# 7.28: Untitled Page 173

## Chapter 7

### Here we have used

(o)

( M

)

to identify the first assumed value for the

C2H4Cl2 5

molar flow rate of dichloroethane in the tear stream. Moving on to Control Volume II, we obtain

Control Volume II

C H :

( M

)

 ( M

)  R

(4a)

2

6

C2H6 3

C2H6 2

C2H6

C H Cl :

( M

)

 ( M

)  R

(4b)

2

4

2

C2H4Cl2 3

C2H4Cl2 2

C2H4Cl2

HCl :

( M

)

 ( M

)  R

(4c)

HCl 3

HCl 2

HCl

C H Cl :

( M

)

 ( M

)  R

(4d)

2

3

C2H3Cl 3

C2H3Cl 2

C2H3Cl

The global net rates of production are constrained by Axiom II that provides the relations

Axiom II: R

 0 , R

  R

, R

  R

(5)

C H

HCl

C H Cl

C H Cl

C H Cl

2

6

2

4

2

2

3

2

4

2

which can be expressed in terms of the conversion (see Eq. 1) to obtain R

 0 ,

R

( M

) ,

R

( M

C

C

) (6)

C2H6

HCl

C2H4Cl2 2

C2H3Cl

C2H4Cl2 2

These representations for the global net rates of production can be used with Eqs. 4 to obtain

Control Volume II

C H :

( M

)

 ( M

)

(7a)

2

6

C2H6 3

C2H6 2

C H Cl :

( M

)

 ( M

) 1

(  C)

(7b)

2

4

2

C2H4Cl2 3

C2H4Cl2 2

HCl :

( M

)

 ( M

) 

( M

C

)

(7c)

HCl 3

HCl 2

C2H4Cl2 2

C H Cl :

( M

)

 ( M

) 

( M

C

)

(7d)

2

3

C2H3Cl 3

C2H3Cl 2

C2H4Cl2 2

This completes the first two steps in the sequence, and we are ready to move on to Control Volume III shown in Figure 7.7. Application of Eq. 2

to Control Volume III provides the following mole balances:

Material Balances for Complex Systems

323

Control Volume III

C H :

( M

)

 ( M

)

(8a)

2

6

C2H6 4

C2H6 3

(1)

C H Cl :

( M

)

 ( M

)

(8b)

2

4

2

C2H4Cl2

C

5

2H4Cl2 3

HCl :

( M

)

 ( M

)

(8c)

HCl 4

HCl 3

C H Cl :

( M

)

 ( M

)

(8d)

2

3

C2H3Cl 4

C2H3Cl 3

Here we have indicated that the molar flow rate of dichloroethane ( C H Cl ) leaving the separator in Stream #5 is the first approximation 2

4

2

based on the assumed value entering the mixer. This assumed value is given in Eq. 3b as ( M

)(o) . Because Stream #5 contains only a single

C2H4Cl2 5

component, this problem is especially simple, and we only need to make use of Eqs. 3b, 7b and 8b to obtain a relation between (1)

( M

)

and

C2H4Cl2 5

( M

)(o) . This relation is given by

C2H4Cl2 5

C H Cl :

( 1

(

) )

M

 1 

C ( M

)  ( M

 (13)

C H Cl

)(o)

2

4

2

2

4

2

C

5

2H4Cl2 1

C2H4Cl2 5 

Neither of the two molar flow rates on the right hand side of this result are known; however, we can eliminate the molar flow rate of dichloroethane ( M

) by working in terms of a dimensionless molar flow rate C2H4Cl2 1

defined by

( M

)

C H Cl

2

4

2 5

M

5

( M

(14)

)

C2H4Cl2 1

This allows us to express Eq. 13 as

(1)

(o)

M

1 C1

 M

,

C  0 . 30

(15)

5

5

(o)

(1)

and we can assume a value of M

in order to compute a value of M

.

5

5

(1)

On the basis of this computed value of M

, the analysis can be repeated

5

(2)

to determine M

that is given by

5

(2)

(1)

M

1 C 1

 M

,

C  0 3

. 0

(16)

5

5

324