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