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20
Analytical Heat Transfer
T ∞ h
T ∞ h
T 0
T s
T s
r 0
r
q
.
FIGURE 2.6
Cylindrical rod heat conduction with internal heat generation.
In 1-D steady-state cylindrical medium with heat generation, the heat
conduction equation is (from Equation 1.18)
1 d
dT
q ˙
r
+ = 0
(2.24)
r dr
dr
k
d
dT
q ˙
r
= − r
dr
dr
k
The general solution is
dT
q ˙
r
= − r
2
+ c 1
dr
2k
dT
q ˙
c 1
= − r +
dr
2k
r
q ˙
T(r) = − r
2
+ c 1 ln r + c 2
4k
with boundary conditions as shown in Figure 2.6:
dT
at r = 0,
= 0 = c 1
dr
q ˙
at r = r 0 , T = T s = − r 0
2
+ c 2
4k
Solve for c 1 (c 1 = 0) and c 2 , one obtains the temperature distribution
2
2
˙
r
0
T(r) = T s +
qr
1 − 2
(2.25)
4k
r 0
At the centerline of the cylindrical rod, the temperature is
2
qr
T 0 = T s +
˙ 0
(2.26)
4k
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20
Analytical Heat Transfer
T ∞ h
T ∞ h
T 0
T s
T s
r 0
r
q
.
FIGURE 2.6
Cylindrical rod heat conduction with internal heat generation.
In 1-D steady-state cylindrical medium with heat generation, the heat
conduction equation is (from Equation 1.18)
1 d
dT
q ˙
r
+ = 0
(2.24)
r dr
dr
k
d
dT
q ˙
r
= − r
dr
dr
k
The general solution is
dT
q ˙
r
= − r
2
+ c 1
dr
2k
dT
q ˙
c 1
= − r +
dr
2k
r
q ˙
T(r) = − r
2
+ c 1 ln r + c 2
4k
with boundary conditions as shown in Figure 2.6:
dT
at r = 0,
= 0 = c 1
dr
q ˙
at r = r 0 , T = T s = − r 0
2
+ c 2
4k
Solve for c 1 (c 1 = 0) and c 2 , one obtains the temperature distribution
2
2
˙
r
0
T(r) = T s +
qr
1 − 2
(2.25)
4k
r 0
At the centerline of the cylindrical rod, the temperature is
2
qr
T 0 = T s +
˙ 0
(2.26)
4k