Heat Chap07 052

7/30/2019 Heat Chap07 052

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Chapter 7 External Forced Convection

7-52 A steam pipe is exposed to a light winds in the atmosphere. The amount of heat loss from the steamduring a certain period and the money the facility will save a year as a result of insulating the steam pipeare to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The plantoperates every day of the year for 10 h a day. 4 The local atmospheric pressure is 1 atm.

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

(Ts + T∞)/2 = (75+5)/2 = 40°C are (Table A-15)

7255.0Pr

/sm10702.1

CW/m.02662.0

25-

=×=

°=

υ

k

Analysis The Reynolds number is

[ ] 4

2510632.1

/sm10702.1

m)(0.1m/s1000/3600)(10Re ×=

××

=υ

=−

∞ DV

The Nusselt number corresponding this Reynolds number is determined to be

[ ]

[ ]19.71

000,282

10632.11

)7255.0/4.0(1

)7255.0()10632.1(62.03.0

000,282

Re1

Pr)/4.0(1

Pr Re62.03.0

5/48/54

4/13/2

3/15.04

5/48/5

4/13/2

3/15.0

=

×+

+

×+=

+

++==

k

hD Nu

The heat transfer coefficient is

C.W/m95.18)19.71(m1.0

CW/m.02662.0 2 °=°

== Nu D

k h

The rate of heat loss by convection is2m77.3m)12)(m1.0( === π π DL A s

W5001=C5)-)(75mC)(3.77.W/m95.18()( 22 °°=−= ∞T T hAQ s s

The rate of heat loss by radiation is

[ ] W1558)K 2730()K 27375().K W/m10)(5.67m(3.77)8.0()(

4442-82

44

=+−+×=−= surr s srad T T AQ σ ε

The total rate of heat loss then becomes

W655915585001total =+=+= rad conv QQQ

The amount of heat loss from the steam during a 10-hour work day is

kJ/day102.361 5×=×=∆= )s/h3600h/day10)(kJ/s559.6(t QQ total

The total amount of heat loss from the steam per year is

kJ/yr 10619.8)days/yr 365)(kJ/day10361.2()daysof no.( 75 ×=×== daytotal QQ

Noting that the steam generator has an efficiency of 80%, the amount of gas used is

therms/yr 1021

kJ105,500

therm1

80.0

kJ/yr 10619.8

80.0

7

=

×== total

gas

QQ

Insulation reduces this amount by 90 %. The amount of energy and money saved becomes

therms/yr 919=)therms/yr 1021)(90.0()90.0(savedEnergy == gasQ

$496== erm))($0.54/ththerms/yr (919=energy)of costtsaved)(UniEnergy(savedMoney

7-36

Wind

V = 10 km/hT ∞

= 5°C

Steam pipe

T s= 75°C

D = 10 cm

ε = 0.8

7/30/2019 Heat Chap07 052



7-53 A steam pipe is exposed to light winds in the atmosphere. The amount of heat loss from the steamduring a certain period and the money the facility will save a year as a result of insulating the steam pipesare to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The plantoperates every day of the year for 10 h. 4 The local atmospheric pressure is 1 atm.


(Ts + T∞)/2 = (75+5)/2 = 40°C are (Table A-15)

7255.0Pr

/sm10702.1

CW/m.02662.0

25-

=×=

°=

υ

k


[ ] 4

2510632.1

/sm10702.1

m)(0.1m/s1000/3600)(10Re ×=

××

=υ

=−

∞ DV

The Nusselt number corresponding this Reynolds number is determined to be

[ ]

[ ]19.71

000,282

10632.11

)7255.0/4.0(1

)7255.0()10632.1(62.03.0

000,282

Re1

Pr)/4.0(1

Pr Re62.03.0

5/48/5

4

4/13/2

3/15.04

5/48/5

4/13/2

3/15.0

=

×+

+

×+=

+

++==

k

hD Nu


C.W/m95.18)19.71(m1.0

CW/m.02662.0 2 °=°

== Nu D

k h

The rate of heat loss by convection is2m77.3m)12)(m1.0( === π π DL A s

W5001=C5)-)(75mC)(3.77.W/m95.18()( 22 °°=−= ∞T T hAQ s s

For an average surrounding temperature of 0 C° , the rate of heat loss by radiation and the total rate of heat

loss are

[ ] W1558)K 2730()K 27375().K W/m10)(5.67m(3.77)8.0(

)(

4442-82

44

=+−+×=

−= surr s srad T T AQ σ ε

W655915885001total =+=+= rad conv QQQ

If the average surrounding temperature is − °20 C , the rate of heat loss by radiation and the total rate of heat loss become

[ ]W1807

)K 27320()K 27375().K W/m10)(5.67m(3.77)8.0(

)(

4442-82

44

=+−−+×=


W680818075001 =+=+= rad convtotal QQQ

which is 6808 - 6559 = 249 W more than the value for a surrounding temperature of 0 °C. This correspondsto

3.8%=×=×=°

100W6559

W249100%change

Ctotal,0

difference

Q

Q

(increase)

If the average surrounding temperature is 25°C, the rate of heat loss by radiation and the total rate of heatloss become

7-37

Wind

V = 10 km/hT ∞

= 5°C

Steam pipe

T s= 75°C

D = 10 cm

ε = 0.8

7/30/2019 Heat Chap07 052



[ ]W1159

)K 27325()K 27375().K W/m10)(5.67m(3.77)8.0(

)(

44442-82

44

=+−+×=


W616011595001 =+=+= rad convtotal QQQ

which is 6559 - 6160 = 399 W less than the value for a surrounding temperature of 0°C. This corresponds to

6.1%=×=×=°

100W6559W399100%change

Ctotal,0

difference

QQ

(decrease)

Therefore, the effect of the temperature variations of the surrounding surfaces on the total heat transfer isless than 6%.

