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# ch05

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For fixed k, increasing alpha corresponds to a reduction in the thermal capacitance per unit volume (\u3c1cp)
of the material and hence to a more pronounced reduction in temperature at both surface and interior
locations. Similarly, for fixed \u3b1, decreasing k corresponds to a reduction in \u3c1cp and hence to a more
pronounced decay in temperature.
COMMENTS: In part (a) recognize that Fig. 5.8 could also be used to determine the required
temperatures.
PROBLEM 5.81
KNOWN: Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to
combustion products at 800°C with a convection coefficient of 20 W/m2\u22c5K.
FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (b)
Temperature distribution at time t = 325s.
SCHEMATIC:
ASSUMPTIONS: (1) Oak wall can be treated as semi-infinite solid, (2) One-dimensional conduction,
(3) Constant properties, (4) Negligible radiation.
PROPERTIES: Table A-3, Oak, cross grain (300 K): \u3c1 = 545 kg/m3, c = 2385 J/kg\u22c5K, k = 0.17 W/m\u22c5K,
\u3b1 = k/\u3c1c = 0.17 W/m\u22c5K/545 kg/m3 × 2385 J/kg\u22c5K = 1.31 × 10-7 m2/s.
ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is
given by Eq. 5.60 or by Figure 5.8. Using the figure with
( ) i
i
T 0, t T 400 25 0.48
T T 800 25
\u221e
\u2212
\u2212
= =
\u2212 \u2212
and
( )1/ 2
x 0
2 t\u3b1
=
we obtain h(\u3b1t)1/2/k \u2248 0.75, in which case t \u2248 (0.75k/h\u3b11/2)2. Hence,
( ) 21/ 22 7 2t 0.75 0.17 W m K 20 W m K 1.31 10 m s 310s\u2212\uf8eb \uf8f6\u2248 × \u22c5 \u22c5 × =\uf8ec \uf8f7\uf8ed \uf8f8 <
(b) Using the IHT Transient Conduction Model for a Semi-infinite Solid, the following temperature
distribution was generated for t = 325s.
0 0.005 0.01 0.015 0.02 0.025 0.03
Distance from the surface, x(m)
25
100
175
250
325
400
Te
m
pe
ra
tu
re
, T
(C
)
The temperature decay would become more pronounced with decreasing \u3b1 (decreasing k, increasing \u3c1cp)
and in this case the penetration depth of the heating process corresponds to x \u2248 0.025 m at 325s.
COMMENTS: The result of part (a) indicates that, after approximately 5 minutes, the surface of the
wall will ignite and combustion will ensue. Once combustion has started, the present model is no longer
appropriate.
PROBLEM 5.82
KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall.
Incident radiant flux and duration of radiant heating. Maximum allowable surface temperatures at the
end of heating.
FIND: If maximum allowable temperatures are exceeded.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite medium
approximation, (3) Negligible convection and radiative exchange with the surroundings at the
irradiated surface, (4) Negligible heat transfer from the back surface, (5) Constant properties.
ANALYSIS: The thermal response of the wall is described by Eq. (5.60)
( ) ( )
1/ 2 2
o o
i
2 q t / q xx xT x, t T exp erfc
k 4 t k 2 t
\u3b1 \u3c0
\u3b1 \u3b1
\uf8eb \uf8f6\u2032\u2032 \u2032\u2032
\u2212 \uf8eb \uf8f6\uf8ec \uf8f7= + \u2212 \uf8ec \uf8f7\uf8ec \uf8f7 \uf8ed \uf8f8\uf8ed \uf8f8
where, 7 2pk / c 6.92 10 m / s\u3b1 \u3c1
\u2212
= = × and for ( )1/ 2ot 30 min 1800s, 2q t / / k 284.5 K.\u3b1 \u3c0\u2032\u2032= = = Hence,
at x = 0,
( )T 0,30 min 25 C 284.5 C 309.5 C 325 C= ° + ° = ° < ° <
At ( ) ( )2 1/ 2ox 0.25m, x / 4 t 12.54, q x / k 1, 786K, and x / 2 t 3.54.\u3b1 \u3b1\u2032\u2032= \u2212 = \u2212 = = Hence,
( ) ( ) ( )6T 0.25m, 30 min 25 C 284.5 C 3.58 10 1786 C ~ 0 25 C\u2212= ° + ° × \u2212 ° × \u2248 ° <
Both requirements are met.
