ch05
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ch05


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and conductivity on
thermal response.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties.
ANALYSIS: (a) For k = 15 W/m\u22c5K, the Biot number is
( ) ( )2oh r 3 300 W m K 0.05m 3Bi 0.333
k 15 W m K
\u22c5
= = =
\u22c5
.
Hence, the lumped capacitance method cannot be used. From Equation 5.50a,
( ) ( )*121 1 *i 1
sin rT T C exp Fo
T T r
\u3b6
\u3b6
\u3b6
\u221e
\u221e
\u2212
= \u2212
\u2212
.
At the surface, r* = 1. From Table 5.1, for Bi = 1.0, \u3b61 = 1.5708 rad and C1 = 1.2732. Hence,
( )260 75 sin 900.30 1.2732exp 1.5708 Fo25 75 1.5708\u2212 = = \u2212\u2212 $
exp(-2.467Fo) = 0.370
2
o
tFo 0.403
r
\u3b1
= =
( )22o
5 2
0.05mr
t 0.403 0.403 100s
10 m s\u3b1 \u2212
= = = <
(b) Using the IHT Transient Conduction Model for a Sphere to perform the parametric calculations, the
effect of \u3b1 is plotted for k = 15 W/m\u22c5K.
Continued...
PROBLEM 5.67 (Cont.)
0 50 100 150 200 250 300
Time, t(s)
25
35
45
55
65
75
Su
rfa
ce
 te
m
pe
ra
tu
re
, T
(C
)
k = 15 W/m.K, alpha = 1E-4 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-6m^2/s
0 50 100 150 200 250 300
Time, t(s)
25
35
45
55
65
75
Ce
nt
er
 te
m
pe
ra
tu
re
, T
(C
)
k = 15 W/m.K, alpha = 1E-4 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-6 m^2/s
For fixed k and increasing \u3b1, there is a reduction in the thermal capacity (\u3c1cp) of the material, and hence
the amount of thermal energy which must be added to increase the temperature. With increasing \u3b1, the
material therefore responds more quickly to a change in the thermal environment, with the response at
the center lagging that of the surface.
The effect of k is plotted for \u3b1 = 10-5 m2/s.
0 50 100 150 200 250 300
Time, t(s)
25
35
45
55
65
75
Su
rfa
ce
 te
m
pe
ra
tu
re
, T
(C
)
k = 1.5 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k = 150W/m.K, alpha = 1E-5 m^2/s
 
0 50 100 150 200 250 300
Time, t(s)
25
35
45
55
65
75
Ce
nt
er
 te
m
pe
ra
tu
re
, T
(C
)
k = 1.5 W/m.K, alpha = 1E-5 m^2/s
k = 15 W/m.K, alpha = 1E-5 m^2/s
k =150 W/m.K, alpha = 1E-5m^2/s
With increasing k for fixed alpha, there is a corresponding increase in \u3c1cp, and the material therefore
responds more slowly to a thermal change in its surroundings. The thermal response of the center lags
that of the surface, with temperature differences, T(ro,t) - T(0,t), during early stages of solidification
increasing with decreasing k.
COMMENTS: Use of this technique to determine h from measurement of T(ro) at a prescribed t
requires an interative solution of the governing equations.
PROBLEM 5.68
KNOWN: Properties, initial temperature, and convection conditions associated with cooling of glass
beads.
FIND: (a) Time required to achieve a prescribed center temperature, (b) Effect of convection coefficient
on center and surface temperature histories.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties, (3) Negligible
radiation, (4) Fo \u2265 0.2.
ANALYSIS: (a) With h = 400 W/m2\u22c5K, Bi \u2261 h(ro/3)/k = 400 W/m2\u22c5K(0.0005 m)/1.4 W/m\u22c5K = 0.143 and
the lumped capacitance method should not be used. From the one-term approximation for the center
temperature, Eq. 5.50c,
( )* 2oo 1 1iT T 80 15 0.141 C exp FoT T 477 15\u3b8 \u3b6\u221e\u221e\u2212 \u2212\u2261 = = = \u2212\u2212 \u2212
For Bi \u2261 hro/k = 0.429, Table 5.1 yields \u3b61 = 1.101 rad and C1 = 1.128. Hence,
( )
*
o
2 211
1 1 0.141Fo ln ln 1.715
C 1.1281.101
\u3b8
\u3b6
\uf8eb \uf8f6 \uf8eb \uf8f6\uf8ec \uf8f7= \u2212 = \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7 \uf8ed \uf8f8\uf8ed \uf8f8
( )
3p 22
o
c 2200kg m 800J kg K
t 1.715r 1.715 0.0015m 4.85s
k 1.4 W m K
\u3c1 × \u22c5
= = =
\u22c5
<
From Eq. 5.50b, the corresponding surface (r* = 1) temperature is
( ) ( ) ( )* 1o i o
1
sin 0.892T r , t T T T 15 C 462 C 0.141 67.8 C
1.101
\u3b6\u3b8 \u3b6\u221e \u221e= + \u2212 = + =
$ $ $ <
(b) The effect of h on the surface and center temperatures was determined using the IHT Transient
Conduction Model for a Sphere.
