Heat 4e SM Chap10

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Solutions Manual

for

Heat and Mass Transfer: Fundamentals & Applications

Fourth Edition

Yunus A. Cengel & Afshin J. Ghajar

McGraw-Hill, 2011

Chapter 10 BOILING AND CONDENSATION

PROPRIETARY AND CONFIDENTIAL

This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and

protected by copyright and other state and federal laws. By opening and using this Manual the user

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Boiling Heat Transfer

10-1C Boiling is the liquid-to-vapor phase change process that occurs at a solid-liquid interface when the surface is heated

above the saturation temperature of the liquid. The formation and rise of the bubbles and the liquid entrainment coupled with the large amount of heat absorbed during liquid-vapor phase change at essentially constant temperature are responsible for the very high heat transfer coefficients associated with nucleate boiling.

10-2C The different boiling regimes that occur in a vertical tube during flow boiling are forced convection of liquid, bubbly

flow, slug flow, annular flow, transition flow, mist flow, and forced convection of vapor.

10-3C Both boiling and evaporation are liquid-to-vapor phase change processes, but evaporation occurs at the liquid-vapor interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature, and it involves no

bubble formation or bubble motion. Boiling, on the other hand, occurs at the solid-liquid interface when a liquid is brought into contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid.

10-4C Boiling is called pool boiling in the absence of bulk fluid flow, and flow boiling (or forced convection boiling) in the

presence of it. In pool boiling, the fluid is stationary, and any motion of the fluid is due to natural convection currents and the motion of the bubbles due to the influence of buoyancy.

10-5C The boiling curve is given in Figure 10-6 in the text. In the natural convection boiling regime, the fluid motion is

governed by natural convection currents, and heat transfer from the heating surface to the fluid is by natural convection. In the nucleate boiling regime, bubbles form at various preferential sites on the heating surface, and rise to the top. In the

transition boiling regime, part of the surface is covered by a vapor film. In the film boiling regime, the heater surface is

completely covered by a continuous stable vapor film, and heat transfer is by combined convection and radiation.

10-6C In the film boiling regime, the heater surface is completely covered by a continuous stable vapor film, and heat

transfer is by combined convection and radiation. In the nucleate boiling regime, the heater surface is covered by the liquid. The boiling heat flux in the stable film boiling regime can be higher or lower than that in the nucleate boiling regime, as can be seen from the boiling curve.

10-7C The boiling curve is given in Figure 10-6 in the text. The burnout point in the curve is point C. The burnout during

boiling is caused by the heater surface being blanketed by a continuous layer of vapor film at increased heat fluxes, and the resulting rise in heater surface temperature in order to maintain the same heat transfer rate across a low-conducting vapor film. Any attempt to increase the heat flux beyond will cause the operation point on the boiling curve to jump suddenly from point C to point E. However, the surface temperature that corresponds to point E is beyond the melting point of most heater materials, and burnout occurs. The burnout point is avoided in the design of boilers in order to avoid the disastrous explosions of the boilers.

&max

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10-8C Pool boiling heat transfer can be increased permanently by increasing the number of nucleation sites on the heater

surface by coating the surface with a thin layer (much less than 1 mm) of very porous material, or by forming cavities on the surface mechanically to facilitate continuous vapor formation. Such surfaces are reported to enhance heat transfer in the nucleate boiling regime by a factor of up to 10, and the critical heat flux by a factor of 3. The use of finned surfaces is also known to enhance nucleate boiling heat transfer and the critical heat flux.

10-9 Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.

10-10 The power dissipation per unit length of a metal rod submerged horizontally in water, when electric current is passed

through it, is to be determined.

Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible.

Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9

kg/m3_{(Table A-9).}

The properties of vapor at the film temperature of

Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16,

ρv = 0.3831 kg/m3 cpv = 1997 J/kg·K

µv = 2.045 × 10−5 kg/m·s kv = 0.04345 W/m·K

Analysis The excess temperature in this case is ∆T = Ts

− Tsat = 400°C, which is much larger than 30°C for water from Fig. 10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 2 5 4 / 1 5 3 3 sat 4 / 1 sat sat 3 film film W/m 10 152 . 1 ) 400 ( ) 400 )( 002 . 0 )( 10 045 . 2 ( )] 400 )( 1997 ( 4 . 0 10 2257 )[ 3831 . 0 9 . 957 )( 3831 . 0 ( ) 04345 . 0 ( 81 . 9 62 . 0 ) ( ) ( )] ( 4 . 0 )[ ( × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × + × − = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − + − = − T T T T D T T c h gk C q _s s v s pv fg v l v v µ ρ ρ ρ &

The radiation heat flux is determined from

2 4 4 4 4 2 8 4 4 ₎ ₍₀_.₅₎₍₅_.₆₇ ₁₀ _W/m _K ₎₍₇₇₃ ₃₇₃ ₎_K ₉₅₇₃_W/m ( − = × ⋅ − = = _T _T − q& εσ

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10-11 The nucleate pool boiling heat transfer rate per unit length and the rate of evaporation per unit length of water being

boiled by a rod that is maintained at 10°C above the saturation temperature are to be determined.

Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible.

Properties The properties of water at the saturation temperature of 100°C are σ = 0.0589 N/m (Table 10-1) and, from Table

A-9,

ρl = 957.9 kg/m3 hfg = 2257 × 103 J/kg

ρv = 0.5978 kg/m3 µl = 0.282 × 10−3 kg/m·s

Prl = 1.75 cpl = 4217 J/kg·K

Also, Csf = 0.0130 and n = 1.0 for the boiling of water on platinum

surface (Table 10-3).

