Chapter 14 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)

/s m 10 562 . 1
kg/m 184 . 1
2 5
3

kg/m.s 10 759 . 1
kg/m.s 10 849 . 1
5
K 282 @
,
5

At 1 atm and the film temperature of (25+9)/2 = 17C = 290 K, the kinematic viscosity of air is, from Table
A-11, /s m 10 488 . 1
2 5
, while the mass diffusivity of water vapor in air is, Eq. 14-15,
D D
T
P
AB


H O-air
2
2
K
atm
m s 187 10 187 10
290
1
2 37 10
10
2 072
10
2 072
5
. .
( )
. /
. .
Analysis The weight of the raindrop before any evaporation occurs is
N 10 38 . 1 ) m/s 8 . 9 (
6
m) (0.003
) kg/m 1000 (
4 2
3
3

1
]
1

Vg mg F
D
The drag force is determined from F C A
u
D D N

2
2
where drag coefficient C
D
is to be determined
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
1536
/s m 10 562 . 1
m) m/s)(0.003 (8
Re
2 5

D V
The corresponding drag coefficient from Fig. 10-20 is 0.5. Then,
4
2 3 2 2
10 34 . 1
2
) m/s 8 )( kg/m 184 . 1 (
4
) m 003 . 0 (
) 5 . 0 (
2


1
]
1

u
A C F
N D D
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
628 . 0
/s m 10 2.37
/s m 10 488 . 1
Sc
2 5
2 5
AB

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

[ ]
( ) ( ) [ ]( ) 9 . 21
10 759 . 1
10 849 . 1
628 . 0 1536 06 . 0 1536 4 . 0 2
Sc Re 06 . 0 Re 4 . 0 2 Sh
4 / 1
5
5
4 . 0 3 / 2 2 / 1
4 / 1
0.4 3 / 2 2 / 1

,
_

+ +

,
_

+ +

s AB
mass
D
D h

Then the mass transfer coefficient becomes

m/s 0.173

m 0.003
) /s m 10 37 . 2 )( 9 . 21 ( Sh
2 5
mass
D
D
h
AB
14-64
Air
25C
1 atm
Raindrop
9C
D = 3 mm
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
m
2
/s
= 1.194 kg /m
3
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
( )
/s m 10 46 . 2
atm 1
K 5 . 295
10 87 . 1 10 87 . 1
2 5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
Analysis The Reynolds number for flow over the flat plate is
964 , 103
/s m 10 539 . 1
m) m/s)(0.4 4 (
Re
2 5

L V
which is less than 500,000, and thus the air flow is laminar
over the entire plate. The Schmidt number in this case is
626 . 0
/s m 10 2.46
/s m 10 539 . 1
Sc
2 5
2 5

AB
D

Therefore, the Sherwood number in this case is determined from Table 14-13 to be
( ) ( ) 1 . 183 626 . 0 964 , 103 664 . 0 Sc Re 0.664 = Sh
3 / 1 5 . 0 1/3 0.5

L
Using the definition of Sherwood number, the mass transfer coefficient is determined to be
m/s 0.0113
m 0.4
/s) m 10 46 . 2 )( 1 . 183 ( Sh
2 5
mass

L
D
h
AB
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,
( )
01433 . 0
kmol kg/ 29
kmol kg/ 18
kPa 101.325
kPa 339 . 2
, ,

,
_

air
A sat A
s A s A
M
M
P
P
M
M
y w
and
w
A,
0
Then the rate of mass transfer to the air becomes
kg/s 10 6.19
5
3
, , mass evap.
) 0 01433 . 0 )( m 4 . 0 m 4 . 0 2 )( kg/m 194 . 1 )( m/s 0113 . 0 (
) (

