.6. COEFFICIENTS FOR FLAT PLATE-TURBULENT FLOW
L.
.\" x (.\" )05 dx
(!l)0.5
(P
v
r
[ x 0.332 x k x
)0333
x
IL.
x-I
I
x xo 5 dx
o
QlE
L
The heat transfer coefficient for turbulent flow can be derived
by using Colburn analogy,
(U)O.5 x (Pr )0333 x IX-05
L.
x kx ~
From colburn analogy, we know that,
dx
sr, ( Pr )21
o
0.332
( !d )0.5
L
x k x
x ( PI' )03JJ
x
y
LX,
x
Qlll
(k)L
0.5
0.664
III
Avcrugc
.
[X- 0.5 I ] L.
(~Y' x ( Pr )033J x [ L 05]
(UI)O
s
..
(Pr)o
(C) ( Rc )o.~ ( PI'
)0,.133
x
x
-:
"
J
x [ pl"]213
0.0296
(Rex
Nux
x ( Pr 'r 1/3
Rex
_
"L
-
0.664
>-1/3
Local Nusselt
1
=
NlI~,
(t) ( Rc
)0.5 ( PI' )U m
x L
k
Nu •
0.664 ( Rc )05
Scanned by CamScanner
)-0.2
ex
r 0.2
>- 0,2
r 0.2
0.0296
x ( Rex) ( Re,f
0.0296
(Re.{)O 8 ( Pr )J/J
0.0296 (Re
)08 ( Pr )OJJJ
.. , (2. J J)
We know that,
k
Nu
ex
0.0296 ( Rex r 0.2
Nux
Number,
Avel'U~C Nussclt N urn bcr, N II.
2
(R
)om
... (2.9)
We know thllt,
= 0.0592 (R
0.0592
2
Nux (PI'
)0.5 (PI'
2
/3
NUr
--'_
Rc = UL
v
~
sr, ( PI' ) 2
Rex PI"
(t) ( Re
- 0.664
0
0,5
[.,'
heat tmnsfor
coc:llil.:icnl,
m
=
[From I1MT data book, 'Page No.113 (Sixth Edition)]
+
- 0.5 + J
0332
3
( PI' )O,JJ3
h,
0.0296
(Rex )0,8 ( Pr )0 JJJ
0.0296
P )0.333 x k
( R,:e:!..x
:.-)O_'8 _x_(_r_
-
x
Convective Heat
2.14
Heal and Mass Transfer
h; = 0.0296
::::)
,...-__
Local he~t .lran~fer)
coefficient.
II .1
.__
_._
e icrent
l.
__.
.
'
0.037
(Re r )08
= 0.0296 (~)..\'
v ( R e)O
The average heat transfer co· t't- .
h
=
(~)
:r
( Pr )0 JJJ
, .. (2,13)
8 ( P r )0.333
'" (2.12)
._-._- -
II IS given
.
by
LI j' hI' d x
we kllow that,
Average Nusselt}
Number, Nu
hL
k
(Pr)03J3xL
0,037 (t)(Re)OR
(l
NlI
1
k
L
L IO.0296 (~)
t f°
( Re., )0.8 ( I'r )0 J)J d x
I.
0.0296 x (.~)
(u.:r)o.8
(P °
v
r)
J<
0,037
Average NlIsselt }
Number, NlI
Fronlequation (2.12)
333 dx
(Re)08
(Pr)O.JJ3
,,, (2.14)
and (2.13),
we know that,
Average heat transfer}
coeffie ient, h
1,25 x fix
["-17--1-,2-5 -h•
= 0.0296
x k
L
U
0.'
x ( ~ )
L [ '.:
( Pr )0 33J
Re - ~x
J' (1.)x x dx
(!)
L
x
= 0.02% x
L
= 00"'7
. J
)0.8
(U
~
(!)
(U)0.8
L
(!) (U)O
= 0.0296 x
0.037
x
x
(!)
(UL)O
L
-;(. k )
L
(Rc)08
~
..< (Pr)(I333
x (Pr)0.3JJ
8
~
x (Pr)03J3
(Pr)OJ3J
(Pr)OJJJ
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Heat Transfer Coefficient for combination of Laminar
and Turbulent Flow
Heat transfer
Ix-L o.
° 02 1
[x-
+
coefficienl
for laminar-turbulent
cOJllbined flow is
given by
dx
2
.r
]
- 0.2 + I
[LOS]
0.8
8
2,6.1.
O.S
x
o
= 0.0296 x
1
L
h
~f
hxdx
°
(Turbulent)
(Lammar)
0
t [j
°
I
I
O.JJ2
U) (R<)"
L
f
x
0.0296
(Pr)' JJJ dx +
(!)(Re)OS(pr)OJ3)
X
dXJ
2.16
Heal and Ma!,'S Transfer
t [J
=
Conveclive Heal Transfer
U)(~)O.~
0.332
o
I
(,x i-
(Pr)OJ3J
~!cpr)O.333
/1 _. L
2. J 7
(Rex)Oj + 0.037. (Re )o.8
L
[0.664
I.
f 0 0296 (~) ( U X)O.8 I'r
.,'
X
.
::: Ii (PI')O.33,J[ 0.332
-;;--
["R .
v
e::::_
U.t]
V
.'
occurs at critical R.eynolds number, Re,. = 5 x 105,
TranSitIOn
. floW IS
. Ian iinar upto Re = 5 x 105, after that flow IS turbulent.
/.t'.,
. t e Re c = Re x = 5 x J 05 .
SubstJtu
x
dx-+-
o
h
(~)C.8
ft ~ ds.]
v
0.0296
]
)°,)]3 d X ]
(
(~)(J..~IX ~
L
- 0.037 (Rex)08
+
= ~ (Pr)O 333 [ 0.664 (5 x J05)O.5
X
0.037 (ReL)O.8 - 0.037 (5 x 105)0.8 ]
x
f
°
00296
!::
L
.r
(~)"
Lk (Pr)03J] [ 0.332 (!l)O.5[~]
V
0.0296
f
(PrJO]]] [ 0.332 (~)"'
-
°
-.~
[ ~05']
+
M
,
(Pr)0333
0.8
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871 ]
]
-k (Pr) 0333
. [ 0 .037 (Re L )0.8 - 871
- L
... (2. 15)
.
ilL
Average Nusselt Number,
:::>
x08 ]]
T
Nu
!!.. (Pr)OJ33 [ 0.037 (ReL)08 - 871]X
_ L
Nu -
~~~Nusselt
L
k
_ Prom [0.03
Number, Nu -
)OB- 87 J 1
7 (R
eL
.. , (2.16)
. v.
_ 0.029Q (' U X)0.8
O.H
-
We know that,
!l)0.8 [ 0.8
U
0.0296 ( v
- 0.8
[O(N-~ (~)O.5 . + ~ (UL)O.8
\
h -
.r
J8
E
+
(!l)0.8 [ x~ ~2++: ] L ]
V
(Re,f8
[ 0.037
rage heat transfer coefficient,
dx ]
x-02
x
0.5 + I
C
(Pr)OJ33
v
]
-
Convective Heal Transfer
L f 63 0 - 56&
~
~J 8
I
AND-
uation for b 'II
equati
We know that, Von Karman momentum
layer flow is
7 do
72 dx
£,
Substitute,
!!..U
_-
;x [[ ~ (I - ~)
P t~
U2 do
'to :::: 72 P
dx
(?)In
,\14
We knoW that,
u
~
'to :::: 0.0225
;x [[ (i)'" [ (i)'" ]
I -
dy ]
Equating equation
7
=>
~ ;x {[[ WI" - (i )211 ] dy I
s
(.JLU0)
(.JL72 dx ::: 0.0225
U8)
do
J +,
,7+ L{
0 117 (
~
.-l!- ,\114
u
0114 do ~ 0.0225 ( ~)
I2 )
_'_L(7+ J
0 -
0
217
~
+,
11/4
do ::: 0.0225 ( p U 8 )
0
/;
,1/4
p
p
0
II
.. (2.18)
(2.17) and (2.18),
0
o
dx
_jL.
(
P U2 P U & )
72 p lJ2 dx ~ 0.0225 P tJl
1.. do
;x { o~n J yin dy - 0:" J y211 dy I
= __d
... (2.17)
.J_
dy ]
s
~
1
:::----
oundary
~
72
~~ l ~o 1
2.7. BOUNDARY LAYER THICK"NESS SHEAR
SKIN FRICTION COEFFICIENT FOR TURB~TLRESS
ENTFlQ
pt~
l
:::: dx
Heal and Mass Transfer
0\14
0
do ~ 0.2314
72
x -:; dx
,\"4 72
x 7" dx
(;0)'" dx
Integrating
~ ;x { o~"(~):dx
[I (0
_- !!....
dx
[78
817
!!.... 8 8ii7
)
I) -
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0:" (~):
_. <)7 (BB217917 ) ]
7 ]
9 I)
)
2.19
=>
J 0'/4 do
.l.- ,\1/4 + C
(
0.2314 pU) x
r
,
Convective Ileal Transfer
_.---.-----....:..:..:..:...:..:....:..:.:::.:.:~:2~::-.~21
~""
u, )11,1
( plJ
5
~
~- ~'J~.
_
S I'
OW thut,
IVc kll
shear stress, to
,
+
I)U.\'
Assumilll.!, boundarv • 11\\'(:r
I'S t III 'b u I~lIt ov
J
the plate.
fo:
\",e"f
.\' ...
( J!..)II-I
0.2314
I
er t ie entire length of
0.0225 p U
==
'tUting () valut.!,
Sllbstl
to
z:
2
(_l!.
)1/4
. pU8
[
0.0225
] 1/4
P U2
pU x 0.370 (Re ,_-0.2 x x
?
So. at x = O. 0 = 0 ~
~
~
4
-5 0 .'/~
OS!~
=
=
~
O. 022- :>
(0.370)1/4,
C =0
(l!_)114
pU
x
0.2314
(_g_)1/4
x 0.2314
U2 [ L x
P
pUx
0.0225
U2
(0.370)114 p
r
pU
.
0.0225
s = [~4 x 0.2314 (_g_)1/4
pU
=
[
0.289 x
x
(~)"4
(_g_)~)(
~
pU
=
(0.289)415
x
=
0.370 x
(ptt.)'/5
(0.370)114 pU
] 4/5
rl5
X
x x
0.0225
(0.370)1/4
0.370 x
= 0.370 x
- 0.370 x
(_l:.!_)115
pUx
(~xY'5
( Boundary layer thickness,
x
(ieYl5
5 = 0.370 (Ret
5
02
x
x x
4/5
x
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1/4
1/4
[.,'v=~J
p
J 1/4
[.,' Re=~
pU2 [(Re)1I5 x (Re)-I]
]
1/4
115
~~=~~o
..__ -~ 2
~--~ pU1
2
Local Skin Friction coefficietll• Cf(:
[ .,' v = ~ ]
We know that,
[ .,' Re = ~
J
xx
03. 70 (RetO.2
2 [~e)O.2
Re
J
0.05769 pU2 [Re 1-01
X
x
Rc 0.2
)
[Y-Ux (Re)02]
0.02884 pU2 [Re]-
l/5 x x~/5
X
(
0.02884 pU2 [( Re r 4/5 ]1/4
~
=
2
0.05769
Shear stress,
Also, we know
x x
I .. , (2.19)
to
to
CfX
x
eY.:
er02
2 (K
eY.:
2
.. ' (2.20)
Convective Heat Transfer
2. 22
Heal and Moss Transfer
-----
Equating both equations,
~
0.05769
pU2
2 (Re)-O.2
C
fx
pU2
2
= 0.05769 (Re}: 0.2
Local friction coefficient, C x = 0 . 057 69 (Re )- 0.2
Avullge Friction Coefficient (elJ :
~
x L-I x
::: 0.072 (~Y'/S
x L- lIS
... (2.21)
We know that,
::: 0.072 (~L
)-115
::: 0.072 (Ret
lIS
Average friction codficient,
L
(U)-I/S
I
~
x L
(~t/S
0.072
CIx
A verage friction coefficient,
y
::: 4 . 0.05796
X
L4/S
L4/5
1
C L == 0.072 (Ret0
elf.fL = L Clx dx
2.23
... (2.22)
[ .: Re == u~ ]
o
L
tf
2.8. HEAT TRANSFER
0.05769 (Re)-O.2 dx
USED
I. If velocity is given, that types of problems are considered
o
as forced convection problems.
TIP + Tao
2. Film temperature, Tf == ~,
where
T; _ Plate surface temperature, oC,.
L
tf
0.05769 (Re)-1I5 dx
o
tf
FROM FLAT SURFACES-FORMUlAE
L
0.05769
(~x }~/5 dx
T
_
o
3.
IL x-
L1 x 0.05769 x (U)-1/5
-;
liS
dx
•
flow is laminar.
Re
==
x 0.05769 x
-'- 0.05769
LXv
- +1
5
(!I)-1/5 L4
Scanned by CamScanner
ater t h an 5 X
If Reynolds number value IS gre
.
is turbulent flow.
0
==
Re
415
5
an
where
u -
. _
1_
v .-
'
!Lh < 5 x 105 -+ Laminar floW
v
(~r/~[~:+1JL
5 x 105 then the
th
If Reynolds number value IS less
o
t
Fluid temperature, °C.
r1J
105 then the floW
'
!:!h:;> 5 x 105 -+ Turbulent floW
v
Velocity, mis,
Length, rn, . sit)' m2/ s·
Kinematic VISCO '
1.24
Hear and Mass Transfer
i
For Flat Plale Laminar Flo", :
IFronlllMT
l
---
data book. Page No.112 (Sixth Edition)!
I. Local Nusselt Number,
Nux
where
0.332 (Re)05 (Pr)0333
Pr
Prandtl number.
-
2. Local Nusselt Number,
\
Convective Heat Transfer
flat plate,
Turbulent Flow (Fully Turbulent from leading
lFrom HMT data book. Page No.113 (Sixth Edition)\
(If
l,Jge)
. \ Nusse\t Number,
LO~a
Nux:::
0.0296 (Re)o.s (Pr)0.333
\.
1Nusse\t Number,
2. LPca
h L
_.1_
Nux:::
k
h~L
=
NUt
where
3.
h
where
k
Length, 01,
k
Thermal conductivity, W/mK.
-
3. Average
Average heat transfer coefficient,
h =
4.
II -
Average heat transfer coefficient, W/m2K
A -
Area, m2,
T" -
Plate surface temperature, 0(',
T." -
Fluid temperature, °C.
5. Hydrodynamic boundary layer thickness,
S x X x (RerO.5
0hx
=
0h.r(Pr)-o.m
=
0.664 (Ret
=
Thermal conductivity, WImK.
heat transfer coefficient,
TIV
-
T", -
Film temperature, °C.
5. Boundary layer thickness
-0.2
8 ::: 0.37 x x x (Re) .
For Flat Plate. Turbulent Flow:
1.328 (Re)-O.5
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k _
Heat transfer Q ::: h A rr,- T~)
4.
h _ Average h~at transfer coefficient W/m2K.,
where
A - Area, m2,
Plate surface temperature, °C,
No 114 (Sixth Edition»)
MT data boOk., Page
.
)
lFrom H
t wall temperature
0.5
8. Average friction coefficient,
efL
Length, m,
x
7. Local friction coefficient,
C/.r
L -
6. Local skin friction coefficient
C/ ,:: 0.0576 (Re)-02
6. Thermal boundary layer thickness,
O-r..
Local heat transfer coefficient. WIm K.
h ::: \.25 hx
2xhr
Heat transfer Q = hA(T,,-T,.,)
where
2
_
x
hx - .:-Local heat transfer coefficient, W/m2K,
L -
2.25
(Laminar, Turbulent com
\.
bined - constan
Average Nusselt Number, ,
037 (Re)o.s - 871 )
Nu ::: prOJ3J lO.
Conve tive Heal Tran
2.26
----
Heal and Mass Transfer
2. Average Nusselt Number,
Nu
where
h -
=
hL
k
Average heat transfer coefficient,
L
Length, m,
k -
Thermal conductivity,
rropt ",t.'
. . ./,,;, at 40 ~ :
oJ
I r mHMTdlJ
Tlwrl11 a I
ol1du ti iry
Kinematic
prandtl
.
.33
0.02756
k
16.96
iry, v
"I C
2.17
I.I 28 kg/m
en iry, p
W/rn2K,
W/mK.
K, Pa e
r
number.
Pr
Re n II.!' number
Re
W/mK,
10-6 m-/s
0.699
:::
We knOW that,
3. Average friction coefficient,
elL
=
0.074 (Re)-O.2 - 1742 (Re)-I.O
2.8.1. Problems on Flat Surfaces
- Forced Convection
I Example 1 I Air at 20 "C~at a pressure of / bar is flowing
over aflat plate at a velocity of 3 m/s. If the plate is maintained til
601:(', calculate tile heat transfer per unit width 0/ the plait.
Assuming the length of the plate along the flow of air is 2 m.
Given: Fluid temperature,
Teo
200e,
Pressure,
P
I bar,
Velocity,
U
3 mis,
Plate surface temperature,
Tw
so-c,
Tofind:
Width, W
1m,
Length,' L
2m.
Heat transfer (Q).
4
5 lOs
)5.377 x 10
So this i laminar
than 5 x I 0
'
Re
ReynoldS numb·r
i le
vahle
flow.
For Flat plate.
Local
II
ell
.
flow
Lanllllar
I
•
:::
0.332 (Rd
u...
,
. book, Page
IF r m II cil data
umber
0.332 (35.377 )(
We knov
that,
_!.--
TI
Scanned by CamScanner
Tw + Teo
2
60 + 20
2
:::
Local Nusselt Number
Nux
k
(Pr
)0 J
. 112 ( ixth Edition)!
o.
IQ4)O 5 x (0.699)
h )( L
Solution: We know that,
Film temperature,
:::
0) J
Convective Heal Transfer
2.28
Heat and Mass Transfer
----
We know.
A veragc heat transfer
coeffic ient
Heat transfer
h
~fa;rat80°C:
oner I
Pr r
[From HMT data book, Page No. 33 (Sixth Edition))
p
1 kglm3
h
2 x 2.415
v
21.09 x IQ-6 m2/s
[/7
4.83 W/m2K]
Pr
0.692
Q
h A (Til" - T a:»
k
0.03047 W/mK
UL
4.83 x 2 (60 - 20)
Res"lt:
Heat transfer
Re
Reynolds Number,
5 x O.S
21.09 x 10"-6
0 = 386.4 Watts
I Re _
I Example 2 I Air at 25'1:' flows over aflat plate at a speed of
5 mls and heated 10 135 '1:'. The plate is 3 m long and 1.5 m wide.
Calculate the local heat transfer coefficient at x = 0.5 In and the
heat transferred from the first 0.5 m of the plate.
Fluid temperature,
T ff)
Plate surface temperature,
Til'
Given:
25°C
[x = L = O.S m]
v
= Width x Length = I x 2::= 2]
[Q = 386.4 Watts J
[.: Area
2.29
1.18 x J05
1.18 x 105 < 5 x lOS
I
.
R < 5 x lOS flow is laminar.
Since
e
'
For flat plate, laminar flow,
Local Nusselt Number Nux
= 0.332 (Re)Os (Pr)0.333
..
[from HMT data book, Page No. 112 (Sixth Edl~lon))
0.332 (1.18 x 105)0.5 (0.692)°33>
Velocity,
U
5 rn/s
Length,
L
3m
Wide,
W
1.5 III
Distance,
x
0.5 III
-[N-u.- _-_ ~10;:-:-0.~9
]
x
We know,
100.9
Tofind:
I.
Local heat transfer coefficient
2.
Heat transferred
Solution:
Nux =
(hx) at x
= 0.5 Ill,
(0) at x = 0.5 m.
h xL
2..k
h x 0.5
...!---
ee
[.,' x= L :::0.5 m]
0.03047
Local heat transfer coefficient,
-c
coefficient,
Average heat nansrer
hx :::
h:::
2xhx
h ::: 2 x 6.14
We know that,
~
TII'+Tff)
Film temperature,
Tf
2
135 + 25
2
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Heat transfer,
Q
_
_
::=
[~
T)
h A (T\I' (I ~)5_ 25)
S 0 5) x
12.29 x (I. x . .
C/)
Convective Heat Transfer
2.30
---
Heal and Mass Transfer
Result:
1.
Local heat transfer coefficient,
hx
6.14 W/m2K,
2.
Heat transferred,
Q
1013.9 W.
I Example 3 I Air at 20°C at atmospheric presSure /1ows
. We know
so/lIlio" .
. tel11perature
Flltll
.
~
rtle
4. Averagefriction coefficient,
5. Local hea~ transfer coefficient,
latic viscosity,
V
prandtl Number,
Pr
0.698
l conductivity,
Therm a
We knoW that,
k
0.02826 W/mK
ReynoldS Number,
Re
=
V
JxOJ
20°C
Velocity, U
3 mls
Wide, W
1m
Tw
Distance, x
Tofind:
[':x=L=O.3m]
17.95 x 10-6
T co
Surface temperature,
1.093 kg/Ill}
17.95 x 10-6 m2/s
UL
7. Heat transfer.
Fluid temperature,
2
[From HMT data book. Page No. 33 (Sixth Edition)!
6. A-verage heal transfer coefficient,
Given:
80 + 20
p
.
3. Local friction coefficient,
liL
T", + T a:
2
~
= SO:C I
Density,
i(lnen
2. Thermal boundary layer thickness,
T!
. s of air aI50°C:
prope
over aflat plate at a velocity of 3 mls. If the plate is 1 m wide and
80 'C, calculate the following at x = 300 mm.
1. Hydrodynamic boundary layer thickness,
.
Since
Re < 5 x 105 , flow is laminar.
For FIL-t plate, laminar now,
80°C
300mm
)
N
112 (Sixth Edition)]
[Refer HMT data book, Page o.
I. Hydrodynamic
= 0.3 m
I
I iyer thicklless :
bounc ary. (
05
5 x x x (Re):"
5
I. Hydrodynamic boundary layer thickness,
2. Thermal boundary layer thickness,
3. Local friction coefficient,
4. Average friction coefficient,
S. Local heat transfer coefficient,
6. Average heat transfer coefficient,
7. Heat transfer.
Scanned by CamScanner
2.3/
5 x OJ x (5.01 x 104t0
[}ltf =__ 6.7 x - 10-3 tn]
2. Thermal boundary IOJ'tier thickness: 3JJ
S (PrtO
"x
(6.7 x 10-3) (0.698t
[~x = 7.5xit!iJ
0 JJJ
----
Heal and Mass Transfer.
3. Local Friction Coefficient:
Cfx
0.664 (Re)-05
r:-__
0_._66_4_,(,
_5,.01 x 104)- 0 5
Ie
2.96 x 10-3
fr
l
Convective Heal Transfer
I~ansfer:
\\ e knoW that.
II A (T'I" - TaJ
Q
I
12.41 x(1
0.3)(80-20)
(Q
223.38 Watts
I
\.
0,
6.7 x 10-3 m,
2.
Tx
3.
f
4.
elL
5.
hx
6.20 W/m2K,
6.
h
12.41 W/m2K,
7.
Example 4
Q
223.38 W.
4. Average friction coefflcient :
Result :
1.328 (Re)-05
C/L
1.328 (5.01 x l<PtO:
5.9 x 10-3
5. Local heat transt;
OJ er coefficient (hx) -:
Local Nusselt Number
0.332 (Re)0.5 (Pr)OJ33
I
. (0.698)0.333
0_.33.2(5.01
x 104)06~
1:"7
____
7.5 x 10-3 m,
2.96 x 10-3,
5.9 x 10-31
I ell
NUl"
2.33
)__:.9:__
We know,
I
5.9 x \0-3,
I Air at 10 ~ (It atmospheric pressure flows over
aflat plate at a velocitv of 3.5 m/s. If the plate is 0.5 m wide and
at 60 ac, calculate tire following at x = 0.400 m.
Local Nusselt Number
n,» L
(i) Boundary layer thickness.
=t:
(ii) Local friction coefficient.
(iii) A verage friction coefficient.
(iv) Shearing stress due to friction.
(v) Thermal boundary layer thickness.
(vi) Local convective lIeat transfer coefficient
(vii) A verage convective heat transfer coefficient.
hx x 0.3
65.9
0:02826
l hx ~-;6-:::-.20::-:W-I-m-2K-1
6. A verage heat transsfer coefficient (II):
h
Lh
==
2xh
x
[.: x == L == 0.3 m J
(viii) Rate of hea! trlrnsfer by convection.
2 x 6.20
(h:)
12.41 W/m2K]
(x)
Scanned by CamScanner
,...
Total drag force 3n the pillte.
Total mass flow rate through the boundary.
\ el
iry,
Wide,
Plate
urfa e temperature,
3.5 n s
W
0.5
T '"
60°C
x
Distance.
To find:
(I) Boundary
Local friction
(iiI)
Average
(iv) Shearing
0.400
coefficient
III
5 flow is laminar.
Re < 5 x 10 ,
IRefer H n data book, Page No. 112 (Sixth Edition)]
(i)
Boult(
.
,,,;ckness, 8 or 8trx (It x == 0.400 m :
Local heat transfer coefficient
(vii)
Average heat transfer
coefficient,
h .
(viii)
Rate of heat transfer
by convection,
Q.
(x) Total mass flow rate through
6.96 x 10-3 m )
h x'
,
Local friction coefficient,
(ii)
T",+Too
:-0
60 + 20
~~
2
•
•
(IV) SIIC'(/rlllg
P,op~rties of air at 40 ~ :
.
[From HMT data book, Page No. 33 (Sixth Edition)1
1.128 kg/m?
16.96 x 10-6
0.02756
Scanned by CamScanner
1112/5
W ImK
___"'_'-J.
stress or j'C(/
We know that,
k
0.664 (8.25 x 104)-05
coefficient,. Cit :
efL == 1.328 (Re)-os
"
\.328 (8.2S x 104)-°.
Tf = --2-
0.699
•
l-.S_-fx-:-=-2~-\-~.
m.
(iii) A ver(lge friction
Pr
CIX':
0.664 (RetO)
fPJ
FD.
the boundary,
=-
C[x
We know that,
=
5 x 0.400 x (8.25 x 104)-0.5
layer thickness , 0 Tr:
Total drag force on the plate,
v
5 x x x (Re)- 0.5
=
oor&hxJ'
(VI)
p
5 x 105
8.25 x 10~<
Since
t;.
stress due to friction,
.I
F I m temperature,
v
Boundar)' layer thiCkness,.\.
Cfl.
Thermal boundary
e So/lIIion:
Re
3.5 x 0.400
16.96 x I Q-{l
, Cf x :
(v)
(ir)
2.35
lar)' laver t/rickness or Hy(lrodynamic bOllndary layer
0
friction coefficient,
Heat Transfer
Ux
Reynolds number,
I Re
layer thickness,
(il)
----
20°C
T'll
GiwIf : Fluid temperature.
Convective
J;lIear stress, f~:
.
Crt =,
.
"
2
2
2.36
Heal and Mass Transfer
Convective Heat Transfer
M Thermal boundary layer thtckness,
0rx
-
0T.\".' •
. )-
6.96 x 10-3 x (0699
Orr
7.84 x 10M) Local heat transfer coefficient, ".1:.'
3 IrQ
I
4.623 x 10--3 x 1.128 x (3.5i
0333
2
~ge
Local Nusselt Number. Nu,.\ == 0.)"2
. .) (R e·))0· (Pr)0333
r+:__
I Nu
0.0319
shear stress,
't
Drag force,
FD
I
1Xusse I 'umber.
Nu
==
I
Drag force on two}
sides of the plate
2 x 6.38 x 10-3
[ Total drag force, FD
l'iu
k
--2__
m
Here,
0lx == Q., 02x
m
:=)
0_5 x 0.400
[.:
r----
T(,,,j;
W> k
" orce
e no\\ lha!.
~
==
9~
0" the plflle, F "
D
cragt fn Ii n C cft-·
IClcnl
_
, C IL.
==
_'
eJE
2
Scanned by CamScanner
5
'8 p U [ 82.r - s
lr ]
8 x = 6.96 x 10---3m
i
3
x 1.128 x 3.5 x [6.96 x 10--- ]
0.017 kg/s I
( i)
I "'0
(ii)
c.,
2.31 x 10---3
(60 - 20)
(iii)
CfL
4.6:!3 x 10---3
.r == L == 0.400 rn]
( i\')
tx
(v)
°Tx
( vi)
hT
5.83 W/m2K
( vii)
h
I 1.66 W/0l2K
(vi ii)
Q
93.28 W
(ix)
FD
0.0127 N
(x)
111
Result,'
(L\") TU(u/d
0.0127 N I
(x} Total mass flow rate, m :
L
11.66
Area x Average shear stress
0.5 x 0.4 x 0.Q319
6.38 x 10-3 N
" .. L
k
s
N/m2 I
Wx Lx t
O_..)_'
3~_ (8.25 x 104)0.5 x (0.699)0333
84.
We know that,
L
e.!E
elL x 2
x (Pr)-- 03J3
(> hx
2.37
6.96 x 10---3m
0.0159 N/m2
7.84 x 10---301
0.017 kgls
(
I?
Convective
2.38
239
4x0.4
Heat and Mass Transfer
\ x \0-4
[ Example 5 , A flat plate -measuring 0.8 m x 0;;-;;;---'
longitudinally in a stream of crude oil which /lows witl Placed
of 4 mls. Calculate tirefollowing:
' a veloCity
(i) Boundary layer thickness at the middle of plate.
(ii) Shear stress at tire middle of plate.
(iii) Friction drag on one side of the plate.
Take
Specific gravity of oil = 0.8
Kinematic viscosity
1 stroke
Given:
Heal Transfer
\< 5 x \05
= \.6 x 104
lRe
.
Re -: 5 )( \ 0\ flow is laminar.
. _ ..
Since
{Refer HMT data book, Page No. 112 (SIxth l:dltlon)1
&
layer thickness,
&
'Boundary
\':x=L=OAm1
[8
:::::)
Length,
L
0.8 m
-:?
:::::)
Width,
W
0.25 m
Velocity,
U
4 m/s
("") S/tear s
:lction
Loca \ f \
Specific gravity oil
Density of oi I, p
= 5 x 0.4 x (\.6 x 104)- 05
-:?
0.8 m x 0.25 m
Plate dimensions
5 x X x (Re)- 0.5
0.0\58
ni]
ess at tile middle of plate, 'l'x:
tr
.
coefficlc.nt,
=
CJ x
0.664 (Reto.
~
0.8
0.8 x 1000
800 kglm3
Kinematic
viscosity,
V
I stroke
=
We know that,
1 x 10- 4 m2/s
Tofind:
(i) Boundary
~
layer thickness
L x 800 x (4)2
at the middle of plate, 8
(i;) Shear stress at the middle of plate , 1: x
(iii) Friction drag on one side of the plate,
FD
2
\
33.54 NIln2
© Solution:
1.
Boundary layer thickness at the middle of plate,
0.8
L =2 =0.4m:
Reynolds number,
Re
UL
\
V
\
Scanned by CamScanner
'tx
..,
2.40
Com ecti
I
Heal and Mass Transfer
t. l:::
I x 10-4
I
Re
= 3.2 x IO~ < 5 x 105
Since Re < 5 x 105, flow is laminar.
Average friction coefficient , -C JL --
1.328 (Re)-o.s
1
1_.3_28_{3_.2
x 104)-0.5
I
7.42x I!rJ ]
elL
lJ.5 .,,';5:
I ({OWIng·
.
L. all, of plale over which
the boundary la) er is laminar.
1[0
/. e"~
Thickness of the boundary layer.
1. Shear streSS (II lite location where boundary layer is
J.
lamillar.
where
toto! drag force on both sides of IIIe plate
4.
boulldar) layer is laminar.
rake, p == 1.205 kg/m3; v= 15.06 x 1tJ-flnt]/J
D
r
We know that,
is 5 m long {l1If1 2 m wide. Calculate the
~/(Jle
0.8 Ill)
pU2
Gil'en: Fluid temperature,
Too
20 e
Velocity,
U
3.5 n s
2
Length, L
5m
Wide,
W
2m
Density,
p
K ineJ113tic viscosity,
V
7.42 x 10-3
=:>
•
47.488
i/m2
I Average shear stress,
t
47.488
N/m2
Drag force on }
one side of the plate
FD
==
0.25 x 0.8 x 47.488
r.::----(i)
(ii)
(iii)
Tofind:
9.49 N
I
0.0158
III
33.54
N/m2
1.205 ks m}
15.06
\{ laminar condition,
Length of the plate, L.
(ii) Thicknes
fthe boundary layer, 8.
Shear tres , • x .
Total drag force on both ides, FD·
(iii)
(iv)
Solutiou : \
that,
e knov
Reynold
!&
number,
Re s:
v
~
1506)(
10-6
~5)(IO~
~
9.49 N
ince
R eynold
Scanned by CamScanner
10
(i)
.Are a x A verage shear stress
W x Lx.
Result:
I
2." /
Air at 20 't'flows over a flttl plate {II a velocity
~
~
Heal Transfer
R
number
5
105. nO\\
. vallie
I
))(
. I rbulenl
i.e.,
IS II
. now
105 after Ihal
i.laminar
is Ilirbulent.
uplO
"-rewm
2.42
Hreal and Mass Transfer
(i) Length of lite plate, L (AlIa
Reynolds number,
!~.----=-~
.
ml11l1r COllditi
!:_Ii
Re
we knOW that,
011) ..
CJ L
=
~
v
I~
L
.
.:.2.15
IIiL~e~n-:gt-;-h-o-=-ft-h-epl-at-.I
(ii)
III
e, ~ - 2 ~
of II b
------.~
.
Ie Olllldary laver
.
II, Tille/mess
BOllndary
1.878 x 10-3
_. 3.5 x L
5 x 105
~~-;I~ear
I
layer thi c k ness, 0 .,:::: 5 0 (AI/lIlllin ar cOlldit'
.
x x x (R )_ _
I()II):
[From HM"I'
stress,
Drag force, F D
data hook I)
x 105)-
.
r;--"_
L~~_ j~:20
.
SS,
LoCClI frict ion
iO-3 ~
(5 x 105)-
~
__~.664
~~
..~2~~
Drag force on both}
sides of the plate
z;'.
(i1~
T,otat draofi
e
orce Oil b
I'
-
Result:
T\
2' x 1.205 x (3.5)2
.
-
----
z. .. ---.
v. 9 x 10-3 N/,V
'~
f.
COI/ditioll) .
011 Slfle!!' of III
I
Averaoc f.' . .
e J1 file, FD (At laminar
~. "ellon eo'ffi
.
\: IClen!,
-C'
1.32R (Re)-U.-\
11.
•
~\
\
__
.. ...I..~~.~ __(~ x W)-- ()'i
~-~
· .._1.87~
/
Scanned by CamScanner
0.0138 N/m~
2 x 0.0593
FD
[F
(J 'i
We know that,
L
ocal shear stress. -T - ..
t
0.0593 N
Ill_J
-1-'
0.0138 N/m2
. (.: At laminar condition, L = 2.15 Ill]
== 2.151111
cotu 111011) ..
coe rcient, C
zr:
.fl
0.664 (Re)-05
=.:
t
2 x 2.15 x 0.0138
II 1011
ffici
0.939 x 10-3
1.205 x (3.5)2
2
Area x Average shear stress
O.S
[.. X - I
x
I~
r.\. (AI/al11il1ar'
t
eO)
.
o :::: 5 ,x 2.ageI 5No.x (5112 (Sixth
Ed .
1I
(iii) Sh(!ar stre
2
0.1186 N I
D
(i)
L
2.15 m
(ii)
0
I S.20 x 10-3 m
(iii)
Tx
6.9 x 10-3 N/m2
(iv)
F[)
0.1186 N
I Example
7
I Castor oil at 30 °C flows over a jl(1t plate at
{I
velocity of 1.5 mls. The length of tire plate is 4 m. rite plale is
heated
uniformly
ami
maintained
(II
90 'C. Calclliate tile
following.
1.
Hydro{lynamic
2.
Thermal bUlIIlllary layer tllickne:;'s,
Total drag force per unit widtlr on one side ofille plate,
3.
4.
boundary layer tlricklless,
Heal transfer rate .
2.44
Heat and Mass Tramfer
Convective Heat Transfer
90 +30
T/ -_ ~::::
At the mean film temperature
properties are taken as follows,'
p= 956.8 kg/m3;
V= 0.65 x lO-4m2/s;
k = 0.213 W/mK,'
a= 7.2 x 1()-8m2/s.
Given:
Fluid temperature
T CtJ
Velocity,
U
Length,
Plate surface temperature,
L
At Tf
= 6QoC ,
60 °
2
5 x 4 x (9.23 x 104t0.5
c,PL,., .
"-"slefl/
0.065 m
3Qoe
2.
Therl1l{l I boundary layer thickness:
s:
= 8 x (Pr): 0333
UTx
hx
~
Til'
4m
9QoC
p
956.8 kglm2
6.74 x 10 3 m
a
Tofind:
0.213
=
4
0.65 x 10a = 7.2 x 10-8
W/mK
0.65 x 10- 4 m2/s
OJ33
= 902.77 J
I
force on one side
I
3. Total {.rag
. of the plate:
Average s kiIn friction coefficient,
7.2 x 10-8 m2/s
1.328 (Ret 0.5
I. Hydrodynamic boundary layer thickness,
1.328 x (9.23 x 104)-05.
2.
Thermal boundary layer thickness,
3.
4.
Total drag force per unit width on one side of the plate,
Heat transfer rate.
4.37 x 10-3 ]
t
We know that,
We know
~ U2
2
R eyno Id' s N urn b er, Re
__ U L
v
1.5 x 4
0.65 x 10-4
[Re
9.23 x 104 < 5 x 105
5
Since Re < 5 x 10 , flow is laminar.
For flat plate, laminar flow,'
4.37 x 10-3
I
=>
r
Average shear streSS r
Drag force, F D
{Refer HMT data book, Page No. 1 12 (Sixth Edition)]
1. Hydrodynamic boundary layer thickness,'
("\\
I
0.065 x (902.77)-
v
Solution,'
[.: x= L = 4 m]
1.5 rn/s
k
°ltx
==
5 x x x (Re)- 0.5
Scanned by CamScanner
2.45
l
Convective
2.46
Heal
and Mass Transfer
-------
4. Heat transfer rate :
We know that,
Local Nusselt Number
0.332 x (Re)o.S (Pr)O.333
..- __
0.332 x (9.23 x 104)0: x (902.77)0
\ Nux
~
3)
le 8 Air at 30°C flows over a flat plaIt at II velocity
. Calculate tht
plate IS'. 2 nI long and 1.5 m Wide.
of 2
r: :
e
folli,,,"II~ydrodynamic and thermal boumlary layer thickntss at
1· . trailing edge of the plale,
the
rotal drag force,
2. rota nWH flow rate Ihroug/. II.t boundary layer ~twtt"
l
J.
_ ~O em and x = 85 em.
Given:
Local Nusselt Number
Nu x
h~.L
k
0.213
Local heat transfer coeffil.: ient
T
30°C
Ve\ocity,
U
2 m/s
Length,
L
2m
Wide. W
To Jintl :
I.
h~ - 5 I .7 W /111- K
A verage heat transfer coeffic ient
2 x hr
Heat transfer,
F\lIid temperature,
h; x 4
972.6
Hydr
dynamic
2. Tota\ drag f rce.
tal rna s flow rate through the boundary layer between
3. T
r == 1\ rn and x == 85 em.
Solulion:
Pr perties of air at 30°C.
HMT dilta book, Pag~ No. }3 tS
fA
"
103.58 W/m2l{]
p
Q
It A (T," - T )
v
" x L x W (T\II - T, )
Pr
\Fro
m
Result:
~\x
0.065 Ill,
2.
0Tx
6.74 x 10-3 Ill,
3.
Drag force , F IJ
18.8 N,
4.
Heat transfer , ()"
24.859 kW.
Scanned by CamScanner
0.701
0.02675 W/mK
We know that,
Reyn
I.
1.165 kglm)
16)( 10-6m2/s
k
103.58 x 4 x I (90 - 30)
I
\.s m
.
and thermal boundary layer thickness.
2x51.7
24.859 kW
2.47
£_to'" rh
x- ..
We know,
Heal Transfer
Ids Number.
Re
v
~
&
10\ floW is laminar.
Since Re
5
ixth
..
Edl\lon)J
Convective Heat Transfer
2 .J8
0.036 N
Heal and Mass Transfer
--------
For flat plate, laminar flow,
(From HMl data book. Page No. 112 (S'
Hydrodynamic
""-1
Ixth Edi .
'fotal rn
aSS f1 0
w rate between x == 40 cm and x = 85 em.
Am
n
lJo ))
boundary layer thickness
bhx
2.49
5 x x x (Ret 0.5
=
.c boundary
5 x 2)< (2.5 x 105)-05
b- --O.-02-m___;1
BY
drodynalnl
r-..,0J
5
"8 p U [0hx=85
rf'
- bhT=40
1 ... (1)
layer thickness
~
== 5 x r x (Ret 0.5
uhx == 0.85
hx
5 xO.85x
Thermal boundary layer thickness,
&rx
bhx x (Pr): 0.333
u x X J-:-O.5
[~
r 2 x 0.85 15 x 0.85 x L 16
0
= 0.02 x (0.701)-0333
==
I &rx = 0.0225 I~J
.5
X }O-6
[.:' x == 85 em,= 0.85 ml
A verage friction coefficient,
CfL
Ghx
1.328 (Re)-os
r------
I SL
0.0130
== 0.85
==
1.328 x (2.5 x 105)- 0.5
Ohx
0:
OJ
5 x x x (Re
0.40
== 5 x 0.40 x
2.65 x 10:]
r 0.5
(u ) -
_?';
0.5
2 x 0.40 )-0.5
We know,
==
5 x 0.40 x ( 16 x 1Q=6
t
e.!!:
2
~
2.65 x 10-3 =
~
t
(l)~
1.165 x (2)2
2
~ I Average shear stress,
Drag force
= 6.1 x 10-3 N/m2
't
I
Area x Average shear stress
2 x 1.5 x 6.1 x 10-3
[.: L = 2m; W = 1.5 m]
I Drag force.
0.018 N
I
Drag force on two sides of the plate
;: 0.018 x 2 = 0.036 N
Scanned by CamScanner
3
5
[00130 - 8.9 x 10- ]
_ x 1 165 x 2 .
.
~
.
..
I:
:::: 0.02 m,
thicknesS,Uhx
\ . Hydrodynamic boundary layer . k"ess Orr:::: 0.0225 m,
layer thlC ,., , .
ry
Therma\ boun d a
_ 0036 N,
2. Drag force, F 0 _.
_
97 x \0-3 kg/so
f).m - 5.
3. T ota\ mass floW rate,
R esult :
Convective Hear Transfer
2.50
Heal and Mass Transfer
[Example 9
\.060 kg/m3
I Air at 30.oC,flows over aflat Plate--;;;-;;----:-
eiOCIty
'J
rerature
90 'C. If the transition occurs at a critical Reynolds nUlllb of
5 x 105, calculate the thickness at which tire boundary lerOf
.
changes from laminar
to tur b uI ent. A t that location fi daYer
,
' n lire
following:
(i) Hydrodynamic boundary layer thickness.
(ii) Thermal boundary layer thickness.
(iii) Local heat transfer coefficient.
(iv) Average heat transfer coefficient.
.
.
.
01" 4 mls and tireplate IS maintatne
\8.97 x \ 0-6 m2/s
d at
at a
a uniform
unl
temo
(v) Heat transfer from both sides for unit width of the plale.
(vi) Mass flow rate.
(vii) The skin friction coefficient.
Given: Fluid temperature, T eo
Velocity, U
Plate surface temperature, TlI'
Pr ::: 0.696
k ::: 0.02896 W ImK
UL
ids number,
Re ::: -v
R~~
5 x \05
4xL
\8.97 x \0-6
:::
~
[L:::
2.37 m~ 237 m. After that flow
. r uplO the length, L - .
F\oW is \am\Oa
turbulent.
. lfrom HMT data
book Page 'No. , ,
'
5 x 105
Tofind : At, Re = 5 x 105
(i) Hydrodynamic boundary layer thickness,
(ii) Thermal boundary layer thickness,
(iii) Local heat transfer coefficient,
0 hx .
(iv) Average heat transfer coefficient,
~Iu
\ 05)- 0.5
237><(5)<
.
==
5><
::::
0.0\67 ~
["x::::L::::2J7m.
\a er thickness .
0Tx'
(ii) Therma\
hx'
2 (Sixth Edition»
At L == 2.37 m .
\ yer thickness:
od
mic boundary a
05
(i) Hydr yna
&:::
5)(. x x (Re t .
h.'(
Critical Reynolds number, Re
2.5/
boundary
Y:::: 0 Iu (pr
OTx
h.
r 0.333
::: 0.0\67 (0.
696t 0.333
(v) Heat transfer from both sides for unit width of the plate, Q.
(vi) Mass flow rate, m.
(v i i) The skin friction coefficient,
~
CIx .
Tf
\ :::: 0.332 (Re)
f
Tw+T<Xl
2
90+30
2
I
:::: 0.332 (S )(
~
---
Scanned by CamScanner
o (pr),J)]
.Loca \ N usse\t
Number, NU
Solution: We know that,
Film temperature,
(iii)
We know that,
h_!--L
'Nux::::
k
6)0 333
105)0.5(0.69
r
~
Convective Heat Transfer
2.51 Healand Mass Transfer
-----
hx x 2.37
Ohx
~
-
0.0167 111
(ii)
°Tx
0.0188 m
(iii)
h;r;
2.54 W/m2K
(iv)
h
5.08 W/m2K
= 2.54 W/m2K
(v)
Q
1444.75 W
(iv) Average heat t~ansfer} _
coefficient, h - 2 x h x
(vi)
III
0.04425 kg/s
(vii)
c.,
0.939 x 10-3
208.07
0.02896
=
::) h, = 2.54 W/m2K
Local heat tranSfer}
coefficient, h,
2 x 2.54
!
I h - 5.08 W/m2K J
- 2
-
x h x A x L\ T
= 2 x h x W'x Lx (T .- TUJ)
11'
2 x 5.08 x I x 2.37 x (90-30)
Mass flow rat'e, III
5
= - pU [
8
Here OJ/IX
= 0,
.
5
°
°
211,r -
T"fJ
30°C
Velocity, U
4 m/s
Plate temperature,
0, IIx]
2hx = 0hx = 0.0167
III
Tofind :
= 8" pU x 0llx
m
10] Air at.30 °C, at II pressure of 1bar is flowin
perI m width of the plate.
Given: Fluid temperature,
['," W = 1 m]
I Q = 1444.75 W 1
(vi)
~mple
overaflat plate at {I velocity of 4 m/s. If the plate is maintained
a uniform temperature of 130 'C, calculate the average he
transfer coefficient over the 1.5 III length of the plate. Al
calculate the rate of heat transfer between the plate and the a
(v) Heat .transfer from both}
sides for unit width
of the plate, Q
~tSIl
Tw
130°C
Length, L
1.5 m
Width,
lm
W
I. Average heat transfer coefficient, h.
2. Heat transfer, Q.
Solution:
5
We know that,
TIP + T",
= 8" x 1.060 x 4 x 0.0167
(vii)
.
Film temperature,
[il = 0.04425 k 1
Skin friction coefficient}
.. or
ocal fnctlon coefficient
L
Tf
2
130 + 130
2
gls
C'fx
= 0.664
2
[fL
(Re )- 0.5
80°C]
Properties of air at 80'C :
[From HMT Jata book, Page No. 33 (Sixth Edition:
p
Scanned by CamScanner
=
I kgllnJ
Heat and Mass Transfer
2. 4
1~
'\I
Pr
k
Reynolds number,
c::
21.09)(
0.692
Re
hxWxL(T
10~~
=
= 0.03047
'
n2/
s'
1.:'__
IQ =
W/Il1I(
/(1!.\lIll:
UL
v
4 x 1.5
[ Re
For flat plate, laminar flow.
Local NUSSeIt}
Number, Nu,
5
:J < :; x 10
6._3 _S8_x_, I x 1.5 x (13 0 - 30
953.7 WJ
)
1.
h = 6.358 W/m2K
2.
Q = 953.7 W
of 4 m/s. Tile plate measures 50 x 30 emm~il .
I
,,'IIi/lt/tilled (II (I uniform temperature of 90°C Compare lite heat
romthe plate when the air flows
/oJ sfi
.
(II)
.
O.332(Re)05(Pr)0333
=
-T '1')
'1'(1)
~~J.
2.84 x 1051
S rnce Re < 5 x I 05 , flow i'5 I am rnar
.
1t'
-----., Air (II 30°C, flows over a flat plate (tl (t
EXlIm
21.09 x 10-6
.
__ ----------~~:('~on:I~'e~C/~il~~~~~e~a/~T.~~,
ran.",er2.55
I From J IM'r data book, Page No. I J 2 S·
Nu, = 0.332 (2.84 x 105)05 x (0.6~~;~~~:itiOnjl
Paral/ello
50 em,
(b) Paral/ello
30 em.
Also calclIlale the percentage of heat loss,
Givell: Fluid temperature,
T:1,) - 30°C
U
Velocity,
I Nux = 156.51 I
Plate dimensions
We know that,
4 rn/s
50 em x 30 ern
0.50 x 0.30 m2
Local Nusselt Number,
Nu
hx L
.r
156.51
hx
Local heat transfer}
coefficient
h
, "
We know that,
.
Surface
k
hx x 1.5
0.03047
3.179
W/m2K
3.179
W/m2K
temperature,
T;
To find :
I. Heat loss when the flow is parallel to 50 em, QI'
2. Heat loss when the flow is parallel to 30 em, Q2'
3. Percentage
of heat loss.
T •• + Too
Solution : Film temperature,
TI
h
[rr
2 x 3.179
Ih
We know that,
Heat transfer,
Q
Scanned by CamScanner
2
90+ 30
2
-
A verage heat tranSfer}
coefficient,
90°C
6.358
I
Properties
of air at 60°C,
..
[From HMT data book, Page No. 33 (SixthEdillon)]
p :::
1.060 kg/rn3
,
Convective
~
2.56
-----_
Heat and Mass Transfer
v
18.97 x 10-6 m2/s
=
PI'
Q\
~-
case"(
0.696
.
~:
Reynolds number, Re
=
v
ReyllO
4xO.3
(':L==30em=O.30m\
18.97 x Io-t'
Re == 6.3 x I O~ < 5 x 16~J
r
UL
\I
. S . lOs flow is laminar.
._~..9.2Q__
18.97x
10-6
[.: L = 50 em = 0 50
-i~o~'x 105 < 5 x 105 1 " III]
~-:-
,
UL
Re ==
Case (i): When the flow is parallel to 50 ern.
I
I
99.36\\
==
2ji
,
ihe flo\\ is parallel 10 30 em side.
..) . \\ hell
Id.; Nurnhe,r
0.02896 W/mK
k
I
Heal Trallsfer
t: <:
"
,
Since R
.I te laminar
for flat p u ,
flow,
Local Nil st;;\1 Number
0.332 (Rer
NUf
.
Since Re < 5 x 105, flow is laminar.
(Pr)o J3J
•
.. ..'} (632 x 104)0) (0.6
0.))-
For flat plate, laminar now,
Local Nusselt Number, Nu, ~ 0.332 (Re)O 5 (Pr)OJ.H
96)0333
.
74.008]
h.L
..2-
k
[From HMT data book. Page No.112 (Sixth blillUlI1!
Nu,
=
0.332 (1.05 x
105)05
x (0.696)0
'.:=~~3T]
-L-oc-a-I-N-l-Is-se-I-t
-N-lI-n-lb-e-r,-N-lix
'I
0.02896
7t008
:::
We know,
k
95.35
=
_!2_x__ ~:~~_
28.96
~I§ill
----------~
II~
5.52 W/1I1 K
Ilcallr;II\$fcr,
coefficient.
>
It
2
h,
It "
2
5.~
lit
I kill truustcr,
f
10 J
r------------------L.ocal hcnl transfer coefficient.
We know rhar,
Average hent transfer
0
1
" t\ ('1'"
~\
"1', .. )
(I.
11.0 I
(OJ
II.W
\')
OJ)
_
O2 '"
\~'~
II.n·1 WhwK
11.0 I'
Scanned by CamScanner
7.141 W/1ll1K
hx
\41 WIln2K
. I /, :::7.
coefliclen,
s
Local heal \ran er
.
h::=
2)( h,
r coefficient.
e Ilcal tr3nSler
h ::= 7- x 7 • 14
A era"
::::>
"x L
NlI.,.
I
h. x 0.30
~
3H
(1"",
(f)()
'J',J
\0)
(ii/) :
l'
n 41
\'IKIII
loSS
11' - T:f})
A x \ \I'
1 W (T\I'~ T,,)
/,)( ~
.
6) - ..0 )
O.
(
O
.
14"8
"
W'~
2. 58
C 011\ ective Heal Tr(Jllsj'er
Hear and Mass Tran~rer
128.5-99.36
99.36 -
[oX of heat loss
Result:
I.
QI
29.3%
99.36 \V
2.
Q2
128.5 W
J
Case (i) :.
3. % of heat loss == 29 ....
.-,-E-x"-u-"p-'-e-1-2-'1 Air at 30°C flows (J1'''r
.
-
~
~
100
?\
.
..
1.060 kg/m?
v
18.97 x 10-0 1112/s
PI'
0.696
k
0.02896 W/mK
For first halfofthe
x
a
m Plate
tit (/
plate,
•
U xL
\\e knoW.
Reynold
number,
Re
v
4 x 0.45
First hut]:of tile plate,
18.97 x 1~
I Re - 9.4 x J04 < 5 x 10 I
2. Full plate,
5
3. Next half of the plate.
Given:
Fluid temperature,
5 x 10\
Since Re
T rn
I
For flat pate,
Velocity, U
Plate surface temperature, T".
Local Nus
Plate' dimension
It
flow is laminar.
laminar flow,
1.
First half of the plate, i. e., x
Full plate,
3.
Next half of the plate.
Solution:
I Nu .•
0.45 rn,
x
405
==
I
90.21
L oca I Nu sell Number, Nux
k
0.90 rn,
90.21
=
[!ix ==
T". + T~.
T
f
2
~------z.:-;;
efficient,
Local heat transfer co
2
Average heat trans
~
[..._T...__/
fer coefficient
h == 2 x hr
_~~~
Properties of air at 600C :
[From HMT dala bod
Scanned by CamScanner
~
v •
Page No, 33 (Sixth Ediuonj]
Heat transfer,
Q1
112 ( iXlh Edillon
0333
(Pr)
h_x_
xL
We know that,
Film temperature,
0,
0.332 (9.4 x 10)
90 X 30 ern?
2.
" )1
N
umber. Nux'
Heat transfer for
i. e.,
.).
[From HMT data book, I age
== 0332 (Re)O.5
0.90 x 0.30 1112
Tofind:
2.59
L == 0.4) m
==
fl
velocity of " m/s. Tile plate is nUlilllllilled at 90 t The pltl/e
dimension is 90 x 30 cmt, Calculale the heat transfer for the
following conditio"
1.
P
hx A x (T
- T",)
II'
(0696)0.333
.
____
2.60
Heat and Mass Transfer
_
2.61
//~ (lise "(iii':
Heat lost from the nextxv half
I
,
nau of
ot tl
tne pate
h x L x W X (Til" - Tn)
[~I
Convective Heat Transfer
----
\_x=~~~
11.61 x 0.45 x 0.30 x (90
.. _
-·30)
[ . x - L = 0.45 Ill' W= 94.04W
-0.30111]
I
~\
\-0,-\_0
'
Case (ii) :
Reynolds Number,
Q3 =
L = 0.90 m
For full plate. ~
\33.48-94.04
v
, Q3
4 x 0.90
18.97 x 10-0
1
Re
1.89 x 105 < 5 x 105
-
Since Re < 5 x 105, flow is laminar.
J
-
For flat plate, laminar flow
Local Nusselt Number , Nu x ' 0.332 (Re)o.s (Pr)0.333
0_._33_2......:(~1.89x
105)05
r:-;- __
I Nux
128.18
I
x (0.696)0333
We know,
=
1.
Heat lost for first half of the plate
Q,
2.
Heat lost for entire plate
O2
3.
Heat lost for next half of the plate
03
O~xample
=
=
Ih =
J.
Overall drag coefficient,
2.
Average shear stress,
Compare tire llverl'ge slrear aress withloeal shear stress
Velocity,
1'0 find:
2 x hx = 2 x 4.12
I
8.24 W/m2K
h x A x (T 11' - T co )
Scanned by CamScanner
133.48 W
I
U ::: 3 m/s
\. Overall drag coefficient,
2 . Averaoe0 shear streSS,
3. Compare
Solutlon :
. loc
hear stress with
the average s
shear stress.
.
f ir at 40°(. :
Propel11es 0 a
e No. 33lSixth
HMT data bOO\;, pag.
IFroll1
8.24 x 0.90 x 0.30 x (90 - 30)
(91 -
39.44 W
a velocity of 3 m/s. ClIlculllte tirefollowing:
Heat transfer for entire plate
Q2
l33.48W
13J Air at 40 °Cflo ws over a flat plate of 0.9 m at
4.12 W/m2K
A verage heat transfer coefficient
h
94.04 W
(shear stresS at tire trailing edge).
Given: Fluid temperature, Ten
40°C
Length, L ::: 0.9 m
hx x 0.90
0.02896
Local heat transfer coefficient
hx
I
39.44 W
Result :
3.
128.18
Q2-Q,
UxL
Re
3-
:::
P
1ll3
1.128 kg/
..
EdItIO
\
t
!,
2.62
v
I
t'i
----
Heal and Mass T ransjer
.c
=
Pr
0.02756 W/mK
Reynolds Number, Re
.
Since Re<5x
we ~now that.
_t_.t
tpCs' skin friction coefficient CIt
'.66)( ,0-3
I Re =
16.96 x 10'
105 , flow
i lami
ow IS
J
5
1.59 x 10 <5
2
S.4 x 10-3 N/m2
nar.
Local shear stress, 'tx
[From HMT data bo .
Local shear stress, 't x
A "erage shear stress, 't
rag coefficient (or) A verage skin. friction
ok, Page No. 112 (S' . h ..
c ffici rxt Edllionll
_
oe IClent
1.328 x (Re)-O,5
r
.:..:.1.3~2~08
x (1.59 x 105)- 05
\ C/l
3.3 x 10-3 \
0.52
3
= 3.3 )( 10-
Average skin friction coefficient, ell
1
A\'erage shear streSS, 't == 0.016 N/tn
2.
1:
Average friction coeffiicient,
.
C- fl
P U2
. ~, '
~
t
=
-
C
fl
x ~
r
lfl
-
Exampl.
Given:
Fluid temperature, T:f)
61llfs
Ve\ocit)', lJ
Length, L
Wide, W
= 0.664 x (Ret05
I III
0.5 III
6 kN/1ll2
6'1. \03 N/II'!
Pressure of air, P
(From HMT
.
0.664
(1 data book, P age No. 112 (Sixth
Edition)}
- x .59 x 105)- O.S
'
Plate surface temperature,
30
Scanned by CamScanner
290°C
pllJl~.
Local skim frinction
. coeffiicient
.
I
p'''''
'.mp.,.,."
I
1-=-.66- x 10-3
141 Air aI 290'C flow' 0'" • fI'" plo" '"
p""."
1.128 x (3)2
Average shear ~tress t
2
w~e~k~n~o~w~,~~~~,~~JO~.0~1~6~N~/m~2 \
___
:; 052
't
0
."ocity of 6 ,,;;.. r•• pial' is I .. long on4 0.5 .. wide. TIl'
of I.' .1, Is 6 ANI"". If IA.
is "",in,.i.tII '" •
of 10 'C, .. ,i..... Ibt ,."t ., h•• ' ,. ... .""
2
3.3 x 10-3 x
't.1'
3.
2
______
S.4 )( 10-3 N/m2
0.0\6 N/m2
Result:
t. Drag coeffICient
or
We kno~' that,
Cjx
\.\28x (3f
x \O~
CIl.
_
~
2
3 x 0.9
UL
For flat plate , la mmar
.
flow,
D
2.63
__ ------~C~o"~v~ec~t~ive~H~ea~/!.~a~1'{.e,
n.~er
_::~
.t
70°C
r
lI'
,ro" ,••
~2~.6!4~R~e~a~ta~n~d~~~m~S~T~~a~m~~_r
~
________
Convective Heat Transfer
Tofind: Heat removed from the plate.
Solution: We know that,
Film temperature,
--..._
TI =
Tw+T~
2
=
70+2~
-----...........--'nee Re
SI'
late. laminar flow,
F r flat p
,
o
Local
,
[From HMT data book. Page No. 112 (Sixth.
Ed"
)1
ilion
Nusselt Number
Nu = 0.332 (Re)05 (Pr)0.333
2
x
Properties of air at ISO°C (A t atmospheric
2.65
< 5 x 105, flow is laminar .
= 0.332 ( 1.10 x 104)05 (0.681 )0333
~=
preSSure) :
30.631
We know that,
[From HMT data book, Page No. 33 (Sixth Edition)]
P
= 0.779 kg/m)
v
Pr
= 32.49 x I~
Loca
m2/s
Nu,
0.037S0 W/mK
Note: The given pressure
=
_x_
k
is not atmospheric
presSure. The
properties of air such as k, Pr and Cp do not change mUch with
V
We know,
.
I at transfer coefficient,
re
,
h
It = 2 x I. 15
PO/III
va/m x~
Pgiven
l1iiO:_ill-W-/m-2K--'1
I bar
32.49 x 1Q-6 x 6 x 103 N/m
::::
Heat transferred,
105 N/m2
6
32.49 x 10- x 6 x 103 N/m2
-.![:....:••_• ..:_I
.:.b=ar_=_1 x 105 N/m2]
5.415 x 10- 4 m /s
2
Reynolds Number, Re
_ 6x 1
5.415 X 10-4
1.10 x 1Q4 < 5 x
IO'J
It A (Too - T
lI,)
254.1 W I
Q
.
b thI sides
of the plate
Heat transfer from 0
::0
::0
2 x 254. J
508.2 W
Q = 508.2 W
Heat transfer,
J 5 nrl area and -# mm
[ EXlImpleJ 5: I A sq uare glass
.
ir at 10 'r'
~ plate
d it is cooled ~, a
_,
I
to
90
\...
an
l,
Ilrick is heated uniformy
.-/ paral/ello lite pate at 3 "VS.
.
bot" slues
Which is flowing over
ling the plate.
.
..
I
te
Calculate tire initta ra oJcoo I
'''_?
1500
kglmp
Take for glass :
Result:
C
p
Scanned by CamScanner
Q
2.31 x (I x 0.5) x (563 - 343)
2
r',' Atmospheric preSSure:::: I bar]
, v
I
A verage
Kinematic viscosity, v
[v Lr= l m]
= 37.S0 x 10-3
----~----~~~~~~~h~~1.15_w./m2K
Local heat tr~nsfer coefficient,
x
30.63
pressure. But the kinematic viscosity will vary with pressure.
~e
h L
0.681
k
r.:-:--~:--:-Kinematic viscosi
I Nusselt Number,
= 0.67 KJlkgK
",'
l
I
I
~
~,
\
66 Heal and Mass Transfer
~2.~~~~~~~~~
Convective Heat Transfer
_______
V ~,'I'"l':
For air at mean tenrperature.55 '(' :
0.132 (1.63x W)
~
= 1.076' kg/m3
Cp = 1008 J/kgK
p
.
.
:::: 0.332 (L63
k = 0.0286 W/mK
x 105)0.5
2.67
Pr ~ ~ ~P
J
\ 19 8 x 1000{)x 100810.333
0.0286
l .
\
I
p
Given:
= 19.8 x 1(j-(J N-S/m2
G lass plate area, A
1.5 m2
Thickness of the plate, t
Plate surface temperature, Til'
4 mm
Fluid temperature,
20°C
For air,
p
4 x 10-3 In
90°C
Ten
Velocity, U
For glass,
=:
hx x I
3 rn/s
h
2500 kg/rn!
Cp
0.67 KJ/kgK
p
Cp
1.076 kg/m!
= 0.67 x 103 J/kgK
We know that,
= pUL
v
~
1.076 x 3 x I
19.8 x 10-6
[ .: v
=; ]
[.: L = I m]
1.63 x 105 J < 5 x 105
Since Re < 5 x lOS, flow is laminar.
[Refer HMT data book, Page No. 112 (Sixth Edition)]
Local heat transfer coefficient for the air flow parallel to the
plate is given by
Nux· = 0.332 (Re)05 (Pr)OJ33
Scanned by CamScanner
t transfer coefficient,
Loca I h e a
h :> 3 40 WIm2K
x
.
.
[h = 6.80 W/m2KJ
To find: Initial rate of cooling the plate.
UL
=, 3.40 W/m2K
x.
We know that,
A verage heat transfer l = 2 x h x
coefficient, h J
= 2 x 3.40
~ = 19.8 x 10-6 N-S/m2
Reynolds number, Re
::::>
(.: L = 1 m]
= 0.0286
\\8.90
= 1008 J/kg-K
k = 0.0286 W ImK
Solution:
\
~
VI e knoW that,
Heat transfer from both l ::::
2 x hA (Tw - T<t)
sides of the plate, Q J
680 x 1 x (90-20)
:::: 2 x .
[Q ::::952WJ
\ .
Convective Heat Transfer
2.68
Heal and Mass Transfer
Reynolds
I Example /6 I Air at 300C
flows over a jI--;;;--:----_,
.
..
P ale ell
number,
velocity. 01/ 3 m/s. TIre plate IS maintained CIt 90 ar-.",
\... lire I ('
dimension is 900 nrnr x 600 mm x 30 mm, lt Ilr I P tile
'J
e I 'er"
conductivity of II,e plate is 27 WlnrK, find,
't"
IRe-'= \.42x\OS\<5x\05
Since
Re <)~ x \ O'i-, flow is laminar.
For fla
t plate laminar now
'
[Refer HMl data book, Page No. "2 (Sixth Edition)l
Velocity, U
Plate surface temperature,
Til'
Plate dimension
Length,
I = 0.332 (Re)Oj (Pr)o.m
"Nusselt num b er, N ty
= 0.332 (1.42 x lQ5)05 (0.696)°.333
900 111m = 0.9 III
Width, W
600 mm
=>
Thickness, t
Thermal conductivity}
of the plate, k
Tofind :
Loca \
900 rum x 600 nun x 30 nun
L
~ Nux
= 0.6 III
30 mm = 0.03 III
27 W/mK
I. Heat loss, Q.
2. Bottom temperature
of the plate for the steady
We know that,
~'l~_:~
Tf =
Tw+T;fJ
h x 0.9
2--I 10.88 :; 0.02896
II
Properties of air CIt 60'C :
1.060 kg/m!
v = 18.97 x 10-6 m2/s
Pr
k
Scanned by CamScanner
Local heat transfer
coetTlcient, hx
f:; 3.567 W/m K
2
L __---
A verage heat transfer \ :; 2 x h.t
= 60°C I
p
:; 3.567 W/m2K
2
90 + 30
2
~
hx L
k
.f
We know that,
Film temperature,
=
Nux
state condition.
Solution:
-;
3 x 0.9
18.97x\(T6
I. Heat lost by the plate.
2. Bottom temperature of the plate for lire
steaely Slate
condition.
Given:
Fluid temperature, T eo
Rt: =
2.69
UL
0.696
0.02896 WlmK
coefficient, h j
:; 2 x3.567
[[~~
Heat losS, Q
:; h A~T
r)
(r It xWx L " II'
09:< (90 - 30)
oJ
7.\34)( 0.6 x .
.'fl
Heal t-« 11sJer
.r.
__ -------c::..:o::n:..:v~eC:"'-lllr·vV((.'
2.70
""\e1\2.th of the plate is turbulent
WhO'"\
...
.
Heat and Mass Transfer
TO
:
I. ft"tl
'thickness
of the boundary layer, I).
We know that.
]}
Heat flow by conduction,
Q = ().RT
where
R
L
kA
Q =
{)'T
LlkA
(
T,
2-
dean value of heat transfer coefficient h
I"·
' .
solution:
properties
0
of air at 20 e :
lFrom HMT data book, Page No. 33 (Sixth Edition)1
\
~
\_
p ::: \ .205 kglm3
v ::: 15.06 x \0-6 ml/s
[From HMT datakA
book.
'. I:.ditionll
.
().T Page No. 43 (SIxth
\
~
Q = _-
Pr ::: 0.703
k ::: 0.02593 W/mK
L
\
We knoW that,
Reynolds
where
L - Thickness
of the plate
Number,
Bottom temperatu~e-}
of the plate,!,
1.
2
.~6~.6_4_X_l~06_>~5_X_l_0~_5
~~e
0.03
Since Re > 5 x \ 05, flow is turbulent.
1
f or flat plate, turbulent flow. [fully turbulent - given
(T, - 90)
90.47SoC or 363.47;1
Q
23-1-14--\. \
.
W
IE'
T, = 90.475°C
. xample 17 \ Air at 20°C'
.
1
O.S m wide at a velocity 0' ;ofloowmg over aflat plate of m
Local Nusse\t Number
,Nux:::
0.0296 (Re)O:8 (Pr)o.m
[From HMT da;' bOok. P"; No. I t J lSixth Ediiio
0.0296 (6.64 x IO'i" (0.103)°333 '
t:
engtll
'J
m/s.
I of tile
. P "ate IS made turbulent
C I Tile fl ow over tile wllOl,
. Thickness of the bounda
. a culate tile flowing
Q"-N-u- -7§J
We know
"
hx xL
k
Nux:::
h x 1
Length, L
1m
\l__
Wide, W
Velocity, U
0.5 m
~~~~-!----~
100 m/s
' ,"
x
2. Mean valu ,£
ry layer.
G'iven :
e OJ heat transijer
I:
Fluid
coefficient.
temperature '<Xlr
20°C
Scanned by CamScanner
Re
\00 x 1
\5.06x lQ-6
231.14
__:0,..4,75
rr;--:-:- __
\JL
v
0.03 m
27 x 0.9 x 0.6 x (T, -90)
~
Result:
2.71
7552
~fi
~
==
0.02593
.
1~--:;\9s~
1
ConvectiveHeat Transfer 2.73
I
I
~1~.7~2~~H~e=w~an~d~~~a~s~s~f,_ra_'~~fi~e_,.___________________
For nat plate, turbulent
..
tlow,
Average heat transfer
-----
coefficient
h
I 25 I
.
Mean heat transfer coefficient,
1x
h
1.25 x 195.8
h
244.75 W/m2K
h
244.75 W/11l2K
Boundary layer thickness:
Boundary
I
~
/'foft"t1:
I
.
for.
(i) Entire
is considered
l-{
.
.
eat translerred
laminar
plate
and turbulent
flow .
. ' E ti e plate is considered
(II) n rr
2.
Percentage
as combination of both
as turbulent flow.
error.
. . We know that,
SolutIOn.
layer thickness
Film temperature,
0.37 x x x (Re)-02
T'F +T'fJ
TJ
2
300 + 40 ::::443 K
2
0.37 x I x (6.64 x 106)-02
[.: x = L = I rn]
IL....:O~_=o.-=-O 1:..:.59n~
Result:
I.
Properties
Boundary
layer thickness
o =
2.
p
0.0159
m
v
Mean heat transfer coefficient
h
=
of air at 170°C:
Pr
244.75 W/m2K
k
I Example 18 I Air at 40°C flows over flat plate, 0.8 m long
0.790 kgltn3
3 1.1 0 x 1Q-6 ro2/s
0.6815
0.037 W/roK
II
._
at a velocity of 50 ntis. The plate surface is maintained at 300°C
Determine the heat transferred from the entire plate length to air
taking into consideration both laminar and turbulent portion of
the boundary layer. Also calculate the percentage error if the
boundary layer is assumed to be turbulent nature from the very
leading edge of the plate.
Given:
Fluid temperature,
T'"
Length,
L
0.8 111
U
SO m/s
Til'
300°C
Velocity,
Plate surface temperature,
Scanned by CamScanner
We know
Reynolds Number,
UL
Re ::::
v
~::::
:::: 31.10xl~
~,~D
~--
t.26x106
_.
.
bulent floW.
tlow is
s
this IS tur
[It llIeans,
.
Re > 5 x 10', so
ombilled.
that floW IS
. _turbulent c.
x 10;, after
Case (i): Lall\ltlar
ber value IS 5
Ids num
laminar upto Reyno
turbulent.1
Convective Heal Transfer
2.74
----
Heat and Mass Transfer
Average Nusselt
Number
} Nu
=
(Pr)OJ33 [0.037 (R )0.8
e
- 871]
(From HMT data book, Page No. 114 (S·
.
. rxth
Edit"
0'"
Nu
=
(.6815)
O
Nu
x
=
1746.09]
QI
80.75
2010.15
x
W/m2K
1_16_:.2:...:_0_W_/m:..:_:-
h x A x (T w ~ 1co)
Percentage
2.
error
J
QI
Q2~
QI
h x A x (Til' - To)
24169.60 - 16796 x 100
16796
n x L x W x (Tw - Too)
43.90
r-:-
80_.7_5_x_0~.8x I x (300 - 40)
I QI
16796W
I
0.0296 x (Re)0.8 x (Pr)0.33
= 0.0296 x (1.286x I06)0.8 x (0.6815)0.333
hx xL
'k
=--
=
O2
L-.
24169.60 W
I Nux = 2010.15 I
Nu
Heat transfer,
I
[From HMT data book, Page No. 113 (Sixth Edition)]
We know
Ih
heat tran~fer}
coeffiCient
116.20 x 0.8 x I x (300 - 40)
Case (ii) : Entire plate is turbulent flow'
Nux
1.25 x 92.96
h x 0.8
0.037
Average heat }
transfer coefficient
h
1.25 x hx
h x L x W x (1 w - Tco)
80.75 W/m2K
=
x
A verage heat transfer }
(for fully turbulent flow)
h
Average
106)08 -871]
h
coefficient,
k
1746.09
Local N usselt }
Number
Nux
I heat transfer
hL
Nu
Heat transfer,
coe
.
f{iclent
IOn))
..).).) [0.037(1.286
[AVerage Nusselt Number
We know
~~a
2.75
92.96 W/m2K
hx x 0.8
0.037
I hx = 92.96 W/m2K I
Scanned by CamScanner
Result:
\
Heat transfer
bi d)
(Laminar-Turbulent
com me
QI == 16796 W
Heat transfer
(Fully turbulent)
.
2
.
3.
Q2 :::: 24169.60
.
W
Percentage error == 43.90 fl tplate at a speed of
"1
0 OCflows over a a
\ Example 19 J Air at
, 60 em long and 75 cm
0 °C Tile plate IS
I ce al
90 mls and heated to' 10
dary layer take P a
'( n of boun
wide. Assuming t/le transl 10
,
,
Re = 5 x 105, Calculate the followmg ,
, ' coefjicient,
1. AveragefrlctlOn
'.JJ
,
.fer coeffiCient,
2. Average Ileat trans),
' sipation,
3. Rate of energy ciIS
Convective Heat Transfer
27~.1~6
__ ~H~e~a,~a~n~d~A~la~s~S~Ti~ra~n~sfi~e_r-:
__ ~~~
.: Given: Fluid temperature, T a: - O°C
__
--------
~inar-turbulent
~
Speed, U = 90 rn/s
Surface temperature, T
H,
loooe
Length, L
60 em
0.60 In
Wide, W
75cm
= 0.75m
flow
for
[From HMT ~ata book, Page No. 114 (Sixth Editionj]
friction } C J L.
Average fficient
coe
::::>
CfL
0,074 (Ret 0.2 - 1742(Ret 10
~
0.074 [10 x 106]-0.2 1742 [3.0 x 106]-10
ToJind:
1. Average friction coefficient,
C IL = 3 .16 x 10-3
A~e
2. Average heat transfer coefficient,
friction}
coefficient
3. Rate of energy dissipation.
Averag e Nusselt } N u
Number
Solution : We know that,
::!>+T
C L = 3.16 x 10-3
f
= (Pr)0.333 [0.037 (Re)08 - 871]
k P No 114 (Sixth Edition»
[From HMT data boo. age .
oo
Film temperature, T/
2.77
2
100
---+ 0
2
(Pr)0.333 [0.037 (3 x 106)0.8 - 871]
.
(3 106)0.8 - 871]
(0.698)0.333 [0,037
x
[N~1215J
Properties of air at 50°C :
We know,
[From HMT data book, Page No. 33 (Sixth Edition))
p
v
1,093 kg/m'
=
17.95
X
h
10-6 m2/s
Pr
0.698
k
0.02826 W/mK
Average heat tran~fer
h
coefficient
T )
'------;Q == h A (TIY - eo
Rate of energy dissipatIOn,
= h » L)( W (T", - T <Xl)
We know,
Reynolds Number, Re
Average Nusselt Number, Nu
UL
::: 198.5)( 0.60){
v
90 x 0.60
17.95 x 10-6
3.0 x 106 > 5 x
Since Re > 5 x 105, flow is turbulent.
~
iQ£J
[Note: Transition Occurs means flow is combination of laminar an~
turbulent flow. i. e., the flow is said to be laminar upto Re value is 5 x 10 ,
after that flow is turbulent.]
Scanned by CamScanner
Result:
1. eft.
2. h
3. Q
== 3.16)( 10==
075(100-0)
.
3
2
198.5 W Im J(
== 8932.5 W
(
I
2.78
Convective Heat Transfer
Heat and Mass Transfer
I
, Example 20 Air at 40 C(' flows over a jl;;;-::----'_,
.
.
.
Pate Qt
velocity of 2 mls. The p Iate IS maintained at 100 cc. Th
Q
.
e length
the plate is 2.5 m. Calculate the heat transfer per unit:..I
of
W",th Us'
(a) Exact method,
llig
(b) Approximate method.
Given:
Fluid temperature,
~(')
. Using exact solution,
ClIst , .
.n
for I,a
tplate,
laminar flow.
Local Nllsselt. \ Nux
Nutnbel I
Too
.
Nu ... = 0.332 (2.49 x. \05)05
~~
We knoW,
Width = 1 m
146.6_}
hx x 2.5
Tofind :
1. Heat transfer (Ql) using exact method.
Heat transfer (Q2) using approximate
x (0.694)°333
hx L
Nux = T
Length L = 2.5 m
\46.6
method.
~-hea.t
tran~fer
= 0.02966
l hx = \.74 W/m2K
L- coefficmet 1
Solution:
Film temperature,
= 0.332 (Re)05 x (Pr)om
[From HMT data book. Page No. 112 (Si'>..1h·Editiont!
Velocity, U
Plate surface temperature, T w = 100°C
2.
2.79
.
Average heat ~ h = 2 x h
transfer coefficient I
x
= 2 x \.74
Tf
QiiOl48 W/m2g
~
Q
Heat transler.
Properties of air at 70°C:
p
= II x A x (1", - 1<1:)
\
::::. 11)( L x W )((1
IV -
1.029 kg/m!
3,48 xl.5 x \ )( (\00 - 40)
v = 20.02 x 10-6 m2/s
Pr
0.694
k
0.02966
~2Y-J
W/mK
We know that,
Case (ii):
Reynolds Number, Re
UL
v
2 x 2.5
20.02 x 10-6
[Re
2.49 x lOs < 5 x 105 I
Since Re < 5 x 105 , flow I'S Iammar
.
Scanned by CamScanner
1co)
t solution:
.
Approxllua e
Local Nusselt
Number Nux
r ;::.O.m
~J
'05
J3J
x (pd
x (Re)
.. ')
.
. 06C)4)ll)J
9 x \05)0) ;< ( .
;::. 0.323 x (2.4
{
Convective Heat Transfer
2.81
Specific heat, Cp = 1.005 KJlkg-K
2.80
Heal and Mass Transfer
hx xL
We know that, Nux
k
= 1005 J/kg-K
Thermal
conductivity,
hx x 2.5
I hx = 1.69 W/m K I
2
roJind:
h = 2 x hx
Average heat}
.
transfer coe ffiicient
We know that,
Ih
3.38 W/m2K
Q2
h x A X (TIV - T
~C .
e
Prandtl number, Pr =
I
2.29 x 10-5 x 1005
;::;
3.38 x 2.5 x 1 x (100-40)
Q-2--50-7-W
[!>r = 0.676J
hx
Exact solution,
Qt
522 W
2. Approximate solution,
Q2
507 W
I
Stanton numbe(, St = Cp p U
N III (Sixth Edition))
HMT data book. Page o.
[F rom n
h
I
Example 21 Air flows over a flat plate at a speed lif
60 mls. If the local skin friction coefficient on a plate is 0.005,
calcukue the local heat transfer coefficient at that point.
Take for air:
l'
5t ;::; 1005 x 0.89 x 60
p = 0.89 kg/m3
IJ = 2.29 x J o-s kg-m/s
Cp
1.005 KJ/kgK
k
Use
Given:
0.034
-'1
Result:
1.
k :
[From HMT data book. Page No. III (Sixth Edi~ion))
<1;»
h x L x W x (Til' - T co)
""-1
2
Local heat transfer coefficient. hx'
Solution:
h = 2 x 1.69
Heat transfer,
~
St Prl13.=
142.7 = 29.66 x 10-3
:::)
k = 0.034 WImK
0.034 W/mK
sr Pr2f);::;
We know that,
~
~
St p,.vJ
Sa2
~
53,667
2
Velocity,
Local friction coefficient,
U
C [x = 0.005
Density,
p
Viscosity,
Il
Scanned by CamScanner
60 m/s
)( (0.676~]
~
0.89 kg/rn!
= 2.29 x 10-5 kg-rn/s
Result:
Local heat transfer
;::; 174.19
}
coefficient, h%
WI 21(
III
) I
..-_-------~COOlnvecl;veH
<orflat plute. tamlnar _ t.rbol
ea'T""",,,
1 Heal and Mass Transfer
I
11 Air at JO~ an~~1
plate• til• a velocity 0"
50 nv.\.
.~. Tile plate
..OI.J] bur Ilow s O"e .
'J
maurlallled
al
70
-c.
C
I
"
. a cutate the heat e I.'. 1.5 n, 10,,· ' Q /l,III
II:.':mmp/~
enl combintdfl
T'
Average Nusselt \
nw:
Number. Nu J = (Pr)0.333 [0 .O~7
J
(Re)08 - 871
I
I e plme, II,king into consider ,.
Irtlll·'ill!rlor It I I( ""d ,.
portion ollhe boundar'"J Iayer. II ton bOlll IU"';'lttr (II1t1" 1IIIklitlll,'I/S
Gi"en •.
PI
FI'd
UI temperature
''F.
T
p
1.013 bar
Velocity,
U
50 m/s
___
'h",'':'"
300(
Pressure.
[From HMT databook. Page No. 114 ( .
