TITLE.PM5

510 ENGINEERING THERMODYNAMICS

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\M-therm\Th11-2.pm5

Also carbon in the fuel = 0.78 kg.

∴ Weight of dry flue gas per kg of fuel

= Weight of carbon in 1 kg of fuel

Weight of carbon in 1 kg of flue gas

=^078

0 0424

.

.

= 18.4 kg. (Ans.)

(ii)Weight of excess air per kg of fuel :

Weight of excess oxygen per kg of flue gas = 0.0832 –

4

7 × .0019

22

56 32 88

1 4

7

11

7

CO O 22 CO

kg kg kg

+=

+=

L

N

M

M

M

M

M

O

Q

P

P

P

P

P

= 0.0832 – 0.0011 (allowing for unburnt carbon monoxide)

= 0.0821 kg.

Weight of excess O 2 per kg of fuel = 18.4 × 0.0821 = 1.51 kg

∴ Weight of excess air per kg of fuel =

151 100

23

. ×
= 6.56 kg. (Ans.)
Example 11.6. A single cylinder was supplied with a gas having the following percentage
volumetric analysis ; CO = 5, CO 2 = 10, H 2 = 50, CH 4 = 25, N 2 = 10. The percentage volumetric
analysis of dry gases was CO 2 = 8, O 2 = 6 and N 2 = 86. Determine the air-fuel ratio by volume.
Solution. Combustion equations are :
2H 2 + O 2 = 2H 2 O
1 vol + 1/2 vol = 1 vol
2CO + O 2 = 2CO 2
1 vol + 1/2 vol = 1 vol
CH 4 + 2O 2 = CO 2 + 2H 2 O
1 vol + 2 vol = 1 vol + 2 vol.

Gas Vol (m^3 )O 2 needed (m^3 ) Products (m^3 )

CO 2 N 2

CO 0.05 0.025 0.05 —

CO 2 0.10 — 0.10 —

H 2 0.50 0.25 — —

CH 4 0.25 0.50 0.25 —

N 2 0.10 — — 0.10

Total 1.0 0.775 0.4 0.10

Volume of air required = 0.775 × 100/21 = 3.69 m^3
Volume of nitrogen in the air = 3.69 × 79/100 = 2.92 m^3
Dry combustion products of 1 m^3 of gases (V) contain 0.4 m^3 of CO 2 + 0.1 m^3 of N 2 (as given
in the table) + 2.92 m^3 of N 2 (from air supplied for complete combustion) = 3.42 m^3.