heat chap07 073
TRANSCRIPT
-
7/30/2019 Heat Chap07 073
1/21
Chapter 7External Forced Convection
7-63
Special Topic: Thermal Insulation
7-73C Thermal insulation is a material that is used primarily to provide resistance to heat flow. It differsfrom other kinds of insulators in that the purpose of an electrical insulator is to halt the flow of electriccurrent, and the purpose of a sound insulator is to slow down the propagation of sound waves.
7-74C In cold surfaces such as chilled water lines, refrigerated trucks, and air conditioning ducts,insulation saves energy since the source of coldness is refrigeration that requires energy input. In thiscase heat is transferred from the surroundings to the cold surfaces, and the refrigeration unit must now workharder and longer to make up for this heat gain and thus it must consume more electrical energy.
7-75C The R-valueof insulation is the thermal resistance of the insulating material per unit surface area.Forflatinsulation theR-value is obtained by simply dividing the thickness of the insulation by its thermalconductivity. That is,R-value =L/k. Doubling the thicknessL doubles theR-value of flat insulation.
7-76C The R-value of an insulation represents the thermal resistance of insulation per unit surface area (orper unit length in the case of pipe insulation).
7-77C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers inan evacuated space. Radiation between two surfaces is inversely proportional to the number of sheets usedand thus heat loss by radiation will be very low by using this highly reflective sheets. Evacuating the spacebetween the layers forms a vacuum which minimize conduction or convection through the air space.
7-78C Yes, hair or any other cover reduces heat loss from the head, and thus serves as insulation for thehead. The insulating ability of hair or feathers is most visible in birds and hairy animals.
7-79C The primary reasons for insulating are energy conservation, personnel protection and comfort,maintaining process temperature, reducing temperature variation and fluctuations, condensation andcorrosion prevention, fire protection, freezing protection, and reducing noise and vibration.
7-80C The optimum thickness of insulation is the thickness that corresponds to a minimum combined costof insulation and heat lost. The cost of insulation increases roughly linearly with thickness while the cost ofheat lost decreases exponentially. The total cost, which is the sum of the two, decreases first, reaches aminimum, and then increases. The thickness that corresponds to the minimum total cost is the optimumthickness of insulation, and this is the recommended thickness of insulation to be installed.
-
7/30/2019 Heat Chap07 073
2/21
Chapter 7External Forced Convection
7-64
7-81 The thickness of flatR-8 insulation in SI units is to be determined when the thermal conductivity of thematerial is known.
AssumptionsThermal properties are constant.
Properties The thermal conductivity of the insulating material is given to be k= 0.04 W/mC.
Analysis The thickness of flat R-8 insulation (in m2.C/W) is determinedfrom the definition ofR-value to be
R L
kL R kvalue value
2m . C / W)(0.04 W / m. C) (8 0.32 m
7-82E The thickness of flat R-20 insulation in English units is to be determined when the thermalconductivity of the material is known.
AssumptionsThermal properties are constant.
Properties The thermal conductivity of the insulating material is given to be k= 0.02 Btu/hftF.
Analysis The thickness of flat R-20 insulation (in hft2F/Btu) is determinedfrom the definition ofR-value to be
ft0.4 F)Btu/h.ft.2F/Btu)(0.0.h.ft20(2
valuevalue kRLk
L
R
R-8
L
R-20
L
-
7/30/2019 Heat Chap07 073
3/21
Chapter 7External Forced Convection
7-65
7-83 A steam pipe is to be covered with enough insulation to reduce the exposed surface temperature to
30C . The thickness of insulation that needs to be installed is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transferis one-dimensional since there is thermal symmetry about the centerline and no variation in the axial
direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible.
Properties The thermal conductivities are given to be k = 52 W/mC for cast iron pipe and k= 0.038
W/mC for fiberglass insulation.Analysis The thermal resistance network for this problem involves 4 resistances in series. The inner radiusof the pipe is r1 = 2.0 cm and the outer radius of the pipe and thus the inner radius of insulation is r2 = 2.3cm. Letting r3 represent the outer radius of insulation, the areas of the surfaces exposed to convection for aL = 1 m long section of the pipe become
A r L
A r L r r r
1 1
3 3 3 3 3
2 2 0 02
2 2 2
( . m)(1 m) = 0.1257 m
(1 m) = m ( in m)
2
2
Then the individual thermal resistances are determined to be
C/W9944.0)mC)(0.1257.W/m(80
1122
1conv,1
AhRR
ii
C/W00043.0)mC)(1W/m.(522
)02.0/023.0ln(
2
)/ln(
1
12pipe1
Lk
rrRR
R Rr r
k L
rr2
3 2
2
33
2
00234188 0 023
insulation
2 (0.038 W/ m. C)(1 mC / W
ln( / ) ln( / . )
). ln( / . )
R Rh A r r
oo
conv,2 2 21
(22 W / m . C)(2 mC / W
1 1
13823 3 3 ) .
