TITLE.PM5

514 ENGINEERING THERMODYNAMICS

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\M-therm\Th11-2.pm5

The percentage composition of products of combustion

Products of Weight Molecular Proportional Percentage

combustion x weight volume

yz = x

y

= z

Σz

× 100

CO 2 1.708 44 0.0388 7.678 per cent (Ans.)

CO 0.884 28 0.0316 6.254 per cent (Ans.)

H 2 O 1.394 18 0.0774 15.317 per cent (Ans.)

N 2 10.01 28 0.3575 70.750 per cent (Ans.)

Σz = 0.5053

Example 11.10. Find an expression for the weight of air supplied per kg of fuel when
carbon content of the fuel and the volumetric analysis of the fuel gas is known.

Show that carbon burnt to CO per kg of fuel is (CO)(C)

CO CO+ 2

, where CO and CO 2 represent

percentage monoxide and carbon dioxide in dry flue gas by volume and C represents percentage
by weight of carbon that is actually burnt.
Solution. Let C = Percentage of carbon, by weight, in the fuel burnt,
CO 2 = Percentage by volume of carbon dioxide in the dry flue gas,
CO = Percentage by volume of carbon monoxide in the dry flue gas, and
N 2 = Percentage by volume of nitrogen in the dry flue gas.
Consider 100 m^3 of the gas.
The actual weight of CO 2 = CO 2 × a × b × 44 in 100 m^3 of gas
The actual weight of CO = CO × a × b × 28 in 100 m^3 of gas
The actual weight of N 2 = N 2 × a × b × 28 in 100 m^3 of gas
The molecular weight of CO = 28
The molecular weight of N 2 = 28
Total mass of carbon per 100 m^3 of flue gas

= [a × b × 44 × CO 2 ] ×

12

44 + [ a × b × 28 × CO] ×

12

28

The reaction producing CO 2 and CO are

C + O 2 = CO 2

12 32 44

∴ C per kg of CO 2 =^12

44

Also 2C + O 2 = 2CO

24 32 56

∴ C per kg of CO =
12
28
All the carbon available in carbon dioxide and carbon monoxide comes from carbon present
in the fuel. Also it can be safely assumed that all nitrogen comes from air (this analysis being not
applicable to the fuels containing nitrogen).