D
OWNSTREAM
P
ROCESSING
Problem Set #5
Problem 1
Consider the following diagram of a countercurrent extraction process in which a solute is extracted
from a heavy phase with flow rate H and concentration X and into a light phase with flow rate L
and solute concentration Y:
a)
Starting with a material balance, derive an equation showing that the recovery (R) of the solute
in the light phase can be expressed as follows:
32
1
R1
EEE
=−
1
+
++
where
d
YL
L
EK
XH
H
==
and E is the extraction factor and K
d
is the distribution coefficient.
Note the light phase initially contains no solute.
(
Hint: Begin your derivation around stage 3
)
3
1
2
Y
2
Y
1
Y
0
Y
1
L
H
X
3
X
1
X
2
X
0
1
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The sum of the geometric sequence 1 + X + X
2
+ X
3
+ X
4
+ ….
. + X
n
can be restated as follows:
n1
1X
S
+
−
=
−
Use this formula to show that a general solution to part a for the recovery of a solute from the
heavy phase (H) using extraction with a particular number of stages (n) is as follows:
EE
R
1E
+
+
−
=
−
c)
Plot recovery as a function of H/L ratio for the case in which K
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 Spring '03
 RogerD.Kamm
 Chemical Engineering, pH, Solubility, 0.2 g, light phase

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