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7/30/2019 Heat Chap08 061
1/20
Chapter 8Internal Forced Convection
Review Problems
8-61 Geothermal water is supplied to a city through stainless steel pipes at a specified rate. The electricpower consumption and its daily cost are to be determined, and it is to be assessed if the frictional heatingduring flow can make up for the temperature drop caused by heat loss.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the
flow is fully developed. 3 The minor losses are negligible because of the large length-to-diameter ratio andthe relatively small number of components that cause minor losses. 4 The geothermal well and the city areat about the same elevation. 5 The properties of geothermal water are the same as fresh water. 6 The fluidpressures at the wellhead and the arrival point in the city are the same.
Properties The properties of water at 110C are = 950.6 kg/m3, = 0.255 10-3 kg/m s, and Cp =4.229kJ/kg C (Table A-9). The roughness of stainless steel pipes is 2 10-6 m (Table 8-3).
Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point ofdelivery at the city, and the entire piping system as the control volume. Both points are at the sameelevation (z2 = z2) and the same velocity (V1 = V2) since the pipe diameter is constant, and the samepressure (P1 =P2). Then the energy equation for this control volume simplifies to
22
upump,turbine2
222
upump,1
211
LL hhhhzgg
Phz
gg
P=++++=+++
VV
That is, the pumping power is to be used to overcome the head lossesdue to friction in flow. The mean velocity and the Reynolds numberare
7
3
3
2
3
2
10186.1skg/m10255.0
m)m/s)(0.60305.5)(kg/m6.950(VRe
m/s305.54/m)(0.60
/sm1.5
4/V
=
=
=
=
=
==
D
D
V
A
V
m
cm
which is greater than 10,000. Therefore, the flow is turbulent. Therelative roughness of the pipe is
1033.3
m60.0
m102/ 6
6
=
=D
The friction factor can be determined from the Moody chart, but to avoid the reading error, we determineit from the Colebrook equation using an equation solver (or an iterative scheme),
+
=
+=
fff
D
f 7
6
10187.1
51.2
7.3
1033.3log0.2
1
Re
51.2
7.3
/log0.2
1
It givesf= 0.00829. Then the pressure drop, the head loss, and the required power input become
kPa2218kN/m1
kPa1
m/skg1000
kN1
2
m/s)305.5)(kg/m6.950(
m0.60
m000,1200829.0
2 2
232
V
=
== mD
LfP
kW5118=
=
== /smkPa1
kW1
0.65
)kPa2218)(/sm(1.5
3
3
motor-pumpmotor-pump
upump,
elect
PVW
W
Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow.
(b) The daily cost of electric power consumption is determined by multiplying the amount of power usedper day by the unit cost of electricity,
kWh/day832,122h/day)kW)(245118(Amount inelect, === tW
$7370/day=== 0.06/kWh)kWh/day)($832,122(costUnitAmountCost
8-42
L = 12 km
D = 60 cmWater
1.5 m3/s
21
7/30/2019 Heat Chap08 061
2/20
Chapter 8Internal Forced Convection
(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventuallydissipated as heat due to the frictional effects. Therefore, this problem is equivalent to heating the waterby a 5118 kW of resistance heater (again except the heat dissipated by the motor). To be conservative, weconsider only the useful mechanical energy supplied to the water by the pump. The temperature rise ofwater due to this addition of energy is
C0.55=
===
)CkJ/kg229.4(/s)m5.1)(kg/m6.950(
kJ/s)(51180.65
33
inelect,motor-pumpelect
p
pCV
WTTCVW
Therefore, the temperature of water will rise at least 0.55C, which is more than the 0.5C drop intemperature (in reality, the temperature rise will be more since the energy dissipation due to pumpinefficiency will also appear as temperature rise of water). Thus we conclude that the frictional heatingduring flow can more than make up for the temperature drop caused by heat loss.
Discussion The pumping power requirement and the associated cost can be reduced by using a largerdiameter pipe. But the cost savings should be compared to the increased cost of larger diameter pipe.
8-43
7/30/2019 Heat Chap08 061
3/20
Chapter 8Internal Forced Convection
8-62 Geothermal water is supplied to a city through cast iron pipes at a specified rate. The electric powerconsumption and its daily cost are to be determined, and it is to be assessed if the frictional heating duringflow can make up for the temperature drop caused by heat loss.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus theflow is fully developed. 3 The minor losses are negligible because of the large length-to-diameter ratio andthe relatively small number of components that cause minor losses. 4 The geothermal well and the city areat about the same elevation. 5 The properties of geothermal water are the same as fresh water. 6 The fluidpressures at the wellhead and the arrival point in the city are the same.
Properties The properties of water at 110C are = 950.6 kg/m3, = 0.255 10-3 kg/m s, and Cp =4.229kJ/kg C (Table A-9). The roughness of cast iron pipes is 0.00026 m (Table 8-3).
Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point ofdelivery at the city, and the entire piping system as the control volume. Both points are at the sameelevation (z2 = z2) and the same velocity (V1 = V2) since the pipe diameter is constant, and the samepressure (P1 =P2). Then the energy equation for this control volume simplifies to
22
upump,turbine2
222
upump,1
211
LL hhhhzgg
Phz
gg
P=++++=+++
VV
That is, the pumping power is to be used to overcome the head lossesdue to friction in flow. The mean velocity and the Reynolds numberare
7
3
3
2
3
2
10187.1skg/m10255.0
m)m/s)(0.60305.5)(kg/m6.950(Re
m/s305.54/m)(0.60
/sm1.5
4/
=
==
====
D
D
V
A
V
m
cm
V
V
which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is
1033.4m60.0
m00026.0/ 4==D
The friction factor can be determined from the Moody chart, but to avoid the reading error, we determineit from the Colebrook equation using an equation solver (or an iterative scheme),
+
=
+=
fff
D
f 7
4
10187.1
51.2
7.3
1033.4log0.2
1Re
51.2
7.3
/log0.2
1
It givesf= 0.01623 Then the pressure drop, the head loss, and the required power input become
kPa4341kN/m1
kPa1
m/skg1000
kN1
2
m/s)305.5)(kg/m6.950(
m0.60
m000,1201623.0
2
V2
232
=
=
= mD
LfP
kW10,017=
=
=
=/smkPa1
kW1
0.65
)kPa4341)(/sm(1.5
3
3
motor-pumpmotor-pump
upump,elect
PVWW
Therefore, the pumps will consume 10,017 W of electric power to overcome friction and maintain flow.
(b) The daily cost of electric power consumption is determined by multiplying the amount of power usedper day by the unit cost of electricity,
kWh/day429,240h/day)kW)(24017,10(Amount elect,in === tW
y$14,426/da=== 0.06/kWh)kWh/day)($429,240(costUnitAmountCost
(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventuallydissipated as heat due to the frictional effects. Therefore, this problem is equivalent to heating the waterby a 5118 kW of resistance heater (again except the heat dissipated by the motor). To be conservative, weconsider only the useful mechanical energy supplied to the water by the pump. The temperature rise ofwater due to this addition of energy is
8-44
L = 12 km
D = 60 cmWater
1.5 m3/s
21
7/30/2019 Heat Chap08 061
4/20
Chapter 8Internal Forced Convection
C1.08=
=
==
)CkJ/kg229.4(/s)m5.1)(kg/m6.950(
kJ/s)(10,0170.65
33
inelect,motor-pumpelect
p
pCV
WTTCVW
Therefore, the temperature of water will rise at least 1.08C, which is more than the 0.5C drop intemperature (in reality, the temperature rise will be more since the energy dissipation due to pumpinefficiency will also appear as temperature rise of water). Thus we conclude that the frictional heatingduring flow can more than make up for the temperature drop caused by heat loss.
Discussion The pumping power requirement and the associated cost can be reduced by using a largerdiameter pipe. But the cost savings should be compared to the increased cost of larger diameter pipe.
8-45
7/30/2019 Heat Chap08 061
5/20
Chapter 8Internal Forced Convection
8-63 The velocity profile in fully developed laminar flow in a circular pipe is given. The radius of thepipe, the mean velocity, and the maximum velocity are to be determined.
AssumptionsThe flow is steady, laminar, and fully developed.
Analysis The velocity profile in fully developed laminar flow in a circular pipe is
=
2
2
max 1)(R
rr VV
The velocity profile in this case is given by
)1001(6)(V 2rr =
Comparing the two relations above gives the pipe radius, themaximum velocity, and the mean velocity to be
m0.10== RR 100
12
Vmax = 6 m/s
m/s3===2
m/s6
2
VV maxm
8-46
Rr
0
Vmax
V(r)=Vmax
(1-r2/R2)
7/30/2019 Heat Chap08 061
6/20
Chapter 8Internal Forced Convection
8-64E The velocity profile in fully developed laminar flow in a circular pipe is given. The volume flowrate, the pressure drop, and the useful pumping power required to overcome this pressure drop are to bedetermined.
Assumptions 1 The flow is steady, laminar, and fully developed. 2 The pipe is horizontal.
Properties The density and dynamic viscosity of water at 40F are = 62.42 lbm/ft3 and = 3.74 lbm/ft h= 1.039 10-3 lbm/ft s, respectively (Table A-9E).
