14 Chapter Mass Transfer 14-134 A circular pan filled with water is cooled naturally. The rate of evaporation of water, the rate of heat transfer by natural convection, and the rate of heat supply to the water needed to maintain its 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 25 C). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Radiation heat transfer is negligible. 4 Both air and water vapor are ideal gases. Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 = (15+20)/2 = 17.5 C = 290.5 K. The properties of dry air at 290.5 K and 1 atm are, from Table A-15,
k
0.0251W/m C, Pr 2.04 10 m /s
5 2
0.731 1.49 10 m /s
5 2
1 atm 20 C 30% RH
Evaporation
The mass diffusivity of water vapor in air at the average temperature of 290.5 K is, from Eq. 14-15,
D AB D H 2O-air 1.87 10 1.87 10
10 10
T 2.072 P 2.37 10
5
Water 15 C
290.5 K 1atm
2.072
m/s
The saturation pressure of water at 20 C is Psat@20 C
h fg 2466 kJ / kg and Pv
2.339 kPa. Properties of water at 15 C are
17051 kPa (Table A-9). The specific heat of water at the average temperature .
of (15+20)/2 = 17.5 C is Cp = 4.184 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). Analysis (a) 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 (1.7051 kPa at 15 C). The vapor pressure of air far from the water surface is determined from
Pv , Psat@T (0.30) Psat@20 C (0.30)(2.339 kPa) 0.7017 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
17051 kPa . (0.4615 kPa m 3 / kg K)(15 + 273) K (101325 17051) kPa . . (0.287 kPa.m / kg K)(15 + 273) K
a ,s 3
Rv Ts Pa ,s Ra Ts
v ,s
0.01283 kg / m3 12052 kg / m3 .
At the surface:
a ,s
s
0.01283 12052 121803 kg / m3 . .
and
Pv ,
v,
0.7017 kPa (0.4615 kPa m 3 / kg K)(20 + 273) K (101325 0.7017) kPa . (0.287 kPa m3 / kg K)(20 + 273 K)
a,
Rv T Pa , Ra T
v,
0.00520 kg / m3 11966 kg / m 3 .
Away from the surface:
a,
0.0052 11966 12018 kg / m 3 . .
Note that water As
ro2
s
, and thus this corresponds to hot surface facing down. The area of the top surface of the
2 ro . Therefore, the characteristic length is
and its perimeter is p
14-95
Chapter 14 Mass Transfer
L As p ro2 2 ro ro 2 0.15 m 2 0.075 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 )(1.2180 1.2018 kg/m3 )(0.075 m)3 [(1.2180 1.2018) / 2 kg/m3 ](1.49 10 5 m2 / s)2
2.53 105
Recognizing that this is a natural convection problem with cold horizontal surface facing up, the Nusselt number and the convection heat transfer coefficients are determined to be (Eq. 14-13)
Nu 0.27(Gr Pr)1 / 4 0.27(2.53 105 0.731)1 / 4 5.60
and
hconv Nuk L (5.60)(0.0250 W/m C) 0.075 m 1.87 W/m 2 C
Then the rate of heat transfer from the air to the water by forced convection becomes
Qconv hconv As (T Ts ) (1.87 W/m 2 C)[ (0.15 m) 2 ](20 15) C 0.66 W (to water)
(b) 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
1.49 10 D AB 2.37 10
5 5
m 2 /s m 2 /s
0.629
Therefore, the Sherwood number in this case is determined from Table 14-13 to be
Sh = 0.27(GrSc)1/ 4 0.27(2.53 10 5 0.629)1/ 4 5.39
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB L (5.39)(2.37 10 0.075m
5
m 2 /s)
0.00170 m/s
Then the evaporation rate and the rate of heat transfer by evaporation become
mv hmass As ( 9.17 10
v,s 7 v,
)
(0.00170 m/s)[ (0.15 m) 2 ](0.01283 0.00520 ) kg/m 3
kg/s = 0.0033 kg/h
and
Qevap mv h fg (917 10 .
