8/4/2019 KremserEquation (1)

1/3

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

8/4/2019 KremserEquation (1)

2/3

2

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)

8/4/2019 KremserEquation (1)

3/3

3

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