3ª Lista TCM II

9-6

9-19 A horizontal hot water pipe passes through a large room. The rate of heat loss from the pipe 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 local atmospheric pressure is 1 atm. 4 The temperature of the outer surface of the pipe is constant. Properties The properties of air at 1 atm and the film temperature of (Ts+T)/2 = (73+27)/2 = 50C are (Table A-15)

Air T = 27C

Pipe Ts = 73C = 0.8

L=10 m

D = 6 cm

1-

25

K 003096.0K)27350(

117228.0Pr

/sm 10798.1

C W/m.02735.0

=+===

==

fT

k

Analysis (a) The characteristic length in this case is the outer diameter of the pipe, Then, m. 06.0== DLc 5

225

3-12

2

3

10747.6)7228.0()/sm 10798.1(

)m 06.0)(K 2773)(K 003096.0)(m/s 81.9(Pr

)( ===

DTTgRa s

( )[ ] ( )[ ] 05.137228.0/559.01 )10747.6(387.06.0Pr/559.01 387.06.02

27/816/9

6/152

27/816/9

6/1=

++=

++= RaNu

2

2

m 885.1)m 10)(m 06.0(

C. W/m950.5)05.13(m 06.0

C W/m.02735.0

======

DLANu

Dkh

s

W516=== C)2773)(m 885.1)(C. W/m950.5()( 22TThAQ ss&(b) The radiation heat loss from the pipe is

[ ] W533=++= = 44428244

)K 27327()K 27373().K W/m1067.5)(m 885.1)(8.0(

)( surrssrad TTAQ &

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-13

9-25 Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical side surface of the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with 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 = (98+25)/2 = 61.5C are (Table A-15)

Vapor 2 kg/h

Water 100C

Pan Ts = 98C = 0.80

Air T = 25C

1-

25

K 00299.0K)2735.61(

117198.0Pr

/sm 10910.1

C W/m.02819.0

=+===

==

fT

k

Analysis (a) The characteristic length in this case is the height of the pan, Then m. 12.0== LLc

6225

3-12

2

3

10299.7)7198.0()/sm 10910.1(

)m 12.0)(K 2598)(K 00299.0)(m/s 81.9(Pr

)( ===

LTTgRa s

We can treat this vertical cylinder as a vertical plate since

4/14/164/1

35 thusand 0.25< 07443.0)7198.0/10299.7(

)12.0(3535Gr

LDGr

L ==

Therefore,

60.28

7198.0492.01

)10299.7(387.0825.0

Pr492.01

Ra387.0825.0

2

27/816/9

6/16

2

27/816/9

6/1=

+

+=

+

+=Nu

2

2

m 09425.0)m 12.0)(m 25.0(

C. W/m720.6)60.28(m 12.0

C W/m.02819.0

======

DLANu

Lkh

s

and

W46.2=== C)2598)(m 09425.0)(C. W/m720.6()( 22TThAQ ss&(b) The radiation heat loss from the pan is

[ ] W47.3=++= = 44428244

)K 27325()K 27398().K W/m1067.5)(m 09425.0)(80.0(

)( surrssrad TTAQ &

(c) The heat loss by the evaporation of water is W940kW 9404.0)kJ/kg 2257)(kg/s 3600/5.1( ==== fghmQ &&Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water then becomes

9.9%==+= 099.0940

3.472.46f


9-40

9-47 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 local atmospheric 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 the Rayleigh 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 170C for the evaluation of the properties 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

