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# Respostas_Livro FT

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```oxygen.
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 C183)( 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 -183°C
O2
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

183)/1.205.1(7.516C)183(1.2
2
1.2
)m 1.2)(C W/m25(
C W/m18
2
1.21
C)20183(
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 oxygen are determined from
negative) since tank the(to
)m 1.2)(C W/m25(
C W/m18
2
1.21
C)20183(m) 1.2()C W/m18(4
1
)(
44)4(
2
21
2
12
12
12
W245,450\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 1.15===
J/kg 000,213
J/s 450,245
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.
2-65
2-131 A large plane wall is subjected to convection, radiation, and specified temperature on the right
surface and no conditions on the left surface. The mathematical formulation, the variation of temperature
in the wall, and the left 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 and emissivity are given to be k = 8.4 W/m\u22c5°C and \u3b5 = 0.7.
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left
surface, and the mathematical formulation of this problem can be expressed as
02
2
=
dx
Td
and ])273[(][])([])([)( 4surr4224surr4 TTTThTLTTLThdx
LdTk \u2212++\u2212=\u2212+\u2212=\u2212 \u221e\u221e \u3b5\u3c3\u3b5\u3c3
C45)( 2 °== TLT
45°C
\u3b5
x
h
T\u221e
L
Tsurr(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
kTTTThC
TTTThkC
/]})273[(][{
])273[(][
4
surr
4
221
4
surr
4
221
\u2212+\u3b5\u3c3+\u2212\u2212=\u2192
\u2212+\u3b5\u3c3+\u2212=\u2212
\u221e
\u221e
Temperature at x = L: LCTCTCLCLT 122221 )( \u2212=\u2192=+×=
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
)4.0(23.4845
m )4.0(
C W/m4.8
]K) 290()K 318)[(K W/m100.7(5.67+C)2545)(C W/m14(C45
)(])273[(][)()()(
444282
4
surr
4
22
212121
x
x
xL
k
TTTThTCxLTLCTxCxT
\u2212+=
\u2212°\u22c5
\u2212\u22c5×°\u2212°\u22c5+°=
\u2212\u2212+\u3b5\u3c3+\u2212+=\u2212\u2212=\u2212+=
\u2212
\u221e

(c) The temperature at x = 0 (the left surface of the wall) is
C64.3°=\u2212+= )04.0(23.4845)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.
2-66
2-132 The base plate of an iron is subjected to specified heat flux on the left surface and convection and
radiation on the right surface. The mathematical formulation, and an expression for the outer surface
temperature and its value are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3
There is no heat generation. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity and emissivity are given to be k = 18 W/m\u22c5°C and \u3b5 = 0.7.
Analysis (a) Noting that the upper part of the iron is well insulated
and thus the entire heat generated in the resistance wires is transferred
to the base plate, the heat flux through the inner surface is determined
to be
\u3b5
x
h
T\u221e
L
Tsurr
q 2
24
base
0
0 W/m667,66
m 10150
W1000 =×== \u2212A
Q
q
&
&
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
0
2
2
=
dx
Td
and 20 W/m667,66
)0( ==\u2212 q
dx
dTk &
])273[(][])([])([)( 4surr
4
22
4
surr
4 TTTThTLTTLTh
dx
LdTk \u2212++\u2212=\u2212+\u2212=\u2212 \u221e\u221e \u3b5\u3c3\u3b5\u3c3
(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
x = 0:
k
q
CqkC 0101
&& \u2212=\u2192=\u2212
x = L: ])273[(][ 4surr
4
221 TTTThkC \u2212++\u2212=\u2212 \u221e \u3b5\u3c3
Eliminating the constant C1 from the two relations above gives the following 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/m667,66]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 = 759°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.
2-67
2-133 The base plate of an iron is subjected to specified heat flux on the left surface and convection and
radiation on the right surface. The mathematical formulation, and an expression for the outer surface
temperature and its value are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3
There is no heat generation. 4 Heat loss through the upper part of the iron is negligible.
Properties The thermal conductivity and emissivity are given to be k = 18 W/m\u22c5°C and \u3b5 = 0.7.
Analysis (a) Noting that the upper part of the iron is well insulated
and thus the entire heat generated in the resistance wires is transferred
to the base plate, the heat flux through the inner surface is determined
to be
\u3b5
x
h
T\u221e
L
Tsurr
q 2
24
base
0
0 W/m000,100
m 10150
W1500 =×== \u2212A
Q
q
&
&
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
0
2
2
=
dx
Td
and 20 W/m000,100
)0( ==\u2212 q
dx
dTk &
])273[(][])([])([
)( 4
surr
4
22
4
surr
4 TTTThTLTTLTh
dx
LdTk \u2212++\u2212=\u2212+\u2212=\u2212 \u221e\u221e \u3b5\u3c3\u3b5\u3c3
(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
x = 0:
k
q
CqkC 0101
&& \u2212=\u2192=\u2212
x = L: ])273[(][ 4surr
4
221 TTTThkC \u2212++\u2212=\u2212 \u221e \u3b5\u3c3
Eliminating the constant C1 from the two relations above gives the following```