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# Respostas_Livro FT

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```expression for the outer
surface temperature T2,
0
4
surr
4
22 ])273[()( qTTTTh &=\u2212++\u2212 \u221e \u3b5\u3c3
(c) Substituting the known quantities into the implicit relation above gives
2442
428
2
2 W/m000,100]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =\u2212+\u22c5×+\u2212°\u22c5 \u2212 TT
Using an equation solver (or a trial and error approach), the outer surface temperature is determined from
the relation above to be
T2 = 896°C
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-68
2-134E The concrete slab roof of a house is subjected to specified temperature at the bottom surface and
convection and radiation at the top surface. The temperature of the top surface of the roof and the rate of
heat transfer are to be determined when steady operating conditions are reached.
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the roof
area is large relative to its thickness, and the thermal conditions on both sides of the roof are uniform. 3
Thermal properties are constant. 4 There is no heat generation in the wall.
Properties The thermal conductivity and emissivity are given to be k = 1.1 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.8.
Analysis In steady operation, heat conduction through the roof must be equal to net heat transfer from the
outer surface. Therefore, taking the outer surface temperature of the roof to be T2 (in °F),
])460[()( 4sky
4
22
21 TTATThA
L
TT
kA \u2212++\u2212=\u2212 \u221e \u3c3\u3b5 x T\u221e
h
T1
L
Tsky
Canceling the area A and substituting the known quantities,
444
2
428
2
22
R]310)460)[(RftBtu/h 101714.0(8.0
F)50)(FftBtu/h 2.3(
ft 0.8
F)62(
)FftBtu/h 1.1(
\u2212+\u22c5\u22c5×+
°\u2212°\u22c5\u22c5=°\u2212°\u22c5\u22c5
\u2212 T
T
T

Using an equation solver (or the trial and error method), the outer surface temperature is determined to be
T2 = 38°F
Then the rate of heat transfer through the roof becomes
Btu/h 28,875=°\u2212×°\u22c5\u22c5=\u2212=
ft 0.8
F)3862()ft 3525)(FftBtu/h 1.1( 221
L
TT
kAQ&
Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside.
Therefore, the house is losing heat as expected.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-69
2-135 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be
determined.
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-
dimensional since this two-layer heat transfer problem possesses symmetry about the center line and
involves no change in the axial direction, and thus T = T(r) . 3 Thermal conductivities are constant. 4 Heat
generation in the wire is uniform.
Properties It is given that C W/m18wire °\u22c5=k and C W/m8.1plastic °\u22c5=k .
Analysis Letting denote the unknown interface temperature, the mathematical formulation of the heat
transfer problem in the wire can be expressed as
IT
01 gen =+\u239f\u23a0
\u239e\u239c\u239d
\u239b
k
e
dr
dTr
dr
d
r
&

r2
r
r1
egen
T\u221e
h
with and ITrT =)( 1 0)0( =dr
dT
Multiplying both sides of the differential
equation by r, rearranging, and integrating give
r
k
e
dr
dTr
dr
d gen&\u2212=\u239f\u23a0
\u239e\u239c\u239d
\u239b \u2192 1
2
gen
2
Cr
k
e
dr
dTr +\u2212= & (a)
Applying the boundary condition at the center (r = 0) gives
B.C. at r = 0: 0 0
2
)0(
0 11
gen =\u2192+×\u2212=× CC
k
e
dr
dT &
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,
r
k
e
dr
dT
2
gen&\u2212= \u2192 22gen4)( Crk
e
rT +\u2212= & (b)
Applying the other boundary condition at 1rr = ,
B. C. at : 1rr = 21gen2221gen 4 4 rk
e
TCCr
k
e
T II
&& +=\u2192+\u2212=
Substituting this relation into Eq. (b) and rearranging give 2C
)(
4
)( 221
wire
gen
wire rrk
e
TrT I \u2212+=
&
(c)
Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as
0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d
with and ITrT =)( 1 ])([)( 22 \u221e\u2212=\u2212 TrThdr
rdTk
The solution of the differential equation is determined by integration to be
1Cdr
dTr = \u2192
r
C
dr
dT 1= \u2192 21 ln)( CrCrT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
r = r1: 112211 ln ln rCTCTCrC II \u2212=\u2192=+
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-70
r = r2: ])ln[( 221
2
1 \u221e\u2212+=\u2212 TCrChr
Ck \u2192
21
2
1
ln
hr
k
r
r
TTC I
+
\u2212= \u221e
Substituting C1 and C2 into the general solution, the variation of temperature in plastic is determined to be