Analysis The velocity profile in fully developed laminar flow in a circular pipe is
=
2
2
max 1)(R
rr VV
The velocity profile in this case is given by
)6251(8.0)( 2rr =V
Comparing the two relations above gives the pipe radius, themaximum velocity, and the mean velocity to be
ft04.0625
12 == RR
Vmax = 0.8 ft/s
ft/s4.02ft/s0.8
2max === VVm
Then the volume flow rate and the pressure drop become
/sft0.00201VV 3==== ]ft)(0.04ft/s)[4.0()( 22 RAV mcm
=
=
lbf1
ft/slbm2.32
ft)s)(80lbm/ft10039.1(128
ft)(0.08)(/sft0.00201
128
2
3
43
4
horiz
P
L
DPV
It gives
psi0.0358== 2lbf/ft16.5P
Then the useful pumping power requirement becomes
ft/slbf0.737
W1)lbf/ft16.5)(/sft00201.0( 23upump, W0.014=
== PVW
Checking The flow was assumed to be laminar. To verify this assumption, we determine the Reynoldsnumber:
1922slbm/ft10039.1
ft)ft/s)(0.084.0)(lbm/ft42.62(Re
3
3
=
==
DmV
which is less than 2300. Therefore, the flow is laminar.
Discussion Note that the pressure drop across the water pipe and the required power input to maintainflow is negligible. This is due to the very low flow velocity. Such water flows are the exception in practicerather than the rule.
8-47
Rr
0
Vmax
V(r)=Vmax
(1-r2/R2)
7/30/2019 Heat Chap08 061
7/20
Chapter 8Internal Forced Convection
8-65 A compressor is connected to the outside through a circular duct. The power used by compressor toovercome the pressure drop, the rate of heat transfer, and the temperature rise of air are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermalresistance of the duct is negligible. 4 Air is an ideal gas with constant properties.
Properties We take the bulk mean temperature for air to be 15C since the mean temperature of air at theinlet will rise somewhat as a result of heat gain through the duct whose surface is exposed to a highertemperature. The properties of air at this temperature and 1 atm pressure are (Table A-15)
/sm10568.1
CW/m.02476.0
kg/m225.1
25-
3
=
==
k
7323.0Pr
CJ/kg.1007
=
=pC
The density and kinematic viscosity at 95 kPa are
P= =95 kPa
101.325 kPaatm0938.
/sm101.673=)/s)/(0.938m10568.1(
kg/m149.1)938.0)(kg/m225.1(
25-25-
33
=
==
AnalysisThe mean velocity of air is
m/s594.8
/4m)(0.2
/sm27.0
2
3
===c
m
A
VV
Then5
25100275.1
/sm10673.1
m)m/s)(0.2(8.594Re =
=
=
hmDV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
m2m)2.0(1010 == DLL th which is shorter than the total length of the duct. Therefore, we assume fully developed flow in a smoothpipe, and determine friction factor from
[ ] 01789.064.1)100275.1ln(790.0)64.1Reln790.0( 2.052 === f The pressure drop and the compressor power required to overcome this pressure drop are
kg/s3101.0)/sm27.0)(kg/m149.1( 33 === Vm
W11.3
V
==
=
===
3
2
2232
kg/m149.1
)N/m74.41)(kg/s3101.0(
N/m74.412
)m/s594.8)(kg/m149.1(m)2.0(m)11()01789.0(
2
PmW
DLfP
pump
m
(b) For the fully developed turbulent flow, the Nusselt number is
5.207)7323.0()100275.1(023.0PrRe023.0 4.08.054.08.0 ====k
hDNu
and C.W/m69.25)5.207(m2.0
CW/m.02476.0 2 =
== NuD
kh
h
Disregarding the thermal resistance of the duct, the rate of heat transfer to the air in the duct becomes2m912.6m)m)(11(0.2 === DLAs
W497.5=+
=+
=
)912.6)(10(
1
)912.6)(69.25(
1
1020
11
21
21
ss AhAh
TTQ
(c) The temperature rise of air in the duct is
C1.6=== TTTCmQ p C)J/kg.kg/s)(1007(0.3101W5.497
8-48
Air10C, 95 kPa
0.27 m3/s
Indoors20C
L = 11 m
D = 20 cm
7/30/2019 Heat Chap08 061
8/20
Chapter 8Internal Forced Convection
8-66 Air enters the underwater section of a duct. The outlet temperature of the air and the fan powerneeded to overcome the flow resistance are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermalresistance of the duct is negligible. 4 The surface of the duct is at the temperature of the water. 5 Air is anideal gas with constant properties. 6 The pressure of air is 1 atm.