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 10°C,

which is less than 30°C for water from Fig. 10-6. Therefore, nucleate boiling will occur. The heat flux in this case can be determined from the Rohsenow relation to be 2 5 3 3 2 / 1 3 3 3 sat 2 / 1 nucleate W/m 10 408 . 1 ) 75 . 1 )( 10 2257 )( 013 . 0 ( ) 10 ( 4217 0589 . 0 ) 5978 . 0 9 . 957 ( 81 . 9 ) 10 2257 )( 10 282 . 0 ( Pr ) ( ) ( × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − n l fg sf s pl v l fg l h C T T c g h q σ ρ ρ µ &

Finally, the nucleate pool boiling heat transfer rate per unit length is

W/m 4420 = × = = (0.010m)(1.408 10 W/m ) / _nucleate 5 2 boiling L π qD π Q& &

The rate of evaporation per unit length is

m kg/s 10 1.96× 3 ⋅ = × ⋅ = = − J/kg 10 2257 m J/s 4420 ) / ( 3 boiling n evaporatio h L Q L m fg & &

Discussion The value for the rate of evaporation per unit length indicates that 1 m of the platinum-plated rod would boil

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10-12 Water is boiled at Tsat = 120°C in a mechanically polished stainless steel pressure cooker whose inner surface

temperature is maintained at Ts = 130°C. The heat flux on the surface is 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 120°C are (Tables 10-1 and A-9)

44 . 1 Pr C J/kg 4244 c N/m 0550 . 0 s kg/m 10 232 . 0 kg/m 121 . 1 J/kg 10 2203 kg/m 4 . 943 3 3 3 3 = ° ⋅ = = ⋅ × = = × = = − l pl l v fg l h σ µ ρ ρ Water 120°C 130°C

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.

=

sf

C

Heating

Analysis The excess temperature in this case is ∆T =Ts−Tsat =130−120=10°Cwhich is relatively low (less than 30°C).

Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be

2 kW/m 228.4 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − 2 3 3 1/2 3 3 3 sat , 2 / 1 nucleate W/m 400 , 228 44 . 1 ) 10 2203 ( 0130 . 0 ) 120 130 ( 4244 0550 . 0 1.121) -9.81(943.4 ) 10 )(2203 10 232 . 0 ( Pr ) ( ) ( n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

10-13 Water is boiled at the saturation (or boiling) temperature of Tsat = 90°C by a horizontal brass heating element. The

maximum heat flux in the nucleate boiling regime is 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 90°C

are (Tables 10-1 and A-9)

Heating element Water, 90°C 96 . 1 Pr C J/kg 4206 c N/m 0608 . 0 s kg/m 10 315 . 0 kg/m 4235 . 0 J/kg 10 2283 kg/m 3 . 965 3 3 3 3 = ° ⋅ = = ⋅ × = = × = = − l pl l v fg l h σ µ ρ ρ qmax

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10-14 Water is boiled at a saturation (or boiling) temperature of Tsat = 120°C by a brass heating element whose temperature

is not to exceed Ts = 125°C. The highest rate of steam production is 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_s−T_sat =125−120=5°C which is in the nucleate boiling range of 5 to 30°C for water.

44 . 1 Pr N/m 0550 . 0 kg/m 12 . 1 kg/m 4 . 943 3 3 = = = = l v l σ ρ ρ Ts=125°C Water 120°C Heating element C J/kg 4244 m/s kg 10 232 . 0 J/kg 10 2203 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ

Also, 0.0060 and n = 1.0 for the boiling of water on a brass 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. =

sf C

Analysis Assuming nucleate boiling, the heat flux in this case can be determined from Rohsenow relation to be

2 3 3 1/2 3 3 3 sat , 2 / 1 nucleate W/m 300 , 290 44 . 1 ) 10 2203 ( 0060 . 0 ) 120 125 ( 4244 0550 . 0 ) 12 . 1 9.81(943.4 ) 10 )(2203 10 232 . 0 ( Pr ) ( ) ( = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

The surface area of the heater is

2 m 04084 . 0 m) m)(0.65 02 . 0 ( = = =πDL π A_s

Then the rate of heat transfer during nucleate boiling becomes

W 856 , 11 ) W/m 300 , 290 )( m 04084 . 0 ( 2 2 nucleate boiling =A q = = Q& _s&

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

kg/h 19.4 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = = h 1 s 3600 J/kg 10 2203 J/s 856 , 11 3 boiling n evaporatio fg h Q m& &

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10-15 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100°C in a mechanically

polished stainless steel pan whose inner surface temperature is maintained at Ts = 110°C. The rate of heat transfer to the water and the rate of evaporation of water are to be determined.

75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ P = 1 atm 110°C 100°C Water C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ Heating

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.

=

sf C

Analysis The excess temperature in this case is which is relatively low (less than 30°C).

C 10 100 110 sat = − = ° − = ∆T T_s T 2 3 3 1/2 3 3 3 sat , 2 / 1 nucleate W/m 700 , 140 75 . 1 ) 10 2257 ( 0130 . 0 ) 100 110 ( 4217 0589 . 0 0.60) -9.8(957.9 ) 10 )(2257 10 282 . 0 ( Pr ) ( ) ( = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

The surface area of the bottom of the pan is

2 2

2 _/₄₌ ₍₀_.₃₀_m) _/₄₌₀_.₀₇₀₆₉_m

=πD π

A_s

Then the rate of heat transfer during nucleate boiling becomes

W 9945 = = = _nucleate (0.07069m2)(140,700 W/m2) boiling Aq Q& _s&

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

kg/s 0.00441 = × = = J/kg 10 2257 J/s 9945 3 boiling n evaporatio fg h Q m& &

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10-16 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100°C 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

Properties The properties of water at the saturation temperature of

100°C are (Tables 10-1 and A-9)

Ts= ? Heating element Water, 100°C 75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ qmax

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

=

sf C

cr = 0.12 (Table 10-4). (It can be shown that L* = 3.99 >

1.2 and thus the restriction in Table 10-4 is satisfied).

