Gas Absorption Calculations

CALCULATIONS: 1. Material Balance: Experiment 1. (a) Calculating the mass rate of absorption of CO2 in the amine in absorber using ideal gas law: PV = nRT P = 80.2 kPa + 101.325 kPa = 181.525 kPa = 181525 Pa V (volumetric flow rate) = 3.8 m3/h = 3.8 m3/h / 3600 = 1.05555 x 10-3 m3/s R = 8.314 J/ mol K T = 19.7 0C = 19.7 + 273 K = 292.7 K n = PV/RT = 181525 Pa x 1.05555 x 10-3 m3/s 8.314 J/ mol K x 292.7 K = 7.8926416 x 10-2 mol/s CO2 to absorber = 20% x 7.8926416 x 10-2 mol/s = 0.01579 mol/s Molar flow rate (mol/s) = Mass flow rate (g/s) / RMM Mass flow rate (g/s) = molar flow rate (mol/s) x RMM RMM of CO2 = 12 + 2(16) = 44 g/mol Mass flow rate = 0.01579 mol/s x 44 g/mol = 0.6946 g/s (mass rate of CO2 to absorber)

(b) Calculating the mass flow rate of CO2 in the sweet gas leaving the absorber: PV = nRT P = 80.2 kPa + 101.325 kPa = 181.525 kPa = 181525 Pa V = 5.0 m3/h = 5.0 m3/h / 3600 = 1.388888 x 10-3 m3/s R = 8.314 J/mol K T = 18.2 0C = 18.2 + 273 K = 291.2 K n = PV/RT = 181525 Pa x 1.388888 x 10-3 m3/s 8.314 J/ mol K x 291.2 K = 0.10414 mol/s CO2 leaving absorber = 4% x 0.10414 mol/s = 0.0041656 mol/s Molar flow rate (mol/s) = Mass flow rate (g/s) / RMM Mass flow rate (g/s) = molar flow rate (mol/s) x RMM RMM of CO2 = 12 + 2(16) = 44 g/mol Mass flow rate = 0.0041656 mol/s x 44 g/mol = 0.1833 g/s (mass rate of CO2 in sweet gas leaving the absorber)

(c) Net rate of CO2 absorption: Mass flow rate to absorber – mass flow rate leaving absorber = 0.6946 g/s – 0.1833 g/ s = 0.5113 g/s

(d)Rate of stripping Room temperature recorded was 21o C. Room pressure = atmospheric pressure R = 8.314 J/ mol K T = 21 + 273 = 294K P = 101.325 kPa For Lean Amine: Volume of CO2 evolved = 7.4cm3 Using a basis of 1 second V = 7.4 cm3/s = 7.4 cm3/s x 10-6 = 7.4 x 10-6 m3/s n = PV/RT = 101325 Pa x 7.4 x 10-6 m3/s 8.314 J/ mol K x 294 K = 3.06 x 10-4 mol/s Molar flow rate (mol/s) = Mass flow rate (g/s) / RMM Mass flow rate (g/s) = molar flow rate (mol/s) x RMM RMM of CO2 = 12 + 2(16) = 44 g/mol Mass flow rate of CO2 = 3.06 x 10-4 mol/s x 44 g/mol = 1.3464 x 10-2 g/s (mass of CO2 in 2mL (2 x 10-6 m3) lean amine)

Calculating the mass of CO2 in 4.5 L/min of lean amine: 4.5L/min = 4.5 L/min x 1 m3/h 16.67L/min = 0.26995 m3/h = 7.4985 x 10-5 m3/s Therefore mass of CO2 present in 7.4985 x 10-5 m3 = 1.3464 x 10-2 g x 7.4985 x 10-5 2 x 10-6 = 5.048 x 10-1 g For Rich Amine: Volume of CO2 evolved = 13.8 cm3 Using a basis of 1 second V = 13.8 cm3/s = 13.8 cm3/s x 10-6 = 13.8 x 10-6 m3/s n = PV/RT = 101325 Pa x 13.8 x 10-6 m3/s 8.314 J/mol K x 294 K = 5.72 x 10-4 mol/s Molar flow rate (mol/s) = Mass flow rate (g/s) / RMM Mass flow rate (g/s) = molar flow rate (mol/s) x RMM RMM of CO2 = 12 + 2(16) = 44 g/mol Mass flow rate of CO2 = 5.72 x 10-4 mol/s x 44 g/mol = 2.517 x 10-2 g/s (mass of CO2 in 2mL (2 x 10-6 m3) rich amine) Calculating the mass of CO2 in 4.5L/min of rich amine: 4.5L/min = 4.5 L/min x 1 m3/h 16.67L/min = 7.4985 x 10-5 m3/s

Therefore mass of CO2 present in 7.4985 x 10-5 m3 = 2.517 x 10-2 g x 7.4985 x 10-5 2 x 10-6 = 9.437 x 10-1 g

Therefore, Using a basis of 1 second: Rate of stripping of CO2 from amine in the regenerator = 9.437 x 10-1 g/s - 5.048 x 10-1 g/s = 4.389 x 10-1 g/h

(e) Comparing the net rate of CO2 absorption (c) with the rate of stripping in (d): Rate of stripping - Net rate of absorption = 5.4709 x 10-4 g/s - 4.13756 x 10-4 g/s = 1.33334 x 10-4 g/ s

Percentage difference = 1.33334 x 10-4 g/ s × 100 % 4.13756 x 10-4 g/s = 32.2%