Heat Chap09 032

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Chapter 9Natural Convection

9-32 A fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The thickness of theinsulation needed to reduce the losses by 85% and the money saved during 10-h are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.

Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature,and possible below it because of the very low sky temperature for radiation heat loss. For convenience, we

use the properties of air at 1 atm and 5C (the anticipated film temperature) (Table A-15),

1-

25

K003597.0K)2735(

11

7350.0Pr

/sm10382.1

CW/m.02401.0

=+

==

==

=

fT

k

AnalysisThe rate of heat loss in the previous problem wasobtained to be 29,094 W. Noting that insulation will cut downthe heat losses by 85%, the rate of heat loss will be

W4364W094,2915.0)85.01( insulationno === QQ

The amount of energy and money insulation will save during a 10-h period is simply determined fromkWh3.247h)kW)(10094.2985.0(, === tQQ savedtotalsaved

$22.26== )kWh/09.0)($kWh3.247(=energy)ofcosttsaved)(UniEnergy(savedMoney

The characteristic length in this case is the outer diameter of the insulated pipe,

insulinsulc ttDL 23.02 +=+= where tinsul is the thickness of insulation in m. Then the problem can be

formulated forTs and tinsul as follows:

)7350.0()/sm10382.1(

)23.0(K])273)[(K003597.0)(m/s81.9(Pr

)(225

3-12

2

3

+=

= insulscs

tTLTTgRa

( )[ ] ( )[ ]

2

27/816/9

6/12

27/816/9

6/1

7350.0/559.01

387.06.0

Pr/559.01

387.06.0

+

+=

+

+=RaRa

Nu

m))(10023.0(

CW/m.02401.0

0 insuls

cc

tLDA

NuL

NuL

kh

+==

==

The total rate of heat loss from the outer surface of the insulated pipe by convection and radiation becomes

])K27330()[.KW/m1067.5()1.0(+)273(4364

)()(

44428

44

+=

+=+=

ssss

surrssssradconv

TAThA

TTATThAQQQ

In steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from theexposed surface of the insulation by convection and radiation, which must be equal to the heat conductedthrough the insulation. Therefore,

]3.0/)23.0ln[(

K)m)(298C)(100W/m.035.0(2W4364)/ln(

)(2 tank

insul

s

o

sinsulation

t

T

DD

TTkLQQ +

=

==

The solution of all of the equations above simultaneously using an equation solver gives Ts = 281.5 K =

8.5C and tinsul = 0.013 m = 1.3 cm.Note that the film temperature is (8.5+0)/2 = 4.25C which is very close to the assumed value of

5C. Therefore, there is no need to repeat the calculations using properties at this new film temperature.9-33E An industrial furnace that resembles a horizontal cylindrical enclosure whose end surfaces are wellinsulated. The highest allowable surface temperature of the furnace and the annual cost of this loss to theplant are to be determined.

9-26

Asphalt

L = 100 m

D + 2tins

25C

Insulation

= 0.1

Tsky

= -30C

T

= 0C

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Properties The properties of air at 1 atm and the anticipated film

temperature of (Ts+T)/2=(140+75)/2=107.5F are (Table A-15)

1-

23

R001762.0R)4605.107(

11

7249.0Pr

/sft101851.0

FBtu/h.ft.01546.0

=+

==

=

=

=

fT

k

Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We

start the solution process by guessing the surface temperature to be 140F for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.

The characteristic length in this case is the outer diameter of the furnace, ft.8==DLc Then,

10

223

3-12

2

3

10996.3)7249.0()/sft101851.0(

)ft8)(R75140)(R001762.0)(ft/s2.32(Pr

)(=

=

=

DTTgRa s

( )[ ] ( )[ ]9.376

7249.0/559.01

)10996.3(387.06.0

Pr/559.01

387.06.0

2

27/816/9

6/1102

27/816/9

6/1

=

+

+=

++=

RaNu

2

2

ft7.326)ft13)(ft8(

F.Btu/h.ft7287.0)9.376(ft8

FBtu/h.ft.01546.0

===

=

==

DLA

NuD

kh

s

The total rate of heat generated in the furnace is

Btu/h10936.3Btu/therm)100,000(therms/h)48)(82.0( 6==genQ

Noting that 1% of the heat generated can be dissipated by natural convection and radiation ,

Btu/h360,39Btu/h)10936.3)(01.0( 6 ==Q

The total rate of heat loss from the furnace by natural convection and radiation can be expressed as

])R46075()[.RBtu/h.ft101714.0)(m7.326)(85.0(

)]R46075()[ft7.326(F).Btu/h.ft7287.0(Btu/h360,39

)()(

444282

22

44

++

+=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is

F141.8== R8.601sT

which is very close to the assumed value. Therefore, there is no need to repeat calculations.

