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7/30/2019 Heat Chap09 032
1/23
Chapter 9Natural Convection
9-32 A fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The thickness of theinsulation needed to reduce the losses by 85% and the money saved during 10-h are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature,and possible below it because of the very low sky temperature for radiation heat loss. For convenience, we
use the properties of air at 1 atm and 5C (the anticipated film temperature) (Table A-15),
1-
25
K003597.0K)2735(
11
7350.0Pr
/sm10382.1
CW/m.02401.0
=+
==
==
=
fT
k
AnalysisThe rate of heat loss in the previous problem wasobtained to be 29,094 W. Noting that insulation will cut downthe heat losses by 85%, the rate of heat loss will be
W4364W094,2915.0)85.01( insulationno === QQ
The amount of energy and money insulation will save during a 10-h period is simply determined fromkWh3.247h)kW)(10094.2985.0(, === tQQ savedtotalsaved
$22.26== )kWh/09.0)($kWh3.247(=energy)ofcosttsaved)(UniEnergy(savedMoney
The characteristic length in this case is the outer diameter of the insulated pipe,
insulinsulc ttDL 23.02 +=+= where tinsul is the thickness of insulation in m. Then the problem can be
formulated forTs and tinsul as follows:
)7350.0()/sm10382.1(
)23.0(K])273)[(K003597.0)(m/s81.9(Pr
)(225
3-12
2
3
+=
= insulscs
tTLTTgRa
( )[ ] ( )[ ]
2
27/816/9
6/12
27/816/9
6/1
7350.0/559.01
387.06.0
Pr/559.01
387.06.0
+
+=
+
+=RaRa
Nu
m))(10023.0(
CW/m.02401.0
0 insuls
cc
tLDA
NuL
NuL
kh
+==
==
The total rate of heat loss from the outer surface of the insulated pipe by convection and radiation becomes
])K27330()[.KW/m1067.5()1.0(+)273(4364
)()(
44428
44
+=
+=+=
ssss
surrssssradconv
TAThA
TTATThAQQQ
In steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from theexposed surface of the insulation by convection and radiation, which must be equal to the heat conductedthrough the insulation. Therefore,
]3.0/)23.0ln[(
K)m)(298C)(100W/m.035.0(2W4364)/ln(
)(2 tank
insul
s
o
sinsulation
t
T
DD
TTkLQQ +
=
==
The solution of all of the equations above simultaneously using an equation solver gives Ts = 281.5 K =
8.5C and tinsul = 0.013 m = 1.3 cm.Note that the film temperature is (8.5+0)/2 = 4.25C which is very close to the assumed value of
5C. Therefore, there is no need to repeat the calculations using properties at this new film temperature.9-33E An industrial furnace that resembles a horizontal cylindrical enclosure whose end surfaces are wellinsulated. The highest allowable surface temperature of the furnace and the annual cost of this loss to theplant are to be determined.
9-26
Asphalt
L = 100 m
D + 2tins
25C
Insulation
= 0.1
Tsky
= -30C
T
= 0C
7/30/2019 Heat Chap09 032
2/23
Chapter 9Natural Convection
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2=(140+75)/2=107.5F are (Table A-15)
1-
23
R001762.0R)4605.107(
11
7249.0Pr
/sft101851.0
FBtu/h.ft.01546.0
=+
==
=
=
=
fT
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 140F for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The characteristic length in this case is the outer diameter of the furnace, ft.8==DLc Then,
10
223
3-12
2
3
10996.3)7249.0()/sft101851.0(
)ft8)(R75140)(R001762.0)(ft/s2.32(Pr
)(=
=
=
DTTgRa s
( )[ ] ( )[ ]9.376
7249.0/559.01
)10996.3(387.06.0
Pr/559.01
387.06.0
2
27/816/9
6/1102
27/816/9
6/1
=
+
+=
++=
RaNu
2
2
ft7.326)ft13)(ft8(
F.Btu/h.ft7287.0)9.376(ft8
FBtu/h.ft.01546.0
===
=
==
DLA
NuD
kh
s
The total rate of heat generated in the furnace is
Btu/h10936.3Btu/therm)100,000(therms/h)48)(82.0( 6==genQ
Noting that 1% of the heat generated can be dissipated by natural convection and radiation ,
Btu/h360,39Btu/h)10936.3)(01.0( 6 ==Q
The total rate of heat loss from the furnace by natural convection and radiation can be expressed as
])R46075()[.RBtu/h.ft101714.0)(m7.326)(85.0(
)]R46075()[ft7.326(F).Btu/h.ft7287.0(Btu/h360,39
)()(
444282
22
44
++
+=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is
F141.8== R8.601sT
which is very close to the assumed value. Therefore, there is no need to repeat calculations.
