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# Respostas_Livro FT

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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.
2-22
2-59 EES Prob. 2-58 is reconsidered. The rate of heat transfer as a function of the thermal conductivity of
the rod is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.

&quot;GIVEN&quot;
L=0.15 [m]
D=0.05 [m]
T_1=20 [C]
T_2=95 [C]
k=1.2 [W/m-C]

&quot;ANALYSIS&quot;
A=pi*D^2/4
Q_dot=k*A*(T_2-T_1)/L

k [W/m.C] Q [W]
1 0.9817
22 21.6
43 42.22
64 62.83
85 83.45
106 104.1
127 124.7
148 145.3
169 165.9
190 186.5
211 207.1
232 227.8
253 248.4
274 269
295 289.6
316 310.2
337 330.8
358 351.5
379 372.1
400 392.7
0 50 100 150 200 250 300 350 400
0
50
100
150
200
250
300
350
400
k [W/m-C]
Q
[
W
]

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.
2-23
2-60 The base plate of a household iron is subjected to specified heat flux on the left surface and to
specified temperature on the right surface. The mathematical formulation, the variation of temperature in
the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base 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. 4 Heat loss through the upper part of
the iron is negligible.
Properties The thermal conductivity is given to be k = 20 W/m\u22c5°C.
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
2
24
base
0
0 W/m000,50
m 10160
W800 =×== \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
02
2
=
dx
Td
and 20 W/m000,50
)0( ==\u2212 q
dx
dTk &
C85)( 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
x = 0:
k
q
CqkC 0101
&& \u2212=\u2192=\u2212
x = L:
k
Lq
TCLCTCTCLCLT 022122221 )(
&+=\u2192\u2212=\u2192=+=
Substituting into the general solution, the variation of temperature is determined to be 21 and CC

85)006.0(2500
C85
C W/m20
m)006.0)( W/m000,50(
)(
)(
2
2
00
2
0
+\u2212=
°+°\u22c5
\u2212=
+\u2212=++\u2212=
x
x
T
k
xLq
k
Lq
Tx
k
q
xT
&&&

(c) The temperature at x = 0 (the inner surface of the plate) is
C100°=+\u2212= 85)0006.0(2500)0(T
Note that the inner surface temperature is higher than the exposed surface temperature, 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.
2-24
2-61 The base plate of a household iron is subjected to specified heat flux on the left surface and to
specified temperature on the right surface. The mathematical formulation, the variation of temperature in
the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base 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. 4 Heat loss through the upper part of
the iron is negligible.
Properties The thermal conductivity is given to be k = 20 W/m\u22c5°C.
x
T2 =85°CQ=1200 W
A=160 cm2
L=0.6 cm
k
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
2
24
base
0
0 W/m000,75
m 10160
W1200 =×== \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
02
2
=
dx
Td
and 20 W/m000,75
)0( ==\u2212 q
dx
dTk &
C85)( 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
x = 0:
k
q
CqkC 0101
&& \u2212=\u2192=\u2212
x = L:
k
Lq
TCLCTCTCLCLT 022122221 )(
&+=\u2192\u2212=\u2192=+=
Substituting into the general solution, the variation of temperature is determined to be C1 and C2

85)006.0(3750
C85
C W/m20
m)006.0)( W/m000,75(
)(
)(
2
2
00
2
0
+\u2212=
°+°\u22c5
\u2212=
+\u2212=++\u2212=
x
x
T
k
xLq
k
Lq
Tx
k
q
xT
&&&

(c) The temperature at x = 0 (the inner surface of the plate) is
C107.5°=+\u2212= 85)0006.0(3750)0(T
Note that the inner surface temperature is higher than the exposed surface temperature, 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.
2-25
2-62 EES Prob. 2-60 is reconsidered. The temperature as a function of the distance is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.

&quot;GIVEN&quot;
Q_dot=800 [W]
L=0.006 [m]
A_base=160E-4 [m^2]
k=20 [W/m-C]
T_2=85 [C]

&quot;ANALYSIS&quot;
q_dot_0=Q_dot/A_base
T=q_dot_0*(L-x)/k+T_2 &quot;Variation of temperature&quot;
&quot;x is the parameter to be varied&quot;

x [m] T [C]
0 100
0.0006667 98.33
0.001333 96.67
0.002 95
0.002667 93.33
0.003333 91.67
0.004 90
0.004667 88.33
0.005333 86.67
0.006 85

0 0.001 0.002 0.003 0.004 0.005 0.006
84
86
88
90
92
94
96
98
100
x [m]
T
[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.
2-26
2-63 Chilled water flows in a pipe that is well insulated from outside. The mathematical formulation and
the variation of temperature in the pipe 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.
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical
formulation of this problem can be expressed as
Water
Tf
L
Insulated
r1r2
0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d
and )]([
)(
1
1 rTTh
dr
rdT
k f \u2212=\u2212
0
)( 2 =
dr
rdT

(b) Integrating the differential equation once with respect to r gives
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 = r2: 00 1
2
1 =\u2192= C
r
C

r = r1:
ff
f
TCCTh
CrCTh
r
C
k
=\u2192\u2212=
+\u2212=\u2212
22
211
1
1
)(0
)]ln([

Substituting C1 and C2 into the general```