heat chap07 035

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Chapter 7External Forced Convection

Flow Across Cylinders And Spheres

7-35C For the laminar flow, the heat transfer coefficient will be the highest at the stagnation point which

corresponds to 0 . In turbulent flow, on the other hand, it will be highest when is between90 and 120 .

13-36C Turbulence moves the fluid separation point further back on the rear of the body, reducing thesize of the wake, and thus the magnitude of the pressure drag (which is the dominant mode of drag). As aresult, the drag coefficient suddenly drops. In general, turbulence increases the drag coefficient for flatsurfaces, but the drag coefficient usually remains constant at high Reynolds numbers when the flow isturbulent.

13-37C Friction drag is due to the shear stress at the surface whereas the pressure drag is due to thepressure differential between the front and back sides of the body when a wake is formed in the rear.

13-38C Flow separation in flow over a cylinder is delayed in turbulent flow because of the extra mixingdue to random fluctuations and the transverse motion.

7-39 A steam pipe is exposed to windy air. The rate of heat loss from the steam is to be determined.Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties.

Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (90+7)/2 = 48.5C are(Table A-15)

7232.0Pr

/sm10784.1

CW/m.02724.0

25-

==

=

k

AnalysisThe Reynolds number is

4

25 10228.6/sm10784.1

m)(0.08]s/h)0m/km)/(3601000(km/h)(50[

Re === DV

The Nusselt number corresponding to this Reynolds number is

( )[ ]

( )[ ]1.159

000,282

10228.61

7232.0/4.01

)7232.0()10228.6(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

The heat transfer coefficient and the heat transfer rate become

C.W/m17.54)1.159(m08.0

CW/m.02724.0 2

=

==Nu

D

kh

length)m(per=C7)-)(90mC)(0.2513.W/m17.54()(

m0.2513=m)m)(108.0(

22

2

W1130==

==

TThAQ

DLA

ssconv

s

7-20

AirV = 50 km/h

T

= 7C

PipeD = 8 cmT

s= 90C

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7-40 A hot stainless steel ball is cooled by forced air. The average convection heat transfer coefficient andthe cooling time are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 The outer surface temperature of the ball is uniform at all times.

Properties The average surface temperature is (350+250)/2 = 300C, and the properties of air at 1 atmpressure and the free stream temperature of 30C are (Table A-15)

7282.0Pr

kg/m.s10934.2

kg/m.s10872.1

/sm10608.1CW/m.02588.0

5

C300@,

5

25-

=

=

=

==

s

k


4

2510597.5

/sm1057.1

m)m/s)(0.15(6Re =

=

=

DV

The Nusselt number corresponding this Reynolds number is determined to be

[ ]

[ ] 6.14510934.2

10872.1)7282.0()10597.5(06.0)10597.5(4.02

PrRe06.0Re4.02

4/1

5

54.03/245.04

4/1

4.03/25.0

=

++=

++==

sk

hDNu

Heat transfer coefficient is

C.W/m25.122 =

== )6.145(

m15.0

CW/m.02588.0Nu

D

kh

The average rate of heat transfer can be determined from Newton's law of cooling by using averagesurface temperature of the ball

W479.5=C30)-)(300mC)(0.07069.W/m12.25()(m0.07069=m)15.0(

22

222

== == TThAQ

DA

ssave

s

Assuming the ball temperature to be nearly uniform , the total heat transferred from the ball during thecooling from 350 C to 250 C can be determined from

Q mC T T ptotal = ( )1 2

where kg23.146

m)(0.15)kg/m8055(

6

33

3

====

D

Vm

Therefore, J683,249=C250)-C)(350J/kg.kg)(48023.14()( 21total == TTmCQ p

Then the time of cooling becomes

min23.75==== s1425J/s5.479J249,683

Q

Qt

7-21

AirV = 6 m/s

T

= 30C

D = 15 cmT

s= 350C

D

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7-41"!PROBLEM 7-41"

"GIVEN"D=0.15 "[m]"

T_1=350 "[C]"T_2=250 "[C]"T_infinity=30 "[C]"P=101.3 "[kPa]"

"Vel=6 [m/s], parameter to be varied"rho_ball=8055 "[kg/m^3]"C_p_ball=480 "[J/kg-C]"

"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_infinity)Pr=Prandtl(Fluid$, T=T_infinity)rho=Density(Fluid$, T=T_infinity, P=P)mu_infinity=Viscosity(Fluid$, T=T_infinity)nu=mu_infinity/rhomu_s=Viscosity(Fluid$, T=T_s_ave)

T_s_ave=1/2*(T_1+T_2)

"ANALYSIS"Re=(Vel*D)/nuNusselt=2+(0.4*Re^0.5+0.06*Re^(2/3))*Pr^0.4*(mu_infinity/mu_s)^0.25h=k/D*NusseltA=pi*D^2Q_dot_ave=h*A*(T_s_ave-T_infinity)Q_total=m_ball*C_p_ball*(T_1-T_2)m_ball=rho_ball*V_ballV_ball=(pi*D^3)/6time=Q_total/Q_dot_ave*Convert(s, min)

Vel [m/s] h [W/m2.C] time [min]

1 9.204 64.831.5 11.5 51.86

2 13.5 44.22.5 15.29 39.01

3 16.95 35.21

3.5 18.49 32.27

4 19.94 29.92

4.5 21.32 27.99

5 22.64 26.36

5.5 23.9 24.96

6 25.12 23.756.5 26.3 22.69

7 27.44 21.74

7.5 28.55 20.98 29.63 20.14

8.5 30.69 19.44

9 31.71 18.81

9.5 32.72 18.24

10 33.7 17.7

7-22

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1 2 3 4 5 6 7 8 9 10

5

10

15

20

25

30

35

10

20

30

40

50

60

70

Vel [m/s]

h

[W/m

2-C]

time

[min]

h

time

7-23

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7-42E A person extends his uncovered arms into the windy air outside. The rate of heat loss from the armis to be determined.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 The arm is treated as a 2-ft-long and 3-in.-diameter cylinder with insulatedends. 5 The local atmospheric pressure is 1 atm.

Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (86+54)/2 = 70F are(Table A-15E)

7306.0Pr

/sft101643.0

FBtu/h.ft.01457.0

23-

==

=

k


[ ] 423

10463.4/sft101643.0

ft(3/12)ft/s/3600)5280(20Re =

=

=

DV

The Nusselt number corresponding this Reynolds number is determined to be

6.129000,282

10463.41

7306.0

4.01

)7306.0()10463.4(62.03.0

000,282

Re1

Pr4.01

PrRe62.03.0

5/48/5

4

4/13/2

3/15.04

5/48/5

4/13/2

3/15.0

=

+

+

+=

+

+

+==k

hDNu

Then the heat transfer coefficient and the heat transfer rate from the arm becomes

F.Btu/h.ft557.7)6.129(ft)12/3(

FBtu/h.ft.01457.0 2 =

== NuD

kh

Btu/h379.8=F54)-)(86ftF)(1.571.Btu/h.ft557.7()(

ft1.571=ft)ft)(212/3(

22

2

==

==

TThAQ

DLA

ssconv

s

7-24

AirV = 20 mph

T = 54F

ArmD = 3 inTs = 86F

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7-43E"!PROBLEM 7-43E"

"GIVEN"T_infinity=54 "[F], parameter to be varied""Vel=20 [mph], parameter to be varied"

T_s=86 "[F]"L=2 "[ft]"D=3/12 "[ft]"

"PROPERTIES"Fluid$='air'k=Conductivity(Fluid$, T=T_film)Pr=Prandtl(Fluid$, T=T_film)rho=Density(Fluid$, T=T_film, P=14.7)mu=Viscosity(Fluid$, T=T_film)*Convert(lbm/ft-h, lbm/ft-s)nu=mu/rho

T_film=1/2*(T_s+T_infinity)"ANALYSIS"Re=(Vel*Convert(mph, ft/s)*D)/nuNusselt=0.3+(0.62*Re^0.5*Pr^(1/3))/(1+(0.4/Pr)^(2/3))^0.25*(1+(Re/282000)^(5/8))^(4/5)

h=k/D*NusseltA=pi*D*LQ_dot_conv=h*A*(T_s-T_infinity)

T [F] Qconv [Btu/h]

20 790.2

25 729.4

30 668.7

35 608.2

40 547.9

45 487.7

50 427.755 367.9

60 308.265 248.6

70 189.2

75 129.9

80 70.77

Vel [mph] Qconv [Btu/h]

10 250.6

12 278.914 305.7

16 331.3

18 356

20 379.8

22 40324 425.6

26 447.7

28 469.330 490.5

32 511.4

34 532

36 552.2

38 572.2

40 591.9

7-25

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20 30 40 50 60 70 80

0

10 0

20 0

30 0

40 0

50 0

60 0

70 0

80 0

T

[F]

Qconv

[Btu/h

]

10 15 20 25 30 35 40

25 0

30 0

35 0

40 0

45 0

50 0

55 0

60 0

Vel [mph]

Qco

nv

[Btu/h]

7-26

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7-44 The average surface temperature of the head of a person when it is not covered and is subjected towinds is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 One-quarter of the heat the person generates is lost from the head. 5 The headcan be approximated as a 30-cm-diameter sphere. 6 The local atmospheric pressure is 1 atm.

Properties The properties of air at 1 atm pressure and the free stream temperature of 10C are (Table A-15)

7336.0Pr

kg/m.s10802.1

kg/m.s10778.1

/sm10426.1

CW/m.02439.0

5

C15@,

5

25-

=

=

=

=

=

s

k


[ ] 525

10045.2/sm10426.1

m)(0.3m/s1000/3600)(35Re =

=

=

DV

The proper relation for Nusselt number corresponding to this Reynolds number is

[ ]

[ ] 7.34410802.1

10778.1)7336.0()10045.2(06.0)10045.2(4.02

PrRe06.0Re4.02

4/1

5

54.03/245.05

4/1

4.03/25.0

=

++=

++==

sk

hDNu

The heat transfer coefficient is

C.W/m28.02)7.344(m3.0

CW/m.02439.0 2 =

== NuD

kh

Then the surface temperature of the head is determined to be

C12.7 =

=+====

)mC)(0.2827.W/m02.28(

W(84/4)+C10)(

m0.2827=m)3.0(

22

222

ssss

s

hA

QTTTThAQ

DA

7-27

AirV = 35 km/h

T

= 10C

HeadQ = 21 W

D = 0.3 m

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7-45 The flow of a fluid across an isothermal cylinder is considered. The change in the drag force and therate of heat transfer when the free-stream velocity of the fluid is doubled is to be determined.

Analysis The drag force on a cylinder is given by

2

2

1=

VNDD ACF

When the free-stream velocity of the fluid is doubled, the drag

force becomes

2

)2(2

2=

VNDD ACF

Taking the ratio of them yields

4==

2

2

1

2 )2(

V

V

D

D

F

F

The rate of heat transfer between the fluid and the cylinder is given by Newton's law of cooling. Weassume the Nusselt number is proportional to the nth power of the Reynolds number with 0.33 < n

heat chap07 035

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