14 Chapter Mass Transfer 14-107 A raindrop is falling freely in atmospheric air. The terminal velocity of the raindrop at which the drag force equals the weight of the drop and the average mass transfer coefficient are to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 The raindrop is spherical in shape. 3 The reduction in the diameter of the raindrop due to evaporation when the terminal velocity is reached is negligible. Properties Because of low mass flux conditions, we can use dry air properties for the mixture. The properties of air at 1 atm and the free-stream temperature of 25C (and the dynamic viscosity at the surface temperature of 9C) are (Table A-15)
1.184 kg/m 3 1.562 10
5
1.849 10
5 5
kg/m.s kg/m.s
m 2 /s
s , @ 282 K
1.759 10
At 1 atm and the film temperature of (25+9)/2 = 17C = 290 K, the kinematic viscosity of air is, from Table A-11, 1.488 10 5 m 2 /s , while the mass diffusivity of water vapor in air is, Eq. 14-15,
T 2.072 (290 K) 2.072 187 10 10 . P 1 atm Analysis The weight of the raindrop before any evaporation occurs is DAB DH 2 O-air 187 10 .
10
2.37 10
5
m2 / s
FD
mg
Vg
(1000 kg/m 3 )
(0.003 m) 3 (9.8 m/s 2 ) 1.38 10 6
CD AN
4
N
u 2 where drag coefficient C D is to be determined 2 using Fig. 10-20 which requires the Reynolds number. Since we do not know the velocity we cannot determine the Reynolds number. Therefore, the solution requires a trial-error approach. We choose a velocity and perform calculations to obtain the drag force. After a couple trial we choose a velocity of 8 m/s. Then the Reynolds number becomes V D (8 m/s)(0.003m) Re 1536 1.562 10 5 m 2 /s The corresponding drag coefficient from Fig. 10-20 is 0.5. Then, Air 2 3 2 2 25 C u (0.003 m) (1.184 kg/m )(8 m/s) FD C D AN (0.5) 1.34 10 4 1 atm 2 4 2
The drag force is determined from FD
which is sufficiently close to the value calculated before. Therefore, the terminal velocity of the raindrop is V = 8 m/s. The Schmidt number is
Sc 1.488 10 D AB 2.37 10
5 5
Raindrop 9 C D = 3 mm
m 2 /s m 2 /s
0.628
Then the Sherwood number can be determined from the forced heat convection relation for a sphere by replacing Pr by the Sc number to be
Sh hmass D D AB 2
1/ 4
2
0.4 Re 1/ 2 0.06 Re 2 / 3 Sc 0.4
s 1/ 2 2/3 0 .4
0.4 1536
0.06 1536
0.628
1.849 10 1.759 10
5 5
1/ 4
21.9
Then the mass transfer coefficient becomes
hmass ShD AB D (21.9)(2.37 10 0.003m
5
m 2 /s)
0.173 m/s
14-64
Chapter 14 Mass Transfer 14-108 Wet steel plates are to be dried by blowing air parallel to their surfaces. The rate of evaporation from both sides of a plate is to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 The plates are far enough from each other so that they can be treated as flat plates. 4 The air is dry so that the amount of moisture in the air is negligible. Properties The molar masses of air and water are M = 29 and M = 18 kg/kmol, respectively (Table A-1). Because of low mass flux conditions, we can use dry air properties for the mixture. The properties of the air at 1 atm and at the film temperature of (20 + 25) = 22.5C are (Table A-15) = 1.53910-5 m2/s = 1.194 kg /m3 Cp = 1007 J / kg K Pr = 0.7303
The saturation pressure of water at 20C is 2.339 kPa (Table A-9). The mass diffusivity of water vapor in air at 22.5C = 295.5 K is determined from Eq. 14-15 to be
D AB DH 2O-air 1.87 10
10
T 2.072 P
1.87 10
10
295.5 K 1atm
2.072
2.46 10
5
m 2 /s
Analysis The Reynolds number for flow over the flat plate is
Re VL (4 m/s)(0.4 m) 1.539 10
5
m 2 /s
103,964
which is less than 500,000, and thus the air flow is laminar over the entire plate. The Schmidt number in this case is
Air 25 C 4 m/s
Sc
1.539 10 D AB 2.46 10
5 5
m 2 /s m 2 /s
Brass plate 20 C
0.626
Therefore, the Sherwood number in this case is determined from Table 14-13 to be
Sh = 0.664 Re L 0.5Sc1/3 0.664 103,964
0.5
0.626
1/ 3
183.1
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB L (183.1)(2.46 10 0.4 m
5
m 2 /s)
0.0113m/s
Noting that the air at the water surface will be saturated and that the saturation pressure of water at 20C is 2.339 kPa, the mass fraction of water vapor in the air at the surface of the plate is, from Eq. 14-10,
w A,s
and
y A,s
MA M
Psat M A P M air
2.339 kPa
18 kg/ kmol
101.325 kPa 29 kg/ kmol
0.01433
wA,
0
Then the rate of mass transfer to the air becomes
mevap. hmass A( wA, s 6.19 10 5 kg/s wA, ) (0.0113 m/s)(1.194 kg/m 3 )(2 0.4m 0.4 m )(0.01433 0)
Discussion This is the upper limit for the evaporation rate since the air is assumed to be completely dry.
