3. Steady state heat transfer in a slab

Heat and Mass Transfer
Module 1: Conduction Heat Transfer
Lecture: 02
Topic: Steady State Heat Transfer in Slab
Recap
• General heat conduction equation in Cartesian
coordinate
Learning Outcomes
• Develop equation for one dimensional steady state heat
conduction in slab, composite slab
• Explain the electrical analogy for heat conduction
• Define thermal resistance
• Solve problems in composite slab
2

One Dimensional Steady State Heat Transfer in Slab
Without Internal Heat Generation
t
T
k
q
z
T
y
T
x
T













1
2
2
2
2
2
2
General heat conduction equation for slab
0
dx
d
)(0 2
2
2
2


 T
or
x
T
Governing Equation
Boundary conditions
T = T1 at x = 0
T = T2 at x = L 3

Integrating the equation
21
1
CxCT
C
dx
dT


Substituting the boundary
conditions
L
TT
C
TLCT
CLCT
12
1
112
212




21211 0 CTCCT 
Temperature distribution
2
12
Tx
L
TT
T 


Rate of heat transfer
 
L
TTkA
Q
L
TT
kAkACQ
dx
dT
kAQ
21
12
1
Lat x







 


kA
L
TT
Q 21 

  212 T
L
x
TTT 
4

Electrical Circuit
T2
Q
R
T1
ResistanceElectrical
DifferencePotential21



R
VV
I
Electrical Analogy of heat transfer
ResistanceThermal
DifferenceeTemperatur21



kA
L
TT
Q
Rth
I
V2V1
C/WK/W or
kA
L
R o
th 
Thermal Resistance
5

Composite Slab
321
41
RRR
TT
Q



Intermediate Temperatures
3
43
2
32
1
21
R
TT
R
TT
R
TT
Q






6

Composite slab with convection on both sides
4321
21
RRRRR
TT
Q
o 

 
4
24
3
43
2
32
1
2111
R
TT
R
TT
R
TT
R
TT
R
TT
Q
o











7

Composite Slab with parallel resistance
Equivalent resistances
765567
3223
1111
111
RRRR
RRR


5674231
51
RRRR
TT
Q



567
54
4
43
23
32
1
21
R
TT
R
TT
R
TT
R
TT
Q








8

Reflection Spot
• Write the governing equation for one
dimensional steady state heat transfer in a
slab
• What is thermal resistance?
9
C/WK/W or
kA
L
R o
th 
0
dx
d
2
2

T

Example 1: A furnace wall is made up of three layers of thicknesses 25 cm,
10 cm, and 15 cm with thermal conductivities of 1.65, k and 9.2 W/mK
respectively. The inside is exposed to gases at 1250 oC, with convection
coefficient of 25 W/m2K and the inside surface is at 1100 oC, the outside
surface is exposed to air at 25 oC with convection coefficient of 12
W/m2K. Determine (i) the unknown thermal conductivity (ii) the overall
heat transfer coefficient (iii) all the surface temperatures (AU-May 2012)
ho = 12 W/m2K
To=25 oC
hi = 25 W/m2K
Ti=1250 oC
Given data
k1 = 1.65 W/mK
k2 = k W/mK
k3 = 9.2 W/mK
L1 = 0.25 m
L2 = 0.1 m
L3 = 0.15 m
T1 = 1100 oC
10