Properties We assume the bulk mean temperature for air to be 20C since the mean temperature of air atthe inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower
temperature. The properties of air at 1 atm and this temperature are (Table A-15)
7309.0Pr
CJ/kg.1007
/sm10516.1
CW/m.02514.0
kg/m204.1
25-
3
=
==
==
pC
k
AnalysisThe Reynolds number is
4
2510959.3
/sm10516.1
m)m/s)(0.2(3Re =
=
=
hmDV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
m2m)2.0(1010 == DLL th
which is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number from
75.99)7309.0()10959.3(023.0PrRe023.0 3.08.043.08.0 ====k
hDNu
h
and
C.W/m54.12)75.99(m2.0
CW/m.02514.0 2 =
== NuD
kh
h
Next we determine the exit temperature of air,
kg/s0.1135=4
m)(0.2m/s))(3kg/m204.1(
m9.425=m)m)(152.0(
23
2
==
==
cm
s
AVm
DLA
and
C18.6===
)1007)(1135.0()425.9)(54.12(
)/()2515(15)( eeTTTT ps
CmhAisse
The friction factor, pressure drop, and the fan power required to overcome this pressure drop can bedetermined for the case of fully developed turbulent flow in smooth pipes to be
[ ] 02212.064.1)10959.3ln(790.0)64.1Reln790.0( 2.042 === f
Pa992.8N/m1
Pa1
m/skg1
N1
2
m/s)3)(kg/m204.1(
m0.2
m1502212.0
2
V22
232
=
=
= m
D
LfP
W1.54=
==
=
=/smPa1
W1
55.0
)Pa992.8)(/sm1135.0(3
3
motor-pumpmotor-pump
upump,fan
PVWW
8-49
Air25C3 m/s
River water15C
L = 15 m
D = 20 cm
7/30/2019 Heat Chap08 061
9/20
Chapter 8Internal Forced Convection
8-67 Air enters the underwater section of a duct. The outlet temperature of the air and the fan powerneeded to overcome the flow resistance are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermalresistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1atm.
Properties We assume the bulk mean temperature for air to be 20C since the mean temperature of air atthe inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower
temperature. The properties of air at 1 atm and this temperature are (Table A-15)
7309.0Pr
CJ/kg.1007
/sm10516.1
CW/m.02514.0
kg/m204.1
25-
3
=
==
==
pC
k
AnalysisThe Reynolds number is
4
2510959.3
/sm10516.1
m)m/s)(0.2(3Re =
=
=
hmDV
which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
m2m)2.0(1010 == DLL th
which is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number and h from
75.99)7309.0()10959.3(023.0PrRe023.0 3.08.043.08.0 ====k
hDNu h
and
C.W/m54.12)75.99(m2.0
CW/m.02514.0 2 =
== NuD
kh
h
Next we determine the exit temperature of air,
kg/s0.1135=4
m)(0.2m/s))(3kg/m204.1(
m9.425=m)m)(152.0(2
3
2
==
==
cm
s
AVm
DLA
The unit thermal resistance of the mineral deposit is
RL
kmineral
2m
W / m. Cm C / W= =
=
0 0015
30 0005
.. .
which is much less than (under 1%) the unit convection resistance,
C/W.m0797.0C.W/m54.12
11 22conv
=
==h
R
Therefore, the effect of 0.15 mm thick mineral deposit on heat transfer is negligible.
Next we determine the exit temperature of air,
C18.6=== )1007)(1135.0(
)425.9)(54.12(
)/()2515(15)( eeTTTT p
CmhAisse
The friction factor, pressure drop, and the fan power required to overcome this pressure drop can bedetermined for the case of fully developed turbulent flow in smooth pipes to be
[ ] 02212.064.1)10959.3ln(790.0)64.1Reln790.0( 2.042 === f
8-50
Water25C3 m/s
River water15C
L = 15 m
D = 20 cm
Mineral deposit0.15 mm
7/30/2019 Heat Chap08 061
10/20
Chapter 8Internal Forced Convection
Pa992.8N/m1
Pa1
m/skg1
N1
2
m/s)3)(kg/m204.1(
m0.2
m1502212.0
2
V22
232
=
=
= m
D
LfP
W1.54=
==
=
=/smPa1
W1
55.0
)Pa992.8)(/sm1135.0(3
3
motor-pumpmotor-pump
upump,fan
PVWW
8-51
7/30/2019 Heat Chap08 061
11/20
Chapter 8Internal Forced Convection
8-68E The exhaust gases of an automotive engine enter a steel exhaust pipe. The velocity of exhaust gasesat the inlet and the temperature of exhaust gases at the exit are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth. 3 The thermalresistance of the pipe is negligible. 4 Exhaust gases have the properties of air, which is an ideal gas withconstant properties.