Nu
[
= (0.698)0333[003
. 7 (4.17Sx 1(6)0.8
!'!_~=
5728.28 \
hL
k
Nu
We know that,
- 871]
Length L
ate surface temperature , 'T II' --
700(
~
5728.28
Width, W
I. ~Ieat transfer Q
I III
~
h
107.92 W/m2K
Average heat transfer 1.
coeffic ient, h J
107.92 W/m
Tofind:
1.5 III
SolUlioll .. W e k now that, ,.
Fihn temperature,
T, = Til' + T.~.
-.'
Heat transfer,
2
. .]
Sixth [dlhon)l
h x 1.5
0.02826
Q
= h x W x L (Tw -T )
C1)
~
')
=
Propert]res of air at 500
~6475.2
e:
[From HMTd ala book , Pago: No. 33'
P _
v
.
1.093 kg/m!
=
17.95 x 10-6
m 215
Scanned by CamScanner
Heat transfer, Q = 6475.2 W
Convection
. \ Example 1 \ Air tl' atmospheric pressure and 200'C flo
~ver tf plate wi,h a velocity of 5 m/s. The plate is 15 mm wide a
at a temperature of 120 'C. Ca/culale the thickn
IS" taintained
•
Of hydrodynamic
and thermal boundary layers and lite local h
transfer coefficient at a distance of 0.5·m from the leading ed
W/mK
Reynolds numb er, Re :::: -...!::
U
v
:::: ~1.5
17.95 x 1tr6
ro:--4"'-1
-~
~_:_Z_8_~>
I
2.8.2• Solved U'nlvarsity
. ProblemS on Flat SUrtKII - Forc
Pr :::: 0.698
I
W
(Sixth Edllionll
-
k :::: 0.02826
Result:
107.92 x 1 x 1.5 x (70 - 30)
Ass ume t'IU' the flow ls on Me side of the plate.
5 x 10;
1 84
Convective Heat Transfer 2.85
Heal and MasJ Transfer
p. 0.815 ",1",1, u= U .s» /(t-(I NsI",1 ;
~",;c
/f!ydfodynal
~
Pr .. 0 7, k = 0.0364 WlnrK
Gil'e,,:
.
{May 2004 An
• n« {ln'
Fluid temperature, T eo = 200°C
,ve,s;ty/
Velocity, U
boundary layer thickness :
5hx = 'S x x x (Re)-0.5
I·
5 x 0.5 x (8.32 x 104)-0.5
,=
= 5 m/s
= 18.667 x 1(}-3 m
Wide of the plate, W = IS mm :: 0.015 m
Plate surface temperature, Tw = 120°C
Distance, x
I 0hx = 8.667 x 10- m I
3
= 0.5 m
~pu
p = 0.815 kglm3
",al boundary layer thickness:
8.667 x 10-3 x (0.7)- OlB
Pr = 0.7
I OTx = 9.76 x IQ-J m I
k = 0.0364 W ImK
1. Hydrodynamic boundary layer thickness 0
,
h:r
2. Thermal boundary layer thickness , 0 Tx :
3. Local heat transfer coefficient, hx.
0.332 (8.32 x 10")05)( (0,7)°.333
v
5 x 0.5
[ .: v =~ ]
!:!.
'p
8.32.x
hx)( L
85.03
104
~.l'
-r
h x 0.5
0.0364
.2---
=
= 6.19
Result:
1.
Ohx
< 5xlOS]
2.
For flat plate, laminar flow,
3.
. [Refer HMT data book, Page No. 112 (Sixth Edition))
= 8.667 x 10- m
= '9.'76 x 10-3 m
h.;
!:::
[.:x::::L::::O.5m]
w/m2ig
3
Since Re < 5 x'IOs , flOW'IS. laminar,
.'
Scanned by CamScanner
Nu, =
We know that,
5 x 0.5
24.5 x 10-6
0 -,815
·IRe
= 85.03
Nux
[.: x = L = 0.5 m]
v
. "
Weknow that,
UL
5 x 0.5
U
3, Local/reat transfer coefficient, It % :
Local NUSSelt} N = 0.332 (Re)os (Pr)0.333
number
"x
Solution: We know that,
Reynolds number, Re
.
= 011%x (Pr):' 0.)33
0Tx
~ = 24.5 x 10-6 Ns/m2
Tofind:
.
6.19 W/m2K
Convective Heal Transfer
2.86
_ Heal and Mass Trall.~fc.'r
I Example 2 I Air
III
,I
1
----
J 34 't' a' a velocity of 3 m/s. The plate is 2 m Inllg c Pierre a,
wide. Calculate .the ..thickness of the hvdrod
. (111(1
.
YIlallllC
b 1.5 sn
ayer
ami
the
skill
friction
coefficient
at
40
c
fi
OUII(/a
...,
m
rom
the
l
..)'
I
edge of 'he
plate.
The
kinematic
viscositv
.£'
. eerdi"D
r...J.
,
•
oJ til r lit 20
eo
IDee . ...
'005 , AmICI lJ . 't' iS
.
.
J 5• 06 X. J U - nrls.
ver
Given : Fluid temperature, T <J'.' = 200C
", silJ,1
Plate surface temperature; T",
134°C
Velocity, U
Length of the plate, L
Wide, W
Distance, x
Kinemati~ viscosity
air at 20°(,"
3 m/s
2 III
1.5 m
= 40 em = 0.40111
~f} = 15.06 ,10-
To find:
x
6
m2/s
Solution: We know that,
or
.
1. Hydrodynall'l1c boundary layer thickness,
3
bhx ::::'7.08 x 10- m
3
2. Skin friction coefficient, CIx = 2.35 x 10- .
.. ~
Air at 25~flows over J m x3 fit (3 mlong)
'zontal plate maintained at 2·00~ at 10 mls. Calculate the
M" e I,eat transfer coefficients.for both laminar and turbulent
averag
.
'\
'ons
"'ake
Re
(critical)
=
3.5
x
lOS.
r~
.1'·
. '
[Dec. 2004, Anna
University}
Fluid temperature, T
local friction
v
.
[Re
7:96
Since Re < 5 x lOS , flow I'S Ialllll1ar.
.
For flat plate, laminar flow,
H d.
[',: L = x = 0.40 m]
< 5 x 105
Plate temperature, T", == 200°C
Velocity, U == 10 m/s
Re(critiCal)= 3.5 x lOS
?_!~= 5 x 0.40 x (7.96 x 104) OJ
[5lrx == _7.08 x 10-3 an I
Soilition:
We know that,
'T1I'+ Tci:l
Film temperature, T/ := ~
200 + 25
:=~
. [From HMT data book, Page No. 112 (Sixth EditiQIl)]
rodynamic boundary layer thickness:
Sir.\' :: 5 x x x (Re)-o.s
Scanned by CamScanner
= 25°C
ToJind:
. fl
1. Average heat transfer coefficient (h) for lammar ow.
· t (h) for turbulentflow.
2. Average heat transfer coeffilClen
Reynolds number, Re == UL
x 1041
ci:l
Length, L == 3 m
.
3 x 0.40
15.06 x 10-6
0.664 (7.96 x \()4,05
2.35 x \0-3 \
llesll1t:
Given:
'
2. Skin friction coefficient
coefficient, C (x .
y
\ Cp
1. Thickness
of the hydrodynamic bmm~~~
d
5
hx :
1.
cllefficient or local friction coeffic~nt :
Cj:f = 0.664(Re)-Oj
~,",i(lft
20'r is flowing (Ilollg a "eclte1 .-----
2.B7
'
~
Properties of air at 112.5 C(' :
»MT datil boOk, Pa~e
lfrom
_
p - Q.
No. 33 (Sixth Edition)]
922 kglro3
It
Im.o)'
Z,)
r'
~
I
I
i
\I
Pr
... 24.29
10-~'~
I
k
= 0.03274 W/rnK
UL
Reyn ld number,
Re ..... \I
Con\lecti'Ve Heat Tt'Cmift'
For turbulent
,~.
,
Sf (il)
'1.1
~
},89
flow,
l/Ca 1 NlI~~t.:.It number, Nux = 0,0296 (Re)O,I(Pr)O
I
0.687
:c:
-. _
'
J)
[From IIMT data book, Page No, 113 (Si, th Ediu nH
Nux
0,0296 (1.23 x l06fl.8 (0,687)0 ).
[Nux
1945 I
1945
= 0.03274
10 x 3
24.29 x 1Q-6
I
IH~w
Rc,(~~ 'I)
number value is 3.5
..
lot;
Re = 1.23 x J 06 1
3,5 x lOS, i.e. flow is laminar
105,
upto Reynolds
after that flow, is turbulent,
~
h r = 21.22 W/m2K
Case (I) : For laminar flow.
Local Nusselt
Number,
0.332
Nul'
(Re )0,5 (Pr)OJJ3
[From HMT data book" Page No, 112 (Sixth Edition))
We kno« that.
~
Nux
0.332 (3.5 x J 05)05 x (0,687)0 m
I Nux
173.33
Nux
hx L
k
173.33
hx L
k
173.33
hx x 3
0.03274
h l'
Result :
~ oefficient for laminar flow,
er c
I . Average heat trans
2K
h == 3.78 W/m
. f rturbulent flow,
sfer coefficient 0
1.89 W/m2K
Local heal transfer coefficient, hx = 1.89 W/m2K
Average heat transfer coefficient, h
2 x hx
2 x 1.89
h
3.78 W/m2K
Average heal transfer coefficient}
for laminar flow If
3.78 W/m2K
Scanned by CamScanner
2 Average heat tran
2
. h - 26525 W/m K
d1 '" long is 10 be
.
'de an 5.
0"C
---:--:-:11
flat late 1 ", WI
eralure of J '
[ Example 4.J A
. P, , h a/ree strea'" temp over flat plale
90'C in air Wit
. ",ust floW
the
maintained at
'with
which air
dissipalio" fro'"
Determine tile velOCity the rate 0/ energy. of air 0150"{;,
. I SO that
along 1.5 m su e,
fi /low;ng pr opertleS 007 KJ/kg "C',
UI -rake t/le
0
Cp ... J.
'1tJ/
plate is 3, 75 k"· I'
8 WI'" I{' ;
Anna Vnb,ers"J
mJ' k ::::0.02 [Mar 20M,
p = 1.09 kg~ 'S' n= 0.7.
'
JJ = 2.03 x 10-5 kgl",- ,
r
2.90
Heal and Mass Transfer
Wide. W = I m
Given :
_----~(_·~O::.t7\:::'(!~CI~il~.e.!.H~e~(I!...1!}.T/~·a~/l;!!J.sfi~i!l:_· ~2?!.!_
_~_L2 0 ......2 ( pLU )0.5
.91
___ .
0.028
.-'-'
~l
x (Pr)OJJ3
I'~S
-:;:J
Length. L = 1.5
Plate surface temperature,
Tw
90°C
Fluid temperature,
IT
00
ro-c
Heat transfer or Energy dissipation,
Q
3.75 kW
[.,'
o .-'-'~x
....., [1.092.03x x1.510-x UJ OJ x (0.7)OJ3J
837.0S
83.66 x (U)05
5
[_~ e-'ocity of air, U = 100.10 m/s
1.007 kJ/kgOC
To find:
Velocity
0.7
/ Pr
Velocity of air, U.
=>
~age
= h (1.5 x 1)(90-
h = 31.25
heat transfer coefficient,
10)
W/m2K
h
31.25
We know that,
Local heat transfer}
coefficient, h,
h
[June 2006, Ann« Universityl
h
= 2
Given:
31.25
x
2
~--:-)~5.-62-5-W-/m-2-K-1
Local NUSSelt}
number, Nux
Flu id temperature,
=
=
0.332 (Re)05
Velocity,
(Pr)O.333
Critical
T
To find:
0.332 (Re)O.5 (Pr)0333
[ .: Nu~~=
hie L ]
T",
= 275 K = 2°e
U = 20 nvs
Length, L = 1.5 m
= 325 K == S2°e
Plate surface temperature, T If
Width, W == 1 m
h~ (From HMT data b 00,k P age No. 112 (Sixth Edition))
Reynolds
number, Re,
= 2 x lOs
I. Average heat transfer coefficient. hi
Boundary layer is laminar]
[
'. t, h
transfer coefficten
I
2. Average h ea t
[Entire length of the plate]
3.
Scanned by CamScanner
air at 2 75 K and a free stream
velocity of 20 nrls flows over a flat plate 1.5 m long tI,nt is
nraintai"ecl at a uniform temperature of 325 K. Calculate the
average heat transfer coefficient over tl,e region where the
boundary layer is laminar, the average lIeat transfer coefjiciellt
over the ell lire length of tire plate und tire lotll/lreat transfer rate
front the plate to tire air over the lengtll 1.5 nt and width I m.
5
Assume trm,sitim, occurs at Re, = 2 x 10 ,
We know that,
Heat transfer, Q
hA (T 111 .- T ex> )
3.75 x 103
I
of air, U = 100.10 m/s
Result:
a-i-x-,,-n-,p-'-eS"""']
Atmospheric
Solution:
=>
100.10 Illls
U
0.028 W/moC
2.03 x 10-5 kg/m-s
7J
817.0S
3.75 x 103 W
1.09 kg/m!
Rc =
'fotal heat transfer rate, Q.
Convective Heat Transfer
2.92
Heat and Mass Transfer·
Solulion:
~
Film temperature,
T,,+T-..::
2
52 + 2
T,
hx = 22.42 W/m2K
_._.
\Lbc31 heat transfer
We know that,
~oefficient,
2 x hx
2 x 22.42
Average h~at tra.risfer
coefficient for
laminar flow, , hi
1.185 kg/m!
v = 15.53 x 1Q-{i m2/s
Pr = 0.702
= 0~02634 W/mK
k
., UL
Reynolds number,' Re
v'
= 2 x 105
Transition occurs at Re,
i.e., flow is laminar upto Reynolds number value is 2 x lOS,
after that flow is turbulent.
20 x L
2 x 105
15.53 x 10-6
0.155
m
I
For flat plate, laminar flow,
Local Nusselt number, Nux = 0.332 (Re)0.5 (Pr)O 333
[F;om HMT data book, Page No. 112 (Sixth Edition)]
Nux
0.332 (2 x 105)0.5(0.702)°·333
N-u-x--13-1.-97~1
hxxO.155
. 0.02634
='
44.84 W/m2K
l
20 x 1.5
15.53 x ]0-6
1.93 x 106 >, 5 x 1Q5
~eL,;; '., .
.
5 105 flow is tur~ulent.
.',
..,
'
.
ReL> x .'
SInce
..
,. . .: ..,. 'b··l t combined flow,.
For flat plate, lammar-tur u en
!
Itl'
Av~rage, N US~~
number, Nu
.,
Nu
~1I
We kn0r-:that,
'.
(pr )0'" [0,031 (ReLl"~~ 87:J ns _ 871J
.. i. .. 0.3)3 (0.037 (1.93 x 10 )
== (0.702).".
..
.
d,'
:=. 2737J!].
,
"
b Nu· ==
, Nusselt-num er,
,..
.
0
v
:
hL
k
,
~4
, ..
2737.18 == 0.0263
.
2"
. , h == 48.06, '1!/m ~
Local Nusselt number , Nu It
=
1
Case (ii) :
,
Reynolds number, ReL
(For entire, lengthlJ
We know that,
131.97
44.84 W/m2K
h
p
Scanned by CamScanner
=
= 27°e
{From HMT data book, Page No. 33 (Sixth Edition)]
r-I
= 22.42 W/m K
Average heat transfer}
coefficient, h
Properties of air at 27'(' 1:1 25 '(' :
Case (i):
2
hx
VI e knoW that,
2
T,
2.93
.I.,
Convective Heal Transfer
2 94
~d
Heal and Mass Transfer '
Average heat tranSfer}
coefficient for
turbulent flow, hi
I
e
hi A Il T
I
2. Average heat transfer coefficient
[For entire length of the plate]
h, = 48.06 W/m2K
3. Total heat transfer rate, Q = 3604.5 W
270C anti I bar flows over (I plate (,t u
speed of 2 mls.
Calculute the boundary layer thickness at 400 mm from
tile leading edg« of tile plate. Find tile mass flow rate
per unit widtt: of the plate.
For air p = 19.8 x 1()-6 kglms at 27OC.
If tile plate is maintained
at 60 't', calculate the IIeat
per hour. The properties of air tit mean
(27 + 60)
temperature of
2
= 43.5 OCare given below.
C
Cp = 1006 JlkgK;
R = 287 Jlkg-K; Pr = 0.7.
[Madras Ulliversi(V, 98/
Scanned by CamScanner
.,
= 1006 J/kgK
p
R '= 28'7' J/kg-K
Pr == 0.7
To find;
Case (i) :
1, B ou
ndary layer thickness, ~\ . ,
rate
'flow
per unit width, 1/1,
2. M ass
Q
.\ Heat transferre~ per hour, .
case (ii) :
Solution:
,,'
Case (;) :
.J!_
Density, P = RT
~
287 x (27 + 273)
g~,~
Reyno\ds number,
Re -
-
transferred
k = 0.02749 WlmK;
'
elise (II '.
'
ties of air at 43.5'(' :
proper, '
k = 0.02749 W/mK
I
Result :
I. Average heat transfer coefficient
[Boundary layer is laminar]
h, = 44.84 W/m2K
(2)
,
..) . Plate surface temperature, Til! = 60°C
48.06 x I x 1.5 x (52 -- 2)
3604.5W
IQ
= 0,400 m
ffiIC"'1"11tof viscosity, ~l = 19.8 x 10-6 kg/m-s
Co
",xWxLx(TII-TT.)
(II
-
= I bar = 1 x 105 N/m2
Pressure. p
.
Distance, x. I.= " 400,mm
. I
Total heat transfer}
. rate, Q
(1)
Tz. = 27°(
Velocity, U = 2 m/s
We know that.
I Example 6 I Air
temperature,
~G'tJvee"":' 1:' lUI
!:!h
V
UL
:s:
~
p
I
••
2.95
Heal and Mass Transfer
2. 96
Convective Heal Tra
2 x 0.4
19.8x 10-6
1.16
[h
~
Heat transfer, Q
I Re =; 4.686 x 1O~ < 5 x 105
.
8.772 Wlrn2K ~
h x W x L (Tv - T,.J
Since Re < 5 x 105 flow I'S la .
.
'
mmar.
8.712 x 1 x 0.4 (60 - _
[Refer HMT data book P
Boundary layer thickness
~
=:
'x
5 ' age No. J 12 (Sixth Ed' .
Xx x (Re j-us
IliOn))
Q
in
Here,
115.79 Jls
5
115.79 x 3600
8" x p x U [ ~2x - 0
~ Ix =: 0 , U2x
~
~x
[in
Local Nusselt number
:::::>
:::::>
N
'
Q
Nu
=:
x
Nux
=:
Nusselt number
I Q = 416.84 x 10 Jib I
3
= 9.23 x 10-3 m
Result:
Case (iJ :
0.0133 kglJ'
_
Ux - 0.332 (Re)o.s (Pr)Om
0332 (4 68
.
.
. 6 x 104)0.5 x (0.7)0.333
63.8D
We know that ,
9.23 x 10-3 m
m
0.0133 kg/s
I Example 7 I Air at 25'(' flows over aflat plale at a ~
Fluid temperature,
Too = 25°C
Velocity, U = 7 mls
k
Plate surface temperature,
T; = 85°C
Distance, x = 20 cm = 0.2 m
To find:
Solution:
Local heat transfer coefficient (h;r)'
We know that,
Film temperature,
Average h
eat tranSfer}
coefficient, h
Scanned by CamScanner
== 2 x hx == 2 x 4.386
-I
7 mls and heated to' 85 '('. Calculate the local lIeat tTtoGfor
coefficient at a distance of 20 em.
IOct. 98, 2000, MU (EEE)/
N _ hx L
ux-
~x
Q = 416.84 x 103 Jib
Case (iiJ :
Given:
,
J
h
]
5
- x I 16
8
. x 2- [9.23 x lO-3J
In
Case (ii):
Ix
)j
115.79 W
5 x 0.4 X (4 686
.
x 104)'-0
9.23 x rO-3
.S
iii]
Mass flow rate,
hA (Tw- T..r.)
T w + Leo
Tf =
8S + 25
---r- -z
=
Conveclive Heal Transfe,.
1.98
Heal and Mass Transfer
Air at 20'(' flows over aflat plate at 60 or with
£~ ", l·(!lllci(V of 6 m/s. Determine tire value of the
sirea
,
.
(ret
. 'e('/ive heat transfer coefficiem upto tllength of I nr
I't, 8
'{llfl
Properties
of air at 55°C:
. [From HMT data hook,·Page No. 33 (S'
Ixth Edit'
Kinematic
Density,
•
p
viscosity,
v
Prandtl Number,
Pr
Thermal conductivity,
k
We know that,
1.075 kglm3
=
1011)]
18.41 x 10-6 m2/s
'.
[x::: L::: 0.2 m]
v
0
rF.l
rojincl :
'
T
Plate temperature,
600e
II'
U
6 m/s
Length, L
1m
A verage heat transfer coefficient
Solution:
Tf
Film temperature,
2
~
7.6 x 104 < 5 x 1051
x lOS, flow is laminar.
For flat plate laminar flow,
. '.., . Local Nusselt .,} .'
,,', ~,'
.
,.
\\
.... Number
,NLt.l'.\:=. 0.332 (Re)0 ..5 (Pr)03~3'
"r
Sundamnar University, April 97/
T
20 e
Fluid temperature,
Velocity,
UL
Re
'
directIOn.
;~(/leflow
IManonma~ium
0.02857 W/mK
~:::
a
{From HMT (fat book, Page Nd. 112 (Sixth Eahibn)]
0.332 (7.6 x 104)0.5 (0.697)0:333
jr--'N-u-x -8-1-.15-,
We know,
111 .
,.
, ,Pr€)pertles
=
}' ,
Nux
hx xL
k
81.15
hx x 0.2
0.02857
_ hx
11.59 W/m2K
Local heat transfer,coefficient,
Scanned by CamScanner
r.
v •
f ai t40oe:'
.
air a
"
, k P 'No 33 (Sixlh Edition)]
'From
HMT
data
boo,
age
.
. . '
(
":"
k m3,
.: "
Density, P
1.128 gJ
" .,
Ie - 'Q.02756 W/mK
Thermal conductiVIty,
II\.J. m21s'
.
. v'
16:96)( 'V Kinematic VISCOSity,
,. ,
'.
.'
' ._'., 0 699 "
Prandtl Number; Pr ,~
.
0
Reynolds
!:Lh
Number,
Re
,\
'
v
6)(1_
"
[.: x = L = 0.2 m]
Local heat tran~fer } L
coefficient
2
40°C 1 .
We know that,
Local Nusselt
Number
Result:
oe com'
avertltJ
Give" :
0.697
7 x 0.2
18.41 x 10--6 ::: 7.6 X 104
Since Re < 5
#)'
.
.
Reynolds Number,
2.99'
,
hx ::: 11.59 W/m2K
=1~_,
___
Re '"
3.53)(
IO~ < 5 x 10
.
,
is laminar.
'
.
105 floW I
Smce Re < S x
,
. th [dilion)]
. arflow
'" ...., ))?(51"
For flat plate, lamm,
book. Page ,.0, , HMT data
IFrom
1.100
Heat and Mass Transfer
0.332 x (Re)o.s x (Pr)0.333--------
LocalNNussebelt } Nu
Ul~
r
x
I Nux
Convective Heal Transfer
I
Number
175.27
j
1~
Local Nusselt
Number
}
Nu
Local heat transfer coefficient.
film temperature,
k
x
4.
'o ' We know that,
solll't n .
hxx L
Local Nusselt } N u
Local friction coefficient,
3. Thermal boundary layer thickness,
•
.- 0.332 x (3.53 x 105)0.5 x (0.699)0.333
175.27
~
75 +25
0.02756
2
[From HMT data book, Page No. 33 (Sixth Edition»)
2 x hx
Density,
9.66 W/m2K
Average heat transfer coefficient
I
h = 9.66 W/m2K
I Example 9 I Air at 25'r:' at the atmospheric pressure is
flowing over a flat plate at 3 m/s. If the plate is 1 m wide and llu
temperature T'III= 75 'r:'. Calculate the following at a location of
leading edge.
Hydrodynamic boundary layer thickness,
Local friction coefficient,
Thermal boundary layer thickness,
LOCIll heat transfer coefficient.
(April, 97, MU/
Fluid temperature,
Velocity,
Distance,
Prandtl Number, Pr =
Thermal conductivity,
0.698
k =
0.02826 W/mK
Wekoow,
Reynolds Number,
{.: x=L=
Re
3x 1
= 1.67 x 105
17.95 x 10-0
1.67 x 105 < 5 x lO~
. laminar
Since Re < 5 x lOS, flow IS
•
U
3 m1s
For flat plate, laminar flow,
MT data book, Page
[From H
1m
r;
75°C
x
1m
I m]
v
25°C
Hydrodynamic boundary layer thickness,
Scanned by CamScanner
17.95 x 1()-O m2/s
.
Kinematic viSCOSity, V ,=
1. Hydrodynamic
boundary
s
-
Uhx
-
TofUld:
I.
1.093 kglm3
p =
T co
Wide, W
Plate surface temperature,
= 323 K = 50°C
Properties of air at 50°C:
Ih
Given:
2
hx x I
2 x -4.83
1 IIIfrom
(i)
{ii)
(iii)
(iv)
Til' +T<I)
Tf
4.83 W/m2K
x
Average heat } h
transfer coefficient
Result:
1.101
~
_
No 112 (Sixth Edition))
.
layer thickness,
x (Rer°.5
5 xot
5 x I x (1.6
•
7" toStO'
.--::::-----=-:;-\..
:
(f~"" I ,Ie J 0
"tn,,>spherlc
, /5
I /
2. J 02
----
Heal and Mass Tnlllsjer
2. Local friction coeffil:ient
ef,
0.664 (Ret
05
i
~~.
\
"
from L = 0 III L ::; 2m. AIsa fiI.d
",,,I'•r codJiciCllt
..
3
1.62 x 10-
0.0122 x (0.698)- o.m
0.01375
surfa'c
I
Total length, L
Width, W
2m
lm
tcmperature,
400 K
4, Local heat transfer coefficient (/1.) :
Til Pllc/ :
\.
\ .ocai h at iran
We know that,
2.
t\\'era~
3.
Heat iran
Local Nusselt } .
Number
NUl'
0.332 (Re)0.5 (Pr)0333
_ 0.332 (1.67 x \05)05
(0.698)0333
120.415
'II'
Nux
f r Q.
Sol Iltion :
Ctlse' (i): Locallreat
transfer coefficient til L == 1m.
1",+1",
Film temperature,
T,
~
, 400 + 300
~
==
. 'II
350\(
k
=
~
['.: x = L == I 01)
0.02826
Local heat-}
transfer coefficient
Properue
770C ~ SO°C·.
of air at \ <om HM1 d," ",,,,. p,g< No. II l""" Ediu,,'1
2
hx = 3.4 W/m K
\ kg,ltn3
p
2 1.09 y.. \ o-b 1\,2/5
Result:
I.
s;
2.
<
1.62 x I O-~
3.
bT.\
O.0137~ m
h\
3.4 W/m2K
4.
=
==
hI. x I
120.415
,~th,.,
fer coefficient at L == \ m,
heat uansfer coeffIcient at L == 2 m.
h; xL
We know,
0('
~~~ill~
300 K
2.S m/s
Fluid tClnpera\\lre,
T""
Velocity, U
(;;l't'/r:
8", x (Prt0333
==
with a vr/o ~.
.
,I "" -;
3. Thermal boundary layer thickness.
[};'.,
,,' ]00 K'
'h' I ,
,,'",II'"''"(,, "",ffie".' (I' J .. lellg,h and' •" cu'''''
.
"'''ag'
h,..
~t
81x
""
"fi'"
'""
O"er
pi ..te ~(I.n.,h
L = 2 m and wid,h Wcrty oj
l/~ 1,{lK..
."
I'; rl . .(1 (It "IIiform tc",pL'rlrtllre oj 400 K C I
-1m
0.664 (1.67 x I O~)· 05
Le"
COllvecrj,e Hear T·
.
"
'."1" IIOJ
0.0122 m
I
Scanned by CamScanner
Pr
k
We know that,
0.692
,
0.03047 W/n \(
.-v
\\L
,/
~
\
'I, \
Healand Mass Transfe,.
2.104
2.5 xl
21.09 x lQ-6
.
I
----
H 8539 45 <
. ' 5 x 1O~
:._j
\ Re
Since Re < 5 x ,()S , flo W .IS lami
ammar.
1
{Refer HMT data boo .
k, Page No. \ 12 (Sixth: Ldillon)l
'.
N
Number
Ux
=
0.332 (R e )0 '.'\ (Pr)0,333
_
_:_\O::_:iJD:.:..l~8
,
•
hxx1
\ hx
Local heat }
transfer coefficient
hx
1.10;
x
h
2 x 2.17
\ h
4.35 W/m2K \
4.35 W/m2K
h A(T",- Too)
.--_-~l:-,·:L = 2 m ; W = 1 m]
0.03047
3.0832 W/m2K
Re
UL
v
Re
2.5 x 2
21.09 x 1~
~Q
\
870 W]
-
Result :
1. Local heat transfer coefficient, at L = \ m = 3.08 W/m2K
at L ~ 2nL
2. Average heat transfer coefficient at L = 2 m = 4.35 W/m2K
3.08 W Im2K
CIIS~ R(i;'
"" : Average heal transfer coefficient
S.
\ Re = 237Q79
mce Re < 5 x lOs fl
.
.18 < 5 x lOs
For flat
' ow IS laminar.
plate, laminar flow
3. Heat transfer, Q = 870 W
\ Example 11] Air at 20 ~ and one almosphere flows overlll II
}hit pI.te at 35 mis, The platt is 75 cOl long "nd is "",UrtBintd
I
600C. Calculate the I.eat transfer per unit width of the platt. Also
calculate th« turbulent boundary /tryer thickness IIIt/o< tnd oftltt
pl.te assuming
of tltt pllJU.
University, Apr. '}7
it to develop fro .. the /eOdiJlg ..".
/Bharathidasall
0.~32 (Re)O.5 (Pr)OJ33
0.332 (237079 .18)°,5 (0.692)°,333
r:\N~-U:r---=1~4'3_'\
Scanned by CamScanner
.c
ansjer
4.35 x 2 x 1 (400 - 300)
101.18
Nux
2xh
Ih
Average heat l
transfer coefficient J h
Cast (iii): Heat transfer, Q
J Nux
eynolds Number,
HealTr
h, xk L
'VI e know that,
A verage heat l
0.692)0\33
1.
=>
Nu,
transfer coeffICient
We know,
Local Nusselt
Number
we know Ihal.
0....332 (1185395)0,5 (
..--- __
_--,--
-------~(·~o~nv~eecclive
.'
Local heat '\..
2
transfer coefficient i h."( = 2.\7 WIm K
For flat plate, laminar flow , '
Local Nusselt
/
::: 20°C
Given:
fluid temperature, Tao
35 rnJs
Velocity, U ::: 15 cro ::: 0.75 m
Length, L :::
2 J 06
Convective Heat Transfer
Heat and Mass Tramier
Plate temperature,
Width,
Tofind:
-----
T",
W
1.
Heat transfer.
2.
Boundary
1111
~
Local Nusselt
. Number
2107
hI xL
l
I
k
hx x 0.75
2341.6
layer thickness.
Solutlon :
0,02756
Local heat
T" + T'l
T}
Film temperature,
2
For flat plate, turbulent
60 + 20
A verage heat } I
transfer l:oeffciellt
'
2
Properties
flow,
1.25 hx
"
of air at 40°C:
p
1.128 kg/m '
k
0.02756
v
=
Pr
W/mK
Heat transfer,
16.96 x 10-6 1112/s
1.25 x 86.04
107.55 W/m2~1
0
h x A x (T" - T Ul)
Q
h x A X (T", - T'/J)
h x L x W x (T
Re
[0
UL
v
.
16.96 x 10-6
Boundary
layer thickness
flow,[Fully,turbulent
Local Nusselt}
,
,NulJ\ber
0.0296 (Re)08
,
=
,
'
,',
I N HI' = 2341.6 I
Scanned by CamScanner
2
0.37)( 0.75)( (1.54 x 106t0
[.:x=L=0.751
- given]
~
x (Pr)om
I From HMT data book. Page No, 113 (Si:\lh Edition)1
, 0.0296 x (1.54
WJ
2
,\ " '\1 '--'_R_e_~I.:..:..5_.:_4_x__;I:...::0_6 _>....::5:....:x~10::._5.....J1
..
"Sinc,~ Re.> 5 x ,lOs: flow is turbulent.
,',for flat plate" turbulent
3226,50
0.37 x x x (Ret0
:
NUl
11)
Bount/ary layer thickness:
35 x 0.75
"
T
107.55 x 0.75 x 1(60 - 20)
We know that,
Reynolds Number,
II' -
0.699
x 106)0,8 x (0,699)0333
,W
f 0 - 3726.50
.
Heat trans er
- === (I,01601\l.
s
h'cknes 8
Boundary layer t I
Result:
1.
2.
2. J 08
Heal and Mass Transfer
I. Example 12 t For a particular engine, the und erslde
.
0"
crank case can be idealised as a flat plate
:J the
.
"'ellS
80 em x 20 em: rile engine runs lit 80 km/h» and the
UI';lIg
.
..
' crank
is cooled by air flowing past It at tile same speed. Cal
clIse
culate th
loss of heat from the crank case surface of temperature 75't e
the ambient air temperature 25 'C. Assume the bounda
to
becomes turbulentfrom the leading edge itself.
IApril ~ laye,
Given:
Area, A
80 em x 20'cm
' M(Jj
1600'cm2
U
Velocity,
Convective Heal Transfer 1.109
22.22 x 0.8
17.95 x 1ij-6
[",' L = 0.8 m]
'[iii-e-:--9.-9 X-I-O'-]
== 0.16 m2
= 9.9 x 105 > 5 x J()5
Re
105,
Flow is turbulent.
Since Re > 5 x
For flat plate, turbulent flow,
[Fully turbulent
Local Nusselt
Number
from leading edge - given]
} Nu.
=
80 km/hr
[From HMT data book, Page No. 113(Sixth Edition)]
80 x 103 m
0.0296 [9.9 x WJO.8 (0.698)033
N-u- --16-:-:4~5.4-:-11
3600 s
"-1
x
22.22 m/s
Surface temperature,
Ambient air temperature,
h;rxL
Til' .= 75°C
T co
We know that,
k
= 25°C
hx x 0.8
o:o2s26
Flow is turbulent from theleading edge, i.e., flow is fully turbulent.
To find: 1. Heat loss.
Solution:
Film temperature,
0.0296 (Re)D.8 (Pr)OJ33
l
75 + 25
=-2-
T,
IT,
[.,' L = O. 8 m J
58.12 W/m2K ]
Local heat } h =
transfer coefficient
x
For turbulent
58.12 WIm2K
flow, flat plate
A verage heat } h
transfer coefficient
Properties of air at ·50°C :
h
Or
[From HMT data book, Page No. 33 (Sixth Edition)]
P
1.093 kg/m-'
v
17.95 x 10--6 m2/s
0.698
Pr
k
We know that,
Reynolds Number,
Re
Scanned by CamScanner
We know,
Heat loss, Q
0.02826 W/mK
!d....h
"
~
Result:
=
h A (Til' - Tao)
=
72.65 x 0.16 (75-
_
581.2 W
Heat Ioss, Q -
25)
,I
~t,
II
I'
2.110
__
----
__
CO"\lecti\)e Heal Transfer
lind ambient comlltlons are pressure 760 mm tI!
IlrIs
te",perature is 15 'C. The plate is maintained at 85 'tHg an.
lengtll of the plal£ is 100 em .Iong the flow 0'
tl".· tr, '"'t/
.. 11 Ihe
',I
h
hea! lost by 50 em of the piette wlticlt is measured ~
, e
Since Re < ~ x \ 0\ flow is laminar.
trailing edge. Plate width IS 50 em.
/BlllIrc,tltidasan
Local Nusselt
Number
, ,
I
'
.
}ron, Ih
'
e
Unlversit J.1 N-",',96/
Re
_.
lJ N u~ --
Length, L
Width, W
=
1m
SO em
0.50 III
I. Heat lost by 50 em of the plate which is measured from the
Solution:
T". + Too
Tf
2
85 + IS
2
Properties of air at 50°C:
.
I
{from HMT data book, Page No. 33 (Sixth Edition)l
Density, p =
Local Nusselt
Number
=:.>
=
17.95 x 10-6 m2/s
Prandtl Number, Pr
=
0.698
Thermal conductivity,
k =
0.02876 W/mK
Reynolds Number, Re
hx L
k
\20.36
hx x I
0,02826
3_.4_W_'n_~_K~1
3.4 W'm2K
2 x II.~
II
2 x 3.4
[F-:::-6-.S-W-'-11l-2K--']
Heat transfer
~
(For entire plate, Lc., L:= l m]
Q2
11 A (1 ...- TaJ
:: hxLxWx(TII,-Ta)
::: 6.S x \ x 0.5 x (85 - \ 5)
~
.
Heat transfer for ftrst half of the plate, 1.(1·,
UL
V
Scanned by CamScanner
r Nux
Local heat} h
transfer coetTlcient
.t
A verage heat \
transfer coefflcitnt I I,
Similarly,
We know that,
l
~h_x
1.093 kg/m)
V
Kinematic viscosity,
0,332 x (\.67 x lQ5)05 (0.698)0333
I N\I.~ = \ 20,36 \
[TZ=
(
0,332 x (Re)05 (Pr)OJ33
=
=:.>[
.
Film temperature,
x \ 0'. < 5 x 105
We know that,
100 em
To flnd :
trailing edge.
.
{From HMT data b 00",l. P age. No. t \2 (SIxth
. Editionj]
Velocity, U = 3 m/s
\
760 mill of H.g = I bar
Pressure
ISoC
Fluid temperature, Too
!II' ,= 8SoC
\ 67
=
For flat plate. laminar flow,
Give" :
Plate temperature,
2. J IJ
3x \
17.95 x 1Q-6
<Example
,'---L!
_ liJ_" Air flows over a flat plate of velo {"Yo!3
!
I
~
Heat and Mass Transfer
V 0 SO 11\
.
I
,,
I
Convective Heal Transfer
I
I
!
2.112
I
Heal and Mass Transfer
UL
Reynolds number. Re ::
v
3 x 0.5
17.95 x I()-6
-----
Q2 (entire plate) - QJ (for first half of the plate)
Q :::: 238 - 168.35
~
I Re :: 0.835 x lOs < 5 x 105 I. -. L'ammar
. flo
For flat plate, laminar flow ,
Local Nusselt }
Number Nux
/lesult :
}-Ieattransfer from 50 em length from trailing edge = 69.65
W
[]!Pnple
=
r:-:- __
I Nux =
0.332 x (Re)O.5 x (Pr)0.333
0_.3_3_2
x (0.835 x 105)0.5 x (0 69
85.1
8)0333
I
.
I
We know that,
Nux =
hxL
k,
Here
L = 0.50 m
~
85.1 = hx x 0.50
0.02826
~
2.1
14] Air
pressure
T
of' 8 kNlm2 and
250°C
C1.)
Wide, W
OJ m
Length, L
1m
Velocity,
U
8 mls
Plate temperature,
Til'
78°C
2
Average heat transfer coefficient h
(I
te",perature of 250°C flows over (I flat plate 0.3 m wide and I
long at a velocity of 8 m/s. If the plate is to be maintained a
temperature of 78°C, estimate the rate of heat to be remov
cOlltinuouslyfrom the plate.
/Bharathiyar University, Apr.
Given:
Pressure, p
8 kN/m2 = 8 x' 103 N/m2
Fluid temperature,
I hx - 4.81 W/m K I
(It
Tofind : Heat transfer.
h
Solution:
Film temperature,
Tf =
Tw+T<tl
2
78 + 250
2
Properties of air at 164°C: (At atmospheric pressure)
I
[From HMT data book. Page No. 33 (Sixth Editio
p :: 0.810 kg/m3
I
v :: .30.08 x 10-6 ro21s
I
k :: 0.03645 W/roK
Pr :: 0.682
/
Scanned by CamScanner
Convective Heat Tramfer
Average heat transfer coefficient, h
2.114
Heal and Mass Transfer
r a pressures
p
Kinematic viscosity, V = va1m X ~
Ih
Ie
0
Heat transfer,
3.08 W/m2K
h x A (Too - T
00 -
Tw)
3.08 x 1 x OJ x (250 - 78)
= 30.08 x I ~
x
30.08 x I~
x
I bar
8 x 1()3 N/~
10
~~~[·~:~I~b~ar~-lxIOSNI
m
= 3.76 x 10-4 m2/s 1 ]
158.9 W I
)-Ieat transfer from both side of the plate
o = 2 x 158.9
lOs N/m2
8 x I ()3 NI;;;i
IQ = 317.85WI
2
Result:
Heat transfer,
0 = 317.85 W
We know that,
UL
Reynolds Number Re
2.9. FLOW OVER CYLINDERS AND SPHERES
V
The flow over a cylinder is shown in Fig.2J.
8xl
3.76
.
x 10-4
I Re - 2.1 x 10
4 <
The flow field can be divided into two regions. They are:
5 x 10
5
I
Since Re < 5 x lOs, flow is laminar.
For flat plate, laminar flow,
Local Nusselt}
Number
1.
Boundary layer region near the surface.
2.
An inviscid region away from the surface.
[From HMTdata book, Page No. 112 (Sixth Edition))
Nux
0.332 (Re)05 (Pr)O.333
0._33:..:2~(.2.1
x 104)0.5 x (0.682)0.333
42.35/
r:-:-
I Nux
We know,
42.35
=
hx x I
0.03645
Local heat}
transfer coefficient
hx = 1.54 W/m2K
For flat plate, laminar flow,
Scanned by CamScanner
I
1r)
h x L x W (T
.
Pgiven
V
2 x h,
2x 1.54
Note: Given pressure is not atmospheric presSure
viscositywill vary with pressure. Pr, k; C are same fo II· So, kinernar
~~==~~~
[Kinematic viscosity,
2.115
Stagnation point
.
Fig. 2.3.
CI w over c)'lim/ers
no
.~
Convective Heat Tramlfer
2.116
u~
Heal and Mass Transfer
'---;;prob'ems
- Flow Over Cylind,,,.
.
-:
soNe01
%.9.2~
Air at 15 OC, 30 IrmI1r flot4ls over a cylinder of
Th Pressure gradient along the surface of the c~
e
.
'.
er IS no
and infaci this pressure gradient IS responsible"
t
'.
I~~
th:
development of a separated flow region 011 the back side of
· der . The separation of flow affects the drag force on a cUrve<!
cy IIn
surface to a great extent.
I
I
l}!~
and 1500 mm height witll surface temperature
", d,a
I
J,o '" «: calculate the heat oss.
\ of 45
Given.
• Fluid temperature,
T'"l
\ SoC
U
30 kmlh
Velocity,
30 x lQ3 m
3600 s
2.9.1. Formulae Used for Flow Over Cylinders and SPhere.
r, + Too
l.
Film temperature,
TI
where
T 00 -
Fluid temperature
T", -
Plate surface temperature "C
2
where
3.
U
Nusselt Number,
Diameter,
-c,
UD
2. Reynolds Number, Re
v
- Velocity,
mls
D -
Diameter,
v -
Kinematic viscosity.
Nu
m
Nusselt Number,
5.
Nu
Heat transfer, Q
where,
A
U
8.33 mts
D
400 mm = 0.4m
Length, L
Plate surface temperature,
T If
Tofind: Heat loss.
I t 'on' . We know that,
Sou'
Film temperature,
T,
= U
1 +1",
45+15
7 = -2-
hD
k
h x A x rr,- T "J
1t
DL
Propertie
f air at 30°C:
)II
n (Si~th Edition)}
[From HMT data boo~, Pase o.
.
= I I 65kg/m3
Density. P
.
v viSCOSity,
be Pr ==
Pr'andtl Num r,
Therma\ con4uctivity. k
Kinematic
.
n...J. 2/
16 \< Iv - m s
0.701
0.02675 W/n,K
We know,
Nusselt Number,
Nu =
0.37 (Re)O.6
Reynolds Number, Re
[From HMT data book, Page No. 119 (Sixth Edilion))
Heat transfer,
Q
A
Scanned by CamScanner
m
e (Re y" (Pr)O.333
For sphere :
where
ISOOmm
4SoC
-,T-I--30-oC~]
m2/s
[From HMT data book, Page No. 115 (Sixth Edil.ionll
4.
2.117
v
~
\6)( 10-0
h A (Tit' - Too)
~
Convective Heat Transfer
2. J J 8
2.119
Heat and Mass Transfer
Nusselt Number,
C(Reyl/(~
Nu
~
[From HMT data book. Page No. 115 (S' h
•
105,
ReD value is 2.08 x
·IXI
correspondmg
.,
Editio
C value is 00266
n)]
. ) and I1J
I.
fl01
temperature,
TJ
130 + 30
2
value is 0.805.
I T = 80°C \
Nu = 0.0266 x (2.08 x 105)0805 x (0.701 )03])
~
f
~lEi-1I-=-4-S1-.3--'1
rties of air at 80°C:
We know that,
prope
NusseJt number,
NlI
p
I kglm3
V
21.09 x 10-6 1112/s
0.02675
PI"
0.692
30.18 W/m2K
k
0.03047 W/mK
Re
UD
v
hxO.4
451.3
~
[From HM I data book, Page No. 33 (Sixth Edition))
hD
k
h
I Heat transfer coefficient,
h
30.18 W/m2K
Heat transfer,
Q
h A (Til' - Too)
We knoW that,
1
Reynolds
Number
h x n x D x L x (Til' - T",)
0.2 x 0.070
21.09 x lo-tJ
[.: A = nDll
30.18 x n x 0.4 x 1.5 x (45 - is)
1
Q
I Example 2 I Air
1706.6 W
_W°C, 0.2 I1Ils flows across a now
electric bulb at 130°C. Find heat transfer ami power lost due to
convection if bulb diameter is 70 min.
Given :
Tofind:
(II
Fluid temperature,
TO")
U
0.2 m/s
Heat energy,
0,
120W
Surface temperature,
Til'
Diameter,
D
70 mm
I. Heat transfer,
Power lost due to convection.
Scanned by CamScanner
We know that, for sphere,
Nu
Nusselt Number,
.
6
:: 0.37 (Re)o
N
119 (Sixth Editionl]
IFrom HMT data book, Page o.
0.37 (663.S2)06
[Bu :: Is.iD
30°C
Velocity,
2.
663.S2 ]
IRe
1706.6 W 1
Heat loss, Q =
Result:
= 663.S2
Nusselt Number,
~
!!.Q
k
Nu::
IS.25
::
h x 0.070
:.:...:..:-:-.:
0.03047
~
~~ :: 7.94 W/m2K
0.070 m
Heat transfer coe
~
fficient, h
Convective Heat Transfer
2. J 20
~air
Heal and Mass Transfer
We know
Heat transfer,
[From HMT data book. Page 0.33 (Sixth Edition)1
p
1 kg/m'
v
21.09 x 10-6 m21s
Pr
0.692
k
0.03047 W/mK
h A (T, - Ten)
hx47tr2[T
u
.-T]
["A-
'"
0.070)
7.94x 4 x 7t x ( -2-
•
-
41[ r2
2
J
x (130-30)
~---------------------,
I Heat transfer, Q 12.22 W I
2
2.
at 80°C:
prope
Q2
Q2
0
% of heat lost
x
1
2. 121
100
12.22
120 x 100
a . Tube is considered as square of side 6 em.
'IOl:t (I, .
CI"'"
L = 6 em = 0.06 m
t.e.,
UL
Reynolds Number Re
v
30 x 0.06
21.09 x I(}-6
10.18%
[ Re
Result:
I.
Heat transfer
=
12.22 W
2.
Percentage of heat lost
=
10.18%
I Example 3 I Air at 40 't" flows over a tube with a velocity of
Nusselt
Number
30 m/s. The tube surface temperature is J 20 't", Calculate tile
heat transfer coefficient for the following cases.
I.
Velocity, U
Tube surface temperature, Tw
30 m/s
1200e
Tofind: Heat transfer coefficient,
Solution: We know that,
Film temperature,
T,
0.675
e
Tube could be square with a side of 6 em.
2. Tube is circular cylinder of diameter 6 em.
Given: Fluid temperature, T",
400e
~
Nu
~
I Nu
0.092
[From HMT dala book. Page No. 118 (Sixth Edilion))
°.333
0.092 (0.853 x I OS)O.67S x (0.692)
173.3]
hL
k
Nu
We know that
(h).
T",+T",
2
120 + 40
2
173.3
Heat transfer
==
h:..:..----x 0.06
0.03047
h == 88 W/rn2K
.
coeffiCIent,
Case (ii) :
Tube diameter,
6 cm == 0.06 In
D
@
ReynoldS Number,
Scanned by CamScanner
I
Nu == C x (Re)" (Pr)o 3H
n
For square
0.853 x lOs
Re
v
~
•. 1.1
Heat alld Mass Transfer
21.09
of
[From HMT data book. PUg'
~t: N o. I rs- (S·
10'
R e value is 085"'
O.O~66 and 0 80-'
.) X.
"
corresponding
. :> respectively,
I
C and
ill)lIlC'
_j_
,
r.l-i-~:_:_'l_t U::_~~2i
-to _l_~.
rr
,
-'f T
:~
- -$-' - - .; - ~'Ldrt
"
u~
IXlh Edit
A"
1011)]
III valu es are
Nu
0.0266 x (0.853 x )05)O.R05x
I"N:-:-u-=--2-19-.3-1
(0.692)0333
i
as shown in Flg.2A,
MOl.
c (ReD)'" (Plf333
__SI\
A ~
\
.:41T-~-¢"
\ .
-
~
II 0
k
Nu
2)9.3
=
hxO.06
0.03047
Result:
I. Heat transfer coefficient for square tube
II =
88 W/m2K
2. Heat transfer coeff icienr
. for circular tube
II =
I 11.3 W 1m2 K
I~~~~
(b) 51aggered
(a) In-line
Fig. 2.4. rltbe Banks
The confIguration
of banks of tubes is characterised by the
tube diameter D, transverse pitch, S" and longitudinal pitch S,
measured between tube centres. The diagonal pitch SD' between
the centres of the tubes in the diagonal row is also sometimes used
for the staggered arrangements. The Reynolds Number is based on
the largest velocity of the fluid tlowing through the bank of tubes.
D
U",ox
~
ReD
v
S,
U
X
U",ax
s=o
,
U
where
-
S, 2.10. FLOW OVER BANK OF TUBES
Heat transfer in tl
nume rous industrial
ow
'. over a b an k or bundl
or air conditio»:
appIrcatlons such as t
e of tubes has
I ioning cooling coil. In thi
seam generation in boiler
s case , one tl UI id moves over
I 5,
\5D:$-'\--~-
- -{\f-t--(\)-i-$ j_
\~\-~~
I
We know,
2.12 J
fluid at . different t emperatures passes
. "s~cond
•
'he 'lube rows ot a bank mavJ be etitl ier staggered
. tl
,
lbcs
Ihe It; I1 the tubes
10-('
0.853 x :O~J
Nussclt Number. Nu
Convective H eat r ramie,.
.
30 x 0.06 __________
.,
0
2.10.1. Formulae
Velocity of fluid, mIs,
Transverse pitch, Ill,
- Diameter, 1\1.
used for Flo'll Over Bank of TU~S
,
UX-S:O
I.
Max.imum velocity, UIIIlIX
where
S, -
'
Transverse pitch, til.
I
.til"'"
Scanned by CamScanner
Convective Heat Transfer
12~./~24~~H~e~a~l~and~M~~=s~~~ra=n~sfi~e_r ~
.:.
_
Umax x 0
2.
Reynolds Number,
Re
3.
Nusselt Number,
Nu
V
=
----
18.97 x I~ m2/s
Pr
0.696
0.02896 WImK
k
1.13 x (Pr)O.33 [C ReI!]
[From HMT data book, Page No.122 (Sixth Ed'r
v =
nOW
We k
:=
that,
s,
lion))
Ulllax = U x S,-D
. urn velocity,
MaxiOl
2.10.2. Solved Problem
0.020
I Example 1 lIn a sur/lice condenser, water flows through
staggered tubes while the air is passed ill cross flow over the
tubes. TIle temperature and velocity 0/ air are 30°C and 8 nrls
respectively. TIle longitudinal and transverse pitches are 22 mm
and 20 mm respectively. TIle tube outside diameter is 18 mm and
tube sur/ace temperature is 90 'C. Calculate the heat transfer
2. J 25
Umax
= 8 x 0.020 - 0.018
[Uma:c
= 80 m/s]
Umo.T X D
Reynolds
=
v
D
==
O.O~
0.018
~
= \.I~
Re
Number,
coefficient.
Given:
Fluid temperature,
T <0
Velocity, U
Longitudinal pitch, S,
Transverse
300e
pitch, S,
20 mm
= 0.020 m
Diameter, D
18 mm
=
Tube surface temperature,
Til'
Tofind:
I. Heat transfer coefficient.
Solution:
We know that,
Film temperature,
Tf
[l?e
8 mls
22 mm = 0.022 m
~.
90 e
o
Til' + T rn
2
[From HMT data book, Page No. 33 (Sixth Edition))
1.060 kg/m3
Scanned by CamScanner
==
1.22
Q:.Qll
S,
Properties of air at 600e :
=
l.ll
0.Q18 m
0
==
Q.018
[3
90 + 30
2
p
==
S,
S,
- 1 1\ O - . 'D
0.556 respectively.
==
e051Sand
. C n values ar .
\ 22 corresponding ,
s: .th Editionli
.
,
i-
rFrom~
~
~
MT data boO"
Page No. 122 ( L"
Convective Heat Transfer
1 1)6
Heat and Mass Transfer
Nusselt Number,
Nu
1.13 (Pr)o m [C (Re)" 1
=>
Irroll1 IIMT
data book. Page No. 122 (Sixth Ed'
, 1(1011)1
Nu
r.n
[0.518 x (7.5 x 104)0 ~561
Nil
266.3
=
Nussclt Number.
Nu
I
liD
k
~kness
of the boundary layer is limited to the pipe
of the flow being within a confined passage .
el.
.The bet'luse
radilis
layerS from the pipe walls meet at the centre of the pipe
sollodar)' I' re flow acquires the characteristics of a boundary layer.
d th~ en I
aO
boundary layer thickness becomes equal to the radius of
once the there will not be any further change in the velocity
Ihe lube.
This invariant velocity distribution is called fully
. tribul Ion.
.'
dts
ed velocity profile. I.e., Poiseulle flow.
develoP
ulae used for Flow through Cylinders
211,1. F or m
,
(Internal flow)
_66.3
Heat transfer
2./27
em .ient, I,
I. Bulk mean temperature
RdUil:
Heal transfer coefficient.
"
428.6 W/m2K
T m;
where
2.11. FLOW THROUGH A CYLINDER -INTERNAL
T mo -
FLOW
Similar I the flow ver a flat plate, a fluid funiform velocu,
entering a tube is retarded near the walls and the boundary layer
begin to develop as sh \ II in Fig.2.S b doned Iii es.
Fully developed
esiabiished flow
Inlet temperature °C,
-
2. Reynolds
Outlet temperature "C.
UD
Re =
Num b er,
v
.
h 2300 flow is laminar.
number value IS less tan,
2300 flow is
. reater than
,
If Reynolds
number values IS g
If Reynolds
turbulent.
3. Laminar
Flow:
Nusselt
Num
be
r,
Nu
=
3.66
123lSixth Editioo))
[From HMT data boOk. Page No.
I Equation)
Flow (Genera
_
023 (Re)os (PrY'
be Nu - O.
Nusselt Num r,
4 _ Heating process
n == O.
== OJ - Cooling process .'
4. Turbulent
n
[From
ng. 2.5. F/ow ""oug"
Scanned by CamScanner
u cylinder
1;. Page No.12
HMT data boO .
5 (Sixth Edition))
2.128
I .
I
Heal and Mass Transfer
This equation is valid for
~
. 0.6 <
PI' < /601
Re >
10000
L
D
>
~I
'1
60
';.-..:.._~~=-~C~o~n~ve=C~/i~ve~JI,~e~a'!
~Tr~a~n
~"
Tube wall temperature °C,
1/11
-
I
Mean temperature 0C ,
T""
-
Inlet temperature °C,
I
J"1II0
Outlet temperature -c
I
I
For turbulent flow,
Nu
Mass flow rate
8.
::: 0.036 (Re)O 8 (Pr)033
(tD)O.OSS
where
This equation is.valid for
i5 < 400, Re < 10,000
6.
"
e
4A ::: 4(LxW)
P
2 (L -i W)
where
A
Area, 1112,
p -
Perimeter,
2.1 1..2
Solved
Problems
Flow)
Gil/ell:
7.
1
Outer diameter,
D;
J nner diameter.
Heat transfer
20 mm = 0.020 m
Length, L
3m
Velocity, U
0.03 mls
Inner temperature of water, Till;
Outer temperature
of water, T
IIIO
Wall temperature, Tit'
To find: Heat transfer (Q).
Solution : We know that,
Q
h A (1~t'- Till) where A ::: 7r x D x L
(or)
Q
- Flow through Cylinders
Diameter of tube, 0
p
7r [ Do + D;
Velocity, m/s,
[ EXfIIlIple J Water flows inside a tube of 20 /11mdsam
(111(13m long at a velocity of 0.03 m/s. The water gets heated
-10"('to 120°C while passing through the tube. The tube w
maintained at COIISlflll1 temperature of 160°C. Find Iitattrans
7r
Do
Area, 4'1[ 02 ,m,2
(Internal
4A
4 x 4' l D~ - DJ 1
where
-
kg/s
I
111,
L - Length, Ill,
W - Width, m.
Equivalent diameter, for hollow cylinder
D,,(or) D" :::
Density, kglm3,
U -
5.. Equivalent diameter for rectangular section,
D (or)D
pxAxU
p A
L
10 <
III
III CJl
Bulk mean temperature, Till
40 + 120
~
2
(T1110- l' .)
111/
TI/I
80°C
[
Scanned by CamScanner
Convective Heal Transfer
2. J 30
Heat and Mass Transfer
-----
Properties of water at 80°C:
I
lFrom HMT data book. Page No. 21 (Sixth E ..
, .
dillon)]
p
974 kglm3
v
0.364 x 10-6 m2/s
Pr
2.22
~
When 0.6 kg of water per minute is passed
~I
2 cm diameter, it is found to be heatedfrom
Ih,olllh a
The heating is .achieved b~ condensingsteamon
ZOIlC to 6 if the tube and subsequently the surfacetemperature
e sur/ace ~ aintained at 90°C. Determine the length of the
th
be IS m
of the "". dfior lully developedjlow.
requIre
. 0.6
;~c.
Mass,
I"be
m = 0.6 kg/min = 60 kgls
Given:
0.6687 W/mK
k
0.01 kg/s
2 em = 0.02 m
Let us first determine the type of flow.
Diameter, D.
UD
Inlet temperature, Tm;
=
outlet temperature, T mo
=
Reynolds Number,
Re
=
v
0.03 x 0.020
Tube surface temperature, Tit'
d : Length oftije tube, (L).
0.364 x 1"0-6
1648.35
IRe
ToJi n .
solution:
I
Since Re < 2300, flow is laminar
For laminar flow,
. Nusselt Number, Nu
r +Tmo
_!!!!.---
Bulk mean temperature, 'r m
2
=
=. 3.66
Nu
3.66
Ih
=>
Heat transfer, Q
hD
k
h x 0.02
.
PropertIes 0
f water 'at 40°C:
I
h A (Til' - Tnr)
h x 1t x D x L X (Til' - Tm)
[.: A = ltDL]
p
v
Result:
Heat transfer,
1845.29 Watts
Q
1845.29 W
Scanned by CamScanner
I
995 kglro3
Mass floW rate, in
U
21s
. 0.657 x 10-6 ro
Pr ::::: 4.340
k
,0.628 W/roK
:::: 4178J~g K
Cp
122.37 x 1t x 0.02 x 3 (160 - 80)
IQ
boOk, Page No.
(From HMT data
668.7 x 10-3
122.39 W/m2K
20+60
.=.:;.....2
40iJ
[From HMT data book, Page No. 123 (Sixth Edition))
We know that,
2.131
:::: pAU
i!pA
21 (Sixth Edition))
2}~.J~3~2~H~ea~l~m~l~d~M~a~s~s~~~ra~lI~sfi~e_,.
__ ~~
.:
0.0 I
Convective Heat Trans'r,
'Jer 2/J3
114.9 x 7'C x 0.02 x Lx (90-40)
__
_________
I
[-L--4.-62-n--',
1t
995 x 4" (0.02)2
l10th of the tube, L = 4.62 m.
. L eng
eSll It •
~
Wuter (It 50 °C enters 50 111m diam~ter and 4 m
I Velocity. U = 0.031 I11ls I
Let us first determine the type of flow
!
I
Re
=
~"elocitv
of 0.8 m/s. r"e lube wall if maintained
be wlI I
·.r
90.0,,", D
'
...., elernllne tlrt heallransfer
/u~gIII s/(ll,1 temperatllre oJ
UD
Iv
.1
II'
Re =
0.031 x 0.02
0.657 x IO~
CI
cotffi ture is 70°C.
...nerO
/tll'l'
Give" :
I
943.6
'ent 111,1 t u« total (/111011111 of "eallralls/erred if exit water
a CO"
Inner tel
nperatllre
Since Re < 2300, now is laminar.
For laminar flow,
f
water,
T
=
3.66
50 mm
Length, L
4m
Tube wall temperature,
[From HMT data book, Page No. 123 (Sixth Editioll)1
liD
k
Nu =
We know that
3.66
II x 0.02
=
[I =
=>
Heat transfer, Q
114.9 W/m
2K
I
11/
Cp dT
11/
Cp (T/IIO - T/II)
1671.2 W
900e
IV
70 e
Exit temperature of water, T",o
t. Heat transfer coefficient, (II).
Tofi"t/ :
2. Heat transfer, (Q).
0
Tm; + T",o
Bulk mean temperature,
2
T",
2
I
60°C
Properties of water at 60°C:
v
h x 1t x D x L x (T". _. 1'/11)
.
(From IIMl data
p
Q
Pr
k
Scanned by CamScanner
0.8 m/s
= 0.05 m
50 + 70
::..----
0.01 x 4178 x (60 - 20)
[Q
U
T
So/ul;oll :
0.628
soae
ml
Diameter, D
Velocity,
Nusselt Number, Nu
We know that,
0
J
N 'I (Sixth Editionli
boOk I'a~~ I 0..-
985 k~tnJ
0.478)( W61112/s
3.020
0.6513 WhnK
Convective Heat Transfer
2. J 34
----
Heat and Mass Transfer
Let us first detennine the type of flow:
UD
_.
Re::
v
0.8
x 0.05
0.478 x ,10-:<>
eRe
o
5, at trmrsfer coefficient.
Diameter of tube, 0
GIve" .
I'
80
Length,
8.36 x 104> 10,000
Tube \ all temperature,
Pr
3.020 => 0.6 < Pr < 160
solution:
Pr perties of water at 40°C.
p
995 kglro3 ,
2
'
= 0.023 x (8.36 x 104)0.8 x (3.020)04
I Nu = 310 I
'
'
Nu
that,
' Nu
310
=
h
..
.
We know,
t
__
Pr
4.340
k
0.628 W/mK
1\'.'
un
Re
h.
2
t
4039.3 W 1m K
o
I.
10 <
h x ~ x 0 x L x (T w - T m )
h
4093.. 3 x 7t x 0.,05 x 4 x (90 - 60)
D ratio
Scanned by CamScanner
.'
v
0.65 x O.O_Q!
0.657)( 10-6
Since Re > 2300 , flow
is turbulent.
.
A (T 1\1 - T m )
76139,W
0.657 x 10-6 m /s ,
ro '
0.6513
,
::
~
= h x 0.05
H eat tr ans fer coefficient h
Heat transfer Q
h' ' .
T;
Heat transfer c:oeffieient, (h).
[Inlet temperature 50·G Exit t
.
Ed "''')1
Process, So, n = 0.4]
'. XI emperature 70·C => Heating
-,
9.65 m/s
To find:
V
.[From HMT data book, Page No 12S'CSixth
~now
140°C
Re
Nusselt Number, Nu :: 0.023 (Re)08 (pr)"
=>
40°C
80> 60,
Pr value is in between 0.6 and 160.'So·,
\
3m
o
oL ratio is greater than 60. Re value is greater than 10,000 and
0.8 em :: 0.008 m
L
Average temperature, T m
Velocity, U
L
2
4039.3 W/m K
.
transfer Q :: 76139 W.
Meat te 4 Water
'
z.,_~
floWS through 0.8 em diameter 3
, m
, t all average temperature of 40 "C. Tireflow velocityis
""gIllbt
",! aOIld [ube wal I temperature IS. UO'f:. C.Ic.I.tt tht
~~
a' !trag•t ,e •
.
n=
I.
~65
8.36 x 104]
Since Re > 2300, flow is turbulent.
L
,4'
0.05::
~esll/I: transfer coefficient
I
2. J 35
.'
_1-::
375
0.008
h .:::'
400
o
d 400 Re <::. IOOOq, SO"
IS in between
10 all
'
.
Convective Heat Transfer
2.136
2./37
Heal and Mass Transf_er.
Nusselt Number,
Nu
D)O.O~
== 0.036 (Re )0.11 (Pr)O 33 (
L
'.
[From HMT data hook. Page No. 125 (S'IXl. h Edir
~
0.036 (7914.76)0.1( (4.340)0.33
Nu
~
=1
x(
o.~~)
lonll
rties of water at 40°C:
prope
(From HMT data book. Page No. 21 (Sixth Edilion)1
O.O~5
v
N=u~===5'=- .4=4=J
We know that,
Nusselt Number,
Nu
55.44
Heat transfer coefficient,
Pr
4.340
k
0.008
628 x 10-3
Cp
0.628 W/mK
4178 j/kg K
rst determine
2
4352.3 W/m K
Let uS fi
the type of flow.
Reyno\ds Number,
Re
Jr = 4352.3 W/m2K
straight tube of 60 mm diameter. TIre IIIbe surface is mailllained
a1 70
and outlet temperatllre of water is 50°C. Find the "eat
traIISfe.r cfHfficient from the tube surface to the water, "eat
ac
transferred and Ihe tube length.
Tofind:
.
of water,
T""
30°C
\ Velocity,
U
20 m/s
Diameter,
D
Tube surface temperature,
Tit,
60 mm = 0.060 m
700C
t:
T",o
50°C
Outlet temperature
of water,
I. Heat transfer coefficient, (h).
2. .Heat transferred,
3.
Tube length, (L).
Solution:
Bulk mean temperature
Scanned by CamScanner
T III
UD
=
(Q).
v
20 x O.Q60
0.657 x \Q-6
I Example 5 I Wattr at 30°C, 20 m/s flows throug" a
Inlet temperature
0.657 x I~ m2/s
=
hD
k
"x
=
h ==
Result:
Heat transfer coefficient,
Giv6r:
995 kg/m3
p
eRe
\.8 x \O~
·
Re > 2300 flow is turbulent.
S tllce'
\ uation is (Re > \0 000) .
For turbulent flow, genera eq
as (P )"
Nu == 0.023 (Re) .
r
...
Pa e No. 125{Sixth Edl~n)1
(From HMT data boO", g
L.
.
So n::: 0.4·
4
This is heatUlg process.,
8 x \06)0.8 (4.340)0
Nu == 0.023 (\..
~
We know that,
hD
Nu
k
~
4\77.7
0.628
4\77.7)(~
~60
J
ConveClive Heal Transfer
2 J 36
I
:
2./37
Heal and Mass Transfor
Nussclt Number, Nu
0.036 (Re)O.R ~(pr)OJ3 (Q,L)0.055
[From HMT datil hook ..Page No. 125 (Si>;th ' '.
. s of water at 40°C:
propertle
EdlltOl\l1
0.036 (79l4.76)0.t( (4.340)0.33 x (~)
=>
Nu
=>
~I N-u--S-S .-44-:-11
[From HMT data book. Page No. 21 (Sixth Edition)l
OOS~
= 995 kglm3
p
We know that,
Nusselt Number, Nu
hD
k
h = 4352.3
0.657 x lQ-6 m2/s
Pr
4.340
k
0.628 W/mK
4l781/kgK
Cp
h x 0.008
55.44 = 628 x lO-3
I Heat transfer coefficient,
v
f
Let us ir st
1
W/m2K
determine the type of flow.
Reynolds Number,
Result:
UO
Re
v
20 x O.~60
0.657 x \0-0
Heat transfer coefficient, h = 4352.3 W/m2K
I Example 5 I Water at se-c. 20 nrls flows
throug"
II
straight tube of 60 mill diameter. The tube surface is nUlintained
at 700 and outlet temperature of water is 50°C. Find the "elll
transfer coefficient from the: tube surface to the water, heo:
transferred and the tube length.
e
Given : Inlet temperature of water, T nil
V~locity, U
Diameter, D
Tube surface temperature, Til'
Outlet temperature of water, T IIIU
To find :
\.8 x 106 \
{ Re
.
R > 2300 flow is turbulent.
Smce e,
. . R > \0000)
for turbulent flow, general equation is ( e.
.
Nu = 0.023 (Re)0.8 {PrY'
30°C
20 m/s
60 mm = 0.060 m
700C
J--,
.
lFrom HMf data
[Bu
We know that,
4\771]
hO
Bulk mean temperature , T
Scanned by CamScanner
",
T
+T
-l!!!__.2!!!!
2
0.060
~62S
O.
4\77.7x~
11)(
4\77:7
Solution :
.
T
Nu
2. .Heat transferred, (Q).
3. Tube length, (L).
..
.
So n = OA.
This is heatlng process.
,
x \06)0.8 (4.340)
Nu
0.023 (\.~
50°C
I. Heat transfer coefficient, (11).
book. Page No. \25 (Six.thEd
=
'I :::: ~60
3
Convective Heat Transfer
d' Hea
2.138 Heal and Mass Transfer
~~~~~~~~=~4:37:2~6.:59~W:=/m~2:K--------
Mass flow rate,
,i,
T mi + T mo
5011l1iO" ,
Mean
temperature, T m
xD2xU
995 x '4 x (0.060)2 x 20
56.2 kg/s 1
Q
m Cp (1"'0 - T",I)
Q
56.2 x 4,178 (50 - 30)
IQ
4.69 x 106
30°C j
propertle
1.165 kglm3
16 x 10-6 m2/s
p
v
Pr
4.69 x 106 W I.
II A (Til'':'' Till) .
k
43726.59 x 1t x D x L x'(70 - 40)
0.701
0.02675 W/mK
=
.
E uivalent diameter
Hydrauhc or q
[.,' Surface area, A = 7t DL)
4.69 x 106
1t
2
4 x:1 [Do 1t [Do +
43726.59 x 7t x 0.060 x Lx (70 - 40)
I L =. 18.96 m I
Result :
1. Heat transfer coefficient,
..
[From HMT data book, Page No. 33 (Sixth EdItion)]
. 5 of air at 30°C:
1m
Q
2
LTm =
1t
We know that,
.
'
px A x U
px%
Heat transfer,
t transfer coefficient, (h).
fOP" . ,
2
Heat tranSl1r"ercoefficient , II = 43726.59 W Im j(-]
-
==
(Do + OJ) (Do -1j2
(Do + OJ>
o
'
Heat transfer, Q .= 4.69 x 106 W.
=
3.
Length, L = 1,8.96 m.
0o-0· '
::: 0.06 - 0.04
cylinder of 4 em inner diameter and 6 em outer diameter and
leaves at 45 CC. Tube wall is maintained lit 60 cC Calculate the
heat transfer coefficient between the air and tile inner tube.
Inlet temperature of air, T mi'
Velocity, U
Imler diameter, DI
Out~r diameter,
Do
15°C
'Tube wall temperature,
600C
Scanned by CamScanner
v
Reynolds Number, Re
35)( 0.02
:::~
6 em = 0.06 m
450C
-
U De
35 m/s
E~'it temperature of air, T mo
Til'
o~
4 ern = 0.04 m
~
Since
j
[02 o _02]'
--0 + O·
,h = 43726.5 W/m2K.
I Example 6 I Air at J 5 CC, 35 m/s, flows tilroul;II a 1r~lIow
02]
Od
2.
Given:
2. J 39
.s turbulent.
se > 2300, now 1
1
2 /40
Heal and Mass Transfer
____
[ De
0.023 (Re)0.8 (Pr)"
IIlon)1
I Nu
0.436 m
we know that,
::;::J
[From HMT data book. Page No. 125 (Sixth Ed' .
This is heating process. So, 11 == 0.4.
=>
Nu
0.023 x (43750)08
Reynolds
Re
Number,
x (0.701)04
6 x 0.436
16 x JO-6
Wi9]
16.3 x
h De
We know,
=>
Result:
Nu
k
102.9
h x 0.02
26.75 x 10-3
Ih
137.7 W/m2K
Heat transfer coefficient,
[From HMT data book, Page No. 125 (Sixth Edition))
h = 137.7 W/m2K
section of size 300 x 800 mm. Calculate tile heat leakage per
metre length per unit temperature difference.
Air temperature,
Velocity,
Till
30°C
U
6 m/s
Area, A
A
Nusselt
length
per
16x 10-6r02/s
Pr
0.701
k
0.02675 W/mK
Number,
Nu
294.96
unit
I.165 kg/m!
=
294.96 ]
0.24 m2
Solution : Properties of air at 30°C:
p
==
We know,
300 x 800 mm?
I. Heat leakage per metre
temperature difference.
v
. g the pipe wall temperature to be higher than air
Assumtn
temperature. So, heating process => n = 0.4.
04
Nu
0.023 (16.3 x 104)0.8 (0.701) .
[Nu
OJ x 0.8 m2
Tofind:
wi
e > 2300 flow is turbulent.
Since R
'
nt flow general equation is (Re > 10000),
le
Fortur b u
Nu == 0.023 (Re)08 (Pr)"
I Example 7 I Air at 30°C, 6 m/s flows ill a rectangular
Given:
2.141
P Perimeter = 2 (L + W)
--------.1
.
where
is (Re > 10000).
For turbulent flow, general equation
Nu =
~
Convective Heal Transfer
Equivalent diameter for 300 x 800 mm? cross sec tiIon IS
. given
.
by
D = 4 A = 4 x (OJ x 0.8)
e
P
2 (0.3 + 0.8)
==
h - 1809 W/m2K
.
.
,
't temperature dIfference.
Heat leakage per unit length per unl.
.
Heat transfer
.
.
coefficIent,
Q
hP
18.09 x [ 2 x (OJ + 0.8) ]
lQ
39.79
WJ
Result : Heat leakage,
Q == 39.79 W.
~.~
....". ""7
=Z;~"'\
Scanned by CamScanner
c;
2.142
Heal and Mass Transfer
Convective Heal""Iransjer
.r.
I
Example i1ln
condenser,
water flows ~
hundred thin walled circular lubes having inner dia",
11110 /
eter 20
and lengtlt 6 m. The mass flow rate of water is 160 k
trr",
water enters at 30°C and leaves at 50°C. Calculale Iii g/s. rhe
Ie aVer
heat transfer coefficient.
Rge
Inner diameter, 0
Given:
20 mm = 0.020 m
It"
Length, L
Inlet water temperature,
T mi
Outlet water temperature,
Tmo
To find:
=
995 x 1txl)2
4
1t
[u
r-I
(h).
2.55 mls
U 0 = 2.55 x 0.020
v
0.657 x I~
R-e--7-7-62-5.-57--.1
Since Re > 2300, flow is turbulent.
Bulk mean temperature,
Tm
For turbulent flow, general equation is (Re > 10000).
Nu = 0.023 x (Re)0.8 (PrY'
TIII;+Tmo
2
30 + 50
2
{From HMT data book, Page No. 125 (Sixth Edition)1
This is heating process. So, n = 0.4
=>
Properties of water at 40°C:
0.023 x (77625.57)0.&x (4.340)0.4
1 Nu
337.8 I
v
= 0.657 x 10-6 m2/s
Pr
4.340
k
0.628 W/mK
Cp
4178 J/kg K
UD
v
Reynolds Number, Re
m
Velocity, U
pAU
m
pA
Scanned by CamScanner
=>
... (1)
hD
Ie
Nu
We know that,
995 kglm3
p
[.,' T mO > T."I]
Nu
[From HMT data book, Page No. 21 (Sixth Edition))
~
I
Re _
(1) ::::)
= 50°C
Solution:
We know that,
No. of tubes = 200]
995 x 4 x (0.020)2
160 kg/s
30°C
Heat transfer coefficient,
[ '.
160
200
6m
m
Mass flow rate,
160
200
337.8
=
h x 0.020
0.628
Heat transfer coefficient, h := 10606.9 W/m2K
Result:
Heat transfer coefficient, n= 10606.9 W/m2K
b r ressure,flowthrough 12
Example 9 Air at 333K, 1.5 a P
if the tube is
," e temperature 0
em diameter tube. The surJac
te is 75 kg/hr.Calculalethe
maintained at 400K and mass flow ra
th 0/ the tuucheat transfer rate for 1.5 m I eng
600C
::: 333 K :::
Given: Air temperature, Tnr
I
I
&.~
Surfa
Dramcrcr.
IJ
C tcmpcrnurre.
T"
( !!!!..v('('/;Vt! Ilc~.
I:! nil
_
.1001\
- '"'7 C
7-J..g.
rr
.
_
__
____
Tnfind:
Nil'.
11115 I.
~~
75b
I Nil
=
Nil
-
WC~O".
1k11 Iransfcr r.uc (0).
J.
0.021 '/ (Re)o.• x (PrY'
. .
3(,00 ~
J . .) III
32.9
Since lit,· prc.,slIrc i~ 1101 Illllch I .
h '.
a )()ve aim
p y~1 111pr pcrncs ( fair 111;1\ he taken ill atmo 'J .'
OSPheric.
).
."
. plenec
IIdi,'
t mpcJ1tCs (If air al (,(1' r ' .
II
IOn.
SII/lllion:
J .U60 kglmJ
\I
11I.97xJ()-Ilm2Is
Pr
0.696
/(
().O:!896 W/rnK
J{e~'lIoJds Nllllllwr.
J I)
"If
0.o1R9()
hAn
It'll!:'"ofthe
1.060
III
Tile' tub« surfuc« ;.5 mainlu;n~tIal80 'r. If
(~rwater increases from JOOC
10 JIJ(Cp",J
lube.
]0 em "' 0:'0 rn
bO
1f
4 ' W.I-f
60 kg/min - 6U k ,I'
I)
I kg!
]
l'ip« xur la
I<l'
l:
(Jill let tvmpvrauu
Me:
f·
Ill.
f]
(Si.\lh
Til'
80"(,
LIe water. '1 ".,
10"
temperature.
c vI' water. Tm"
1'0 Jill":
I. Lt:II.!!lh oftlu: lube. L.
SOlllliolt:
W\.· 1..1I0\\' that.
T
~
10000)
Scanned by CamScanner
-1')III
7~?~~!1l x 0.12>- l.5)y(127-(,oJ
1.060
-
II'
I ,,'tller flow» 1I1",ugll 3(1 em dianln" IlIb~al
mil' of MJ"l:llIIi".
,h,. tempertuurc
IlIlel kill perature
SillLl'
~~
I leal Iran fer rare. () == 301).82 W
[ EmiliI'll'
'" (I
J .MS Ill/s
(1)
liD
k
[__!l_ ~_1(~·X2 ~
Rr.m/I:
'"
0.0 0
(O.()%)() I
h'
v
0.0 (I
0
___
We lno\\' Ihal,
M:I"'~ 11 \\ r;llc.
).
(J0551.3\()~
J(
II /. (11 ~ D x L) / (T - T
!.!_Q
I<c
--
002
3_2.1}
7.94 W!m~K-
Heal Iran. fer rate.
rom J J 1 J ,filla Oo.ll Pa., •
. ~c N(l, J J
.
p
_
np proce. s. S(I, n= 04
. c; heilll
n.ozn _J..W.:.J
I
J.clI;!lh.
.
~
Bulk mean
Ecillinnil
h:lOperalUrc.
T",
T
2
,,"
II~
_--------~~C=o=n~v~ec~t;~ve~H~e~a~tT~~.(.~r
Iran.',er 2.147
I
2.146
~
Heat and Mass Transfer
~
r,
------
2
20°C
r Nu -\ Nu =
'I
we knoW that, .
Nu
~
39.50
Heat transfer,
1.006 x 10-6 m2/s
Pr
7.020
k
0.597 W/mK
4178 J/kgK
Cp
=
h x 0.30
0.597
78.60 W/m2K
Q
m Cp~T
m Cp (Tmo - Tm;)
\Q
... (1)
v
83.56 x )03 w \
Q
We know that,
h x 1[ DL (T w - T m)
We know that,
Mass flow rate,
=>
83.56 x 1O~
m
pAU
m
pX'4D2xU
Result:
1[
1000 x '4 (OJO}2 x U
ju
0.014 m/s
UD
(1) => Re
I
v
0.014 x OJO
1.006 x.] 0-6.
.
j Re
4174
Since Re > 2 , 300 , fl ow IS
. turbulent.
(lnternaljlow)
Nusselt number, Nu
=
I > 2300
Length of the tube required, L = 18.79 m
I Example 11 \ Air at 2 bar pressure and 60'(' is heatedas it
flows through a tube of diameter 25 mm at a velocityof 15mls.lf
the wall temperature is maintained at 100'(', find tl,e heat
transfer per unit length of tl,e tube. How much wouldbe the bulk
temperature increase over one metre length of the tube.
Given: Pressure, p = 2 bar
2 x lOs N/m2
0
Inlet temperature
of air, T m;
Diameter
of tube, D
Velocity,
:
0.023 (Re)0.8 (Pr)"
Tube wall temperature,
=>
Nu = 0.023 (4174)°·8 (7.020)"
ng process. So, n = 0.4
Scanned by CamScanner
60 e
;: 25 mm ;: 0.025 m
U
T". ;:
Length, L ;:
[From HMT d at a b ook, Page No. 125 (Sixth Editionll
This is heati
78.60 x 1t x 0.30 x L x (80 - 20)
~ I L = 18.79 m \
1[
=> 1 kg/s
For turbulentjlow
I
\h
1 x 4178 (30 - 10)
UD
Reynolds number, Re
hD
k
1
[From HMT data book, Page No . 21 (Sixth
. E ..
p = 1000 kg/m!
dillon)!
=
39.50J
I
Propert;e.~of water at 20'(' :
v
002
. 3(4174)0.8(7.020),1,4
To find:
15 m/s
1000e
1m
. I th of the tube, Q.
1. Heat transfer per unit eng . T - T ).
2. Rise in bulk temperature of air, (,.0
1ft'
I
; \
Convective /I al Transfer
2.148
Nu
Heat and Ma'IS Transfer
SollltilJII:
(From IIMT data book, Page No 33 (Sixth
. Ed' .