Noting that all resistances are in series, the total resistance is
C/W)2.138/(1)023.0/ln(188.400043.009944.0 3321total rrRRRRR oi
Then the steady rate of heat loss from the steam becomes
C/W])2.138/(1)023.0/ln(188.400043.009944.0[
C)22110(
33total
rrR
TTQ oi
Noting that the outer surface temperature of insulation is specified to be 30C, the rate of heat loss can alsobe expressed as
( )
QT T
R rro
o
3
33
30 221106
C
1/ (138.2 ) C / W
Setting the two relations above equal to each other and solving for r3 gives r3 = 0.0362 m. Then theminimum thickness of fiberglass insulation required is
t= r3 - r2 = 0.0362 - 0.0230 = 0.0132 m = 1.32 cm
Therefore, insulating the pipe with at least 1.32 cm thick fiberglass insulation will ensure that the outer
surface temperature of the pipe will be at 30C or below.
RiTi
Rinsulation RoTo
R i e
T1 T2 T3
-
7/30/2019 Heat Chap07 073
4/21
Chapter 7External Forced Convection
7-66
7-84"!PROBLEM 7-84"
"GIVEN"
T_i=110 "[C]"
T_o=22 "[C]"
k_pipe=52 "[W/m-C]"
r_1=0.02 "[m]"
t_pipe=0.003 "[m]""T_s_max=30 [C], parameter to be varied"
h_i=80 "[W/m^2-C]"
h_o=22 "[W/m^2-C]"
k_ins=0.038 "[W/m-C]"
"ANALYSIS"
L=1 "[m], 1 m long section of the pipe is considered"
A_i=2*pi*r_1*L
A_o=2*pi*r_3*L
r_3=r_2+t_ins*Convert(cm, m) "t_ins is in cm"
r_2=r_1+t_pipe
R_conv_i=1/(h_i*A_i)
R_pipe=ln(r_2/r_1)/(2*pi*k_pipe*L)
R_ins=ln(r_3/r_2)/(2*pi*k_ins*L)R_conv_o=1/(h_o*A_o)
R_total=R_conv_i+R_pipe+R_ins+R_conv_o
Q_dot=(T_i-T_o)/R_total
Q_dot=(T_s_max-T_o)/R_conv_o
Ts, max [C] tins [cm]
24 4.45
26 2.489
28 1.733
30 1.319
32 1.055
34 0.871
36 0.7342
38 0.6285
40 0.5441
42 0.4751
44 0.4176
46 0.3688
48 0.327
-
7/30/2019 Heat Chap07 073
5/21
Chapter 7External Forced Convection
7-67
20 25 30 35 40 45 500
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Ts,max [C]
tins
[cm]
-
7/30/2019 Heat Chap07 073
6/21
-
7/30/2019 Heat Chap07 073
7/21
Chapter 7External Forced Convection
7-69
To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat thecalculations above for 2, 3, . . . . 15 cm thick insulations, and list the results in the table below.
Insulationthickness
Rate of heat lossW
Cost of heat lost$/yr
Cost savings$/yr
Insulation cost$
0 cm 133,600 12,157 0 01 cm 15,021 1367 10,790 2828
5 cm 3301 300 11,850 3535
10 cm 1671 152 12,005 9189
11 cm 1521 138 12,019 9897
12 cm 1396 127 12,030 10,604
13 cm 1289 117 12,040 11,310
14 cm 1198 109 12,048 12,017
15 cm 1119 102 12,055 12,724
Therefore, the thickest insulation that will pay for itself in one year is the one whose thickness is 14 cm.
-
7/30/2019 Heat Chap07 073
8/21
Chapter 7External Forced Convection
7-70
7-86 A cylindrical oven is to be insulated to reduce heat losses. The optimum thickness of insulation and theamount of money saved per year are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulation is one-dimensional.3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 Thesurface temperature of the furnace and the heat transfer coefficient remain constant. 6 The surfaces of thecylindrical oven can be treated as plain surfaces since its diameter is greater than 1 m.
Properties The thermal conductivity of insulation is given to be k= 0.038 W/mC.
