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# Respostas_Livro FT

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```2 Thermal conductivity is
constant. 3 Heat is generated uniformly.
Analysis The differential equation and the boundary conditions for this heat conduction problem can be
expressed as
01 gen
2
2
=+\u2202
\u2202+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202
k
e
z
T
r
Tr
rr
&

h
T\u221e
egen
qH
z
ro

Hqz
HrTk
z
rT
&=\u2202
\u2202
=\u2202
\u2202
),(
0)0,(

]),([
),(
0),0(
\u221e\u2212=\u2202
\u2202\u2212
=\u2202
\u2202
TzrTh
r
zrT
k
r
zT
o
o

2-126E A large plane wall is subjected to a specified temperature on the left (inner) surface and solar
radiation and heat loss by radiation to space on the right (outer) surface. The temperature of the right
surface of the wall and the rate of heat transfer are to be determined when steady operating conditions are
reached.
Assumptions 1 Steady operating conditions are reached. 2 Heat
transfer is one-dimensional since the wall is large relative to its
thickness, and the thermal conditions on both sides of the wall are
uniform. 3 Thermal properties are constant. 4 There is no heat
generation in the wall.
qsolar
x
520 R
L
T2
Properties The properties of the plate are given to be k = 1.2
Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.80, and 60.0=s\u3b1 .
Analysis In steady operation, heat conduction through the wall must
be equal to net heat transfer from the outer surface. Therefore, taking
the outer surface temperature of the plate to be T2 (absolute, in R),
solar
4
2
21 qATA
L
TTkA ssss &\u3b1\u3b5\u3c3 \u2212=\u2212
Canceling the area A and substituting the known quantities,
)ftBtu/h 300(60.0)RftBtu/h 101714.0(8.0
ft 0.8
R) 520(
)FftBtu/h 2.1( 242
4282 \u22c5\u2212\u22c5\u22c5×=\u2212°\u22c5\u22c5 \u2212 TT
Solving for T2 gives the outer surface temperature to be T2 = 553.9 R
Then the rate of heat transfer through the wall becomes
2ftBtu/h 50.9 \u22c5\u2212=\u2212°\u22c5\u22c5=\u2212=
ft 0.8
R )9.553520()FftBtu/h 2.1(21
L
TT
kq& (per unit area)
Discussion The negative sign indicates that the direction of heat transfer is from the outside to the inside.
Therefore, the structure is gaining heat.
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.
2-61
2-127E A large plane wall is subjected to a specified temperature on the left (inner) surface and heat loss
by radiation to space on the right (outer) surface. The temperature of the right surface of the wall and the
rate of heat transfer are to be determined when steady operating conditions are reached.
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the wall
is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 3
Thermal properties are constant. 4 There is no heat generation in the wall.
Properties The properties of the plate are given to be k = 1.2
Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.80.
Analysis In steady operation, heat conduction through the wall must
be equal to net heat transfer from the outer surface. Therefore, taking
the outer surface temperature of the plate to be T2 (absolute, in R),
42
21 TA
L
TT
kA ss \u3b5\u3c3=\u2212
Canceling the area A and substituting the known quantities,
42
4282 )RftBtu/h 101714.0(8.0
ft 0.5
R) 520(
)FftBtu/h 2.1( T
T \u22c5\u22c5×=\u2212°\u22c5\u22c5 \u2212
Solving for T2 gives the outer surface temperature to be T2 = 487.7 R
x
520 R
L
T2
Then the rate of heat transfer through the wall becomes
2ftBtu/h 77.5 \u22c5=\u2212°\u22c5\u22c5=\u2212=
ft 0.5
R )7.487520()FftBtu/h 2.1(21
L
TT
kq& (per unit area)
Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside.
Therefore, the structure is losing heat 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.
2-62
2-128 A steam pipe is subjected to convection on both the inner and outer surfaces. The mathematical
formulation of the problem and expressions for the variation of temperature in the pipe and on the outer
surface temperature are to be obtained 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 steady and one-
dimensional in the radial r direction, the mathematical
formulation of this problem can be expressed as
r2
r
r1Ti
hi
To
ho
0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d
and )]([)( 11 rTThdr
rdTk ii \u2212=\u2212
])([)( 22 oo TrThdr
rdTk \u2212=\u2212
(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 = r1: )]ln([ 211
1
1 CrCTh
r
Ck ii +\u2212=\u2212
r = r2: ])ln[( 221
2
1
oo TCrChr
Ck \u2212+=\u2212
Solving for C1 and C2 simultaneously gives
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212
++
\u2212\u2212=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212\u2212=
++
\u2212=
1
1
211
2
0
1
112
211
2
0
1 ln
ln
ln and
ln rh
kr
rh
k
rh
k
r
r
TTT
rh
krCTC
rh
k
rh
k
r
r
TTC
i
oi
i
i
i
i
oi
i
Substituting into the general solution and simplifying, we get the variation of temperature to
be
21 and CC

211
2
11
1
111
ln
ln
)(lnln)(
rh
k
rh
k
r
r
rh
k
r
r
T
rh
krCTrCrT
oi
i
i
i
i
++
+
+=\u2212\u2212+=
(c) The outer surface temperature is determined by simply replacing r in the relation above by r2. We get

211
2
11
2
2
ln
ln
)(
rh
k
rh
k
r
r
rh
k
r
r
TrT
oi
i
i
++
+
+=
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.
3-1
Chapter 3

Steady Heat Conduction in Plane Walls

3-1C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod
is the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the
rod are insulated, the heat transfer area of the rod is the lateral surface area of the rod,
4/2DAs \u3c0=
DLA \u3c0= .

3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out
of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the
wall does not change during steady heat conduction. However, the temperature along the wall and thus the
energy content of the wall will change during transient conduction.

3-3C The temperature distribution in a plane wall will be a straight line during steady and one dimensional
heat transfer with constant wall thermal conductivity.

3-4C The thermal resistance of a medium represents the resistance of that medium against heat transfer.

3-5C The combined heat transfer coefficient represents the combined effects of radiation and convection
heat transfers on a surface, and is defined as hcombined = hconvection + hradiation. It offers the convenience of
incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in
heat transfer calculations.

3-6C Yes. The convection resistance can be defined as the inverse of the convection heat transfer
coefficient per unit surface area since it is defined as )/(1 hARconv = .

3-7C The convection and the radiation resistances at a surface are parallel since both the convection and