¥
Note:
ki
Given
pressure
....
IIlCI1HIIIC VISCOSity,
v an
•
P
==
1.060 kglmj
V
_;
18.97 x '0-6 m2/ S
is above
d densi
Itlon)1
. . lleatinc process. So, n = 0.4,
IS
Nu ~ == 0.023 (39.53 x I03)OM , (0.696)04
,hiS
~
0.696
0.02896
W/IIlK
atmospheric
il
pre'
.
.
94.70
. r, I.
.
v
V,111I
x--
Ma - now rate,
P!_:i\'Cll
I09.70=~i~2K]
Q!
/1[11111
viscosity.
" x 0,025
0,02896
.SlIrc. So
ensuy. p WI 1 vary with pres ure I' '
p
Killcmalic
"0k
Nil
C are same for all pressures.
,!
0.023 (Re)08(Pr)"
=
IFrom IIMT data bUllk. PJgc ! 0, 125 (Sixth EdilionH
Properties of air at 60°C:
Pr
k
2./49
pAU
III
\
\
18.97
l',' Atmuspheric
18.97'
Density,
p
. 10-(, x 1 bar
2ba~
pressure
z-
I bart
1 10·
10-6 x --.
_ x 10
r
We kllnW that.
u.
Ileat iran fer. Q
RT
r)
I;' C" ("")
o
,,11
, (T - 60)
0' 5 '/. 10)0
1".1
,
.'
. C :; 1005 J/kgKJ
[,.' ~or all fI
." (I)
Q
We know that,
We k.now that.
Heat transfer,
UD
Reynolds number, Rc
Q
v
hA (Til' - T,,')
.
T)
" ;<nDL" t I" - "'
0'
)( I
10,).70;< n 0
J(
8.615{IOO-
(100 - T",)
,.' (2
,~
T.,)
:! 00
.,quatill!? I andt2),
Since
Re > 2300,
For turbulent
now is turbulent.
internal
now, general equation
Scanned by CamScanner
i~ Rc
10.000).
15.075 (T".o -
6U) "" 8.615(IOO_T
.. )
~2.IJ5~O~~R~ea~/~an~d~M~~~s~~~a~m~~:r~~~~~
~
1.749 (T mo - 60)
100
Tm
~
T
1.749 (T mo - 60) = 100 - (
~
1. 749 T mo - 104.94 = 100- (
~
1.749 Tmo - 104.94 = 100-30-
n., +T
2 mo )
60 + Tmo)
2.249 Tmo
174.94
~
Tmo
77.78°C
Rise in bulk temperature of air, ~ T
Tmo
2
\m
= 0.056 kgls
erature of air, T mi = 100°C
Mean
-
77. 78°C
I
temperature,
=
ity , v
=
Kinematic
.
VISCOSI
:::
0.015 x 1005 (I7.78°C)
[v
I Q = 268.03 W I
!
:::
1t
I Example I I 205 kg/hr of air are cooled from 100'(' to
~
in to a helix of 0.6 m diameter. Calculate the value of air side
heat transfer coefficient if the properties of air at 65°C are
Re:::
-
~
(0035)2 x U
4
UD
v
557 x 0.035
~
= 1.044 kg/m!
to« 97, Madras University/
Scanned by CamScanner
(1) ~
p = 0.003 kg/hr-m
P
3600
1.044 kglm3
7.98 x 10-7 m2/~
0.056 ::: 1.
30't' byflowing through a 3.5 em inner diameter pipe coil bent
= 0.0298 W/mK;
= 0.7;
l!
P
044x-x.
2.11.3. Solved University Problems - Internal Flow
k
... (I)
v
::: pAU
Mass flow rate, m
1t
{)2 x U
0.056 ::: 1.044 x 4 x
268.03 W
Ii
Pr
UD
Q.:QQl kgls - m
p
.
2
Solution:
Reynolds Number, Re
I ~ T = 17.78°C I
Heat transfer, Q = m C (Tmo - Tm;)
Q
I
T m; + T rno = 650C
Tm =
Tmo ~ Tm i
77.78 - 60
Result : , 1.
205 kglhr
205
3600 k~s
=
t nsfer coefficient, (h).
roJind: Heat ra
~
0
m =
10
1.749Tmo +2 = 100 - 30 + 104.94
I Outlet temperature of air, Tm
floW rate,
let telllP
. T ature of air, ma- 30°C
telllper
Outlet
Diameter, D = 3.5 em = 0.035 m
2
Tmo
=>
~
~e" : MaS;)
Com ective Heal Tran.ifer
~~~~~~~~----------2.152
2./5J
Heal and Mass Transfer
Since Re > 2300. tlow is turbulent.
For turbulent now. general equation is (Re > 10000).
Nu = 0.023 x (Re)08 x (Pr)«
(From HMT data book. Page No. 125 (Sixth Edltlon)1
This is cooling process. So n = 0.3.
[.: T,I/(J<: T
.
Nli
0.023 x (2.44 x 106)0.8 x (0.7)0.3
[Nli
:!661.71
We know that.
hO
k
Nu
2661.7
=
I Heat transfer coctlicient,
Result:
~
1
I,
" x 0.035
OW
v
15.53 x 10-6 m2/s
Pr
0.702
k
0.02634 W/mK
that.
We"n .
Equivalent
HydrallhC or .
diiameter
0.0298
h = 2266.2 W/m2K
Heat transfer coefficient,
4A
P
D"
the air is healed by maintaining the tempera/lire of the outer
surface 0/ inner til he at 5f) 'C. The air enters (II 16 'C utul leaves
at 32 'C. Ill' flow rate is 30 III/so Estimate the heat transfer
coefficient between air and the inner tube.
Inner diameter,
D;
[Apr. 20(J0, Madras University/
3.125 em = 0.03125 m
Outer diameter,
Do
5 cm = 0.05 rn
Tube wall temperature,
T",
50 e
Inner temperature of air, T mi
IGoe
Outer temperature of air, T rna
sz-c
30 rn/s
Toflnd : Heat transfer coefficient. (lJ).
Solution :
Tm;+ T",o
~~..••
!..
"L__
Scanned by CamScanner
Do-D,
0.05 - 0.03125
0.01875 m
Reynold
Number,
Re
1
v
30 x 0.01875
15.53 x 10-6
0
Flow rate, U
Mean temperature, TII/
D,I
[Do + 0,1 (Do - 0,1
[Do + OJ
h = 2266.2 W/m2K
I Example 2 I III a long annulus (3.125 em ID and S em OD)
Give" :
1t rOo +
..,
36.2)( 10D
[ Re
...00 fl
',s turbulent.
Re : 2.> , ow
. . e > 10000).
Iequation IS {R
For turbulent flow, g,t:nera
Re)os (PrY'
Nu == 0.023 {
') I til .·,bll,'
.
Since
p-.rOJH IIM['dJlabll~'"
Thi
he.uinu
PI' cess. S0 /I
.
Nu
:::0
0 4.
.
l'u~C: . u. 1- (
. T
l':
"'" r I
Ill.
.., 1
_ 0 0'23 {6.-
.
loJ)08 {.
c1"
,q
--.
2.154
Heat and Mass Transfer
We know that,
Convective Heal Ttan.t[er
I Nu -
88.591
Nu -
hDh
Ie
~300
~tlce
SOcmls = O.SOm/s
200°C
Length, L
2m
Tofind: Average heat transfer coefficient, (h).
Solution: Properties of engine oil at 147°C.
[From HMT data book, Page No. 24 (Sixth Edition)1
p
v
=
Pr
k
816 kg/m3
8 x 10-6 m2/s
......
116
Q.133SW/mK
We know that,
Reynolds Number, Re =
I Re
Scanned by CamScanner
UD
v
0.8 x 0.05
8 x 10-6
5000 I
IFrom HMT datil book, Page No. 12S(Sixth F.ditionll
Glnele,
'ube at an av~rag~ 'emp~ra'ur~ of U7°C. The flow velocityb
80 cm/s. Calcula'~ the average heat transfer coefficient If Iht
'ube wall Is maintained at a temperature of 200°C and it Is 2 1ft
Ion,.
IOct. 2002,MU/
Given:
Diameter. D
SOmm = 0.050 m
Average temperature. T m
147°C
Velocity, U
Tube wall temperature, Til'
0L < 400
b lent flow, (Re < 10000)
for tur u
(D)0.055
Nu
==
0.036
(Re)O.8
(Pr)OJ3
Nusselt Num b er,
L
h = 124.4 W/m2K
\ §Xamp/~ 3 , Engln~ oil flows 'hrough a SO mm ttl
o
\0 <
I h = 124.4 W/m2K 1
R~sul': Heat transfer coefficient,
flow is turbulent.
L
2
= 0.050 = 40
Re:>. •
88.59 ... h x 0.01875
26.34 x 10-3
2. J 5.5
.
(0.050)0.055
'Nu :::: 0.036 (5000)08 x (116)0.33 x -2-
Gu ::: 128.42J
Nu
We knoW that,
=
hD
k
h x 0.050
\28.42 =
~
-o.m'8
QJ == 343.65 W/m2g
~
Result:
.
fti'
h == 34365 W/m2K
fifom alf 1"let
heati", waler
fi
or
lure 0/ 40 I(' I""ol",s
[Example 4] A0C system
to all outlel tempera
I , Thepip'
lemperature of 20
5 ", diamelersteelp P .
m
passing tl.e waler '1"~U!:i:,!'i";d al 110°C by co"d:;:~"~~~~h'
surface temperature IS
floW ral' 0/0.5 kl
'
F
waltf ",ass
on its surface. or a
Oct 2002}
lenolh ot tl.e tube desired.
N " 97 MadrasU"I"., .
'J
U I" 0.
IBlwratllitlasall
n ., T :; 200e
erature.
""
Given: inlet tem P
:; 400e
Outlet temperature. T "'~ :; 2.5 em := 0.025 m
Diameter,
:; IIOoe
erature, Til'
.
Piper surface temP
m :; 0.5 kv)l1l1n3 ~/S
oW rate,
:; 8 .33 )( 10- klY
M ass fl
Heat tran
sfer coe tcten
t
I
•
L..;_~__''__-
&
I
2.156
Heal and Mass Transfer
Tofind:
~-We knOW
Length of the tube (L).
Convective Heal T.ransfer
.
hD .
Nu
that,
k
Solution :
Bulk mean temperature,
~tr~l1sfer
20+40
~
p
nlC(T·-T)
p
mo
0.857 x 10-6 m2/s
k
0.610 W/mK
Cp
4178 J/kg K
Reynolds Number,
IQ
Heat transfer,
Re
6%'r-0_5__
IL
::.:: U D
... (I)
We know that ,
pAU
1t
8.33 x 10-3
P x 4" 02 y U
8.33 x 10-3
997 x 4" x (0.025)2 x U
1t.
[U ~
=
__
·(foI7 m/s I
UD
v
0.017 x 0.025
0.857 x J 0-(,
~Re == _i2?]
, flow IS 1 .
For laminar flow,
. amllJar.
Nu
== 3.66
(From HMT I
( ala hook. Pane
N.).
t:
,I ~3 (Sixlh Edilion)1
Scanned by CamScanner
I
Re.HIII:
8_9._3_x~J(
x 0.025 x Lx (110-30)
I
.t.24 III
Length of the tube, L == 1.24 m
I EXlImple 5 I Lubricating oil (II
III
Re
Q
696.05 W
h A (T_,- Tm)
h x 1t X D x L (T IV - Tm )
v
Mass flow rate,
1111
8.33 x 1(,.3 x 4178 (40 - 20)
5.5
Nusseh Number
m Cp' AT
Q
997 kg/m!
Pr
Since Re < 2300
89.3 W/m2K]
-._
v
(1) ~
0.610
coefficient, h
Heat transfer,
[L, ::-~ooe]
Properties of water at 300e :
h x 0.025
3.66
TI/1 ==
(I
temperetsre
0/
60 l('
enters / em diameter tube wit" a velocity (1/ 3 m/s. The tube
sur/ace is maintained at 40 (C Assuming that the oil has the
following average properties, ealcil/llte tile tube length required
to cool tile coil to 45 'C.
p= 8M kglllr3; k = 0.140 WlmK;
<: 1.78kJIk:'t'
Assume laminar allll/IIII;V developedjlow.
IBlllIratllitinsan V,,;versity, 97/
Given :
Inlet temperature of} T
lubricating oil
ml
Diameter, D
Velocity,
Tube surface temperature,
6GoC
I em =: 0.01 III
U
3 m/s
I, ·lI':= 40°C
Out let temperature of} T
lubricating oil
",0
==
45°(
1
_-----------------c~o='~I\~Je~ct~;v~e~/~le~>a~t~T.~/u~ru~ifI~e~r2.158
Heat and Mass Transfer
p
865 kg/m3
k
0.140 W/mK
1.78 kJ/kgOC = 1.78 x 103 J/kgoC
Cp
----
~
T -40 ]
J
[}xtlmple
6 Air (It :! bar pressure and bulk temperature of
ZOO'(";.'1 heated (IS it flows through a tube wit/r (I dil,meter of 25.4
pAU
I
1t
pX4"D2xU
tl,e wall
1t
0.204 kg/s
For cooling process,
Q
0.204 x 1.78 x 103 x (60 -:-45)
We know that,
For laminar, internal flow
Nu
and
temperllture
the
wall temperalllre
Given :
•
T
L~,
[ ': Nu = h~ ]
Case (ii) :
To find :
h x 0.01
h = 51.24 W/m2K
I Average heat transfer coefficient, h 51.24 W/m2KJ
We know that,
Heat transfer, Q
hA (T",- Tm)
200
==
2U:=
nODe
w
Length. L :::: 3 m
1. Heat transfer per unit length of the lube.
2. Increase ill bull-. ternperalufC over a 3 01 length
of the tube.
C
=::)
Scanned by CamScanner
200°C
Length, L == I III
--o.i4O = 3.66
hA (Tm- T",)
'Y'
2 x 105 Nlrn1
2 bar'
Case (i) :
Pressure. p
Air bulk temperature, 'I'm
.
= 3.66
3.66
above the
illcrc!ll.5cover II J m lengtl! of the tube.
IMtIIlrllJ Unb'ers;ty, 96/
[From HMT data book, Page No. 123 (Sixth Edition)1
h~
i.., 10't'
1111along tile tength of tire tube. flow much would
25.4 11\111 =- 0.025 m
Diameter of tube. D
10 m/s
Velocity, U
0
Wall temperature is 20 e above the air It:nJp~rature.
5446.8 W I
IQ
of 10 m/s. Calculate th« heut transfer per unit
the bllik temperature
I
m Cp (Tmo - T mJ
m Cp (Tm; - Tmo)
Heat transfer, Q
",m at a velocity
letlgth of IIIbe if el cOIutmrt/retll 1111x condition is maintained at
865 x 4" (0.0 1)2 x 3
1m
Length of the tube, L = 270.69 In
Result:
Solution: We know that,
m
....
60 + 4 51.24 x It x 0.0 I x L x [
5446.8
'-L------:-.-l
?
~_.
__
--2-70-.-69--m~
I
1. Length of the tube, L.
Mass flow rate,
+T
]
2 1110 - T
"'I
L
Flow is laminar and fully developed.
Tofind:
iT
h x 1t x D x L x I
Solution:
. (i)· Properties
{Ue
of air at _UOO( :
•
.
.
11\11
hta
bOIl".
l[-ronl'
'
0.746 kglll1
v ==
34.85)(
P
6 . )/
I(}
N· J \ ( 'l\th Fdilillnll
P.lgc . O.
In-"
.
'om' clive Heal Iran ifer
2.160
----
Heal and Mass Transfer
25.99 x 10"-6 Ns/rn?
1.1
Pr
0.680
0.03931
~I
Nu
W/mK
Note:
Given
pressure
is above
kinematic viscosity, v and density,
C are same for all pressures.
atm
p will
pheric
pres ure
ary \\ ith pre
p
Density,
p =
170
:=
k
h 0.Q25
0.03931.
41 ....8:=
1026 J/kgK
"
216/
64.90 W/m~K
:=
S
ure 'Pr .
.k
hA (T .. - T III)
f.r
h
('1',., - T III)
It DL
(220 - 200)
Ip
Case (ii) :
We know that,
Reynolds number.
o
Re
lie. I iran fer
v
o
We kn
v,
-----~
rh I,
!!_
Heal tr ncr,
p
[... m =pA
~
J..l
1.473
10
I Re
Since Re> 2300, flow is turbulent.
T", _ 'I
For turbulent internal flow, general equari n is (Re
Nu
=
~ (O.025i
2300
0.023(Re)
41.20
I
10,000).
(Pr)n
I From HMT data bo k. Pa I: o. 12 - ( ixth EdIlIOn)]
This is heating process.
~
So,
11
= 0.4.
Nu
0.023 ( 14.17
[Nu
41.28
I
Scanned by CamScanner
10')0
(0.680)04
Result __ I.
Q ;;;:
T
1026
[T",I)-T""I
1./61
HIt(/1
----
"lid Mass Transfer
2.12. FREE CONVECTION
If the fluid motion
.
.
is produced
due
.
resulting trom ICIIlJJCmlure gradients,
the
,
.
said 10 be free or natural convection,
'1 his 1110 Ic of hcilt transfer
ivcn be Iow.
exnmp Ics arc urvcn
in d
to change
1II0dl:
.
cns""
Convective He(J1Transfer
4.
I. The hearing of rooms by use of radiators.
lines, electric
transforllls
of heat transfer is calculated
convection equation given below.
Q
-
Heat transfer in W,
A
-
Area in 1112,
T",
-
Pipe surface temperature
rex:> -
Thermal conductivity,
G r ::;;
L Sf
t Ire general
5.
II A er", .- T co
Q
where
.
using
k -
W/m2K
W/mK.
g x P x J) x .1T
v2
(From HMT data hook, Page No. 134 (~'Ixlh f:.dilum)J
where,
rate
Length, m,
GrashofNumber}
for vertical plate
The hca~ tran fer from the pipe carrying steam from the
wall of turnaces, from the wall of air conditioninu e IIUuse
[rom the condenser of some refrigeration
units.
'
The
L -
and
rcct If icrs.
3,
Heallransfer coefficient;
and So
Ine
Thl', ~o ling of transmission
h -
where,
of heat tr'l I' I
• n~ er i
very cummonly
K
J.
S
occurs
Nu = hI.
Number,
~c;e'1
-
Length of the plate,
TII'- T"",
m2/s.
v -
Kinematic viscosity,
p -
Coefficient of thermal expansion.
If Grl'r value is less than 109, now is laminar. If (JrPr
value is greater than 109, now is turbulent.
.
i.e.,
Gr Pr < 109, -+ Laminar flow
Gr Pr > f09, -+ Turbulent flow.
Fluid temperature
in DC,
6.
in "C.
For laminar
Nusselt
now (Vcr1ical plate) :
Number,
This expression
2.12.1. Formulae Used for Free Convection
Nu :: 0.59 (Gr Pr)1)25
is valid for.
104 < GrPr<I09
I.
Film temperature.
where
2,
T
.
b k 1':lI!f No. 135 (SI~11! [dilion)j
[From JIM r data 00.
~
f
Til' -
Surface temperature
T <T.l Fluid temperature
OefticiCllt of thermal expansion
p =
in DC,
7.
in "C.
For turbulent
Nussclt
flow (Vertical plate):
Number,
- 010 IGr Prj(.lm
Nu -
.
S' 'II! EdilionlJ
k I' gc No 11.5(. IX
[From HMT tlala boll .. a
R.
Heat transfer (Vertical plate) :
Q
Scanned by CamScanner
2.J6J
::
,.
T
'ddT" - "')
1
Heal and Mass' Transfer
liM
9.
l
I
~sPhere,
15.
GrashofNumber for Horizontal Plate:
g x p x L~ x .1T
Gr
where
v2
Lc -
Characteristic length
W -
Width of the plate.
:::-
W
2 '
10.
Convective Heal T"
Iransfer
Nusselt Number,
[From HMT data book P
, age No. 137 (Sixth E
Heat transfer, Q ::: h x A x (T - T
\I'
Boundary layer thickness
16.
Ox = [3.93 x (Pr): 0.5(0.952 + Pr)025 x (Gr)-02S]
Nusselt Number, Nu ::: 0.54 [Gr Pr]0.2S
[From HMT data book, Page No. 134 (Sixth E
This expression is valid for
Maximum velocity,
17.
Gr Pr < 8 x 106
(From HMT data book, Page No. 135 (Sixlh E ..
dillon)]
Nusselt Number, Nu = O. 15 [Gr Pr]OJ3J
Umax
= 0.766 x v x (O.952+Pr)-'12
x
[g p (:; - Ta»]
112
Mass flow rate,
18.
This expression is valid for. 8 x 106 < Gr Pr < 10' I
II.
(I)
A ::: 4 n ,2
where
For horizontal plate, upper surface heated,
2 x 104 <
Nu ::: 2 + 0 .43 [G r Pr]025
"1 =
For horizontal plate, lower surface heated.
G
] 0.25
[
I.7 x P x v (Pr}2 (pr : 0.952)
Nusselt Number, Nu::: 0.27 [Gr Pr]02S
2.12.2. Solved Problems
Convection
T~is expression is valid for 105 < Gr Pr < 10' I.
12.
Heat transfer (Horizontal plate)
where
·13.
-
Upper surface heated, heat transfer
coefficient W/m2K,
hI -
Lower surface heated, heat transfer
hll
I Example J I A vertical plate 01 O.75 m heigl" is at J
(hu + hI) x A x rr, - Tco)
Q
and is exposed to air at a temperature 01 105'(' alld
atmosphere. Calculate:
1. Mean heat transfer coefficient,
2.
coefficient, W/m2K.
For horizontal cylinder,
Rate of heal transfer per unil widthollile plale.
Given:
Nusselt Number, Nu ::: C [Gr PrJ'"
14.
[From HMT data book, Page No. 137 (Sixth Edition))
Tofind:
Heat transfer, Q
0.75 m
T",
170 e
Fluid temperature, T
105°e
IX>
0
I. Heat transfer coefficient, (h).'
2. Heat transfer (Q) per unit width.
nOL
./
..
Scanned by CamScanner
Length, L
Wall temperature,
For horizontal cylinder,
where, A
on Free Convection (or) N
2.166
l
Heal and Muss Transfer
Solution: Velocity
convection type problem.
(U) is not given.
_-===:-::-:~====:::::~C::...:().:::"::Ve~Clive
Heat T,
~
.: 8.35 x lOK]
ran_ifi_e_,_
So this-:-:;-Utal
Tj
Film temperature,
I
= 8.35 x 108 x 0.684
Gr Pr
T".+ T'l.
2
~.~~
5.71
x
J
IOS
Since Gr Pr <. 1()9. flov is laminar.
,-I
Properties of air at
170 + 105
2
GrPrvalue
is in between IO",md 109 i.e., 1000<GrPr<
T-L.j__
137.5~;~
So. Nussel,
Number
Tj
137.5°C ~ 1400C
Nil
-
0.59 (Gr Pr)02:;
[From 11"11' data book. Page No. 33 fSixlh Edi .
I~
Density,
p
0.854 kg/m!
K incmatic viscosity,
V
27 .80 x I ()-6 m2/s
Prandil Number, Pr
Thermal conductivity,
[From IIMT datu book. Page No. 135 (Sixth Edilionll
'lIon)1
0.59 (5.7 I x 1011)0.25
CHiC = 91.2 I]
0.684
Ie
We know thai.
0.03489 W/mK
We know that,
Coefficienl
N us: ell Number,
expansron
I) J.2 I
137.5 + 273
r-- -
-
----
------
II JI Ira us fer,
()
(if
4.24 W/m2gj
"A (1'., - Tel)
o. 134 (Srxrh Erlilion)/
1).81:<2.4:.:103
(0.75) /(170
(n.RO
10 6)2
105)
Re.\ult :
I.
I le<ll transfer
co ffiril'nl.
h
Heat transfer. Q
Scanned by CamScanner
II
4.24 x 1 x 0.75 x (170- 105)
[.,' W= J m]
g x {3 x LJ x ~T
v2
,FrOIllIlMT dal;} hool;. I'a 'C
4.24 W/m2K
hxWxLx(T.,-T'I))
We know Ihal,
=
ItxO.75
0.03489
We kn "
2.4 x 10-3 ~
Gr
hI.
~
Ii
.-
He at rail fer coefficient,
1
410.5
Grashof Number,
Nil
f3
()fthefJ~lal}
109
4.2.' Wlm K
206.H \\.
2.J 68
Heal and Mass Transfer
I Example 2 I A vertical plate of 0.7 m wide (lml ~. m he; l
maintained at a temperature of 90't' ill a room.
Calculate the convective hem Ion.
If It
\...
__ ----------~~~C:nn~v~eC:'I~·v~e~H~eq~I~U~a~ns~~~r~3
b
g x P x L3 x ~ T
GrashofNum
er, Gr
Height (or) Length,
L
1.2 m
Wall temperature,
T\II
90°C
Room temperature,
Too
30°C
v2
3() Qr,
[From HMT data book, ~age No. 134 (Sixth Edition»
0.7 m
Wide, W
Given:
(If
~
~
Ir'-1
9.81 x 3 x 10-3 x (1.2)3 x (90 30)
(IS.97 x 1~)2
-r--S-.4-x-,-09-'1
G
Gr Pr
S.4 x 109 x 0.696
G-r-P-r--5.9-X)Q9J
To fillll: Convective heat loss (Q).
Solution:
Velocity
convection type problem.
(U) is not given.
So, this
IS
Since Gr Pr > 109, flow is turbulent.
natural
For turbulent flow,
Nusselt Number,
We know that,
Film temperature,
Tw + Too
Tj
Nu
=
0.10 (Or Pr)0.333
{From HMT data book, Page No. 135 (Sixth Edition)]
2
Nu
90 + 30
2
1 Nu
0.10 [5.9 x 109]0.333
179.3 I
We know that,
Nusselt Number,
Properties of air at 60°C :
Nu
hL
k
179.3
= 'Q.02896
h x 1.2
[From HMT data book, Page No. 33 (Sixth Edition))
p
1.060 kg/m-
v =
18.97 x 10-6 m2/s
Pr
0.696
k
0.02896 W/mK
Convective heat
transfer coefficient
Heat loss, Q
We know,
l
, .
[Q = 218.16 ~
~
60 + 273
[p
3 x '0-3 K-I
/
Scanned by CamScanner
h A (AT)
h x W x Lx (T1I'- T.o)
4.32 x 0.7 x 1.2 x (90 - 30)
Coefficient of }
1
thermal expansion
~ = T inK
j
1
h = 4.32 W/m2K
= 3 x 10-3 K-I
Result:
Convective heat loss,
Q == 218.16 W
2./70
1
Heat and Mass Transfer
____
I £wllnple 3 I A vertical pipe of 12 em oute» diameler
----
.--------~C~Omnveclive
9 81 x
.
'''',g, at a surface temperamre (If 120't' is iN a room I h' 2.5 IJI
I' ere ~
Calculate lite hem loss per melre leltC/110 lle
pipe.
If Ihe
Diameter,
D
12 em
Length,
L
2.5 III
Surface temperature, Til'
120°C
Ten
Room temperature,
0.12m
2.91 x 1Q-3X(25)3
. x ( 120 - 2Q1
(20 .0 2 x 1~)2
--____
[Gr
1.11 x 10"]
Gr Pr
1.11 x 10" x 0.694
nil' is at 20't'.
Given :
Heal t- ,/:
ansJer 2.17/
[GrPr
7.72 x 1010J
Since Gr PI' > 109, flow is turbulent.
For turbulent
20°C
flow,
= 0.10 (Gr Pr)OJ33
Nu
Toflnd : Heat loss (Q) per metre length of the pipe.
Solutio« "
[From HMT data book, Page No. 135]
Film temperature,
TI
TII'+TIYj
Nu
2
I Nil
120 + 20
2
0.10 [7.72 x 10 °]0333
'
422.3
I
We know that,
700C]
Nusselt
Number,
Nu
hL
k
422.3
h x 2.5
0.02966
Properties of air at 70°C:
[From '-IMJ data hook. 1';J~e No, .1] (Sixlh Etlilinn)1
p
1.029 kg/lllJ
v
= 20.02
Pr
0.694
Ie ==
O.021)(j() W/ml<
[~Jleat
;" 10-6 11J2/s
transfer
Heal loss per}
metre length
IQ
== 2.9J
x JO-.1 K' J
[f3 == ~2.c)1 x 10-.1 K·-t]
==
gxOxl)x.1T
v2
IFmmllM
r daril honk. Page Nn. I.H (Sixlh EdilillU)f
Scanned by CamScanner
I
h A!!.T
5.01 x1txO.12x
f3 = T inK
J
70 + 273
5.01 Wlm2K
II
h x 1t X D x L X (Til' - T .,,)
Result:
[ EX(lmple.J
. -
I x(120-20)
I
188.8 Wlm
-'
Q= 1888 WIlli
Heat loss per metre length ot pipe.
.
J
Number, Gr
Q
I
We know that,
Grashof
coefficient,
_.
if 800
Ion". 70 111m
! A hortzouto! plate
r
ITeof 140 'C ill Itlrgr I{/II" OJ
/11/11
(I
".
I.•
(I
"'/(Ie is maintained (It {/ temperatl
I . firo/lllhr plate.
,
. the 101(11 he(ll 015
iltll oJ water at 6(} 't'. Determ",e
L
800111111= O.S III
Gil'e",' Horizontal plate length,
2. J 72
Heal and Mass Transfer
Wide,
Tofind:
C onveclive Heal Transfer
____
W
70 rnm
Plate temperature,
Til'
140°C
Fluid temperature,
T
60°C
<I)
-- 0::----0
. 70 rn
(I):::>
Gr:=
9.81 xO.76x
_.
I Gr = 0.297 x I09_]
Gr Pr
Total heat loss from the plate.
Tw+T~
2
Tj
2./7
60
1
x 109 x 1.740
= 0.297
~-~
Solution:
Film temperature,
IO-3X(0.035)3x(140
(0.293 >< I~)2
0.518x 109J
Gr Pr value is in between 8 x 106 and lOll,
i.e., 8x J06<GrPr<lOll.
140 + 60
2
So, for horizontal
Nusselt
plate, upper surface heated ,
Nu
Number,
= 0.15 (Gr Pr)0.333
[From HMT data book, Page No. 135 (Sixth Editionj]
Properties
of water at 100°C :
[From HMT data book, Page No. 21 (Sixth Edition)1
p
v
961 kg/rn-'
=
I Nu
119.66
I
We know that
Nusselt
1.740
k
0.6804 W/mK
/3 (water)
0.76 x 1O-3K-1
gx/3xL~
=
Number,
Nu
119.66 =
[From HMT data book, Page No. 29 (Sixth Editionj]
Gr
0.15 [0.5 I 8 x 109j0.333
0.293 x 10-6 m2/s
Pr
GrashofNumber,
Nu
hu x 0.Q35
0.6804
Heat transfer coefficient for}
= 2326.19W/m2K
upper surface heated, hu
xl\T
v
... (I)
2
For horizontal
plate,
Lower surface heated,
[From HMT data book, Page No. 134 (Sixth Editiom]
For horizontal
plate,
Lc
Lc
I Lc
[From HMT data book, Page No. 136(Sixth Edition))
Nusselt Number,
Characteristic
length
W
2
[Nu
0.070
2
We know that,
O.oJ5 m
O.oJ5 m
Scanned by CamScanner
Nu = 0.27 [Gr PrjU25
Nu = 0.27 [0.518 x 109j025
I
Nusselt Number,
Nu
40.73 ]
Convective Heal Transfer
1. / 74
Heal and Mass Transfer
h, J( 0.035
40.73
0.6804
Heat transfer coefficient for}
lower surface heated, h,
"---
Total heat transfer,
~ 791.79 W/m'K
I
----
(hll + hi) A ~T
Q
(hll + h,) x W x L x (T +T
u-
Q
[2326.19 + 791.79] x [0.070 x 0.8] x [140-60
IQ
13,968.55
I
Tf in K
We know
wi)
I
30 + 273 - 303
Total heat loss, Q = 13,968.55 W.
Result:
I Example 5 I Air flow through
long
(I
recltlllgular
0
300 mm heigh: x 800 111mwidth air-conditioning duct mainla;!s
the. outer duct sur/ace temperature at 20°C. If the duel ;1'
un~nsulated anti exposed to air tit 4(J0C. Calculate the heal
gamed by the duct. Assuming duct to he horizontal.
Given:
0.02675 W/mK
k
"')
L
Length (or) Height,
30e mm
Since the duct
convection
from
W
is laid horizontally,
the vertical
gx!3xL3x~T
Gr
800 mm
==
v2
[From HMT data book, Page No. 134 (Sixth Editionll
0.8 111
Surface temperature,
T",
20°C
Fluid temperature,
T ""
40°C
the heat gain is by free
and the horizontal top and bottom
sides.
Free convection from the vertical sides:
OJ 111
Width,
3J x 10-3 K-l I
[ 13
9.81 x3.3 x 10-3 x (OJ)3 x (40-2,Ql
(16 x 10-6)2
[0 == 6.8 x 10~
To flnd : Heat gained per metre length (Q/L).
7
6.8 X 107 x 0.701 :: 4.7 X 10
Gr Pr
Solutio" :
4.7 x
[Gr Pr
Film temperature,
Tf
TII'+T""
9
Gr Pr < 109 , floW is laminar.
. 104 <: Gr Pr <: 10 .
IQ4 and 109 i.e.,
Gr Pr values is in between
025
:: 0.59 (Gr Pr) . .
.
So, Nusselt Number, Nu
No 135 (Sixth Edlllon)1
Since
2
20 + 40
2
..
[From
37
(",
Scanned by CamScanner
IQD
t.1MT data ~ok, Page
r
.
2.1 6
HCQt and Mass
Transfer
Nu
=
Nu
,:I
48.85
:;;
)
11{,~"ttrans lcr from
verticul side
I'
,I
;
k
0.15 (1.13 x 108]0331
\ Nu
72.17J
We knoW that,
'Nusselt Number,
I
,,,
4. 5 x 0.8 x 0') )' (40
I~~- -20.88 W
J
I
:\
I
X
II
-
\:
l(),
Ileal transferfrom
Fur horizontal
Upper
-
hll = 4.82 W/m2K
surface heated:
Nusselt Number,
41.76 W
I
41.76W
surface heated, heat transfer coefficient
1
\i,
I
20)
lower
f - 2
hu x 0.4
= 0.02675
h" = 4.82 W/m2K
l
l lcat transfer from
both side of
vertical sides
k
II'
II x W x I. ('I' 00 - T w )
I
h" Lc
Nu
72.17
It t\ (T - T )
<
I
Nu
::?
4.35 W/1ll2K
-
!'
.
' x 0 < Gr Pr < Ie
0.15lGr Pr)OJ33
'Nussel' Number, Nu =
lU)2675
I,
II'
I',
-izontal plate, upper surface heated 8 1 6
1
" x OJ
"I: j
.
7'
I
for hor
hL
48.85
.I
/'
0.59 [4.7 x 107JO!;------'_
~u
We know that.
_------...::C:.:::'o~nVvteClive Heal ransfer 2
..------
Nu
0.27 [Gr PrJO.25
= 0.27 [1.13 x 108]0.25
I"-N-u-ill]
'" (I)
We know that,
IlOrlzollllll.\·itlt!!I·"
Nusselt Number, Nu
plate.
hi x 0.4
Characteristic
length , I"c
W
0.8
27.8
2" = T = 0.4
5 \.85 W/m2K ]
=
,r-I,-c --0.4-1
Gr
9.81 x 3.3 x 10-3 x (OA}l x (40 -- 20)
(16 x 10-6)2
I'- G.::...:r_ _.:...:1.:::._6..:.:_x
iQ!]
Gr Pr
1.6
I Gr Pr
1.13 x IOQ£]
x 108 x 0.701
L
Scanned by CamScanner
= Q.0267s
Lower surface heated, heat transfer coefficient
hi Heat transfer from
I
horizontal pate
\.85 W/m2K
f Q H = (h + hi) A sr
II
= (hll + 11() x W x
L t,T
x
_
1 I 8
----------~--~(~·o~nv~e~c~/iI~,e~H~e~a~/~Tr~an~s2if1~er~~2
k - 0.02675 W/mK
Ilealalld Mass Transfer
[QH
(4.82
J2.~
:=
-'20)
". (2)
J Heat tran~fer 1 +
f
from vertical
sides
l
1 otal heat transfer
~
1.85) x 0.8 x O~
Heat
{
transfer
from
hori.zontal
side,
I
l
We knoW that,
coefficient
of}
thermal expansion
~
T, in K
f
73.8 W I
Re!-;u/t: Heat transfer,
Q
:=
I E:mlllp/e 6 I A plate of 6
gx~xL3
Grashof
73.8 W
8 em x 14 em site '''lIill_
tuined tIl (I temperature of 60'(' (Inti heat lost to lite air is at (J't.
Tire vertical
dimension
is
CIII
14
Number,
Determine
heat
... ( 1')
Characteristic
where
length
LII x Lv
Irlllr,~fer
LH + Lv
6 em x 8 cm x 14 em
Plate size
0.06 rn x 0.08 m x 0.14 111
0.08 x 0.14
0.08 + 0.14
0.0509 m I
Plate temperature,
r,
60°C
_Fluid temperature,
To"
O°C
1'0find : Heat transfer coefticient,
(h).
Film temperature,
Tf
T,,,+T'.L)
2
=
60 + 0
2
I Tf
9.81 x 3.3 x 10-3 y (0.0509)3
(16 x 10-6)2
Gr
(I) =>
1 x 106 1
Gr Pr
I x 106 x 0.701
[ G-r-P-r --7-.0-1-x-1-'05:11
x (60 -
0)
I Gr
Sotution : We know that,
Properties
A~T
[From HMT data book, Page No. 1341
X
CIII,
c
Gr
coefficient.
Given :
1
303
3.3 x 10-3 K-Ij
41.76 + 32.05
IQ
1
30 + 273
30°C
5
7.01 x 10
Since Gr Pr < 109, flow is laminar.
9
I
For laminar flow, 104 < Gr Pr < 10 ,
Nusselt Number Nu := 0.59 (Gr Pr)025
(From HMT data book. Page No. 135 (Sixth Edition)]
of air at JO°C :
0.59 x [7.01 x 105]025
[From HMT data book, Page No 33 (Sixth EditionJl
p
1.165 kg/m!
v
16 x 10-6 1112/s
Pr
0.701
~[N-t-I-=-1--7-::-:.07:1J
We know that,
Nusselt
Scanned by CamScanner
Number,
Nu
2.180
Heal and Mass Transfer
h x 0.0509
26.75)( 10-3
17.07
.---_--_.
Ih
Result:
8.97 Wlrn2K
Heat transfer coefficient,
I
----
.~~~~~::C~on~v~e~C/~iV~e~lf,~ea~/~'l~ra~Il~'(,0:r~_2
(l
~
.~
2.55 x 10-3 K-T]
Grashof
number.
Given:
9.81 x 2.55 x 10-3 x (0.15)3 x (200 - 37)
(25.45 x 1()-{>)2
Wall temperature,
r,
200°C
Ambient air temperature,
Ton
37°C
E
0.92
Emissivity,
Tofind:
. lJ
o air Qt
if emissivity 0
1/
15 em == 0.15 m
Diameter of pipe, D
~
·-2-.1-2-x-lO-7-'1
Gr Pr
2.12 x J01 x 0.686
[ Gr Pr
1 .45 x 107
For horizontal
Nusselt
cylinder,
Number,
Nu
=
C [Gr Pr]"
Gr Pr == 1.45 x 107, corresponding
Solution:
~
Tf
I
[From HMT data book, Page No. 137 (Sixth Editionj]
I.Heat loss per metre length.
Film temperature,
g x p x D3 x ~T
=
[From HMT data book. Pace
No . 134 IS'
. h Edi
. 1I
\ ixt
.d I!IOn
I Example 7 I A horizontal pipe of 15 em dian 'eter '.
maintained at wall temperature of 200°C and is exposed t
Gr
2 I, I
,,2
h = 8.97 W Im2K
37°C. Calculate the heat loss per metre length
pipe is 0.92.
').
II
Til' + Ton
2
200 + 37
__
-
We know that,
2
Nusselt
C = 0.125, and In == 0.333.
Nu
0.125(1.45xI07)OJJ3
I Nli
30.31
hD
Number,
Nu
k
h x 0.15
30.31
Properties of air at 118.5°C ~ 120°C,
[From HMT data book, Page No. 33 (Sixth Editionl]
We know,
p
0.898 kg/Ill]
v
25.45 x 10-6 m2/s
Pr
0.686
k
0.03338 W/mK
Scanned by CamScanner
lIi].74
~
Heat tran. fer .by } Q
convectiOn
o.oms
W/m2KJ
hA sr
h x 1t x D x L X (Til' - T"')
6.74xnxO.1Sx 1x(200-37)
:l (I) ["L=lm]
517.7 W/~
1
TJ in K
1
118.5+273
=
Heat tn~ll'i.fcr }
by radtaUOll
391.5
o.
.. '
4
E x A x <1 [T .. -
.
T4]
~~~~~~~~~--------2. J 82
tion thickness
Insu \ a
1 diameter of)
Actua
.
the p'pe, D
Heal and Mass Transfer
where
Emissivity
A - Area, m2
E
-
o
Emissivity,
temperature,
surface
be
111
Air temperature,
Fluid temperature, K
Too
200 + 273
Til'
473 K
Qr
I
rClI1.~fer
30 mm
::: 0.Q30 rn
= 0.080
+ 2 x 0.030
0.\4 m
Stefen Boltzmann constant
5.67 x 10-8 W/m2K4
TIl" - Surface temperature, K
Too
Convective Heal T
0.94
E
Til
T
rL
37+273
I,---T_oo
_310
KJ
x [ T411' - T4 ]
E x 1t x 0 x L x o
CSJ
0.92 x 1t x 0.15 x 1 x 5.67 x 10-8
x [(473)4-(310)4]
IQ
r
I
1003.4118 W 1m
'" (2)
1. Heat loss from 5 m length of the pipe,Q.
2. Overa\\ heat transfer coefficient, hI'
Tofind:
Total heat transfer per metre length
Q
Heat trans~er } + { Heat tr~n~fer }
{ by convection
by radiation
517.7 + 1003.41
IQ
1521.12 W/m
I
3. Heat transfer coefficient due to radiation,hr'
Solution:
We know that.
Film temperature,
TJ
-
85 + \5
2
Result: Total heat transfer per metre length == 1521.12 W/m
I Example 8 I A steam pipe 80 mm in diameter is covered
with 30 mm thick layer of insulation which has a surface
emissivity of 0.94. The insulation surface temperature is 850C
and the pipe is placed ill atmospheric air lit 15 'C If the heat is
lost both by radiation and free convection, find tile following:
1. The "eat lossfrom 5 m length of tirepipe.
2.
The overall "eat transfer coefficient.
3. Heat transfer coefficient due to radiation.
Given:
Diameter of pipe
80 mm
==
(
Scanned by CamScanner
0.080 m
s 'm Edition)\
Propertiesof air tit 50"C :
p No 33 ~ IX
{From HMT data book, age .
p ==
\ .093 kg!m'
v ==
\ 7 .95 )( \ 0-6 m
2/s
Pr == 0.698
k
==
0.02826 W/mK
__
2. J 84
-----C-o...:.n:...::_vective
Heal and Mass Transfer
If,
eat Transfer
h x 7t D x Lx (T
Coefficient of thermal }
expansion, p
Tf in K
[L-Q_c:..;:_on_v__
I
50 + 273
3.095 x
K-I]
fleat loS
Qrad
E(JA[~v-~]
E
Emissivity
A
Area- m2
(J
Stefen Boltzmann constant
where,
=
gxf3xD3x.1T
v2
[From HMT data book, Page No. 134 (Sixth Edition))
5.67 x 10-8 W/m2K4
9.81 x 3.095 x 10-3 x (0.14)3 x (85 - 15)
(17.95 x 10-{j)2
Ir-G-r--18-.1-0-x-,-06-',
Tw
Surface temperature, K.
Too
Fluid temperature, K.
GrPr
18.10x'06xO.698
T II'
85+273
Tet):::
[ Gr Pr
1.263 x 1071
G
358 K ]
I T et) ::: 288 K I
x (J x 7t DL x [T~
-'J!]
L!I_'~I'_----'_
For horizontal cylinder,
::> Q rad ==
Nusselt number, Nu = C [ Gr Pr Jill
:=:
[From HMT data book. Page No. 137 (Sixth Edition))
o- ':,
[ Qrad
1.263 x 107,
C = 0.125,
corresponding
and m = 0.333
E
We know that,
Nu
1118.90 W ]
~
28.952
~
h
~ective
h x 0.14
0.02826
5.84 W/m2K
heat transfer coefficient, he == 5.84 W/n~
Heat lost by convection,
Qconv ::: h A (.1 T)
Scanned by CamScanner
Qconv + Qrad
Total heat loss, Qt
898.99+ 1118.90
::=
1
hD
k
15+273
0.94 x 5.67 x 10-8 x 7t x 0.14 x 5 x [3584_288]
Nu = 0.125 [ 1.263 x 107 JO 333
l:HiC::: 28.952
(85 -15)
t by radiation,
10-3
We know that,
Grashof number, Gr
w- TaJ
5.84x1t xO.14x5x
8_9_8._99_W]
Total heat transfer,
Qt
::=
::=
2017.89
2017.89 ~
4
Convective Heal Transfer
2. J 86
Heat and Mass Transfer
13.108 - 5.84
=
W/m7K]
e&_-;;':-_7.268
----
Result:
(ii)
By radiation,
Or
:=0
:::
h , = 7.268
Heat transfer coefficient,
2.
Maximum current. Take resistance of wire is Zohm/m.
Horizontal wire diameter, 0
Surface temperature,
Til'
Air temperature,
T'"
Resistance of the wire, R
I.
Heat transfer coefficient,
2.
Maximum current, (I).
Film temperature,
3 x 10~J0l
g x p x 03 x IlT
ofNumber,
Gras 11
v2
Gr
9 81 x 3 >< 10-3 >< (3 >< 10-3)3 x (100 - 20)
Or =='
=>
(18.97 x 10--6)2
176.64
~==
Gr Pr ==
3 x 10-3 m
Cili_1!_ == JtiiJ
J
176.64 x 0.696
For horizontal cylinder,
Nusselt Number,
7 ohm/Ill
(h),
Nu = C [Gr Pr]"
(From HMT data book, Page
Nusselt Number,
~lsfer
coefficient,
Heat transfer,
Scanned by CamScanner
137 (Sixth E inon
C - 0 85 and m == 0.188.
Nu
0.85 [122.9]°·188
I Nu
2.IJ
.-
.
2.1
hO
k
h x 3 x 10-3
0.02896
h
20.27 W/t;:KJ
Nu
[From HMT data book, Page No. 33 (Sixth Edition)]
~.
o.
We know that,
Properties of air at 60°C:
J
d ..
N
.
Gr Pr = 122.9, correspondins
Tf
1.060 kg/m!
..
ook Page No. 134 (Sixth Editiorn]
,
[From 1-1 MT d a ta b
3 mrn
100 + 20
2
p =
333
3 x 10-3 K-l
W/m2K
I.
I
1
60 + 273
h , = 13.108 W /m2K
maintained at J 00'(' and is exposed to air at 20 '('. Calcillate the
following:
Solution :
0.02896 W/mK
1
T in K
f
:::
:::
I Example 9 I A horizontal wire of 3 mm diameter is
Tofind:
~
1118.90 W
Radiative heat transfer coefficient,
Given:
0.696
Ol" ,.= 898.99 W
2. Overall heat transfer coefficient,
3.
k
,
18.97 x 10--6m2/s
:::
Pr
we knoW,
1. Heat loss from 5111length of pipe
(i) By convection,
v
I
2.187
Q = h A ~T
)]
_------C.:..o::n:..:v~ec~tl;ve HearT,
2,188
Heat and Mass Transfer
h x 1t x 0 x L x (T - T rn )_______
II
IQ
20.27x1tx3
15.2 W/m
I
ransfer
. 5 of air at I 62.5°C
ertle
proP
x 10-3x I x(IOO
-20)
.
p
=
0.815 kg/m3
v
=
30.09 x 10-6 m2/s
Pr
0.682
k
0.03640 W/mK
We know that,
Heat transfer,
~ 160°C·
Q
1
Tf in K
We knoW,
I
'-1
OrashofNumber,
1.47 Amps]
I.
Heat transfer coefficient,
2.
Maximum current,
h
=
I =
20.27 W/m2K
immersed in air at 25 't:'. Calculate tile convective heat loss.
Diameter of sphere, 0
20 mm
Surface temperature,
TlI'
300°C
Fluid temperature,
T
25°C
a)
Gr
::::>
1.47 Amps
I EXtlmple 10 I A sphere of diameter 20 mm is at 300°C is
0.020 m
=
1 Gr
Gr Pr
1 Gr
Pr
Gr
Tf
I
54734.2 x 0.682
37328.7
I
N
{From HMT data book, Page o.
Tw + Too
2
162.5°C
2 + 0.43 [37328.7]025
I Nu
7.97]
Nusselt Number, Nu
162.5°C
I
137 (Sixth Edition)]
Nu
We know that,
300 + 25
2
Scanned by CamScanner
54734.2
[1 < Gr Pr < lOS]
Nusselt Number, Nu = 2 + 0.43 [Or Pr]025
We know that,
Film temperature,
9.81 x 2.29 x 10-3 x (0.020)3 x (300- 25)
(30.09 x I~Y
For sphere,
Tofind : Convective heat loss, (Q).
Solution:
I
{From HMT data book, Page No. 134(Si>.1hEdition)!
Result:
Given:
I
162.5 + 273
435.5
f3---2.-29-x-I-0--3
-K-~I
ill
==
k
,,-----
_
--------------~(~·~o'~'v~e:c/~{V~c:H.~e~a/~~~r~an~~~e:r~l~.
80 + 22
""~
4",.1,
h
14: I ,4
( r..
T, )
n
O.~~O )
( 00 - 2 )
pro/",,.",;e~·
. (II {Ii, lit J / cr' III JO '(' :
[ Q -~~_.01 W
[From HMT data book. Page No.3) (Sixth Edition)!
he3t I,) s, 0
nveciivc
p
5.01 W
I £.m"'r/~ II I .-4 vertica! plate of 40 em 1,,111: is mainlailltd
If' 60 ~ (Ind if c..\:ptl.fNI IIi,
III
I.
Bound",)'
2.
Calculate
1I~)'t" thickness tI"IIe
tailing
the fol/owing
:
TI,t' sa",t' plait' is placed ill tI wind tunnel
1111(1
boundary
L
40 em = 0.40
Plate temperature,
r,
80°
Fluid temperature,
T
22°C
Length,
0.02826 W/mK
I
Tf in K
III
m2/s
I
51 + 273
3.086 x 10-3 K-I I
Ip
Case (i) : For free convection,
g x p x L3 x L\T
Gr
v2
=
[From HMT data book. Page:No. 134 (Sixth Edition)j
9.81 x3.086
.. Cast' (i) ..
i) Boundary
x 10-3 x (0.4)3 x (80-22}
(17.95 x 1~)2
layer thickness (Natural convection).
3.48 x 108 ]
Case (ii):
(i) Boundary
layer thickness
at velocity
U = 5 m/s (Forced
convection).
Cast' (iii) :
erage heat transfer coefficient tor natural convection,
(ii) Average heat transfer coefficient
Solutio"
k
II,icb,~n·.
G;'~" ..
(i) A
17.95 x I~
0.698
air is
,-4 vemge ''''(1' transfer coefficient for natural (Inti forctd
convection for th« above mentioned data.
Ttl find
v =
Pr
edge of the pl(lle.
blo"'n over ;1 al a velocity of 5 m/s. Cnlclilme
In),t"
J.
(I' 11°C.
1.093 kg/m!
: We know that,
for forced convection,
h.
h.
Gr Pr
3.48 x 108 x 0.698
I Gr Pr
2.43 x
IOU < 109
S·mce G r P r < 109• flow is laminar.
. I mino' flow:
For free convectIOn, 0
Boundary layer thickness,
b.T '"'
025
_ 05 x (0.952 + Pr)'
[3.93 x (Pr)
1.
Film temperature,
T..,+T
1, =
2
Scanned by CamScanner
[From'
LIMT data boO...
Page No.
(Grt 0.25] x X
x
134 (Si.xthEdilion»
Ox ==
=>
_
----------------~~C~o~n~ve=c~fi~ve~H~ea~f~T~r~~m~s(l~e
~
hL
.
We knoW that.
Nu
k
Heal and Mass Transfer
2.192
lox
==
[3.93 x (0.698'-
0.0156
+ 0.698)0.25 x ----
0 5 x (0.952
(3048 x I 08)- 0.2~J
. )( 0-4
l. x == L c- 040
. In)
m
Ih
z»
Case (ii) : For forced convection.
Reynolds number,
23.29
Re =
==
UL
[From
Ox
Nux
I. I I x I O~
I Nux
'" (I )
0.332 ( 1.11 x JOS)05x (0.698)0333
==
98.13
h, L
k
<5 x or
°
IIx
== 5 x x x ( Re )" 0 5
98.13
IIMT data book. Page No. 112 (Sixth Fditionj]
Local heat tran~fer} h
coeffiCient
x
5 x 0.40 x (1.11 x 10)-05
10-3
[.:
x = L = 0.40 m]
mi···
(2)
equation (I) and (2), we know that, boundary layer
in forced convection is less than that in free convection.
Case (iii): A verage heat transfer
convection, I, :
coefficient
for
nat"ral
Nusselt number.
Nu
Hr"n data book.
I
Page No. 135 ( ixth Edilion))
0.59 (Gr Pr)O 2:
0.5'-) (2.43
~23.~
Scanned by CamScanner
A veraae
106)<W
heat tran~fer} h
coeffic lent
II
[II
II x x 0.4
0.02826
6.932 W/m2K
2 x 6.932
13.86 W/m2KJ
.. , (4)
. C) and (4) we know that heat transfer
From equation)
. "s much larger than that in free
coefficient in forced convection I
con ecti
For free convection, laminar flow, vertical plate :
[From
... (3)
We know that,
I <5x == 6.003 x
From
thickness
I
[From HMT data book. Page No. 112 (Sixth Editionj]
For forced convection, laminar flow:
layer thickness
1.645 W/m2K
for forced convection, laminar flow.fla: plate :
Local Nusselt number, Nux = 0.332 (Re)05 (Pr)OJ33
v
Since Re < 5 x 105, flow is laminar.
Boundary
0.02826
A lIerage heat transfer coefficient for forced convection,h :
5 x 0040
17.95xI0-6
Re
~
n,
Result:
Case (i) :
I.
b s ( alu/lIl cOll~ec(loo)
Case (ii):
I.
Ox (FOIc.ed c.()n~ewoo)
Case (iii):
1.
h ( '&Iura! OOIlVt\.··UOIl)
2.
h (FOIU'd conveCtloo)
0.0156 m
6.003 X 10-3 In
).645
W/m2K
13.86 W/m2K
I
I
]./94
Heal and Mass Transfer
-----------------~~~~~
11
Convective Heat Transfer
2.12.3.SolvedUniversity Problems - Free ConVectio;--GErample I ] A large vertical plate 5 In height is Ill'
f
I
I
",
I,
i
.
t 100'(' and exposed
a
htattransfer coefficient.
10
_
Height or length, L -
ersltyJ
5m
6.68 x 1011 x 0.695
[Gr Pr
4.64 x 1011
T...
for turbulent flow,
Fluid temperature,
Teo
Nusselt number,
h.
~
T,. + T",
2
100 + 30
2
Nu = 0.10 [Gr PrJ0333
(From HMT data book, Page No. 135 (Sixth Edition))
We know that,
T,
Nu
0.10 [4.64 x 101lJ0333
I Nu
767.27
= 1.~5
"
=
Pr
= 0.695
19A95
kg m3
1 Q-6 m-/s
~
k
=
-,'M of rhermal expansion, f3 = I ~ K
h = 4.49 W/m2K
I Heat transfer coefficient, h = 4.49 W/m2K I
Result:
Convective heat transfer coefficient, h = 4.49 W/m1K
horizontally in a room at 23 DC Tale the olllSiIk surface
temperature of pipe as 165't: DettTmW the Ileal loss per metre
Itngth of the pipe.
{D«. 2004, Anna UnMniIy/
Given :
Diameter of the pipe D = 10 cm = O.J 0 m
Ambient air temperature, Ix
f3
fa
hx5
0.02931
I Emmple 2 I A steam pipe 10 em olltSiIk diamdeT TUIIS
J.: = 0.0_931 W/mK
Coe
hL
Nusselt number, Nu
Propenies of air Q/ 65'(' :
p
I
We know that,
767.77
[From H.\IT dam bo k, Page No. 33 (Sixth Edition)!
I
Since Gr Pr > 109, flow is turbulent.
Surface temperature,
Film temperature,
. Gr
=
23'T
\\ all temperature, T... = 165°C
To find:
Solution:
Heat loss per metre length .
We know that
I ... + Tx
Film temperature, If
I II
Scanned by CamScanner
2.195
6.68 x 1011
Gr Pr
.f
n~~
{June 2006, Anna Unl» . e
Tofind: Convectiv e heat transfer coefficient,
SOIUlif1n:
.
""lIa", d
air at 30ac. Calculate II'e co
.
Given:
Gr
2
=
94°C
i65 + 23
= --2-
I
.
2.196
-----
Heal and Mass Tran:..fer
Properties of air 0194 DC R: 95 DC:
I From HMT data book l'agc No. 33 ( ..
P
__ 0 .959
IXth~.....
....llIon)l
v ==
22.615 / 10-6 m2/s
Pr
0.689
k
0.03169
W/mK
We know that,
Coefficient of thermal 1
expansion J J3
==
94 + 273
~
7.22x
1t
x 0.10 x (165 - 23)
322.08 W/m
I
2.72 y 10-3 K-I
Grashof number,
Gr
2.72 x 1O-iJ0]
Result:
gxpxIYxLlT
v2
[ExlImple 3 A steam pipe 10 em OD runs 1I0rizontallyin a
10-3 x (0.10)3 x(165
(22.615 x 10-6)2
=>
23)
7.40 x 106
Gr
7.40 x 106 x 0.689
I Gr Pr
5.09 x 106 I
For horizontal
cylinder,
elise (i) :
Diameter of the pipe, D
Ambient air temperature, T so
Nusselt number,
322.08 W/m
room at 23 CC. Take outside temperature of pipe as J 65 cr'.
Determine tile heat loss per unit length of the pipe. If pipe
surface temperllture reduces to 80 cr' witlll.5 em insulation, what
is tile reduction in heat loss?
{Dec.2005, Anna University]
Given:
Gr Pr
==
I
[From HMT data book. Page No. 134 (Sixth Edition)]
=> Gr == 9.81 x2.72x
Heat loss per metre length, ~
Nu == C [ Gr Pr ]"
Surface temperature of the pipe, Til'
10 cm == 0.10 m
23°C
165°C
[From HMT data book, Page No. 137 (Sixth Edition))
=>
C == 0.48, and 1/1 == 0.25
5.09 x 106, corresponding
Gr Pr
r:N__
-:-u 0._48__:l:...,5.09
x 106 ]0.25
I Nu
22.79
Insulation thickness, t
I
We know that,
Nusselt number,
Nu
(
Scanned by CamScanner
Case (ii) :
Surface temperature of the pipe, T".
hD
k
BO°C
\.Scm
Tojind:
1. Heat loss per unit length of the pipe, Q.
2. % of reduction in heat loss.
==
O.OISm
2.198 Heat and Mass Transfer
Solution: Case (i): We know that,
Tf
Film temperature,
Tw + Too
2
=
----
2
94°C
f
----
~
r-I
Nu
0.48 [ 5.09 x 106 ]025
N-u--2-2.-79----,'
We know that,
165 + 23
IT
Convective Heat Transfer
Nusselt
number,
I
22.79
Properties of air at 94'[' 1:195'[' :
(From HMT data book, Page No. 33 (Sixth Ed' .
~
Ilion)]
p
== 0.959
v
= 22.615 x 10--6 m2/s
..
L..'
k
0.03169
L
Heat loss per} Q
unit length
L
Tf in K
h x 0.10
0.03169
h A ~T
7.22 x 7t x 0.10 x (165 - 23)
322.08
W/m
Case (ii) :
1
New diameter,
- 367
Ci.:--= 2.72 x 10-3 K-I
k
h x 7t x 0 x L x (T .. - Tao)
W/mK
1
94 + 273
hD
h~__:_7=.22=---.:.W:....:..:/m=2..::..;K~'
Q
Q
0.689
Grashof number, Gr
Heat loss,
kg/m''
Pr
Coefficient ofthennal }
expansion
P
Nu
0+2
0,
r
0.10 + 2 (0.015)
1
0.13 m I
= g x p x D3 x L\ T
v2
(From HMT deta book, Page No. 134 (Sixth Edition»)
=>
Gr
==
9.8Ix2.72x1O-3x(0.10)3x(l65
(22.615 x 10--6)2
Gr
==
7.40 x 106
23)
Gr Pr == 7.40 x 106 x 0.689
[Gr Pr
==
5.09 x 106]
Surface temperature
For horiZOntal cylinder
Nusselt number N
'
, u == C [ Gr Pr ]m
G
r Pr == 5.09 x 106,
Scanned by CamScanner
= 80aC
T .. +Too
Film temperature,
[From HMT data book, Page No.137 (Sixth Edition»)
corresponding
of the pipe, T..
C = 0.48, and 111 == 0.25
Tf =
2
~
2
2.199
1_}(){) Heal and Mass Transfer
Proputia of air at 5/.S 't:" .. so t :
p =
1.093 kg/m3
=
17.95 x I~
y
Pr
Coefficient
of the~al}
---- --
( 'onveclive Heal Transfer
Ileat
0.02826
II x 7t 0, L x (T .. - T"')
f-l
I
T/ in K
~
273
E
129.66 W/m
perunit}Q,
length
L
129.66 W/m
Q
0.1' reduction}
111 heat 10 s
Percentage
[p - 3.08 x 10-3 K-i_]
gxf-lxrY,
r=-
9.81 x 3.08 x 10-3 /. (013)2 / (80 _ 2:>
--.(17.95 x IO~)2
~
IGr
1.17xl071
Gr Pr
1.17 x 107 X 0.698
8.16 x 106
:I
Result :
2. % of reduction
= 8.16 x 106 ,
C = 0.48 and m = 0.25
corresponding
IFrom IIMT data book, Page No.1) 7 (Sixth Editionj]
Nu
= 0.48[8.16x
[ Nu
25.65 ]
Nu
25.65
~
[h
in heat loss
=
59.74%
pkue is maintained at a temperature of /30'(' in large lank full of
water {II 70 'e Estimate Ihe rate of heal inpul into Ihe plale
necessary 10 maintai« the temperature of lJO '(: IMIIY 2005, A VI
Given:
Horizontal
106)0.25
We know that,
Nusselt number,
= 322.08 Wlm
I Exumple 4 I A thin 80 em /ang and Il em wide horiZ/Jnlal
Nu = C [ Gr Pr )m
Gr Pr
59.74%
I. Heal loss per unit length of the pipe. ~
For honzomal cylinder,
Nusselt number,
L
-Q--X 100
L
322.08 - 129.66
322.08
x 100
yL\T
y2
[ Gr Pr
Q
L
W/mK
I
Gr
2 2U I
II A L\T
5.57 x 7t x 0.13 x L x (80- 23)
Ie
51.5
QI
m21s
0.698
expansion
Grahof number,
1(155.
To find:
= h 0,
T If'
130 e
Fluid temperature.
T",
7Uoe
0
Film t,!:npcrature,
T/
-2--
.!1_0 + 70
2
h x 0.13
Scanned by CamScanner
Plate temperature,
Rate of heat input into the plate. Q.
Solutlon :
= 0.02826
5.57 W/m2K
80 em = 0.80 m
8 ern = 0.08 m
T .. +T",
Ie
-
plate length. L
Wide. W
J
2.202
Convective Heat Transfer
Heal and Mass Transfer
-------
Properties of water 01 100 'r' :
[From HMT data book, Page No 2 I (S'
.
Ixth Ed' .
P = 961 kg/m!
- IIlon)1
Pr
hll X 0.04
0.6804
124.25
[From HMT data book, Page No. 29 (S' h ..
ixt Edlllon)j
Heat transfer coefficient}
for upper surface heated hll = 2113.49 W/m2K
For horizontal plate, Lower surface heated:
g x f3 x L x t1 T
J
Grashof number Gr
,
=
y2
Nusselt number, Nu
c
'" (I)
L,
=
L
=
Characteristic length = W
2
0.08
2
c
=
0.27 [Gr Pr ]0.25
[From HMT data book, Page No. 136 (Sixth Edition))
(From HMT data book, Page No. 134 (Sixth E '.
For horizontal plate:
dUlon)j
(J) ::::;>
hll Lc
k
Nusselt number, Nu
1.740
k = 0.6804 W/mK
f3waler = 0.76 x 1O-3K-1
I
We know that,
= 0.293 x 1 ()-6 m2/s
Y
We know that,
124.25
2.203
=
0.27 [0.580 x )09]
I Nu =' 4i061
We know that,
hi Lc
Nusselt number, Nu
k
0.04 m
L,
=
Gr
= 9.81 >:0.76x 10-3(0.04)3>:(130 .....70~
hi x 0.04
42.06
0.6804
(0.293 x 1~)2
715.44 W/m2K
Gr = 0.333 x 109
Gr Pr = 0.333 x I ()9 x l.740
[Gr Pr - 0.580 % IO')]
Gr Pr valve is in betw"'hn 8 I""
'.
"'"
/ v-and 10".
t.e., 8 / If)') <: Gr Pr <: 10"
.
face heated,
. So, for honzontal plate, upper
ussclt numL~r,
!J\:
Nu = (.15
) (Gr PrfJ.JJJ
nook p.
(hom II.\1T d;;r.;,
.' .
. u
' .g,e
.,
o. J 35 (S'Y.lh Edition»
= O. J 5 {O.580 / lO'lJO JJ3
Scanned by CamScanner
Heat transfer coefficient
for lower surface heated
hi = 715.44 W/m2K
Total heat transfer, Q = (hll + hi) A t1 T
(hll + hi) .;.W x Lx (T .. - T,,,,)
(2113.49 + 7 J 5.44) x (0.08 x 0.8) x (130 - 70)
[0
J 0.86 '" 10J W
I
Result: Rate of heat input into the plate, Q:o 10.86 x 10J W
·lO.J
Heat and Mass Transfer
Convective Heat Trun.ifer
I Example 5 I A hot plate 1.2 m wide, (J.35
III I,(.;/:-;;;;--
.
'11'
lid at
/l5"C is exposed 10 I h e am btent
sti tur (II 25 't: nile l
.
" ale tI'e
following:
,;1 !tftl\';mum velocity III 180 mm from lite lellding edo .1'
fteOJII'e
1'1
plate.
(ii) The boundary layer thickness at J 80 mm fro", the le d'
a
edge of the plate.
'"g
---------) Total mass flow through the boundary,
(v
.) Heat loss from the plate. Q.
2.205
in.
(VI
..) Rise
in temperature
of the air passing
through the
(VII
boundary,
tl T.
Solution: We know that,
T",+T."
TJ
Fluid temperature,
(iii) Local htat transfer coefficient at J 80 mm from lite leadillg
edge of the plate.
2
115 + 25
2
(iv) Average heal transfer coefficient over lite surface of the
plate.
(v) Total massflow through the boundary.
Properties of air al 70'(' :
(vi) Heat lossfrom the plate.
(vii) Rise in temperature of the air passing t"roug" the
boundary. Use approximate solution.
{May 2004, AIII,a Universityj
Given:
Wide,
W
1.2 m
Height or length,
L
0.35 m
T..
115°C
T
rf)
25°C
x
180 mm = 0.180
Plate surface temperature,
Fluid temperature,
Distance,
We know that,
Coefficient ofthen~al}
expanSIon
p
1.029 kg/mJ
v
20.02 x 10-6 m2/s
Pr
0.694
k
0.02966 W/mK
J3
I
TJ in K
I
In
I
70 + 273 = 343
Tilflnd :
(i)
Maximum velocity
the plate, "",ax'
(/I)
The boundary layer thickness
edge of the plate, 0) ..
(Ill)
Local heltt transfer coefficient
lending edge of the plate,
at
Average
over the surface
(III)
..
(From HMT data book, Page No. 33 (Sixth Edlllon))
at 180 mill from the leading
".r'
heat transfer
2.91 x )O--J K--I
at 180 mrn from the leading edge of
coefficient
plutc, h.
Scanned by CamScanner
180 mill from
the
of the
8. x p)( Xl)( ~ T
Grashof llumber.
Gr
v2
bo k Page No. 13
..
4 (Sixlh EdltlOn)1
[From ','MT "qat ax ~~l x (0.18)3 (115 - 2~
9 8 x 2. l
n.-/,\2
.
(20.02 x Iv-r
[9r .. 37.4 x 12£J
2.206
Heal and Mass Transfer
(i) Maximum velocity al180 mm from the leading edge, u
= 0.766 x v (0.952 + Prtl/2
u
x
[g f3 (Til' - ~]
0.766 x 20.02
X
[ GrL Pr = 1.90 x 1081
)( .\"112
Since GrL Pr < 109, now is laminar.
JO-6 [0.952 + 0.694J-112 x
J
GrL Pr value. is in between 104and 104.
9.81 x 2.91 x 10-3 (115 - 251 1/2
(20.02 x 10-6)2
x (O.18)IQ
-_-=-_-0-.4-0-6 -ml-s--',
[
~I u-
____------------=~~~~C~on~v~ec~(/~·ve~~~e~a~/~~~a~m~~~
=> GrL Pr = 27.5 x 107 x 0.694
1/2"'ox.
v2
max
~
.
i.e.,
So, Nusselt number, Nu
max
(ii) The boundary layer thickness at 180 mm from the t.
104 < GrL Pr < 109
(From HMT data book, Page No. m (Sixth Edition»
.
eadlng
edge of the plate, 0 :
Ox = 3.93 x x x Pr 0.5 (0.952 + Pr)0.25 (Gr): 0.25
[From HMT data book. Page No . 134 (Sixth Ed'itlonll
.
x
i
I
I
4.82 W/m2K
(iv) Average heat transfer coefficient
plate, I, :
Grashofnumber}
(for entire plate)
Gr L
L
Scanned by CamScanner
().()'2%6
We know that,
over the surface of the
_ g x p x L3 x ~ T
v2
_
27.5 x 107
I
m =
I
9.81 x 2.91 x JO-3 x (0.35)3 x (115 - 2~
(20.02 x 10-6)2
I Gr
h x 0.35
(v) Total mass flow through the boundary (,;,):
2 x 29.66
0.01229
!
Ie
x
x 10-3
! I
Ie
hL
I h = 5~86 W/m2K I
[From HMT data book, Page No. 134 (Sixth Edition)]
I
hL
2k
°
x
69.26
69 .26
edge of the plate, h x :
h
Nu
69.26
(iii) Local heat transfer coefficient at 180 mm from the leading
Local heat transfer}
.
coe ffi cient
0.59 (1.90 x 108)0.25
Nusselt number, Nu
(0.952 + 0.694)0.25 x (37.4 x 106)-025
~"--x -=-0.-0 1-22--'9-m--',
I
Nu
We know that,
3.93 x 0.180 x (0.694)-0.5
i '
= 0.59 (Gr Pr)025
[m
GrL
] 0.25
1.7 x P x v [ (Pr)2 (Pr + 0.952)
27.5 x 107
»25
1.7 x 1.029 x 20.02 x l~x [ (0.694)2 (0.694 + 0.952)J
0.00478
kg/s
I
(vi) Heat loss from the piale, Q:
Q
h A (T",- T..,)
For both sides, Q
2 x h x A (T," - T..,)
2 x 5.86 x (0.35 x 1.2) x (115 - 25)
[Q
443.01
WJ
n
"
i~
,
2.208
Convective Heat Transfer
Heat and Mass Transfer
I:
-(viii Rise in temperature of the air passing---;;;;;;--
I
We know that,
II
,
I
g
I
boundary (J T)
Heat lost, Q
=> 443.01
0.00478 x lOllS x 1\ T
Result:
of air at 356°( ~ 350°C:
Iht
[From HMl data hook, Page No. 33 (Sixth Editionl]
p = 0.566 kg./mJ
mCp~T
92.21 K
GT
=:>
----;;-erties
Pr -= 0.676
k = 0.04908 W/mK
0.406 mi.
umax
(ii)
Dx
0.01229
(iii)
hx
4.82 \\ Im·'K
(iv)
h
5.86 W/Ill 'K
(v)
m
0.00478
kg s
(VI)
Q
443.01
\V
( vii)
~T
92.21 K
Coefficient of } p = _1_
thermal expansion
Tf in K
III
1
356 + 273 - 629
I p = 1.58. io?J0J
Grashof Number.
I Example 6 I A large vertical plate 4 m height is maintainel
Given: Vertical plate length (or) Height,
9.81
Wall .ernperarure,
Til'
606°C
Air te nperature,
T
106°e
W
10m
I Gr
Solution ,
Heat transfer, t 0).
1.61 x 1011 x 0.676
Film temperature,
T
f
2
I.08xI01i]
I Gr Pr
Sin c
rPr > 109. flow is turbulent.
1- r turbulent
T +T
__
_1_"
flow,
==
Nu
eh Number
Nu
.
lI'r III
606 + 106
2
Nu
~
Scanned by CamScanner
,,2
1.6IxIO"]
Gr Pr
Wide
g x (~x I) x ~T
1.58 x 10-3 x (4)3 x (606-I06)
(55.46 x 10-6)2
Gr
4m
L
=
Gr
[From IIMT data book. Page No. 134 (Sixth Edition))
at 606'C (II1t1 exposed to atmospheric air tit J 06 'C Calculate II.t
heat transfer if the plate is 10 In wide.
I Oct. 2001, MUI
• I
= 55.46 ' 10-6 m2/s
v
I
(i)
To find:
2.209
0.10 [Gr PrJO333
IIMT data boO~, P ag
_ 01011.08
.
e
1(1
.
135 l ixth
10"J·
diti n)]
1
I
2.110
------
Heal and Mass Transfer
We know that,
Nusselt Number,
hL
Nu
Ie
hx4
0.04908
472.20
I
[ Heat transfer coefficient,
Heat transfer,
h
S.78 W/m2K]
Q
hA~T
I!
____-----------~~~::c~o~~~~t'~·ve~H~e~a~t~~r~a~~~~er~22~.2~l
v 0.264 x 1~ m2/s
Pr
1.55
Ie
0.683 W/mK
0.8225 x 10-3K-I
P(for water)
[From HMT data book, Page No. 29 (Sixth Edition)!
h x W x L x (T w - T
Q»
GrashofNumber,
g x P x L3c x ~T
Gr
'" (1)
I
Result:
F or horizontal
plate,
Q
S.78 x 10 x 4 x (606-106
IIS600 W
)
IQ
IIS.6
x
103
WJ
Characteristic
length, L"
Q
IIS.6
x
103 W
Heat transfer,
I Example 71 A thin 100 em long and 10 em wide horiz.ontal
plllte is maintained at a uniform temperature of 150 CC in a large
tIlnkfull of water at 75 'C. Estimate the rate of heat to be supplied
to the plate to maintain constant plate temperature as heat is
dissipaud from either side of plate.
[Oct. 2000, MU/
Given: Length of horizontal
Tofind:
Solution:
plate, L
100 cm = l rn
Wide,
W
10cm
Plate temperature,
Tw
ISO°C
Fluid temperature,
Too
7SoC
= 0.10m
T/
= 0.10
2
2
I
0.05 m
(1) ~ Gr
9.81 x 0.8225 x 10-3 x (0:05)3 x (150 -75)
(0.264 x 1O-~)2
"-1 G-r--l-.0-8-53-x-l-09---.1
1
Gr Pr
1.0853 x 109 x 1.55
Gr Pr
1.682 x 109 I
Gr Pr value is in between 8)( 106 and 1011.
i.e., 8 x 106<GrPr<
1011.
For horizontal plale, uppersurface healed:
Nusselt Number, Nu = 0.15 (Gr Pr)0.333
Heat loss (Q) from either side of plate.
Film temperature,
W
[From HMT data book, Page No. 135 (Sixth Editiom]
Nu = 0.15 [1.682 x 109)0.333
=
150 + 75
2
[!\Ju = 177.13]
~
We know that,
hll t,
Properties of water at I 12.5°C :
Nusselt Number, Nu =
[From HMT data book, Page No. 21 (Sixth Edition))
p
=
9S1 kglm3
177.13 ==
T
h x 0.05
~
0.683
/
Scanned by CamScanner
j :.
2.2 J 2
Convective Heat Transfer
Heal and Mass TramJer
8
~{Imple
Upper sun,face heated. ' heat transfer coefficient
hu = 2419.7 W/m2K
For horizontal plate, lower surface heated:
Nusselt Number Nu
I ~ hot ~/ate 20 em in /reig/" lind 60 em wide is
. osed to tile ambient a" at 30 't'. Assuming the temperature of
eXP late IS. mainuunec
.,
I at 110CV"\". F'lnd the heat loss from bot"
tire P
-tace of the plate. Assume horizontal plttte. IApril2003, MU/
sur)'
Given:
Height (or) Length of the plate, L
20 cm
0.20m
0.27 [Gr Pr]O 2:i
Wide, W
60cm
[From HMT data book, Page No. 136 (Sixth
. Ed't'
lion)!
I
,
=
0.60 m
I
i
Nu
0.27 [1.682 x 1091°25
[NU
54.68
"
I
To find:
We know that,
2.2 J J
Fluid temperature,
T <'J'J
30°C
Plate surface temperature,
T",
110°C
Heat loss from both the surface of the plate (Q).
T", + Too
Solutio" :
Nusselt Number, Nu
Tf
Film temperature,
=
--2-
110 + 30
hlx Lc
54.68
2
k
\r,
hi x 0.05
54.68
{From HMT data book, Page No. 33 (Sixth Edition)!
p = 1.029 kg/m3
heat transfer coefficient
hi =
Total heat transfer,
of air at 70°C:
W/m2K
746.94
Lower surface heated,
Properties
0.683
746.94 W/m2K
v
= 20.02 x 10-{' m2/s
Pr
= 0.694
k
We know,
Q = (h'l + h,) x A x ~T
= (h" + hi) x W x L X (Tw - Too)
Coefficient of l
thermal expansion J ~
0.02966 W ImK
I
I
T I in K == 70 + 273
\
= (2419.7 + 746.94)
x 0.10 x (150 -75)
343
:::: 2.9 \ x 10--3 K--1
UL 23,749.8 W I
==
Result:
Heat transfer,
Q
=
23,749.8
W
We know,
Scanned by CamScanner
1,.~}~/4~~#~oa~/~a~nd~U~~~s~TI~ro~m~fi_er
:-~~~
~
g x 13 x L~ x.1T
GrashofNul1lber. Gr =
v2
_
__
---------'" (I)
Convective Heat Transfer
For horizontal plate, lower sUrface heated:
Nusselt Number,
[From HMT data book, Page No. 134 (Sixth E ..
Characteristic
where
length ==
i
2.
Nu =
0.27 (Gr Pr)025
0.277 [1.06 x IOS]0.25
I'-N7"u--2-8-. 1
dlhon))
5-1
We know that,
~
= OJOm
0.30m
(I)~
Nusselt Number,
I
9.81 x 2.91 x 1(}-3 x (0.30)3 x (I 10 _ 30
(20.02 x 1Q-6)2
:JQl
I"-G-r--.l.-53-8-4-x-I-Os--'1
1.5384 x ]OS x 0.694
/ Gr Pr
1.0676 x lOS
i.e., 8x 106<GrPr<
2.78 W/m2K
Lower surface heated, heat transfer coefficient
I
Gr Pr value is in between 8 x 106 and 1011.
hi
Total heat transfer, Q
]0".
0.15 [1.0676 x IOSJ0.333
/Nu
=
70.72
Nusselt Number, Nu
=
hu Lc
I
We know that,
70.72
k
hu x 0.30
0.02966
6.99 W/m2K
Upper surface heat d' h
e, eat transfer coefficient
-
hu ==
Scanned by CamScanner
2.78 W/m2K
(hu + hi) A.1T
= (6.99 + 2.78) x 0.60 x 0.20 x (110-30)
I Q = 93.82 W I
Q
0.15 (Gr Pr)0.333
[From HMT data book, Page No. 135 (Sixth Edition)]
NI,J
=
(hu + hi) x W x Lx (T; - TaJ
For horizontal plate, Upper surface heated,
Nusselt Number, Nu =
hi Lc
k
hi x 0.30
0.02966
28.15
Gr
Gr Pr
Nu ==
6.99 W/m2K
Result:
Heat transfer from both surface of the plate == 93.82 W
I Example 9 I A horizontal plate 1 m x 0.8 m is kept in a
water tank with the top surface at 60°C providing heat to warm
stagnant water at 20°C Determine the value of convection
coefficient.
IBharathiyar University, Nov. 96/
[Procedure
is same as Example 7J
I Example 10 I A vertical pipe 80 mm diameter and 2 '"
height is maintained at a constant temperature of 120'C. The
pipe is surrounded by still atmospheric air at 30 'C. Find heat loss
by natural convection. '
IManonmanium Sundaranar Univ~rsity,Nov. 97/
2.216
Heal and Mass Transfer
Given:
---
~turbulent
Vertical pipe diameter. ~80
. h
D = RO rnm -0 -----2 rn
- .080 m
Herg t (or) Length. L
Surface temperature.
T II'
.
120 e
Air temperature,
T a:
30 e
for
Convective II
flow,
Nu
[From HMT data b 00k , Page No. 135 .
== 0.IO(3.32x 1010]0333
(Sixth Editionll
ffiu.__
0
31:..:...:.8 . .:::___j8\
We knoW that,
Solution : We know that ,
hL
k
Nusselt Number, Nu
Film temperature,
r, + T<:IJ
T/
2
120 + 30
2
[17-::=-75oe
~
3 \8.8
0.03006
h
4.79 W/m2K
h x A x ~T
h x 'It x D x L x (T
Heat transfe! coefficient,
Heat loss, Q
I
I\' -
Properties of air at 75°e :
1.0145 kg/rn-'
v ._
20.55 x 10-6 m2/s
Pr
0.693
k
0.03006 x 10-3 W /mK
1
Tf in K
1
75 + 273
1(3
Gr
=
Q
Result:
1:::-__
Gr
Gr Pr
transfer rate from
I. First half of the plate.
3 -IJ
g x P x L3 x ~T
v2
3.
x 1~)2
4.80 x 1010 x 0.693
Scanned by CamScanner
Full plate.
Next half of the plate.