Analysis We treat the surfaces of this cylindrical furnace as plain surfaces since its diameter isgreater than 1 m, and disregard the curvature effects. The exposed surface area of the furnace is
222sidebase m69.70m)m)(65.1(2m)5.1(2222 rLrAAAo
The rate of heat loss from the furnace before the insulation is installed is
W101,794=C)2775)(mC)(70.69.W/m30()( 22 TTAhQ soo
Noting that the plant operates 5280 = 4160 h/yr, the annual heat lost fromthe furnace is
Q Q t ( . 101794 kJ / s)(4160 3600 s/ yr) =1.524 10 kJ / yr9
The efficiency of the furnace is given to be 78 percent. Therefore, to generate thismuch heat, the furnace must consume energy (in the form of natural gas) at a rate of
Q Qin oven9 910 kJ / yr) / 0.78 =1.954 10 kJ / yr =18,526 therms/ yr / ( . 1524
since 1 therm = 105,500 kJ. Then the annual fuel cost of this furnace before insulation becomes
Annual Cost Q Unit cost therm / yr)($0.50 / therm) = $9,263 / yrin ( ,18526
We expect the surface temperature of the furnace to increase, and the heat transfer coefficient to decreasesomewhat when insulation is installed. We assume these two effects to counteract each other. Then the rateof heat loss for 1-cm thick insulation becomes
W445,11
C.W/m30
1CW/m.038.0
m01.0
C)2775)(m69.70(
1
)(
2
2
ins
insconvinstotalins
o
soss
hk
t
TTA
RR
TT
R
TT
Q
Also, the total amount of heat loss from the furnace per year and the amount and cost of energyconsumption of the furnace become
Q Q t
Q Q
Q
ins ins8
in,ins ins oven8 8
in,ins
kJ / s)(4160 3600 s / yr) =1.714 10 kJ / yr
10 kJ / yr) / 0.78 = 2.197 10 kJ / yr = 2082 therms
Annual Cost Unit cost therm / yr)($0.50 / therm) = $1041/ yr
( .
/ ( .
(
11445
1714
2082
Cost savings = Energy cost w/o insulation - Energy cost w/insulation = 9263 - 1041 = $8222/yr
The unit cost of insulation is given to be $10/m2 per cm thickness, plus $30/m2 for labor. Then the total cost
of insulation becomes
Insulation Cost Unit cost)(Surface area) =[($10/ cm)(1 cm) +$30/ m m ) = $28282 2 ( ]( .7069
RoT
RinsulationTs
-
7/30/2019 Heat Chap07 073
9/21
Chapter 7External Forced Convection
7-71
To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat thecalculations above for 2, 3, . . . . 15 cm thick insulations, and list the results in the table below.
InsulationThickness
Rate of heat lossW
Cost of heat lost$/yr
Cost savings$/yr
Insulation cost$
0 cm 101,794 9263 0 01 cm 11,445 1041 8222 2828
5 cm 2515 228 9035 3535
9 cm 1413 129 9134 8483
10 cm 1273 116 9147 9189
11 cm 1159 105 9158 9897
12 cm 1064 97 9166 10,604
Therefore, the thickest insulation that will pay for itself in one year is the one whose thickness is 9 cm. The10-cm thick insulation will come very close to paying for itself in one year.
-
7/30/2019 Heat Chap07 073
10/21
Chapter 7External Forced Convection
7-72
7-87E Steam is flowing through an insulated steel pipe, and it is proposed to add another 1-in thick layerof fiberglass insulation on top of the existing one to reduce the heat losses further and to save energy andmoney. It is to be determined if the new insulation will pay for itself within 2 years.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transferis one-dimensional since there is thermal symmetry about the centerline and no variation in the axial
direction. 3 Thermal properties are constant. 4 The heat transfer coefficients remain constant. 5 The thermalcontact resistance at the interface is negligible.
Properties The thermal conductivities are given to be k = 8.7 Btu/hftF for steel pipe and k = 0.020
Btu/hftF for fiberglass insulation.
Analysis The inner radius of the pipe is r1 = 1.75 in, the outer radius of the pipe is r2 = 2 in, and the outerradii of the existing and proposed insulation layers are r3 = 3 in and 4 in, respectively. Considering a unitpipe length ofL = 1 ft, the individual thermal resistances are determined to be
R Rh A h r L
ii i
conv,1 21
(30 Btu / h.ft . F)[2 (1.75/ 12 ft)(1 ft)h. F / Btu
1 1
20 0364
1 1( ) ].
R Rr r
k L1
2 1
12
2 175000244
pipe
2 (8.7 Btu / h.ft. F)(1 fth. F / Btu
ln( / ) ln( / . )
).