Properties We take the bulk mean temperature for exhaust gases to be 700 C since the mean temperatureof gases at the inlet will drop somewhat as a result of heat loss through the exhaust pipe whose surface is
at a lower temperature. The properties of air at this temperature and 1 atm pressure are (Table A-15)
/sft105902.0
FBtu/h.ft.0280.0
lbm/ft03422.0
23-
3
=
==
k 694.0Pr
FBtu/lbm.2535.0
=
=pC
Noting that 1 atm = 14.7 psia, the pressure in atm is
P= (15.5 psia)/(14.7 psia) = 1.054 atm. Then,
/sft100.5598=)/s)/(1.054ft105902.0(
lbm/ft03608.0)054.1)(lbm/ft03422.0(
23-23-
33
=
==
Analysis(a) The velocity of exhaust gases at the inlet of the exhaust pipe is
( ) ft/s82.97==== 4/ft)(3.5/12)lbm/ft03608.0( lbm/s0.2 23cmcm AmVAVm
(b) The Reynolds number is
231,43/sft105598.0
ft)12ft/s)(3.5/(82.97Re
23=
==
hmDV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
ft917.2ft)12/5.3(1010 == DLL thwhich are shorter than the total length of the duct. Therefore, we can assume fully developed turbulentflow in the entire duct, and determine the Nusselt number from
4.105)694.0()231,43(023.0PrRe023.0 3.08.03.08.0 ====k
hDNu h
and F.Btu/h.ft12.10)4.105(ft)12/5.3(
FBtu/h.ft.03422.0 2 ==== NuDkhhh
i
2ft7.33=ft)ft)(812/5.3( == DLAsIn steady operation, heat transfer from exhaust gases to the duct must be equal to the heat transfer fromthe duct to the surroundings, which must be equal to the energy loss of the exhaust gases in the pipe. Thatis,
Q Q Q E = = =internal external exhaust gases
Assuming the duct to be at an average temperature ofTs , the quantities above can be expressed as
:Qinternal
=
==
800ln
F800)ft33.7)(F.Btu/h.ft12.10(
ln
22ln
s
es
e
is
es
iesisi
T
TT
TQ
TT
TT
TTAhTAhQ
:Qexternal F)80)(ftF)(7.33.Btu/h.ft3()(22 == sosso TQTTAhQ
:Eexhaust gases F)F)(800Btu/lbm.535lbm/h)(0.236002.0()( == eiep TQTTCmQ
This is a system of three equations with three unknowns whose solution is
F1.609and,, === se TTQ F736.3Btu/h11,635
Therefore, the exhaust gases will leave the pipe at 865F.8-69 Hot water enters a cast iron pipe whose outer surface is exposed to cold air with a specified heattransfer coefficient. The rate of heat loss from the water and the exit temperature of the water are to bedetermined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth.
8-52
Exhaust800F
0.2 lbm/s
80F
L = 8 ft
D = 3.5 in
7/30/2019 Heat Chap08 061
12/20
Chapter 8Internal Forced Convection
Properties We assume the water temperature not to drop significantly since the pipe is not very long. Wewill check this assumption later. The properties of water at 90 C are (Table A-9)
96.1Pr
CJ/kg.4206/s;m10326.0/
CW/m.675.0;kg/m3.965
26-
3
=
===
==
pC
k
Analysis(a) The mass flow rate of water is
kg/s0.9704m/s)8.0(4
m)(0.04)kg/m3.965(
23 === VAm c
The Reynolds number is
062,98/sm10326.0
m)m/s)(0.04(0.8Re
26=
=
=
hmDV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
m4.0m)04.0(1010 == DLL thwhich are much shorter than the total length of the pipe. Therefore, we can assume fully developedturbulent flow in the entire pipe. The friction factor corresponding to Re = 98,062 and /D = (0.026 cm)/(4 cm) = 0.0065 is determined from the Moody chart to be f= 0.034. Then the Nusselt number becomes
6.52196.1062,98034.0125.0PrRe125.03/13/1
==== fkhD
Nuh
and C.W/m8801)6.521(m04.0
CW/m.675.0 2 =
=== NuD
khh
hi
which is much greater than the convection heat transfer coefficient of 15 W/m2.C. Therefore, theconvection thermal resistance inside the pipe is negligible, and thus the inner surface temperature of thepipe can be taken to be equal to the water temperature. Also, we expect the pipe to be nearly isothermalsince it is made of thin metal (we check this later). Then the rate of heat loss from the pipe will be thesum of the convection and radiation from the outer surface at a temperature of 90C, and is determined tobe
A D Lo = = 0 0046( . m)(15 m) = 2.168 m2
W2601=C)10)(90mC)(2.168.W/m15()( 22 == surrsooconv TTAhQ
[ ] W942K)27310(K)27390().KW/m1067.5)(m168.2)(7.0()( 444282440 =++== surrsrad TTAQ W3543=942+6012=+= radconvtotal QQQ
(b) The temperature at which water leaves the basement is
C89.1=
===)CJ/kg.4206)(kg/s9704.0(
W3543C90)(
pieeip
Cm
QTTTTCmQ
The result justifies our assumption that the temperature drop of water is negligible. Also, the thermalresistance of the pipe and temperature drop across it are
C06.0)C/W1065.1)(W3543(
C/W1065.1m)C)(15W/m.52(4
)4/6.4ln(
4
)/ln(
5
512
===
=
==
pipetotalpipe
pipe
RQT
kL
DDR
which justifies our assumption that the temperature drop across the pipe is negligible.8-70 Hot water enters a copper pipe whose outer surface is exposed to cold air with a specified heattransfer coefficient. The rate of heat loss from the water and the exit temperature of the water are to bedetermined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth.
