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# Respostas_Livro FT

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```2
)/ln(
oconv,pipetotal
2oconv,
2
12
pipe
°=+=+=
°=°==
°=°==
RRR
Ah
R
kL
rr
R
o \u3c0
\u3c0\u3c0

Water pipe
Tair = -5°C
Soil
W658.1
C/W 3.0153
C)]5(0[
total
21 =°
°\u2212\u2212=\u2212= \u221e\u221e
R
TT
Q&
kJ 57.83)s 3600J/s)(14 658.1( =×=\u394= tQQ &
The amount of heat required to freeze the water in the pipe completely is
kg 157.0)m 5.0()m 01.0()kg/m 1000( 232 ==== \u3c0\u3c1\u3c0\u3c1 Lrm V
kJ 4.52kJ/kg) 7.333kg)( 157.0( === fgmhQ
The water in the pipe will freeze completely that night since the amount heat loss is greater than the
amount it takes to freeze the water completely . )4.5257.83( >

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.
3-109
3-167E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an
environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is
wrapped completely in a towel are to be determined.
Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2
The thermal contact resistance at the interface is negligible. 3 The heat transfer coefficients for wrapped
and unwrapped potatoes are the same.
Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h\u22c5ft\u22c5°F. We take the
properties of potato to be those of water at room temperature, \u3c1 = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm\u22c5°F.
Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the
potato cools down and the temperature difference between the potato and the surroundings decreases.
However, we can solve this problem approximately by assuming a constant average temperature of
(300+200)/2 = 250°F for the potato during the process. The mass of the potato is

lbm 5089.0
)ft 12/5.1(
3
4)lbm/ft 2.62(
3
4
33
3
=
=
==
\u3c0
\u3c0\u3c1\u3c1 rm V

Rconv
T\u221e
RtowelTs
Potato
The amount of heat lost as the potato is cooled from 300 to 200°F is
Btu 8.50F200)-F)(300Btu/lbm. lbm)(0.998 5089.0( =°°=\u394= TmcQ p
The rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings
are
Btu/h 6.609
h) 60/5(
Btu 8.50 ==\u394= t
QQ&
F.Btu/h.ft 17.2 2 °=°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e F)70250()ft (3/12
Btu/h 6.609
)(
)(
2\u3c0TTA
QhTThAQ
so
so
&&
When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be

F/Btuh 6012.12539.03473.1
F/Btuh. 2539.0
ft )12/24.3(F).Btu/h.ft 2.17(
11
F/Btuh 3473.1
ft)12/5.1(ft]12/)12.05.1(F)[Btu/h.ft. 035.0(4
ft)12/5.1(ft]12/)12.05.1[(
4
convtoweltotal
222conv
21
12
towel
°=+=+=
°=°==
°=+°
\u2212+=\u2212=
RRR
hA
R
rkr
rr
R
\u3c0
\u3c0\u3c0

Btu/h 4.112
F/Btuh 1.6012
F)70250(
total
=°
°\u2212=\u2212= \u221e
R
TT
Q s&
min 27.1====\u394 h 452.0
Btu/h 4.112
Btu 8.50
Q
Qt &
This result is conservative since the heat transfer coefficient will be lower in this case because of the
smaller exposed surface temperature.
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.
3-110
3-168E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an
environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is loosely
wrapped completely in a towel are to be determined.
Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2
The heat transfer coefficients for wrapped and unwrapped potatoes are the same.
Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h\u22c5ft\u22c5°F. The thermal
conductivity of air is given to be k = 0.015 Btu/h\u22c5ft\u22c5°F. We take the properties of potato to be those of water
at room temperature, \u3c1 = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm\u22c5°F.
Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the
potato cools down and the temperature difference between the potato and the surroundings decreases.
However, we can solve this problem approximately by assuming a constant average temperature of
(300+200)/2 = 250°F for the potato during the process. The mass of the potato is
lbm 5089.0
)ft 12/5.1(
3
4)lbm/ft 2.62(
3
4
33
3
=
=
==
\u3c0
\u3c0\u3c1\u3c1 rm V