. 2 158 W; 3. 46.3W 1
(Ans: I. 111.79 W, . 25 mls. Tbe plate
at
2. Air at 250C floWS past a flat plate. .' d at a uniform
d i malotaUle
measures 600 mrn x 300 mm an IS
frolll the plate if the
5
temperature of 950C. Calculate the heat 1055 h this heat \05 be
air flows . parallel to the 600 mm side. llltotel.·
HoWIllhuc3noI11Ill side?
2.
9.81 x 2.87 x 10-3 x (2)3 x (120
1
FOR PRACTICE
of2 m1s over a plate maintained at 100°C. The lengthandwidthof
the plate are 800 rnrn and 400 mm respectively. Calculateheat
2.87 x 10- K
Gr Pr == 3.32 x 1010 J
Since Gr Pr > 109, fl'ow IS turbulent..
216.7 W
216.7 W
e and at atmospheric pressure flowSat a velocity
= 2.87 x 10-3 K-I
__:___~(20.55
4.80 x LOIOI
Heat loss, Q
2.13. PROBLEMS
l. Air at 200
[From HMT data book, Page No. 134 (Sixth Edition))
=
T co)
= 4.79 x 'It x 0.080 x 2 x (120 - 30)
p
We know,
2.2/7
0.10 (Gr Pr]Om
0
Tofind: Heat loss (Q).
eat Transfer
3Q2
affected if the flow of air is made para e [AIlS: 100.5 W; l42 Wl
2.218
Heat and Mass Transfer
~RK
3. A thin plate of length 1 m is placed longitudinally i
stream flow of water. Calculate the mean heat transfe
n a free
r coeffi .
and the rate of heat flow from the plate, if it is kept at S00C. Clent
[Ans:
~.14.
I·
3 kW/m2K 23
'
• X 103 W
4 Air at atmospheric pressure and at a temperatur
f
]
.
e 0 3SoC
flows over a heated cylinder of 50 mm diameter whose SUrf:
.
°
D
.
h
I
f
h
aCe
IS
maintained at 150 C. etermme t e oss 0 eat from the I'
.
..
cy IOder
if the air velocity IS 50 mls.
[Ans: 3260 W/ ]
•
Ill
5. Water at 10°C with a free stream velocity of 1.524 IllIs
flows across a cylinder of 2.54 ern diameter whose sUrface is kept
at 65.6°C. Compute the average heat transfer coefficient.
[Ans:
7275 W/m2K]
6. Air at 27°C flows across a heated 30 mrn diameter pipe at
77°C with a velocity of 1 m/s. Compute the heat transfer rate per
unit length of pipe.
[Ans: 84.5 W/m]
7. Find the convective heat loss from a radiator 0.5 m wide
and I m high maintained at a temperature of 84°C in a room at
20°C. Consider the radiator as a vertical plate.
[Ans: 110 W]
, ,
. I;
, I
I
8. A horizontal steam pipe of 0.1 m diameter is placed
horizontally in a room at 20°C. The outside surface temperature is
80°C and the emissivity of the pipe material is 0.93. Estimate the
total heat loss from the pipe per metre length due to free
. an d ra diration,
.
[Ans' . 2617 W]
convection
. 20 em x 30 em IS
. use d as a wa ter heater-. in a
9. A plate of size
process plant. The temperature of water is 20°C, while the heater
. the heatt
plate is maintained at a temperature of 120°C. Determme
'
"
id
f
heat(fr
is kW]
kep
transfer rate by free convection when 20 em Sl e 0
vertical.'
[Ans : 20
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QUESTION
Jf'hat
is dimensional analysis ?
. .
engltl
. d resistances,
of f1UI
.
thermodynamics.
2.
{Nov. 96, MUj
. sional analysis IS a mathematical h
.
Du:tletl
d
f
.
met od which mak
of the stu y 0 the dimensions for I'
es
use
bl
ThiIS method can be ap so
. eering pro ems.
redVlng several
'
State B
uckingham
P I toall types
heat flow problemsin fluidm han'
ec ICS and
1r theorem.
lAp,. 97, MUj
kingham 1t theorem states as follows: "If thereare n
Suc
.,
II h
.
. bles in a dimensiona y omogeneous equationand if
varIa ntain m fundamenta I d'unensions,
. then th e variables
.
are
these cO
.
.
arrange d into (n - m) dimensionless terms. These
.
sl'onless terms are called 1t terms.
dlmen
3. What are a II tire advantages of dimensionalanalysis?
1. It expres ses the functional relationship betweenthe
variables in dimensional terms.
. .
·
up a theoretical solutionin a simplified
2. It enables ge tt 109
red to a large
. f tests can be app I
3 The results of one senes 0
·th the help of
.
. ilar problems WI
number of other Simi
dimensionless
form.
dimensional
analysis.
.
. alanalysis?
if dimension
4. What are all the limitatIOns 0
,
'ded by dimensional
hiIp
ation IS. not proVI
relations
1. The complete mlOrm
th t there is some
. dicates a
analysis. It on IY 10
.
f
eters.
echanlSOl 0
between the param
t the internal01
. . iven abou
2. No informatlo~ IS g
dingthe
enon.
.
I
h
nom
y clueregar
h
P ysica p e
f.~ "
t givean
'~iS'-doesno
,
3. Dimensional analy I
selection of variables.
. e
2.220
5.
Convective
Heat and Mass Tran:-,fer
Deline Reynolds number (Re).
'./'
.'
IMay 2005, June ;----006
.
,All}
It is defined as the ratio of inertra force to viscous fore e.
Inertia force
Re == Viscous force
6. Define Prandtl number (Pr).
{May 2005 A V, June 2006 A V, Oct. 98, Apr. 2002, MUI
It is the ratio of the momentum
diffusivity
to the thermal
Heat Transfer
2.221
10. What is meant by Newtonian and non-newtonionfluids ?
The fluids which obey the Newton's law of viscosity are called
Newtonion fluids and those which do not obey are called nonnewton ion fluids.
11. What is meant J,ylaminar flow and turbulentflow ?
Laminar flow: Laminar flow is sometimes called stream line
flow. In this type of flow, the fluid moves in layers and each
fluid particle follows a smooth continuous path. The fluid
particles in each layer remain in an orderly sequence without
mixing with each other.
diffusivity.
Pr ==
Momentum diffusivity
Thermal diffusivity
Turbulent flow
~
7. Define Nusselt Number (Nu).
Laminar flow
IDec. 200J A U, Apr. 97, 98, MUI
~,
I
r
It is defined as the ratio of the heat flow by convection
process
under an unit temperature gradient to the heat flow rate by
conduction under an unit temperature gradient through a
I
.
;
stational)' thickness (L) of metre.
qconv
Nusselt Number (Nu)
I
t
j
i I
\
I
, f
8. Defme Grashof number (Gr).
It is defined as the ratio of product
qcolld
IOct. 97, 99, MUI
of inertia force and
buoyancy force to the square of viscous force.
Gr == Inertia force x Buoyancy force
(Viscous forceY
9. Define Stanton number (SI).
IDee. 2005, /.U/
It is the ratio of Nusselt number to the Jlr~duct of Reynolds
I
,I
number and Prandtl number.
Nu
,
Re x Pr
St == --
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Time
Turbulent flow: In addition to the laminar type of flow, .a
distinct irregular flow is frequently observed in nature. This
type of flow is called turbulent flow. The. path of any
individual particle is zig-zag and irregular. FIg. shows the
instantaneous velocity in laminar and turbulent flow.
12. Whl.' is lIydrodynamic boundary 'layer ?
fl .d i I
In hydrodynamic boundary layer, velocity of the UI IS ess
than_990/0offree stream velocity.
13. Wha! is tllermal bountlary layer?
. . I than
l',lu.,cl.hem1albO.undary layer, temperat\lre ofthp- fluid IS ess
lOci. 98, Mu,r,I
14. Define convectIOn.
f that will occur between
Convection is a process o.fheatd~:s :en they are at different
a solid surface and,a fluid me IU
. '~9% of free stream temperature.
. I.
.
temperatures.
•
.
Heat transfer from the mov ing fluid to solid surface i
.........
.
u~eqUitlOO
----
.
s gIven by
Q = h A (T; - T.J
This equation is referred to as Newton's
wbere
___
Convective Heat Transfer
convection
law of Cooling.
Local heat transfer coetlicient
Surface area in m2
in W Im2K
Surface (or) Wall temperature
in K
TGO -
Temperature
of fluid in K
16. Wiat is Iftealll by free or natural convection?
(May 2004, Dec. 2004, June 2006, May 2004 AU,
IOct.1999, MU/
1. . Reynolds number (Re).
3.
T t10 is given by
,
IDec. 2005, Dec. 2004, June 2006, AU/
In the boundary layer concept the flow field over a body is
divided into two regions:
• A thin region near the body called the boundary layer
where the velocity and the temperature gradients are large.
• The region outside the boundary layer where the veloci~
and the temperature gradients 'are very nearly equal to their
free stream values.
23. An electrically heated plate dissipates heat by convection at a
rate of 8000 Wlm2 into the ambient air at 25,\:'./fthe surface
of the hot plate is at 125~, calculate the
transfer
r:
coefficient for convection belween the plate and air_
,
{May 2005, May 2006, AU/
_
(Nov. 1994, MUj
of equation
used to calculate
for flow through cylindrical pipes?
Nu
n =
n =
Surface temperature, T w :::
[Oct: 1999, MUj
Tofind':' Heat"tl'flos-fer--coeffic'ielrt,
We know that,
- hA(T
Heat transfer, Q -
Solution:
0.4 for heating of fluids.
0.3 for cooling of fluids.
40
.
250C + 273 :::298 K
l250C + 273 :::39~ K '
Ambient temperature, Ta:;
heat transfer
0.023 (Re)O.8 (PrY'
Scanned by CamScanner
8000 W/m2
Given: Heat dissipation, Q
Ans : Q = hA (T w - Too)
19. What is the form
{May 2004" AU/
22. Indicate the concept or significance of boundary layer.
convection.
18. According to Newton's
law of cooling the amount of heat
transfer from a solid surface of area A at a temperature Tw to
bOllndary layer thickness.
The thickness of the boundary layer has been defined as the
distance from the surface at which the local velocity or
temperature reaches 99% of the external velocity or
temperature.
If the fluid motion is produced due to change in density
resulting from temperature gradients, the mode of heat transfer
is said to be free or natural convection.
(May 2004, Dec. 2004,. June 2006 AU, Nov. 96, Apr. 98, MUJ
If the fluid motion is artificially created by means of an
external force like a blower or fan, that type of heat transfer is
known as forced convection.
Prandtl number (Pr).
11. Define
Nov. 96, Oct. 97, MUJ
aJluid at a temperature
1
2. Nusselt number (Nu).
It A T... -
17. What is forced
2.213
,0. "'hilt are the dimensionless parameters used in forced
(h
-Ta:;)
IV
r ",.'
[
I
2.224
Heat and Mass Transfer
Convective Heat Transfer
h x 1(398 -298)
8000
:::)
= h x 100
h = 80 W Im2K
Result: Heat transfer coefficient,
2.225
26. Define displacement thickness.
h = 80 W Im2K
U. Write down the momentum
equation for a stead
.
·bl
[Y, two
dimensional flow 0if an mcompressi
e, constant pf',
. Jl·d·
operty
newtonwn
UI
m th e rec tId·
angu ar coor mate system
mention the physical significance of each term.
flIrd
Th~ displacement thickness is the distance, measured
perpendicular to the boundary, by which the free stream is
displaced on account off ormation of boundary layer.
27. Define momentum thickness.
The momentum thickness is defined as the distance through
which the total loss of momentum per second be equal to if it were
passing a stationary plate.
{June 2006, Anna University}
28. Define energy thickness.
Momentum equation,
I
au
\
The energy thickness can be defmed as the distance, measured
perpendicular to the boundary of the solid body, by which the
boundary should be displaced to compensate for the reduction in
kinetic energy of the flowing fluid on account of boundary layer
au )
(
. (au
au ]
where, P u ax + ay
p u ax + V "By
V
Inertia forces.
formation.
Body force.
ap
ax
00
Pressure force.
a- 2u + -a2u = Viscous forces.
ox2 ay2
25. Sketch the boundary development of aflow.
!~ ::
•
~:
!:
:
~
I
I
~
I
I
I
I
I
I
~
I
~
'y
~U!
,
Laminar
I Tranii- I
boundary layar---71Ion"""__ Turbulent
boundary
layer
--I
l-U
~_~I-=~~
:
\~
~I
u;;'fJl '.
1
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,
, CHAPTER~III
3. PHASE CHANGE HEAT TRANSFER
AND HEAT EXCHANGERS
3.1 Boiling and condensation
3.1.1 Introduction
In the last chapter of convective heat transfer, we have
considered the fluid as a homogeneous single phase system. But, in
many situations, the fluid changes its phase during convective heat
transfer process. Boiling and condensation are such convective heat
transfer process that are associated with change in phase of liquid.
3.1..2 Boiliag
The change of phase from liquid to vapour state is known as
boiling
3.1.3 Condensation
The change of phase from vapour to liquid state is known as
condensation.
3.1.4 Applications
Boiling and condensation process finds wide applications as
mentioned below.
1. Thermal and Nuclear power plant.
2. Refrigerating systems.
3. Process of heating and cooling
4. Heating of metal in furnaces
5. Air conditioning systems.
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' ."
-----------------------------
~
----
J. 2 Heal and Mass Transfer
3.l.5 8oili~g beat tran~fer phenomena
Boiling is a convection process involving a change f
O
from liquid to vapourstate, This is possible only when the tem phase
of the surface (Tw) exceeds the saturation temperature o~~tu~
{TsaV'
.
qUid
Boiling and Condensation. 13
c
S,g
~e
Nucleate
boiling
c~
->
Filmbollin9
Ql
I
II
III
107
IV
V
VI
According to convection law,
Q = hA (Tw- TsaV
Q=hA(~T)
where
~T = (TwTsat> is known as excess temperature.
If heat is added to a liquid from a submerged solid surface
the boiling process is referred to as pool boiling. In this case th~
liquid above the hot surface is essentially stagnant and its motion
near the surface is due to free convection and mixing induced by
bubble growth and detachment.
Fig. 3.1 shows the temperature distribution in saturated pool
boiling with a liquid - vapour interface.
.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:y~~~.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:
..
.....................................................
.
~.~.~ ~
Vapour~_-
~'<i' _ ~
--_
-
_- s .:_-__
-_
- -~.:-
_-_-§_-_- .r:0._- .D_- _-9..- _- _- _- _- _.0_-_- ----13.....: Liquid
~-~---------------~---bubbles -_-_-_-_-_-_-_-_0_-_-_-_ -<:_-:_-_-2- .z:': -_0_-_- _-_-_-_
___
-0
-0-
0_
-
-
/;?;?~;;;;;?;;;;?;?;);;;;;;;;?;;?;;;
Solid surface
.
Fig 3.1 Pool Boiling
I .
I '
The different regions of boiling. are indicated in figJ.2.
This specific curve has been obtained from an electrically heated
platinum wire submerged in a pool of water by varying its surface
temperature and measuring the surfac,e heat flux. (q).
Scanned by CamScanner
10L_----~--------L-----~----~-100
50
10
1
Excess Temperature
150
6 T e = T w - T sal
- Free convection
11 - Bubbles condense in super heated liquid
III
- Bubbles raise to surface
IV
- Unstable film
V
- Stable film
VI
- Radiation
coming into play
F;g~ s.: Pool Boiling Curve for Water
1. Interface evaporation
,
.
ocess with no bubble
Interface evaporation i.e., evaporation pr
.
I.
. the excess temperature
formation exists in region l. In tms region
"' ..
.
small (SoC). Here the liquid near the surface tS super
~ T IS very
.
h r id surfllce.
heated slightly, and evaporation takes p'ace at t e iqu .
3.4 Heat and Mass Transfer
2. Nucleate Boiling
j
I
I
----
This type of boiling exists in regions II and III Th
.
. e nucle
boiling begins at region II. As the excess temperature is f ate
increased, bubbles are formed more rapidly and rapid eva un~et
. d'icate d .10 region
. III . N ucleate boilinPOtation
takes place. This. .IS 10
.
.
g eXIsts
upto L\T = 50OC. The maximum heat flux, known as critic I h
.
a eat
flux, occurs at point A.
I~
. t
I'
[From HMT dOlo book
page No. I 42(Sixlh edilion)]
Q = ~I hfg \ g x (PI -Pv)\o.s x \
neatflUX
a.
l
A
Cp, x AT \~
.
lCsf hfgP;l ".(3.\)
c
x
;
= q = heat flux,
Wlrn2
Film boiling exists in regions IV, V and VI.
In region IV the vapour film is not stable and collapses and
reforms rapidly. With further increase in L\T (excess temperature)
the vapour film is stabilised as indicated in region V.
'
The surface temperature required to maintain a stable film are
high and under these conditions a sizeable amount of heat is lost by
the surface due to radiation. This is indicated in VI.
From fig.3.2 it is clear that high heat transfer rates are
associated with small values of the excess temperature in nucleate
boiling region.
I,
V
leate Pool Boiling
where
3. Film Boiling
I:
1.J-luc
11,- Dynamic viscosity of liquid,Ns/ml
hfg - enthalpy of evaporation, l/kg
g _ Acceleration due to gravity,9.81 rn/s2
P, - Density of liquid, kglm3
Pv - Density of vapour, kglm3
o - Surface tension for \iquid vapour interface,N/m
specific heat ofliquid, J/kg K
Cpl-
CsJ - Surface fluid constant
3.1.6 Flow Boiling
Flow boiling or forced convection boiling may occur when
a fluid is forced through a pipe or over a surface which is
maintained at a temperature higher than the saturation temperature
of the fluid.
This type of boiling occurs in water tube boilers
forced convection.
involving
3.l.7 Boiling Correlations
It is obvious from the boiling curve that various physical
mechanisms are involved in different regions and there will be
correspondingly many types of correlations for the boiling procesS .
. Some of them are given below.
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P r - Prandtl Number
~ T _ Excess temperature = T w - Tsat
T _ Surface temperature, °C
w
T
_ Saturation temperature, °C
sat
d 1 7 for other fluids.
n = \ for water an .
b. Critical heat Flux
9_= 0.\8
A
hfg Pv
\~O 25
~a)( g(P/- PvJ
l---P;-
•• , (3.2)
3.6 Heal and Mass Transfer
c. Excess temperature
.1T = Tw - Tsat < 50°C for Nucleate
pool boiling
-----
,:::----------;here
cr - Stefan Boltzmann constant = 5.67 x 10--8 W/m2K4
E -
d. Heat transfer, Q = m x hfg
... (3.3)
2. Film Pool boiling
_:B:o:il/~·ng~a~nd~C~o~n~~~m~a~t/~·M
emissivity
Tw - Surface temperature, °C
T sat - Saturation temperature, °C
[From HMT data book page No. 142 (Sixth edition)J
b. Excess temperature
a. Heat transfer co efficient
h = hconv + 0.75 hrad
...
(3.4)
3.1.8 Solved Problems
III Water is 10 be boiled (II atmospheric pressure in a polished copper
h
cony
= 0 62 [k~
.
x Pv x (p,-
p)x g x (hfg + 0.4 Cpv ~T) ]025
pan by means of an electric healer. TI,e diameter of the pan is
D.1T
0.38 m and is kept at I 15" C. Calculate the following
~IV
I..
where
I. Power required 10 boil the water
(3.5)
of vapour, W ImK
k, - Thermal conductivity
2. Rate of evaporation
3. Critical heat flux.
Pv - Density of vapour, kglm3
Given:
P,- Density of liquid, kg/rn-'
Diameter, d = 0.38 m;
g - Acceleration
Surface temperature, T w = 1150 c.
due to gravity, 9.81 m/s-'
"Jg - Enthalpy of evaporation J/kg
Tofind:
Cpv - Specific heat of vapour at constant
Jlv - Dynamic viscosity of vapour,
D - Diameter,
pressure
Ns/m?
2. Rate of evaporation, (m)
Q
111
6 T - Excess temperature
1. Power required, (P)
3. Critical heat flux, (A)
= 1'w - T sal
... (3.6)
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!
1
IlolI/nR and ('und,n.wIIQn
100'"
Specific volume of vapour v
• R
Dcnsily of vapour, I)"
.9
1.673 mJ/kg
..L
v
il
_11.673
copper
pan
Ii0RICl'
~T·
Excess tempermure • Tw - TSIII • II sn - 100 • IS° C
16'1'. 15° I 50° . So this is Nucleate pool boiling process.
Fig 3.3
1. Power required to boil the water
We know that. saturation temperature of water is 100° C.
For Nucleate pool boiling
i.e·1 T sal = 100° C ,
Heal flux,
t
= 111 x hfg
[g x (~/-PV)]O~
Properties of water at 100° C.
[From HMT data book page No. I 42 (Sixth edition)]
[From IIMT data book page No.21.
Sixth edition]
Density, P, = 961 kg/m!
Kinematic viscosity, v = 0.293 x I O~ m2/s
Prandtl Number, P, = 1.740
Dynamic viscosity, J.l, = PI x v = 961 x 0.293 x 1~
= 281.57 x 10-6 Ns/m2
[R.S. Khurmi Steam table page No.4}
At JOO°c.
I cr = 0.0588 N/m I
n = 1 for water
"Jg = 2256.9 kJ/kg
hfg = 2256.9
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[From HMf data book page No. 144]
x
103 J/kg
[From HMT data book page No. J4J]
Substitute
~"hfg, PI. PV' o, Cpb ~T, Csf hfg' nand Pr values in Equn (I)
Q
(1) ::::) -
A
AI JOO°C
Enthalpy of evaporation,
where c = surface lension for liquid vapour interface
For water - copper::::) Csf = surface fluid constant = 0.013
Specific heat, Cpl = 4216 J/kg K
From Steam Table
[ Cpl x ~T ]3 ... (I)
Csfxhfg P;
= 281.57 x 10-6 x 2256.9 x 103 x
[9.81x (961-0.597)] 0.5
0~~8
4216 x 15
[
x
0.013 x 2256.9 x 103 x (1.74)1
]3
l3J,/~O~H~ea~t~a~n~d~U~as~s~~~a~n~~~e~r
Heat flux,
-t
____
Boili"g and COfldemation J. II
= 4.83 x lOS W/ll12
= 0.18 x 2256.9 x 103 x 0.597
Heat transfer, Q = 4.83 x lOS x A
x[ 0.0588 x 9,81 x (961 _ 0.597>]0,25
f-d
= 4.83 lOS f (0.38)2
=4,83
Q =54.7
Q
2
x 105 x
x
x
x
103 W
(0.597)2
Critical Heal flux, q =
-t
= 1.52 x 106 W/m2
= 54.7 x 103 = P
Result:
I Power = 54.7 x 103 Wi
I. P = 54.7 x 103 W
2. Rate of evaporation,
(,;,)
2.
We know that,
3. _2. = q = 1.52 x 106 W/m2.
Heat transferred, Q =';, x
~
.
m = 0.024 kg/s
A
Jrg
[l) Water is boiled at the rate 0/ ].I kg/h in a polished copper
Q
m=-
pan, 300mm in diameter, at atmospheric pressure. AsslUninl
hlg
nucleate boiling conditions, calculate th« temperature of the
54.7 x 103
bottom sur/ace 0/ the pan.
2256.9 x 103
Given:
~.024kglsl
Mass flow rate, ,;, = 24 kglh
3. Critical heat flux
_ 24 kg
- 3600 s
For Nucleate pool boiling, critical heat flux,
A = 0.18 hI"
I,;. 6.6 10- kgls I
:t'''') j
=
O.25
Q
P, a' g
[
Diameter, d = 300mm
3
l'
c
0.3m
Toflnd:
[From HMT data book page No /42}
Surface Temperature. Tit'
41
Scanned by CamScanner
I
3.12 Heal and Mass Transfer
Boiling and Condensation 3.1J
Solution:
For Nucleate boiling
We know that,
Heat flux,
Saturation temperature of water is 100° C,
.2_= 11 xh
i.e. Tsat = 100° C.
A
I
[gX(PI-PII)]0.5x
'fg
[ Cp/x LiT]3
C
h
0'
Heat transferred, Q =m x h_rg
0.293 x 10--{)m2/s
Kinematic viscosity, \I
(I)
[From HMT data book page No. 142 (Sitth editiOlt)J
{From HMT dolo book page No. 21 (Sixth edition)}
961 kg/m!
...
sf x ,/g Pr
Properties of water at 100° C.
Density, PI
.
II
g_=~
A
A
=>
Prandtl number, P, = 1.740
6.6 x 10-] x 2256.9
Q
Specific heat, Cpl = 4216 J/kg K
A
x
103
.!!.. d2
4
Dynamic viscosity, III
= PI x V
6.6 x 10-] x 2256.9 x 103
= 961 ~ 0.293 x 10-6
.!!.. (0.3)2
4
\ .~I
~
From Steam table
2~ 1.57 x 10-6 Nszm21
-t-
[R.S. Khurmi Steam table page No. 4J
= 210 x 10] W/m2 \
At 100°C
Enthalpy of evaporation,
hfg
I }Jfg = 2256.9
Specific volume of vapour,
I
c = Surface tension for liquid vapour interface
= 2256.9 kJ/kg
x 103 J/kg
Vg = 1.673 m3/kg
I
I
At
100° C
[From HMT data book page No.144J
10 == 0.0588 N/m \
For water - copper => Csi == surface ~uid constant = 0.013
,
\'
Density of vapour,
\Csi <= 0.013 \
Pv
(From HMT dota book page No.143]
"g
I
n == I for water
1.673
b
Substitute
== 0.597~
Scanned by CamScanner
Equation (I)
11/,
hfg> QI' Pv' 0, Cpl' hlg,
t,
nand Pr values in
~3.iI4~H~w~t~and~M~~~s~~~mu~ife~r
---------------(1)::::)
210 x 103
-= 281.51 x I~
x 2256.9 x 103
Tofind:
x {9.81 x (961-0.591)1°·5
l
Voltage, (V)
0.0588
Sollltion:
4216 x &T
,[
x
Boiling and Condensation 115
____
]3
We know that, saturation temperature of water is 100° C.
0.013 x 2256.9 x 103 x (1.14)
i.e., Tsat = 100° C
I
il·
4216 x 6T)3 = 0.825
( 5105 l.l
Properties of water at 100° C.
::::) [0.0825 6T]3
= 0.825
V
0.0825 6T
= 0.931
Pr=1.740
!6T=11.35°CI
Excess temperature, 6 T = T w - T sat
1l.35 = T w - 100° C.
I Tw= III.JSOC I
Result :
Surface temperature, T w = 111.35° C
o A nickel wire carrying electric current 0/1.5 mm diameter
and 50 cm long, is submerged in a water bath w/rich is open to
atmospheric pressure. Calculate the voltage at the burn out
point, if at this point the wire carries a current 0/200A.
Given:
d = 1.5 mm = 1.5 x 10-3 m; L = 50 cm = 0.50 m ;
Current, I= 200 A
Scanned by CamScanner
(Sixth edition))
= 0.293 x 10-6 m2/~
Cpt = 4216 JlkgK
11,=P,xv=961
We know that,
{From HMT data book page No.21.
P, = 961 kg/m3
X
0.293 x 10-6
11,= 281.57 x 10-6 Nslm2
From Steam Table At 100° C
{R. S. Khurmi
Steam table. page No.4]
hfg = 2256.9 kJlkg
hfg = 2256.9 x loJ Jlkg
"s = 1.673 m3lkg
Pv =
I
I
Vi
= 1.673 = 0.591 kglm3
o = Surface tension for liquid - vapour interface.
At 100° C
Icr = 0.0588 N/m I
{From HMT data book page No. 144]
"
J /6 Heat and Mass Transfer
Boiling and Condensation 3.17
For Nucietlle Pool Boiling Critical heatjlux
(At burn out)
Given:
Diameter, D = 8 mm '" 8 )( 10-3 m ;
".
(I) .
[From HMT data book page No. 142]
Substitute hit:' PI> P o, values in Equation (I)
2...= 0.18)( 2256.9 )( 103 )( 0.597
A
Power dissipation
025
x
[0.0588
x
Surface temperature, T w = 260°C.
Tofind:
'
V
(I) =>
Emissivity, E = 0.92
9.81 (961 - 0.597) ]
(0.597)2
Solution:
We know that, saturation temperature of water is 100 c.
0
I
i.e. Tsat = 1000Cl
i
Excess temperature, 11 T '" T w - T sat
i
I'lT=
=>
Q=
A
=>
1.52 )( 106
=>
1.52)( 106
=>
IV
Vx[
A
V x 200
[':A=1tdL1
xdl,
V x 200
It )( 1.5 )( 10-3 x 0.50
[7.9 Volts
260-100
I~T '" 1600 C I> 500 C
Heat transferred, Q = V x [
So, this is Film pool boiling.
Film temperature,
Tf
=
Tw +- Tsat
2
260 t 100
2
I
Properties of water vapour at 1800 C. (Saturated Steam)
Result:
Voltage, V =
[From HUT data book page No. 39
17.9 Volts.
(Sixth edition))
I!l A Ileating element c/added wit/I metal is 8 mm diameter and of
Py
= 5.16 kglm3
emissivity is 0.92. The element is horizontally immersed in a
ky
= 0.03268 W/mK
waer bath. Tile susface temperature of the metal is 260"C under
steady state boiling conditions. Calculate the power dissipation
per unit length of the heater;
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Cpv = 2709 J/kg K
Ily
= 15.10 x
10-6 Ns/m2
118 Heat and Mass Transfer
.
Properlle5 0
f saturated water at 100· C.
.
[From HMT data book page No.21 (SIXth edition)]
I
BOiling and CorrcklUation
I
I
hrad= 20 W/m2K
1.19
• .. (3)
Substitute (2), (3) in (I)
3
PI '" 961 kglm
(I) => h = 421.02 + 0.75(20)
From steam table At /OO·c.
[R.S. Khurmi Steam table. page No.4]
Ih = 436.02 W/m2K I
"Jg'" 2256.9 kJlkg
I hlg '" 2256.9)( 103 JlkgJ
Heat transferred, Q
'1' g heat is transferred due to both convection and
In film poo I b01 ID ,
h " It )( 0 )( L (T", - T sat>
radiation.
436.02)(
••• (I)
Heat transfer co-efficienl, h '" hconv + 0.75 hrad
h
k3)( P )( (P/-P »( g)( [hlg +0.4 (CPv .1T»)
'" 0.62
v
v
v
[
COny
I'v
0 .1T
j
" [2256.9 )( 103 + (0.4)( 2709 )( 160)]
hconv '" 0.62
1
15.10)( 10-6" 8)(
10-3)(
8 )( 10-3)( 1 )( (260--100)
[.: L'" 1m)
(or)
Power dissipation, P'" 1753.34 W/m
O.2S
(32.68" 10-3)3)( 5.16)( (961 - 5.16»( 9.81 0.25
It "
1753.34 W/m
Q
ResulJ :
Power dissipation, P = 1753.34 W/m.
page No. 142]
[FromHMfdatabook
I
hA(T",-Tsat)
(B Water is boiling
maintained
on a horiz.ontal
tube whose waU temperatllre
at J S·C above the saturation
temperature
Calculate the nue/eate boiling heat transfer c~ff1clertL
Is
of water.
Assume the
water to be at a pressure of J 0 atm. And also jlnd the change in vallie
160
of heat transfer c~fficient
4.10" 106 ]0.25
hconv '" 0.62 [-1.93 )( 10-5
I. The temperature
w"en
difference
is increased to 10· C at a presJllre
of lOatm.
I hconv '" 421.02 W/m2K I
1. The pressure is raised to 10 atm at ..:iT = 15"(,.
. . • (2)
Given:
[From HMf data book page No. 142]
hrad = 5.67"
10..,11
)( 0.92
x (260
[
+ 273)4 - (100 + 273)4]
(260 + 273) - (100 + 273)
C·: Stefan boltzman constant, a = 5.67"
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10-8 W/m2~]
Wall temperature
temperature.
is maintained
T", = 115°C.
P = 10 atm = 10 bar
at 15°C above the saturation
[ .: Tsat = 100· C; Tw = 100 + 15 = 115°C)
\
3.20 Heat and Mass Transfer
t
Case (i)
Boiling and Condensation ),11
Case (ii)
~T=300C;p=
10atm= lObar
p = 20 bar; ~ T = 15° C
Heat transfer co-efficient, h = 5.56 (~T)3 = 5.56 (15)3
Case (ii)
Ih= 18765 W/m2K I
P = 20 atm = 20 bar; ~T = 15° C
Solution:
Heat transfer co-efficient other than atmospheric pressure
We know that, for horizontal surface, heat transfer co-efficient
h
h
P
= 5.56 (~T)3
= 18765 (20)0.4
[From HMT data book page 11'0.143
(Sixth edition)]
h = 5.56 (Tw- Tsat P
=>
lhp = 62.19 x 103 W/m2K
Nucleate boiling heat transfer co-efficient
= 18765 W/m2K r
hp=47.13
Heat transfer co-efficient other than atmospheric pressure
hp = hpO.4
=
[FromHMTdatabook
page No. 143]
hp = 47.13 x 103 W/m2K
Case (i)
p= IObar;~T=.30"C
[From HMt data book page 11'0.143]
Heat transfer co-efficient, h = 5.56 (~T)3 = 5.56 (30)3
x
103 W/m2K
Case (i)
hp = 377 x 103 W/m2K
Case (ii)
!8765:x dO]04
Heat transfel' ico-efficient,
I
Result:
= 5.56 (115 - ~.oW
Ih
= h p04
hp = 62.19 x 103 W/m2K
I!J A electric wire of 1.5 mm diameter and 100 mm long is laid
/.orizontally and submerged In water at atmospheric prnsure. The
wire has all applied '!oltage of I(J V and carries a current of 41
ampture». Determine heat flux and excess temperature.
Tile
followlng correlation for wster boiling on I,orizolftai submerged
surface IIolds good.
h = 1.54
Given:
(g/14 = 5.58 (.!inJ Wlm2K
A
Heat transfer co-efficient other than atmospheric pressure
Diameter, D = 1.5 mm = 1.5 x 10-3m;
hp = h p04
Length, L = 200 nun = 0.2 m ;
= 150 x 10J ( (0)°.4
I = 377 103
hp
x
W/mlK
Voltage, V = 16 V; Current I = 42 amps;
I
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. h = 1.54 (~)3/4
= 5.58 (~T)3
",S.
r'
:J.:2:!2H~t~at~a~nd~u~au~~~a:·m~~~u
----------------
'fIF:;'tI-WdVW1.J'P·;TTiSf#,
___
BOiling and Condensation1.2J
3.1.9 Solved Anna Univenity Problems
To/flld:
o
I. Heat flux. ( A)
IIIAn aluminium pan of 15cm diameter is lUed to boil waler and
2. Excess temperature, (~n
SDbltlt'" :
We know that, heat transfer
the water depth at the time of boiling is 2.5 em. The pan is
placed on an electric stove and the heating element raises the
temperature of the pan to 110-C. Calculate the power Input/or
boiling and the rale of evaporation. Take Cs/= 0.0132.
O==V"I
{Dec.2005. Anna Univ]
==16" 42
Given:
[O==672W]
Diameter, d = 15 em = 0.15 m
Surface Area, A = nOL
Distance, x = 2.5 em = 0.025 m
= n )( 1.5 x 10'-3 )( 0.2
T w = 110° C.
Surface temperature,
!A ~ 9.42 " 10-' m21
=> _g_ ='
672
- 713.3 )( 103
A 9.42)( 10-'
Heat flux.
t=
713.3 )( 103 W/m2
Csf= 0.0132
Tojind:
I. Power input, (P)
2. Rate of evaporation,
Solution:
We know that,
h = 1.54
_____
L
~sat
(.2.r = 5.5S (~T)3 (Given)
= 1000(
A
~ 1.54 (713.3 )( 103)3/4
= 5.5S(~T)3
(~T)3 = 6773.92
I~T IS.')O C I
=
IExcess temperature, ~ T = IS. 9° C I
Aluminium
pan
~===~=:t--....
Electric stove
ReslIlJ:
I.
(m)
.2..= 713.3)( 103 W/m2
A
2.~T= IS.C)OC
I
I
l.•
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Fig 3.4
Boiling
3.24 Heal and Moss Transfer
We know that,
0
Saturaridn temperature of water is 100 C.
A
.
I
j
0.5 [
cr
)(
.. (I)
n = I for water
Prandtl Number, P, == 1.740
Specific heat, Cpl == 4216 J/kg K
I cr - 0.0588 N/m I
[R.S Khurmi Steam table, page No.4)
[From HMT data book page No. 144]
Vo" hfg> PI'
.' values in Equn (I)
, P... o, ep"I ~T Csf h'fg> n a~d P,
Q
"
..
' ,:
(~)~ A '" 2,81.57. x I <i6)( 2256.9 x. 103 x [9.81
AI JOO°C
0.013
Specific volume of vapour, Vg ==1.673 m3/kg
t
=_11.673
x
] 3
103 x 1.740
= 1.43 x 105 W/m2!
'
Heat transfer, Q = 1.43 x 105 x A
!p" '" 0.597 kg/m31
= 1.43 )( 105 x
~T =Excess temperature
. ==T w - T sat
= 1.43 x
'" IIOoe - lOooe
pool boiling.
105)(
2!. d2
4
f (0.15)2
Q = 2527 W = P
I ~T'" looe I
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x 2256.9
r::::------
Density of vapour, p" = _!_
Vg
I Power input for boiling, P 2527 wi
_1
(961--0.597')1 0.5 '
0.OS88
4216 x 10
x [
g
, Nucleate
. S0, thoIS IS
X'
. .
Enthalpy of evaporation, h/g == 2256.9 kJ/kg
h/ == 2256.9 x 103 J/kg
,1T== looe <50oe
editioni]
Substitute,
11, = 281.57 x 10-6 Nslm2
~
r;
At 1000 C.
Dynamic viscosity, 11, == P, x v
= 961 x 0.293 x 10--0
From Steam Table
~T ]3
[From HMT dolo book page No.142(Sixth
cr = Surface tension for liquid vapour interface
i"
'I
(P/-P,,)]
Kinematic viscosity. v = 0.293 x 10--0m /s
I
'j
[g)(
s/xlyg
Where
(Sixth edition)j
2
\
= Vol x h
'fg
[From HMT data book page No.2 I
Density. P, == 961 kg/m3
1
en'ep/)(
1. Power iDput for boiliDg
Heat flux,.9..
,
i\
1.25
For Nucleate pool boiling
.
T'sat-~ 10000C
i.e.,
Properties of water at 1000C.
and Condensation
u
3. 26 Heat and MasS
TransLl"ifij~er:_
-----_
:
Boiling and Condensation 3.27
-
Saturation
I·
1. Rate of evaporation,
(';')
::m)(
Properties of water at 100° C.
[FromHMTdatabook
hr.g
n
Q
=>
of water is 100° C.
i.e. Tsat = 100° C
We know thaI,
Heat transferre d, Q
temperature
page No.2/,
(Sixth editioni]
Density, PI = 961 kglm3
,;,::hi
Kinematic viscosity, v = 0.293 x 10-6 m2/s
2527
Prandtl Number, P, = 1.740
:: 2256.9 )( 103
Specific heat, Cpl = 4216 l/kg K
Dynamic viscosity, iii = P/)(
Resllll:
I. p:: 2527 W
2.
1.11 x 1(J3 kgls
m ::
t
. . d' b 'il' water at atmospheric pressure on a COppe,
m/t is desire to 0
W
. h' l ctrically heated. Estimate ti,e heatfluxfrom
surface whlc IS e e
'.,
. d
•
h
ter. lifthe sur/ace IS malntalne at llO C
the surface to I e w~ ,
.
and also the peak hea.tfl~
[June. 2006, Anna Univ]
[R.S. Khurmi Steam table, page No.4]
From Steam Table
At
ioo-c
Enthalpy of evaporation,
hlg = 2256.9 kJ/kg
hfg = 2256.9
Specific volume of vapour,
Given:
Surface temperature,
V
= 961 x 0.293 x 10-6
iii = 281.57 x 10--6 Ns/m2
x
"s = 1.673 m3/kg
Density of vapour, Pv = _!_
0
Vg
T w = 110 C.
=_11.673
Tofmd:
Q
1. Heat flux, A
I Pv
= 0.597 kg/m31
aT = Excess temperature
2. Critical heat flux, ~'
= T w - T sat
= II O°C- 1000
e
I aT= 10 e I
0
Solution:
aT = 100 e < 50° e. So, this is Nucleate pool boiling process.
We know that,
42
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\03 l/kg
-------------jt,22·~8~H~~~I~m~ld~A~,,~~.~~S~~~o='~~ifi~er
- Nuclcare pool boiling
For
Q
0.5
[8)( (P/-PI')1 x
xh
Heal flux, A = 111
___
'fg
----
Boiling and Condensation 1. 29
a
- 0.18)( 2256.9 )( 103)( 0.597
[From HMT data book page No. 142 (Sixth edt«'on))
x [ 0.0588 )( 9.81 )( (961- 0.597») 0.25
(0.597)2
Where
n = I for water
= 1.52 )( 106 W/m2
a = surface tension for liquid vapour interface
Critical heal flux,
At JOO°c.
@ 0.0588 N/m I
[From HMT data book page No. 144)
0=
Result:
I. Heat flux,
For water - copper => Csf = surface fluid constant = 0.013
t
*
= 1.52 )( 106 W1m2
= 142.83 x 103 W/m2
2. Critical heat flux,
[From HMT data book page No. 143)
t
= 1.52 x 106 W 1m2
Substitute,
11/, hfg> PI>PV' a, Cpb £\T, Csp hfg, nand P, values in Equation (I)
(\)~ -t
I
I
= 281.57 x I~
3
9.8Ix
x 2256.9 x 10 x [
(961-0.597)10.5
0.0588
4216 x 10
J3
x [ 0.013 x 2256.9 x 103 x 1.74
I
I
Heat flux, ;
3.1.10 Condensation
The change of phase from vapour to liquid state is known as
condensation.
3.1.11 Modes of condensation
There are two modes of condensation
1.Filmwise
= 142.83 x 103 W/m2
condensation"
2. Dropwise condensation.
For Nucleate pool boiling
Critical heat flux,
t
3.1.12 Filmwise condensation
= 0.18 h
fg
x P
v
[a x g x (P/_pv)]0.25
Pv2
The liquid condensate wets the solid surface, spreads out and
forms a continuous film over the entire surface is known as filmwise
condensation.
[From HMT data book page No. J 42 (Sixth edition)]
Film condensation
\.
I'
L
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occurs when a vapour free from impurities.
I'I
J.I.1 3
Dropwise condensation
J-~"'''';(''' condensation,
an UIVP""'· .. '
•
Bofling and Condensation J J I
the vapour condenses
into
sl11alll'
. us sizes which fall down the surface in a rand
IqUid
droplets 0 f vano
0111 rash,'
.
d
.
sfer rates in dropwlse con ensatron may be as
.
Heat rran
.'
mUch as
.
h' h than in tilmwlse condensatIOn.
10
tnnes Ig er
x ... Distance along the surface. m
0"
Tsal ... Saturate temperature. OC
T w ... Surface temperature. OC
g ... Acceleration due to gravity. 9.81 mlsl
4 Nusselt's Tbeory for film condensation
3..1 1
hlg ... Enthalpy of evaporation, J/kg
e mathematical solution given by Nusselt
is described OVerhere
. .
Th
The following assumption are made for derivation.
.
I.
The plate is maintained at a uniform temperature TWIn "'h'Ichis
less than the saturation temperature T sat' of the vapour
2.
Fluid properties are constant.
3.
The shear stress at the liquid vapour interface is negligible.
4.
The heat transfer across the condensate
conduction and the temperature distribution
5.
p ... Density of fluid, kglmJ
b. Local heat transfer co-4ficient
(h~ for vertical Jllrfllce, laminar flow
h = !...
x
Ox
c. Average heat transfer c~fflClt!IIt
... (3.8)
(II)for vertical Jllrfau, laminar flow
layer is by pure
is linear.
. •. (3.9)
The condensing vapour is entirely clean and free from gases,
The factor 0.943 may be replaced by 1.13 for more accurate result
air and non condensing impurities.
as suggested by Mc adams.
3.1.1S Correlation
for filmwise condensing
process
••• (3.10)
[From HMT data book page No. 148 (Sixth edition)]
II. Film
s, = [4 ~ k x (Tsat - T w) 1 0.25
,!
,i
r
d. Average heat transfer co-efflcient for Horizontal surface, laminar flow
thickn6s for laminar flow vertical surface,
... (3.7)
•.. (3.11)
g x hlg x p2
where
e. Average heat transfer co-efficient
!'
'1.
for bank o!tubes, laminar flow
Ox - Boundary layer thickness - m
... (3.12)
J.l - Dynamic viscosity of fluid, Ns/m2
I
k - Thermal conductivity
of the liquid, W/mK
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l
CL:WZ
&&ilME
ea&!S&4ZE&ZLW
.__
3.32 Heat and Mass Transfer
--:--Where a, '"~
f. For La",lnar flow, Re < 1800.
PIl
g. For turblilant floW Re > 1800
h. Average heatlransfer
P - Perimeter
co-efflclent
for vertical sur/ace, turbula"t/l
k) p2 g
h = 0.0077 (Re)o 4 [ -11-2 -
]0,)))
Boiling and Condensation 3.33
We know that,
F'II m temperature, T
Tw+Tsal
J = --...::.:::.
2
0",
110 + 133.5
2
" • (3.13)
I TJ = 121.75 C I
0
Properties of saturated water at 121.75 C == 120 C
0
3.1.16 Solved Problems on Laminar flow, Vertical surfaces
o
0
[From HMf data book page No.2l
p = 945 kglm3
Dry saturated stea", at a pressure of 3 bar, condenses
0/ a vertical tube of heightl
atIlO"c'
on the s urfact
m. TI,e tube sur/ace temperature'
15 Ie tpt
(Sixth editions]
v = 0.247 x 10--6m2/s
k = 0.685 W/mK
Calculate thefol/owing
J.I '" p x V = 945 x 0.247 x 10-6
1. Thickness o/the condensatefllm
2. Local heat transfer co-efflcient
at a distance 0/0.25 m:
Assume Laminar flow
Given,'
I J.I = 2.33 x 10-"' Nslm21
For vertical surfaces,
Pressure, p = 3 bar
0
Surface temperature, T w = 110 C
Thicknes, Ox=
[4 J.I k x (Tsal - T w) 1°·25
x
x
g x hfg x p2
Distance, x = 0.25 m
[From HMf data book page No.l48 (Sixth editiont]
Toftnd,'
I. Ox
4 x 2.33 x 10-"' x 0.685 x 0.25 x [133.5 - 110]
[
2. hx at x = 0.25 m
Sollltiolf "
Properties of steam at 3 bar
From steam table,
I
[R.S. Khurmi steam table. page No.JO]
hfg = 2163.2 kJ/kg = 2163.2 x 103 J/kg
L1
I
t
i
Scanned by CamScanner
9.81 x 2163.2 x 103 x (945)2
-1
rIT-h-i-Ckn-e-ss-,o-x-=-I.-I-8X-1-0-4m
Local heat transfer coefficient , h x
-
1°.25
3.34
Heat and Mass Transfer
0.685
hx -1.I8x 10-4
[hx
Boiling and Condensation 3.35
= 5805.08 W/m2 KJ
We know that,
Result:
Ft'I m temperature, T = Tw +T sat
f
2
Ox =).)8 x 10-4 m
hx = 5805.08 W/m2 K
r:;l A vertical tube of 65 mm outside diameter and 1.5
~
60+ 100
=
.'"
exposed'to steam at atmospheric pressure. The out
2
ITj
IO"g~
er SU'./IlCt
of the tube is maintained at a temperature 0/ 600C by
circulating cold water through
el
0
= 80
Properties of saturated water at 800 e
[From HMF data book page No.2 J]
tire tube. Calemate tht
P = 974 kglm3
following:
v = 0.364 x 10-{; m2/s .
1. The rate of heat transfer to the coolant.
2.
k = 0.6687 W/mK
The rate of condensation of steam.
65 mm = 0.065 rn;
Il = P x v = 974 x 0.364 x 10-6
Length, L
1.5 m
III 354.53
Surface temperature, T
60°C
Diameter, D
Given:
\I'
Tofind:
=
x
10-{;Nslm21
Assuming that the condensate film is laminar
For laminar flow, vertical surface heat transfer co-efficient
I. The rate of heat transfer to the coolant (0)
l-r
h=O.943
2. The rate of condensation of steam (/;1)
Solution:
{From HMT data book page No. 148 (Sixth edition)]
We know, saturation temperature of water is 100DC.
i.e..
I Tsat
= 100°C
I
h
] 025
~
.
Il L (Tsat - T w)
k3 P 2
The factor 0.943 may be replaced by 1.13 for more accurate result
as suggested by Mc Adams
Properties of steam at IOODC
=
[From R.S.Khurllli steam table. page 110. 4}
Enthalpy of evaporation, hfg
I II = 4684 W/m~
1
J
x (974)2 x 9.81 x 2256.9 x 103
354.53 x 10-{; x 1.5 x (100 - 60)
= 2256.9 kJ/kg
= 2256.9 x 103 J/kg
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1.13 [(0.6687)3
0.25
J 36 Heal and Mass Transfer
I. H_ "."sfer
BOiling and Condensalion 3.37
Q
Rtsllh:
Q= 57,389 W
m = 0.0254 kgls
4,684 x 1t x 0.065
jQ
57,389
x
1.5 x (100 - 60)
Wi
[II A vertical flal plale in Ihe /0':'" 0/ fill is 500 """ ill "elg'" aIId is
exposed to steam al atmospheric pressllre. If slIr/tlce 0/ tile pi tile is
maintained til 60· C, calcllltlle Ihe /ollowing
ii) TIre rate of cOlldtlfSatiolf of steam (,;,)
I. The Jilm thickness tlllhe Irtlilinl edge
2. Overall hetlllrtlns/er co-ejfic;elft
We know that,
Heat transfer, Q =
=>
m=
3. Heat trtlllS/er rete
m hfg
4. The condenstlte
Q
Assllme laminar flow conditions and IInit width o/the pltlle..
~g
m=
mtlSs flow rate:
Given:
57,389
2256.9
x
lIP
Height (or) Length, L = 500 mm = 0.5 m
Surface temperature, Tw = 60° C
1m = 0.0254 kgls I
Let us check the assumption of laminar film condensation
Toflnd :
1. Ox
We know that,
2. h
,
II
Reynolds Number, Re = 4m
PJl
where
3.Q
4.m
I
P = Perimeter'" ltD = It x 0.065 = 0.204 m
=> Re =
4 x .0254
0.204 x 354.53 x 10-6
Soilltion:
We know that, saturation temperature of water is 100" C
i.e.,
I Tsat 100° C I
=
[·'R-e-=-14-0-6~...L3
< 1800
Properties of steam at 100° C
So OUrassumption (laminar flow) .
IS correct.
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[From R.S.Khurmi steam table, page No. 41
3. 38 Heat and Mass Transfer
hfg = 2256.9 kJ/kg
Boiling and Condensation J. 39
hfg = 2256.9 x 103 J/kg
2. Average heat transfer co-efficient, (h)
For vertical surface, Laminar flow
We know that,
Film temperature,
=
Tf
Tw+Tsat
2
h = 0.943'
60 + 100
{From HMT data book page No.21}
v = 0.364 x 10-ti m2/s
I h = 6164.3 W/m2K·1
Heat transfer, Q = h A (Tsat - Tw )
11 = P x v = 974 x 0.364 x 10-ti
= h »
111 = 354.53 x 10-ti Ns/m2
IQ
(ox)
We know, For vertical plate,
4 u k x (T
r-
[
g x
where
sat
1
w
- T )
"lg x p2
Q
(Sixth edition)}
4 x 354.53 x 10-6 x 0.6687 x 0.5 x (100-60)
9.81 x 2256.9 x 103 x (974)2
~
m
0.25
1
m
1m
I
-
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1,23,286 W
We know that,
'.
x = L = 0.5 m
=
4. Condensate mass flow rate, (,;,)
0.25
{From HMT data book page No. 148
[
Lx »:» (Tsat-Tw)
-= 6164.3 x 0.5 x 1 x (100 - 60)
1
~ Ox =
w)
1 0.25
J
3. Heat transfer rate, (Q)
k = 0.6687 W/mK
Film thickness, Ox =
l~
x Lx (Tsat - T
'u
x (974)2 x 9.81 x 2256.9 x _103J 0.25
h = 1.13 (06687)3
_:_.
[ 354.53 x 10-6 x 0.5 x (100 - 60)
.
0
.
p2 x g xh(v
as suggested by Mc Adams
Properties of saturated water at 80 C
1. Film thickness
x
The factor 0.943 may be replaced by 1.13 for more accurate result
2
p = 974 kglm3
r k3
. m x hJg
Q
hfg
1,23,286
2256.9 x 103
0.054 kg/s I
I
'l1f11l4/JlI
f III '/PII' i/I,d Mil"
"'11' ~'J~1iIJ Vilit ~
Ii
/1',
''J,).?') I) I II) IIYif,.
)'1/1
'{".I'I'
~1l>lJitlA'''';, 1/4IJi!Q>i/ t ','II
W ¥IIIIWIII"I,
~¢/
'I Wi '/.,:1/
1'11111 ",,,1))<'1 tIl'",;, 'I,
17· 1)'/-1'~ I
.'
Ill",/i
'iI:
,l,r
! Ji / I, ' (J ~III J '11)
j
'J h'i r" u)r ,fJ 94') II ij/l)~ r<lipiawJ 1'1 J ) ') {'Jf " 'II '. a<
{IIi I IIJO
Z
(O,fiIi~7)3
11-1,11.
I 354053
water nt WI" C
425f).9
_.
I I() I, I OJ / (100
(0)
I
(974P / ,),IjJ
/
i
/ rr()l11 IIM1' data baok p(lg~ N~,2IJ
\I ~
0.364 x 10 r, tn2/S
J. Iteut tronsfer rate, (Q)
Q = It A (Tsol - Tw )
Heat rransfer,
k = 0,(,687 W/",K
l' = P x \I C 974 x 0.364
10-6
x
~ 6164,)
1,1 ~ 354.53 x 10-6 NS/m2
x
0,5 x 1 x (100 - 60)
1
I. FIt""lJic/"'I!.H
I Q = 1,2),286 W I
(oJ
4. Condensate man flow role, (,;,)
We know, For vertical plate,
We know that,
<
[41lk
x (TS<lt-Tw)]Q,25
F'It m t hick
Ie' ness, Ur =
.
g x hJg x p2
N IjJ
[From /-IMT data book page 0
(Sixth editiM)}
where
r
= L = 0.5 m
::::> Ox
=[ 4 354,53 10x
x
6 x
0,6687 x 0,5 x (100-6Q2] 0,25
::::>
Q
m l'Jg
m
Q
m
9.81 x 2256.9 x 103 x (974)2
1m
1
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JliOJJi
I'~ "I~i4'i'I,J(J by M~ AtJ"JI1,
['I[ - 1l()"C]
Pl'Op~rtles 01' saturated
I'
x
hJg
1,23,286
2256,9 x 103
0,054 kg/s I
<)2
j()lj
.
-
r b'J
3~1Y ~# ,,_<1~' =':.1'. y/
~~
'v#
¢ ~
'fi'
.t~ <;:¢
I' ~"1"'" Wi,
'fO __
.(;
~
ill tfi~
~
/
!#,~,
~
1~ --4.,.:L
UlJ-Jl_ #H 11IJv.'-~
~AI
,.
Sf; ~J!"~
-.
~"e
If-
i
~iI!t#~iJf
~
I'/,PJ..,
j;yt;t
S'fi UftI~!'
'"W
""'1If~
~
l6-_4MM ~JI«frr ~
~f_
tilt
*
·~~r1<l'~.
Jl"t(e:!'iit!YI7(.f
• ~ If!tke,t4lfe..
~.
~ .. -.,__
/f'N_ Hr-ri tf£.vN//,k
VA/ '.-"1:/J
j;,.;;.t ,,~~1I!:iottI
L
e.
1iwI ~,,-ifd
lMt11 __
t
Pfb( ItIWJI
H lite P--IIlflOt$fU
w-qFurior11j Ole
'**"
~.:
Pr=~,"': 1}.1ia
a)
bar
M<2, A: '::O-cm / 50 em:
050 ' 0.'::0; a.25m)
Film thidlU1f
We know, FIlf vertical surfaces
Serface temperamre, T... = 2ft C
Distance, x = 25 em = (US m
Iix =
4 I' k x (T sat - T...) 025
g""lg"p2
(From H,..rr data book page No. /48J
Tafoul:
a) Ii..
c) h
0x =[4><82751'
9.81 x 2403.2
d) Q
j) h at 3D·
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10--6>< O.612xO.25:<{41.53-20)
Iii..
= 1.46 x 1Q-4 m
I
><loJ " (997))
025
F
Boiling and Condensation J.43
3. 42 Heat and Mass Transfer
b) Local heat transfer co-efficient (h::J
[Assuming Laminar q Ow]
e) Tolal sleam condensalion
h = !....
x
Ox
Heal transfer,
Q
~
m
0.612
hx""
rale (,;,)
=nlxh/g
g__
l'fg
1.46)( 10-4
[&=4,191 W/m2gJ
c) Average heat transfer co-efficient (h)
E~~-o.o I
[Assuming Laminar flow]
125 kgls
k3)( p2)( g x hr.] 025
h ""0.943
30,139.8
2403.2)( 10J
m
jg
[ 11x L x (T sat - T w)
f) /ft"e plate is inclined al ONlil" IlOri'l.onla:
hinclined
hYl!rtical x (sin 0)\4
~
hinclined
hYertical ~ (sin 30)\4
~
hinclined
The factor 0.943 may be replaced by 1.13 for more accurate result
as suggested by Mc Adams
. [k3 p2 g hr. ]0.25
~ h = 1.13
jg
11 L (T sat - T w)
=
5599.6 x (Yl)\4
I hinclined =
4,708.6 W/m2
K]
Let us check the assumption of laminar film condensation
where L = 50 em = 0.5 m
h = 1.13 [(0.612)3 x (997)2 x 9.81 )(2403.2 x 103] 0.25
We know that.
827.51 x 10-6 x .5 x (41.53 - 20)
Reynolds Number, R
e
Ih 5599.6 W/m K·1
=
2
where
W = width of the plate = 50 cm = 0.50 m
d) Heat transfer (Q)
~
We know that,
Q = h A (T sat - T w )
=
4 x 0.0125
0.50 x 827.51 x 10-6
1800
So our assumption (laminar flow ).IS correct
(5599.6) x 0.25 x (41.53 - 20)
I Q = 30,139.8 W I
43
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Re
~io.~<
= h x A x (Tsat - T w)
=
= 4WI1
~
$.44 Heal and Mass Transfer
-----
Result:
a. 5.\" = 1.46 x 10-4 m
b. h.t =4191
Properties of steam at 1.7 bar
[From R.S.Khurmi steam table, page No.9]
W/m2K
c. h = 5599.6 W/m2K
hfg = 2215.8 kJ/kg = 2215.8 x ](P J/kg
d. Q = 30,139.8 W
We know that,
e. ,;, = .0125 kg/s
Film temperature,
f. hinclined = 4708.6
m
Boiling and Condensation 1.45
Solulion:
Tw+Tsat
Tf = _.:___:.:=-
2
W/m2K.
85+115.2
2
Tire outer surface of a cylindrical vertical drum huvlng 25c",
diameter is exposed te saturated steam at 1.7 bar for condensation.
The surface temperature of the drum is maintained at 85"C.Calculate
tirefol/owing
Properties of saturated water at 100° C
[From HMT data book page No.2/ J
p = 961 kglmJ
v = 0.293 x 10--6m2/s
I. Lengtlr of the drum
k = 0.684 W/mK
2. ThicA"nessof condensate layer to condense 65 kg/h of steam.
f.I = p x V = 961 x 0.293 x 10--6
Givm:
I f.I 281.57 x 10--6Nslm21
=
Diameter, D = 25 em = 0.25 rn; Pressure, p = 1.7 bar
For vertical surfaces, .
Surface temperature,
Tw = 85° C
•
65
[Assuming Laminar flow]
Average heat transfer coefficient
Mass, m = 65 kglh = 3600 kg/s
h = 0.943
1m O.OI8(J kg/sl
=
Tofuul :
LL
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1
kJ p2 g x h.fJ
0.25
g
[ f.I L (T sat - T ....)
[From HMJ data
book page No. 148J
Using Mc Adam correlation, 0.943 is replaced by 1.13
J. 46 Heo: and Mass Transfer
(0.6804)3)(
h-1.I3
I
[
(961)2)( 9.81 )( 2215.8)(
281.S7)(
h = 5900 L- 0.2S
10-6)( L)(
I
103] OH
Boiling andCondefUation
(115.2 - 85)
0.%- [4
X
9.81 "2215.8)(
~ ... =
Heat transfer, Q
X!Q-{> x 0.6804xO.18X(l15.2-S5)]O.2S
281.57
... (I)
1.20)( 1()-4 m
103)( (961)2
I
Let us check the assumption
~ Xhfg
1.47
of laminar flow
We know that,
0.0180 kg/s x 2215.8
x
10J J/kg
Reynolds number,
R
e
39.8 x 1031/5
Q
39.8)(
103 W
I
= 4';'
PJI
where
P = Perimeter = 1t0 = 1t )( 0.25 = 0.785
We know that,
R = _ __:_4_x-=0..:..:.0...;_I..:..SO.;__~
e
0.785 x 2SI.57 x IQ-{>
i
Q
, ~i
39.8 x 103
h » 1tDL)( (Tsat - T w )
39.8 x 103
h x 1t x .25 x L (115.2 - 85)
I,j
IR" = 325.71<
So our assumption
(laminar flow) is correct
Result:
Substitue h value
I. L = O.IS m
=>
39.8)( 103
2.0 x = 1.20 x 10-4 m
=>
0.278
(5900 L-0.25) )( 1t )( .25 )( L )((115.2 - 85)
LO.7S x (115.8 - S5)
=>
L
O.ISm
~fthe
drum,
L = O.IS m
ISOO
!11 Saturated steam at tsat = 1oo·e condenses on the outer JUrface of
1.4 m long, 2m outer diameter
temperature
folloHling.
TK, = 60·C Assuming
film condensation,
find the
i) Local heat transfer co-efficient at the bottom of the tube.
ii) Average heat transfer co-efficient over th« entire length of
tile tube.
2. Film thickness
.t
1
4 " k x (T sat - T w ) 0.25
~=,..
[
g x
IX=L=0.18m
"tg)( p2
1
Scanned by CamScanner
Given:
T sal = 100°C
Saturation temperature,
Length, L = 1.4 m
Outer diameter, 0 = 2m
Surface temperature,
II
venice! tube maintained at allnl/orm
T K' = 60° C
~~~~~~~-------------------J -# Helll and Mass Transfer
Tojlltl:
I. Local heat transfer co-c·fficient,
h;K
2. Average heat transfer co-efficient,
Boiling and ComkJUolion J.49
= [4)( 354.53 )( I~)(
0.6687)( 1.4)( (100
60>] 0.25
9.81 " 2256.9 )( loJ x (974)2
h
[.: x=L=
SoIlIlio" :
10.1'=2.24)(
of steam at I OO"C
Properties
[From R.S.Khurmi
Enthalpy
l.4m)
l0-4ml
steam table. page NOA}
Local heat transfer co-efficient
of evaporation,
h =
.1'
"tg = 2256.9 kJlkg
= 2256.9
h
x loJ Jlkg
=
.t
(hx)'
k
[From HMT data book
0.1'
page No. 148]
0.6687
2.24)( 10-4
I h;K = 2985.26 W/m2K I
We know that,
Tw+ Tsat
Tf =
2
Film temperature,
Average heat transfer co-efficient
60 + 100
h = 0.943
2
k3)(
(h),
p2)( g )( h£. ] 0.25
JS
[ J.l)( L)( (Tsat-T >
w
[From HMT data book page No.J48]
Properties
of saturated
water at 80° C
[From HMT data book page No.21
p = 974 kglm3
The factor 0.943 may be replaced
(Sixth edition)}
v = 0.364 )( 10-{i m2/s
by 1.13 for more accurate result
as suggested by Me Adams
=:>h=1.13
k3
[
p2 g h;
] 0.25
J8
J.l L (Tsat - T IV)
k so 0.6687 WImK
Il = P)( 'II = 974 )( 0.364 )( 10-6
Il = 354.53 )( I~
Film thickness
surfaces,
0.1' =
film is laminar.
laminar flow,
4 Ilk x (T
[
- T ) ]0.25
sat
IV
g x hfg x p2
[From HMT data book page No. 148 (Sixth edition)]
Scanned by CamScanner
(974)2 x 9.81 x 2256.9 )( 103] 0.25
354.53 x 1~
N~m2
Assuming that the condensate
For vertical
h = 1.13 [(0.6687)3)(
I
x 1.4)(
h = 4765.58 W/m2K
Let us check the assumption
(100-60)
I
of laminar film condensation.
3. 50 Heat and Mass Transfer
Boiling and Condensation J 51
We know that,
Reynolds number,
R.. '"
4 IIi
p;-
... (I)
[1) A I'ertlcal plate 0.4 m heIgh and 0.3 m wide, at 40-C, Is expoud to
saturated steam at atmospheric preuure. Find the following
I) Film thldne.u
Heat transfer, Q - hA 6 T
- II
It
/II) Total heat fluX to the plate.
D L x (TSIl1 - T",)
2 x 1.4
IQ-1.67
at the bottom of the pln'~
II) Maximum velocity at the bottom of the ptate
)t
(100 - 60)
Given:
Height or Length, L - 0,4 m
106wl
Wide, W - 0.3 m
We know that,
Surface temperature, T w ~ 40°C
Q "" ';1 II'}.'g
1.67 x 106 =
Toflnd:
m (2256.9" 103)
I. Film thickness,
1m = 0.739 kg/s I
...
(2)
s,
2, Maximum velocity, umax
3, Total heat flux, Q
Perimeter, P = nD
Solution:
=nx2
We know that, Saturation temperature of water is loooe
I P 6.283 m I
i.e. TSBI = 100° C
..• (3)
=
Properties of water at 100° C
Substitute P, ~, Jl values in equation ( I)
(I) ~
[From R.SKhllrmi steam table, page No,4}
R =
4 x 0,739
e
6.283 x 354.53 x 10-6
"Jg = 2256.9 kJ/kg
= 2256.9 x 103 J/kg
We know that,
So our assumption (laminar flow) is correct
Film temperature,
Result:
Til'
Tf = --2--
+ TSBI
1. Local heat transfer co-efficient,
hI' = 2985.26 W Im2K
40 + 100
2. Average heat transfer co-efficient,
h = 4765.58 Wlm2K
2
Scanned by CamScanner
~J~.5~2~R~e~.m~an~d~U~~~~~~~a~m~ife~r~~~::~--------. of saturated water at 70° C
~
Properlles
[From HMT data book p
. age No.(/
(SIXthedit'
p = 979.5 kg/m3
Boiling and Condensation J.53
Average heat transfer co-efficient (h),
10")J
v = 0.421 x 10-0 m2/s
.
h = 0.943
[
kJ x p2 x g X hr. ] 0.25
jg
.
Il )( L x (Tsat - T w)
k =0.66 W/mK
[From HMT data book page No. 148]
u= pxv
The factor 0.943 may be replaced by 1.13 for more accurate result
= 979.5 x 0.421 x 10-6
as suggested by Mc Adams.
Il = 4.12 x 10-4 Nslm2
h = 1.13 [ (0.66)3 x (979.5)2 x 9.8 I x 2256.9
Assuming tha,t the condensate film is laminar.
Film thickness, .sx =
[41l k x (Tsat - Tw)]
I
h = 5633.22 W/m2K
Total heat flux is given by
Q = hA (Tsa! - Tw)
[From HMT data book page No.ua
(Sixthedition)]
= h x (L x W) x (Tsa! - Tw)
=
= [4 x 4.12 x 10-4 x 0.66 x 0..4 x (100 - 40)] 0.25
5633.22
=
Result:
[.: x = L = O.4m)
I. c\. = 1.87 x 10-4 m
2. umax = 0.407 m/s
pg (ox)2
21l
979.5 x 9.81 (1.87 x 10-4)2
2 x 4.12 x 10-4
= 0.407 rnIs /
Scanned by CamScanner
x (0.4 x 0.3) x (100 -
I Q 40, 559 wi
9.81 x 2256.9 x 103 x (975.9)2
fmax
I
0.25
g x hfg x p2
Maximum velocity , umax =
103]°.25
4.12 x 10-4 x 0.4 x (100-40)
For laminar flow, vertical surface,
.,
x
:. Q = 40, 559 W
40)
3. 54 Heat and Mass Transfer
3.1.17 Solved problems on Laminar now, Horizontal sur;;---a
----------~--------------------
fl)
~~
Boiling and Condensation J jj
c~
for horizontal tubes, heat transfer co-efficient
A horizontal tube of outer diameter 2.2 em is exposed to dry stea",
tit 100· C. Tile pipe surface is maintained at 62· C by CircUlating
water through it. Calculate tile rate of formation Of condensate
per metre length of the pipe.
h = 0.728 [
h = 0.728[
(0.6687)J
[From HMT data book page
No. 148 (Sixth edition)/
x (974)2 x 9.81 x 2256.9 x
10J jO.25
354.53 x 10-6 x 2.2 x 10-2 x (100 - 62)
Diameter, D = 2.2 cm = 2.2 x 10-2 m
Surface temperature,
] 0.25
g
Il 0 (TS81- Tw)
Given:
Dry steam temperature,
kJ pl g ".h
lr-h-=-8-78-3.-4 -W-'m-=-2K-',
Tsat = 100° C
T w = 62° C
Heat transfer, Q = h A (Tsat - T w )
1'0find:
h x nDL x (Tsat - T w)
m
8783.4 x
It x 2.2
x 10-2 x I (100 - 62)
Solution:
[':L=lml
Properties
IQ = 23,068.5 W I
of steam at 100° C
{From R.S.Khurmi steam table page No.4}
hrg = 2256.9
I jg
h
= 2256.9
We know that,
kJ/kg
Q=';' ~g
x 10J J/kg /
=> ,;,=!L
~g
We know that,
Film temperature,
Tf =
Tw + Tsal
m=
62 + 100
I Ii, 0.010 kgls I
2
)I =
Result:
o
of saturated water at 80° C
p = 974 kg/rn!
0.010 kgls
=
2
Properties
23,068.5
2256.9 x 10J
{From HMT data book page No.21
(Sixth edition)]
I
0.364 x 10-6 m2/s
A steam condenser consisling of II square Ilrray of 900 I,orizolllil/
tubes each 6mm in diameter. The tubes are exposed to sllturaled
steu", at a pressure of 0./8 bar and II,e tube surface temperalllre
is maintained at 23· C, calculate
k = 0.6687 W/mK
I
Il = P x I' = 974 x 0.364 x 10-6
I fJ
/. Heal transfer co-efficielll
2. The rate at whiclt steam is condensed
= 354.53
x
10-6 Ns/m2/
1
I
I
Scanned by CamScanner
_..-(
J56 Heal and Moss Transfer
1.57
Boiling and Condensation
G/~n:
Horizontal tubes = 900
With 900 tubes, a 30 )( 30 tube of square array could be formed
3
Diameter, D = 6mm = 6 )( 10- m
N = .j9Oo = 30
i.e.
Pressure, p = O. 18 bar
Surface temperature, T w = 23° C
Toflnd:
For horizontal bank of tubes, heat transfer co-efficient
I. Heat transfer co-efficient, (h)
{From HMT data book
2. The rate at which steam-is condensed, (m)
page No.J48]
Sol"t;on:
Properties of steam at, p = O. 18 bar
{From R.SKhurmi
steam table, page No.8]
(0.628)3 x (995)2 x 9.8 I x 2363.9 x loJ
h = 0.728 [ 653.7 x 10-ti x 30 x 6 x 10-3 x (57.83 - 23)
T sal = 57.83° C
hfg = 2363.9
0'2.S
I h = 4443 W/m2K I
"Jg = 2363.9 kJlkg
I
1
I
)( 103 J/kg
Heat transfer,
Q = h A (T sat - T w )
We know that,
Tw+ Tsal
4443 x 1t x 6x 10-3 x 1(57.83 - 23)
Film temperature, Tf = ----=.::::....
2
[':L= 1m]
23 + 57.83
IQ
2
I Tf = 40.41° C 1== 40° C
Properties of saturated water at 40° C
p = 995 kglm3
(From HMT data book page No. 21]
Q = ,;, x hfg
~~=_g_ h
fg
m = _2_9_1_6._9_
2363.9 x 103
, J,'
11 = p x V = 995 x 0.657 x 10;-6
I,n 1.23 x 10-3 kgls I
=
111-
2916.9 W
We know that,
. v = 0.657)( 10-ti ~2/s
k = 0.628 W/mK
=
653.7 x lo-tiNslm21 .:
Scanned by CamScanner
I
-
.
('
3.58 Heat and Mass Transfer
for complete array, the rate of condensation
is
Boiling and Condensation J. 59
10-3
,;, = 900 x 1.23 x
Solution:
Properties of steam at 0.12 bar
,;, = II 07 x 10-3 kgls
[From R.S.Khurmi steam table page No. 7J
Tsat = 49.45° C
,;, = I. I kg/s
I
Irg
IIrg
Result:
h = 4443 W/m2K
= 2384.3
=
kJ/kg
2384.3 x 103 J/kg
I
We know that,
Film temperature, T = Tw+Tsat _
f
2
"' = 1.1 kg/s
30 + 49.45
II) A condlmser Is to be Ilesiglled to condense 600 kgn, of dry saturated
2
steam lit a pressure of 0./1 bar. A square array of 400 tubes, each 01
8 nrm diameter Is to be used. The lube surface Is mointained at I
ITf=
30· C. Calculate the I,eat transfer co-efficient and the lellgf/, 01
Properties of saturated water at 40° C
39.72°C 1=40°C
P = 995 kg/m3
each tube.
[From HMT data book page No. III
v = 0.657 x 10-6 m2/s
Given
,;, = 600 kg/h = ~
3600
m-' =-0-.1-66-k-gJ-s
I~
k = 0.628 W/mK
kg/s = 0.166 kg/s
I
)l
= P x v = 995 x 0.657 x 10-6
l)l 653.7 10-6 Ns/m21
=
Pressure, p = 0.12 bar
x
No. of tubes = 400,
With 400 tubes, a 20 x 20 tube of square array could be formed
Diameter, D = 8mm = 8 )( 10-3 m
i.e.
Surface temperature,
N = .j4Oo = 20
IN = 201
T w = 30° C
Toftnd:
For horizontal bank of tubes, heat transfer co-ellicient
I. h
k3
h = 0.728
[
2. L
44
Scanned by CamScanner
P
2
g
h
)l N D (Tsat -
'fg
T w)
] 0.25
[From HMT date
book page No. 148
/Ju111nKlind ('fJndenHol/on
=""""'-=="=-"~'~~--
. 'I
.
"
0.728
[
h.
(1),628)3 / (99~)2 I. 9.81 'I 2384,3 ,. .I.OJ
--_._}
65),7" 10 (j r 20 ~ IJ Yo 10~3)( (<19.45
.jO)
j
(12
>_
__....
J. 61
~
==~~--==-==----,h.
3.1..8 Annll UnlvU8lty Solved p,,)blem~
-----=====~== ..
fD Dry,.,.ru,e;} " ..... t « p"".re 0/1.41 bur c•• J..... on
,fur/tlCe 0/ a verllco/lube
o/lle/K/Il
lm. The lube ,fur/flU
a'e
lemperature I" !ltpl fll J/7'C. £."Im
II,e 11,lckneu 0/ II"
5304.75 Wlm2K
condenftllefl
We know that,
.'
{ May 2005, Anna Untv]
.
1m
Given:
Pressure, p ,. 2.45 bar
Heat transfer, Q. h A (T sat - Til')
Distance or height, .r '" 1m
Surface temperature,
No. of tubes '" 400
::::> Q '" 400'
h x 11 Yo D y L x (T sat - Til')
::) Q =: 400
5304.75 Yo 11 Yo 8 Yo 10-3 Yo L Yo (49.45 - 30)
Yo
IQ 1.05 10 LJ
=:
6 Yo
x
Til' '"' 117° C
roflnd:
film, ox'
Thickness of the condensate
Solulion:
... ( I )
Properties of steam at 2.45 bar.
{From R.s.Khllrmi steam table. page No.fO]
We know that,
Q = ;" x hlg
hlg = 2183 kJlkg
~ = 0.166 x 23843 x 103
IQ '" 0.3957 10 W
x
6
1
... (2)
We know that,
Film temperature,
TI = T
Equating (I) and (2)
Result:
h = 5304.75 W/m2K
L= 0.37 m
Scanned by CamScanner
-t T sal
2
117 -! 127
2
~ 0.3957 x 106 = 1.05 x x 106 L
::)IL = 0.37 m I
\I'
I TI = ]220 C I
Properties of saturated water at 122° C '" 120°C
{From HMT data book page No.2/
(Sixth edition)]
p = 945 kg/m!
\I =
0.247 x 10-{' m2/s
k = 0.6850 W/mK
_13~.6~]~H~e(.~1f~a~nd~~~a~·~~f!1;~a~n~.if.~e~r
--------------
~
Boiling and Condensation
IJ - P" v
'- 945"
0.247 )( 10-6
Saturated steam temperature, Tsal = 100° C
Tube surface temperature, T w = 92° C
1" - 2JJ )( 10-4 Ns/m2 )
Tojlnd:
For vcr,.icnl surlirccs, (Assuming condensate
'I I' k ~. ( 'I'.,.
sill -
.5 _
,I
[
Y )( hIll
II'
film is laminur)
)] U,2$
p2
10 I
co-efficient. h
2. Rate of condensation,
m
Properties of steam at 100° C
[From R.S.Khurmi
/1M"/, data book page No, 148
(.5'lxlhedlllon)}
><
I. Average heatlransfer
So/ul/on:
/FI'OIII
'2,3
0.61l50)(
I )( (127-
117)
"Jg - 2256.9
I oj J/kg
X
We know rha ••
Film temperature,
Tw'" Tsal
2
Tf =
.. [6. 84" IO-Jlo.2~
92'" 100
1.912)( 10'3
2
1.35 )( 10-4111
steam table, page No.4}
hfg a 2256.9 kJ/kg
l·2$
!1.I1I " 21113 " 10·) x (945)2
IOx •
J. 63
I
Properties of saturated water at 96° C
p ~ 965 kg/mJ
Rel'ull:
Thickness of the condensate
film 0x -
1.35
x
10-4 m
v = 0.310 x 10-6 m2/s
k = 0.677 W/mK
IIJ
A lube a/2m lellglll u/l(l 25 mm outer diameter is If}be condense
saturated steam 01 I ()o'e willie the lube surface is malnlalned al
,,= pxv=965xO.310x
92'C Estimate the average heat transfer co-efflclent and II,e fale
I"
0/ condensation 0/ steam If Ille lube if kept korizontal. Tiresteam
For horizontal
condenses on the outside of the lube,
Given:
Tube length, L = 2m
Diameter, D = 25 mm = 0.025 m
Scanned by CamScanner
10-6
= 2.99 x 1Q-4 Nslm2
J
tubes, heat transfer co-efficient
[June 2006, Anna Univ)
k3
II = 0.728
p2
g h
] 0.25
'fg
[
"D (Tsal- Tw)
[From HMT data book page No. 148
(Sixth editionl]
1 J Hi,]' an.1.",us Tr,m,~f~'r
[{0.677),) )( (965)2 )(9.81 x 2256.~]
---~-
Boilmg and Condensation J. 65
2.99 x 10-" x 0.025 x (100 -92)' °11
It - O. 28
3.1.19 Problems (or practice
r h = I .166.08 W/m K I
1
I.
A wire of I mm diameter and 150 II1Ill length is submerged
horizontally
in water at 7 bar, The wire carries a current of I J 1.5 A with an applied
voltage of 2. I 5 V. If the surface of the wire is mainrained
Heal transfer.
(i) The heat flux and (ii) The boiling hear transfer coefficient.
[Ans : (i) 0.6 MWln,1, (#) 199]0 W/",J'q
h x 1t x D x L x (Tsar - TIl')
2,
13,166.08 x 1t x 0.025 x 2 x (100
IQ
We knov
16544.98 W
I
- 92)
=:>11/
11/
A electric wire vi' 1.5 rnm diameter and 200 mm long is laid horizontally
and submerged in water at atmospheric pressure. The wire has an applied
voltage of 16 V and carries a current of 40 amperes, Calculate
(i) The heat flux, and (ii) The excess
111al,
Q
ar 180° C,
calculate
temperature
[Ans : 0.679 MW/mJ, (ii) 18.5]"
= ,;, x !JIg
J,
=Q.
A metal
lad healing element is of8 nun diameter and emissivity 0,95.
The element is horizontally
immersed in a water bath. Ihe surface
temperature ofrhe metal is 260 C under steady state boiling conditions.
alculate the power dissipation per unit length for the healer if water is
exposed to atmospheric pressure and is at uniform temperature.
0
II
165-R9S
~_'-6 9 '( 10·;
{Am' : I. 75 K WI"'I
4,
11/
A heated brass plate at 150 C is submerged horizontally
in water at a
pressure corresponding to a saturation temperature of 1250 • Whal is
the heat transfer per unit area? Calculate also the heat transfer coefficient
in boiling.
0
[Ans : 1.15 x 1(J6WI",l, 900 KWlmlKI
A heated pol.ished copper plate j.; umncrsed in a pool of water boiling
at atll10S~heflc pressure. If the 5l1. ;':1ce temperature or the copper plate
I' maintainer] at tempcrauvc
of 113, <)0 C. determine
the urtace heal
flux and the evaporanon rate per unit .rea of the plate.
" = 13,166.08 W1m2 K
11/
== 7.33 x
a
10-J kg/s
I .,I/U':
6,
Water at atmospheric
IIJ KWlml, 49.9] Itg/mlll
p"e~-;lIrc is bodo:d ill a .dle lIlade ofcopper.
I
bottom or kettle i5 Ilat, 30 CIIl ill diauictcr
and is IlIa:lllaillCd
The
at a
temperature 01'118° " '"kulatc the rate ofhcat required to boil water,
Also estimate the rate uf evaporaliull of water from the kettle.
II",,:
Scanned by CamScanner
17.1 K'II. 17.41 Itlll"f
'.
fr'
_j~.~6~6~f~/~oo~t~al~I(:/~A~~a~~~1}~·(~"'~~fi~o~r
__
~~==~---------~
~
3.2 Heat Exchnngers
Heat Exchangers
3.2. t Introduction
t Non condensable
A heat exchanger is defined as an equipment
-_I --_-_---__-_--"----L
which trnllSfe
'l
the hellt from II hot fluid to 11 cold fluid.
