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ch05


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(2) Constant properties.
ANALYSIS: Prescribed surface temperature is analogous to h ® ¥ and T¥ = Ts. Hence, Bi = ¥.
Assume validity of one-term approximation to series solution for T (x,t).
(a) At the midplane,
( )2o so 1 1
i s
T T
0.50 C exp Fo
T T
q z* -= = = -
-
1 1 1tan Bi /2.z z z p= = ¥ ® =
Hence
( )
1
1
1 1
4sin 4
C 1.273
2 sin 2
z
z z p
= = =
+
( )o 1
2
1
ln / C
Fo 0.379
q
z
*
= - =
( )22
7 2
0.379 0.01 mFoL
t 63 s.
6 10 m /sa -
= = =
´
<
(b) With ( )21 11C exp Fo cos xq z z* *= -
( ) ( ) ( )i s i s 21 1 11T T T T T C exp Fo sin x x L L x
¶ ¶q z z z
¶ ¶
*
*
*
- -
= = - -
 
max
4
x 1
300 C
 T/ x T/ x 0.5 2.36 10 C/m.
0.01 m 2
p¶ ¶ ¶ ¶ *== = - = - ´
o
o <
COMMENTS: Validity of one-term approximation is confirmed by Fo > 0.2.
PROBLEM 5.41
KNOWN: Thickness and properties of rubber tire. Convection heating conditions. Initial and final
midplane temperature.
FIND: (a) Time to reach final midplane temperature. (b) Effect of accelerated heating.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)
Negligible radiation.
ANALYSIS: (a) With Bi = hL/k = 200 W/m2\u22c5K(0.01 m)/0.14 W/m\u22c5K = 14.3, the lumped capacitance
method is clearly inappropriate. Assuming Fo > 0.2, Eq. (5.41) may be used with C1 = 1.265 and \u3b61 \u2248
1.458 rad from Table 5.1 to obtain
( ) ( )* 2oo 1 1iT T C exp Fo 1.265exp 2.126 FoT T\u3b8 \u3b6\u221e\u221e\u2212= = \u2212 = \u2212\u2212
With ( ) ( )*o o iT T T T\u3b8 \u221e \u221e= \u2212 \u2212 = (-50)/(-175) = 0.286, ( ) 2fFo ln 0.286 1.265 2.126 0.70 t L\u3b1= \u2212 = =
( )2
f 8 2
0.7 0.01m
t 1100s
6.35 10 m s\u2212
= =
×
<
(b) The desired temperature histories were generated using the IHT Transient Conduction Model for a
Plane Wall, with h = 5 × 104 W/m2\u22c5K used to approximate imposition of a surface temperature of 200°C.
0 200 400 600 800 1000 1200
Time, t(s)
0
50
100
150
200
Te
m
pe
ra
tu
re
, T
(C
)
x = 0, h = 200 W/m^2.K
x = L, h = 200 W/m^2.K
x = 0, h = 5E4 W/m^2.K
x = L, h = 5E4W/m^2.K
The fact that imposition of a constant surface temperature (h \u2192 \u221e) does not significantly accelerate the
heating process should not be surprising. For h = 200 W/m2\u22c5K, the Biot number is already quite large (Bi
= 14.3), and limits to the heating rate are principally due to conduction in the rubber and not to
convection at the surface. Any increase in h only serves to reduce what is already a small component of
the total thermal resistance.
COMMENTS: The heating rate could be accelerated by increasing the steam temperature, but an upper
limit would be associated with avoiding thermal damage to the rubber.
PROBLEM 5.42
KNOWN: Stack or book comprised of 11 metal plates (p) and 10 boards (b) each of 2.36 mm
thickness and prescribed thermophysical properties.
FIND: Effective thermal conductivity, k, and effective thermal capacitance, (\u3c1cp).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistance
between plates and boards.
PROPERTIES: Metal plate (p, given): \u3c1p = 8000 kg/m3, cp,p = 480 J/kg\u22c5K, kp = 12
W/m\u22c5K; Circuit boards (b, given): \u3c1b = 1000 kg/m3, cp,b = 1500 J/kg\u22c5K, kb = 0.30 W/m\u22c5K.
ANALYSIS: The thermal resistance of the book is determined as the sum of the resistance of
the boards and plates,
tot b pR NR MR\u2032\u2032 \u2032\u2032 \u2032\u2032= +
where M,N are the number of plates and boards in the book, respectively, and i i iR L / k\u2032\u2032 =
where Li and ki are the thickness and thermal conductivities, respectively.
( ) ( )
( ) ( )tot p p b btot
3 2 2
tot
R M L / k N L / k
R 11 0.00236 m/12 W/m K 10 0.00236 m/0.30 W/m K
R 2.163 10 7.867 10 8.083 10 K/W.\u2212 \u2212 \u2212
\u2032\u2032 = +
\u2032\u2032 = \u22c5 + \u22c5
\u2032\u2032 = × + × = ×
The effective thermal conductivity of the book of thickness (10 + 11) 2.36 mm is
tot
-2
0.04956 mk L/R 0.613 W/m K.
8.083 10 K/W
\u2032\u2032= = = \u22c5
×
<
The thermal capacitance of the stack is
( ) ( )( ) ( )tot p p p b b b3 3tot
4 4 5 2
tot
C M L c N L c
