Effect of pressure on distillation

M. J. McCready, 4/21/21

Fenske equation

For a simple situation, the Fenske equation gives the minimum number of trays  (even Wikipedia has this...) as a function of the relative volatility.  The distillate mole fraction is xd and the bottoms mole fraction is xb.

fenske = Log[ ( xd/(1-xd) (1-xb)/xb)]/Log[α]

Plot[fenske/.{xd->.99,xb->.01},{α,1.1,4}, AxesLabel->{" α -- relative volatility","minimum trays"}]

Relative volatility, α: toluene, hexane

T is in C.

Psat1[T_]:=(1/760)*10^(6.95087-(1342.31/(T+219.187)));(*toluene*)
Psat2[T_]:=(1/760)*10^(6.91058-(1189.64/(T+226.28))); (*n hexane*)

Raoult’s law

For Raoult’s law, the relative volatility is just the ratio of vapor pressures.   Note that the total pressure divides out, but as P increases, T must increase and stay above the bubble point.

Plot[Psat2[T]/Psat1[T],{T,25,250}, AxesLabel->{"T - Celsius","α"}]

We see that as T increases, the relative volatility decreases significantly.   Does this occur all of the time... ???  Yes!

Alpha as a function of T.

Try this as Log[Psat2]/Log[Psat1]

lnalfa = (A2 - B2/( T + C2))/(A1 - B1/( T + C1))

Since we have a C1 and C2, we should be able to expand around T = 0.

Series[lnalfa,{T,0,1}]

ExpandAll[Normal[%]]

With only some loss of generality....

%/.{C1->C2, A1->A2}

FullSimplify[%]

Note from the numbers above, the reason that Psat 2 is larger, is that B2 subtracted less than B1 from the “A” constant.   So as T is increased, the value of the ratio gets less.

Apart[%]

This expansion is consistent with the general trend that the higher the heat of vaporization, the higher the boiling point. For our range of comparison, α>1 would be a situation where the vapor pressure of the more volatile component is in the numerator.  Hence the heat of vaporization for this component, is expected to be smaller  -- meaning “B” for this component is smaller.

The result shown above is general.  As T increases, the relative volatility is expected to decrease.  Hence as P increases, the same is true and so distillations will be harder and take more energy as at higher pressures.

Back to Fenske

We will do a bubble point calculation below, but the Fenske equation tells us that the minimum number of trays will vary as

Plot[fenske/.{α-> Psat2[T]/Psat1[T],xd->.99, xb->.01},{T,25,250}, AxesLabel->{"T-- Celsius","minimum trays"}]

Since the actual trays will be some factor, we see that the number of trays would be expected to double over this temperature range.

Bubble point, Raoult’s Law

We will pick a total pressure and find the bubble point T.  Use a x = 0.5 feed.

bpeq= ptotal == .5 Psat1[T] + .5 Psat2[T]

Plot[T/.FindRoot[bpeq,{T,25}],{ptotal,1,10}, AxesLabel->{"P (ATM)","Bubble point T"}]

Or check the boiling point of toluene

bp1 = ptotal == Psat1[T]

Plot[T/.FindRoot[bp1,{T,25}],{ptotal,1,10}, AxesLabel->{"P (ATM)","Toulene Boiling point T"}]

Now back to distillation scaling

We know Raoult’s law is a poor approximations as P increases past a couple of ATM.  However, save fixing it for another day...

Temperature range to check...

Plot[T/.FindRoot[bp1,{T,25}],{ptotal,1,25}, AxesLabel->{"P (ATM)","Toluene Boiling point T"}]

Minimum Trays... (Fenske eq.)

Plot[fenske/.{α-> Psat2[T]/Psat1[T],xd->.99, xb->.01},{T,110,280}, AxesLabel->{"T-- Celsius","minimum trays"}]

So the minimum trays would vary from ~8 to ~13 at the highest pressure.

Plot[(fenske/.{α-> Psat2[T]/Psat1[T],xd->.99, xb->.01})/.FindRoot[bp1,{T,25}][[1]],{ptotal,1,25}, AxesLabel->{"P (ATM)","minimum trays"}]

Minimum reflux...

The minimum reflux directly reflects energy costs.  How does this change?

The simplest relation for minimum reflux is the “Underwood Equation”.   A reference for this is King’s book:  “Separation Processes”, McGraw Hill, 1971, but there is probably a 2nd edition and the Underwood equation is probably in many texts...

((((Subscript[x, D]) D/((Subscript[x, F])F) - α (1-(Subscript[x, D])) D/((1-(Subscript[x, F]))F)))/(α-1))

We pick a saturated liquid feed for simplicity, a 50 - 50 initial mixture, .99 xd and 0.01 xb

LminslashF = (.99 - α .01)/(α-1)

L is the liquid flow in the upper part of the column, LminslashF is (Subscript[L, min])/F, where F is the (liquid) Feed.

Plot[LminslashF/.α-> Psat2[T]/Psat1[T],{T, 110,280},AxesLabel->{"Temperature, C","L/F - top section"}]

Perhaps the standard “R” reflux ratio, which is “L” from here divided by D, which is the distillate product flow rate, would be easier.  F = 2 D for this idealized case...

minRflux = (.99 - α .01)/(α-1) 2

Plot[minRflux/.α-> Psat2[T]/Psat1[T],{T, 110,280},AxesLabel->{"Temperature, C","minimum reflux ratio"}]

Plot[(minRflux/.α-> Psat2[T]/Psat1[T])/.FindRoot[bp1,{T,25}][[1]],{ptotal,1,25}, AxesLabel->{"P (ATM)","minimum reflux ratio"}]

So the reflux ratio and hence the energy requirements will vary by about a factor of 2.   For commodity chemical processing, even a few % of energy savings is important!