# 7.33: Untitled Page 178

## Chapter 7

### N :

( M

)

 (1  )

 ( M )

(17c)

2

N2 5

N2 4

and we will use these conditions in our analysis of ethylene (C H ) , 2

4

carbon dioxide ( CO ) , nitrogen ( N ) . Directing our attention to the 2

2

ethylene flow rate leaving the reactor, we note that the use of Eq. 4 in Eq. 11a leads to

( M

)

  1  C CY

(18)

C H

3

2

4

Moving from the reactor to the absorber, we use Eq. 12a with Eq. 18 to obtain

( M

)

  1  C CY

(19)

C H

4

2

4

and when this result is used in the first splitter condition given by Eq. 17a we obtain

( M

)

 (1  )

  1  C CY

(20)

C H

5

2

4

Here we must remember that  is the parameter that we want to determine by means of an iterative process. At this point we move on to the mole balances for nitrogen ( N ) and make use of Eqs. 10c, 11c, 12c, 16

2

and 17c to obtain

1   

5

3 

Y

2

( M

)

(21)

N2 5

 Y

Finally we consider the mole balances for carbon dioxide ( CO ) and 2

make use of Eqs. 7b, 10b, 11b, and 17b to obtain the following result for the molar flow rate of carbon dioxide in the recycle stream.

1   2 1 

( M

Y

)

(22)

CO2 5

Y

At this point we consider a total molar balance for the mixer M

M

M

(23)

1

5

2

and note that the total molar flow rate of Stream #2 was given earlier by Eq. 6b. This leads to the mole balance around the mixer given by Mixer:

M

  

M

C Y

(24)

1

5

Material Balances for Complex Systems

333

and we are ready to develop an iterative solution for the recycle flow and the parameter  .

At this point we begin our analysis of the tear stream, i.e., Stream #5

that can be expressed as

M

 ( M

)  ( M

)  ( M

)

(25)

5

C2H4 5

CO2 5

N2 5

Given these values, we make use of Eq. 24 to express the molar flow rate entering the mixer in the form

M

  CY   

( M

) 

( M

) 

( M

)

(26)

1

C2H4 5

CO2 5

N2 5 

Representing the total molar flow rate of Stream #1 in terms of the three species molar flow rates leads to

M

( M

) 

( M

) 

( M

) )

1

C2H4 1

O2 1

N2 1

in which the molar flow rates of oxygen ( O ) and nitrogen ( N ) are 2

2

specified by Eqs. 15 and 16. Use of those results provides

5

M

 ( M

)  (3  Y)  1  

Y

(27)

1

C H

1

2

4

2

and substitution of this result into Eq. 26 yields

5

( M

)

  CY  (3  Y)  1   Y

C H

1

2

2

4

(28)

 ( M

)  ( M

)  ( M

) 

C

2H4 5

CO2 5

N2 5 

Here we can use Eqs. 4b and 10a to obtain

 ( M

)  ( M

CY

)

(29)

C2H4 1

C2H4 5

which allows us to express Eq. 28 as

5

  

 (3 

) 1     ( M

)  ( M

CY

CY

Y

Y

)  (30)

CO

2

2 5

N2 5 

At this point we return to Eqs. 20, 21 and 22 in order to express the molar flow rates of carbon dioxide ( CO ) and nitrogen ( N ) in the tear stream 2

2

as

334