Analysis The maximum or critical heat flux is determined from

2 W/m 1,017,000 = − × × × = − = 4 / 1 2 3 4 / 1 2 max )] 60 . 0 9 . 957 ( ) 6 . 0 ( 8 . 9 0589 . 0 )[ 10 2257 ( 12 . 0 )] ( [ _v _l _v fg crh g C q& σ ρ ρ ρ

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0130 . 0 ) 100 ( 4217 0589 . 0 0.60) -9.8(957.9 ) 10 )(2257 10 282 . 0 ( 000 , 017 , 1 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = − Ts It gives C 119.3° = s T

Therefore, the temperature of the heater surface will be only 19.3°C above the boiling temperature of water when burnout occurs.

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10-17 Prob. 10-16 is reconsidered. The effect of local atmospheric pressure on the maximum heat flux and the

temperature difference Ts –Tsat is to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" D=0.002 [m] P_sat=101.3 [kPa] "PROPERTIES" Fluid$='steam_IAPWS' T_sat=temperature(Fluid$, P=P_sat, x=1) rho_l=density(Fluid$, T=T_sat, x=0) rho_v=density(Fluid$, T=T_sat, x=1) sigma=SurfaceTension(Fluid$, T=T_sat) mu_l=Viscosity(Fluid$,T=T_sat, x=0)

Pr_l=Prandtl(Fluid$, T=T_sat, P=P_sat+1[kPa]) c_l=CP(Fluid$, 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=0.0130 "from Table 10-3 of the text"

n=1 "from Table 10-3 of the text"

C_cr=0.12 "from Table 10-4 of the text"

g=9.8 [m/s^2] “gravitational acceleraton" "ANALYSIS" q_dot_max=C_cr*h_fg*(sigma*g*rho_v^2*(rho_l-rho_v))^0.25 q_dot_nucleate=q_dot_max 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))^3 DELTAT=T_s-T_sat Psat

[kPa] _[kW/mq&max 2_] ∆T [C] 70 871.9 20.12 71.65 880.3 20.07 73.29 888.6 20.02 74.94 896.8 19.97 76.59 904.9 19.92 78.24 912.8 19.88 79.88 920.7 19.83 81.53 928.4 19.79 83.18 936.1 19.74 84.83 943.6 19.7 885 920 955 990 1025 19.5 19.6 19.7 19.8 19.9 20 20.1 20.2 qmax [ k W /m 2 ] ∆ T [ C ] Heat Temp. Dif.

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10-18E Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 212°F by a horizontal

polished copper heating element whose surface temperature is maintained at Ts = 788°F. The rate of heat transfer to the water per unit length of the heater is to be determined.

Properties The properties of water at the saturation temperature of 212°F are ρl =59.82lbm/ft3 and hfg =970Btu/lbm

(Table A-9E). The properties of the vapor at the film temperature of are (Table

A-16E) F 500 2 / ) 788 212 ( 2 / ) ( _sat + = + = ° = _s f T T T P = 1 atm F ft Btu/h 02267 . 0 F Btu/lbm 4707 . 0 h lbm/ft 0.04561 s lbm/ft 10 267 . 1 lbm/ft 02571 . 0 5 3 ° ⋅ ⋅ = ° ⋅ = ⋅ = ⋅ × = = − v pv v v k c µ ρ Heating element Water, 212°F

Also, g = 32.2 ft/s2_{= 32.2×(3600)}2_ft/h2_{. Note that we expressed the properties in units that will cancel each other in boiling} heat transfer relations. Also note that we used vapor properties at 1 atm pressure from Table A-16E instead of the properties of saturated vapor from Table A-9E since the latter are at the saturation pressure of 680 psia (46 atm).

Analysis The excess temperature in this case is , which is much larger than 30°C or 54°F.

Therefore, film boiling will occur. The film boiling heat flux in this case can be determined to be

F 576 212 788 sat = − = ° − = ∆T T_s T 2 4 / 1 3 2 sat 4 / 1 sat 3 film ft Btu/h 600 , 18 ) 212 788 ( ) 212 788 )( 12 / 5 . 0 )( 04561 . 0 ( )] 212 788 ( 4707 . 0 4 . 0 970 )[ 02571 . 0 82 . 59 )( 02571 . 0 ( ) 02267 . 0 ( ) 3600 ( 2 . 32 62 . 0 ) ( ) ( )] ( 4 . 0 )[ ( 62 . 0 ⋅ = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − × + − = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − + − = T T T T D T T c h gk q _s s v sat s pv fg v l v v µ ρ ρ ρ &

[

]

2 4 4 4 2 8 4 sat 4 rad ft Btu/h 7 . 761 R) 460 212 ( R) 460 788 ( ) R ft Btu/h 10 1714 . 0 )( 2 . 0 ( ) ( ⋅ = + − + ⋅ ⋅ × = − = − T T q& εσ _s

Note that heat transfer by radiation is very small in this case because of the low emissivity of the surface and the relatively low surface temperature of the heating element. Then the total heat flux becomes

2 rad film total ₄ 761.7 19,171Btu/h ft 3 600 , 18 4 3 ⋅ = × + = + =q q

q_& _& _&

Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat transfer surface area,

Btu/h 2509 = ⋅ × × = = = ) ft Btu/h ft)(19,171 1 ft 12 / 5 . 0 ( ) ( 2 total total total π πDL q q A

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10-19E Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 212°F by a horizontal

polished copper heating element whose surface temperature is maintained at Ts = 988°F. The rate of heat transfer to the water per unit length of the heater is to be determined.

Properties The properties of water at the saturation temperature of 212°F are ρl =59.82lbm/ft3 and hfg =970Btu/lbm

(Table A-9E). The properties of the vapor at the film temperature of T_f =(T_sat +T_s)/2=(212+988)/2=600°F are, by interpolation, (Table A-16E)

P = 1 atm F ft Btu/h 02640 . 0 F Btu/lbm 4799 . 0 h lbm/ft 0.05099 s lbm/ft 10 416 . 1 lbm/ft 02395 . 0 5 3 ° ⋅ ⋅ = ° ⋅ = ⋅ = ⋅ × = = − v pv v v k c µ ρ Heating element Water, 212°F

Also, g = 32.2 ft/s2_{= 32.2×(3600)}2_ft/h2_{. Note that we expressed the properties in units that will cancel each other in} boiling heat transfer relations. Also note that we used vapor properties at 1 atm pressure from Table A-16E instead of the properties of saturated vapor from Table A-9E since the latter are at the saturation pressure of 1541 psia (105 atm).