Current Case:
Rr r
k Linsulation
ins 2 (0.020 Btu / h.ft. F)(1 fth. F / Btu
ln( / ) ln( / )
).3 2
2
3 23227
R Rh A h r
oo o
conv,2 21
(5 Btu / h.ft . F)[2 ft)(1 fth. F / Btu
1 1
2 3 1201273
3 3( ) ( / )].
Then the steady rate of heat loss from the steam becomes
( )
. . . . ) ..Q
T
R
T T
R R R R
i o
i ocurrent
total pipe ins
F
( h F / BtuBtu / h
400 600 0364 0 00244 3 227 01273
1002
Proposed Case:
Rr r
k Linsinsulation
2 (0.020 Btu / h.ft. F)(1 fth. F / Btu
ln( / ) ln( / )
).3 2
2
4 25516
R Rh A h r
oo o
conv,2 21
(5 Btu / h.ft . F)[2 ft)(1 fth. F / Btu
1 1
2 4 1200955
3 3( ) ( / )].
Then the steady rate of heat loss from the steam becomes
( )
. . . . ) ..Q
T
R
T T
R R R R
i o
i oprop
total pipe ins
F
( h F / BtuBtu / h
400 600 0364 0 00244 5 516 0 0955
602
Therefore, the amount of energy and money saved by the additional insulation per year are
yr/504.3$)Btu1000/01.0)($Btu/yr400,350()costUnit(=savedMoney
Btu/yr400,350)h/yr8760)(Btu/h0.40(
Btu/h0.402.602.100
saved
savedsaved
currentpropsaved
Q
tQQ
QQQ
or $7.01 per 2 years, which is barely more than the $7.0 minimum required. But the criteria is satisfied, andthe proposed additional insulation isjustified.
Ri
Ti
Rinsulation Ro
To
R i e
T1 T2 T3
-
7/30/2019 Heat Chap07 073
11/21
Chapter 7External Forced Convection
7-73
7-88 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air.The pipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe. Thethickness of insulation that will protect the water from freezing under worst conditions is to be determined.
Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions. 2 Heat transferis one-dimensional since there is thermal symmetry about the centerline and no variation in the axial
direction. 3 Thermal properties are constant. 4 The water in the pipe is stationary, and its initial temperature
is 15C. 5 The thermal contact resistance at the interface is negligible. 6 The convection resistance inside
the pipe is negligible.
Properties The thermal conductivities are given to be k = 0.16 W/mC for plastic pipe and k = 0.035
W/mC for fiberglass insulation. The density and specific heat of water are to be = 1000 kg/m3 and Cp =
4.18 kJ/kg.C (Table A-15).
Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the innerradius of insulation is r2 = 3.3 cm. We let r3 represent the outer radius of insulation. Considering a 1-m
section of the pipe, the amount of heat that must be transferred from the water as it cools from 15 to 0 C isdetermined to be
kJ177.3=C0)-C)(15kJ/kg.kg)(4.18827.2(
kg827.2m)](1m)(0.03)[kg/m1000()(
total
2321
TmCQ
LrVm
p
Then the average rate of heat transfer during 60 h becomes
,
(.Q
Q
tave
total J
s)W
177 300
60 36000821
The individual thermal resistances are
R Rr r
k L1
2 1
2
0 033 0 0300948
pipe
pipe 2 (0.16 W/ m. C)(1 mC / W
ln( / ) ln( . / . )
).
Rr r
k L
rrinsulation
2 (0.035 W/ m. C)(1 mC / W
ln( / ) ln( / . )
). ln( / . )3 2
2
33
2
00334 55 0 033
R R h A r ro
o conv 2 2
1
(30 W / m . C)(2 m C / W
1 1
18853 3 3 ) .
Then the rate of average heat transfer from the water can be expressed as
[ . ( )]
. . ln( / . ) / ( . ).
,Q
T T
R r rr
i ave o
total
0.821 WC
[ ] C / Wm
7 5 10
0 0948 4 55 0 033 1 1885350
3 33
Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is
t= r3 - r2 = 3.50 - 0.033 = 3.467 m
which is too large. Installing such a thick insulation is not practical, however, and thus other freezeprotection methods should be considered.
Ri 0
Ti
Rinsulation RoTo
R i e
T1 T2 T3
-
7/30/2019 Heat Chap07 073
12/21
Chapter 7External Forced Convection
7-74
7-89 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air. Thepipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe. Thethickness of insulation that will protect the water from freezing more than 20% under worst conditions is tobe determined.
Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions. 2 Heat transferis one-dimensional since there is thermal symmetry about the centerline and no variation in the axial
direction. 3 Thermal properties are constant. 4 The water in the pipe is stationary, and its initial temperature
is 15C. 5 The thermal contact resistance at the interface is negligible. 6 The convection resistance insidethe pipe is negligible.
Properties The thermal conductivities are given to be k = 0.16 W/mC for plastic pipe and k = 0.035
W/mC for fiberglass insulation. The density and specific heat of water are to be = 1000 kg/m3 and Cp =
4.18 kJ/kg.C (Table A-15).
Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the inner radiusof insulation is r2 = 3.3 cm. We let r3 represent the outer radius of insulation. The latent heat of freezing ofwater is 33.7 kJ/kg. Considering a 1-m section of the pipe, the amount of heat that must be transferred from
the water as it cools from 15 to 0C is determined to be
kJ0.3667.1883.177
kJ7.188)kJ/kg7.333)(kg827.2(2.02.0
kJ177.3=C0)-C)(15kJ/kg.kg)(4.18827.2(
kg827.2m)](1m)(0.03)[kg/m1000()(
freezingcoolingtotal
freezing
total
2321
QQQ
mhQ
TmCQ
LrVm
if
p
Then the average rate of heat transfer during 60 h becomes
,
(.Q
Q
tave
total J
s)W
366 000
60 36001694
The individual thermal resistances are
R Rr r
k L1
2 1
2
0 033 0 0300948
pipe
pipe 2 (0.16 W/ m. C)(1 mC / W
ln( / ) ln( . / . )
).
R r r
k L
r rinsulation2 (0.035 W/ m. C)(1 m
C / W
ln( / ) ln( / . ))
. ln( / . )3 2
2
33
20033 4 55 0 033
R Rh A r r
oo
conv 2 21
(30 W / m . C)(2 mC / W
1 1
18853 3 3 ) .
Then the rate of average heat transfer from the water can be expressed as
[ . ( )]
. . ln( / . ) / ( . ).
,Q
T T
R r rr
i ave o
total
1.694 WC
[ ] C / Wm
7 5 10
0 0948 4 55 0 033 1 18850312
3 33
Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is
t= r3 - r2 = 0.312 - 0.033 = 0.279 m
which is too large. Installing such a thick insulation is not practical, however, and thus other freezeprotection methods should be considered.
Ri 0
Ti
Rinsulation RoTo
R i e
T1 T2 T3
-
7/30/2019 Heat Chap07 073
13/21
Chapter 7External Forced Convection
7-75
Review Problems
7-90 Wind is blowing parallel to the walls of a house. The rate of heat loss from the wall is to bedetermined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3
Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties Assuming a film temperature ofTf= 10C for theoutdoors, the properties of air are evaluated to be (Table A-15)
7336.0Pr
/sm10426.1
CW/m.02439.0
25-
k
AnalysisAir flows along 8-m side. The Reynolds number in this case is
625
10792.7/sm10426.1
m)(8m/s)3600/100050(Re
LVL
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow.Using the proper relation for Nusselt number, heat transfer coefficient is determined to be
C.W/m78.30)096,10(m8
CW/m.02439.0
096,10)7336.0(871)10792.7(037.0Pr)871Re037.0(
2
3/18.063/18.00
NuL
kh
k
LhNu
o
L
The thermal resistances are
2m24=m)m)(83( wLAs
C/W0014.0)mC)(24.W/m78.30(
11
C/W1408.0
m24
C/W.m38.3)38.3(
C/W0052.0)mC)(24.W/m8(
11
22
2
2
22
so
o
s
valueinsulation
sii
AhR
A
RR
AhR
Then the total thermal resistance and the heat transfer rate through the wall are determined from
W122.1
C/W1474.0
C)422(
C/W1474.00014.01408.00052.0
21
total
oinsulationitotal
R
TTQ
RRRR
Ri Rinsulation Ro
T1 T2
Air
V = 50 km/h
T2 = 4C
L = 8 m
WALL
T1 = 22 C
-
7/30/2019 Heat Chap07 073
14/21
Chapter 7External Forced Convection
7-76
7-91 A car travels at a velocity of 60 km/h. The rate of heat transfer from the bottom surface of the hotautomotive engine block is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Air is
an ideal gas with constant properties. 4 The pressure of air is 1 atm. 5 The flow is turbulent over the entire
surface because of the constant agitation of the engine block. 6 The bottom surface of the engine is a flatsurface.
