SOLUTIONS MANUAL
1-1
,Solutions Manual for Engineering Heat Transfer, 3rd Ed
Chapter 1
1.1 Equation 1.7 is the integrated form of Fourier’s Law of Conduction in one dimension for
constant thermal conductivity and steady-state operation. If a more accurate model is
desired, the variation of thermal conductivity with temperature may be accounted
for by using
k = k0(1 + C1(T – To))
where k is thermal conductivity, k0 is thermal conductivity at some reference
temperature T0, C1 is a constant, and T is temperature. Determine k0 and C1 for copper
if T0 = 650 K. Use the graphs in this chapter to obtain the necessary data.
Solution:
k = k0(1 + C1(T – To)) to fit data on copper. At T0 = 650 K, k0 = 360 W/(m·K) from Figure
1.3. At 200 K, k0 = 415 W/(m·K) also from Figure 1.3. Second point is needed and
selected arbitrarily. Substitute into the equation:
k = k0(1 + C1(T – To))
415 = 360 (1 + C1(200 – 650))
(415/360 – 1)
C1 = = – 3.4 x 10-4 so the equation becomes
–450
k = 360(1 – 3.4 x 10-4(T – 650)) [SI units]
1.2 Repeat Problem 1 for glycerin if T0 = 170°F.
Solution:
k = k0(1 + C1(T – To)) to fit data on glycerin. At T0 = 170 + 460 = 630°R, k0 = 0.17
BTU/(hr·ft2·°R). At 590°R, k0 = 0.18 BTU/(hr·ft2·°R) selected arbitrarily. Both
data points from Figure 1.4. Substitute into the equation:
k = k0(1 + C1(T – To))
0.18 = 0.17 (1 + C 1(590 – 630))
(0.18/0.17 – 1)
C1 = = – 1.47 x 10-3 so the equation becomes
(590 – 630)
k = 0.17(1 – 1.47 x 10-3(T – 630)) [Engr units]
1-1
,1.3 Repeat Problem 1 for hydrogen if the reference temperature is 350 K.
Solution:
k = k0(1 + C1(T – To)) to fit data on H2. At T0 = 350 K, k0 = 0.2 W/(m·K) from Figure
1.5. At 600 K, k0 = 0.315 W/(m·K) also from Figure 1.5. Second point is needed and
selected arbitrarily. Substitute into the equation:
k = k0(1 + C1(T – To))
0.315 = 0.2 (1 + C1(600 – 350))
(0.315/0.2 – 1)
C1 = = – 2.3 x 10-3 so the equation becomes
600 – 350
k = 0.2(1 – 2.3 x 10-3(T – 350)) [SI units]
1-2
, 1.4 Equation 1.7 was obtained by integration of Equation 1.6, assuming a constant thermal
conductivity and steady-state operation. If thermal conductivity in Equation 1.6 is not
a constant but instead is written as
k = k0(1 + C1(T – To))
where k is thermal conductivity, k0 is thermal conductivity at some reference
temperature T0, C1 is a constant, and T is temperature, how is Equation 1.7 affected?
Derive the new expression.
Solution:
L
T2
qx
dx = -kdT Substitute for k = k0(1 + C1(T – To))
A T1
0
L
T2 T2
qx
dx = -k0(1 + C1(T – To))dT = (-k0 – k0C1T + k 0 C 1 T o )dT)
A
T1 T1
0
For constant qx and A,
qx
L = – k0(T2 – T1) – k0C1(T22 – T12)/2 + k0C1To(T2 – T1)
A
qx k 0C 1
L = (– k0 + k0C1T o)(T 2 – T1) – (T2 – T1)(T2 + T1)
A 2
qx k 0C 1
L = [– k0 + k0C 1To – (T2 + T1)](T2 – T1)
A 2
qx C1
L = – k0[1 – C 1To + (T2 + T1)](T2 – T1) or
A 2
k0 A C1
qx = – (T2 – T1) [1 – C1To + (T2 + T1)]
L 2
1-3
1-1
,Solutions Manual for Engineering Heat Transfer, 3rd Ed
Chapter 1
1.1 Equation 1.7 is the integrated form of Fourier’s Law of Conduction in one dimension for
constant thermal conductivity and steady-state operation. If a more accurate model is
desired, the variation of thermal conductivity with temperature may be accounted
for by using
k = k0(1 + C1(T – To))
where k is thermal conductivity, k0 is thermal conductivity at some reference
temperature T0, C1 is a constant, and T is temperature. Determine k0 and C1 for copper
if T0 = 650 K. Use the graphs in this chapter to obtain the necessary data.
