254 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
or: D
dCA
dy
{
D
dCA
dy
C
d
dy
(
D
dCA
dy
)
dy
}
Dk 2 C^2 A
dy
D
d^2 CA
dy^2
Dk 2 C^2 A
d^2 CA
dy^2
k 2
D
C^2 AD 0
Putting: qD
dCA
dy
, then:
d^2 CA
dy^2
D
dq
dy
D
dq
dCA
Ð
dCA
dy
Dq
dq
dCA
Substituting: q
dq
dCA
k 2
D
C^2 AD 0
qdqD
k 2
D
C^2 A
Integrating:
q^2
2
D
k 2
D
C^3 A
3
CK
WhenyD1,qD0,CAD0 and:KD 0
Thus:
1
2
(
dCA
dy
) 2
D
k 2
D
C^3 A
3
dCA
dy
Dš
√
2
3
k 2
D
C^3 A/^2
AsNADDdCA/dy
is positive, negative root must apply and:
C
3 / 2
A dCAD
√
2
3
k 2
D
dy
Integrating: 2 CA^1 /^2 D
√
2
3
k 2
D
yCK
WhenyD0,CADCAsand:KD 2 CAs^1 /^2
Thus: C
1 / 2
A C
1 / 2
As D
1
2
√
2
3
k 2
D
y
or: CAs^1 /^2
[(
CA
CAs
) 1 / 2
1
]
D
√
1
6
k 2
D
y
C
1 / 2
As
{(
CAs
CA
) 1 / 2
1
}
D
√
1
6
k 2
D
y.
(i) For the conditions given:
√
1
6
k 2
D
D
√{(
1
6
)(
9. 5 ð 103
1. 8 ð 10 ^9
)}
D 0. 938 ð 106 (m/kmol)^0.^5