7-38

7/30/2019 Heat Chap07 052



7-54E An electrical resistance wire is cooled by a fan. The surface temperature of the wire is to bedetermined.

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

Properties We assume the film temperature to be 200 °F . The properties of air at this temperature are (Table A-15E)

7124.0Pr

/sft102406.0

FBtu/h.ft.01761.0

23-

=×= °=

υ

k


8.692/sft102406.0

ft)12ft/s)(0.1/(20Re

23=

×==

−∞

υ

DV

The proper relation for Nusselt number corresponding this Reynolds number is

[ ]

[ ] 34.13000,282

8.6921

)7124.0/4.0(1

)7124.0()8.692(62.03.0

000,282

Re1

Pr)/4.0(1

Pr Re62.03.0

5/48/5

4/13/2

3/15.0

5/48/5

4/13/2

3/15.0

=

+++=

+

++==

k

hD Nu


F.Btu/h.ft19.28)34.13(ft)12/1.0(

FBtu/h.ft.01761.0 2 °=°

== Nu D

k h

Then the average temperature of the outer surface of the wire becomes

2ft3142.0ft)12)(ft12/1.0( === π π DL A s

F662.9°=°

×°=+= → −= ∞∞

)ftF)(0.3142.Btu/h.ft19.28(

Btu/h3.41214)(1500+F85)(

22hA

QT T T T hAQ s s s

Discussion Repeating the calculations at the new film temperature of (85+662.9)/2=374°F gives

T s=668.3°F.

7-39

Air V∞ = 20 ft/s

T ∞ = 85°F

Resistance wir D = 0.1 in

7/30/2019 Heat Chap07 052



7-55 The components of an electronic system located in a horizontal duct is cooled by air flowing over theduct. The total power rating of the electronic device is to be determined.



(Ts + T∞)/2 = (65+30)/2 = 47.5°C are (Table A-15)

7235.0Pr

/sm10774.1

CW/m.02717.025-

=×=

°=υ

k


[ ] 4

2510758.3

/sm10774.1

m)(0.2m/s(200/60)Re ×=

×=

υ=

−∞ DV

Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be

2.112)7235.0()10758.3(102.0Pr Re102.0 3/1675.043/1675.0 =×===k

hD Nu


C.W/m24.15)2.112(m2.0

CW/m.02717.0 2 °=°

== Nu D

k h

Then the rate of heat transfer from the duct becomes

2m2.1m)5.1)(m2.04( =×= s A

W640.0=C30)-)(65mC)(1.2.W/m24.15()( 22 °°=−= ∞T T hAQ s s

7-40

Air

30°C200 m/min

20 cm

65°C

7/30/2019 Heat Chap07 052



7-56 The components of an electronic system located in a horizontal duct is cooled by air flowing over the

duct. The total power rating of the electronic device 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 (T s + T∞)/2 = (65+30)/2 = 47.5°C are(Table A-15)

7235.0Pr

/sm10774.1

CW/m.02717.025-

=×=

°=υ

k

For a location at 4000 m altitude where the atmospheric pressure is61.66 kPa, only kinematic viscosity of air will be affected. Thus,

/sm10915.2)10774.1(66.61

325.101 255kPa66.61@

−− ×=×

=υ


[ ] 4

2510287.2

/sm10915.2

m)(0.2m/s(200/60)Re ×=

×

=υ

=−

∞ DV

Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be

21.80)7235.0()287.2(102.0Pr Re102.0 3/1675.03/1675.0 ====k

hD Nu


C.W/m90.10)21.80(m2.0

CW/m.02717.0 2 °=°

== Nu D

k h

Then the rate of heat transfer from the duct becomes

2m2.1m)5.1)(m2.04( =×= s A

W457.7=C30)-)(65mC)(1.2.W/m90.10()( 22 °°=−= ∞T T hAQ s s

7-41

Air

30°C200 m/min

20 cm

65°C

7/30/2019 Heat Chap07 052


Q

Resistor 0.4 W

D = 0.3 cm


7-57 A cylindrical electronic component mounted on a circuit board is cooled by air flowing across it. Thesurface temperature of the component is to be determined.


Properties We assume the film temperature to be 50°C. The properties of air at 1 atm and at this temperature are (Table A-15)

7228.0Pr

/sm10798.1CW/m.02735.025-

=×=

°=υ k


1.417/sm10798.1

m)m/s)(0.003(150/60Re

25=

×== −

∞

υ

DV

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

[ ]

[ ] 43.10000,282

1.417

1)7228.0/4.0(1

)7228.0()1.417(62.0

3.0

000,282

Re1

Pr)/4.0(1

Pr Re62.03.0

5/48/5

4/13/2

3/15.0

5/48/5

4/13/2

3/15.0

=

+++=

+

++==

k

hD Nu


C.W/m09.95)43.10(m003.0

CW/m.02735.0 2 °=°

== Nu D

k h

Then the surface temperature of the component becomes2m0001696.0m)018.0)(m003.0( === π π DL A s

C64.8°=°

°=+= → −= ∞∞)m96C)(0.00016.W/m09.95(

W0.4+C40)(

22hA

QT T T T hAQ s s s

7-42

Air

V∞ = 150 m/min

T ∞ = 40°C

7/30/2019 Heat Chap07 052



7-58 A cylindrical hot water tank is exposed to windy air. The temperature of the tank after a 45-mincooling period is to be estimated.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 The surface of the tank is at the same temperature as the water temperature. 5

The heat transfer coefficient on the top and bottom surfaces is the same as that on the side surfaces.

