# 7.33: Untitled Page 178

- Page ID
- 18311

## 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