Solution:
k = k0(1 + C1(T – To)) to fit data on copper. At T0 = 650 K, k0 = 360 W/(m·K) from Figure
1.3. At 200 K, k0 = 415 W/(m·K) also from Figure 1.3. Second point is needed and
selected arbitrarily. Substitute into the equation:
k = k0(1 + C1(T – To))
415 = 360 (1 + C1(200 – 650))
(415/360 – 1)
C1 = = – 3.4 x 10-4 so the equation becomes
–450
k = 360(1 – 3.4 x 10-4(T – 650)) [SI units]
1.2 Repeat Problem 1 for glycerin if T0 = 170°F.
Solution:
k = k0(1 + C1(T – To)) to fit data on glycerin. At T0 = 170 + 460 = 630°R, k0 = 0.17
BTU/(hr·ft2·°R). At 590°R, k0 = 0.18 BTU/(hr·ft2·°R) selected arbitrarily. Both
data points from Figure 1.4. Substitute into the equation:
k = k0(1 + C1(T – To))
0.18 = 0.17 (1 + C 1(590 – 630))
(0.18/0.17 – 1)
C1 = = – 1.47 x 10-3 so the equation becomes
(590 – 630)
k = 0.17(1 – 1.47 x 10-3(T – 630)) [Engr units]
1-1
,1.3 Repeat Problem 1 for hydrogen if the reference temperature is 350 K.
Solution:
k = k0(1 + C1(T – To)) to fit data on H2. At T0 = 350 K, k0 = 0.2 W/(m·K) from Figure
1.5. At 600 K, k0 = 0.315 W/(m·K) also from Figure 1.5. Second point is needed and
selected arbitrarily. Substitute into the equation:
k = k0(1 + C1(T – To))
0.315 = 0.2 (1 + C1(600 – 350))
(0.315/0.2 – 1)
C1 = = – 2.3 x 10-3 so the equation becomes
600 – 350
k = 0.2(1 – 2.3 x 10-3(T – 350)) [SI units]
1-2
, 1.4 Equation 1.7 was obtained by integration of Equation 1.6, assuming a constant thermal
conductivity and steady-state operation. If thermal conductivity in Equation 1.6 is not
a constant but instead is written as
k = k0(1 + C1(T – To))
where k is thermal conductivity, k0 is thermal conductivity at some reference
temperature T0, C1 is a constant, and T is temperature, how is Equation 1.7 affected?
Derive the new expression.
Solution:
L
T2
qx
dx = -kdT Substitute for k = k0(1 + C1(T – To))
A T1
0
L
T2 T2
qx
dx = -k0(1 + C1(T – To))dT = (-k0 – k0C1T + k 0 C 1 T o )dT)
A
T1 T1
0
For constant qx and A,
qx
L = – k0(T2 – T1) – k0C1(T22 – T12)/2 + k0C1To(T2 – T1)
A
qx k 0C 1
L = (– k0 + k0C1T o)(T 2 – T1) – (T2 – T1)(T2 + T1)
A 2
qx k 0C 1
L = [– k0 + k0C 1To – (T2 + T1)](T2 – T1)
A 2
qx C1
L = – k0[1 – C 1To + (T2 + T1)](T2 – T1) or
A 2
k0 A C1
qx = – (T2 – T1) [1 – C1To + (T2 + T1)]
L 2
1-3