Properties The properties of air at 1 atm and the film temperature of
(Ts + T)/2 = (75+5)/2 = 40C are (Table A-15)
7255.0Pr
/sm10702.1
CW/m.02662.0
25-
k
AnalysisThe Reynolds number is
525
10855.6/sm10702.1
m)(0.7m/s)3600/100060(Re
LV
L
which is less than the critical Reynolds number. But we will assumeturbulent flow because of the constant agitation of the engine block.
C.W/m97.58)1551(m7.0
CW/m.02662.0
1551)7255.0()10855.6(037.0PrRe037.0
2
3/18.053/18.0
NuL
kh
k
hLNu L
W1734=C5)-(75m)m)(0.7(0.6C).W/m97.58()( 2 ssconv TThAQ
The heat loss by radiation is then determined from Stefan-Boltzman law to be
W181K)273+(10K)273+(75).KW/m10(5.67)m7.0)(m6.0)(92.0()(
4442-8
44
surrssrad TTAQ
Then the total rate of heat loss from the bottom surface of the engine block becomes
W1915 1811734radconvtotal QQQ
The gunk will introduce an additional resistance to heat dissipation from the engine. The total heat transferrate in this case can be calculated from
W1668=
m)0.7m6.0)(CW/m.3(
)m002.0(
m)]m)(0.7C)[(0.6.W/m97.58(
1
C5)-(75
12
ss
s
kA
L
hA
TTQ
The decrease in the heat transfer rate is
1734-1668 = 66 W
Ts = 75C
= 0.92
Air
V = 60 km/h
T = 5C
L = 0.7 m
Engine block
Ts = 10C
-
7/30/2019 Heat Chap07 073
15/21
Chapter 7External Forced Convection
7-77
7-92E A minivan is traveling at 60 mph. The rate of heat transfer to the van is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3
Radiation effects are negligible. 4 Air flow is turbulent because of the intense vibrations involved. 5 Air is
an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties Assuming a film temperature of Tf = 80F, theproperties of air are evaluated to be (Table A-15E)
7290.0Pr
/sft101697.0
FBtu/h.ft.01481.023-
k
AnalysisAir flows along 11 ft long side. The Reynolds number in this case is
6
2310704.5
/sft101697.0
ft)(11]ft/s)3600/528060[(Re
LV
L
which is greater than the critical Reynolds number. The air flow is assumed to be entirely turbulent becauseof the intense vibrations involved. Then the Nusselt number and the heat transfer coefficient are determinedto be
F.Btu/h.ft39.11)8461(ft11
FBtu/h.ft.01481.0
8461)7290.0()10704.5(037.0PrRe037.0
2
3/18.063/18.0
NuL
kh
k
LhNu
o
L
o
The thermal resistances are
2ft8.240ft)ft)(116(+ft)ft)(112.3(+ft)ft)(62.3(2 sA
F/Btuh.0004.0)ftF)(240.8.Btu/h.ft39.11(
11
F/Btuh.0125.0)ft(240.8
F/Btu.h.ft3)3(
F/Btuh.0035.0)ftF)(240.8.Btu/h.ft2.1(
11
22
2
2
22
soo
s
valueinsulation
si
i
AhR
A
RR
AhR
Then the total thermal resistance and the heat transfer rate into the minivan are determined to be
Btu/h1220
F/Btuh.0164.0
F)7090(
F/Btuh.0164.00004.00125.00035.0
21
total
oinsulationitotal
R
TTQ
RRRR
Air
V = 60 mph
T = 50F
L = 11 ft
Minivan
Ri Rinsulation Ro
T1 T2
-
7/30/2019 Heat Chap07 073
16/21
Chapter 7External Forced Convection
7-78
7-93 Wind is blowing parallel to the walls of a house with windows. The rate of heat loss through thewindow is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3
Radiation effects are negligible. 4 Air flow is turbulent because of the intense vibrations involved. 5 The
minivan is modeled as a rectangular box. 6 Air is an ideal gas with constant properties. 6 The pressure of airis 1 atm.
