PROBLEM 3.95
KNOWN: Dimensions and thermal conductivity of a spherical container. Thermal conductivity and
volumetric energy generation within the container. Outer convection conditions.
FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperature
distribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energy
generation rate.
SCHEMATIC:
500
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radial
conduction.
ANALYSIS: (a) For a control volume which includes the container, conservation of energy yields
−q
E g − E out = 0 , or qV
conv = 0 . Hence
( )
(
q ( 4 3 ) π ri3 = h4π ro2 Ts,o − T∞
5
)
3
and with q = 10 W/m ,
Ts,o = T∞ +
3
i
qr
3hro2
105 W m 2 ( 0.5 m )
3
= 25$ C +
3000 W m ⋅ K ( 0.6 m )
1500
2
2
<
$
= 36.6
48.2C .
(b) Performing a surface energy balance at the outer surface, E in − E out = 0 or q cond − q conv = 0 .
Hence
4π k ss Ts,i − Ts,o
= h4π ro2 Ts,o − T∞
(1 ri ) − (1 ro )
(
)
(
)
(
)
500 W m 2 ⋅ K
1000
 ro 
$
$
23.2
141 $ C .
48.2C +
Ts,i = Ts,o +
C = 129.4
− 1  ro ( Ts,o − T∞ ) = 36.6
( 0.2 ) 0.6 m 11.6

k ss  ri
15 W m ⋅ K

h
<
(c) The heat equation in spherical coordinates is
d  dT 
2
k rw  r 2
= 0.
 + qr
dr  dr 
Solving,
r2
dT
=−
3
qr
+ C1
dr
3k rw
Applying the boundary conditions,
dT
=0
and
dr r = 0
C1 = 0
and
and
T (r ) = −
2
qr
C
− 1 + C2
6k rw
r
T ( ri ) = Ts,i
2
C2 = Ts,i + qr
i 6k rw .
Continued...
PROBLEM 3.95 (Cont.)
Hence
ri2 − r 2 )
(
6k rw
q
T ( r ) = Ts,i +
<
At r = 0,
T ( 0 ) = Ts,i +
2
i
qr
6k rw
= 129.4$ C +
105 W m3 ( 0.5 m )
2
6 ( 20 W m ⋅ K )
<
= 337.7$ C
(d) The feasibility assessment may be performed by using the IHT model for one-dimensional, steadystate conduction in a solid sphere, with the surface boundary condition prescribed in terms of the total
thermal resistance
( )
ri2
2
R ′′tot,i = 4π ri R tot = R ′′cnd,i + R ′′cnv,i =
[(1 ri ) − (1 ro )] + 1  ri 2
k ss


h  ro 
where, for ro = 0.6 m and h = 1000 W/m2⋅K, R ′′cnd,i = 5.56 × 10-3 m2⋅K/W, R ′′cnv,i = 6.94 × 10-4 m2⋅K/W,
Center temperature, T(0) (C)
and R ′′tot,i = 6.25 × 10-3 m2⋅K/W. Results for the center temperature are shown below.
675
625
575
525
475
0
2000
4000
6000
8000
10000
Convection coefficient, h(W/m^2.K)
ro = 0.54 m
ro = 0.60 m
Clearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2⋅K (a practical upper limit), T(0) > 475°C and
the desired condition can not be met. The corresponding resistances are R ′′cnd,i = 2.47 × 10-3 m2⋅K/W,
R ′′cnv,i = 8.57 × 10-5 m2⋅K/W, and R ′′tot,i = 2.56 × 10-3 m2⋅K/W. The conduction resistance remains
dominant, and the effect of reducing R ′′cnv,i by increasing h is small. The proposed extension is not
feasible.
COMMENTS: A value of q = 1.79 × 105 W/m3 would allow for operation at T(0) = 475°C with ro =
0.54 m and h = 10,000 W/m2⋅K.