The total amount of heat loss and its cost during a-2800 hour period is
Btu10102.1h)2800)(Btu/h360,39( 8 === tQQ totaltotal
$716.== therm)/65.0)($therm000,100/10102.1(Cost 8
9-27
L = 13 ft
Furnace = 0.1
Air
T
= 75F
D = 8 ft
7/30/2019 Heat Chap09 032
3/23
Chapter 9Natural Convection
9-34 A glass window is considered. The convection heat transfer coefficient on the inner side of thewindow, the rate of total heat transfer through the window, and the combined natural convection andradiation heat transfer coefficient on the outer surface of the window are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (5+25)/2 = 15C are (Table A-15)
1-
25
K003472.0K)27315(
11
7323.0Pr
/sm10471.1
CW/m.02476.0
=+
==
==
=
fT
k
Analysis(a) The characteristic length in this case is the height of the window, m.2.1==LLc Then,
9
225
3-12
2
3
10986.3)7323.0()/sm10471.1(
)m2.1)(K525)(K34720.0)(m/s81.9(Pr
)(=
=
=
0
cs LTTgRa
7.189
7323.0
492.01
)10986.3(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/19
2
27/816/9
6/1
=
+
+=
+
+=Nu
2m4.2m)m)(22.1(
)7.189(m2.1
CW/m.02476.0
==
=
==
sA
NuL
kh C.W/m3.915 2
(b) The sum of the natural convection and radiation heat transfer from the room to the window is
W9.187C)525)(m4.2)(C.W/m915.3()( 22convection === ss TThAQ
W3.234])K2735()K27325)[(.KW/m1067.5)(m4.2)(9.0(
)(444282
44radiation
=++==
ssurrs TTAQ
W422.2=+=+= 3.2349.187radiationconvectiontotal QQQ
(c) The outer surface temperature of the window can be determined from
C65.3)m4.2)(CW/m.78.0(
)m006.0)(W346(C5)(
2
total,,,,total =
===
s
isososiss
kA
tQTTTT
t
kAQ
Then the combined natural convection and radiation heat transfer coefficient on the outer window surface becomes
C.W/m20.35 2 =
=
=
=
C)]5(65.3)[m4.2(
W346
)(
or
)(
2
,,
totalcombined
,,combinedtotal
ooss
ooss
TTA
Qh
TTAhQ
Note that T QR= and thus the thermal resistance R of a layer is proportional to the temperature dropacross that layer. Therefore, the fraction of thermal resistance of the glass is equal to the ratio of thetemperature drop across the glass to the overall temperature difference,
4.5%)(or045.0)5(25
65.35
total
glass
total
glass =
=
=
TR
T
R
R
which is low. Thus it is reasonable to neglect the thermal resistance of the glass.
9-28
Q
Outdoors
-5C
Glass
Ts= 5C
= 0.9
L = 1.2 m
Room
T
= 25C
7/30/2019 Heat Chap09 032
4/23
Chapter 9Natural Convection
9-35 An insulated electric wire is exposed to calm air. The temperature at the interface of the wire and theplastic insulation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (50+30)/2 = 40C are (Table A-15)
1-
25
K003195.0K)27340(
11
7255.0Pr
/sm10702.1
CW/m.02662.0
=+
==
==
=
fT
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 50C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.The characteristic length in this case is the outer diameter of the insulated wireLc =D = 0.006 m. Then,
3.339)7255.0()/sm10702.1(
)m006.0)(K3050)(K003195.0)(m/s81.9(Pr
)(225
3-12
2
3
=
=
=
DTTgRa s
( )[ ] ( )[ ]101.2
7255.0/559.01
)3.339(387.06.0
Pr/559.01
387.06.0
2
27/816/9
6/12
27/816/9
6/1
=
++=
++=
RaNu
2
2
m2262.0m)m)(12006.0(
C.W/m327.9)101.2(m006.0
CW/m.02662.0
===
=
==
DLA
NuD
kh
s
The rate of heat generation, and thus the rate of heat transfer is
(Q VI= = =8 80V)(10 A) W
Considering both natural convection and radiation, the total rate of heat loss can be expressed as
])K27330()273)[(.KW/m1067.5)(m2262.0)(9.0(
C)30)(m226.0)(C.W/m327.9(W80
)()(
444282
22
44
+++
=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is
C6.52 =sT
which is close to the assumed value of 50C. Then the temperature at the interface of the wire and theplastic cover in steady operation becomes
C57.5=
=+==)m12)(CW/m.15.0(2
)3/6ln()W80(+C6.52
2
)/ln()(
)/ln(
2 12
12
kL
DDQTTTT
DD
kLQ sisi
9-29
Air
T
= 30C
Ts
= 0.9
L = 12 m
D = 6 mm
Resistanceheater
7/30/2019 Heat Chap09 032
5/23
Chapter 9Natural Convection
9-36 A steam pipe extended from one end of a plant to the other with no insulation on it. The rate of heatloss from the steam pipe and the annual cost of those heat losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (170+20)/2 = 95C are (Table A-15)
1-
25
K002717.0K)27395(
11
7121.0Pr
/sm10252.2
CW/m.0306.0
=+
==