Properties The properties of water at 80°C are (Table A-9)

CJ/kg.4197

kg/m8.971 3

°==

pC

ρ

The properties of air at 1 atm and at the anticipated film temperature of 50°C are (Table A-15)

7228.0Pr

/sm10798.1

CW/m.02735.0

25-

=×=

°=

υ

k


015,309

/sm10798.1

m)(0.50m/s3600

100040

Re25

=

×

×

=

υ

=−

∞ DV

The proper relation for Nusselt number corresponding to this

Reynolds number is

( )[ ]

( )[ ]9.484

000,282

015,3091

7228.0/4.01

)7228.0()015,309(62.03.0

000,282

Re1

Pr /4.01

Pr Re62.03.0 Nu

5/48/5

4/13/2

3/15.0

5/48/5

4/13/2

3/15.0

=

+

++=

+

++=


C.W/m.5326)9.484(

m50.0

CW/m.02735.0 2 °=°

== Nu

D

k h

The surface area of the tank is

222

m885.14/)5.0(2)95.0)(5.0(4

2 =+=+= π π π π D

DL A s

The rate of heat transfer is determined from

C182

80)mC)(1.885.W/m53.26()( 222 °

−

+°=−= ∞

T T T hAQ s s

(Eq. 1)

where T 2 is the final temperature of water so that (80+T 2)/2 gives the average temperature of water duringthe cooling process. The mass of water in the tank is

kg27.181m)/4(0.95m)(0.50)kg/m8.971(4

232

=π=ρπ=ρ= L D

V m

The amount of heat transfer from the water is determined from

C)-C)(80J/kg.kg)(419727.181()( 212 °°=−= T T T mC Q p

Then average rate of heat transfer is

s6045

C)-C)(80J/kg.kg)(419727.181( 2

×°°

=∆

=T

t

QQ (Eq. 2)

Setting Eq. 1 to be equal to Eq. 2 we obtain the final temperature of water

7-43

Air

V =40 km/h

T ∞

= 18°C

Water tank D =50 cm L = 95 cm

7/30/2019 Heat Chap07 052



C69.9°= →

×°°

=°

−

+°=

2

2222

s6045

C)-C)(80J/kg.kg)(419727.181(C18

2

80)mC)(1.885.W/m53.26(

T

T T Q

7-44

7/30/2019 Heat Chap07 052



7-59 "!PROBLEM 7-59"

"GIVEN"D=0.50 "[m]"L=0.95 "[m]"T_w1=80 "[C]"T_infinity=18 "[C]"Vel=40 "[km/h]"

"time=45 [min], 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/rhoT_film=1/2*(T_w_ave+T_infinity)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)T_w_ave=1/2*(T_w1+T_w2)

"ANALYSIS"Re=(Vel*Convert(km/h, m/s)*D)/nu Nusselt=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*L+2*pi*D^2/4Q_dot=h*A*(T_w_ave-T_infinity)m_w=rho_w*V_wV_w=pi*D^2/4*LQ=m_w*C_p_w*(T_w1-T_w2)Q_dot=Q/(time*Convert(min, s))

time [min] Tw2 [C]

30 73.06

45 69.86

60 66.83

75 63.96

90 61.23

105 58.63

120 56.16

135 53.8

150 51.54

165 49.39

180 47.33

195 45.36

210 43.47

225 41.65240 39.91

255 38.24

270 36.63

285 35.09

300 33.6

7-45

7/30/2019 Heat Chap07 052



0 50 100 150 200 250 300

30

35

40

45

50

55

60

65

70

75

time [min]

T w 2

[ C ]

7-46

7/30/2019 Heat Chap07 052



7-60 Air flows over a spherical tank containing iced water. The rate of heat transfer to the tank and the rateat which ice melts are to be determined.


Properties The properties of air at 1 atm pressure and the free stream temperature of 25°C are (Table A-15)

7296.0Pr

kg/m.s10729.1

kg/m.s10849.1

/sm10562.1

CW/m.02551.0

5

C0@,

5

25-

=

×=

×=×=

°=

−°

−∞

s

k

µ

µ

υ


658,806/sm10562.1

m)m/s)(1.8(7Re

25=

×=

υ=

−∞ DV

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

[ ]

[ ] 1.79010729.1

10849.1)7296.0()658,806(06.0)658,806(4.02

Pr Re06.0Re4.02

4/1

5

54.03/25.0

4/1

4.03/25.0

=

××

++=

µ

µ++==

−

−

∞

sk

hD Nu


C.W/m.2011)1.790(m8.1

CW/m.02551.0 2 °=°

== Nu D

k h

Then the rate of heat transfer is determined to be

W2850=°−°=−=

==

∞ C)025)(mC)(10.18.W/m20.11()(

m10.18=m)8.1(

22

222

T T hAQ

D A

s s

s

π π

The rate at which ice melts is

kg/min0.512== → = → = kg/s00854.0kJ/kg)7.333(=kW850.2 mmhmQ fg

7-47

Air V = 7 m/s

T ∞

=25°C

Iced water 0°C

D = 1.8 m

7/30/2019 Heat Chap07 052



7-61 A cylindrical bottle containing cold water is exposed to windy air. The average wind velocity is to beestimated.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gaswith constant properties. 4 Heat transfer at the top and bottom surfaces is negligible.

