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Chemical Engineering 150B Fall 2005
Problem Set #4 Solutions
Problem 1.
(15 points) Consider liquid A evaporating into initially pure gaseous B in a closed
vessel shown below:
This vessel initially has 0.4 liter of A with 75 cm
2
of surface area in a total volume of 20 liters.
After 4 minutes, B is five percent saturated with A.
(a)
What is the mass transfer coefficient of A? Note that the vessel is isothermal at 298 K,
and the equilibrium vapor pressure of A at 298K is 24 torr.
Both A and B are ideal in the
gas phase.
The flux at 4 minutes can be found directly from the values given:
( )( )
( )( )
1
vapor concentration
air volume
N
liquid area
time
=
( )
( )( )
2
1
2
1
24
273
0 05
19 6
760
22 4
298
75
240
mol
.
.
. L
N
mol / cm s
cm
sec
=
=
8
7.0 x 10
The concentration difference is that at the interface between A and B minus that in the solution.
That at the surface is the value at saturation; that in the bulk at short times is essentially zero.
Thus,
( )
1
A,i
A
N
k c
c
=

2
3
3
1
24
273
0
760
298
22 4 10
mol
mol / cm s
k
. *
cm
k
=

=
8
7.0 x 10
0.054 cm/s
This value is lower than that commonly found for transfer in gases.
(b)
How long will it take to reach ninety percent saturation of A in B?
The time required for 90 percent saturation can be found from a mass balance:
accumulation in gas phase =
evaporation rate
B (
g
)
A (
l
)
2 pt
2 pt
3 pt
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1
1
1
1
1
d
Vc
AN
dt
d
Vc
Ak c ( sat ) c
dt
=
=

B is initially pure, so t=0, c
1
=0.
We use this condition to integrate the mass balance:
1
1
1
( kA /V )t
c
e
c ( sat )

= 
( )
( )
1
1
3
3
2
1
20
0 4 10
1 0 9
0 054
75
11052
3 1
c
V
t
ln
kA
c ( sat )
.
cm
t
ln
.
.
cm / s*
cm
t
s
t
. hr
= 


= 

=
=
[
Note
that credit will be given to answers that use vapor concentration of A at 4 minutes as 5%
of total pressure (1 atm), only because the statement “five percent A” may be interpreted in that
manner.
Answers for that situation are as follows (a) k=1.72 cm/s and (b) t=349 s.
Since the
mol fraction of A at saturation is y
A
= 24/760=0.03 < 0.05 , that interpretation of 5% is not
physically possible.]
Problem 2.
(15 points) A 1m square, thin plate of napththalene is oriented parallel to an air
stream flowing at 30 m/s.
The air is at 310 K and 1 atm.
The plate is at 300 K.
Determine the
rate of sublimation from the top of the plate.
[The following properties for naphthalene are known: Diffusivity of naphthalene is at 298 K =
6.11 x 10
6
m
2
/s ;
vapor pressure at 300 K = 25 Pa.]
Note that diffusivity changes with temperature, so we must first use equation 2441 at the
average temperature to calculate the appropriate diffusivity.
Next we check whether this flow is laminar or turbulent:
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 Spring '08
 Bell
 Fluid Dynamics, Aerodynamics, Mass Transfer, Naphthalene

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