14-65
Chapter 14 Mass Transfer 14-109E Air is blown over a square pan filled with water. The rate of evaporation of water and the rate of heat transfer to the pan to maintain the water temperature constant are to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 80 F). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Water is at the same temperature as the air. Properties The molar masses of air and water are M = 29 and M = 18 lbm/lbmol, respectively (Table A1E). Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 80 F and 1 atm, for which = 0.17 10-3 ft2/s, and = 0.074 lbm/ft3 (Table A-15E). The saturation pressure of water at 80F is 0.5073 psia, and the heat of vaporization is 1048 Btu/lbm. The mass diffusivity of water vapor in air at 80F = 540 R = 300 K is determined from Eq. 14-15 to be
D AB DH 2O-air 1.87 10
10
T 2.072 P
1.87 10
10
300 K
2.072
1atm
2.54 10
5
m 2 /s = 2.74 10
4
ft 2 /s
Analysis The Reynolds number for flow over the free surface is
Re VL (10ft/s)(15 / 12 ft) 0.17 10
3
ft 2 /s
73,530
which is less than 500,000, and thus the flow is laminar over the entire surface. The Schmidt number in this case is Air 0.17 10 3 ft 2 /s Sc 0.622 80 F D AB 2.734 10 4 ft 2 /s Saturated 1 atm air 10 ft/s Therefore, the Sherwood number in this 30% RH case is determined from Table 14-13 to be
Sh = 0.664 Re L0.5Sc1/3 0.664 73,530
0.5
Evaporation
0.622 1 / 3
153.7
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB L (153.7)(2.734 10 15/12 ft
4
Water 80 F
ft 2 /s) 0.0336 ft/s
Noting that the air at the water surface will be saturated and that the saturation pressure of water at 80F is 0.5073 psia (= 0.0345 atm), the mass fraction of water vapor in the air at the surface and at the free stream conditions are, from Eq. 14-10,
w A,s w A,
y A,s y A,
MA M MA M air
Psat M A P M air Psat M A P M air
wA,
(0.3)(0.5073 psia) 18 lbm/ lbmol 14.7 psia 29 lbm/ lbmol (1.0)(0.5073 psia) 18 lbm/ lbmol 14.7 psia 29 lbm/ lbmol
0.00643
0.02142
Then the rate of mass transfer to the air becomes
mevap hmass As wA,s 0.0336 ft/s 0.074 lbm/ft 3 15 / 12ft2 0.02142 0.00642 5.83 10 5 lbm/s
Noting that the latent heat of vaporization of water at 80F is hfg = 1048 Btu/ lbm, the required rate of heat supply to the water to maintain its temperature constant is Q m h (583 10 5 lbm / s)(1048 Btu / lbm) 0.0611 Btu / s = 220 Btu / h .
evap. fg
Discussion If no heat is supplied to the pan, the heat of vaporization of water will come from the water, and thus the water temperature will have to drop below the air temperature.
