7.8: Untitled Page 153

Chapter 7

rate which provides 10% excess oxygen, and the composition of Stream #2

is assumed to be 79% nitrogen ( N ) and 21% oxygen ( O ) .

2

2

In this example we want to determine: (I) the molar flow rate of air relative to the molar flow rate of the synthesis gas, and (II) the species molar flow rates of the components of the flue gas relative to the molar flow rate of the synthesis gas. We assume complete combustion so that all hydrocarbon species in Stream #1 are converted to carbon dioxide ( CO ) 2

and water ( H O ) . In addition, we assume that the nitrogen is inert so 2

that no NOX appears in the exit stream. This means that the molecular species in the flue gas are N , O , CO and H O .

2

2

2

2

Figure 7.3 Combustion of synthesis gas

In this example, the development leading to Eq. 7‐29 is applicable for the combustion of synthesis gas and that result takes the form

i  3 4 mn 2 p

 

( M

)

O

( M

2 Theoretical

 

 

4

)

C H O

m n p 1 

i

i

1

i

(1)

4  4

2

4 

( M

) 

( M

) 

2 

CH

H

( M CO)

4 1

2 1

1

4

4

4

Material Balances for Complex Systems

283

The species molar flow rates in Stream #1 can be expressed in terms of the total molar flow rate, M

 according to

1

( M

)  0 004

.

M

, ( M )  0 528

.

M

, ( M )  0 383

.

M

 (2)

CH4

1

H

1

CO

1

1

2 1

1

and these results can be used in Eq. 1 to obtain the molar flow rate of theoretical air in terms of the molar flow rate of synthesis gas.

( M

)

O

Theoretical

2

 4  4

2

4  2

0 . 004  0 . 528 

0 . 383  M

(3)

1

 4

4

4

 0 . 4635 M1

Taking into account that Stream #1 contains oxygen, the condition of 10%

excess oxygen requires

( M

) 

( M

)

1 1

. 0 ( M

)

(4)

O

O

  

2

2

O

1

2

2 Theoretical

and this allows us to express the molar flow rate of oxygen in Stream #2 as

( M

)

1 1

. 0 ( M

)

( y

) 

M

(5)

O

  

2 2

O2 Theoretical

O2 1 1

Given that the mole fraction of oxygen ( O ) in the synthesis gas 2

(Stream #1) is 0.001, we can use this result along with Eq. 3 to obtain ( M

)

 0 509

.

M

(6)

O2 2

1

We are given that the mole fraction of oxygen ( O ) in the air (Stream #2) 2

is 0.21 and this leads to the total molar flow rate for air given by M

 2 4

. 2 M

(7)

2

1

Knowing the molar flow rate of the air stream relative to the molar flow rate of the synthesis gas that is required for complete combustion is crucial for the proper operation of the system. Knowing how much air is required to achieve complete combustion will generally be determined experimentally, thus the ratio 2.42 will be adjusted to achieve the desired operating condition.

In addition to knowing the required flow rate of air, we also need to know the molar flow rates of the species in the flue gas since this gas will

284