Respostas_Livro FT
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Respostas_Livro FT


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solution, the variation of temperature is determined to be 
 fTrT =)(
This result is not surprising since steady operating conditions exist. 
 
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. 
 2-27
2-64E A steam pipe is subjected to convection on the inner surface and to specified temperature on the 
outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat 
loss are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its 
thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There 
is no heat generation in the pipe. 
Properties The thermal conductivity is given to be k = 7.2 Btu/h\u22c5ft\u22c5°F. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical 
formulation of this problem can be expressed as 
 0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d 
and )]([
)(
1
1 rTTh
dr
rdT
k \u2212=\u2212 \u221e 
 F160)( 22 °== TrT
(b) Integrating the differential equation once with respect to r gives 
Steam 
250°F 
h=12.5
L = 30 ft 
 T =160°F 
 1Cdr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
r
C
dr
dT 1= 
 21 ln)( CrCrT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: )]ln([ 211
1
1 CrCTh
r
C
k +\u2212=\u2212 \u221e 
r = r2: 22212 ln)( TCrCrT =+=
Solving for C1 and C2 simultaneously gives 
 2
11
2
2
22122
11
2
2
1 ln
ln
ln and 
ln
r
hr
k
r
r
TT
TrCTC
hr
k
r
r
TT
C
+
\u2212\u2212=\u2212=
+
\u2212= \u221e\u221e 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
F160
in 4.2
ln74.24F160
in 4.2
ln
)ft 12/2)(FftBtu/h 5.12(
FftBtu/h 2.7
2
4.2ln
F)250160(
ln
ln
)ln(lnlnln)(
2
2
2
11
2
2
2212121
°+\u2212=°+
°\u22c5\u22c5
°\u22c5\u22c5+
°\u2212=
+
+
\u2212=+\u2212=\u2212+= \u221e
rr
T
r
r
hr
k
r
r
TT
TrrCrCTrCrT
 
 (c) The rate of heat conduction through the pipe is 
 
Btu/h 33,600=
°\u22c5\u22c5
°\u22c5\u22c5+
°\u2212°\u22c5\u22c5\u2212=
+
\u2212\u2212=\u2212=\u2212= \u221e
)ft 12/2)(FftBtu/h 5.12(
FftBtu/h 2.7
2
4.2ln
F)250160(F)ftBtu/h 2.7ft)( 30(2
ln
2)2(
2
11
2
21
\u3c0
\u3c0\u3c0
hr
k
r
r
TT
Lk
r
C
rLk
dr
dTkAQ&
 
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. 
 2-28
2-65 A spherical container is subjected to specified temperature on the inner surface and convection on the 
outer surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are 
to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there 
is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. 
Properties The thermal conductivity is given to be k = 30 W/m\u22c5°C. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical 
formulation of this problem can be expressed as 
02 =\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d 
r1 r2
T1k 
T\u221e
h and C0)( 11 °== TrT
])([)( 22 \u221e\u2212=\u2212 TrThdr
rdTk 
(b) Integrating the differential equation once with respect to r gives 
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
2
1
r
C
dr
dT = 
 21)( Cr
CrT +\u2212= 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 12
1
1
1 )( TCr
C
rT =+\u2212= 
r = r2: \u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212+\u2212=\u2212 \u221eTCr
Ch
r
Ck 2
2
1
2
2
1 
Solving for C1 and C2 simultaneously gives 
 
1
2
21
2
1
1
1
1
12
21
2
12
1
1
 and 
1
)(
r
r
hr
k
r
r
TT
T
r
C
TC
hr
k
r
r
TTr
C
\u2212\u2212
\u2212+=+=
\u2212\u2212
\u2212= \u221e\u221e 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
)/1.205.1(63.29C01.2
2
1.2
)m 1.2)(C W/m18(
C W/m30
2
1.21
C)250(
1
11)(
2
1
2
1
2
21
2
1
1
1
1
1
1
1
1
r
r
T
r
r
r
r
hr
k
r
r
TTT
rr
C
r
CT
r
CrT
\u2212=°+\u239f\u23a0
\u239e\u239c\u239d
\u239b \u2212
°\u22c5
°\u22c5\u2212\u2212
°\u2212=
+\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212
\u2212\u2212
\u2212=+\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212=++\u2212= \u221e
 
(c) The rate of heat conduction through the wall is 
 
 W23,460=
°\u22c5
°\u22c5\u2212\u2212
°\u2212°\u22c5\u2212=
\u2212\u2212
\u2212\u2212=\u2212=\u2212=\u2212= \u221e
)m 1.2)(C W/m18(
C W/m30
2
1.21
C)250(m) 1.2()C W/m30(4
1
)(44)4(
2
21
2
12
12
12
\u3c0
\u3c0\u3c0\u3c0
hr
k
r
r
TTrkkC
r
Crk
dr
dTkAQ&
 
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. 
 2-29
2-66 A large plane wall is subjected to specified heat flux and temperature on the left surface and no 
conditions on the right surface. The mathematical formulation, the variation of temperature in the plate, 
and the right surface temperature are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its 
thickness, and the thermal conditions on both sides of the wall are uniform. 2 Thermal conductivity is 
constant. 3 There is no heat generation in the wall. 
Properties The thermal conductivity is given to be k =2.5 W/m\u22c5°C. 
Analysis (a) Taking the direction normal to the surface of the wall 
to be the x direction with x = 0 at the left surface, the 
mathematical formulation of this problem can be expressed as 
 02
2
=
dx
Td 
and 20 W/m700
)0( ==\u2212 q
dx
dTk & 
 C80)0( 1 °== TT
 (b) Integrating the differential equation twice with respect to x yields 
x 
q=700 W/m2
T1=80°C 
L=0.3 m 
k 
 1Cdx
dT = 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Heat flux at x = 0: 
k
qCqkC 0101 
&& \u2212=\u2192=\u2212 
Temperature at x = 0: 12121 0)0( TCTCCT =\u2192=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 80280C80
C W/m5.2
 W/m700)(
2
1
0 +\u2212=°+°\u22c5\u2212=+\u2212= xxTxk
q
xT
&
 
 (c) The temperature at x = L (the right surface of the wall) is 
 C-4°=+×\u2212= 80m) 3.0(280)(LT
Note that the right surface temperature is lower 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. 
 2-30
2-67 A large plane wall is subjected to specified heat flux and temperature on the left surface and no 
conditions on the right surface. The mathematical formulation, the variation of temperature in the plate, 
and the right surface temperature are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its 
thickness, and the thermal conditions on both sides of the wall are uniform. 2 Thermal conductivity is 
constant. 3 There is no heat generation in the wall. 
Properties The thermal conductivity is given to be k =2.5 W/m\u22c5°C. 
Analysis (a) Taking the direction normal to the surface of the wall 
to be the x direction with x = 0 at the left surface, the 
mathematical formulation of this problem can be expressed as 
 02
2
=
dx
Td 
and 20 W/m1050
)0( ==\u2212 q
dx
dTk & 
 C90)0( 1 °== TT
 (b) Integrating the differential equation twice with respect to x yields 
x 
q=1050 W/m2
T1=90°C 
L=0.3 m 
k 
 1Cdx
dT = 
 21)(