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Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Membrane flux 10 lm2 h with a separation factor of

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Unformatted text preview: psi) / (22,400 x 454 cm3 (STP)/lbmol) = 4.724 x 10-14 lbmol/ft2-s-psi Therefore, the permeances are: PM CO2 = 266,000 4.724 × 10-14 = 1.26 × 10-8 lbmol / ft 2 - s - psi Using the given selectivities, PM H2S = 1.26 × 10-8 lbmol / ft 2 - s - psi PM CH4 = 2.52 × 10-10 lbmol / ft 2 - s - psi Now make the crossflow calculations, letting A = H2S, B = CO2, C = CH4, and n = local molar flow rate on the feed-retentate side. Local mole fractions are x on the feed-retentate side and y on the permeate side. From a rearrangement of Eq. (14-45), the differential component material balances are as follows, including the incremental Euler form for making the numerical calculations, Exercise 14.21 (continued) Analysis: (continued) dx A y A − x A = , dn n ( ( ( ( xAk +1) − xAk ) yAk ) − xAk ) = navg ∆n (1) dxB yB − xB = , dn n ( ( ( ( xBk +1) − xBk ) yBk ) − xBk ) = ∆n navg (2) ( ( ( with xCk +1) = 1 − xAk +1) − xBk +1) (3) In the absence of film mass-transfer resistances on either side of the membrane, the ratios of mass transfer rates through the membrane are given by Eq. (14-39) combined with Eq. (14-40), which are as follows, including the Euler form for making the numerical calculations, yA xA PF − yA PP = α* , AC yC xC PF − yC PP yA yC yB xB PF − yB PP = α* , BC yC xC PF − yC PP yB yC Since, yA + yB + yC = 1, ( yCk +1) = =α * AC ( k +1) y ( k +1) A yA yC =y ( k +1) + ( k +1) C yA yC ( ( xAk +1) PF − yAk ) PP ( ( xCk +1) PF − yCk ) PP ( ( xBk +1) PF − yBk ) PP ( ( xCk +1) PF − yCk ) PP = α* BC yA y B 1 + +1 = yC yC yC 1 1+ with ( k +1) (4) (5) and, therefore, yB yC ( k +1) (6) ( k +1) (7) ( k +1) yB and y =y (8) yC For a differential molar transfer rate, dn, the differential membrane area required is given by Eq. (14-50), which is also given in Euler incremental form for numerical calculation. If we base this calculation on H2S, ( k +1) B d AM = PM A yA dn , xA PF − yA PP ( k +1) C ∆AM = PM A ( yAk +1) ∆n ( ( xAk +1) PF − yAk +1) PP (9) The calculations are made with spreadsheet, using Eqs. (1) to (9), with the following values that remain constant: PP = 20 psia, α * = 50, α * = 50, PM A = 126 × 10−8 lbmol / ft 2 - s - psi . . AC BC Exercise 14.21 (continued) Analysis: (continued) The pressure on the feed-retentate side, PF, decreases from 1,000 to 980 psia. The incremental change in this pressure will have to be adjusted by trial and error so that when the pipeline gas composition is reached on the feed-retentate side, the pressure will have decreased to 980 psia. It is not possible to obtain the exact specified pipeline gas composition with respect to H2S and CO2. So the specification (retentate) is changed to a mol% for H2S of not more than 0.04 and a mol% for CO2 of not more than 2.0. Either H2S or CO2 will control. The initial value of n is 0.44 lbmol/s. Choose an increment, ∆n = -0.005 lbmol/s. The initial values of x(0) are 0.1 for A, 0.2 for B, and 0.7 for C. The following is a set of calculations for...
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