Heat 4e SM Chap10

10-1

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

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10-2

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 q&max 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.

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10-3

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

−5

µv = 2.045 × 10 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 q& film

⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4c pv (Ts − Tsat )] ⎤ ⎥ = C film ⎢ µ v D(Ts − Tsat ) ⎢⎣ ⎥⎦

1/ 4

(Ts − Tsat )

⎡ 9.81(0.04345) 3 (0.3831)(957.9 − 0.3831)[2257 × 10 3 + 0.4(1997)(400)] ⎤ = 0.62 ⎢ ⎥ (2.045 × 10 −5 )(0.002)(400) ⎣⎢ ⎦⎥

1/ 4

(400)

= 1.152 × 10 5 W/m 2

The radiation heat flux is determined from 4 q& rad = εσ (Ts4 − Tsat ) = (0.5)(5.67 × 10 −8 W/m 2 ⋅ K 4 )(773 4 − 373 4 ) K 4 = 9573 W/m 2

Then the total heat flux becomes q& total = q& film +

3 3 q& rad = 1.152 × 10 5 W/m 2 + (9573 W/m 2 ) = 1.224 × 10 5 W/m 2 4 4

Finally, the power dissipation per unit length of the metal rod is Q& total / L = πDq& total = π (0.002 m)(1.224 × 10 5 W/m 2 ) = 769 W/m

Discussion The contribution of radiation to the total heat flux is about 8%.

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10-4

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 q& nucleate

⎡ g (ρl − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

c pl (Ts − Tsat ) ⎤ ⎥ ⎢ ⎢⎣ C sf h fg Prln ⎥⎦

1/ 2 ⎡

3

⎡ 9.81(957.9 − 0.5978) ⎤ = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/ 2

⎡ ⎤ 4217(10) ⎢ ⎥ 3 ⎢⎣ (0.013)(2257 × 10 )(1.75) ⎥⎦

3

= 1.408 × 10 5 W/m 2

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

Q& boiling / L = πDq& nucleate = π (0.010 m)(1.408 × 10 5 W/m 2 ) = 4420 W/m The rate of evaporation per unit length is m& evaporation L

=

(Q& boiling / L) h fg

=

4420 J/s ⋅ m 2257 × 10 3 J/kg

= 1.96 × 10 − 3 kg/s ⋅ m

Discussion The value for the rate of evaporation per unit length indicates that 1 m of the platinum-plated rod would boil water at a rate of about 2 grams per second.

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10-5

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)

ρ l = 943.4 kg/m 3 ρ v = 1.121 kg/m 3 σ = 0.0550 N/m

h fg = 2203 × 10 3 J/kg

µ l = 0.232 × 10 −3 kg/m ⋅ s

120°C

c pl = 4244 J/kg ⋅ °C

Water

Prl = 1.44

130°C

Also, C sf = 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.

Heating

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 130 − 120 = 10°C which 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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(943.4 - 1.121) ⎤ = (0.232 × 10 )(2203 × 10 ) ⎢ ⎥ 0.0550 ⎣ ⎦ −3

3

1/2

⎞ ⎛ 4244(130 − 120) ⎟ ⎜ ⎜ 0.0130(2203 × 10 3 )1.44 ⎟ ⎠ ⎝

3

= 228,400 W/m 2 = 228.4 kW/m 2

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)

ρ l = 965.3 kg/m 3 ρ v = 0.4235 kg/m 3 σ = 0.0608 N/m

h fg = 2283 × 10 3 J/kg

µ l = 0.315 × 10 c pl

Water, 90°C

−3

kg/m ⋅ s = 4206 J/kg ⋅ °C

qmax Heating element

Prl = 1.96

Also, C sf = 0.0060 and n = 1.0 for the boiling of water on a brass heating (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, Ccr = 0.12 (Table 10-4). (It can be shown that L* = 1.58 > 1.2 and thus the restriction in Table 10-4 is satisfied). Analysis The maximum or critical heat flux is determined from

q& max = C cr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4 = 0.12(2283 × 10 3 )[0.0608 × 9.81× (0.4235) 2 (965.3 − 0.4235)]1 / 4 = 873,200 W/m 2 = 873.2 kW/m 2

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10-6

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 − Tsat = 125 − 120 = 5°C which is in the nucleate boiling range of 5 to 30°C for water. Properties The properties of water at the saturation temperature of 120°C are (Tables 10-1 and A-9)

ρ l = 943.4 kg/m 3 ρ v = 1.12 kg/m 3 σ = 0.0550 N/m Prl = 1.44 h fg = 2203 × 10 3 J/kg

µ l = 0.232 × 10

−3

Ts=125°C

Water 120°C

Heating element

kg ⋅ m/s

c pl = 4244 J/kg ⋅ °C

Also, C sf = 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. Analysis Assuming nucleate boiling, the heat flux in this case can be determined from Rohsenow relation to be q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(943.4 − 1.12) ⎤ = (0.232 × 10 )(2203 × 10 ) ⎢ ⎥ 0.0550 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4244(125 − 120) ⎜ ⎟ ⎜ 0.0060(2203 × 10 3 )1.44 ⎟ ⎝ ⎠

3

= 290,300 W/m 2

The surface area of the heater is As = πDL = π (0.02 m)(0.65 m) = 0.04084 m 2

Then the rate of heat transfer during nucleate boiling becomes Q& boiling = As q& nucleate = (0.04084 m 2 )(290,300 W/m 2 ) = 11,856 W

(b) The rate of evaporation of water is determined from m& evaporation =

Q& boiling h fg

=

⎛ 3600 s ⎞ ⎜ ⎟ = 19.4 kg/h 2203 × 10 J/kg ⎝ 1 h ⎠ 11,856 J/s 3

Therefore, steam can be produced at a rate of about 20 kg/h by this heater.