------------0- - - -
_
-_ =- =- =- =- =_p=- =- =- =- ='0=-
3.2.2 Types of Heat Exchanger
There nre silvern I types of heat exchangers
which
3.67
gas
-
HOI water
Illay be
classified on the basis of
I. Nature of heat exchange
11. Relative direction
process
III. Design and constructional
=- -- =- =_p=- =- =- =- =- =b-- =---~
-----_
----.-:.:
Steum-·
of fluid motion
I
features
i
Cold water
IV. Physical state of fluids,
Fig 1. of Direct
I. Nature of hC1l1 exchange
On the basis of the
exchangers
nrc classified
II. Direct
contact
process
nature
h. Indirect
of heat exchange
process, helll
as
heat exchangers
b. Indirect contact
or Open heat exchangers
heat exchangers.
II. Direct contact "eat exclumgcrs
or Ope" "cal e:~:cll(I"gers
takes place
by direct mixing of hot and cold fluids. This heat transfer
is usually
accompanied
by mass transfer.
Examples: Cooling
towers, Direct contact
CX(:/IIIIIKcr
c.\'c1I1111gcr.\·
In this type of heat exchangers. the transfer of heat between two
fluids could be carried out by transmission through u wall which separates
the two fluids.
It may be classi tied as
i. Regenerators
the heal exchange
In direct contact heat exchanger,
(,I}tIIIICI/WIII
COIIIIU'I//{!III
ii. Rccuperators
(or) Surface hem exchangers
i. RegcllerlllllfJ
In the type of heat exchangers,
through the slime space.
Examples : IC engine
feed heaters
II. Rccuperutors
hot and cold fluids now alternately
. gas turbines.
(or) Surface
This is the most common
Ileal exchungers
type of heat exchanger
and cold fluid do nOI come into direct contact
separated by a tube wall or a surface.
I
Scanned by CamScanner
in which the hot
with each other but are
(
•••
~
~:
• .,..,
3.68 Heal and Mass Transji!r.
.~
les
Automobile
radiators.
Air
pre heaters, Economlsers . ttc.
a",p C.I.
.
-';
~~
...
c
•••
Heal Exchangers
1:"_
r;..l
~
••
J 69
b. Counter flow h~u' ~xchang~r
Af/vtl"'tlg~.f
I. Easy construction
In this type, hot and cold fluids move in parallel but opposite:
directions.
2. More economical
Cold fluid
). More surface area for heat transfer
DI.Jn(lvu"'tlg~.,
-
I, Less heat transfer co·eflicient
2. Less generating capacity
II. Relative direction
Hot
fluid
of nuid motion
..
-
This typo of heat exchangers arc classified as follows
D,
1
Parallel flow heat exchanger
b. .ourucr flow heat exchanger
e.
II, "lIrllllel
'rU88
flllw
Fig. J. 7 Counter flow "eal excltanger
flow heat exchanger
111.1111
flxcllllltllcr
III I.hl~type, hot uud cold
c. Cros» flow keut exchanger
fluldtl move In the UUllle dircctlon.
In Ihlo type, the hot and cold fluids move at right angles
other,
old l1uld
1101
~l
j
C
j
a
-~
~
I?
'.
thlld
. 1
II
II(Jr,
IlLlld
1
'~Il'
,(j 1'1""II,,/lI1W
11(/111~ dulltller
1111/, ). fJ
Scanned by CamScanner
'ft/IIM//ow /111111
e. dla",,,
10 each
•
.....
J. 0 "'qot and Mass TrollSfer
Heat Exchangers
III. Design and constructional features
On the basis of design and constructional
features. the heat c"changers
are clas itied a follows.
tubes
than one time. This type of exchanger
manufacture,
b. Shell and lube
c. Multiple
shell and tube passes
d. Compact
heat exchangers
In thi type, two concentric
more
is preferred due to its low cost of
and easy to repair.
d. Compact h~at ~xcl'QnguJ
There are many special purpose heat exchangers called compact heat
tubes
pipes, each carrying
exchangers.
They are generally employed when convective
one of the fluids are
co.efficient
associated
associated
with the other fluid.
used as a heat exchanger.
The direction
of flow may be parallel or counter.
b. S"~II lind tube
In this type of
heat exchanger,
one of the fluids move through a
heat transfer
with one of the fluids is much smaller than that
IV. Physical state of fluids
bundle of tubes enclosed
by a shell. The other fluid is forced through the
Tubes
are classified
state of fluids inside the exchanger,
heat
as
a. Condensers
Hot lluid
(001)
Ba"'e plate
Based on the physical
exchangers
shell and it moves over the outside surface of the tubes.
SIleII
J/It/l and tub~ paSl~J
In order to increase the over all heat transfer. multiple shell and rube
passes arc used. In this type. the two fluids traverse the exchanger
a. Concentric
II. COlfulft,lc
c. Multlplt
J 71
b. Evaporators.
t
a. COnlJl!nsers
In a condenser.
Ihroughoullhe
the condensing
fluid remains at constant temperature
exchanger while the temperature of the colder fluid gradually
increased from inlet to outlet. It is shown in fig 3.10.
In other words, the hot fluid loses latent heat which is accepted by
the cold fluid.
<).-
HoI tIuid
(In)
b. Evaporators
In a evaporator,
the temperature
shown in fig 3.11.
Fig J. 9 S/.~II and tube I.tat txChlllfgt,
Scanned by CamScanner
the cold fluid remains at constant temperature while
of hot fluid gradually decreases from inlet to outlet.
11 is
J
1 HUll and MmJ TTamfeT
~~
~
'~.cn
~
ttc.:pa"::1Il'
c5fruma, ~
~ r.ruJ heza ~A:"
IBCXI
~
is tx 0t.2I.e7~
D
~as
~-------------------l
3.1_" ..us..ptiom
zn dcm-e oq:n:srioo far unD (-or \'IIrious ~pes 0( bc3I
~
...~~.-e~
o.~~~
I. .Bo!' is. 5l~'
.~
2.
Tbe p\-=u1 bar t:ransfer ~fficicm
3.
l'be ~ilX
~.
Tbe r:nots5 flo ...· l'1IR of bod! fluids are C'OIISQD(
$.
A.oo rondu~:rioo along ~
is ctIOSWJI
bats ofbodl fluids are c:oomm..
~
is negligible.
The change in kinetic, and potential ~~
of the fluids are
negligihle..
j .,
"1
3.1.5
~l'=t..:
,~------------_L--L
A single pass p3l'1I)JeI now heat e.xcbangtt'5 is sbo"''1l in fi&. 3.12.
L
Scanned by CamScanner
Logaridlmic Men Temperahl~ Dirre~."
for Parallel Flo,,'
1. 74 /,!.W( and M(ISS Transfer
_,
....
-. -.
-.-. -.-.
Cold fluid
lIot fluid
Cold Iluid
-. --..
--. -.-.
-. -.-.
from (3. I~).
dQ e III" 'pedt
dO
dt ..
III,
e" c
fig 3.12 Flow Ilrrtlllgelllcllt
~
Let
"'II - Mass now rate of hot fluid
dT-dt..
"'e - Mass flow rate of cold fluid
Cpll -
Specific heat of hot fluid
r',' C,"""'u " C,JCI
-dO
dO
c;;- - c,
.. -dO
[...!_.., ...!_]
C"
. Cpc - Specific heat of cold fluid
T I - Entry temperature of hot fluid
C,
de - - eo [...!_ + ...!_]
C"
c,
T 2 - Exit temperature of hot fluid
r ·,'de .. dT-cit]
t2 - Exit temperature of cold fluid
(3.IS)~
C
"
Let us consider an elemental area dA of the heat exchanger. The
heat flow rate is given by
[','9- T-tJ
~
'" (3.14)
We know that,
~
dQ = -mh (ph dT
~
dT
.. , (3.15)
I I]
.!!!L = _ UdA [
e·.,
Ch +
C
,
cI
+
C
Integrating
=~
'~ide
I '9 = -
[I
. '2
[ I.
C
h
I]
U IdA
c
mhCph
~
L_5]
I]
[In B] = - U + I
_ CII
Cc
. ,,(3.16)
45
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(J.I B)
[_'_+ ~]c,
de - -UdA(T-t)
U - Overall heat transfer co-efficient.
dQ = -mh Cph dT = me Cpe dt
...
Substituting dQ value from Equn. (3.14) in Equn. (3.IS)
tl - Entry temperature of cold fluid
dQ = UdA (T-I)
... 13.17)
A
1/;,01 };xcltO",lIrt J
7
I
I
'" (J.I~
w. know ,h.,.
0-
nt"
'ph Cr, - T1)'"
"'e C/~ (12-I,)
I
••• (3.20))
fJ
, ':11
e
I
I
···(l.21
From equn (3.20).
(Qr)
Q- C, (It -I,)
o .. VA (M)",
~II..It-'I)
Cc
where (An", - 'o~rirhmjc
••• (J.21l
Q
I
(AT)",
..
UTI
~n
T,-T2
_•
It-t'l
3.2.6
-I'])J
-I,) - (T1
In [~:
92)
In ( 8, .. -UA
lempt.nlure d,trermu
= ::1
Loga,.itllmic mn. temper.tue
tor co•• ter n01t'
00.
cmd fl.uid
=
-
A
iT, - T1 - It - 1
~
"J
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dirrereue
,
'I
~~r:
~ :..... _.,.
J 78 Heal and Mass Transfer
':.
. ...
.
I
Heal Exchangers
Lei
/II Ir _ Mass
J 79
flow rate of hot flu id
... (3.27)
//I
_
e
Mass flow rete of cold fluid
(.,'c, = me • Cpcl
Cph - Specific heal of hot flui~
Cpc _ Specific heal of cold fluid
T 1 - Entry lemperalure o~ hoi fluid
T 2 - Exillemperalure
,i
"
of hot fluid
t1 - Entry temperature of cold fluid
...
t2 - Exit temperature of cold fluid
U _ Overall heat transfer co-efficient.
Substituting dQ value from Equn(3.24).
Let us consider an elemental area dA of the heat exchanger.
(3.28)~
The heat flow rate is given by
dQ = UdA (T-t)
d6
= - UdA (T _ t)
in Equn (3.28)
[J_- _!_]
c,
C/
• " (3,24)
[':6 = T-t]
We know that,
dQ = -nih Cph (dT) = -me Cpe (dt)
• " (3.25)
..!!! = _ UdA [ 1
6
C" -
1]
Ce
Integrating
jI2da
--0 = - [11]
CI, - C
~
[___§J
'I'.
From Equn. (3.2S).
dQ -
,r
(3.28)
[','d6 = dT -dt]
-"'e CfX' dt
dt.~
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e
j
U dA
• •• (3.26)
[.: C"
="''' x Cplrl
[In
all2 = _ UA [ CI,1
1]
Cc
Heal Exchangers
.•. (3.29)
UA (TI
J. 8/
-12) - (Tr II)]
Q=
We know !bat,
T2)= m,Cpc (12-1,)
Q= mhCph(Tl~Q=
C, (12-1,)
C/.(TI-T2)=
... (3.30)
[.: C = m x
~Q=
C~I
[':92=T2-11
9,=TI-12J
Ch(T,-T2)
~IIc, =TI-T'I
...
(3.31)
I
Q
from equn (3.30)
Q= CC (12-11)
I~t =¥I
Substitute
-ci and
h
.•. (3.33)
Q=UA(~n",
...
(3.32)
f
where
(~T)m
- logarithmic
mean temperature
difference
values in Equn (3.29)
c
3.2.7 Fouling Factors
We know, the surfaces of a heal exchangers
after it has been in use for some time. The surfaces
do nOI remain clean
become
fouled with
scaling or deposits. The effect of these deposits affecting the value of overall
heat transfer co-efficient (U). This effect is taken care of by introducing
additional thermal resistance
given by as follows.
Scanned by CamScanner
called the fouling resistance
an
(R) which is
['
Ilttll ttl hliffglffl
.--,;;;,
- - ----~----"Iltl/iitl JlIIi¥
1'01 ('1/1111'" flOHI
).~.8Il:tfll(\II"4lIl"~.
lIy 11_1111 NUlllhlol"
otTl'jUl~r"J'rJuJ11i (NTI))
A hQnl oS hlln~ol' 01111hi) clo~I",110d hy tho l,olllll'ltIJmlc
(t.MTI») when Inlet and olltl~t cj)lIclhloll~
Tallil't\I'IlIUI'~ I)IIlQI\'nu
til''' ~Iledni.id, I1l1t when Ihu prohlel1l INto dUlcnnlne
h.\Il1IWI'tlllIl'e of helll es.chllnl:!lll'. effccllvelle~s
The hem oxchnnger
ef(ecllvellLlsN
where
Ihe Inlot or e_1t
method IH usod,
Is deflned
ns rho mIlo of
actunl hem transfor to the maximum possible hen I unnsfer.
Actual hC1l1 transfer
Effecilvcncss •
Me"1J
Maximum possible heat transfer
or I - P,nlry h:mperll,"re
or hoI nuld "C
'1'2 - r.xh temperature
orJlIJ' iJuld "C
I. _ En.ry temperature of cold fluid PC
'2 - Exl. temperature of cold fluid "C
..!.L
«:
2.
Ileullo.fl by 11111
fluid"
ber o·j'·r'rails f'er U'"lilts (N rU) = --UA
N 11111
O,,"Oc
c.,
= m"Cp,,(TI-T2)"mCCpc(~-tl)
3.2.9 Problems on Parallel now lind
where
Counter now heat exchangers
F(}rmulfle
';'11 - Mass flow rate of hot fluid, kg/s
used
';'C - Mass flow rate of cold fluid, kg/s
I From IiMT data book page no. I 51 (Sixth edition)]
l.Heat tmnsfer Q = VA (LJT)nr
"f
U - Over~1I heat transfer co-efficient,
~n'
m
Cp" - Specific heat of hot fluid, J/kg K
I
Cpc - Specific heatof cold fluid, J/kg K
where,
,A
/leul gullied by cold fluid
W/m2 K
- Area, m2
-
L
" ,
.
ogariihmic Mean Temperature
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3. Surface urea of lube
A = n DI L
.
Difference '(LMTD)
where DI - Inner diameter
_-
J If)
_-.-~
3.84 Heat and Mass Transfer
I
4. Q=,;, x hlg
!I
Heat Exchanger!
where hfg - Enthalpy of evaporation,
Jlkg K
1,85
Specific heat of water, Cpc = 4180 J/kg K
I' ,
S. Mass flow rate
Overall heat transfer co-efficient. U = 280 W/m2K
,~=pAC
Toflnd:
III In a counter flow do"ble pipe I,eal exchanger; oil Is COoledfiro",
8S-C 10 SS·C by water entering al 2S· C. TI,e IIIass flow rat« 0,/
I. Heat exchanger area, (A)
2. Heallransfer rate, (Q)
all
ls 9,800 kgll' and specific I,eal of oil is 2000 Jlkg K. TI'e "'113
flow rate of water is 8,000 kgll' alld specific Ileal of waler ;
4180 Jlkg K. Determine II,e heat exchanger area alld "eallranSltr
rate for all overall I,eallrallsfer co-ejJlcielll of 280 WI",lK.
Solulloll :
We know that,
Heat lost by oil (Hot tluid)
=
Heat gained by water (cold fluid)
Q" = Qe
Give" :
Hot fluid - oil,
(TI' T2)
Cold tluid - water
,;,}, Cpl. (T I - T 2) = ,i'e Cpe (t2 -II)
:::)
Water
(tl' t2)
Entry temperature of oil, T I = 85° C
Oil
Exit temperature of oil, T 2 = 55° C
Water
Fig.l.U
Entry temperature of water, tl = 25° C
:::)
2.72 x 2000 [85 - 55)
:::)
163.2 x 103
:::)
t2
= 2.22 x 4180 x [t2 - 25)
= 9279.6 t2 - (231.9 x 103)
= 42.5° C
Exit temperature of water, t2 = 42.5° C
Mass flow rate of oil (Hot fluid), ';'11 = 9,800 kg/h
9,800 kg/s
3600
=
111111 = 2.72 kgls
I
';'c = 8,000 kg/h
= 8,000 kg/s
3600
I me 2.22 kgls I
=
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mh Cpl. (T 1- T 2)
Q = 2.22 x 4180 x (42.5 - 25)
Q = 162 x 103 W
We know that,
Specific heat of oil, Cpll = 2000 J/kg K
Mass flow rate of water (cqld-fluid),
Heal transfer, Q = ';'C Cpc (t2 - tl) (or)
:::)
Heat transfer,
Q = UA (~T)III
... (I)
[From HMT data book page No. 151
(Sixth editionl]
where
(~T)m - Logarithmic Mean Temperature Difference. (LMTD)
3.86 Heal and Moss Transfer
For Counter flow,
Heal Exchanger« J.87
Given:
Hot fluid - oil,
Cold fluid - water
(TI, T2)
(II' t2)
Mass now rate of water (cold fluid), ';'C = 65 kg/min
65
(85 - 42.5) - (55 - 25)
=
85 -42.5]
In [ 55 - 25
[(6T)m
Substitute
(I)
Imc
= 35.8° C
(6 T)m ' U andQ values in Equn (I)
162 x 103
= 1.08 kg/s
I
Ii = 500 C
Entry temperature
of 'water,
Exit temperature
of water, t2' = 75° C
Specific heat of oil (Hot fluid), Cph = 1.780 kJ/kg K
= 1.780 x 103 J/kg K
Q
~
60 kg/s
280 x A x 35.8
16.16 m21
Entry lemperature
of oil, T I = 115° C
Exit temperature
of oil, T 2 = 70· C
Overall heat transfer co-efficient,U
= 340
W/m2K
toflnd:
Resull:
I. Heat exchanger area, A = 16.16 m2
2. Heat transfer,
Q = 162 x 103 W
I. Heat exchanger
area: (A)
2. Heat transfer rate, (Q)
Solution:
o
We know that,
Water flows 01 the rate of 65 kg/min through a double pi .
counter flow Ileal exchanger: Water is heated from 50'C 10 7
by an oil flowing tllrougll tile lube. Tire specific Ireal of tl,e oU
1.780 kJlkg K. Tire oil enters at 115°C and leaves 11170"(. TI,eovt
hea: transfer co-efficient is 340 WI",1 K. Calcuulte II,e following
I. Heat exchanger area
2. Role of lIeallransfer
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Heat transfer,
Q = ';'e Cpe (t2 - tl)
~Q=
l/~eCpe(t2-tl)
~Q=
1.08x4186x(75-50)
I'
(or)';'h
Cph (Tt - T2)
[.: Specific heat of water, Cpc = 4186 J/kg K]
Q-=-11-3 x-I-03-w~1
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.• "'r'
3. 90 Ileal and Mass Transfer
(380-210),(6T)m'1:'
"In
,'I
(300-25)
[380-210]
'I
Heal ExchQng",~ J. 91
) ,
We know that,
300 - 25
•
I
\
I (6T)nr = '2'1~:~O~
,"
.;
,\
•
Heat transfer,
I
Heat transfer,
IV
(
~ 184' 103
'I'
,,,,
..
=
750. 'l' A,'I,.[:Z18.3)
A' '=
~
7S0·A'>«193.1)
=
1.27 m2
Case (iii)
Percentage of increase in area = 1.27 - 1,12
1.12
I. 1'2m2
.,
= 13.3'% '
Resull:
Case (il)
For Parallel flow,
r :
f.:
"
,
"
I. Area required for parallel tlow = 1.27 m2
,,1,
2. Area required for counter flqw
.
I
I
.I \,
Area for counter flo~
~ U • A (i.\!)",
IArea for parallel tlow A
(6 T)(li
QUA
~
184'103
=>
\
,',
We know that,
Q
(6T)", .
J,
= 1.12 m2
»,
Percentage of increase in area = 13.3 %
It) /11 a counter flow single pass I,eal exchanger
Is used 10 cool Ihe
engille oil from J 50"C 10 55"C will, water; available at 23" C as Ihe
cooling medium. The specific Ileal of 011/s 2125 Jlkg K. The flow
(380 - 25) - (300'- 210)
(6T)nr -
rete 0: coollllg water I("ougl, the hiller lube of 0.4m diameter is
.
[ 380 ~ 25 ]
I
2.2 kgls. Tile flow rate of all througt: II,e outer lube of 0.75 m
300-2LO
, (, (.1 .,/~
dlameler
I (6T)ni - ~93.lo)C f
co-efficient
is 2.4 kgls. /f II,e value of
10 meet tts coollllg requlremenl?
Givell :
Hot tluid - oil.
(TI• T2)
,
= I x 2300 x [380 - 3001
IQ
Cold tluid - water
(11.12)
Entry temperature of oil. T I = ISO· C
wl
Exit temperature of oil. T 2 = 55° C
184xI03
Entry temperature of water. t I - 23' C
46
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the overall Ileal transfer
is 140 Wlml K, how 10llg must II,e Ileal exchanger be
J. 92 Heal and Mass Transfer
Specific heat of oil (hot fluid). Cph = 2125 Jlkg K
Heat Exchangers J.93
Inner diameter. 01 = 0.4 m
Flow rate of water (cooling fluid).
We know that.
mc = 2.2 kgls
'Heal transfer ,Q = U A (~T)m
... (1)
Outer diameter. O2 = 0.75 m
where
Flow rate of oil (Hot fluid). ;,,, = 2.4 kgls
[From HMT data book page Na.151J
(.1 T)m - Logarithmic Mean Temperature Difference. (LMTD)
Overall heat transfer co-efficient. U = 240 W/m2K
For Counter flow.
Toflnd:
(.1T)m
Length of the heal exchanger. L
SO/lit/Oil
B
[(TI - t2) - (T2 - tl)J
:
In [~~ ~ :~
1
We know that.
Heat lost by oil (HOI fluid)
= Heal gained by water (Cold
::)
=
Qc
=
2.2 x 4186x(t2-23)
(.1T)m _ (ISO -75.6)
fl4
- (55 - 23)
In [ISO - 75.6]
Q"
55 - 23
=
= 2.4x2125(150-55)
[.: Specific heal of water. Cpc = 4186J
=
484.5x 103
=
9209.2t2-(211
Substitute (.1T)m ' U and Q values in equn (1)
(I) ::)
::)
x 103)
484.4 x 103
::)
=
I Exit temperature of water. t2 75.6° C I
=
a
240
X
A x 50.2
IA = 40.20 m21
We know that,
Area. A = It x DI <L
40.20 = It x 0.4 x L
::)
::)
Q = 2.2 x 4186 x (75.6 - 23)
I Qz484.4 x 103 wi
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IL=31.9ml
Result .'
Length of the heat exchanger, L = 31.9 m.
3. 94 Heal and Mass Transfer
!II Salllraled sleam .,/16- C b condmsing on Ihe OilIer'''be
0" • single
Heat Exchangers J 95
p.ss heal exchanger.
Tire hea, txc/r"
""It,
1050 kgllr 01 ttI.,er Irom 10"C '0 95- C. The OVerallIre.,
ctH/flciml is 1690 W/,.JK, C.lcul.'e Ihelol/ottlbt,
'J
We know that,
Heat transfer,
I. Area 01"tIIl exd.lller
Q '" ';'h x hlg
91 x 1()3= nIh x 2185 x 1()3
2 R.1t 01 cOlltitludOlf 01steam.
I
rde"/r = 1115 Ulkg
Rate of condensation of steam, n;h= O.0416kg1s
I
We know that,
Gha:
Hot fluid - steam
Cold fluid - water
(TI'
(11,12)
T2
Heat transfer, Q = U A (~T)",
fFronrHA"dolohool.
pDg1!No.151J
where
Saturated steam temperature, T I = T. = 126"C
Mass tlow rate oh"3ler,"
.... (I)
(~T)", - Logarithmic Mean Temperature
'" 10:0 kg h
1050 kg
3600 S
Difference. (LMTD)
For Parallel now
[ (T,-t,) - (Tl-IV]
(~T)",= ......._-[-T-I---II-]--~
Ew..~ ~ofwlter
.. I, =:!O"'C
E.l",c~of"aler.
(~'" 95~C
()\
&"""l':'
ha.llr3:ll3:-er co-efficiem,
'''~~h
In -TZ -12
(126 - 20) - (126 - 95)
'" 1800 W m2K
In [ 126- 20
126 - 95
"'_ISjUkg
'" _IS5
I~
Jk"
r-,
(~-n-",-=-6-1. C--'I
r•.foM :
..Ala oflxa1 a~.
(A)
Substitute (.~T)", Q, U values in equn (I)
2.. Rz:t of mMensatioo of sseam, tit
(I)
s..tm..:
Hcz tn::l5!a:. Q'"
mc Cpr:
Q'" 0.19
f
- I,)
4186" (95 - 20)
loJ W
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Q=U
:::::>
91 x I<P= 1800
:::::>
I. A '" 0.828 m2
I
2.
A (dT)",
A x 61
1Area, A '" 0.828 mll
Itesab:
(.: Specjjj beal of water Cpc = ~ 186 J ~
.Q=91"
:::::>
mh = 0.0416 kgls
.
I.'
J.96 1111(11
anti Moss 7rcm.lfer
In An ,II co,le, of tl" fom
W
,,,,.p,,.,u,,
of tubula, htal txchungt, cools 011t:
.
"10,,,
0190"(' 10 _'j-t by a largt pool 01 Slain 0", '" I
.$$"",,11111 "o"s'a""tmptrtl'llft
""" dI,,_lt,
H,al ExchanglTs
of 21- C. TIlt IlIbt It"IIh Is j"f'
Is 11 ",m. Tilt sptdf'" htal and Sptclflc gravity 'f
Mass flow rate of oil.mh
2",
O"'.
011." 1.45 /(Jlkl K and 0.8 rtsptCllvtly. Tilt vtloclty o/Ihtoll&f
61 cmls. Ca/clIl.'t Iht ove'fllI httll I"mslt,
.. Po)( A )( C
'" 800)(
~
(02»( 0.62
c~fflcltnL
800 )(t(O.028)2
GIvrtJf:
Cold fluid - water
Hot fluid - oil
(TI'
T2)
0.305 legis
)(0.62
!
(11.12)
Heat transfer, Q
Entry temperature of oil, T, :: 90· C
Exit temperature of oil,
';'h)( Cph (TI -T2)
0.305)(
T2 = 35· C
Entry and Exit temperature of water, t, = t2 = 28· C
Tube length, L = 32 m
Cph = 2.45 x loJ J/kg K
41x103W
\L
Q'::
U A (.1nm
Velocity of oil. C = 62 cm/s = 0.62 mls
(.1T>~,- Logarithmic
Mean Temperature Difference, (LMTO).
[(TI - tl> - (T2 - '2) J
In[~::::]
Tofmd:
\
Overall heat transfer co-efficient U
(90 - 28) - (35 - 28)
Solution:
Specific gravity of oil
=
Density of oil
Density of water
In [~=~:
]
jr(-.1-T)-m-C'-=
-2-s.-2"-c-'1
_&
Pw '
0.8
=~
1000
[Density of oil. Po = 800 kglmJ
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' ~ •• '(1)'
[Fro';' HMT data book page No, HI]
where
For Parallel flow.
Specific gravity of oil = 0.8
103 x (90 - 35) .
.r
Heat transfer; .
Specific heat of oil. Cph = 2.45 kJ/kg K
2.45)(
!
/Q
" We know that,
Diameter, D = 28 mm = 0.028 m
~
J 97
Substitule (.1T)", Q. values in equn (I)
3. 98 Heal anti Meiss Transfer
(I)
=>
Q
U A (6T)m
u " 11 0 Lx (6T)nr
41 • 103
StlbilkM! .
Wtknowtflat,
U )( 11 x 0.028 x 32 x 25.2
Heat transfer, Q .. 'Ir{c C~ (c, - tl)(or) mh Cp" (11 - T1)
57.7.9
U
Overall heat transfer co-efficient,
~
U = 577.9 W/m2 K
Resllh:
U = 577.9W/m2
!Z)ln
II
m .Cpc(I:2-.lr) ~ MACp,,(Tr-·Tl)
mc C~ (80 - 30) - mh Cph (220 -100)
mc C~ (50) - ni" Cph (120)
Co
__ --
K
pIIrllll~1flow dOllb/e pipe helll exchanger waler flows III .
Ihe Inner pipe and is healed from JO"C 10 80"C. OflflOwing t
lJ:e ",ini","",
120
nih Cph
50
';'cCpe
.
.
---2.4
';'h Cpli
tile IInnlllllS is cooled fro", 220"C 10 IOO"C. II is desired 10 COOl
exchllllger. /)dermi"e
_"';'cC~
Let't' is"the lowest temPerature 10 wbich the oil is cooled.
terrrperlltll.re 10 whldl Ii,
"'"Y be cooled.
So,
GIwn:
"
.
~
';'cC~
(1- 30) - "'h Cph (220 -t)
I'
Hot fluid - oil,
Cold fluid -: water
..
~
(TI, T2)
';'cC~
-.--)(
,
I
(t -30) -(220 -t)
o/II~CpIr
Entry temperature of water, tl = 30· C
~
2.4 x (t - 30).- (22Q r: t)
Exit temperature of water, t2 = 80· C
~
2.4 x 1-.72
Entry temperature of oil, T I = 220· C
~
Exit temperature of oil, T2 = 100· C
~
~
z(~O:-t)
I
I
\
2.4 t + t ~ 220 + 72
n.
Tuflnd:
~
t (2.4 +1) = 292
.. 3.4t - 292
""
Ii - 85.:88·C I
Minimum temperature to which the oil is cooled, (t)
RDIIlt:
Mioirnum lempermln to wilich ibe oil may be coOled is IS:"O C
' .. _JIII!lp'.......
" ....._--
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!) I•• ,.,.1Id fl-..ItHI txjM,~r,ItOl_I~r is coo/~d
I~
4" C
6, ~
~"'~riIt,
., 10" C. n~"'.ss flo .. ""~
_trr
I
Htlll £.xchangers 1.101
0 ~ III
_trr is 1.1 j,1s ."d IIt~IfHISS flo.' ,." 01 cold..tI/~ris O.S t IfA.,
Irusf~r co-~fJ1ci~,,'s
on bOl" :Sidlit If
rs I,
IIt~ i.di"id .. 1 h.,
680 w,.,z K. fllld tlI~ ana of tIt~ IIHI urhtuf,~r.
~
r
Q=0.S.4186·(36-20)
IQ
I
... (I)
Heat transfer, Q = U A (.:\n",
... (2)
GiwII:
= 33,488 W
We know that,
Hot water (T I' T2),
Entry temperature of hot warrer, T 1 = 80· C
Exit temperature of hot watrer. T2 = 40· C
Entry temperature of cold water, tl = 20· C
(.:\T)", - Logarithmic Mean Temperature Difference. (LMTD)
Mass flow rate of hot water,';'" = 0.2 kgls
Mass flow rate of cold water,
{From HMT data book page No. lSI
(Sixlh edition))
Where
For parallel flow,
me =.O.S kgls
(.:\T)", - [(TI
Heat transfer co-efficients on both sides = hi = ho = 600 W/m2)(
Toflnd:
-II)
-
(T2 -12>]
In [~~ ~ :~]
Heat exchanger area, A
(80 - 20) - (40 - 36)
(.:\1)", - -------
Sol,,'lon:
We know thai,
In [80 - 20]
40- ]6
Heat 1051 by hot water = Heal gained by cold water
Q/.
mIl Cpl. (TI - T2>
0.2'
~
(11-20)
[.: Specific heat of water, Cp
33,488
12
1~llemp"'lure
~ 2093 12- 41.860
-
)6· C
of cold WIler. 12~ 36· C
Scanned by CamScanner
I
me Cpc (12 - tl)
- O.S' 4186.
4186(80-40)
-~
~ Qc
2.708
(.:\T)", - 20.67· C
I
We know that,
4186 J/kg 1<1
Overall heal Iransfer co-efficient
I
U
s
!.+.!..
hi
ho
!. _ ho+
"I
U
hi
"0
... (3)
Heal Exchangers
u=ho+ hj
Inside diameter of the tube, d, = 0,06 m
Outside diameter of the tube, do = 0.08 m
600 x 600
Heal transferred, Q = 1.6 x JOS KJIhr
600 + 600
= 1.6 x lOS x loJ lIs
[u = 300 W/m2 K I
(2)
3600
.•• (4)
Q = 44, 444.4 W
Substitute, Q, U and (~nmvalue in equation (2)
Toflnd:
Length of the tube, L
~
Q=UA(~T)m
~
33,488 = 300 x A x 20.67
~
IA = 5.40 m21
So/ulio" :
We know that,
{From HMT data book page No. 151
(Sixth edition)]
Heal exchanger area, A = 5.40 m2
I!l /"
... (I)
Heal transfer, Q = U A (LH)m
ResIIIf :
0
mltn
where
/rot liquid ~/"trS III 4fWC
JHlroll~1flow ~Ol uc"onf~r,
l~tflIeSal15fYC. Coldfluld
190 Wfm1K rapectfp~1y.
(~T)/II - Logarithmic Mean Temperature Difference. (LMTD)
.
IIIS(J"C lind leav~.f lit I/(/'f,
I"sld~ lind outsld~ htOI tl'flsnsfer cDtffici~nlS
Tht'lhsldt
IIr~ 110 WIm21
lind outsidt
tulH ort 0.06m lind 0.08 m rtsp~cti",Iy.IfI"t
hour Is 1.6 x IOSk), find tht Itngt" oflht
dillmdtfSll/
,
For Parallel flow
[(Tt-tt)
htat transfmS
tub« rtqulrtd.
In [~~
- (T2-tV]
~:J
GiPtn:
(400 - 50) - (250 ~ 110)'
,,
Entry temperature of hot fluid, T 1 = 400" C
I [400-50]
n 250- 110
'
Exillemperature of hot fluid, T2 ',:"250 C
0
I
Entry lemperature of cold fluid, 11;" 50" C
210
0.916
Exit temperature of cold fluid, 12 = I 10· C
I
I
I
1,101
hjhO
Inside heal transfer co-efficient, hi = 120 W/m2K
Outside heat transfer co.efficient, ho = '190 W/m2K
L ..n,.
_
Scanned by CamScanner
I (M)m = 229,25· C I
...
(2)
('
We know that,
Overall heat transfer co-efficient
I
Glvtn:
Hoi fluid - oil (TI, T2),
I
I
-+-
ro
-U
r;
h;
Entry temperalure of oil, T I - 200" C
hO
Exit lemperalure of oil, T2 - 120" C
_~.
0.03
_1_+_1120
190
Enlry temperature ofwaler, II - 25· C
Exillemperalure of waler, 12 - 70" C
_ 0.0111 + 5.26 • 10-3
Specific heal of oil, CPh - I.S kJlkg K
- I.S • 103 J/kg-K
.1. - 0.0163
U
I U -61.35
W/m2
KI
Substitute, Q, (AT) .. , U values in equn (I)
(I)
~
~
Overall heat transfer co-efflcienr, U - 400 W/m2K
Toflnd:
I. Rare of heallrlnsfer,
Q=U A(6nm
Q = U • II • do • L • (AT)m
~
Mass flow rate of oil, ';'h - 0.8 kg/s
44,444.4 = 61.35 • II x 0.08 • L • 229.25
IL=
12.57ml
Q
2. Mass flow rate of water,
me
3. Area of heal exchanger, A
Solllllon:
We know that.
Heat transfer, Q = mh CpIJ (TI - T2)
Raid, :
= 0.8'
Length of the tube, L = 12.57 m
[!!lIn a cOllnl~rflow dOllb/~ plp~ heal exchanger Is IIs~d 10 cool lite
/ro1ll 20frC to 12frC wilh waUr ava;lab/~ al 15"C as tht c
~dllllfL TIlt exh I~mp~ratllr~ 0/ waUr Is 70~ Th~ sptclflc hm ollis 1.5 tJllg K and Iht ",assflow ret« 0/011 Is 0.8 /igls.1f
ol'Uall hea, lrans/t, c~fflcitnl
Is 400 WI",zK,find Iht /ollow
I) Ratt 0/ .tat trans/tr
1) Mas flow ralt 0/ wal~r
J) Ana
0/ Ittal excllangt,
~:__--
Scanned by CamScanner
I
1.5' 103 • (200- 120)
Q = 96,000 lIs
I
We know Ihat,
Heal lost by oil (hot fluid)
= Heal gained by water (cold fluid)
~
= ';'eCpc(12-11)
~
mhCph(TI-T2)
0.8'
1.5 x 103 x (200-120)
=
me 4186 • (70 - 25)
x
[.: Specific heat of water Cpc = 4186 J/kg KJ
EM
M
1. 106 Hem and MUll {rumltr
Ilwl t,u:It{j"y"'~
I
Man
now rile o( willer, me
rm
O,S(ifj kgl, 1
In /III (III t(lt/lt'I"'
III
For Counter now
fIre by utili/( u CIJIIIIIIIIIII/llt' Ill'w III ]1"(: 111f~"""'I /hiw '"k
(1/1111 I, 'JIJ(J 1If1111 lind Ille mil" //(1'" 'lilt
[OI-tV
(I''' ,",
T2 -II
(i/ Wille, I. ?IHI 1If/1I,
GiveY(ill' chotc« fl" II plI'lIl1el/III'"II' Wlllllt' fluw helll e~hllnKlu,
- (T1-11)]
I"[~I
J_t,,:_
I,J/em, (1111. t(lq4:d /",m ?lre
II 'ub,kIlIWII
[From IIMT dato bOtit,
wilit ",atO/U', If II,e (Ive'lIl1Iteal "alll/"
tfI·el/lcleII,l.
Jln" lite ore« ul II", Itelll a.t:lIIIII/(".
Talle ,pecl/Ie I'wa/
]11 WI",J K.
,,111,
1 kJllI'(C
(luge M.I j I (Shih edlllo,,)}
HOI Huia - oil ('1'" '121.
(2()() - 7(1) - (12() - 25)
I" [ 200 -70
1
'-,nlry ltmpc:rawre
120 - 25
(IiTI", - 111.112' C
of oil. '1'1 - 7(1) C
Exit temperarure of oil. '1'2 .. 40" C
Emr)' temperature
I
Cold Huid - ....ater (I,. Iz)
01' water, II .. 2$° C
:n
'1 he rmm flow rate of oil.
I
h ..
90() kglh
9()O
.. --
k~;'
3600
We koow that,
- 0.25 kg/s
The mass 1101'1rate of water,
';'c" 701) kg/h
Substnute, 0, U, and (6'1')"1 values.
;:,
701)
.. - 3600
96,000 - 400 < A " 111.82
I
kg1s
.. 0.194 kg/s
A - 2,146 rn2
Overall heat transfer co-etflcient,
Rtfllll:
Specific
heat of oil, Cph"
I, 0 - %,000 J/s
2. ';'C - 0.509 kg/s
2 kJlkgo C
n 2 x 10J J/kgo C
rQ/llld:
I. Choice of heat exchanger
3.A-2.146I11l
2. Area of heat exchanger.
47
Scanned by CamScanner
U = 20 W/m2K
(Whether
parallel flow or counter now)
SlIiIIIiM :
Heal £TchangC'Ts 3.109
w~know that.
\\~ blow thai,
H~
Heat transfer. 0 = n~1, CpnI (TI - T,)_ or me
• CP" (t 2 - tl )
klst b)' oil (Hot fluid : Heat gained by water (Cold
Q;.
:::>
Qe
0 = IIi"~
:::::-
Cph (TI - T!)
= 0.25 x :2 • 103
(70 - -10)
10 = 15.000 J/s I
[.: Specific heat of water ~:
15,000 .= 812.08t112
... (3)
Substitute. O. U. and (6T)m values in equation (I).
41 86 Jik&~
20.302.10
15.000 = 20 • A x (20.26)
= 43.47" C
IA = 37.02 m21
[Exit temperature of water, t2 = 43.47" Cl > T2
Result:
Since 12 > T 2, counter flow arrangement should be used.
I. Choice of heat exchanger - counter flow arrangement
We know that,
Heat transfer,
Q = U A (6 T)m
... (1)
2. Surface area, A = 37.02 ml
3.2.10 Problems on cross now heat exchangers
For Counter flow
(or)
Shcllandtubeheatexchangers
Formulae used
{From HMT data book
I. Q = F U A (LJT)
page No. } 5} (Sixth edit;
where
F - Correction
(70~43.47) - (40 -25)
In [ 70 - 43.47]
page No. 151 (Sixth editiont]
m [counter flow]
factor - (From data book)
U - Overall heat transfer co-efficient,
W1m2 K
(6T)",- Logarithmic
difference.
mean temperature
40 - 25
For Counter flow
11.53
(6T)",
0.569
I
(6T)m = 20.26° C
Scanned by CamScanner
I
[(T I - (2) - (T 2 - tl)
In [~~ ~ :~
... (2)
1
1
J 1/(/ /lila/
m,d MII,u 'l'ril/l.I!m'
where
T, ~ 1'.,11' y llilliplinlh,r~
or hlit IllIllI, "C
Tl - Exit tVIIII'~rlltllf' (,floul 1111111,
"C
I, - Elllry IClllrCl'lllurC uf collllluld,
"C
(A'I)",
Ex II IClilpCfhlUrc uf' cold Iluhl, ,.C
12·
LUIIlH'ltlllolc
'"V,II]
IC'"11crUlurcdlllere'ltt
1'01' eout1lcrll,.w,
For 'ounrcr llow,
],
/lilil/lo.f/
by "01 PIIIII • /lillIl/lIIIIWII
Q/,
•
"'I, Cpl,(T,
n
b)' 1:11111
Jill III
Qr
- 1'2) e "tc
/)('(12 -I,)
!I1 III UcrossJIIII~ I"!III I'.H'I"IIIKers, hoi" fli,lll.f u","/xell, 11111flllid 1111/.
II
(380-210)
In 80 - 2 10
JOO - 25
Calli flulds enters III 15' C und teuves at 110' C. Cult'lllllfe th,
requircil SIIr/II"1! areu II/ heat t!xcllanger, Take aVl!rtllll'l'lIllr""J/fl
co-t!J!1cil!lIl is 750 WI",} K, MIISI flow rail' 0/ hot flllill
- (300-25)
[2
IpeclJl.: IWII 0/1JOO Jlkg K tillers ut JIIO' C 1lIIlllell,'eJ at JOO'C,
2 I 8 J. C
is I kg!.!,
1
I
Given :
Heat transfer, Q ~ "'" C"" (T I -. T2)
Specific heat of hot fluid, Cph ~ 2300 J/kg K
~
Entry temperature of hot fluid. T I~ 380· C
Q = I • 2300 (380 - 300)
I Q ~ 184 • 0 I
Exit temperature of hot fluid, T2 ~ 300· C
I 3 W
Entry temperature of cold fluid, t, ~ 25· C
Tofind
Exit temperature of cold fluid, t2 ~ 210· C
correction factnr F. refer IIMT data bookpag.
110
16 I (Sixt/, edition}
{Single pass cross flow heal e.rchanger - Both fluids unmixed]
Overall heat transfer co-efficient, U ~ 750 W/m2K
From graph.
Mass flow rate of hot fluid, IIi" = 1 kg/so
12-11
210-25
X.xis value P = -= --~ 0.52
T I - II 380 - 25
Tolilld:
Heat exchanger area (A)
T I - T,
Curve value R ~ ----
Solution :
12 - tl
380 - 300
~ -::-- __
210 _ 25 ~ 0.432
This is cross flow, both fluids unmixed type heat exchanger.
For cross flow heat exchanger,
Q ~ F U A (6,.)
m [counter flow J
... (1)
{From HMT cia/a book page No. 151 (Sixth dition)
Scanned by CamScanner
X.XIS value is 0.52, curve value is 0.432 corresponding
value is 0.97,
'IX"
i.e.
IF 0.971
=
Y
.
,I 11-: II,' II ,,,,,I ""L~,'
~--.-----------------
tr'"L'J""
G/lltlII
:
1101 l1uid
water
Cold fluid - brine SOlulion
(TI' '1'2)
O,()
Enlry temperature of water, T I - 20" C
0.8
0.7
r
(II' (2)
Exil temperature
of Wilier, 1'2 - 7° C
Entry temperature
of brine solution,
Exil temperature
of brine solution, t2 • 3° C
II •
_2° C
Heat load, Q - 5500 W
0.6
Overall hcut transfer co-efficient,
0.5
U - 800 W/m2K
Ttl fillll :
Area required (A)
0
P - 0.52
So/uliml:
Shell and lube heal exchanger
FlK' ,US
- One shell pass and two lube passes
For shell and tube heal exchanger (or) cross now heat exchanger.
Q
Substitute Q, F, (6 T)", and U value in Equn (I)
Q -
(I)
III
where
F U A (6T)",
A"
Correct ion factor
m21
Logarithmic mean temperature dilTerence for counter now.
For Counter now,
R~.full :
(ATJ",
Surface area, A • 1.15 1112
iII lit II rtlr/KulIl/IIK
.fIIllIl/OIII:tller/IIK
pllltll
wilier if ('lIoled IfI"" 1"" C III 7" C by IIr/lI~
(20
111-1" C 1I11111elll,IIII(III"" C 1111! tle.fll(lI Itellll'llld
Is 55(JO W 111111
lite Ilveflllllll!lI(
WIIII( arc« re"u/rl1ll
trunsfer
"II-IIU'c1elll
when ".J/III( II ,f/wllllll,lluhe
wilit lite wilier milk/III(
I.~ROO W/",1 K,
Scanned by CamScanner
I"
3)
(7 I 2)
[1.Q__l]
7 2
I
heat excl"II/Ktif
one ,Jllell/ ,I(IU ,,,,,111,,: IIr/1I1! ",11111111(
IW(I
Illhe IJlI,f:fI!,~,
.• , (I)
218.3
(6'1')",
A - 1.15
[counter tlow]
{From /-1M,/,data book page No.151 }
F
184 x 10) - 0.97 x 750"
F U A x (AT)
[(AT)",
e
12.S2:iJ
"'I
11.1111.,,;11 Mill'
III/lilil,
1I"II\llil
' mil
1111, 111111 1III /111 I',III /I"
/ltllil l'II/IIIIIW"t
,1"/11 IIIIuA /111,i Ilfl /lll
/11111' 11111/1 /iini 11II11111'fl11I111J
1'1'"' UIIII'Ii,
'",,_ I'"hl , I' •
I"
"I I 1\ IAI JI/I
I I'
"i ~(I(I ,
1I'liI, .1,1, I 1\ • ') ~7
III'
II ',II 111/ J
HjI~1I111
II ~:
i "II I'll ¥hlll~,II -
I' , ' •
J
01
II
II
111/ 1IIIIIp (~, Jill' (), '/ ~III' " tt~'IIF III ,,",,1111 (')
/IIINliJ1/
II
II.
111'111lit h!llil "Xi h~I1I!"I,
'I, 'Il
,I\I?
I~ I I
~ , 1,
11
Xft"i
value Iii 0,22, curve
1'111111)
Is 0,94.
II- fI,~.11I1i/
I.!J ,\'11111'11111(1 fill1l11l1,,, /111' (_' II (IImlllllllflll,'"
",I'('IUlllfler,
n,e (Jiml/"fI wnter
", lUI" C. ClIlI'lIlf1ll1ll1i! '''1/1'''1111111,·
III tI '''''IIIflelllelll I~'
IIIltl""""P"lIluflllll//'''IIflll!
(II) counte« flow
(t) Cflllf JlIIW.
(II) 1"""l/ul/lIlIY
fluid - steam
"01
Cold (Iuld - water
(T,. T2)
(I" '2)
0.94
Saturated
0,9
Entry temperature
of water,
tl = 25° C
Ex it temperature
of water,
t2 = 80° C
0.8
R = 2.6
0.7
d,"11 111111 IlIh, ,"",
,1"'lIf. 1/," 1111"III J ,. (' "",1 II"",_'
GII'c" :
l.e.!I~;;~~J
F
I1/'
steam temperature,
T, ~ T 2 ~ 120° C
Tojitlll :
(tlT)III for paraliel flow, counter flow and cross flow.
Solution :
0.6
Case (i)
0.5
For Parallel flow,
(t1T)", =
0
P = 0.22
Fig.. U6
Scanned by CamScanner
[FromHMTdatabook
page No. 151]
3.116 Heal and Mass Transfer
Heal Exchangers JI17
(120 - 25) - (120 - 80)
In [
XaxlS va Iue P=~
120 - 25 ]
120-80
I
= 80-25
TI-tt
120-25
P = 0.5781
· ., (I)
0
(~T)m for parallel flow = 63.5 C
T t - T2
120 - 120
Curve value R = --= --t2-tl
80-25
Case (ii)
I R=O I
For Counter flow,
>:axis value is 0.578, curve value is 0,
So corresponding
~
[correction
Yaxis value is I
Factor
F = I
I
(120 - 80) - (120 - 25)
In [
(3)~(~nm=Fx63.5°C=
120-80]
""i"2O-2s
I
(~T)m
I x 63.5
for cross flow = 63.5° C
I
... (4)
· .. (2)
0
(~T)mfor counter flow = 63.5 C
From (I), (2) and (4) we came to know when one of the fluids in a
heat exchanger changes phase. the logarithmic mean temperature difference
Case (iii)
and rate of heat transfer will remain same for parallel flow, counter flow
For cross flow
(~T)m = F x (~T)m for counter flow
I (~T)", = F 63.5 I
x
· .. (3)
0
and cross flow.
3.2.11 Anna
University
Solved Problems
III In a double pipe counter flow I.eal exchanger, 10,000 IIg1hr of an
where
oil huving a specific "eal of 1095 J/lIg-K Is cooled from 80'C 10
Fe correction factor
(Refer flMT data book l'oge No. 160)
[Correction factor for sillRle pass cross flow heal exchangerone fluid mixed, other unmixed]
50'C by 8000 kglhr of water enlering al 25'C Deurmine
exchanger
area lor all overall heal trensfe» co-efflclent
JOO WlmlK.
Take Cpfor waler as 4180 Jlkg-K.
IDee-ZOU4.
Scanned by CamScanner
Ihe heal
of
Anna Unlv]
3. 118 Heal (111<1Mew '/'rIl1l.'1~cr:_
.__
-------
_----------
!.fI'.!!!wl/ £'rchan/(lfr., J 11'1
Ilent trnnster. 0 .. ';'1 C
GII'I.'n :
,
Cold lluid - water (t I' t2)
UI
111
I' 2
.
'.
10 000 kg/hI'
The mass flow rate (If oil (Hot fluid), "'''
'
10,000 kg
1
1,ot
fl'd
'1 ('1' '1')
=-
• 3600s
We know Ihal,
:-2 :27 kgl~
Hcallrallsfcr,
G",
'd)
e
Q. 2.22 • 418() , (43.85 _ 25)
Q
17492.
iOl3
Q - UA (~T)",
(~T)I/I- Logarithmic
,
,"'e-
- 8000 kglh
= .~
.•. (1)
page No. 151 (.')/xlh ed/l/olI) /
Mean Temperature Diff
, erence. (LMTD)
For Counter flow,
kg/s
3600
Gne = 2.22 kglS]
Entry temperature of water, II = 25· C
2
Overall heat transfer co-e ffici
icient, u -- 300 W/m K
(SO - 43.85) - (SO- 25)
In [ 80 - 43.85]
Specific heal of water, Cpc = 41S0 J/kg-K
50 - 25
Toflnd:
Heal exchanger area, A
I (~T)m = 30.23· C I
~~:
I
Substitute
Heat lost by oil (Hot fluid) = Heal gained by water (Cold f1u~ql.I
Qit
=
Qc
(I) =>
I
=>
=>
2.777 x 2095 (SO- 50)
174.53'
103
t2
=
103t2-
U and Q value in equn (I)
Q= UA(~T)m
174.92 x 103 = 300 x A x 30.23
I Heat exchanger area, A 19.287 m21
231.99 x 103
=
43.SSo C
Result:
I
l
Scanned by CamScanner
(~T)m'
A = 19.287 m2
2.22 x 41S0 x (tz - 25)
= 9.27 x
=
'I)
Where,
of oil, T 2 = 50· C
Mass flow rate of water, (Cold flui
lIlt
'1')
.--~.
1 - 2 or ,;," l)" (12
[From flMT dau, hook
Specific heat of oil, Cph = 2095 J/kg-K
'1 T 1-- SO· C
Entry temperature 0 f 01,
Exitlemperature
I
('I'
Heat exchanger area, A = 19.2S7 m2
3.120 Heat and Mass Transfer
III In a counter flow double pipe heat txcl,anger, water Is heated /ro",
Heat Exchangerl 1.12/
25"(' to 6S'C by an 011 with a specific I,eat of /.45 KJ/Kg K Gild
Q = UA (.1T)",
Heattransfc:r,
mass flow rate is 0.9 Kgls. The ollis cooled from 230·C to 160· C.
... (1)
If tl,e overall heat tronsfer co-elJicient is ,120 WI",2 ·C, calcu~
[From HMT data book
.
the following.
(Sf)m- Loganthmic
J) Tire rate of heat transfer
I
3) The surface area of tIre heat txchanger
I
page No./S/ (Sixth edition)]
Mean ...
•emperature Difti
.
2) TI,e mass flow rate of water
.
ereuce (lMTD)
For Counter flow,
I
Given:
(6.1)", = [(TI-t2)-
[May-2004, Anna Un~J
In
(T2-tl)]
rTI - t2l
lT2 l
-t
Hot fluid - oil (T I' T 2)
Entry temperature
Cold fluid - water (tl'~)
(230 - 65) - (160 - 25)
0
of water, tl = 25 C
In [230 -65]
_;_____.._:___160 - 25
Exit temperature of water, t2 = 65 C
.-:- __
Specific heat of oil. Cph = 1.45 kJlkg
1(6.1)", = 1·49.4~ C I
0
= 1.45 x 103 Jlkg
Mass flow rate of oil,
mh = 0.9 kgls
Substitute
(I)
Entry temperature of oil, T I = 2300 C
(6.1) '" Q and U val'ues ID equation
. (I)
=>
=>
Q=U A (.1T)m
91.35 x 1()3= 420" A )c 149.49
Exit temperature of oil, T 2 = 1600 C
Overall heat transfer co-efficient,
IA ~ 1.455 m21
U = 420 W/m2 0 C
We know that, For cold fluid,
Toflnd:
Q = mc Cpc (t2 - tl)
I. The rate of heat transfer, Q
2. Mass flow rate of water,
me
3. Surface area of the heat exchanger, A
[.,' Specific heat of water, Cpc = 4 r~JA& I()
Solution:
Heat transfer,
Q = mh \ph (T I ....:T 2)
=
0.9 x 1.45 x 103)( (230- 160)
IQ = 91.35
x
103wI
I mc
= 0.545 kgls
I
.
Result;
1. Heat transfer, Q = 91.35 )( 103 W
2. Mass flow rate of water,
me = 0.545 kgls
3. Surface area of the heat exchanger, /Ii. = ··1;45.fm2
Scanned by CamScanner
--
lin
Heal and Mass Trallsfer
\]
,4 ('o,mterflow cOllcelltric tube IIeat excllallger is used 10 cool engille
)( 2130.(160-60)
25.(' as tl,e coolillg mediuIII. Tl,eflow rate of cooling water tllrougl,
42.6 x 104 == 8372
,,: . .'
(rr 25)
tl,e i""er tube 0/0.5111 is 2kgls wllile tl,eflow rate of oil tllroug/, tl't
Ollter diameter
42.6 x 104 - 8'37
= 0.7 ", is also 2 kg/so If U is
I • ~.
250 Wlm1 K, "ow long must tile IIeat t.'(cilallger be 10 meet its
cooling requirement?
t2
(11.12)
(tl,t2)
~
"
Specific heat of oil (Hot fluid). Cph = 2130 J/kg K
75.88 C
0
==
.
(or.).ni"cph (T 1- T2)
Q = 2)( 4186 x·(15.88 -2'5)
t:
I Q -'425.96)( '10 W r --'
~~~-:--_.:..._
__
3
Entry temperature of oil, 1 I = 160° C
We know that,
Exit temperature of oil. 12 = 60° C
Q ,;; 'u A (,1T) m ,
Heat transfer
Entry temperature of water. tl = 25° C
Where,
Inner diameter. DI = 0.5 m
Flow rate of water (Cooling fluid). mc = 2 kg/s
.
t2 - 209.3 x 10J
~
Heat transfer , Q == ni c cpc (trtl)'
Give" :
Hot fluid - oil
'
Exit temperature ofwarer • t2-- 75.:8~oC
(May-200S, Anna Univj
Cold fluid - water
..
[ .. S
2,
'; pecitic
hear of water Cpc -- 4/86 l/kg KJ
:
oil (C == 2130 Jlkg K) frolll 160· C to 60· C willi waler available at
outer a"nulus,
_
. , 2·X4186)( (r. -25)
S
I
I'
f,From H MJ. daf(1 b qDf"ol- pfJgB,no., [1$11
'
(,1T)m-Logarith~'ii;M~an'Te;':;pl
,
'.
"
,,','.
I
','
For Counter flow,
Outer diameter. D2 = 0.7 m
••• (I)
"'.
.,'
"f
erature Difference. (LMrD)
•
'
\' .
t ..
Flow rate of oil (Hot fluid). ,nh = 2 kg/s
Overall heat transfer co-efficient, U = 250 W/m2 K
"
Tofind :
Length of the heat exchanzer
to , L
(,1 T)m
Solutio" :
= (160 '- 75,88) - (60 - 25)
In [160 ~ 75.88]
We know that,
,
Heat lost by oil (Hot flUI.d) -- Heat gamed
.
by water (Cold fluid)
49.12
=>
0.8768
Q/r
Qc
60 - 25
,
, ,
48
Scanned by CamScanner
'
3.124 Heal and Mass Transfer
Substitute (~T)m' U and Q values in equation (I)
r
Q'
Heat Exchangers 3.125
Given:
Mass flow rate or hot liquid'';'h = 4.2 kg/min
(1)
Q =U A(~T)m
=>
=>
I = 0.07 kg/s I
mh
425.96 x UP = 250 x A x 56.02°C
=>
~
Specific heat of hot liquid,
= 30.415 m~
Cph = 3.5 IUllegK
[CPh = 3.S x loJ Jlleg K
We know that,
Inlet temperature of hot liquid, T, = 1300C
Area, A = 1t. D, . L
Specific heat of water,
30.415 = 1t x 0.5 x L
=>
Cpc = 4.18 lUIkg K
Mass flow rate of cooling water,
Length of the heat exchanger, L = 19.36 m
'0 de.ermint th« inlet or exil
,empera'ures of heat exc/ranger
III A parallel flOH/:hea. exchanger is used '0 cool 4.2 kglmin Of 1101
liquid of specifiC hea' 3.S kJlkg K a. 130' C. A cooling wa.er of'
.empera.ure
kJlkg K is used for cooling purpose a••
of IS·C. Tire mass flow rate of cooling water is
Overall heat transfer co-efficient, U =
3. Effec.iveness of heat exchanger
Take
Overall heat transfer co-efficient is HOO Wlm2K.
Heat exchanger area is 0.30 m2
Scanned by CamScanner
I
1100 w/m2K
Tofind:
I. Outlet temperature of liquid, (T 2)
2. Outlet temperature of water, (t2)
3. Effectiveness of heat exchanger, (e)
Solution:
Capacity rate of hot liquid, C = mh)( Cph
= 0.07 x 3.S x 103
of liquid
2. Outlet temperature of water
= 0.28 kg/s
Area, A = 0.30 m2
17 kglmin. calculate .he following.
I. Ou.let ttmpera.urt
mc = 17 kg/min
Imc
ResuU:
specific heat 4.18
.
Inlet temperature of cooling water, tl = 15°C
Length of the heat exchanger, (Single pass), L = 19.36 m
INote) NTU me.hod is ustd
I
ICpc - 4.18 x, loJ·Jlleg K!
L=19.36m
3.2.12 Solved problems on NTU
[Number Transfer Units) method
=,
IC=24SWIK!
Capacity rate of water,
... (1)
C· = me x Cpe
= 0.28 x 4.18 x 103
IC 1170.4 W ! ... (2)
=
IK
7
J~.~/l~6~H!,a~,~a~n~d~A~la~s~s~~~a~m~~~fr~
----------------
__
-
From (I) and (2).
Cl1Ii~= 245 W/K
CmIX;
CmiR = ~=
C'ma.~
1170.4
CmiR
64%
1170.4 W/K
0.209
.
Effecliveness
E
.•• (3)
= 0.209
CmIX
NTU
Number of transfer units.
UA
=_
CmiR
NTU=
Maximumpossible
1100 x 0.30
245
.. ' Qmax
.' ."
. •• (4)
[NTU = Ij~r
Tofind effectiveness E, refer HMT data book page no 161
(Parallel flow heat exchanger)
I
",,' Cmin'(T, -I,)
. -
Qmax
245 ( 130 - 15) \ '.
= 28,175 W
I
Actual heallransfe! rate
Q
C·
Curve -+ ~
Cmax
•!
'I~''.
. '.
= 0.209
"
EX
Qmax
0.64 ~ 28,175
18,032 'w
I
We know thai,
Heal transfer,
Corresponding Yaxis value is 64%
i.e.,
hellf Iransf~r
\
From graph.
Xaxis -+ NTU = 1.34
1.34
NTU
[From HMT data book page no. 151}'
IE =0.64
1
.'
.
Q
;',; C;(12 -I,)
18,032
0.28' x4.18 x 103 (12 - 15)
18,032
.11 70.4 12 - 17556
30.40° C
Outlel lemperalure of cold waler,
Scanned by CamScanner
12 = 30.40·
C
J. 128 Heat and Mass Transfer
We know that,
Mass flow rate of oil,
Heat transfer Q
mh Cph (T, - T2)
=>
0.07 x 3.5 x 103 (130 - T2)
18,032
18,032
=>
T2
=
mit .. 520 kglh
'"
k
gts
mit '" 0.144 kgts
31850 - 245 T2
Inlet temperature of oil,
= 56.4° C
IOutlet temperature of hot r 'd T2 - 56 .4° C I
T, .. 95" C
U'"' 1000 W/m2 K
Overall heat transfer co-efficient,
IqUl,
Heat exchanger area, A '" 1m2
Resllil :
I. T2" 56.4°C
Toftnd:
I. Total heat transfer, (Q)
2. t2 '" 30.40° C
2. Outlet temperature of water, (t )
3. e= 0.64
2
Ill/"• COll"'~rflo"" "tal t.t:cllangtr, water all0" Cflowing .llltt I'GIt
01/100 k,l1I. It is "taltd by oil 01Sptcific "tat 1100 Jilt, K/10""u.,
.1 '''t rtlI~01510 kglll al inltl I~"",~ralUrt 0195" C Dtlt""I,,~ lit,
lollo""i",
3. Outlet temperature of oil, (T )
2
Sollilian :
Capacity rate of hot oil, C ,.
..
0.144"2100
[c ,. 302.4 W~
1. Olld~ It'fffptrtlhlrt 01",.,,,
J. Dulin tt"'l'trtlhlrt qf oil
Capacity nile of water,
C·
t!.\·C".",u.MJ is Iwfl.
GI\wt:
HOI Iluid - oil
... (I)
';'e" Cpt.
,.
OI~rtllI"Ht ''''"sltrr CtNffiritlfl if 1000 H~:A'.
ol..! Iluid - "'liter
,;,,,,, C
plt
I. Tot.llu., lTa"sltr
HH'
520
3600
0.33"4116
[C
1381.3 W/~
(SfHcijic
Ir"at o/""Qttrr Cp.,. -
... (2)
1/86 Jlt, IV
From Equn (I) and (2)
em In - 30•. 4 WII(
1381.3 WI"
o JJ ~ s
l'1
e~I(IOJ~1\
-0:.1.4
UIU
.. 0,
O••dl
I'
...( l
-
Scanned by CamScanner
---------
Number of transfer units,
, ,{Fi;om. HMr t/QtQ
'NTU"'~'
book page no lSI}
Cmin
1000.)( 1
.' i
302.4
[BTl)''''
}1l '.
'II
I
: .....,.
,(4)
: T(I),fi;,d.ejf~ctiv~n~s ~j':te/P' HMf: t!.qIa· bOok Page' no 163
(COU.n,{~"jlpw, 'h~at,~C#J(;lnger)
f'rOIll graph,
Xaxis -+ NTU
.~
•
I
.
."
\
Heat trapsfer, Q
",q
= 3.3
I:;
2J,5~
I
: .
I ~, .
!.
Corresppnding YlI?'isvab,ae is 0•.95
i.e.,
Jl
.. ';'cCpc{~-tl)
'. ~
0.31 ~ 4~'86..{~':'~Q.")
i-
.:
"
.l'~-:'pc.'rffllM,Jlkg KJ
~='
r
,[ e = 0:.95J
,I.
=> .:' 2I"S46 = IJB}'.38;.~""','J.7,6'1f/4~
r::::=--=>-----:-""":':,~I( ===- =3=S.S='C~)~ , "",' -:
Curve-+ __..:..;...=··Q;~':l8
1 ; t;
. \ ',"
'
I"
Cmin
CmIX
.... : -.' ,'.'\
\\\'\
We knQW, m.t,
1~~~~~eOf~~rit2'
•
Cmin
___.;. = 0.218
Cmax
:I
."
.,,1
'CI·" .../',
='; 35.5
0
() ~. I
,
•
'I
\
1
We know that,
H~at tl;~fer,
Q
95%
. : : ?;I~~,~~~~,(~~-
21,5~",
,~,,2;~,?:2,8:-:-,.Wi1T~.,.
IT2
EtfecQ~eness
I o"tlot .... _ofoil.
e
\ . :'{ L .:
',I'
I. Q = 21.546 W'
3.>
'.' NTU
Maximum possible peat transfer~
('
,.'
. Qmax;"
..
'
2. T2 = 23.751)C
3. ~ = 35S"C
~~in ("(1,;;-11)
302.4 (95 -,20)
I Qmax
. 22,680
:W, I,
Scanned by CamScanner
;:
!
I
I ~
=
23.75o
:'
t ". ,: 1,
. ,'.; .'
i,
1.
'I·
I ," ":.""
';1
I
to';
.
, J I •I
'/
I,'
cl
T, ~ ~1S"C
;.: ~ _ t ,',.' ! :..
Res"lt:
~2)\
21,5~
I
,
1 I •~ :".
/ I'
(,
.;
j.,
(, ,I
',
1
ll32
-
Heat and Mass Transfer
ill 111 • cross flow both fluids unmLted heat exchanger, water at 6' C
flowing .tthe rate 0//.15 legis.It Is used to cool 1.1kgls 0/ air that
Is initially at a temperature 0/50' C. Calculate the /ollowing
Heat Exchangers 3.111
Capcity rate of air
C = mh x Cph
= 1.2 x 1010
I. Exit temperature 0/ air
2. £Xii temperature 0/ water
Assume overaU heattrans/er co-efficient Is 130 Wlm1 K and area Is
IC=1212WIK!
... (2)
From Equn ( I) and (2), We know that,
23m2.
Cmin = 1212 W/K
Givell :
Cmax = 5232.5 WIK
Hot' fluid - air
Cold fluid - water
Cmin _ 1212
C
- 5232.5
max
Inlet temperature of water, tl = 6° C
Mass flow rate of water,
Mass flow rate of air,
me'= 1.25 kgls
= 0.23
Cmin
= 0.23
Cmax
mh = 1.2 kg/s
.• : (3)
Initial temperature of air, T I = 50° C
Overall heat transfer co-efficient, U = 130 W/m2 K
Number of transfer units, NTU = VA
Cmin
Surface area, A = 23 m2
= 130 x 23
Tojlnd:
[From HMJ'data book
page no.151J
1212
I. Exit temperature of air, (T2)
INTU=2.46
2. Exit temperature of water, (t2)
Tofind effectiveness
I
... (4)
E • reier
HMT UIdata b 00k page
'./,
SollItion
no 165J
(Cross flow. both fluids unmixed)
We know that,
From graph,
Specific heat of water, Cpc = 4186 J/kg K
Specific heat of air, Cph = 10 I0 J/kg K (constant)
Curve -+ ~
Corresponding
= 1.25 x 4186
=
Scanned by CamScanner
'"' 0.23
Cmax
x Cpc
IC 5232.5 W/K!
= 2.46
C·
we know
Capacity rate of water
C =mc
Xaxis -+ NTU
Ya.xis value is 0.85
i.e.,
... (1)
Ie" 0.85 I
r
Heal exchangers 3.11.5
3. 13.4- ·H.«r a"d M(JssTransftr
-- We knoW Ihat.
Maximum heat transfer
_ Cmin (T, - tl)
Q
Heal Iransfer, Q •
I,
mex
45,328
• 1212(SO-6).
[ Qmax ... S3,328
45.328
w] '\
T2
'
,
1.2)( 1010(50-T2)
= 60,600,-
12.6°
=
12.\2,T2
c
I
Cm'ln
-.
0.23
emlX
8S%
Outlellcmperalure
I
of air, T 2 '" 12.6° C
Result:
I. T2 == 12.6° C
2., t2 =, !~.6° C
Effectiveness
:'
.
o /n a counter flow heat exchanger; "'aler Is Ileatedfrom 20·C BO-C
10
e
by an oil HlI~/, a specific Ileat of 2. 5 kJ/kg-K and nlass flow
rate of
0.5 kg/so Tile oil is cooled from 1/o·C 10 40"C. If IIIeoverall Ileal
Iransfer co-efflc/~nt is UfIO wl':"2 K,flhd thefollowing by using NTU
'2.46
",_
"
method
, "
I. Mass flo HI rate of water
Actual heat transfer . -rate
,
• I .
"
. , ,
Q,'i'j
•
e><Qmax
3. Surface area
\= 0.85 >< 51,3~8 .
\'
..v:
2. Effectiveness, of he,lltexcllt~ng~r
.,
Given :
Hot fluid r: oil ,
IL.:Q:.____45_,3_28_W
.......
1
Heat transfer, Q
,'I,'; ,
me Cpc (t2 - tl)
•
.
I
i
'I',
I
Cold fluid - water
(T(, T2)
(t(, t2)
Inlet temperature of water, t( = 20° C
•
ti == 80° C
45,328
"1.25 x 4186 (t2 - 6)
Outlet temperature of water,
45,328.
'5232.5 ~'-,31·,195
Specific heat of oil, Cph == 2.5 kJ/kg - K
.~
."
I
= 2.5 x 103 J/kg - K
'.';
Outlet temperature of water, t2 = 14.6° C
',\"
')
"j'
,
I
Mass flow rate of oil, mh = 0.5 kg/s
Inlet temperature of oil, T (= 110° C
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,
I
I
-....,\ ,
1 J 36 Heat and Mass Transfer
Outlet temperature of oil. T 2 = 40° C
From equation (I) and (2).
2
Overall heat transfer co-efficient, U = 1400 W/m K
Toji"d:
I. Mass flow rate of water,
Cmin'"
1250 W/K
Cmax = 1456.73 W/K
me
Cmin
2. Effectiveness of heat exchanger, e
_
C
max
3. Surface area, A
Cmin
Cmax
Solll 1;0" :
1250
-"""i456J3 = 0.858
= 0.858
'" (3)
We know that,
Heat lost by the oil
Qh
= Heat gained by the water
We know that,
.
o,
T1-T2
T, _
Effectiveness.
t···
. mllc",. -- Cmin1
t,
E = _-
mh Cph (T I - T 2) = me Cpc (t2 - tl)
[From HMfd ata boak, page 110 1S 1J
110-40
0.5x2.5x
103(110-40)
= mcx
4186 x (80-20)
110-20
[ .: Specific heat of water Cpe = 4186 llkg
me
0.348 kg/s
Mass flow rate of water, me
0.348 kgls
From graph,
Yaxis -+ E
103
Ie = 1250 W/K I
= 0.77
C·
Curve -+ ~
Cmax
••• (1)
Capacity rate of water (Cold fluid), C = ~c Cpe
= 0.348 x 4186
Ie = 1456.73 WIK \ •.. (2)
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IE = 0.71\
[To find NTU, refer H ut data book page no 163J I'«:ower fl ow)
Capacity rate of oil (Hot fluid), C = mh Cph
= 0.5 x 2.5 x
.
= 0.858
Corresponding X axiS
. value is 3 •4 , .'..e , NTU = 3 .4
We know that,
NTU = UA
Cmin
3.4
=
1400 x A
1250
(FromJlMT data book,
page no.l51 J
3.138 Heat and Mass Transfer
Inlet temperature of water t _ 2
, ,- O°C
0.858
0.77
Heat ExchangerJ
Mass flow rate of Water. ';'c '" 10 kg/s
Overall heat transfer co-efficie I Un , - 600 W 1m2 K
Heat exchanger area. A '" 6 m2
Effectiveness,
Toflnd:
E
I. Exit lemperature of oil, (T 2)
2. Exit temperature
I.
';'c = 0.348 kgls.
2.
E = 0.77
C =,;,,, x Cph
=3x2.lxIOJ
IC = 6300 W IK I
3. A = 3.03 1112
apcit
=
Q] It is desired to use double pipe counter flow heat exchunger to cool
J kg/s of oil (Cp = 2.1 kllkg K)from 120
10 x 4186
C=4186~
e. Cooling water at 10'(
D
... (2)
(._. Specific heal of water, C{X; = 4186 J/k
enters the heat exchanger at a rate of /0 kg/so The overall Iteat
transfer co-efficient of the heat exchanger is 600 WI",} K and tht
heat transfer area is 6",1. Calculate the exit temperatures of oil and
water.
[JlIl1e-2006. Anna
niv]
Fr m Equn (I) and (2),
Cmlll = 6300 W/K
Given :
max = 41860
c
water u.j t-)
Mass flow rate of oil,
Specific heal of oil
HOltluid-oil
W/K
T"T2
,i,,, = 3 k s
=
6300
41,860
C/)IJ = 2.1 kJ/kgK
=2.1
Inlet temperature
'" (I)
rate of water,
C = ';'c x Cpc
3.2.13 Anna University Solved Problems
Cold fluid
2
Solution:
Capacity rate of hOI oil.
3.4
NTU
Result :
of water, (1 )
=>
10 J gK
max
of oil, T, = 120°C
49
Scanned by CamScanner
Cl11m
= 0.150
···c
.,
----~:-----IF~;h~~--[From HMT data
3~.~14~O~H~ea~t~a~n~d~M~~~s~v~rwu~g.~e~r
__
UA
NTU= -Cmin
Number of transfer units,
b00 k page no 15/]
___ ------~-----:--------------~H~e~a~t£r~c~h~an~g~e~n~J~.~/4~/~
Actual heat transfer rate
Q
600 x 6
=6300
U
[NTU =0.57
0.42
..• (4)
[Tofind effectiveness
E,
0
mc Cpc (t2 - II)
2,64,600
Curve --+ C
max
I
2,64,600 W
10 x 4186 (12 -20)
= 0.571
Cmin
104
10
Heat transfer,
From graph,
X axis --+
x 63 x
We know that,
refer HMT data book page no J6J} (Counter flow)
NTIl
E x Oma.x
26.32° C
= 0.150
[Exit temperature
t2 = 26.32° C
ofwaler,
I
We know'that,
Corresponding
Yaxis value is 42%
i.e.,
Heat transfer, Q
I = 0.42 I
E
;"" Cph (TI - T2)
3 x 2.1 x 103 (120 - T 2)
78°C
Effectiveness
E
1 Ex it temperature
of oil, T 2
78° C I
Result:
I. T2 = 78° C
2. t2 = 26.32° C
III A parallel flow heat exchanger lias IlOt and cold water stream
0.571
NTU
running through u.theflow rates are 10 and 1S kg/min respectillel),.
Maximum possible heat transfer,
Omax
Cmin (TI - tl)
6300,(120.-
lOmax
= 63 x 104
20)
WI
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IIIlettemperatures
are 7S·C and 15"(' on hot and cold sides. rhe
exit temperature
on tile hot side sllould not exceed SO-C.. usume
"i "0
=
E -
= 600 Wlm1K. Calculate tile area of heat exchanger using
NTU approach.
{Dec-Z005. Anna Univ]
GOlf''' .I \) k~ "ill
MtlSS l10w rol( of Iml WI\I~·r. "'II
~ k~/s
6(l
M:lSS 11,1\\1111113 of cold water.
0,166 "'SIs
,i"." ~ k ' 111111
• 60 k~/s • 0.·116 kg/s
')
We know Ihnl.
E 111,)\;1 j Vlln~lSS.
In!cllemj)¢t1IIUI\l of hOI water, T I • 75"
I',' ';'/', ph .. Crnln )
/Fm", IIM1' ItII/a hoo4, flaRO 110, I J/
Inllll h.'IIIj)crnlure of c,)ld W[lIe,"',II • .5'
I'}xil ICIIIJlllnlIUI\l of hoI water, T2• 50"
75 - 25
Overall hcnllrnnster l'l)·ctl1cicnl, "0 • ", • 600 W1m2 K
III fi"il :
Helll Ilxch:lIIgcr area, A
(1'()find N7V. refer N MT d.
paJ.:() no 161 (Par /I.
From graph.
- 0.5
-';'11 x Cph
-0.166
C,
Curve -~ C
~
x 4186
• 0.399
fIIllJI
[£394,87
ata book
a el flow heat exchanger/
Solulill/'
Capacily rate of'ho: fluid.
C
(Sixth iJdltlOll)J
• 75 - 50
W/K!
... (I)
Corresponding
t: Specific heal of water. Cp = 4 J 86 J/Kg KI
X,nxis value rs
. 0.84 , "' e " N'ru '" 0,84
0,5
Capcity rate of cold fluid,
c- ,;,c x Cpc
··1
= 0.416 x 4186
EoorrwlK]
... (2)
Effectiveness,
E
From Equn (I) and (2)
Cmin = 694.87 W/K
Cmu = 1741.37W/K
NTU
Scanned by CamScanner
0.84
.r.
3144 Heat and Mass Trans,er
Heal Exchangers 1.145
UA
We know that,
NTU (Number
---------------
-
0f trans
fer units)
== -
•.. (4)
Cmin
{From
3.2.14 Problems for Practice
I.
HMT data book page no /5/ J
The specific heats of exhaust gases and water may be taken as 1.13
~
_efficient,
Overall heat tranSler co
I
..!....+..!....
U
h,
I
ho+ h;
Exhaust gases flowing through a tubular heat exchanger at a rate of
0.4 kg/s are cooled from 450° C to 150° C by water initially at 150C.
and 4.19 kllkgOe
efficient
respectively.
area required for the following cases, when the cooling water flow is
ho
0.5 kg/s; (i) parallel flow (ii) counter flow.
[Ans : (i) 4.84 m2 (ii) 4./5 ",1/
'" _.;;..---
U
h;ho
1
600 + --600
2.
16.67 kg/s of the product at 700° C (Cp = 3.6 kJlkg "C) in a chemical
plant, are to be used to heat 20 kg/s of the incoming fluid from
100° C (Cp = 4.2 kl/kg "C), If the overall heat transfer coefficient is
U - 600" 600
1 k W 1m2 "C and the installed heat transfer surface is 42 m2, calculate
the fluid outlet temperatures
Eow/m2K]
. and U values
Substitute NTU , e mill
=>
(4)
NTU"'-
and parallel flow
arrangements.
in equation
[Ans : th] = 438.4
(4)
3.
·c, te] = 186.8 6C, 17M' CJ
8000 kg/h of air at 105° C is cooled by passing it through a counter
flow heat exchanger.
emin
at 150 C and flows at a rate of 7500 kg/h. The heat exchanger has
A
694.87
Result:
2
area, A = 1 .945 m
Find the exit temperature of air if water enters
heat transfer
area equal to 20 m2 and the overall heat transfer
coefficient
to this area is 145 W/m2 °C.
corresponding
Take Cp (air) = IkJlkg 6C and Cp(water)
[A = \.945 m~
Heat exchanger
for the counter-flow
UA
300"
0.84 '"
and the overall heat transfer co-
from gases to water is 140 W/m2 "C. Calculate the surface
== 4.18 kJlkg DC
[Ans : 76.1 -C/
4.
A shell-and-tube
type of heat exchanger is designed to cool 1.51~
kg/s of oil (Cp == 2093 J/kg K) from 65.56· C to 42.220 C by using
1.008 kg/s of water at a inlet temperature
of26.67 "C. Assuming an
overall heat transfer coefficient of681.6
W/m2K and a single-shell.
2 tube pass type of heat exchanger determine the required heal transfer
area. use the effectiveness
method.
IAns: 7.9m2/
Scanned by CamScanner
3. 146 Heat and Mass Transfer
(C
1000 ~
. han er hot exhaust gases
p
g ~~
In a cross flow heat exc
g.
1000 C are used to heat Water
.
3000 C and leavmg at
,
entenng at
1250 C The overall heat
0
35 C to'
"
flowing at I kg/s rom
.
face area has been found to be
.
.
the oas side sur
,
coefficient based on
'"
h d estimate the required gas
100 W Im2K. Using the NTU met 0 ,
5.
WI,al is meant by Filmwise condensation?
6.
transr""l
fi
{April 2000. Oct 2000 MUj
. .
The liquid condensate
Si~1
3.2.15 Two mark
1.
In dropwise
droplets
"'IIat is meant by condensation?
"'
The change
of phase
to liquid
from vapour
state
boiling and cOlldensatioll.
".
Boiling and condensation.
proc ess finds wide applications as menu
•
•
.1"
In dropwise
directly
below
4.
1.
Thermal and nuclear power plant
2.
Refrigerating
3.
Process of heating and cooling
4.
Air conditioning
sizes which fall down the surface in a random
condensation,
exposed
9.
systems
10
{April 1999 MUj
a large portion of the area of the plate is
vapour.
The heat transfer
rate in dropwise
is 10 times higher than in film condensation.
condensation
Write tIll!force balance equation on a I'D/umeelementfor fllmwise
on a vertical plane surface.
condensation
systems
"'"at is meant by pool boiling?
[May-2004.
Where,
AnnaUn·.
Bx - Body force in x direction
. a dd e d to a liquid
from a submerzed
solid surface, the boll
If heat IS
I
e
.
~
d
to
as
pool
boiling
In
this
case
process IS re.erre,
.
. the liquid above
hot surface is essentially
stagnant
and its monon
due to free convection
and mixing
induce!
Op - Pressure gradient
ax
near the surface
by bubble
growth
/0.
Draw different regtous 0/ bollill/l alld wtuu ts Nucleate boil/nil'!
[Apri//99Y
detachment.
5.
condensation,
of various
Give the merits of drop wise condensation.
8.
Give tile appllClltlOnoJ
{Dec2004 . 2005 & June 2006 A UJ
the vapour condenses into small liquid
fashion.
is known.
condensation.
3.
is known as film wise
[April 2000 , Oct 2000 MUJ
The change of phase from rqui
2.
surface
What is meant by Dropwise condensation?
7.
r .d to vapour state is known as boil"
Define boiling
[Dec 2004 . 2005 & June 2006 A Uj
wets the solid surface, spreads out and forms a
continuous
film over the entire
condensation.
surface area.
.
and Answers
QuestIOns
Heal EJchangers 3.147
WlllIt are tlu: modes tlf condensation?