1
2
plastic
1
2
111plastic ln
ln
lnln)(
r
r
hr
k
r
r
TT
TrCTrCrT III
+
\u2212+=\u2212+= \u221e
We have already utilized the first interface condition by setting the wire and plastic layer temperatures
equal to at the interface . The interface temperature is determined from the second interface
condition that the heat flux in the wire and the plastic layer at
IT 1rr = IT
1rr = must be the same:

1
2
plastic
1
2
plastic
1gen1plastic
plastic
1wire
wire
1
ln
2

)()(
r
hr
k
r
r
TT
k
re
dr
rdT
k
dr
rdT
k I
+
\u2212\u2212=\u2192\u2212=\u2212 \u221e&
Solving for and substituting the given values, the interface temperature is determined to be IT
C97.1°°+\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
°\u22c5
°\u22c5+°\u22c5
×=
+\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b += \u221e
=C25
m) C)(0.007 W/m(14
C W/m1.8
m 0.003
m 007.0ln
C) W/m2(1.8
m) 003.0)( W/m105.1(
ln
2
2
236
2
plastic
1
2
plastic
2
1gen T
hr
k
r
r
k
re
TI
&

Knowing the interface temperature, the temperature at the center line (r = 0) is obtained by substituting the
known quantities into Eq. (c),
C97.3°=°\u22c5×
×°=+=
C) W/m(184
m) 003.0)( W/m105.1(+C1.97
4
)0(
236
wire
2
1gen
wire k
re
TT I
&

Thus the temperature of the centerline will be slightly above the interface temperature.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-71
2-136 A cylindrical shell with variable conductivity is
subjected to specified temperatures on both sides. The rate of
heat transfer through the shell is to be determined.
r2
T2
r
r1
T1
k(T)
Assumptions 1 Heat transfer is given to be steady and one-
dimensional. 2 Thermal conductivity varies quadratically. 3
There is no heat generation.
Properties The thermal conductivity is given to be
. )1()( 20 TkTk \u3b2+=
Analysis When the variation of thermal conductivity with
temperature k(T) is known, the average value of the thermal
conductivity in the temperature range between is
determined from
21 and TT

( ) ( )
( )\u23a5\u23a6\u23a4\u23a2\u23a3\u23a1 +++=
\u2212
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 \u2212+\u2212
=\u2212
\u239f\u23a0
\u239e\u239c\u239d
\u239b +
=\u2212
+
=\u2212=
\u222b\u222b
2
121
2
20
12
3
1
3
2120
12
3
0
12
2
0
12
avg
3
1
33
)1()(
2
1
2
1
2
1
TTTTk
TT
TTTTk
TT
TTk
TT
dTTk
TT
dTTk
k
T
T
T
T
T
T
\u3b2
\u3b2\u3b2\u3b2

This relation is based on the requirement that the rate of heat transfer through a medium with constant
average thermal conductivity equals the rate of heat transfer through the same medium with variable
conductivity k(T).
avgk
Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be
( )
)/ln(3
12
)/ln(
2
12
212
121
2
20
12
21
avgcylinder rr
TT
LTTTTk
rr
TT
LkQ
\u2212
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 +++=\u2212= \u3b2\u3c0\u3c0&
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part
of Eq. 2-77, and performed the indicated integration.

2-137 Heat is generated uniformly```