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10-7

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. 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 100°C are (Tables 10-1 and A-9)

ρ l = 957.9 kg/m 3

P = 1 atm

3

ρ v = 0.60 kg/m σ = 0.0589 N/m

100°C Water

Prl = 1.75

110°C

h fg = 2257 × 10 3 J/kg

µ l = 0.282 × 10 −3 kg ⋅ m/s

Heating

c pl = 4217 J/kg ⋅ °C

Also, C sf = 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. Analysis The excess temperature in this case is ∆T = T s − Tsat = 110 − 100 = 10°C which 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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠

3

⎡ 9.8(957.9 - 0.60) ⎤ = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

1/2

3

⎛ ⎞ 4217(110 − 100) ⎜ ⎟ ⎜ 0.0130(2257 × 10 3 )1.75 ⎟ ⎝ ⎠

3

= 140,700 W/m 2

The surface area of the bottom of the pan is As = πD 2 / 4 = π (0.30 m) 2 / 4 = 0.07069 m 2

Then the rate of heat transfer during nucleate boiling becomes Q& boiling = As q&nucleate = (0.07069 m 2 )(140,700 W/m 2 ) = 9945 W

(b) The rate of evaporation of water is determined from m& evaporation =

Q& boiling h fg

=

9945 J/s 2257 × 10 3 J/kg

= 0.00441 kg/s

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

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10-8

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)

ρ l = 957.9 kg/m 3

Water, 100°C

3

h fg = 2257 × 10 J/kg

ρ v = 0.60 kg/m 3 σ = 0.0589 N/m

Ts = ? qmax

µ l = 0.282 × 10 −3 kg ⋅ m/s

Heating element

c pl = 4217 J/kg ⋅ °C

Prl = 1.75

Also, C sf = 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, Ccr = 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

q& max = C cr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4 = 0.12(2257 × 10 3 )[0.0589 × 9.8 × (0.6) 2 (957.9 − 0.60)]1 / 4 = 1,017,000 W/m 2 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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.8(957.9 - 0.60) ⎤ 1,017,000 = (0.282 × 10 )(2257 ×10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0130(2257 ×10 3 )1.75 ⎟ ⎝ ⎠

3

It gives Ts = 119.3°C

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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10-9

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

[kW/m ] 871.9 880.3 888.6 896.8 904.9 912.8 920.7 928.4 936.1 943.6 951.1 958.5 965.8 973 980.1 987.2 994.1 1001 1008 1015

∆T [C] 20.12 20.07 20.02 19.97 19.92 19.88 19.83 19.79 19.74 19.7 19.66 19.62 19.58 19.54 19.5 19.47 19.43 19.4 19.36 19.33

1025

20.2 20.1

990

20 Heat 19.9

955

19.8 19.7

920 Temp. Dif.

∆ T [C]

2

2

q& max

qmax [kW/m ]

Psat [kPa] 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

19.6 19.5

885

19.4 850 70

75

80

85

90

95

100

19.3 105

Psat [kPa]

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10-10

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. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. Properties The properties of water at the saturation temperature of 212°F are ρ l = 59.82 lbm/ft 3 and h fg = 970 Btu/lbm

(Table A-9E). The properties of the vapor at the film temperature of T f = (Tsat + Ts ) / 2 = (212 + 788) / 2 = 500°F are (Table A-16E)

P = 1 atm

ρ v = 0.02571 lbm/ft

3

Water, 212°F

µ v = 1.267 × 10 −5 lbm/ft ⋅ s = 0.04561 lbm/ft ⋅ h c pv = 0.4707 Btu/lbm ⋅ °F

Heating element

k v = 0.02267 Btu/h ⋅ ft ⋅ °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 ∆T = Ts − Tsat = 788 − 212 = 576°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 to be q& film

⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4c pv (Ts − Tsat )] ⎤ ⎥ = 0.62 ⎢ µ v D(Ts − Tsat ) ⎢⎣ ⎥⎦

1/ 4

(Ts − Tsat )

⎡ 32.2(3600) 2 (0.02267) 3 (0.02571)(59.82 − 0.02571)[970 + 0.4 × 0.4707(788 − 212)] ⎤ = 0.62 ⎢ ⎥ (0.04561)(0.5 / 12)(788 − 212) ⎦⎥ ⎣⎢

1/ 4

(788 − 212)

= 18,600 Btu/h ⋅ ft 2

The radiation heat flux is determined from 4 ) q& rad = εσ (Ts4 − Tsat

[

= (0.2)(0.1714 × 10 −8 Btu/h ⋅ ft 2 ⋅ R 4 ) (788 + 460 R) 4 − (212 + 460 R) 4 = 761.7 Btu/h ⋅ ft

]

2

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 q& total = q& film +

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

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, Q& total = As q& total = (πDL)q& total = (π × 0.5 / 12 ft × 1 ft)(19,171 Btu/h ⋅ ft 2 ) = 2509 Btu/h

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10-11

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. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. Properties The properties of water at the saturation temperature of 212°F are ρ l = 59.82 lbm/ft 3 and h fg = 970 Btu/lbm

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

P = 1 atm

ρ v = 0.02395 lbm/ft 3

Water, 212°F

µ v = 1.416 ×10 −5 lbm/ft ⋅ s = 0.05099 lbm/ft ⋅ h c pv = 0.4799 Btu/lbm ⋅ °F

Heating element

k v = 0.02640 Btu/h ⋅ ft ⋅ °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 = Ts − Tsat = 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 q& film

⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4C pv (Ts − Tsat )] ⎤ ⎥ = 0.62 ⎢ µ v D(Ts − Tsat ) ⎢⎣ ⎥⎦

1/ 4

(Ts − Tsat )

⎡ 32.2(3600) 2 (0.0264) 3 (0.02395)(59.82 − 0.02395)[970 + 0.4 × 0.4799(988 − 212)] ⎤ = 0.62 ⎢ ⎥ (0.05099)(0.5 / 12)(988 − 212) ⎣⎢ ⎦⎥

1/ 4

(988 − 212)

= 25,147 Btu/h ⋅ ft 2

The radiation heat flux is determined from 4 ) q& rad = εσ (Ts4 − Tsat

[

= (0.2)(0.1714 × 10 −8 Btu/h ⋅ ft 2 ⋅ R 4 ) (988 + 460 R) 4 − (212 + 460 R) 4

]

= 1437 Btu/h ⋅ ft 2 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 q& total = q& film +

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

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, Q& total = As q& total = (πDL)q& total = (π × 0.5 / 12 ft × 1 ft)(26,225 Btu/h ⋅ ft 2 ) = 3433 Btu/h

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10-12

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 q& nucleate

⎡ g (ρl − ρv ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

c pl (Ts − Tsat ) ⎤ ⎥ ⎢ ⎢⎣ C sf h fg Prln ⎥⎦

1/ 2 ⎡

3

⎡ 9.81(957.9 − 0.5978) ⎤ = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/ 2

⎡ ⎤ 4217(10) ⎢ ⎥ 3 ⎢⎣ (0.013)(2257 × 10 )(1.75) ⎥⎦

3

= 1.408 × 10 5 W/m 2

Using the Newton’s law of cooling, the boiling heat transfer coefficient is q& nucleate = h(Ts − Tsat )

h=

→

h=

q& nucleate Ts − Tsat

1.408 × 10 5 W/m 2 = 14,100 W/m 2 ⋅ K (110 − 100) K

Discussion Heat transfer coefficient on the order of 104 W/m2·K can be obtained in nucleate boiling with a temperature difference of just 10°C.

