Exercise 07 Solution ENG

Pagina 1

Assignment #7: Results 1. Absorption of H2S Assuming a total molar flow rate for the inlet gas stream (containing H2S and N2) of 1 [kmol/h], the molar flow rate of the non-migrant gas component is G = 0.997 [kmol/h]. a) Minimum water flow rate required Under the assumption of dilute solutions: 483.38 L min G kmolL G kmol     . The minimum water flow rate required is 481.93 kmol per kmol of inlet gas. b) Number of equilibrium stages using (L/G) = 1.2·(L/G)min N = 11. c) Composition of the outlet streams 9.75 5NYe  , 1 5.02 6Xe  . 2. Non-linear absorption The equilibrium curve is represented in this case by a branch of hyperbole having K < 1. By solving the equation: 2( ) 4 0A B D     the following two roots are obtained for (L/G)min: 1) (L/G)min = 0.0934 2) (L/G)min = 0.1840. These two values can be used to calculate Ytg: 1) Ytg = 0.027198 2) Ytg = -0.025525. The first solution is acceptable because it is such that YN < Ytg < Y0. The oil flow rate flowing within the gas-absorption system is: 8.9045L  [kmol/h].  Determination of the number of ideal stages by the graphical solution of the Riccati equation Starting from:  1 1 0 0.344371NN GX X Y Y L     it is possible to find that the number of ideal stages required to reach the desired purity level in the outlet gas stream is 6.  Determination of the number of ideal stages by the analytical solution of the Riccati equation The analytical solution of the Riccati equation is given by: 0 11 2 11 2 n n i i i i ii Y Y A B Y A Y BY Y Y A B Y           where: 2( ) ( ) 4 2 i A B A B DY     and A, B and D are calculated with the actual (L/G). When 2( ) 4 0A B D     , the ana lytical solution of the Riccati equation can be rearranged to determine the number of ideal stages,…