Respostas_Livro FT
1537 pág.

Respostas_Livro FT


DisciplinaFenômenos de Transporte I13.217 materiais112.689 seguidores
Pré-visualização50 páginas
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