TITLE.PM5

520 ENGINEERING THERMODYNAMICS

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and b = 2 – a = 2 – 1.3 = 0.7

Thus the combustion equation becomes :

C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5)

79

21

F

HG

I

KJ N^2

→ 1.3CO 2 + 0.7CO + 3H 2 O + (0.9) (3.5)

79

21

F

HG

I

KJ N^2

Total number of moles of dry products of combustion

= 1.3 + 0.7 + (0.9) (3.5)

79

21

F

HG

I

KJ

= 1.3 + 0.7 + 11.85 = 13.85 moles/mole of fuel

Volumetric analysis of dry products of combustion is as follows :

CO 2 =^13

13 85

.

.

× 100 = 9.38%. (Ans.)

CO =

07

13 85

.

.

× 100 = 5.05%. (Ans.)

N 2 =

11 85

13 85

.

.

× 100 = 85.56%. (Ans.)

+ Example 11.18. Methane (CH 4 ) is burned with atmospheric air. The analysis of the

products on a ‘dry’ basis is as follows :

CO 2 = 10.00%, O 2 = 2.37%, CO = 0.53%, N 2 = 87.10%.

(i)Determine the combustion equation ; (ii)Calculate the air-fuel ratio ;

(iii)Percent theoretical air.

Solution. (i) Combustion equation :

From the analysis of the products, the following equation can be written, keeping in mind

that this analysis is on a dry basis.

x CH 4 + y O 2 + z N 2 → 10.0 CO 2 + 0.53 CO + 2.37 O 2 + a H 2 O + 87.1 N 2

To determine all the unknown co-efficients let us find balance for each of the elements.

Nitrogen balance : z = 87.1

Since all the nitrogen comes from the air,

z

y =

79

21 ; y =

871

79 21

.
(/) = 23.16
Carbon balance : x = 10.00 + 0.53 = 10.53
Hydrogen balance : a = 2x = 2 × 10.53 = 21.06
Oxygen balance. All the unknown co-efficients have been solved for, and in this case the
oxygen balance provides a check on the accuracy. Thus, y can also be determined by an oxygen
balance

y = 10.00 +^053

2

. + 2.37 + 21 06
2
. = 23.16

Substituting these values for x, y, z and a, we have,

10.53 CH 4 + 23.16 O 2 + 87.1 N 2 → 10.0 CO 2 + 0.53 CO + 2.37 O 2 + 21.06 H 2 O + 87.1 N 2

Dividing both sides by 10.53, we get the combustion equation per mole of fuel,

CH 4 + 2.2 O 2 + 8.27 N 2 → 0.95 CO 2 + 0.05 CO + 2H 2 O + 0.225 O 2 + 8.27 N 2. (Ans.)