The total amount of heat loss and its cost during a-2800 hour period is

Btu10102.1h)2800)(Btu/h360,39( 8 === tQQ totaltotal

$716.== therm)/65.0)($therm000,100/10102.1(Cost 8

9-27

L = 13 ft

Furnace = 0.1

Air

T

= 75F

D = 8 ft

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9-34 A glass window is considered. The convection heat transfer coefficient on the inner side of thewindow, the rate of total heat transfer through the window, and the combined natural convection andradiation heat transfer coefficient on the outer surface of the window are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The local atmospheric pressure is 1 atm.

Properties The properties of air at 1 atm and the film temperature of

(Ts+T)/2 = (5+25)/2 = 15C are (Table A-15)

1-

25

K003472.0K)27315(

11

7323.0Pr

/sm10471.1

CW/m.02476.0

=+

==

==

=

fT

k

Analysis(a) The characteristic length in this case is the height of the window, m.2.1==LLc Then,

9

225

3-12

2

3

10986.3)7323.0()/sm10471.1(

)m2.1)(K525)(K34720.0)(m/s81.9(Pr

)(=

=

=

0

cs LTTgRa

7.189

7323.0

492.01

)10986.3(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/19

2

27/816/9

6/1

=

+

+=

+

+=Nu

2m4.2m)m)(22.1(

)7.189(m2.1

CW/m.02476.0

==

=

==

sA

NuL

kh C.W/m3.915 2

(b) The sum of the natural convection and radiation heat transfer from the room to the window is

W9.187C)525)(m4.2)(C.W/m915.3()( 22convection === ss TThAQ

W3.234])K2735()K27325)[(.KW/m1067.5)(m4.2)(9.0(

)(444282

44radiation

=++==

ssurrs TTAQ

W422.2=+=+= 3.2349.187radiationconvectiontotal QQQ

(c) The outer surface temperature of the window can be determined from

C65.3)m4.2)(CW/m.78.0(

)m006.0)(W346(C5)(

2

total,,,,total =

===

s

isososiss

kA

tQTTTT

t

kAQ

Then the combined natural convection and radiation heat transfer coefficient on the outer window surface becomes

C.W/m20.35 2 =

=

=

=

C)]5(65.3)[m4.2(

W346

)(

or

)(

2

,,

totalcombined

,,combinedtotal

ooss

ooss

TTA

Qh

TTAhQ

Note that T QR= and thus the thermal resistance R of a layer is proportional to the temperature dropacross that layer. Therefore, the fraction of thermal resistance of the glass is equal to the ratio of thetemperature drop across the glass to the overall temperature difference,

4.5%)(or045.0)5(25

65.35

total

glass

total

glass =

=

=

TR

T

R

R

which is low. Thus it is reasonable to neglect the thermal resistance of the glass.

9-28

Q

Outdoors

-5C

Glass

Ts= 5C

= 0.9

L = 1.2 m

Room

T

= 25C

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9-35 An insulated electric wire is exposed to calm air. The temperature at the interface of the wire and theplastic insulation is to be determined.



temperature of (Ts+T)/2 = (50+30)/2 = 40C are (Table A-15)

1-

25

K003195.0K)27340(

11

7255.0Pr

/sm10702.1

CW/m.02662.0

=+

==

==

=

fT

k


start the solution process by guessing the surface temperature to be 50C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.The characteristic length in this case is the outer diameter of the insulated wireLc =D = 0.006 m. Then,

3.339)7255.0()/sm10702.1(

)m006.0)(K3050)(K003195.0)(m/s81.9(Pr

)(225

3-12

2

3

=

=

=

DTTgRa s

( )[ ] ( )[ ]101.2

7255.0/559.01

)3.339(387.06.0

Pr/559.01

387.06.0

2

27/816/9

6/12

27/816/9

6/1

=

++=

++=

RaNu

2

2

m2262.0m)m)(12006.0(

C.W/m327.9)101.2(m006.0

CW/m.02662.0

===

=

==

DLA

NuD

kh

s

The rate of heat generation, and thus the rate of heat transfer is

(Q VI= = =8 80V)(10 A) W

Considering both natural convection and radiation, the total rate of heat loss can be expressed as

])K27330()273)[(.KW/m1067.5)(m2262.0)(9.0(

C)30)(m226.0)(C.W/m327.9(W80

)()(

444282

22

44

+++

=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is

C6.52 =sT

which is close to the assumed value of 50C. Then the temperature at the interface of the wire and theplastic cover in steady operation becomes

C57.5=

=+==)m12)(CW/m.15.0(2

)3/6ln()W80(+C6.52

2

)/ln()(

)/ln(

2 12

12

kL

DDQTTTT

DD

kLQ sisi

9-29

Air

T

= 30C

Ts

= 0.9

L = 12 m

D = 6 mm

Resistanceheater

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9-36 A steam pipe extended from one end of a plant to the other with no insulation on it. The rate of heatloss from the steam pipe and the annual cost of those heat losses are to be determined.



