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8/4/2019 KremserEquation (1)
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The Kremser Equation
Tore Haug-Warberg
Dept. of Chemical Engineering
October 1st, 2004
The Kremser Equation represents an analytical solution to a classical sepa-
ration problem ofN ideal equilibrium stages connected with countercurrent
gas and liquid flows:
V1
1
V2
Vn
n
Vn+1
VN
N
VN+1
L0
L1
Ln1
Ln
LN1
LN
It is assumed that one (single) chemical component takes part in the separa-
tion, or, if the system is multicomponent, that all the components are being
separated independently. Moreover, the equilibrium relation is assumed to
be linear (Henrys law). The component balance is written:
Lnxn + VN+1yN+1 = LNxN+ Vn+1yn+1
Constant molar flow assumption:
Ln = Ln+1 = = LN = LConstant
and Vn+1 = Vn+2 = = VN+1 = VConstant
Combine the equations above into a simplified component balance:
L (xn xN) = V(yn+1 yN+1) , n [0,N]
The compositions (xn+1,yn+1) for n [0,N are assumed to represent true
equilibrium pairs whereyn+1 = mxn+1 is a linear equilibrium relation (Henrys
law). The mass balance becomes L (xn xN) = V(mxn+1 yN+1), or on a
slightly modified form (operating line):
xn+1 L
mVxn =
yN+1
m
L
mVxN
A L/mV is the so-called absorption factor (problem dependent). The
final balance equation reads
xn+1 Axn =yN+1
mAxN
Constant c
, n [0,N]
These are the N+ 1 equations that must be solved simultaneously to fix
the concentration profile of the column. The equations constitute a set of
1
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coupled algebraic equations ideally suited for matrix algebra. Let us start
by visualizing the coupling in matrix terms:
AA...A
1 0 0
A 1 0
0 A 1. . .
A 1
x0x1x2...
xN
=
x0c
c...
c
The first (and only) stage in the elimination process is indicated to the left.
The outcome of the elimination is
1 0 0
0 1 00 0 1
. . .
0 1
x0
x1x2...
xN
=
x0
c + Ax0c + A(c + Ax0) = c + cA + A
2x0...
cN
k=1 Ak1
+ANx0
The vector elements on the right hand side are recognized as members of
the geometric seriesNk=1
Ak1 =1 AN
1 A
The solution ofxN can thereby be written
xN = cNk=1
Ak1 + ANx0 = c1 AN
1 A+ ANx0
This equation constitutes The Solution to the original problem, and the
remaining part of the note shows simply(?) how to tweak it into something
recognizable from Geankoplis. Substitutes the definition ofc yN+1m
AxN
xN =
yN+1
m AxN
1 AN
1 A+ ANx0
and isolates an expression for xN (the right hand side is now in terms of
column input parameters only)
xN =
yN+1m
1AN1A
+ANx0
1 + A
1AN
1A
Multiplication on both sides by 1A
1Agives
xN = xN 1 A
1 A1
=
yN+1m
1 AN
+ ANx0 (1 A)
(1 A) + A (1 AN)
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Add and subtract the term1 AN
x0 in the nominator
xN =
yN+1m
1 AN
+ ANx0 (1 A) +
0
1 AN
x0
1 AN
x0
1 AN+1
=
yN+1m
x0
1 AN+ x0
1 AN+1
1 AN+1
=
1 AN
1 AN+1
yN+1
m x0
+ x0
Collect all the concentration variables on the left hand side and the absorp-
tion parameters to the right
x0 xN
x0 yN+1m
=1 AN
1 AN+1=
1 AN
1 AN+1A(N+1)
A(N+1)1
=
1A
N+1
1A
1A
N+1 1
Eq. 10.3-21
=S N+1 S
S N+1 1
Here, S A1 mV/L is the so-called stripping factor. Rather than
asking for xN we can ask for y1. The same procedure repeated leads to:
yN+1 y1
yN+1 mx0= A
N+
1 AAN+1 1
Eq. 10.3-24
The equations above express the concentration differences in a column of
fixed size (number of trays N). We may occasionally want to invert the
relationship and calculate Ngiven a separation specification :
x0 xN
x0 yN+1m
=1 AN
1 AN+1=
1 AN
1 A AN
Rewrites the last equation into 1 AN
= AAN
which yields
AN =1
1 A= N=
ln
11A
lnA