Analysis The excess temperature in this case is ∆T =T_s−T_sat =988−212=776°F, which is much larger than 30°C or 54°F.

Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from

2 4 / 1 3 2 sat 4 / 1 sat 3 film ft Btu/h 147 , 25 ) 212 988 ( ) 212 988 )( 12 / 5 . 0 )( 05099 . 0 ( )] 212 988 ( 4799 . 0 4 . 0 970 )[ 02395 . 0 82 . 59 )( 02395 . 0 ( ) 0264 . 0 ( ) 3600 ( 2 . 32 62 . 0 ) ( ) ( )] ( 4 . 0 )[ ( 62 . 0 ⋅ = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − × + − = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − + − = T T T T D T T C h gk q _s s v sat s pv fg v l v v µ ρ ρ ρ &

[

]

2 4 4 4 2 8 4 sat 4 rad ft Btu/h 1437 R) 460 212 ( R) 460 988 ( ) R ft Btu/h 10 1714 . 0 )( 2 . 0 ( ) ( ⋅ = + − + ⋅ ⋅ × = − = − T T q& εσ _s

Note that heat transfer by radiation is very small in this case because of the low emissivity of the surface and the relatively low surface temperature of the heating element. Then the total heat flux becomes

2 rad film total ₄ 1437 26,225Btu/h ft 3 147 , 25 4 3 ⋅ = × + = + =q q

q& & &

Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat transfer surface area,

=

= _total ( ) _total

total A q πDL q

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10-20 The nucleate pool boiling heat transfer coefficient of water being boiled by a horizontal platinum-plated rod is to be

determined.

Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible.

Properties The properties of water at the saturation temperature of 100°C are σ = 0.0589 N/m (Table 10-1) and, from Table

A-9,

ρl = 957.9 kg/m3 hfg = 2257 × 103 J/kg

ρv = 0.5978 kg/m3 µl = 0.282 × 10−3 kg/m·s

Prl = 1.75 cpl = 4217 J/kg·K

Also, Csf = 0.0130 and n = 1.0 for the boiling of water on platinum

surface (Table 10-3).

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 10°C,

which is less than 30°C for water from Fig. 10-6. Therefore, nucleate boiling will occur. The heat flux in this case can be determined from the Rohsenow relation to be 2 5 3 3 2 / 1 3 3 3 sat 2 / 1 nucleate W/m 10 408 . 1 ) 75 . 1 )( 10 2257 )( 013 . 0 ( ) 10 ( 4217 0589 . 0 ) 5978 . 0 9 . 957 ( 81 . 9 ) 10 2257 )( 10 282 . 0 ( Pr ) ( ) ( × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − n l fg sf s pl v l fg l h C T T c g h q σ ρ ρ µ &

Using the Newton’s law of cooling, the boiling heat transfer coefficient is → ) ( _sat nucleate hT T q& = _s − sat nucleate T T q h s − = & K W/m 14,100 2⋅ = − × = K ) 100 110 ( W/m 10 408 . 1 5 2 h

Discussion Heat transfer coefficient on the order of 104 W/m2·K can be obtained in nucleate boiling with a temperature

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10-21 The nucleate boiling heat transfer coefficient and the value of Csf for water being boiled by a long electrical wire are to

be determined.

Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible.

Properties The properties of water at the saturation

temperature of 100°C are σ = 0.0589 N/m (Table 10-1) and, from Table A-9,

ρl = 957.9 kg/m3 hfg = 2257 × 103 J/kg

ρv = 0.5978 kg/m3 µl = 0.282 × 10−3 kg/m·s

Prl = 1.75 cpl = 4217 J/kg·K

Also, n = 1.0 is given.

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 28°C, which is less than 30°C for water from Fig. 10-6.

Therefore, nucleate boiling will occur. The nucleate boiling heat transfer coefficient can be determined using → ) ( _sat boiling hT T q& = _s − sat boiling T T q h s − = & Also, we know W/m 4100 / _boiling boiling L= Dq = Q& π & 2 6 boiling = 4100_πD W/m =_π4100₍₀_.₀₀₁ W/m_m₎ =1.305×10 W/m q&

Hence, the nucleate boiling heat transfer coefficient is

K W/m 46,600 2⋅ = − × = K ) 100 128 ( W/m 10 305 . 1 6 h

The value of the experimental constant Csf can be determined from the Rohsenow relation to be

3 sat 2 / 1 boiling Pr ) ( ) ( ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s pl v l fg l h C T T c g h q σ ρ ρ µ & or 3 / 1 3 sat 2 / 1 boiling Pr ) ( ) ( ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg s pl v l fg l sf h T T c g q h C σ ρ ρ µ & ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × × = − 3 / 1 3 3 2 / 1 6 3 3 ) 75 . 1 )( 10 2257 ( ) 100 128 ( 4217 0589 . 0 ) 5978 . 0 9 . 957 ( 81 . 9 10 305 . 1 ) 10 2257 )( 10 282 . 0 ( sf C

(14)

10-22 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100°C 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).

75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ Coffee maker Water, 100°C 1 L P = 1 atm C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ

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.