7
kg / s)(2466 kJ / kg)
0.00226 kW
2.26 W
(c) The net rate of heat transfer to the water needed to maintain its temperature constant at 15 C is
Qnet
Qevap Qconv
2.26 ( 0.66) 1.6 W
Discussion Note that if no heat is supplied to the water (by a resistance heater, for example), the temperature of the water in the pan would drop until the heat gain by convection equals the heat loss by evaporation.
14-135 Air is blown over a circular pan filled with water. The rate of evaporation of water, the rate of heat transfer by convection, and the rate of energy supply to the water to maintain its temperature constant are to be determined.
14-96
Chapter 14 Mass Transfer 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 25 C). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Radiation heat transfer is negligible. 4 Both air and water vapor are ideal gases. Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 = (15+20)/2 = 17.5 C = 290.5 K. The properties of dry air at 290.5 K and 1 atm are, from Table A-15,
k
0.0251W/m C, Pr 2.04 10 m /s
5 2
0.731 1.49 10 m /s
5 2
The mass diffusivity of water vapor in air at the average temperature of 290.5 K is, from Eq. 14-15,
D AB DH 2 O-air 187 10 . 187 10 .
10 10
1 atm 20 C 30% RH 3 m/s Water 15 C
5
Evaporation
T 2.072 P 2.37 10 m / s
(290.5 K) 2.072 1 atm
The saturation pressure of water at 20 C is Psat@20 C
h fg 2466 kJ / kg and Pv
3
2.339 kPa. Properties of water at 15 C are
17051 kPa (Table A-9). Also, the gas constants of water is Rwater = 0.4615 .
kPa.m /kg.K (Table A-1). Analysis (a) Taking the radius of the pan r0 = 0.15 m to be the characteristic length, the Reynolds number for flow over the pan is
Re VL (3m/s)(0.15 m) 1.49 10 5 m 2 /s 30,201
which is less than 500,000, and thus the flow is laminar over the entire surface. The Nusselt number and the heat transfer coefficient are
Nu = 0.664 Re L 0.5 Pr 1/3 0.664 30,201
0.5
0.731 1 / 3
103.9
hheat
Nuk L
(103.9)(0.0250 W/m C) 0.15m
17.3 W/m 2
C
Then the rate of heat transfer from the air to the water by forced convection becomes
Qconv hconv As (T Ts ) (17.3 W/m 2 C)[ (0.15 m) 2 ](20 15) C 6.1 W (to water)
(b) 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
1.49 10 D AB 2.37 10
5 5
m 2 /s m 2 /s
0.629
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 30,201
0.5
0.629
1/ 3
98.9
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB L (98.9)(2.37 10 5 m2 / s) 0.15m 0.0156 m / s
14-97
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 (1.7051 kPa at 15 C). The vapor pressure of air far from the water surface is determined from
Pv , Psat@T (0.30) Psat@20 C (0.30)(2.339 kPa) 0.7017 kPa
Treating the water vapor and the air as ideal gases, the vapor densities at the water-air interface and far from the surface are determined to be At the surface:
Pv ,s
v ,s
17051 kPa . (0.4615 kPa m3 / kg K)(15 + 273) K
0.7017 kPa (0.4615 kPa m3 / kg K)(20 + 273) K
Rv Ts
Pv ,
0.01283 kg / m3
Away from the surface:
v,
Rv T
0.00520 kg / m3
Then the evaporation rate and the rate of heat transfer by evaporation become
mv hmass As ( 8.41 10
6 v,s v,
)
(0.0156 m/s)[ (0.15 m) 2 ](0.01283 0.00520) kg/m 3
kg/s = 0.0303 kg/h
and
Qevap
mv h fg
(8.41 10
6
kg / s)(2466 kJ / kg)
0.0207 kW
20.7 W
(c) The net rate of heat transfer to the water needed to maintain its temperature constant at 15 C is
Qnet
Qevap Qconv
20.7 ( 6.1) 14.6 W
Discussion Note that if no heat is supplied to the water (by a resistance heater, for example), the temperature of the water in the pan would drop until the heat gain by convection equals the heat loss by evaporation. Also, the rate of evaporation could be determined almost as accurately using mass fractions of vapor instead of vapor fractions and the average air density from the relation mevap hmass A( w A,s w A, ) .
