Distillation

To reduce load times, this material is divided into seven files, corresponding to the numbered points below. The present file (distill2.html) contains point 2 only.
  1. Distillation Principles
  2. Distillation Modeling
  3. Distillation Operating Equations
  4. Distillation Calculations
  5. Distillation Enthalpy Balances
  6. Enthalpy-Concentration Method
  7. Equipment & Column Sizing

Distillation II: Modeling

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When modeling a distillation column, one can draw balances on the entire system, the column base, the accumulator, each tray (or a group of trays). Mass and component balances are always required. In special cases, the energy balances can be neglected.

The overall material balance on a distillation column with NT trays is:

'deriv(M;B+M;D+'Sum(M;n,i,1,NT),t)=F-D-B
where F,D, and B are mass flow rates, MB is the mass in the column base, MD the mass in the accumulator, and Mn the mass on tray n. Often, the mass of the vapor in the column is much smaller than the mass of the liquid in the column; in such cases, it is often a reasonable approximation to write the mass accumulation terms for the liquid phase only.

The overall component balance is:

'deriv(M;B*x;B+M;D*x;D+'Sum(M;n*x;n,i,1,NT),t)=F*x;F-D*x;D-B*x;B
This can also be written with accumulation terms representing the liquid alone, if the appropriate assumptions are made.

The balances required for the accumulator are:

'deriv(M;D,t)=V;1-L-D
where V1 is the vapor flow off the top tray to the condenser, and
'deriv(M;D*x;D,t)=V;1*y;1-L*x;D-D*x;D
for a total condenser or
'deriv(M;D*x;D,t)=V;1*y;1-L*x;D-D*y;D
for a partial condenser with a vapor product.

The base and reboiler balances are

'deriv(M;B,t)=L;NT-V;B-B
for a partial reboiler with boilup rate VB (LNT is the liquid flow off the bottom tray of the column), and
'deriv(M;B*x;B,t)=L;NT*x;NT-V;B*y;B-B*x;B
Don't forget that xB and yB are in equilibrium with each other.

For each tray, we can write:

deriv(M;n,t)=L;(n-1)+V;(n+1)-L;n-V;n+F;n-P;n
'deriv(M;n*x;n,t)=L;(n-1)*x;(n-1)+V;(n+1)*y;(n+1)-L;n*x;n-V;n*y;n+F;n*x;Fn-P;n*x;n
in a general form that allows one feed (Fn) and one liquid product (Pn) on each tray. Usually, only one tray will have a feed stream, and there will be only a few, if any sidedraw products. You can also have vapor products, which will require replacing Pnxn in the component balance with Pnyn.

Steady-State Model

The column mass and component balances in their steady state forms are:

0=F-D-B
0=F*x;F-D*x;D-B*x;B
These are equivalent to Equations 21.3 and 21.4 in McCabe, Smith, & Harriott.

These two equations can be solved simultaneously to eliminate any one variable. For example:

0=F*x;F-D*x;D-(F-D)*x;B
0=F*(x;F-x;B)-D*(x;D-x;B)
'quot(D,F)='quot(x;F-x;B,x;D-x;B)
Similar rearrangements yield:
'quot(B,F)='quot(x;D-x;F,x;D-x;B)
and
'quot(D,B)='quot(x;B-x;F,x;F-x;B)='quot(x;F-x;B,x;D-x;F)
Some authors like to use these ratios in their calculations. I usually find it makes more sense to work directly from the balance equations.

Feed Tray

It also may be useful to examine the steady state balances on a feed tray. The total material balance on feed tray n is

0=F+L;(n-1)+V;(n+1)-L;n-V;n
To capture all necessary detail, one must recall that the feed can be both vapor and liquid. As described by the feed variable q, the fraction liquid is Fq, so the vapor rate above the feed tray and the liquid rate below the feed tray will change. The new values will be
V;n=V;(n+1)+(1-q)*F
L;n=L;(n-1)+q*F
An equivalent analysis can be made for draw trays. It will be necessary to specify whether the draw is made from the liquid space or vapor space on the tray. This defines a q value for the draw. The equations will then be essentially the same, but keep in mind that the effect of a draw is to reduce the traffic in the column while a feed increases traffic.


R.M. Price
Original: 7 February 1997
Updated: 19 February 1997, 22 Jan 1998, 14 Feb 2003

Copyright 1997, 1998, 2003 by R.M. Price -- All Rights Reserved