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# Respostas_Livro FT

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```CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
Heat flux at x = 0:
k
q
CqkC 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
90420C90
C W/m5.2
W/m1050)(
2
1
0 +\u2212=°+°\u22c5\u2212=+\u2212= xxTxk
q
xT
&

(c) The temperature at x = L (the right surface of the wall) is
C-36°=+×\u2212= 90m) 3.0(420)(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.
2-31
2-68E A large plate is subjected to convection, radiation, and specified temperature on the top surface and
no conditions on the bottom surface. The mathematical formulation, the variation of temperature in the
plate, and the bottom surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is
constant. 3 There is no heat generation in the plate.
Properties The thermal conductivity and emissivity are
given to be k =7.2 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.7.
x
T\u221e
h
Tsky
L
75°F
\u3b5
Analysis (a) Taking the direction normal to the surface
of the plate to be the x direction with x = 0 at the bottom
surface, and the mathematical formulation of this
problem can be expressed as
02
2
=
dx
Td
and ])460[(][])([])([)( 4sky
4
22
4
sky
4 TTTThTLTTLTh
dx
LdTk \u2212++\u2212=\u2212+\u2212=\u2212 \u221e\u221e \u3b5\u3c3\u3b5\u3c3
F75)( 2 °== TLT
(b) Integrating the differential equation twice with respect to x yields
1C
dx
dT =
21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
Convection at x = L:
kTTTThC
TTTThkC
/]})460[(][{
])460[(][
4
sky
4
221
4
sky
4
221
\u2212++\u2212\u2212=\u2192
\u2212++\u2212=\u2212
\u221e
\u221e
\u3b5\u3c3
\u3b5\u3c3
Temperature at x = L: LCTCTCLCLT 122221 )( \u2212=\u2192=+×=
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be

)3/1(2.2075
ft )12/4(
FftBtu/h 2.7
]R) 480()R 535)[(RftBtu/h 100.7(0.1714+F)9075)(FftBtu/h 12(F75
)(
])460[(][
)()()(
44428-2
4
sky
4
22
212121
x
x
xL
k
TTTTh
TCxLTLCTxCxT
\u2212\u2212=
\u2212°\u22c5\u22c5
\u2212\u22c5\u22c5×°\u2212°\u22c5\u22c5+°=
\u2212\u2212++\u2212+=\u2212\u2212=\u2212+= \u221e \u3b5\u3c3

(c) The temperature at x = 0 (the bottom surface of the plate) is
F68.3°=\u2212×\u2212= )03/1(2.2075)0(T
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.
2-32
2-69E A large plate is subjected to convection and specified temperature on the top surface and no
conditions on the bottom surface. The mathematical formulation, the variation of temperature in the plate,
and the bottom surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is
constant. 3 There is no heat generation in the plate.
Properties The thermal conductivity is given to be k =7.2 Btu/h\u22c5ft\u22c5°F.
Analysis (a) Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the
bottom surface, the mathematical formulation of this problem can be expressed as
0
2
2
=
dx
Td x T\u221e
h
L
75°F
and )(])([)( 2 \u221e\u221e \u2212=\u2212=\u2212 TThTLThdx
LdTk
F75)( 2 °== TLT
(b) Integrating the differential equation twice with respect to x yields
1Cdx
dT =
21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
Convection at x = L: kTThCTThkC /)( )( 2121 \u221e\u221e \u2212\u2212=\u2192\u2212=\u2212
Temperature at x = L: LCTCTCLCLT 122221 )( \u2212=\u2192=+×=
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be

)3/1(2575
ft )12/4(
FftBtu/h 2.7
F)9075)(FftBtu/h 12(F75
)()()()()(
2
2
212121
x
x
xL
k
TThTCxLTLCTxCxT
\u2212\u2212=
\u2212°\u22c5\u22c5
°\u2212°\u22c5\u22c5+°=
\u2212\u2212+=\u2212\u2212=\u2212+= \u221e

(c) The temperature at x = 0 (the bottom surface of the plate) is
F66.7°=\u2212×\u2212= )03/1(2575)0(T

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.
2-63
2-129 A spherical liquid nitrogen 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
evaporation of nitrogen 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 of the tank is given to be k = 18 W/m\u22c5°C. Also, hfg = 198 kJ/kg for
nitrogen.
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
and C196)( 11 °\u2212== TrT
])([)( 22 \u221e\u2212=\u2212 TrThdr
rdTk
(b) Integrating the differential equation once with respect to r gives
r1
r2
h
T\u221e
r -196°C
N2
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 = \u2192 21)( Cr
C
rT +\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
196)/1.205.1(8.549C)196(1.2
2
1.2
)m 1.2)(C W/m25(
C W/m18
2
1.21
C)20196(
1
11)(
2
1
2
1
2
21
2
1
1
1
1
1
1
1
1
\u2212\u2212=°\u2212+\u239f\u23a0
\u239e\u239c\u239d
\u239b \u2212
°\u22c5
°\u22c5\u2212\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
r
r
T
r
r
r
r
hr
k
r
r
TT
T
rr
C
r
C
T
r
C
rT

(c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from
negative) since tank the(to W 261,200
)m 1.2)(C W/m25(
C W/m18
2
1.21
C)20196(m) 1.2()C W/m18(4
1
)(
44)4(
2
21
2
12
12
12
\u2212=
°\u22c5
°\u22c5\u2212\u2212
°\u2212\u2212°\u22c5\u2212=
\u2212\u2212
\u2212\u2212=\u2212=\u2212=\u2212= \u221e
\u3c0
\u3c0\u3c0\u3c0
hr
k
r
r
TTr
kkC
r
C
rk
dx
dTkAQ&

kg/s 1.32===
J/kg 000,198
J/s 200,261
fgh
Qm
&
&
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.
2-64
2-130 A spherical liquid oxygen 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 evaporation of oxygen 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 of the tank is given to be k = 18 W/m\u22c5°C. Also, hfg = 213 kJ/kg for```