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# Respostas_Livro FT

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2-44
2-89 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature
of the wire is to be determined using the applicable relations directly and by solving the applicable
differential equation.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer
is one-dimensional since there is thermal symmetry about the center line and no change in the axial
direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform.
Properties The thermal conductivity is given to be k = 15.1 W/m\u22c5°C.
T\u221e
h
r
k
egen
T\u221e
h
ro0
Analysis (a) The heat generation per unit volume of the wire is
38
22
gen
wire
gen
gen W/m10061.1
m) (6m) 001.0(
W2000 ×==== \u3c0\u3c0 Lr
EE
e
o
&&
&
V

The surface temperature of the wire is then (Eq. 2-68)
C323°=°\u22c5
×+°=+= \u221e
C) W/m175(2
m) 001.0)( W/m10061.1(
C20
2 2
38
gen
h
re
TT os
&

(b) The mathematical formulation of this problem can be expressed as
01 gen =+\u239f\u23a0
\u239e\u239c\u239d
\u239b
k
e
dr
dTr
dr
d
r
&

and ])([)( \u221e\u2212=\u2212 TrThdr
rdTk oo (convection at the outer surface)
0)0( =
dr
dT (thermal symmetry about the centerline)
Multiplying both sides of the differential equation by r and integrating gives
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 line,
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 boundary condition at , orr =
B. C. at : orr = 2gengen222gengen 42 42 o
o
o
o r
k
e
h
re
TCTCr
k
e
h
k
re
k
&&&& ++=\u2192\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212+\u2212=\u2212 \u221e\u221e
Substituting this relation into Eq. (b) and rearranging give C2

h
re
rr
k
e
TrT oo 2
)(
4
)( gen22gen
&& +\u2212+= \u221e
which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the
surface (r = ro ) is determined by substituting the known quantities to be
C323°=°\u22c5
×+°=+=+\u2212+= \u221e\u221e
C) W/m175(2
m) 001.0)( W/m10061.1(
C20
22
)(
4
)(
2
38
gen0gen22gen
0 h
re
T
h
re
rr
k
e
TrT ooo
&&&

Note that both approaches give the same result.
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.
2-45
2-90E Heat is generated uniformly in a resistance heater wire. The temperature difference between the
center and the surface of the wire is to be determined.
Assumptions 1 Heat transfer is steady since there is no change
with time. 2 Heat transfer is one-dimensional since there is
thermal symmetry about the center line and no change in the
axial direction. 3 Thermal conductivity is constant. 4 Heat
generation in the heater is uniform.
r
Ts
ro
Heater
0
Properties The thermal conductivity is given to be
k = 5.8 Btu/h\u22c5ft\u22c5°F.
Analysis The resistance heater converts electric energy
into heat at a rate of 3 kW. The rate of heat generation
per unit length of the wire is
38
22
wire
gen
gen ftBtu/h 10933.2
ft) (1ft) 12/04.0(
Btu/h) 14.34123( \u22c5×=×=== \u3c0\u3c0 Lr
EE
e
o
gen
&&
&
V

Then the temperature difference between the centerline and the surface becomes
F140.5°=°\u22c5\u22c5
\u22c5×==\u394
F)ftBtu/h 8.5(4
ft) 12/04.0)(ftBtu/h 10933.2(
4
2382
gen
max k
re
T o
&

2-91E Heat is generated uniformly in a resistance heater wire. The temperature difference between the
center and the surface of the wire is to be determined.
Assumptions 1 Heat transfer is steady since there is no change
with time. 2 Heat transfer is one-dimensional since there is
thermal symmetry about the center line and no change in the
axial direction. 3 Thermal conductivity is constant. 4 Heat
generation in the heater is uniform.
Properties The thermal conductivity is given to be
k = 4.5 Btu/h\u22c5ft\u22c5°F.
r
Ts
ro
Heater
0
Analysis The resistance heater converts electric energy
into heat at a rate of 3 kW. The rate of heat generation
per unit volume of the wire is
38
22
gen
wire
gen
gen ftBtu/h 10933.2
ft) (1ft) 12/04.0(
Btu/h) 14.34123( \u22c5×=×=== \u3c0\u3c0 Lr
EE
e
o
&&
&
V

