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# Respostas_Livro FT

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```no total,
21 =°
°\u2212=\u2212= \u221e
R
TT
Q&
(a) The rate of heat transfer after insulation is
W9.9966615.015.0 ins noins =×== QQ &&
The total thermal resistance with the foam insulation is

C) W.m/42.0(
C/W 02253.0
)m C)(16.8 W/m.025.0(
C/W 02253.0
4
2
4
conv,2foam321total
°+°=
°+°=
++++=
L
L
RRRRRR

The thickness of insulation is determined from
cm 5.4m 0.054 ==\u23af\u2192\u23af
°+°
°\u2212=\u23af\u2192\u23af\u2212= \u221e 4
4total
21
ins
C) W.m/42.0(
C/W 02253.0
C)823(
W9.99 L
LR
TT
Q&
The outer surface temperature of the wall is determined from
C8.3°=\u23af\u2192\u23af°
°\u2212=\u23af\u2192\u23af\u2212= \u221e 22
conv
22
ins C/W 00350.0
C)8(
W9.99 T
T
R
TT
Q&
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.
3-121
(b) The total thermal resistance with the fiberglass insulation is
C) W.m/6048.0(
C/W 02253.0
)m C)(16.8 W/m.036.0(
C/W 02253.0 4
2
4
conv,2glassfiber 321total
°+°=°+°=
++++=
LL
RRRRRR

The thickness of insulation is determined from
cm 7.7m 0.077 ==\u23af\u2192\u23af
°+°
°\u2212=\u23af\u2192\u23af\u2212= \u221e 4
4total
21
ins
C W.m/6048.0(
C/W 02253.0
C)823(
W9.99 L
LR
TT
Q&
The outer surface temperature of the wall is determined from
C8.3°=\u23af\u2192\u23af°
°\u2212=\u23af\u2192\u23af\u2212= \u221e 22
conv
22
ins C/W00350.0
C)8(
9.99 T
T
R
TT
Q&
Discussion The outer surface temperature is same for both cases since the rate of heat transfer does not
change.

3-177 Cold conditioned air is flowing inside a duct of square cross-section. The maximum length of the
duct for a specified temperature increase in the duct is to be determined.
Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are
constant. 4 Steady one-dimensional heat conduction relations can be used due to small thickness of the duct
wall. 5 When calculating the conduction thermal resistance of aluminum, the average of inner and outer
surface areas will be used.
Properties The thermal conductivity of aluminum is given to be 237 W/m\u22c5°C. The specific heat of air at the
given temperature is cp = 1006 J/kg\u22c5°C (Table A-15).
Analysis The inner and the outer surface areas of the duct per unit length and the individual thermal
resistances are

2
22
2
11
m 0.1m) 1(m) 25.0(44
m 88.0m) 1(m) 22.0(44
===
===
LaA
LaA
Ri Ralum
T\u221e1 T\u221e2
Ro
[ ]
C/W 09214.007692.000007.001515.0
C/W07692.0
)m 0.1(C). W/m13(
11
C/W 00007.0
m 2/)188.0(C) W/m.237(
m 015.0
C/W01515.0
)m 88.0(C). W/m75(
11
oalumitotal
22
2
o
2alum
22
1
i
°=++=++=
°=°==
°=+°==
°=°==
RRRR
Ah
R
kA
LR
Ah
R

The rate of heat loss from the air inside the duct is
W228
C/W09214.0
C)1233(
total
12 =°
°\u2212=\u2212= \u221e\u221e
R
TT
Q&
For a temperature rise of 1°C, the air inside the duct should gain heat at a rate of
W805C)1)(CJ/kg. kg/s)(1006 8.0(total =°°=\u394= TcmQ p&&
Then the maximum length of the duct becomes
m 3.53===
W228
W805total
Q
Q
L &
&

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.
3-122
3-178 Heat transfer through a window is considered. The percent error involved in the calculation of heat
gain through the window assuming the window consist of glass only is to be determined.
Assumptions 1 Heat transfer is steady. 2 Heat transfer is
one-dimensional. 3 Thermal conductivities are constant.
4 Radiation is accounted for in heat transfer
coefficients.
Ri Rglass
T\u221e1 T\u221e2
Ro
Properties The thermal conductivities are given to be
0.7 W/m\u22c5°C for glass and 0.12 W/m\u22c5°C for pine wood. Ri Rwood
T\u221e1 T\u221e2
Ro
Analysis The surface areas of the glass and the wood
and the individual thermal resistances are
2wood
2
glass m 45.0m) 2(m) 5.1(15.0 m 55.2m) 2(m) 5.1(85.0 ==== AA
C/W 41433.117094.092593.031746.0
C/W 08787.003017.000168.005602.0
C/W17094.0
)m 45.0(C). W/m13(
11
C/W03017.0
)m 55.2(C). W/m13(
11
C/W 92593.0
)m 45.0(C) W/m.12.0(
m 05.0
C/W 00168.0
)m 55.2(C) W/m.7.0(
m 003.0
C/W31746.0
)m 45.0(C). W/m7(
11
C/W05602.0
)m 55.2(C). W/m7(
11
woodo,woodwoodi,woodtotal,
glasso,glassglassi,glass total,
22
wood2
woodo,
22
glass2
glasso,
2
woodwood
wood
wood
2
glassglass
glass
glass
22
wood1
woodi,
22
glass1
glassi,
°=++=++=
°=++=++=
°=°==
°=°==
°=°==
°=°==
°=°==
°=°==
RRRR
RRRR
Ah
R
Ah
R
Ak
L
R
Ak
L
R
Ah
R
Ah
R