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10-13

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

q& boiling = h(Ts − Tsat )

→

h=

q& boiling Ts − Tsat

Also, we know Q& boiling / L = πDq& boiling = 4100 W/m

q& boiling =

4100 W/m 4100 W/m = = 1.305 × 10 6 W/m 2 πD π (0.001 m)

Hence, the nucleate boiling heat transfer coefficient is h=

1.305 × 10 6 W/m = 46,600 W/m 2 ⋅ K (128 − 100) K

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

⎡ g(ρl − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎡

c pl (Ts − Tsat ) ⎤ ⎢ ⎥ ⎢⎣ C sf h fg Prln ⎥⎦

3

or 1/ 3

C sf

3⎫ ⎧ 1/ 2 ⎪ µ l h fg ⎡ g ( ρ l − ρ v ) ⎤ ⎡ c pl (Ts − Tsat ) ⎤ ⎪ ⎥ ⎬ =⎨ ⎥ ⎢ ⎢ n σ ⎥⎦ ⎪ ⎦ ⎢⎣ h fg Prl ⎪ q& boiling ⎣ ⎩ ⎭

1/ 3

C sf

3⎫ 1/ 2 ⎧ −3 3 ⎪ (0.282 × 10 )(2257 × 10 ) ⎡ 9.81(957.9 − 0.5978) ⎤ ⎡ 4217(128 − 100) ⎤ ⎪ =⎨ ⎥ ⎬ ⎢ ⎥ ⎢ 0.0589 1.305 × 10 6 ⎣ ⎦ ⎣⎢ (2257 × 10 3 )(1.75) ⎦⎥ ⎪ ⎪⎩ ⎭ = 0.0173

Discussion The boiling heat transfer coefficient of 46,600 W/m2·K is within the range suggested by Table 1-5 for boiling and condensation (2500 to 100,000 W/m2·K).

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10-14

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

ρ l = 957.9 kg/m 3

Coffee maker

3

ρ v = 0.60 kg/m σ = 0.0589 N/m Prl = 1.75

P = 1 atm

Water, 100°C 1L

3

h fg = 2257 × 10 J/kg

µ l = 0.282 × 10 −3 kg ⋅ m/s c pl = 4217 J/kg ⋅ °C

Also, C sf = 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. 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 Q = Q& ∆t = mh fg → Q& =

mh fg ∆t

=

(0.5 kg)(2257 kJ/kg) = 0.5878 kW (32 × 60 s)

As = πDL = π (0.004 m)(0.30 m) = 0.003770 m 2 q& = Q& / As = (0.5878 kW)/(0.003770 m 2 ) = 155.92 kW/m 2 = 155,920 W/m 2

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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(957.9 − 0.60) ⎤ 155,920 = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0130(2257 × 10 3 )1.75 ⎟ ⎝ ⎠

3

It gives Ts = 110.3°C

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

Q = Q& ∆t = mc p ∆T → ∆t =

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

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10-15

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)

ρ l = 957.9 kg/m 3

Coffee maker

ρ v = 0.60 kg/m 3 σ = 0.0589 N/m

P = 1 atm

Water, 100°C 1L

Prl = 1.75 h fg = 2257 × 10 3 J/kg

µ l = 0.282 × 10 −3 kg ⋅ m/s c pl = 4217 J/kg ⋅ °C

Also, C sf = 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. 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 Q = Q& ∆t = mh fg → Q& =

mh fg ∆t

=

(0.5 kg)(2257 kJ/kg) = 0.5878 kW (32 × 60 s)

As = πDL = π (0.004 m)(0.30 m) = 0.003770 m 2 q& = Q& / As = (0.5878 kW)/(0.003770 m 2 ) = 155.92 kW/m 2 = 155,920 W/m 2

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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(957.9 − 0.60) ⎤ 155,920 = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0130(2257 × 10 3 )1.75 ⎟ ⎝ ⎠

3

It gives Ts = 110.3°C

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

Q = Q& ∆t = mc p ∆T → ∆t =

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

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10-16

10-24 Water is boiled at Tsat = 90°C in a brass heating element. The surface temperature of the heater is to be determined. 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 90°C are (Tables 10-1 and A-9)

ρ l = 965.3 kg/m 3 ρ v = 0.4235 kg/m 3 σ = 0.0608 N/m

Water, 90°C

h fg = 2283 × 10 3 J/kg

qmin

µ l = 0.315 × 10 −3 kg/m ⋅ s

c pl = 4206 J/kg ⋅ °C

Heating element

Prl = 1.96

Also, C sf = 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 ⎡ σg ( ρ l − ρ v ) ⎤ q& min = 0.09 ρ v h fg ⎢ ⎥ 2 ⎢⎣ ( ρ l + ρ v ) ⎥⎦

1/ 4

⎡ (0.0608)(9.81)(965.3 − 0.4235) ⎤ = 0.09(0.4235)(2283 × 10 ) ⎢ ⎥ (965.3 + 0.4235) 2 ⎢⎣ ⎥⎦

1/ 4

3

= 13,715 W/m 2

The surface temperature can be determined from Rohsenow equation to be q& nucleate

⎡ g(ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(965.3 - 0.4235) ⎤ 13,715 W/m = (0.315 × 10 )(2283 × 10 ) ⎢ ⎥ 0.0608 ⎣ ⎦ 2

−3

3

1/2

⎛ ⎞ 4206(Ts − 90) ⎜ ⎟ ⎜ 0.0060(2283 × 10 3 )1.96 ⎟ ⎝ ⎠

3

Ts = 92.3°C

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10-17

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. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. Properties The properties of water at the saturation temperature of 100°C are (Tables 10-1 and A-9)

ρ l = 957.9 kg/m 3 ρ v = 0.5978 kg/m σ = 0.0589 N/m

h fg = 2257 × 10 3 J/kg

3

µ l = 0.282 × 10 −3 kg ⋅ m/s c pl = 4217 J/kg ⋅ °C

Prl = 1.75

Also, C sf = 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 Tf = (Ts+Tsat )/2 of 1000°C (will be checked) (Table A-16) c pv = 2471 J/kg ⋅ °C ρ v = 0.1725 kg/m 3 P = 1 atm k v = 0.1362 W/m ⋅ °C µ v = 4.762 × 10 −5 kg/m ⋅ s Water, 100°C

Analysis (a) For a horizontal heating element, the coefficient Ccr is determined from Table 10-4 to be ⎛ g (ρ l − ρ v ) ⎞ L* = L⎜ ⎟ σ ⎝ ⎠

1/ 2

⎛ 9.81(957.9 − 0.5978) ⎞ = (0.002)⎜ ⎟ 0.0589 ⎠ ⎝

Ts

qmax

1/ 2

= 0.7986 < 1.2

Heating element

C cr = 0.12 L *−0.25 = 0.12(0.7986) −0.25 = 0.1269

Then the maximum or critical heat flux is determined from q& max = C cr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4 = 0.1269( 2257 × 10 3 )[0.0589 × 9.81 × (0.5978) 2 (957.9 − 0.5978)]1 / 4 = 1,074,000 W/m 2 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

q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(957.9 - 0.5978) ⎤ 1,074,000 = (0.282 × 10 )(2257 × 10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0060(2257 × 10 3 )1.75 ⎟ ⎝ ⎠

