SEPARATION POWER
 The degree of separation which can be obtained with any
particular separation process is indicated by “ the
separation power or separation factor”.
 The objective of a separation device is to produce
products of differing composition. So it is logical to define
the separation factor in terms of product composition.
ChE 334: Separation Processes
Dr Saad Al-Shahrani
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Product P1
Xi,1, Xj,1
Feed
Components
(i, j)
(e.g L-L Extraction)
Separation
device
(liq.)
Product P2
Xi,2, Xj,2
(liq.)
Ci1 / Ci2
SPi , j  1 2
Cj /Cj
Where C= Concentration (mole fraction, mass fraction or mole/ vol., mass/ vol.)
SPi , j 
mole fraction of i in P1 / mole fraction of j in P1
mole fraction of i in P2 / mole fraction of j in P2
ChE 334: Separation Processes
Dr Saad Al-Shahrani
COMPONENT RECOVERIES
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SPi , j 
x i,1/x j,1
x i,2 /x j,2

Ratio of mole fraction in P1
Ratio of mole fraction in P2
P1
yi,, yj,
Separation
Feed
device
Components
i,j
(e.g Distillation or evaporation)
(vap.)
P2
Xi,, Xj,
(liq.)
SPi , j 

Ratio of mole fraction in P1
Ratio of mole fraction in P2
(y i / y j ) P1
(x i / x j ) P2
ChE 334: Separation Processes
xj
yi
 ( ) P1 * ( ) P2
yj
xi
Dr Saad Al-Shahrani
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
If
SPi , j >1: component i in product (1) more than component j and
component j in product (2) more than component i.

If SPi , j <1: component j in product (1) more than component i and
component i in product (2) more than component j.

If SPi , j =1: No separation takes place between i and j.
ChE 334: Separation Processes
Dr Saad Al-Shahrani
COMPONENT RECOVERIES
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SFi / SF j
SRi
Ratio of mole fraction in P1
SPi , j 


SR j (1  SFi )(1  SF j ) Ratio of mole fraction in P2
Where SFi, SFj = split fraction of component i and j respectively.
ChE 334: Separation Processes
Dr Saad Al-Shahrani
COMPONENT RECOVERIES
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Example
Referring to example 1.
Key component split
Column (separator)
nC4H10/ iC5H12
C1
C3H8/ iC4H10
C2
iC4H10/ nC4H10
C3
ChE 334: Separation Processes
Separation factor, SPi,j
226.6 / 0.7
 137
28.1 / 11.9
54.8 / 2.2
SP 
 7103
0.6 / 171.1
SP 
SP  377.6
Dr Saad Al-Shahrani
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Example: A feed, F, of 100 kmol/h of air containing 21 mol% O2 and 79 mol%
N2 is to be partially separated by a membrane unit according to each of
the following four sets of specifications. For each case, compute the
amounts, in kmol/h, and compositions, in mol%, of the two products
(retentate, R, and permeate, P). The membrane is more permeable to
the oxygen.
Case 1: 50% recovery of O2 to the permeate and 87.5% recovery of N2
to the retentate.
Case 2: 50% recovery of O2 to the permeate and 50 mol% purity of O2 in
the permeate.
Case 3: 85 mol% purity of N2 in the retentate and 50 mol% purity of O2
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.
ChE 334: Separation Processes
Dr Saad Al-Shahrani
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Solution
Th feed is:
50% recovery of O2 to the permeate and
87.5% recovery of N2 to the retentate
F
Membrane
Case 1: This is the simplest case to
calculate because two recoveries
are given:
?
?
O2
N2
Permeate (P)
fast
slow
Retentate (R)
ChE 334: Separation Processes
Dr Saad Al-Shahrani
?
?
COMPONENT RECOVERIES
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Case 2: With the recovery for O2 given, calculate its distribution into the two products:
50% recovery of O2 to the permeate and
50 mol% purity of O2 in the permeate
Using the purity of O2 in the permeate,
the total permeate is:
?
?
Permeate (P)
Membrane unit
By a total permeate material balance,
F
Retentate (R)
By an overall N2 material balance,
ChE 334: Separation Processes
Dr Saad Al-Shahrani
?
?
COMPONENT RECOVERIES
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Case 3: With two purities given, write two simultaneous material balance equations,
one for each component, in terms of the total retentate and total permeate.
For nitrogen, with a fractional purity of 1.00 − 0.50 = 0.50 in the permeate,
For oxygen, with a fractional purity of
1.00 − 0.85 = 0.15 in the retentate,
85 mol% purity of N2 in the retentate and 50
mol% purity of O2 in the permeate
F
Membrane unit
Solving (1) and (2) simultaneously for
the total products gives
Permeate (P)
?
?
Retentate (R)
ChE 334: Separation Processes
Dr Saad Al-Shahrani
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?
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Therefore, the component flow rates are
Case 4: First compute the O2 flow
rates using the split ratio and an
overall O2 material balance,
85 mol% purity of N2 in the retentate and a
split ratio of O2 in the permeate to the
retentate equal to 1.1.
F
Membrane unit
Permeate (P)
?
?
Retentate (R)
ChE 334: Separation Processes
Dr Saad Al-Shahrani
?
?
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Solving these two equations simultaneously gives
Since the retentate contains 85 mol% N2 and, therefore, 15 mol% O2, the flow
rates for the N2 are
ChE 334: Separation Processes
Dr Saad Al-Shahrani