Chapter 7
Heat Transfer by Conduction
Theory
As with all transport phenomena, the rate of the transferred quantity is propor-
tional to the driving force and inversely proportional to the resistance. For heat
transfer by conduction, the driving force is the temperature difference T and
the resistance R ¼ x=kA, where x is the wall thickness, k is the thermal
conductivity and A is the surface area perpendicular to the direction of transfer.
For a cylindrical wall, A is equal to the logarithmic mean surface area, while for
a spherical wall, A is equal to the geometric mean surface area. Thus:
Rate of heat transfer
For a single wall T
q¼
R
For a composite wall T
q¼P
R
Resistance to heat transfer
Resistance, R Surface area, A
Plane wall x=kA A
A1 A2
Cylindrical wall r=kALM ALM ¼
ln A1 =A2
pffiffiffiffiffiffiffiffiffiffiffiffi
Spherical wall r=kAG AG ¼ A1 A2
Review Questions
Which of the following statements are true and which are false?
1. Heat is conducted in solids, liquids, and gases by the transfer of the energy of
motion from one more energetic molecule to an adjacent less energetic one.
S. Yanniotis, Solving Problems in Food Engineering. 55
Ó Springer 2008
56 7 Heat Transfer by Conduction
2. Fourier’s law is the basic relationship for heat transfer by conduction.
3. Resistance to heat transfer is proportional to thermal conductivity.
4. Air has low thermal conductivity.
5. Metals have higher thermal conductivity than non-metals.
6. Ice has a thermal conductivity much higher than water.
7. The thermal conductivity of gases is higher than the thermal conductivity of
solids.
8. Thermal conductivity is a weak function of temperature.
9. In all cases, thermal conductivity varies with temperature gradient.
10. At steady state, the rate of heat transfer is always zero.
11. At steady state, the temperature at various points in a system does not
change with time
12. At steady state, the temperature at various points in a system may change
with position.
13. The temperature gradient is positive.
14. For the same heat transfer rate, the slope of the temperature gradientin
insulating materials is smaller than in non-insulating materials.
15. The temperature distribution in a plane wall varies linearly with distance in
the wall if there is no heat generation in the wall and the thermal conduc-
tivity is constant.
16. The temperature distribution in a cylindrical wall varies logarithmically
with the distance in the wall if there is no heat generation in the wall and the
thermal conductivity is constant.
17. The arithmetic mean area differs from the logarithmic mean area by less
than 1.4% if A2/A1 < 1.5.
18. In a composite wall at steady state, the heat transfer rate in each layer
depends on the thermal conductivity of the layer.
19. The temperature drop in a plane wall is inversely proportional to the
resistance.
20. The slope of the temperature gradient in each layer of a composite plane
wall depends on the thermal conductivity of the layer.
Examples
Example 7.1
Calculate the rate of heat transfer through a glass window with 3 m2 surface
area and 5 mm thickness if the temperature on the two sides of the glass is 14 8C
and 15 8C respectively and the thermal conductivity of the glass is 0.7 W/m 8C.
The system is at steady state.
Solution
Examples 57
Step 1
Draw the process diagram:
T1
T2
Δx
Step 2
Calculate the resistance of the glass to heat transfer:
x 0:005 m
R¼ ¼ ¼ 0:00238 8C=W
kA ð0:7 W=m8CÞð3 m2 Þ
Step 3
Calculate the rate of heat transferred:
T1 T2 15 14 8C
q¼ ¼ ¼ 420 W
R 0:00238 8C=W
Example 7.2
Hot water is transferred through a stainless steel pipe of 0.04 m inside diameter
and 5 m length. The inside wall temperature is 90 8C, the outside surface
temperature is 88 8C, the thermal conductivity of stainless steel is 16 W/m 8C,
58 7 Heat Transfer by Conduction
and the wall thickness is 2 mm. Calculate the heat losses if the system is at
steady state.
