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 wallq ¼ �T

R

For a composite wallq ¼ �T

P

R

Resistance to heat transfer

Resistance, R Surface area, A

Plane wall �x=kA A

Cylindrical wall �r=kALM ALM ¼ A1 ÿA2

ln A1=A2

Spherical wall �r=kAG AG ¼ffiffiffiffiffiffiffiffiffiffiffiffi

A1A2

p

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.

Ó Springer 2008

55

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

56 7 Heat Transfer by Conduction

Step 1

Draw the process diagram:

T1

T2

∆x

Step 2

Calculate the resistance of the glass to heat transfer:

R ¼ �x

kA¼ 0:005 m

0:7 W=m8Cð Þ 3 m2ð Þ ¼ 0:00238 8C=W

Step 3

Calculate the rate of heat transferred:

q ¼ T1 ÿ T2

R¼ 15ÿ 14 8C

0:00238 8C=W¼ 420 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,

Examples 57

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:

r2r1

A1 A2

T1

T2

Step 2

Calculate the logarithmic mean area of the wall:

i) A1 ¼ 2p r1L ¼ 2p 0:02mð Þ 5mð Þ ¼ 0:6283 m2

ii) A2 ¼ 2p r2L ¼ 2p 0:022mð Þ 5mð Þ ¼ 0:6912 m2

iii) ALM ¼ A1 ÿA2

lnA1

A2

¼ 0:6283ÿ 0:6912

ln 0:62830:6912

¼ 0:6592 m2

Step 3

Calculate the resistance of the metal wall to heat transfer:

R ¼ r2 ÿ r1

kmALM

¼ 0:002m

16W=m 8Cð Þ 0:6592m2ÿ � ¼ 0:00019 8C=W

Step 4

Calculate the rate of heat transfer:

q ¼ �T

R¼ T1 ÿ T2

R¼ 90ÿ 88 8C

0:00019 8C=W¼ 10526 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

58 7 Heat Transfer by Conduction

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

T3

T4

∆x3∆x2∆x1

Step 2

State your assumptions.

The system is at steady state.

Step 3

Calculate the resistance to heat transfer:

i) Inner metal wall

R1 ¼�x1

k1A¼ 0:002m

16W=m8Cð Þ 1m2ð Þ ¼ 0:00013 8C=W

ii) Insulation:

R2 ¼�x2

k2A¼ 0:05m

0:055W=m8Cð Þ 1m2ð Þ ¼ 0:90909 8C=W

iii) Outer metal wall:

R3 ¼�x3

k3A¼ 0:002m

16W=m8Cð Þ 1m2ð Þ ¼ 0:00013 8C=W

Examples 59

iv) Total resistance:

�R ¼ R1þR2þR3 ¼ 0:00013þ 0:90909þ 0:00013 ¼ 0:909358C=W

Step 4

Calculate the heat transfer through the wall:

q ¼ �TP

R¼ T1 ÿ T4

P

R¼ 200ÿ 50 8C

0:90935 8C=W¼ 165 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:

r2r1

A1 A2

T1

T2

r3

A3

T3

60 7 Heat Transfer by Conduction

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:

q ¼ �TP

R¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: W

Step 6

Calculate temperature T2.

The temperature drop is proportional to the resistance:

T1 ÿ T2

T1 ÿ T3

¼ RwP

R

where Rw is the metal wall resistance.

Therefore,

T2¼::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::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

Exercises 61

of the outer surface is 20 8C. The thermal conductivity of the metal wall is 16W/

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

T3T4

∆x3∆x2∆x1

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

62 7 Heat Transfer by Conduction

Step 4

Calculate the thickness of insulation:Since

X

R ¼ �T

q¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::

:::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=W

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

q ¼ �T

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 15W/m 8C respectively. If 100W/m2

are transferred through the wall at steady state, calculate the temperature

gradient in the two layers.

Exercises 63

Solution

Step 1

State your assumptions:

......................................................................................................................

Step 2

From Fourier’s law:

dT

dx¼ :::::::::::::::::::::::::::::::::::::::::::::::::

Therefore,

dT

dx

� �

A

¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m

and

dT

dx

� �

B

¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m

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

A¼ ÿk

dT

dx¼ ÿ koþaTð Þ dT

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.

64 7 Heat Transfer by Conduction

Hint:

The relationships that give the temperature variation as a function of distance in

a wall are:

i) Plane wall:

T ¼ T1 ÿxÿ x1

kAq ¼ T1 ÿ

x

x2 ÿ x1�T

ii) Cylindrical wall:

T ¼ T1 ÿln r=r1ð Þ2pkL

q ¼ T1 ÿln r=r1ð Þln r2=r1ð Þ�T

iii) Spherical wall:

T ¼ T1 ÿ1r1 ÿ

1r

4pkq ¼ T1 ÿ

1ÿ r1r

1ÿ r1r2

�T

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.

Exercises 65