A s A
w w A h m
Discussion This is the upper limit for the evaporation rate since the air is assumed to be completely dry.
14-65
Air
25C
4 m/s
Brass
plate
20C
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 80F). 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 A-
1E). Because of low mass flux conditions, we can use dry air properties for the mixture at the specified
temperature of 80F and 1 atm, for which = 0.17 10
-3
ft
2
/s, and = 0.074 lbm/ft
3
(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
( )
/s ft 10 2.74 = /s m 10 54 . 2
atm 1
K 300
10 87 . 1 10 87 . 1
2 4 2 5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
Analysis The Reynolds number for flow over the free surface is

530 , 73
/s ft 10 0.17
) ft 12 / 15 )( ft/s 10 (
Re
2 3

L V
which is less than 500,000, and thus the flow is laminar over the entire surface. The
Schmidt number in this case is

622 . 0
/s ft 10 2.734
/s ft 10 17 . 0
Sc
2 4
2 3

AB
D

Therefore, the Sherwood number in this

case is determined from Table 14-13 to be
( ) ( ) 7 . 153 622 . 0 530 , 73 664 . 0 Sc Re 0.664 = Sh
3 / 1 5 . 0 1/3 0.5
L

Using the definition of Sherwood number, the
mass transfer coefficient is determined to be
ft/s 0.0336
ft 15/12
/s) ft 10 734 . 2 )( 7 . 153 ( Sh
2 4
mass

L
D
h
AB
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,

00643 . 0
lbmol lbm/ 29
lbmol lbm/ 18
psia 14.7
psia) 5073 . 0 )( 3 . 0 (
, ,

,
_

air
A sat A
s A s A
M
M
P
P
M
M
y w
02142 . 0
lbmol lbm/ 29
lbmol lbm/ 18
psia 14.7
psia) 5073 . 0 )( 0 . 1 (
, ,

,
_

air
A sat
air
A
A A
M
M
P
P
M
M
y w

Then the rate of mass transfer to the air becomes
( ) ( )( )( )( ) lbm/s 10 5.83
5

00642 . 0 02142 . 0 ft 12 / 15 lbm/ft 074 . 0 ft/s 0336 . 0

2 3
, , mass evap A s A s
w w A h m
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
evap fg

583 10 1048
5
lbm/ s Btu / lbm 0.0611 Btu / s = 220 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.
14-66
Air
80F
1 atm
10 ft/s
30% RH
Evaporation
Saturated
air
Water
80F
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 60F). 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 A-
1E). Because of low mass flux conditions, we can use dry air properties for the mixture at the specified
temperature of 60F and 1 atm, for which = 0.15910
-3
ft
2
/s, and = 0.076 lbm / ft
3
(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

( )
/s ft 10 2.53 = /s m 10 35 . 2
atm 1
K 9 . 288
10 87 . 1 10 87 . 1
2 4 2 5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
Analysis The Reynolds number for flow over the free surface is

620 , 78
/s ft 10 0.159
) ft 12 / 15 )( ft/s 10 (
Re
2 3

L V
which is less than 500,000, and thus the
flow is laminar over the entire surface.
The Schmidt number in this case is
628 . 0
/s ft 10 2.53
/s ft 10 159 . 0
Sc
2 4
2 3

AB
D

Therefore, the Sherwood number in this case

is determined from Table 14-13 to be
( ) ( ) 9 . 158 622 . 0 620 , 78 664 . 0 Sc Re 0.664 = Sh
3 / 1 5 . 0 1/3 0.5
L

Using the definition of Sherwood number, the
mass transfer coefficient is determined to be
ft/s 0.0322
ft 15/12
/s) ft 10 53 . 2 )( 9 . 158 ( Sh
2 4
mass

L
D
h
AB
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,
00325 . 0
lbmol lbm/ 29
lbmol lbm/ 18
psia 14.7
psia) 2563 . 0 )( 3 . 0 (
, ,

,
_

air
A sat A
s A s A
M
M
P
P
M
M
y w
01082 . 0
lbmol lbm/ 29
lbmol lbm/ 18
psia 14.7
psia) 2565 . 0 )( 0 . 1 (
, ,