Properties Assuming a film temperature of 5C, the propertiesof air at 1 atm and this temperature are evaluated to be (TableA-15)
7350.0Pr
/sm10382.1
CW/m.02401.0
25-
k
AnalysisAir flows along 1.2 m side. The Reynolds number in this case is
625
10447.1/sm10382.1
m)(1.2m/s)3600/100060(Re
LVL
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow.Using the proper relation for Nusselt number, heat transfer coefficient is determined to be
C.W/m93.40)2046(m2.1
CW/m.02401.0
2046)7350.0(871)10447.1(037.0Pr)871Re037.0(
2
3/18.063/18.0
NuL
kh
k
hLNu L
The thermal resistances are
2m5.4=m)m)(1.52.1(3sA
C/W0045.0)mC)(5.4.W/m93.40(
11
C/W0012.0)mC)(5.4W/m.(0.78
m005.0
C/W0231.0)mC)(5.4.W/m8(
11
22,
2
22,
so
oconv
s
cond
siiconv
AhR
kA
LR
AhR
Then the total thermal resistance and the heat transfer rate through the 3 windows become
W833.3
C/W0288.0
C)]2(22[
C/W0288.00045.00012.00231.0
21
,,
total
oconvcondiconvtotal
R
TTQ
RRRR
AirV = 60 km/h
T2 = -2C
L = 1.2 m
WINDOW
T1 = 22 C
-
7/30/2019 Heat Chap07 073
17/21
Chapter 7External Forced Convection
7-79
7-94 A fan is blowing air over the entire body of a person. The average temperature of the outer surface ofthe person is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The
pressure of air is 1 atm. 4 The average human body can be treated as a 30-cm-diameter cylinder with anexposed surface area of 1.7 m2.
Properties We assume the film temperature to be 35C. The
properties of air at 1 atm and this temperature are (Table A-15)
7268.0Pr
/sm10655.1
CW/m.02625.0
25-
k
AnalysisThe Reynolds number is
4
2510063.9
/sm10655.1
m)m/s)(0.3(5Re
DV
The proper relation for Nusselt number corresponding to this Reynolds number is
6.203
000,282
10063.91
7268.0/4.01
)7268.0()10063.9(62.03.0
000,282
Re
1Pr/4.01
PrRe62.0
3.0
5/48/5
4
4/13/2
3/15.04
5/48/5
4/13/2
3/15.0
khD
Nu
Then
C.W/m02.18)6.203(m3.0
CW/m.02655.0 2
NuD
kh
Considering that there is heat generation in that person's body at a rate of 90 W and body gains heat byradiation from the surrounding surfaces, an energy balance can be written as
Q Q Qgenerated radiation convection
Substituting values with proper units and then application of trial & error method yields the averagetemperature of the outer surface of the person.
C36.2K309.2
s
ss
ssssurrs
T
TT
TThATTA
)]27332()[7.1)(02.18(])27340)[(1067.5)(7.1)(9.0(90
)()(W90
448
44
V = 5 m/s
T = 32C
Person, Ts90 W
= 0.9
D = 0.3 m
-
7/30/2019 Heat Chap07 073
18/21
Chapter 7External Forced Convection
7-80
7-95 The heat generated by four transistors mounted on a thin vertical plate is dissipated by air blown overthe plate on both surfaces. The temperature of the aluminum plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3
Radiation effects are negligible. 4 The entire plate is nearly isothermal. 5 The exposed surface area of the
transistor is taken to be equal to its base area. 6 Air is an ideal gas with constant properties. 7 The pressureof air is 1 atm.
Properties Assuming a film temperature of 40C, theproperties of air are evaluated to be (Table A-15)
7255.0Pr
/sm10702.1
CW/m.02662.0
25-
k
AnalysisThe Reynolds number in this case is
425
10386.5/sm10702.1
m)(0.22m/s)60/250(Re
LV
L
which is smaller than the critical Reynolds number. Thus we have laminar flow. Using the proper relationfor Nusselt number, heat transfer coefficient is determined to be
C.W/m75.16)5.138(m22.0
CW/m.02662.0
5.138)7255.0()10386.5(664.0PrRe664.0
2
3/15.043/15.0
NuL
kh
khLNu L
The temperature of aluminum plate then becomes
C50.0
])m22.0(2)[C.W/m75.16(
W)124(C20)(
22s
ssshA
QTTTThAQ
Discussion In reality, the heat transfer coefficient will be higher since the transistors will cause turbulencein the air.