C 11 8000 kg/m 0.00236 m 480 J/kg K 10 1000 kg/m 0.00236 m 1500 J/kg K
C 9.969 10 3.540 10 1.35 10 J/m K.
\u3c1 \u3c1\u2032\u2032 = +
\u2032\u2032 = × × \u22c5 + × × \u22c5
\u2032\u2032 = × + × = × \u22c5
The effective thermal capacitance of the book is
( ) 5 2 6 3p totc C / L 1.351 10 J/m K/0.04956 m 2.726 10 J/m K.\u3c1 \u2032\u2032= = × \u22c5 = × \u22c5 <
COMMENTS: The results of the analysis allow for representing the stack as a homogeneous
medium with effective properties: k = 0.613 W/m\u22c5K and \u3b1 = (k/\u3c1cp) = 2.249×10-7 m2/s. See
for example, Problem 5.38.
PROBLEM 5.43
KNOWN: Stack of circuit board-pressing plates, initially at a uniform temperature, is subjected by
upper/lower platens to a higher temperature.
FIND: (a) Elapsed time, te, required for the mid-plane to reach cure temperature when platens are
suddenly changed to Ts = 190°C, (b) Energy removal from the stack needed to return its temperature
to Ti.
SCHEMATIC:
PROPERTIES: Stack (given): k = 0.613 W/m\u22c5K, \u3c1cp = 2.73×10
6
 J/m3\u22c5K; \u3b1 = k/\u3c1cp = 2.245×10
-7
m
2/s.
ANALYSIS: (a) Recognize that sudden application of surface temperature corresponds to h \u2192 \u221e, or
Bi-1 = 0 (Heisler chart) or Bi \u2192 \u221e (100, Table 5.1). With Ts = T\u221e,
( ) ( )
( )
s
o
i s
T 0,t T 170 190 C
0.114.
T T 15 190 C
\u3b8\u2217 \u2212 \u2212= = =
\u2212
\u2212
$
$
Using Eq. 5.41 with values of 1 11.552 and C 1.2731 at Bi 100\u3b6 = = = (Table 5.1), find Fo
( )2o 1 1C exp Fo\u3b8 \u3b6\u2217 = \u2212
( ) ( ) ( )o 12 21
1 1Fo ln / C ln 0.114/1.2731 1.002
1.552
\u3b8
\u3b6
\u2217
= \u2212 = \u2212 =
where Fo = \u3b1t/L2,
( )232 3
7 2
1.002 25 10 mFoL
t 2.789 10 s 46.5 min.
2.245 10 m / s\u3b1
\u2212
\u2212
×
= = = × =
×
<
The Heisler chart, Figure D.1, could also be used to find Fo from values of -1o and Bi 0.\u3b8\u2217 =
(b) The energy removal is equivalent to the energy gained by the stack per unit area for the time
interval 0 \u2192 te. With oQ\u2032\u2032 corresponding to the maximum amount of energy that could be transferred,
( )( ) ( )( )6 3 -3 7 2.o iQ c 2L T T 2.73 10 J/m K 2 25 10 m 15 190 K 2.389 10 J/m\u3c1 \u221e\u2032\u2032 = \u2212 = × \u22c5 × × \u2212 = \u2212 ×
Q\u2032\u2032 may be determined from Eq. 5.46,
( )1
o
o 1
sin 1.552radQ sin1 1 0.114 0.795Q 1.552rad
\u3b6 \u3b8\u3b6
\u2217\u2032\u2032
= \u2212 = \u2212 × =
\u2032\u2032
We conclude that the energy to be removed from the stack per unit area to return it to Ti is
7 2 7 2
oQ 0.795Q 0.795 2.389 10 J/m 1.90 10 J/m .\u2032\u2032 \u2032\u2032= = × × = × <
PROBLEM 5.44
KNOWN: Car windshield, initially at a uniform temperature of -20°C, is suddenly exposed on its
interior surface to the defrost system airstream at 30°C. The ice layer on the exterior surface acts as an
insulating layer.
FIND: What airstream convection coefficient would allow the exterior surface to reach 0°C in 60 s?
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, transient conduction in the windshield, (2) Constant properties,
(3) Exterior surface is perfectly insulated.
PROPERTIES: Windshield (Given): \u3c1 = 2200 kg/m3, cp = 830 J/kg\u22c5K and k = 1.2 W/m\u22c5K.
ANALYSIS: For the prescribed conditions, from Equations 5.31 and 5.33,
 
( ) ( ) ( )
( )
o
i i i
0,60s T 0,60s T 0 30 C
0.6
T T 20 30 C
\u3b8 \u3b8
\u3b8 \u3b8
\u221e
\u221e
\u2212 \u2212
= = = =
\u2212
\u2212 \u2212
$
$
( )2 23
kt 1.2 W m K 60Fo 1.58
cL 2200kg m 830J kg K 0.005m\u3c1
\u22c5 ×
= = =
× \u22c5 ×
The single-term series approximation, Eq. 5.41, along with Table 5.1, requires an iterative solution to
find an appropriate Biot number. Alternatively, the Heisler charts, Appendix D, Figure D.1, for the
midplane temperature could be used to find
1Bi k hL 2.5\u2212 = =
2h 1.2 W m K 2.5 0.005m 96 W m K= \u22c5 × = \u22c5 <
COMMENTS: Using the IHT, Transient Conduction, Plane Wall Model, the convection coefficient
can be determined by solving the model with an assumed h and then sweeping over a range of h until the
T(0,60s) condition is satisfied. Since the model is based upon multiple terms of the series, the result of h
= 99 W/m2\u22c5K