==
=
fT
k
AnalysisThe characteristic length in this case is the outer diameter of the pipe, m0.0603== DLc . Then,
6
225
3-12
2
3
10231.1)7121.0()/sm10252.2(
)m0603.0)(K20170)(K002717.0)(m/s81.9(Pr
)(=
=
=
DTTgRa
s
( )[ ] ( )[ ] 42.157121.0/559.01)10231.1(387.0
6.0Pr/559.01
387.0
6.0
2
27/816/9
6/162
27/816/9
6/1
=
+
+=
++=
Ra
Nu
2
2
m37.11m)m)(600603.0(
C.W/m823.7)42.15(m0603.0
CW/m.0306.0
===
=
==
DLA
NuD
kh
s
Then the total rate of heat transfer by natural convection and radiation becomes
kW27.4==+++
=
+=
W
TTATThAQ surrssss
27,388
])K27320()K273170)[(.KW/m1067.5)(m37.11)(7.0(
C)20170)(m37.11)(C.W/m823.7(
)()(
444282
22
44
The total amount of gas consumption and its cost during a one-year period is
therms/yr496,10s/h)3600h/yr8760(kJ105,500
therm1
78.0
kJ/s388.27=
=
=tQ
Qgas
$5647/yr== therm)/538.0)($therms/yr496,10(Cost
9-30
Steam
L = 60 m
D =6.03 cm
Ts= 170C
= 0.7
Air
T
= 20C
7/30/2019 Heat Chap09 032
6/23
Chapter 9Natural Convection
9-37"!PROBLEM 9-37"
"GIVEN"L=60 "[m]"D=0.0603 "[m]"
T_s=170 "[C], parameter to be varied"T_infinity=20 "[C]"
epsilon=0.7T_surr=T_infinityeta_furnace=0.78UnitCost=0.538 "[$/therm]"time=24*365 "[h]"
"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_film)Pr=Prandtl(Fluid$, T=T_film)rho=Density(Fluid$, T=T_film, P=101.3)mu=Viscosity(Fluid$, T=T_film)nu=mu/rho
beta=1/(T_film+273)T_film=1/2*(T_s+T_infinity)sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant"g=9.807 "[m/s^2], gravitational acceleration"
"ANALYSIS"delta=DRa=(g*beta*(T_s-T_infinity)*delta^3)/nu^2*PrNusselt=(0.6+(0.387*Ra^(1/6))/(1+(0.559/Pr)^(9/16))^(8/27))^2h=k/delta*NusseltA=pi*D*LQ_dot=h*A*(T_s-T_infinity)+epsilon*A*sigma*((T_s+273)^4-(T_surr+273)^4)Q_gas=(Q_dot*time)/eta_furnace*Convert(h, s)*Convert(J, kJ)*Convert(kJ, therm)
Cost=Q_gas*UnitCost
9-31
7/30/2019 Heat Chap09 032
7/23
Chapter 9Natural Convection
Ts [C] Q [W] Cost [$]
100 11636 2399
105 12594 2597
110 13577 2799
115 14585 3007
120 15618 3220
125 16676 3438130 17760 3661
135 18869 3890
140 20004 4124
145 21166 4364
150 22355 4609
155 23570 4859
160 24814 5116
165 26085 5378
170 27385 5646
175 28713 5920
180 30071 6200
185 31459 6486
190 32877 6778195 34327 7077
200 35807 7382
10 0 120 14 0 160 180 200
10000
15000
20000
25000
30000
35000
40000
2000
3000
4000
5000
6000
7000
8000
Ts
[C ]
Q
[W]
Cost[$]
Q
Cost
9-32
7/30/2019 Heat Chap09 032
8/23
Chapter 9Natural Convection
9-38 A steam pipe extended from one end of a plant to the other. It is proposed to insulate the steam pipefor $750. The simple payback period of the insulation to pay for itself from the energy it saves are to bedetermined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (35+20)/2 = 27.5C are (Table A-15)
1-
25
K003328.0K)2735.27(
11
7289.0Pr
/sm10584.1
CW/m.0257.0
=+
==
==
=
fT
k
AnalysisInsulation will drop the outer surface temperature to a value close to the ambient temperature. Thesolution of this problem requires a trial-and-error approach since the determination of the Rayleigh numberand thus the Nusselt number depends on the surface temperature which is unknown. We start the solution
process by guessing the outer surface temperature to be 35C for the evaluation of the properties and h.We will check the accuracy of this guess later and repeat the calculations if necessary. The characteristic
length in this case is the outer diameter of the insulated pipe,m.1603.0==DL
c Then,6
225
3-12
2
3
10856.5)7289.0()/sm10584.1(
)m1603.0)(K2035)(K003328.0)(m/s81.9(Pr
)(=
=
=
DTTgRa s
( )[ ] ( )[ ]23.24
7289.0/559.01
)10856.5(387.06.0
Pr/559.01
387.06.0
2
27/816/9
6/162
27/816/9
6/1
=
+
+=
++=
RaNu
2
2
m22.30m)m)(601603.0(
C.W/m884.3)23.24(m1603.0
CW/m.0257.0
===
=
==
DLA
NuD
kh
s
Then the total rate of heat loss from the outer surface of the insulated pipe by convection and radiationbecomes
W2039
])K27320()K27335)[(.KW/m1067.5)(m22.30)(1.0(+
C)2035)(m22.30)(C.W/m884.3(
)()(
444282
22
44
=++
=+=+=
surrssssradconv TTATThAQQQ
In steady operation, the heat lost from the exposed surface of the insulation by convection and radiation must be equalto the heat conducted through the insulation. This requirement gives the surface temperature to be