,
_

air
A sat
air
A
A A
M
M
P
P
M
M
y w

Then the rate of mass transfer to the air becomes
( ) ( )( )( )( ) lbm/s 10 2.35
5

00325 . 0 01082 . 0 ft 12 / 15 lbm/ft 076 . 0 ft/s 0322 . 0

3 3
, , mass evap A s A
w w A h m
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 m h
fg

evap
lbm/ s Btu / lbm 2 35 10 1060
5
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
Air
60F
1 atm
10 ft/s
30% RH
Evaporation
Saturated
air
Water
60F
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 0C. 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 T
s
+ 2 which cannot be determined at this point because of the unknown
surface temperature Ts. We know that T T
s
<

and, for the purpose of property evaluation, we take Ts to be
20C. Then, the properties of water at 20C and the properties of dry air at the average temperature of 25C
and 1 atm are (Tables A-9 and A-15)

/s m 10 141 . 2 , C kJ/kg 007 . 1 : C 25 at air Dry
kPa 4.25 = C, 30 at Also, kPa. 34 . 2 , kJ/kg 2454 : C 20 at Water
2 5
C 30 @

p
sat v fg
C
P P h
Also, the mass diffusivity of water vapor in air at 25C is
D
H O-air
2
2
m s

2 50 10
5
. /
(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

T T
h
C
M
M
P P
P
s
fg
p
v v s v

Le
2 3 /
, ,
where the Lewis number is
856 . 0
/s m 10 2.50
/s m 10 141 . 2
Le
2 5
2 5

AB
D

Note that we could take the Lewis number to be

1 for simplicity, but we chose to incorporate it
for better accuracy.
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
P P P
v T ,
( . ) ( . )( . .

sat@ sat@30 C
kPa) kPa 0 35 0 35 4 25 1488
Noting that the atmospheric pressure is 1 atm = 101.3 Pa, substituting the known quantities gives
C 15.9

kPa 3 . 101
kPa ) 488 . 1 34 . 2 (
kg/kmol 29
kg/kmol 18
C)(0.856) kJ/kg. 007 . 1 (
kJ/kg 2454
C 30
2/3
s
T
Therefore, the temperature of the drink can be lowered to 15.9C 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 = 18C and water properties at 16.0C. But the improvement will be minor.
14-69
Hot dry air
30C
35% RH
Water that
leaks out
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
Ts [C]
0.1 12.72
0.15 14.05
0.2 15.32
0.25 16.53
0.3 17.68
0.35 18.79
0.4 19.85
0.45 20.87
0.5 21.85
0.55 22.8
0.6 23.71
0.65 24.58
0.7 25.43
0.75 26.25
0.8 27.05
0.85 27.82
0.9 28.57
0.95 29.29
1 30
14-70
Chapter 14 Mass Transfer
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
10
14
18
22
26
30

T
s


[
C
]

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 80F). 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 T
s
+ 2
which cannot be determined at this point because of the unknown
surface temperature Ts. We know that
T T
s
<

and, for the purpose of property evaluation, we take Ts to be
60F. Then the properties of water at 60F and the properties of dry air at the average temperature of
(60+80)/2 = 70F and 1 atm are (Tables A-9E and A-15E)

/s ft 10 2.25 = /h ft 8093 . 0 , F Btu/lbm 24 . 0 : F 70 at air Dry
psia 0.5073 = F, 80 at Also, psia. 2563 . 0 , Btu/lbm 1060 : F 60 at Water
2 4 2
F 80 @

p
sat v fg
C
P P h
Also, the molar masses of water and air are 18 and 29 lbm/lbmol, respectively (Table A-1E), and the mass
diffusivity of water vapor in air at 80F (= 294.4 K) is
( )
ft/s 10 2.63 = m/s 10 44 . 2
atm 1
K 4 . 294
10 87 . 1 10 87 . 1
4 5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
Analysis The surface temperature of the jug can be determined by rearranging Chilton-Colburn equation as