Properties We assume the water temperature not to drop significantly since the pipe is not very long. Wewill check this assumption later. The properties of water at 90 C are (Table A-15)
8-53
Water90C
0.8 m/s
10C
L = 15 m
Di= 4 cm
Do= 4.6 cm
Water90C
0.8 m/s
10C
L = 15 m
Di= 4 cm
Do= 4.6 cm
7/30/2019 Heat Chap08 061
13/20
Chapter 8Internal Forced Convection
96.1Pr
CJ/kg.4206/s;m10326.0/
CW/m.675.0;kg/m3.965
26-
3
=
===
==
pC
k
Analysis(a) The mass flow rate of water is
kg/s0.9704m/s)8.0(4
m)(0.04)kg/m3.965(
23 =
== VAm c
The Reynolds number is
062,98/sm10326.0
m)m/s)(0.04(0.8Re
26=
=
=
hmDV
which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
m4.0m)04.0(1010 == DLL th
which are much shorter than the total length of the pipe. Therefore, we can assume fully developedturbulent flow in the entire pipe. Assuming the copper pipe to be smooth, the Nusselt number isdetermined to be
1.27796.1062,98023.0PrRe023.03.08.03.08.0 ====
k
hDNu
h
and C.W/m4676)1.277(m04.0
CW/m.675.0 2
=
=== NuDk
hh hi
which is much greater than the convection heat transfer coefficient of 15 W/m2.C. Therefore, theconvection thermal resistance inside the pipe is negligible, and thus the inner surface temperature of thepipe can be taken to be equal to the water temperature. Also, we expect the pipe to be nearly isothermalsince it is made of thin metal (we check this later). Then the rate of heat loss from the pipe will be thesum of the convection and radiation from the outer surface at a temperature of 90C, and is determined tobe
A D Lo = = 0 0046( . m)(15 m) = 2.168 m2
W2601=C)10)(90mC)(2.168.W/m15()( 22 == surrsooconv TTAhQ
W942]K)27310(K)27390)[(.KW/m1067.5)(m168.2)(7.0(
)(
444282
440
=++=
=
surrsrad TTAQ
W3543=942+6012=+= radconvtotal QQQ
(b) The temperature at which water leaves the basement is
C89.1=
===)CJ/kg.4206)(kg/s970.0(
W3544C90)(
pieeip
Cm
QTTTTCmQ
The result justifies our assumption that the temperature drop of water is negligible. Also, the thermalresistance of the pipe and temperature drop across it are
C007.0)C/W1092.1)(W3543(
C/W1092.1m)C)(15W/m.386(4
)4/6.4ln(
4
)/ln(
6
612
===
=
==
pipetotalpipe
pipe
RQT
kL
DDR
which justifies our assumption that the temperature drop across the pipe is negligible.
8-71 Integrated circuits are cooled by water flowing through a series of microscopic channels. Thetemperature rise of water across the microchannels and the average surface temperature of themicrochannels are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the microchannels are smooth. 3Entrance effects are disregarded. 4 Any heat transfer from the side and cover surfaces are neglected.
Properties We assume the bulk mean temperature of water to be the inlet temperature of 20C since themean temperature of water at the inlet will rise somewhat as a result of heat gain through the microscopicchannels. The properties of water at 20C and the viscosity at the anticipated surface temperature of 25Care (Table A-9)
8-54
L = 1 cmMicro-channel
0.3 mm 0.05 mm
Water20C
7/30/2019 Heat Chap08 061
14/20
Chapter 8Internal Forced Convection
01.7PrC;J/kg.4182
/sm10004.1/
CW/m.598.0
kg/m998
26-
3
====
==
pC
k
Analysis(a) The mass flow rate of water is
kg/s00998.0)/sm10)(0.01kg/m998(3-33
=== Vm
The temperature rise of water as it flows through the micro channels is
C1.2=
===)CJ/kg4182)(kg/s00998.0(
J/s50
pp
Cm
QTTCmQ
(b) The Reynolds number is
1.569/sm10004.1
m)1057.8m/s)((6.667Re
m10571.8)m103.0+m1005.0(2
)m103.0)(m1005.0(44
m/s667.6100)m103.0)(m1005.0(
/sm1001.0
26
5
5
33
33
33
33
=
=
=
=
==
=
==
hm
ch
cm
DV
P
AD
A
VV
which is less than 2300. Therefore, the flow is laminar, and the thermal entry length in this case is
m0.0171=m)10571.8)(01.7)(1.569(05.0PrRe05.0 5== ht DL
which is longer than the total length of the channels. Therefore, we can assume thermally developingflow, and determine the Nusselt number from (actually, the relation below is for circular tubes)
[ ]224.5
)01.7)(1.569(m01.0
m10571.804.01
)01.7)(1.569(m01.0
m10571.8065.0
66.3PrRe)/(04.01
PrRe)/(065.066.3
3/25
5
3/2=
+
+=+
+==
LD
LD
k
hDNu
and C.W/m445,36)224.5(m10571.8
CW/m.598.0 25
=
==
Nu
D
kh
h
Then the average surface temperature of the base of the micro channels is determined to be
C22.6 =
+
+=+=
=
=+==
)m107)(C.W/m445,36(
W)100/50(C
2
2.2120
)(
m10701.010)05.03.0(2
262,,
,,
263
savemaves
avemavess
s
hA
QTT
TThAQ
pLA
8-55
7/30/2019 Heat Chap08 061
15/20
Chapter 8Internal Forced Convection
8-72 Integrated circuits are cooled by air flowing through a series of microscopic channels. Thetemperature rise of air across the microchannels and the average surface temperature of the microchannelsare to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the microchannels are smooth. 3Entrance effects are disregarded. 4 Any heat transfer from the side and cover surfaces are neglected. 5 Airis an ideal gas with constant properties. 6 The pressure of air is 1 atm.