)m60)(CW/m.038.0(2
)03.6/03.16ln(
C)170(W2039
2
)/ln( 12
..insulation
=
=
==
ssis
ins
sis T
kL
DD
TT
R
TTQQ
It gives 30.8C for the surface temperature, which is somewhat different than the assumed value of 35 C.Repeating the calculations with other surface temperatures gives
W1988andC3.34 == QTs
Heat loss and its cost without insulation was determined in the Prob. 9-36 to be 27.388 kW and $5647.Then the reduction in the heat losses becomes
kW40.25988.1388.27saved =Q or 25.388/27.40 = 0.927 (92.7%)Therefore, the money saved by insulation will be 0.921 ($5647/yr) = $5237/yr which will pay for the costof $750 in $750/($5237/yr)=0.1432 year = 52.3 days.
9-33
Steam
L = 60 m
D =16.03 cm
170C, = 0.1
Air
T
= 20C
Insulation
= 0.1
7/30/2019 Heat Chap09 032
9/23
Chapter 9Natural Convection
9-39 A circuit board containing square chips is mounted on a vertical wall in a room. The surfacetemperature of the chips is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withconstant properties. 3 The local atmospheric pressure is 1 atm. 4 The heattransfer from the back side of the circuit board is negligible.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (35+25)/2 = 30C are (Table A-15)
1-
25
K0033.0K)27330(
11
7282.0Pr
/sm10608.1
CW/m.02588.0
=+
==
==
=
fT
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 35C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The characteristic length in this case is the height of the board, m.3.0== LLc Then,
7
225
3-12
2
3
10463.2)7282.0()/sm10608.1(
)m3.0)(K2535)(K0033.0)(m/s81.9(Pr
)(=
=
=
LTTgRa
s
57.40
7282.0
492.01
)10463.2(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/17
2
27/816/9
6/1
=
+
+=
+
+=Nu
22
2
m09.0m)3.0(
C.W/m50.3)57.40(m3.0
CW/m.02588.0
==
=
==
sA
NuL
kh
Considering both natural convection and radiation, the total rate of heat loss can be expressed as
])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(
C)25)(m09.0)(C.W/m50.3(W)05.0121(
)()(
444282
22
44
+++
=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is
Ts = 33.5 C
which is sufficiently close to the assumed value in the evaluation of properties and h. Therefore, there is noneed to repeat calculations by reevaluating the properties and h at the new film temperature.
9-34
Air
T
= 25CT
surr= 25C
PCB, Ts
= 0.7121 0.05W
L = 30 cm
7/30/2019 Heat Chap09 032
10/23
Chapter 9Natural Convection
9-40 A circuit board containing square chips is positioned horizontally in a room. The surface temperatureof the chips is to be determined for two orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The heat transfer from the back side of the circuit board is negligible.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (35+25)/2 = 30C are (Table A-15)
1-
25
K0033.0K)27330(
11
7282.0Pr
/sm10608.1
CW/m.02588.0
=+
==
==
=
fT
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 35C for the evaluation of theproperties and h. The characteristic length for both cases is determined from
m.075.0
m)]3.0(+m)3.02[(
m)3.0( 2===
p
AL
sc
Then,
5
225
3-12
2
3
10848.3)7282.0()/sm10608.1(
)m075.0)(K2535)(K00333.0)(m/s81.9(Pr
)(=
=
=
cs LTTgRa
(a) Chips (hot surface) facing up:
45.13)10848.3(54.054.0 4/154/1 === RaNu
22
2
m09.0m)3.0(
C.W/m641.4)45.13(m075.0
CW/m.02588.0
==
=
==
s
c
A
NuL
kh
Considering both natural convection and radiation, the total rate of heat loss can be expressed as
])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(
C)25)(m09.0)(C.W/m641.4(W)05.0121(
)()(
444282
22
44
+++
=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is Ts = 32.5 Cwhich is sufficiently close to the assumed value. Therefore, there is no need to repeat calculations.(b) Chips (hot surface) facing up:
725.6)10848.3(27.027.0 4/154/1 === RaNu
C.W/m321.2)725.6(m075.0
CW/m.02588.0 2 =
== NuL
kh
c
Considering both natural convection and radiation, the total rate of heat loss can be expressed as
])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(
C)25)(m09.0)(C.W/m321.2(W)05.0121(
)()(
444282
22
44
+++=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is Ts = 35.0 Cwhich is identical to the assumed value in the evaluation of properties and h. Therefore, there is no need torepeat calculations.