K 002699.0K)2735.97(

117116.0Pr

/sm 10279.2

C W/m.03077.0

=+===

==

fT

k

Analysis The characteristic length in this case is Lc = D = 0.08 m. Then,

6225

31-2

2

3

10694.2)7116.0()/sm 10279.2(

)m 08.0)(K 25170)(K 002699.0)(m/s 81.9(

Pr)(

==

=

DTTgRa s

Air T = 25C

Lamp 60 W = 0.9

D = 8 cm Light 6 W

( )[ ] ( )[ ] 42.207116.0/469.01 )10694.2(589.02Pr/469.01 589.02 9/416/94/16

9/416/9

4/1=

++=

++= RaNu

Then

222

2

m 02011.0m) 08.0(

C.W/m 854.7)42.20(m 08.0

CW/m. 03077.0

======

DANu

Dkh

s

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

])K 27325()273)[(.KW/m 1067.5)(m 02011.0)(9.0(

C)25)(m 02011.0)(C.W/m 854.7(W )6090.0(

)()(

444282

22

44

+++=

+=

s

s

surrssss

T

T

TTATThAQ &

Its solution is Ts = 169.4C

which is sufficiently close to the value assumed in the evaluation of properties and h. Therefore, there is no need to repeat calculations.


9-72

9-82 A vertical plate in water is considered. The forced motion velocity above which natural convection heat transfer from the plate is negligible is to be determined. Assumptions 1 Steady operating conditions exist. Properties The properties of water at the film temperature of (Ts+T)/2 = (60+25)/2 = 42.5C are (Table A-9)

1-3

26

K 10396.0

/sm 10630.0/

===

Analysis The characteristic length is the height of the plate Lc = L = 5 m. The Grashof and Reynolds numbers are

VVLV

LTTgGr s

626

13226

3-132

2

3

1094.7/sm 10630.0

)m 5(Re

1028.4)/sm 10630.0(

)m 5)(K 2560)(K 10396.0)(m/s 81.9()(

===

===

Water T = 25C

V

Plate, Ts = 60C

L = 5 m

and the forced motion velocity above which natural convection heat transfer from this plate is negligible is

m/s 2.61=== V

VGr 1.0

)1094.7(1028.41.0

Re 2613

2

9-83 Thin square plates coming out of the oven in a production facility are cooled by blowing ambient air horizontally parallel to their surfaces. The air velocity above which the natural convection effects on heat transfer are negligible is to be determined.

18C Hot plates

270C

2 m

2 m

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The atmospheric pressure at that location is 1 atm. Properties The properties of air at 1 atm and 1 atm and the film temperature of (Ts+T)/2 = (270+18)/2 = 144C are (Table A-15)

1-

25

K 002398.0K)273144(

11/sm 10791.2

=+===

fT

Analysis The characteristic length is the height of the plate Lc = L = 2 m. The Grashof and Reynolds numbers are

VVVL

LTTgGr s

425

10225

3-12

2

3

10166.7/sm 10791.2

)m 2(Re

1009.6)/sm 10791.2(

)m 2)(K 18270)(K 002398.0)(m/s 81.9()(

===

===

and the forced motion velocity above which natural convection heat transfer from this plate is negligible is

m/s 10.9=== V

VGr 1.0

)10166.7(1009.61.0

Re 2410

2


10-5

10-15 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100C by a mechanically polished stainless steel heating element. The maximum heat flux in the nucleate boiling regime and the surface temperature of the heater for that case are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible.

P = 1 atm

qmax Ts = ? Water, 100C

Heating element

Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9)

75.1PrN/m 0589.0

kg/m 60.0

kg/m 9.9573

3

====

l

v

l

CJ/kg 4217m/skg 10282.0

J/kg 1022573

3

==

=

pl

l

fg

c

h

Also, 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note that we expressed the properties in units specified under Eqs. 10-2 and 10-3 in connection with their definitions in order to avoid unit manipulations. For a large horizontal heating element, C

=sfC

cr = 0.12 (Table 10-4). (It can be shown that L* = 5.99 > 1.2 and thus the restriction in Table 10-4 is satisfied). Analysis The maximum or critical heat flux is determined from

2 W/m1,017,000=

==

4/123

4/12max

)]60.09.957()6.0(8.90589.0)[102257(12.0

)]([ vlvfgcr ghCq &

The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into the Rohsenow relation together with other properties gives

3sat,

2/1

nucleatePr

)()(

=nlfgsf

slpvlfgl

hC

TTcghq

&

3

3

1/233

75.1)102257(0130.0)100(4217

0589.00.60)-9.8(957.9

)10)(225710282.0(000,017,1

= sT

It gives C119.3=sTTherefore, the temperature of the heater surface will be only 19.3C above the boiling temperature of water when burnout occurs.


10-4

10-14 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100C in a mechanically polished stainless steel pan whose inner surface temperature is maintained at Ts = 110C. The rate of heat transfer to the water and the rate of evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9)

75.1PrN/m 0589.0

kg/m 60.0

kg/m 9.9573

3

====

l

v

l

Heating

P = 1 atm

110C

100C Water

CJ/kg 4217m/skg 10282.0

J/kg 1022573

3

==

=

pl

l

fg

c

h

Also, 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note that we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit manipulations.

=sfC

Analysis The excess temperature in this case is C10100110sat === TTT s which is relatively low (less than 30C). Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be

2

3

3

1/233

3sat,

2/1

nucleate

W/m700,140

75.1)102257(0130.0)100110(4217

0589.00.60)-9.8(957.9)10)(225710282.0(

Pr

)()(

=

=

=

nlfgsf

slpvlfgl

hC

TTcghq

&

The surface area of the bottom of the pan is

222 m 07069.04/m) 30.0(4/ === DAsThen the rate of heat transfer during nucleate boiling becomes

W9945=== ) W/m700,140)(m 07069.0( 22nucleateboiling qAQ s &&(b) The rate of evaporation of water is determined from

kg/s 0.00441=== J/kg 102257J/s 99453

boilingnevaporatio

fghQ

m&

&

That is, water in the pan will boil at a rate of 4.4 grams per second.


10-9

10-19 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C in a mechanically polished AISI 304 stainless steel pan placed on top of a 3-kW electric burner. Only 60% of the heat (1.8 kW) generated is transferred to the water. The inner surface temperature of the pan and the temperature difference across the bottom of the pan are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is nucleate boiling (this assumption will be checked later). 4 Heat transfer through the bottom of the pan is one-dimensional. Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9)

75.1PrN/m 0589.0

kg/m 60.0

kg/m 9.9573

3

====

l

v

l

Electric burner, 3 kW

P = 1 atm

100C Water

CJ/kg 4217m/skg 10282.0

J/kg 1022573

3

==

=

pl

l

fg

c

h

Also, ksteel = 14.9 W/mC (Table A-3), =sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3 ). Note that we expressed the properties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit manipulations. Analysis The rate of heat transfer to the water and the heat flux are

22

222

W/m25.46=)m 69 W)/(0.0701800(/

m 07069.04/m) 30.0(4/

W1800=kW 8.1kW 360.0

=====

==

s

s

AQq

DA

Q

&&

&

Then temperature difference across the bottom of the pan is determined directly from the steady one-dimensional heat conduction relation to be

C3.10 ====

C W/m9.14m) )(0.006 W/m460,25(

2

steelsteel k

LqTLTkq

&&

The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to determine the surface temperature when the heat flux is given. Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be

3sat,

2/1

nucleatePr

)()(

=nlfgsf

slpvlfgl

hC

TTcghq

&

3

3

1/233

75.1)102257(0130.0)100(4217

0589.00.60)9.81(957.9

)10)(225710282.0(460,25

= sT

It gives C105.7=sTwhich is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is valid.


10-11

10-21 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C by a stainless steel heating element. The surface temperature of the heating element and its power rating are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the coffee maker are negligible. 3 The boiling regime is nucleate boiling (this assumption will be checked later). Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9)

75.1PrN/m 0589.0

kg/m 60.0

kg/m 9.9573

3

====

l

v

l

P = 1 atm

1 L Water, 100C

Coffeemaker

CJ/kg 4217m/skg 10282.0

J/kg 1022573

3

==

=

pl

l

fg

c

h

Also, 0.0130 and n = 1.0 for the boiling of water on a stainless steel surface (Table 10-3 ). Note that we expressed the properties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit manipulations.

=sfC

Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18C is nearly 1 kg. The rate of energy transfer needed to evaporate half of this water in 25 min and the heat flux are

222

2

W/m29,940=kW/m 29.94=)m 13kW)/(0.025 7523.0(/

m 02513.0m) m)(0.2 04.0(

kW 7523.0s) 60(25kJ/kg) kg)(2257 5.0(

=====

=====

s

s

fgfg

AQq

DLA

tmh

QmhtQQ

&&

&&

The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to determine the surface temperature when the heat flux is given. Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be

3sat,

2/1

nucleatePr

)()(

=nlfgsf

slpvlfgl

hC

TTcghq

&

3

3

1/233

75.1)102257(0130.0)100(4217

0589.00.60)9.81(957.9

)10)(225710282.0(940,29

= sT

It gives C106.0=sTwhich is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is valid. The specific heat of water at the average temperature of (14+100)/2 = 57C is cp = 4.184 kJ/kgC. Then the time it takes for the entire water to be heated from 14C to 100C is determined to be

min 7.97=s 478kJ/s 0.7523

C14)C)(100kJ/kg kg)(4.184 1( =====Q

TmctTmctQQ pp &

&


10-21

10-28 Water is boiled at a temperature of Tsat = 150C by hot gases flowing through a mechanically polished stainless steel pipe submerged in water whose outer surface temperature is maintained at Ts = 165C. The rate of heat transfer to the water, the rate of evaporation, the ratio of critical heat flux to current heat flux, and the pipe surface temperature at critical heat flux conditions are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is nucleate boiling since

which is in the nucleate boiling range of 5 to 30C for water.

C15150165sat === TTT s

Water, 150C

Boiler

Hot gases

Vent

Ts,pipe = 165C

Properties The properties of water at the saturation temperature of 150C are (Tables 10-1 and A-9)

16.1PrN/m 0488.0

kg/m 55.2

kg/m 6.9163

3

====

l

v

l

CJ/kg 4311m/skg 10183.0

J/kg 1021143

3

==

=

pl

l

fg

c

h

Also, 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note that we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit manipulations.

=sfC

Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be

2

3

3

1/233

3sat,

2/1

nucleate

W/m000,383,1

16.1)102114(0130.0)150165(4311

0488.0)55.29.8(916.6

)10)(211410183.0(

Pr

)()(

=

=

=

nlfgsf

slpvlfgl

hC

TTcghq

&

The heat transfer surface area is

2m 854.7m) m)(50 05.0( === DLAsThen the rate of heat transfer during nucleate boiling becomes

W10,865,000=== ) W/m000,383,1)(m 854.7( 22nucleateboiling qAQ s &&(b) The rate of evaporation of water is determined from

kg/s 5.139===kJ/kg 2114

kJ/s 865,10boilingnevaporatio

fghQ

m&

&

(c) For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be

cylinder) large thusand 1.2 > * (since 12.0

1.2 >7.100488.0

)55.26.916(8.9)025.0(

)(*

2/12/1

LC

gLL

cr

vl

==

=

=

Then the maximum or critical heat flux is determined from


10-22

2

4/123

4/12max

W/m1,852,000

)]55.26.916()55.2(8.90488.0)[102114(12.0

)]([

==

= vlvfgcr ghCq &

Therefore, 1.34==000,383,1000,852,1

current

max

qq&&

(d) The surface temperature of the pipe at the burnout point is determined from Rohsenow relation at the critical heat flux value to be

C166.5=

=

=

crs

crs

nlfgsf

crslpvlfgl

T

T

hC

TTcghq

,

3

3,

1/233

3sat,,

2/1

crnucleate,

16.1)102114(0130.0

)150(43110488.0

)55.29.8(916.6)10)(211410183.0(000,852,1

Pr

)()(

&


10-35

10-49 Saturated steam at atmospheric pressure thus at a saturation temperature of Tsat = 100C condenses on a vertical plate which is maintained at 90C by circulating cooling water through the other side. The rate of heat transfer to the plate and the rate of condensation of steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv

10-37

10-51 Saturated steam condenses outside of vertical tube. The rate of heat transfer to the coolant, the rate of condensation and the thickness of the condensate layer at the bottom are to be determined. Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 3 The tube can be treated as a vertical plate. 4 The condensate flow is wavy-laminar over the entire tube (this assumption will be verified). 5 Nusselts analysis can be used to determine the thickness of the condensate film layer. 6 The density of vapor is much smaller than the density of liquid, lv

10-46

10-59 Saturated steam at a pressure of 4.25 kPa and thus at a saturation temperature of Tsat = 30C (Table A-9) condenses on the outer surfaces of 100 horizontal tubes arranged in a 1010 square array maintained at 20C by circulating cooling water. The rate of heat transfer to the cooling water and the rate of condensation of steam on the tubes are to be determined. Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. Properties The properties of water at the saturation temperature of 30C are hfg = 2431103 J/kg and v = 0.03 kg/m3. The properties of liquid water at the film temperature of =+= 2/)( sat sf TTT (30 + 20)/2 = 25C are (Table A-9),

C W/m607.0

CJ/kg 4180/sm10894.0/

skg/m10891.0

kg/m 0.997

26

3

3

==

===

=

l

pl

lll

l

l

k

c

Analysis (a) The modified latent heat of vaporization is

n = 100 tubes 20C

L = 8 m

P = 4.25 kPa

Cooling water

Saturated steam

J/kg 102,459=C0)2C(30J/kg 41800.68+J/kg 102431

)(68.033

sat*

=+= splfgfg TTchh

The heat transfer coefficient for condensation on a single horizontal tube is

C. W/m8674

m) C(0.03)2030(s)kg/m 10891.0()C W/m607.0)(J/kg 102459)(kg/m 03.0997)(kg/m 997)(m/s 8.9(

729.0

)()(

729.0

2

4/1

3

33332

4/1

sat

3*

horizontal

=

=

==

DTTkhg

hhsl

lfgvll

Then the average heat transfer coefficient for a 10-pipe high vertical tier becomes

C W/m4878C) W/m8674(10

11 224/1 tube1 horiz,4/1 tubesN horiz,

=== hN

h

The surface area for all 100 tubes is

2total m 75.40= m) m)(8 03.0(100 == DLNAsThen the rate of heat transfer during this condensation process becomes

kW 3678==== W3,678,000C)2030)(m 40.75)(C. W/m4878()( 22sat ss TThAQ&(b) The rate of condensation of steam is determined from

kg/s 1.496=== J/kg 102459J/s 000,678,3

3*oncondensatifgh

Qm&

&


10-60

10-82 Steam at a saturation temperature of Tsat = 40C condenses on the outside of a thin horizontal tube. Heat is transferred to the cooling water that enters the tube at 25C and exits at 35C. The rate of condensation of steam, the average overall heat transfer coefficient, and the tube length are to be determined. Assumptions 1 Steady operating conditions exist. 2 The tube can be taken to be isothermal at the bulk mean fluid temperature in the evaluation of the condensation heat transfer coefficient. 3 Liquid flow through the tube is fully developed. 4 The thickness and the thermal resistance of the tube is negligible. Properties The properties of water at the saturation temperature of 40C are hfg = 2407103 J/kg and v = 0.05 kg/m3. The properties of liquid water at the film temperature of

(50+20)/2 = 35C and at the bulk fluid temperature of

=+= 2/)( sat sf TTT=+= 2/)( outin TTTb (25

+ 35)/2 = 30C are (Table A-9),

C W/m623.0

CJ/kg 4178skg/m10720.0

kg/m 0.9943

3

==

==

l

pl

l

l

k

c

:C35At

5.42=PrC W/m615.0

CJ/kg 4178/sm10801.0/

kg/m 0.99626

3

==

===

l

pl

lll

l

k

c

:C30At

Steam 40C

Condensate

25C

Cooling water

35C

Analysis The mass flow rate of water and the rate of heat transfer to the water are

W58,820=C25)C)(35J/kg kg/s)(4178 408.1()(

kg/s 408.1]4/m) 03.0(m/s)[ )(2kg/m 996( 23water==

===inoutp

c

TTcmQ

VAm&&

&

The modified latent heat of vaporization is

J/kg 102435=C0)3C(40J/kg 41780.68+J/kg 102407

)(68.033

sat*

=+= splfgfg TTchh

The heat transfer coefficient for condensation on a single horizontal tube is

C W/m9292

m) C(0.03)3040(s)kg/m 10720.0()C W/m623.0)(J/kg 102435)(kg/m 05.0994)(kg/m 994)(m/s 8.9(

729.0

)()(

729.0

2

4/1

3

33332

4/1

sat

3*

horizontal

=

=

==

DTTkhg

hhsl

lfgvllo

The average heat transfer coefficient for flow inside the tube is determined as follows:

C W/m7357m 0.03

359C) W/m615.0(Nu359)42.5()906,74(023.0PrRe023.0Nu

906,74100.801

m) m/s)(0.03 2(Re

2

4.08.04.08.0

6-avg

======

===

Dkh

DV

i

Noting that the thermal resistance of the tube is negligible, the overall heat transfer coefficient becomes

C. W/m4106 2 =+=+= 9292/17357/11

/1/11

oi hhU


10-73

10-94 Saturated steam at a saturation temperature of Tsat = 95C (Table A-9) condenses on a canned drink at 2C in a dropwise manner. The heat transfer coefficient for this dropwise condensation is to be determined.

Steam 95C

Drink 2C

Assumptions The heat transfer coefficient relation for dropwise condensation that was developed for copper surfaces is also applicable for aluminum surfaces. Analysis Noting that the saturation temperature is less than 100C, the heat transfer coefficient for dropwise condensation can be determined from Griffiths relation to be C W/m245,284 2 =+=+== 952044104,512044104,51 satdropwise Thh 10-95 Water is boiled at 1 atm pressure and thus at a saturation temperature of Tsat = 100C by a nickel electric heater whose diameter is 2 mm. The highest temperature at which this heater can operate without burnout is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water are negligible. Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9)

75.1PrN/m 0589.0

kg/m 60.0

kg/m 9.9573

3

====

l

v

l

CJ/kg 4217m/skg 10282.0

J/kg 1022573

3

==

=

pl

l

fg

c

h

Ts= ?

Heating wire

1 atm

Water

Also, 0.0060 and n = 1.0 for the boiling of water on a nickel surface (Table 10-3).

=sfC

Analysis The maximum rate of heat transfer without the burnout is simply the critical heat flux. For a horizontal heating wire, the coefficient Ccr is determined from Table 10-4 to be

151.0)399.0(12.0*12.0

1.2 < 399.00589.0

60.09.957(8.9)001.0()(

*

25.025.0

2/12/1

===

=

=

=LC

gLL

cr

vl

Then the maximum or critical heat flux is determined from

2

4/1234/12max

W/m1,280,000

)]60.09.957()6.0(8.90589.0)[102257(151.0)]([

=== vlvfgcr ghCq &

Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into Rohsenow relation together with other properties gives

3sat,

2/1

nucleatePr

)()(

=nlfgsf

slpvlfgl

hC

TTcghq

&

3

3

1/233

75.1)102257(0060.0)100(4217

0589.00.60)-9.8(957.9)10)(225710282.0(000,280,1

= sT

It gives the maximum temperature to be: C109.6=sT


10-61

The logarithmic mean temperature difference is:

C10.9)5/15ln(

515)/ln(ln

===

oi

ei

TTTT

T

The tube length is determined from

m 16.7==== C)10.9)(m 03.0()C W/m4106( W820,58

)(

2ln

ln TDhQLThAQ s&&

Note that the flow is turbulent, and thus the entry length in this case is 10D = 0.3 m is much shorter than the total tube length. This verifies our assumption of fully developed flow.


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3ª Lista TCM II