For rectification I am given the following information

$x_e$ = 0.95

$x_f$ = 0.50

$x_b$ = 0.15

$v$ = 8

$q$ = 1.17

**VLE Data**    
x       y
0.00    0.00
0.10    0.17
0.20    0.31
0.30    0.44
0.40    0.55
0.50    0.645
0.60    0.73
0.70    0.81
0.80    0.88
0.90    0.94
0.95    0.97
1.00    1.00

The VLE data is derived from a given diagram. There may be some inaccuracies there. However drawing with pen and paper in the diagramm gives me the same error I'll explain later on.

With all this data I developed the McCabe Thiele Diagramm for this problem.

Putting everything together in Excel

enter image description here

The grey dotted line indicates my $x_b$ and in orange I drew the theoretical stages. My Problem is: Usually you would come up with a whole number of stages, in this case I would get $9\frac{5}{7}$.

There is also an $E_k = 0.7$ given.

$$n=\frac{n_{th}}{E_k} = 13.87$$

My question is: Why do I not get whole numbers? I am under the impression that there always have to be whole numbers. However in Absorption we where supposed to give answers like $n = 1.2$.

I thought maybe the given mole fractions took $E_k$ into account and therefore I would get a whole number after dividing $n_{th}$ by 0.7.

Is this just an error in my VLE Data and actually I would end up with n = 14? 13+1 to be accurate. It's hard to use the given diagramm since it's very small and I couldn't draw accurate enough in it. Also on what stage is the feed? I wouldn't get a whole number the either.

  • $\begingroup$ Can you explain what the different x are? Some textbooks/teachers use different names for the same. $\endgroup$
    – mart
    Commented Sep 20, 2015 at 20:14
  • $\begingroup$ also what's the blue line in your diagram? $\endgroup$
    – mart
    Commented Sep 20, 2015 at 20:16
  • $\begingroup$ Please use this diagram for reference. The blue line is the stripping section. x_b is the bottom composition, x_f the feed and x_e the distillate composition. v is the reflux ratio. $\endgroup$
    – idkfa
    Commented Sep 20, 2015 at 21:16
  • $\begingroup$ It's been ages since I learened this and I don't work with columns but I don't recall ever arriving at an even number of stages. $\endgroup$
    – mart
    Commented Sep 21, 2015 at 8:13
  • $\begingroup$ Then I'll take it we only got whole numbers for convenience during the course. Appreciated! $\endgroup$
    – idkfa
    Commented Sep 21, 2015 at 8:15

1 Answer 1


Well a stage is an equilibrium stage ie. given some feed composition z, temperature and pressure, vapour of composition y goes up and liquid composition x goes down.

Whether you are calculating actual or theoretical stages, you will hardly ever get a whole number. If you look at the graph, last stage (where xb = 0.15), the 5/7 stage probably comes from the (Stage N-1 composition - xb)/(Stage N-1 composition - Stage N composition). You can verify this yourself. Looks about right from eyeballing it.

Another way to see it is if your column had 10 theoretical stages, then xb will be around 0.12 (from the graph).

Ek just accounts for the fact that in an actual column the stages will not be at equilibrium. Meaning on each stage n y will be less than what is predicted on the VLE curve.

To conclude, you didn't do anything wrong. If this were an actual design and you come up with 13.87 actual stages required you would then select 14 stages. Then if you really want to meet xb=0.15 you can do that by playing with the bottom or top temperature.

  • $\begingroup$ Thank you! I didn't need to verify since I had all x,y data for each stage calculated in order to plot the diagram. $\endgroup$
    – idkfa
    Commented Sep 24, 2015 at 15:39

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