14-66
Chapter 14 Mass Transfer 14-110E Air is blown over a square pan filled with water. The rate of evaporation of water and the rate of heat transfer to the pan to maintain the water temperature constant are to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 60 F). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Water is at the same temperature as air. Properties The molar masses of air and water are M = 29 and M = 18 lbm/lbmol, respectively (Table A1E). Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 60 F and 1 atm, for which = 0.15910-3 ft2/s, and = 0.076 lbm / ft3 (Table A-15E). The saturation pressure of water at 60F is 0.2563 psia, and the heat of vaporization is 1060 Btu/lbm. The mass diffusivity of water vapor in air at 60F = 520 R = 288.9 K is determined from Eq. 14-15 to be
D AB DH 2O-air 1.87 10
10
T 2.072 P
1.87 10
10
288.9 K 1atm
2.072
2.35 10
5
m 2 /s = 2.53 10
4
ft 2 /s
Analysis The Reynolds number for flow over the free surface is
Re VL (10ft/s)(15 / 12 ft) 0.159 10
3
ft 2 /s
78,620
which is less than 500,000, and thus the flow is laminar over the entire surface. The Schmidt number in this case is
Sc
0.159 10 3 ft 2 /s D AB 2.53 10
4
ft 2 /s
0.628
Air 60 F 1 atm 10 ft/s 30% RH
Saturated air
Evaporation
Therefore, the Sherwood number in this case is determined from Table 14-13 to be
Sh = 0.664 Re L 0.5 Sc 1/3 0.664 78,620
0.5
Water 60 F
1/ 3
0.622
158.9
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB L (158.9)(2.53 10 15/12 ft
4
ft 2 /s)
0.0322 ft/s
Noting that the air at the water surface will be saturated and that the saturation pressure of water at 60F is 0.2563 psia, the mass fraction of water vapor in the air at the surface and at the free stream conditions are, from Eq. 14-10,
w A,s w A,
y A,s y A,
MA M MA M air
Psat M A P M air Psat M A P M air
wA,
(0.3)(0.2563 psia) 18 lbm/ lbmol 14.7 psia 29 lbm/ lbmol (1.0)(0.2565 psia) 18 lbm/ lbmol 14.7 psia 29 lbm/ lbmol
0.00325
0.01082
Then the rate of mass transfer to the air becomes
mevap hmass A wA, s 0.0322 ft/s 0.076 lbm/ft 3 15 / 12ft3 0.01082 0.00325 2.35 10 5 lbm/s
Noting that the latent heat of vaporization of water at 60F is hfg = 1060 Btu/ lbm, the required rate of heat supply to the water to maintain its temperature constant is Q mevap h fg (2.35 10 5 lbm / s)(1060 Btu / lbm) 0.0249 Btu / s = 89.5 Btu / h Discussion If no heat is supplied to the pan, the heat of vaporization of water will come from the water, and thus the water temperature will have to drop below the air temperature.
14-67
Chapter 14 Mass Transfer
Simultaneous Heat and Mass Transfer
14-111C In steady operation, the mass transfer process does not have to involve heat transfer. However, a mass transfer process that involves phase change (evaporation, sublimation, condensation, melting etc.) must involve heat transfer. For example, the evaporation of water from a lake into air (mass transfer) requires the transfer of latent heat of water at a specified temperature to the liquid water at the surface (heat transfer).
14-112C It is possible for a shallow body of water to freeze during a cool and dry night even when the ambient air and surrounding surface temperatures never drop to 0 C. This is because when the air is not saturated ( < 100 percent), there will be a difference between the concentration of water vapor at the water-air interface (which is always saturated) and some distance above it. Concentration difference is the driving force for mass transfer, and thus this concentration difference will drive the water into the air. But the water must vaporize first, and it must absorb the latent heat of vaporization from the water. The temperature of water near the surface must drop as a result of the sensible heat loss, possibly below the freezing point.
14-113C During evaporation from a water body to air, the latent heat of vaporization will be equal to convection heat transfer from the air when conduction from the lower parts of the water body to the surface is negligible, and temperature of the surrounding surfaces is at about the temperature of the water surface so that the radiation heat transfer is negligible.
14-68
Chapter 14 Mass Transfer 14-114 Air is blown over a jug made of porous clay to cool it by simultaneous heat and mass transfer. The temperature of the water in the jug when steady conditions are reached is to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 Radiation effects are negligible. Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 which cannot be determined at this point because of the unknown surface temperature Ts. We know that Ts T and, for the purpose of property evaluation, we take Ts to be 20 C. Then, the properties of water at 20 C and the properties of dry air at the average temperature of 25 C and 1 atm are (Tables A-9 and A-15)
Water at 20 C : h fg Dry air at 25 C : C p 2454 kJ/kg, Pv 1.007 kJ/kg C, 2.34 kPa. Also, at 30 C, Psat @ 30 C = 4.25 kPa 2.141 10
5
m 2 /s
Also, the mass diffusivity of water vapor in air at 25 C is DH 2O-air
2.50 10
5
m2 / s (Table 14-4), and
the molar masses of water and air are 18 and 29 kg/kmol, respectively (Table A-1). Analysis The surface temperature of the jug can be determined by rearranging Chilton-Colburn equation as
Ts
T
h fg C p Le
2/ 3
M v Pv ,s Pv , M P
5 5
where the Lewis number is
Le
2.141 10 D AB 2.50 10
m 2 /s m /s
2
0.856
Note that we could take the Lewis number to be 1 for simplicity, but we chose to incorporate it for better accuracy.