III
There arc two modes of condensation
arc formed more rapidly and rapid evaporation
I. Filmwise condensation
Scanned by CamScanner
MU. April 200] MUj
Nucleate boiling exists in regions II and III. The nucleate boiling begins
2. Dropwise
condensation
region II. As the excess temperature
indicaled
is further increased. bubbles
takes place. This is
in region III. Nucleate boiling exists upto 6T - 50· C.
~~~----3.148 Heal and Mass Transfer
Heat Exchangers 3.149
c::
0
u ';::
5. Counter flow heat exchangers
CI)
~...
~
0
-
!:! c.
r:: '";.
Nucleate
boiling
6. Cross flow heat exchangers
Filmboiling -
IU
II
7. Shell and tube heat exchangers
v
VI
8. Compact heat exchangers
7
10
/3.
B
10
In direct contact heat exchanger, the heat exchange takes place by
direct mixing of hot and cold fluids.
6
10
U.
5
E
Whal is meant by lndirec: c,?ntaciheal exchanger?
In'this type ofheatexchangers,
,....
M
W/tal is meant by Direct heal exchanger (or) open heat exchanger?
~fh;at between two fluids
could be carried out by transmission through a wall which separates
104
the two fluids.
~
'-'
~ 103
15. J!,~a~is!!Iea'!.lby Regeneralors?,
In th is type of heat exchangers, hot and cold fluids flow alternately
through the same space.
(;
10
the t~;fer
2
-
...
t
Examples : IC engin~:...gas turbines.
10
100
50
10
Excess Temperature ~ Te = Ts - Tsat
150
I - Free convection
II - Bubbles condense in super heated liquid
IV - Unstable film
III - Bubbles raise to surface
VI
- Radiation coming into play
V - Stable film
l l,
What are the types of helll exchangers?
The types of heat exchangers are as follows
I. Direct contact heat exchangers
2. Indirect contact heat exchangers
3. Surface heat exchangers
4. Parallel flow heat exchangers
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[Dec 2005,AU]
Whal is meant byJltCUperalOTf;(or) Surface Ileal excuangen»
.....
- ..
This is the most common type of heat exchangers in which the hot and
cold fluid do not come into direct contact with each other but are
separated by a tube wall or a surface.
-
Examples:
/7.
Whut is heat exchanger?
A heat exchanger is defined as an equipment which transfers the heat
from a hot fluid to a cold fluid.
/2.
/6.
Automobile radiators, Air preheaters, Economisers etc.
What is meant by parallel flow Ileal exchanger? [May-05, AU)
In this type of heat exchanger, hot and cold fluidUDoye in the same
direction.
/B. Wluu is meant by counter flow Ileal exclla~ger? {May-05. AU}
In this type of heat exchanger, hot and Gold fluids move i'!.E_arallelbut
opposite directions.
/""
19.
Wluu is meant by cross flow heal exchanger?
In thi~ type of heat-exchanger, hot and coJd floids move at right angles
to each other.
\
,.
3.150 Heat and Mass Tralls[e'
20. Wlrat is meant by SI,ell alld tube I,eat exchanger?
------:~~==~~-:=------------U'''at is meant by Effectiveness r
Exchangers 151
]~.
In this type of heat exchanger, one onhe fluids move through a
Heat
The heat exchanger effectivene ss ISdefined
.
as th
.
transfer to the maximum possibl e heat transfer. e ratio of actual he·at
of tubes enclosed by a shell. The other fluid is forced through the
and it moves over the outside surface of the tubes.
Effectiveness
E =
21. Wlrat is ",eant by compactl,eat e..'(c"angers?
There are many special purpose heat exchangers called compact
associated with the other fluid.
Actual heat transfer
Maximum possible heat transftr
Q
s.;
exchangers They are generally employed when convective heat
co-efficient associated with one of the fluids is much smaller than
J
25. Sketc" tI,e temperature variatio ns In
. parallel flo
d
"eat exchangers.
"' an counter flo"'
(Dec-O". AU]
22. WI,at is meant by LMTD'!
We know that the temperature difference between the hot and
fluids in the heat exchanger varies from point to point. In
various modes of heat transfer are involved. Therefore based on
}fOt
..
alJicj
II
of appropriate mean temperature difference, also called
a
C!
II
mean temperature difference, the total heat transfer rate in the
c.
exchanger is expressed as
~
Ih2
92
9)
IC2
E
Q~UA(~T)m
where
U _ Overall heat transfer co-efficient, W/m2K
Cold fluid
Ie)
Area
Temperature distribution - rara
n
l",ei flo"'
A - Area, m2
(~T)m - Logarithmic mean temperature difference.
21.
What is meant by Fouling factor?
[ Nov.96
We
know,
the
surfaces
ofa
heat
exchangers
do
not
rem'
.'
~
~c~ I
It has been In use for some time. The surfaces become fouled
scaling or deposits. The effect of these deposits affecting the
overall
heat
transfer co-efficient. This effe C t iIS ta k en care of
.
.
introducing an additional thermal resistance called th r.
resistance.
e rou
Scanned by CamScanner
Area
Temperature distribution - r...ounter flo"'
9) = Ih
92=lh
I
2
-I
-I
CI
c2
CHAPTER-IV
4. RADIATION
4.1. INTRODUCTION
The heat is transferred from one body to another without any
transmitting medium is known as radiation. It is an electromagnetic
wave phenomenon.
All types of electromagnetic waves are classified in terms of
wavelength
and are propagated
at the speed of light,
i.c., 3 x 10M m/s.
4.2. EMISSION PROPERTIES
The rate of emission of radiation by a body depends upon the
following f"etors.
I.
The wavelength or frequency of radiation.
2.
The temperature of surface.
J.
The nature of the surface.
4.3. EMISSIVE
POWER [EtJ
The emissive power is defined as the total amount of radiation
emitted by a body per unit time and unit area.
._
It is expressed
Scanned by CamScanner
in YiJJJ12.
4.2
Heal and Mass Transfer
4.4. MONOCHROMATIC
The energy
time
EMISSIVE POWER (EbJ
emitted
by the surface
per unit area in all directions
emissive
is known
length erp
as monochr Olllat~
un'
When
the radiant
energy
A part is reflected
surface, and (he remainder
AND TRANSMISSION
falling
I
at a given
power.
4.5. ABSORPTION, REFLECTION
happen.
Radiation
on a body,
back, a part is transmitted
three
=
4.3
a+p+'[
where
p and t are known as absorptivity
..
.
. a,
.
IVI
,re fl ecnvity
and
transmissIVity of the surface.
i.e., Absorptivity,
a
Radiation absorbed
Incident radiation
Reflectivity,
p
Radiation reflected
Incident radiation
Transmissivity,
t
Radiation transmitted
Incident radiation
thin I
throUgh t~
is absorbed.
Q
4.6. CONCEPT OF BLACK BODY
Black body is an ideal surface having the following properties.
Fig. 4.1.
If the incident
Fig.4.I,
energy
Qa is absorbed,
Q is falling on a body as shown in
Qr is reflected
energy balance yields,
Dividing
and Q, is transmitted,
then
).
A black body absorbs all incident radiation, regardless of
wave length and direction.
2.
For a prescribed temperature and wave length, no surface
can emit more energy than black body.
A black body is regarded as a perfect
absorber of incident radiation. A black body
condition can be approached in practice by
forming a cavity in a material as shown in
Fig.4.2. Radiation passing through the hole
into the cavity is repeatedly absorbed and
reflected at the cavity walls until it all
absorbed.
the above equation
by Q
Q
Q
Scanned by CamScanner
Fig. 4.2.
A black body is a perfect emitter. This is a fact which can be
proved as follows. Consider a black body at a uniform temperature,
placed inside an arbitrarily shaped, perfectly insulated enclosure
composed of another black body whose temperature is also
4.4
r
Heal and Mas.' ),a",}"
-ii&'~'
uniform but different from that of the former. The bla~
the enclosure will reach a common equilibrium temperature Y.and'
. d ..
alter'
peno
ot time due to heat transfer.
a,
Radiation
AnlfLt
Enclosure at
uniform
temperature
T
-
4.5
28981lmK
IFrom HMT data book, Page No. 81(Sixth Edition)1
I Amax T
2.9 x 10-3 m~_]
... (4,2)
[':p=IO-om1
4.9. STEFAN-BOLTZMANN
Fig. 4.3.
The emissive
LAW
power of a black body is proportional to the
fourth power of absolute temperature.
4.7. PLANCK'S DISTRIBUTION LAW
The relationship between the monochromatic emissive power
of a black body and wave length of a radiation at a -particular
temperature is given by the following expression, by Planck
c1 1.-5
J~~
]_
[l-rom IIMl data "link. !'agl' No. KI(Sixth Editiun)!
, .. (4.3)
where
... (4.1)
E
"
a
I
where
EbA
Stefan-Boltzmann
constant
5.67 x 10 x W/1I12 K4
r
[From IIl'vlT data book. Page No. 81(Sixth EditionJl
Monochromatic
Emissive power - W/m2
Temperature
-. K
emissive power W/m2
Wavelength - m
4.10. MAXIMUM EMISSIVE POWER, (EllA)max
0.374 x 10-15 W-m2
A combination of Planck's law and Wicns dispiacelllcill law
yields the condition for the maximum monochromatic emissive
14.4 x 10-3 mK
I
power
1'01' a
black body.
(4
4.8. WIEN'S DISPLACEMENT LAW
The Wien's law gives the relationship between temperat~re
maxi
and wavelength corresponding. to t hee maximum
spec t·ra I e miSSive
power of the black body at that temperature.
Scanned by CamScanner
\\ here
T5
1.307 >. I () ~
II,-adial ion ('oll,lant
I
I
. (-I ~)
bLCLl1
2~
1.307 x 10 ' I ;
4.6
Heat and Mass Transfer
4.11. EMISSIVITY
h
It is defined as the ability of the surface of a body to rad'
I .
late
ea. t IS also defined as the ratio of the emissive power of
body to the emissive power of a black body of equal temperatu any
reo
t
Emissivity,
E
=
It states that the total emissive
. any directi
.
surface rn
IrectlOn IS directl POWerEb
. froma radIating
pi
. .
y proportIonal
ane
angle of emission.
tothecosineof the
E
Eb
[Eb
C(
cos~
'" (4.6)
4.12. GRAY BODY
4.16. FORMULAE
If a body absorbs a definite percentage of incident radiation
irrespective of their wave length, the body is known as gray body.
The emissive power of a gray body is always less than that of the
black body.
4.13.
KIRCHOFF'S
USED
[From HMT data b~ok, P
J.
Emissive Power (or) Total Emissive Power:
E,
LAW OF RADIATION
where
This law states that the ratio of total emissive power to the
absorptivity is constant for all surfaces which are in thermal
equilibrium with the surroundings. This can be written as
IT
0 T4
=
W/m2
StefanBoltzmannconstant
5.67 x 10-8 W/m2 K4
2.
Wien's Law:
T =
Amax
3.
It also states that the emissivity of the body is always equal to
2898,.LlnK = 2.9 x Woo) mK
Monochromatic Emissive Power (or)
Spectral Emissive Power:
its absorptivity when the body remains in thermal equilibrium with
its surroundings.
u,
4.14.
INTENSITY
=
E,;
OF RADIATION
u2 = E2 and so on.
C2
It is defined as the rate of energy leaving a surface in a given
In
=
Scanned by CamScanner
0.374 x 10-15 W-m2
where
(Ib>
direction per unit solid angle per unit area of the emitting surface
normal to the mean direction in space.
... (4.5)
age No. 8 I(Sixth Edition))
4.
Maximum
14.4 x 10-3 mK
Emissive Power (Eb;)1II/lX
:
where
c4
4.8
5.
Hear andM ass Transfer
Intensity
of Radiation (/ t) :
--------------------2. According to Stefan-Bolt
zman law
Emissive power, Eb :: a T4
Eb
7t
6.
Absorptivity,
a
Radiation absorbed
Incident radiation
p
Radiation reflected
Incident radiation
[From HMT dal~ book P
. age No 81(Sixlh Ed' .
E
_
"b -
{ .:
Reflectivity,
Transmissivity,
I
A
=
0.5 Jl == 0.5 x 10-6 III
[ '.: I ~l = 10-6 m]
To find : I. Surface temperature,
2.
Surface temperature,
llillmple
to Wien's displacement
[From
0.5 x 10-6 x T
=
HMT data book.
=
T
Page 1\(\. X I (~i.\lh
Scanned by CamScanner
64.1 x 106 W/m2
J A black body at 3000 K emits radiation.
Wave length at .~'lricllel"issio" is mtu:imilm.
3.
Maximum
4.
5.
Tottll emissive I'llwer,
Calculate the tot(ll emissive of tI,e [umace if it is
assumed (IS u 1'('(11surface huving emissivity equal to
(1.85.
IMtlClras Ulliversity, April 96/
Editiollli
\.
Surface temperature,
T
3000 K
==
l\1onochlOmatic emissive power EbA at
A -c. \ P .~ \ x \ 0 -(, J1l.
2,
Maximum
wave length, (A max ).
-,
Maximum
emls~)VC
).
\
emissive power.
To find :
2.9 x 10-3
5800 K
2
Given :
law,
2.9 x 10-3 rnK
I T =_ 5~0Q_ []
Surface temperature,
5800 K
2.
Emissive power.
Amax T
T
Calculate tile following:
1. MOlloc/rromatic emissive power (It 1 pm wave lengtll,
Solution :
1. According
x 10-8 W/m2 K4)
64.1 x 106 W/m21
Emissive power, Eb
~xamp_le 1 Assuming
SUII
to be black body emiUillg
radiation with maximum intensity at A. = 0.5 u; calcukue its
surface temperature and emissive power.
Wave length
Stefan-Bohzman
constant
Re,.",lt:
I.
Given:
=
=_5.67
\t,
4.17. SOLVED PROBLEMS
_I
a
r-
Radiation transmitted
Incident radiation
t
IIlon)1
5.67xl(}&"(5800)4
.'
Po\\ 'eI'" (F··)
'1".
1111
/\
.'
4.10
Heat and Mass Transfer
4.
e;
Total emissive power,
J. Maximum
emissive power (E
Radialion
.1
IINnta.t
5.
Emissive
power of real surface at E = 0.85.
Maximum
emissive power
Solution:
\.307 x 10--5 TS
1. Monochronuuie Emissive Power:
1.307 x 10-5 x (3000)5
From Planck's
3.17 x 1012 W/m2]
distribution
law. we know that,
ci A-5
4. Total emissive power (E,) :
From Stefan-Bohzmann
Eb
law, we know that
OT4
=
[From HMT data book, Page no. 811
0.374 x
where
Cz
10-15
W m2
{From HMT data book. Page no. 811
where
5.67 x 10-8 W/m2 K4
[Given]
1 xl~m
0.374 x 10-15 [I x 10-6]-5
I EhI.
2.
Stefan-Boltzman constant
a
14.4 x 10-3 mK
e
14.4)( 10-3 ]
[ I )( 10-6 x 3000
- I
Eb
(5.67 x 10-8) x (3000)4
I s,
4.59 x 106 W/m2\
em;ssivJ!.P0wer of a real surface /
5. Total
I'
2
3.10 x 1012 W/m
where
E
Emissivity
-
'1
II. max
3.90 x 1()6 W/m2]
law, we know that,
T
2.9 x 10-3 mK
2.9 x JO-3
A max
I Amax
Result:
1.
EH
=::
3.10 x 1012 W/m2
2.
A max
=::
0.966 x I~
3000
0.966 x 10-6 m
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= 0.85
0.85 x 5.67 x 10-8 x (3000)4
Maximum wave length, (A.max) :
From Wien's
41/
:
m
4.12
Heat and Mass Transfer
3.
--------------------From Wien's displacement la
3.17 x 1012 W/m2
4.
4.59x
5.
3.90
Amax T
106 W/m2
X
Rudiulioll
w, We know that
2.9 x 10--3 mK
106 W/m2
2.9 x 10-3
I Example 3 I A gT(~Vsill/ace is maintained at temperature
2.4xl~m
(I
emissive power at tlmt tempermure is
IA x 10 "'/",1. Calculate the emissivity of the body and ti'e
wavelength corresponding to the maximum intensity olmdialion.
of 90(1't' IIlId maximum
[_A_max--,-__
10
Given : Surface temperature,
T
1173 K
Maximum
To find:
emissive
power.
(Eh),
)1II0X
1.4 x 1010 W/m2
I.
Emissivity
of the body (s).
2.
Wave
length
corresponding
intensity of radiation (A ilia)'
to
maximum
I.
Emissivity of the body, E
2.
Maximum wave length , Amar
power,
0.48
2.41l
I
[ Example 4 A black body of 1200 cml emits radiatio
1000 K. Calculate tile following:
1.
Total rate of energy emission.
2.
Intensity
3.
of normal radiation.
Wavelength
of
maximum
monochromatic emi
power.
4.
emissive
[.: I Il = I~
Result:
Solution : We know that,
Maximum
2._4 ~IlJI
(Eh)IIIGX
of radiation along a direction at 60° to
normal.
,
1.307 x 10-5 W[m2 K5
where
Intensity
== c4 T)
Give" :
Area, A
1200 x 10-4 m2
1.307 x"-IO-5 x (I) 73)5
Surface temperature, T
2.90 x 1010 W/m2
1200 cm2
1000 K
To flnd :
).4 x 1010 W/m2
So,
/'
Emissivity,
f.
Scanned by CamScanner
==
1.4 x 1010
2.90 x 1010
rGiven]
1. Total rate of energy emission, Eb·
2.
Intensity of normal radiation, In'
4. !-I
Heat and Mass Transfer
3.
Wave length
power, A maX"
4.
Intensity
of maximum
monochromatic
4.
of radiation
415
at 60°, Ie.
Solution:
::: 18,048 W 1m2
From Stefan-Boltzmann
law,
1.
Eb
Result :
c T4
1.
Energy emission E
[From HMT data book, Page no. 811
2.
Intensity of normal radiation, In
Eb
5.67 x 10-8 x (1000)4
3.
Maximum wave length, A. max
\ e,
56.7 x 103 W/m2
I
4.
Intensity of radiation at 60°, Ie
Here
Area
1200 x 10-4 m2,
~
Eb
56.7 x 103 x 1200 x 10-4
I Eb
6804 W
Energy emission,
=
2. Intensity of normal radiation
In
=
'
.7t
A max
A max
I
Scanned by CamScanner
In == 18,048 W/m2
2.
The emission received per m2 just outside the earth's
3.
atmosphere.
The total energy received by the eartl. if no radiationis
blocked by the earth's atmosphere.
t:
The energy received by a 2 x 2 m solar collector
normal is inclined at 45 to the sun. The energy_~ss
. 50% d the diffuse radiatIOn
through the atmosphere IS
0 an
is 20% of direct radiation.
Surface temperature, T == 6000 K
12 x 1010 m
Distance between earth and sun, R
Given:
2.9 x 10-3
1000
Diameter of the sun, DI
2.9 x 10-6 m
[.:
==
0
2.9 x 10-3 mK
2.9 Il I
2.9 ~
Total energy emitted by the sun.
4.
3. From Wien's law, we know that
T
18,048 W/m2
1.
Eb
7t
Amax
6804 W
I
,.--
18,048 W/m2
==
[ Example 5 Assuming sun to be black bo.dy emitting
radiatiolt at 6000 K at a mean distance of 12 x 1010 m from the
earth. 'tu« diameter of the sun if1.5 x'lOl) m and th~tofthe'earth
is 13.2 x 1(J6m. Calculate thefollowing.
56.7 x 103 W/m2
\ In
b
1.5 x 109m
13.2 x 106 m
1 Il = 10-6 m]
Diameter of the earth, D2
" 16
Solution : I. Total ellcrJ:Y emitted :
Energy emitted by sun, E b ==
-1/7
5.67 x 10-8 x (6,000)4
[.: o
2855.S W/m2
Stefen-Boltzman constant
3.
5.67 x 10-8 W/m2 K4]
IrE-b-----73-.4--x-10-6-W--/m-2~1
Energy received by the earth:
Earth area
Area of sun, AI
~ (D )2
2
1t
4 x [13.2 x 106]2
47t x (
1.5 x 109) 2
2
I Earth area
I
Energy received by the earth
7 x 1018 m2
3.88 x 1017 W
73.4 x 106 x 7 x 1018
5. 14 x 1026 W
I
Tile emission received per ml just
atmosphere :
4.
outside
tile earth's
The energy received by a 2 x 2 m solar col/ector:
Energy loss through the atmosphere is 50%. So ener
reaching the earth
100-50
The distance between earth and sun
Energy received by the earth
41t R2
Area, A
0.50 x 2855.5
4 x 7t x (12 x 1010)2
1.80 x 1023 m2
= 50%
0.50
12 x 1010 m
R
1.36 x 1014 m2 I
2855.5 x 1.36 x 1014
=> Energy emitted by the sun
2.
=
I
=> The radiation received outside the earth atmosphere per m
1427.7 W/m2
2
.,. (
Diffuse radiation is 20%.
Eb
=>
0.20 x 1427.7
285.5 W/m2
A
I Diffuse radiation
285.5 w!ffi2]
.,. (2
51
eUIMP.!&IIiJiiC
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-1./8
---
Heat and Mass Transfer
Total radiation reaching the collection
1427.7 + 285.5
1713.2
Plate area
Radiation
Incident radiation
Given:
800 W/m2
Absorbed energy
300 W/m2
Reflected energy
W/1112
100 Wlm2
Transmitted energy
A x cos El
800 - (300 + 100)
2 x 2 x cos 45°
400 W/m2
Tofind:
2.82 Jl12
Energy received by the collector
2.82 x 1713.2
I.
2.
'.,
-'.
4.
I
2.
Reflectivity, p.
3.
Transmissivity,
We know that,
1.
Absorptivity,
2855.5
Total energy received}
by the earth
300
3.88x
=
Wlm2
IOJ7W
2.
3.
Transmissivity
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0.375
p
Radiation reflected
Incident radiation
4831.2 W
I
100
800
Ie
3.
Transmissivity,
t
Ahsorptivity
Reflectivity
la
Reflectivity,
I
2.
Radiation absorbed
Incident radiation
a
800
The radiation received}
outside the earth's atmosphere
Example 6 800 Wlml of radiant energy is incident upon a
surface, out 'of which 300 Wlm2 is absorbed, 100 Wlml is
reflected and the remainder is transmitted through tile surface:
Calculate thefollowing:
1.
t.
5.14 x 1026 W
Energy emitted by the sun, Eb
Energy received by the}
solar collector
Absorptivity, a.
Solution:
4831.2 W
Result:
1.
0.125/
Radiation transmitted
Incident radiation
400
800
It
0.5)
4./9
4.20
Heat and Mass Transfer
____-------~~~----------------~R~a~~~a/~io~n--i4.~21
Result:
\.
Absorptivity,
a
0.375
2.
Reflectivity,
p
0.125
Transmissivity,
t
0.5
3.
Eb (0 - A.z T)
c 1'4
0.6195
(From HMT data book, Page no. 82J
E, (0- '''2
L T)
Eb (0-11.11 T)
o T4
o T4
/\.
I ExamplklA
... (2)
black body is kept at a temperature Of
9491:('. Estimate the fraction of thermal radiation emitted hy the
0.6195 - 0.0025
surface in the wave length band lu and 4J.L
0.617
Given:
Surface temperature,
T
1222 K
IT
1222 K I
A.)
I ~l
Final wave length, A.2
4~
Initial wave length,
:::>
E b (/") T - 1..2T)
o T4
0.617
:::>
E b (A) T - A2 T)
0.617 x c x T4
0.617 x 5.67 x 10-8 x (1222)4
:::>
To find:
Radiation emitted by the surface [E b (A.) T - ~ T) ].
Solution:
I x 1222 ~K
I Eb(A) A
Result:
T- 2T)
78 x 103 W/m21
Energy emitted E b (A) T _ A2 T) = 78 x 10J W1m2
I EXlImple 8 I A surface emits radiation as a black body at
3000 K. Calculate
1222 ~K
the emission from
the surface in the
wavelength interval Zum L l L 5 pm:
A) T = 1222 ilK, corresponding fractional emission,
Given:
3000 K
Initial wave length, AI
... (1)
= . 0.0025
Surface temperature, T
Final wave length, A2
[From HMT data book, Page no. 82 (Sixth edition)J
Tofind:
4 Il x 1222 K
4888 ilK
A2 T = 4888 ilK,
l,,-
corresponding fractional emission
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Solution:
I. Emission from the surface E b (I. I T -1..2 T) .
2 x 3000 ilK
6000 ilK ]
4.22
•
Heal and Mass Transfer
A IT
6000 ilK, corresponding
fractional emission
1. Emissive power.
2.
The wave length A. b I
.,.
J
e Ow W!riclt 20
emiSSIOn IS cOlleen/rated d
percent 0,/ th
an tlte W
e
which 20 percent of the em!
. aile length A.} abo lie
3.
The maximum
5 x 3000 JlK
4.
Spectral emissive POWer.
I
5.
The irradiation incident.
0.7378
'" (I)
[From HMT data book, Page no. 82)
}.2 T
15,000 JlK
}.2 T = 15,000 JlK,
corresponding
lSSIOn IS conce
Given:
fractional emission
walle length.
ntrated
Surface temperature, T
3000K
Solution:
0.9699
'" (2)
1.
Emissive power, Eb
5.67 x 10-8 x (3000)4
[From HMT data book, Page no. 82J
E b (0 - A2 T)
c T4
E b (0 - AI T)
c T4
0.9699 - 0.7378
[Eb
-
4.59 x 106 W/m2]
2. The wave length 1.( corresponds to the Upper I'un It,
.
containing 20% of emitted radiation.
Eb (0 - 1.( T)
0.2321
0.20, corresponding
o T4
,
AIT
o x T4 x 0.2321
[From HMT data book, Page no. 821
5.67 x 10-8 x (3000)4 X 0.2321
AIT
1.06 x 106 W/m2
==>
Result:
Energyemitted
I Example
em perature
9
Eb(AI
T-A2 T)
1.06 x 106 W/m2
I A large enclosure is maintained at a uniform
of 3000 K. Calculate
the following:
Scanned by CamScanner
2666 ilK
==>
I
2666 ilK
AI
2666
3000
AI
0.88 Il
I
T he wave length 1.2 corresponds to t he Iower limit,containing
20% of emitted radiation
4.24
Heat and Mass Transfer
(1-0.20)
where
ci
c2
0.80, corresponding
Eb}.
@374 x 10-15) x (0.96 x 1()-6 5
6888 JlK
[(e 096 "",0-'
) ] to
10-6 3000 _ I
x
[From HMT data book, Page no. 82)
6888 JlK
So,
law,
4.18. SOLVED UNIVERSITY PROBLEMS
AIIIOX T
2.9 x 10-3 mK
10-3
A max
2.9 x
3000
9.6 x 10-7 rn
I
Amox
I Example 1 I Ti,e
emits maximum radiation' at
A. = 0.52 J..L Assuming tire sun to be a black body, calculatethe
surface temperature of tire sun. Also calculate th«
monochromatic emissive power o/tlre sun's surface.
SUIl
/April 98, MUj
0.96 x 10-6 m
4. Spectral Emissive Power:
distribution
3.1 x 1012 W/m2
The irradiation incident on a small ob' t I
'.
[ec paced wlthm the
enclosure m~ be treated as equal to emissionfrom a black body at
the enclosure surface temperature.
3. Maximum wave tength (A.ma.J :
From Planck's
EbA
5. Irradiation:
6888
3000
From Wien's
=>
x
A lIIax --
Tofind:
1. Surface temperature, T.
2.
law, we know that,
0.S2j.l
= 0.52 x 10-{) m
Given:
Monochromatic emissive power, Eb)"
Solution:
1. From Wien's law,
A
T
max
[From HMT data book. Page no. 811
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= 2.9 x 10-3 mK
[From HMT data book, Page no.
81 (Sixth edition)]
4.26
Heat and Mass Transfer
T
=
----~~-;::==~=-:---~--=
2.9 x 10-3
0.52 x 10--6
Given: Temperature,
IT = 5576 K I
Solulion:
/. Monochromatic
2. Monochromatic emissive power (Eb;) :
emissive pOHler (Eu) :
From Planck's distribution law, we know that
law,
From Planck's
c1 A-5
_
Eb}. -
[J:~)- J
[From HMT data book, Page no. 81]
[From HMT data book, Page no. 81
0.374 x 10-15 W m2
where
where
14.4.x 10-3 mK
0.52 x 10--6 m
T
[Given]
5576K
[.(052': ;0"J~-:576)6.9 x 1013 W/m2
Result:
1.
T
I
5576K
14.4 x 10-3 mK
A
1 urn
lEbA
=
4".-3)
J4
1 x 10--6 x 2000
Tora! emissivepower.
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[Given]
_
]
1
2.79 x 1011 W/m2/
A,nax T
=
2.9 x 10-3
I
Monochromatic radiant flux
length.
m
0.374 x 10-15 [I x I~I-5
e
=>
= 1 x I~
From Wi en 's law, we know that,
density at J pm wave
2. Wave lenglll at which emission is maximum and tne
corresponding emissive power.
J.
c2
[(
Example 2 A furnace wall emits radiation at 2000 K.
Treatingit as btack hody radiation, calculate
1.
0.374 x 10-15 Wm2
2. Maximum WaveLength (AmaJ :
6.9 x 1013 W/m2
2.
I
J
c,
EbA
0.374 x 10-15 [0.52 x 10--61-5
=>
~R~a~$~m~iO
l um = I x lQ-6
2000 K ; A
T
IApril98, MU/
(From HMT data book, Page no. 811
Amax
2.9 x 10-3
T
2.9 x 102000
1.45 Il 1
3
1.45 x IQ-6 m
4.28
Heat and Mass Transfer
Corresponding
emissive power
--
The wavelength of maxi",u",
. power.
JA :T)_1
Given:
0.374 x 10-15 x [1.45 x 10-6]-5
Tojind:
144 x JO-3)
Solution:
m
[ e ( 1.45 x JO-6 x 2000
=
]
-
1
1.
T
650 + 273 == 923 K
2. In;
where
o -
Stefan-Boltzmann
Eb
l s,
41151.8 W/m2]
Here,
Area
0.25 m2
=>
Eb
41151.8 W/m2 X 0.25 m2
I s,
10.28 X IQ3 Watts I
Intensity, In
2.
5.67 x 10-8 x (2000)4
10.28 x 103
907.2 x 103 W/m2/
7t
Result:
1.
Eb}.
2.79 x 1011 W/m2
2. (i)
A max
1.4511
(ij)
Eb}.
4.09 x 1011 W/m2
Eb
907.2 x 103 W/m2
3.
I Example 3 I Tile temperature of a black surface 0.25 ml of
area is 650't: Calculate,
1. Tile total rate of energy emission.
2. The intensity of normal radiation.
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5.67 x 10-8 (923)4
I s,
constant
5.67 x 10-8 W/m2 K4
3. A max .
Emissive power, Eb
From Stefan-Boltzmann law, we know that,
I In
3274.7 W
I
T
2.9 x 10-3 mK
3. From Wien's law,
Amax
A max
429
-
noc'''O",atic
emissive
/'0
I· ct. 96 EEE, MUj
0.25 m2
3. Total emissive power (E J) :
o T4
I
A
4.09 x 1011 W/m2
s,
Radiation
"'0
3
2.9 x 10-3
923
3.13 x lQ-6m 1
Result:
I.
Eb
10.28 x 103 W
2.
In
3274.7 W
3.
A max
3.13 X 10-6 m
4.30
Heal and Mo." Transfer
_______
[&amele ., 1 Assuming
sun 10 be black body enritr;n,
radiaJi()n with maximum intensity at .A. = 0.5 J1, calculate the
temperature of the surface of the sun and the heat flux at its
Surface.
/ApriI97, MU, EEE}
Given:
"-max
Radiation
4.3/
0.5 11
0.5 X 10~ m
Tofind :
I.
Surface temperature,
T.
Radiation and reflection process are assumed to be diffuse.
2. Heat flux, q.
The absorptivity of a surface is taken equal to its
emissivity and independent of temperature of the source of
the incident radiation.
Solution :
1. From Wien's Jaw, we know that,
2.9 x 10-3 mK
"-max T
RADIATION EXCHANGE BETWEEN TWO BLACK
4.20. SURFACES SEPARATED BY A NON-ABSORBING
2.9 X 10-3
T
MEDIUM
"-max
2.9 X 10-3
0.5 X 10~
IT
2.
q
Heat flux,
q
Result:
5800 K
.
Let us consl ider two black bodies separated by a non-absorbing
. .
.
medIUm.
T h e pro blem is to determine the net radiation heat
h
between them.
exc Consider
ange
area e Iemen t s dA 1 and dA 2 on the two surfaces. The
I
. r and the angles , the normals to the two
.
between t Ilem IS
distance
k
with
the line joining them are 91 and 92,
area elements rna e
Q = Eb = IT T4
A
5.67 X 10-8 (5800)4
/q
64.16 x 106 W/m2
T
5800 K
q
64.16 X 106 W/m2
I
Normal to dA2
Normal 10 dA1
co
rig.
4.4. Rat/ilttion-lte(lt
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exclw/lg e. betweell two
bla£'ksurfaces
4.32
Heal and Mass Transfer
Radiation
FigAA shows the projection of d A I normal to the line between
the centres. The projected area is ciA I cos
dQ2_1
eI .
'" cos 9, cos 8, dA, dA,
Energy leaving d A I in direction 9 I
I dAI cos 91
"I
'''1 -
where
I
... (4.11)
,2
'" (4.7)
The net rate of heat transfer between d A I an d A2 is
Intensity of radiation at surface AI
dOl2 = dQI_2
We know that,
- dQ2_1
'''1 cos 01 cos 02 ciA I dA2
Intensity of radiation,
1"1
,2
= Eol
1[
dQlz
Radiation arriving at any area normal to r will depend on the
solid angle subtended by it.
Let dw I be subtended
dAz by dAI.
So,
4.33
at d A I by d Az and d(J)z subtended
[
cos 01 cos 92 "AI dAz ]
,2
We know
al
I"
dAz cos (:)z
,.z
dWI
dwz
= (1"1 -1"2)
=
dAI cos
°
1
,.2
... (4.8)
From Stefan-Boltzmann
.. , (4.9)
s,
The rate of radiant energy leaving dAI and striking on dA2 is
given by
law, we know
0' T4
, 4
(0'
IQ 12
(. 'r4I - 'r4)
2
Co
4
r I - 0' T 2 )
0'
[
[cos 91 cos 0z dAI clAz ]
1[
,.2
cos 9( cos 92 dAI clA2 ]
1[
,.2
... (4.12)
." (4.10)
The rate of energy radiated by ciA 2 and absorbed
.
given by
by dAI is
The rate of total net heat transfer for the total areas A( and A2
by
IS given
QI'c
52
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==
j' IQ 12
(i
)J2
4.34
Heat and Mass Transfer
II
4
( TI -
Radiation
4
T2 )
[creosol
cos 02 dAI dA2]
QI-2
QI
1t r2
II
I
AI
AI A2
cos 91 cos 92 dAI dA2
1t r2
4.35
... (4.16)
AI A2
QI-2
QI
where
FI2
FI2 -
... (4.17)
Shape factor (or) configuration factor
From equation (4.10)
(or) View factor
n,
I cos 91 cos 92 dAI dA2
° °
cos 1cos 2dAI dA2
Shape factor is defined as "The fraction of radiative energy
that is diffused from one surface and strikes the other surface
directly with no intervening reflections."
r2
E:I
QI-2
JJ ° °
ff °
::::)I QI-2
cos 1cos 2dAI dA2
Similarly,
cos 91 cos 2dAI dA2
a[.T:J
... (4.19)
r2
... (4.14)
The total energy radiated by A2 is given by
Total energy radiated by AI is given by
QI = AI o
... (4.18)
r2
Q2
Ti
... (4.15)
Q2-1
Q2
° °
cos 1cos 2 dAI dA2
4
A2 c T2
1
A2
1t r2
AI A2
1t r2
Q2-1
Q2
where
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If
cos 91 cos 92 dAI dA2
F21 -
F2_1
Shape factor of A2 with respect to AI·
4.36
Heat and Mass Transfer
~
Radiation
4.23. HEAT EXCHANGE BETWEEN TWO NON BLA
F2_ I Q2
Q2 -I
I Q2-1
=
PARALLEL PLANES
... (4.20)
F2_1 A2aT~
=
. CK (GRAY)
Consider two very large parallel gray surf
f
.
rraces 0 areas A I and
A2 , at a small distance apart 'raand exchangl'ng diratiIon as sh own In
.
Fig.4.5.
From equation (4.18), (4.20), we know that,
AIFI_2
.'
4.37
A2F2-1
This is known as reciprocity theorem.
Thus the net rate of heat transfer between two surfaces AI and
A2 is given by
-,Q-12-=-A- -F- -a-[ T-4-=-1
-_-TA;I]
I 12
I
'" (4.21)
This equation is applicable to black surfaces only. If surface
having emissivity,
Q-12-- - -I- -12- -[
T-4----T-~-]-I
gA F a
'" (4.22)
1
I~
E -
Emissivity of surface
4.21. SHAPE FACTOR
Shape factor is defined as "The fraction of radiative energy
that is diffused from one surface and strikes the other surface
directly with no intervening reflections."
4.22.
Fig. 4.5.
Let T I' (XI and E I be the temperature, absorptivity and
emissivity of the surface I.
SHAPE FACTOR ALGEBRA (OR) VIEW FACTOR
ALGEBRA
Similarly T2, (X2 and E2 be the temperature, absorptivity and
emissivity of the surface 2.
In order to compute the shape factor for certain geometric
arrangements for which shape factors charts or equations are not
available, the concept of shape factor as a fraction of intercepted
energy and the reciprocity theorem can be used.
The following assumptions are made for the analysis.
I.
2. There is no absorbing medium in between the surfaces.
The shape factors for these geometries can be derived in termS
of known shape factors of other geometries. The interrelation
between various factors is called shape factor algebra.
3. The emissive and reflective properties are constant for
over all surfaces.
",
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The configuration factor of either surface is unity.
__ ~H~ea~/~a~nd~M~a~ss~~~r~an~~~e~r
__ ---------------------~4~.3~8
--
-.......
The surface 1: emits radiant energy E( which falls on the
surface 2. Out of this, a part of a2 E( is absorbed by the surface 2
and the remainder (I - (2) E( is reflected back to surface I.
Radiation
E( [I - 1-(I-t()(I" (I t-2),,)
Q(
1 -(1- t()(1 -~)
E,[I-[l-:-~,
continuing.
1
The rate of radiant energy leaving surface 1 is given by
=
(1 - a()2 (1 - (2)2 + ...... ]
E( - a.<1 - (2) Edl + P + p2 + ...... ] ... (4.23)
Q(
El t2
=
£(
where P
(1 - al) (J - (2)
Since (l( and a2 are less than unity, P will be less than unity.
. fini
In
imty
Q(
when extended to
The rate of radiant energy leaving surface 2 is given by
Q
1
2
El - a1 (1 - (2) EI x 1 _ p
El .'
E2 tl
=
... (4.25)
£1 + £2 - £1 £2
The net radiative heat exchange from surface I to 2 is given by
al (1- (2) El
I-P
Q(2
= \Q1 -Q2
E1 £2
~
£( + £2 - £1 £2
From Kirchoffs law, we know that,
absorptivity of a surfaces are equal.
... (4.24)
+ t2 - tl t2
Similarly,
gives 1 _1 P .
(4.23) ~
~-t(+t(t2]
EI t2
·t( + t2 -tl t2
E( - a( (I - (2) E( [1 + (I - a()(1 - (2) +
As P < 1, the series 1 + Y + p2 +
+',"]-', (1-<,) ]
[1
E,[' -1+"
+.,-', "-', +" "J
1 - 1 + t2 + t( - tl t2
E('_[a((l-a2)E(+al(l-a()(l-a2)2E(+
a( (1 - a()2 (I - (2)3 E( + ...... ]
Q(
emissivity
and
I
E1 £2-~
EI
£1 + £2 - E1 ~
E1
£1 + £2 - E1 E2
From Stefan-Boltzmann law, we know that,
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]
E,[1-(1-.,)(1-,,)_., (1-,,) ]
On reaching surface 1, a part a( (1 - (2) E! is absorbe~ and the
remainder (1 .- a() (1 - (2) E( is reflected. This process will go on
Q(
439
.,. (4.26)
4,40
Heal and Mass Transfer
Radiation
o r-
Heat exchange
(considering Area).
,4
cr 1 ,
0'2
=>
E2
£2 cr
4
.
cr r, £2 - £2 cr
E o A [T~ - T;]
... (4.28)
[From equation (4.27)]
where
E
r42 £ ,
£,
+ £2 - £, £2
£, £2 cr
[ri - r; ]
£,
between two para'lle'l surface is given by
4
r2
Substitute E, and E2 values in equation (4.26),
£,
=
4.41
4.24. HEAT EXCHANGE BETWEEN TWO LARGE
CONCENTRIC CYLINDERS OR SPHERES
+ £2 - £, £2
Consider two large concentric cylinders of areas A( and A2
exchanging radiation as shown in FigA.6 .
... (4.2n
-
where,
£
-
Fig. 4.6.
£'
Let
rl, (11 and £1 be the temperature,
absorptivity
and
emissivity of the Inner cylinder.
£
=
Similarly
T 2' (12 and £2 be the ,temperature, absQrpti,vity and
emissivity of the outer cylinder.
We can use the technique
£
=
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except
as we have used in parallel plates
4.42
Heat and Mass Transfer
--....._/
:'0
Radiation
I
4.43
'" (4.29)
Considering the energy emitted by the inner cylinder,
1. Inner cylinder emits the energy = El
2. Outer cylinder absorbs energy = u2 E.
E2
3. Outer cylinder reflects energy
4. Inner cylinder absorbs energy
s,
r. <X2== ~
El (1 - E2)
El (1 - E2) F2l <Xl
Al
El (I - E2) A2 EI
[.: F" ~ ~~ • al,"Ell
... (4.30)
E, (I - E,) [ 1 - E, ~; ]
6. Energy absorbed by the inner cylinder on the second
reflection
E, (1- E,>, E, ~; [ 1 - ~;..
~
]
This absorption and reflection go on indefinitely. So we ~
find the net energy lost by the inner. cylinder, considering infiniteI
times absorption and reflections.
Heat lost by the inner cylinder per unit area is given by
AI
01 = EI - EI (I - £2) EI A 2
Edl -£2)2
AI
£1
A ['AI]
I - A2 £1
2
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+ "..... .. ,'
... (4.31)
The net radiation heat transfer between the inner and outer
concentric cylinders is given by QI2 = QI - Q2
AI
~
£1 A2
4.44
---
Heal and Mass Transfer
Considering
area
and A2
AI
445
'" (4.33)
where
For cylinders,
For sphere,
AI EI E2 _ AI E2 EI
=>
QI2
=
AI
-A
EI
2
+ E2 _ A
EI E2
2
From Stefan-Boltzmann
law, we know that,
Eb
a T4
E
EI
ElaTI
=>
E2
E2 a
EI and E2 values
AI EI a
Ti
T;
in the equation (4.32),
E2 _ AI E2 a
AI
AI
2
2
A EI + E2 _- A
Aa
I
[ ~:
E I E2
EJ Eo
4.25. RADIATION SHIELD
Radiation shields constructed from I
.
.I I .
ow emissivity (h'
reflective) materia s. t IS used to reduce th e net radiat,
.
igh
between two surfaces.
on transfer
Let us consider two parallel planes I and 2
d Ts resnecti
each of area A t
temperatures T I an T 2 respectively. A radiatl'on S hiie ld ISi placed'a
between them as shown in Fig.4.7.
In
4
=>
Substituting
... (4.32)
AI
T;
EJ
EI E2
[Ti _ T; ]
(t, -
I) ] + E,
2
Radiation
shield
Fig. 4. 7. Radiation shield
The net heat exchange
radiatio 11 S I'
.
lie Id IS. given
by
between parallel plates without
A a
QI2 =
(Ti _ Ti)
... (4.34)
1 +l_1
EI
·E2
[From equationno. (4.28)J
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4.46
Heal and Mass Transfer
where
A _ Area in m2,
a
_ Stefan Boltzmann constant
= 5.67 x
10-8 W/m2 1(4
s I' £2
_
T1, Tz
_ Temperature of surface 1 and 2 respectively.
Emissivities of surface I and 2 respectively.
Under equilibrium condition
Heat exchange between 1 and 3 is
A a (T~ _ T;)
QIJ
=
I
1
-+-_
£1
Heat exchange
'" (4.35)
£3
=> QJ3 [
(~3+ 1) +( t +~ - J
I)
£~ _
= A c (T~ - Ti)
between 3 and 2 .is
QI3 =
... (4.37)
... (4.36)
Dividing the equation (4.37) by equation (4.34),
From equation (4.35),
QI3
QI2
(1+1_1)
(1+1_1)+(1+1_1)
EI
£3
If £1
£2
QI3
I
2
QI2
Substitute T; value in equation (4.36)
=>
I
QI3
£2
E2
EI
E3
= £3
= 2" QI2 (or)
I
Q32 = 2"
012
..
. b
tw parallel surfaces,the
Thus by msertmg one shield etween 0
direct radiation heat transfer between them is halved.
4.36) =:> Q32
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4.48
Heat and Mass Transfer
FORMULAE
USED
1. Helll e.'(challge between two large parallel plate is given by
4
4
E O'A(TJ -T2)
QJ2
-
Where emissivity,
I
----
E
I
I
- +--1
EJ
~
2.
Emissivity of surface 1
E2 -
Emissivity of surface 2
TJ -
Temperature of surface 1 - K
T2 -
Temperature of surface 2 - K
-
QJ2
where
4
-
3.
4
-
where
Area, A
A2
E
(I- )
E2-
1
27t r L
Ti ]
AO' [T; I
211
- + - + - - (n + I)
E, E2 Es
a.649 x o x A x [(900)4 - (400)4]
Stefan-Boltzmann constant
5.67 x 10-8W/m2 K4
0.649 x 5.67 x 10-8 x A [(90W-(400)4]
0'
Q
Q
I
-
Es -
Emissivity of shield.
[~
Result:
-
23.20 x 103 W/m2
A
Number of shields.
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I
+ 0.7 - I
Q
where
Heat transfer willi n shield is given by
II
I
~
0.6491
:::)
47t r2
Area, A
where
[From equation no.(4.28)]
1
E
I
I
E
-+-AI
For sphere,
Q = E o A [T: - T~ J
E AI 0' [ T J - T 2 ]
EJ
For cylinder,
:::: 900K
Cold plane temperature , T 2 :::: 400 K TI •
Emissivity of hot plane, E)
0.9
Emissivity of cold plane, E2
0.7
Tofind: Net radiant heat exchange per
.
square meter.
Solution: The heat exchange between tw I
o arge parallel plate
is given by
two large concentric cylinder (or)
Heat exchange between
sphere is given by
j
Given: Hot plane temperature , T )
-
EJ
I
[§xample 1 Calculate the n
.
et radiant .
",. for two large planes at a temo
Interchanoe
sq·
."erature of 9
II
per
esnectively. Assume tflat tile enziss;v',"
00 K and 400 K
~ r
.
l·oT of hOI l
that 0/ cold plane IS O.7.
_.
Pane is 0.9 and
E2
Stefan-Boltzmann constant
5.67 x 10-8 W/m2 K4
0'
4.26. SOLVED PROBLEMS
23.201 kW/m2]
Heat exchange,
Q
A
:::
23.20 kW/'l12
4.50
Heat and Mass Transfer
I Example 2 I Estimate
square
meter from
the net radiant
heat eXcllm~
a very large plate at a temperature
and 320°C. Assume
0/ hot plate is 0.8 and t
that emissivity
cOld
plate is 0.6.
Given:
T,
550 + 273 = 823 K
T2
320 + 273 = 593 K
c\
0.8
Solution:
3] Two large parallel
Radial"
Ion
4.5
of 900 K and 500
plclleslITeIII'
.
2
K res'
a",tQfnedat
area 0/6 m . ,Compare II,e net heal pectlvely. Eac" plate has a
for the following cases:
exchange hetweenth
an
e plllles
1. Both plats are hlack.
[Example
temperature
2.
{~,
Plates have an emissivity .1'
oJ 0.5.
Given:
0.6
c2
To find
Pel'
of 550
: Heat exchange
Heat exchange
T,
Tl = 900 K
900 K
T2=SOOK
500 K
per square meter, (Q/A).
A
between two large parallel plate is
given by
E (J A [T~ - T; ]
Q
[From equation no.( 4.28)]
where
To find:
Heat exchange for
1.
Both plates are black.
2.
Plates have an emissivity 0[0.5.
Fig. 4.8.
Solution:
This is heat transfer between two large parallel
plates problem.
I
I
Heat transfer, Q12
0.8 + 0.6 - 1
IE
Q
Q
A
I~
Result:
E a A (Ti - T~)
... (I)
Case 1: For black surface,
0.521
0.52 x 5.67 x 10-8 x A [ (823)4 - (593)4]
[.:
=>
::
(J =
Emissivity,
E
Q 12
5.67 x 10-8 W/m2 K4]
4
4
a A (T, - T 2 )
5.67 x 10-8 x 6 x [ (900)4 - (500)4]
9880.6 W/m2
lli~:__- 201.9 x 10 W I
3
9.88 kW/m21
Heat exchange,
~
= 9.88 kW/m2
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Case 2:
Emissivity,
EI
4.52
Heal and Mass Transfer
Radiation
In equation ( 1).
E
Tofind:
4.53
Heat exchange. (Q).
Heat exchange between two large concentric
Solution:
cylinder is given by
1
1
0.5 + 0.5 -1
... (I)
[From equation no.(4.33)]
0.331
0.33 x 5.67 x 10-8 x 6 x [(900)4 - (500~1
-
where
E
1. + AI
EJ
66.6 x 103 W ]
(1. _I)
~
1
Result:
1. Case 1:
2. Case 2:
Q'2
Q'2
=
201.9 x 103 W
=
66.6 x 103 W
I Example 4 I Calculate
the heat
rJ
0.6
exchange
IE
(I) ~
QI2
0.46 x 5.67 x 10-8 x 1t x 0, x L ~ [(403)4 - (303)4]
0.46 x 5.67 x 10-8x1txO.12x
60 mm
IQ
Result:
120mm
130° C + 273
E2
403
0.6
T2
30°C + 273
~
0.5
=
Heat exchange, QI2
1 x [(403)4-(303)4]
176.47 W
I
= 176.47 W
I Example 5 I A liquid oxygen is stored in double walled
0.12 m
£1
O.S-
0.461
12
TI
1t 02 ~
1)
_10.6 + 0.12
(_1
1)
0.24 O.S-
by radiation
0.060 m
r2
[.: A=1tOL]
1t 0, L, ( -1
-1 +---
between the surfaces 0/ two long cylinders having radii 120 mIll
and 60 mm respectively. Tile axis 0/ the cylinders are petrallelto
each other. The inner cylinder is maintained at a temperature of
130'C and emissivity 0/0.6. Outer cylinder is maintained at II
temperature of 30'C and emissivity 0/0.5.
Given:
A2
spherical vessel. Inner wall temperature is - 160'C and outer
watt temperature is 30 'C. Inner diameter of sphere is 20 em and
outer diameter is 32 em. Calculate the fol/owing :
if emissivity of spherical surface is 0.05.
2. Rale of evaporation of liquid oxygen if its rate of
1. Heallrans/er
303 K
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Fig. 4.9.
vapourizalion of latent heat is 200 kJ/kg.
4.54
Heat and Mass Transfer
_______ ------------------------~R~a~d~ia~lio~n--~4.~jj
Given:
Inner wall temperature,
Heat transfer,
- 160°C + 273
TI
Q12
o AI [T~ - Ti 1
E
... (1)
[From equation no.(4.33)1
113 K
where
-E
2
1
41t' I [I --+-0.05
41t ,2
2
[.:
0.05-
I
j
Area A = 41t,2;
E,
I
_1_ + 41t (0.10)2 [_1_
0.05
41t (0.16)2
0.05 - I
Fig. 4.10.
Inner diameter, DI
303 K
20cm
Inner radius, 'I
0.20 m
0.10 m
Emissivity,
£1
Latent heat
To jbld :
0.036
(I) =>
[( 113)4 - (303)41
[Q
=
200 x 103 J / kg
2.
This is heat exchange
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Rate of evaporation
Heat transfer
Latent heat
2.12 Jls
200· x 103 J/kg
.
here
between large concentnv sp
problem.
I
2.12 W
2. Rate of evaporation.
1.
W
200 x 103 J/kg
I. Heat transfer, Q12·
Sotution :
-2.12
[-ve sign indicates heat is transferred from outer surface to
inner surface 1
0.05 = E2
200 kJ / kg
0.036 x 5.67 x 10-& x 41t'~ x
[ (113)4 - (303)41
= 0.16 m
=
=
Q12
\
. 0.036 x 5.67 x 10-& x 4 x 1t x (0.10)2 x
32 cm = 0.32 m
Outer diameter, D2
r2
J
30°C + 273
Outer wall temperature, T 2
Outer radius,
= E2 = 0.051
= \ x \0-5 kg/s
4.56
Heat and Mass Transfer
Result:
l.
Heat transfer,
Q12
=
2.12W
2.
Rate of evaporation
=
1 x 10-5 kg/s
Tofind:
I Example 6 \ Two concentric spheres 30 em and 40 CIII '
Rate of evaporation.
Solution: This is heat exchange betw
sphere problem.
een largeconcentric
diameter
witt. , the space between them evacuate d are Usedto Sl 'II
tameter wu
liquid air at - 130"C in a room at 25 "C. The surfaces Of ;;e
spheres are flushed with aluminium of emissivity E ==
Calculate the rate of evaporation of liquid air if the latent heat0'
vapourisation of liquid air is 220 kJ/kg.
if
Given:
Inner diameter, DI
30 em
0.30 m
0,0;
Inner radius,
rl
0.15 m
Outer diameter,
D2
40 em
r2
0.40 m
0.20m
TI
- 130°C + 273
Outer radius,
143 K
25°C + 273
298 K
E
Latent heat of vapourisation
0.05
220 kJ/kg
220 x 103 J/kg
41tr~ [(143)L(298tl
0.032 x 5.67 x 10-8 x 41t x (0.15)2 x
[(143)4- (298t 1
I Q12 =
-3.83 W]
.
f ed from outer surface to
[- ve sign indicates heat IS trans err
inner surface]
Fig. 4./1.
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4.58
Heat and Mass Transfer
Heat transfer
Latent heat
Rate of evaporation
3.83
220 x 103
--
rofind:
I.74 x 10-5 kg/s
Result:
I Example 7 I A pipe of outside diameter JO em hav;"
Heat exchange, QI2
emissivity 0.6 and at a temperature of 600 K runs centrally ill :
brick duct of 40 em side square section having emissivity 0.8 a/fd
at a temperature of JOOK. Calculate the foilowing :
I.
Heat exchange per metre length.
2.
Convective "eat transfer coefficient when surrounding Of
duct is 280 K.
2. Convective heat t
Teo ::::: 280 K
ransfer coefficient
where,
30cm
DI
1t DI L
1t x
0.30 x 1
0.942 m2
1
Surface area, A2
0.942
Ir-£'--0-.5---'5 I
T42]
." (I)
(I0.8-1 )
4"l~
Fig. 4./1.
0.55 x 5.67 x 10-8 x 0.942 x
[ (600)4 - (300)4]
! Heat exchange. Q
I2
= 3569.2 W/m
Case (ii) :
Heat transfer by convection, Q
600 K
40 cm
-
I
0.6 + ~
". (2)
hA(T(J)-T..,)
h x A x (T 2 - T..,)
QI2
h x I x (300 - 280)
IQ
I2
20h
3569.2
20h
I
Equating (2) and (3).
= 0.40 m
(0.4 x I) x 4
[length L = 1m;
I
Q12
[.: L = 1m]
0.6
Brick duct side
When
(I) ~
0.30 m
Surface area, AI
x c x AI [T41
E
£'
Given:
Pipe diameter,
(h)
Solution:
Case I: We know that
1.74 x 10-5 kg/s
Rate of evaporation
I. Heat exchange, (Q).
Heat transfer coefficient,
No. of sides::: 4 J
h
178.46 W/m2 K
Result:
I A2
1.6 m2/
I.
Heat exchange, QI2
E2
0.8
2.
Heat transfer coefficient. h
T2
300 K
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3569.2 W/m
178.46 W/mlK
... (3)
I
4.60
Heal and Mass Transfer
4.27 SOLVED PROBLEMS ON RADIATION SHIED
I Example 1 I Emisslvities
------
-
0/ two
Given:
TI
~I
-+
OJ OJ-I
0.230 1
0.230 x o x A [T4 _ T4 ]
I
2
0.230 x 5.67 x 100a x A x [(1073)4 _ (573)4]
Radiation
shield
800°C + 273
1
E
where,
large paral/el pl
Qles
maintained at 800 'r and 300 'r are 0.3 and 0.5 respectivel" ",.
v- r'''d
net radiant heat exchange per square metre/or these plate,.. ",.
". r'''d
the percentage reduction in heat transfer when a polish
aluminium radiation shield 0/ emissivity 0.06 is placed hettv ed
tell
them. Also find the temperature 0/ the shield.
15,880.7 W/m2
1073 K
= 15.88 kW/m2
Heat transfer per square metre without radiation shield
300°C + 273
[_QAI2
573 K
0.3
Shield emissivity,
0.06
Plate 2
T,
where,
E
3.
in heat transfer
Temperature
of the shield (T3).
1
EI
1::3
cr x A [Ti - T~ ]
1
1
... (A)
- +--1
1::1
due to radiation
1::3
Heatexchange between radiation shield 3 and plate 2 is given by
Q32
Solution: Heat exchange between two large parallel plates
without radiation shield is given by
Ql2
1
- +--1
1. Net radiant heat exchange per square metre. (Q/A)
reduction
... (1)
-
Tofind:
Percentage
shield.
.
by
Fig. 4.13.
2.
15.88 kW/m2!
Heat exchange between plate 1 and radiation shield 3 is given
Plale 1
E)
=
=
E c A [Ti -
T1 ]
[From equation no.( 4.28)J
where,
e o A [T; - Ti ]
1
1 +1_1
E3
1::2
cr A (T; - T~ ]
1 +1_1
E3
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E2
... (B)
4,62
H!!CI/and Mass Transfer
We know that,
Radia/ion
013 = 032
~ransfe~_with
(j
'"
A I T; - T~]
I
I
- +--1
E3
[T; - T; ]
I
I
OJ + 0.06 - I
I
I
0.06 + 0.5 - I
(1073)4 - (T3)4
T; - (573)4
19
17.6
4
T3
Owithout shield
I Radiation shield temperature,
Substituting
I. Heat exchange per square metre without radiation shield
012 = 15.88 kW/m2
2. Percentage reduction in heat transfer
3. Temperature of radiation shield T3
I
1.33 x 1012
6.90 x lOll
I
T3 = 911.5 K
I
r2
900 K
E,
0.4
E2
0.7
E3
0.05
T3 value in equation (A) (or) equation (8),
5.67 x 10-8 x A x [(1073)4 - (911.5)4]
I
I
OJ + 0.06 - I
A
= 88%
= 911.5 K
[Example 2 Two large parallel plates are maintained at a
temperature of 600 K and 900 K and emissivities of 0.4 and 0.7
respectively. Determine heat transfer by radiation and also
calculate percentage of reduction in heat transfer and shield
temperature when another plate of emissivity 0.05 introduced in
betweenthem.
Radialion
shield
Given:
TI
600 K
Heat transfer with radiation shield
013
012 - 013
012
-
Result:
1.33 x 1012
911.5K
shield
0.88 = 88 %
0.926 x (1073)4 - 0.926 x (T3)4 + (573)4
(T)4
'" (2)
15.88 - 1.89
15.88
0.926 [ (1073)4 - (T3)4 ] + (573)4
(1.926)(T3)4
1.89 kW/m']
_. Owith shield
Owithoul
17.6 [(1073)4 - (T3)4]
19
+(573)4
(T3)4 + 0.926 (T3)4
[()~J ~
Reduction in heat transfer due to radiation shield
E2
[T~ - T;]
4,63
radiation shield
Plalel-
Tofind:
I. Heat transfer
T,
2. % of reduction in heat transfer
1895.76 W/m2
3. Shield temperature
..
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Plale2
(1'3)
Fig. 4.14.
WT
4.64
Heat and Mass Transfer
-£
Solutio" : Heat transfer between two large parallel plat
without radiation shield is given by
.
es
E o A [T~ - Ti ]
-e
where,
=
1
We know that,
1
OA + 0.7 -I
I
0.341
=>
Q12
0.341 x 5.67 x 10-8 x A x [ (600)4 - (900)4 ]
I
.
~
Heat transfer without}
radiation shield
Q12
_
•
A
-
-10,179.6W/m
2
I
1
- +--1
£3
£3
1
£1
£3
- +--1
[ (600)4 - (T3)4 ]
1
1
0.4 + 0.05 - 1
£,
=
I
I
- +--1
£,
=
£2
T; -(900)4
1
1
0.05 + 0.7 -I
Tj -(900)4
21.5
2Q.42
0.949 [ (600)4- Tj ] + (900)4
7.79 x 1011
.. , (A)
(1.949) T;
£3
:::).
I T3
7.79 x lOll
795.1 K
[ Shield temperature, T 3
E c A [Tj - T~ ]
54
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£3
20.42
21.5 [(600)4 - Tj ] + (900)4
Heat transfer between radiation shield 3 and plate 2 is given by
Q32
1
£3
cr A [T~ - Tj]
Q13
1
(600)4 - Tj
I
I
1
- +--1
£
(T; - T~)
- +--1
E c A [Ti - Tj ]
=>
£2
... (I)
Heat transfer between plate I and radiation shield 3 is given by
where,
_!_+_!_-1
1
". (8)
Q32
cr A (Tj - T~)
(T~ - T;)
0.341 x o x A x (Ti - T~ )
- 10,179.6 W/m2
QI3
-+1
£3
£2-1
a A (T~ - T;)
£1
~
\. QA12
c A (Tj _ T4 )
~
1
...~
I
795.1 KJ
FFtCZ'M"f'? ..
4.66
Heal and Mass Transfer
Substituting T3 value in equation (A) or Equation (B),~
A r T~ - T43 ]
(J
Heat transfer wit h
0 ==
}
radiation shield
13
_!_ + _!_ - 1
EI
013
013
A
II-leattransferwith
I
radiation shield
•
What would be the I
•
OSS 0/
which ts enclosed
in (I 55 c
Given :
E3
Case J :
5.67 x 10-8 x A x ( (600)4 - (795.!tJ
1
1
0.4 +0.05 - 1
W/m2
Diameter of pipe, D
30 em = 0.30 m
J
Surface temperature , T 1
}013 =
-712.13W/m2
300°C + 273
~,
Emissivity of the pipe , £1
_J
==
012-013
Case 2 :
Emissivity,
+ 712.13
Toflnd :
Solution :
93%
Case J :
in heat transfer
Shield temperature,
T3
Reduction in heat loss.
25'C
~
0
- 10,179.6 W/m2
o A [Ti - T; 1
£1
£1 x o x n
D L [Ti - T; 1
[.: A = nDLl
93%
o
795.1 K
J
0.8 x 5.67 x 10-8 x n x 0.30 x
'
L x [ (573)4 - (298)4 1
I Example 3 I A pipe of diameter 30 em, carrying steamrulll
in a large room ami is exposed to air at a temperature of 25,:1
Tile surface temperature of tile pipe is 300 'C. CalcuMe ti,e ~
of heat to surrounding per meter length of pipe due to ther I
radiation. Tile emissivity of tile pipe surface isJ).8.
L
Scanned by CamScanner
0.91
£2
W
Heat transfer,
Result:
55 em = 0.55 m
I. Loss of heat per metre length, (Q/L).
2.
0.93
0.8
Outer diameter , D2
012
Heat transfer without } 012
radiation shield
A
25°C + 273
298 K
'" (2)
-10.179.6
3.
ue to radialion 01' th
.
m ~"'meter hrick 0
'.
'J
e pipe
if enllsSll1ity 0.91 ?
Air temperature , T 2
-10,179.6
2. % of reduction
467
.
Radiation
heat d
573 K
-712.13
Reduction in heat tranSfer}
due to radiation shield
1.
ti.
lOlL
4271.3 W/m I
Heat loss per metre length = 4271.3 W/m
ASPEiM- ,
.M·"'·
.n
Radialion
4.68
4.69
Heal and Mass Transfer
Case 2: When the 30 cm diameter pipe is enclosed ~
diameter
~
pipe, heat exchange between two large concentrjlll
c
cylinder is given by
. ivity E value in equation (I),
stituting emlssl
sub
Q::::
0.76 x 5.67 x 10-8 x 7t x DI x L,
<I)
x
z»
Q
l (573)4 - (298)4 ]
0.76 x 5.67 x 10-8 x 7t x 0.30
L
x [ (573)4
[~
- (298)4 ]
4057.8 W/m \
Reduction in heat loss
4271.3 - 4057.8
Fig. 4.15.
213.4
'" (I)
where
-E =
[From equation no. 4.331
Result:
I. Heat loss per metre length
4271.3
2. Reduction in heat loss
213.4 W 1m
W/m
I Example 4 I Tire outlet header of a higl. pressure steam
1
L (1-
7t 01
-+-0.8 7t 02 L
1
-+-01
0.8 D2
0.91-
(10.911)
1
)
superheater consists of pipe (e = 0.8) of diameter 27.5 em. Its
surfacetemperature is 500 CC. Calculate tire loss of heat per unit
lengthby radiation if it is placed in an enclosure at 30 CC.
If the header is now enveloped in II steel screen of diameter
32.5 em and emissivity of 0.7 and the temperature of the screen is
340 "(:,/ind the reduction in heat by radiation due to provision of
thisscreen.
Given:
_1 + 0.30 (_1_
0.8 0.55 0.91-
/E = 0.76/
Scanned by CamScanner
I)
Case 1:
Emissivity,
£1
0.8
Diameter,
0,
27.5 em = 0.275 m
T,
500°C + 273 = 7 3 K
T2
30°C + 273 = 303 K
-.I. 70
Heal and Mass Transfer
Case 2:
32.5 em = 0.325 In
Screen diameter, D2
Emissivity,
0.7
£2
340°C + 273
Screen temperature, Ts
Tofind:
= 613 K
I. Loss of heat per unit length.
2. Reduction in heat loss.
@
So/ulilln:
Fig. 4.16.
30'C
C
Cast! J: We know that,
E
1
1t D( L
~ + 1t D2 L
Heat transfer, Q
Q
£(
x o x 1t D L x [T4
(
-
T4]2
(1 )
£2 -
1
1
[.: A = 7t D LI
0.8 x 5.67 x 10-8 x 1t x 0.275 x
L x [ (773)4 - (303)41
~
13661.41 W/m
=
I~ -
1;~6kW/m
I
Substituting
Case 2: Heat exchange between two large concentric cylinder
is given by
Q =
Here
~
where,
T2 =
Q =
E
I
=
I
£1
AI ( I
+ A2
£2
Scanned by CamScanner
-I)
Q
0.62xcrxAx[T~
-T~]
0.62 x 5.67 x \0-8 x 1t x D( x Lx [T~ - T~ ]
£' o A [T~ - T~ ]
T s = Screen temperature.
£' a A [T~ - T~]
(I) ~
E value in equati. lIi (1),
=
0.62 x 5.67 x 10-3 x 1t x 0.275 x L.x
{ (773)4-(613)4]
." (I)
4.72
Radiation
Heal and Mass Transfer
Reduction in heat transfer due to screen
13.66-~
7.11 kW/m
Result:
roJind:
1.
Heat loss per metre length,
~
13.7 kW/m
Temperature,
T2
50°C + 273
Emissivity,
£2
323 K
0.9
).
Heat lost by radiation.
2.
Reduction in heat loss.
4.73
solution:
2.
Reduction in heat}
transfer due to screen
=
7.1 kW 1m
I Example 5 I Calculate the heat lost by radiation per "'elre
length of 8 em diameter pipe at 400 'C and emissivity of 0.7, whell
E, = 0.7
(a) It is located in a large room with a red brick walls
maintained at a temperature of 35 DC.
(b) It is enclosed in a 20 em diameter of red brick pipe
maintained at a temperature of 50 DCand emissivity oJ
0.9.
Fig. 4.17.
Case 1 :
Heat exchange
Q1
£1 o A [T~
0.7 x 5.67 x 10-8 x 7t x Dl x L x
Also find reduction in heat loss.
Given:
Case 1:
Length, L
Diameter of pipe, DI
[(673)4"": (308)4]
0.7 x 5.67 x
1m
8 ern = 0.08 m
1956.5 W \
Temperature,
TI
Temperature,
T2
is given by
0.7
35°C + 273
308K
Case 2:
Diameter,
D2
20cm
0.20m
Fig. 4.18.
Scanned by CamScanner
10-8 x 7t x 0.08 x 1 x
[ (673)4 - (308)4]
... (1)
Case 2 : Heat exchange between two large concentric cylinder
673 K
Emissivity, EI
- Ti ]
4.74
r
Heat and Mass Transfer
--------------~
E
crAdTI
4
4
...........
-T2]
-[Example
". (2)
6
I Emissivitles
Of tw
Radialion
large paraUel plata
",aintained at T, K and T2 K are 06
. and 0 6 respe,. I
,Fer is reduced 75 tl
.
C Ive!y. Heat
trans,.
'mes When a
I' h
adia/ion shields of emissivity 0 04
I po IS ed aiulflilliwn
r
. are P aced ill 6etw . h
Calculate the number of shields required.
eell t em.
E
Given:
EI = 0.6
E2
Heat transfer reduced
= 0.6
= 75 times
Emissivity of radiation shield
E = E
'
J
_I + 0.08 (_1 I)
0.7
0.2
3 -
~_~
Q2
004
.
Radiation
0.9-
shields
£3
(2) =>
4.75
0
-
where
= 0.04
,0.67 \
=
0.67x5.67xlO-8xltxD1xLx
[ (673)4 - (323)41
=
0.67 x 5.67 x 10-8 x It x 0.08 x Ix
Fig. 4.19.
[(673)4 - (323t1
IQ
\854.7 W
2 - =
Reduction
in heat loss,
Q1 - Q2
I
To find:
...(3)
Solution:
A o (T~ - T~]
'" (I)
101.8 W
Result:
2.
Heat loss,
Reduction
Q1 =
Q2
in heat loss
Scanned by CamScanner
Heat transfer with n shield is given by
1956.5 - 1854.7
=
I.
Number of shields required.
=
Heat transfer without shield, i.e., n = O.
1956.5 W
1854.7 W
101.8 W
\
(I) =>
\
Q12
=
Acr(TI 4 -T2 4 1
I
I
EI
£2
- +- - I
... (:
,f,7(1
Ileal and Mass Transfer
Radiation
Heat transfer is reduced 75 times,
Qwithout shield
~",Z7lle
71 Two large parallel plates with e = 0.5 each,
~ intained at different temperatures and are exchanging
are trIaI by radiation. Two equally large radiation shields with
"eat on ~"'issivity 0.05 are introduced in parallel to the plates.
sur/ache percentage of reduction in net radiative heat transfer.
find t e
.
. n : Emissitivlty of plate 1, EI = O.S
G,ve .
Emissivity of plate 2, E2
O.S
75
, Qwilh sh.eld
QI2
o.,
75
A cr [T~ - T~]
1
1
EI
E2
- +--1
o
(1) =>
4,77
Emissivity of shield,
7S
Es
Number of shields, n
2
Plate. 2
Plate. 1
7S
I
I
2n
0.6 + 0.6 + Q04 - (n + 1)
I
1
0.6 + 0.6 - 1
3.33 + 50 n - n - 1
2.33
50n-n-l
171.67
49n-l'
171.67
49 n
Result:
7S
3.52 ~ 4
In
4 I
Number of shields required, n = 4 nos.
shields
Fig. 4.20.
Tofind: Percentage of reduction in net radiative heat transfer.
75
Solution:
Case 1 : Heat transfer without radiation shield:
Heat exchange
between
radiation shield is given by
172.~7
n ,=
Scanned by CamScanner
Radiation
where,
-E
two large parallel plates without
478
Heal and Mass Transfer
~ED
1
1
o.s + O.S - I
\E
0.333 \
Case 2 : Heal transfer
0.333 (J A [T4I - T42 1
with radiation
•. 28.
~
Colculate the net r d'
~
alanthet
, ea for IWO large parallel plates at te
a exchange per
"r ar
. I
mperature of 4270C
7 respecl"'e 'Yo &(ho, pia',) = 0.9 anll e
_
a"d
0C
1 I'
aluminium
shield is placed b ::'~" platt) - 0.6. If a
o ,s ti
t. wee" them fi
",age of rel/u('lion in II.e heat transl'e
• ",d the
P
perce
')1 r. £(.11;",,) :; 0.4.
/May 2004, A""a Uni\lf!rSity/
". (I)
s"ield :
Given:
427°C + 273
TI
We know that,
700 K
Heat transfer with 11 shield
27°( + 273
A (J [T~ - T~ 1
Qwi1h shield
4.79
PROBLEMS
"e
0.333 (J A [T4I - T41
2
Q",ithout shield
UNIVERSITY
3
Plate. 1
Plcite.2
[I
T,
300 K
=
2
0.9
0.6
A (J (T~ - T~ 1
I
1
0.4
2 x 2
0.5 + 0.5 + 0.05 - (2 + I)
fig. 4.21.
A (J [T~ - T~ 1
Tofind:
I.
et radiant
81
... (2)
0.0123 A (J (T~ - T~)
QWilh shield
heat exchange.per
Percentage
of reduction
m'! area.
in the heat transfer.
Solution :
Case 1 : Heat transfer without radiation shield:
Heat ex hange between 1\ 0 large parallel
We know that,
Reduction in heat transfer l
due to radiation shield J
Qwithoul
shield - Qwith
Qwithout
shield
radiation shield i given by,
hield
QI_
0.333 (J A (T~ - T~] - 0.0123 A (J [T~ - T~ 1
0.333 (J A {T~ - T~ 1
0.333 -0.0123
.....,..,
0.
= 096
.
3
where
'£ (JA[T~ -T~l
I
'£
96.3%
..)..)..)
Remit: Percentage of reduction in l = 96 ...
net radiative heat transfer J
Scanned by CamScanner
.J
plates without
I
I
0.9
0.6 - I
[Fromequationno.t4.281
·y.,oitr=!
WE
~5
4.80
Heat and Mass Transfer
\E
Radiation
= 0.5625 \
_______
0.5625 x 5.67 x lO-8 x A x [(7
00)4- (31\1\
-z»
V\I~l
cr A [T~ - T~ )
cr A tT~ - 1~ 1
l.+_!__1
- +--1
1
1
E,
E3
E2
E)
= 7.39 x 103 W/m2
Heat tran~f~r Without}
radiation shield
Q12
A
3
7.39 x 10 W/m2
"'(\1
Case 2 : Heat transfer with radiation shield:
Heat exchange
between
14 _ 14
I
I
I
1
- +--1
EI
E)
£)
£2
plate 1 and radiation shield 3' IS gIVen
. by
T~ - (300)4
1
1
0.9 + 0.4 - 1
1
\
0.4 + 0.6 - \
T;
T~ - (300)4
(700)4 -
where,
3.166
2.611
=> 7.60 X io'
-t
2.611 T~ - 2.11 x \010
- 3.166 T~
5.77 T~
7.81 x 1011
1.353 x lO'l
T4
)
cr A [T~ - Tj]
'" (2)
1
1
- +--1
t1
Heat exchange
2
(700)4 - T;
E o A [T~ - Tj ]
Q13
3
- +--1
_
-
[T~ - T~ 1
Radiation shield temperature,
T) =
606.55 K
Substitute T) value in equation (2) or (3),
t3
between radiation shield 3 and plate 2 is given by
Heat transfer with \.
=>
radiation shield J Q\3
_
-
4 T4)
cr A lT I - 3
l. + _!__ 1
EI
where,
E3
5.67 x 10-8 x A [(70W - (606.55)j
\
I
0.9 +0.4 - 1
s
Q\3
cr A [Tj - T~]
1
1
- +--1
t3
£2
2.27 x 103 W/m2
A
.. , (3)
Reduction in heat loss \.
due to radiation shield J
We know that,
55
Scanned by CamScanner
4,81
Qwithout shield - Qwith shield
Qwithout shield
..
4.82
Heal and Mass Transfer
QI2 - QI3
Radiation
Q12
Solution .'
7.39 x 103 - 2.27 x 103
7.39 x 103 -
Case J .' Heat transfer without radiation shield :
Heat exchange between two parallel plates without radiation
= 69.2%
I. Net radiant heat exchange} QI2
0.692
Result:
(without
shield)
shield is given by
= 7.39
A
103 II·
012
I'I'I~
2.
Percentage of reduction in !lIe}
heat transfer due to hield
= 69.2
£ a A [T~ - T~]
I::
I
-+
0.5
have emissivities of 0.5 (1/1(1O.B respectively. A rat/illtionSA~
havillg (111 emissivity of 0.1 011 one side 0/1(/ (III emiSS;";tyo/tf
011 the other side hi placed between the plates. C(tlell/me tht 6e
IDec.2005,
Given :
TI
800 K
T2
600 K
EI
0.5
E2
0.8
EJa
0.1
E3b
0.05
3a
Plane.
with and lVil~
VnivtrJi
3b
Plaro;l
1
£,
T,
£2
E3a
::::>
E3b
::::>
0.444
012
0.444 x 5.67 x 10-8 x A
A
=
7.048
Heat transfer without}
radiation shield
I
-0.8
I
1£
012
AIIIUI
[From equation no.(4.28)]
-
where.
1'-E-.\-'1I-'-I1P-le-2---',
Two large parallel planes {It BOOK and 6ft
transfer rate hy radiation per square meter
radiation shield. Comment 011 the results.
4.83
r(800)4 - (600)4]
103
W/Il12
=
7.048 x 103 Whn2
QI2
A
... (I)
Case 2 : Heat transfer with radiation shield "
T2
Heat exchange between plate I and radiation shield 3a is given by
T3
where,
Radiation
E
shield
Fig. 4.22.
Tojilld:
I.
2.
witlt
Heat transfer rate per square metre
radiation shield.
~~
Heat transfer rate per square metre
radiation shield.
_,
J.
Comment
on the re LIlt
Scanned by CamScanner
a A (T~ - T;]
I
EI
I
+- E a
... (2
4.84
Heat and Mass Transfer
Heat exchang e b etween_ radiation
Q 3b, 2 =
where,
E
-.........
shield 3b an d pate
I 2 IS'
.---given b
o A [T~ - T~ 1
)
II stitllte T 3 value
in equation (2) Or (3),
-;:::l
}-leattransfer with}
radiation shield QI,3a
:::
-£1 +l..
£
-\
30
~[(800'j4
_ r,u 'h ...
l..~
'" (3)
1
1
-+-_
Q 1,3a
}-leattransfer With} QI, 3a
radiation shield
A
Q3b,2
cr A [Tj _ T~)
cr A [T~ _ Tj)
1
-+-El
509.74 W/m2
E2
We know that,
1
E3a
1
-+
1
-_
E3b
E2
Reduction in heat transfer}
due to radiation shield
3
1
1
El
E3a
2
1
o.os
1
20.25 [(800)4 - Tj ) ==
4
8.29 x 1.012 - 20.25 T 3
11 T~ _1.42x
gg
12
2.
10
significantly.
3 ==
Heat transfer with l~ = 509.74 W/m2
radiation shield J A
The presence of radiation shield reduces tbe beat transfer
3
Scanned by CamScanner
transfer rate significantly.
Result:
1. Heat transfer without l
= 7.048 x \03 W/m2
radiation shield 1 A
11 [T~ - (600)4]
3.1072 x lOll
T4
T
The presence of radiation shieldreducesihe h
31.25 Tj
9.71 x 1012
shield temperature,
Comment:
2025
11
::::>
Radiation
]a
92.7%
1
+ 0.8 - 1
Tj - (600)4
(800)4 - T~
:::>
shield
0.927
Tj - (600)4
o.s+0.1-1
:::>
shield - Q\I;1h shield
Qwilhout
7.048 x \OL 509.74
7.048 x IQ3
(800)4 - T~
:::>
Qwilhout
Qu
-+--
:::>
509.74 W/m2
QI2-QI
T4 _ T4
1
OJ - \
0.5
cr A [Tj _ Ti]
E3b
4.85
a A lTi - 141
~
1
E
R adialion
5 b
746.60 K
4.86
Heat and Mass Transfer
I
[ Example 3 Two very large parallel plates with em' '.
ISSIIII('
0.5 exchange IIeat. Determine tile percentage reduction' le,
"eat transfer rate if a polished aluminium radiation shiel/: Iht
Ife~
0.04 is placed in between t lte p Ia tes.
G
of plate 1, El
Emissivity
of plate 2, E2 = O.S
Emissivity
of radiation
shield,
E)
rSI~J
= O.S
Emissivity
= 0.04 = Es
'
J-T--+
0.5
[June 2006, Anna Unille .
Given:
I
==
==
::::>
Q12
==
::::>
Qwithoul shield
=
OJ - I
~
0.333 (j A [T~ - T; 1
0.333 (j A [T~ - T; 1
... (1)
Case 2 : Heat transfer with radiation shield:
We know that,
Heat transfer with n shield,
Qwith shield
=
where,
-
Radiation shield
Es
Emissivityof radiationshield.
n -
Numberof radiationshield.
A (j [T~ - T; 1
Fig. 4.23.
Qwi1h shield
To find:
radiation
Percentage
of reduction
radiation
exchange
0.04
A cr [T~ - T;]
between
two
shield is given by,
where,
_i_+li!1_(I+I)
0.5 + 0.5
shield.
Solution :
Case 1 .. Heat transfer without radiation shield:
Heat
_1
in heat transfer due to
large parallel
plates witho~
52
Qwi1h shield
=
We know that,
Reduction in heat tran~fer }
due to radiation shield
=
Scanned by CamScanner
0.0192 A o [T~ - T;]
,- QWIith shield
Q without shield
Qwithout
shield
". (2)
4.88
~
Heal and Mass Transfer
.>
0.333 A (J [Ti - Ti] - 0.0192 A (J [Ti - ~
4
4
0.333 A (J [T ( - T 2 ]
Radiation
Result:
=
[
~
15.8 x 1Q3 W/m2]
Q
Percentage of reduction in heat transfer rate = 94.2%
of
two
large
parallel
planes
at 800 '(' and 300'(' are 0.3 and 0.5 respectively. Find
the net radiant heat exchange per square metre for these plates.
15.8 x 103 W/m2
]lesu1t:
A
~
Find
~
lite relative heat Iransl'.er b
'J'
etween two
1000 K ancl500 K wit en lit
I nes at temperature
large p a
1.
2.
ey are
Black bodies.
