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ch05


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.
Since Bic > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder,
( ) ( ) ( )* * 2 *1 0 11r , Fo C exp Fo J r\u3b8 \u3b6 \u3b6= \u2212 × (1)
Solving for Fo and setting r* = 1, find
( )
*
2 1 0 11
1Fo ln
C J
\u3b8
\u3b6\u3b6
\uf8ee \uf8f9
= \u2212 \uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
where ( ) ( ) ( )( )
o o*
i
T r , t T 500 350 K
1,Fo 0.231
T T 1000 350 K
\u3b8 \u221e
\u221e
\u2212 \u2212
= = = =
\u2212 \u2212
.
From Table 5.1, with Bi = 0.441, find \u3b61 = 0.8882 rad and C1 = 1.1019. From Table B.4, find J0(\u3b61) =
0.8121. Substituting numerical values into Eq. (2),
( ) [ ]2
1Fo ln 0.231 1.1019 0.8121 1.72
0.8882
= \u2212 × = .
From the definition of the Fourier number, Fo = 2ot r\u3b1 , and \u3b1 = k/\u3c1c,
2
2o
o
r c
t Fo Fo r
k
\u3c1
\u3b1
= = \u22c5
( )2 3t 1.72 0.015m 400kg m 1600J kg K 1.7 W m K 145s= × × \u22c5 \u22c5 = . <
(b) Using the IHT Transient Conduction Model for a Cylinder, the following surface temperature
histories were obtained.
Continued...
PROBLEM 5.49 (Cont.)
0 50 100 150 200 250 300
Time, t(s)
300
400
500
600
700
800
900
1000
Su
rfa
ce
 te
m
pe
ra
tu
re
, T
(K
)
h = 250 W/m^2.K
h = 50 W/m^2.K
Increasing the convection coefficient by a factor of 5 has a significant effect on the surface temperature,
greatly accelerating its approach to the oil temperature. However, even with h = 250 W/m2\u22c5K, Bi = 1.1
and the convection resistance remains significant. Hence, in the interest of accelerated cooling,
additional benefit could be achieved by further increasing the value of h.
COMMENTS: For Part (a), note that, since Fo = 1.72 > 0.2, the approximate series solution is
appropriate.
PROBLEM 5.50
KNOWN: Long pyroceram rod, initially at a uniform temperature of 900 K, and clad with a thin
metallic tube giving rise to a thermal contact resistance, is suddenly cooled by convection.
FIND: (a) Time required for rod centerline to reach 600 K, (b) Effect of convection coefficient on
cooling rate.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Thermal resistance and capacitance of
metal tube are negligible, (3) Constant properties, (4) Fo \u2265 0.2.
PROPERTIES: Table A-2, Pyroceram ( T = (600 + 900)K/2 = 750 K): \u3c1 = 2600 kg/m3, c = 1100
J/kg\u22c5K, k = 3.13 W/m\u22c5K.
ANALYSIS: (a) The thermal contact and convection resistances can be combined to give an overall heat
transfer coefficient. Note that t,cR\u2032 [m\u22c5K/W] is expressed per unit length for the outer surface. Hence,
for h = 100 W/m2\u22c5K,
( ) ( )
2
2t,c
1 1U 57.0 W m K
1 h R D 1 100 W m K 0.12m K W 0.020 m\u3c0 \u3c0
= = = \u22c5
\u2032+
\u22c5 + \u22c5 ×
.
Using the approximate series solution, Eq. 5.50c, the Fourier number can be expressed as
( ) ( )2 *o 11Fo 1 ln C\u3b6 \u3b8= \u2212 .
From Table 5.1, find \u3b61 = 0.5884 rad and C1 = 1.0441 for
( )2oBi Ur k 57.0 W m K 0.020m 2 3.13W m K 0.182= = \u22c5 \u22c5 = .
The dimensionless temperature is
( ) ( ) ( )( )
*
o
i
T 0, t T 600 300 K
0, Fo 0.5.
T T 900 300 K
\u3b8 \u221e
\u221e
\u2212 \u2212
= = =
\u2212 \u2212
Substituting numerical values to find Fo and then the time t,
( )2
1 0.5Fo ln 2.127
1.04410.5884
\u2212
= =
2
2o
o
r c
t Fo Fo r
k
\u3c1
\u3b1
= = \u22c5
( )2 3t 2.127 0.020 m 2 2600kg m 1100J kg K 3.13W m K 194s= × \u22c5 \u22c5 = . <
(b) The following temperature histories were generated using the IHT Transient conduction Model for a
Cylinder.
Continued...
PROBLEM 5.50 (Cont.)
