A sample of coal was found to contain C= 80%, H=5%, O= 1% N= 2%and ash =12%. Calculate
the minimum amount of air required for complete combustion of 1Kg of coal Sample (CO6)
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Let us calculate the 02 required for 1kg of coal first
Weight of carbon = 81 / 100 * 1 = 0.81 kg
Weight of hydrogen = 5 / 100 * 1 = 0.05 kg
Weight of sulphur = 1 / 100 * 1 =0.01 kg
Weight of oxygen = 8 / 100 * 1 = 0.08 kg
Calculation of 02 needed for 1 kg of coal
CO2 = 0.81 * 32 /12 = 2.16 kg
2H20 = 0.05 * 32 / 12 = 0.4 kg
SO2 = 0.01 * 32 / 12 =0.01 kg
Total 02 required = 2.57 kg
Less 02 available= -0.08
Net 02 required = 2.49 kg
Weight of air required = weight of 02 / 23 *100
= 2.49 / 23 *100
= 10.82 kg of air
Volume of air:
:.28.94 kg of air = 22400 ml volume at NTP
:. 10.82 kg of air = 22400 * 10.82 / 28.94
= 837484 m!
air = 8.375 litres of air
Weight of air requires
= 10.28 kg
Volume of air required = 8.375 Litres.
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