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 + C****O****2**

(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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