14-98
Chapter 14 Mass Transfer 14-136 A spherical naphthalene ball is hanged in a closet. The time it takes for the naphthalene to sublimate completely is to be determined. Assumptions 1 The concentration of naphthalene in the air is very small, and the low mass flux conditions exist so that the Chilton-Colburn analogy between heat and mass transfer is applicable (will be verified). 2 Both air and naphthalene vapor are ideal gases. 3 The naphthalene and the surrounding air are at the same temperature. 4 The radiation effects are negligible. Properties The molar mass of naphthalene is 128.2 kg/kmol. Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 298 K and 1 atm, at which 2.14 10 5 m 2 /s (Table A-15). 1.18 kg/m 3 , C p 1007 J/kg K , and Analysis The incoming air is free of naphthalene, and thus the mass fraction of naphthalene at free stream conditions is zero, wA, = 0. Noting that the vapor pressure of naphthalene at the surface is 11 Pa, the mass fraction of naphthalene on the air side of the surface is PA,s M A 11Pa 128.2 kg/kmol w A,s 4.8 10 4 P M air 101,325Pa 29 kg/kmol Normally we would expect natural convection currents to develop around the naphthalene ball because the amount of naphthalene near the surface is much larger, and determine the Nusselt number (and its counterpart in mass transfer, the Sherwood number) from Eq. 14-16,
Nu 2 0.589 Ra 1/ 4
Closet 1 atm 25 C
Sublimation
Naphthalene 25 C
[1+ (0.469 / Pr) 9/16 ]4/9 But the mass fraction value determined above indicates that the amount of naphthalene in the air is so low that it will not cause any significant difference in the density of air. With no density gradient, there will be no natural convection and thus the Rayleigh number can be taken to be zero. Then the Nusselt number relation above will reduce to Nu = 2 or its equivalent Sh = 2. Then using the definition of Sherwood number, the mass transfer coefficient can be expressed as ShD AB 2 D AB hmass D D
The mass of naphthalene ball can be expressed as m
naphV
1 6
naph (
D3 ) . The rate of change of the
mass of naphthalene is equal to the rate of mass transfer from naphthalene to the air, and is expressed as dm hmass air A( w A, s w A, ) dt 2 D AB d 1 3 2 w A, ) naph ( D ) air ( D )(w A, s dt 6 D 3 2 dD 2 D AB ( D ) air ( w A,s w A, ) naph D 6 dt 4 air DAB DdD ( w A,s w A, )dt Simplifying and rearranging,
naph
Integrating from D
t 8
Di
0.03 m at time t = 0 to D = 0 (complete sublimation) at time t = t gives
2 naph Di air D AB ( w A, s
w A, )
Substituting, the time it takes for the naphthalene to sublimate completely is determined to be 2 (1100 kg / m3 )(0.01 m) 2 naph Di t 3.95 106 s 45.7 days 3 4 2 8 air D AB ( w A,s w A, ) 8(1.19 kg / m )(4.80 10 0) m / s)
14-99
Chapter 14 Mass Transfer 14-137E A swimmer extends his wet arms into the windy air outside. The rate at which water evaporates from both arms and the corresponding rate of heat transfer by evaporation 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 arm can be modeled as a long cylinder. Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the average temperature of (40 + 80)/2 = 60 F and 1 atm, for which = 0.15910-3 ft2/s , and = 0.077lbm / ft3 (Table A-15E). The saturation pressure of water at 40F is 0.1217 psia. Also, at 80F, the saturation pressure is 0.5073 psia and the heat of vaporization is 1048 Btu/lbm (Table A-9E). The molar mass of water is R = 0.5956 psia.ft3/lbm.R (Table A-1E). 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 a cylinder is VD (20 5280 / 3600ft/s)(3 / 12 ft) Re 46,120 0.159 10 3 ft 2 /s The Schmidt number in this case is Wet arm