(Ts+T)/2 = (170+20)/2 = 95C are (Table A-15)

1-

25

K002717.0K)27395(

11

7121.0Pr

/sm10252.2

CW/m.0306.0

=+

==

==

=

fT

k

AnalysisThe characteristic length in this case is the outer diameter of the pipe, m0.0603== DLc . Then,

6

225

3-12

2

3

10231.1)7121.0()/sm10252.2(

)m0603.0)(K20170)(K002717.0)(m/s81.9(Pr

)(=

=

=

DTTgRa

s

( )[ ] ( )[ ] 42.157121.0/559.01)10231.1(387.0

6.0Pr/559.01

387.0

6.0

2

27/816/9

6/162

27/816/9

6/1

=

+

+=

++=

Ra

Nu

2

2

m37.11m)m)(600603.0(

C.W/m823.7)42.15(m0603.0

CW/m.0306.0

===

=

==

DLA

NuD

kh

s

Then the total rate of heat transfer by natural convection and radiation becomes

kW27.4==+++

=

+=

W

TTATThAQ surrssss

27,388

])K27320()K273170)[(.KW/m1067.5)(m37.11)(7.0(

C)20170)(m37.11)(C.W/m823.7(

)()(

444282

22

44

The total amount of gas consumption and its cost during a one-year period is

therms/yr496,10s/h)3600h/yr8760(kJ105,500

therm1

78.0

kJ/s388.27=

=

=tQ

Qgas

$5647/yr== therm)/538.0)($therms/yr496,10(Cost

9-30

Steam

L = 60 m

D =6.03 cm

Ts= 170C

= 0.7

Air

T

= 20C

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9-37"!PROBLEM 9-37"

"GIVEN"L=60 "[m]"D=0.0603 "[m]"

T_s=170 "[C], parameter to be varied"T_infinity=20 "[C]"

epsilon=0.7T_surr=T_infinityeta_furnace=0.78UnitCost=0.538 "[$/therm]"time=24*365 "[h]"

"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_film)Pr=Prandtl(Fluid$, T=T_film)rho=Density(Fluid$, T=T_film, P=101.3)mu=Viscosity(Fluid$, T=T_film)nu=mu/rho

beta=1/(T_film+273)T_film=1/2*(T_s+T_infinity)sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant"g=9.807 "[m/s^2], gravitational acceleration"

"ANALYSIS"delta=DRa=(g*beta*(T_s-T_infinity)*delta^3)/nu^2*PrNusselt=(0.6+(0.387*Ra^(1/6))/(1+(0.559/Pr)^(9/16))^(8/27))^2h=k/delta*NusseltA=pi*D*LQ_dot=h*A*(T_s-T_infinity)+epsilon*A*sigma*((T_s+273)^4-(T_surr+273)^4)Q_gas=(Q_dot*time)/eta_furnace*Convert(h, s)*Convert(J, kJ)*Convert(kJ, therm)

Cost=Q_gas*UnitCost

9-31

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Ts [C] Q [W] Cost [$]

100 11636 2399

105 12594 2597

110 13577 2799

115 14585 3007

120 15618 3220

125 16676 3438130 17760 3661

135 18869 3890

140 20004 4124

145 21166 4364

150 22355 4609

155 23570 4859

160 24814 5116

165 26085 5378

170 27385 5646

175 28713 5920

180 30071 6200

185 31459 6486

190 32877 6778195 34327 7077

200 35807 7382

10 0 120 14 0 160 180 200

10000

15000

20000

25000

30000

35000

40000

2000

3000

4000

5000

6000

7000

8000

Ts

[C ]

Q

[W]

Cost[$]

Q

Cost

9-32

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9-38 A steam pipe extended from one end of a plant to the other. It is proposed to insulate the steam pipefor $750. The simple payback period of the insulation to pay for itself from the energy it saves are to bedetermined.



temperature of (Ts+T)/2 = (35+20)/2 = 27.5C are (Table A-15)

1-

25

K003328.0K)2735.27(

11

7289.0Pr

/sm10584.1

CW/m.0257.0

=+

==

==

=

fT

k

AnalysisInsulation will drop the outer surface temperature to a value close to the ambient temperature. Thesolution of this problem requires a trial-and-error approach since the determination of the Rayleigh numberand thus the Nusselt number depends on the surface temperature which is unknown. We start the solution

process by guessing the outer surface temperature to be 35C for the evaluation of the properties and h.We will check the accuracy of this guess later and repeat the calculations if necessary. The characteristic

length in this case is the outer diameter of the insulated pipe,m.1603.0==DL

c Then,6

225

3-12

2

3

10856.5)7289.0()/sm10584.1(

)m1603.0)(K2035)(K003328.0)(m/s81.9(Pr

)(=

=

=

DTTgRa s

( )[ ] ( )[ ]23.24

7289.0/559.01

)10856.5(387.06.0

Pr/559.01

387.06.0

2

27/816/9

6/162

27/816/9

6/1

=

+

+=

++=

RaNu

2

2

m22.30m)m)(601603.0(

C.W/m884.3)23.24(m1603.0

CW/m.0257.0

===

=

==

DLA

NuD

kh

s

Then the total rate of heat loss from the outer surface of the insulated pipe by convection and radiationbecomes

