# 7.6: Untitled Page 151

## Chapter 7

excess air to displace the reaction equilibrium. Matters are complicated further by the presence of water in the air stream. In this simplified treatment of combustion, we will ignore the complexities associated with NOX in the exit stream and water vapor in the entrance stream.

Theoretical air

The first step in analyzing a combustion process is to determine the rate at which air must be supplied in order to achieve complete combustion. This is called the theoretical air requirement for the combustion process and it is based on the fuel composition. Most fuels consist of many different hydrocarbons; however, in this simple example we consider the fuel to consist of a single molecule represented by C H O . Examples are propane ( m  3 , n  8 , p  0 ), m

n

p

carbon monoxide ( m  1 , n  0 , p  1), and methanol ( m  1 , n  4 , p  1 ). Total combustion is the conversion of all the fuel to CO and H O . Our tools for this 2

2

analysis are Axioms I and II as represented by Eqs. 7‐4 and 7‐5 with the global rate of production defined by Eq. 7‐3. We first express the atomic matrix as Molecular species 

C H O

O

m

n

p

2

CO2 H2O

carbon

m

0

1

0

(7‐16)

hydrogen

n

0

0

2

oxygen

p

2

2

1

and make use of this representation to express Axiom II in the form

R

 C H O

m n p

A N

m 0 1 0

0

RO

 

Axiom II:

N

R

n

0 0 2 

2

 

JA A

0

(7‐17)

 R

A  1

p 2 2 1

CO2

0

 

 

 RH O 

2

The atomic matrix can be expressed in row reduced echelon form according to

2

1

0

0

n

4 m n

2 p

N

0

1

0

(7‐18)

JA

2 n

2 m

0

0

1



n



Material Balances for Complex Systems

279

This indicates that the rank is three, and all the species production rates can be expressed in terms of a single pivot species that we choose to be water. Use of Eq. 7‐18 in Eq. 7‐17 allows us to express Axiom II in the form

2

1

0

0

R

C H O

m n p

n

0

4 m n  2 p

RO

 

2

0

1

0

 

0

(7‐19)

2 n

 

 R

CO

2

0

2 m

 

0

0

1

  RH



 

2O

n

This equation can be partitioned according to the development in Example 6.5 or the development in Sec. 6.4 in order to obtain

2

 RC H O

1 0 0

n

 

m n p

0

 4 m n  2 p

 

0 1 0 

R

R

0

(7‐20)

O

 

2

H

 

2

2 n

O 

0 0 1

0

 

R

 

CO

2 m

2



n



This immediately leads to a special case of the global pivot theorem (see Eq. 6‐81)

2

RC H O

n

m n p

4 m n

2 p

Global Pivot Theorem:

R

(7‐21)

O

R

2

H

 

2

2 n

O 

 R

CO

2 m

2



n



which provides the species global net rates of production in terms of the global net rate of production of water as given by

2

4 m n

R

  R

R

 

2 p R

R

 2 m

,

,

C H O

H

R

(7‐22)

m n p

2O

O2

H2O

CO2

H O

n

2 n

2

n

In this particular problem, it is convenient to represent the global net rate of production of fuel (which is negative) in terms of the global net rate of production of oxygen (which is negative), thus we use the first two of Eqs. 7‐22 to obtain

4 m n  2 p

R

(7‐23)

O

R

2

C

4

m H n O p

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