Properties We assume the bulk mean temperature for air to be 60C since the mean temperature of air atthe inlet will rise somewhat as a result of heat gain through the microscopic channels whose base areasare exposed to uniform heat flux. The properties of air at 1 atm and 60C are (Table A-15)
0.7202Pr
CJ/kg.1007
/sm10895.1
CW/m.02808.0
kg/m060.1
25-
3
=
==
==
pC
k
Analysis(a) The mass flow rate of air is
kg/s10298.5)/sm10)(0.5kg/m060.1( 43-33 === Vm
The temperature rise of air as it flows through the micro channels is
C93.7=
=== )CJ/kg.1007)(kg/s10298.5(J/s50
4p
pCm
QTTCmQ
(b) The Reynolds number is
1508/sm10895.1
m)1057.8m/s)((333.3Re
m10571.8)m103.0+m1005.0(2
)m103.0)(m1005.0(44
m/s3.333)m103.0)(m1005.0(
/sm/100)105.0(
25
5
5
33
33
33
33
=
=
=
=
==
=
==
hm
ch
cm
DV
P
AD
A
VV
which is smaller than 2300. Therefore, the flow is laminar and the thermal entry length in this case is
m0.004653m)10571.8)(7202.0)(1508(05.0PrRe05.0 5 === ht DL
which is 42% of the total length of the channels. Therefore, we can assume thermally developing flow,and determine the Nusselt number from (actually, the relation below is for circular tubes)
[ ]174.4
)7202.0)(1508(m01.0
m10571.804.01
)7202.0)(1508(m01.0
m10571.8065.0
66.3PrRe)/(04.01
PrRe)/(065.066.3
3/25
5
3/2=
+
+=+
+==
LD
LD
k
hDNu
and C.W/m1368)174.4(m10571.8
CW/m.02808.0 25
=
==
Nu
D
kh
h
Then the average surface temperature of the base of the micro channels becomes
C119.1=
+
+=+=
=
=+==
)m107)(C.W/m1368(
W)100/50(C
2
7.11320
)(
m10701.010)05.03.0(2
262,,
,,
263
savemaves
avemavess
s
hA
QTT
TThAQ
pLA
8-56
L = 1 cm
Micro-channel0.3 mm 0.05 mm
Air0.5 L/s
7/30/2019 Heat Chap08 061
16/20
Chapter 8Internal Forced Convection
8-73 Hot exhaust gases flow through a pipe. For a specified exit temperature, the pipe length is to bedetermined.
Assumptions 1 Steady operating conditions exist. 2 The inner surface of the pipe is smooth. 3 Air is anideal gas with constant properties. 4 The pressure of air is 1 atm.
Properties The properties of air at 1 atm and the bulk mean temperature of (450+250)/2 = 350 C are(Table A-15)
6937.0Pr
CJ/kg.1056
/sm10475.5
CW/m.04721.0kg/m5664.0
25-
3
=
==
==
pC
k
AnalysisThe Reynolds number is
9864/sm10475.5
m)m/s)(0.15(3.6Re
25=
==
DmV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
m1.5=m)15.0(1010 = DLL thwhich is probably much shorter than the total length of the duct. Therefore, we can assume fullydeveloped turbulent flow in the entire duct, and determine the Nusselt number from
31.32)6937.0()9864(023.0PrRe023.0 3.08.03.08.0 ====k
hDNu
Heat transfer coefficient is
C.W/m17.10)31.32(m15.0
CW/m.04721.0 2 =
== NuD
kh
The logarithmic mean temperature difference is
C2.148
450180
250180ln
450250
ln
ln =
=
=
is
es
ie
TT
TT
TTT
The rate of heat loss from the exhaust gases can be expressed as
[ ] LLThAQ s 25.710)C2.148(m)15.0()C.W/m17.10(2
ln ===
whereL is the length of the pipe. The rate of heat loss can also be determined from
kg/s0.03603=/4m)(0.15m/s))(3.6kg/m5664.0( 23 == cVAm
W7612C)250450)(CJ/kg.kg/s)(105603603.0( === TCmQ p
Setting this equal to rate of heat transfer expression above, the pipe length is determined to be
m10.72=== LLQ W761225.710
8-57
L
D = 15
Exhaust
gases450C3.6 m/s
250C
Ts
= 180
C
7/30/2019 Heat Chap08 061
17/20
7/30/2019 Heat Chap08 061
18/20
Chapter 8Internal Forced Convection
8-75 Cold-air flows through an isothermal pipe. The pipe temperature is to be estimated.
