Cascades

1
Cascades
Last lecture We discussed leaching techniques
and an algebraic and graphical approach to
solving leaching problems.
This lecture will focus on extending our analysis
of separations techniques to include use of
various configurations of separators.
Thus far we have only discussed separations which
involve using a single unit. However, combining
several separations units together into a cascade
is often desirable.
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Cascades

• Cascades are sequences of separators.
• Sequences of separators can be used to achieve
  higher purities or to recover more solute from a
  product stream.
• Cascades can be configured in various ways.

Three single section cascade configurations are
S
P1
Countercurrent
F
P2
S
Crosscurrent
F
P1
P2
P3
P4
S
F
P6
2-D diamond
P5
P4
P1
P2
P3
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Cascades
Multiple section cascades can achieve high
product purities. A typical example of a two
section cascade is simple distillation.
S1
P1
Example Two Section Cascade

• Two Sections consist of a
• stripping section below the feed and an
• adsorbtion section above the feed

F
S1
P1
P2
S2
One way to think of this two section cascade is
as two countercurrent cascades with the feed
split into two feed streams (one for each
section). The solvent S1 is a liquid which
absorbs less volatile species from the upward
passing vapor stream. The solvent stream S2 is a
vapor which strips the counter-flowing liquid of
volatile species.
F
S2
P2
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Countercurrent Solid-Liquid Cascade Leaching
Xn the underflow liquid concentration of B
(solute-free basis) Yn the overflow liquid
concentration of B (solute-free basis)

• Solid feed consists of carrier A (insoluble) and
  solute B (soluble)
• Pure solvent C is supplied to the Nth stage and
  can completely dissolve B, but not A
• All of B is dissolved in first stage, but is
  transferred to the overflow in subsequent stages.
• The underflow consists of solid A with a
  constant flow rate FA and liquid (B and C).
• The liquid in the underflow slurry has the same
  composition as the liquid in the overflow.
• The ratio of solvent mass (in the underflow) to
  insoluble mass is R and is constant.

R MC/MA (in the underflow)
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Countercurrent Solid-Liquid Cascade Leaching
Xn the underflow liquid concentration of B
(solute-free basis) Yn the overflow liquid
concentration of B (solute-free basis)

• The flow rate of solvent in the underflow out of
  stage 1 is
• RFA (since slurry completely flows in the
  underflow carrying all of A)
• The flow rate of solvent in the overflow out of
  stages 2 to N is S.
• The flow rate of solvent in the overflow out of
  stage 1 is
• S RFA (since RFA is transferred to the slurry)

Solid Feed Insoluble A (FA) Soluble B (FB)
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Countercurrent Leaching Material Balances
Xn the underflow liquid concentration of B
(solute-free basis) Yn the overflow liquid
concentration of B (solute-free basis) R MC/MA
(in the underflow)
Yn
Yn1
A material balance of B on any interior stage
gives
n
Xn-1
Xn
B entering stage n
B leaving stage n
A material balance of B on stage 1 gives
P1
Y2
Y1
1
B entering stage 1
B leaving stage 1
F
X1
A material balance of B on stage N gives
YN
S
B entering stage N
B leaving stage N
N
XN-1
XN
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Countercurrent Leaching Equilibrium and Washing
Factor
Since the liquid trapped in the slurry and the
liquid in the overflow have the same compositions
Equilibrium condition
Defining the washing factor to be the ratio of
the solvent flow rate S to the solvent flow rate
in the underflow RFA
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Countercurrent Leaching Algebra
1) Use the equilibrium condition XnYn to
eliminate Y from the mass balances. 2) Rearrange
and substitute in the washing factor W. 3) Use
Gaussian elimination to solve set of N linear
equations (shown below) which result from 2. The
N linear equations are
For n2 to N-1
The solution is
Back substitution into the above equations
results in
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Configuration Efficiencies
Cocurrent
Crosscurrent
Countercurrent
FA,XB(F)
S
FA,XB(F)
FA,XB(F)
Extract
YB(1)
Extract 1
1
S/2
1
1
YB(1)
XB(1)
YB(1)
XB(1)
XB(1)
YB(2)
2
2
2
S/2
Extract 2
XB(2) XB(R)
YB(2)
XB(2) XB(R)
XB(2) XB(R)
YB(2)
S
Raffinate
Extract
Raffinate
Raffinate
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Configuration Efficiencies Cocurrent Stage 1
For a feed of carrier A and solute B (A
insoluable in C) the separation in the first
stage is found by combining a mass balance of the
solute
With the equilibrium condition between the
raffinate and extract streams
To obtain
Substituting in the extraction factor
We obtain
Using the equilibrium condition we find
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Configuration Efficiencies Cocurrent Stage 2 to N
The second stage mass balance of the solute B
Combined with the equilibrium condition between
the exiting streams
Results in
Adding the second stage (and consequently any
number of additional stages) in a cocurrent
configuration leads to No additional separation!
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Configuration Efficiencies Crosscurrent
The only difference between the separation that
occurs in the first stage in the crosscurrent
configuration and the separation that occurs in
the 1st stage of the cocurrent separator comes
from using 1/N times as much solvent so
The product of the N ratios is
XB(2)
For any intermediate stage we can just multiply
the first n ratios to find
The equilibrium condition between the extract and
raffinate streams gives
So even with an infinite number of stages there
is a finite amount of B left in the raffinate.
With an infinite number of stages
Note that with infinite stages you only
introduce an infinitesimal amount of solvent at
each stage.
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Configuration Efficiencies Countercurrent
Extract
FA,XB(F)
The countercurrent cascade has two streams
flowing in opposite directions. This means that
the intermediate concentrations are going to
depend on the number of stages in the cascade
(this was also the case for the crosscurrent
cascade, but downstream coupling was only due to
distributing the solvent over N stages).
YB(1)
1
XB(1)
YB(2)
Material balance around the first stage
2
The equilibrium condition
XB(2) XB(R)
S
Raffinate
Material balance around the second stage
Extract
FA,XB(F)
YB(1)
The equilibrium condition
1
XB(1)
YB(2)
Combining these equations we can obtain
XB(1)
YB(2)
And for N stages
2
XB(2) XB(R)
S
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Cascade Efficiencies
Cocurrent
Crosscurrent
Countercurrent
S
FA,XB(F)
FA,XB(F)
FA,XB(F)
Extract
YB(1)
Extract 1
S/2
1
1
1
YB(1)
XB(1)
XB(1)
YB(2)
XB(1)
YB(1)
2
2
S/2
2
Extract 2
YB(2)
XB(2) XB(R)
XB(2) XB(R)
YB(2)
XB(2) XB(R)
Raffinate
S
Raffinate
Extract
Raffinate
Increased separation, but cannot obtain arbitrary
purity because the solvent is divided into N
stages.
No additional separation. Added stages only
useful for increasing residence time.
If E gt 1 then arbitrary high purity can be
obtained. If Elt1 then 1-E times as much B will
remain in the raffinate.
Countercurrent
Crosscurrent
Extraction
Cocurrent
Number of Stages
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Complex Configurations
Example Various cascade configurations canbe
combined to make a more complex
separationssystem. To analyze such a cascade,
break down the combination into its component
configurationsand use the analysis weve
developed for simpleconfigurations.
YB(4)
YB(5)
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Summary
This lecture We learned how multiple stages
could be combined into a cascade to achieve
higher degrees of separation. We described
several different sorts of cascade
configurations.  We learned how to analyze
cascades using combinations of equilibrium and
mass balance relationships.  We analyzed the
separation efficiency of several cascade
configurations.
Next lecture well describe group methods for
solving cascade separations problems.