9-35
Air
T
= 25C
Tsurr
= 25CPCB, Ts = 0.7121 0.05W
L = 30 cm
7/30/2019 Heat Chap09 032
11/23
Chapter 9Natural Convection
9-41 It is proposed that the side surfaces of a cubic industrial furnace be insulated for $550 in order toreduce the heat loss by 90 percent. The thickness of the insulation and the payback period of the insulationto pay for itself from the energy it saves are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withconstant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (110+30)/2 = 70C are (Table A-15)
1-
25
K002915.0K)27370(
11
7177.0Pr
/sm10995.1
CW/m.02881.0
=+
==
==
=
fT
k
AnalysisThe characteristic length in this case is the height of the furnace, m.2==LLc Then,
10
225
3-12
2
3
10301.3)7177.0()/sm10995.1(
)m2)(K30110)(K002915.0)(m/s81.9(Pr
)(=
=
=
LTTgRa s
2.369
7177.0
492.01
)10301.3(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/110
2
27/816/9
6/1
=
+
+=
+
+=Nu
22
2
m16)m2(4
C.W/m318.5)2.369(m2
CW/m.02881.0
==
=
==
s
c
A
NuL
kh
Then the heat loss by combined natural convection and radiation becomes
W119,15
])K27330()K273110)[(.KW/m1067.5)(m16)(7.0(
C)30110)(m16)(C.W/m318.5(
)()(
444282
22
44
=+++
=
+=
surrssss TTATThAQ
Noting that insulation will reduce the heat losses by 90%, the rate of heat loss after insulation will be
W1512W119,151.0)9.01(
W607,13W119,159.09.0
insulationnoloss
insulationnosaved
===
===
The furnace operates continuously and thus 8760 h. Then the amount of energy and money the insulationwill save becomes
therms/yr5215s/yr)3600(8760kJ105,500
therm1
78.0
kJ/s607.13savedEnergy =
== tQsaved
$2868)therm/55.0)($therms5215(=energy)ofcosttsaved)(UniEnergy(savedMoney ==
Therefore, the money saved by insulation will pay for the cost of $550 in
550/($2868/yr)=0.1918 yr = 70 days.
9-36
Hot gases
T
= 30C
Furnace
Ts= 110C = 0.7
2 m
2 m
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
Insulation will lower the outer surface temperature, the Rayleigh and Nusselt numbers, and thusthe convection heat transfer coefficient. For the evaluation of the heat transfer coefficient, we assume the
surface temperature in this case to be 50C. The properties of air at the film temperature of ( Ts+T)/2 =(50+30)/2 = 40C are (Table A-15)
1-
25
K003195.0K)27340(
117255.0Pr
/sm10702.1
CW/m.02662.0
=+
===
=
=
fT
k
Then,
10
225
3-12
2
3
10256.1)7255.0()/sm10702.1(
)m2)(K3050)(K003195.0)(m/s81.9(Pr
)(=
=
=
LTTgRa s
0.272
7255.0
492.0
1
)10256.1(387.0825.0
Pr
492.0
1
Ra387.0825.0
2
27/816/9
6/110
2
27/816/9
6/1
=
+
+=
+
+=Nu
C.W/m620.3)0.272(m2
CW/m.02662.0 2 =
== NuL
kh
m)2m)(22(4 insuls tA +=
The total rate of heat loss from the outer surface of the insulated furnace by convection and radiationbecomes
])K27330()K273)[(.KW/m1067.5()7.0(+C)30(C).W/m620.3(W1512
)()(
444282
44
++=
+=+=
ss
surrssssradconv
TATA
TTATThAQQQ I
n steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from theexposed surface of the insulation by convection and radiation, which must be equal to the heat conductedthrough the insulation. Therefore,
insul
ss
ins
ssinsulation
t
TA
t
TTkAQQ
C)(110C)W/m.038.0(W1512
)( furnace =
==
Solving the two equations above by trial-and error (or better yet, an equation solver) gives
Ts = 48.4C and tinsul = 0.0254 m = 2.54 cm
9-37
7/30/2019 Heat Chap09 032
13/23
Chapter 9Natural Convection
9-42 A cylindrical propane tank is exposed to calm ambient air. The propane is slowly vaporized due to acrack developed at the top of the tank. The time it will take for the tank to empty is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 Radiation heat transfer is negligible.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (-42+25)/2 = -8.5C are (Table A-15)
1-
25
K003781.0K)2735.8(
11
7383.0Pr
/sm10265.1
CW/m.02299.0
=+
==
==
=
fT
k