3

It gives Ts = 109.1°C (b) Heat transfer in the film boiling region can be expressed as q& total = q& film

⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4c pv (T s − T sat )] ⎤ 3 ⎥ + q& rad = 0.62 ⎢ 4 µ v D (Ts − Tsat ) ⎥⎦ ⎢⎣

1/ 4

(T s − Tsat ) +

3 4 εσ (Ts4 − Tsat ) 4

Substituting, ⎡ 9.81(0.1362) 3 (0.1725)(957.9 − 0.1725)[2257 × 10 3 + 0.4 × 2471(Ts − 100)] ⎤ 1,074,000 = 0.62⎢ ⎥ (4.762 × 10 −5 )(0.004)(Ts − 100) ⎣⎢ ⎦⎥

[

1/ 4

]

3 (0.3)(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (Ts + 273) 4 − (100 + 273) 4 4 Solving for the surface temperature gives Ts = 2200°C. Therefore, the temperature jump of the wire when the operating point jumps from nucleate boiling to film boiling is ∆T = Ts, film − Ts,crit = 2200 − 109 = 2091°C Temperature jump: × (Ts − 100) +

Note that the film temperature is (2200+100)/2=1150°C, which is close enough to the assumed value of 1000°C for the evaluation of vapor paroperties.

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10-18

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. Analysis The problem is solved using EES, and the solution is given below. "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" n=1 "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_filmT_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

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10-19

q& max

2

[kW/m ] 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762 1073762

∆T [C] 2760 2535 2380 2262 2168 2091 2025 1967 1917 1872 1831 1794 1761 1730 1701 1674 1649 1626 1604

1.080x106

2180

1.060x106 2160

1.040x106 1.020x106

2140

1000000

2120

980000

∆ T [C]

2

qmax [W/m ]

[kW/m ] 925656 934417 943050 951559 959948 968222 976385 984439 992389 1000237 1007987 1015642 1023205 1030677 1038062 1045363 1052581 1059719 1066778 1073762

∆T [C] 2073 2079 2084 2090 2096 2101 2107 2112 2117 2122 2127 2132 2137 2142 2146 2151 2155 2160 2164 2168

2100

960000 940000

2080

920000 70

75

80

85

90

95

100

105

P [kPa]

1.5x10 6

2800 2600

1.3x10 6 2400

2

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 1

2

qmax [W/m ]

ε

q& max

∆ T [C]

P [kPa] 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

1.0x10 6

2200 2000

7.5x10 5 1800 1600 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

5.0x10 5 1

ε

PROPRIETARY MATERIAL. © 2011 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.

10-20

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

ρ l = 957.9 kg/m 3

P = 1 atm

3

ρ v = 0.60 kg/m σ = 0.0589 N/m

100°C Water

Prl = 1.75

Ts 3

h fg = 2257 × 10 J/kg

µ l = 0.282 × 10 −3 kg ⋅ m/s c pl = 4217 J/kg ⋅ °C

Heating

Also, C sf = 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. Analysis The rate of heat transfer to the water and the heat flux are m& evap =

m evap

=

∆t Q& = m& evap h fg

(957.9 kg/m 3 )(π × (0.2 m) 2 /4 × 0.10 m) = 0.003344 kg/s 15 × 60 s ∆t = (0.03344 kg/s)(2257 kJ/kg) = 7.547 kW

ρ∆V

=

As = πD 2 / 4 = π (0.20 m) 2 / 4 = 0.03142 m 2 q& = Q& / A = (7547 W)/(0.03142 m 2 ) = 240,200 W/m 2 s

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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.8(957.9 − 0.60) ⎤ 240,200 = (0.282 × 10 )(2257 ×10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0058(2257 × 10 3 )1.75 ⎟ ⎝ ⎠

3

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.

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10-21

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

ρ l = 957.9 kg/m 3

P = 1 atm

3

ρ v = 0.60 kg/m σ = 0.0589 N/m

100°C Water

Prl = 1.75

Ts 3

h fg = 2257 × 10 J/kg

µ l = 0.282 × 10 −3 kg ⋅ m/s c pl = 4217 J/kg ⋅ °C

Heating

Also, C sf = 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. Analysis The rate of heat transfer to the water and the heat flux are m& evap =

m evap

=

∆t Q& = m& evap h fg

(957.9 kg/m 3 )(π × (0.2 m) 2 /4 × 0.10 m) = 0.003344 kg/s 15 × 60 s ∆t = (0.03344 kg/s)(2257 kJ/kg) = 7.547 kW

ρ∆V

=

As = πD 2 / 4 = π (0.20 m) 2 / 4 = 0.03142 m 2 q& = Q& / A = (7547 W)/(0.03142 m 2 ) = 240,200 W/m 2 s

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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.8(957.9 − 0.60) ⎤ 240,200 = (0.282 × 10 )(2257 ×10 ) ⎢ ⎥ 0.0589 ⎣ ⎦ −3

3

1/2

⎛ ⎞ 4217(Ts − 100) ⎜ ⎟ ⎜ 0.0130(2257 ×10 3 )1.75 ⎟ ⎝ ⎠

3

It gives Ts = 111.9°C

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

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10-22

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

As = πDL = π (0.002 m)(0.50 m) = 0.003142 m 2

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 Q& = hAs (Ts − Tsat ) → h =

Heating wire, 3.8 kW

Q& 3800 W = = 40,320 W/m 2 ⋅ °C As (Ts − Tsat ) (0.003142 m 2 )(130 − 100)°C

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. 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)

ρ l = 943.4 kg/m 3 ρ v = 1.121 kg/m 3 σ = 0.0550 N/m

h fg = 2203 × 10 3 J/kg

µ l = 0.232 × 10 −3 kg/m ⋅ s

120°C

c pl = 4244 J/kg ⋅ °C

Water

Prl = 1.44

128°C

Also, C sf = 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.