Hot dry air 30 C 35% RH
Water that leaks out
The air at the surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (2.34 kPa). The vapor pressure of air far from the surface is determined from
Pv , Psat@T (0.35) Psat@30 C (0.35)(4.25 kPa) 1488 kPa .
Noting that the atmospheric pressure is 1 atm = 101.3 Pa, substituting the known quantities gives
Ts 30 C 2454 kJ/kg (1.007 kJ/kg. C)(0.856)
2/3
18 kg/kmol (2.34 1.488) kPa 29 kg/kmol 101.3 kPa
15.9 C
Therefore, the temperature of the drink can be lowered to 15.9 C by this process. Discussion The accuracy of this result can be improved by repeating the calculations with dry air properties evaluated at (30+16)/2 = 18 C and water properties at 16.0 C. But the improvement will be minor.
14-69
Chapter 14 Mass Transfer 14-115 "!PROBLEM 14-115" "GIVEN" P=101.3 "[kPa]" T_infinity=30 "[C]" "phi=0.35 parameter to be varied" "PROPERTIES" Fluid$='steam_NBS' h_f=enthalpy(Fluid$, T=T_s, x=0) h_g=enthalpy(Fluid$, T=T_s, x=1) h_fg=h_g-h_f P_sat_s=Pressure(Fluid$, T=T_s, x=0) P_sat_infinity=Pressure(Fluid$, T=T_infinity, x=0) C_p_air=CP(air, T=T_ave) T_ave=1/2*(T_infinity+T_s) alpha=2.18E-5 "[m^2/s], from the tables in the text" D_AB=2.50E-5 "[m^2/s], from the text" MM_H2O=molarmass(H2O) MM_air=molarmass(air) "ANALYSIS" Le=alpha/D_AB P_v_infinity=phi*P_sat_infinity P_v_s=P_sat_s T_s=T_infinity-h_fg/(C_p_air*Le^(2/3))*MM_H2O/MM_air*(P_v_s-P_v_infinity)/P
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
Ts [C] 12.72 14.05 15.32 16.53 17.68 18.79 19.85 20.87 21.85 22.8 23.71 24.58 25.43 26.25 27.05 27.82 28.57 29.29 30
14-70
Chapter 14 Mass Transfer
14-71
Chapter 14 Mass Transfer 14-116E In a hot summer day, a bottle of drink is to be cooled by wrapping it in a wet cloth, and blowing air to it. The temperature of the drink in the bottle when steady conditions are reached is to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 80 F). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 Radiation effects are negligible. Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 which cannot be determined at this point because of the unknown surface temperature Ts. We know that Ts T and, for the purpose of property evaluation, we take Ts to be 60 F. Then the properties of water at 60 F and the properties of dry air at the average temperature of (60+80)/2 = 70 F and 1 atm are (Tables A-9E and A-15E)
Water at 60 F : h fg Dry air at 70 F : C p 1060 Btu/lbm , Pv 0.24 Btu/lbm F, 0.2563 psia. Also, at 80 F, Psat @ 80 0.8093 ft /h = 2.25 10
2 4 F
= 0.5073 psia
ft /s
2
Also, the molar of masses water and air are 18 and 29 lbm/lbmol, respectively (Table A-1E), and the mass diffusivity of water vapor in air at 80 F (= 294.4 K) is
D AB DH 2O-air 1.87 10
10
T 2.072 P
1.87 10
10
294.4 K 1atm
2.072
2.44 10
5
m/s = 2.63 10
4
ft/s
Analysis The surface temperature of the jug can be determined by rearranging Chilton-Colburn equation as
Ts
T
h fg C p Le
2/ 3
M v Pv ,s Pv , M P
Wrapped with a wet cloth
where the Lewis number is
Le
2.25 10 4 ft 2 /s D AB 2.63 10
4
ft 2 /s
0.856
Note that we could take the Lewis number to be 1 for simplicity, but we chose to incorporate it for better accuracy.
Air 80 F 30% RH
2-L drink
The air at the surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (0.2563 psia). The vapor pressure of air far from the surface is determined from
Pv , Psat @T (0.3) Psat@80F (0.3)(0.5073 psia) 0152 psia .