The rate of heat gain through the glass and the wood and their total are

W4.1933.111.182
W3.11
C/W41433.1
C)2440(
W 1.182
C/W08787.0
C)2440(
woodglasstotal
woodtotal,
12
wood
glasstotal,
12
glass
=+=+=
=°
°\u2212=\u2212==°
°\u2212=\u2212= \u221e\u221e\u221e\u221e
QQQ
R
TT
Q
R
TT
Q
&&&
&&

If the window consists of glass only the heat gain through the window is
2
glass m 0.3m) 2(m) 5.1( ==A

C/W 07469.002564.000143.004762.0
C/W02564.0
)m 0.3(C). W/m13(
11
C/W 00143.0
)m 0.3(C) W/m.7.0(
m 003.0
C/W04762.0
)m 0.3(C). W/m7(
11
glasso,glassglassi,glass total,
22
glass2
glasso,
2
glassglass
glass
glass
22
glass1
glassi,
°=++=++=
°=°==
°=°==
°=°==
RRRR
Ah
R
Ak
L
R
Ah
R

W2.214
C/W07469.0
C)2440(
glasstotal,
12
glass =°
°\u2212=\u2212= \u221e\u221e
R
TT
Q&
Then the percentage error involved in heat gain through the window assuming the window consist of glass
only becomes
10.8%=×\u2212=\u2212= 100
4.193
4.1932.214Error %
with wood
with woodonly glass
Q
QQ
&
&&

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.
3-123
3-179 Steam is flowing inside a steel pipe. The thickness of the insulation needed to reduce the heat loss by
95 percent and the thickness of the insulation needed to reduce outer surface temperature to 40°C are to be
determined.
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 centerline and no variation in the axial
direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is
negligible.
Properties The thermal conductivities are given to be k = 61 W/m\u22c5°C for steel and k = 0.038 W/m\u22c5°C for
insulation.
Analysis (a) Considering a unit length of the pipe, the inner and the outer surface areas of the pipe and the
insulation are

2
3333
2
2
2
1
m 1416.3m) 1(
m 3770.0m) 1(m) 12.0(
m 3142.0m) 1(m) 10.0(
DDLDA
LDA
LDA
o
i
===
===
===
\u3c0\u3c0
\u3c0\u3c0
\u3c0\u3c0 Ri
T\u221e1
R2 Ro
T\u221e2
R1
The individual thermal resistances are
C/W 02274.0
0.23876
)12.0/ln(
03079.0
02274.0
0.23876
)12.0/ln(
00048.003031.0
C/W 22026.018947.000048.003031.0
C/W 02274.0
)m 1416.3(C). W/m14(
11
C/W 18947.0
)m 3770.0(C). W/m14(
11
C/W
0.23876
)12.0/ln(
)m 1(C) W/m.038.0(2
)12.0/ln(
2
)/ln(
C/W 00048.0
)m 1(C) W/m.61(2
)5/6ln(
2
)/ln(
C/W 03031.0
)m 3142.0(C). W/m105(
11
3
3
3
3
insulationo,21insulation total,
steelo,1insulation no total,
3
2
3
2insulationo,
22steelo,
33
2
23
insulation2
1
12
pipe1
22
°++=
+++=+++=
°=++=++=
°=°==
°=°==
°=°===
°=°===
°=°==
D
D
D
D
RRRRR
RRRR
DDAh
R
Ah
R
DD
Lk
rr
RR
Lk
rr
RR
Ah
R
i
i
oo
oo
ii
i
\u3c0\u3c0
\u3c0\u3c0

Then the steady rate of heat loss from the steam per meter pipe length for the case of no insulation becomes
W976.1
C/W 0.22026
C)20235(21 =°
°\u2212=\u2212= \u221e\u221e
totalR
TT
Q&
The thickness of the insulation needed in order to save 95 percent of this heat loss can be determined from

C/W 02274.0
0.23876
)12.0/ln(
03079.0
C)20235( W)1.97605.0(
3
3insulationtotal,
21
insulation
°\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b ++
°\u2212=×\u23af\u2192\u23af\u2212= \u221e\u221e
D
DR
TT
Q&
whose solution is
cm 10.78===\u23af\u2192\u23af=
2
12-33.55
2
-
thicknessm 3355.0 233
DD
D
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution```