Heat Chap07 099

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QTsky = 100 K

Chapter 7External Forced Convection

7-99 Wind is blowing over the roof of a house. The rate of heat transfer through the roof and the cost of thisheat loss for 14-h period are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Air isan ideal gas with constant properties. 4 The pressure of air is 1 atm.

Properties Assuming a film temperature of 10C, the properties ofair are (Table A-15)

7336.0Pr

/sm10426.1CW/m.02439.0

25-

==

=k

AnalysisThe Reynolds number is

[ ] 725

10338.2/sm10426.1

m)(20m/s)3600/100060(Re =

=

=

LVL

which is greater than the critical Reynolds number. Thus we havecombined laminar and turbulent flow. Then the Nusselt number andthe heat transfer coefficient are determined to be

C.W/m0.31)10542.2(m20

CW/m.02439.0

10542.2)7336.0](871)10338.2(037.0[Pr)871Re037.0(

24

43/18.073/18.0

===

====

NuLkh

k

hLNu L

In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal tothe heat transfer from the roof to the surroundings (by convection and radiation), which must be equal tothe heat transfer through the roof by conduction. That is,

Q Q Q Q= = =room to roof, conv+rad roof, cond roof to surroundings, conv+rad

Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out, respectively, the quantitiesabove can be expressed as

[ ]4,44282,

224,

4,rad+convroof,toroom

K)273(K)27320().KW/m1067.5)(m300)(9.0(

C))(20mC)(300.W/m5()()(

+++

=+=

ins

insinsroomsinsroomsi

T

TTTATTAhQ

m15.0

)m300)(CW/m.2( ,,2,,condroof, outsinsoutsinssTT

L

TTkAQ

=

=

[ ]44,4282,

2244,,rad+convsurr,toroof

K)100(K)273().KW/m1067.5)(m300)(9.0(

C)10)(mC)(300.W/m0.31()()(

++

=+=

outs

outssurroutsssurroutsso

T

TTTATTAhQ

Solving the equations above simultaneously gives

C5.3andC,6.10,W025,28 ,, ==== outsins TTQ kW28.03

The total amount of natural gas consumption during a 14-hour period is

therms75.15kJ105,500

therm1

85.0

)s360014)(kJ/s03.28(

85.085.0=

=

==

tQQQ totalgas

Finally, the money lost through the roof during that period is

$9.45== )therm/60.0$therms)(75.15(lostMoney

7-84

AirV = 60 km/h

T = 10C

Tin = 20C

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7-100 Steam is flowing in a stainless steel pipe while air is flowing across the pipe. The rate of heat lossfrom the steam per unit length of the pipe is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Thepressure of air is 1 atm.

Properties Assuming a film temperature of 10C, the properties of air are (Table A-15)

7336.0Prand/s,m10426.1C,W/m.02439.0 2-5 ===k

AnalysisThe outer diameter of insulated pipe isDo= 4.6+2 3.5=11.6 cm = 0.116 m. The Reynoldsnumber is

4

2510254.3

/sm10426.1

m)m/s)(0.116(4Re =

=

=

oDV

The Nusselt number for flow across a cylinder is determined from

( )[ ]

( )[ ]0.107

000,282

10254.31

7336.0/4.01

)7336.0()10254.3(62.03.0

000,282

Re1

Pr/4.01

PrRe62.03.0

5/48/5

4

4/13/2

3/15.04

5/48/5

4/13/2

3/15.0

=

+

+

+=

+

++==

k

hDNu o

and CW/m50.22)0.107(m116.0

CW/m0.02439 2 === NuD

kh

oo

Area of the outer surface of the pipe per m length of the pipe is

2m3644.0)m1)(m116.0( === LDA ooIn steady operation, heat transfer from the steam through the pipe and the insulation to the outer surface (byfirst convection and then conduction) must be equal to the heat transfer from the outer surface to thesurroundings (by simultaneous convection and radiation). That is,

Q Q Q= =pipe and insulation surface to surroundings

Using the thermal resistance network, heat transfer from the steam to the outer surface is expressed as

[ ]

C/W874.3)m1)(CW/m.038.0(2

)3.2/8.5ln(

2

)/ln(

C/W0015.0)m1)(CW/m.15(2

)2/3.2ln(

2

)/ln(

C/W0995.0

)m1(m)04.0()C.W/m80(

11

23

12

2,

=

==

=

==

=

==

kL

rrR

kL

rrR

Ah

R

insulation

pipe

iiiconv

andC/W)874.30015.00995.0(

C)250(

,

1insandpipe ++

=

++

= sinsulationpipeiconv

s T

RRR

TTQ

Heat transfer from the outer surface can be expressed as

[ ]4442822244

rad+convsurr,tosurface

K)2733(K)273().KW/m1067.5)(m3644.0)(3.0(

C)3)(mC)(0.3644.W/m50.22()()(

+++

=+=

s

ssurrsosurrsoo

T

TTTATTAhQ

Solving the two equations above simultaneously, the surface temperature and the heat transfer rate per mlength of the pipe are determined to be

length)m(perandC9.9 W60.4== QTs

7-85

Steel pipeD

i= D

1= 4 cm

D2= 4.6 cm

Steam, 250C

Do

Di

Air

3C, 4 m/s

Insulation

= 0.3

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7-101 A spherical tank filled with liquid nitrogen is exposed to winds. The rate of evaporation of the liquidnitrogen due to heat transfer from the air is to be determined for three cases.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 The pressure of air is 1 atm.

Properties The properties of air at 1 atm pressure and the free stream temperature of 20C are (Table A-15)

7309.0Pr

kg/m.s10023.5

kg/m.s10825.1

/sm10516.1

CW/m.02514.0

6

C196@,

5

25-

=

=

=

==

s

k

Analysis(a) When there is no insulation,D =Di = 4 m,and the Reynolds number is

[ ] 625

10932.2/sm10516.1

m)(4m/s1000/3600)(40Re =

=

=

DV

The Nusselt number is determined from

[ ]

[ ] 233310023.5

10825.1)7309.0()10932.2(06.0)10932.2(4.02

PrRe06.0Re4.02

4/1

6

54.03/265.06

4/14.03/25.0

=

++=

++==

sk

hDNu

and C.W/m66.14)2333(m4

CW/m.02514.0 2 =

== NuD

kh

The rate of heat transfer to the liquid nitrogen is

[ ] W200,159C)196(20(]m)(4C)[.W/m66.14(

))(()(

22

2

==

==

TTDhTThAQ sss

The rate of evaporation of liquid nitrogen then becomes

kg/s0.804====kJ/kg198

kJ/s2.159

ifif

h

QmhmQ

(b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to

evaluate dynamic viscosity at a new surface temperature which we will assume to be -100 C. At -100C,kg/m.s10189.1 5= . Noting thatD =D0 = 4.1 m, the Nusselt number becomes

[ ] 625

10005.3/sm10516.1

m)(4.1m/s1000/3600)(40Re =

=

=

DV

[ ]

[ ] 191010189.110825.1)7309.0()10005.3(06.0)10005.3(4.02

PrRe06.0Re4.02

4/1

5

54.03/265.06

4/1

4.03/25.0

=

++=

++==

sk

hDNu

and C.W/m71.11)1910(m1.4

CW/m.02514.0 2 =

== NuD

kh


7-86

Nitrogen tank

-196C

Do

Di

Wind

20C40 km/h

Insulation

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W7361

)m81.52)(C.W/m71.11(

1

m)m)(2C)(2.05W/m.(0.0354

m)205.2(

C)]196(20[

1

4

m81.52)m1.4(

22

21

12

tan,tan,

222

=

+

=

+

=

+

=

===

s

ks

convinsulation

ks

s

hArkr

rr

TT

RR

TTQ

DA


kg/s0.0372====kJ/kg198

kJ/s361.7

ifif

h

QmhmQ

(c) We use the dynamic viscosity value at the new estimated surface temperature of 0C to bekg/m.s10729.1 5= . Noting thatD =D0 = 4.04 m in this case, the Nusselt number becomes

[ ] 625

10961.2/sm10516.1

m)(4.04m/s1000/3600)(40Re =

=

=

DV

[ ]

[ ] 172410729.1

10825.1)7309.0()10961.2(06.0)10961.2(4.02

PrRe06.0Re4.02

4/1

5

54.03/265.06

4/1

4.03/25.0

=

++=

++==

sk

hDNu

and C.W/m73.10)1724(m04.4

CW/m.02514.0 2 =

== NuD

kh


W4.27

)m28.51)(C.W/m73.10(

1

m)m)(2C)(2.02W/m.(0.000054

m)202.2(

C)]196(20[

1

4

m28.51)m04.4(

22

21

12

tan,tan,

222

=

+

=

+

=

+

=

===

s

ks

convinsulation

ks

s

hArkr

rr

TT

RR

TTQ

DA


kg/s101.38 4-====kJ/kg198

kJ/s0274.0

if

ifh

QmhmQ

7-87

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7-102 A spherical tank filled with liquid oxygen is exposed to ambient winds. The rate of evaporation ofthe liquid oxygen due to heat transfer from the air is to be determined for three cases.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 7 The pressure of air is 1 atm.

Properties The properties of air at 1 atm pressure and the free stream temperature of 20C are (Table A-15)

7309.0Pr

kg/m.s10127.6

kg/m.s10825.1

/sm10516.1

CW/m.02514.0

5

C183@,

5

25-

=

=

==

=

s

k

Analysis(a) When there is no insulation,D =Di = 4 m,and the Reynolds number is

[ ] 625

10932.2/sm10516.1

m)(4m/s1000/3600)(40Re =

=

=

DV

The Nusselt number is determined from

[ ]

[ ] 22201005.1

10825.1)7309.0()10932.2(06.0)10932.2(4.02

PrRe06.0Re4.02

4/1

5

54.03/265.06

4/1

4.03/25.0

=

++=

++==

sk

hDNu

and C.W/m95.13)2220(m4

CW/m.02514.0 2 =

== NuD

kh

The rate of heat transfer to the liquid oxygen is

[ ] W372,142C)183(20(]m)(4C)[.W/m95.13())(()( 222 ==== TTDhTThAQ sss

The rate of evaporation of liquid oxygen then becomes

kg/s0.668====kJ/kg213

kJ/s4.142

ifif

h

QmhmQ

(b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to

evaluate dynamic viscosity at a new surface temperature which we will assume to be -100C. At -100C,kg/m.s10189.1 5= . Noting thatD =D0 = 4.1 m, the Nusselt number becomes

[ ] 625

10005.3/sm10516.1

m)(4.1m/s1000/3600)(40Re =

=

=

DV

[ ]

[ ] 191010189.1

10825.1)7309.0()10005.3(06.0)10005.3(4.02

PrRe06.0Re4.02

4/1

5

54.03/265.06

4/1

4.03/25.0

=

++=

++==

sk

hDNu

and C.W/m71.11)1910(m1.4

CW/m.02514.0 2 === NuDkh


7-88

Oxygen tank

-183C

Do

Di

Wind

20C40 km/h

Insulation

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W6918

)m81.52)(C.W/m71.11(

1

m)m)(2C)(2.05W/m.(0.0354

m)205.2(

C)]183(20[

1

4

m81.52)m1.4(

22

21

12

tan,tan,

222

=

+

=

+

=

+

=

===

s

ks

convinsulation

ks

s

hArkr

rr

TT

RR

TTQ

DA


kg/s0.0325====kJ/kg213

kJ/s918.6

ifif

h

QmhmQ

(c) Again we use the dynamic viscosity value at the estimated surface temperature of 0C to bekg/m.s10729.1 5= . Noting thatD =D0 = 4.04 m in this case, the Nusselt number becomes

[ ] 625

10961.2/sm10516.1

m)(4.04m/s1000/3600)(40Re =

=

=

DV

[ ]