Air, 1 atm 40 F, 50% RH 20 mph
Sc
0.159 10 3 ft 2 /s D AB 2.53 10
4
ft 2 /s
0.628
80 F
46,120 28200
5/8 4/5
Then utilizing the analogy between heat and mass convection, the Sherwood number is determined from Eq. 10-32 by replacing Pr number by the Schmidt number to be
Sh 0.3 0.62 Re 0.5 Sc1 / 3 1 (0.4 / Sc) 2 / 3
1/ 4
1
Re 28200
5/8 4/5
= 0.3
0.62(46,120)0.5 (0.628)1 / 3 1 (0.4 / 0.628) 2 / 3
1/ 4
1
198
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
4 2 ShD AB (198)(2.53 10 ft /s) 0.2004 ft/s D 3/12 ft 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 (0.5073 psia at 80 F). The vapor pressure of air far from the water surface is determined from Pv , Psat@T (0.50) Psat@40 F (0.50)(01217 psia) 0.0609 psia .
hmass
Treating the water vapor as an ideal gas, vapor the densities at the water-air interface and far from the surface are determined to be Pv ,s 0.5073 psia At the surface: 0.00158 lbm / ft 3 v ,s Rv Ts (0.5956 psia ft 3 / lbm R)(80 + 460) R Away from the surface:
Pv ,
v,
0.0609 psia (0.5956 psia ft 3 / lbm R)(40 + 460) R
Rv T
0.000205 lbm / ft 3
Then the evaporation rate and the rate of heat transfer by evaporation become mv hmass As ( v,s (3/12 ft)(2 ft)](0.00158 0.000205) lbm/ft 3 v, ) (0.2004 ft/s)[2 and
Qevap
8.66 10 4 lbm/s = 3.12 lbm/h mv h fg (8.66 10 4 lbm / s)(1048 Btu / lbm)
0.907 Btu / s
Discussion The rate of evaporation could be determined almost as accurately using mass fractions of vapor instead of vapor fractions and the average air density from the relation mevap hmass A( w A,s w A, ) .
14-100
Chapter 14 Mass Transfer 14-138 A nickel part is put into a room filled with hydrogen. The ratio of hydrogen concentrations at the surface of the part and at a depth of 2-mm from the surface after 24 h is to be determined. Assumptions 1 Hydrogen penetrates into a thin layer beneath the surface of the nickel component, and thus the component can be modeled as a semi-infinite medium regardless of its thickness or shape. 2 The initial hydrogen concentration in the nickel part is zero. Properties The molar mass of hydrogen H2 is M = 2 kg/kmol (Table A-1). The solubility of hydrogen in nickel at 358 K (=85C) is 0.00901 kmol/m.bar (Table 14-7). The mass diffusivity of hydrogen in nickel at 358 K is DAB =1.2 10-12 m2/s (Table A-3b). Also, 1 atm = 1.01325 bar. Analysis This problem is analogous to the onedimensional transient heat conduction problem in a semi-infinite medium with specified surface temperature, and thus can be solved accordingly. Using mass fraction for concentration since the data is given in that form, the solution can be expressed as
w A ( x , t ) w A, i wA, s wA,i erfc x 2 DABt
H2 3 atm 85C
The molar density of hydrogen in the nickel at the interface is determined from Eq. 14-20 to be
CH 2 , solid side (0) S PH 2 , gas side
Nickel part
(0.00901 kmol / m.bar )(3 101325 bar) . 0.0274 kmol / m
The argument of the complementary error function is
x 2 D AB t 2 (12 . 10 2
12
10
2
3
m
3105 .
m / s)(24 3600 s)
The corresponding value of the complementary error function is determined from Table 4-3 to be
erfc x 2 DABt erfc(3.105) 0.000015
Substituting the known quantities,
C A ( x, t ) 0 0.0274 0 0.000015 C A ( x, t ) 4.1 10
7
kmol / m 3
Therefore, the hydrogen concentration in the steel component at a depth of 2 mm in 24 h is very small.
