We know that, Lime stone is nothing but Calcium carbonate (CacO3)
Lime stone (CaCo3) on heating gets converted into slaked lime and carbon dioxide
i.e,
CaCo3 + Heat = Cao + Co2
Shall calculate, quantity of CacO3 required for converting into Cao.
Molecular weight of Ca = 40 g/mol
Molecular weight of Oxygen = 16 g/mol
Molecular weight of Carbon = 12 g/mol
Molecular weight of Sulphur = 32 g/mol
Sulphur on heating gets converted into Sulphur di-oxide (SO2)
Therefore we have,
CaCo3 + Heat = Cao + CO2
(40+12+3X16) + Heat = (40+16) + (12+2X16)
100 + Heat = 56 + 44
1 + Heat = 0.56 + 0.44-----------I
This implies, 1 kg of calcium carbonate (CaCO3) on heating gets converted into 0.56 kg of slaked lime & 0.44 kg of CO2
Further,Sulphur present in coal, on combustion gets converted into sulphur dioxide (SO2)
Shall calculate, quantity of SO2 generated on combustion of 1 kg of sulphur
S + O2 = SO2
32 + 2X16 = 32+2X16
32 + 32 = 64
1 + 1 = 2-----II
This means, 2 kg of SO2 will produce on combustion of 1 kg of sulphur.
Further,Slaked lime (Cao) reacts with Sulphur dioxide (SO2) & converts into Calcium sulphate (CaSO4)
i.e
Cao + SO2 + O2 = CaSO4
(40+16) + (32 + 2X16) + 16 = (40+32+4X16)
56 +(32 + 2X16) + 16 = 136
56 + 64 + 16 = 136
0.875 + 1 + 0.25 = 2.12------III
This means, 1 kg of So2 needs 0.875 kg of Cao to produce 2.42 kg of Calcium sulphate
i.e, On burning 1 kg of sulphur there produces 2 kg of SO2 (refer equation-II), hence Cao required to reduce sulphur from 1 kg So2 = 2 X 0.875 = 1.75 kg
Similarly referring equation-I, 1 kg of calcium carbonate (CaCO3) on heating gets converted into 0.56 kg of slaked lime(Cao)
Therefore lime stone required for 1 kg sulphur to convert it into CaSO4 is;
1.75/0.56 = 3.125 kg
Considering 95% efficiency for above combustion, total lime stone required for 1 kg sulphur to get converted it into CaSo4 is;
3.125 X 105% = 3.28 kgs-------IV
Demonstration with example.
A 50 MW coal based power plant has specific fuel consumption 0.75 kg/kwh.The sulphur content in the coal is 0.75%, the maximum permissible limit of sulphur in the coal is 0.5%.Calculate the amount of lime stone required to reduce SO2 emission.
Assuming SO2 emission at 0.5% sulphur in coal is normal.
Extra % of sulphur in coal is = 0.75-0.5 = 0.25%
Coal consumption per day
Specific fuel consumption = 0.75 kg/kwh
Total power generation in a day = 50 X 1000 X 24 = 1200000 kw
Total coal consumption = 1200000 X 0.75 / 1000 = 900 MT/day
Extra Sulphur burned = 900 X 0.25% = 2.25 MT or 2250 kg/day
Lime stone required to reduce SO2 emission is 2250 X 3.28 = 7380 kg (refer equation-IV)