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ch05


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of the spheres and
convection coefficients.
FIND: (a) Show in a qualitative manner, on T-t coordinates, temperatures at the center and the outer
surface for each sphere; explain features of the curves; (b) Time required for the outer surface of each
sphere to reach 415K, (c) Energy gained by each bath during process of cooling spheres to a surface
temperature of 415K.
SCHEMATIC:
Sphere A Sphere B
ro (mm) 150 15
\u3c1 (kg/m3) 1600 400
c (J/kg\u22c5K) 400 1600
k (W/m\u22c5K) 170 1.7
h (W/m2\u22c5K) 5 50
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Uniform properties, (3) Constant
convection coefficient.
ANALYSIS: (a) From knowledge of the Biot number and the thermal time constant, it is possible to
qualitatively represent the temperature distributions. From Eq. 5.10, with Lc = ro/3, find
( )
( )
( )
2
3
A
o
2
B
5 W/m K 0.150m/3
 Bi 1.47 10 
170 W/m K
h r / 3
Bi 
k
50 W/m K 0.015m/3
 Bi
1.
 (1)\u2212\u22c5= = ×
\u22c5
=
\u22c5
= 0.147 
7 W/m K
(2)=
\u22c5
The thermal time constant for a lumped capacitance system from Eq. 5.7 is
( ) ( )
3
A 2
s
1600 kg/m 0.150m 400 J/kg K1
Vc 6400s
hA 3 5 W/m K
\u3c4 \u3c1 \u3c4 × \u22c5= = =
× \u22c5
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb (3)
( )3o
B 2
400 kg/m 0.015m 1600 J/kg K
 r c
 64s
3h 3 50 W/m K
\u3c1
\u3c4 \u3c4
× \u22c5
= = =
× \u22c5
(4)
When Bi << 0.1, the sphere will cool in a
spacewise isothermal manner (Sphere A).
For sphere B, Bi > 0.1, hence gradients will
be important. Note that the thermal time
constant of A is much larger than for B;
hence, A will cool much slower. See sketch
for these features.
(b) Recognizing that BiA < 0.1, Sphere A can be treated as spacewise isothermal and analyzed
using the lumped capacitance method. From Eq. 5.6 and 5.7, with T = 415 K
( )
i i
T T
exp t/
T T
\u3b8
\u3c4
\u3b8
\u221e
\u221e
\u2212
= = \u2212
\u2212
(5)
Continued \u2026..
PROBLEM 5.64 (Cont.)
A A
i
T T 415 320
t ln 6400s ln 10,367s 2.88h.
T T 800 320
\u3c4 \u221e
\u221e
\uf8ee \uf8f9
\u2212 \u2212\uf8ee \uf8f9
= \u2212 = \u2212 = =\uf8ef \uf8fa \uf8ef \uf8fa
\u2212 \u2212\uf8f0 \uf8fb\uf8f0 \uf8fb
<
Note that since the sphere is nearly isothermal, the surface and inner temperatures are
approximately the same.
Since BiB > 0.1, Sphere B must be treated by the Heisler chart method of solution beginning
with Figure D.8. Using
( )2
-1o
B B
50 W/m K 0.015mhrBi 0.44 or Bi 2.27,
k 1.7 W/m K
\u22c5 ×
\u2261 = = =
\u22c5
find that for r/ro = 1,
( ) ( ) ( )o
o o o
1, t T r , t T 415 320
0.8.
\u3b8
\u3b8 \u3b8 \u3b8
\u221e\u2212 \u2212
= = = (6)
Using Eq. (6) and Figure D.7, find the Fourier number,
( )( ) ( )
( )
oo
2i i o
T r , t T / 0.8 415 320 K/0.8
 t0.25 Fo 1.3.
T T 800 320 K r
\u3b8 \u3b1
\u3b8
\u221e
\u221e
\u2212
\u2212
= = = = =
\u2212 \u2212
( )22o
B 6 2
1.3 0.015mFo r
t 110s 1.8 min
2.656 10 m / s\u3b1 \u2212
= = = =
×
<
where \u3b1 = k/\u3c1c = 1.7 W/m\u22c5K/400 kg/m3 × 1600 J/kg\u22c5K = 2.656×10-6m2/s.
(c) To determine the energy change by the spheres during the cooling process, apply the
conservation of energy requirement on a time interval basis.
Sphere A:
( ) ( )in out AE E E Q E E t E 0 .\u2212 = \u2206 \u2212 = \u2206 = \u2212
( ) ( ) ( ) [ ]33A iQ cV T t T 1600kg/m 400J/kg K 4/3 0.150m 415 800 K\u3c1 \u3c0\uf8ee \uf8f9= \u2212 = × \u22c5 × \u2212\uf8f0 \uf8fb
6
AQ 3.483 10 J.= × <
Note that this simple expression is a consequence of the spacewise isothermal behavior.
Sphere B: ( ) ( )in out BE E E Q E t E 0 .\u2212 = \u2206 \u2212 = \u2212
For the nonisothermal sphere, the Groeber chart, Figure D.9, can be used to evaluate QB.
With Bi = 0.44 and Bi2Fo = (0.44)2×1.3 = 2.52, find Q/Qo = 0.74. The energy transfer from
the sphere during the cooling process, using Eq. 5.44, is
( )B o iQ 0.74 Q 0.74 cV T T\u3c1 \u221e\uf8ee \uf8f9= = \u2212\uf8f0 \uf8fb
( ) ( ) ( )33BQ 0.75 400kg/m 1600J/kg K 4/3 0.015m 800 320 K 3257 J.\u3c0= × × \u22c5 \u2212 = <
COMMENTS: (1) In summary:
Sphere Bi = hro/k \u3c4 s\ufffd \ufffd t(s) Q(J)
 A 4.41×10-3 6400 10,370 3.48×106
 B 0.44 64 110 3257
PROBLEM 5.65
KNOWN: Spheres of 40-mm diameter heated to a uniform temperature of 400°C are suddenly
removed from an oven and placed in a forced-air bath operating at 25°C with a convection coefficient
of 300 W/m2\u22c5K.