0 4 8 12 16 20
Time, t(s)
0
100
200
300
400
500
Ce
nt
er
 te
m
pe
ra
tu
re
, T
(C
)
h = 100 W/m^2.K, r = 0
h = 400 W/m^2.K, r = 0
h = 1000 W/m^2.K, r = 0 
0 4 8 12 16 20
Time, t(s)
0
100
200
300
400
500
Su
rfa
ce
 te
m
pe
ra
tu
re
, T
(C
)
h = 100 W/m^2.K, r = ro
h = 400 W/m^2.K, r = ro
h = 1000 W/m^2.K, r = ro
Continued...
PROBLEM 5.68 (Cont.)
The cooling rate increases with increasing h, particularly from 100 to 400 W/m2\u22c5K. The temperature
difference between the center and surface decreases with increasing t and, during the early stages of
solidification, with decreasing h.
COMMENTS: Temperature gradients in the glass are largest during the early stages of solidification
and increase with increasing h. Since thermal stresses increase with increasing temperature gradients, the
propensity to induce defects due to crack formation in the glass increases with increasing h. Hence, there
is a value of h above which product quality would suffer and the process should not be operated.
PROBLEM 5.69
KNOWN: Temperature requirements for cooling the spherical material of Ex. 5.4 in air and in a
water bath.
FIND: (a) For step 1, the time required for the center temperature to reach T(0,t) = 335°C while
cooling in air at 20°C with h = 10 W/m2\u22c5K; find the Biot number; do you expect radial gradients to be
appreciable?; compare results with hand calculations in Ex. 5.4; (b) For step 2, time required for the
center temperature to reach T(0,t) = 50°C while cooling in water bath at 20°C with h = 6000 W/m2\u22c5K;
and (c) For step 2, calculate and plot the temperature history, T(x,t) vs. t, for the center and surface of
the sphere; explain features; when do you expect the temperature gradients in the sphere to the largest?
Use the IHT Models | Transient Conduction | Sphere model as your solution tool.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the radial direction, (2) Constant properties.
ANALYSIS: The IHT model represents the series solution for the sphere providing the temperatures
evaluated at (r,t). A selected portion of the IHT code used to obtain results is shown in the Comments.
(a) Using the IHT model with step 1 conditions, the time required for T(0,ta) = T_xt = 335°C with r =
0 and the Biot number are:
at 94.2 s Bi 0.0025= = <
Radial temperature gradients will not be appreciable since Bi = 0.0025 << 0.1. The sphere behaves as
space-wise isothermal object for the air-cooling process. The result is identical to the lumped-
capacitance analysis result of the Text example.
(b) Using the IHT model with step 2 conditions, the time required for T(0,tw) = T_xt = 50°C with r = 0
and Ti = 335°C is
wt 3.0 s= <
Radial temperature gradients will be appreciable, since Bi = 1.5 >> 0.1. The sphere does not behave
as a space-wise isothermal object for the water-cooling process.
(c) For the step 2 cooling process, the temperature histories for the center and surface of the sphere are
calculated using the IHT model.
Continued \u2026..
PROBLEM 5.69 (Cont.)
At early times, the difference between the center and surface temperature is appreciable. It is in this
time region that thermal stresses will be a maximum, and if large enough, can cause fracture. Within 6
seconds, the sphere has a uniform temperature equal to that of the water bath.
COMMENTS: Selected portions of the IHT sphere model codes for steps 1 and 2 are shown below.
/* Results, for part (a), step 1, air cooling; clearly negligible gradient
Bi Fo t T_xt Ti r ro
0.0025 25.13 94.22 335 400 0 0.005 */
// Models | Transient Conduction | Sphere - Step 1, Air cooling
// The temperature distribution T(r,t) is
T_xt = T_xt_trans(&quot;Sphere&quot;,rstar,Fo,Bi,Ti,Tinf) // Eq 5.47
 T_xt