written 8.7 years ago by | • modified 8.7 years ago |
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
written 8.7 years ago by | • modified 8.7 years ago |
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
written 8.7 years ago by |
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+O2→CO2
12 32 44
12kg of Carbon requires 32 kg of Oxygen and forms 44kg of Carbon Dioxide
Therefore 78 kg of Carbon requires (3212×78)=208kg of Oxygen to form 286 kg of Carbon Dioxide.
For Hydrogen, the chemical reaction is given as:
2H2+O2→2H2O
4 32 36
4 kg of Hydrogen requires 32 kg of Oxygen to form 36 kg of Water
Therefore 3 kg of Hydrogen requires (324×3)=24kg of Oxygen and forms 27 kg of Water.
For Sulphur, the chemical reaction is given as:
S+O2→SO2
32 32 64
32kg of Sulphur requires 32 kg of Oxygen and forms 64kg of Sulphur Dioxide
Therefore 1 kg of Carbon requires (3232×1)=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 = (10023×230)=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 (CO2+H2O+SO2)
= 2.86 + 0.27 + 0.02
= 3.15 kg /kg of fuel