Chapter Two

Conduction Heat Transfer

Fourier’s Law of Heat Conduction

dx

dTk

A

Qq x

x

dx

dTkAQx

Heat Flux(Wm-2)

Heat Transfer (W)Per unit Area (m2)(W/m2)

Thermal Conductivity (Wm-1K-1)

dT = temperature difference (K)dx = distance across section (m)

• Heat is conducted in the direction of decreasing temperature• Thus, the temperature gradient is negative when heat is

conducted in +ve x-direction• -ve sign to ensure that the heat transfer in +ve x-direction is +ve

quantity

dx

dTkAQx

Hot face Cold face

Temp.gradient

Hot face temp.

Cold face

temp.

Conduction

Through a Flat Slab or Wall Through a Hollow Cylinder Through a Hollow Sphere Through Solids in Series

Plane walls in series Multilayer Cylinders

Combined Convection and Conduction and Overall Coefficients

Conduction with Internal Heat Generation

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Plane wall

12

21

1212

][

][][

2

1

2

1

xx

TTkAQ

TTkAxxQ

kAdTdxQ

dx

dTkAQ

x

x

T

T

x

x

x

x

T1 T2

Qx

x2x1

* T1> T2

From Fourier’s Law:

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dxdx

dTkAdxQ

dx

dTkAQ

xatT

xatT

x

x

x

x

2

1

2

1

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The rate of heat transfer :

Note that T1 > T2

Thermal Resistance, RTH:

(conduction resistance)

Therefore, RTH = x/kA

12

21

xx

TTkAQx

T1 T2

RTH

Calculate the heat loss per square meter of surface area for an insulating wall composed of 25.4 mm thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K.

From App A.3, the thermal conductivity of fiber insulating board is 0.048 W/m.K. The thickness x2-x1=0.0254 m:

352.7 K(Inside)

297.1 K(Outside)

25.4 mm on fiber insulating board

x

2

2112

W/m1.105

)1.2977.352(0254.0

048.0)(

TTxx

k

A

q

rLA 2

kL

r

r

R2

)ln(1

2

Consider the hollow cylinder with an inside radius of r1 ,

1T

2T

1r2r

q

where

R1T 2Tq

)/ln( 12

12

AA

AAAlm

R

TT

kLrr

TTq 21

1/2

21

2/)ln(

12

21

rr

TTkAq lm

kL

rr

kA

rrR

lm 2

)/ln( 1212

)22

ln(

)22(

1

2

12

LrLr

LrLrAlm

Length of Tubing for Cooling Coil

A thick-walled cylindrical tubing of hard rubber having an inside

radius of 5mm and an outside radius of 20 mm is being used as a

temporary cooling coil in a bath. Ice water is flowing rapidly inside

and the inside wall temperature is 274.9K. The outside surface

temperature is 297.1K, A total of 14.65W must be removed from the

bath by cooling coil. How many m of tubing are needed?

From Appendix A.3, k = 0.151 W/m.K

m 02.01000

2

m 005.01000

5

2

1

r

r

)ln(

2

1/2

21

rr

kLTTq

mmW

Wlength

W

rr

TTkAq

AA

AAA

lm

lm

964.0/2.15

65.14

2.15

)005.002.0

1.2979.274)(0682.0(151.0

m 068.0)0314.0/1257.0ln(

0314.01257.0

)/ln(

12

21

2

12

12

The calculation will be done first for a length of 1 m tubing:

The negative sign indicates that the heat flow is from r2 on the outside to r1 on the inside.Since 15.2 W is removed for 1 m length, the needed length is:

R

TT

krr

TTq 21

21

21

4/)/1/1(

k

rrR

4

)11

(21

1r

2r

q

1T

2T

where

R1T 2Tq

24 rA

R

TT

RRR

TT

Ak

x

Ak

x

Ak

xTT

qCBA

C

C

B

B

A

A

414141

Ak

x

Ak

x

Ak

xR

C

C

B

B

A

A

1T2T

3T

4T

q

AR1T 2Tq

BR3T CR

4T

A B C

where

Electrical analogy

EXAMPLE 4.3-1 Heat Flow Through an Insulated Wall of a Cold Room

A cold-storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board, and an outer layer of 76.2mm of concrete. The wall surface temperature is 255.4K inside the cold room and 297.1K at the outside surface of the concrete. Use conductivities from Appendix A.3 for pine, 0.151; for cork board, 0.0433; and for concrete, 0.762 W/m.K. Calculate the heat loss in W for 1 m2 and the temperature at the interface between the wood and cork board.

The resistance for each material are

0841.0151.0

107.12

346.20433.0

106.101

100.0762.0

102.76

3

3

3

Ak

xR

Ak

xR

Ak

xR

C

CA

B

BB

A

AC

W16.48

530.2

41.7-

0.12.3460.084

1.2974.255

4141

CBA

C

C

B

B

A

A RRR

TT

Ak

x

Ak

x

Ak

xTT

q

To calculate the temperature T2,

K 79.256084.0

4.25548.16

2

2

21

T

T

R

TTq

A

R

TT

RRR

TT

Lk

rr

Lk

rr

Lk

rr

TTq

CBA

CBA

4141

342312

41

2

)/ln(

2

)/ln(

2

)/ln(

Lk

rr

Lk

rr

Lk

rrR

CBA 2

)ln(

2

)ln(

2

)ln( 3/42/31/2

q

ABC1T

2T3T

4T

3r

2r

1r

4r

EXAMPLE 4.3-2 Heat Loss from an Insulated Pipe

A thick-walled tube of stainless steel (A) having a k = 21.63 W/m.k with dimensions of 0.0254m ID and 0.0508m OD is covered with a 0.0254m layer of asbestos (B) insulation, k = 0.2423 W/m.k. The inside wall temperature of the pipe is 811K and the outside surface of the insulation is at 310.8K. For a 0.305m length of pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation.

BA RR

TTq

31

The resistances are

K/W 01673.0

)305.0)(63.21(2

)0127.0

0254.0ln(

2

)ln(

2

)ln( 1/21/2

Lk

dd

Lk

rrR

AAA

K/W 493.1

)305.0.0)(2423.0(2

)0508.0

1016.0ln(

2

)ln(

2

)ln( 1/21/2

Lk

dd

Lk

rrR

BBB

The heat transfer rate is

BA RR

TTq

31

W7.331 493.101673.0

8.310811

q

K 5.80501673.0

8117.331

2

2

21

T

T

R

TTq

A