T T
h
C
M
M
P P
P
s
fg
p
v v s v

Le
2 3 /
, ,
where the Lewis number is

856 . 0
/s ft 10 2.63
/s ft 10 25 . 2
Le
2 4
2 4

AB
D

Note that we could take the Lewis number

to be 1 for simplicity, but we chose to
incorporate it for better accuracy.
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
P P P
v T , @
( . ) ( . )( . ) .

sat sat@80 F
psia psia 0 3 0 3 05073 0152
Noting that the atmospheric pressure is 1 atm = 14.7 psia, substituting the known quantities gives

( )( )
( )
F 58.4

,
_

psia 7 . 14
psia 152 . 0 2563 . 0
lbm/lbmol 29
lbm/lbmol 18
856 . 0 F Btu/lbm. 24 . 0
Btu/lbm 1060
F 80
3 / 2
s
T
Therefore, the temperature of the drink can be lowered to 58.4F by this process.
Discussion Note that the value obtained is very close to the assumed value of 60F for the surface
temperature. Therefore, there is no need to repeat the calculations with properties at the new surface
temperature of 58.7F
14-72
Air
80F
30% RH
2-L drink
Wrapped
with a wet
cloth
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
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 The entire water body and the metal
container are maintained at a uniform temperature
of 55C. 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
effects.
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 T
s
+ 2 =
(25+55)/2 = 40C = 313 K. The properties of dry air at
40C and 1 atm are, from Table A-15,

/s m 10 70 . 1 /s m 10 35 . 2
726 . 0 Pr , C W/m 0266 . 0
2 5 2 5



k
The mass diffusivity of water vapor in air at the average temperature of 313 K is determined from Eq. 14-
15 to be
( )
m/s 10 77 . 2
atm 1
K 313
10 87 . 1 10 87 . 1
5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
The saturation pressure of water at 25C is
P
sat@25 C
kPa.

3169 .
Properties of water at 55C are
h P
fg v
2371 1576 kJ / kg and kPa .
(Table A-9). The specific heat of water at the average temperature
of (15+55)/2 = 35C is Cp = 4.178 kJ/kg.C.
The gas constants of dry air and water are Rair = 0.287 kPa.m
3
/kg.K and Rwater = 0.4615 kPa.m
3
/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
m m
bottle bottle
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 55C is
( )( )( ) W 0,000 6 C 25 55 C kJ/kg. 1 kg/s 2
bottle bottle
T C m Q
p

The amount of water removed by the bottles is

( )( )
( )( ) kg/s 10 8 = g/min 0 48 g/bottle 0.6 min / bottles 800
bottle per removed Water bottles of rate Flow
3
out water,

m
Noting that the water removed by the bottles is made up by fresh water entering at 15C, the rate of heat
removal by the water that sticks to the bottles is

( )( )( ) Q m C T
p water removed water removed
3
10 kg / s J / kg C C W

8 4178 55 15 1337
Therefore, the total amount of heat removed by the wet bottles is

, Q Q Q
total, removed glass removed water removed
+ + 60 000 1337 61,337 W
14-73
Air, 25C
1 atm
50% RH
Resistance heater
Water
bath
55C
Heat
supplied
Q
evap
Q
rad
Q
conv
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

( ) ( . )( )( . )[( ) ( ) ] Q A T T
s rad,top surr
4 2 2 4
m W/ m K K K W + +

4 8 4 4
0 95 8 567 10 55 273 15 273 2023
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 55C). The vapor pressure of air far from the
water surface is determined from
P P P
v T ,
( . ) ( . )( . .

sat@ sat@25 C
kPa) kPa 050 050 3169 1585
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 water-
air interface and far from the surface are determined to be
At the surface:

v s
v s
v s
a s
a s
a s
s v s a s
P
R T
P
R T
,
,
,
,
, ,
.
( .
.
( . . )
( .
.
. . .

