# 7.7: Untitled Page 152

## Chapter 7

At this point we have completed our analysis of Axiom II and we are ready to apply Axiom I using the control volume illustrated in Figure 7‐3. Application of the steady state form of Eq. 7‐5 for both the fuel ( C H O ) and the oxygen m

n

p

2

Fuel:

 ( M

)  ( M

)

 R

(7‐24)

C H O

1

C H O

3

C H O

m

n p

m

n p

m n p

Oxygen:

 ( M )  ( M )

 R

(7‐25)

O2 2

O

3

O

2

2

In order to determine the theoretical air needed for complete combustion, we assume that all the fuel and all the oxygen that enter the combustion chamber are reacted, thus there is no fuel in Stream #3 and there is no oxygen in Stream #3.

Under these circumstances Eqs. 7‐24 and 7‐25 reduce to

Fuel ( complete combustion):

 

( M

)

 R

(7‐26)

C H O

1

C H O

m n p

m

n p

Oxygen ( complete combustion):

 ( M )

 R

(7‐27)

O

2

O

2

2

At this point we represent R

in terms of R

using Eq. 7‐23, and this

O2

C H O

m

n p

allows us to express the theoretical oxygen required for complete combustion as 4 m n  2 p

( M

)

 ( M )

( M

)

(7‐28)

O2 Theoretical

O2 2

C

4

m H n O p 1

If the fuel in Stream #1 contains N molecular species that can be represented as C H O , the theoretical oxygen required for complete combustion is given by m

n

p

i N 4 mn2 p

 

( M

)

(

O

M

)

2 Theoretical

 

 

4

 

C m H n O

(7‐29)

p 1 

i  1

i

i

This analysis for the theoretical oxygen can be extended to other fuels that may contain sulfur provided that the molecules in the fuel can be characterized by C H O S . In this case the complete combustion products would be CO , H O

m

n

p q

2

2

and SO .

2

EXAMPLE 7.2: Determination of theoretical air

A fuel containing 60% CH , 15% C H , 5% CO and 20% N (all 4

2

6

2

mole percent) is burned with air to produce a flue gas containing CO , 2

H O , and N . The combustion process takes place in the unit illustrated 2

2

Material Balances for Complex Systems

281

in Figure 7‐3, and we want to determine the molar flow rate of theoretical air needed for complete combustion. The solution to this problem is given by Eq. 7‐29 that takes the form

i  3 4 mn 2 p

 

( M

)

( M

)

O2 Theoretical

 

 

4

C H O

m n p 1 

1

i

i

i

 4  4 

 8  6 

 4  2 

( M

) 

( M

) 

( 

M

)

(1)

CH

C H

 4 

4

 4 

2 6

 4 

CO

1

1

1

2 ( M

)  7 

( M

)  1 

( M

)

CH4

C2H6

CO

1

1

1

2

2

If we let M be the total molar flow rate of Stream #1, we can express 1

Eq. 1 as

7

1

( M

)

2 0 60

.

( 0 15

.

)

( 0 05

.

)

M

O

Theoretical

 

2

1

2

2

(2)

 1 . 75 M1

This indicates that for every mole of feed we require 1.75 moles of oxygen for complete combustion. Taking the mole fraction of oxygen in air to be 0.21, we find the molar flow rate of air to be given by

 1 

( M

)

1 7

. 5 M

8 3

. 3 M

(3)

air Theoretical

1

1

 0 2

. 1 

The determination of the theoretical molar flow rate of air in Example 7.2 was relatively simple because the complete combustion process involved a feed stream that could be described in terms of a single molecular form given by C H O . In the next example, we consider a slightly more complex process in m

n

p

which the excess air is specified.

EXAMPLE 7.3: Combustion of residual synthesis gas

In Figure 7.3 we have illustrated a synthesis gas (Stream #1) consisting of 0.4% methane ( CH ) , 52.8% hydrogen ( H ) , 38.3% carbon monoxide 4

2

( CO ) , 5.5% carbon dioxide ( CO ) , 0.1% oxygen ( O ) , and 2.9% nitrogen 2

2

( N ) . The synthesis gas reacts with air (Stream #2) that is supplied at a 2  282