=

sf C

Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18°C is nearly 1

kg. The rate of energy transfer needed to evaporate half of this water in 32 min and the heat flux are

2 2 2 2 W/m 155,920 = kW/m 155.92 = ) m 770 kW)/(0.003 5878 . 0 ( / m 003770 . 0 m) m)(0.30 004 . 0 ( kW 5878 . 0 s) 60 (32 kJ/kg) kg)(2257 5 . 0 ( = = = = = = × = ∆ = → = ∆ = s s fg fg A Q q DL A t mh Q mh t Q Q & & & & π π

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0130 . 0 ) 100 ( 4217 0589 . 0 0.60) 9.81(957.9 ) 10 )(2257 10 282 . 0 ( 920 , 155 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = − Ts It gives C 110.3° = s T

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

The specific heat of water at the average temperature of (14+100)/2 = 57°C is cp = 4.184 kJ/kg⋅°C. Then the time it

takes for the entire water to be heated from 14°C to 100°C is determined to be

612s=10.2min kJ/s 0.5878 C 14) C)(100 kJ/kg kg)(4.184 1 ( → ∆ = ∆ = ⋅° − ° = ∆ = ∆ = Q T mc t T mc t Q Q _p p & &

(15)

10-23 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100°C by a

copper 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 100°C are

(Tables 10-1 and A-9)

Coffee maker Water, 100°C 1 L P = 1 atm 75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ

Also, 0.0130 and n = 1.0 for the boiling of water on a copper 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.

=

sf C

Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18°C is nearly 1

kg. The rate of energy transfer needed to evaporate half of this water in 32 min and the heat flux are

2 2 2 2 W/m 155,920 = kW/m 155.92 = ) m 770 kW)/(0.003 5878 . 0 ( / m 003770 . 0 m) m)(0.30 004 . 0 ( kW 5878 . 0 s) 60 (32 kJ/kg) kg)(2257 5 . 0 ( = = = = = = × = ∆ = → = ∆ = s s fg fg A Q q DL A t mh Q mh t Q Q & & & & π π

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = _n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0130 . 0 ) 100 ( 4217 0589 . 0 0.60) 9.81(957.9 ) 10 )(2257 10 282 . 0 ( 920 , 155 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = − Ts It gives C 110.3° = s T

(16)

10-24 Water is boiled at Tsat = 90°C in a brass heating element. The surface temperature of the heater is to be determined.

Properties The properties of water at the saturation temperature of 90°C

are (Tables 10-1 and A-9)

Heating element Water, 90°C 96 . 1 Pr C J/kg 4206 c N/m 0608 . 0 s kg/m 10 315 . 0 kg/m 4235 . 0 J/kg 10 2283 kg/m 3 . 965 3 3 3 3 = ° ⋅ = = ⋅ × = = × = = − l pl l v fg l h σ µ ρ ρ qmin

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

Analysis The minimum heat flux is determined from

2 4 / 1 2 3 4 / 1 2 min W/m 715 , 13 ) 4235 . 0 3 . 965 ( ) 4235 . 0 3 . 965 )( 81 . 9 )( 0608 . 0 ( ) 10 2283 )( 4235 . 0 ( 09 . 0 ) ( ) ( 09 . 0 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − = v l v l fg v g h q ρ ρ ρ ρ σ ρ &

The surface temperature can be determined from Rohsenow equation to be

C 92.3° = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − s s n l fg sf s l p v l fg l T T h C T T c g h q 3 3 1/2 3 3 2 3 sat , 2 / 1 nucleate 96 . 1 ) 10 2283 ( 0060 . 0 ) 90 ( 4206 0608 . 0 0.4235) -9.81(965.3 ) 10 )(2283 10 315 . 0 ( W/m 715 , 13 Pr ) ( ) ( σ ρ ρ µ &

(17)

10-25 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100°C by a horizontal

nickel plated copper heating element. The maximum (critical) heat flux and the temperature jump of the wire when the operating point jumps from nucleate boiling to film boiling regime are to be determined.

75 . 1 Pr N/m 0589 . 0 kg/m 5978 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ

Also, 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 specified under Eqs. 10-2 and 10-3 in connection with their definitions in order to avoid unit

manipulations. The vapor properties at the anticipated film temperature of T =

sf C

f = (Ts+Tsat )/2 of 1000°C (will be checked) (Table A-16) s kg/m 10 762 . 4 C J/kg 2471 C W/m 1362 . 0 kg/m 1725 . 0 5 3 ⋅ × = ° ⋅ = ° ⋅ = = − v pv v v c k µ ρ _P_{= 1 atm} Ts Heating element Water, 100°C

Analysis (a) For a horizontal heating element, the coefficient Ccr is

determined from Table 10-4 to be _q

max 1269 . 0 ) 7986 . 0 ( 12 . 0 * 12 . 0 1.2 < 7986 . 0 0589 . 0 ) 5978 . 0 9 . 957 ( 81 . 9 ) 002 . 0 ( ) ( * 25 . 0 25 . 0 2 / 1 2 / 1 = = = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − − L C g L L cr v l σ ρ ρ

Then the maximum or critical heat flux is determined from

2 W/m 1,074,000 = − × × × = − = 4 / 1 2 3 4 / 1 2 max )] 5978 . 0 9 . 957 ( ) 5978 . 0 ( 81 . 9 0589 . 0 )[ 10 2257 ( 1269 . 0 )] ( [ _v _l _v fg crh g C q& σ ρ ρ ρ

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = _n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0060 . 0 ) 100 ( 4217 0589 . 0 0.5978) -9.81(957.9 ) 10 )(2257 10 282 . 0 ( 000 , 074 , 1 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = − Ts It gives T_s =109.1°C

(b) Heat transfer in the film boiling region can be expressed as

) ( 4 3 ) ( ) ( )] ( 4 . 0 )[ ( 62 . 0 4 3 ₄ sat 4 sat 4 / 1 sat 3 rad film total _D_T _T T T T T T T c h gk q q q _s _s s v sat s pv fg v l v v ₋ ₊ ₋ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − + − = + = εσ µ ρ ρ ρ & & & Substituting,

(18)

10-26 Prob. 10-25 is reconsidered. The effects of the local atmospheric pressure and the emissivity of the wire on the

critical heat flux and the temperature rise of wire are to be investigated.

