7.18: Untitled Page 163

Chapter 7

We can utilize Eqs. 2 of Example 7.4 to express the molar flow rates according to

At Entrances & Exits:

M

x M ,

A  1 , 2 , 3 , 4

A

A

(9a)

At Entrances & Exits:

M

M  M  M  M

(9b)

A

B

C

D

and in many cases it will be convenient to use Eq. 9a in the form

M

At Entrances & Exits:

x

A ,

A

A

1 , 2 , 3 , 4

(9c)

M

Use of Eq. 8 allows us to express the species mole balances for Control Volume I in the form

Control Volume I

C H :

 ( M

)  ( M

)

 R

(10a)

2

6

C2H6 1

C2H6 4

C2H6

C H Cl :

 ( M

)  ( M

)

 R

(10b)

2

4

2

C2H4Cl2 1

C2H4Cl2 4

C2H4Cl2

:

 ( M

)  ( M

HCl

)

 R

(10c)

HCl 1

HCl 4

HCl

C H Cl :

 ( M

)  ( M

)

 R

(10d)

2

3

C2H3Cl 1

C2H3Cl 4

C2H3Cl

Since there are no chemical reactions taking place in the mixer illustrated in Figure 7.5c, the mole balances for Control Volume II take the simple forms given by

Control Volume II

C H :

 ( M

)  ( M

)  ( M

)

 0

(11a)

2

6

C2H6 1

C2H6 5

C2H6 2

C H Cl :  ( M

)  ( M

)  ( M

)

 0

(11b)

2

4

2

C2H4Cl2 1

C2H4Cl2 5

C2H4Cl2 2

HCl :

 ( M

)  ( M

)  ( M

)

 0

(11c)

HCl 1

HCl 5

HCl 2

C H Cl :

 ( M

)  ( M

)  ( M

)

 0

(11d)

2

3

C2H3Cl 1

C2H3Cl 5

C2H3Cl 2

We begin our analysis with Control Volume I and simplify the mole balances in terms of conditions on the molar flow rates. Some of those molar flow rates are zero and when we make use of this information the mole balances for Control Volume I take the form

Material Balances for Complex Systems

303

Control Volume I ( constraints on molar flow rates)

C H :

 ( M

)  ( M

)

 R

(12a)

2

6

C2H6 1

C2H6 4

C2H6

C H Cl :

 ( M

)

 R

(12b)

2

4

2

C2H4Cl2 1

C2H4Cl2

:

( M

HCl

)

 R

(12c)

HCl 4

HCl

C H Cl :

( M

)

 R

(12d)

2

3

C2H3Cl 4

C2H3Cl

We now make use of the results from Axiom II given by Eqs. 7 and impose those constraints on global rates of production to obtain Control Volume I ( constraints on global net rates of production) C H :

 ( M

)  ( M

)

 0

(13a)

2

6

C2H6 1

C2H6 4

C H Cl :

 ( M

)

 R

(13b)

2

4

2

C2H4Cl2 1

C2H4Cl2

HCl :

( M

)

  R

(13c)

HCl 4

C2H4Cl2

C H Cl :

( M

)

  R

(13d)

2

3

C2H3Cl 4

C2H4Cl2

At this point we can use Eq. 1 to express the global net rate of production in terms of the conversion leading to

Control Volume I ( global net rates of production in terms of the conversion) C H :

 ( M

)  ( M

)

 0

(14a)

2

6

C2H6 1

C2H6 4

C H Cl :

( M

)

( M

C

)

(14b)

2

4

2

C2H4Cl2 1

C2H4Cl2 2

HCl :

( M

)

( M

C

)

(14c)

HCl 4

C2H4Cl2 2

C H Cl :

( M

)

( M

C

)

(14d)

2

3

C2H3Cl 4

C2H4Cl2 2

Moving on to Control Volume II, we note that all global rates of production are zero and that only ethane and dichloroethane are present in this control volume. This leads to the two non‐trivial species mole balances given by

Control Volume II ( constraints on molar flow rates)

C H :

 ( M

)  ( M

)

 0

(15a)

2

6

C2H6 1

C2H6 2

304