7/30/2019 Heat Chap07 099
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QTsky = 100 K
Chapter 7External Forced Convection
7-99 Wind is blowing over the roof of a house. The rate of heat transfer through the roof and the cost of thisheat loss for 14-h period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Air isan ideal gas with constant properties. 4 The pressure of air is 1 atm.
Properties Assuming a film temperature of 10C, the properties ofair are (Table A-15)
7336.0Pr
/sm10426.1CW/m.02439.0
25-
==
=k
AnalysisThe Reynolds number is
[ ] 725
10338.2/sm10426.1
m)(20m/s)3600/100060(Re =
=
=
LVL
which is greater than the critical Reynolds number. Thus we havecombined laminar and turbulent flow. Then the Nusselt number andthe heat transfer coefficient are determined to be
C.W/m0.31)10542.2(m20
CW/m.02439.0
10542.2)7336.0](871)10338.2(037.0[Pr)871Re037.0(
24
43/18.073/18.0
===
====
NuLkh
k
hLNu L
In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal tothe heat transfer from the roof to the surroundings (by convection and radiation), which must be equal tothe heat transfer through the roof by conduction. That is,
Q Q Q Q= = =room to roof, conv+rad roof, cond roof to surroundings, conv+rad
Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out, respectively, the quantitiesabove can be expressed as
[ ]4,44282,
224,
4,rad+convroof,toroom
K)273(K)27320().KW/m1067.5)(m300)(9.0(
C))(20mC)(300.W/m5()()(
+++
=+=
ins
insinsroomsinsroomsi
T
TTTATTAhQ
m15.0
)m300)(CW/m.2( ,,2,,condroof, outsinsoutsinssTT
L
TTkAQ
=
=
[ ]44,4282,
2244,,rad+convsurr,toroof
K)100(K)273().KW/m1067.5)(m300)(9.0(
C)10)(mC)(300.W/m0.31()()(
++
=+=
outs
outssurroutsssurroutsso
T
TTTATTAhQ
Solving the equations above simultaneously gives
C5.3andC,6.10,W025,28 ,, ==== outsins TTQ kW28.03
The total amount of natural gas consumption during a 14-hour period is
therms75.15kJ105,500
therm1
85.0
)s360014)(kJ/s03.28(
85.085.0=
=
==
tQQQ totalgas
Finally, the money lost through the roof during that period is
$9.45== )therm/60.0$therms)(75.15(lostMoney
7-84
AirV = 60 km/h
T = 10C
Tin = 20C
7/30/2019 Heat Chap07 099
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Chapter 7External Forced Convection
7-100 Steam is flowing in a stainless steel pipe while air is flowing across the pipe. The rate of heat lossfrom the steam per unit length of the pipe is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Thepressure of air is 1 atm.
Properties Assuming a film temperature of 10C, the properties of air are (Table A-15)
7336.0Prand/s,m10426.1C,W/m.02439.0 2-5 ===k
AnalysisThe outer diameter of insulated pipe isDo= 4.6+2 3.5=11.6 cm = 0.116 m. The Reynoldsnumber is
4
2510254.3
/sm10426.1
m)m/s)(0.116(4Re =
=
=
oDV
The Nusselt number for flow across a cylinder is determined from
( )[ ]
( )[ ]0.107
000,282
10254.31
7336.0/4.01
)7336.0()10254.3(62.03.0
000,282
Re1
Pr/4.01
PrRe62.03.0
5/48/5
4
4/13/2
3/15.04
5/48/5
4/13/2
3/15.0
=
+
+
+=
+
++==
k
hDNu o
and CW/m50.22)0.107(m116.0
CW/m0.02439 2 === NuD
kh
oo
Area of the outer surface of the pipe per m length of the pipe is
2m3644.0)m1)(m116.0( === LDA ooIn steady operation, heat transfer from the steam through the pipe and the insulation to the outer surface (byfirst convection and then conduction) must be equal to the heat transfer from the outer surface to thesurroundings (by simultaneous convection and radiation). That is,
Q Q Q= =pipe and insulation surface to surroundings
Using the thermal resistance network, heat transfer from the steam to the outer surface is expressed as
[ ]
C/W874.3)m1)(CW/m.038.0(2
)3.2/8.5ln(
2
)/ln(
C/W0015.0)m1)(CW/m.15(2
)2/3.2ln(
2
)/ln(
C/W0995.0
)m1(m)04.0()C.W/m80(
11
23
12
2,
=
==
=
==
=
==
kL
rrR
kL
rrR
Ah
R
insulation
pipe
iiiconv
andC/W)874.30015.00995.0(