Then the temperature difference between the centerline and the surface becomes
F181.0°=°\u22c5\u22c5
\u22c5×==\u394
F)ftBtu/h 5.4(4
ft) 12/04.0)(ftBtu/h 10933.2(
4
2382
gen
max k
re
T o
&

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.
2-46
2-92 Heat is generated uniformly in a spherical radioactive material with specified surface temperature.
The mathematical formulation, the variation of temperature in the sphere, and the center temperature are to
be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat transfer is steady since there is no indication of any changes with time. 2 Heat transfer
is one-dimensional since there is thermal symmetry about the mid point. 3 Thermal conductivity is
constant. 4 Heat generation is uniform.
Properties The thermal conductivity is given to be k = 15 W/m\u22c5°C.
Analysis (a) Noting that heat transfer is steady and one-
dimensional in the radial r direction, the mathematical
formulation of this problem can be expressed as
ro0
Ts=80°Ck
egen
r constant with 01 gen
gen2
2
==+\u239f\u23a0
\u239e\u239c\u239d
\u239b e
k
e
dr
dTr
dr
d
r
&
&

and 80°C (specified surface temperature) == so TrT )(
0)0( =
dr
dT (thermal symmetry about the mid point)
(b) Multiplying both sides of the differential equation by r2 and rearranging gives
2gen2 r
k
e
dr
dTr
dr
d &\u2212=\u239f\u23a0
\u239e\u239c\u239d
\u239b
Integrating with respect to r gives
1
3
gen2
3
Cr
k
e
dr
dTr +\u2212= & (a)
Applying the boundary condition at the mid point,
B.C. at r = 0: 0 0
3
)0(
0 11
gen =\u2192+×\u2212=× CC
k
e
dr
dT &
Dividing both sides of Eq. (a) by r2 to bring it to a readily integrable form and integrating,
r
k
e
dr
dT
3
gen&\u2212=
and 2
2gen
6
)( Cr
k
e
rT +\u2212= & (b)
Applying the other boundary condition at r r= 0 ,
B. C. at : orr = 2gen222gen 6 6 osos rk
e
TCCr
k
e
T
&& +=\u2192+\u2212=
Substituting this relation into Eq. (b) and rearranging give 2C
)(
6
)( 22gen rr
k
e
TrT os \u2212+=
&

which is the desired solution for the temperature distribution in the wire as a function of r.
(c) The temperature at the center of the sphere (r = 0) is determined by substituting the known quantities to
be
C791°=°\u22c5×
×°=+=\u2212+=
C)m W/ 15(6
)m 04.0)( W/m10(4+C80
6
)0(
6
)0(
2372
gen22gen
k
re
Tr
k
e
TT osos
&&

Thus the temperature at center will be about 711°C above the temperature of the outer surface of the
sphere.
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.
2-47
2-93 EES Prob. 2-92 is reconsidered. The temperature as a function of the radius is to be plotted. Also, the
center temperature of the sphere as a function of the thermal conductivity is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.

&quot;GIVEN&quot;
r_0=0.04 [m]
g_dot=4E7 [W/m^3]
T_s=80 [C]
k=15 [W/m-C]

&quot;ANALYSIS&quot;
T=T_s+g_dot/(6*k)*(r_0^2-r^2) &quot;Temperature distribution as a function of r&quot;
T_0=T_s+g_dot/(6*k)*r_0^2 &quot;Temperature at the center (r=0)&quot;

r [m] T [C]
0 791.1
0.002105 789.1
0.004211 783.2
0.006316 773.4
0.008421 759.6
0.01053 741.9
0.01263 720.2
0.01474 694.6
0.01684 665
0.01895 631.6
0.02105```