ch02
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ch02


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One-dimensional heat transfer through samples
and reference material, (3) Negligible thermal contact resistance between materials.
PROPERTIES: Table A.2, Armco iron T = 350 K k W / m K.r\ufffd \ufffd: .= \u22c569 2
ANALYSIS: (a) Recognizing that the heat rate through the samples and reference material, all of the
same diameter, is the same, it follows from Fourier\u2019s law that
k
T
x
k T
x
k
T
x
t
t,1
r
r
t
t,2\u2206
\u2206
\u2206
\u2206
\u2206
\u2206
= =
k k T
T
 W / m K 2.49 C
3.32 C
 W / m K.t r r
t
= = \u22c5
°
°
= \u22c5
\u2206
\u2206
69 2 519. . <
We should assign this value a temperature of 350 K. <
(b) If the test samples are identical in every respect, \u2206Tt,1 \u2260 \u2206Tt,2 if the thermal conductivity is highly
dependent upon temperature. Also, if there is heat leakage out the lateral surface, we can expect
\u2206Tt,2 < \u2206Tt,1. Leakage could be influential, if the thermal conductivity of the test material were less
than an order of magnitude larger than that of the insulating material.
PROBLEM 2.19
KNOWN: Identical samples of prescribed diameter, length and density initially at a uniform
temperature Ti, sandwich an electric heater which provides a uniform heat flux \u2032\u2032qo for a period of
time \u2206to. Conditions shortly after energizing and a long time after de-energizing heater are
prescribed.
FIND: Specific heat and thermal conductivity of the test sample material. From these properties,
identify type of material using Table A.1 or A.2.
SCHEMATIC:
ASSUMPTIONS: (1) One dimensional heat transfer in samples, (2) Constant properties, (3)
Negligible heat loss through insulation, (4) Negligible heater mass.
ANALYSIS: Consider a control volume about the samples
and heater, and apply conservation of energy over the time
interval from t = 0 to \u221e
E E E = E Ein out f i\u2212 = \u2212\u2206
P t Mc T To p i\u2206 \u2212 = \u221e \u22120 \ufffd \ufffd
where energy inflow is prescribed by the Case A power condition and the final temperature Tf by
Case B. Solving for cp,
c
P t
M T T
 W 120 s
2 3965 kg / m m m 33.50 - 23.00 Cp
o
i
3 2=
\u221e \u2212
=
×
× × × °
\u2206
\ufffd \ufffd \ufffd \ufffd
15
0 060 4 0 0102\u3c0 . / .
c J / kg Kp = \u22c5765 <
where M = \u3c1V = 2\u3c1(\u3c0D2/4)L is the mass of both samples. For Case A, the transient thermal response
of the heater is given by
Continued \u2026..
PROBLEM 2.19 (Cont.)
T t T q t
c k
k = t
c
q
T t T
o i o
p
p
o
o i
\ufffd \ufffd
\ufffd \ufffd
\u2212 = \u2032\u2032
\ufffd
\ufffd
\ufffd
\ufffd
\ufffd
	
\u2032\u2032
\u2212
\ufffd
\ufffd
\ufffd
\ufffd
	
2
2
1 2
2
\u3c0\u3c1
\u3c0\u3c1
/
k = 30 s
3965 kg / m J / kg K
 W / m
24.57 - 23.00 C
 W / m K3
2
\u3c0 × × \u22c5
×
°
\ufffd
\ufffd
\ufffd
\ufffd
	
 = \u22c5765
2 2653 36 0
2
\ufffd \ufffd
. <
where
\u2032\u2032 = = =
×
=q P
2A
P
2 D
 W
2 0.060 m
 W / mo
s
2 2 2
2
\u3c0 \u3c0/ /
.
4
15
4
2653
\ufffd \ufffd \ufffd \ufffd
With the following properties now known,
\u3c1 = 3965 kg/m3 cp = 765 J/kg\u22c5K k = 36 W/m\u22c5K
entries in Table A.1 are scanned to determine whether these values are typical of a metallic material.
Consider the following,
\u2022 metallics with low \u3c1 generally have higher thermal conductivities,
 
\u2022 specific heats of both types of materials are of similar magnitude,
 
\u2022 the low k value of the sample is typical of poor metallic conductors which generally have
much higher specific heats,
 
