Tutorial Session for Unit – I

Conduction Heat Transfer

Presented by
V.Sivaramkumar
AP(Sr.Gr)/Mech
KLNCE
One Dimensional Steady State Heat
Conduction Through Plane walls,
Cylinders and Spheres
Formulas Used
V
We Know that Ohms Law I
R
dT
Compare to Ohms Law We can get Heat flow Equation Q
R
L
Thermal Resistance for Walls R
kA

1 r
Thermal Resistance for Cylinders R ln 2
2kL r1

1 1 1
R   
Thermal Resistance for Spheres 4k  r1 r2 

1
Convection Thermal Resistance R
hA
Problems in Heat Conduction Through walls
Solved Problem: 1

The wall of a room is composed of three layers .The outer layer

is brick 20cm thick, the middle layer is sand 10cm thick, the
inside layer is cement 5cm thick. The temperature of the
outside air is 250c and inside air is -200c. The heat transfer
Coefficient for outside air and brick is 45.4w/m2K and for
inside air and cement is 17 w/m2K. Assume thermal
conductivity for brick is 3.45w/mK, for sand is 0.043w/mK and
for cement is 0.294w/mK . Determine the Total thermal
resistance, heat flow rate and Interface temperatures.
Solution:
Rtot=R1+R2+R3+R4+R5.
1 L L L 1
Rtot   1  2  3 
h o A k 1A k 2 A k 3 A h i A

1 0.2 0.1 0.05 1

Rtot     
45.4 1 3.45 1 0.043 1 0.294 1 17 1

K  m2
Rtot  2.6344
w

dT 298  253
Q 
Rtot 2.6344

w
Q  17.08
m2
 TO  T1 TO  T1 298  T 1
Q  
R1  1   1 
   
 ho A   45 .4  1 

298  T1
17.08 
0.02203

T1  297.62K

T1  T2 297.62  T2 297.62  T2
Q  
R2  L1   0.2 
   
 k1 A   3.45  1 

297.62  T2
17.08 
0.05797

T2  296.63K
 T2  T3 T2  T3 298  T1
Q  
R3  L2   0.1 
   
 k2 A   0.043  1 

296.63  T3
17.08 
2.3256

T3  256.9 K

T3  T4 T3  T4 256.9  T4
Q  
R4  L3   0.05 
   
 k3 A   0.294  1 

256.9  T4
17.08 
0.17

T4  253.9 K
Exercise Problem: 1

A furnace wall is made up of three layers, inside layer with

thermal conductivity 8.5w/mK, the middle layer with
conductivity 0.25w/mK, the outer layer with conductivity
0.08w/mK. The respective thickness of the inner , middle and
outer layers are 25cm, 5cm and 3cm respectively. The inside
and outside wall temperatures are 6000c and 500c
respectively. Determine the Total thermal resistance, heat
flow rate and interface temperatures.
Ans:
2
Rtot=0.604 K  m T2=846.23K
w

w
Q=909.97 m2
T3=664.23K
Exercise Problem: 2
A wall is constructed of several layers. The first layer consist of sand
20cm thick of thermal conductivity 0.66w/mK, the second layer consist
of 3cm thick cement of thermal conductivity 0.6w/mK, the third layer
consist of 8cm thick stone of thermal conductivity 0.58w/mK and outer
layer consist of 1.2cm thick brick of thermal conductivity 0.6w/mK. The
heat transfer coefficients on the interior and outside of the wall are
5.6w/m2K and 11w/m2K respectively. Interior room temperature is 220c
and outside air temperature is
-50c. Calculate Overall thermal resistance, rate of heat transfer,
Temperature at the junction between cement and stone and overall
heat transfer coefficient.