Solution
Step 1
Draw the process diagram:
r2
T2 r1
T1
A1 A2
Step 2
Calculate the logarithmic mean area of the wall:
i) A1 ¼ 2p r1 L ¼ 2pð0:02 mÞð5 mÞ ¼ 0:6283 m2
ii) A2 ¼ 2p r2 L ¼ 2pð0:022 mÞð5 mÞ ¼ 0:6912 m2
0:6283 0:6912
iii) ALM ¼ A1 A2 ¼ ¼ 0:6592 m2
ln A 1 ln 0:6283
A2 0:6912
Step 3
Calculate the resistance of the metal wall to heat transfer:
r2 r1 0:002 m
R¼ ¼ ¼ 0:00019 8C=W
km ALM ð16 W=m 8CÞ 0:6592m2
Step 4
Calculate the rate of heat transfer:
T T1 T2 90 88 8C
q¼ ¼ ¼ ¼ 10526 W
R R 0:00019 8C=W
Example 7.3
The wall of an oven consists of two metal sheets with insulation in between. The
temperature of the inner wall surface is 200 8C and that of the outer surface is
50 8C. The thickness of each metal sheet is 2 mm, the thickness of the insulation
Examples 59
is 5 cm, and the thermal conductivity is 16 W/m 8C and 0.055 W/m 8C respec-
tively. Calculate the total resistance of the wall to heat transfer and the heat
transfer losses through the wall per m2 of wall area.
Solution
Step 1
Draw the process diagram:
T1 T2
T4
T3
Δx1 Δx2 Δx3
Step 2
State your assumptions.
The system is at steady state.
Step 3
Calculate the resistance to heat transfer:
i) Inner metal wall
x1 0:002 m
R1 ¼ ¼ ¼ 0:00013 8C=W
k1 A ð16 W=m8CÞð1 m2 Þ
ii) Insulation:
x2 0:05 m
R2 ¼ ¼ ¼ 0:90909 8C=W
k2 A ð0:055 W=m8CÞð1 m2 Þ
iii) Outer metal wall:
x3 0:002 m
R3 ¼ ¼ ¼ 0:00013 8C=W
k3 A ð16 W=m8CÞð1 m2 Þ
60 7 Heat Transfer by Conduction
iv) Total resistance:
R ¼ R1 þ R2 þ R3 ¼ 0:00013 þ 0:90909 þ 0:00013 ¼ 0:909358C=W
Step 4
Calculate the heat transfer through the wall:
T T1 T4 200 50 8C
q¼P ¼ P ¼ ¼ 165 W
R R 0:90935 8C=W
Comments:
1) The main resistance to heat transfer (99.97%) is in the insulation layer.
2) The slope of the temperature gradient is steeper in the layer where the
resistance is higher.
Exercises
Exercise 7.1
If an insulation of 2 cm thickness with thermal conductivity equal to 0.02 W/
m 8C is wrapped around the pipe of Example 7.2 so that the outside surface
temperature of the insulation is 35 8C, while the inside wall temperature is still
90 8C, what would be the heat loss? What will be the outside surface metal wall
temperature T2?
Solution
Step 1
Draw the process diagram:
T3
r2
T2 r1
r3
T1
A1 A2 A3
Exercises 61
Step 2
Calculate the logarithmic mean area of the insulation:
iÞ A3 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m2
iiÞ ALMi ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m2
Step 3
Calculate the resistance of the insulation layer to heat transfer:
R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W
Step 4
Calculate the total resistance:
X
R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::
Step 5
Calculate the rate of heat transfer:
T
q ¼ P ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: W
R
Step 6
Calculate temperature T2.
The temperature drop is proportional to the resistance:
T1 T2 Rw
¼P
T1 T3 R
where Rw is the metal wall resistance.
Therefore,
T2 ¼::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C
Exercise 7.2
The wall of a refrigerator of 4 m2 surface area consists of two metal sheets with
insulation in between. The temperature of the inner wall surface is 5 8C and that
62 7 Heat Transfer by Conduction
of the outer surface is 20 8C. The thermal conductivity of the metal wall is 16 W/
m 8C and that of the insulation is 0.017 W/m 8C. If the thickness of each metal
sheet is 2 mm, calculate the thickness of the insulation that is required so that
the heat transferred to the refrigerator through the wall is 10 W/m2.