+ +
1576
0 4615
01041
101325 1576
0 287
0 9090
01041 0 9090 10131
kPa
kPa. m / kg K)(55+273 K)
kg / m
kPa
kPa. m / kg K)(55+273 K)
kg / m
kg / m
3
3
3
3
3
and
Away from the surface:

v
v
v
a
a
a
v a
P
R T
P
R T
,
,
,
,
, ,
.
( .
.
( . . )
( .
.
. . .

+ +
1585
0 4615
00115
101325 1585
0 287
11662
00115 11662 11777
kPa
kPa m / kg K)(25+273 K)
kg / m
kPa
kPa m / kg K)(25+273 K)
kg / m
kg / m
3
3
3
3
3
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 m
2
and its perimeter is p = 2(2 + 4) = 12 m. Therefore, the characteristic
length is
m 667 . 0
m 12
m 8
2

p
A
L
s
Then using densities (instead of temperatures) since the mixture is not homogeneous, the Grashoff number
is determined to be

Gr
m/ s )(1.1777 kg / m )(0.667 m)
kg / m ](1.70 m s
ave
2 3 3
3 2

g L
s
( ) ( . .
[( . . ) / / )
.


3
2 5 2
9
9 81 10131
11777 10131 2 10
151 10
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
155 ) 726 . 0 10 51 . 1 ( 15 . 0 Pr) Gr ( 15 . 0 Nu
3 / 1 9 3 / 1

and C W/m 17 . 6
m 667 . 0
C) W/m 0266 . 0 )( 155 ( Nu
2
conv



L
k
h
Then the natural convection heat transfer rate becomes
W 1480

C 25) )(55 m C)(8 W/m 17 . 6 ( ) (
2 2
conv
conv
T T A h Q
s s

(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
m s
m s
2
2

D
AB
170 10
2 77 10
0 614
5
5
. /
. /
.
The Sherwood number and the mass transfer coefficients are determined to be
Sh GrSc 015 015 151 10 0 614 146
1 3 9 1 3
. ( ) . ( . . )
/ /
14-74
Chapter 14 Mass Transfer
h
D
L
AB
mass
2
Sh m / s)
m
m/ s

( )( .
.
.
146 2 77 10
0 667
000606
5
Then the evaporation rate and the rate of heat transfer by evaporation become
kg/h 16.1 = kg/s 00448 . 0
)kg/m 0116 . 0 )(0.1041 m m/s)(8 00606 . 0 (
) (
3 2
, , mass

v s v s
v
A h m
and
kg / s)(2371 kJ / kg) kW 10,600 W
evap

( . . Q m h
v fg
0 00448 10 6
Then the total rate of heat loss from the open top surface of the bath to the surrounding air and surfaces is
W 14,103 + + + + 600 , 10 1480 2023
evap conv rad top total,
Q Q Q Q

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
9
2 2 5
3 2
2
3
10 25 . 3
) s / m 10 (1.70
m) K)(1 25) K)(55 )(1/313 m/s 81 . 9 ( ) (
Gr

L T T g
s
133 ) 726 . 0 10 25 . 3 ( 1 . 0 Pr) Gr ( 1 . 0 Nu
3 / 1 9 3 / 1

C W/m 54 . 3
m 1
C) W/m 0266 . 0 )( 133 ( Nu
2
conv



L
k
h
W 1275 C 25) )(55 m 1 C)(12 W/m 54 . 3 ( ) (
2 2
conv
side conv,


T T A h Q
s s

The radiation heat loss from the side surfaces of the bath to the surrounding surfaces is
W 2498 ] ) K 273 15 ( ) K 273 55 )[( K W/m 10 67 . 5 )( m 1 m 12 )( 61 . 0 ( ) (
4 4 4 2 8 4
surr
4
side rad,
+ +

T T A Q
s s

and
W 3773 + + 2498 1275
rad conv side total,
Q Q Q

(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,
. . m m m
make-up removed evap
+ + 000800 000448 0.01248 kg / s = 44.9 kg / h

