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some concept of Novel Separation Processes are Asymmetric Membrane, Centrifugal Separation Processes, Cloud Point Extraction, Colloidal Particles, Common Stationary Phase.Main points of this lecture are: Gas Separation, High Diffusivity, Concentration Gradient, Membrane Surface, Membrane Resistance, Gas Phase, Direction Normal, Stream Flows Parallel, Permeate, Cross Flow
Typology: Study notes
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Novel Separation Processes
Separation processes, membranes, electric field assisted separation, liquid
membrane, cloud point extraction, electrophoretic separation, supercritical fluid
extraction
Novel Separation Processes
Gas Separation
In case of gas separation by membranes, high pressure feed gas is supplied to one
side of the membrane and permeate comes out normal to the membrane to the low
pressure side. Due of high diffusivity in gases, concentration gradient in the gas phase
normal to the membrane surface is small. So, gas film resistance is neglected compared to
membrane resistance. This means concentration in gas phase in a direction normal to
membrane is uniform whether gas stream flows parallel to the surface or not.
There are various types of gas separation processes depending upon the flow
characterizations. Since the permeate comes normal to the flow direction of the feed, this
is known as simple cross flow (Fig. 5.1a). If there is complete mixing of the feed and
permeate by an external agent (stirrer or mixer), then the configuration is complete
mixing (Fig. 5.1b). If feed and permeate are in the same direction, then the flow is
cocurrent flow (Fig. 5.1c). If they are in opposite direction, then it is counter current flow
(Fig. 5.1d).
Novel Separation Processes
In the following section, the working principles and calculations involved in complete
n.
p in feed and permeate side.
In ;
mixing mode are considered. This case is like a continuous stirred tank reactor (CSTR).
The assumptions involved are:
(i) Isothermal conditio
(ii) Negligible pressure dro
(iii) Permeability of each component is constant.
qp = θqf
Fig. 5.2: Schematic of a complete mixing configuration with the process condition
Pl, y (^) p Low pressure
High pressure side Ph , x (^0) q 0 =( 1 - θ)qf Feed q (^) f, xf
s
the above figure,
3 q (^) f is total feed flow rate (in m /s);
3 q 0 (^) is outlet reject flow (m /s)
3 q (^) p is outlet permeate flow (m /s) θ is fraction of feed permeate
p
f
q
q
Overall material balance yields the following relation.
q (^) f = q 0 + qp (5.1)
Rate of diffusion/ permeation of species A (in a binary mixture of A and B) is given as,
( )
' qA q y p p PA P x P y
h 0 l p Am Am t
Novel Separation Processes
where,
' P A is permeability of A in membrane
3
2
cm cm
s cm cmHg
⎟ ;^ q^ A is the flow rate of A in
permeate; is the membrane area; is the membrane thickness; is feed side total
pressure ( cm.Hg );
A m t Ph
x 0 is mole fraction of A in reject; x (^) f is mole fraction of A in feed;
y (^) p is mole fraction of A in permeate; is partial pressure of A in reject gas phase.
Rate of permeation of species B is given as,
P xh 0
( ) ( ) ( )
'
0
B p^ p B h l m m
q q^ y P P x P y A A t
⎣ p^
Where, is permeability of B. Dividing Eq.(5.2) by (5.3), the following expression is
obtained.
' P B
( ) ( )
0
0
l p p h
p (^) l p h
x y y P
y (^) P x y P
Where,
'
'
A
B
An overall balance of component A results into the following equation.
q xf f = q x 0 0 + q yp p (5.5)
Rearrangement of above equation results,
0 0 p^ p f f f
q x q y x q q
Defining, ;
p
f
q
q
0 1 f
q
q
Novel Separation Processes
Case 2: , , ,
l f h
x P
∗ are given and y (^) p , x 0 , A m to be calculated
If all the feed is permeated, then θ =1 and feed composition xf = yp
For all values of θ < 1, yp > xf
Substitute, xf = yp in Eq. (5.3).
x 0 (^) m = Minimum rejection component for a given xf
( ) ( )
( )
l f f h
f f
x x P
x x
∗
∗
So, a feed component xf cannot be stripped lower than x0m even with an infinitely large
membrane area for a completely mixed system. To do this cascade may be used.
qf, xf
Feed in high pressure (^) dA m
q, x (^) q – dq
x -dx
Permeat
qp = θqf
Ph
Pl y, dq
Reject x 0 , q 0 = (1-θ)qf
Plug Flow
Low pressure
Fig. 5.3: Schematic of a cross flow model
Novel Separation Processes
Longitudinal velocity in high pressure or reject stream is high. So that gas is in plug flow
and flows parallel to membrane. Low pressure side, permeate stream is almost pulled into
vacuum. So, flow is essentially perpendicular to membrane. No mixing is assumed. So
that composition varies as length. Over a different membrane area dAm at any point, local
permeation rates are presented below.
