# 7.22: Untitled Page 167

## Chapter 7

In row reduced echelon form Eq. 9 can be expressed as

 R

C2H4

1

1

0

0

1

0   R

O2

 

2

0

 

 

3

1

 0

1

0

0   RCO

0

2

2

2

 

 

  

(10)

R

0

 0

0

1

1

0

0  

H

2O

 

 

R

0

0

0

0

0

0

1

  C2H4O

RN

2

and the stoichiometric constraints on the global rates of production are given by application of the global pivot theorem (see Sec. 6.4) 1

R

  R

C

R

(11a)

2H4

H2O

C2H4O

2

3

1

R

  R

(11b)

O

R

2

H2O

C

2

2

2H4O

R

 R

(11c)

CO2

H2O

R

 0

(11d)

N2

Degree of Freedom Analysis

To gain some insight into a strategy for solving this problem, we need a degree‐of‐freedom analysis for the control volumes illustrated in Figure 7.6c. In this case we will use molar flow rates and global rates of production as our independent variables. Of the eight streams illustrated in Figure 7.6c, only seven are cut by the surface of a control volume.

Keeping in mind that we are dealing with six molecular species, we see that we have 42 stream variables in terms of species molar flow rates. In addition, we have six global rates of production so that the total number of generic variables is 48. The three control volumes illustrated in Figure 7.6c, coupled with the six molecular species, gives rise to 18 generic constraints associated with Axiom I. In addition, Axiom II provides 4 stoichiometric conditions (see Eqs. 11), and the physics of the splitter provides A  1  5 constraints on the composition (see Eq. 7‐44) that we list here as

Splitter condition:

( x )

 ( x ) ,

A

6

4

1 , 2 ,... , 5

A

A

(12)

Material Balances for Complex Systems

311

This provides us with 27 generic constraints and we can move on to the matter of particular constraints.

Control Volume I

We begin our exploration of the particular constraints with Control Volume I and direct our attention to Stream #1. The problem statement indicates that only ethylene ( C H ) , oxygen ( O ) and nitrogen ( N ) are 2

4

2

2

present in that stream, thus we have

Stream #1:

( M

)

 ( M

)

 ( M

)

 0

(13)

C2H4O 1

H2O 1

CO2 1

In addition, we are given that the oxygen and nitrogen in Stream #1 are supplied by air, thus we have a constraint on the composition given by Steam #1:

( M

)

 ( M ) ,

 

(14)

O

1

N

1

21 79

2

2

Moving on to the other stream entering Control Volume I, we draw upon information about the splitter to conclude that Stream #5 contains no ethylene oxide ( C H O ) , water ( H O ) or oxygen ( O ) , thus we have 2

4

2

2

Stream #5:

( M

)

 ( M

)

 ( M )

 0

(15)

C2H4O 5

H2O 5

O2 5

It is obvious that there is no ethylene oxide ( C H O ) or water ( H O ) in 2

4

2

Stream #2. However, we must be careful not to impose such a condition as a particular constraint, since this condition is “obvious” on the basis of the species mole balances and the conditions given by Eqs. 13 and 15. At this point we conclude that there are 7 particular constraints associated with Control Volume I and we list this result as

Control Volume I:

particular constraints  7

(16)

Moving on to Control Volume II, we note that the problem statement contained the comment “…the mole fraction of ethylene in Stream #2

entering the reactor must be maintained at 0 . 05 for satisfactory catalyst operation.” This provided the basis for Eq. 5 and we list this particular constraint as

Stream #2:

( x

)

  ,

  0 . 05

(17)

C2H4 2

There are no particular constraints imposed on Stream #4, thus we move on to Stream #7 and Stream #8 for which the following particular constraints are imposed:

312