Noting that the entire make-up water enters the bath 15C, the rate of heat supply to preheat the make-up
water to 55C is

( . )( )( ) Q m C T
p preheating water make-up water
kg / s J / kg C C W 001248 4178 55 15 2086
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,
( ) ( )
W 79,962 + + +
+ + + + + +
2086 3773 103 , 14 000 , 60
water makeup side conv rad
top
evap conv rad bottle total
Q Q Q Q Q Q Q Q

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
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 The entire water body and the metal container
are maintained at a uniform temperature of 50C. 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 effects.
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 T
s
+ 2 =
(25+50)/2 = 37.5C = 310.5 K. The properties of dry air at
310.5 K and 1 atm are, from Table A-15,

/s m 10 68 . 1 /s m 10 31 . 2
726 . 0 Pr , C W/m 0264 . 0
2 5 2 5



k
The mass diffusivity of water vapor in air at the average temperature of 310.5 K is, from Eq. 14-15,
( )
m/s 10 72 . 2
atm 1
K 5 . 310
10 87 . 1 10 87 . 1
5
072 . 2
10
072 . 2
10
air - O H
2


P
T
D D
AB
The saturation pressure of water at 25C is
P
sat@25 C
kPa.

3169 .
Properties of water at 50C are
h P
fg v
2383 12 35 kJ / kg and kPa .
(Table A-9). The specific heat of water at the average temperature
of (15+50)/2 = 32.5C is Cp = 4.178 kJ/kg.C.
The gas constants of dry air and water are Rair = 0.287 kPa.m
3
/kg.K and Rwater = 0.4615
kPa.m
3
/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
m m
bottle bottle
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 55C is
( )( )( ) W 0,000 6 C 25 55 C kJ/kg. 1 kg/s 2
bottle bottle
T C m Q
p

The amount of water removed by the bottles is

( )( )
( )( ) kg/s 10 8 = g/min 0 48 g/bottle 0.6 min / bottles 800
bottle per removed Water bottles of rate Flow
3
out water,

m
Noting that the water removed by the bottles is made up by fresh water entering at 15C, the rate of heat
removal by the water that sticks to the bottles is

( )( )( ) Q m C T
p water removed water removed
3
10 kg / s J / kg C C W

8 4178 55 15 1337
Therefore, the total amount of heat removed by the wet bottles is

, Q Q Q
total, removed glass removed water removed
+ + 60 000 1337 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

W 1726 ] ) K 273 15 ( ) K 273 50 )[( K W/m 10 67 . 5 )( m 8 )( 95 . 0 ( ) (
4 4 4 2 8 2 4
surr
4
top rad,
+ +

T T A Q
s s

14-76
Air, 25C
1 atm
50% RH
Resistance heater
Water
bath
50C
Heat
supplied
Q
evap
Q
rad
Q
conv
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 50C). The vapor pressure of air far from the
water surface is determined from
P P P
v T ,
( . ) ( . )( . .

sat@ sat@25 C
kPa) kPa 050 050 3169 1585
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 water-
air interface and far from the surface are determined to be
At the surface:

v s
v s
v s
a s
a s
a s
s v s a s
P
R T
P
R T
,
,
,
,
, ,
.
( .
.
( . . )
( .
.
. . .

+ +
12 35
04615
0 0829
101325 12 35
0287
09598
0 0829 0 9598 10427
kPa
kPa. m / kg K)(50 +273 K)
kg / m
kPa
kPa. m / kg K)(50 +273 K)
kg / m
kg / m
3
3
3
3
3
and
Away from the surface:

v
v
v
a
a
a
v a
P
R T
P
R T
,
,
,
,
, ,
.
( .
.
( . . )
( .
.
. . .