Noting that the atmospheric pressure is 1 atm = 14.7 psia, substituting the known quantities gives
Ts
80 F
1060 Btu/lbm 0.24 Btu/lbm. F 0.856
2/ 3
18 lbm/lbmol 29 lbm/lbmol
0.2563 0.152 psia 14.7 psia
58.4 F
Therefore, the temperature of the drink can be lowered to 58.4 F by this process. Discussion Note that the value obtained is very close to the assumed value of 60 F for the surface temperature. Therefore, there is no need to repeat the calculations with properties at the new surface temperature of 58.7 F
14-72
Chapter 14 Mass Transfer 14-117 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat loss from the top and side surfaces of the bath by radiation, natural convection, and evaporation as well as the rates of heat and water mass that need to be supplied to the water are to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn analogy between heat Air, 25 C and mass transfer is applicable since the mass 1 atm Qconv fraction of vapor in the air is low (about 2 percent Qevap Qrad 50% RH for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 The entire water body and the metal container are maintained at a uniform temperature of 55 C. 4 Heat losses from the bottom surface are negligible. 5 The air motion around the bath is negligible so that there are no forced convection Water effects. bath Heat Properties The air-water vapor mixture is assumed to be 55 C supplied dilute, and thus we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 = (25+55)/2 = 40 C = 313 K. The properties of dry air at 40 C and 1 atm are, from Table A-15, Resistance heater k 0.0266 W/m C, Pr 0.726
2.35 10 5 m 2 /s
1.70 10 5 m 2 /s
The mass diffusivity of water vapor in air at the average temperature of 313 K is determined from Eq. 1415 to be
D AB DH 2O-air 1.87 10
10
T 2.072 P
1.87 10
10
313 K
2.072
1atm
2.77 10
5
m/s
The saturation pressure of water at 25 C is Psat@25 C
h fg 2371 kJ / kg and Pv
3169 kPa. Properties of water at 55 C are .
15.76 kPa (Table A-9). The specific heat of water at the average temperature
of (15+55)/2 = 35 C is Cp = 4.178 kJ/kg. C. The gas constants of dry air and water are Rair = 0.287 kPa.m3/kg.K and Rwater = 0.4615 kPa.m3/kg.K (Table A-1). Also, the emissivities of water and the sheet metal are given to be 0.61 and 0.95, respectively, and the specific heat of glass is given to be 1.0 kJ/kg. C. Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is mbottle mbottle Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s Then the rate of heat removal by the bottles as they are heated from 25 to 55 C is Q m C T 2 kg/s 1kJ/kg. C 55 25 C 60,000 W
bottle bottle p
The amount of water removed by the bottles is m water, out Flow rateof bottles Water removed per bottle
800 bottles / min 0.6 g/bottle 480 g/min = 8 10
3
kg/s
Noting that the water removed by the bottles is made up by fresh water entering at 15 C, the rate of heat removal by the water that sticks to the bottles is Qwater removed mwater removed C p T (8 10 3 kg / s)(4178 J / kg C)(55 15) C 1337 W Therefore, the total amount of heat removed by the wet bottles is Q Q Q 60,000 1337
total, removed glass removed water removed
61,337 W
14-73
Chapter 14 Mass Transfer (b) The rate of heat loss from the top surface of the water bath is the sum of the heat losses by radiation, natural convection, and evaporation. Then the radiation heat loss from the top surface of water to the surrounding surfaces is
Qrad,top A (Ts4
4 Tsurr )
(0.95)(8 m2 )(567 10 .
8
W / m2 K4 )[(55 273 K) 4
(15 273 K) 4 ] 2023 W
The air at the water surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (15.76 kPa at 55 C). The vapor pressure of air far from the water surface is determined from
Pv , Psat@T (0.50) Psat@25 C (0.50)(3169 kPa) 1585 kPa . .
Treating the water vapor and the air as ideal gases and noting that the total atmospheric pressure is the sum of the vapor and dry air pressures, the densities of the water vapor, dry air, and their mixture at the waterair interface and far from the surface are determined to be
Pv , s
v ,s
15.76 kPa (0.4615 kPa.m 3 / kg K)(55 + 273 K) (101325 15.76) kPa . (0.287 kPa.m 3 / kg K)(55 + 273 K)
a ,s
Rv Ts Pa , s Ra Ts
v ,s
01041 kg / m 3 . 0.9090 kg / m3
At the surface:
a ,s
s
01041 0.9090 10131 kg / m 3 . .
and
Pv ,
v,
1585 kPa . (0.4615 kPa m / kg K)(25 + 273 K) (101325 1585) kPa . . (0.287 kPa m / kg K)(25 + 273 K)
a, 3 3
Rv T Pa , Ra T
v,
0.0115 kg / m 3 11662 kg / m3 .
