Question

Sulphur dioxide is to be scrubbed from an air stream in a small packed tower by contacting it with an organic amine. The feed gas 3% SO2, by volume, and 95% of it is to be absorbed. The total gas rate is 150 m^{3}/h. at 20^{

Answer 1

Feed gas: $150 m^3/h, 20^0 C; 1.1 bar \implies G_1= 150 * \frac{273}{293}*\frac{1.1}{1.013}*\frac{1}{22.4}$

$Inlet \space SO_2 \space conc, y_1 = 0.03* G_s= 6.775 (1-0.03) = 6.572 \frac{k mol}{h}$

= 0.1930 kmol/ k ; SO₂ leaving = 0.0102 kmol/h

$\implies y_2 = \frac{0.0102}{0.0102+6.572}=0.00155 \\ Liquid \space rate = 1.4 kmol /h = L_s= L_2\\ x_1= \frac{0.1930}{0.1930+1.4}=0.1212; x_2=0 (feed \space liquid \space SO_2-free) )$