Ts
12 W
V = 250 m/min
T = 20C
L= 22 cm
-
7/30/2019 Heat Chap07 073
19/21
Chapter 7External Forced Convection
7-81
7-96 A spherical tank used to store iced water is subjected to winds. The rate of heat transfer to the icedwater and the amount of ice that melts during a 24-h period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Thermal resistance of the tank is negligible. 3 Radiation
effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties The properties of air at 1 atm pressure and the free stream temperature of 30C are (Table A-15)
7282.0Pr
kg/m.s10729.1
kg/m.s10872.1
/sm10608.1
CW/m.02588.0
5
C0@,
5
25-
s
k
Analysis(a) The Reynolds number is
625
10304.1/sm10608.1
m)(3.02m/s1000/3600)(25Re
DV
The Nusselt number corresponding to this Reynolds number is determined from
105610729.1
10872.1)7282.0()10304.1(06.0)10304.1(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
sk
hDNu
and C.W/m05.9)1056(m02.3
CW/m.02588.0 2
NuD
kh
The rate of heat transfer to the iced water is
W7779 C)030(]m)(3.02C)[.W/m05.9())(()(222
TTDhTThAQ sss
(b) The amount of heat transfer during a 24-hour period is
kJ079,672s)3600kJ/s)(24779.7( tQQ
Then the amount of ice that melts during this period becomes
kg2014kJ/kg7.333
kJ079,672
ifif
h
QmmhQ
1 cmIced water
Di = 3 m
0C
Q
Ts = 0CV = 25 km/hT = 30C
-
7/30/2019 Heat Chap07 073
20/21
Chapter 7External Forced Convection
7-82
7-97 A spherical tank used to store iced water is subjected to winds. The rate of heat transfer to the icedwater and the amount of ice that melts during a 24-h period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 7 Thepressure of air is 1 atm.
Properties The properties of air at 1 atm pressure and the free stream temperature of 30C are (Table A-15)
kg/m.s10872.1
/sm10608.1
CW/m.02588.0
5
25-
k
7282.0Pr
kg/m.s10729.1 5
C0@,
s
Analysis(a) The Reynolds number is
625
10304.1/sm10608.1
m)(3.02m/s1000/3600)(25Re
DV
The Nusselt number corresponding to this Reynolds number is determined from
105610729.1
10872.1)7282.0()10304.1(06.0)10304.1(4.02
PrRe06.0Re4.02
4/1
5
54.03/265.06
4/1
4.03/25.0
sk
hDNu
and C.W/m05.9)1056(m02.3
CW/m.02588.0 2
NuD
kh
In steady operation, heat transfer through the tank by conduction is equal to the heat transfer from the outersurface of the tank by convection and radiation. Therefore,
( ) ( ), ,
, ,
Q Q Q
QT T
Rh A T T A T T
s out s in
sphereo o surr s out o surr s out
through tank from tank, conv+rad
4 4
where C/W10342.2m)m)(1.50C)(1.51W/m.15(4
m)50.151.1(
4
5
21
12
rkr
rrRsphere
222 m28.65m)02.3( DAo
Substituting,
]K)273(K)27315)[(.KW/m1067.5)(m65.28)(9.0(
C))(30mC)(28.65.W/m05.9(C/W1034.2
C0
4,
44282
,22
5
,
outs
outsouts
T
TT
Q
whose solution is
kW9.63 W9630andC23.0 QTs
(b) The amount of heat transfer during a 24-hour period is
kJ032,832s)3600kJ/s)(2463.9( tQQ
Then the amount of ice that melts during this period becomes
kg2493kJ/kg7.333
kJ032,832
ifif
h
QmmhQ
1 cmIced water
Di = 3 m
0C
Ts, outV = 25 km/h
T = 30C 0C
-
7/30/2019 Heat Chap07 073
21/21
Chapter 7External Forced Convection
7-98E A cylindrical transistor mounted on a circuit board is cooled by air flowing over it. The maximumpower rating of the transistor is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas
with constant properties. 4 The pressure of air is 1 atm.
Properties The properties of air at 1 atm and the film temperatureof F1502/)120180( fT are (Table A-15)
7188.0Pr
/sft10210.0
FBtu/h.ft.01646.023-
k
AnalysisThe Reynolds number is
5.727/sft10210.0
ft)/12ft/s)(0.22(500/60Re
23
DV
The Nusselt number corresponding to this Reynolds number is
72.13
000,282
5.7271
7188.0/4.01
)7188.0()5.727(62.03.0
000,282
Re1
Pr/4.01
PrRe62.03.0
5/48/5
4/13/2
3/15.0
5/48/5
4/13/2
3/15.0
k
hDNu
and F.Btu/h.ft32.12)72.13(ft)12/22.0(
FBtu/h.ft.01646.0 2
NuD
kh
Then the amount of power this transistor can dissipate safely becomes
Btu/h)3.412=W(1
C)120180(ft)2ft)(0.25/1(0.22/12F).Btu/h.ft32.12(
)()(
2
W0.26=Btu/h0.887
TTDLhTThAQ sss
Air500 ft/min
120F
Powertransistor
D = 0.22 in