# 7.9: Untitled Page 154

- Page ID
- 18287

## 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

leads to the general expression

( *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