Away from the surface:
a,
0.0115 11662 11777 kg / m3 . .
Note that s , and thus this corresponds to hot surface facing up. The area of the top surface of the water bath is As = 2 m 4 m = 8 m2 and its perimeter is p = 2(2 + 4) = 12 m. Therefore, the characteristic length is As 8 m 2 L 0.667 m p 12 m Then using densities (instead of temperatures) since the mixture is not homogeneous, the Grashoff number is determined to be
Gr g(
ave 3 s)L 2
(9.81 m / s2 )(1.1777 10131 kg / m3 )(0.667 m) 3 . [(11777 10131) / 2 kg / m3 ](1.70 10 . .
5
m2 / s) 2
151 109 .
Recognizing that this is a natural convection problem with hot horizontal surface facing up, the Nusselt number and the convection heat transfer coefficients are determined to be
Nu 0.15(Gr Pr)1 / 3 0.15(1.51 109 0.726)1 / 3 155
Nuk (155)(0.0266 W/m C) 6.17 W/m 2 C L 0.667 m Then the natural convection heat transfer rate becomes Qconv hconv As (Ts T ) (6.17 W/m 2 C)(8 m 2 )(55 25) C 1480 W
and
hconv
(c) Utilizing the analogy between heat and mass convection, the mass transfer coefficient is determined the same way by replacing Pr by Sc. The Schmidt number is determined from its definition to be
Sc 170 10 . D AB 2.77 10
5 5
m2 / s m2 / s
0.614
The Sherwood number and the mass transfer coefficients are determined to be
Sh
015(GrSc)1/ 3 .
015(151 109 0.614)1/ 3 . .
146
14-74
Chapter 14 Mass Transfer
ShD AB (146)(2.77 10 5 m2 /s) 0.00606 m / s L 0.667 m Then the evaporation rate and the rate of heat transfer by evaporation become mv hmass As ( v,s v, ) hmass
(0.00606 m/s)(8 m 2 )(0.1041 0.0116 )kg/m 3 0.00448 kg/s = 16.1 kg/h
and
Qevap
mv h fg
(0.00448 kg / s)(2371 kJ / kg)
10.6 kW
10,600 W
Then the total rate of heat loss from the open top surface of the bath to the surrounding air and surfaces is Q Q Q Q 2023 1480 10,600 14,103 W
total, top rad conv evap
Therefore, if the water bath is heated electrically, a 14 kW resistance heater will be needed just to make up for the heat loss from the top surface. (c) The side surfaces are vertical plates, and treating the air as dry air for simplicity, heat transfer from them by natural convection is determined to be
Gr g (Ts T ) L3
2
(9.81 m/s2 )(1/313 K)(55 25) K)(1 m)3 (1.70 10 5 m2 / s)2
3.25 109
Nu
0.1(Gr Pr)1 / 3
0.1(3.25 109 0.726)1 / 3
133
hconv
Nuk L
(133)(0.0266 W/m C) 1m
hconv As (Ts T )
3.54 W/m 2 C
C)(12 1 m 2 )(55 25) C 1275 W
Qconv, side
Qrad, side
(3.54 W/m 2
The radiation heat loss from the side surfaces of the bath to the surrounding surfaces is
4 As (Ts4 Tsurr )
(0.61)(12 m 1 m )(5.67 10
8
W/m 2 K 4 )[(55 273 K ) 4
(15 273 K ) 4 ]
2498 W
and
Q total, side
Qconv Q rad
1275 2498
3773 W
(d) The rate at which water must be supplied to the maintain steady operation is equal to the rate of water removed by the bottles plus the rate evaporation, mmake-up mremoved mevap 0.00800 0.00448 0.01248 kg / s = 44.9 kg / h Noting that the entire make-up water enters the bath 15 C, the rate of heat supply to preheat the make-up water to 55 C is Qpreheating water mmake-up water C p T (0.01248 kg / s)(4178 J / kg C)(55 15) C 2086 W Then the rate of required heat supply for the bath becomes the sum of heat losses from the top and side surfaces, plus the heat needed for preheating the make-up water and the bottles, Q Q Q Q Q Q Q Q
total bottle rad conv evap top rad conv side makeup water
60,000 14,103 3773 2086 79,962 W
Therefore, the heater must be able to supply heat at a rate of 80 kW to maintain steady operating conditions
14-75
