Problem 6.2.5

Calculate the liquid and vapor flow rates in a fractionation column that is producing 84.5 moles per hour when it is being fed at 216.8 moles per hour. The column is operating with a reflux ratio of 0.915. The feed quality can be represented by q-value of 1.04. Assume equi-molar flow and vaporization.

Solution:

Flow rates in the rectification section are represented by subscript n.

Equi-molar assumption implies that reflux and vapor flow rates on all plates in the rectification section are equal. i.e., L_n = L_n+1 = L_n+2 = L_n+3 = …=L_n

And V_n = V_n+1 = V_n+2 = V_n+3 = …= V_n.

By basis definition, L_n is equal to D times R. Thus L_n is 84.5 times 0.915 or 77.32 moles per hour.

Vapor flow rate in the rectification section, V_n, is given by D times (R plus one). So Vn is equal to 84.5 times (0.915 plus 1) or 161.82 moles per hour.

Similarly, the liquid and vapor flow rates on all plates in the stripping section are equal. i.e., L_m = L_m+1 = L_m+2 = L_m+3 = …=L_m

And V_m = V_m+1 = V_m+2 = V_m+3 = …= V_m.

Liquid flow rate in the stripping section, L_m is equal to q times F plus L_n. So L_m is equal to 1.04 times 216.8 plus 77.32 or 302.79 moles per hour.

Vapor flow rate in the stripping section, V_m is equal to (q minus 1) times F plus V_n. So V_m is equal to (1.04 minus 1) times 216.8 plus 161.82 or 170.49 moles per hour.