Question

Air containing 1.6 vol% sulphur dioxide is scrubbed withpure water in a packed column of 1.5 m2 cross sectional area and 3.5 m heightpacked with no.2 plastic Intalox saddles, at a pressure of 1 atm. Total gasflow rate is 0.062 kmol/s, the liquid flow rate is 2.2 kmol/s, and the outlet gassulphur dioxide concentration is y = 0.004. At the column temperature, theequilibrium relationship is given by y = 40x. (a) What is L/Lmin (b) Calculate NOG. (c) Determine HOG. (d) Calculate KGa from the data

Answer 1

Gas flow rate, V = 0.062 kmol/s

Sulphur dioxide flow rate in gas = 0.062 * 1.6 / 100 = 0.001 kmol/s

Air flow rate, V' = 0.062 - 0.001 = 0.061 kmol/s

Y'_N+1 = 0.001 / 0.061 = 0.016

Liquid flow rate, L = L' = 2.2 kmol/s

X'₀ = 0

Outlet gas sulphur dioxide concentration, Y₁ = 0.004

Y'₁ = Y₁ / (1 - Y₁) = 0.004/ (1 - 0.004) ~ 0.004

Mass Balance:

L' (X'_N - X'₀) = V' (Y'_N+1 - Y'₁)

2.2 X'_N = 0.061 (0.016 - 0.004)

X'_N = 0.00034

Refer operating and equilibrium lines drawn below for the system:

Slope of operating line (at min condition) = (40 * 0.00034 - 0.004) / (0.0034 - 0) = 28.172

L' min / V' = slope = 28.172

L' min = 28.172 * 0.061 = 1.72 kmol/s

a)

L / Lmin = 2.2 / 1.72 = 1.28

b)

NOG = (Y_N+1 - Y₁) / [(Y_N+1 - Y*_N+1) - (Y₁ - Y*₁)] * ln [(Y_N+1 - Y*_N+1) / (Y₁ - Y*₁)]

Y*_N+1 = 40 * X'_N = 40 * 0.00034 = 0.0136

Y*₁ = 40 * X'₀ = 40 * 0 = 0

NOG = (0.016 - 0.004) / [(0.016 - 0.0136) - (0.004 - 0)] * ln[(0.016 - 0.0136) / (0.004 - 0)]

NOG = 3.7

c)

Height of tower, Z = NOG * HOG = 3.5 m

HOG = Z / NOG = 3.5 / 3.7 = 0.94 m

d)

KGa = G_m / (HOG . P)

P = 1 atm

G_m = V / Area = 0.062 / 1.5 = 0.041 Kmol /m²-s

KGa = 0.041 / (0.94 * 1) = 0.044 Kmol/m³-s-atm