14-101
Chapter 14 Mass Transfer 14-139 A 0.1-mm thick soft rubber membrane separates pure O2 from air. The mass flow rate of O2 through the membrane per unit area and the direction of flow are to be determined. Assumptions 1 Steady operating conditions exist. 2 Mass transfer through the membrane is onedimensional. 3 The permeability of the membrane is constant. Properties The mass diffusivity of oxygen in rubber at 298 K is DAB = 2.1 10-10 m2/s (Table 11-3). The solubility of oxygen in rubber at 298 K is 0.00312 kmol / m.bar (Table 14-7). The molar mass of oxygen is 32 kg / kmol (Table A-1). Analysis The molar fraction of oxygen in air is 0.21. Therefore, the partial pressure of oxygen in the air is
yO 2 PO 2 ,2 P PO 2 ,2 yO 2 P 0.21 (12 atm) . 0.252 atm
Rubber membrane O2 1 atm 25 C Air 1.2 atm mO2 L
The partial pressure of oxygen on the other side is simply PO2 ,1 1 atm . Then the molar flow rate of oxygen through the membrane by diffusion can readily be determined to be
N diff, A,wall D AB S PA,1 L
10
PA,2 1 0.252 atm 1.01325 bar 1 atm 0.1 10 3 m
(2.1 10 4.97 10
m/s ) 0.00312 kmol/m.ba r
9
kmol/m s
Then the mass flow rate of oxygen gas through the membrane becomes
mdiff MN diff (32 kg / kmol)(4.97 10
9
kmol / m.s)
1.59 10
7
kg / m.s
The direction of the flow will be from the pure oxygen inside to the air outside since the partial pressure of oxygen is higher inside.
14-102
Chapter 14 Mass Transfer 14-140E The top section of a solar pond is maintained at a constant temperature. The rates of heat loss from the top surface of the pond by radiation, natural convection, and evaporation are 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 The water in the pool is maintained at a uniform temperature of 80 F. 4 The critical Reynolds number for flow over a flat surface is 500,000. Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 = (70+80)/2 = 75 F. The properties of dry air at 75 F and 1 atm are, from Table A15E,
k Pr 0.0147 Btu/h ft 0.73 0.824 ft 2 /h 0.167 10
3
Air, 70 F 1 atm 100% RH 40 mph
Tsurr =60 F Qevap Qrad Qconv
F
ft 2 /s
The saturation pressure of water at 70 F is Psat@70 F 0.3632 psia. Properties of water at 80 F are h fg
Pond 80 F Heating fluid
1048 Btu/lbm
and Pv 0.5073 psia (Table A-9). The gas constant of water is Rwater = 0.5956 psia.ft3/lbm.R (Table A-1E). The emissivity of water is 0.95 (Table A-15). The mass diffusivity of water vapor in air at the average temperature of 75 F = 535 R = 297.2 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
297.2K 1atm
2.072
2.49 10
5
m 2 /s
2.68 10
4
ft 2 /s
Analysis (a) The pond surface can be treated as a flat surface. The Reynolds number for flow over a flat surface is
Re VL (40 5280 / 3600ft/s)(100 ft) 0.167 10
3
ft 2 /s
3.51 10 7