Ans: K  m2
Rtot=0.78 w T3=276.5K
w w
Q=34.56 m 2
U=1.28 m2 K
Problems in Heat Conduction Through Cylinders
Solved Problem: 1

A steel pipe of 120mm inner diameter, 140mm outer

diameter with thermal conductivity 55w/mK is covered with
two layers of insulation each having a thickness of 55mm. The
thermal conductivity of the first insulation material is
0.05w/mK and that of second is 0.11w/mK. The temperature
of inside tube surface is 2400c and that of outside surface of
the insulation is 600c. Calculate the heat loss, interface
temperature between two layers of insulation and Overall
heat transfer coefficient.
 dT
Q
Rtot
513  333
Q
Rtot
Rtot  R1  R2  R3
1 r  1 r  1 r 
Rtot  ln 2   ln 3   ln 4 
2k1 L  r1  2k 2 L  r2  2k3 L  r3 

1  0.07  1  0.125  1  0.18 

Rtot  ln  ln  ln 
2  55 1  0.06  2  0.05 1  0.07  2  0.11  1  0.125 

K m
Rtot  2.4236
w

513  333
Q
2.4236
w
Q  75.83
m
 T1  T2 513  T2 513  T2
Q  
R1 1 r  1  0.07 
ln 2  ln 
2k1 L  r1  2  55  1  0 . 06 

513  T2
75.83 
4.4607 10  4

T2  512.97 K

T2  T3 512.97  T3 512.97  T3
Q  
R2 1  r3  1  0.125 

ln   ln  
2k 2 L  r2  2  0. 05  1  0 . 07 

512.97  T3
75.83 
1.8456

T3  372.7 K
Q  UAdT Overall

75.83  U  2r4l  dT Overall

75.83  U  2    0.18 1 513  333

w
U  0.372
m2 K
Exercise Problem: 1

A steel pipe of 170mm inner diameter and 190mm outer diameter

with thermal conductivity 55w/mK is covered with two layers of
insulation. The thickness of first insulation is 25mm(k=0.1w/mk)
and the second layer thickness is 40mm(k=0.18w/mk). The
temperature of the inner fluid is 3200c and ambient air
temperature is 250c. The heat transfer coefficients for inside and
outside surfaces are 230w/m2K and 6w/m2K respectively.
Determine the heat loss, interface temperature at the junction
between two insulations (assume inner fluid temperature is equal
to inner surface temperature) and Overall heat transfer coefficient.
Ans:
w
Q= 368.5 T3=455.88K
m
w
U=1.24 m2 K
Exercise Problem: 2

An insulated steel pipe carrying a hot liquid. Inner diameter of

the pipe is 25cm, wall thickness is 2cm, thickness of insulation
is 5cm, temperature of hot liquid is 1000c, temperature of
surrounding is 200c, inside heat transfer coefficient is
730w/m2K and outside heat transfer coefficient is 12w/m2K.
Calculate the heat loss per meter length of pipe. Take
ksteel=55w/mk, kinsulating material=0.22w/mK.
Ans:
Q=281.18 w/m
Problems in Heat Conduction Through Spheres

Solved Problem: 1

A hollow sphere (k=65w/mk) of 120mm inner diameter and

350mm outer diameter is covered with 10mm layer of
insulation sphere (k=10w/mk). The inside and outside Surface
temperatures are 5000c and 500c respectively. Calculate the
rate of heat flow, overall heat transfer Coefficient and
interface temperatures.
 dT
Q
Rtot
773  323
Q
Rtot
Rtot  R1  R2
1 1 1 1 1 1
Rtot        
4k1  r1 r2  4k 2  r2 r3 

1  1 1  1  1 1 
Rtot       
4  65  0.06 0.175  4  10  0.175 0.185 

K
Rtot  0.01585
w

773  323
Q
0.01585

Q  28361w
Q  UAdT Overall

 
28361  U  4r3  dT Overall
2

 
28361  U  4    0.1852  773  323

w
U  146.54
m2 K

T1  T2 773  T2 773  T2
Q  
R1 1 1 1 1  1 1 
     
4k1  r1 r2  4  65  0 . 06 0 . 175 

773  T2
28361 
0.0134

T2  392.7 K
Exercise Problem: 1

A steel sphere (k=43.26w/mK) of 5.08cm inner diameter and

7.62cm outer diameter is covered with 2.5cm layer of
insulation sphere (k=0.208w/mK), the inside fluid
temperature is 3160c with heat transfer coefficient 28w/m2K.
While the outer surface exposed to the ambient air at 300c
with heat transfer coefficient 17w/m2K. Calculate heat loss
and interface temperatures.
Ans:
Q= 29.8433 w T2=456.8K