+ +
1585
0 4615
00115
101325 1585
0 287
11662
00115 11662 11777
kPa
kPa m / kg K)(25+273 K)
kg / m
kPa
kPa m / kg K)(25+273 K)
kg / m
kg / m
3
3
3
3
3
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 m
2
and its perimeter is p = 2(2 + 4) = 12 m. Therefore, the characteristic
length is
m 667 . 0
m 12
m 8
2

p
A
L
s
Then using densities (instead of temperatures) since the mixture is not homogeneous, the Grashoff number
is determined to be

9
2 2 5 3
3 3 2
2
ave
3
10 27 . 1
) s / m 10 ](1.68 kg/m 2 / ) 0427 . 1 1777 . 1 [(
m) )(0.667 kg/m 0427 . 1 )(1.1777 m/s 81 . 9 ( ) (
Gr
+

L g
s
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
146 ) 726 . 0 10 27 . 1 ( 15 . 0 Pr) Gr ( 15 . 0 Nu
3 / 1 9 3 / 1

and C W/m 78 . 5
m 667 . 0
C) W/m 0264 . 0 )( 146 ( Nu
2
conv



L
k
h
Then the natural convection heat transfer rate becomes
W 1156 C 25) )(50 m C)(8 W/m 78 . 5 ( ) (
2 2
conv
conv


T T A h Q
s s

(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
618 . 0
s / m 10 72 . 2
s / m 10 68 . 1
Sc
2 5
2 5

AB
D

The Sherwood number and the mass transfer coefficients are determined to be
138 ) 618 . 0 10 27 . 1 ( 15 . 0 ) GrSc ( 15 . 0 Sh
3 / 1 9 3 / 1

m/s 00564 . 0
m 667 . 0
/s) m 10 72 . 2 )( 138 ( Sh
2 5
mass

L
D
h
AB
Then the evaporation rate and the rate of heat transfer by evaporation become
14-77
Chapter 14 Mass Transfer
kg/h 11.6 = kg/s 00323 . 0
)kg/m 0116 . 0 )(0.0829 m m/s)(8 00567 . 0 (
) (
3 2
, , mass

v s v s
v
A h m
and
kg / s)(2383 kJ / kg) kW 7670 W
evap

( . . Q m h
v fg
0 00323 7 67
The total rate of heat loss from the open top surface of the bath to the surrounding air and surfaces is
W 10,552 + + + + 7670 1156 1726
evap conv rad top total,
Q Q Q Q

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
9
2 2 5
3 2
2
3
10 83 . 2
) s / m 10 (1.68
m) K)(1 25) K)(50 )(1/310.5 m/s 81 . 9 ( ) (
Gr

L T T g
s
127 ) 726 . 0 10 83 . 2 ( 1 . 0 Pr) Gr ( 1 . 0 Nu
3 / 1 9 3 / 1

C W/m 36 . 3
m 1
C) W/m 0264 . 0 )( 127 ( Nu
2
conv



L
k
h

W 1007 C 25) )(50 m 1 C)(12 W/m 36 . 3 ( ) (
2 2
conv
side conv,


T T A h Q
s s

The radiation heat loss from the side surfaces of the bath to the surrounding surfaces is
W 1662 ] ) K 273 15 ( ) K 273 50 )[( K W/m 10 67 . 5 )( m 1 m 12 )( 61 . 0 ( ) (
4 4 4 2 8 4
surr
4
side rad,
+ +

T T A Q
s s

a
nd
W 2669 + + 1662 1007
rad conv side total,
Q Q Q

(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,

. . m m m
make-up removed evap
+ + 0 00800 000323 0.01123 kg / s = 40.4 kg / h

Noting that the entire make-up water enters the bath 15C, the rate of heat supply to preheat the make-up
water to 50C is

( . )( )( ) Q m C T
p preheating water make-up water
kg / s J / kg C C W 001123 4178 50 15 1642
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,
( ) ( )
W 74,863 + + +
+ + + + + +
1642 2669 552 , 10 000 , 60
water makeup side conv rad
top
evap conv rad bottle total
Q Q Q Q Q Q Q Q

Therefore, the heater must be able to supply heat at a rate of 75 kW to maintain steady operating conditions

14-78