C)250(
,
1insandpipe ++
=
++
= sinsulationpipeiconv
s T
RRR
TTQ
Heat transfer from the outer surface can be expressed as
[ ]4442822244
rad+convsurr,tosurface
K)2733(K)273().KW/m1067.5)(m3644.0)(3.0(
C)3)(mC)(0.3644.W/m50.22()()(
+++
=+=
s
ssurrsosurrsoo
T
TTTATTAhQ
Solving the two equations above simultaneously, the surface temperature and the heat transfer rate per mlength of the pipe are determined to be
length)m(perandC9.9 W60.4== QTs
7-85
Steel pipeD
i= D
1= 4 cm
D2= 4.6 cm
Steam, 250C
Do
Di
Air
3C, 4 m/s
Insulation
= 0.3
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Chapter 7External Forced Convection
7-101 A spherical tank filled with liquid nitrogen is exposed to winds. The rate of evaporation of the liquidnitrogen due to heat transfer from the air is to be determined for three cases.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 The pressure of air is 1 atm.
Properties The properties of air at 1 atm pressure and the free stream temperature of 20C are (Table A-15)
7309.0Pr
kg/m.s10023.5
kg/m.s10825.1
/sm10516.1
CW/m.02514.0
6
C196@,
5
25-
=
=
=
==
s
k
Analysis(a) When there is no insulation,D =Di = 4 m,and the Reynolds number is
[ ] 625
10932.2/sm10516.1
m)(4m/s1000/3600)(40Re =
=
=
DV
The Nusselt number is determined from
[ ]
[ ] 233310023.5
10825.1)7309.0()10932.2(06.0)10932.2(4.02
PrRe06.0Re4.02
4/1
6
54.03/265.06
4/14.03/25.0
=
++=
++==
sk
hDNu
and C.W/m66.14)2333(m4
CW/m.02514.0 2 =
== NuD
kh
The rate of heat transfer to the liquid nitrogen is
[ ] W200,159C)196(20(]m)(4C)[.W/m66.14(
))(()(
22
2
==
==
TTDhTThAQ sss
The rate of evaporation of liquid nitrogen then becomes
kg/s0.804====kJ/kg198
kJ/s2.159
ifif
h
QmhmQ
(b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to
evaluate dynamic viscosity at a new surface temperature which we will assume to be -100 C. At -100C,kg/m.s10189.1 5= . Noting thatD =D0 = 4.1 m, the Nusselt number becomes
[ ] 625
10005.3/sm10516.1
m)(4.1m/s1000/3600)(40Re =
=
=
DV
[ ]
[ ] 191010189.110825.1)7309.0()10005.3(06.0)10005.3(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
=
++=
++==
sk
hDNu
and C.W/m71.11)1910(m1.4
CW/m.02514.0 2 =
== NuD
kh
The rate of heat transfer to the liquid nitrogen is
7-86
Nitrogen tank
-196C
Do
Di
Wind
20C40 km/h
Insulation
7/30/2019 Heat Chap07 099
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Chapter 7External Forced Convection
W7361
)m81.52)(C.W/m71.11(
1
m)m)(2C)(2.05W/m.(0.0354
m)205.2(
C)]196(20[
1
4
m81.52)m1.4(
22
21
12
tan,tan,
222
=
+
=
+
=
+
=
===
s
ks
convinsulation
ks
s
hArkr
rr
TT
RR
TTQ
DA
The rate of evaporation of liquid nitrogen then becomes
kg/s0.0372====kJ/kg198
kJ/s361.7
ifif
h
QmhmQ
(c) We use the dynamic viscosity value at the new estimated surface temperature of 0C to bekg/m.s10729.1 5= . Noting thatD =D0 = 4.04 m in this case, the Nusselt number becomes
[ ] 625
10961.2/sm10516.1
m)(4.04m/s1000/3600)(40Re =
=
=
DV
[ ]
[ ] 172410729.1
10825.1)7309.0()10961.2(06.0)10961.2(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
=
++=
++==
sk
hDNu
and C.W/m73.10)1724(m04.4
CW/m.02514.0 2 =
== NuD
kh
The rate of heat transfer to the liquid nitrogen is
W4.27
)m28.51)(C.W/m73.10(
1
m)m)(2C)(2.02W/m.(0.000054
m)202.2(
C)]196(20[
1
4
m28.51)m04.4(
22
21
12
tan,tan,
222
=
+
=
+
=
+
=
===
s
ks
convinsulation
ks
s
hArkr
rr
TT
RR
TTQ
DA
The rate of evaporation of liquid nitrogen then becomes
kg/s101.38 4-====kJ/kg198
kJ/s0274.0
if
ifh
QmhmQ
7-87
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Chapter 7External Forced Convection
7-102 A spherical tank filled with liquid oxygen is exposed to ambient winds. The rate of evaporation ofthe liquid oxygen due to heat transfer from the air is to be determined for three cases.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 7 The pressure of air is 1 atm.
