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Calculate the mass of air supplied. Also calculate individual and total mass of products of combustion per kg of fuel if 30% of excess air is supplied for combustion.

The ultimate analysis of a solid fuel is as follows:- - C = 78%, O2 = 3%, H2 = 3%, S = 1%, moisture= 5% and ash content = 10% - Mumbai university > MECH > SEM 3 > THERMO

Marks: 10M

Year: Dec 2013

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Element Percentage present in fuel
Carbon (C) 78
Hydrogen (H) 3
Sulphur (S) 1
Oxygen (O) 3
Ash 10
Moisture 5

Consider 100 kg of fuel.

For Carbon, the chemical reaction is given as:

$C + O_2 → CO_2$

$12 \ \ \ \ 32 \ \ \ \ \ 44$

12kg of Carbon requires 32 kg of Oxygen and forms 44kg of Carbon Dioxide

Therefore 78 kg of Carbon requires $(\frac{32}{12} × 78) = 208kg$ of Oxygen to form 286 kg of Carbon Dioxide.

For Hydrogen, the chemical reaction is given as:

$2H_2 + O_2 → 2H_2O$

$ 4 \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ 36$

4 kg of Hydrogen requires 32 kg of Oxygen to form 36 kg of Water

Therefore 3 kg of Hydrogen requires $(\frac{32}{4} × 3) = 24kg$ of Oxygen and forms 27 kg of Water.

For Sulphur, the chemical reaction is given as:

$S + O_2 → SO_2$

$32 \ \ \ \ 32 \ \ \ \ \ 64$

32kg of Sulphur requires 32 kg of Oxygen and forms 64kg of Sulphur Dioxide

Therefore 1 kg of Carbon requires $(\frac{32}{32} × 1) = 1$ kg of Oxygen and forms 2 kg of Sulphur Dioxide.

Total Oxygen Required = 208 + 24 + 1 = 233 kg

Total Theoretical Oxygen to be supplied = 233 – 3 = 230 kg

Therefore Total Theoretical Air to be supplied = $(\frac{100}{23} × 230) = 1000kg$ for 100kg of fuel

= 10kg / kg of fuel

Actual Air to be supplied = Theoretical Air x Excess Factor

= 10 x 1.3

= 13 kg /kg of fuel

Now Total Mass of Combustion products = Mass of $(CO_2 + H_2O + SO_2)$

= 2.86 + 0.27 + 0.02

= 3.15 kg /kg of fuel

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