"GIVEN" L=0.3 [m] D=0.004 [m] epsilon=0.3 P=101.3 [kPa] "PROPERTIES" Fluid$='steam_IAPWS' T_sat=temperature(Fluid$, 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+1)

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 10-3 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=9.81 [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 10-4 of the text"

C_cr=0.12*L_star^(-0.25) L_star=D/2*((g*(rho_l-rho_v))/sigma)^0.5 q_dot_max=C_cr*h_fg*(sigma*g*rho_v^2*(rho_l-rho_v))^0.25 q_dot_nucleate=q_dot_max q_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_nucleate

q_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

(19)

P

[kPa] q&max [kW/m2_] ∆T [C] 70 71.65 73.29 74.94 76.59 78.24 79.88 81.53 83.18 84.83 86.47 88.12 89.77 91.42 93.06 94.71 96.36 98.01 99.65 101.3 925656 934417 943050 951559 959948 968222 976385 984439 992389 1000237 1007987 1015642 1023205 1030677 1038062 1045363 1052581 1059719 1066778 1073762 2073 2079 2084 2090 2096 2101 2107 2112 2117 2122 2127 2132 2137 2142 2146 2151 2155 2160 2164 2168 70 75 80 85 90 95 100 105 920000 940000 960000 980000 1000000 1.020x₁₀6 1.040x₁₀6 1.060x₁₀6 1.080x₁₀6 2080 2100 2120 2140 2160 2180 P [kPa] qma x [ W /m 2 ] ∆ T [C ] ε q&max [kW/m2_] ∆T [C] 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 2760 2535 2380 2262 2168 2091 2025 1967 1917 1872 1831 1794 1761 1730 1701 1674 1649 1626 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1600 1800 2000 2200 2400 2600 2800 5.0x₁₀5 7.5x₁₀5 1.0x₁₀6 1.3x₁₀6 1.5x₁₀6 ε ∆ T [ C ] qma x [ W /m 2 ]

(20)

10-27 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100°C in a

teflon-pitted stainless steel pan placed on an electric burner. The water level drops by 10 cm in 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 boiling regime is nucleate

boiling (this assumption will be checked later).

75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ P = 1 atm Ts 100°C Water C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ Heating

Also, 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 their definitions in order to avoid unit

manipulations. =

sf C

Analysis The rate of heat transfer to the water and the heat flux are

2 2 2 2 2 evap 2 3 evap evap W/m 240,200 = ) m 42 W)/(0.031 7547 ( / m 03142 . 0 4 / m) 20 . 0 ( 4 / kW 547 . 7 kJ/kg) kg/s)(2257 03344 . 0 ( kg/s 003344 . 0 s 60 15 m) 0.10 /4 m) 0.2 ( )( kg/m 9 . 957 ( = = = = = = = = = × × × = ∆ ∆ = ∆ = s s fg A Q q D A h m Q t V t m m & & & & & π π π ρ

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0058 . 0 ) 100 ( 4217 0589 . 0 0.60) 9.8(957.9 ) 10 )(2257 10 282 . 0 ( 200 , 240 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = − Ts It gives Ts = 105.3°C

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

(21)

10-28 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100°C in a

polished copper pan placed on an electric burner. The water level drops by 10 cm in 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 boiling regime is nucleate

boiling (this assumption will be checked later).

75 . 1 Pr N/m 0589 . 0 kg/m 60 . 0 kg/m 9 . 957 3 3 = = = = l v l σ ρ ρ P = 1 atm Ts 100°C Water C J/kg 4217 m/s kg 10 282 . 0 J/kg 10 2257 3 3 ° ⋅ = ⋅ × = × = − pl l fg c h µ Heating

Also, 0.0130 and n = 1.0 for the boiling of water on a copper 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.

=

sf C

Analysis The rate of heat transfer to the water and the heat flux are

2 2 2 2 2 evap 2 3 evap evap W/m 240,200 = ) m 42 W)/(0.031 7547 ( / m 03142 . 0 4 / m) 20 . 0 ( 4 / kW 547 . 7 kJ/kg) kg/s)(2257 03344 . 0 ( kg/s 003344 . 0 s 60 15 m) 0.10 /4 m) 0.2 ( )( kg/m 9 . 957 ( = = = = = = = = = × × × = ∆ ∆ = ∆ = s s fg A Q q D A h m Q t t m m & & & & & π π π ρ V

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

3 sat , 2 / 1 nucleate Pr ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = _n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ & 3 3 1/2 3 3 75 . 1 ) 10 2257 ( 0130 . 0 ) 100 ( 4217 0589 . 0 0.60) 9.8(957.9 ) 10 )(2257 10 282 . 0 ( 200 , 240 _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − × × = − Ts It gives Ts = 111.9°C

(22)

10-29 Boiling experiments are conducted by heating water at 1 atm pressure with an electric resistance wire, and measuring

the power consumed by the wire as well as temperatures. The boiling heat transfer coefficient is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water

are negligible. _{1 atm}

Analysis The heat transfer area of the heater wire is

Ts=130°C Heating wire, 3.8 kW 2 m 003142 . 0 m) m)(0.50 002 . 0 ( = = =πDL π A_s

Noting that 3800 W of electric power is consumed when the heater surface temperature is 130°C, the boiling heat transfer coefficient is determined from Newton’s law of cooling to be

C W/m 40,320 2⋅° = ° − = − = → − = C ) 100 )(130 m (0.003142 W 3800 ) ( ) ( ₂ sat sat h _A _T Q_T T T hA Q s s s s & &

10-30 Water is boiled at Tsat = 120°C in a mechanically polished stainless steel pressure cooker whose inner surface

temperature is maintained at Ts = 128°C. The boiling heat transfer coefficient is to be determined.

44 . 1 Pr C J/kg 4244 N/m 0550 . 0 s kg/m 10 232 . 0 kg/m 121 . 1 J/kg 10 2203 kg/m 4 . 943 3 3 3 3 = ° ⋅ = = ⋅ × = = × = = − l pl l v fg l c h σ µ ρ ρ 128°C 120°C Water

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.