Value of thermal resistances
C/W083.0
121
11
C/W0163.0
2.91
15.0
C/W
1.0
1
1.0
C/W1515.0
65.11
25.0
C/W04.0
251
11
o
o
3
3
3
o
2
2
2
o
1
1
1
o















o
o
i
i
Ah
R
Ak
L
R
kkAk
L
R
Ak
L
R
Ah
R
W3750
04.0
110012501





i
i
R
TT
Q
Intermediate temperatures
11
C8.531T
1515.037501100T
o
2
2
112
1
21




 QRTT
R
TT
Q
C4.337T
0833.0375025T
o
4
4
4
4




 oo
o
o
QRTT
R
TT
Q
C5.398T
0163.037504.337T
o
3
3
343
3
43




 QRTT
R
TT
Q

12
Unknown thermal
conductivity
W/mK82.2
0355.0
1.0
0355.0
3750
5.3988.5311.0 32
2






k
Q
TT
k
R
Overall heat transfer Coefficient
KW/m06.3
3263.0
11
3263.0
3263.0083.00163.00355.01515.004.0
2
321




RA
U
RA
RRRRRR oi
Answer
i) The unknown thermal conductivity k2 = 2.82 W/mK
ii) The overall heat transfer coefficient U = 3.06 W/m2K
iii) Surface temperatures: T2 = 531.8 oC, T3 = 398.5 oC,
T4 = 337.4 oC

Example 2: A furnace wall consists of three layers. The inner layer of 10
cm thickness is made of fire brick (k=1.04 W/mK). The intermediate layer
of 25 cm thickness is made of masonry brick (k=0.69 W/mK) followed by
a 5 cm thick concrete wall (k=1.37 W/mK). When the furnace is in
continuous operation the inner surface of the furnace is at 800oC while
the outer concrete surface is at 50 oC. Calculate the rate of heat loss per
unit area of the wall, the temperature at the interface of the firebrick and
masonry brick and the temperature at the interface of the masonry brick
and concrete. (May 2006, Nov 2012)
13
Given data
k1 = 1.04 W/mK
k2 = 0.69 W/mK
k3 = 1.37 W/mK
L1 = 0.1 m
L2 = 0.25m
L3 = 0.05 m
T1 = 800 oC, T4 = 50 oC

C/W0365.0
37.11
05.0
C/W362.0
69.01
25.0
C/W096.0
04.11
1.0
o
3
3
3
o
2
2
2
o
1
1
1









Ak
L
R
Ak
L
R
Ak
L
R
W7.1516
0365.0362.0096.0
50800
321
41







RRR
TT
Q
14
C4.654T
096.07.1516800T
o
2
2
112
1
21




 QRTT
R
TT
Q
C36.105T
0365.07.151650T
o
3
3
343
3
43




 QRTT
R
TT
Q
Answer
i) Rate of heat transfer = 1516.7 W
ii) Surface temperatures: T2 = 654.4 oC, T3 = 105.36 oC

15
Reference Books
1. Y. A. Cengel - Heat transfer A Practical
approach
2. Incropera and Dewitt - Fundamental of Heat
and Mass Transfer
3. Osizik - Heat transfer A Basic approach
4. A. Bejan - Heat Transfer
5. K. Kannan - Heat and Mass Transfer

Features of the book
 Lucid explanation of
subject content
 More solved problems
from Anna University
Question Papers
 Two mark questions with
answers
Publisher
Anuradha Publications
Chennai
16

Dr. K. Kannan
Professor Mechanical Engineering
Anjalai Ammal Mahalingam Engineering College
Kovilvenni – Tamil Nadu
k.kannan@aamec.edu.in
17

Heat and Mass Transfer
Module 1: Conduction Heat Transfer
Lecture: 03
Topic: Problems in composite slab
Recap
• Heat transfer in composite slab
Learning Outcomes
• Solve problems in composite slab
19

Example 3: Determine the steady state heat transfer through a double
pane window, 0.8 m high, 1.5 m wide consisting of two 4 mm thick
glass layers (k=0.78 W/mK) separated by a 10 mm thick stagnant layer
of air (k=0.26 W/mK). Inside temperature of room air is maintained at
20 oC with a convective heat transfer coefficient of 10 W/m2K, outside
air temperature is -10 oC and convective heat transfer coefficient on
the outside is 40 W/m2K. Also determine the overall heat transfer
coefficient. (AU-Nov 2010)
20
ho = 40 W/m2K
To= - 10 oC
hi = 10 W/m2K
Ti=20 oC
Given data
k1 = k3 = 0.78 W/mK
k2 = 0.26 W/mK
L1 = L3 = 0.004 m
L2 = 0.01 m
A = 0.8 x 1.5 = 1.2 m2