Properties The properties of water at the average temperature of ( T 1 + T 2)/2 = (3+11)/2 = 7°C are (Table A-9)

CJ/kg.4200

kg/m8.999 3

°==

pC

ρ

The properties of air at 1 atm and the film temperature of (T s + T ∞)/2 = (7+27)/2 = 17°C are (Table A-15)

7317.0Pr

/sm10489.1

CW/m.02491.0

25-

=×=

°=

υ

k

Analysis The mass of water in the bottle is

kg2.356m)/4(0.30m)(0.10)kg/m8.999(4

232

=π=ρπ=ρ= L D

V m

Then the amount of heat transfer to the water isJ 79,162=C3)-C)(11J/kg. kg)(4200 356.2()( 12 °°=−= T T mC Q p

The average rate of heat transfer is

W32.29s6045

J162,79=

×=

∆=

t

QQ


C.W/m55.15C7)-)(27m(0.09425W32.29)(

m0.09425=m)m)(0.3010.0(

22conv

2

°= → °= → −=

==

∞ hhT T hAQ

DL A

s s

s

π π

The Nusselt number is

42.62CW/m.0.02491

m)C)(0.10.W/m55.15( 2

=°

°

== k

hD Nu

Reynolds number can be obtained from the Nusselt number relation for a flow over the cylinder

( )[ ]

( )[ ]856,12Re

000,282

Re1

7317.0/4.01

)7317.0(Re62.03.042.62

000,282

Re1

Pr /4.01

Pr Re62.03.0

5/48/5

4/13/2

3/15.0

5/48/5

4/13/2

3/15.0

= →

+

++=

+

++= Nu

Then using the Reynolds number relation we determine the wind velocity

m/s1.91VVV

= → ×

= → υ

= ∞−∞∞

/sm10489.1

)m10.0(856,12Re

25

D

7-48

Air

V

T ∞

= 27°CBottle

D =10 cm L = 30 cm

7/30/2019 Heat Chap07 052



Flow Across Tube Banks

7-62C In tube banks, the flow characteristics are dominated by the maximum velocity V max that occurs

within the tube bank rather than the approach velocity V . Therefore, the Reynolds number is defined onthe basis of maximum velocity.

7-63C The level of turbulence, and thus the heat transfer coefficient, increases with row number because of the combined effects of upstream rows in turbulence caused and the wakes formed. But there is nosignificant change in turbulence level after the first few rows, and thus the heat transfer coefficient remainsconstant. There is no change in transverse direction.

7-64 Combustion air is preheated by hot water in a tube bank. The rate of heat transfer to air and the pressure drop of air are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of hot water.

Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air

properties at the assumed mean temperature of 20°C (will be checked later) and 1 atm (Table A-15):

k = 0.02514 W/m-K ρ = 1.204 kg/m3

C p =1.007 kJ/kg-K Pr = 0.7309

µ = 1.825× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7132

Also, the density of air at the inlet temperature of 15°C (for use in the mass flow rate calculation at the

inlet) is ρi = 1.225 kg/m3.

Analysis It is given that D = 0.021 m, S L = S T = 0.05 m, and V = 3.8 m/s.Then the maximum velocity and the Reynolds number based on the maximum velocity become

m/s552.6m/s)8.3(021.005.0

05.0

max =−=−=VV

DS

S

T

T

9075skg/m10825.1

m)m/s)(0.021552.6)(kg/m204.1(Re

5

3max =

⋅×=

µρ

=−

D D

V

The average Nusselt number is determined usingthe proper relation from Table 7-2 to be

59.75)7132.0/7309.0()7309.0()9075(27.0

)Pr (Pr/Pr Re27.0 Nu

25.036.063.0

25.036.063.0

==

= s D D

This Nusselt number is applicable to tube banks with N L > 16. In our case the number of rows is N L = 8, andthe corresponding correction factor from Table 7-3 is F = 0.967. Then the average Nusselt number and

heat transfer coefficient for all the tubes in the tube bank become

1.73)59.75)(967.0( Nu Nu , === D N D F L

CW/m5.87m0.021

C)W/m02514.0(1.73 2, °⋅=°⋅

== D

k Nuh L N D

The total number of tubes is N = N L × N T = 8× 8 = 64. For a unit tube length ( L = 1 m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are

2m .2224m)m)(1021.0(64 =π=π= DL N A s

7-49

S L

S T

V=3.8 m/s

T i=15°C

T s=90°C

D

7/30/2019 Heat Chap07 052



kg/s862.1m)m)(105m/s)(8)(0.8.3)(kg/m225.1()( 3 ==ρ== LS N mm T T ii V

Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become

C42.28C)J/kgkg/s)(1007(1.862

C)W/m5.87)(m222.4(exp)1590(90exp)(

22

°=

°⋅°⋅

−−−=

−−−=

p

si s se

C m

h AT T T T

C07.68)]42.2890/()1590ln[(

)42.2890()1590(

)]/()ln[(

)()(ln °=−−

−−−=−−

−−−=∆

e si s

e si sT T T T

T T T T T

W25,148=°°⋅=∆= )C07.68)(mC)(4.222W/m5.87( 22lnT hAQ s

For this square in-line tube bank, the friction coefficient corresponding to Re D = 9075 and S L/ D = 5/2.1 =