T1=457.5K T3=338.08K
Fins or Extended surfaces
Problems in Fins
Solved Problem: 1

A long fin 5cm diameter its base is connected to a furnace

wall at 1500c, Surrounding air temperature is 200c. The
temperature of the fin at a distance 20cm apart from its base
is 600c. The conductivity of the fin material is 200w/mK.
Determine Convective heat transfer coefficient and heat
transfer rate.
Tx  T
 e  mx
Tb  T

333  293
 e  m0.2
423  293

 333  293 
ln   ln e 
 m0.2

 423  293 

 1.17865   m  0.2

m  5.8932 m 1

hP
 5.8932
kA

P  2r  2    0.025  0.157 m

A  r 2    0.025 2  1.963  10 3 m 2
 h  0.157
3
 5.8932
200  1.963 10

w
h  86.85
m2 K

Q  (Tb  T )  (hPkA) 0.5

Q  (423  293)  (86.85  0.157  200 1.963  10 3 ) 0.5

Q  300.8w
Exercise Problem: 1

An aluminium alloy rectangular fin of 7mm thick 50mm side

protrudes from a wall, which is maintained at 1200c. The
ambient air temperature is 220c. The heat transfer coefficient
and conductivity of the fin material are 140w/m2K and
55w/mK respectively. Assume length of the fin is 50mm.
Determine the temperature at the end of the fin, temperature
at the middle of the fin and heat transfer rate.
Ans:
TL=338.98K TL/2=350.88K

Q=48.54 w
Exercise Problem: 2

A square aluminium fins of 0.5mm side and 12mm long are

attached to a plane wall which is maintained at 800c.
Surrounding air temperature is 220c. Calculate the number of
fins required to generate 35*10-3w of heat. Take conductivity
of the fin material is 165w/mK and heat transfer coefficient is
10w/m2k. Assume no heat loss from the tip of the fin.
Ans:
N=2.57≈3 fins required.
Unsteady State Heat Conduction
Problems in Unsteady State Heat Conduction

Solved Problem: 1

A copper plate (30*30 cm), 2mm thick is heated up to 4000C

and then immersed into water at 300C. Find the time required
for the plate to reach the temperature of 500C. Heat transfer
Coefficient is 100 w/m2K.
 hL
Bi 
k

V l bt t
L  SignificantLength   
As 2  l  b 2

0.002
L  0.001m
2

100  0.001
Bi   2.5  10  4 0.1
386

 hAs 
T  T   cV  
 e 
TO  T

 h 
T  T    
 cL 
e
TO  T

323  303 

100 
 
e  3830.0018954 

673  303
0.05405  e 0.02916

ln 0.05405  ln(e 0.02916 )

 2.9178  0.02916

  100.1s
Solved Problem: 2

A long steel cylinder 12cm in diameter and initially at a

temperature of 200c is placed into a furnace 8200c with the
local heat transfer coefficient 140w/m2K. Calculate the time
required for the axis temperature to reach 8000c. Take
thermal diffusivity for steel is 6.11*10-6m2/s and thermal
conductivity for steel is 21w/mK.
 hL
Bi 
k

V r 2 L r 0.06
L     0.03m
As 2rL 2 2

140  0.03
Bi   0.2  0.1
21

we can' t use Lumped analysis

from Heislers chart

To  T 1073  1093
yaxis    0.025
Ti  T 293  1093

hR0 140  0.06

Curve    0.4
k 21
Project y axis value horizontaly, it intersect curve value at one point
project intersection point verticaly

coresponding x axis value is 2
 5.2
R0

6.11 10-6 
 5.2
0.6 2

  3064 s  51 min
Exercise Problem: 1

An aluminium sphere weighing 5.5kg and initially at a

temperature of 2900c is suddenly immersed in a fluid at 150c.
The convective heat transfer coefficient is 58w/m2K. Estimate
the time required to cool the aluminium to 950c.
Ans:
Ʈ= 1357 sec.
Exercise Problem: 2

A aluminium slab of 5cm thick initially at a temperature of

4000c. It is suddenly immersed in a water at 900c. Calculate
the mid plane temperature after 1 minute. Take heat transfer
coefficient is 1800W/m2k.
Ans:
To= 421.9K.