C/W0208.0
402.1
11
C/W00427.0
78.02.1
004.0
C/W032.0
26.02.1
01.0
C/W00427.0
78.02.1
004.0
C/W083.0
102.1
11
o
o
3
3
3
o
2
2
2
o
1
1
1
o















o
o
i
i
Ah
R
Ak
L
R
Ak
L
R
Ak
L
R
Ah
R
21
14434.0
0208.000427.0032.000427.0083.0
321



R
R
RRRRRR oi
Total resistance

22
W84.207
14434.0
1020





R
TT
Q oi
Overall heat transfer coefficient
KW/m773.5
2.114434.0
11 2



RA
U
Answer
i. Rate of heat transfer = 207.84 W
ii. Overall heat transfer coefficient = 5.773 W/m2K

Example 4: The wall of an oven consists of 3 layers of brick. Inside one is
built of 20 cm of fire bricks surrounded by 10 cm of insulating brick and
outside layer is binding bricks of 12 cm thick. The oven operates at 900oC,
such that the outside surface of the oven is maintained at 60oC.
Calculate the heat loss per m2 surface area, the interfacial temperature.
Given the thermal conductivity of fire brick, insulating brick and binding
brick are 1.2, 0.26 and 0.68 respectively in W/mK (AU-Nov. 2009)
23
Given data
k1 = 1.2 W/mK
k2 = 0.26 W/mK
k3 = 0.68 W/mK
L1 = 0.2 m
L2 = 0.1 m
L3 = 0.12 m
T1 = 900 oC, T4 = 60 oC

C/W176.0
68.01
12.0
C/W625.0
16.01
1.0
C/W167.0
2.11
2.0
o
3
3
3
o
2
2
2
o
1
1
1









Ak
L
R
Ak
L
R
Ak
L
R
W77.867
176.0625.0167.0
60900
321
41







RRR
TT
Q
24
C1.755T
167.077.867900T
o
2
2
112
1
21




 QRTT
R
TT
Q
C7.212T
176.077.86760T
o
3
3
343
3
43




 QRTT
R
TT
Q
Answer
i) Rate of heat transfer = 867.77 W
ii) Surface temperatures: T2 = 755.1 oC, T3 = 212.7 oC

Example 5: A composite wall consists of 10 cm thick layer of brick k=0.7
W/mK and 3 cm thick plaster k=0.5 W/mK. An insulating material of
k=0.08 W/mK is to be added to reduce the heat transfer through the wall
by 40%. Find its thickness. (Nov. 2005, Nov. 2004, Nov. 2009)
25
Given data
k1 = 0.7 W/mK
k2 = 0.5 W/mK
k3 = 0.08 W/mK
L1 = 0.1 m
L2 = 0.03 m
2
2
1
1
k
L
k
L
T
A
Q



3
3
2
2
1
1
'
k
L
k
L
k
L
T
A
Q



A
Q
A
Q
 4.0
'

26
cm2.44m0244.0
305.006.0143.0508.0
08.0
08.0
06.0143.0
97.1
1
97.1
06.0143.0
4.0
08.0
06.0143.0
1
5.0
03.0
7.0
1.0
1
4.0
08.05.0
03.0
7.0
1.0
1
4.0
3
3
3
3
3
2
2
1
1
3
3
2
2
1
1















