CHEMICAL ENGINEERING

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