Chapter 14 Mass Transfer 14-118 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat loss from the top and side surfaces of the bath by radiation, natural convection, and evaporation as well as the rates of heat and water mass that need to be supplied to the water are to be determined. Assumptions 1 The low mass flux conditions exist so that Air, 25 C the Chilton-Colburn analogy between heat and mass 1 atm Qconv Qevap Qrad transfer is applicable since the mass fraction of vapor in 50% RH the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 The entire water body and the metal container are maintained at a uniform temperature of 50 C. 4 Heat losses from the bottom surface are negligible. 5 The air motion around the bath is negligible so that there are no Water forced convection effects. bath Heat Properties The air-water vapor mixture is assumed to be 50 C dilute, and thus we can use dry air properties for the supplied mixture at the average temperature of (T Ts ) / 2 = (25+50)/2 = 37.5 C = 310.5 K. The properties of dry air at 310.5 K and 1 atm are, from Table A-15,
k
0.0264 W/m C, 2.31 10 m /s
5 2
Pr
0.726
5 2
Resistance heater
1.68 10 m /s
T 2.072 P 310.5 K 1atm
2.072
The mass diffusivity of water vapor in air at the average temperature of 310.5 K is, from Eq. 14-15,
D AB DH 2O-air 1.87 10
10
1.87 10
10
2.72 10
5
m/s
The saturation pressure of water at 25 C is Psat@25 C
h fg 2383 kJ / kg and Pv
3169 kPa. Properties of water at 50 C are .
12.35 kPa (Table A-9). The specific heat of water at the average temperature
of (15+50)/2 = 32.5 C is Cp = 4.178 kJ/kg. C. The gas constants of dry air and water are Rair = 0.287 kPa.m3/kg.K and Rwater = 0.4615 3 kPa.m /kg.K (Table A-1). Also, the emissivities of water and the sheet metal are given to be 0.61 and 0.95, respectively, and the specific heat of glass is given to be 1.0 kJ/kg. C. Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is mbottle mbottle Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s Then the rate of heat removal by the bottles as they are heated from 25 to 55 C is Q m C T 2 kg/s 1kJ/kg. C 55 25 C 60,000 W
bottle bottle p
The amount of water removed by the bottles is m water, out Flow rateof bottles Water removed per bottle
800 bottles / min 0.6 g/bottle 480 g/min = 8 10
3
kg/s
Noting that the water removed by the bottles is made up by fresh water entering at 15 C, the rate of heat removal by the water that sticks to the bottles is Qwater removed mwater removed C p T (8 10 3 kg / s)(4178 J / kg C)(55 15) C 1337 W Therefore, the total amount of heat removed by the wet bottles is Q Q Q 60,000 1337
total, removed glass removed water removed
61,337 W
(b) The rate of heat loss from the top surface of the water bath is the sum of the heat losses by radiation, natural convection, and evaporation. Then the radiation heat loss from the top surface of water to the surrounding surfaces is
Qrad, top
4 As (Ts4 Tsurr )
(0.95)(8 m2 )(5.67 10 8 W/m 2 K 4 )[(50 273 K )4 (15 273 K )4 ] 1726 W
14-76
Chapter 14 Mass Transfer The air at the water surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (12.35 kPa at 50 C). The vapor pressure of air far from the water surface is determined from
Pv , Psat@T (0.50) Psat@25 C (0.50)(3169 kPa) 1585 kPa . .
Treating the water vapor and the air as ideal gases and noting that the total atmospheric pressure is the sum of the vapor and dry air pressures, the densities of the water vapor, dry air, and their mixture at the waterair interface and far from the surface are determined to be
Pv ,s
v ,s
12.35 kPa (0.4615 kPa.m / kg K)(50 + 273 K) (101325 12.35) kPa . (0.287 kPa.m 3 / kg K)(50 + 273 K)
a ,s 3
Rv Ts Pa ,s Ra Ts
v ,s
0.0829 kg / m 3 0.9598 kg / m 3
At the surface:
a ,s
s
0.0829 0.9598 10427 kg / m3 .
and
Pv ,
v,
1585 kPa . (0.4615 kPa m3 / kg K)(25 + 273 K) (101325 1585) kPa . . (0.287 kPa m 3 / kg K)(25 + 273 K)
a,
Rv T Pa , Ra T
v,
0.0115 kg / m 3 11662 kg / m3 .