W2039

])K27320()K27335)[(.KW/m1067.5)(m22.30)(1.0(+

C)2035)(m22.30)(C.W/m884.3(

)()(

444282

22

44

=++

=+=+=

surrssssradconv TTATThAQQQ

In steady operation, the heat lost from the exposed surface of the insulation by convection and radiation must be equalto the heat conducted through the insulation. This requirement gives the surface temperature to be

)m60)(CW/m.038.0(2

)03.6/03.16ln(

C)170(W2039

2

)/ln( 12

..insulation

=

=

==

ssis

ins

sis T

kL

DD

TT

R

TTQQ

It gives 30.8C for the surface temperature, which is somewhat different than the assumed value of 35 C.Repeating the calculations with other surface temperatures gives

W1988andC3.34 == QTs

Heat loss and its cost without insulation was determined in the Prob. 9-36 to be 27.388 kW and $5647.Then the reduction in the heat losses becomes

kW40.25988.1388.27saved =Q or 25.388/27.40 = 0.927 (92.7%)Therefore, the money saved by insulation will be 0.921 ($5647/yr) = $5237/yr which will pay for the costof $750 in $750/($5237/yr)=0.1432 year = 52.3 days.

9-33

Steam

L = 60 m

D =16.03 cm

170C, = 0.1

Air

T

= 20C

Insulation

= 0.1

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9-39 A circuit board containing square chips is mounted on a vertical wall in a room. The surfacetemperature of the chips is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withconstant properties. 3 The local atmospheric pressure is 1 atm. 4 The heattransfer from the back side of the circuit board is negligible.



1-

25

K0033.0K)27330(

11

7282.0Pr

/sm10608.1

CW/m.02588.0

=+

==

==

=

fT

k


start the solution process by guessing the surface temperature to be 35C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.

The characteristic length in this case is the height of the board, m.3.0== LLc Then,

7

225

3-12

2

3

10463.2)7282.0()/sm10608.1(

)m3.0)(K2535)(K0033.0)(m/s81.9(Pr

)(=

=

=

LTTgRa

s

57.40

7282.0

492.01

)10463.2(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/17

2

27/816/9

6/1

=

+

+=

+

+=Nu

22

2

m09.0m)3.0(

C.W/m50.3)57.40(m3.0

CW/m.02588.0

==

=

==

sA

NuL

kh


])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(

C)25)(m09.0)(C.W/m50.3(W)05.0121(

)()(

444282

22

44

+++

=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is

Ts = 33.5 C

which is sufficiently close to the assumed value in the evaluation of properties and h. Therefore, there is noneed to repeat calculations by reevaluating the properties and h at the new film temperature.

9-34

Air

T

= 25CT

surr= 25C

PCB, Ts

= 0.7121 0.05W

L = 30 cm

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9-40 A circuit board containing square chips is positioned horizontally in a room. The surface temperatureof the chips is to be determined for two orientations.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The heat transfer from the back side of the circuit board is negligible.



1-

25

K0033.0K)27330(

11

7282.0Pr

/sm10608.1

CW/m.02588.0

=+

==

==

=

fT

k


start the solution process by guessing the surface temperature to be 35C for the evaluation of theproperties and h. The characteristic length for both cases is determined from

m.075.0

m)]3.0(+m)3.02[(

m)3.0( 2===

p

AL

sc

Then,

5

225

3-12

2

3

10848.3)7282.0()/sm10608.1(

)m075.0)(K2535)(K00333.0)(m/s81.9(Pr

)(=

=

=

cs LTTgRa

(a) Chips (hot surface) facing up:

45.13)10848.3(54.054.0 4/154/1 === RaNu

22

2

m09.0m)3.0(

C.W/m641.4)45.13(m075.0

CW/m.02588.0

==

=

==

s

c

A

NuL

kh


])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(

C)25)(m09.0)(C.W/m641.4(W)05.0121(

)()(

444282

22

44

+++

=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is Ts = 32.5 Cwhich is sufficiently close to the assumed value. Therefore, there is no need to repeat calculations.(b) Chips (hot surface) facing up:

725.6)10848.3(27.027.0 4/154/1 === RaNu

C.W/m321.2)725.6(m075.0

CW/m.02588.0 2 =

== NuL

kh

c


])K27325()K273)[(.KW/m1067.5)(m09.0)(7.0(

C)25)(m09.0)(C.W/m321.2(W)05.0121(

)()(

444282

22

44

+++=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is Ts = 35.0 Cwhich is identical to the assumed value in the evaluation of properties and h. Therefore, there is no need torepeat calculations.

9-35

Air

T

= 25C

Tsurr

= 25CPCB, Ts = 0.7121 0.05W

L = 30 cm

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9-41 It is proposed that the side surfaces of a cubic industrial furnace be insulated for $550 in order toreduce the heat loss by 90 percent. The thickness of the insulation and the payback period of the insulationto pay for itself from the energy it saves are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withconstant properties. 3 The local atmospheric pressure is 1 atm.


(Ts+T)/2 = (110+30)/2 = 70C are (Table A-15)

1-

25

K002915.0K)27370(

11

7177.0Pr

/sm10995.1

CW/m.02881.0

=+

==

==

=

fT

k

AnalysisThe characteristic length in this case is the height of the furnace, m.2==LLc Then,

10

225

3-12

2

3

10301.3)7177.0()/sm10995.1(

)m2)(K30110)(K002915.0)(m/s81.9(Pr

)(=

=

=

LTTgRa s

2.369

7177.0

492.01

)10301.3(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/110

2

27/816/9

6/1

=

+

+=

+

+=Nu

22

2

m16)m2(4

C.W/m318.5)2.369(m2

CW/m.02881.0

==

=

==

s

c

A

NuL

kh

Then the heat loss by combined natural convection and radiation becomes

W119,15

])K27330()K273110)[(.KW/m1067.5)(m16)(7.0(

C)30110)(m16)(C.W/m318.5(

)()(

444282

22

44

=+++

=

+=

surrssss TTATThAQ

Noting that insulation will reduce the heat losses by 90%, the rate of heat loss after insulation will be

W1512W119,151.0)9.01(

W607,13W119,159.09.0

insulationnoloss

insulationnosaved

===

===

QQ

QQ

The furnace operates continuously and thus 8760 h. Then the amount of energy and money the insulationwill save becomes

therms/yr5215s/yr)3600(8760kJ105,500

therm1

78.0

kJ/s607.13savedEnergy =

== tQsaved

$2868)therm/55.0)($therms5215(=energy)ofcosttsaved)(UniEnergy(savedMoney ==

Therefore, the money saved by insulation will pay for the cost of $550 in

550/($2868/yr)=0.1918 yr = 70 days.

9-36

Hot gases

T

= 30C

Furnace

Ts= 110C = 0.7

2 m

2 m

7/30/2019 Heat Chap09 032

12/23


Insulation will lower the outer surface temperature, the Rayleigh and Nusselt numbers, and thusthe convection heat transfer coefficient. For the evaluation of the heat transfer coefficient, we assume the

surface temperature in this case to be 50C. The properties of air at the film temperature of ( Ts+T)/2 =(50+30)/2 = 40C are (Table A-15)

1-

25

K003195.0K)27340(

117255.0Pr

/sm10702.1

CW/m.02662.0

=+

===

=

=

fT

k

Then,

10

225

3-12

2

3

10256.1)7255.0()/sm10702.1(

)m2)(K3050)(K003195.0)(m/s81.9(Pr

)(=

=

=

LTTgRa s

0.272

7255.0

492.0

1

)10256.1(387.0825.0

Pr

492.0

1

Ra387.0825.0

2

27/816/9

6/110

2

27/816/9

6/1

=

+

+=

+

+=Nu

C.W/m620.3)0.272(m2

CW/m.02662.0 2 =

== NuL

kh

m)2m)(22(4 insuls tA +=

The total rate of heat loss from the outer surface of the insulated furnace by convection and radiationbecomes

])K27330()K273)[(.KW/m1067.5()7.0(+C)30(C).W/m620.3(W1512

)()(

444282

44

++=

+=+=

ss

surrssssradconv

TATA

TTATThAQQQ I

n steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from theexposed surface of the insulation by convection and radiation, which must be equal to the heat conductedthrough the insulation. Therefore,

insul

ss

ins

ssinsulation

t

TA

t

TTkAQQ

C)(110C)W/m.038.0(W1512

)( furnace =

==

Solving the two equations above by trial-and error (or better yet, an equation solver) gives

Ts = 48.4C and tinsul = 0.0254 m = 2.54 cm

9-37

7/30/2019 Heat Chap09 032

13/23


9-42 A cylindrical propane tank is exposed to calm ambient air. The propane is slowly vaporized due to acrack developed at the top of the tank. The time it will take for the tank to empty is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 Radiation heat transfer is negligible.


