TITLE.PM5

512 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th11-2.pm5

+Example 11.8. A sample of fuel has the following percentage composition : Carbon = 86
per cent ; Hydrogen = 8 per cent ; Sulphur = 3 per cent ; Oxygen = 2 per cent ; Ash = 1 per cent.
For an air-fuel ratio of 12 : 1, calculate :
(i)Mixture strength as a percentage rich or weak.
(ii)Volumetric analysis of the dry products of combustion.
Solution.
Element, wt. (kg) O 2 reqd. (kg)

C = 0.86 0.86 ×^83 = 2.29

H 2 = 0.08 0.08 × 8 = 0.64

S = 0.03 0.03 ×^11 = 0.03

O 2 = 0.02

Total O 2 = 2.96

Weight of oxygen to be supplied per kg of fuel = 2.96 – 0.02 = 294 kg.

Weight of minimum air required for complete combustion =

2 94 100

23

. ×
= 12.78 kg

Hence “correct” fuel-air ratio =^1

12 78.

: 1

But actual ratio is

1

12 : 1.

(i)Mixture strength =

12 78

12

.

× 100 = 106.5%

This show that mixture is 6.5% rich. (Ans.)

Deficient amount of air = 12.78 – 12 = 0.78 kg

Amount of air saved by burning 1 kg of C to CO instead of CO 2

= Oxygen saved ×

100

23

=

8

3

4

(^23)
L (CO )− (CO)
NM
O
QP
×
100
23
= 5.8 kg
Hence^078
58
.
.
= 0.134 kg of carbon burns to CO and as such 0.86 – 0.134 = 0.726 kg of carbon
burns to CO 2.
∴ CO formed = 0.134 ×^7
3
= 0.313 kg
CO 2 formed = 0.726 ×^11
3
= 2.662 kg
N 2 supplied = 12 × 0.77 = 9.24 kg
SO 2 formed = 0.03 × 2 = 0.06 kg.