Component A balance:
[ ]
' A h l
ydq P x P y dAm t
Component B balance:
( ) ( ) ( )
' 1 1 1
B h l
y dq P x P y dAm t
dq = total flow rate perpendicular to dAm. Dividing Eq.(5.12) by (5.13),
( ) ( )
l
h
l
h
x y y P
y (^) P x y P
Permeate composition y as a function of reject composition x at a point along the length.
Analytical solution:
The design equation is presented below:
( )(^ )
( )
R S (^) T f (^) f f
f
x u u F u F D
x E^ u^ F^ u^ F u D
∗ (^) ∗
∗
Where, 1
f
q
q
θ
∗ = − , ; 1
x i x
2 2 2 u = − Di + D i + 2 Ei + F ;
Novel Separation Processes
q
’ , y yi^ = 0
q, x q 0 , x 0
Am (^) Am = 0
Fig. 5.5: Schematic of balance over a small element
The schematic of .4 and the small
= Total material out
Total A out
In the above differential volum
the counter current flow model is presented in Fig. 6
element is shown in Fig. 5.5.
Overall material balance:
Total material in
' q = q 0 + q
Overall A balance:
Total A in =
' qx = q x 0 0 + q y
Species A balance over a differential element,
' d qx ( ) = d q y ( )
q, x
(q - dq)
(x - dx)
dq, y
dAm
e, species A balance provides,
qx = (^) ( q − dq (^) )( x − dx (^) )+ ydq
Novel Separation Processes
ydq = d (^) ( qx ) (5.20)
Local flux of A across the membrane is presented,
[ ]
' A h l m
− ydq = P x − P y dA t
For species B, the following balance equ ation is provided:
( ) ( ) ( )
' 1 1 1
B h l m
− − y dq = ⎡⎣ P − x − P − y ⎤⎦ dA t
Combining Eqs. (5.21) and (5.22), the following e xpression is obtained.
( ) ( )
'
' 1 1 1
x P^ l y
A^ h
B (^) l
h
y P P
y P (^) P x y P
Eliminate by using equations (5.17) and (5.18),
' q
qx = q x 0 0 (^) + (^) ( q − q 0 (^) ) y
The above equation can be rearranged as
( )
( )
0 0
x y q q x − y
Bu using this equation
substitute q in equation (5.21) then we get,
( )
( ) [ ]
x 0 y d q 0 ' A h l m
x y (^) P y P x P y dA t
[ ]
0 '
0
A h l m
x y d x y (^) P yq P dA t
− = x − P y
By derivating this equation and by rearranging it finally we get it as,
Novel Separation Processes
f^ (^1 )^0 p
x x y
iven θ , then
y (^) p from equation (5.32)
ary differential equations.
Solved Problems
For a value of g
(i) Guess x 0
(ii) Solve for
(iii) Check value of yp for solving ordin
(iv) Iterate.
A membrane is used t
4 3 q (^) f = 10 cm ( STP ) / s and feed composition of A, xf =0.5 ; The desired composition of
the reject is x 0 = 0.25. The membrane thickness, t = 3*
80 cm Hg and Pl = permeate side pressure = 20 cm Hg. The permeabilities are,
3 ' 10 2
cm STP cm p
− = × and
' 10 pB 6 10 s cm cmHg..
− = × of above units. Assuming complete
mixing model, calculate permeate co ncentration, y
p , fraction permeated^ θ^ and membrane
area ( Am ) required?
x (^) f , x 0 ,
2 4
2
p
b b ac y a
From Eq.(), =
Where,
( )
0 0
; h^ 1 1 h l l
b x x P P
Novel Separation Processes
0
h
l
c x P
' 10
' 10
A
B
p
p
−
−
( )
(^1 0 )
h h
l l
b x x P P
( )
( )
0
h
l
c x P
2 4
2
p
b b ac y a
0 1
x (^) f yp x
( )( ) ( )
4
' 10 0 3
f p m A h l p
q y A p t P x P y
−
−
8 2 Am = 2.7 × 10 cm
cm thickness with oxygen permeability
3 ' 10 (^300 10 )
..
p (^) A s cm cmHg
− = × and
permeability ratio of oxygen to nitrogen. F P ) / s a tion
cm ( STP cm ).
eed rate, q (^) f = × cm ST
6 3 2 10 ( nd frac
Novel Separation Processes
. C. J. Geankoplis, Transport Processes and Unit Operations, Prentice Hall of India,
New Delhi,1997.