Away from the surface:
a,
0.0115 11662 11777 kg / m3 . .
Note that
s
, and thus this corresponds to hot surface facing up. The area of the top surface of the
water bath is As = 2 m 4 m = 8 m2 and its perimeter is p = 2(2 + 4) = 12 m. Therefore, the characteristic length is As 8 m 2 L 0.667 m p 12 m Then using densities (instead of temperatures) since the mixture is not homogeneous, the Grashoff number is determined to be
Gr g(
ave 3 s )L 2
(9.81 m/s2 )(1.1777 1.0427 kg/m3 )(0.667 m)3 [(1.1777 1.0427) / 2 kg/m3 ](1.68 10 5 m2 / s)2
1.27 109
Recognizing that this is a natural convection problem with hot horizontal surface facing up, the Nusselt number and the convection heat transfer coefficients are determined to be
Nu 0.15(Gr Pr)1 / 3
Nuk L
0.15(1.27 109 0.726)1 / 3
146
C
and
hconv
(146 )(0.0264 W/m C) 0.667 m
5.78 W/m 2
Then the natural convection heat transfer rate becomes Qconv hconv As (Ts T ) (5.78 W/m 2 C)(8 m 2 )(50 25) C 1156 W (c) Utilizing the analogy between heat and mass convection, the mass transfer coefficient is determined the same way by replacing Pr by Sc. The Schmidt number is determined from its definition to be
Sc 1.68 10 D AB 2.72 10
5 5
m2 / s m2 / s
0.618
The Sherwood number and the mass transfer coefficients are determined to be
Sh
hmass
0.15(GrSc)1 / 3
0.15(1.27 109 0.618)1 / 3
5 2
138
ShD AB (138)(2.72 10 m /s) 0.00564 m/s L 0.667 m Then the evaporation rate and the rate of heat transfer by evaporation become
14-77
Chapter 14 Mass Transfer
mv hmass As (
v,s v,
)
(0.00567 m/s)(8 m 2 )(0.0829 0.0116 )kg/m 3 0.00323 kg/s = 11.6 kg/h
and
Qevap
mv h fg
(0.00323 kg / s)(2383 kJ / kg)
7.67 kW
7670 W
The total rate of heat loss from the open top surface of the bath to the surrounding air and surfaces is Q Q Q Q 1726 1156 7670 10,552 W
total, top rad conv evap
Therefore, if the water bath is heated electrically, a 10.55 kW resistance heater will be needed just to make up for the heat loss from the top surface. (c) The side surfaces are vertical plates, and treating the air as dry air for simplicity, heat transfer from them by natural convection is determined to be
Gr g (Ts T ) L3
2
(9.81 m/s2 )(1/310.5 K)(50 25) K)(1 m)3 (1.68 10 5 m2 / s)2
2.83 109
Nu
0.1(Gr Pr)1 / 3
0.1(2.83 109 0.726)1 / 3
127
hconv
Qconv, side
Qrad, side
Nuk L
(127)(0.0264 W/m C) 1m
T ) (3.36 W/m 2
3.36 W/m 2 C
C)(12 1 m 2 )(50 25) C 1007 W
hconv As (Ts
The radiation heat loss from the side surfaces of the bath to the surrounding surfaces is
4 As (Ts4 Tsurr )
(0.61)(12 m 1 m )(5.67 10
8
W/m 2 K 4 )[(50 273 K ) 4
(15 273 K ) 4 ] 1662 W a
nd
Q total, side Qconv Q rad
1007 1662
2669 W
(d) The rate at which water must be supplied to the maintain steady operation is equal to the rate of water removed by the bottles plus the rate evaporation, mmake-up mremoved mevap 0.00800 0.00323 0.01123 kg / s = 40.4 kg / h Noting that the entire make-up water enters the bath 15 C, the rate of heat supply to preheat the make-up water to 50 C is Qpreheating water mmake-up water C p T (0.01123 kg / s)(4178 J / kg C)(50 15) C 1642 W Then the rate of required heat supply for the bath becomes the sum of heat losses from the top and side surfaces, plus the heat needed for preheating the make-up water and the bottles, Q Q Q Q Q Q Q Q
total bottle rad conv evap top rad conv side makeup water
60,000 10,552 2669 1642 74,863 W
Therefore, the heater must be able to supply heat at a rate of 75 kW to maintain steady operating conditions
14-78