2.38 is, from Fig. 7-27a, f = 0.22. Also, χ = 1 for the square arrangements. Then the pressure drop acrossthe tube bank becomes

Pa 45.5=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)552.6)(kg/m204.1()1)(22.0(8

2

V f N P L

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (15 + 29.1)/2 = 22.1°C, which is fairly

close to the assumed value of 20°C. Therefore, there is no need to repeat calculations.

7-50

7/30/2019 Heat Chap07 052



7-65 Combustion air is preheated by hot water in a tube bank. The rate of heat transfer to air and the pressure drop of air are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of hot water.



k = 0.02514 W/m-K ρ = 1.204 kg/m3

C p =1.007 kJ/kg-K Pr = 0.7309

µ = 1.825× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7132



Analysis It is given that D = 0.021 m, S L = S T = 0.05 m, and V = 3.8 m/s.Then the maximum velocity and the Reynolds number based onthe maximum velocity become

m/s552.6m/s)8.3(021.005.0

05.0max =

−=

−= VV

DS

S

T

T

since 2/)( DS S T D +>

9075skg/m10825.1

m)m/s)(0.021552.6)(kg/m204.1(Re

5

3max =

⋅×=

µρ

=−

D D

V

The average Nusselt number is determined using the proper relation from Table 7-2 to be

55.74)7132.0/7309.0()7309.0()9075()05.0/05.0(35.0

)Pr (Pr/Pr Re)/(35.0 Nu

25.036.06.02.0

25.036.06.02.0

==

= s D LT D S S

This Nusselt number is applicable to tube banks with N L > 16. In our case the number of rows is N L = 8, andthe corresponding correction factor from Table 7-3 is F = 0.967. Then the average Nusselt number and

heat transfer coefficient for all the tubes in the tube bank become

09.72)55.74)(967.0( Nu Nu , === D N D F L

CW/m29.86m0.021

C)W/m02514.0(09.72 2, °⋅=°⋅

== D

k Nuh L N D


2m .2224m)m)(1021.0(64 =π=π= DL N A s



C25.28C)J/kgkg/s)(1007(1.862

C)W/m29.86)(m222.4(exp)1590(90exp)(

22

°=

°⋅°⋅

−−−=

−−−=

p

si s se

C m

h AT T T T

C16.68)]25.2890/()1590ln[(

)25.2890()1590(

)]/()ln[(

)()(ln °=

−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W24,834=°°⋅=∆= )C16.68)(mC)(4.222W/m29.86( 22lnT hAQ s

7-51

S L

S T

V=3.8 m/s

T i=15°C

T s=90°C

D

7/30/2019 Heat Chap07 052



For this staggered tube bank, the friction coefficient corresponding to Re D = 9075 and S T / D = 5/2.1 = 2.38

is, from Fig. 7-27ba, f = 0.34. Also, χ = 1 for the square arrangements. Then the pressure drop across thetube bank becomes

Pa 70.3=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)552.6)(kg/m204.1()1)(34.0(8

2

V f N P L

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (15 +28.3)/2 = 21.7°C, which is fairly


7-66 Combustion air is heated by condensing steam in a tube bank. The rate of heat transfer to air, the pressure drop of air, and the rate of condensation of steam are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of steam.



k = 0.02625 W/m-K ρ = 1.145 kg/m3

C p =1.007 kJ/kg-K Pr = 0.7268

µ = 1.895× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7111


inlet) is ρi = 1.204 kg/m3. The enthalpy of vaporization of water at 100°C is h fg = 2257 kJ/kg-K (Table A-9).

Analysis (a) It is given that D = 0.016 m, S L = S T = 0.04 m, and V = 5.2 m/s.Then the maximum velocity and the Reynolds number based onthe maximum velocity become

m/s667.8m/s)2.5(016.004.0

04.0max =

−=

−= VV

DS

S

T

T

since 2/)( DS S T D +>

8380skg/m10895.1

m)m/s)(0.016667.8)(kg/m145.1(Re

5

3max =

⋅×=

µρ

=−

D D

V

The average Nusselt number is determined using the proper relation from Table 7-2 to be

88.70)7111.0/7268.0()7268.0()8380()04.0/04.0(35.0


25.036.06.02.0

25.036.06.02.0

==

= s D LT D S S

Since N L =20, which is greater than 16, the average Nusselt number and heat transfer coefficient for all thetubes in the tube bank become

88.70 Nu Nu , == D N D L

CW/m3.116m0.016

C)W/m02625.0(88.70 2, °⋅=°⋅

== D

k Nuh L N D


2m .0510m)m)(1016.0(200 =π=π= DL N A s

kg/s504.2m)m)(1.04m/s)(10)(02.5)(kg/m204.1()( 3 ==ρ== LS N mm T T ii V

7-52

S L

S T

V=5.2 m/s

T i=20°C

T s=100°C

D

7/30/2019 Heat Chap07 052




C68.49C)J/kgkg/s)(1007(2.504

C)W/m3.116)(m05.10(exp)20100(100exp)(

22

°=

°⋅°⋅

−−−=

−−−=

p

si s se

C m

h AT T T T

C01.64)]68.49100/()20100ln[(

)68.49100()20100(

)]/()ln[(

)()(ln °=

−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W74,836=°°⋅=∆= )C01.64)(mC)(10.05W/m3.116( 22lnT hAQ s