Analysis The tank gains heat through its cylindrical surface as well as its circular end surfaces. Forconvenience, we take the heat transfer coefficient at the end surfaces of the tank to be the same as that of itsside surface. (The alternative is to treat the end surfaces as a vertical plate, but this will double the amountof calculations without providing much improvement in accuracy since the area of the end surfaces is muchsmaller and it is circular in shape rather than being rectangular). The characteristic length in this case is the
outer diameter of the tank, m.5.1==DLc Then,
10
225
3-12
2
3
10869.3)7383.0()/sm10265.1(
)m5.1](K)42(25)[(K003781.0)(m/s81.9(Pr
)(=
=
=
DTTgRa s
( )[ ] ( )[ ]1.374
7383.0/559.01
)10869.3(387.06.0
Pr/559.01
387.06.0
2
27/816/9
6/1102
27/816/9
6/1
=
+
+=
++=
RaNu
222
2
m38.224/m)5.1(2)m4)(m5.1(4/2
C.W/m733.5)1.374(m5.1
CW/m.02299.0
=+=+=
=
==
DDLA
NuD
kh
s
and
W8598C)]42(25)[(m38.22)(C.W/m733.5()(
22
=== ss TThAQ
The total mass and the rate of evaporation of propane are
kg/s02023.0kJ/kg425
kJ/s598.8
kg4107)m4(4
)m5.1()kg/m581(
4
23
2
===
=
=
==
fgh
Qm
LD
Vm
and it will take
hours56.4==== s996,202kg/s02023.0
kg4107
m
mt
for the propane tank to empty.
9-38
D = 1.5 m
L = 4 m
Propane tank
0T
s= -42C
Air
T
= 25C
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
9-43E The average surface temperature of a human head is to be determined when it is not covered.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The head can be approximated as a 12-in.-diameter sphere.
Properties The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 120F for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The properties of air at 1 atm and the anticipated film temperature of (Ts+T)/2 = (120+77)/2 = 98.5F are(Table A-15E)
1-
23
R001791.0R)4605.98(
11
7262.0Pr
/sft10180.0
FBtu/h.ft.01525.0
=+
==
==
=
fT
k
AnalysisThe characteristic length for a spherical object isLc =D = 12/24 = 0.5 ft. Then,
6
223
3-12
2
3
10943.6)7262.0(
)/sft10180.0(
)ft5.0)(R7795)(R001791.0)(ft/s2.32(Pr
)(=
=
=
DTTgRa s
39.25
7262.0
469.01
)10943.6(589.02
Pr
469.01
589.02
9/416/9
4/16
9/416/9
4/1
=
+
+=
+
+=Ra
Nu
222
2
ft7854.0)ft5.0(
F.Btu/h.ft7744.0)39.25(ft1
FBtu/h.ft.01525.0
===
=
==
DA
NuD
kh
s
Considering both natural convection and radiation, the total rate of heat loss can be written as
])R46077()R460)[(.RBtu/h.ft101714.0)(m7854.0)(9.0(
F)77)(ft7854.0(F).Btu/h.ft7744.0(Btu/h)4/287(
)()(
444282
22
44
+++=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is
Ts = 125.9 F
which is sufficiently close to the assumed value in the evaluation of the properties and h. Therefore, there isno need to repeat calculations.
9-39
Air
T
= 77F
HeadQ = 287 Btu/h
D = 12 in
= 0.9
7/30/2019 Heat Chap09 032
15/23
Chapter 9Natural Convection
9-44 The equilibrium temperature of a light glass bulb in a room is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The light bulb is approximated as an 8-cm-diameter sphere.
Properties The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by guessing the surface temperature to be 170 C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The properties of air at 1 atm and the anticipated film temperature of ( Ts+T)/2 = (170+25)/2 = 97.5C are(Table A-15)
1-
25
K002699.0K)2735.97(
11
7116.0Pr
/sm10279.2
CW/m.03077.0
=+
==
==
=
fT
k
AnalysisThe characteristic length in this case isLc =D = 0.08 m. Then,
6
225
3-12
2
3
10694.2
)7116.0(
)/sm10279.2(
)m08.0)(K25170)(K002699.0)(m/s81.9(Pr
)(
=
=
=
DTTgRa s
( )[ ] ( )[ ]
42.20
7116.0/469.01
)10694.2(589.02
Pr/469.01
589.02
9/416/9
4/16
9/416/9
4/1
=+
+=
++=
RaNu
Then
222
2
m02011.0m)08.0(
C.W/m854.7)42.20(m08.0
CW/m.03077.0
===
=
==
DA
NuD
kh
s
Considering both natural convection and radiation, the total rate of heat loss can be written as
])K27325()273)[(.KW/m1067.5)(m02011.0)(9.0(
C)25)(m02011.0)(C.W/m854.7(W)6090.0(
)()(
444282
22
44
+++
=
+=
s
s
surrssss
T
T
TTATThAQ
Its solution is
Ts = 169.4 C
which is sufficiently close to the value assumed in the evaluation of properties and h. Therefore, there is noneed to repeat calculations.