Solution
Step 1
Draw the process diagram:
T1 T2
T3 T4
Δx1 Δx2 Δx3
Step 2
State your assumptions
....................................................................................................................
Step 3
Calculate the resistance to heat transfer:
i) Inner metal wall:
R1 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W
ii) Insulation:
R2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W
iii) Outer metal wall:
R3 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W
iv) Total resistance:
X
R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=W
Exercises 63
Step 4
Calculate the thickness of insulation:Since
X T ::::::::::::::::::::::::::::::::::::::::::::::::::::::
R¼ ¼ 8C=W
q :::::::::::::::::::::::::::::::::::::::::::::::::::::::
the thickness of the insulation will be
x2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m
Exercise 7.3
A layer of fat 5 mm thick underneath the skin covers a part of a human body. If
the temperature of the inner surface of the fat layer is 36.6 8C and the body loses
heat at a rate of 200 W/m2, what will be the temperature at the surface of the
skin? Assume that the thermal conductivity of fat is 0.2 W/m 8C.
Solution
Step 1
State your assumptions:
..................................................................................................................
Step 2
Calculate the resistance to heat transfer:
R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=W
Step 3
Calculate the surface temperature:
Since
T
q¼
R
the surface temperature of the skin will be
T2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C
Exercise 7.4
A composite plane wall consists of two layers A and B. The thermal conductiv-
ity of layers A and B are 0.02 W/m 8C and 15 W/m 8C respectively. If 100 W/m2
are transferred through the wall at steady state, calculate the temperature
gradient in the two layers.
64 7 Heat Transfer by Conduction
Solution
Step 1
State your assumptions:
......................................................................................................................
Step 2
From Fourier’s law:
dT
¼ :::::::::::::::::::::::::::::::::::::::::::::::::
dx
Therefore,
dT
¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m
dx A
and
dT
¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m
dx B
Exercise 7.5
Find an analytical expression to calculate the heat flux in a plane wall if the
thermal conductivity varies with temperature according to the equation
k=koþaT.
Solution
Step 1
Use the expression for k in Fourier’s law:
q dT dT
¼ k ¼ ðko þaTÞ
A dx dx
Step 2
Separate the variables and integrate from x1 to x2 and T1 to T2:
Exercise 7.6
Develop a spreadsheet program to plot the temperature distribution in a plane
wall, a cylindrical wall, and a spherical wall of 4 cm thickness if the outer surface
of the wall is at 20 8C and the inner surface is at 200 8C. The inner wall radius for
the cylinder and the sphere is 0.1 m, the outer wall radius is 0.14 m, and the
thermal conductivity of the wall is 0.02 W/m 8C. Assume steady state.
Exercises 65
Hint:
The relationships that give the temperature variation as a function of distance in
a wall are:
i) Plane wall:
x x1 x
T ¼ T1 q ¼ T1 T
kA x2 x1
ii) Cylindrical wall:
lnðr=r1 Þ lnðr=r1 Þ
T ¼ T1 q ¼ T1 T
2pkL lnðr2 =r1 Þ
iii) Spherical wall:
1 1 1 rr1
T ¼ T1 r1 r q ¼ T1 T
4pk 1 rr1 2
Exercise 7.7
Develop a spreadsheet program to plot the temperature distribution in a plane
wall of 4 cm thickness and inner wall surface temperature of 200 8C, at steady
state, when a) the thermal conductivity of the wall in W/m 8C varies according
to the equation k = 0.0325þ0.0004T and 250 W/m2 of heat are transferred
through the wall; and b) the thermal conductivity is constant and equal to the
average value for the temperature range of the wall calculated in case a.
Calculate the heat transfer rate in case b and compare it to the heat transfer
rate of case a.
Hint:
Calculate the temperature as a function of distance in the wall using the
expression developed in Exercise 7.5.