(Ts+T)/2 = (-42+25)/2 = -8.5C are (Table A-15)

1-

25

K003781.0K)2735.8(

11

7383.0Pr

/sm10265.1

CW/m.02299.0

=+

==

==

=

fT

k

Analysis The tank gains heat through its cylindrical surface as well as its circular end surfaces. Forconvenience, we take the heat transfer coefficient at the end surfaces of the tank to be the same as that of itsside surface. (The alternative is to treat the end surfaces as a vertical plate, but this will double the amountof calculations without providing much improvement in accuracy since the area of the end surfaces is muchsmaller and it is circular in shape rather than being rectangular). The characteristic length in this case is the

outer diameter of the tank, m.5.1==DLc Then,

10

225

3-12

2

3

10869.3)7383.0()/sm10265.1(

)m5.1](K)42(25)[(K003781.0)(m/s81.9(Pr

)(=

=

=

DTTgRa s

( )[ ] ( )[ ]1.374

7383.0/559.01

)10869.3(387.06.0

Pr/559.01

387.06.0

2

27/816/9

6/1102

27/816/9

6/1

=

+

+=

++=

RaNu

222

2

m38.224/m)5.1(2)m4)(m5.1(4/2

C.W/m733.5)1.374(m5.1

CW/m.02299.0

=+=+=

=

==

DDLA

NuD

kh

s

and

W8598C)]42(25)[(m38.22)(C.W/m733.5()(

22

=== ss TThAQ

The total mass and the rate of evaporation of propane are

kg/s02023.0kJ/kg425

kJ/s598.8

kg4107)m4(4

)m5.1()kg/m581(

4

23

2

===

=

=

==

fgh

Qm

LD

Vm

and it will take

hours56.4==== s996,202kg/s02023.0

kg4107

m

mt

for the propane tank to empty.

9-38

D = 1.5 m

L = 4 m

Propane tank

0T

s= -42C

Air

T

= 25C

7/30/2019 Heat Chap09 032

14/23


9-43E The average surface temperature of a human head is to be determined when it is not covered.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The head can be approximated as a 12-in.-diameter sphere.

Properties The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We

start the solution process by guessing the surface temperature to be 120F for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.

The properties of air at 1 atm and the anticipated film temperature of (Ts+T)/2 = (120+77)/2 = 98.5F are(Table A-15E)

1-

23

R001791.0R)4605.98(

11

7262.0Pr

/sft10180.0

FBtu/h.ft.01525.0

=+

==

==

=

fT

k

AnalysisThe characteristic length for a spherical object isLc =D = 12/24 = 0.5 ft. Then,

6

223

3-12

2

3

10943.6)7262.0(

)/sft10180.0(

)ft5.0)(R7795)(R001791.0)(ft/s2.32(Pr

)(=

=

=

DTTgRa s

39.25

7262.0

469.01

)10943.6(589.02

Pr

469.01

589.02

9/416/9

4/16

9/416/9

4/1

=

+

+=

+

+=Ra

Nu

222

2

ft7854.0)ft5.0(

F.Btu/h.ft7744.0)39.25(ft1

FBtu/h.ft.01525.0

===

=

==

DA

NuD

kh

s

Considering both natural convection and radiation, the total rate of heat loss can be written as

])R46077()R460)[(.RBtu/h.ft101714.0)(m7854.0)(9.0(

F)77)(ft7854.0(F).Btu/h.ft7744.0(Btu/h)4/287(

)()(

444282

22

44

+++=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is

Ts = 125.9 F

which is sufficiently close to the assumed value in the evaluation of the properties and h. Therefore, there isno need to repeat calculations.

9-39

Air

T

= 77F

HeadQ = 287 Btu/h

D = 12 in

= 0.9

7/30/2019 Heat Chap09 032

15/23


9-44 The equilibrium temperature of a light glass bulb in a room is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The light bulb is approximated as an 8-cm-diameter sphere.

Properties The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We

start the solution process by guessing the surface temperature to be 170 C for the evaluation of theproperties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.

The properties of air at 1 atm and the anticipated film temperature of ( Ts+T)/2 = (170+25)/2 = 97.5C are(Table A-15)

1-

25

K002699.0K)2735.97(

11

7116.0Pr

/sm10279.2

CW/m.03077.0

=+

==

==

=

fT

k

AnalysisThe characteristic length in this case isLc =D = 0.08 m. Then,

6

225

3-12

2

3

10694.2

)7116.0(

)/sm10279.2(

)m08.0)(K25170)(K002699.0)(m/s81.9(Pr

)(

=

=

=

DTTgRa s

( )[ ] ( )[ ]

42.20

7116.0/469.01

)10694.2(589.02

Pr/469.01

589.02

9/416/9

4/16

9/416/9

4/1

=+

+=

++=

RaNu

Then

222

2

m02011.0m)08.0(

C.W/m854.7)42.20(m08.0

CW/m.03077.0

===

=

==

DA

NuD

kh

s

Considering both natural convection and radiation, the total rate of heat loss can be written as

])K27325()273)[(.KW/m1067.5)(m02011.0)(9.0(

C)25)(m02011.0)(C.W/m854.7(W)6090.0(

)()(

444282

22

44

+++

=

+=

s

s

surrssss

T

T

TTATThAQ

Its solution is

Ts = 169.4 C

which is sufficiently close to the value assumed in the evaluation of properties and h. Therefore, there is noneed to repeat calculations.