9-40
Air
T
= 25C
Lamp60 W
= 0.9
D = 8 cm
Light,6 W
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
9-45 A vertically oriented cylindrical hot water tank is located in a bathroom. The rate of heat loss from thetank by natural convection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The temperature of the outer surface of the tank is constant.
Properties The properties of air at 1 atm and the film temperature
of (Ts+T)/2 = (44+20)/2 = 32C are (Table A-15)
1-
25
K003279.0K)27332(
11
7276.0Pr
/sm10627.1
CW/m.02603.0
=+
==
==
=
fT
k
AnalysisThe characteristic length in this case is the height of the cylinder,
m.1.1==LLc Then,
9
225
3-12
2
3
10883.3)/sm10627.1(
)m1.1)(K2044)(K003279.0)(m/s81.9()(Gr =
=
=
LTTg s
A vertical cylinder can be treated as a vertical plate when
m1542.0)10883.3(
m)1.1(35
Gr
35m)4.0( 4/191/4 ===
LD
which is satisfied. That is, the Nusselt number relation for a vertical plate can be used for the side surfaces.For the top and bottom surfaces we use the relevant Nusselt number relations. First, for the side surfaces,
99 10825.2)7276.0)(10883.3(GrPrRa ===
2.170
7276.0
492.01
)10825.2(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/19
2
27/816/9
6/1
=
+
+=
+
+=Nu
2
2
m382.1)m1.1)(m4.0(
C.W/m027.4)2.170(m1.1
CW/m.02603.0
===
===
DLA
NuLkh
s
W6.133C)2044)(m382.1)(C.W/m027.4()( 22side === TThAQ ss
For the top surface,
m1.04
m4.0
4
4/2=====
D
D
D
p
AL
sc
6
225
3-12
2
3
10123.2)7276.0()/sm10627.1(
)m1.0)(K2044)(K003279.0)(m/s81.9(Pr
)(Ra =
=
=
cs LTTg
61.20)10123.2(54.0Ra54.0Nu 4/164/1
===
222
2
m1257.04/)m4.0(4/
C.W/m365.5)61.20(m1.0
CW/m.02603.0
===
=
==
DA
NuL
kh
s
c
W2.16C)2044)(m1257.0)(C.W/m365.5()( 22top === TThAQ ss
For the bottom surface,
31.10)10123.2(27.0Ra27.0Nu 4/164/1 ===
9-41
AirT
= 20CTankT
s= 44C
= 0.4L = 1.1 m
D = 0.4 m
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
C.W/m683.2)31.10(m1.0
CW/m.02603.0 2 =
== NuL
kh
c
W1.8C)2044)(m1257.0)(C.W/m683.2()( 22bottom === TThAQ ss
The total heat loss by natural convection is
W157.9=++=++= 1.82.166.133bottomtopsideconv QQQQ
The radiation heat loss from the tank is
[ ]W101.1=
++++=
= 444282
44rad
)K27320()K27344().KW/m1067.5)(m1257.01257.0382.1)(4.0(
)( surrss TTAQ
9-42
7/30/2019 Heat Chap09 032
18/23
7/30/2019 Heat Chap09 032
19/23
Chapter 9Natural Convection
C11.7== K7.2842T
We could repeat the solution using air properties at the new film temperature using this value to increasethe accuracy. However, this would only affect the heat transfer value somewhat, which would not havesignificant effect on the final water temperature. The average rate of heat transfer can be determined fromEq. 2
W34.3== 2)-(11.753976.3Q
9-44
7/30/2019 Heat Chap09 032
20/23
Chapter 9Natural Convection
9-47"!PROBLEM 9-47"
"GIVEN"height=0.28 "[m]"L=0.18 "[m]"w=0.18 "[m]"
T_infinity=24 "[C]"
T_w1=2 "[C]"epsilon=0.6T_surr=T_infinity"time=3 [h], parameter to be varied"
"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_film)Pr=Prandtl(Fluid$, T=T_film)rho=Density(Fluid$, T=T_film, P=101.3)mu=Viscosity(Fluid$, T=T_film)nu=mu/rhobeta=1/(T_film+273)
T_film=1/2*(T_w_ave+T_infinity)T_w_ave=1/2*(T_w1+T_w2)rho_w=Density(water, T=T_w_ave, P=101.3)C_p_w=CP(water, T=T_w_ave, P=101.3)*Convert(kJ/kg-C, J/kg-C)sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant"g=9.807 "[m/s^2], gravitational acceleration"
"ANALYSIS"delta=heightRa=(g*beta*(T_infinity-T_w_ave)*delta^3)/nu^2*PrNusselt=0.59*Ra^0.25h=k/delta*NusseltA=2*(height*L+height*w+w*L)Q_dot=h*A*(T_infinity-T_w_ave)+epsilon*A*sigma*((T_surr+273)^4-(T_w_ave+273)^4)
