Heat Chap08 061

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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

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(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

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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


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

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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

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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)

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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)

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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

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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


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


upump,fan

PVWW

8-49

Air25C3 m/s

River water15C

L = 15 m

D = 20 cm

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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


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


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.


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

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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


upump,fan

PVWW

8-51

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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


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

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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


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

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96.1Pr

CJ/kg.4206/s;m10326.0/

CW/m.675.0;kg/m3.965

26-

3

=

===

==

pC

k


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

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01.7PrC;J/kg.4182

/sm10004.1/

CW/m.598.0

kg/m998

26-

3

====

==

pC

k


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


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

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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


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

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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


9864/sm10475.5

m)m/s)(0.15(3.6Re

25=

==

DmV


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

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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


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

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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

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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

Documents

Heat Chap08 061