9-40

Air

T

= 25C

Lamp60 W

= 0.9

D = 8 cm

Light,6 W

7/30/2019 Heat Chap09 032

16/23


9-45 A vertically oriented cylindrical hot water tank is located in a bathroom. The rate of heat loss from thetank by natural convection and radiation is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The temperature of the outer surface of the tank is constant.

Properties The properties of air at 1 atm and the film temperature

of (Ts+T)/2 = (44+20)/2 = 32C are (Table A-15)

1-

25

K003279.0K)27332(

11

7276.0Pr

/sm10627.1

CW/m.02603.0

=+

==

==

=

fT

k

AnalysisThe characteristic length in this case is the height of the cylinder,

m.1.1==LLc Then,

9

225

3-12

2

3

10883.3)/sm10627.1(

)m1.1)(K2044)(K003279.0)(m/s81.9()(Gr =

=

=

LTTg s

A vertical cylinder can be treated as a vertical plate when

m1542.0)10883.3(

m)1.1(35

Gr

35m)4.0( 4/191/4 ===

LD

which is satisfied. That is, the Nusselt number relation for a vertical plate can be used for the side surfaces.For the top and bottom surfaces we use the relevant Nusselt number relations. First, for the side surfaces,

99 10825.2)7276.0)(10883.3(GrPrRa ===

2.170

7276.0

492.01

)10825.2(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/19

2

27/816/9

6/1

=

+

+=

+

+=Nu

2

2

m382.1)m1.1)(m4.0(

C.W/m027.4)2.170(m1.1

CW/m.02603.0

===

===

DLA

NuLkh

s

W6.133C)2044)(m382.1)(C.W/m027.4()( 22side === TThAQ ss

For the top surface,

m1.04

m4.0

4

4/2=====

D

D

D

p

AL

sc

6

225

3-12

2

3

10123.2)7276.0()/sm10627.1(

)m1.0)(K2044)(K003279.0)(m/s81.9(Pr

)(Ra =

=

=

cs LTTg

61.20)10123.2(54.0Ra54.0Nu 4/164/1

===

222

2

m1257.04/)m4.0(4/

C.W/m365.5)61.20(m1.0

CW/m.02603.0

===

=

==

DA

NuL

kh

s

c

W2.16C)2044)(m1257.0)(C.W/m365.5()( 22top === TThAQ ss

For the bottom surface,

31.10)10123.2(27.0Ra27.0Nu 4/164/1 ===

9-41

AirT

= 20CTankT

s= 44C

= 0.4L = 1.1 m

D = 0.4 m

7/30/2019 Heat Chap09 032

17/23


C.W/m683.2)31.10(m1.0

CW/m.02603.0 2 =

== NuL

kh

c

W1.8C)2044)(m1257.0)(C.W/m683.2()( 22bottom === TThAQ ss

The total heat loss by natural convection is

W157.9=++=++= 1.82.166.133bottomtopsideconv QQQQ

The radiation heat loss from the tank is

[ ]W101.1=

++++=

= 444282

44rad

)K27320()K27344().KW/m1067.5)(m1257.01257.0382.1)(4.0(

)( surrss TTAQ

9-42

7/30/2019 Heat Chap09 032

18/23

7/30/2019 Heat Chap09 032

19/23


C11.7== K7.2842T

We could repeat the solution using air properties at the new film temperature using this value to increasethe accuracy. However, this would only affect the heat transfer value somewhat, which would not havesignificant effect on the final water temperature. The average rate of heat transfer can be determined fromEq. 2

W34.3== 2)-(11.753976.3Q

9-44

7/30/2019 Heat Chap09 032

20/23


9-47"!PROBLEM 9-47"

"GIVEN"height=0.28 "[m]"L=0.18 "[m]"w=0.18 "[m]"

T_infinity=24 "[C]"

T_w1=2 "[C]"epsilon=0.6T_surr=T_infinity"time=3 [h], parameter to be varied"

"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_film)Pr=Prandtl(Fluid$, T=T_film)rho=Density(Fluid$, T=T_film, P=101.3)mu=Viscosity(Fluid$, T=T_film)nu=mu/rhobeta=1/(T_film+273)

T_film=1/2*(T_w_ave+T_infinity)T_w_ave=1/2*(T_w1+T_w2)rho_w=Density(water, T=T_w_ave, P=101.3)C_p_w=CP(water, T=T_w_ave, P=101.3)*Convert(kJ/kg-C, J/kg-C)sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant"g=9.807 "[m/s^2], gravitational acceleration"

