Calculate the minimum amount of air required for the complete combustion of 1 kg of coal containing C = 81 %, H2 = 5 %, O2 = 8.5 %, S = 1 %, N2 = 1 %, ash = 3.5 %.
Answers
Answer:This may help you
Explanation:
The quantity of oxygen required for combustion of 1 kg of the fuel is: (2.66C + 8H + S) kg. ∴ Oxygen required from air for the complete combustion of fuel will be (2.66C + 8H + S – O) which can be written as 2.66C + 8 (H – O/8) + S, the term in the bracket being known as the available hydrogen.
Answer:
Weight of air required =10.28 kg
Volume of air required = 8.375 litres
Explanation:
Weight of hydrogen = 5 / 100 1 = 0.0 5 k g
Weight of carbon = 8 1 / 100 1 = 0.8 1 k g
Weight of sulphur = 1 / 1 00*1=0.01 k g
Weight of oxygen = 8 /100*1 = 0.0 8 K g
Calculation of O2 needed for 1 k g of coal CO2 = 0.81 32/12 = 2.16 k g
2 H 2 O= 0.05 32/12 = 0.4 kg SO2 = 0.0132/12 =0.01 k g
Total 0 2 required = 2.57 k g Less O2 available = -0.08
Net 02 required = 2.49 k g
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
= 8374.84 m l
air 8.375 litres of air
Weight of air required =10 . 28 k g
Volume of air required = 8 . 375 litres
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