CHEMICAL ENGINEERING 150B
PROBLEM SET #2
ARJAN SINGH PUNIANI
Problem 1
We know the system is in a pseudosteadystate. Let us assume that
R
A
= 0, steady
state conditions, or
∂c
a
∂t
= 0, and unidirectional flux:
dN
A,z
dz
. Let us assert that
y
A,
2
’
0
and
y
A,
1
=
P
A
P
= 0
.
03737. We can immediately solve for an explicit relation between flux
and height:
N
A,z
=

cD
AB
dy
A
dz
+
y
A
(
N
A,z
+
N
B,z
)
⇒
N
A,z
=

cD
AB
1

y
A
dy
A
dz
Separating differentials:
N
A,z
Z
z
2
z
1
dz
=

cD
AB
Z
y
A,
2
y
A,
1
dy
A
(1

y
A
)
⇒
N
A,z
(
z
2

z
1
) =
cD
AB
ln
1

y
A,
2
1

y
A,
1
A mass balance of the system yields the timedependence of the height:
N
A,z
=
ρ
A,L
M
A
dy
A
dz
=
cD
AB
zy
B,lm
(
y
A,
1

y
A
2
)
⇒
Z
t
t
=0
dt
=
ρ
A,L
(
y
B,lm
/M
A
)
cD
AB
(
y
A,
1

y
A,
2
)
Z
z
(
t
)
z
(0)
z dz
Where
y
B,lm
=
y
A,
1

y
A,
2
ln
h
1

y
A,
2
1

y
A,
1
i
=
0
.
03737
ln
h
1

0
1

0
.
03737
i
= 0
.
9794
To solve for
c
,
c
=
n
V
=
P
RT
=
1 atm
0
.
08206
×
298
= 0
.
0409
kgmol
m
3
Since
D
AB
= 8
.
44
×
10

6
m
2
/s:
N
A,z
(
z
2

z
1
) = 1
.
31472
×
10

7
mol/cms
Solving the time ODE above:
z
2
(
t
)

z
2
(
t
0
) =

2
cD
AB
y
A,
1
ρ
A,L
y
B,lm
/M
A
t
= 2
0
.
0409
×
8
.
44(10

6
)
×
0
.
03737
0
.
866
×
0
.
9794
92
.
2
×
10
6
t
⇒
t
=
ρ
A,L
M
A
1
1
.
472
×
10

8
[(Δ
z
)
2

0
.
01
m
2
]
Since at
t
= 0,
z
2
(
t
0
) = 0
.
01 m
2
. Thus:
t
=
0
.
866
92
.
2
1
1
.
314
×
10

7
[(Δ
z
)
2

100 cm
2
]
1