# 6.11: Untitled Page 134

## Chapter 6

### Axiom I:

c v n dA

R dV ,

A  C H , C H , H

(2)

A A

A

2

6

2

4

2

A

V

Application of this result to the control volume illustrated in Figure 6.2

provides the following three equations:

Ethane:

 ( y

) M

 ( y

) M

 R

(3)

C2H6 1 1

C2H6 2

2

C2 H6

Ethylene:

 ( y

) M

 ( y

) M

 R

(4)

C2H4 1 1

C2H4 2

2

C2 H4

Hydrogen:

 ( y ) M

( y

) M

 R

(5)

H

1

1

H

2

2

H

2

2

2

Here we have used R to represent the global net rate of production for A

species A that is defined by (see Eq. 6‐30)

R

R dV ,

A  C H , C H , H

(6)

A

A

2

6

2

4

2

V

The units of the global rate of production, R , are moles time while the A

units of the rate of production, R , are moles ( time volume) , and one A

must be careful to note this difference.

At the entrance and exit of the control volume, we have two constraints on the mole fractions given by

Stream #1:

( y

)  ( y

)  ( y

)

 1

(7)

C2 H6 1

C2 H4 1

H2 1

Stream #2:

( y

)  ( y

)  ( y

)

 1

(8)

C H

2

C H

2

H

2

6

2

4

2

2

For this particular process, the global form of Axiom II can be expressed as

A N

Axiom II

N R

 0 ,

J  C , H

(9)

JA A

A  1

The visual representation of the atomic matrix is given by Molecular Species  C H

C H

H

2

6

2

4

2

carbon

 2

2

0 

(10)

hydrogen

 6

4

2 

and we express the explicit form of this matrix as

Stoichiometry

247

2 2 0

2 2 0

A

,

or

N  

(11)

6 4 2

JA

6 4 2

Use of this result for the atomic matrix with Eq. 9 leads to

RC

2 H6 

2 2 0

0

R

 

(12)

H2

6 4 2 

0

R

C

2 H4 

At this point we can follow the development in Sec. 6.2.5 to obtain

RC

2 H6 

1

0

1

0

R

H

 

(13)

2

0

1

1 

0

R

C

2 H4 

in which C H has been chosen to be the pivot species (see Sec. 6.4).

2

4

Carrying out the matrix multiplication leads to

R

  R

(14a)

C2 H6

C2 H4

R

 R

(14b)

H

C H

2

2

4

in which R

is to be determined experimentally. A degree of freedom

C2H4

analysis will show that a unique solution is available and we can summarize the various equations as

Ethane mole balance:  100 kmol/min 

( y

) M

 R

(14)

C2H6 2

2

C2 H6

Ethylene mole balance:

( y

) M

 R

(15)

C2H4 2

2

C2 H4

Hydrogen mole balance:

30 kmol/min  R

(16)

H2

Stream #1:

( y

)  1 , ( y

)  0 , ( y

)

 0

(17)

C2 H6 1

C2 H4 1

H2 1

Stream #2:

( y

)  ( y

)  ( y

)

 1

(18)

C2 H6 2

C2 H4 2

H2 2

Axiom II constraint:

R

 R

(19)

C2 H6

C2 H4

Axiom II constraint:

R

 R

(20)

H

C H

2

2

4

The solution to Eqs. 14 through 20 is given by

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