=

sf

C

Heating

Analysis The excess temperature in this case is ∆T =Ts −Tsat =128−120=8°Cwhich is relatively low (less than 30°C).

2 3 3 1/2 3 3 3 sat , 2 / 1 nucleate W/m 900 , 116 44 . 1 ) 10 2203 ( 0130 . 0 ) 120 128 ( 4244 0550 . 0 1.121) -9.81(943.4 ) 10 )(2203 10 232 . 0 ( Pr ) ( ) ( = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

The boiling heat transfer coefficient is

C kW/m 14.6 2⋅ = ° ⋅ = ° − = − = ⎯→ ⎯ − = 14,610 W/m C C ) 120 128 ( W/m 900 , 116 ) ( 2 2 sat nucleate sat nucleate _T _T q h T T h q s s & &

(23)

10-31 The initial heat transfer rate from a hot metal sphere that is suddenly submerged in a water bath is to be determined.

Assumptions 1 Steady operating condition exists. 2 The metal sphere has uniform initial surface temperature.

Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9

kg/m3_{(Table A-9). The properties of vapor at the film temperature of T}

f = (Tsat + Ts)/2 = 400°C are, from Table A-16,

ρv = 0.3262 kg/m3 cpv = 2066 J/kg·K

µv = 2.446 × 10−5 kg/m·s kv = 0.05467 W/m·K

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 600°C, which is much larger than 30°C for water from Fig.

10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from

2 5 4 / 1 5 3 3 sat 4 / 1 sat sat 3 film film W/m 10 052 . 1 ) 600 ( ) 600 )( 02 . 0 )( 10 446 . 2 ( )] 600 )( 2066 ( 4 . 0 10 2257 )[ 3262 . 0 9 . 957 )( 3262 . 0 ( ) 05467 . 0 ( 81 . 9 67 . 0 ) ( ) ( )] ( 4 . 0 )[ ( × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × + × − = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − − + − = − T T T T D T T c h gk C q s s v s pv fg v l v v µ ρ ρ ρ &

2 4 4 4 4 4 2 8 4 sat 4 rad W/m 10 729 . 3 K ) 373 973 )( K W/m 10 67 . 5 )( 75 . 0 ( ) ( × = − ⋅ × = − = − T T q& εσ _s

Then the total heat flux becomes

2 5 2 4 2 5 rad film total ₄(3.729 10 W/m ) 1.332 10 W/m 3 W/m 10 052 . 1 4 3 × = × + × = + =q q

q_& _& _&

Finally, the initial heat transfer rate from the submerged metal sphere is

W 167 = × = = 2 5 2 2 total total q πD (1.332 10 W/m )π(0.02m) Q& &

(24)

10-32E Water is boiled at a temperature of Tsat = 250°F by a nickel-plated heating element whose surface temperature is

maintained at Ts = 280°F. The boiling heat transfer coefficient, the electric power consumed, and the rate of evaporation of water are to be determined.

nucleate boiling since ∆T T T= _s− _sat=280 250 30− = °F which is in the nucleate boiling range of 9 to 55°F for water.

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

43 . 1 Pr lbm/s 1208 . 0 lbf/ft 003755 . 0 lbm/ft 0723 . 0 lbm/ft 82 . 58 2 3 3 = = = = = l v l σ ρ ρ Ts=280°F Water 250°F Heating element F Btu/lbm 015 . 1 h lbm/ft 0.556 s lbm/ft 10 544 . 1 Btu/lbm 946 4 ° ⋅ = ⋅ = ⋅ × = = − pl l fg c h µ

Also, g = 32.2 ft/s2_and _{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 heat transfer relations.

=

sf C

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

2 3 1/2 3 sat , 2 / 1 nucleate ft Btu/h 221 , 475 , 3 43 . 1 ) 946 ( 0060 . 0 ) 250 280 ( 015 . 1 1208 . 0 ) 0723 . 0 32.2(58.82 ) )(946 556 . 0 ( Pr ) ( ) ( ⋅ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

Then the convection heat transfer coefficient becomes

F ft Btu/h 115,840 ⋅ 2⋅° = ° − ⋅ = − = → − = F ) 250 280 ( ft Btu/h 3,475,221 ) ( 2 sat sat _T _T q h T T h q s s & &

(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from Btu/h) 3412 = kW 1 (since = Btu/h 905 , 454 ) ft Btu/h 221 ft)(3,475, 2 ft 12 / 25 . 0 ( ) ( 2 kW 133.3 = ⋅ × × = = = =Q qA πDL q π

W&_e & & _s &

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

lbm/h 480.9 = = = Btu/lbm 946 Btu/h 905 , 454 boiling n evaporatio fg h Q m& &

(25)

10-33E Water is boiled at a temperature of Tsat = 250°F by a platinum-plated heating element whose surface temperature is

maintained at Ts = 280°F. The boiling heat transfer coefficient, the electric power consumed, and the rate of evaporation of water are to be determined.

nucleate boiling since ∆T =T_s−T_sat =280−250=30°F which is in the nucleate boiling range of 9 to 55°F for water.

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

43 . 1 Pr lbm/s 1208 . 0 lbf/ft 003755 . 0 lbm/ft 0723 . 0 lbm/ft 82 . 58 2 3 3 = = = = = l v l σ ρ ρ Ts=280°F Water 250°F Heating element F Btu/lbm 015 . 1 h lbm/ft 0.556 s lbm/ft 10 544 . 1 Btu/lbm 946 4 ° ⋅ = ⋅ = ⋅ × = = − pl l fg c h µ

Also, g = 32.2 ft/s2_and _{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 heat transfer relations.