(b) For this staggered tube bank, the friction coefficient corresponding to Re D = 7713 and S T / D = 4/1.6 =

2.5 is, from Fig. 7-27b, f = 0.33. Also, χ = 1 for the square arrangements. Then the pressure drop across thetube bank becomes

Pa 283.9=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)667.8)(kg/m145.1()1)(33.0(20

2

V f N P L

(c) The rate of condensation of steam is

kg/s0.03316=°⋅

== → =

°

°CkJ/kg2257

kW836.74

C100@condC100@cond

fg fg

h

QmhmQ

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (20 + 49.7)/2 = 34.9°C, which is very


7-53

7/30/2019 Heat Chap07 052



7-67 Combustion air is heated by condensing steam in a tube bank. The rate of heat transfer to air, the pressure drop of air, and the rate of condensation of steam are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of steam.



k = 0.02625 W/m-K ρ = 1.145 kg/m3

C p =1.007 kJ/kg-K Pr = 0.7268

µ = 1.895× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7111


inlet) is ρi = 1.204 kg/m3. The enthalpy of vaporization of water at 100°C is h fg = 2257 kJ/kg-K (Table A-9).

Analysis (a) It is given that D = 0.016 m, S L = S T = 0.05 m, and V = 5.2 m/s.

Then the maximum velocity and the Reynolds number

based on the maximum velocity become

m/s647.7m/s)2.5(

016.005.0

05.0max =

−=

−= VV

DS

S

T

T

7394skg/m10895.1

m)m/s)(0.016647.7)(kg/m145.1(Re

5

3max =

⋅×=

µρ

=−

D D

V

The average Nusselt number is determined usingthe proper relation from Table 7-2 to be

26.66)7111.0/7268.0()7268.0()7394(27.0


25.036.063.0

25.036.063.0

==

= s D D

Since N L =20, which is greater than 16, the average Nusselt number and heat transfer coefficient for all thetubes in the tube bank become

26.66 Nu Nu , == D N D L

CW/m7.108m0.016

C)W/m02625.0(26.66 2, °⋅=°⋅

== D

k Nuh L N D


2m .0510m)m)(1016.0(200 =π=π= DL N A s

kg/s130.3m)m)(1.05m/s)(10)(02.5)(kg/m204.1()( 3 ==ρ== LS N mm T T ii V


C44.43C)J/kgkg/s)(1007(3.130

C)W/m7.108)(m05.10(exp)20100(100exp)(

22

°=

°⋅°⋅−−−=

−−−= p

si s se

C m

h AT T T T

C6.67)]44.43100/()20100ln[(

)44.43100()20100(

)]/()ln[(

)()(ln °=

−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W73,882=°°⋅=∆= )C6.67)(mC)(10.05W/m7.108( 22lnT hAQ s

7-54

S L

S T

V=5.2 m/s

T i=20°C

T s=100°C

D

7/30/2019 Heat Chap07 052



(b) For this in-line arrangement tube bank, the friction coefficient corresponding to Re D = 6806 and S L/ D =

5/1.6 = 3.125 is, from Fig. 7-27a, f = 0.20. Also, χ = 1 for the square arrangements. Then the pressure dropacross the tube bank becomes

Pa 133.9=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)647.7)(kg/m145.1()1)(20.0(20

2

V f N P L

(c) The rate of condensation of steam is

kg/s0.03273=°⋅

== → =°

°CkJ/kg2257

kW882.73

C100@condC100@cond

fg fg

h

QmhmQ

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (20 + 43.4)/2 = 31.7°C, which is fairly


7-55

7/30/2019 Heat Chap07 052



7-68 Water is preheated by exhaust gases in a tube bank. The rate of heat transfer, the pressure drop of exhaust gases, and the temperature rise of water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of steam. 3 For exhaust gases, air properties are used.



k = 0.04104 W/m-K ρ = 0.6746 kg/m3

C p =1.033 kJ/kg-K Pr = 0.6946

µ = 2.76× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7154



Analysis (a) It is given that D = 0.021 m, S L = S T = 0.08 m, and V = 4.5 m/s.Then the maximum velocity and the Reynolds number based on the maximum velocity become

m/s102.6m/s)5.4(021.008.0

08.0max =

−=

−= VV

DS

S

T

T

3132skg/m1076.2

m)m/s)(0.021102.6)(kg/m6746.0(Re

5

3max =

⋅×=

µρ

=−

D D

V

The average Nusselt number is determined using

the proper relation from Table 7-2 to be

46.37)7154.0/6946.0()6946.0()3132(27.0


25.036.063.0

25.036.063.0

==

= s D D

Since N L =16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become

46.37 Nu Nu , == D N D L

CW/m2.73m0.021

C)W/m04104.0(46.37 2, °⋅=°⋅

== D

k Nuh L N D


2m .4458m)m)(1021.0(128 =π=π= DL N A s



C0.237C)J/kgkg/s)(1033(1.773C)W/m2.73)(m445.8(exp)30080(80exp)(

22

°=

°⋅ °⋅−−−=

−−−=

p

si s seC m

h AT T T T

C7.186)]23780/()30080ln[(

)23780()30080(

)]/()ln[(

)()(ln °=

−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W115,425=°°⋅=∆= )C7.186)(mC)(8.445W/m2.73( 22lnT hAQ s

(b) For this in-line arrangement tube bank, the friction coefficient corresponding to Re D = 3132 and S L/ D =

8/2.1 = 3.81 is, from Fig. 7-27a, f = 0.18. Also, χ = 1 for the square arrangements. Then the pressure dropacross the tube bank becomes

7-56

S L

S T

V=4.5 m/s

T i=300°C

T s=80°C

D

7/30/2019 Heat Chap07 052



Pa 36.2=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)102.6)(kg/m6746.0()1)(18.0(16

2

V f N P L

(c) The temperature rise of water is

C4.6°=°⋅

==∆ → ∆=C)kJ/kg18.4)(kg/s6(

kW425.115

,water

,water

water p

water water water pC m

QT T C mQ

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (300 + 237)/2 = 269°C, which is

sufficiently close to the assumed value of 250°C. Therefore, there is no need to repeat calculations.

7-57

7/30/2019 Heat Chap07 052



7-69 Water is heated by a bundle of resistance heater rods. The number of tube rows is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the rods is constant.

Properties The properties of water at the mean temperature of (15°C +65°C)/2=40°C are (Table A-9):

k = 0.631 W/m-K ρ = 992.1 kg/m3

C p =4.179 kJ/kg-K Pr = 4.32

µ = 0.653× 10-3 kg/m-s Pr s = Pr @ Ts = 1.96

Also, the density of water at the inlet temperature of 15°C (for use in the mass flow rate calculation at the

inlet) is ρi =999.1 kg/m3.

Analysis It is given that D = 0.01 m, S L = 0.04 m and S T = 0.03 m, and V = 0.8 m/s.Then the maximum velocity and the Reynolds number based on the maximum velocity become

m/s20.1m/s)8.0(01.003.0

03.0max =

−=

−= VV

DS

S

T

T

232,18skg/m10653.0

m)m/s)(0.0120.1)(kg/m1.992(Re

3

3max =

⋅×=

µρ

=−

D D

V



3.269)96.1/32.4()32.4()232,18(27.0


25.036.063.0

25.036.063.0

==

= s D D

Assuming that N L > 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become

3.269 Nu Nu , == D N D L

CW/m994,16

m0.01

C)W/m631.0(3.269 2, °⋅=°⋅

==

D

k Nuh L N D

Consider one-row of tubes in the transpose direction (normal to flow), and thus take N T =1. Then the heattransfer surface area becomes

L Ltube s N N DL N A 1257.0m)m)(401.0()1( =×== π π

Then the log mean temperature difference, and the expression for the rate of heat transfer become

C51.45)]6590/()1590ln[(

)6590()1590(

)]/()ln[(

)()(ln °=

−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

L L s N N T hAQ 220,97)C51.45()C)(0.1257W/m994,16( 2ln =°°⋅=∆=

The mass flow rate of water through a cross-section corresponding to N T =1 and the rate of heat transfer are

kg/s91.95m/s) )(0.8m0.03)(4kg/m1.999( 23 =×== V Am c ρ

W10004.2C)1565(J/kg.C)9kg/s)(41791.95()( 7×=°−=−= ie p T T C mQ

Substituting this result into the heat transfer expression above we find th e number of tube rows

206=→=×→∆= L L s N N T hAQ 220,97W10004.2 7ln

7-58

S L

S T

V=0.8 m/s

T i=15°C

T s=90°C

D

7/30/2019 Heat Chap07 052



7-70 Air is cooled by an evaporating refrigerator. The refrigeration capacity and the pressure drop acrossthe tube bank are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of refrigerant.


properties at the assumed mean temperature of -5°C (will be checked later) and 1 atm (Table A-15):

k = 0.02326 W/m-K ρ = 1.316 kg/m3

C p =1.006 kJ/kg-K Pr = 0.7375

µ = 1.705× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7408

Also, the density of air at the inlet temperature of 0 °C (for use in the mass flow rate calculation at the inlet)

is ρi = 1.292 kg/m3.

Analysis It is given that D = 0.008 m, S L = S T = 0.015 m, and V = 4 m/s.Then the maximum velocity and the Reynolds number based on the maximum velocity become

m/s571.8m/s)4(008.0015.0

015.0max =

−=

−= VV

DS

S

T

T

5294skg/m10705.1

m)m/s)(0.008571.8)(kg/m316.1(Re

5

3max =

⋅×=

µρ

=−

D D

V



61.53)7408.0/7375.0()7375.0()5294(27.0


25.036.063.0

25.036.063.0

==

= s D D

Since N L > 16. the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become

61.53 Nu Nu , == D N D F L

CW/m8.155m0.008

C)W/m02326.0(61.53 2, °⋅=°⋅

== D

k Nuh L N D

The total number of tubes is N = N L × N T = 30× 15 = 450. The heat transfer surface area and the mass flowrate of air (evaluated at the inlet) are

2m .5244m)m)(0.4008.0(300 =π=π= DL N A s

kg/s4651.0m)m)(0.4.015m/s)(15)(04)(kg/m292.1()( 3 ==ρ== LS N mm T T ii V

Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer (refrigeration capacity) become

C57.15C)J/kgkg/s)(1006(0.4651

C)W/m8.155)(m524.4(exp)020(20exp)(

22

°−=

°⋅°⋅−−−−−=

−−−= p

si s se

C m

h AT T T T

[ ]

C33.10)]57.1520/()020ln[(

)57.15(20)020(

)]/()ln[(

)()(ln °=

+−−−−−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W7285=°°⋅=∆= )C33.10)(mC)(4.524W/m8.155( 22lnT hAQ s

7-59

S L

S T

V=4 m/s

T i=0°C

T s=-20°C

D

7/30/2019 Heat Chap07 052



For this square in-line tube bank, the friction coefficient corresponding to Re D = 5294 and S L/ D = 1.5/0.8 =

1.875 is, from Fig. 7-27a, f = 0.27. Also, χ = 1 for the square arrangements. Then the pressure drop acrossthe tube bank becomes

Pa 391.6=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)571.8)(kg/m316.1()1)(27.0(30

2

V f N P L

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (0 -15.6)/2 = -7.8°C, which is fairly close

to the assumed value of -5°C. Therefore, there is no need to repeat calculations.

7-60

7/30/2019 Heat Chap07 052



7-71 Air is cooled by an evaporating refrigerator. The refrigeration capacity and the pressure drop acrossthe tube bank are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to thetemperature of refrigerant.


properties at the assumed mean temperature of -5°C (will be checked later) and 1 atm (Table A-15):

k = 0.02326 W/m-K ρ = 1.316 kg/m3

C p =1.006 kJ/kg-K Pr = 0.7375

µ = 1.705× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7408

Also, the density of air at the inlet temperature of 0 °C (for use in the mass flow rate calculation at the inlet)

is ρi = 1.292 kg/m3.


m/s571.8m/s)4(008.0015.0

015.0max =

−=

−= VV

DS

S

T

T

5294skg/m10705.1

m)m/s)(0.008571.8)(kg/m316.1(Re

5

3max =

⋅×=

µρ

=−

D D

V



73.53)7408.0/7375.0()7375.0()5294()015.0/015.0(35.0


25.036.06.02.0

25.036.06.02.0

==

= s D LT D S S

Since N L > 16. the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become

73.53 Nu Nu , == D N D F L

CW/m2.156m0.008

C)W/m02326.0(73.53 2, °⋅=°⋅

== D

k Nuh L N D

The total number of tubes is N = N L × N T = 30× 15 = 450. The heat transfer surface area and the mass flowrate of air (evaluated at the inlet) are

2m .5244m)m)(0.4008.0(300 =π=π= DL N A s

kg/s4651.0m)m)(0.4.015m/s)(15)(04)(kg/m292.1()( 3 ==ρ== LS N mm T T ii V

Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer (refrigeration capacity) become

C58.15C)J/kgkg/s)(1006(0.4651

C)W/m2.156)(m524.4(exp)020(20exp)(

22

°−=

°⋅°⋅−−−−−=

−−−= p

si s se

C m

h AT T T T

[ ]

C32.10)]58.1520/()020ln[(

)58.15(20)020(

)]/()ln[(

)()(ln °=

+−−−−−−−−−

=−−−−−

=∆e si s

e si s

T T T T

T T T T T

W7294=°°⋅=∆= )C32.10)(mC)(4.524W/m2.156( 22lnT hAQ s

7-61

S L

S T

V=4 m/s

T i=0°C

T s=-20°C

D

7/30/2019 Heat Chap07 052



For this staggered arrangement tube bank, the friction coefficient corresponding to Re D = 5294 and S L/ D =

1.5/0.8 = 1.875 is, from Fig. 7-27b, f = 0.44. Also, χ = 1 for the square arrangements. Then the pressuredrop across the tube bank becomes

Pa 638.2=

⋅=

ρχ=∆

2

232max

m/skg1

N1

2

m/s)571.8)(kg/m316.1()1)(44.0(30

2

V f N P L

Discussion The arithmetic mean fluid temperature is (T i + T e)/2 = (0 -15.6)/2 = -7.8°C, which is fairly close

to the assumed value of -5°C. Therefore, there is no need to repeat calculations.

7-72 Air is heated by hot tubes in a tube bank. The average heat transfer coefficient is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is constant.


properties at the assumed mean temperature of 70°C and 1 atm (Table A-15):

k = 0.02881 W/m-K ρ = 1.028 kg/m3

C p =1.007 kJ/kg-K Pr = 0.7177

µ = 2.052× 10-5 kg/m-s Pr s = Pr @ Ts = 0.7041




m/s50.10m/s)7(02.006.0

06.0max =

−=

−= VV

DS

S

T

T

524,10skg/m10052.2

m)m/s)(0.0250.10)(kg/m028.1(Re

5

3max =

⋅×=

µρ

=−

D D

V



33.82)7041.0/7177.0()7177.0()524,10(27.0


25.036.063.0

25.036.063.0

==

= s D D

Since N L > 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become

33.82 Nu Nu , == D N D L

CW/m118.6 2 °⋅=°⋅

==m0.02

C)W/m02881.0(33.82,

D

k Nuh L N D

S L

S T

V=7 m/s

T i=40°C

T s=140°C

D

Heat Chap07 052

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