Heating

Analysis The excess temperature in this case is ∆T = Ts − Tsat = 128 − 120 = 8°C which 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 q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(943.4 - 1.121) ⎤ = (0.232 × 10 −3 )(2203 × 10 3 ) ⎢ ⎥ 0.0550 ⎣ ⎦

1/2

⎛ ⎞ 4244(128 − 120) ⎜ ⎟ ⎜ 0.0130(2203 × 10 3 )1.44 ⎟ ⎝ ⎠

3

= 116,900 W/m 2

The boiling heat transfer coefficient is ⎯→ h = q& nucleate = h(Ts − Tsat ) ⎯

q& nucleate 116,900 W/m 2 = = 14,610 W/m 2 ⋅ °C = 14.6 kW/m 2 ⋅ C (128 − 120)°C Ts − Tsat

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10-23

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

ρv = 0.3262 kg/m3 −5

µv = 2.446 × 10 kg/m·s

cpv = 2066 J/kg·K 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 ⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4c pv (Ts − Tsat )] ⎤ ⎥ q& film = C film ⎢ µ v D(Ts − Tsat ) ⎢⎣ ⎥⎦

1/ 4

(Ts − Tsat )

⎡ 9.81(0.05467) 3 (0.3262)(957.9 − 0.3262)[2257 × 10 3 + 0.4(2066)(600)] ⎤ = 0.67 ⎢ ⎥ (2.446 × 10 −5 )(0.02)(600) ⎦⎥ ⎣⎢

1/ 4

(600)

= 1.052 × 10 5 W/m 2

The radiation heat flux is determined from 4 ) q& rad = εσ (Ts4 − Tsat

= (0.75)(5.67 × 10 −8 W/m 2 ⋅ K 4 )(973 4 − 373 4 ) K 4 = 3.729 × 10 4 W/m 2 Then the total heat flux becomes q& total = q& film +

3 3 q& rad = 1.052 × 10 5 W/m 2 + (3.729 × 10 4 W/m 2 ) = 1.332 × 10 5 W/m 2 4 4

Finally, the initial heat transfer rate from the submerged metal sphere is Q& total = q& total πD 2 = (1.332 × 10 5 W/m 2 )π (0.02 m) 2 = 167 W

Discussion The contribution of radiation to the total heat flux is about 21%, which is significant and cannot be neglected.

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10-24

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. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is nucleate boiling since ∆T = Ts − Tsat = 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)

ρ l = 58.82 lbm/ft 3 ρ v = 0.0723 lbm/ft 3

Water 250°F

σ = 0.003755 lbf/ft = 0.1208 lbm/s 2 Prl = 1.43

Ts=280°F

Heating element

h fg = 946 Btu/lbm

µ l = 1.544 × 10 − 4 lbm/ft ⋅ s = 0.556 lbm/ft ⋅ h c pl = 1.015 Btu/lbm ⋅ °F

Also, g = 32.2 ft/s2 and C sf = 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. Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

⎡ 32.2(58.82 − 0.0723) ⎤ = (0.556)(946) ⎢ ⎥ 0.1208 ⎣ ⎦

1/2

3

⎛ 1.015(280 − 250) ⎞ ⎜⎜ ⎟⎟ ⎝ 0.0060(946)1.43 ⎠

3

= 3,475,221 Btu/h ⋅ ft 2

Then the convection heat transfer coefficient becomes q& = h(Ts − Tsat ) → h =

q& 3,475,221 Btu/h ⋅ ft 2 = = 115,840 Btu/h ⋅ ft 2 ⋅ °F (280 − 250)°F Ts − Tsat

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

W& e = Q& = q&As = (πDL)q& = (π × 0.25 / 12 ft × 2 ft)(3,475,221 Btu/h ⋅ ft 2 ) = 454,905 Btu/h = 133.3 kW

(since 1 kW = 3412 Btu/h)

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

Q& boiling h fg

=

454,905 Btu/h = 480.9 lbm/h 946 Btu/lbm

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10-25

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. 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 − Tsat = 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)

ρ l = 58.82 lbm/ft 3 ρ v = 0.0723 lbm/ft 3

Water 250°F

σ = 0.003755 lbf/ft = 0.1208 lbm/s 2 Prl = 1.43

Ts=280°F

Heating element

h fg = 946 Btu/lbm

µ l = 1.544 × 10 − 4 lbm/ft ⋅ s = 0.556 lbm/ft ⋅ h c pl = 1.015 Btu/lbm ⋅ °F

Also, g = 32.2 ft/s2 and C sf = 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. Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be q& nucleate

⎡ g (ρ l − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠

1/ 2 ⎛

⎡ 32.2(58.82 − 0.0723) ⎤ = (0.556)(946) ⎢ ⎥ 0.1208 ⎣ ⎦

1/2

3

⎛ ⎞ 1.015(280 − 250) ⎜ ⎟ ⎜ 0.0130(0.1208 × 10 3 )1.43 ⎟ ⎝ ⎠

3

= 341,670 Btu/h ⋅ ft 2

Then the convection heat transfer coefficient becomes q& = h(Ts − Tsat ) → h =

q& 341,670 Btu/h ⋅ ft 2 = = 11,390 Btu/h ⋅ ft 2 ⋅ °F (280 − 250)°F Ts − Tsat

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

W& e = Q& = q&As = (πDL)q& = (π × 0.25 / 12 ft × 2 ft)(341,670 Btu/h ⋅ ft 2 ) = 44,724 Btu/h = 13.1 kW

(since 1 kW = 3412 Btu/h)

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

Q& boiling h fg

=

44,724 Btu/h = 47.3 lbm/h 946 Btu/lbm

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10-26

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. Analysis The problem is solved using EES, and the solution is given below. "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" n=1 "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)

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10-27

12919 18603 25321 33073 41857 51676 62528 74413 87332 101285 116271 132290 149343 167430 186550 206704 227891 250111 273366 297653 322974

m& evap

2250

[lbm/h] 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

2000 1750 1500 1250 1000 750 500 250 0 260

265

270

300000

600

250000

500

200000

400

2

700

h

300

We

100000

285

290

295

200

50000 0 260

280

Ts [F]

350000

150000

275

We [kW]

260 262 264 266 268 270 272 274 276 278 280 282 284 286 288 290 292 294 296 298 300

W& e [kW] 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

mevap [lbm/h]

h [Btu/h.ft2.F]

h [Btu/h-ft -F]

Ts [F]

100

265

270

275

280

285

290

295

0 300

Ts [F]

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300

10-28

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 c pl = 4.226 kJ/kg ⋅ °C (Table A-9). 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

Steam 200°C

Steam generator

Water, 200°C

q preheating = c pl ∆T = (4.226 kJ/kg ⋅ °C)(200 − 15)°C = 782 kJ/kg

and

q total = q boiling + q preheating = 1941 + 782 = 2723 kJ/kg

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

q preheating q total

=

Water, 15°C

782 = 0.287 (or 28.7%) 2723

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.

