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Problem 1 Unsteady state equimolar counter diffusion in
gases
Problem Statement:
Two bulbs A (V1) and B (V3), which initially contain pure gases
A and B at 1 atm, are connected by a tube (V2) of length 5 cm and
the internal diameter of 0.4 cm. After 1 hr, the partial pressure
of gas A in bulb A (V1) drops to 0.75 atm while its corresponding
partial pressure in bulb B (V3) is 0.3 atm. The temperature is
constant at 25
oC. If the volume of bulb A is 1 liter, determine:
A) The partial pressure of A in bulb B after very long time
B) The diffusivity of A in B
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Problem 1 Unsteady state equimolar counter diffusion in
gases
Known:
Two pure gases A and B at 1 atm placed in two separate bulbs
connected by a tube of known length and diameter
Initial partial pressure of gas A in bulb 1 (1 atm) and 3 (0
atm), and partial pressure of gas A in bulb 1 (0.75 atm) and 3 (0.3
atm) after 1 hour
Volume of bulb 1 (1 L) and constant temperature (25 oC)
Diameter of the connecting tube (0.4 cm) and length (5 cm)
Find:
A) Partial pressure of A in bulb 3 after very long time (pA at
t=)
B) Diffusivity of A in B (DAB)
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Assumptions:
1) Applicability of Ideal Gas Law
2) Equimolar counter diffusion at any time
3) Pseudo steady state diffusion in the tube
4) Negligible tube volume compared to bulb volumes
5) Constant total concentration, temperature, and pressure
6) Constant diffusivity DAB 7) Well mixed V1 and V3
Problem 1 Unsteady state equimolar counter diffusion in
gases
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Schematic:
Problem 1 Unsteady state equimolar counter diffusion in
gases
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A) The partial pressure of gas A in bulb 3 after a very long
time.
Moles are conserved throughout the system.
1 = 1|=1 + 2|=1 + 3|=1
1 1
=1|=11
+
2|=12
+3|=13
1 1 = 1|=11 + 3|=13
Check difference in volumes:
1 = 1 2 = 2
4 3 =?
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A) The partial pressure of gas A in bulb 3 after a very long
time.
Was it safe to neglect tube volume?
2 = (0.004)2
40.05
1000
13
2 = 6.28 104
Since V2
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A) The partial pressure of gas A in bulb 3 after a very long
time.
At = the system will have reached equilibrium, therefore will be
constant throughout.
1|= = 2|= = 3|= = |=
1 1 = |=1 + |=3
1 1 = |=(1 + 3)
|= =1 1
(1 + 3)
|= =(1)(1)
(1 + 0.83)
|= = 0.55
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A) The partial pressure of gas A in bulb 3 after a very long
time.
Figure A.1 - Partial pressure profile gradient along z at
various times
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B) Diffusivity of A in B (DAB).
= , , (. )
3, 3, = 3,
3, = 3,
= 3,3,2 = 3,2 (. )
3, = 3,3,2
2 = 3,3, 22
= 3,3,2 = 3,2 (. )
General transient mole balance of species i inside
CV j, no gen/cons:
3
=(33)
= 3
3
Knowing that:
Integrating B.3 over the constant CV area of the inlet cross
sectional area of tube outlet:
Knowing that: 3 = 33
And that: 3 ()
Then:
Knowing that in equimolar counter diffusion: 3, = 3,
Then substituting B.5 back into B.4:
3, 3, + 3,3,
= 3,
+ 3, 3, + 3,
= 3,
(. )
3, = 3,
2
33
= 3,
2(. )
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2 = 0 = 1
0 + 2 2= 2 = 0 = 1
2
= 0 =(
2 )
2= 1 =
2
2= 1
+ 2
2
+2
= 0
33
= 3,
2 = 3,2(. )
Problem: we cannot solve this PDE simply
Solution: perform new balance with CV of the tube and solving
for the flux
Recall the Continuity Equation from Transport Phenomena with a
1-D
cylinder (axial diffusion only):
Another Problem: we cannot solve this PDE simply neither!
Another Solution: HOWEVER, the tube volume V2 is so
insignificant relative to the bulb volumes, one may assume
pseudo-steady state and thus neglect any accumulation of species in
the tube!
2
= 0 Since , then is constant! 2
Integrating and replacing our flux for Ficks First Law to get
the concentration as a function of position z:
Solving for both boundary conditions given after 1 hour and
using the assumption of well mixed bulb volumes:
BC 1:
-
2 = = 1
+ 2 1=
2 = 1
= (3 1)
33
= 3,2 = 12 = (3 1) 2
33
= 3 1 2
3
3
= 3 1 2
3 1 2
=
3 3 3
12
=
3(1 + 31
) 2
BC 2:
33
= 3,
2 = 3,2 (. ) Recall:
Substituting K1 into B.6:
Realizing =
and that T is constant:
Yet another problem: cannot integrate both & since both are
time dependent 3 1
Yet another solution: find in terms of 1 3
Noting that: = 11 + 33 Then: 1 = 3 3
1
Thus:
33
= 3
1 + 31
2
(. )
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3
3(1 + 31
)
3
3
= 23
0
= 3
11 + 32
ln3
1 + 31
33
= 3
1 + 31
2
(. )
Equation B.7 is now a simple separable variable/first-order DE
ready to be solved. Knowing the BCs corresponding to the respective
time, and after rearranging for DAB:
11 + 3
ln 3
1 + 31
3 1 + 3
1
= 23
= (0.83
1 3
1000 )(0.05 )1
1 + 0.83
0.004 2
4 (3600 )ln
(0.3 )1 + 0.83
1 1
1
= 3.99 104
2
THE END!