Assumptions 1 Steady operating conditions exist. 2 The inner surface of the duct is smooth. 3 Air is anideal gas with constant properties. 4 The pressure of air is 1 atm.
Properties The properties of air at 1 atm and the bulk mean temperature of (5+19) / 2 = 12 C are (TableA-15)
7331.0Pr
CJ/kg.1007
/sm10444.1
CW/m.02454.0
kg/m238.1
25-
3
=
==
=
=
pC
k
AnalysisThe rate of heat transfer to the air is
m/s03499.0m/s)5.2(4
m)12.0()kg/m238.1(
23 === mcAm V
W1.493C)519)(CJ/kg.kg/s)(100703499.0( === TCmQ p
Reynolds number is
775,20/sm10444.1
m)m/s)(0.12(2.5
Re 25 === DV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case areroughly
m1.2=m)12.0(1010 = DLL thwhich is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number from
79.57)7331.0()775,20(023.0PrRe023.0 4.08.04.08.0 ====k
hDNu
Heat transfer coefficient is
C.W/m82.11)79.57(
m12.0
CW/m.02454.0 2 === NuD
kh
The logarithmic mean temperature difference is determined from
[ ] C535.5m)20(m)12.0()C.W/m82.11(W1.493 lnln2
ln === TTThAQ s
Then the pipe temperature is determined from the definition of the logarithmic mean temperaturedifference
C3.8=
=
= s
s
s
is
es
ie T
T
T
TT
TT
TTT
5
19ln
519C535.5
ln
ln
8-59
20 m
Air
5C2.5 m/s
12 cm
Ts
19C
7/30/2019 Heat Chap08 061
19/20
Chapter 8Internal Forced Convection
8-76 Oil is heated by saturated steam in a double-pipe heat exchanger. The tube length is to bedetermined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces of the tube are smooth. 3 Air is an idealgas with constant properties.
Properties The properties of oil at the average temperature of (10+30)/2=20 C are (Table A-13)
08.2Pr
CJ/kg.1880CW/m.145.0
kg/m888 3
=
==
=
pCk
AnalysisThe mass flow rate and the rate of heat transfer are
kg/s5022.0m/s)(0.84
m)03.0()kg/m888(
23 === mcAm V
W881,18C)1030)(CJ/kg.1880)(kg/s5022.0()( === iep TTCmQ
The Nusselt number is determined from Table 8-4 atDi/Do =3/5=0.6 to be Nui = 5.564. Then the heattransfer coefficient, the hydraulic diameter of annulus, and the logarithmic mean temperature difference
are
C.W/m34.40)564.5(m02.0
CW/m.145.0 2 =
== ih
i NuD
kh
C58.79
10100
30100ln
3010
ln
m02.0m03.0m05.0
ln =
=
=
===
is
es
ei
ioh
TT
TT
TTT
DDD
The heat transfer surface area is determined from
2
2ln
ln m881.5)C58.79)(C.W/m34.40(
W881,18=
=
==
Th
QAThAQ ss
Then the tube length becomes
m62.4====)m(0.03
m5.8812
2
i
ss
D
ALDLA
8-77 . 8-79 Design and Essay Problems
8-60
L
Oil
10C0.8 m/s 5 cm
Ts
= 100C
30C
3 cm
7/30/2019 Heat Chap08 061
20/20
Chapter 8Internal Forced Convection
8-79 A computer is cooled by a fan blowing air through the case of the computer. The flow rate of the fanand the diameter of the casing of the fan are to be specified.
Assumptions 1 Steady flow conditions exist. 2 Heat flux is uniformly distributed. 3 Air is an ideal gaswith constant properties.
Properties The relevant properties of air are (Tables A-1 and A-15)
/kg.KkPa.m287.0
CJ/kg.1007
3=
=
R
Cp
AnalysisWe need to determine the flow rate of air for the worst case scenario. Therefore, we assume theinlet temperature of air to be 50C, the atmospheric pressure to be 70.12 kPa, and disregard any heattransfer from the outer surfaces of the computer case. The mass flow rate of air required to absorb heat ata rate of 80 W can be determined from
kg/s007944.0C)5060)(CJ/kg.1007(
J/s80
)()( =
=
==
inoutp
inoutpTTC
QmTTCmQ
In the worst case the exhaust fan will handle air at 60C. Then the density of air entering the fan and thevolume flow rate becomes
/minm0.64973====
===
/sm01083.0kg/m0.7337
kg/s007944.0
kg/m7337.0273)K+/kg.K)(60kPa.m287.0(
kPa70.12
3
3
3
3
mV
RT
P
For an average velocity of 120 m/min, the diameter of theduct in which the fan is installed can be determined from
cm8.3====== m083.0)m/min120(
)/minm6497.0(44
4
32
V
VDV
DVAV c
8 61
Coolingair