How to calculate the quantity of lime stone required to reduce SO2 emission ??

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)

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