E3.10 Dry Scrubbing

Example 3.10: A flue gas containing 700 lb/h of SO₂ and 500 lb/h of HCl enters a dry scrubber where lime feed rate is 1.6 times the stoichiometric requirement. Dry scrubber is 80% efficient in removing SO₂ and 90% efficient in removing HCl. How much is the lime addition-rate? What is the mass flow rate of CaO, SO₂, HCl, CaSO₃, CaCl₂ and H₂O in the flue gas, leaving scrubber. Assume that CO₂ does not react with the lime (CaO).

Solution:

Mass of SO₂ = 700 lb = 700/64 = 10.938 lbmol

Moles of SO₂ converted = 0.8(10.938) = 8.75 lbmol

Mass of HCl = 500 lb = 500/36.5 = 13.7 lbmol

Moles of HCl converted = 0.9(13.7) = 12.33 lbmol

Stoichiometric requirement of CaO = 1(10.938) + 0.5(13.7) = 17.79 lbmol

Excess of lime to be used = 60%

Actual requirement of lime = 1.6(17.79) = 28.46 lbmol = 28.46(56) = 1593.7 lb

Actual consumption of CaO = 1(8.75) + 0.5(12.33) = 14.92 lbmol = 14.92(56) = 835.24 lb

Total mass = 700 + 500 + 1593.7 = 2793.7 lb

CaO in the product = 1593.7 - 835.24 = 758.46 lb

CaSO₃ formed = 8.75 lbmol = 8.75(120) = 1050 lb

CaCl₂ formed = 0.5(12.33) = 6.165 lbmol = 6.165(111) = 684.27 lb

Water formed = 0.5(12.33) = 6.165 lbmol = 6.165(18) = 110.96 lb

SO₂ unconverted = 0.2(10.938) = 2.188 lbmol = 2.188(64) = 140 lb

HCl unconverted = 0.1(13.7) = 1.37 lbmol = 1.37(36.5) = 50 lb

Total products rate = 758.46 + 1050 + 684.27 + 110.96 + 140 + 50 = 2793.69