which is much larger than the critical Reynolds number of 500,000. Therefore, the air flow over the pond surface is turbulent, and the Nusselt number and the heat transfer coefficient are determined to be
Nu = 0.037 Re L 0.8 Pr 1/3 0.037(3.51 10 7 ) 0.8 0.73
1/ 3
36,215
hheat
Nuk L
(36,215)(0.0147 Btu/h ft F) 100ft
5.32 Btu/h ft 2
F
Then the rate of heat transfer from the air to the water by forced convection becomes
Qconv hconv As (T Ts ) (5.32 Btu/h ft 2 F)(10,000 ft 2 )(80 70) F 532,000 Btu/h (to water)
(b) Noting that the emissivity of water is 0.95 and the surface area of the pool is As (100 ft)(100 ft) = 10,000 ft 2 , heat transfer from the top surface of the pool by radiation is
14-103
Chapter 14 Mass Transfer
Qrad
4 As (Ts4 Tsurr )
(0.95)(10,000 ft2 )(0.1714 10
8
Btu/h ft2 R 4 )[(540 R )4
(520 R )4 ]
194,000 Btu/h
(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
0.167 10 3 ft 2 /s D AB 2.68 10
4
ft 2 /s
0.623
Then utilizing the analogy between heat and mass convection, the Sherwood number is determined by replacing Pr number by the Schmidt number to be
Sh = 0.037 Re L 0.8 Sc1/3 0.037(3.51 10 7 ) 0.8 0.623
1/ 3
34,350
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass ShD AB D (34,350)(2.68 10 100 ft
4
ft 2 /s)
0.0921ft/s
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 (Pv,s = 0.5073 psia at 80 F). The humidity of air is given to be 100%, and thus the air far from the water surface is also saturated. Therefore, Pv, Psat@70 F 0.3632 psia. Treating the water vapor as an ideal gas, the vapor densities at the water-air interface and far from the surface are determined to be At the surface:
Pv ,s
v ,s
0.5073 psia (0.5956 psia ft 3 / lbm R)(80 + 460) R 0.3632 psia
Rv Ts Pv ,
0.00158 lbm / ft 3
Away from the surface:
v,
Rv T
(0.5956 psia ft 3 / lbm R)(70 + 460) R
0.00115 lbm / ft 3
Then the evaporation rate and the rate of heat transfer by evaporation become
mv hmass As (
v, s v,
)
(0.0921 ft/s)(10,000 ft 2 )(0.00158 0.00115) lbm/ft 3
0.396 lbm/s = 1426 lbm/h
and
Qevap mv h fg (1425 lbm / h)(1048 Btu / lbm) 1,493,000 Btu / h
Discussion All of the quantities calculated above represent heat loss for the pond, and the total rate of heat loss from the open top surface of the pond to the surrounding air and surfaces is
Qtotal, top Qrad Qconv Qevap
194,000 532,000 1,493,000 2,219,000 Btu/h
This heat loss will come from the deeper parts of the pond, and thus the pond will start cooling unless it gains heat from the sun or another heat source. Note that the evaporative heat losses dominate. Also, the rate of evaporation could be determined almost as accurately using mass fractions of vapor instead of vapor fractions and the average air density from the relation mevap hmass As (wA, s wA, ) .