=

sf C

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

2 3 3 1/2 3 sat , 2 / 1 nucleate ft Btu/h 670 , 341 43 . 1 ) 10 1208 . 0 ( 0130 . 0 ) 250 280 ( 015 . 1 1208 . 0 ) 0723 . 0 32.2(58.82 ) )(946 556 . 0 ( Pr ) ( ) ( ⋅ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ₋ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = n l fg sf s l p v l fg l h C T T c g h q σ ρ ρ µ &

Then the convection heat transfer coefficient becomes

F ft Btu/h 11,390 ⋅ 2⋅° = ° − ⋅ = − = → − = F ) 250 280 ( ft Btu/h 341,670 ) ( 2 sat sat h _T q_T T T h q s s & &

(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from Btu/h) 3412 = kW 1 (since = Btu/h 724 , 44 ) ft Btu/h 0 ft)(341,67 2 ft 12 / 25 . 0 ( ) ( 2 kW 13.1 = ⋅ × × = = = =Q qA πDL q π

W&_e & & _s &

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

lbm/h 47.3 = = = Btu/lbm 946 Btu/h 724 , 44 boiling n evaporatio fg h Q m& &

(26)

10-34E Prob. 10-32E is reconsidered. The effect of surface temperature of the heating element on the boiling heat

transfer coefficient, the electric power, and the rate of evaporation of water is to be investigated.

"GIVEN" T_sat=250 [F] L=2 [ft] D=0.25/12 [ft] T_s=280 [F] "PROPERTIES" Fluid$='steam_IAPWS' P_sat=pressure(Fluid$, 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 10-3 of the text"

g=32.2 [ft/s^2] "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))^3 q_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_fg Q_dot_boiling=W_dot_e*Convert(kW, Btu/h)

(27)

Ts [F] h [Btu/h.ft2_.F] W&e [kW] evap m& [lbm/h] 260 262 264 266 268 270 272 274 276 278 280 282 284 286 288 290 292 294 296 298 300 12919 18603 25321 33073 41857 51676 62528 74413 87332 101285 116271 132290 149343 167430 186550 206704 227891 250111 273366 297653 322974 4.956 8.564 13.6 20.3 28.9 39.65 52.77 68.51 87.11 108.8 133.8 162.4 194.8 231.2 272 317.2 367.2 422.2 482.4 548.1 619.5 17.89 30.91 49.08 73.27 104.3 143.1 190.5 247.3 314.4 392.7 483 586.1 703 834.5 981.5 1145 1325 1524 1741 1978 2236 260 265 270 275 280 285 290 295 300 0 250 500 750 1000 1250 1500 1750 2000 2250

T

s

[F]

m

evap

[

lb

m

/h

]

260 265 270 275 280 285 290 295 300 0 50000 100000 150000 200000 250000 300000 350000 0 100 200 300 400 500 600 700

T

_s

[F]

h

[B

tu

/h

-f

t

2

-F]

W

e

[k

W]

h

W

_e

(28)

10-35 Cold water enters a steam generator at 15°C and is boiled, and leaves as saturated vapor at Tsat = 200°C. The fraction

of heat used to preheat the liquid water from 15°C to saturation temperature of 200°C is 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 200°C is hfg = 1941 kJ/kg and the

specific heat of liquid water at the average temperature of (15+200)/2 = 107.5°C is (Table A-9). C kJ/kg 226 . 4 ⋅° = pl c Water, 200°C Steam generator Steam 200°C Water, 15°C

Analysis The heat of vaporization of water represents the amount of heat needed to

vaporize a unit mass of liquid at a specified temperature. Using the average specific heat, the amount of heat needed to preheat a unit mass of water from 15°C to 200°C is determined to be kJ/kg 782 = C ) 15 C)(200 kJ/kg 226 . 4 ( preheating =c ∆T = ⋅° − ° q pl

and q_total =q_boiling+q_preheating =1941+782=2723kJ/kg

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

) (or 2723 782 preheat o Fraction t total preheating 28.7% 0.287 = = = q q

10-36 Cold water enters a steam generator at 40°C and is boiled, and leaves as saturated vapor at boiler pressure. The boiler

pressure at which the amount of heat needed to preheat the water to saturation temperature 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 of

vaporization hfg and the specific heat of water cp at specified temperatures, and

they can be obtained from Table A-9.

Water, 100°C Steam generator Steam 100°C Water, 40°C

Analysis The heat of vaporization of water represents the amount of heat needed to

vaporize a unit mass of liquid at a specified temperature, and represents the amount of heat needed to preheat a unit mass of water from 40°C to the saturation temperature. Therefore, T c_p∆ sat @ sat , boiling preheating ) 40 ( _fg _T avg p T h c q q = − =

The solution of this problem requires choosing a boiling temperature, reading the heat of vaporization at that temperature, evaluating the specific heat at the average temperature, and substituting the values into the relation above to see if it is satisfied. By trial and error, the temperature that satisfies this condition is determined to be 320.4°C at which (Table A-9)

and T kJ/kg 1235 4 . 320 @ °C = fg h avg = (20+320.4)/2 = 180.2°C → c_p_,_avg =4.41kJ/kg⋅°C Substituting, kJ/kg 1237 C ) 40 4 . 320 )( C kJ/kg 41 . 4 ( ) 20 ( _sat , T − = ⋅° − ° = c_p_avg

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

MPa 11.3

= =P _T ₌ _°_C P_boiler _sat_@_sat ₃₂₀_.₄

(29)

10-37 Prob. 10-36 is reconsidered. The boiler pressure as a function of the cold water temperature is to be plotted. Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" T_cold=40 [C] "ANALYSIS" Fluid$='steam_IAPWS' q_preheating=q_boiling q_preheating=c_p*(T_sat-T_cold) T_sat=temperature(Fluid$, P=P, x=1) c_p=CP(Fluid$, T=T_ave, x=0) T_ave=1/2*(T_cold+T_sat) q_boiling=h_fg h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) 0 5 10 15 20 25 30 9600 9800 10000 10200 10400 10600 10800 11000

T

cold

[C]

P

[k

P

a

]

h_fg=h_g-h_f Tcold [C] P [kPa] 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 9726 9806 9886 9966 10046 10127 10207 10287 10368 10449 10529 10610 10691 10772 10852 10933