[ ] 172410729.1

10825.1)713.0()10961.2(06.0)10961.2(4.02

PrRe06.0Re4.02

4/1

5

54.03/265.06

4/1

4.03/25.0

=

++=

++==

sk

hDNu

and C.W/m73.10)1724(m04.4

CW/m.02514.0 2 =

== NuD

kh


W8.25

)m28.51)(C.W/m73.10(

1

m)m)(2C)(2.02W/m.(0.000054

m)202.2(

C)]183(20[

1

4

m28.51)m04.4(

22

21

12

tan,tan,

222

=

+

=

+

=

+

=

===

s

ks

convinsulation

ks

s

hArkr

rr

TT

RR

TTQ

DA

The rate of evaporation of liquid oxygen then becomes

kg/s101.21 4-====kJ/kg213

kJ/s0258.0

ifif

h

QmhmQ

7-89

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7-103 A circuit board houses 80 closely spaced logic chips on one side. All the heat generated is conductedacross the circuit board and is dissipated from the back side of the board to the ambient air, which is forcedto flow over the surface by a fan. The temperatures on the two sides of the circuit board are to bedetermined.

Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 7 The pressure of air is 1atm.

Properties Assuming a film temperature of 40C, the properties of air are (Table A-15)

7255.0Pr

/sm10702.1

CW/m.02662.0

25-

==

=

k

AnalysisThe Reynolds number is

[ ] 425

10051.7/sm10702.1

m)(0.18m/s)60/400(Re =

=

=

LV

L

which is less than the critical Reynolds number. Therefore,the flow is laminar. Using the proper relation for Nusseltnumber, heat transfer coefficient is determined to be

C.W/m43.23)4.158(m18.0

CW/m.02662.0

4.158)7255.0()10051.7(664.0PrRe664.0

2

3/15.043/15.0

=

==

====

NuL

kh

khLNu L

The temperatures on the two sides of the circuit board are

C39.52

C39.48

=

+=

+==

=

+=

+==

m)m)(0.18C)(0.12W/m.16(

m)W)(0.00306.080(C48.39

)(

m)m)(0.18C)(0.12.W/m43.23(

W)06.080(C30

)(

2121

2

22

s

s

ss

kA

LQTTTT

L

kAQ

hA

QTTTThAQ

7-90

T1

T2

T =30C400 m/min

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7-104E The equivalent wind chill temperature of an environment at 10F at various winds speeds are

V = 10 mph: T T V V equiv ambient = +

= =

914 914 0 475 0 0203 0 304

914 914 10 0 475 0 0203 10

. ( . )( . . . )

. . ( . . (F) mph) + 0.304 10 mph 9 F

V = 20 mph: Tequiv = = 914 914 10 0 475 0 0203 20. . ( . . (F) mph) + 0.304 20 mph 24.9 F



In the last 3 cases, the person needs to be concerned about the possibility of freezing.

7-91

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7-105E "!PROBLEM 7-105E"

"ANALYSIS"T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*Vel+0.304*sqrt(Vel))

Vel [mph] Tambient [F] Tequiv [F]4 20 19.87

14.67 20 -4.383

25.33 20 -15.05

36 20 -20.57

46.67 20 -23.15

57.33 20 -23.77

68 20 -22.94

78.67 20 -21.01

89.33 20 -18.19

100 20 -14.63

4 40 39.91

14.67 40 22.45

25.33 40 14.7736 40 10.79

46.67 40 8.935

57.33 40 8.493

68 40 9.086

78.67 40 10.48

89.33 40 12.51

100 40 15.07

4 60 59.94

14.67 60 49.28

25.33 60 44.59

36 60 42.16

46.67 60 41.02

57.33 60 40.75

68 60 41.11

78.67 60 41.96

89.33 60 43.21

100 60 44.77

7-92

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0 22 44 66 88 110

-3 0

-2 0

-1 0

0

10

20

30

40

50

60

Vel [mph]

Tequiv

[F]

20 F

40 F

60 F

7-106 . 7-110 Design and Essay Problems

7-93

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Heat Chap07 099