Properties The properties of air at 1 atm pressure and the free stream temperature of 20C are (Table A-15)
7309.0Pr
kg/m.s10127.6
kg/m.s10825.1
/sm10516.1
CW/m.02514.0
5
C183@,
5
25-
=
=
==
=
s
k
Analysis(a) When there is no insulation,D =Di = 4 m,and the Reynolds number is
[ ] 625
10932.2/sm10516.1
m)(4m/s1000/3600)(40Re =
=
=
DV
The Nusselt number is determined from
[ ]
[ ] 22201005.1
10825.1)7309.0()10932.2(06.0)10932.2(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
=
++=
++==
sk
hDNu
and C.W/m95.13)2220(m4
CW/m.02514.0 2 =
== NuD
kh
The rate of heat transfer to the liquid oxygen is
[ ] W372,142C)183(20(]m)(4C)[.W/m95.13())(()( 222 ==== TTDhTThAQ sss
The rate of evaporation of liquid oxygen then becomes
kg/s0.668====kJ/kg213
kJ/s4.142
ifif
h
QmhmQ
(b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to
evaluate dynamic viscosity at a new surface temperature which we will assume to be -100C. At -100C,kg/m.s10189.1 5= . Noting thatD =D0 = 4.1 m, the Nusselt number becomes
[ ] 625
10005.3/sm10516.1
m)(4.1m/s1000/3600)(40Re =
=
=
DV
[ ]
[ ] 191010189.1
10825.1)7309.0()10005.3(06.0)10005.3(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
=
++=
++==
sk
hDNu
and C.W/m71.11)1910(m1.4
CW/m.02514.0 2 === NuDkh
The rate of heat transfer to the liquid nitrogen is
7-88
Oxygen tank
-183C
Do
Di
Wind
20C40 km/h
Insulation
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Chapter 7External Forced Convection
W6918
)m81.52)(C.W/m71.11(
1
m)m)(2C)(2.05W/m.(0.0354
m)205.2(
C)]183(20[
1
4
m81.52)m1.4(
22
21
12
tan,tan,
222
=
+
=
+
=
+
=
===
s
ks
convinsulation
ks
s
hArkr
rr
TT
RR
TTQ
DA
The rate of evaporation of liquid nitrogen then becomes
kg/s0.0325====kJ/kg213
kJ/s918.6
ifif
h
QmhmQ
(c) Again we use the dynamic viscosity value at the estimated surface temperature of 0C to bekg/m.s10729.1 5= . Noting thatD =D0 = 4.04 m in this case, the Nusselt number becomes
[ ] 625
10961.2/sm10516.1
m)(4.04m/s1000/3600)(40Re =
=
=
DV
[ ]
[ ] 172410729.1
10825.1)713.0()10961.2(06.0)10961.2(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
=
++=
++==
sk
hDNu
and C.W/m73.10)1724(m04.4
CW/m.02514.0 2 =
== NuD
kh
The rate of heat transfer to the liquid nitrogen is
W8.25
)m28.51)(C.W/m73.10(
1
m)m)(2C)(2.02W/m.(0.000054
m)202.2(
C)]183(20[
1
4
m28.51)m04.4(
22
21
12
tan,tan,
222
=
+
=
+
=
+
=
===
s
ks
convinsulation
ks
s
hArkr
rr
TT
RR
TTQ
DA
The rate of evaporation of liquid oxygen then becomes
kg/s101.21 4-====kJ/kg213
kJ/s0258.0
ifif
h
QmhmQ
7-89
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Chapter 7External Forced Convection
7-103 A circuit board houses 80 closely spaced logic chips on one side. All the heat generated is conductedacross the circuit board and is dissipated from the back side of the board to the ambient air, which is forcedto flow over the surface by a fan. The temperatures on the two sides of the circuit board are to bedetermined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 7 The pressure of air is 1atm.