"ANALYSIS"delta=heightRa=(g*beta*(T_infinity-T_w_ave)*delta^3)/nu^2*PrNusselt=0.59*Ra^0.25h=k/delta*NusseltA=2*(height*L+height*w+w*L)Q_dot=h*A*(T_infinity-T_w_ave)+epsilon*A*sigma*((T_surr+273)^4-(T_w_ave+273)^4)

m_w=rho_w*V_wV_w=height*L*wQ=m_w*C_p_w*(T_w2-T_w1)Q_dot=Q/(time*Convert(h, s))

9-45

7/30/2019 Heat Chap09 032

21/23


time [h] Tw2 [C]

0.5 4.013

1 5.837

1.5 7.496

2 9.013

2.5 10.41

3 11.693.5 12.88

4 13.98

4.5 15

5 15.96

5.5 16.85

6 17.69

6.5 18.48

7 19.22

7.5 19.92

8 20.59

8.5 21.21

9 21.81

9.5 22.3710 22.91

0 2 4 6 8 102. 5

7

11.5

16

20.5

25

time [h]

Tw2

[C]

9-46

7/30/2019 Heat Chap09 032

22/23


9-48 A room is to be heated by a cylindrical coal-burning stove. The surface temperature of the stove andthe amount of coal burned during a 30-day-period are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The temperature of the outer surface of the stove is constant. 5 The heattransfer from the bottom surface is negligible. 6 The heat transfer coefficient at the top surface is the sameas that on the side surface.



1-

25

K002857.0K)27377(

11

7161.0Pr

/sm10066.2

CW/m.02931.0

=+

==

==

=

fT

k

AnalysisThe characteristic length in this case is the height of the cylindir,

m.7.0==LLc Then,

9

225

3-12

2

3

10387.2)/sm10066.2(

)m70.0)(K24130)(K002857.0)(m/s81.9()(Gr =

=

=

LTTg s

A vertical cylinder can be treated as a vertical plate when

m1108.0)10387.2(

m)7.0(35

Gr

35m)32.0(

4/191/4=

==

LD

which is satisfied. That is, the Nusselt number relation for a vertical plate can be used for side surfaces.

99 10709.1)7161.0)(10387.2(GrPrRa ===

2.145

7161.0

492.01

)10709.1(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/19

2

27/816/9

6/1

=

+

+=

+

+=Nu

222

2

m7841.04/)m32.0()m7.0)(m32.0(4/

C.W/m080.6)2.145(m7.0

CW/m.02931.0

=+=+=

===

DDLA

NuL

kh

s

Then the surface temperature of the stove is determined from

C127.6==

+=

+=+=

K6.400

)290)(.KW/m10)(5.67m841(0.85)(0.7)297)(mC)(0.7841.W/m080.6(W1200

)()(

4442-8222

4surr

4radconv

s

ss

ssss

T

TT

TTATThAQQQ

The amount of coal used is determined from

kg3.102==

=

==

kJ/kg30,000

kJ)/0.65(60,480/

kJ60,480=s/h)3600h/daykJ/s)(142.1(

coalHV

Qm

tQQ

9-47

Air

T

= 24CStoveT

s

= 0.85L =0.7 m

D = 0.32 m

7/30/2019 Heat Chap09 032

23/23


9-49 Water in a tank is to be heated by a spherical heater. The heating time is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature of the outer surface of the sphere isconstant.

Properties Using the average temperature for water (15+45)/2=30as the fluid temperature, the properties of water at the film

temperature of (Ts+T)/2 = (85+30)/2 = 57.5C are (Table A-9)

1-3

26

K10501.0

12.3Pr

/sm10474.0

CW/m.6515.0

=

==

=

k

Also, the properties of water at 30C are (Table A-9)

CJ/kg.4178andkg/m996 3 == pC

AnalysisThe characteristic length in this case isLc =D = 0.06 m. Then,

8

226

3-132

2

3

10108.8)12.3()/sm10474.0(

)m06.0)(K3085)(K10501.0)(m/s81.9(Pr

)(Ra =

=

=

DTTg s

( )[ ] ( )[ ]14.89

12.3/469.01

)10108.8(589.02

Pr/469.01

589.02

9/416/9

4/18

9/416/9

4/1=

++=

++= RaNu

222

2

m01131.0m)06.0(

C.W/m9.967)14.89(m06.0

CW/m.6515.0

===

=

==

DA

NuD

kh

s

The rate of heat transfer by convection is

W1.602)30)(85mC)(0.01131.W/m9.967()( 22conv === TThAQ ss

The mass of water in the container is

kg.8439)m)(0.040kg/m996( 33 === Vm

The amount of heat transfer to the water isJ104.994=C15)-C)(45J/kg.kg)(417884.39()( 612 == TTmCQ p

Then the time the heater should be on becomes

hours2.304==

== s8294J/s602.1

J104.994 6

Q

Qt

Water

T,ave

= 30C

Resistanceheater

Ts = 85CD = 6 cmD = 6 cm

Date post:	14-Apr-2018
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Heat Chap09 032

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