Page 55 - Separation process principles 2

20 Chapter 1 Separation Processes

Table 1.8 Main Separation Factors for Hydrocarbon Case I: This is the simplest case to calculate because two recover-
Recove~y Process ies are given:
ng) = 0.50(21) = 10.5 kmolh 1
Key-Component Separation i
I
Split Column Factor, SP ng' = 0.875(79) = 69.1 kmolih
1
ng) = 21 - 10.5 = 10.5 kmolh
ng) = 79 - 69.1 = 9.9 kmolh
I
Case 2: With the recovery for O2 given, calculate its distribution
into the two products:

power; a small value larger than but close to 1.0 corre-
sponds to a low degree of separation power. For example, if
SP = 10,000 and SRi = l/SRj, then, from (1-5), SRi = 100
and SR, = 0.01, corresponding to a sharp separation. How- Using the purity of Oz in the permeate, the total permeate is
ever, if SP = 9 and SRi = l/SRj, then SRi = 3 and SRj = 4, n") = 10.5/0.5 = 21 kmolh I
corresponding to a nonsharp separation.
For the hydrocarbon recovery process of Figure 1.9, the By a total permeate material balance,
values of SP in Table 1.8 are computed from the data in ng' = 21 - 10.5 = 10.5 knlolih
Tables 1.5 or 1.6 for the main split in each of the three sepa-
rators. The separation factor in Column Cl is relatively By an overall N2 material balance,
small because the split for the heavy key, iC5H12, is not ng' = 79 - 10.5 = 68.5 kmol/h
sharp. The largest separation factor occurs in column C2,
where the separation is relatively easy because of the fairly Case 3: With two purities given, write two simultaneous material-
large volatility difference between the two keys. Much more balance equations, one for each component, in terms of the total re-
tentate and total permeate.
difficult is the butane-isomer split in Column C3, where only
a moderately sharp split is achieved. For nitrogen, with a fractional purity of 1.00 - 0.50 = 0.50 in the
When applying the conservation of mass principle to sep- permeate,
aration operations using (1 -I), component specifications in n~, = 0.85n(~) + 0.50n(') = 79 kmolih (1)
terms of component recoveries are easily applied, while
those in terms of split ratios and, particularly, purities are For oxygen, with a fractional purity of 1.00 - 0.85 = 0.15 in the
retentate,
more difficult, as shown in the following example.
no2 = 0.50n(') + 0.15n(~) = 21 kmolih (2)
Solving (1) and (2) simultaneously for the total products gives
EXAMPLE 1.1
n(') = 17.1 kmolih n(R) = 82.9 kmolh
A feed, F, of 100 kmolh of air containing 21 mol% O2 and 79
mol% N2 is to be partially separated by a membrane unit according Therefore, the component flow rates are
to each of the following four sets of specifications. For each case,
compute the amounts, in kmollh, and compositions, in mol%, of the ng' = 0.85(82.9) = 70.5 kmolh
two products (retentate, R, and permeate, P). The membrane is
ng) = 82.9 - 70.5 = 12.4 krnolh
more permeable to the oxygen.
ng) = 0.50(17.1) = 8.6 kmolih
Case 1: 50% recovery of 02 to the permeate and 87.5% recovery
of N2 to the retentate. ng' = 17.1 - 8.6 = 8.5 kmolih
Case 2: 50% recovery of 02 to the permeate and 50 mol% purity
Case 4: First compute the O2 flow rates using the split ratio and an
of 02 in the permeate.
overall 02 material balance,
Case 3: 85 mol% purity of N2 in the retentate and 50 mol% purity
of 02 in the permeate.
Case 4: 85 mol% purity of N2 in the retentate and a split ratio of O2
in the permeate to the retentate equal to 1.1.
Solving these two equations simultaneously gives
ng)=lOkmolh ng)=2l-lO=llkmovh
SOLUTION
Since the retentate contains 85 mol% N2 and, therefore, 15 mol%
The feed is 02, the flow rates for the N2 are
JF)
02 = 0.21(100) = 21 kmolh
ng' = 0.79(100) = 79 krnolh