L
L
L
L
L
k
L
k
L
T
k
L
k
L
k
L
T
Answer:
Thickness of insulation L3 = 2.44 cm

Example 6: A composite wall of thermal insulation has a rectangular
section 2 x 0.5 m and is made from timber 0.15 m, card board 0.3 m and
steel 0.05 m thick. The temperature at outside surface of timber and
steel are 25 oC and 150 oC. How the heat transfer rate would be affected
if aluminium rod of 4 cm diameter were inserted through each square
metre of wall. Thermal conductivity of timber 0.12, steel 45 card board
0.035 and aluminium 205 W/mK.
27
Given data
k1 = 0.12 W/mK
k2 = 0.035 W/mK
k3 = 45 W/mK
k4 = 205 W/mK
L1 = 0.15 m
L2 = 0.3 m
L3 = 0.05 m
T1 = 25 oC, T4 = 150 oC
Area of composite slab A1 = A2 = A3 = A = 2 x 0.5 = 1 m2

C/W1011.1
451
05.0
C/W57.8
035.01
3.0
C/W25.1
12.01
15.0
o3
3
3
3
o
2
2
2
o
1
1
1










Ak
L
R
Ak
L
R
Ak
L
R
without aluminium rod
W72.12
1011.157.825.1
25150
3
321
41








RRR
TT
Q
28
C/W1.94
20504.0
4
05.03.015.0
R
4
o
2
4
4
2
321
4
4
4









kd
LLL
kA
L
R
cy
Equivalent Resistance due to
aluminium rod
621.1617.0
R
1
94.1
1
1011.157.825.1
1
R
1
11
R
1
eq
3
eq
4321eq








eqR
RRRR

29
Answer
i) Rate of heat transfer without aluminium rod = 12.72 W
ii) Rate of heat transfer with aluminium rod = 77.11 W
iii) Percentage increase in heat transfer = 83.5 %
Rate of heat transfer with aluminium rod
W11.77
621.1
25150
' 41





eqR
TT
Q
Increase in heat transfer
%5.83100
11.77
72.1211.77
100
'
'





Q
QQ

30
Example 7: To defrost ice accumulated on the outer surface of a car
windshield, warm air is blown over the inner surface of the windshield.
Consider windshield thickness is 5 mm and its thermal conductivity is
1.4 W/mK. The outside ambient temperature is – 10 oC and the
convection heat transfer coefficient is 200 W/m2K, while the ambient
temperature inside the car is 25 oC. Determine the value of the
convection heat transfer coefficient for the warm air blowing over the
inner surface of the windshield necessary to cause the accumulated ice to
begin melting (AU-May 2019)
Outside
T∞1=– 10 oC
h1=200 W/m2K
Inside
T∞2=25oC
h2 W/m2K
Wind shield k = 1.4,
L = 0.005 m
2
2221
1
11
11
h
TT
k
L
TT
h
TT  




T1 = 0 oC for melting of ice

31
2
22
1
25
4.1
005.0
0
200
1
010
h
TT 




143.7
280
2000
2802000
4.1
005.0
0
200
1
010
2
2
2





T
T
T
KW/m112
86.17
2000
86.172000
1
25143.7
4.1
005.0
143.70
1
25
4.1
005.0
0
2
2
2
2
2
22








h
h
h
h
TT
Answer
Heat transfer coefficient inside = 112 W/m2K
Taking first two terms
Taking last two terms

32
Reference Books
1. Y. A. Cengel - Heat transfer A Practical
approach
2. Incropera and Dewitt - Fundamental of Heat
and Mass Transfer
3. Osizik - Heat transfer A Basic approach
4. A. Bejan - Heat Transfer
5. K. Kannan - Heat and Mass Transfer

Features of the book
 Lucid explanation of
subject content
 More solved problems
from Anna University
Question Papers
 Two mark questions with
answers
Publisher
Anuradha Publications
Chennai
33

Dr. K. Kannan
Professor Mechanical Engineering
Anjalai Ammal Mahalingam Engineering College
Kovilvenni – Tamil Nadu
k.kannan@aamec.edu.in
34

3. Steady state heat transfer in a slab

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