\u2022 more than likely, the material is nonmetallic.
From Table A.2, the second entry, polycrystalline aluminum oxide, has properties at 300 K
corresponding to those found for the samples. <
PROBLEM 2.20
KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a given
instant of time.
FIND: Regions where the temperature changes with time.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties of infinite medium and (2) No internal heat generation.
ANALYSIS: The temperature distribution throughout the medium, at any instant of time, must
satisfy the heat equation. For the three-dimensional cartesian coordinate system, with constant
properties and no internal heat generation, the heat equation, Eq. 2.15, has the form
\u2202
\u2202
\u2202
\u2202
\u2202
\u2202 \u3b1
\u2202
\u2202
2 2 2 1T
 x
T
 y
T
 z
 T
 t2 2 2
+ + = . (1)
If T(x,y,z) satisfies this relation, conservation of energy is satisfied at every point in the medium.
Substituting T(x,y,z) into the Eq. (1), first find the gradients, \u2202T/\u2202x, \u2202T/\u2202y, and \u2202T/\u2202z.
\u2202
\u2202
\u2202
\u2202
\u2202
\u2202 \u3b1
\u2202
\u2202 x
x - y
 y
y - x + 2z
 z
z + 2y T
 t
2 4 2 1\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd+ \u2212 + = .
Performing the differentiations,
2 4 2 1\u2212 + =
\u3b1
\u2202
\u2202
 T
 t
.
Hence,
\u2202
\u2202
 T
 t
= 0
which implies that, at the prescribed instant, the temperature is everywhere independent of time.
COMMENTS: Since we do not know the initial and boundary conditions, we cannot determine the
temperature distribution, T(x,y,z), at any future time. We can only determine that, for this special
instant of time, the temperature will not change.
PROBLEM 2.21
KNOWN: Diameter D, thickness L and initial temperature Ti of pan. Heat rate from stove to bottom
of pan. Convection coefficient h and variation of water temperature T\u221e(t) during Stage 1.
Temperature TL of pan surface in contact with water during Stage 2.
FIND: Form of heat equation and boundary conditions associated with the two stages.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in pan bottom, (2) Heat transfer from stove is
uniformly distributed over surface of pan in contact with the stove, (3) Constant properties.
ANALYSIS:
Stage 1
Heat Equation:
2
2
T 1 T
tx \u3b1
\u2202 \u2202
=
\u2202\u2202
Boundary Conditions: ( )oo 2x 0
qTk q
x D / 4\u3c0=
\u2202
\u2032\u2032
\u2212 = =
\u2202
( ) ( )
x L
Tk h T L, t T t
x
\u221e
=
\u2202 \uf8ee \uf8f9\u2212 = \u2212\uf8f0 \uf8fb\u2202
Initial Condition: ( ) iT x,0 T=
Stage 2
Heat Equation:
2
2
d T 0
dx
=
Boundary Conditions: o
x 0
dTk q
dx
=
\u2032\u2032
\u2212 =
( ) LT L T=
COMMENTS: Stage 1 is a transient process for which T\u221e(t) must be determined separately. As a
first approximation, it could be estimated by neglecting changes in thermal energy storage by the pan
bottom and assuming that all of the heat transferred from the stove acted to increase thermal energy
storage within the water. Hence, with q \u2248 Mcp d T\u221e/dt, where M and cp are the mass and specific
heat of the water in the pan, T\u221e(t) \u2248 (q/Mcp) t.
PROBLEM 2.22
KNOWN: Steady-state temperature distribution in a cylindrical rod having uniform heat generation
of \ufffd .q W / m1
3
= ×5 107
FIND: (a) Steady-state centerline and surface heat transfer rates per unit length, \u2032qr . (b) Initial time
rate of change of the centerline and surface temperatures in response to a change in the generation rate
from \ufffd \ufffd .q to q = 10 W / m1 2
8 3
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the r direction, (2) Uniform generation, and
(3) Steady-state for \ufffd .q = 5 10 W / m1 7 3×
ANALYSIS: (a) From the rate equations for cylindrical coordinates,
\u2032\u2032 = \u2212q k T
 r
 q = -kA T
 r
r r
\u2202
\u2202
\u2202
\u2202 .
Hence,
q k 2 rL T
 r
r = \u2212 \u3c0
\u2202
\u2202\ufffd \ufffd
or
\u2032 = \u2212q kr T
 r
r 2\u3c0
\u2202
\u2202
where \u2202T/\u2202r may be evaluated from the prescribed temperature distribution, T(r).
At r = 0, the gradient is (\u2202T/\u2202r) = 0. Hence, from Eq. (1) the heat rate is
\u2032 =qr 0 0\ufffd \ufffd . <
At r = ro, the temperature gradient is
\u2202
\u2202
\u2202
\u2202
 T
 r
K
m
r m
 T
 r
 K / m.
r=r
2 o
r=r
o
o
\ufffd
\ufffd\ufffd
= \u2212 ×
\ufffd
\ufffd\ufffd
\ufffd
\ufffd\ufffd
= \u2212 ×
\ufffd
\ufffd\ufffd
= \u2212 ×
2 4 167 10 2 4167 10 0 025
0 208 10
5 5
5
. . .
.
\ufffd \ufffd 	 
\ufffd \ufffd
Continued \u2026..
PROBLEM 2.22(Cont.)
Hence, the heat rate at the outer surface (r = ro) per unit length is
\u2032 = \u2212 \u22c5 \u2212 ×q r W / m K m K / mr o\ufffd \ufffd \ufffd \ufffd2 30 0 025 0 208 105\u3c0 . .
\u2032 = ×q r W / m.r o\ufffd \ufffd 0 980 105. <
(b) Transient (time-dependent) conditions will exist when the generation is changed, and for the
prescribed assumptions, the temperature is determined by the following