Steam generator

Steam 100°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 c p ∆T represents the

amount of heat needed to preheat a unit mass of water from 40°C to the saturation temperature. Therefore,

Water, 100°C

q preheating = q boiling c p ,avg (Tsat − 40) = h fg @ Tsat 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) h fg @ 320.4°C = 1235 kJ/kg

Water, 40°C

and Tavg = (20+320.4)/2 = 180.2°C → c p ,avg = 4.41 kJ/kg ⋅ °C

Substituting, c p ,avg (Tsat − 20) = (4.41 kJ/kg ⋅ °C)(320.4 − 40)°C = 1237 kJ/kg

which is practically identical to the heat of vaporization. Therefore, Pboiler = Psat @ Tsat =320.4°C = 11.3 MPa

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10-29

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) h_fg=h_g-h_f

P [kPa] 9726 9806 9886 9966 10046 10127 10207 10287 10368 10449 10529 10610 10691 10772 10852 10933

11000 10800 10600

P [kPa]

Tcold [C] 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

10400 10200 10000 9800 9600 0

5

10

15

20

25

30

Tcold [C]

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10-30

10-38 Water is boiled at a temperature of Tsat = 150°C by hot gases flowing through a mechanically polished stainless steel pipe submerged in water whose outer surface temperature is maintained at Ts = 160°C. 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 ∆T = Ts − Tsat = 160 − 150 = 10°C which is in the nucleate boiling range of 5 to 30°C for water. Properties The properties of water at the saturation temperature of 150°C are (Tables 10-1 and A-9) ρ l = 916.6 kg/m 3 h = 2114 × 10 3 J/kg fg

ρ v = 2.55 kg/m 3 σ = 0.0488 N/m

µ l = 0.183 × 10 −3 kg ⋅ m/s

c pl = 4311 J/kg ⋅ °C Prl = 1.16 Also, C sf = 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. Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be q& nucleate

⎡ g (ρl − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎣ ⎦

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(916.6 − 2.55) ⎤ = (0.183 × 10 )(2114 × 10 ) ⎢ ⎥ 0.0488 ⎣ ⎦ −3

3

Vent Boiler 1/2

⎛ ⎞ 4311(160 − 150) ⎜ ⎟ ⎜ 0.0130(2114 × 10 3 )1.16 ⎟ ⎝ ⎠

3

= 410,090 W/m 2 The heat transfer surface area is As = πDL = π (0.05 m)(25 m) = 3.927 m 2 Then the rate of heat transfer during nucleate boiling becomes Q& boiling = As q& nucleate = (3.927 m 2 )(410,090 W/m 2 ) = 1,610,400 W = 1610 kW

Water, 150°C Ts,pipe = 160°C

(b) The rate of evaporation of water is determined from Q& boiling 1610 kJ/s = = 0.762 kg/s m& evaporation = 2114 kJ/kg h fg

Hot gases

(c) For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 1/ 2

⎛ g (ρ l − ρ v ) ⎞ ⎛ 9.8(916.6 − 2.55) ⎞ L* = L⎜⎜ ⎟⎟ = (0.025)⎜ ⎟ 0.0488 σ ⎝ ⎠ ⎝ ⎠ C cr = 0.12 (since L * > 1.2 and thus large cylinder) Then the maximum or critical heat flux is determined from q& max = C cr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4

1/ 2

= 10.7 > 1.2

= 0.12( 2114 × 10 3 )[0.0488 × 9.8 × ( 2.55) 2 (916.6 − 2.55)]1 / 4 = 1,852,000 W/m 2

Therefore, q& max 1,852,000 = = 4.52 410,090 q& current (d) The surface temperature of the pipe at the burnout point is determined from Rohsenow relation at the critical heat flux value to be q& nucleate,cr

⎡ g(ρl − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

⎞ ⎜ c p ,l (Ts ,cr − Tsat ) ⎟ n ⎟ ⎜ C h Pr sf fg l ⎠ ⎝

1/ 2 ⎛

3

⎡ 9.8(916.6 − 2.55) ⎤ 1,852,000 = (0.183 × 10 )(2114 × 10 ) ⎢ ⎥ 0.0488 ⎣ ⎦ −3

3

1/2

⎞ ⎛ 4311(Ts ,cr − 150) ⎟ ⎜ ⎜ 0.0130(2114 × 10 3 )1.16 ⎟ ⎠ ⎝

3

Ts ,cr = 166.5°C

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10-31

10-39 Water is boiled at a temperature of Tsat = 155°C by hot gases flowing through a mechanically polished stainless steel pipe submerged in water whose outer surface temperature is maintained at Ts = 160°C. 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 ∆T = Ts − Tsat = 160 − 155 = 5°C which is in the nucleate boiling range of 5 to 30°C for water. Properties The properties of water at the saturation temperature of 155°C are (Tables 10-1 and A-9) ρ l = 912.0 kg/m 3 h = 2099 × 10 3 J/kg fg

ρ v = 2.901 kg/m 3 σ = 0.0477 N/m

µ l = 0.177 × 10 −3 kg ⋅ m/s

c pl = 4326 J/kg ⋅ °C Prl = 1.125 Also, C sf = 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. Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be q& nucleate

⎡ g (ρl − ρ v ) ⎤ = µ l h fg ⎢ ⎥ σ ⎦ ⎣

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(912.0 − 2.901) ⎤ = (0.177 × 10 )(2099 × 10 ) ⎢ ⎥ 0.0477 ⎣ ⎦ −3

3

Vent 1/2

⎞ ⎛ 4326(160 − 155) ⎟ ⎜ ⎜ 0.0130(2099 × 10 3 )1.125 ⎟ ⎠ ⎝

3

Boiler

= 56,197 W/m 2 The heat transfer surface area is As = πDL = π (0.05 m)(25 m) = 3.927 m 2 Then the rate of heat transfer during nucleate boiling becomes Q& boiling = As q& nucleate = (3.927 m 2 )(56,197 W/m 2 ) = 220,700 W

Water, 155°C Ts,pipe = 160°C

(b) The rate of evaporation of water is determined from Q& boiling 220.7 kJ/s = = 0.105 kg/s m& evaporation = 2099 kJ/kg h fg

Hot gases

(c) For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 1/ 2

⎛ g (ρl − ρ v ) ⎞ ⎛ 9.81(912.0 − 2.901) ⎞ L* = L⎜ ⎟ = (0.025)⎜ ⎟ σ 0.0477 ⎝ ⎠ ⎝ ⎠ C cr = 0.12 (since L * > 1.2 and thus large cylinder) Then the maximum or critical heat flux is determined from q& max = Ccr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4

1/ 2

= 10.8 > 0.12

= 0.12( 2099 × 10 3 )[0.0477 × 9.81 × ( 2.901) 2 (912.0 − 2.901)]1 / 4 = 1.948 × 10 6 W/m 2

Therefore, q& max 1.948 × 10 6 W/m 2 = = 34.7 q& current 56,197 W/m 2

(d) The surface temperature of the pipe at the burnout point is determined from Rohsenow relation at the critical heat flux value to be ⎡ g (ρl − ρ v ) ⎤ 1.948 × 10 = µ l h fg ⎢ ⎥ σ ⎦ ⎣ 6