COMMENTS: The foregoing analysis is conservative since heat transfer at the irradiated surface
due to convection and net radiation exchange with the environment have been neglected. If the
emissivity of the surface and the temperature of the surroundings are assumed to be \u3b5 = 1 and Tsur =
298K, radiation exchange at Ts = 309.5°C would be ( )4 4 2rad s surq T T 6, 080 W / m K,\u3b5\u3c3\u2032\u2032 = \u2212 = \u22c5
which is significant (~ 60% of the prescribed radiation).
PROBLEM 5.83
KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties of
finger.
FIND: Whether copper or glass feels cooler to touch.
SCHEMATIC:
ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constant
properties, (3) Negligible contact resistance.
PROPERTIES: Skin (given): r = 1000 kg/m
3
, c = 4180 J/kg×K, k = 0.625 W/m×K; Table A-1
(T = 300K), Copper: r = 8933 kg/m
3
, c = 385 J/kg×K, k = 401 W/m×K; Table A-3 (T = 300K),
Glass: r = 2500 kg/m
3
, c = 750 J/kg×K, k = 1.4 W/m×K.
ANALYSIS: Which material feels cooler depends upon the contact temperature Ts given by
Equation 5.63. For the three materials of interest,
( ) ( )
( ) ( )
( ) ( )
1/2 1 / 2 2 1/2
skin
1/2 1/2 2 1/2
cu
1/2 1/2 2 1/2
glass
k c 0.625 1000 4180 1,616 J/m K s
k c 401 8933 385 37,137 J/m K s
k c 1.4 2500 750 1,620 J/m K s .
r
r
r
= ´ ´ = × ×
= ´ ´ = × ×
= ´ ´ = × ×
Since ( ) ( )1/2 1/2cu glassk c k c ,r r>> the copper will feel much cooler to the touch. From Equation
5.63,
( ) ( )
( ) ( )
1/2 1/2
A,i B,iA B
s 1/2 1/2
A B
k c T k c T
T
k c k c
r r
r r
+
=
+
( )
( ) ( )
s cu
1,616 310 37,137 300
T 300.4 K
1,616 37,137
+
= =
+
<
( )
( ) ( )
s glass
1,616 310 1,620 300
T 305.0 K.
1,616 1,620
+
= =
+
<
COMMENTS: The extent to which a material\u2019s temperature is affected by a change in its thermal
environment is inversely proportional to (krc)
1/2
. Large k implies an ability to spread the effect by
conduction; large rc implies a large capacity for thermal energy storage.
PROBLEM 5.84
KNOWN: Initial temperatures, properties, and thickness of two plates, each insulated on one
surface.
FIND: Temperature on insulated surface of one plate at a prescribed time after they are pressed
together.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible
contact resistance.
PROPERTIES: Stainless steel (given): r = 8000 kg/m
3
, c = 500 J/kg×K, k = 15 W/m×K.
ANALYSIS: At the instant that contact is made, the plates behave as semi-infinite slabs and, since
the (rkc) product is the same for the two plates, Equation 5.63 yields a surface temperature of
sT 350K.=
The interface will remain at this temperature, even after thermal effects penetrate to the insulated
surfaces. The transient response of the hot wall may therefore be calculated from Equations 5.40
and 5.41. At the insulated surface (x* = 0), Equation 5.41 yields
( )2o s 1 1
i s
T T
C exp Fo
T T
z- = -
-
where, in principle, h ® ¥ and T¥ ® Ts. From Equation 5.39c, Bi ® ¥ yields 1z = 1.5707, and
from Equation 5.39b
( )
1
1
1 1
4sin
C 1.273
2 sin 2
z
z z
= =
+
Also,
( )
( )
6 2
2 2
3.75 10 m / s 60st
Fo 0.563.
L 0.02 m
a -´
= = =
Hence, ( )2oT 350 1.273exp 1.5707 0.563 0.318400 350
-
= - ´ =
-
oT 365.9 K.= <
COMMENTS: Since Fo > 0.2, the one-term approximation is appropriate.
PROBLEM 5.85
KNOWN: Thickness and properties of liquid coating deposited on a metal substrate. Initial temperature
and properties of substrate.
FIND: (a) Expression for time required to completely solidify the liquid, (b) Time required to solidify
an alumina coating.
SCHEMATIC:
ASSUMPTIONS: (1) Substrate may be approximated as a semi-infinite medium in which there is one-
dimensional conduction, (2) Solid and liquid alumina layers remain at fusion temperature throughout
solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact
resistance at the coating/substrate interface, (4) Negligible solidification contraction, (5) Constant
properties.
ANALYSIS: (a) Performing an energy balance on the solid layer, whose thickness S increases with t,
the latent heat released at the solid/liquid interface must be balanced by the rate of heat conduction```