# 7.9: Untitled Page 154

## Chapter 7

often be discharged into the atmosphere. Determination of these molar flow rates requires the application of Axioms I and II. We begin with the atomic matrix

Molecular Species  CH

H

CO CO

H O O

4

2

2

2

2

carbon

 1

0

1

1

0

0 

(8)

hydrogen

4

2

0

0

2

0

oxygen

 0

0

1

2

1

2 

which leads to the following form of Axiom II:

R

CH4

 R

H2

 

A  6

1 0 1 1 0 0

0

  RCO 

 

Axiom II:

N R

4 2 0 0 2 0

 

0

(9)

JA

A

 

 RCO

A  1

2

0 0 1 2 1 2

0

RH

2O 

RO

2 

This axiom can be expressed in row reduced echelon form according to

R

CH4

 R

H2

1 0 0 1 1

2

  

0

  RCO 

 

0 1 0

2

3

4

 

0

(10)

 RCO 

2

0 0 1

2

1

2

0

 

RH

2O 

RO

2 

Following the development in Sec. 6.4 indicates that the solution for the three non‐pivot global net rates of production is given by R

 R

 R

 2

CH

R

4

CO2

H2O

O2

R

  2 R

 3R

 4

H

CO

H O

R

(11)

O

2

2

2

2

R

  2 R

 R

 2

CO

CO

R

2

H2O

O2

This completes our analysis of Axiom II, and we move on to the steady state form of Axiom I that is given by

Material Balances for Complex Systems

285

Axiom I:

c

dA  R , A  1 , 2 , ..., N

(12)

v n

A

A

A

A

Application of this result to the control volume illustrated in Figure 7.3

 ( M )  ( M )  ( M )

 R ,

A  1 , 2 , ... , 7

(13)

A 1

A 2

A 3

A

while the individual species balances take the forms given by CH :

 ( M

)

  0 004

.

M

 R

(14)

4

CH4 1

1

CH4

H :

 ( M )

 0 528

.

M

 R

(15)

2

H2 1

1

H2

CO :

 ( M )

 0 383

.

M

 R

(16)

CO 1

1

CO

CO :

 ( M

)  ( M

)

 R

(17)

2

CO2 1

CO2 3

CO2

N :

 ( M )  ( M )  ( M )

 R

(18)

2

N2 1

N2 2

N2 3

N2

O :

 ( M )  ( M )  ( M )

 R

(19)

2

O2 1

O2 2

O2 3

O2

H O :

( M

)

 R

(20)

2

H2O 3

H2O

At this point we need to simplify the balance equations for carbon dioxide, nitrogen and oxygen. Beginning with carbon dioxide, we obtain CO :

( M

)

 R

 0 055

.

M

(21)

2

CO2 3

CO2

1

Moving on to the nitrogen balance given by Eq. 18, we make use of the conditions

N :

( M

) 

0 0

. 29 M ,

( M

) 

0 7

. 9 M ,

R

 0

(21)

2

N2 1

1

N2 2

2

N2

to express the molar flow rate in the flue gas as

N :

( M

)

 ( M )  ( M )

 0 0

. 29 M

 0 . 79 M (22)

2

N2 3

N2 1

N2 2

1

2

Finally, the use of Eq. 7 leads to the following result for the molar flow rate of nitrogen:

N :

( M

)

 1 . 94 M

(23)

2

N2 3

1

286