1/ 2 ⎛

⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ ⎜ C h Pr n ⎟ ⎝ sf fg l ⎠

3

⎡ 9.81(912.0 − 2.901) ⎤ = (0.177 × 10 −3 )(2099 × 10 3 ) ⎢ ⎥ 0.0477 ⎣ ⎦

1/2

⎞ ⎛ 4326(Ts ,cr − 155) ⎟ ⎜ ⎜ 0.0130(2099 × 10 3 )1.125 ⎟ ⎠ ⎝

3

Ts ,cr = 171.3°C

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10-32

10-40 The electrical current at which a nickel wire would be in danger of burnout in nucleate boiling 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 Analysis The danger of burnout occurs when the heat flux is at maximum in nucleate boiling, which can be determined using q& max = C cr h fg [σgρ v2 ( ρ l − ρ v )]1 / 4

Using Table 10-4, the parameter L* and the constant Ccr are determined to be

⎡ g (ρl − ρ v ) ⎤ L* = L ⎢ ⎥ σ ⎣ ⎦

1/ 2

⎡ 9.81(957.9 − 0.5978) ⎤ = (0.0005) ⎢ ⎥ 0.0589 ⎣ ⎦

1/ 2

= 0.1997

which correspond to C cr = 0.12 L *−0.25 = 0.12(0.1997) −0.25 = 0.1795

Hence, the maximum heat flux is

q& max = (0.1795)(2257 × 10 3 )[(0.0589)(9.81)(0.5978) 2 (957.9 − 0.5978)]1 / 4 = 1.519 × 10 6 W/m 2 We know that q& =

I 2R I 2R = As πDL

Thus, the electrical current at which the wire would be in danger of burnout is

⎡ q& πD ⎤ I = ⎢ max ⎥ ⎣ ( R / L) ⎦

1/ 2

⎡ (1.519 × 10 6 W/m 2 )π (0.001 m) ⎤ =⎢ ⎥ 0.129 Ω/m ⎢⎣ ⎥⎦

1/ 2

= 192 A

Discussion The electrical current at which burnout could occur will decrease if the resistance of the wire increases.

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10-33

10-41 The initial heat transfer rate from a hot steel rod that is suddenly submerged in a water bath is to be determined. Assumptions 1 Steady operating condition exists. 2 The steel rod 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 Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16,

ρv = 0.3831 kg/m3 −5

µv = 2.045 × 10 kg/m·s

cpv = 1997 J/kg·K 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 ⎡ gk v3 ρ v ( ρ l − ρ v )[h fg + 0.4c pv (Ts − Tsat )] ⎤ ⎥ q& film = C film ⎢ µ v D(Ts − Tsat ) ⎢⎣ ⎥⎦

1/ 4

(Ts − Tsat )

⎡ 9.81(0.04345) 3 (0.3831)(957.9 − 0.3831)[2257 × 10 3 + 0.4(1997)(400)] ⎤ = 0.62 ⎢ ⎥ (2.045 × 10 −5 )(0.02)(400) ⎣⎢ ⎦⎥

1/ 4

(400)

= 6.476 × 10 4 W/m 2

The radiation heat flux is determined from 4 ) q& rad = εσ (Ts4 − Tsat

= (0.9)(5.67 × 10 −8 W/m 2 ⋅ K 4 )(773 4 − 373 4 ) K 4 = 1.723 × 10 4 W/m 2 Then the total heat flux becomes q& total = q& film +

3 3 q& rad = 6.476 × 10 4 W/m 2 + (1.723 × 10 4 W/m 2 ) = 7.768 × 10 4 W/m 2 4 4

Finally, the initial heat transfer rate from the submerged steel rod is Q& total = q& totalπDL = (7.768 × 10 4 W/m 2 )π (0.02 m)(0.2 m) = 976 W

Discussion The contribution of radiation to the total heat flux is about 17%, which is significant and cannot be neglected.

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10-34

10-42 Water is boiled at Tsat = 100°C by a spherical platinum heating element immersed in water. The surface temperature is Ts = 350°C. The rate of heat transfer 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 100°C are (Table A-9)

h fg = 2257 × 10 3 J/kg

ρ l = 957.9 kg/m 3 The properties of water vapor at (350+100)/2 = 225°C are (Table A-16)

ρ v = 0.444 kg/m

350°C

3

Water 100°C

µ v = 1.749 ×10 −5 kg/m ⋅ s c pv = 1951 J/kg ⋅ °C k v = 0.03581 W/m ⋅ °C

Analysis The film boiling occurs since the temperature difference between the surface and the fluid. The heat flux in this case can be determined from q& film

[

⎡ gk v3 ρ v ( ρ l − ρ v ) h fg + 0.4c pv (Ts − Tsat ) = 0.67 ⎢ µ v D(Ts − Tsat ) ⎢⎣

]⎤⎥ 1 / 4 (T

[

⎥⎦

s

− Tsat )

⎡ (9.81)(0.03581) 3 (0.444)(957.9 − 0.444) 2257 × 10 3 + 0.4(1951)(350 − 100) = 0.67 ⎢ (1.749 × 10 −5 )(0.15)(350 − 100) ⎢⎣

]⎤⎥ ⎥⎦

1/ 4

(350 − 100)

= 25,207 W/m 2

The radiation heat transfer is

[

]

4 q& rad = εσ (T s4 − Tsat ) = (0.10)(5.67 × 10 −8 ) (350 + 273) 4 − (100 + 273) 4 = 745 W/m 2

The total heat flux is

q& total = q& film +

3 3 q& rad = 25,207 + (745) = 25,766 W/m 2 4 4

Then the total rate of heat transfer becomes Q& total = Aq& total = π (0.15) 2 (25,766 W/m 2 ) = 1821 W

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10-35

Condensation Heat Transfer 10-43C Condensation is a vapor-to-liquid phase change process. It occurs when the temperature of a vapor is reduced below its saturation temperature Tsat. This is usually done by bringing the vapor into contact with a solid surface whose temperature Ts is below the saturation temperature Tsat of the vapor.

10-44C In film condensation, the condensate wets the surface and forms a liquid film on the surface which slides down under the influence of gravity. The thickness of the liquid film increases in the flow direction as more vapor condenses on the film. This is how condensation normally occurs in practice. In dropwise condensation, the condensed vapor forms droplets on the surface instead of a continuous film, and the surface is covered by countless droplets of varying diameters. Dropwise condensation is a much more effective mechanism of heat transfer.

10-45C The presence of noncondensable gases in the vapor has a detrimental effect on condensation heat transfer. Even small amounts of a noncondensable gas in the vapor cause significant drops in heat transfer coefficient during condensation.

