Absorption design with real stages
Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik
1. Murphree Efficiency
Up to now, we always treated a case where the equilibrium is reached in every stage. However, this
assumption is often not correct. Therefore we need to use an efficiency to describe the non-ideal behaviour.
The Murphree efficiency for the gas phase can be defined for every stage i:
MG
Ei

y i 1  y i
y i 1  y i
*
The Murphree efficiency compares the actually absorbed amount in stage i (yi+1 – yi) to the maximum
possible amount absorbed in stage i if the equilibrium was reached (yi+1 – yi*). In the equation, the M in
EiMG stands for Murphree, the G stands for gas phase.
We assume, that the Efficiency is constant during the whole column and doesn’t depend on
the stage i. Therefore we get:
E  const  E i
MG

y i 1  y i
y i 1  y i
*
For the operating line we need the mass balance for the column from stage 1 to stage i:
y i 1 
L
G
xi  ( y1 
L
G
x0 )
The equation for the equilibrium is our third equation to solve the problem. Please note, that
the equation is for yi* and not for yi.
y i  m xi
*
First we draw the equilibrium line and the operating
line into the diagram:
Now we need to draw a third line using the Murphree
efficiency. This line is drawn between equilibrium and
operating line at a fractional vertical distance from the  y i 1  y i 
equilibrium line equal to the Murphree gas efficiency.
y
Thus the value of E is is the ratio of the length of
lines, AB/AC.
E 
y i 1  y i
y i 1  y
*
i

Op. line
y
 yi
*
i 1

A
Eq. line
B
C
AB
AC
x
2. The stage-by-stage construction
Using this line describing the stage compositions of the column, we can apply the usual stage-bystage construction, switching from operating line to the stage composition line.
Since the stage composition line represents the real effluent compositions from the trays, it is
used instead of the equilibrium line to complete the tray construction.
y
Op. line
yn+1
y4
Eq. line
L /G
y=mx
y3
y2
y1
xo x1 x2
x3
x4
x
3. Mathematical solution
Now we want to calculate the number of stages instead of a graphical solution. That’s why we take
the three equations seen before:
y i 1 
L
G
xi  ( y1 
L
G
1 
x0 )
2 
y i  m xi
*
E 
y i 1  y i
y i 1  y
 y i  y i 1 ( 1  E )  Ey i
*
3 
*
i
If we combine the second and the third equation, we get:
E m x i  y i  y i 1 ( 1  E )
Multiplying the first equation with E/A and knowing that A = L / (mG), we get:
E
A
y i 1  E m x i 
E
A
y1  E m x0
Combining the last two equations:
E
A
y i 1  y i  ( 1  E ) y i 1 
E
A
y 1  Ey 0
*
3 a 
Solving for yi+1:
A

y i 1  
 E  A  EA
E



*
y i  
 y 1  Ay 0

 E  A  EA 

A'
B'
We can introduce two new variables A’ and B’:
E


B'  

 E  A  EA 
A


A'  

 E  A  EA 
We therefore get the following equation:
y i  1  A' y i  B ' y 1  Ay 0 
*
If we set the Efficiency E = 1, we get for A’ and B’:
A'  A
B'  1

As a next step, we would like to calculate the number of
real stages. This calculation needs some mathematical
effort. At first, we take the result from the last slide:
However, we need an equation for i instead of i+1:
This equation is a First Order Difference
Equation. For a general equation...
... the solution has the following form:
If we look into our equation we can
identify the coefficients as:
y i  1  A' y i  B ' y 1  Ay 0 
*
y i  A' y i 1  B ' y 1  Ay
*
0

y i   y i 1  
yi  y0 
i

1
i
1 
  B ' y 1  Ay 0 
  A'
*

And using those coefficients in the general
solution:
y i  y 0  A'   B ' y 1  Ay
Furthermore we need equation (3a) for i=1:
y 0  y 1 ( 1  E )  Ey
i
*
0
*
0
 1   A' 
1  A'
i
Combining the last two equations, we get:

y i   y 1 1  E

The fractional absorption α can now be
calculated. After some mathematical
transformations, we finally get:
 

Ey
y n 1  y 1
y n 1  y
*
0
*
0

B'
1  A'
 A' n '

( A' )
n'
y 1 
1

1
A
The number of real stages n’ is then:
n' 
1 / A 
log 

log A'
 1

The Kremser equation for the ideal case
where equilibrium is reached on every stage:
n 
1 / A 
log 

log A
 1

Dividing the second equation by the first
one, we get the overall efficiency E0:
E
It is important to note that the overall
efficiency E0 and the Murphree Efficiency E
are not the same.
0
1
1

n
n'

log A'
log A
Ay
*
0
  A' 

i

B'
1  A'
y
1
 Ay
*
0