14-104
Chapter 14 Mass Transfer 14-141E The top section of a solar pond is maintained at a constant temperature. The rates of heat loss from the top surface of the pond by radiation, natural convection, and evaporation are 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 The water in the pool is maintained at a uniform temperature of 90 F. 4 The critical Reynolds number for flow over a flat surface is 500,000. Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture at the average temperature of (T Ts ) / 2 = (70+90)/2 = 80 F. The properties of dry air at 80 F and 1 atm are, from Table A-15E,
k Pr 0.0148 Btu/h ft F 0.73 0.838 ft /h 0.170 10
3 2
Air, 70 F 1 atm 100% RH 40 mph
Tsurr =60 F Qevap Qrad Qconv
ft 2 /s
The saturation pressure of water at 70 F is Psat@70 F 0.3632 psia. Properties of water at 90 F are h fg
1043 Btu/lbm and Pv
0.6988 psia (Table
Heating fluid
A-9). The gas constant of water is Rwater = 0.5956 psia.ft3/lbm.R (Table A-1E). The emissivity of water is 0.95 (Table A-15). The mass diffusivity of water vapor in air at the average temperature of 80 F = 540 R = 300 K is determined from Eq. 14-15 to be
D AB DH 2O-air 1.87 10
10
Pond 90 F
T 2.072 P
1.87 10
10
(300K ) 2.072 1atm
2.54 10
5
m 2 /s
2.72 10
4
ft 2 /s
Analysis (a) The pond surface can be treated as a flat surface. The Reynolds number for flow over a flat surface is
Re VL (40 5280 / 3600ft/s)(100 ft) 0.170 10
3
ft /s
2
3.45 10 7
which is much larger than the critical Reynolds number of 500,000. Therefore, the air flow over the pond surface is turbulent, and the Nusselt number and the heat transfer coefficient are determined to be
Nu = 0.037 Re L 0.8 Pr 1/3 0.037(3.45 10 7 ) 0.8 (0.73) 1 / 3 35,720
hheat
Nuk L
(35,720)(0.0148 Btu/h ft F) 100ft
5.29 Btu/h ft 2
F
Then the rate of heat transfer from the air to the water by forced convection becomes
Qconv hconv As (T Ts ) (5.29 Btu/h ft 2 F)(10,000 ft 2 )(90 70) F 1,057,000 Btu/h (to water)
(b) Noting that the emissivity of water is 0.95 and the surface area of the pool is As (100 ft)(100 ft) = 10,000 ft 2 , heat transfer from the top surface of the pool by radiation is
Qrad
4 As (Ts4 Tsurr )
(0.95)(10,000 ft2 )(0.1714 10
8
Btu/h ft2 R 4 )[(550 R )4
(520 R )4 ]
299,400 Btu/h
(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
14-105
Chapter 14 Mass Transfer
Sc
0.170 10 3 ft 2 /s D AB 2.72 10
4
ft 2 /s
0.625
Then utilizing the analogy between heat and mass convection, the Sherwood number is determined by replacing Pr number by the Schmidt number to be
Sh = 0.037 Re L 0.8 Sc1/3
ShD AB D
0.037(3.45 107 ) 0.8 (0.625)1/ 3
4
33,920
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
hmass (33,920)(2.72 10 100 ft ft / s) 0.0923 ft / s
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 (Pv,s = 0.6988 psia at 90 F). The humidity of air is given to be 100%, and thus the air far from the water surface is also saturated. Therefore, Pv, Psat@70 F 0.3632 psia. Treating the water vapor as an ideal gas, the vapor densities at the water-air interface and far from the surface are determined to be At the surface: Away from the surface:
Pv ,s
v ,s
0.6988 psia (0.5956 psia ft / lbm R)(90 + 460) R
3
Rv Ts
0.00213 lbm / ft 3
Pv ,
v,
0.3632 psia (0.5956 psia ft 3 / lbm R)(70 + 460) R
Rv T
)
0.00115 lbm / ft 3
Then the evaporation rate and the rate of heat transfer by evaporation become
mv hmass As (
mv h fg
v, s v,
(0.0923 ft/s)(10,000 ft 2 )(0.00213 0.00115) lbm/ft 3
3,396,000 Btu / h
0.905 lbm/s = 3256 lbm/h
and
Qevap
(3256 lbm / h)(1043 Btu / lbm)
Discussion All of the quantities calculated above represent heat loss for the pond, and the total rate of heat loss from the open top surface of the pond to the surrounding air and surfaces is Q Q Q Q 299,400 1,057,000 3,396,000 4,752,400 Btu/h
total, top rad conv evap
This heat loss will come from the deeper parts of the pond, and thus the pond will start cooling unless it gains heat from the sun or another heat source. Note that the evaporative heat losses dominate. Also, the rate of evaporation could be determined almost as accurately using mass fractions of vapor instead of vapor fractions and the average air density from the relation mevap hmass A( w A,s w A, ) .
14-142 .... 14-146 Design and Essay Problems
14-106