Grey bodies with emissivities of eaclt surface is 0.7.
[Oct. 2001, MUI
800 e + 273
1'(
Given:
0
[May 2002, MUI
Given:
T,t t T,
1073 K
300° e + 273
573 K
1>,
0.3
£2
0.5
Q = g (J A (T1 - T;)
£
=
[From equation no.(4.28))
T2
500 K
£,
0.7
E2
0.7
~T'
£,
1.
Heat transfer
for black bodies.
2.
Heat transfer
for grey bodies.
Case
1: Heat 'exchange
0.23 x a x A (T~ - T~ )
Scanned by CamScanner
between
two large parallel plate is
given by
Q
1
1
OJ + 0.5 - 1
£2
Solution:
I
0.23
Q
1000 K
Fig. 4.25.
To find :
Solution : The heat exchange between two large parallel plate
is given by
where
T,
Fig. 4.24.
Heat exchange per square metre.
To find:
- (573)41
94.2%
I Example 4 I Emissivities
maintained
48CJ
2
0.333 - 0.0192
0.333
0.942
0.23 x 5.67 x 10-8 x l(1073)4
QA:::::
For black bodies ,
E
A a (T 4I - T4)
2
=
Q = A a (T~ - Ti )
E
Q
A
5.67 x \0-8 [(IOOW-(SOO)41
4.90
:: 53.15 x 10 W/Il12
£" A (J (T~ - T~ )
Q
Case 1:
-f. =
where
1::2
I
~
Inner temperature,
TI :::
Outer temperature,
T2
O.os
- 183°C + 273 '" 90 K
E
293 K
I
0.7
I
0.7
-+ --
Latent heat of oxygen
0.5381
Toflnd : Rate of evaporation
0.538 x A x 5.67 x 10-8 x [ (1000)4 - (500)41
I~
210 kJ/kg
210 x J03 J/kg
IE
Q
20°C + 273
28.6 x 103
Solution:
w/m'l
Result:
I.
2.
I
Q
A (Black surface)
53.15 x 103 W/m2
AQ (Grey body)
28.6 x 103 W/m2
I
Example 6 The inner sphere of liquid oxygell container is
40 em diameter and outer sphere is 50 em diameter. Both have
emissivities 0.05. Determine the rate at which lite liquid oxygen
would evaporate at - 183'C wizen lite outer sphere at 201(.
Latent heat of oxygen is 210 kJlkg.
{April 99, MUI
Given: Inner diameter,
Inner radius,
D(
40 cm
r(
0.20 m
Outer diameter,
D2
50 ern
Outer radius,
r2
0.25 m
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=
Fig. 4.26.
This is heat exchange between two large concentric spheres
problem.
Heat transfer,
0.50 m
-E 0' Al [T 4I - T4]2
... (I)
[From equation no.(4.33)]
0.40 m
where
=
QJ2
E
4.92
Heat and Mass Transfer
\
47t"1
0.05 + 47t
r;
(\
)
300°C + 273 _
- 573 K
0.05 - \
OJ
O.~5 + :\
(O.~5 - I )
Radiation shield emissivity"
- 0
"3 -
.05
I
(I0.05 - I )
o.osI + (0.20)2
(0.25)2
Radiation
shield E:! :: 0 05
I E -_ 0.031 I
(\) =>
Plate 1
0.031 x 5.67 x 10-8 x 4 x 7t x (0.20)2)(
QI2
Plate 2
[ (90)4 - (293~1
- 6.45 W
[ - ve sign indicates
inner surface.]
heat is transferred
Fig. 4.27.
Heat transfer
Latent heat
Rate of evaporation
2 lOx
Tofilld:
6.45 W
\ 03 1 I kg
I. Net radiant heat exchange per square metre
6.45 lis
2.
210 x 103 l/kg
3.07 x 10-5 kgls
, Rate of evaporation
Result:
Rate of evaporation
=
I
3.07 x 10-5 kg/s
I Example 7 I Emissivities 0/ two large parallel plaltl
maintained at 800'(' and 300'(' are 0.3 and 0.5 respectively.Fin'
the net radiant heat exchange per square metre 0/ the plates.If'
polished aluminium shield (E = 0.05) is placed between the1l'o
Find the percentage 0/ reduction ill heat transfer. [Oct. 99, MW
Given:
TI
=
T1 = BOO'C
from outer surface to
(~2).
Percentage of reduction in heat transfer due to radiationshield.
Solution:
Case 1 : Heat transfer without radiation shield:
Heat exchange between two large parallel plates without
radiation shield is given by
where
E
=
e + 273 = 1073 K
8000
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4.94
Heat and Mass Transfer
1
1
OJ + 0.5 - 1
=>
(J
I
0.230
Q12
0.230 x 5.67 x 10-8 x A x [(1073)4-(57)
EI
E3
~
3
~I
1
I
0.3 + 0.05 - I
=
Q12
A
'" (I)
-
4
E crA[T,
-T3]
(l073)4
=>
4
1
-
... (~
=>
2.78 x 1013 - 21 r4
=>
3.02 x 1013
-EcrA[T - 4 T4]
2
3
_!_+_!_-1
-+1
2
I
-
T; - (573)4
~
-+
0.05
OJ - I
T~ - (573)
21
22.3 Tj -2.4
x 1012
43.3 Tj
I r3:::
913.8K
1
T 3 value in equation (2) or (3),
Heat transfer With} Q _ 5.67 x 10-8 x A x [(1073)4 (913.8)4]
radiation shield
13 1
I
-OJ +0.05 - 1
~J
1594.6
:::
w/m'l
... (4)
Qwilhout shield - Owilh shield
Qwithout shield
012
- Ti]
_!_+_!_-l
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E
012 -013
E2
cr A [Tj
We know that,
=>
% of reduction in heat transfer}
due to radiation shield
1
E3
3
I
Heat exchange between radiation shield 3 and plate 2 is givenby
3
22.3
Substitute
E3
- r:
Shield temperature
E
E
E2
495
(1073)4 - r4
Heat exchange between plate 1 and radiation shield 3 is givenby
where
I
+--1
Radiation
o A [ T4
3 - T4]
£3
Case 2: Heat transfer wit" radiation shield :
where
1
-
\E
Heat transfer Without}
radiation shield
A [T~ - Tj]
I
... (~
I
15.8 x 103 - 1594.6
15.8 x 103
0.899 == 89.9%
I
il
II
~
Heat and Mass Transfer
4.96
Result:
--- roJind:
1.
2.
Heat exchange without} QI2
radiation shield
A
15.8 x 103 W/rn2
Solution:
Heat transfer,
= 89.9%
% of reduction in heat transfer
I Example 8 I Tile amount
250
0"
of radiant energy falling
50 em x 50 em horizontal thin metal plate insulated to the bolto II
is 3600 kJlm2 hr. If the emissivity of the plate surface is 0.8
the ambient air temperature
is 30 'C, find the equilibriuIPJ
temperature of tile plate.
/April97, MUJ
Q"~
Given:
Area,
A
50 em x 50 em
0.5 m x 0.5 m
0.25 ffi2]
Q
Radiant energy,
3600 kl1m2 hr
3600 x 103
1
3600
m2 s
103 lis x m2
m
Area
E
c A [T~ _ T; ]
0.8 x 5.67 x 10-8
250
41
_
(303)4
T
2.2 x 1010
I TI
417.89
Result:
1.13x1Q--8[T4
Plate temperature,
.
)( O.2Sx [T4
1-(03)4]
I - (03)4 ]
TI ::: 417.8K
~u~_.
I
[Example 9 A pipe carrying st
'
ea", havm
diameter of 20 em runs in a large room d ' g an Outside
30 "C. Tire pipe surface temperature is 400Clr>
toQlrat
.
.....Calcu/ateth l
of heat to surroundings per metre length of
' e oss
thermal radiation. Emissivity of the pipe s ."the~,pe,dueto
.
"rlace u 0,8, What
would be the loss of heat if tire pipe is encl d'
..
ose m a 40 Clft
diameter brick conduit of e= 0.9 ?
/MU A '
•
, p,,12001/
{The procedure of this problem is sameas problemno.3 _
(Solved problems 011 radiation shield - Section4.27) J
I Example 10 I The surface of douhlewalledsphericalvessel
1000 ~
Here,
Q
0.25 m2
W
Q
1000 x 2 x 0.25 m2
IQ
2S0W
rn
I
used for storing liquid oxygen are coveredwitha layerofsilver
lining having an emissivity of 0.03. The temperatureof outer
surface of the inlier wall is - 153't' and tl,etemperature
ofthe
inner surface of the outer wall is 27't'. Thesphereare21 em and
30 em diameter
with tire space betwee» them evacuated.
Calculate the salt of evaporation of liquid oxygenduetoradiant
l leat transfer. Latent heat of vapoumatlon
• • °ifl' Iqui'd ot}'oen
'. is 220
kJlkg.
.
as problemno.6
{The procedure of this problem IS same
(Solved university problems - Section 4.28)J
&
Emissivity,
E
Ambient air temperature,
0.8
T2
303 K
56
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,
4.98
Heat and Mass Transfer
I
( Example 11 Two large parallel
I
'
oifO 3
dO'
.
panes hav'
. an
.5 are maintained at a t
'Ilg e",·
K'
emperature »r«
ISsi~'
respectively. A radiation shield havi
'J
00 /( q~
b h
laVing an en .
Q"d
ot sides is placed between two pia
C
IlSSivity 0'0 4~
hi
..
nes.
alculat la 'J .OS
oif .s IIeld, (u) ratio of heat transfer rate
.
e
te"'Per ~
shield.•
Without sh,' e I d I 1Il~.
"1
Radiation
~
,I,
1M
0",.
[The procedure of this problem is
U, April 20'"
/S I d
.
.
same as pr b
~I
I' 0 ve
umverSI
roblems - Section 4.28)J
0 Ie", I!o.~
Exam Ie 12 Liquid nitrogen boil'
.
'.
Ing at - 1960C .
In a 15 litres spherical
container ot 32
.
IS Sto,~
tai
.
'J
em d,am
con atner IS surrounded by a concentric spheri
eler. 1~
.
erlcal shell
d iameter whose inside temperature is mai t .
Of 40 ~
In alned at jOt'
annular space between the two is evacuated. Tl.
. 1'1rt
l
faci
te surfaces 'f
sp teres acing each other is silvered and hav
. oJI~
35 ", ki
e an emlSsivit.,
. . ~a Ing the latent heat of vapourisauan fi
h :' ~
or t e "qllij
oxygen as 200 kI/kg, find the rate at which it evaporat IV,
the thermal resistance offered by the inner surface .~St·he?lta
, II'
JI
OJ
e Iftlltr
sne and by the thickness of the same.
Emissivities
diameter of 20 em runs In a I~rge room and is exposed to air at
temperature of 30~. TI,e pipe surface temperature is 400~.
Calculate the loss of heat to the surroundings per metre length of
pipe due to thermal radiation. TI,e emissivity of the pipe surface
. [Bltarathidasan University, Nov. 9~
[The procedure of this problem is same as problem no.!
r-;,,;;.:...~roblems- Section 4. 28)J
of a double
walled spherkl
vessel used for storing liquid oxygen are covered with a layer~
silver having
an emissivity
of O.03. The temperature
of tile oulll
surface
of the inner wall is - 153 CC and the temperature of innll
surface
of the outer
30 em in diameter.
Calculate
wall is 27CC. The spheres are 21 cm anI
With the space
the radiation
heat transfer
vessel and the rate of evaporation
vapourisution
is 220 kJlkg.
[The procedure
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them evacuated.
the walls intoIhl
if the ratt~
University, Apr. 91)
0
is same as problem 710.
[Bharathiyar
of this problem
(Solved university problems
between
through
of liquid oxygen
- Section 4.28)J
parallel plates
. t ,'ned at 800 ~ and 300 c:c are 0.3 and 0.5 respectively. Find
",a,n a
net radiant heat exchange per square metre between the
the
[Nov. 97, MKU/
plates.
[The procedure of this problem is same as problem no.4
d university problems - Section 4.28)J
(SO Ive
a-x-a-m-p-Ie-l-S"IA pi~e carrying steam having an outside
o
The surface
of .two large
4.99
is 0.8.
What would be the loss of heat due to radiation if the pipe is
enclosed in a 40 em diameter brick conduct of emissivity 0.91.
IBllUratlriyar University, Nov. 96/
[The procedure of this problem is same as problem no. 3
(Solved problems on radiation shield - Section 4.27)J
I Example 16 \ Consider two large parallel plates one at TI
10000[( with emissivity
emissivity
62
6]
= 0.8 and the other at T2 = 500"1( wit
= 0.4. An aluminium
emissivity (botl: sides)
63 =
radiation
shield
with
0.2 is placed between the plate
Calculate the percentage reduction in tile heat transfer rate as
result of the radiation shield.
[Bhorathidasan
University, Nov.
[The procedure of this problem is same as problem n
(Solved university problems - Section 4.28)J
4. J 00
Radiation
Heal and Mass Transfer
I
[Example 17 Two very large parallel pi
•
•
Q~
Illsnvities 0.3 ami 0.8 exchange heat by radiatio
. "'i1h
n, F'''d
P eTCentage reductio II ill heat transfer when a POlis/,ed.,
. 'hI
Ta.uiation shield of emissivity = 0.04 is placed betweell the",. '14",
~
(I
~ we n
.
Absorptiv1t)'
+ Reflectivity
, pr.98/
and radiosity.
(,-a)G+£Eh
a == £
(1-£)G+£Eb
". (4.39)
[Radiosiry.
J - f. Eb
It is defined
incident
unit time per unit area. It is expressed
in W/m2.
upon a surface per
". (4.40)
" 'a surface
The net energy 1 eav ing
radiosil)' (J) and irradiat i n (G).
A
as the total radiation
(I - c) G
G ==
Irradiation,
QI-2
Irradiation (G)
is the difference
J-G
J
_ (J-£Eb)
I _£
J(I-c)-(J-cEb)
(I - c)
Radiosity (J)
J-Jc-J
It is used to indicate
the total ra~iation
unit time per unit area. It is expressed
The rad iosiry (J) consists
So,
I.
Reflected
2.
Emitted
~eaving a surface per
111 W/m-.
by the surface
E Eb
by the surface
J
==
I -
pG
pG + E Eb
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.. ' (4.38)
cEb
1-
E Eb
of two parts.
t=O]
(4.38) ::::>
We knoW that,
An alternate approach for analysing thermal radiation between
gray or black surfaces is called electrical network analogy. This
approach
is more direct, more general and much simpler. The two
terms often used in the electrical analogy approach are irradiation
[.:
a+p
,
-[p--'-_-a--"lj
[Bharathidasan University ....
4.29. ELECTRICAL NETWORK ANALOGY FOR THERMAL
RADIATION SYSTEMS BY USING RADIOSITY AND
IRRADIATION
= I
+ Transmissivity
a+p+t
"n",,·
[The procedure of this problem is same as
bl
pro ell, IIo.j
(Solved University problems - Section 4. 28)J
4.101
-
t;
.J r;
c
between its
4.102
Heat and Mass Transfer
=
Q'-2
=
QI-2
A£(E'-J~
I-E
Radiation
Eb-J
--;his
l-E
AE
figA.29.
I
I
again can be represented by an electric circuit as shownin
0
". (4~
J,
in the form of electric I .
a CltcUQI
This can be represented
shown in Fig.
1 -E .
0
1
J2
Fig. 4.19.
J
w
Fig. 4.28.
1
. kn
.
here --F- IS own.as space resistance.
AI
12
If two surface resistance of the two bodies and space resistance
them is considered, then, the net heat flow can be
e sented by an electric circuit as shown in Fig.4.30.
repre
l-E
where
\10M
A, F'2
oo----~~~----~o
__
b tween
A E is known as surface resistance of the body.
If two bodies which are radiating heat with each other~'
I
the radiating heat of one b0 dy per unit'. area IS not fallingon~
Eb2
other and part of it has gone elsewhere, then, it is takenit
account by a factor which is known as shape factor or view fa~
The heat radiated by the first body }
and received by the second body
Heat radiated from second}
and received by first
Fig. 4.30.
Ebl - Eb2
JIA1FI_2
__
4.103
J2 A2 F2_1
So, net heat lost by the first body,
QI _ 2 =
=
J, AI F, _ 2 - J2 A2 F2 _ I
AI
FI_2 (11 -h)
[.: AI F'2=A2F1J
.. , (4.43)
J, -J2
1·
A, F,'2
I
,,' (4.J
where,
0'
-
c.
B 0ltzmann constant
Stefan
5.67 x JO-8 W/m2 K4
-
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4.104
Heat and Mass Transfer
TI T2
A2 FI2 -
Temperature of surface I, K
Temperature of surface 2, K
Emissivity of surface I
Emissivity of surface 2
Area of surface I, m2
Area of surface 2, m2
Radiation
AI FIJ
J2 -JJ
_L__
A2 F2J
Shape factor.
For black surface,
The values of 012, 013, 02J are determined from the values of
the radiosities (J I' J2 and )3)' Kirchoffs law which states that the
(4.43) ~
sum of the current entering a node is zero, is used to find the
radiosity.
4.30. RADIATION OF HEAT EXCHANGE FOR THREE GRAy
SURFACES
The network for three gray surfaces is shown in Fig.4.31..
this case each of the bodies exchanges heat with the other two.Tt
heat expressions are as follows:
4.31. SOLVED PROBLEMS
I Example 1 I Calculate
configuralions
the
shape
factors
for
shown in Fig.
I. A black body inside a black enclosure.
2. A tube with cross section of an equilateral triangle.
Ebl
4./05
11 - IJ
I
JI
1 - £1
J2
AI FI2
AI £1
Fig. 4.31.
Q/2
=
J1 -J2
1
A, F'2
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1 - £2
A2 £2
3. Hemispherical sur/ace and 1I plane surface.
the
4./06
Hear and Mass Transfer
Solution:
Case J:
F'_2
[All radiation emitted from the black surface 2 .
IS abs
the enclosing surface I.]
oriled~
-
(2) ~
. [F,_2
.
-
p_,-.l
(Since symmetry rriangle1
O·D
f!,-.l
Now considering radial ion fro
We know that,
4107
Radlollon
r.::-
- 0.5J
....
m su"ace 2,
F2-,
=
FI _ 1+ FI -2
I
+ F2_J
0]
2_2
FI_2
F, -2
=
!F
"'(ij
By reciprocity theorem,
AI
+ F2-2
F2 -I + F2-J
=
!F2_J
-
~
I-F2_ ]
1
... (3)
By reciprocity theorem, we know
(1) :::)
FI -I
=
l F,_, =
(3) ~
~:I
I-F2_1
1- FI_2
c·: F
1-0.5
C': F,_2=0.5]
2_1
0.5/
Result:
Case 2:
F/_I
Result:
FI _ I =
We know that,
+ FI-2+FI-3
=
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I
=
0,
FI _2
0.5,
FI -J
0.5
Case 3: We know that,
For flat surface, shape factor FI _ I = O.
I FI - 2 + FI - 3 =
F1_
.J
FI_I +FI_2
F2_1
= FI_2
F2-2
= 0
F2-J
0.5
= 0.5
= F,_2]
4.108
Heal and Mass Transfer
By reciprocity theorem,
A1FI_2
=>
[F'_2
=
0
A2F2-1
t, F,_, I
'" (4)
Since all radiation ermttmg from the black surface 2 are
absorbed by the enclosing surface 1,
F2-1
(4) ~
[':F2_1==1]
AI
FI_2
1t ,.2
21t ,.2
I FI_2
We know that, FI_I +FI_2
FI_I +0.5
Result :
A2
FI_2
=
0.5
-
2
Fig. 4.32.
= 0.5
I
From Fig., we know that,
As
==
AI + A3
A6
=
Az + A4
=
=
1
I FI_I
0.5
FI_I
0.5
FI_2
F2_1
0.5
I
We know that,
AsFs-6
==
AIFI_6+A3F3_6
[.: As=AI
=
Al FI_4+AI
I Example 2 I Find the shape factor FI_2 for the figure
+A);
FS-6=FI-6+F3_6]
FI_2+A3F3_6
[.: FI_6=FI_4+FI-z]
shown below. In the Fig., the areas AI and A2 (Ireperpendicular
but do not share the common edge.
As Fs -4 - A3 F3-4 + AI FI_2 + A3 F3-6
[.:
AI=As-A3;
FI_4=Fs_4-F3-4]
AIFI_2
=
As F5_6-As
AtFI_2
=
As [F5_6-FS_4]+AdF3-4-FJ-6]
A5 [F
AI
S-6-
FS_4+A3F3-4-A/J-6
F
5-4
J+
AJ
[F
-FJ-61
A
3-4
.. ·(I)
I
94 (Sixth edition)]
[Refer HMT data book, Page no.
.
21A'*-
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S""pe }ilClor fior Me area A and
--~~~~~~~~~---------------------.1
4.110
I
Heat and Mass Transfer
~
5
B
1
T
1
. = m ......
B
T
L2
L2 = 2 m
2
~I/(JII
A4 :
1
A4
Ll =4
~'---':::'
Fig. 4.33.
Fig. 4.35.
Shape factor for the area A 5 and A6 :
L2
~
2
B = 2 =1
Z
S = 42 =2
Y
B
IF
0.116431
5-4
[From tables]
Shapefactor for the area A 3 and A4 :
Fig. 4.34.
=
Y =
L2
4
-=-=2
B
2
L(
4
----2
B-2
-
Fig. 4.36.
Z =
Z value is 2, Y value is 2. From that, we can find
corresponding shape factor value is 0.14930 (From tables, Page No. 94).
I F5-6
Y =
0.14930 1
!F
3_ 4
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4".
-l~
L.-.-----'-.
"""'Ll
Z
~
=
L2
2
=2=1
B
..,
Ll
=2=1
B
J
0.20004
4.112
H
---.:.:_eal and Mass Transfer
SIIapefi ac tor for the area A and A .
3
Given: Area, A
6'
TI
I. B::: 2 m .1
T
l~
=
2 x 2 == 4 012
1000°C + 273
1273 K
As
L2::: 4 m
T2 == ,500~C +273
773 K
==
"' L'-- __ --=::::.,
L1::: 2 m
Distance
0.5 III
Fig. 4.37.
L2
z
4
Fig.4.1B.
B = 2" = 2
Tofind:
Solution:
y
0.23285
(I)~
I
Heat transfer
F s _ 6' F 5 _ 4' F3 _ 4 and F3 _ 6 values in equation (I),
Substitute
As
FI_2
AI
Heat transfer, (Q).
[0.14930-0.11643]+
A
A3
AI
QI2
[0.20004-0.23285]
1
l-~
AI EI
Al F12
A2E2
E2 =
2x2
c
= 2 [0.03287] - I [0.03281]
0.03293
Shape factor,
1 -EI
i
For black body
2 x 2 [0.03287] - 2 x 2 [0.03281]
Result:
I
Ti I
I
4x2
FI -2 =
-
[From equationnO.(4.43)]
= AS [0.03287] - A [0.03281]
I
o [Ti
=
--+-+-
A3
I
by radiation general equation is
=
I
[Ti - T; ] x AI FI2
5.67 x 10-8 [ (1273)4 - (773)4] x 4 x FI2
5.14xlOSFlz
F I _ 2 == 0.03293
where
I Example 3 I Two black square plates of size 2 by 2m are
olaced parallel to each other at a distance of 0.5 m. One plate is
maintained at a temperature of 1000 CCand the other at 500
oc.
F 12
-
!
... (1)
Shape factor for square plates
ln order to find shape factor F 12' refer HMT data book, Page
no. 90 (Sixth Edition) .
Find the heat exchange between the plates.
X axis
57
Scanned by CamScanner
1
=:
Smaller side _
Distance between planes
I
·1
I!
j3
4.114
Heat and Mass Transfer
2
0.5
I X axis
Curve
~
~
2
[Since given is squar
X axis value is 4, curve is 2. So, corresponding
TI ::: 750°C + 273 :::
1023K
e Plates]
Y axis VI.
T2 ::: 350°C + 273 :::
a Ue,s
0.62.
i.e.,
£1
I FI2
623 K
OJ
0.621
Distance between discs = 0.2 m.
0.62
Tofind:
Heat exchange between discs, (Q ).
F12
Solution:
Heat transfer by radiation general equation is
4
Fig. 4.39.
[T~ - T;]
1 - £1 + __ I
1 - e2
__
+ __
(J
(I) ~
5.14 x 105 x 0.62
012
1012
Result:
Heat transfer,
I
I
3.18 x 105 W
AI £1
012 = 3.18x
IOsW.
I
T
0.3 m
°1
°2
=
AI = A2
=
0.3 m
§
T,075OC
El =0.3
l_§
T2:: 350'C
E2:: 0.6
1t
4' (0.3)2
Scanned by CamScanner
5.35 x 104
... (I)
42.85 + 0.070 FI2
where
F 12
-
Shape factor for disc
In order to find shape factor F 12' refer HMT data book,Page
no. 90 (Sixth edition).
O.2m
! 02
4
A2 £2
5.67 x IO-S [ (l023t - (623)4)
1 - 0.3
1
1- 06
+
+'
0.070 x 0.3
0.070 FI2 0.070 x 0.6
Example"
Two circular discs of diameter 0.3 m eachart
placed parallel to each other at a distance of 0.2 m: One disc is
maintained at a temperature of 750 't' and the other at 350er and
their corresponding emissivities are 0.3 and 0.6. Calculate heat
exchange between the discs.
Given:
AI FI2
[Fromequationno.(4.33)]
Fig. 4.40.
X axis
=
Diameter
Distance between discs
OJ
0.2
j2
4.116
Heat and Mass Transfer
I X axis
=
1.5
Diameter of disc 2, ~
Curve
-)- I
X axis value is 1.5, curve is I. So, corresponding
is 0.28.
y
.
aXIS valUe
Temperature
0.62 m
Distance
of disc I, TI ::: 125 crn _
Temperature
of disc 2, T·2
10jind:
0.28
=>
Radiation
62 crn :::
[Since given is disc]
4. iJ7
1150 K - 1.25 m
:::
620 K
Heat flow by radiation.
1 . When no other, surfaces
.
:
are present
2.' When the discs rare connected b. .
.
Y non-conducting
surface,
Solulion:
Area,'
~
TV
T,=1150K
AI
'r~
'.v
1.5
= '. ~ ;;< (0.62)2
Fig. 4.41.
5.35 x 104
(I) =>
,
T2~620K
030 in2'/.
42.85 + 0.070 x 0.28
Result:
Heat transfer,
I Example 5 I Two
~Fig. 4.41.
569.9 W /
We know that
QJ2
Heat/transfer
black
= 569.9 W
discs of diameter
by.radrauo.
a [T4 , .:. ~r42 y'
62 em are
arranged directly opposite to each other and separated by a
distance of 125 em. Tile temperature of tile discs are 1150 K and
620 K. Calculate the heat flow by radiation between tile discsfor
the following cases.
When 110 other surfaces are present.
2.
When the discs
I
1-&2
A,E,
A,FI2
A2~
I
are
connected
Emissivity,
by IIoll-colltlucting
£,
E2 = )
cr [ r~ - r;)
I
surface.
Diameter
I-E,
--+--+--
For black surface,
1.
Given:
t;.... :al e411aill.)l1l~
of disc I, D,
Scanned by CamScanner
62 em
0.62 III
A, F'2
iFrom 'equation no.(4.33))
4. / /8
Heat and Mass Transfer
.
5.67 x 10-8 x 0.30 x F12 [(1150)4_
.
..
(620~
3
27.2 x 10 F121
I
=
••. (1)
where
Radiation
X axis value is 0.496, curve is S S . .
. 0, correspo d'
'
n Ing Y axis
value .IS 034
.
•
Shape factor for disc.
F12
4
Case 2 : The dISCS are connected b
' ·1/9
Y non-COndu .
So, choose curve 5.
ctlng,surfaces.
0.34J
In order to find shape factor F 12' refer HMT data book, p
no. 90 (Sixth edition)..
.
age
Diameter
Distance between discs
X axis
0.62
1.25
I X axis
0.496
I
0.496
Case 1: When no other
radiation. So, choose curve 1.
surfaces
are present
I Q12 ::: 9248 W I
i.e., direct
Result:
Y axis value is 0.05.
I F12 =
0.05
27.2 x 1()3 x 0.34
Q12
(1) ::::)
X axis value is 0.496, curve is 1.
So, corresponding
Fig. 4.44.
,
. QI2 (Direct radiation) = 1360 W
Q 12 (planes
I
connected
by non-conducting
surfll:CS) =
9248
I
Example 6 Two parallel rectangul~ surfaces 1 m x 2m
are opposite to each other at a distance of 4 m..1he surfacesare
black and at 300 ~ and 200~. Calculate the heat exchangeby
radiation between two surfaces.
r,
Given:
Area, A
Distance
TI
0.496
lx2=2m2
= 4m
300°C+273
S73K
Fig. 4.43.
(I) =>
Q12
[Q12
:::
:::
27.2 x 103 x 0.05
1360 W I
-1--V
i7.
1m
4m
2m
L
T2
1m
T 2 = 200°C + 273
= 473 K
Fig. 4,45.
..
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4.120
Heat and Mass Transfer
(012), .
Tofind:
Heat exchange
Solution:
Heat transfer by radiation
From graph, we know that,
general equation is
,
(J
QI2
r T~ - T~]
I
EI .
J -
--
+ --
AI EI
,
.
(I):::::>
J - E
+ ---.l
AI FI2
Result:
EI
&2
=>
012
(J [
where
FI2
4
T I - T 2 ] x AI FI2
... (1)
Shape factor for parallel rectangles
In order to find shape factor, refer HMT data book Page nO.91
and 92 (Sixth Edition).
X
=
Longer side
Distanc.e
2
.4
---..--~
1
0.5 '
D__,___=(
y
B
D
-
4.
012
S.67x 10-8[(573),,_(
012
261.9 W
Heat exchange,
012 = 261.9 W
I
4'
-
o~
=
473)4 ~x 2 x 0.04
A2 E2
For Black surface,
L
D
I FI2
,
[Example 7 Two parallel plates of s;ze3
2
m x m areplaced
arallel to each other at a distance or I
0
P
.
'J
III.
lie plate is
maintained at a temperature of 550 C(' and the otheral 250't
the emissit'ilies are 0.35 and 0.55 resnect;ve!u Th land
I,
l'
:.t.
e p ales are
located in a large room whose walls are al 35't: If the plales
excllange Ileal with eac!, other and ",;th tht room,calculale
1.
Heal lost by the plates,
2.
Heat received by the room.
B=1m
m_t!J
0.25
B=1m
Fig. 4.46.
Size of the plates
Distance between plates
TI
550°C + 273 = 823 K
Second plate temperature,
T2
250°C + 273 = 523 K
Emissivity
of first plate, EI
0.35
of second plate, E2
0.55
Room temperature,
TOfind, :
0.5.
X = lID
Fig. 4.'47.
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1m
First plate temperature,
Emissivity
BID = 0.25
3 m x 2m
T3
350C + 273 = 308 K
I.
Heat lost by the plates.
2.
Heat received by the room.
4.122
Heat and Mass Transfer
Solution:
In this problem, heat exchange take Pla~
two plates and the room. So, this is three surface problem and the
corresponding radiation network is given below.
Eb3
Fig. 4.49.
To find shape factor F
92 (Sixth edition).
'2- refer HMT data book, p
age no.91 &:
Fig. 4.48. Electrical network
Area,
A,
07
3 x 2 = 6 m2
D=1m
I A,
=:>
A2 = 6 m2
Since the room is large,
I
I
. ,,''J
__j___"
A
,
B=2m
AJ = 00
Fig.4.50.
From electrical network diagram,
I-E,
I - E2
E2 A2
1 -EJ
EJ AJ
Apply
1 - EJ = 0,
EJ AJ
L
3
x = 5=1=3
=
1-0.35
0.35 x 6
0.309
=
1-0.55
0.55 x 6
= 0.136
=
0
1 - EI = 0.309,
EI AI
ctrical network diagram,
y
B 2
= 5=1=2
X value is 3, curve value is 2. From that, we can find
corresponding shape factor value is 0.47, ie., F12= 0.47.
[From graph]
[.: A3 =CX)j
[F12
IE.
---=-.2 = 0.136 values JII
E2 A2
=
0.47J
We know that ,
Fll+FI2+F13
But,
FII
=
=
0
!&li!1®5~
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Eb2
::
::
I - 0.47
I
Q); = 0.531
Similarly,
=
... (5)
o T4
J
F21 + F22 + F23
5.67x 10-8 [308J4
0
We know that,
F22
=>
F23
=>
F23
=
1- FI2
F23
=
1-0.47
network
=
1-
=
J3 :: 510.25 W/m2]
... (6)
J I and J2 can be calculated by using Kirchotrs
law.
=> The sum of current entering the node J1 is zero.
AI Node
AI FI3
6 x 0.53
A2F23
6 x 0.53
0.314
... (1)
= 0.314
... (2)
r.,
[From diagram]
=
0.354
26.0IxIOJ-JI
0.309
... (3)
J2-J1
+ 0.354
,
J1
=> 84.17 x IOJ - 0.309
law,
=
EbJ
[From diagram]
The radiosities
0.53 )
6 x 0.47
Eb
I
F21
diagram,
From Stefan-Boltzmann
510.25 W/m2-
Eb3
I F23
From electrical
4.24x~
Eb2
EbJ
5.67 x 10-8 [523J4
crT4
=>
+
510.25-J1
0.314
=
0
J2
J1
J1
+ 0.354 - 0.354 + 1625 - 0Ji4 :; 0
-9.24J1+2.82J2
= -85.79xloJ
... (7)
AI Node Jz:
5.67 x 10-8 [823
J4
... (4)
+
Eb3-J2
I
A2F23
br
Scanned by CamScanner
EbJ-J2
+ 0~36
o
-------:------!.Radintioll
= 4.24 x 103 - 4.73 x )OJ
~
J
_1__
0.354.
J2
0354
_2_
510.25
.
+Q3j4 -0.314
2.82 J1 - 13.3 J2
6 x 0.55
4.24 x 103 __2__
+ 0.136
- 0.136 - 0
=
-32.8
x 103
IQ
2
. ". (8)
Tota I
heat lost by the plates
Q
Solving equation (7) and (8),
- 9.24 J1 + 2.82 J2
=
_ 85.79 x 103
". (7)
2.82 J1 - 13.3 J2
=
- 32.8 x 103
'" (8)
- 3.59 x 103 ~
=
= Q1 +Q2
49.36 x 103 - 3.59 x 1()3
=
I Q = 45.76 x 100W]
'" (9)
Heat received by the room
I By solving,
:::)
J2 =
4.73 x 103 W/m2
:::)
J1 =
10.73 x 103 W/m2
11 - 1)
Q
I
=
10.73 x 103 - 510.25 + 4.73 x )(}J-510.25
0.314
, 0.314
Heat lost by plate I is given by
Ebl -J1
Q,
01
~
=
(:,-;:
J
[.: Eb3=J]=512.9]
[Q
26.01 x 103 - 10.73 x 103
1-0.35
0.35
I 01
12- 1)
+~
x 6
49.36 x 103 W
Heat lost by plate 2 is given by
I
45.9 x 10)
wi
... (10)
From equation (9), (10), we came to know heat lost by the
plates is equal to heat received by the room.
[Example
8] T"e water tank of size 2 m x 1 m x l m and
radiates heat from each side. The surface emissivityof tank is 0.8
andthe Surface temperature of tank is 32OC.
Calculate the following:
1. Hear lost by radiation
2. Reduction
if ambient temperatureis 4t:
in heat loss if the tank is coated with an
tlluminium paint of emissivity 0.6.
Scanned by CamScanner
4.128
Heal and Mass Transfer
~~~~~~~~T-al-lk~sl'-ze----~
Given:
Emissivity of tank,
€I
0.8
Surface temperature,
T1
32°e + 273 =: 305 K
Ambient temperature,
T2
4°e + 273
Emissivity of aluminium,
€2
0.6
Reduction in heat loss
=:
Result:
277 K
I.
Heat loss by radiation Q _ ,
2.
Reduction in heat loss
I
1.003 kW
-
_
- O,250kW
4.32. UNIVERSITY SOLVED PROBLEMS
To find :
I. Heat loss by radiation (Q).
I
rexample 1 Determine the view faclor (F lfi
2.
L.;
Reduction in heat loss.
Shown
I.
Solutio" :
1. From Stefan-Boltzmann
law, we know that
1(1 Or the figure
IDee. 2004 & May 2005 Anna u,'
below.
1m
I
.1
1m
E b (or) Q == o T4
. . ity (c)c.' and Area (A) are given. SO,
E ITIISSIVI
Heat transfer, Q
€
x A x o T4
1':1 x A x o [T~
- T; ]
Solution:
0.8 x 8 x 5.67 x 10-8 [ (30S)L (27m
[.: Area
==
2 x 1 x 4 == 8 m2 (4 sides)1
1003.83 W I
[Q
(or)
IQ
1.003 kW
2. Emissivity of aluminium,
I
£2 ==
0.6.
Fig. 4.5/.
£1 - £2
Reduction in heat loss
==
x Q
£1
SA
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.
",verslty)
~1~,/~j~O~H~e~a~/a~n~d~~~~~s~s~~~o~~~fu~
--------------~
From Fig., we know that,
Shapt factor for the area A J and A, :
As ...
AI + A2
A6 '"
A3 + A.
~Bz1m-l
T
Ae
L2 = 2 m
-tJ~
Further.
AsFs-6
==
AIFI_6+A2F2-6
[.: As == AI + A2; FS-6 = FI_6 + F2_ ]
6
=
Flg.4.SJ.
A I F I _ 3 + A I F I - 4 + A2 F2 - 6
t: FI-6=Ft-3+Fl-(]
As FS-3 - A2 F2-3 + AI FI_4 + A2 F2_6
A~ FS-6
[·:AI=As-A2;
==
FI-3=Fs_3-Fd
AsFs_6-A~F~-3+A2F2-3-A2F2-6
L2
2
i.,
2
z
B = T =2
y
B = T =2
Z value IS 2, Y value is 2. From
that, we can find
corresponding shape factor value is 0.14930.
[From tables]
F 5 - 6 = 0.14930
~
I
J
Shape factor for tile area A J and A J :
~ B= 1 m ~
T
(Refer HMT data book, Page nO.94 (Sixth EditiOliI
L2:: 1 m
A3
-ll~
L, = 2 m
~
~'''----
Fig. 4.54.
Fig. 4.52.
Z =
L2
)
B - --)
) -
Y
L,
2
- -- --2
) -
=
B
[FS_3 = 0.116431
Scanned by CamScanner
[From tables]
·132
Hear and Mass Transfer
Shape factor for tIre area A 2 ami A j :
II B = 1 m
1
A5
.1
A; [0.14930
- 0.11643] +
L2 = 1 m
A3
A2
-1J~
AI [0.200IlA
A
-l.
\J<f - 0.23285]
AI [0.03287] _ ~
2
Al [0.032811
Fig. 4.55.
L2
Z
=>
U
I
-
-
B
T [0.03287] _ !
,.--____
FI-4
1 [0.03281]
O.0329
View factor, FI -4 -- 003
. 293
[Example 2 Determine the .
.
VIew factor F
Illefigure shown below.
I'D
/ - 2 alld F2 fi
ec. 2005
<]
Or
--...J
'Alilia Ulliversity}
Result:
Y
!::.L -
F2-3
0.20004
-
B
I
Shape factor for tire area A 2 and A6 :
I_ B= 1 m .1
T1 ..
2
L2.
L1
= 1m
Fig. 4.57.
~~---..::::..
Solution:
I.
Fig. 4.56.
L2
2
z
B=T =2
y
s_
Substitute FS-6'
FS-3'
5m
T
_
B-1
-
0.23285
I
F2-3 and F2-6 values in equation (I),
Fig. 4.58.
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____.-----
4.136 HeatandMa!sTra/lsf~------Result: View factors FI_, ~ 0.0978
F2 _ I ==
Radiation
~
0.0489
D
[ Example
Two pilfallel plates of size J m x J "'lit
spaced 0.5 0' IIp'''' are lacaWI in a very large room, the ""'''. '
whiclt are maintained at a lemperatllre of 27 OC. One P alt' of
maiolainetl III a temperalure of 900"C lind tne other ., 46 •
Tlreir emissivilies are 0.2 111111
0.5 respeclively. If the p1artJ
6t
excltange heat belween tlte",selve!i a/l(l sllrr(Jllfltiillgs,find Ihe n
•
. . Consider (J1111
Y I~II
hea! ITilnsfer 10 eilell
plale anti to lite room.
venlt}1
pltlle .wrft,cesfac",g eac/r olher.
{May 2004, Anna Uni ,
,!
"'2 £2
Fig. 4.62. Electrlcul IItlwor' dl IIgrum
Area, AI
SOllllion:
Size of the plates
==
I J11 X I 111
Distance between plates
==
0.5 III
Room temperatur'!,
==
27 '
foirst plate temperature,
T I ==
900
Second plate temperature,
T2
=
400 '
Emissivity of first plate,
CI
;::
0.2
TJ
0
_/
IA
I
..
1)( I • 1m2
-
"2
Since the room is large, A)·
-I- 27
0
0
-I
., 300 K
27
.. 1173 K
27
.. 673 K
-
2
Im
I
r:/)
From electrical network diagram,
1-0.2
--4
I )( 0.2
1- O.S
Emissivity of second plate , e·2 ==
(;Q.'5 • I
0.5
==
1'0/1/1(1:
I.
o
Net heat transfer to each plaLc.
2. Net heat transfer to room.
s
Salution
: In thi pro bl em, heat exchange take place bet\l'~
om. S 0, this., IS three surface problem and tlt
.wo ,. Iares and the roor
orrcsponding radiation network is given below.
Scanned by CamScanner
I-I:
Apply __
JI
AI
(;1
nctwo r'k (I'iagram.
I-e
I, -:;:;
A3 (;3
0 values in e"':l:trical
4. 138
Neal and Mass Transfer
--------
But,
::::>
::::>
J2
Eb2
Fig. 4.63. Electricat network diagram
To find shape factor FI2 ' refer HMT data book Page nO.91&
92 (Sixth Edition).
Similarly,
We know that,
F22
o
::::>
F23
I-F 21
1-0.41525
0.5847J
From electrical network diagram,
8=1m
Fig.4.M.
X
=
L
0
= 0.5 = 2
y
=
B
0
= 0.5 = 2
FI2
=
==
1.7102
I x 0.5847
==
1.7102
I
I
I
X value is 2, Y value is 2. From that, we can find correspon·
ding shape factor value is 0.41525.
[From table}
i.e.,
I x 0.5847
AI FI2
From Stefan-Boltzmann
I x 0.41525 == 2.408
law ,
0.41525
4
We know that,
o TI
5.67 x 10-8 [II 73J4
107.34 x I 03 W/m~
Scanned by CamScanner
4.140
Heat and Mass Transfer
5.67 x 10-8 [673]4
I Eb2
11.63 x 103
Eb3
o T3
W/~
4
=
Eb3 =
5.67 x 10-8 [300]4
459.27 W/m2
From electrical network diagram, we know that,
I Eb3 =
The radiosities
13 = 459.27
11 and 12 can be calculated
W/m~
by using Kirchoff's
Jaw.
0.415 J1 - 1.4997 J
2
==
- 6.08 x 103 '" (2)
Solving equation (1) and (2),
=> The sum of current entering the node J 1 is zero.
AINodeJ/
- 1.2497 J1 + 0.415 J2
0.415JI-1.4997J2
:
Ebl -1)
[From electrical network diagram]
11
26835 -
12-\11
459.27-1)
+ 2.408 +
1.7102
=
Heat lost by plate ( I), Q1
=
25.35 x 103 W/m2
Ebl -J1
0
[From electricalnetwork diagram]
lz
1)
1)
'4 + 2.408 - 2.408 + 268.54 - 1.7102 =
0
~6835-0.2511
0
+0.41512-0.41511}
::::
107.34 x IOL 25.35x J03
1-0.2
1 x 0.2
+ 268.54 - 0.5847 1)
-1.2497
-27.lOx 103
-6.08 x 103
11.06 x J03 W/m2
J1
107.34 x 103-11
4
==
By solving,
o
4
==
J1 + 0.41512
-27.IOxIOJ
... (1)
b
Scanned by CamScanner
4./42
Heat and Mass Tramjer
Heat lost by plate (2), Q2
~
Eb2- J2
=
1- E2
A2 E2
11.63 x 103 - 11.06 .___
x 103
1-0.5
=
-I x 0.5
[lxomple"
, Two hI" It
Rd'
C 'qll("e
U lOlto"
" 143
'(lettI p{/Tllllelto eaclt olh,
pInt, 0' i
.
P"
.'
al a di
'J sUI
b
"'{Iinlflined tu a lempert/lure "
'Slnlleeof 0.4
~ I", are
oJ 900't:
f1I. Olle pl
.
r:/'nd the net heal exclutnoe 0'
and lire 01"
ate u
T"
It
'J tllergy d.
er at 400 "C.
lie to 'adintio L_
•
t'ltt IHIOplates.
II oelwetll
fl'
Given:
I Q2 = 570 W I
Total heat lost by the} Q
=
plates (1) and (2)
+
A, F'3
=
T2 = 400°C + 273
T2 = 4OO'C
I
A2F23
Fig. 4.65.
Tofind:
Heat exchange, (Q).
Solulion:
Heat transfer by radiation generalequationis
Q12 =
[.,' Eb3 = J) = 459.27 W/m2j
IQ =
1
673 K
12 7" 13
25.35 x 103 - 459.27
11.06 x 103 - 459.27
1.7102
+
1.7102
20.752 x 103 W
~Tl=900'C
O.4m
= 1173 K
IQ = 21.06 x 103 W I
I
1m
T, = 900°C + 273
= 20.49 x 103 + 570
1, -13
I
a [Ii - I;]
1- E,
1
--+-+-2
A, E,
A, F'2
[Nole: Heat lost by the plates is equal to heat received by the
Result:
I. Net heat lost by each plates
Q,
= 20.49 x 103 W
Q2
=
£, =
For black body,
:::::)
2. Net heat transfer to the room
Scanned by CamScanner
A2 £2
[From equationno.(4.43)]
,I
570 W
Q = 20.752 x 103 W
l-~-
I
I
room.]
lOCI. 99, MUj
Distance = 0.4 m
Q, +Q2
Total heat received or} Q =
absorbed by the room
Area A ::: I x I ::: 1 m2
where
QI2 =
£2 = I
a[Ti
-I~] A, FI2
=
5.67 x IO-S[ (I 173)4- (673)41F'2
IQ
I2
=
95.7 x 103 FI21
F'2
-
Shape factor for squareplates
... (I)
,
4. J 44
Heal and Mass Transfer
In order to find shape factor F,2, refer HMT data book, Page
2Qcm ::: O.2m
no.90 (Sixth edition).
0.2m
Smaller side
Distance between planes
X axis
=
8000e + 273
=
1073 K
T,
1
0.4
3000e + 273
T2
I X axis
2.5
I
Curve ~
2 [since given is square plate]
=
X axis value is 2.5, curve is 2. So, corresponding Y axis value
is 0.42.
i.e.,
0.42
I
573 K
E,
0.3
E2 =
0.5
Fig. 4.67.
Tofind:
Heat exchange, (0).
Solution:
1t
= 4 (0.2)2 = 0.031 m2
A,
= 0.031 m2
A2
= 0.031 m1
2.5
Fig. 4.66.
I0
Result:
Heat transfer by radiation generation equation is
95.7 x 103 x 0.42
(I) =>
,2
40 x 10J W
I
(J
[r: ...r; J
Heat exchange, 0,2 = 40 x JOJ W
I Example 5 I Two circular discs of diameter 20 em each ore
placed 2 m apart. Calculate the radiant heat exchong« for these
discs if there are maintained at 800 't:' 0/1(1300't:' respectively and
the corresponding ennssivities are 0.3 and 0.5. IApr. 2000, MOl
_5.67 x 1O-8((107Jt-(573tl
1 - 0.3
I
I - 0.5
0.031 x 0.3 + 0.31 x F'2 . O.OJI x 0.5
+
Scanned by CamScanner
4.146
Heal and Mass Transfor
69 x J()l
~
where F'2
-
[!xample 6 I TlVo black d'
Radian
•
ISc., Of d.'
011 4147
directly opposite at U di.ft(tnce "
IQ"'eler 0 5
'
oJ", Th
. "'artpl
fOOD K and 500 K respectivel
. t discs are .
aced
discs.
lV· Calculate the h "'m"t"i"td al
t.."e
tat /lOti! bettl!te"
J, WI,ell no other surfaces
areprese
2. Whell the discs are
nt.
~on"ected bv
,
surface.
J
non-conducting
'" (lJ
J 07.45 + 0.03 J F'2
Shape factor for disc.
In order to find shape factor, FJ2,
(Refer HMTdata book, Page no.90 (Sixth Ed"
Diameter
"axis =
,Ilionl)
Distance between disc
to« 97, MUI
I X axis =
Curve
-+ 1 (since givel1 is djs~)
.
Temperature, of disc, J
is 0.01.
FJ2
=
0.5m
T~
Distance = ) m
X axis value is 0.1, curve is 1. So, corresponding Y axis value
I
0.5 m
Diameter of disc, 2
0.1 I
\
~
Diameter of disc, )
Givell:
0.2
2
Temperature
-
1m
JOOOK
of disc, 2
5001(
T,=l000K
l~
0.01
Solutio" :
Fig. 4.69.
1t
4' (0.5)2
'-.(r.f
I,
" Fig. 1-68.
.
Heat transfer by radiation general equation is
c [T~ - T~]
012 = 1- t ,\' I
)-&2
.. 69 xJl~;
(l)~
'tel
, 1.1
107.45 + 0~031,x 0.01
rQ~~"~
,:-':2([7
W~":l
Result: Heat exchange, Q = 20.7 Watts
____l + + -:61'
Al FI2
Ai &2
I AI
1
For black surface,
Emissivity, £1 =
.J
~
,i'
t'
012 =
I'
62
.'
~
=
::'4'
,I
,
4
o AI FI2 r T I - T 2 ]
= 5.67 x 1O-8xO.196xFI2x
[(1000)4 - (soW]
Scanned by CamScanner
,
--;:xiS
value is 0.5, curve is 5. S
'
.'1ad mio"
~ 149
~4~./~4~8~~H~e~a/~(~/II~d~A~U~/~~J~T~"~~II~lv~e~,.~==~-------------o cprr .spo t'
. 0 J4.
" n(!Og y axi~ value
, 10'fl2J
@;~~4
where,
FI1
~
,s .
Shape factor for disc.
-
In order to find shape factor f12,
I
z»
:::::>
(I)
(Refer J·IMT dalft book, Page no. 90 (Sixlh Edilir.~
Diameter
Distance between discs
X axis
Q2
I
[x axis
0.5 ]
other surfaces are present i.e., directQ
Case: J Wh en no
.'
.'
S
I
curve I. X axis value IS 0.5, curve IS I,
radiatIon.
0, C ioose
corresponding Y axis value is 0.05.
[F12
==
0.05 ]
FI2
0.3'1 )
QI2
JO.4x IOJxO,1'
3536 W ,
[ 012
Resull:
I.
012 (DireCI radialion)
2.
012 (Planes connected by non-condU(linpurflcc)
= 520.9 W
= 3536 W
[!:xalllple 7 I A long cylintlricallrealer 30 mm in diameleris
mainlabretl til 700°C II has surface emissivilyof 0.8. The healer
is localed in (/ large room whose wall are 351('. Fimlll,e radianl
Ileallrans/er. Find the percentage of reduclionin Ireallransfer if
the heater is complelely covered by radialiollshieltl (s= 0.05) and
diameler 40 mm:
IApril99, MU/
Give":
30 mm = 0.030 m
Diameter of cylinder, DI
700°C + 273 = 973 K
Temperature, TI
0.8
Emissivity, E,
Room temperature, Tz
=
35°C + 273 = 308 K
Room
0.5
Fig. 1.70.
(I) =>
• T 2 [2
10.4 x 103 x 0.05
CaJe 2: The discs are connected
. surft.1
by non-conductJOB
:0, choose curve 5.
Scanned by CamScanner
Fig. 4.71.
4. 150
Heat and Mass Transfer
Radiation S"ield :
Emissivity, E3 =
Diameter, D3
=
----------------Since room is large ~
0.05
Rod iQlion
~
40 mm == 0.040 rn
4.15/
Shapefactor
Small body enclosed by largebod
F 12
Radiation shield
==
Y~FI2'=l
[Refer HMT data
(1) ~
Q
book,p
5.67 x lo-a [ (973
1 - 0.8
age 110.83(Sixth edition)]
t ~(30S)4_J
12
0.094 x O.S + 0.094 x 1 + 0
[Since A '=
ex)
1- ~
'A
2
Heat transfer without shield
I QI2
=
3783.2 W
J
Heat
transfer between heater (1) and ra dilatlon shield
.
.
b
(3) is
given y
Fig. 4.72.
Toflnd : I. Heat transfer.
l-EI
__
2. % of reduction in heat transfer.
AI EI
where
Solutio" :
c [Ti - T~]
1 - EI
I
I - E2
where,
Al
I Al
1t DL
Al FI2
Scanned by CamScanner
== 1t x 0.040 x 1
[Refer HMT data book, Page no.83 (Sixth edition»
5.67 x 10-8[(973)" - Tj I
1 - 0.8 +
1
I - 0.05
0.094 x 0.8 0.094 x I + 0.125 x 0.05
A2 E2
= 1t x 0.030 x 1 = 0.094 m
0.094 m21
A3 &3
... (11
--+-+AI EI
AI FI3
1-& 3
+ __
Shape factor for concentric long cylinder F13 = 1
Heat transfer by radiation general equation is
=
1t D3 L
I
+
I A3 == 0.125 m21
Case 1 : Heat transfer wit/rout shietd :
QI2
=O}
'" (2)
Case 2: Heat transfer with shield:
Diameter 03 = 0.040 m
2~
==
3.43 x 10-10 [(973)4 - T~]
I
.., (3)
Hear and Mass Transfer
... I:JL.
Heat exchange between radiation shield (3) and D
[\OOrn (
2),
given by
a [Tj - T;]
Reduction in heat
loss due to
radiation shield
1
.
==
Radia/io"
Q.
Without shield - Q
Q.
.
"I~
Without shield
QIl-Qn
QI2
Since room is large,
A2
=
3783.2 -154.6
3783.2
1-£2
o
A2 £2
95.9%
I. Heat transfer without radiation shield
Shape factor for small body enclosed by large body
F32
QIl
I
==
3783.2 W
2. % of reduction in heat transfer = 95.9%
[Refer HMT data book,Page noll\
5.67 x 10-8 [ T; - (308)4 ]
=>
==
Result:
1 - 0.05
1
0.125 x 0.05 + 0.125 x 1 + 0
3.54 x 10-10 [T;
- (308)4]
I
'" (4)
I
[Example 8 A disc oj 10 em diameter at 4000C is situated
2m below tile centre oj another disc of I.S m diameter which is
maintained at 200 'C. Find the net radiant energy excl.ange
between tile surfaces if tile emisslvities of smaller and larger discs
are 0.8 and 0.6 respectively.
/Manonmaniunr
Sundaranar Unil1ersity,NOI1. 96/
[The procedure of this problem is same as problem no.5J
We know 013
=>
032
3.43 x 10-10 1(973)4 - T;]
3.54 x 10-10 [Tj -(308~1
307.4 - 3.43 x 10-10 T;
3.54 x 10-IOT; -3.18
6.97 x 10-10 Tj
310.58
817 K I
=>
Substitute T3 value in (3) or (4).
Heat transfer with radiation shield
013
==
3.43 x 10-10 [ (973)4 - (817)4 ]
L§lL =
Scanned by CamScanner
154.6 W
I
4.33. RADIATION FROM GASES AND VAPOURS- EMISSION
AND ABSORPTION
Many gases such as N2, 02' H2, dry air etc., do not emit or
absorb any appreciable amount of thermal radiation. These gases
may be considered as transparent to thermal radiation. On the other
hand, some gases and vapours such as CO2, CO, H20, S02' NH3,
etc., emit and absorb significant amount of radiant energy. As
illustration we shall take up radiation from CO2 and H20, which
are the most common absorbing gases present in atmosphere
industrial furnace, etc.
4./54
Heal and Mass Transfer
4.33.1. Radiation from Gases Differs From Solids
~
The radiation from gases differs from solids in the fOllow'
lilt
ways:
• The radiation from solids is at all wavelengths, whe
gases radiate over specific wavelength ranges or b I'eas
within the thermal spectrum.
iIItds
• The intensity of radiation as it passes through an absorb'
gas decreases with the length of passage through the Illg
volume. This is unlike solids wherein the absorption gas
radiation takes place Wit. hiIn a sma II d'istancs from thtof
surface.
4155
I
r£xantple 1 A gas is en l
I...!::
c oSed in
'''7CC TIle mean bea", leng/~
a bOdy III
N
•
I Of the
a temper t
es~ure of water vapOur is 02
gas body is J
a lire of
' '
pll
. at".
"'- The
.
lit' Calculate the emissivity 01"
and tire total
partIal
a •
'J ",aler Vapo
preSSureis 2
Ur.
Temperature T _
Given:
,
- 727°C +
Mean.beam length L _
273 == 1000 K
,
", -
3m
Partial pressure of water vapou' p
I,
Tojinll: I
H
20
==
0.2 atm.
Total preSSure p :::
. . .
'2atm
E rmssrvity of water vapo
ur, (£H 0).
2
4.34. MEAN BEAM LENGTH
Solution:
PH 0 x
2
Hottel and Egbert evaluated the emissivities of a number of
gases at various temperature and pressures are presented the results
in the form of graphs.
Their results are strictly valid for hemispherical gas volum~sof
radius L, radiating to an elemental surface at the centre of the base
as shown in Fig.
GH~W
L",
0.2 x 3
0.6mat~
From HMT data book, Page no. I 07 we C f d
'"
H 0.
'
an In emissIvity of
2
Fig. 4.73.
However,
calculated by
for other
shapes,
mean
beam
length
lO00K
can be
Fig. 4.74.
From graph,
i;
where
=
3.6 x
AV
V
Volume of gas
A
Surface area of gas
Scanned by CamScanner
Emissivity
of H20
=
OJ
... (1)
4. 156
Heat and Mass Transfer
To fintl correction/actor/or
H]O:
-
0.2 + 2
2
:::: 1.1
2
Partial pressure of CO p
2,
CO
2
Partial pressure of HOp
2 , Ht' '" IOOIe '" 0 10
=
2
Given:
.
PH 0 L
1.1,
m '" 0.6
2
aIm
Total pressure, p '" 2 abn
From HMT data book, Page nO.l08 (Sixth editio )
n, Wt ~
find correction factor for H20.
Temperature, T .. 92""
rC + 273
'" 1200 K
Mean beam length, l. '"OJ m
Tofind:
Emissivity of mixture, (t.-a).
So/lilian: TafindemissivityofCo~
P~
xL.
0.2)( OJ
==
I P~ xL.
P¥+P
__
=1.1
0.06 m-atuiJ
From HMT databook, Page no. I05, we can find emissivityofC~.
2
Fig. 4.75.
From graph,
Correction
factor for H20
IC
=
Emissivity of H20,
(H 0
2
OJ x 1.36
EH20
0.408
E
0.408
H20
So,
I
Result:
1.36
Emissivity of H20,
0
H2
1.36
I
... (.
I
I Example 2 I A gas mixture contains 20% COl and J~
H P by volume. TIle total pressure is 2 atm. The temperattPt
the gas is 927'\:". The mean beam length IS. 0• 3 m. Calculatl emissivity 0/ the mixture.
Scanned by CamScanner
1200K
Fig. 4.76.
From graph,
Emissivity 0(002
I~
== 0.09
==
O.l19J
u
4.158
Heat and Mass Transfer
To find correction factor for CO2 :
~
Total pressure, P =
P C02 Lm =
2 atm
0.06 m-atrn.
From HMT data book, Page no. 106, we can find
factor for CO2,
co~
Fig. 4.71.
From graph,
Emissivity of H20
Tofind correction factor for Hp:
PH20 + P
P=2atm
2
From graph, correction factor for CO2 is ,i .25.
,',
1.,25 I
x Cco2
I EC02' x CC02
=
2
pH 'L0,
2
From HMT data book",Pag~
III,
0:09 x 1.25
.,,(
0.'11251
0.1 x OJ
I 'PH20,L~ = 0.03 m-atrnJ '
,-
. i...
ity d
From HMT data book, Page n~.107, we can find emlss
, P"zO + P ,1,05
2
F(g.'.4.7fJl "
,',
H20.
Scanned by CamScanner
2
= 1.05
1.05,
,
0.03 m-ann
no. 108 (Sixth edition), we can
find correction factor for H20.
Tofind emissivity of H20 :
'J>H20 x Lm
0.1 +2
PH20 + p
Fig. 4.77.
EC0
2
0.048
r
4.158
Heal and Mass Transfer
Tofind correction factor for CO2 :
Total pressure, P =
=
PC02 L",
2 atm
0.06 m-atm.
From HMT data book,' Page no. 106, we can find COrr....,
"~Q~
factor for CO2,
.'
,
Fig. 4.78.
From graph,
Emissivity of H20 = 0.048
0.048 J
Tofind correction/actor/or Hp:.
PH20 +p
"
I
•
I
.'
Fig. 4.77.
I
From graph, correction factor for CO2 is ,j .25.
I
I,Ci~
=
1.25,/
x CC~
=
0:09 x 1.25
EC02,' x CC~
=
Oh'12S
=
0.1 x OJ
,
"
,
! I',
.!
2
,
'"
J ~.
..,(I)
'\ ,
F
I
i
' '.
, '.
. 'ty of
rom HMT data book, Page no.107, we can find emisSIVI
H20.
' ,
Scanned by CamScanner
. ' t,
PHl 0 '.'
L", = 0.03 m-ann
"
',',
'.'
.'
)'~..
'I
From HMT data book, page. no. 108 '(Sixth edition), we can
find correction factor for H2O', '
I
,'PH20 :L';' ~". 0.03rn-a~.J
1.05,'
=
,
To find emissivity of H:P :
J>H20 'x L",
= 1.05
PH20 + p
I
EC~
0.1 +2
= ~
'2
P=2atm
, PHzO+ P
--,,',05
2
Ffg.'·-#. 7.'/1 '
4.160
Heal and Mass Transfer
From graph,
Tota I
,...C- -
1.39
--1.-39-',
x CH20
=
Correction factor for H20
emissivity of gaseous mixtllr
Emtx
Ec~
H 20
EH20
r+=
0.048 x 1.39
I EmU =
0.0661
I EH20 x CH20
6/'
~.,
Ceo + E
2
1i20 CH 0
0.1125 + 0 066
2 - l\E
. - 0.002
~[Fromequal'
0.176U
IOn(I),(2)and(3)]
cr
;'td 01 a temperature of 925 OV~r;t
.
vollI"'eis
~.
s ~nl"t valli
I rtS!iUre of the combustion gases is J
The
lola .~e of water vapour hi O.J atm and that 01" ~O"" .the PII"illl
.
",0111'01
=
~/
Emissivity of gaseous mixture
Jl es ull: .]
, E"'I: ::: 0.1765
[f!a",e1e 3 A furnace of 25 nrl or~a and J2",J .
Correction factor for mu1ure of CO] and H]O:
PH20
ROd"",.
e
0.1
0.1 +0.2
pres!;U
"'e.
.~ . .
'J
~olate II.t emunVity of tile gaseous mixture .
Given:
Area, A = 25 m2
Volume V
12 m)
:lIS 0.]5 IIIIft.
I
=
0.333
Temperature
2
0.09 ,
x L", + PH20 x L",
r·1200K
P~L",+PHzOL",
0.002
0.25 atm.
Solution:
We know, Mean beam length for gaseous mixture.
V
12
Lit,
3.6 x A = 3.6 x 25
PC(
(Emu)'
1.72 m
I
of CO] :
0.25 x 1.72
0.43 rn-atm.
:00.333
From liMT data bo
CO2,
Fig. 4.80.
..' (3)
60
Scanned by CamScanner
cOz
Emis iviry of mixture
TOfind emissivity
I
0.1 atm.
Tofind:
I Lltt
0.002
2
Partial pressure of CO2, P
From HMT data book, Page no. 109 (Sixth edition), we can
find correction factor for mixture of CO2 and H20.
From graph,
,
1198 K
Total pres ure, P
3 atrn
Partial pres ure of water vapour, PH 0
0.06 + 0.03
, Peo
T - 925 + 273
I
105, we can find emissiviry of
4. 162
Heat and Mass Transfer
From graph. we find Cc~
..
I
::: 1.2
I Cc~
I~
te~
x Cc~
0.15 x 1.2
te02
x CC02
::::
O.I~
'" (I)
T = 1198 K
Tofind emissivity of HzO :
PH20 xL",
Fig. 4.81.
IP
From graph,
H20 x
Em issivity of CO2
I
0.15
0.15
EC0
2
To find correction factor for COl:
Total pressure, P
Peo2 t., =
From
::::
0.1 x 1.72
t, :::: O.I72J
From HMT data book, Page no. 107 we can find e ...
'
mlsslvlly of
H2 O .
3 atm.
0.43 m-atm.
HMT data book, Page no. 106, we can find correcta
factor for CO2,
T = 1198 K
Fig. 4.13.
From graph,
P = 3 atm
Fig. 4.81.
Scanned by CamScanner
Emissivity of H20 =
0.15
I
0.15
EH20
=
J
4./64
Heal and Mass Transfer
Tofind correction factor for H20 :
PH20 + P
PH20
0.1 + 3
=
2
~
2
= 1.55
+P
2
From HMT data book, Page no. 108 (Sixth edition)
find correction factor for H20.
' 'We can
From HMT data book, Page no I
.
.
c.
c
. 09 (SIxth di
nd
correctIOn
ractor
lor
mixture
of
CO
e Ilion), We c
fi
2andHO
an
2 .
0.602
PH 0
__2
PH 0
__1_
+P _
-1.55
=0.285
P,,<o+ p~
2
Fig. 4.85.
Fig. 4.84.
From graph, we find
CH20
r-I
From graph, we find
1.58
~E
= 0.045.
I ~E = 0.045\
C-H- --1.-58-,
... (3)
20
=>
EH20 x CH20
L!_EH_:2:_O_x _C_H-=.20
0_.2_3_7__J1
Correction Factor for mixture of CO2 and H20 :
PH 2
°
Total emissivity
0.15 x 1.58 = 0.237
0.1
0.1 + 0.25
of the gaseous mixture is
.,. (2)
EmU'
0.18 + 0.237 - 0.045
[From equation (I), (2) and (3)1
= 0.285
I E""x
Result ; Total emissivity
0.285
Scanned by CamScanner
0.372
1
of gaseous mixture,
En/u = 0.372.
~=-----------___
---.........
__ /~/~Cl~ll~tI~n~d~U~a~~~7r~a=,u~~=~
__
-44~./~6~6
4.36. PROBLEMS FOR PRACTICE
<,
I. 1,wo equ al discs of diameter 200 mill each are arranged' InfII.
nes 400 m apart. The temperature of first d' 0
para IIe I pla
IS(: ~
and that of second disc is 200°e. Determine the radia
heat flux between them, I·t" t Ilese are
-
seo-c
(i) Black
(ii) Grey with emissivities
OJ and 0.5 respectively.
[Ans,' 30 W, 4.5 W)
2.
A steam main (E = 0.79) having an outside diameter of 80 mill
runs in a large room in which the air temperature is 27°C. Tht
surface temperature of the stearn main is 300°e. Calculate tht
loss of heat to surroundings per metre length of pipe due10
radiation.
Calculate also the reduction in heat loss if the above pipeis
enclosed in a brick conduit (at 27°C) of emissivity 0.93.
[Ans,'
3.
4.
1151.3 7 W 1m, 29.075 W/m)
Two large parallel planes of emissivity 0.8 and 0.6 are
maintained at temperature of 560°C and 300°C respectively.
Compute the radiant heat exchange per square metre between
them.
[Ans,' 11.28 kW/ml)
A double-walled
spherical vessel used for storing liquid
oxygen consists of an inner sphere of 30 cm diameter and an
outer sphere of 36 cm diameter. Both the surfaces are covered
with a paint of emissivity 0.5. The temperature of liquid
oxygen stored is - 183°C whereas the temperature of the outer
sphere is 20°e. Calculate the radiation heat transfer throu~
the walls into the vessel and the rate of evaporation of liqUId
oxygen if its latent heat of vapourisation is 2 13.54 kJlkg.
[Ails:
Scanned by CamScanner
3.6 W, 0.0607 kg!llJ
Two parallel plates 0.5 by I
Radiation 4./67
0
5. plate is maintained at 10000C' m are spaced 0.5 a
,,
and th
part. One
emissivltJes of plates are 0.2 d
e other at SOO°C Th
.
an 0.5 res'
. e
are located In a very large
pectlvely,The plates
room,
maintame. d at 27°C . The plates ex
has the w II which are
c ange he t '
and with the room, but only the I
a with each other
be
consi
p
ate
surfaces
f .
other are to e consIdered in the a I .
aCingeach
na YSISFind th
transfer to each plate and to the room.
.
e net heat
[Ans: 14.425kW 2595 kW
' .
,17.02 kW]
Two very large parallel planes with emis ...
6.
",
SIVltlesOJ and 0 8
exchange heat by radiation. Fmd the percenta
.:
.
ge reductIon In
heat transfer when a polished aluminium radI'at' h' I
. .'
Ion s Ie d of
emisSIVIty = 0.04 IS placed between them.
[Ans: 93.6%]
7. Two parallel plates 2 m x I m are placed I m apart facingeach
other. Their temperature and emissivity values are 500°C and
0.8, and 300°C and 0.5 respectively. Estimate the net radiant
heat transfer between the two plates,
If another identical plate (E = 0.6) is introduced between the
two plates equi-distant from each, find its temperature and the
heat gained by the colder plate due to its presence.
[Ans: 335°C, 4.163 kW]
8. Two parallel plates 3 m x 2m, placed I m apart, are maintained
at 500°C and 200°C ; their respective emissivities lbeingOJ
and 0.5. If the temperature of the room in which these plates
are located at 40°C, estimate the heat lost by the hotter plate.
Consider radiation only.
[Ans: 6.629 kW]
9. Two parallel plates each of emissivity 0.8 are maintainedat
temperatures of 400 K an 600 K in an evacuated space. A
screen of emissivity 0.05 is now introduced between these
plates. Determine the temperature of the screen and also the
heat flux per unit area of the screen.
[Ans: 727 K, 146 W/m2]
-I. 168
Heat and Hass Transfer
10. A chamber is filled with a gas mixture at a pressur
and 1000°C. The gas mixture is transparent to radi t~ of 2 ~
CO2 whose partial pressure is 0.3 atm. Assuming a ~~n e~~
length of 1.2 m, estimate
4.37.
of the gas V I
the emissivity
TWO MARK QUESTIONS
an"
o Ul'lle.
[Ails: 0• I 7fi1
AND ANSWERS
'1
from one body to another with
d'
.
k
transrmttmg meorum IS
nown
electromagnetic
wave phenomenon.
as
.
radiation.
OUI~
It is :
iI
2. Define emissive power {Ebl.
{Dec.2005, Anna University, Oct. 97, MU, Oct. 2000,Ml~
The emissive power is defined as the total amount of radiali
emitted by a body per unit time and unit area. It is expressed~
W/m2.
Define monochromatic emissive power. {E b;'/'
The energy emitted by the surface at a given length per UM
time per unit area in all directions is known as monochromalK
emissive power.
4.
Incident~
{APril
·....tation
.
DeC.200S 97, April 99, MU
Black body IS an ideal sf:'
lillie 2006
' Dec.2004,
ur aee h'
,Alllla U •
tn
g
I. A black body absorbs all . a.... the followin nrve1'J~J
wave length and di
. InCIdent Ild"
g propertIes.
Ireclion.
laban, regard I
2. For a prescribed tempe
ess of
.
rature and
can emit more energy th b
wa...e lenm'"
an lack bod
I!>"', no surface
8. State Planck's distribution' ,alii.
y.
to« 97, April 2000
The relationship between the
,MY, May 2004, AU7
monOChromat'
.
of a black body and wave length of
. I~ emissive power
. . .
a radiationat
.
temperature IS given by the folio .
.
a partIcular
wmg expressIOn, by Planck.
c ).-5
E hi,. = ;:::--:,....1_-
where
[Jrt) - J
Absorptivity
is defined as the ratio between radiation absor~
ci
Monochromatic emissive powerW1m2
Wavelength - m
0.374 x 10-15 W m2
and incident
radiation.
c2
14.4 x 10-3 mk
Wltat is meant by absorptivity? {Dec.2004, Anna Un;vers~1
Absorptivity,
.5.
. 'Y trails
•
The heat is transferred
3.
'.
Transmissivity is d "'lss#Vlljl r
"
etiOed
to the mCldent radial'
as the '.
IOn
ratIO of
.'
radiation
T ransmlssivity
~. .
transmitted
, t "" ~tran._:_
.
7. What is hlack hody ,
,_f. Define Radiation.
. .
6. What is "',-ant h
a =
Radiation absorbed
Incident radiation
What is meant by reflectivity?
Reflectivity
incident
is defined
9. State Wien's displacemem to«
{Dec.2004, Alma Un;vers~!
as the ratio of radiation
reflected lotht
radiation.
Reflectivity,
p
EbA
A =
{Dec.2004, June 2006, Anna University}
The Wien's law gives the relationship, between temperature
and wave length corresponding to the maximum spectral
ernissi ve power of the black body at that temperature.
Amax
Rad iation retlected
Incident radiation
where
All/ax
~:
Scanned by CamScanner
T
cJ
c3
2.9 x 10-3
T
2.9 X 10-3 mk
[Radiationconstant]
&aiCfS2&J!!~
4.170
Heal and Mass Transfer
l;':state Stefan-Boltzmann
10". IApr.2002, MU ~
•
, Qy 200
The emissive power of a black body is p
~~
ropOrtional ' ~
fourth power of absolute temperature.
10 ~
ec T4
where
s,
s,
(J
Eb
Emissive
T4
ell == EI;
power,
W/m2
lYDejine
T
and May 2005, Anna Univel1'
It is defined as the ability of the surface of a body to rad('
heat. It is also defined as the ratio of emissive POWerof ate
body to the emissive
power of'a
black body of
tern pera tu re.
eq:
Emissivity,
E
=
E
Eb
15. State Lambert's cosine law.
It states that the total emissive power Eb from a rad' ti
ia Ingplane
surfac.e i~ any direction proportional to the cosine of the angle
of emISSion.
[Apr. 99, Oct. 99, Apr. 2001, MU, May 2004,AUj
Radiation shields constructed from low emissivity (high
reflective) materials. It is used to reduce the net radiation
transfer between two surfaces.
17. Define irradiation
(G).
[Nov. 96, MU/
It is defined
as the total radiation incident upon a surfaceper
unit time per unit area. It is expressed in W/m2.
[Aprit 2001, MD, Dec.2004, JUlie 20(16, Anna Univtn/IyJ
This law states that the ratio of total emissive power to ~
absorptivity
is constant for all surfaces which are in lhennal
equilibrium with the surroundings.
This can be written as
u. Whal is radiosity (J).
IDec.2005, Anna University, April 2001,MU/
It IS. used to indicate the total ra diianon Ieaving a surfaceper
unit time per unit area. It is expressed in W/m2.
E3
A:
Scanned by CamScanner
ex cos 0
16. Wlrat is tire purpose of radiation sllield?
law of radiation.
E2
and Soon.
direction per urnt so I angle per unit
a given
area of the '.
surface normal to the mean direction in spa
emitting
ceo
Eb
In = 7t
MD, Dec.2004, Dec.2005, June 2006 AU]
If a body absorbs a definite percentage of incident radiatiOll
irrespective
of their wave length, the body is known as gray
body. The emissive power of a gray body is always less than
that of the black body.
£(
4 17/
intensity of radiation (I,j.
Eb
J 2. What is meant by gray body?
IApri12000,
10"
IS
INov. 96, Ocl. 98 9
It is defined as the rate of energy leavin
.' 9, MUI
.
I'd
g a space In .
to« 2000, April 2002, MlJ , Dec.llJ04
11. Define Emissivity.
(X2 = E2
Radial"
.
alwaySI'n I
alns In the
-..,ua to
rntaleqUilibrium
I
.
with its surroundings.
Stefan-Boltzmann
constant
5.67 x 10-8 W/m2 K4
Temperature,
K
(J
13. SUIte Kirchoff's
'.
It also states th a t th e emIssIvity of th
'ts absorptivity when the body rem .e ~y
4.172
Heat and Mass Transfer
What are the assumptions made to calcuJ~
19. exchange between the surfaces
r
I.
~
All surfaces are considered to be either black or Ilh..
QO"l
2. Radiation and reflection process a~ assumed to be dl
3.
The absorptivity of a surface IS taken equaJ ~.
emissivity and i~d~pendent of temperature of the so~ ~
the incident radIatIOn.
q
o
What is meant by shape factor and mention its Ph .
2 . signifICance.
.
{May 2005, Anna Un.!n~
OcL 1997, Apr. 98, Oct. 20~'
The shape factor is defined as "The fraction of the rad~
energy that is diffused fro.m one ~urface e!ement an~ strikes~
other surface directly WIth no mterve~m~ reflections". II U
represented by F if • Other names for radiation shape factor an
view factor, angle factor and ~onfigurat~o~ factor. The s~
factor is used in the analysis of radiative heat exchan~
between two surfaces.
21. The heat transfer by radiation takes place by ntellllS
~
{MU, EEE, Nov. 1994j
Ans : Electromagnetic waves.
22. A perfect black body is one which
_
{MU, EEE, April95J
Ans: Absorb heat radiation of all wavelength falling on it.
23. Two plates spaced 150 mm apart are maintained at lOOOf
and 70't: The hetu transfer will take place mainly~
{MU, EEE, Oct 1996/
Ans : Radiation.
24. According to Stefan-Boltzmann law, ideal radiatorstmf
radiant energy at a rate proportional to
•
{MY. EEE, Oct 199~
Ans : Fourth power of absolute temperature.
Scanned by CamScanner
I
•
Raditllio
","en II,e."eal ~ Iransferredfrol1l hot bo
"4.17
J
J5.
aiol,t tine without aJfectino the t
dy to cold60"" .
str "
e
e Interveni
-r, In a
reI''''erredto as heat transfer hy
. ng ItIedi"",',.
..., I l$
Ans : Radiation.
IMU,EEE, Apr.1997/
J6. fhe amount of radiation mainly dependson
,4ns:
Nature of body, temperature of body an-d--
ofbo d y.
.
type of surface
17. fhe heal transfer equation Q = aAT' is knownas
_
{MU, EEE, Apr. 1997}
Ans: Stefan-Boltzmann equation.
carbon d' 'do
18• DiscusS the radiation characteristics or
'J
lOX' e and
watervapour.
{Dec.lOBS, Anna University}
Ans : The CO2 and H20 both absorb and emit radiation ov
.
certain wavelength regions called absorption bands.
er
The radiation in these gases is a volume phenomenon.
The emissivity of C?2 and the emissivity of H20 at a
particular temperature Increases with partial pressure and mean
beam length.
DO
CHAPTER-V
5~TRANSFER
~UCTION
~11I""'-
.
..
f
In a system consisting 0 two or more components whose
ntrationsval)' from point to point, there is a natural tendency
cOnce~ies
(particles) to be transferred from a region of higher
for Sy·. Sl'd)e to a region
. of lower
ntration Slide (hiig her density
, conce•
.
•
tionside
(lower
density
side).
ntra
conce
This process of transfer of mass as a result of the species
concentration
difference in a mixture is known as mass transfer.
Someexamples of mass transfer are
I.Humidification of air in cooling tower.
2. Evaporation of petrol in the carburetter of an Ie engine.
3. The transfer of water vapour into dry air.
4. Dissolution of sugar added to a cup of coffee.
5.2MODES OF MASS TRANSFER
There are basically two modes of mass transfer given below
that are similar to the conduction
transfer.
and convection
I. Diffusion mass transfer
2. Convective mass transfer
SJ DIFFUSION MASS TRANSFER
It may be c I assified
.
into two types.
I. Molecular diffusion
2. Eddy diffusion.
Scanned by CamScanner
modes of heat
~~~-------.....I·
I
I
5.4
The r:ra;lSpol1 of water on a microscopic
.
.
- hi h
PoQell~ J\
Weigh
.
.. t of Co
(iii) Mass/raction
ItlPootnlA.
.
.
IOf
ig er concentranon to a region of 10\\
diffusion from a region of
concentration
diffusion.
P \ - Density of corn
level as a re.sul
1\" '\ - Molecular
in a mixture of liquids or gases is know n as mOlecUI:
The mass fraction
is defi
.
llled as th
.n1"Cies to the total mass density of the . e IllaSs~.
S.S EDDY DIFFUSION
When one of the diffusion fluids is in turbulent mOlion, edd
diffusion takes place. Mass transfer is more rapid by eddy diffusio Y
than by molecular diffusion.
n
)r- .
Mass fraction
.
or Mass density
The mass concentration is defined as the mass of a component
per unit volume of the mixture It is expressed in kg/m '.
Mass of a component
.
Unit volume of mixture
The molar concentration is defined as the number of molecules
of a component per unit volume of the mixture. It is expressed in
kg-rnole/m-.
.
Number of molecules of component
Molar concentration =
.
.
Unit volume of mixture
and molar concentration
rnass den'
.
~.tlon
(If
of as
SII)'
P
(il') Mole fraction
The mole concentration
is d fi
.
f
. .
e tned 8S th
concentratIOn 0 . a species to the total rn I
e ratio of III I
o ar concenll1.
0 e
Mole
liOn.
Mole fract ion
~n
of a srwo.-;••
Totalmolbr,.n~
Its
o ar concentration
are related
Consider a system shown in
Fig.S.1. A partition separates the two
gases,a and b. When the partition is
removed, the two gases diffuses
throughone other until the equilibrium
IS established throughout
the system.
. The diffusion
rate is given by
theFlck'sl
hi
n
aw, w ich states that molar
uxof:lnel
.
di
.
emel1l per un It area is
Ctly
D:cd proportional to concentration
~. rem
Scanned by CamScanner
55 concentration
''''IC~I_
S.8 FJCK'S LAW OF DIFFUSION
(ii) Moltl~ Concentration or Molar density
The mass concentration
by the expression
fe.
rni\==-
5.7 CONCENTRATIONS
=.
Illl:':tu .
PA
Convective mass transfer is a process of mass transfer that
will occur between a surface and a fluid medium when they are at
different concentrations.
.
Mass concentration
=
Ma
TOtal
5.6 CONVECTfVE MASS TRANSFER
(i) Mass concentration
.
a
Fig.S.1
b
5.4 Heat and Mass Transfer
dCa
rna
-oc-A
dx
:::)
rna .