Rconv
T\u221e
Rair Ts
Potato
Rtowel
The amount of heat lost as the potato is cooled from 300 to 200°F is
Btu 8.50F200)F)(300Btu/lbm. lbm)(0.998 5089.0( =°\u2212°=\u394= TmcQ p
The rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings are
Btu/h 6.609
h) 60/5(
Btu 8.50 ==\u394= t
QQ&
F.Btu/h.ft 17.2 2 °=°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e F)70250()ft (3/12
Btu/h 6.609
)(
)(
2\u3c0TTA
QhTThAQ
so
so
&&
When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be

F/Btuh 1195.22477.03134.15584.0
F/Btuh. 2477.0
ft )12/28.3(F).Btu/h.ft 2.17(
11
F/Btuh 3134.1
ft)12/52.1(ft]12/)12.052.1(F)[Btu/h.ft. 035.0(4
ft)12/52.1(ft]12/)12.052.1[(
4
F/Btuh. 5584.0
ft)12/50.1(ft]12/)02.050.1(F)[Btu/h.ft. 015.0(4
ft)12/50.1(ft]12/)02.050.1[(
4
convtowelairtotal
222conv
32
23
towel
21
12
air
°=++=++=
°=°==
°=+°
\u2212+=\u2212=
°=+°
\u2212+=\u2212=
RRRR
hA
R
rkr
rr
R
rkr
rr
R
\u3c0
\u3c0\u3c0
\u3c0\u3c0

Btu/h 9.84
F/Btuh. 2.1195
F)70250(
total
=°
°\u2212=\u2212= \u221e
R
TT
Q s&
min 35.9====\u394 h 598.0
Btu/h 9.84
Btu 8.50
Q
Qt &
This result is conservative since the heat transfer coefficient will be lower because of the smaller exposed
surface temperature.

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.
3-111
3-169 An ice chest made of 3-cm thick styrofoam is initially filled with 45 kg of ice at 0°C. The length of
time it will take for the ice in the chest to melt completely is to be determined.
Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not
change with time. 2 Heat transfer is one-dimensional. 3 Thermal conductivity is constant. 4 The inner
surface temperature of the ice chest can be taken to be 0°C at all times. 5 Heat transfer from the base of the
ice chest is negligible.
Properties The thermal conductivity of styrofoam
is given to be k = 0.033 W/m\u22c5°C. The heat of
fusion of water at 1 atm is . kJ/kg 7.333=ifh
Rconv
T\u221e
Rchest Ts
Ice chest Analysis Disregarding any heat loss through the
bottom of the ice chest, the total thermal resistance
and the heat transfer rate are determined to be

C/W 6678.107508.05927.1
C/W 07508.0
)m 74.0(C). W/m18(
11
C/W 5927.1
)m 5708.0(C) W/m.033.0(
m 03.0
m 74.0)5.0)(4.0()5.0)(3.0(2)4.0)(3.0(2
m 5708.0)06.05.0)(06.04.0()06.05.0)(03.03.0(2)06.04.0)(03.03.0(2
convchesttotal
22conv
2chest
2
2
°=+=+=
°=°==
°=°==
=++=
=\u2212\u2212+\u2212\u2212+\u2212\u2212=
RRR
hA
R
kA
LR
A
A
o
i
o
i
W79.16
C/W 1.6678
C)028(
total
=°
°\u2212=\u2212= \u221e
R
TT
Q s&
The total amount of heat necessary to melt the ice completely is
kJ 685,16kJ/kg) kg)(333.7 50( === ifmhQ
Then the time period to transfer this much heat to the cooler to melt the ice completely becomes
days 11.5==×===\u394 h 276s 10937.9
J/s 79.16
J 000,685,16 5
Q
Qt &

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.
3-112
3-170 A wall is constructed of two large steel plates separated by 1-cm thick steel bars placed 99 cm apart.
The remaining space between the steel plates is filled with fiberglass insulation. The rate of heat transfer
through the wall is to```