TITLE.PM5

526 ENGINEERING THERMODYNAMICS

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O 2 =

075

19 85

.

.

× 100 = 3.78%. (Ans.)

N 2 =

14 1

19 85

.

.

× 100 = 71.03%. (Ans.)

The total dry moles = 2 + 0.75 + 14.1 = 16.85

Hence dry analysis is :

CO 2 =

2

16 85.

× 100 = 11.87%. (Ans.)

O 2 =^075

16 85

.

.

× 100 = 4.45%. (Ans.)

N 2 = 14 1

16 85

.

.

× 100 = 83.68%. (Ans.)

Considering a mixture strength of 130% :

Now, 1.3 = Stoichiometric ratio

Actual A/F ratio

∴ Actual A/F ratio =

896

13

.

.

= 6.89/1. (Ans.)

This means that

1

13.

or 0.769 of the stoichiometric air is supplied. The combustion cannot

be complete, as the necessary oxygen is not available. It is usual to assume that all hydrogen is

burned to H 2 O, since hydrogen atoms have a greater affinity for oxygen than carbon atoms. The

carbon in the fuel will burn to CO and CO 2 , but the relative proportions have to be determined.

Let, a = Number of moles of CO 2 in the products, and

b = Number of moles of CO in the products.

Then the combustion equation is as follows :

C 2 H 6 O + 0.769 3379

(^2221)
F ON+×
HG
I
KJ
→ a CO 2 + b CO + 3H 2 O + 0.769 × 3 ×^79
21
N 2
To find a and b a balance of carbon and oxygen atoms can be made,
i.e., Carbon balance :
2 = a + b ...(i)
and Oxygen balance :
1 + 2 × 0.769 × 3 = 2a + b + 3
or 2.614 = 2a + b ...(ii)
From eqns. (i) and (ii), we get
a = 0.614, b = 1.386
i.e., The products are : 0.614 moles CO 2 + 1.386 moles CO + 3 moles H 2 O + 8.678 moles N 2
The total moles = 0.614 + 1.386 + 3 + 8.678 = 13.678.
Hence wet analysis is :
CO 2 =
0 614
13 678
.
.
× 100 = 4.49%. (Ans.)
CO = 1 386
13 678
.
.
× 100 = 10.13%. (Ans.)