Unformatted text preview: psi) / (22,400 x 454 cm3 (STP)/lbmol)
= 4.724 x 1014 lbmol/ft2spsi
Therefore, the permeances are:
PM CO2 = 266,000 4.724 × 1014 = 1.26 × 108 lbmol / ft 2  s  psi
Using the given selectivities,
PM H2S = 1.26 × 108 lbmol / ft 2  s  psi
PM CH4 = 2.52 × 1010 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 feedretentate side. Local mole fractions are x on the feedretentate side and y on
the permeate side. From a rearrangement of Eq. (1445), 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 masstransfer resistances on either side of the membrane, the ratios of
mass transfer rates through the membrane are given by Eq. (1439) combined with Eq. (1440),
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.
(1450), 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 feedretentate 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 feedretentate 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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 Spring '11
 Levicky
 The Land

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