Properties Assuming a film temperature of 40C, the properties of air are (Table A-15)
7255.0Pr
/sm10702.1
CW/m.02662.0
25-
==
=
k
AnalysisThe Reynolds number is
[ ] 425
10051.7/sm10702.1
m)(0.18m/s)60/400(Re =
=
=
LV
L
which is less than the critical Reynolds number. Therefore,the flow is laminar. Using the proper relation for Nusseltnumber, heat transfer coefficient is determined to be
C.W/m43.23)4.158(m18.0
CW/m.02662.0
4.158)7255.0()10051.7(664.0PrRe664.0
2
3/15.043/15.0
=
==
====
NuL
kh
khLNu L
The temperatures on the two sides of the circuit board are
C39.52
C39.48
=
+=
+==
=
+=
+==
m)m)(0.18C)(0.12W/m.16(
m)W)(0.00306.080(C48.39
)(
m)m)(0.18C)(0.12.W/m43.23(
W)06.080(C30
)(
2121
2
22
s
s
ss
kA
LQTTTT
L
kAQ
hA
QTTTThAQ
7-90
T1
T2
T =30C400 m/min
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Chapter 7External Forced Convection
7-104E The equivalent wind chill temperature of an environment at 10F at various winds speeds are
V = 10 mph: T T V V equiv ambient = +
= =
914 914 0 475 0 0203 0 304
914 914 10 0 475 0 0203 10
. ( . )( . . . )
. . ( . . (F) mph) + 0.304 10 mph 9 F
V = 20 mph: Tequiv = = 914 914 10 0 475 0 0203 20. . ( . . (F) mph) + 0.304 20 mph 24.9 F
V = 30 mph: Tequiv = = 914 914 10 0 475 0 0203 30. . ( . . (F) mph) + 0.304 30 mph 33.2 F
V = 40 mph: Tequiv = = 914 914 10 0 475 0 0203 40. . ( . . (F) mph) + 0.304 40 mph 37.7 F
In the last 3 cases, the person needs to be concerned about the possibility of freezing.
7-91
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Chapter 7External Forced Convection
7-105E "!PROBLEM 7-105E"
"ANALYSIS"T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*Vel+0.304*sqrt(Vel))
Vel [mph] Tambient [F] Tequiv [F]4 20 19.87
14.67 20 -4.383
25.33 20 -15.05
36 20 -20.57
46.67 20 -23.15
57.33 20 -23.77
68 20 -22.94
78.67 20 -21.01
89.33 20 -18.19
100 20 -14.63
4 40 39.91
14.67 40 22.45
25.33 40 14.7736 40 10.79
46.67 40 8.935
57.33 40 8.493
68 40 9.086
78.67 40 10.48
89.33 40 12.51
100 40 15.07
4 60 59.94
14.67 60 49.28
25.33 60 44.59
36 60 42.16
46.67 60 41.02
57.33 60 40.75
68 60 41.11
78.67 60 41.96
89.33 60 43.21
100 60 44.77
7-92
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Chapter 7External Forced Convection
0 22 44 66 88 110
-3 0
-2 0
-1 0
0
10
20
30
40
50
60
Vel [mph]
Tequiv
[F]
20 F
40 F
60 F
7-106 . 7-110 Design and Essay Problems
7-93