FIND: (a) Time the spheres must remain in the bath for 80% of the thermal energy to be removed,
and (b) Uniform temperature the spheres will reach when removed from the bath at this condition and
placed in a carton that prevents further heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction in the spheres, (2) Constant properties, and
(3) No heat loss from sphere after removed from the bath and placed into the packing carton.
PROPERTIES: Sphere (given): \u3c1 = 3000 kg/m3, c = 850 J/kg\u22c5K, k = 15 W/m\u22c5K.
ANALYSIS: (a) From Eq. 5.52, the fraction of thermal energy removed during the time interval \u2206t =
to is
( ) ( )3o 1 1 11
o
Q 1 3 / sin cosQ \u3b8 \u3b6 \u3b6 \u3b6 \u3b6
\u2217 \uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fb (1)
where Q/Qo = 0.8. The Biot number is
2
oBi hr / k 300 W / m K 0.020m /15 W / m K 0.40= = \u22c5 × \u22c5 =
and for the one-term series approximation, from Table 5.1,
1 11.0528 rad C 1.1164\u3b6 = = (2)
The dimensionless temperature o ,\u3b8
\u2217
 Eq. 5.31, follows from Eq. 5.50.
( )2o 1 1C exp Fo\u3b8 \u3b6\u2217 = \u2212 (3)
where 2o oFo t / r .\u3b1= Substituting Eq. (3) into Eq. (1), solve for Fo and to.
( ) ( ) ( )2 31 1 1 11 1
o
Q 1 3 C exp Fo / sin cosQ \u3b6 \u3b6 \u3b6 \u3b6 \u3b6\uf8ee \uf8f9= \u2212 \u2212 \u2212\uf8f0 \uf8fb (4)
oFo 1.45 t 98.6 s= = <
(b) Performing an overall energy balance on the sphere during the interval of time to \u2264 t \u2264 \u221e,
in out f iE E E E E 0\u2212 = \u2206 = \u2212 = (5)
where Ei represents the thermal energy in the sphere at to,
( ) ( ) ( )i o iE 1 0.8 Q 1 0.8 cV T T\u3c1 \u221e= \u2212 = \u2212 \u2212 (6)
and Ef represents the thermal energy in the sphere at t = \u221e,
( )f avgE cV T T\u3c1 \u221e= \u2212 (7)
Combining the relations, find the average temperature
( ) ( )( )avg icV T T 1 0.8 T T 0\u3c1 \u221e \u221e\uf8ee \uf8f9\u2212 \u2212 \u2212 \u2212 =\uf8f0 \uf8fb
avgT 100 C= ° <
PROBLEM 5.66
KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex spheres in packed bed
thermal energy storage system. Convection coefficient and inlet gas temperature.
FIND: Time required for sphere to acquire 90% of maximum possible thermal energy and the
corresponding center temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to or
from a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constant
properties.
ANALYSIS: With Bi \u2261 h(ro/3)/k = 75 W/m2\u22c5K (0.0125m)/1.4 W/m\u22c5K = 0.67, the approximate
solution for one-dimensional transient conduction in a sphere is used to obtain the desired results. We
first use Eq. (5.52) to obtain o .\u3b8\u2217
( ) ( )
3
1
o
o1 1 1
Q1 Q3 sin cos
\u3b6
\u3b8 \u3b6 \u3b6 \u3b6
\u2217 \uf8eb \uf8f6
= \u2212\uf8ec \uf8f7\uf8ee \uf8f9\u2212 \uf8ed \uf8f8\uf8f0 \uf8fb
With Bi \u2261 hro/k = 2.01, 1 2.03\u3b6 \u2248 and C1 \u2248 1.48 from Table 5.1. Hence,
( )
( )
3
o
0.1 2.03 0.837 0.155
5.3863 0.896 2.03 0.443
\u3b8\u2217 = = =\uf8ee \uf8f9\u2212 \u2212\uf8f0 \uf8fb
The center temperature is therefore
( )o g,i i g,iT T 0.155 T T 300 C 42.7 C 257.3 C= + \u2212 = ° \u2212 ° = ° <
From Eq. (5.50c), the corresponding time is
2
o o
2 11
r
t ln
C
\u3b8
\u3b1\u3b6
\u2217\uf8eb \uf8f6\uf8ec \uf8f7= \u2212 \uf8ec \uf8f7\uf8ed \uf8f8
where ( )3 7 2k / c 1.4 W / m K / 2225 kg / m 835 J / kg K 7.54 10 m / s.\u3b1 \u3c1 \u2212= = \u22c5 × \u22c5 = ×
( ) ( )
( )
2
27 2
0.0375m ln 0.155 /1.48
t 1,020s
7.54 10 m / s 2.03\u2212
= \u2212 =
×
<
COMMENTS: The surface temperature at the time of interest may be obtained from Eq. (5.50b).
With r 1,\u2217 =
( ) ( )o 1s g,i i g,i
1
sin 0.155 0.896T T T T 300 C 275 C 280.9 C
2.03
\u3b8 \u3b6
\u3b6
\u2217
×\uf8eb \uf8f6
= + \u2212 = ° \u2212 ° = °\uf8ec \uf8f7\uf8ed \uf8f8 <
PROBLEM 5.67
KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in a
fluid of prescribed temperature and convection coefficient.
FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity