Heat Chap10 021

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Chapter 10Boiling and Condensation

10-21 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100Cby a horizontal nickel plated copper heating element. The maximum (critical) heat flux and thetemperature jump of the wire when the operating point jumps from nucleate boiling to film boiling regimeare to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible.

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

l

v

l

=

=

=

=

957 9

060

00589

175

.

.

.

.

kg /m

kg /m

N / m

Pr

3

3

hfg

l

pl

=

= =

2257 10

0 282 10

4217

3

3

J / k g

kg m /s

C J / kg C

.

Also, Csf = 0.0060 and n = 1.0 for the boiling of water on a nickel plated surface (Table 10-3 ). Notethat 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.The vaporproperties at the anticipated filmtemperature ofTf= (Ts+Tsat)/2 of 1000C (will be checked) (Table A-16)

skg/m10762.4

CW/m1362.0

CJ/kg2471

kg/m1725.0

5

3

=

=

==

v

v

pv

v

k

C

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

136.0)60.0(12.0*12.0

1.2

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)(4

3)(

)(

)](4.0)[(62.0

4

3 4sat

4sat

4/1

sat

3

radfilmtotal TTTTTTD

TTChgkqqq ss

sv

satspvfgvlvv +

+=+=

Substituting,

[ ]44428

4/1

5

33

)273100()273()KW/m1067.5)(5.0(

)100()100)(003.0)(10762.4(

)]100(24714.0102257)[1725.09.957)(1723.0()1362.0(81.962.0000,153,1

+++

+=

s

s

s

s

T

TT

T

Solving for the surface temperature gives Ts = 1871C. Therefore, the temperature jump of the wire whenthe operating point jumps from nucleate boiling to film boiling is

Temperature jump: C1762=== 1091871crit,films, sTTT

Note that the film temperature is (1871+100)/2=985C, which is close enough to the assumed value of1000C for the evaluation of vapor paroperties.

10-15

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10-22"!PROBLEM 10-22"

"GIVEN"L=0.3 "[m]"D=0.003 "[m]""epsilon=0.5 parameter to be varied"P=101.3 "[kPa], parameter to be varied"

"PROPERTIES"Fluid$='steam_NBS'

T_sat=temperature(Fluid$, P=P, x=1)rho_l=density(Fluid$, T=T_sat, x=0)rho_v=density(Fluid$, T=T_sat, x=1)sigma=SurfaceTension(Fluid$, T=T_sat)mu_l=Viscosity(Fluid$,T=T_sat, x=0)Pr_l=Prandtl(Fluid$, T=T_sat, P=P)C_l=CP(Fluid$, T=T_sat, x=0)*Convert(kJ/kg-C, J/kg-C)h_f=enthalpy(Fluid$, T=T_sat, x=0)h_g=enthalpy(Fluid$, T=T_sat, x=1)h_fg=(h_g-h_f)*Convert(kJ/kg, J/kg)

C_sf=0.0060 "from Table 8-3 of the text"n=1 "from Table 8-3 of the text"

T_vapor=1000-273 "[C], assumed vapor temperature in the film boiling region"rho_v_f=density(Fluid$, T=T_vapor, P=P) "f stands for film"C_v_f=CP(Fluid$, T=T_vapor, P=P)*Convert(kJ/kg-C, J/kg-C)k_v_f=Conductivity(Fluid$, T=T_vapor, P=P)mu_v_f=Viscosity(Fluid$,T=T_vapor, P=P)

g=9.8 "[m/s^2], gravitational acceleraton"sigma_rad=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant"

"ANALYSIS"

"(a)""C_cr is to be determined from Table 8-4 of the text"C_cr=0.12*L_star^(-0.25)L_star=D/2*((g*(rho_l-rho_v))/sigma)^0.5q_dot_max=C_cr*h_fg*(sigma*g*rho_v^2*(rho_l-rho_v))^0.25q_dot_nucleate=q_dot_maxq_dot_nucleate=mu_l*h_fg*(((g*(rho_l-rho_v))/sigma)^0.5)*((C_l*(T_s_crit-

T_sat))/(C_sf*h_fg*Pr_l^n))^3"(b)"q_dot_total=q_dot_film+3/4*q_dot_rad "Heat transfer in the film boiling region"q_dot_total=q_dot_nucleateq_dot_film=0.62*((g*k_v_f^3*rho_v_f*(rho_l-rho_v_f)*(h_fg+0.4*C_v_f*(T_s_film-

T_sat)))/(mu_v_f*D*(T_s_film-T_sat)))^0.25*(T_s_film-T_sat)

q_dot_rad=epsilon*sigma_rad*((T_s_film+273)^4-(T_sat+273)^4)DELTAT=T_s_film-T_s_crit

10-16

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P [kPa] qmax [kW/m2] T [C]

70 994227 1865

71.65 1003642 1870

73.29 1012919 1876

74.94 1022063 1881

76.59 1031078 1886

78.24 1039970 1891

79.88 1048741 189681.53 1057396 1900

83.18 1065939 1905

84.83 1074373 1909

86.47 1082702 1914

88.12 1090928 1918

89.77 1099055 1923

91.42 1107085 192793.06 1115022 1931

94.71 1122867 1935

96.36 1130624 1939

98.01 1138294 1943

99.65 1145883 1947

101.3 1153386 1951

qmax [kW/m2] T [C]

0.1 1153386 2800

0.15 1153386 25740.2 1153386 2418

0.25 1153386 2299

0.3 1153386 2205

0.35 1153386 2126

0.4 1153386 2059

0.45 1153386 2002

0.5 1153386 19510.55 1153386 19050.6 1153386 1864

0.65 1153386 1827

0.7 1153386 1793

0.75 1153386 1761

0.8 1153386 1732

0.85 1153386 1705

0.9 1153386 1680

0.95 1153386 1657

1 1153386 1634

10-17

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70 75 80 85 90 95 100 105

9.75x105

1.01x106

1.06x106

1.10x10

6

1.14x106

1.18x106

1860

1880

1900

1920

1940

1960

P [kPa]

qmax

[W/m

2]

Red-Heat

Blue-Temp. Dif.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.0 x106

1.1 x106

1.1 x106

1.1 x106

1.2 x106

1.3 x106

1.3 x106

1600

1800

2000

2200

2400

2600

2800

qmax

[W/m

2]

Heat

Temp. Dif.

10-18

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10-23 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat= 100C in a teflon-pitted stainless steel pan placed on an electric burner. The water level drops by 10 cmin 30 min during boiling. The inner surface temperature of the pan is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the pan are negligible. 3 The boilingregime is nucleate boiling (this assumption will be checked later).

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

l

v

l

=

=

=

=

9579

060

00589

175

.

.

.

.

kg /m

kg /m

N / m

Pr

3

3

hfg

l

pl

=

=

=

2257 10

0 282 10

4217

3

3

J / k g

kg m /s

C J / kg C

.

Also, Csf = 0.0058 and n = 1.0 for the boiling of water on a teflon-pitted stainless steel surface (Table 10-3). Note that we expressed the properties in units specified under Eq. 10-2 connection with theirdefinitions in order to avoid unit manipulations.

AnalysisThe rate of heat transfer to the water and the heat flux are

22

222

evap

3evap

evap

W/m2,402,000=)m2W)/(0.0314470,75(/

m03142.04/m)20.0(4/

kW47.75kJ/kg)7kg/s)(22503344.0(

kg/s03344.0s6030

m)0.10m0.2)(kg/m9.957(

==

===

===

=

=

=

=

s

s

fg

AQq

DA

hmQ

t

V

t

mm

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

be

3

sat,2/1

nucleatePr

)()(

=

nlfgsf

slpvlfgl

hC

TTCghq

3

3

1/233

75.1)102257(0058.0

)100(4217

0589.0

0.60)9.8(957.9)10)(225710282.0(000,402,2

= s

T

It gives

Ts = 111.5 C

which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleateboiling assumption is valid.

10-19

Heating

P= 1 atm

Ts

100 CWater

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10-24 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat= 100C in a polished copper pan placed on an electric burner. The water level drops by 10 cm in 30 minduring boiling. The inner surface temperature of the pan is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the pan are negligible. 3 The boilingregime is nucleate boiling (this assumption will be checked later).

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

l

v

l

=

=

=

=

9579

060

00589

175

.

.

.

.

kg /m

kg /m

N / m

Pr

3

3

hfg

l

pl

=

=

=

2257 10

0 282 10

4217

3

3

J / k g

kg m /s

C J / kg C

.

Also, Csf = 0.0130 and n = 1.0 for the boiling of water on a copper surface (Table 10-3). Note that weexpressed the properties in units specified under Eq. 10-2 connection with their definitions in order toavoid unit manipulations.

AnalysisThe rate of heat transfer to the water and the heat flux are

22

222

evap

3evap

evap

W/m2,402,000=)m2W)/(0.0314470,75(/

m03142.04/m)20.0(4/

kW47.75kJ/kg)7kg/s)(22503344.0(

kg/s03344.0s6030

m)0.10m0.2)(kg/m9.957(

==

===

===

=

=

=

=

s

s

fg

AQq

DA

hmQ

t

V

t

mm

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

be

3

sat,2/1

nucleatePr

)()(

=

nlfgsf

slpvlfgl

hC

TTCghq

3

3

1/233

75.1)102257(0130.0

)100(4217

0589.0

0.60)9.8(957.9)10)(225710282.0(000,402,2

= s

T

It gives

Ts = 125.7 C

which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleateboiling assumption is valid.

10-20

Heating

P= 1 atm

Ts

100 CWater

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10-25 Water is boiled at a temperature of Tsat = 150C by hot gases flowing through a mechanicallypolished 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 tocurrent 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 T T Ts= = = sat C165 150 15 which is in the nucleate boilingrange of 5 to 30C for water.

Properties The properties of water at the saturation temperature of150C are (Tables 10-1 and A-9)

l

v

l

=

=

=

=

9166

2 55

00488

116

.

.

.

.

k g / m

k g / m

N / m

Pr

3

3

hfg

l

pl

=

=

=

2114 10

0183 10

4311

3

3

J / k g

kg m /s

C J / kg C

.

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

manipulations.

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

2

3

3

1/2

33

3

sat,

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

2m854.7m)m)(5005.0( === DLAs

Then the rate of heat transfer during nucleate boiling becomes

W10,865,000=== )W/m000,383,1)(m854.7( 22nucleateboiling qAQ s

(b) The rate of evaporation of water is determined from

,m

Q

hfgevaporation

boiling k J / s

2114 kJ / kg= = =

10 8655.139 kg/s

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

cylinder)largethusand1.2>*(since12.0

0.12>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

[ ( )]

. ( )[ . . ( . ) ( . . )]

max/

/

q C h g cr fg v l v=

=

=

2 1 4

3 2 1 4012 2114 10 0 0488 9 8 2 55 916 6 2 55

1,852,000 W / m2

10-21

Water, 150C

Boiler

Hotgases

Vent

Ts,pipe

= 165C

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Therefore,

, ,

, ,

maxq

qcurrent= =

1852 000

1 383 0001.34

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

C166.5=

=

=

crs

crs

nlfgsf

crslpvlfgl

T

T

hC

TTCghq

,

3

3

,1/2

33

3

sat,,

2/1

crnucleate,

16.1)102114(0130.0

)150(4311

0488.0

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

Pr

)()(

10-22

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10-26 Water is boiled at a temperature of Tsat = 160C by hot gases flowing through a mechanicallypolished 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 tocurrent heat flux, and the pipe surface temperature at critical heat flux conditions are to be determined.


boiling regime is nucleate boiling since T T Ts= = = sat C165 160 5 which is in the nucleate boilingrange of 5 to 30C for water.

Properties The properties of water at the saturation temperature of160C are (Tables 10-1 and A-9)

l

v

l

=

=

=

=

907 4

326

00466

109

.

.

.

.

kg /m

kg /m

N / m

Pr

3

3

hfg

l

pl

=

=

=

2083 10

0170 10

4340

3

3

J / k g

kg m /s

C J / kg C

.

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

manipulations.


2

3

3

1/233

3

sat,2/1

nucleate

W/m359,61

09.1)102083(0130.0

)160165(4340

0466.0

)26.39.8(907.4)10)(208310170.0(

Pr

)()(

=

=

=

nlfgsf

slpvlfgl

hC

TTCghq

The heat transfer surface area is

2

m854.7m)m)(5005.0( === DLAsThen the rate of heat transfer during nucleate boiling becomes

W481,900=== )W/m359,61)(m854.7( 22nucleateboiling qAQ s

(b) The rate of evaporation of water is determined from

.m

Q

hfgevaporation

boiling k J / s

2083 kJ / kg= = =

48190.231 kg/s

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

cylinder)largeand thus1.2>*(since12.0

0.12>9.10

0466.0

)26.34.907(8.9)025.0(

)(*

2/12/1

LC

gLL

cr

vl

=

=

=

=

Then the maximum or critical heat flux is determined from

[ ( )]

. ( )[ . . ( . ) ( . . )]

max/

/

q C h g cr fg v l v=

=

=

2 1 4

3 2 1 4012 2083 10 0 0466 9 8 326 907 4 326

2,034,000 W / m2

10-23

Water, 150C

Boiler

Hot

gases

Vent

Ts,pipe

= 160C

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Therefore,

, ,

,

maxq

qcurrent= =

2 034 000

6135933.2

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

C176.1=

=

=

crs

crs

n

lfgsf

crslpvlfgl

T

T

hC

TTCghq

,

3

3

,1/2

33

3

sat,,2/1

crnucleate,

09.1)102083(0130.0

)160(4340

0466.0

)26.39.8(907.4)10)(208310170.0(000,034,2

Pr

)()(

10-24

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10-27E Water is boiled at a temperature ofTsat = 250F by a nickel-plated heating element whose surfacetemperature is maintained at Ts = 280F. The boiling heat transfer coefficient, the electric powerconsumed, and the rate of evaporation of water are to be determined.


boiling regime is nucleate boiling since T T Ts= = = sat F280 250 30 which is in the nucleate boilingrange of 9 to 55F for water.

Properties The properties of water at the saturation temperature of 250F are (Tables 10-1 and A-9E)

l

v

l

=

=

= ==

5882

00723

0 003755 01208

143

.

.

. .

.

lbm/ ft

lbm / ft

lbf / ft lbm / s

Pr

3

3

2

hfg

l

pl

=

= =

946

0556

1015

Btu / lbm

lbm / h ft

C Btu / lbm F

.

.

Also, g = 32.2 ft/s2 and Csf = 0.0060 and n = 1.0 for the boiling of water on a nickel plated surface(Table 10-3). Note that we expressed the properties in units that will cancel each other in boiling heattransfer relations.


2

31/2

3

sat,2/1

nucleate

ftBtu/h221,475,3

43.1)946(0060.0

)250280(015.1

1208.0

)0723.032.2(58.82))(946556.0(

Pr

)()(

=

=

=

nlfgsf

slpvlfgl

hC

TTCghq

Then the convection heat transfer coefficient becomes

( )

( )q h T T h

q

T Ts

s

= =

=

= sat

sat

2

3,471,670 Btu / h ft

F280 250115,840 Btu / h ft F2

(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from

Btu/h)3412=kW1(since=

Btu/h811,909)ftBtu/h221ft)(3,475,2ft12/5.0()(2

kW266.6

===== qDLAqQW se

(c) Finally, the rate of evaporation of water is determined from

lbm/h961.7===Btu/lbm946

Btu/h811,909boilingnevaporatio

fgh

Qm

10-25

Ts=280 FWater

250 F

Heating element

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10-28E Water is boiled at a temperature of Tsat = 250F by a platinum-plated heating element whosesurface temperature is maintained at Ts = 280F. The boiling heat transfer coefficient, the electric powerconsumed, and the rate of evaporation of water are to be determined.


boiling regime is nucleate boiling since T T Ts= = = sat F280 250 30 which is in the nucleate boilingrange of 9 to 55F for water.

Properties The properties of water at the saturation temperature of 250F are (Tables 10-1 and A-9E)

l

v

l

=

=

= ==

5882

00723

0 003755 01208

143

.

.

. .

.

lbm/ ft

lbm / ft

lbf / ft lbm / s

Pr

3

3

2

hfg

l

pl

=

= =

946

0556

1015

Btu / lbm

lbm / h ft

C Btu / lbm F

.

.

Also, g= 32.2 ft/s2 and Csf = 0.0130 and n = 1.0 for the boiling of water on a platinum plated surface(Table 10-3). Note that we expressed the properties in units that will cancel each other in boiling heattransfer relations.


2

3

3

1/2

3

sat,2/1

nucleate

ftBtu/h670,341

43.1)101208.0(0130.0

)250280(015.1

1208.0

)0723.032.2(58.82))(946556.0(

Pr

)()(

=

=

=

nlfgsf

slpvlfgl

hC

TTCghq

Then the convection heat transfer coefficient becomes

( )

( )q h T T h

q

T Ts

s

= =

=

= sat

sat

2

341,670 Btu / h ft

F280 25011,390 Btu / h ft F2

(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from

Btu/h)3412=kW1(since=

Btu/h450,89)ftBtu/h0ft)(341,672ft12/5.0()(2

kW26.2

===== qDLAqQW se

(c) Finally, the rate of evaporation of water is determined from

lbm/h94.6===Btu/lbm946

Btu/h450,89boilingnevaporatio

fgh

Qm

10-26

Ts=280 FWater

250 F

Heating element

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10-29E "!PROBLEM 10-29E"

"GIVEN"T_sat=250 "[F]"L=2 "[ft]"D=0.5/12 "[ft]""T_s=280 [F], parameter to be varied"

"PROPERTIES"Fluid$='steam_NBS'P_sat=pressure(Fluid$, T=T_sat, x=1)rho_l=density(Fluid$, T=T_sat, x=0)rho_v=density(Fluid$, T=T_sat, x=1)sigma=SurfaceTension(Fluid$, T=T_sat)*Convert(lbf/ft, lbm/s^2)mu_l=Viscosity(Fluid$,T=T_sat, x=0)Pr_l=Prandtl(Fluid$, T=T_sat, P=P_sat+1) "P=P_sat+1 is used to get liquid state"C_l=CP(Fluid$, T=T_sat, x=0)h_f=enthalpy(Fluid$, T=T_sat, x=0)h_g=enthalpy(Fluid$, T=T_sat, x=1)h_fg=h_g-h_f

C_sf=0.0060 "from Table 8-3 of the text"n=1 "from Table 8-3 of the text"g=32.2 "[ft/s^2], gravitational acceleraton"

"ANALYSIS""(a)"q_dot_nucleate=mu_l*h_fg*(((g*(rho_l-rho_v))/sigma)^0.5)*((C_l*(T_s-T_sat))/(C_sf*h_fg*Pr_l^n))^3q_dot_nucleate=h*(T_s-T_sat)"(b)"W_dot_e=q_dot_nucleate*A*Convert(Btu/h, kW)A=pi*D*L"(c)"

m_dot_evap=Q_dot_boiling/h_fgQ_dot_boiling=W_dot_e*Convert(kW, Btu/h)

10-27

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Ts [F] h [Btu/h.ft2.F] We [kW] mevap [lbm/h]

260 12908 9.903 35.74

262 18587 17.11 61.76

264 25299 27.18 98.07

266 33043 40.56 146.4

268 41821 57.76 208.4

270 51630 79.23 285.9

272 62473 105.5 380.5274 74348 136.9 494.1

276 87255 174.1 628.1

278 101195 217.4 784.5

280 116168 267.4 964.9

282 132174 324.5 1171

284 149212 389.2 1405

286 167282 462.1 1667

288 186386 543.4 1961

290 206521 633.8 2287292 227690 733.7 2648

294 249891 843.6 3044

296 273125 964 3479

298 297391 1095 3952300 322690 1238 4467

260 265 270 275 280 285 290 295 300

0

50000

100000

150000

200000

250000

300000

350000

0

20 0

40 0

60 0

80 0

1000

1200

1400

Ts[F]

h

[Btu/h-ft^2-F]

W

kW

h

We

10-28

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260 265 270 275 280 285 290 295 300

0

50 0

1000

1500

2000

2500

3000

3500

4000

4500

Ts

[F]

mevap

[lbm/h]

10-29

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10-30 Cold water enters a steam generator at 15C and is boiled, and leaves as saturated vapor at Tsat =100C. The fraction of heat used to preheat the liquid water from 15C to saturation temperature of 100Cis to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible.

Properties The heat of vaporization of water at 100C is hfg = 2257 kJ/kg and the specific heat of liquidwater at the average temperature of (15+100)/2 = 57.5C is CkJ/kg184.4C =pl (Table A-9).

Analysis The heat of vaporization of water represents the amount of heatneeded to vaporize a unit mass of liquid at a specified temperature. Usingthe average specific heat, the amount of heat needed to preheat a unit massof water from 15C to 100C is determined to be

kJ/kg355.6=C)15C)(100kJ/kg184.4(preheating == TCq pl

and

kJ/kg6.26126.3552257preheatingboilingtotal =+=+= qqq

Therefore, the fraction of heat used to preheat the water is

)(or

2612.6

6.355preheattoFraction

total

preheating13.6%0.136===

q

q

10-30

Water, 100C

Steamgenerator

Water, 15C

Steam100C

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10-31 Cold water enters a steam generator at 20C and is boiled, and leaves as saturated vapor at boilerpressure. The boiler pressure at which the amount of heat needed to preheat the water to saturationtemperature is equal to the heat of vaporization is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible.

Properties The properties needed to solve this problem are the heat ofvaporization hfg and the specific heat of water Cp at specified temperatures,and they can be obtained from Table A-9.

Analysis The heat of vaporization of water represents the amount of heatneeded to vaporize a unit mass of liquid at a specified temperature, and

C Tp represents the amount of heat needed to preheat a unit mass of waterfrom 20C to the saturation temperature. Therefore,

q q

C T hp ave fg T

preheating boiling

sat sat

=

=, @( )20

The solution of this problem requires choosing a boiling temperature,reading the heat of vaporization at that temperature, evaluating the specificheat at the average temperature, and substituting the values into the relationabove to see if it is satisfied. By trial and error, the temperature that satisfiesthis condition is determined to be 315C at which (Table A-9)

hfg C@315 1281 = kJ / kg and Tave = (20+315)/2 = 167.5C Cp ave, .= 4 37 kJ / kg C

Substituting,

C Tp ave, ( ) ( . )( )sat kJ / kg C C kJ / kg = =20 4 37 315 20 1289

which is practically identical to the heat of vaporization. Therefore,

P P Tboiler sat sat= =@ 10.6 MPa

10-31

Water, 100C

Steam

generator

Water, 20C

Steam100C

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10-32"!PROBLEM 10-32"

"GIVEN""T_cold=20 [C], parameter to be varied"

"ANALYSIS"Fluid$='steam_NBS'

q_preheating=q_boiling

q_preheating=C_p*(T_sat-T_cold)T_sat=temperature(Fluid$, P=P, x=1)C_p=CP(Fluid$, T=T_ave, x=0)

T_ave=1/2*(T_cold+T_sat)

q_boiling=h_fgh_f=enthalpy(Fluid$, T=T_sat, x=0)h_g=enthalpy(Fluid$, T=T_sat, x=1)h_fg=h_g-h_f

Tcold [C] P [kPa]

8 10031

9 1007110 10112

11 1015212 10193

13 10234

14 10274

15 10315

16 10356

17 10396

18 10437

19 1047820 10519

21 10560

22 10601

23 10641

24 10682

25 10723

26 10764

27 1080528 10846

29 10887

30 10928

10-32

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0 5 10 15 20 25 30

9600

9800

10000

10200

10400

10600

10800

11000

Tcold

[C]

P

[kPa]

10-33

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10-33 Boiling experiments are conducted by heating water at 1 atm pressure with an electric resistancewire, and measuring the power consumed by the wire as well as temperatures. The boiling heat transfercoefficient is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heatlosses from the water are negligible.

AnalysisThe heat transfer area of the heater wire is

2m003142.0m)m)(0.50002.0( === DLA

s

Noting that 3800 W of electric power is consumed when theheater surface temperature is 130C, the boiling heat transfercoefficient is determined from Newtons law of cooling to be

CW/m40,320 2 =

=

==C)100)(130m(0.003142

W3800

)()(

2sat

satTTA

QhTThAQ

ssss

Ts=130 C

Heating wire, 3.8 kW

1 atm

Date post:	04-Apr-2018
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Heat Chap10 021

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