10-46C The modified latent heat of vaporization h*fg is the amount of heat released as a unit mass of vapor condenses at a

specified temperature, plus the amount of heat released as the condensate is cooled further to some average temperature between Tsat and Ts . It is defined as h *fg = h fg + 0.68c pl (Tsat − Ts ) where cpl is the specific heat of the liquid at the average film temperature.

10-47C During film condensation on a vertical plate, heat flux at the top will be higher since the thickness of the film at the top, and thus its thermal resistance, is lower.

10-48C The condensation heat transfer coefficient for the tubes will be the highest for the case of horizontal side by side (case b) since (1) for long tubes, the horizontal position gives the highest heat transfer coefficients, and (2) for tubes in a vertical tier, the average thickness of the liquid film at the lower tubes is much larger as a result of condensate falling on top of them from the tubes directly above, and thus the average heat transfer coefficient at the lower tubes in such arrangements is smaller.

10-49C In condensate flow, the wetted perimeter is defined as the length of the surface-condensate interface at a crosssection of condensate flow. It differs from the ordinary perimeter in that the latter refers to the entire circumference of the condensate at some cross-section.

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10-36

10-50 The necessary surface temperature of the plate used to condensate saturated water vapor at a desired condensation rate is to be determined. Assumptions 1 Steady operating condition exists. 2 The plate surface has uniform temperature. 3 The film temperature is 90°C. Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρv = 0.5978 kg/m3 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 90°C are, from Table A-9,

ρl = 965.3 kg/m3

cpl = 4206 J/kg·K

−3

µl = 0.315 × 10 kg/m·s −6

kl = 0.675 W/m·K 2

νl = µl / ρl = 0.326 × 10 m /s Analysis The calculation of the modified latent heat of vaporization requires the knowledge of the Ts. Hence, we assume Ts = 80°C, and iterate the solution, if necessary, until good agreement with the calculated value of Ts is achieved: h ∗fg = h fg + 0.68c pl (Tsat − Ts ) = 2257 × 10 3 + 0.68(4206)(100 − 80) = 2314 × 10 3 J/kg

The Reynolds number is Re =

4m& 4(0.016 kg/s) = = 406.3 pµ l (0.5 m)(0.315 × 10 −3 kg/m ⋅ s)

which is between 30 and 1800, and thus the flow is wavy-laminar. The heat transfer coefficient is h = hvert, wavy

⎛ g ⎜ = 1.22 1.08 Re − 5.2 ⎜⎝ ν l2 Re k l

⎞ ⎟ ⎟ ⎠

1/ 3

⎤ (406.3)(0.675 W/m ⋅ K ) ⎡ 9.81 m/s 2 = ⎢ 1.22 2 2⎥ −6 − 5.2 ⎣⎢ (0.326 × 10 m /s) ⎦⎥ 1.08(406.3)

1/ 3

= 7558 W/m 2 ⋅ K Hence, the surface temperature can be calculated using hAs (Tsat − Ts ) = m& h ∗fg

Ts = 100°C −

→

Ts = Tsat −

(0.016 kg/s)(2314 × 10 3 J/kg) (7558 W/m 2 ⋅ K )(0.5 m) 2

m& h ∗fg hAs

= 80.4°C

Discussion The assumed Ts = 80°C is good, thus the solution does not require iteration.

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10-37

10-51 The local heat transfer coefficients at the middle and at the bottom of a vertical plate undergoing film condensation are to be determined. Assumptions 1 Steady operating condition exists. 2 The plate surface has uniform temperature. 3 The flow is laminar. Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρv = 0.5978 kg/m3 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 90°C are, from Table A-9,

ρl = 965.3 kg/m3

cpl = 4206 J/kg·K

−3

µl = 0.315 × 10 kg/m·s −6

kl = 0.675 W/m·K 2

νl = µl / ρl = 0.326 × 10 m /s Analysis The modified latent heat of vaporization is h ∗fg = h fg + 0.68c pl (Tsat − Ts ) = 2257 × 10 3 + 0.68(4206)(100 − 80) = 2314 × 10 3 J/kg

The local heat transfer coefficient can be calculated using ⎡ gρ l ( ρ l − ρ v )h ∗fg k l3 ⎤ ⎥ hx = ⎢ ⎢⎣ 4 µ l (Tsat − Ts ) x ⎥⎦

1/ 4

⎡ (9.81)(965.3)(965.3 − 0.5978)(2314 × 10 3 )(0.675) 3 ⎤ =⎢ ⎥ 4(0.315 × 10 −3 )(100 − 80) x ⎥⎦ ⎢⎣ ⎛1⎞ = 4008⎜ ⎟ ⎝x⎠

1/ 4

1/ 4

W/m 2 ⋅ K

The local heat transfer coefficient at the middle of the plate (x = 0.1 m) is ⎛1⎞ h x = 4008⎜ ⎟ ⎝x⎠

1/ 4

⎛ 1 ⎞ W/m 2 ⋅ K = 4008⎜ ⎟ ⎝ 0.1 ⎠

1/ 4

W/m 2 ⋅ K = 7130 W/m 2 ⋅ K

The local heat transfer coefficient at the bottom of the plate (x = 0.2 m) is ⎛1⎞ h x = 4008⎜ ⎟ ⎝ x⎠

1/ 4

⎛ 1 ⎞ W/m ⋅ K = 4008⎜ ⎟ ⎝ 0.2 ⎠ 2

1/ 4

W/m 2 ⋅ K = 5990 W/m 2 ⋅ K

Discussion The assumption that the flow is laminar is verified to be appropriate: Re ≅

4 gρ l2 ⎛ k l ⎜ 3µ l2 ⎜⎝ h x =L

3

⎞ 4(9.81)(965.3) 2 ⎟ = ⎟ 3(0.315 × 10 −3 ) 2 ⎠

3

⎛ 0.675 ⎞ ⎜ ⎟ = 176 < 1800 ⎝ 5990 ⎠

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10-38

10-52 The hydraulic diameter Dh for all 4 cases are expressed in terms of the boundary layer thickness δ as follows: (a) Vertical plate:

Dh =

4 Ac 4 wδ = = 4δ p w

(b) Tilted plate:

Dh =

4 Ac 4 wδ = = 4δ p w

(c)Vertical cylinder:

Dh =

4 Ac 4πDδ = = 4δ p πD

(d) Horizontal cylinder:

Dh =

4 Ac 4(2 Lδ ) = = 4δ p 2L

(e) Sphere:

Dh =

4 Ac 4πDδ = = 4δ p πD

Therefore, the Reynolds number for all 5 cases can be expressed as Re =

4 Ac ρ l Vl D h ρ l Vl 4δρ l Vl 4m& = = = pµ l pµ l µl µl

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10-39

10-53 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 Nusselt’s 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, ρ v