Apply boundary condition
A =-Dab
dCa
dx
Ca = C IX + C
2
At.
x =0
At,
x = L
••• (5.1)
where
Ca2=Cll+C2
N a = rna
A _ Molar flux - Unit is kg - mole
_
Co2 = Cil + Cal
s - m2
(or)
- Co2 -Cal
C 1i,
Mass flux - Unit - ~
s- m2
Substituting
C I, C2 values in equation (5.2)
Dab - Diffusion co-efficient of species a and b-.!!t
dCa
-- Concentration gradient
dx
s
(5.2) :::)
C = [C
alai
a2 -
c, I1x+C
From Fick's law, we know that,
5.9 STEADY STATE DIFFUSION THROUGH A PLANE
MEMBRANE
rna
dC
Molar flux, A = -Dab dx
Q
Consider a plane membrane of thickness L, containingfl~
'a'. The concentrations of the fluid at the opposite wall face51J!
Cal and Ca2 respectively.
Considering the diffusion is along X axis, then the controll~
equation is
d2C a = 0
__
x
dx2
Integrating above equation
dCa
dx = CI
Again integrating,
TL
1
Membrane
=
Dab
-lC
L
a 2 - C a I]
I
Where,
-
Fig. 5.2
rtlits
Scanned by CamScanner
Mo Iar fl ux -rna
., A
rna
A
kg-mole
- Molar flux - -=--s-m2
••• (5.3)
5.6 Heal and Mass Transfer
Dab -
Diffusion co-efficient -
.
.
~~
'de - ~_kg-rnol~
Cal - ConcentratIOn at inner Sl
Given :
ransfer 5.7
PartIal pressure of 0
m3
Ca2 - Concentration at outer side - ~
Mass 11
.
2, Po
2
::::0 .,)(
"\
iota\
pressure
::::0.2\ x \. \ bar
m3
::::0.21 )( \.\)(
L - Thickness - m
Partial pressure of N
n... _ 0
.19)( iotal
2, 1"1'12 -
For cylinders,
~
10, N/m2
pressure
::::0.79 x \.\ bar
L = r2 - r,
Temperature,
21tL(r2 - r,)
A = --=----'-"-
=: 0.79)( \.\)( I05N/m2
T = 2O"C + 273
= 293 K
Tofind:
For sphere,
I. Molar concentrations
Co
2'
L=r2-r,
2. Mass densities,
P02' PN
A = 41t r, r2
3. Mass fractions,
m~, m
4. Molar fractions,
Xo , XN
2
2
where,
N2
2
N2
We know that,
r2 - Outer radius - m
Molar concentration,
L- Length - m
S.lO SOLVED PROBLEMS
C
Solution:
r, -Inner radius - m
o
,
C = _!_
GT
ON CONCENTRATIONS
A vessel contains a binary mixture of O2 and 'N2 with JHIIIi'
pressures in the ratio 0.21 and 0.79 at 20"e. If the lOll
pressure of the mixture is 1.1 bar, calculate tl,efollowillX:
i) Molar concentrations
0.2\ x 1.\ )( \05
83\4 x 293
(.: Universal gas constant, G =: 83\4 J/kg-mole
ii) Mass densities
iii) Mass fractions
iv) Molar fractions of each species
Scanned by CamScanner
C~
= 9.48 x \0-3 kg - mole 1m
5. 8 Heat and Mass Transfer
Mass fractions:
1.1 x 105
8314 x 293
mo = P0 =~
= 0.79 x
I C = 35.67 x
N2
10-3 kg -
2
mole
1m3]
We know that,
Molar concentration, C = ...e_
M
p=C
__
P
2
1.302
I
I
Im02 = 0.2~
m = PN2
_ 0.9987
P - 1:302
,
"
N2
x M
,I
We know that,
= 9.48 x 10-3 x 32
Total concentration, C = C + C
°2
= 9.48 x 10-3+35.67 x 10-3
[C = 0.045]
[.: Molecular weight of02 is 32]
I P02 = 0.303 kglm31
Mole fractions:
=_2
2
I:
C
= 9.48 x 10--3
= 35.67 x 10-3 x 28
0.045
I PN2 = 0.9987 kglm31
Overall density, P = Po
2
+ o..
I"N2
= 0.303 + 0.9987
I
P = 1.302 kglm31
Scanned by CamScanner
IX02 = 0.210
I
CN
xN =_2
2
I
I'
. Co
Xo
[': Molecular weight ofN2 is 21]
N2
C
35.67 x 10--3
0.045
[XN2 = 0.7921
5.10 Heal and Mass Transfer
Result:
...---:
foft"d:
I. COz = 9.48 x 10-3 kg - mole 1m3
C
Nz
I . Molar concentrations , Co2' CN2
= 35.67 x 10-3 kg - mole 1m3
2. Mass densities, Poz' PN
. 2. POz = 0.303 kg/m
3
3. Mass fractions, ,;,oz'
PN = 0.9987 kg/m'
moz = 0.233
';'NZ
4.
o
= 0.210
xNZ
= 0.792
z
mN
2
Solution:
We know that,
= 0.767
xO
2
4. Average molecular weight, M
z
3.
Mass Transfer 5. J J
.
Molar concentration, C = -
P
GT
C0-2
Po .
2
GT
= 0.21 x I x lOS
A mixture of O2 and N2 with their partial pressures intht
ratio 0.21 to 0.79 is in a container at 25 C CalculatetAt
molar concentration, the mass density, and the IIUlSsjractifJ"
of each species for a total pressure of 1bar: What Hlould lit
the average molecular weight of the mixture?
8314 x 298
D
[.: Universal gas constant, G = 8314 Jlkg-mole-KJ
jC
3
3
O2 = 8.476 x 10- kg - mole Im /
[Dec-2004 & 2005, Anna Univ]
Given:
Partial pressure of 0z, POz = 0.21 x Total pressure
= 0.21 x I bar
=
0.21 x I x 105 Nlm2
Partial pressure of Nz, PNZ = 0.79 x Total pressure
= 0.79 x I bar
= 0.79 x I x 105 N/m2
Temperature,
T = 25°C + 273
= 298 K
Scanned by CamScanner
=
I
0.79 x 1 x lOs
8314 x 298
CNZ = 31.88 x 1(r3 kg - mole 1m3
We know that,
Molar concentration, C = ~
p=C x M
I
1
&
1
~
== 8.476x
10-3 x 32
A"~Jr#O
Molecular weight
M~
M :: p~ ""'2 +p~~
== 0.2] x 32 + 0.79)( 28
[.: Molecular weight of 0 .
21\ J);
:: 0.271
~02
M
k.g/m3
== 28.84
1
JleSu/J:
I.
==
I
31.88 x 10-3 x 28
2.
kglmJI
== 8.476 x 10-3 kg - molelmJ
~2
== 31.88
x 10-l kg - mole/mJ
== 0.27]
kglm3
== 0.893
kg/m!
p~
PN2
[.: Molecular weight ofN2 iJlt;
PN2 == 0.893
CO2
== 0.233
.
== 0.767
Overall density, p::
P02
P 2
... 0.271
0.893
I P - 1.164 kgl
m31
Mas fraction
4.
M
== 28.84
111 The molHular weights of the two COmpoMIIlI A
IIIfd B f1/
a gay mixture are U and 48 re.pective/y. TU MOIecu/4r
weight of a gas mixture ;, found to be JO. If tlu IffIIII
c()ncentration of the mixture U 1.1 kglmJ, detmrrbr.t tit
foll() wing:
O.R93
111
2
1.164
(i)
Density of component A and B
(II)
Molar fractions
(Iii)
Mas« fractions
(Iv)
1'0101 pressure
290 K.
if the temperatureof tht mixlllft;,
[Muy-2004, Anno Univ)
Given:
Molecular weight of component A, MA ~ 24
Molecular weight of component B, Ma == 48
Scanned by CamScanner
5. J 4 Heat and Mass Transfer
Molecular weig.ht of gas mixture, M ::: 30
P = 1.2 kg/m3
Mass concentration,
We know that,
Temperature, T = 290 K
Tofind:
I .Density of component A and B, pA, PB
~
124CA+48CB==~)''1..
2. Molar fractions, x A' and xB
Solving equation (1) and (2)
'3. Mass fractions, m~, and m~
~
CA = 0.03 kg mole/m3
4. Total pressure, p
CB = 0.01 kg mole/m3
Solution:
Molar concentration of the mixture,
(i) Density
Density,
C = _£_
PA = 24 CA
M
= 24 x 0.Q3
1.2
30
IPA = 0.72 kglm3\
I C = 0.04 I
Density,
PB = 48 CA
We know that,
= 48 x 0.01
I PB 0.48 kglm
=
I
CA + CB = 0.041
3 \
,..[I) ii) Mole fractions
CA
xA=-=C
We know that,
x ;:::CB
B
C
0.03
=075
0.04
.
;::: 0.01 = 0 25
0.04
.
iii) Mass fractions
·1
[,: ~lB'
11
Scanned by CamScanner
rnA;:::
PA ;::: 0.72 = 06
P
1.2
.
' •. (2)
5.16 Heat and Mass Transfer
Mass r,.
allsfer 5.17
SOLVED PROBLEMS ON l\1El\1B
1
5.1 .
Elts
iv) Total pressure at 290 K
rt1
rIelium diffuses thTOUgha plane
..d
pi
L!.J At tile tnner
Gas law, pV = mRT
p= m RT
V
IIIenrbran
st e the conce
. e oil "''''tho
3
ntratlO"
lelc.
0025 kg mole/m . At the Outers."
Of "efill", .
.
.
k
Ille the co"
ts
'elium IS 0.007 g mole/mJ. 'WI,li. .
ce"tratioll .f
,I
.., IS tI,e difJi .
oJ
helium through tile membrane. ASSumedi I. IISlo" flllX Of
ifhelium with respect to plastic is 1 )(/'!!IlSIQ" cO-e/Jicklll
o
=pRT
1/
9 ",1Is.
Given:
ThlC. kess
n , L = 2mm = 0.002 m
concentration at inner side,
C
= 1 2 x 8314 x 290
.
30
[ .: Universal gas constant,
kg-mole
m3
•
C
a2
= 0.007
kg-mole
Diffusion co-efficient,
= 0.72 kg/m!
PB = 0.48 kglrn3
2. x A = 0.75
-l
We know that, for plane membrane
Molar flux,
niB = 0.4
Resll/t:
1 x 10-9 [0.025 _ 0.007]
0.002
m ::;:9 x 10-9 --"-__
kg-mole
_a
A
5-m2
D·fti .
.
I us Ion flux of helium,
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[Fromequationno.5.3]
rna
Dab
A
= L [Cal - Ca2]
A
= 0.6
P = 96.442 kN/rn2
-fm
Solution:
xB= 0.25
4.
L
ToJind:
Diffusion flux,
rnA
Dab::;: 1 x 10-9 m2/s
2\
Result:
3.
Ca2
m3
I p 96.442 kN/rn
1. PA
Cal
G = 83 14 J/kg-mol~Kl Concentration at outer side,
p = 96442 N/rn2
=
= 0.025
al
rna = 9 x 10-9
A
kg- mole
S _ m2
5.J 8 Heat and Mass Transfer
r:;")
~
Gaseous hydrogen is stored in a rectangular
cO"lQ;
walls of the container area of steel having 25 IIInr ~. ~
At the inner surface of the cOntainer tJ,~la.
.1
•
'It
"ire' "
concentration of '11'Yurogen
In t e steel is 1.2 Ie
while at the outer surface of the cOn/ainer t ~
concentration is zero. Calculate the molar diff . lire ,
T
'J USIO" Jl
hydrogen through tile steet.l ~'ake diffusion Co-.Il !itA
effie;,.
--,
hydrogen in steel is 0.24 xl 0-11 "r/s.
lilA
111 Hydrogen gases at 3 s
Given: Thickness, L = 25 mm = 0.025 m
Molar concentration
at inner side, Cal
Ca2
== 1.2~
at outer side,
=0
10-12m2/s
=0.24x
Hydrogen
Tofind: Molar diffusion flux, A
Dab
A = T [Cal - Ca2J
A
1.15 x io-
I. Molar concentration on both sides Cal and Ca2
2. Molar flux
12
0.24 X 10- [I 2 - OJ
0.025
.
II
Diffusion co-efficient, Dab = 9.1 x 10-8 m2/s
Tofind :
We know that, for plane membrane
rna
Inside pressure, PI = 3 bar Outsid
'
e pressure, P2 = 1 bar
Thickness, L = 0.25 mm = 0.25 x 10-3 m
Solubility of hydrogen 2.1 x 10-3 kg-mole
m3_ bar
Temperature, T == 20°C
. Solution:
Molar flux,
Sleeiplate
Dill
Ina
iii) Mass flux of hydrogen
Given:
cl1
Diffusion co-efficient,
Dab
71
i) Molar concentration 01'/1d
'J Y rogenon 60th .,.1
'~I1101 l
fl
Slues
II~ mo ar ux of hydrogen
mJ
Molar concentration
Ma
membrane Ita .
ar and I 6ar
ss ransfer 5. J 9
Vllrgthick"e
areseparated6
.
co-efficient 01' Itlld
.ss0,25 mm."",
. ~ aplastiC
'J
J
roge" I" h
I lie6mary dl(r.
•
Tile solubility
01' h'e
pla.flieis 9 J
'.J,USlon
:J
Ydroge'
,)(
J~
",21s
2. J x J 0-1 kg-molel",1 b "In tile "'e"'6
.'
' .
ar; A,
erane IS
con d Ilion of 200 is assu- d.
n uni/or", tem
•..e
mperature
Calculale II,efollowing
kg - mole
2
s-m
3. Mass flux
Solution:
I. Molar concentration
on inner side,
Result:
Cal == Solubility
m
kg- mole
Molar diffusion flux, AU = 1.15 x 10-11 S _ m2
Cal == 2.1 x 10-3 x 3
x Inner pressure
Cal == 6.3 x 10-3 kg-mole
m3
.-
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&_
,.
5.20
Heat and Mass Transfer
Molar concentration on outer side,
'01,\'ED UNIVERSITY
Ca2
= Solubility x Outer pressure
C a2
= 2.1 x 10-3 x
IC =
a2
2.1 x 10-3
1
lila
] .,smm-
Dab
T[Cal-Ca2]
9.1 x 10-8 [6.3 X 10-3 - 2.1 x 10-3
0.25 X 10-3 ~
1.52 x 10-6 kg - mole
A
Mass flux
3)
s-m2
Molar flux X Molecular weight
kg- mole
1.52 x 10-6 s-m 2 x 2/mole
[.: Molecular weight of H2 is21
Mass flux
3.04 x 10-6 __3_
2
.
s-m
I
=.
Give" :
Temperature,
T == 25°C
Inside pressure,
p, == 2 bar
Inner diameter,
d, == 25 mm
Inner radius, r, == 12.5
Thickness,
mm == 0.0125 m ~
t == 2.5 mm == 0.0025 m
Outer radius, r2 = Inner radius + Thickness
== 0.0125 + 0.0025
I r2 == 0.015 ml
Cal
kg- mole
= 6.3 X 10-3
m3
Ca2
= 2.1
x 10-3
kg-mole
m3
Diffusion co-efficient,
Dab == 0.21 x 10-91112/5
Solubility, == 3.12 x 10-3 kg-mole
m3 - bar
TOfind:
2. Molar flux
kg-mole
= 1.52 x 10-6
S _ m2
Loss of 02 by diffusion per metre length
SO/Ulioll :
3.
Mass flux
= 3.04 x 10-6
kg
s- m2
I. M~Iar concentration
on inner side,
Cal == SolUbili~y x Inner
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Q
Tlte diffuSlvily
of 0 l
Hlal/ thick"
9 m 1Is and tile soluhilil 2 1 'rough rUbber ell.
1
x
10]
O.
Yo/ O·
's
2 tn rubbe .
3 kg-mole
.
r 'S
3.12 x 103 b
. Fmd lite loss %
b .
m - ar
2 Y diffusion
per
[AI')"'2000 &
11,elre Iengtt! of pipe.
. Apr' 1998 - Mill
Result:
1.
.
(1] rll bber pipe of inside
dlllmeler 25
floHilngthro
"
111111 and
Ugh
2) We know that,
Molar flux, A
SON
.....-- Oxygen at 25°~ fI~d pr~ssure of 2 hur is
I
kg-mole
m3
PROBLEM Mass Transfer 5.21
S.lZ SPLANE MEMBRANE
pressure
5.22 Heal and Mass Transfer
= 3.12 x 10-3 x 2
Cal
Cal
= 6.24 x 10-3
------------~
rna =4.51
x 10-11 ~
s
kg-mole
m3
ItSP":
_
LOSS of oxygen - 4.51 x 10-11 ~
,
tration on outer side,
Molar concen
x Outer pressure
C
=Sou I b'I'ty
II
s
'LIvdrogen gas at 2 atm and 250(' .
r11 P.'
ISflottJ'
~
tpe o/ID = 25 111111and OD ::::
50 IIrg throllgh a h
PIt,vJrogen t/lroug/I tI,e ruhher .III0nr. The difJll.fi;;tyher
a2
C
=3.12x 10-3 x 0
C
=0
a2
,
IS
J
a. the partial. pressur e of 02 on the outer surface of the
[Assummg
tube is zero 1
Oluhility
F: d
-;--.:..:.:
mJ - hQr '. III tire loss Of
Hydrogen hy diffusion per metre lenalh .~ .
J
OJ
G
[The procedure
Molar flux, A
of
".l/h. The
of hydrogen == 0.053 kg-IIIole
S
We know,
ma
.7)( J fH
Pipe.
[Apr
. '97 - MUJ
prevIOUs problem]
of this. problem is same as
'" (I)
{Ans : 4.46 x Ut·' !!!-rno.!!.}
For cylinders,
5.13 STEADY
Consider
Dab [Cal - Ca2l
STATE
two large chambers a and b connected by a passage
as shown in Fig.5.3.
(r2 -r,)
(1) z» 21r L (r2 -r,)
In(
s
EQUIMOLARCOlINTERDIFFUSION
L
Na and Nb are the steady
components a and b respectively.
;,2)
2 1r L . Dab rCa' - Ca2l
Chamber
.....
/
--'
a
In( ;;)
2 X 1r X I X 0.2 J x 10-9 (6.24 X 10-3 - OJ
0.015 )
In ( 0.0125
[.: Length
= I m]
---_--J
Chamber
+--Nb
,--------./
b
Pb, C
b
Fig.5.3
Equimolar diffusion
is defined as each molecules of 'a' is
replaced by each molecule of 'b' and vice versa. The tot~1pressure
p:::Pa + Pb is uniform throughout
the system.
p = Pa + Pb
Scanned by CamScanner
state molar diffusion rates of
(~
i
1i nsfer
Heal and Mass ra·
5.24
ith respect to x
. .n g WI
DifTerentlaU
I
\
I
__...
d
dp dPa + _!!_!!....
dx
;;; --- dx
press
S·Inee the total
.
steady state co
ure of the system
-----~__:_-----~M~a.~~s..!.Ti~ran5!.sfi~er:..:5~.
2~5
d
A Po
So, Na = -D GT dx
remains Co
nSlanI
nditJons,
dp
7h
d
dpb
'P{/ + _
== 0
== -;;;
dx
Integrating,
dpi,
2
!
N = rna = _ _Q_ dpo
a
A
GT , dx
te conditions, the total molar flux isl!..
Under steady sta
~
Na + Nb == 0
Na == - Nb
A dp., _ D ~
dPb
-Dab GT dr - ba GT dx
A
"·Ii)
A
dPb
GT
{IX
Molar flux, Nb = mb = _Q_ \ Pb' - Pb2 \
A
GT x2-x, 1
••• (5.7)
where,
dPa
Na == -Dab GT -dX
Nb == -Dba
... (5.6)
Similarly,
From Fick's law,
[
Molar flux, No = mAO= _Q_ Ipa, - Pa2\
GT x2-x, j
rna
kg- mole
- Molar flux A
s- m2
J
----=:;---
D - Diffusion co-efficient - m2/s
We know,
[F rom equation
dx
dx
A - Area - m2
Substitute in equation (5.5)
GT -;t; = - Db _GT dx
A
(5.5) ~
z>
-Dab
dPa
IDab= Dba =
01
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G - Universal gas constant - 8314 ----kg-mole - K
A
Pal - Partial pressure of constituent at I in N/m2
dPa
.
.
Pa2 - Partial pressure of constituent at 2 In N/rn
T - Temperature - K
2
\
\i
I
I
.~
.26 Heal and Mass Transfer
.14 SOLVED PROBLEMS ON EQUIMOLAR
COUNTER DIFFUSION
III Ammonia and air are in equimolar counter diffusion ill a
cylindrical lube of 2.5 mm diameter and 15 m lengtlt. rhe
total pressure is J atmosphere and the temperature is 25 C.
One end of the tube is connected to a large reservoir Of
ammonia and the other end of 'the tube is open to atmosphere.
If the mass diffusivity for the mixture is 0.28 x J(J-I ",2/s.
Calcalate the following
0
For equirnolar COunter djA:...
,
h) Mass rate of air in kg/ll
Given:
G - Universal
Diameter, d = 2.S mm = 2.S x 10-3 m
A-Area::::!Id2
Length, (x2 - XI) = ISm
Total pressure, P.= I atm = 1.0 I3 bar
Dab = 0.28 x 10-4 m2/s
Atmospheric
Air
gas consta
nt - 8314
oe-K
3/
I A:::: 4.90 x 10-6m~
rna
(I)~
_
490 x 10-6 r----
.
0.28 x 10-4 [I
x ._:_OI3 x 105-0)
8314 x 298
Molartransfer rate of ammonia
_
TOfind:.
J
~II)I
4
::::f- (2.Sx 10-
Temperature, T = 2SoC + 273 = 298 K
Ammonia
I
D ( Pal - P [F
Molar tlux, - rna == -..!!!..
A
GT ~
rOil)
x2 - .tI eqUation
where,
00.(5.6)J '" (1
a) Mass rate of ammonia in kg/II
Diffusion co-efficient,
IIUSIOn
.
15-
-
-3
k
,rna - .74 x IO-U g-mole
I. Mass rate of ammonia in kg/h
We know
2. Mass rate of air ill kg/h
Masstransfer rate . M
of ammonia
== Ol~ transf~r rate x Molecularweight
o ammoma
ofammonia
So/ution :
We know
that ,
Total pressure
,
p=p
al
+ p a2
::::3.74 x IQ-lJ x 17.03
[Molecula
. h
r welg t of ammonia = 17.03, refer HMTdata,
page no. 182 (Sixth edilion)]
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r
5.28 Heal and Mass Transfer
= 6.36
I
x I 0-12 kg/~
= 6.36 x 10-12 ~
~
!
I~/3~600h
Mass transfer rate of ammonia
Mass Transfer 5.29
~ransferrateofCOl
2.
Mass transfer rate of air
I
= 2.29 x 10-8 kg/h
Given:
Diameter,
We know,
Mass transfer rate = Molar transfer
.
of air
rate of air
x
= -3.74
x
10-13 x 29
=-1.08
x
10-11 kg/s
(
Temperature, T = 273 K
S
Partial pressure of CO2 at one end
Molecul
.
ar Weigh
of air.
I
200
Pal = 200 mm of Hg = 760 bar
Pal = 0.263 bar
I Pa I = 0.263
= -1.08 x 10-11 _k_,;g::___
x
[.: I bar = 760 mm ofHg]
lOs N/m2
I [.: I bar = lOs Nlm2]
Partial pressure of CO2 at other end
1/3600 h
90
Pa2 = 90 mm of Hg = 760 bar
I Mass transfer rate of air :::- 3.88 10- kg/h I
x
d ::: 60 mm = 0.060 m
1.2 m
J..,ength , (X2 -XI):::
Total pressure, p = I atm = I bar
Molar transfer rate of air, mb = -3.74 x IO-I3~
[Due to equimolar diffusion, rna = -mb]
and
8
=>
Pa2 = O.118 bar
=>
I Pa2 = 0.118 x lOs N/m2 I
Result:
I. Mass transfer rate of ammonia = 2.29 x 10-8 kg/h
2. Mass transfer rate of air = - 3.88 x 10-8 kg/h
[II CO2 and air experience equimolar counter diffusionin
Q
cO2
1
I
d=60mm
circular tube whose length and diameter are 1.2 m and60mm
-__
-.JI Xrx 1= 1.2
respectively. TI,e system is at (I total pressure of 1atm andQ
temperature of273 K. Tile ends of the tube are connected
to
TOfind:
large ell ambers. Partial press lire of CO2 at one endis
1. Mass transfer rate of CO
200 mm of Hg while at tile other end ls 90 mm of Hg·
2 M
2
. ass transfer rate of air
Calculate tile following
Scanned by CamScanner
Air
ml
'-----_--....J
5.30
y
Heat and Mass 7}ans/er
Solution:
Mass Tran.ifer 5.31
.
We know that, for equimolar COullter d'ff
Molar flux,
Ill"
=
A
Dnb
GT
[~.a2
.
I] uSIOIl
We knOW,
.
t of air mb = - 1.785 x
Molar transfer ra c
,
x2-xl
where,
DRh - Diffusion
The diffusion
co-efficient
- 1112/s
for CO - A'
co-efficient
rr COlllb'
II 89
111alio
. )( IO"{) n b
2
[From HMT data book page no. 180
I
Dab = 11.89 x 10-6 rn2/s
.
/(101,
:
G - Universal
gas constant - 8314 ~
1t
kg - mole - K
A - Area = - d2
4
(0.060)2
= :
I A = 2.82
x
x 10-6
I =:>
rna
= 11.89
()
2.82 x 10-3
8314 x 273
Molar transfer
rlo.263 x J05-0.118xW'
1.2.
We know,
transfer
x
gb
Molecular wei t
= 1.785 x 10-10 x 44.01
[Molecular
HMTdIil
weight of CO2 :::; 44.01, ref~r hediti~l
page no.182 (Slxt
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Molecular weight
of air
x
x 10-10 x 29
Mass transfer rate of air = -5.176
x
9
10- kg Is
Result:
9
/
I. Mass transfer rate of CO2, :::; 7.85 x 10- kg s
2. Mass transfer rate of air
UNIVERSITY
>
-5.176
x
PROBLEMS
COUNTER
10-9 kg Is
ON
DIFFUSION
ill Two large tanks, maintained at the same temperature
rate of COl> ma = 1.785 x 10- 10 kg - mole
Mass transfer rate·
= Molar
of CO2
=-1.785
EQUlMOLAR
I
x
Molar transfer
of air
WsOLVED
10-3 m2
==
1112/1
(S'X/hec/.
I
MasS transfer
. rate
and
pressure are connected by a circular 0.15m diameter direct,
which is 3 m in length. One tank contains a uniform mixture
of 60 mole % ammonia and 40 mole % air and the other
tank contains a uniform mixture of 20 mole % ammonia
and 80 mole % air. The system is at 273 K and
1.013 x .'05 pa. Determine the rate of ammonia transfer
betweenthe two tanks. Assuming a steady state ~ass transfer.
[Manonmanium Sundaranar Univ - Nov '96, MU - Nov '96J
Given Data:
Diameter, d = 0.15 m
Length, (x2 - xl) = 3 m
5.32 Ileal and Mass Transfer
_ .!!.. )( (0.15)
2
- 4
Pa I =
I~OO = 0.6 bar = 0.6 x lOs N/m2
Pbl =
40
100
= 0.4 bar = 0.4 x 105 N/m2
~go
= ~go=
Pal =
= 0.2 bar = 0.2 x 105 N/m::!
Pb2
0.8 bar = 0.8 x 105 N/m2
~
·ff .Ion co-efficient
;: 21.6 x ) Q-6 m2/s
_01 uS
..
Dab
monia with air
of am
HMT data book page no. 180 (Sixth edition]
[From
T = 273 K
~
)
P = 1.013
x
105N/m2
Tank I
Tank 2
Ammonia
+Air
Ammonia
+ Air
Pal
Pal
Pbl
Pb2
(I) z»
.
Molartransfer rate of ammoma,
Masstransfer rate
of ammonia
'a' - Ammonia
;: Molar transfe.r
of ammonia
=2.15
'b' -Air
rna ;:
x
2.15 x 10-9 kg-mole
S
rate
x Molecular
weight
of ammonia
10-9 x 17.03
[Refer HMJ data book, page no. 182 ]
Tofind :
Rate of ammonia
transfer
Mass transfer
Solution:
rate of ammonia
= 3.66 x 10-8 kg Is
Result:
We know that, for equimolar
counter
diffusion,
1. Rate of ammonia
'M oar
I fl ux -rna = -Dab [ Pal - Pa2]
'A
GT
x2-xl
where,
G - Universal
10-8 kg Is
circular tube whose length ami diameter are lm and 50mm
respectively. Tire system is at a total pressure of 1atm and a
temperature of 25°C. Tire ends of the tube tire connected til
large clrambers in whicl: the species concentrations are
maintained at fixed values. Tire partial pressure of C01 at
J
= 8314 ----=---
kg - mole - K
A - Area = ~ d2
4
63
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= 3.66 x
[I CO2 and air experience equimolar counter diffusion in a
... (I)
gas constant
transfer
I
I'
Mass Transfer 5.35
5.34 Heat and Mass Transfer
'ffusion, we can find
wof d I
one end is 190 mm of Hg while the other end is 95
Estimate the mass transfer rate of CO2 and air th "'Itt Ii..
rO"lIh ".
tube.
[Bharathidasan
Univ-Apr '98, MU-Apr '98 0
'he
frolfl
, c"200
[This problem is same as problem No.2 - Solved Pr hi 2 ]
o elll)
[Ans: 1. Mass transfer rate of CO2 = 5.17 )( 19-9 s
kgls
2. Mass transfer rate of air = - 3.40 x 19-9
kglsi
5.16 ISOTHERMAL EVAPORATION OF WATER INTO AIR
Consider the isothermal evaporation of water from a
waler
surface and its diffusion through the stagnant air layer over'
II
shown in Fig.S.4. The free surface of the water is eXposedto ..
Dab ~
~ ~ -aT (Xl - X I)
Molar flU'" A
(or)
Dab
~~ err
flu",
Molar
\Pal
Pa
In -
1
l
In p- pwl
P-Pwl
~
(Xl-XI)
rna _ Molar flux USI
__
'versal gas constant - 83 \4
Water vapour
G- Uru
/'"
.
J
••• (5.9)
~J
_
kg - mole
s _ ml
---A Di if 'on co-efficient - mlls
as
(5 8)
I
Dab _.
Air==
...
A
\\'here,
air In
the tank.
fick's la
kg _ mole - K
T _ Temperature - K
p _ Total pressure in bar
Water
Pw\.
Tank
Pw2
Fig. 5.4
_ Partial pressure of water vapour corresponding
saturation temperature a t I' III N/m2
to
_ Partial pressure of dry air at 2 in N/m2
For the analysis of this type of mass diffusion, following 5.17SOLVED PROBLEMS ON ISOTHERMAL
assumptions are made,
EVAPORATION OF WATER INTO AIR
1. The system is isothermal and total pressure remains
Determine tile diffusion rate of water from the bottom of a
constant.
test tube of25 mm diameter llml35 mm long into dry air at
2. System is in steady state condition.
ill
3. [here
is slight air movement over the top of the tankto
remove the water vapour which diffuses to that point.
4.
we. Take diffusion
co-efficient
of water
0.28 x lQ-4 m2 Is.
Given:
Both the air and water vapour behave as ideal gases.
Diameter,d::. 25 rnrn ::. 0.025 m
Length, (x2 - xl)::' 35 mrn = 0.035
Scanned by CamScanner
111
in air is
Mass Transfer 5.37
ssure at the top of the test tube. Here, air
5.36 Heal and Mass Transfer
'
rtial pre
.
_
"::":":=-":'T~e:':m:':'pe~ra:':'tu':"'r':"'e,_;_T-=-2-5""::OC~+-2-7-3-=-2-9-8-K---~_________
'\
.... pa
d there IS no water vapour. So, Pw2 - O.
is drY an
Diffusion co-efficient. Dah = 0.28 x 10-4 m2/s
\
I
~~
Dry saturated air
r
A===
Area,
t
i
=== (0.025)2
\
water
2
rna
Dab
GT
(
x
'"
__
kg- mole- K
.::__J __
watervapour
= I atm = 1.0 \ 3 ?~r
At 25° C
Pwl = 0.03166 bar
[From R.S. Khurmi steam table. page no.2J
2
\ Pwl = 0.03166 x IOsN/m
-
Scanned by CamScanner
\
\.0 \3 x 10L 0
\0-\0
x
kg-
~
105 J
mole
s
Molar rate of
water vapour
x
Molecular weight of
water vapour
5.09 x 10-10
x
\8.0\6
I : Molecular
weight of steam = 18.016.
refer HMT data book. page 110.183 I
1.013 x \ 05 N/m2
Pw I = Partial pressure at the bottom of the test tube
corresponding to saturation temperature 25° C
~
r
11 L1.013 x 10S_0.03166
Weknow that,
Mass rate of
~
0.035
(I)
(From equation no.5.91
=
\.0 \3 x \ 05
rna = 5.09 x
p ) In \ p - Pw2 \
x2-x\
lp-pw\)
G - Universal gas constant = 8314
x
x 298
I
where,
p - Total pressure
0.28 )( 10-4
83\4
(\)==' ~
Solution:
We know that, for isothermal evaporation,
Molar flux, mAa =
10-4 m \
~.90)(
\\
Tofind:
Diffusion rate of water
2
d
Masstransfer rate of water vapour
= 9. I 70 x
, 1ts"lt:
lYff
I us ion rate of water
==
9.170
x
10-9kgls
10-9 kgls
5.38
Heat and Mass Transfer
Mass Transfer 5.39
Estimate the rate of diffusion of water vapour fro
water at tire bottom of a well which is 62 _ l ", (IPOol
. .., (eep (I
Of
diameter 10 dry ambient air over lire lop of tire lid 2.2",
entire svstem may be assumed at 30°C and Oil
lVell. 'l'h
•
e (It",
e
pressure. Tile diffusion co-efficient is 0.24)( 1()-4 oSPh
ert
where,
",2Is.
Given:
Diameter,
Deep,
== 8314
kg - mole - K
d == 2.2 m
Partla. I pre ssure at the bottom of the well
pwl - correspon ding
to saturation temperature 30° C
I
== 6.2 m
(x2 -XI)
T == 30°C + 273 == 303 K
Temperature,
Total pressure,
Diffusion
J
.
I gas constant
G - Unlversa
p = 1 atm = 1.013 bar = 1.013 x 105 N/rn2
co-efficient,
Dab = 0.24 x
Dry saturated
d
air
10-4 m2/s
TQ)
_
:::>
G-pw-I ==-0-.0-4-24-2----:1-=-0~5
N~/m~2;!1
X
. pressure
_ Partial
:::>
_l__ (j)
- - - - - - -
Pwl ==0.04242
Pw2
x2-xl
---------
:::>
(l)~
water
bar
[From steam table,
page 110.21
at the top of the well.which
is zero.
IPw2 == 0 I
rna == 0.24 x 10-4
8314 x 303
3.80
x
1.013 x 105
6.2
1 013 x I OL 0
x In [ 1.013 x ; 05 _ 0.04242
ToJind:
rna == 2.53
Diffusion
10-8
kg-
---"'----
Molar rate of water
.
We know that, for isothermal
-rna
flux,
mole
S
Solution:
Molar
x
rate of water
]
x 105
D ab
A == GT (X2-XI)
Area,
ln
[~l ...
(l)
kg- mole
10-8 ---=----
P-Pwl
We know that,
Mass rate of
Watervapour
Molar rate of
water vapour
x
Molecular weight
of stearn
2.53 x 10-8
x
1:::.016
4.55
Scanned by CamScanner
x
S
evaporation,
p
2.53
x
10-7 kg/s
Mass Transfer 5.41
5.40 Heal and Mass Transfer
O
Result:
_
7
~
Diffusion rate of water - 4.55 x 10- kg/s
an 210 mm in diameter and 75 nun ~
An open P
weep co"
at 25 fie and is exposed to dry attnos h '.l.I~
water
P eric
r_'
'ate
the
diffusion
co-efficient
of
Water
in
a'
Q;'.
,--rucu.,
tr, 'l'll/{
rate of diffusion of water vapour is 8.52 x 16-4 kglh. e
'~t
Diameter, d
Given:
knoW' that,
MaSs rate of
water vapour
= 210 mrn = 0.210 ....
,.,
Temperature, T
= 25°C + 273 ::::298 K
where,
= 8.52 x 10-4 kg/h
Area,
Diffusion rate (or) mass rate
= 8.52 x 10-4~
x
Molecular
x
p
(x2-xl)
x In [P-PW2
P-PWIJ
••• ( 1)
A= ~ d2
4
I A = 0.0346 m2 \
= 2.36 x 10-7 kg/s
G - Universal gas constant
I
1--------- _I
1---------
= 8314 ---
J __
kg-mole-
®
p - Total pressure
K
= 1 atm = 1.013 bar
= 1.013 x 10sN/m2
x2 - x1
d
Ix 18.016\
4
= 2.36 x 10-7 kg/s
Dry atmospheric air
weight
of steam
= ~ (0.210)2
3600 S
Mass rate of water vapour
rate of
water vapour
Dab x ~
= 75 rnrn
0.075 III
Molar
2.3 6)()0-7=0
Deep, (x2 -XI)
e
=
we
(j)
I-------:~
- - - - - - - - r-- Water
Pwl - Partial pressure
corresponding
at the bottom of the pan
to saturation
temperature
25° C
At 25° C
I;;.-=.-=.-=.-=.-=.-=.-=.-=.
=::)
Pw: == 0.03166 bar
=::)
Pw: ::::0.03166 x lOS N/m2
Toflnd:
Diffusion co-efficient,
(Dab)
Solution: We know that, molar rate of water vapour,
ma
A
-
::::
Dab
pGT (X2-xl)
-
Scanned by CamScanner
X
In
(P -PW2)
P-Pwl
[From (R..s. Khurrni) steam
table. page no. 2]
Pw2 - Partial pressure at the top of the pan, which is zero.
==>
~W2::::
~
5.42 Heal and Mass Transfer
(I)=>
2.36 x 10-7::
Mass Transfer 5.43
DabXO.~
~
8314 x 298
xln(~
=:
.1 gas constant ==8314 -----
kg - mole - K
ssure == 1 atm = 1.013 bar
5
1.013 x 10 - 0.03166 x 105)(
[nab
G - Unlvers
18'()16
p - Total pre
2.18 x I 0-5 m2/~
= 1.013 x lOs N/m2
- Partla. I pressure at the bottom of the pan
o
pwl
eorres ponding to saturation temperature 30 C
Rf!Sull:
Diffusion co-efficient, Dab == 2.18 x 10-5 m2/s
A pan of 40 mm deep, isfilled with water to a I I
. e.:'Cposed
and t«
to dry air at 30°C. Calculate th eVe
ti 0/20 "''''
e
'lne req
.
for all the water to evaporate. Take, mass diff
. ~"ed
0.25 x 1()-4 mlls.
':JUS'VlIy is
Pw2
Given:
==0.04242 bar
:::>
Pwl
:::>
~w -1-==-0.
-0-42-4-2-X-I
O--:5-:-N-/~m-=2-'\
[From steam table
page no.2J
_ Partla. I pressure at the top of the pan, which is zero.
~ (PW2 ==0 \
Deep, (x2 - xl) == 40 - 20 == 20 rnrn == 0.020 m
Temperature,
T == 300e + 273 =: 303 K
Diffusion co-efficient,
rna _
(Il=>
Dab == 0.25 x 10-4 m2/s
A
0.25 x 10-4
8314 x 303
1.013
x
x
105
0.020
I
1(1)
Tofind :
L
~==2.15
x
10-6
kg-mole
s
A
Time required for all the
water to evaporate, 1.
1
I
1.013 x 1 OS_ 0
x n 1.013 x 105 - 0.04242 x 105 J
Dry atmospheric air
For unit Area, A ::::1m?
water
Solution:
We know that, for isothermal evaporation
Molar flux ~
'A
==
Dab
p
GT (x2-xl)
l
L
Scanned by CamScanner
Molar rate of water
m :::: 2.15
,
10-6 kg - mole
sm2
x
a
We knowthat,
PW2]
x In [p p- Pwl
... (I)
Mass rate of
WatervapOur
Molar rate of
water vapour
x
Molecular weight
of steam
Mass Transfer 5.45
5.44 Heal and Mass Transfer
::: 8.54 x 10--4 kg
3600 s
=2.15XIO-6~
[Molar rate of water vapour
3.87 x 10-5 ~
== 2.37
x
10--7 kgls
The total amount of water to be evaporated per m2 area
= (0.020
Dry atmospheric air
x l ) x 1000
= 20 kglm2 Area
fO
Time required,
I
=~
DiffuSI
'on co-efficient,
= 516.79 x
We knoW that,
103 sJ
Molar rate of water vapour
An open pan
20 em in diameter
and
ISOTH'ERMAl
8 em deep Contailll
water at 25°C and is exposed
to dry atmospheric
rate
vapour
of diffusion
of
the diffusion
water
co-efficient
air.lftht
is 8.54 x /0'-4 kglh,
Diameter,
Length,
(X2 -xI)
Temperature,
Diffusion
d
T
= 20 ern = 0.20 m
=-= 8 em = 0.08
= 25°C
m
+ 273 = 298 K
rate (or)
Mass rate of water vapour
:::>
rna:::
p
Dab x A x
P
GT
(x2-x,)
I \p - Pw2\
n
P-Pw'
Mass rate of
water vapour
x In \p - Pw2\
p-Pw,J
Molar rate of
water vapour
2.37 x 10-7 = _D_u_b_x_A_
x
p
GT
(x2-x,)
x
Molecular weight
of steam
Area,
x In \p - Pw2
Ix 18.016
P-Pwl]
\
... (1)
where,
A = .2!_ d2
4
=
= 8.54 x 10- 4 kg/h
J
We know that,
*
(0.20i
\ A = 0.0314 m
Scanned by CamScanner
x
(x2-X,)
of water in air.
[May '05 -Anna Univj ,
Given:
Dab
A - GT
I == 516.79)( 103 S
SOLVED UNIVERSITY
PROBLEMS
ON
EVAPORATION OF WATER INTO AIR
estimate
========= -L
CD
water
soilltion:
rna -
Time required for all the water to evaporate,
o
Dab
vapour
Result:
5.18
t-a>
Mass rate of Water~
20
3.87 x 10-S
11
find:
2
\
Mass Transfer 5.47
5.46 Heat and Mass Transfer
G - Universal gas constant
-;-----L_----... t.jI"tI :
= 8314
kg - mole - K
p _ Total pressure
= I atm = 1.013 bar
Partial pressure at the bottom of the t
corresponding
to saturation
Pwl = 0.03166 bar
I
table
I
(.%2-X1
T == 2SoC
. n CO-e
r ,
) == 1 5 em
= O. 15m
.
+ 273 = 298 K
fticient, Dab = 0255
IIr
• page n02J
Partial pressure at the top of the pan. He
'.
.
re, air IS d
and there IS no water vapour. So, Pw2 ::: O.
ry
x '10~ rolls
·d.'S10
DIi'"
Dry saturated ~ir
ure 25° C
[From (R S Kh
'"
Urmi) Slea
Pwl = 0.03166 x 105 N/m2
Pw2 -
re(l1P
est tube
temperat
OiaJ11 '
l)logtb,
erature,
::: 1.013 x 105N/m2
Pwl -
v
=::::lOmm==o.OlOm
ter
e d
T~
d'
10ft'··
. n rate of water
Difi'uS10
-.
solution:
We knoW that, for
.....al evaporation,
--------
isothe,,,.
(1) => 2.37 x 10-7 = Dab x 0.0314 x 1.013 x lOS
8314 x 298
0.08
5
1
xI [
1.013 x 10 - 0
n 1.013 x 105-0.03166 x 105 x18.016
2.58 x 10-5 m2/s
I
where,
Area, A = ~ d2
= ~
(0.010)2
Result:
Diffusion co-efficient,
5
\ A = 7.85 x 10- m
Dab = 2.58 x 10-5 m2/s
II] Estimate the diffusion rate of water from tile bottom oja
test tube 10mm in diameter (I/l(1 15cm 100Ig into dry
atmosphere air at 25°C. Diffusion co-efficient of water into
air is 0.255 x 10-4 mt/s.
[Nov '96· MUl
G - Universal gas constant
\
J
= 8314
kg-mole-
K
P - total ores sure = 1 atm = 1 .013 bar
= 1.0i.3 x 105 N/m2
Pwl - Partial pressure
correspoIiding
Scanned by CamScanner
2
at the bottom of the test tube
to ·saturation temperature 250 C
Mass Transfer
5.48
Heal and Mass Transfer
1.5 cm = 0.015 m
15 cm = 0.15 m
Pwl = 0.03 166 bar
{Fro", SI
~
/PWI
.
=0.03166
x
IOSN/m2]
_
=
0.(
,W
hi
lch'
01
7.85 x 10-5
Dry air
IS~
to
0.255 x 1D-4 x
(I)~
0.256 cm2/s = 0.256 x 10-4 m2/s
Page II I
w
IPw2
250C + 273 = 298 K
falll/ubi
P 2 - Partial pressure at the top of the test tube
~
5.49
1.013 x 105
8314 x 298
Molar rate of water vapour
0.15-
= 1.73
ma
Molar flux,
A
p
Dab
GT (X2 -xI)
In(P-PW2)
P =P«,
... (1)
We know that,
Mass rate of
water vapour
=
Molar rate of
water vapour
1.73 x 10-11
x
Molecular weight
of steam
x
18.016
Mass rate of water vapour
= 3.11 x 10-10
Area, A
!!..d2
4
~ (0.015)2
[.: Molecular weight of steam:: 18.0J6
refer HMf data book, page no.J8J]
I
where,
kglsJ
A
G- Universal.gas constant
1.76 x 10-4 m2
8314
Result:
p - Total pressure
Diffusion rate of water = 3.11 x 10-10 kg/s
kg-
J
mole - K
1 atm = 1.013 bar
1.013 x 105 N/m2
I
Estimate the diffusion rate of water vapourfrom tile bonollli
of a test tube 1.5 em diameter and l Scm long into dry air •.
25°C Take D = 0.256 cml/s.
I
[Apr '2001 - MU, Bharathidasan
Univ- Nov'901
I
h
Scanned by CamScanner
Partial
pressure
the
tube
test
saturation
at the bottom
corresponding
temperature
25°C.
Mass Transfer 5.51
5.50 Heal and Mass Transfer
Pw2 -
Pwl
0.03166 bar [From steam
IPwl
0.03166 X 105 N/m2]
b
(1) => 1.76 x 10-5
50%
Relative humidity
ta le, Po
Atmospheric
air 50% RH
ge tJo,21
raJi"d:nOration
. rate of water in grams
Partial pressure at the top of the test tube who .
,
lch IS
Pw2 = 0
l.ero,
I
==>
25°C + 273 = 298 K
~rature,T
~
I
pvllr-
per hour.
0.256 x 10-4 1.013 x 105
8314 x 298 x
0.15x
In[
I"-----------
Molar rate of water vapour
~-----------We know that,
Mass rate of water vapour
1.013 x 105_0
~6\
_:_:__."..J
)( lOS J
[From HMT data book, page no. 180J
==3.899 x 1 0-11 ~
D ab = 25.83 x 10-6 m2/s
:s
We knoW that , for isothermal evaporation,
{~~l:~:~e
vapour
} x
J ~~:;~~ \
l steam
J
ma
3.899 x 10-11 x 18.016
I Mass rate of water vapour
where,
Area, A
P
Dab
Molar flux, A = GT
I
(X2 - XI)
n
(p - Pw2 '\
\.P - Pw\)
... (1)
~d2
4
7.02 x 10-10 kg/s 1
Result: Diffusion rate of water = 7.02 x 10-10 kg/s
An open pan of 150 mm diameter and 75 mm deepcontains
\A==0.0176m2\
water at 25°C and is exposed to atmosphere air at 25°C
and 50% R.B. Calculate the evaporation rate of water in
[Apr '2002-MU]
grams per hour.
G - Universal gas constant
P - Total pressure
Given:
Diameter, d
150 mm == 0.150 m
Deep, (x2 - xl)
75 mm == 0.075 m
= 1 atm = 1.013 bar
= 1.013 x lOS N/m2
Pw\
-
Partial
pressure
corresponding
Scanned by CamScanner
= 8314 kg _ mole _ K
at the bottom
of the test tube
to saturation temperature
25°C
Mass Transfer 5.53
Heal and Mass Transfer
5.52
water
diameter a nd 8 em deep contains
.,
pan 20 em
d.
tmospheric air: Determine
,4" ope nd is exposed to ry ~
vapour in glhr. Take
01]5" C aof diffusion of wa er
[Del '99 _ MV]
At 25° C
lt
=>
Pwl = 0.03166 bar
=>
IL.Pwl~ = 0.03166
{From Slea
x 10 N/m2
5
I1J 1Qb{
Page 110.2]e,
1
-...J
.
P ., - Partial pressure at the top of the pan corres
wz
. Iiurm idi
POndln g to
25°C and 50% re Iative
ity.
bar = 0.03166
:::>
Pw2 = 0.03166
:::>
R.H.= 50 % = 0.50
Pw2 = 0.03166
:::>
I Pw2
x
0.0176
- 321 x 10-/0 kg/so
s : Diffusion rate of water - .
lOs x 0.50
[]-r:
.
"ate oif water from tile hottom of a
tlte diffusIOn"
10 mm in diameter and 15 em long into dry
iest lub.
. t 25°e. Diffusion co-efficient of water
almosplterlc atr a
2
.
.
0255
x ](;--4 m '/s.
inloaIT IS •
An .
= 1583 N/m21
25.83 x 10-6
(I):::>
lOs N/m2
x
0-4 m2/s.
/It role
~,;:;0.259 x 1
ter vapour = 0.855 g/hr
rateofwa
,4"s: MOSS
•
if water from the bottom of a
. te tlte diffUSIOn~ate 0
nd 20 em long into dry
r11 tSII/1l0be10 mm in dlOmeter ~ = 0.26 x 10--4 m2/s.
~ test ta
. t 30°e. Assume
almosplterealf a
[Apr '99 - MUJ
x
1.013 x 105
8314 x 298
0.075
x In [
1.013 x 105_1S83
1.013 x 105_0.03166~
[Nov '96 - Mano'!manium Sundaranar Univ 1
]
Ans : Diffusion rate of water = 3.12 x 10-/0
[Theprocedure of above problems
Molar rate of water vapour, ma = 3.96
Molar rate of
water vapour
x
Molecular weight
of steam
x
18.016
(1125" C and is exposed to dry atmospheric air. If the rate of
diffusionof water vapour is 8.54 x 10-4 kg/It. estimate tile
diffusionco-efficient of water in air.
= 7.13 x 10-8 kgls
= 7. 13 x
Mass rate of water
I 0-8
vapour
1000 g
1/3600 h
= 0.256 g/h
Result:
Evaporation
5.17,
lil An openpan 20 em in diameter and 8cm deep contains water
= 3.96 x 10-9
I
are same as, Section
Problem no. IJ
We know that,
Mass rate of
water vapour
kg/so
rate of water = 0.256 glh
Scanned by CamScanner
Ans:
Dab
[Apr '97 - Manonrnaniu-n Sundaranar Univ
& Apr '98 - Bharathidasan UnivJ
= 2.58 x 10-5 m2/s
[The procedure
Problem no. I ]
of this problems
.
IS
same
as, Section
5.18,
Mass Transfer 5.55
5.54 Heat and Mass Transfer
~x_Distance-m
5.19 CONVECTIVE MASS TRANSFER
Convective mass transfer is a process of mass t
.
.
ransfu
will occur between a surface and a fluid medium when tl
r thai
different concentrations.
ley are al
v - Kinematic viscosity - m2/s
For flat plate,
If Re < 5 x lOs, flow is laminar
S.20 TYPES OF CONVECTIVE MASS TRANSFER
If Re > 5 x lOs, flow is turbulent
I. Free convective mass transfer
·/t Number (Sc)
2.SChttlit
2. Forced convective mass transfer
.
I is defined as the ratio of the molecular
: m to the molecular diffusivity of mass.
S.21 FREE CONVECTIVE MASS TRANSFER
If the fluid motion is produced due to change' I d
n ellS'1
resulting from concentration gradients, the mode of mass t
Iy
.
.
ransfer'
said to be free or natural convective mass transfer.
IS
diffusivity
of
mornen u
SC ==
Molecular
diffusivity of momentum
Molecular diffusivity of mass
Example: Evaporation of alcohol
Sc= - v (or)Sc=-Dab
5.22 FORCED CONVECTIVE MASS TRANSFER
~fthe fluid motion is artificially created by means of an exte~al
force like a blower or fan, that type of mass transfer is known as
forced convective mass trasfer.
where,
v - kinematic
Dab-
Example: The evaporation of water from an ocean whenair
viscosity - 1U2/s
Diffusion co-efficient - m2/s
3. scnerwood Number (Sir)
blows over it.
5.23
11
pDab
SIGNIFICANCE
OF DIMENSIONLESS
GROUPS
It is defined as the ratio of concentration
boundary.
gradients at the
1. Reynolds Number (Re)
It is defined
as the ratio of the inertia force to the viscous
force.
where,
Re ==
Re=
Inertia force
Viscous force
hm - Mass transfer co-efficient - m/s
Ux
Dab- Diffusion co-efficient - m2/s
v
where,
U - velocity - mls
Scanned by CamScanner
x - Length - m
Mass Transfer 5·.57
5.56 Heat and Mass Transfer
5.24 FORMULAE
USED FOR FLAT PLATE P
Reynolds Number,
~c tnbi;ed Laminar - Turbulent flow
./(ii) Cotllu ....
SherWoodNumber; Sh = (0.031 ReO.8- 81l1Sc 0.333
~
ROBLEM
Re = U.x
S
v
Sh = h"r
where,
Dab
U - velocity - mls
~ROBLEMS
ON FLAT PLATE
5.ZSS0
1 Air at 10llC witll a velocity of 3 m/s flows over a ]lat plate.
GJ 1/ the plate is 0.3 m long, calculate the mass transfer
co.efJicient•
x - Distance - m
v - Kinematic viscosity '- m2/s
IfRe < 5 x lOs, flow is laminar
(
If Re > 5 x lOs, flow is turbulent
Given:
Fluid temperature, Too= lODe
For Laminar Flow
[From HMI data book, page no .]75
IS'IXt h edilio))
L
. \'
ocal Sherwood Number, Sh, = 0.332 (Re.x)o.s(Sc)0.333 n
Velocity, U = 3 mls
Length, x = 0.3 m
Average Sherwood Number, Sh = 0.664 (Re)O.S(Sc)0.331
Tofind:
where,
Mass transfer co-efficient, (hm)
Sc = Schmidt Number
solution:
= _v_
[From HMf data book,
page no.33]
Kinematic viscosity, V = 14.16 x \0-6 m2/s
Properties of air at 10 e
D
Dab
v -
kinematic viscosity
Weknow that,
Dab- Diffusion co-efficient
Reynolds Number, Re = Ux
V
Scherwood
Number
,
Sh = hrnX
3 x OJ
14.\6 x \0-6
Dab
hm - Mass transfer co-efficient - m/s
For Turbulent
Flow
Re = 0.63 x 105-< 5 x lOs
[From HMT data book, page no.17o
Since, Re < 5 x \ 05, flow is laminar
j
(i) Fully turbulent from leading edge.
Sherwood
Number,
\,
Scanned by CamScanner
Sh = 0.0296 (Ke)O.8 (Sc)OJ33
ForL ammar
. flow, flat plate,
Sherwood Number (Sh) = 0.664 (Re)u.5 (Sc)o.m
..• (I)
[From HMTdaca book, page no. 175]
Mass Transfer 5.59
5.58 Heal and Mass Transfer
~55nl/s
~e1ocitY,
u -._
'lie
:x:::::: 600 mm = 0.6 m
where,
Sc - Schmidt Number ==
V
D~b
Dab- Diffusion co-efficient (water + Air) at lOa
c , 80C
= 20.58 X 10-6 m2/s
[From HMT data book pag
...--_--[Dab
__
J..,ellgth,
..• (2)
--,
•
= 20.58 x 10-6 m2/s I
e 1I0./80J
14.16 x 10-6
(2) => Sc = 20.58 x 10-6
(.
r'MasS tran sfer co-efficient, (h
1
m)
0
[From HMT data book.
page no. 33]
50lplion:
. s of air at 30°C
propertle
. Vise
- osity , v = 16 x IQ-6 m2/s
. .....atlc
Kille".
We knOw that,
Ux
ids Number, Re =
Reyna
v
55 x 0.6
16 x 10-6
I Sc = 0.6881
Substitute Sc, Re values in equation (I)
Re = 2.06 x 106> 5 x 105
(I) => Sh = 0.664 (0.63 x 105)05 (0.688)0 3JJ
Since, Re > 5 x lOs, flow is turbulant
ISh= 147.151
[Flow is lami~ar upto Re = 5 x 105, after that flow is turbulant
1
We know that,
Sherwood Number, Sh =
hmx
For combined Laminar - Turbulant flow.flat
plate,
Dab
hm x 0.3
147.15 = _....;.;_-20.58 x 10-6
Mass transfer co-efficient,
hm == 0.0 I m/s
Sherwood Number (Sh) = [0.037
Mass transfer co-efficient,
hm = 0.0 I rn/s
Too = JO°C
Scanned by CamScanner
=
Dab- Diffusion co-efficient
[I] Dry air at30 e and one atmospheric pressure flows over a
Fluid temperature,
(1)
where,
D
flat plate of 600 mm long at a velocity of 55 mls. Calculate
the mass transfer co-efficient at the end of tile plate.
I
Given:
••.
[From HMTdata book. page no. 176]
Sc - Schmidt Number
Result:
(Re)O.8 - 871 ]ScO.333
==
~
25.83
x
v
Dab
.•. (2)
(water + Air) at 30° C :::::260C
10-6 m2/s
[From HMJ data book. page no. 180]
-::;:-=-2S=-.-83-x-I-0--6- -2/m
s
I
~~~~~-------~
5.60
Heat and Mass Transfer
(2) ~
16>< 10-6
Sc = 25.83 x 10-6
S/I
(
k
ge no.33]
HMTdata boo, pa
8°C = 30°C
.,,:
111//0
~
rtieS 0
fair at 2
prope
[From
v === 16 x 10-6 m2/s
.
. ViSCOSIty,
ttC
[ Sc = 0.619 ]
. (lla
I(,oe
Substitute Sc, Re values in equation (I)
(I) ~
Sit = [0.037 (2.06 x 106)08 - 871] (0.619)0333
I Sh
that,
ow
we kO
= 2805.131
....lumber, Re ===
~
v
Ids J'"
ReynO
2.5 x 15
We know that,
16 x 10-6
h,nX
Sherwood Number, Sh = -D
Re-_ 2 .34 x 10 > 5 x lOS
6
ab
=>
hm x 0.6
2805.13 = 25.83 x 10-6
Mass transfer co-efficient,
05 flow is turbulant
R /' 5 x 1 ,
.
Since, e
= 5 x lOs, after that flow IS turbulent
[floW is laminar upto Re
hnr = 0.121 m/s
.
_ Turbulantfiow,fillt
mar
eo b 'netl Lam
for ", ,
(SI ) = [0 037 (Re)0.8
h rWood Number
1
.
Result:
Mass transfer co-efficient,
hili = 0.121 m/s
Q] TIre water in II 6m x J 5 m outdoor swimming pool is
maintained at a temperature of 28°C. Assuming a wind speed
of 2.5 m/s in tIre direction of the long side of the pool.
Calculate tile mass transfer co-efficient.
Sc - Schmidt Number
= 25.83
Too = 28°e
rl
Speed, U = 2.5 m/s
Wind speed in the direction
of the long side of pool.
So, x = IS m
Tofind:
Mass transfer co-efficient,
(hm)
=
v
... (2)
Dab
(water + Air) at 28° C :::;26°C
x 10-6 m2/s
(From HMT data book, page no. 180]
O-ab-=-2-5-.8-3-x-1 0---6 2-/
(2)~ Sc =
-m- s-'l
16 x 10-6
25.83 x 10-6
SUbstitute SR.
c, e values In equation (1)
Scanned by CamScanner
(\)
book, page no. 176]
EO.619]
;A
•••
where,
Size = 6m x 15 m
Fluid temperature,
- 871 )ScO.333
{From HMIdata
5e
Dab- Diffusion co-efficient
Given:
plate,
1
Mass Transfer 5.63
5 62 Heat and Mass Transfer
We know that,
Sherwood Number, Sh =
3185.90
hnrX
15+25
Dab
2
= __ h m x 15 _
25.83 x 10-6
. s of air at 20°C
Mass transfer co-efficient, hm =
prO
Result:
Mass transfer co-efficient, hm = 5.486 x 10-3 m/s
Air dr2SOCyrows over a tray full 0/ water wit" a vel .
2.8
Tile tray measures ~ .10 em along tile flow dirOCItyO/.
•
eClion
and 40 em wide. Tile partial pressure 0/ water present' h
In I e
air is 0.007 bar: Calculate tile evaporation rate 01" Wat .
'J
er if
the temperature on tile water sur/ace is J 5°C. Take diffusion
co-efficient is 4.2 x Ifr5 m}/s.
PerUe
uc viscosity,
I(inema I
V
[From HMT data book, page no.33J
x 10-6 m2/s
= 15.06
'lie knoWthat,
"
Ux
Reynolds Number, Re =
v
mls.
Given:
2.8 x 0.30
15.06 x 10-6
Re=
Sherwood Number
Speed, U = 2.8 mls
Flow direction is 30 cm side. So, x = 30 ern = 0.30 m
Area,
A = 30 ern x 40 ern = 0.30 x 0.40 m2
Partial pressure of water,
!
Pw2
Dab = 4.2
Sc - Schmidt Number
x 10-5
N/m2
(Re )0.5 (Sc) 0.3 3\ ... (
=
V
Dab
15.06
4.2x
6
10-.
EOJ58]
Substitut S
e
Scanned by CamScanner
( 'h) = [0.664
[From IIMT data book, page no. 17
Sc:::
Diffusion co-efficient,
flow:
= 0.007 bar
T w = 15°C
5 x 10
where,
Pw2 = 0.007 x 105 N/m21
Water surface temperature,
x 105
Since, Re < 5 x 105, flow is laminar.
Forflat plate, Laminar
Fluid temperature, Too = 25°C
0.557
C
Re value
In
equati
n (I)
Mass Transfer 5.65
5.64
Heat and Mass Transfer
----uNIVERSITY
SOLVED PROBLEMS
/,,~t\ u~
(I) ~
Sh
[0.664 (0.557
I Sh
x
105)0.5(0.358)0.~
5.Z6 Ofll fLt\T~P::L:.:A:.:.T.:....E
I
111.37
20"C /p = 1.2 kg/m3, v = l5 x /0-6 ml/s.
. air 01
)( J(r5 ml/sl flows over aflat plate oj length JO em
l!J D:::4.~ overed wit" a thin layer oj water at a velocity of
h'ch ,s c
11"
Estimate the local mass transfer co-efficient of a
/",Is.
« tncm from tile leading edge anti the ave";ge
.I' lance 0,
.
S
ul
iter co-efficient.
[June 20()~-A"'1Q Univ]
",asstrans,.
~
~t
rJ1 Dry
We know that,
hmx
Sherwood Number, Sh
Dab
~
hm X 0.30
111.37
Mass transfer co-efficient,
Mass transfer co-efficient
4.2 X 10-5
hm
Density, p = 1.2 kg/m)
based on pressure difference I'S .
given
Kinematic viscosity,
hmp
_
0.0155
- R Til'
-
287 x 288
1 Pil'I
Velocity, U = I m/s
Distance, x = 10 em = 0.10 m
Saturation pressure of water at 15°C
(RS
5
= 0.017 x 10 N/m
2
Tofilld:
{From steam table
Khumi) page no I)
I
= hmp x A [Pwl
2. Average mass tran fer co-efficient,
hm for entire length.
co-efficient
at x = 0.10 DI
We know that ,
= 1.88 x 10-7 x (0.30 x 0.40)
'I
h.t at a distance of 0.10 m.
Case(i) : Local mass transfer
- Pw2]
x (0.017 x IOLO.007x
I. Local mass transfer co-efficient,
Solution:
The evaporation rate of water is given by,
mil'
I O~ m2/s
Length, L = 50 em = 0.50 m
= 1.88 x 10-7 mls I
= 0.017 bar
v = 15 x
Diffusion co-effie ient, Dab = 4.2 x 10-5 m2/s
_~
[.,' Til' = 15°C + 273 = 288 K, R = 287 JlkgKj
1 hmp
= 200C
Give" :
Fluid temperature, 1
0.0155 mls
by,
Pwl
--
Re = Ux
Reynolds number,
IOSI
v
l-n-II'--=--2.-2-5-x--IO--~5~k-W~s'l
==
Result:
Evaporation rate of water, mit' = 2.25 x 10-5 kg/s
l Re == 6666.67 < 5 x lOS 1
.
Since Re < 5 )( . 05
I
i
6E
I
1
Scanned by CamScanner
I x 0.1
ISxIO-6
.Tlow is laminar
5.66 Heat and Mass Transfer
For Laminar Flow,jlat plate
Local Sherwood Number, Sh = 0.332 (Re)o.s (S )0
C .333
[From HMJ'data book
where,
.Sc = Schmidt Number
"'(1)
Page ~O.17S1
= -2_
Dab
Re - 3 3 ----::::---:.
- . 3)( 104
5
Since Re < 5 x 10 , flow is 1 .
amlnar.
For flat plate laminar flow
Sherwood Number, Sh « 0
Sc = 15 x 10-6
4.2 x 10-5
Substitute Re and Sc VI'
.664 (Re)O 5
(SC)0.333
a Ues.
'" (2)
Sh == 0.664 (3.33 x 104 )0.5
(0.357)0333
\ Sc = 0.357 \
[Sh == S5.99\
Substitute Sc, Re values in equation (1)
We know that,
(1) ~
Sh = 0.332 (6666.67 )0.5 (0.357)0.333'
- hmL
SI1-Dab
\ Sh = 19.24\
85.~9 ==
We know that,
hxx
Sherwood Number, Sh =
Dab
19.24=
=:>
hx x 0.1
---4.2.x 10-5
\ hx = S.OS·x 10-J m/s.\
Local mass transfer co-efficient
at x = 0.1 m is 8.08 x 10-3mil.
Case (ii): Average mass transfer co-efficient h m' for entirelen~h
We know that,
Reynolds
number,
Re =
=
UL
v
1 x 0.50
15 x 10-6
Scanned by CamScanner
hm x 0.50
4.2 x 10-5
::::>
hm == 0.007 m/s.
Average mass transfer
co-efficient
~d:
1. hx == 8.08 x 10-3 m/s,
2. hm == 0.007 rn/s.
f
.
or entIre length'
IS 0.007 mJ
5.68
Hem U"" .'_
5.17 FORMULAE USED FOR INTERNAL FLOW
(CYLINDERS or PIPES) PROBLEMS
~
UD
1. Reynolds number, Re ==
C\'t,~ ~'t<\ \
MGt's
r1l Air at 30· C
tlltr,~'ll~ .. · T'CDujer 5,69
t;J
Gild
l'Lo
12 mm diameter t I4bQ'IIIo 'p~.
W
e
t
2.5 m/s. rile ;11 . Of I 'rl
•
Side
p""
deposit of naphth SUr/Itt' ""tl~U~f flow! in
",ass transfer co-effie'
,r. ale",. ',,01I~, ""h Q ""ocitu :
UtI
'11"
-, OJ
v
where,
U-
SOLVEDpQ""
Vnlt
(pIPES AND
t.t~
""'r
Velocity - mls
D - Diameter - m
Dab == 0.62 x J(j-S na2
v _ Kinematic viscosity - m2/s
1
IS.
. n .
GIlle
If Re < 2000, flow is laminar
Ie",. 1'1t~_t'IIIi"f e con'ain. a
elil
~
'he
IlIlioll
•
a"erage
co-efficien,
fluid temperature T _ 3Ooe
Velocity, U = 2.5 mls
If Re > 2000, flow is turbulent
•
For laminarflow:
'
'
<J:)-
Diameter, 0 = 12 mm » 0 .0 \2m
Sherwood Number, Sh == 3.66
h
Length, x == 1 m
D
Sherwood Number, Sh == _m_
Diffusion co-efficient , D ab -- 0 .62 x \ I.)-~
Tofind :
Dab
where,
hm - Mass transer co-efficient - mls
Dab - Diffusion co-efficient
- m2/s
ll\11~
Average mass transfer co-eff IClent,
.
h
.
11\
SoI uuon :
Properties of air at 30°C
For turbulentflow:
.'
Kinematic
Sherwood Number, Sh == 0.023 (Re)O.83(Sc)O.44
[From HMT data book, page no. J 76 (Sixth edition))
[From HMT data boo
viscosity , v == 16 X In...J.
k, page no
v m2/s
v
We know that ,
Reynolds number,
where,
Re == UD
v
Sc == Schmidt Number == _v_
Dab
==
2.5 x 0.0\2
16 x 10"-6
Sh = hm D
Dab
.
Re
= 1875 < 2000
Since Re <2000 , fl ow IS
. laminar.
or/ami nar Internaljlow:
F,
Sherwood Number, Sh = 3.66
Scanned by CamScanner
5.70 Heat and Mass Transfer
We know that,
hmD
Sh=Dab
3.66 =
::::' 4
~
hmxO.012
R e ::::10
Since Re > 2000, flow'
3.66=
hm x 0.012
a"., r~Qltlfe
)( IlK
~5.71
,624.1 ~ 2
IS turbU\
For turbulent, Internal flow
0.62 x 10-5
~
0%
tnt
Sherwood Number (Sh):::
0.023 (R \1\
I:r·&3 ~
where,
Result :
Mass transfer co-efficient,
o
[From HI".
Sc - Schmidt Number::: -!_
hm = 1.89 x 10-3 mls.
St)O.44
'''ll data book
... 0)
,Page 110.1761
Dab
Sc = 15.06 x 10-6
0.75 x 10-5
Air at 20° C and atmospheric pressure, containin
quantities of iodine flows witb a velocity of 4 m/: ? s~
.'
•
s mSldeQ
4cm inner diameter tube. Determine the mass transfrr
co-efficient. Assume Dab = 0.75 x 10-5 m2/s.
I
Sc = 2.0081
Substitute Sc, Re values in equation
. (1)
Given:
Fluid temperature, T CXl = 20°C
(1) => Sh = 0.023 (10,624)0.83 (2.008)0.44
Velocity, U = 4 mls
ISh = 68.661
Diameter, D = 4 em = 0.04 m
Diffusion co-efficient, Dab = 0.75 x 10-5 m2/s
We know that ,
Sherwood Number, Sh = hmD
Toflnd :
Dab
Mass transfer co-efficient,
hm
~
Solution:
Properties
[From HMf data book, page no.Jl]
viscosity, v = 15.06 x 10-6 m2/s
We know that,
Reynolds
number,
Scanned by CamScanner
hm x 0.04
0.75 x 10-5
Mass transfer co-efficient, hm = 0.0128 m1s
of air at 20°C
Kinematic
68.66 =
Re = UD
v
ReSUlt:
Mass transfer co-efficient, hm = 0.0128 mls
where,
5.72 Heat and Moss Transfer
\
SC -
5.29 UNIVERSITY SOLVED PROBLEMS
IIIA'
I tm and 25"C containing small quantities of iOdill
I..!J ir at a
2 _/.'
'../ 35
di
e
'/h a velocity of 6. TW S msiae a mm lameter tuh
floww!
. fi 'd'
t.
Calculate mass transfer co-efficient or 10 me. The therltl{j
Schmidt
NUtn\.Ul:r::::
Sc = 15.5 x 1(}-{l
0.82 x lQ=s
"
1)ill
physical properties of air are
v::: 15,5 x 1(J-fI m1/s
substitute
{May 2004 - Anna Un;v]
D::: 0.82 x 10-5 m1/s
( 1)
-;::J
Given:
Pressure, p ::: 1 atm= 1.013 bar
Fluid temperature,
Sc, Re values in
'\
\
84.07 \
Number, Sh:: ~
Dab
Diameter, D::: 35 mrn ::: 0.035 m
Kinematic viscosity, v= 15.5 x 1~
Tofind:
Mass transfer co-efficient,
I
\ .&9()~.44
Sherwood
Vel~city, U ::: 6.2 mls
84.07
m2/s
0.82 x 10-5 m2/s
Dab:::
\
We know that,
Too::: 25°C
Diffusion co-efficient,
.
t<\uatl()1\ t\)
Sh ::: 0.023 (14,000)0.&3 1\
@h:::
=
hm x 0.035
0.82 x \()-5
Mass transfer co-efficient, hm:: O.()\I}(:, mls
hm
Result :
Solution:
Mass transfer co-efficient, hm = 0.0\96 mls
We know that,
Reyno Ids num b er, Re=
\1] Dry air at 27·C and 1 aIm flows OVfr II IIIf' jla. platt SO CIII
UD
long and velocity of 50 mls. Calcullllt 'ht mass "wltr C~
efficient of water vapour in air aI 'he end of .ht platt. Talt
V
6.2 x 0.035
15.5 x 10-6
IRe:::
D = 0.26 x 1()-4 m2/s.
[Ocl'99· Madras Univ }
14,000 I > 2000
Ans:
Since Re > 2000, flow is turbulent.
For turbulent, Internal flow
Sherwood Number (Sh) = 0.023 (Rep·g3
(Sc) 0.44
•• , (1)
{From HMf data book, page no.176}
Scanned by CamScanner
hm
1
! '
= 0.11 mls
[The procedure of this problems is same as, Section- 5.25
Problem no.2)
5. 74 Heal and Mass Transfer
o.
"c and atmospheric pressure flows wit/I a ~
A" at ~5.
JO mm diameter tube of J m length. The' Iy 0/
3 m/s inside a
.
.
I"S;d,
.r the tube con tams a deposit of naphthal e
surface 0,
.r.'
e"e
. the average mass transfer co-effiCient. Talc .
Determtne
. D = 0. 62 x J(i-5 m1/s.
e /0,
Naphtllalene air,
.
{Apr' 2000 - Madras 1
T
.
I.
•
VIl/\! }
•
S
A n·
h
= 2.27 x J(i-3 m/s.
",
[The procedure· of this problems
Problem no. 1]
is same as, Section - 5.28
r7l Air at 20DCflows part a tray full of water with a veloc;".
L:J
.'J
of
2.5 mls. Calculate the evaporatto» rate of water if th
D
temperature on the water surface is 15 C. The tray measur e
25 em along the flow direction and its width is 40 em;
A plastIC membrane 0 2
pressure of water associated with it is 0.0075 bar. The physical
properties of air are Density = 1.205 kg/m', kinematic Viscosity
15.06 x 1~ m2/s and diffusivity = 0.15 m2/llr.
'
[Oct' 98 - Madras Univ}
of this problems
[Ans . (i) 0003
(ii) 76.5 x 1fH kgll ~ 75ando.OlSk
nro e!s-tn1j 153 x 1 g trloitlntl
gas is maintained at
()-I kgls-nr2/
2.
Hydrogen
ite sid
pressuresof3
the OpposIte Sl. e of a 0 .3 rnm thiIek rubbe bar and \ bar on
entire system IS at 25°C. What' th r membraneand the
h
IS e molard'~
hydrogen .t . rough the membrane? Take0 _ I usion flux of
and solubility of H2 in rubber = 1.5 x 1f\...)AB - 8.7 x \0-3 m21s
v kmoVm)bar,
fAns: 1.75 x Ifr6 kgls-nrZ/
is same as, Section - 5.25
[IJ Dry air at 27DCand 1 atm flows over a wet flat plate 50 cm
long at a velocity of 50 mls. Calculate the mass transfer
co-efflceint of water vapour in air at the end of the plate,
D = 0.26 cm2/s.
[Oct' 2001 - Madras UnivJ
Ans: h", = 0.11 mls.
[The procedure
Problem no.2J
of this problems
l
Scanned by CamScanner
..\1(1,I.'j,
""-tft, j
. 5 rn
.75
of2.5 bar
m thic~h
.
and I bash
diffusIon
co-effieie
aronits Ydrogen
8.5 >< J()-8m2/sandthe~:
~~ hYdr:PPosiless:lllaintaif\ed
2b
Ublll",0 gen·
!'lbe bi
kg mo Ie Imar.
Underth
.'1 fhYdr In the
lnary
eUnlforrn ~enlll_ Plastic is
COnd'
""Ibranc .
(i) The molar concentrations
IhonsOf2St ISO.0015
of the membrane, and
of hYdrogen
,Calculate
atthe0
PPositefaces
(ii) The molar and mass diffu '
Sianflu
membrane.
x ofhYdr
agenthrough the
Ans : mw = 1.846 x lrrs kg/so
[The procedure
Problem no.4J
....11(:£
at pressures
es
The moving air has a total pressure of 1. 01 bar and the partial
P~f'ot,.,
is same as, Section-
5.25
3. Estimate
the diffusion rate of wate fr th
.
r om e bottom f
tube 10mm In diameter and IS em 100 . t dry
0 a test
°C T k h '.
g 10 0 atmosphericair
t
25
a
. 1 a e t e diffuSIon co-effieieot of
bra
"
0.255 x 10-4 m2 Is.
watert ughau- IS
fAns: 1.13 x l~ kglll/
4. Air at 3.5°C and I atm flows over a wet flatplate50emlongwith
a velocity of 30 mls. Calculate the mass transfereo-efficeintof
water vapour in air at the end of the plate.
fAns: 0.075 mis/
_ME
5. 76 Heal and Mass Transfer
Compute the rate of evaporation of~ter vapour. from~
5. ofa flat pan filled with water at 150C mtoa~
air SI:reaJn .~
with a velocity of3m1s parallel to water s~rface. The tern~
of air is 200C and the length of the pan LD the flow diftcti 'IIrt
30 ern while its width is 50 cm. Take the total pressllrt of ~~
.' I
f
lIr}'
1.013 x lOs N/m2 and the partia pressure 0 water vapoUr~~
as 800 N/m2.
q
[Ans ; 0.096.
""I
5.31 TWO MARK QUESTIONS
AND ANSWERS
-------
1. What is mass transfer?
The process of transfer of mass as a result of the spec"
concentration difference in a mixture is known as mass
transter.
2. Give the examples of mass transfer.
~
"",al is Eddy diffll.s'
~~~
When one of the diffi .
•
USIOn t1
diffuSion takes place
uids i .
•
~
o-
",haIlS
convective '"
.
CI.t.s t'Q"~1".
. 7
.
mOtion, eddy
•...
It"
•
7. Slate
,
ru« "S law of diffus;oll.
en they are at
{J"~t·06 ~
.'
.
{Oct'97: 99,~~S &. Dtc'04 A VI
The diffusion rate ISgiven by th'
0 &. Ap,'98 MVI
e FICk's I
molar fl ux 0 f an element per un't
. aw, which state th
I area IS d'
S
at
concentration gradient.
lrectly proponional to
ma =-0
dea
A
ab d;"
fMU-Nov'96,Oct'''1
where,
of air in cooling tower
ma
k
- Molar flux _ g - mole
A
s-m2
2. Evaporation of petrol in the carburertor of an IC engine
3. The transfer of water vapour into dry air
3. Wlia!are the modes of mass transfer?
Urbulent
5 7
Convective mass lransfe .
.
II
""t.20 06,A VI
occur between a surfacer IS a pr<>tessof
and a t1 .
mass tra
different concentrations •
Uld med'
nsfer that w'II
IUm wh
I
Some examples of mass transfer are
I. Humidification
S lilt
•
~
asl "'r~fer
fJune - 2006,AUj
There are basically two modes of mass transfer,
Dab -
Diffusion co-efficient of species a and b
dCa
dx
- Concentration gradient, kgtm3
2/
.m s
1. Diffusion mass transfer
2. Convective mass transfer
Lt('Wllat is molecular diffusion?
8. Whal is free
[June - 2006,A~1
The transport of water on a microscopic level as a result of
diffusion from a region of higher concentration to a regiOll
of
lower concentration in a mixture of liquids or gases is known IS
molecular diffusion.
L
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convective mlUs transftr'!
IOe,'97. MU/
If the fluid m o tiIon ISpr
. od uce d due to change ID'. density resulting
from concentration gradients, the mode of mass transfer is said
to be free or natural convective mass transfer.
Example:
Evaporation
of alcohol.
J
5. 78 Heat and Mass Transfer
9. Defineforced convective mass transfer.
{Apr'97
If'th fluid motion is artificially created by means of
.....(;/
e
an e)(te... .
force like a blower or fan, that type of mass transfer is kno -"Qj
convective mass transfer.
\\'n~
Example: The evaporation
blows over it.
of water from an ocean wh
en a~
10. Define Schmidt Number.
{Apr'97, Oct'97 - Mll/
It is defined as the ratio of the molecular diffusivity of rno
.,
f
lllenluJn
to the molecular diffusivity 0 mass.
(iii) Mass fr .
.
action
(IV) Mole fra
.
ctlon
(i) Mass concelilra .
no" 0, At
Mass of a CQrn
lIss Ii
Pon
ttr!il)l
expressed in kg; lent Pet UIl'
rn .
It vOlu
me Of Ih
Mass concentrati
on ""
(ii) Molar Concent,,...
""0/1
Sc =
Molecular diffusivity of momentum
Molecular diffusivity of mass
11. Define Scherwood Number.
/Apr'97& 2001- MU I {May -2004 -AU,
It is defined as the ratio of concentration
Sh=
Dab -
Diffusion co-efficient,
e Of min..
~'ure
oi"'d
N urn b er of molecul
tl!$il)l
.
es of a
the mixture. It is exp
tOtnPolle
tesSed in k
lit Pet Unit
Molar concentration""
&-rnoleJml Volultle of
'NUtnber 0
.
f tnoles of
Dnit volu
COmponent
tne of mixture
(iii) Mass fraction
mls
The mass fraction is d fi
.
e Ined
concentration of species to the t
as the ratio f
otal mass den'
0 mass
Slty ofthe mixture
Mass fraction = ~Mass concentrationof
.
'I'
a species
lotal mass density
m2/s
x- Length, m
(iii). Mole fraction
12. Give two examples of convective mass transfer:
/ May -2004 -AUf
1. Evaporation
of alcohol
2. Evaporation
of water from an ocean when air blows overit.
IJI.'Dejine thefotlowing.
Or At,
gradients at the boundary,
h",x
Dab
hIn - Mass transfer co-efficient,
\.
e mixture I '
IYJasS
' t IS
D' Of a Corn
nit vOlum
Ilelll
The mole fraction is defined as the
.
concentration of a species to the t tal
raho of mol
o molar concentration.
Mole concentration
= Mole concentrationofa species
Total molar concentration
I Dec -04 & 05 -AUf
(i) Mass Concentration
(ii) Molar Concentration
-
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