0 50 100 150 200 250 300
Time, t(s)
300
400
500
600
700
800
900
Su
rfa
ce
 te
m
pe
ra
tu
re
, (K
)
r = ro, h = 100 W/m^2.K
r = ro, h = 500 W/m^2.K
r = ro, h = 1000 W/m^2.K
 
0 50 100 150 200 250 300
Time, t(s)
300
400
500
600
700
800
900
Ce
nt
er
lin
e 
te
m
pe
ra
tu
re
, (K
)
r = 0, h = 100 W/m^2.K
r = 0, h = 500 W/m^2.K
r = 0, h = 1000 W/m^2.K
While enhanced cooling is achieved by increasing h from 100 to 500 W/m2\u22c5K, there is little benefit
associated with increasing h from 500 to 1000 W/m2\u22c5K. The reason is that for h much above 500
W/m2\u22c5K, the contact resistance becomes the dominant contribution to the total resistance between the
fluid and the rod, rendering the effect of further reductions in the convection resistance negligible. Note
that, for h = 100, 500 and 1000 W/m2\u22c5K, the corresponding values of U are 57.0, 104.8 and 117.1
W/m2\u22c5K, respectively.
COMMENTS: For Part (a), note that, since Fo = 2.127 > 0.2, Assumption (4) is satisfied.
PROBLEM 5.51
KNOWN: Sapphire rod, initially at a uniform temperature of 800K is suddenly cooled by a convection
process; after 35s, the rod is wrapped in insulation.
FIND: Temperature rod reaches after a long time following the insulation wrap.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) No heat losses
from the rod when insulation is applied.
PROPERTIES: Table A-2, Aluminum oxide, sapphire (550K): r = 3970 kg/m
3
, c = 1068 J/kg×K, k =
22.3 W/m×K, a = 5.259´10
-5
 m
2
/s.
ANALYSIS: First calculate the Biot number with Lc = ro/2,
( ) ( )2oc h r / 2 1600 W/m K 0.020 m/2h LBi 0.72.
k k 22.3 W/m K
×
= = = =
×
Since Bi > 0.1, the rod cannot be approximated as a lumped capacitance system. The temperature
distribution during the cooling process, 0 £ t £ 35s, and for the time following the application of
insulation, t > 35s, will appear as
Eventually (t ® ¥), the temperature of the rod will be uniform at ( )T .¥ To find ( )T ,¥ write the
conservation of energy requirement for the rod on a time interval basis, in out final initialE E E E E .- = D º -
Using the nomenclature of Section 5.5.3 and basing energy relative to T¥, the energy balance becomes
( )( ) oQ cV T T Qr ¥- = ¥ - -
where Qo = rcV(Ti - T¥). Dividing through by Qo and solving for ( )T ,¥ find
( ) ( )( )i oT T T T 1 Q/Q .¥ ¥¥ = + - -
From the Groeber chart, Figure D.6, with
2
ohr 1600 W/m K 0.020mBi 1.43
k 22.3 W/m K
× ´
= = =
×
( ) ( ) ( )( )2 22 2 2 -6 2oBi Fo Bi t/r 1.43 5.259 10 m /s 35s/ 0.020m 0.95.a= = ´ ´ =
find Q/Qo » 0.57. Hence,
( ) ( ) ( )T 300K 800 300 K 1-0.57 515 K.¥ = + - = <
COMMENTS: From use of Figures D.4 and D.5, find T(0,35s) = 525K and T(ro,35s) = 423K.
PROBLEM 5.52
KNOWN: Long bar of 70 mm diameter, initially at 90°C, is suddenly immersed in a water bath
(T¥ = 40°C, h = 20 W/m
2
×K).
FIND: (a) Time, tf, that bar should remain in bath in order that, when removed and allowed to
equilibrate while isolated from surroundings, it will have a uniform temperature T(r, ¥) = 55°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties.
PROPERTIES: Bar (given): r = 2600 kg/m
3
, c = 1030 J/kg×K, k = 3.50 W/m×K, a = k/rc =
1.31´10
-6
 m
2
/s.
ANALYSIS: Determine first whether conditions are space-wise isothermal
( ) ( )2oc h r / 2 20 W/m K 0.035 m/2hLBi 0.10
k k 3.50 W/m K
×
= = = =
×
and since Bi ³ 0.1, a Heisler solution is appropriate.
(a) Consider an overall energy balance on the bar during the time interval Dt = tf (the time the bar is
in the bath).
( ) ( )
( )
( )
( )
in out
final initial f i
f o
f
o i
E E E
0 Q E E Mc T T Mc T T
Q Mc T T Q
55 40 CQ T T
1 1 0.70
Q T T 90 40 C
¥ ¥
¥
¥
¥
- = D
- = - = - - -
- = - -
--
= - = - =
- -
o
o
where Qo is the initial energy in the bar (relative to T¥; Eq. 5.44). With Bi = hro/k = 0.20 and
Q/Qo = 0.70, use Figure D.6 to find Bi
2
Fo = 0.15; hence Fo = 0.15/Bi
2
 = 3.75 and
( )22 6 2f ot Fo r / 3.75 0.035 m /1.31 10 m / s 3507 s.a -= × = ´ = <
(b) To determine T(ro, tf), use Figures D.4 and D.5 for q(ro,t)/qi (Fo = 3.75, Bi
-1
 = 5.0) and qo/qi
(Bi
-1
 = 5.0, r/ro = 1, respectively, to find
( ) ( ) ( )o oo f i
o i
r , t
T r , t T 40 C 0.25 0.90 90 50 C 49 C.
q q q
q q¥