TITLE.PM5

FUELS AND COMBUSTION 511

dharm

\M-therm\Th11-2.pm5

Excess air supplied =

OV

(O)

2

(^212)
×
− =
60 342
21 6
..
()
×
− =
20 52
15

. = 1.37 m 3

Total quantity of air supplied = 3.69 + 1.37 = 5.06 m^3.

Air fuel ratio =

Volume of air

Volume of fuel =

506

1

.

= 5 (say). (Ans.)

+Example 11.7. The following is the ultimate analysis of a sample of petrol by weight :
Carbon = 85 per cent ; Hydrogen = 15 per cent.
Calculate the ratio of air to petrol consumption by weight if the volumetric analysis of the
dry exhaust gas is :
CO 2 = 11.5 per cent ; CO = 1.2 per cent ; O 2 = 0.9 per cent ; N 2 = 86 per cent.
Also find percentage excess air.
Solution.

Name of gas Volume per m^3 Molecular Relative Weight per kg

of flue gas weight weight of flue gas

(x) (y) z = x × y Σzz

CO 2 0.115 44 5.06 0.1700

CO 0.012 28 0.336 0.0113

O 2 0.009 32 0.288 0.0096

N 2 0.86 28 24.08 0.8091

Σz = 29.76

∴ Weight of carbon per kg of flue gas

= Weight of carbon in 0.17 kg of CO 2 + Weight of carbon in

0.0113 kg of CO

=

3

11

× 0.17 +

3

7

× 0.0113 = 0.0512 kg

∴ Weight of dry flue gas per kg of fuel =

085

0 0512

.

.

= 16.6 kg

Vapour of combustion = 9 × 0.15 = 1.35 kg

Total weight of gas = 16.6 + 1.35 = 17.95 kg per kg of fuel

∴ Air supplied = (17.95 – 1) = 16.95 kg/kg of fuel

∴ Ratio of air to petrol = 16.95 : 1. (Ans.)

Stoichiometric air =^085

8

3

F .(.)× 015 8

HG

I

KJ

+×

L

N

M

O

Q

P ×

100

23

= 15.07 kg per kg of fuel

∴ Excess air = 16.95 – 15.07 = 1.88 kg

∴ Percentage excess air =^188

15 07

.

.

× 100 = 12.47%. (Ans.)