m_w=rho_w*V_wV_w=height*L*wQ=m_w*C_p_w*(T_w2-T_w1)Q_dot=Q/(time*Convert(h, s))
9-45
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
time [h] Tw2 [C]
0.5 4.013
1 5.837
1.5 7.496
2 9.013
2.5 10.41
3 11.693.5 12.88
4 13.98
4.5 15
5 15.96
5.5 16.85
6 17.69
6.5 18.48
7 19.22
7.5 19.92
8 20.59
8.5 21.21
9 21.81
9.5 22.3710 22.91
0 2 4 6 8 102. 5
7
11.5
16
20.5
25
time [h]
Tw2
[C]
9-46
7/30/2019 Heat Chap09 032
22/23
Chapter 9Natural Convection
9-48 A room is to be heated by a cylindrical coal-burning stove. The surface temperature of the stove andthe amount of coal burned during a 30-day-period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The temperature of the outer surface of the stove is constant. 5 The heattransfer from the bottom surface is negligible. 6 The heat transfer coefficient at the top surface is the sameas that on the side surface.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (130+24)/2 = 77C are (Table A-1)
1-
25
K002857.0K)27377(
11
7161.0Pr
/sm10066.2
CW/m.02931.0
=+
==
==
=
fT
k
AnalysisThe characteristic length in this case is the height of the cylindir,
m.7.0==LLc Then,
9
225
3-12
2
3
10387.2)/sm10066.2(
)m70.0)(K24130)(K002857.0)(m/s81.9()(Gr =
=
=
LTTg s
A vertical cylinder can be treated as a vertical plate when
m1108.0)10387.2(
m)7.0(35
Gr
35m)32.0(
4/191/4=
==
LD
which is satisfied. That is, the Nusselt number relation for a vertical plate can be used for side surfaces.
99 10709.1)7161.0)(10387.2(GrPrRa ===
2.145
7161.0
492.01
)10709.1(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/19
2
27/816/9
6/1
=
+
+=
+
+=Nu
222
2
m7841.04/)m32.0()m7.0)(m32.0(4/
C.W/m080.6)2.145(m7.0
CW/m.02931.0
=+=+=
===
DDLA
NuL
kh
s
Then the surface temperature of the stove is determined from
C127.6==
+=
+=+=
K6.400
)290)(.KW/m10)(5.67m841(0.85)(0.7)297)(mC)(0.7841.W/m080.6(W1200
)()(
4442-8222
4surr
4radconv
s
ss
ssss
T
TT
TTATThAQQQ
The amount of coal used is determined from
kg3.102==
=
==
kJ/kg30,000
kJ)/0.65(60,480/
kJ60,480=s/h)3600h/daykJ/s)(142.1(
coalHV
Qm
tQQ
9-47
Air
T
= 24CStoveT
s
= 0.85L =0.7 m
D = 0.32 m
7/30/2019 Heat Chap09 032
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Chapter 9Natural Convection
9-49 Water in a tank is to be heated by a spherical heater. The heating time is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The temperature of the outer surface of the sphere isconstant.
Properties Using the average temperature for water (15+45)/2=30as the fluid temperature, the properties of water at the film
temperature of (Ts+T)/2 = (85+30)/2 = 57.5C are (Table A-9)
1-3
26
K10501.0
12.3Pr
/sm10474.0
CW/m.6515.0
=
==
=
k
Also, the properties of water at 30C are (Table A-9)
CJ/kg.4178andkg/m996 3 == pC
AnalysisThe characteristic length in this case isLc =D = 0.06 m. Then,
8
226
3-132
2
3
10108.8)12.3()/sm10474.0(
)m06.0)(K3085)(K10501.0)(m/s81.9(Pr
)(Ra =
=
=
DTTg s
( )[ ] ( )[ ]14.89
12.3/469.01
)10108.8(589.02
Pr/469.01
589.02
9/416/9
4/18
9/416/9
4/1=
++=
++= RaNu
222
2
m01131.0m)06.0(
C.W/m9.967)14.89(m06.0
CW/m.6515.0
===
=
==
DA
NuD
kh
s
The rate of heat transfer by convection is
W1.602)30)(85mC)(0.01131.W/m9.967()( 22conv === TThAQ ss
The mass of water in the container is
kg.8439)m)(0.040kg/m996( 33 === Vm
The amount of heat transfer to the water isJ104.994=C15)-C)(45J/kg.kg)(417884.39()( 612 == TTmCQ p
Then the time the heater should be on becomes
hours2.304==
== s8294J/s602.1
J104.994 6
Q
Qt
Water
T,ave
= 30C
Resistanceheater
Ts = 85CD = 6 cmD = 6 cm