Chapter 4
Distillation
4.1 Vapor Liquid Equilibrium Relations
The equilibrium in vapor-liquid equilibrium is restricted by the phase rule. For a binary
mixture in two phases, the degree of freedom F is given by
F =m−π+2=2−2+2=2
(4.1-1)
As an example we will use the benzene-toluene, vapor-liquid system. The four variables are
temperature, pressure, the mole fraction of the more volatile component in the vapor and
liquid phases. Benzene is the more volatile component in the benzene-toluene system. If the
pressure is fixed at 200 kPa, only one more variable can be set. If we set the temperature, the
mole fraction of benzene (component 1) can be determined from the following relations:
P sat
P sat
y1
1 − y1
= 1 and
= 2
x1
P
1 − x1
P
(4.1-2a, b)
In these expressions, P = 200 kPa, P1sat and P2sat are the vapor pressures of benzene and
toluene at the specified temperature, respectively. For ideal solution such as benzene-toluene
system, the bubble and dew point temperatures of the mixture will be between the saturation
temperatures of benzene and toluene. At 200 kPa, the saturation temperatures of benzene and
toluene are 377.31 K and 409.33 K, respectively. If we set the mixture temperature at 400 K,
P1sat = 352.160 kPa and P2sat = 157.8406 kPa. Equations (4.1-2a, b) become
y1
352.16
1 − y1 157.841
=
and
=
x1
200
200
1 − x1
(4.1-3a, b)
Solving these two equations, we obtain x1 = 0.2170 and y1 = 0.3820. The vapor-liquid
equilibrium relations for benzene (1)-toluene (2) at a total pressure of 200 kPa are given as a
boiling-point Txy diagram shown in Figure 4.1-1. The upper curve is the saturated vapor
curve (the dew-point curve) and the lower curve is the saturated liquid curve (the bubblepoint curve). The region between these two curve is the two-phase region.
In Figure 4.1-1, if we start with a cold liquid mixture of x1 = 0.217 and heat the mixture, it
will start to boil at 400 K, and the composition of the first vapor in equilibrium is y1 = 0.382.
As we continue boiling, the composition x1 in the remaining liquid mixture will decrease
since y1 is richer in benzene. If we start with a hot gas mixture of y1 = 0.382 and cool the
mixture, it will start to condense at 400 K, and the composition of the first liquid drop in
equilibrium is x1 = 0.217. As we continue condensing, the composition y1 in the remaining
gas mixture will increase since x1 is richer in toluene. The Txy diagram for an ideal mixture
can be calculated from the pure vapor-pressure data as illustrated in Example 4.1-1.
4-1
Figure 4.1-1 Calculated Txy diagram of benzene and toluene at 200 kPa.
Example 4.1-1 ---------------------------------------------------------------------------------Construct a Txy diagram for a mixture of benzene and toluene at 200 kPa. Benzene and
toluene mixtures may be considered as ideal.
Data: Vapor pressure, Psat, data: ln Psat = A − B/(T + C), where Psat is in kPa and T is in K.
Compound
Benzene (1)
Toluene (2)
A
14.1603
14.2515
B
2948.78
3242.38
C
− 44.5633
− 47.1806
Solution -----------------------------------------------------------------------------------------The temperature for the Txy diagram should be between the boiling points of benzene and
toluene given by
T1boil =
2948.78
+ 44.5633 = 377.31 K
14.1603 − log(200)
T2boil =
3242.38
+ 47.1806 = 409.33 K
14.2515 − log(200)
4-2
The simplest procedure is to choose a temperature T between 377.31 K and 409.33 K,
evaluate the vapor pressures, and solve for x and y from the following equations:
sat
yi
P
= i
xi
P
(E-1)
At 400 K, the vapor pressure of benzene and toluene are given by
P1sat = exp(14.1603 − 2948.78/(400 − 44.5633)) = 352.160 kPa
P2sat = exp(14.2515 − 3242.38/(400 − 47.1806)) = 157.8406 kPa
Therefore
y1 352.16
=
⇒ y1 = K1 x1 = 1.7608x1
x1
200
(E-1)
1 − y1 157.841
=
⇒ 1 − y1 = K2 (1 − x1) = 0.7892 (1 − x1)
1 − x1
200
(E-1)
Substituting Eq. (E-2) into (E-1) yields
1 − K1 x1 = K2 (1 − x1)
Therefore
x1 =
1 − K2
1 − 0.7892
=
= 0.2170
K1 − K 2
1.7608 − 0.7892
y1 = K1 x1 = 1.7608x1 = 0.3820
The Matlab program listed in Table E4.1-1 plots the Txy diagram shown in Figure 4.1-1.
Table E4.1-1-------------------------------------------------------------------------% Example 4.1-1, Txy diagram for benzene-toluene mixture at 200 kPa
%
P=200; % kPa
A=[14.1603 14.2515]; B=[2948.78 3242.38]; C=[-44.5633 -47.1806];
% Boling point at 200 kPa
Tb=B./(A-log(P))-C;
fprintf('Boiling point of Benzene at P = %g, Tb = %6.2f C\n',P,Tb(1))
fprintf('Boiling point of Toluene at P = %g, Tb = %6.2f C\n',P,Tb(2))
T=linspace(Tb(1),Tb(2),50);
K1=exp(A(1)-B(1)./(T+C(1)))/P;
K2=exp(A(2)-B(2)./(T+C(2)))/P;
x=(1-K2)./(K1-K2); y = K1.*x;
ymin=round(Tb(1)-1);ymax=round(Tb(2)+1);
4-3
% Solve for x and y at T = 400 K
T1=400;
K1=exp(A(1)-B(1)./(T1+C(1)))/P;
K2=exp(A(2)-B(2)./(T1+C(2)))/P;
x1=(1-K2)/(K1-K2); y1 = K1*x1;
plot(x,T,y,T,':',[0 y1],[T1 T1],'--',[x1 x1],[ymin T1],'--',[y1 y1],[ymin T1],'--')
axis([0 1 ymin ymax]);
grid on
xlabel('x,y');ylabel('T(K)')
legend('Saturated liquid','Saturated vapor')
----------------------------------------------------------------------------------------When deviations from Raoult’s law are large enough, the Tx and Ty curves can go through a
maximum or a minimum. The extreme point (either minimum or maximum) is called
azeotrope where the liquid mole fraction is equal to the vapor mole fraction for each species:
xi = yi
(4.1-1)
A system that exhibits a maximum in pressure (positive deviations from Raoult’s law) will
exhibits a minimum in temperature called minimum boiling azeotrope as shown the top part
of Figure 4.1-1 for a mixture of chloroform and hexane. The Pxy diagram is plotted at 318 K
and the Txy diagram is plotted at 1 atm. This is the case when the like interaction is stronger
than the unlike interaction between the molecules. The mixture will require less energy to go
to the vapor phase and hence will boil at a lower temperature that that of the pure
components.
If the unlike interaction is stronger than the like interaction we have negative deviations from
Raoult’s law and the system will exhibit a minimum in pressure or a maximum in
temperature called maximum boiling azeotrope. A mixture of acetone and chloroform shows
this behavior in the bottom part of Figure 4.1-1. The Pxy diagram is plotted at 328 K and the
Txy diagram is plotted at 1 atm. The data for vapor pressure and Wilson model are from the
Thermosolver program by Koretsky. This program can also plot Pxy and Txy diagrams for
different mixtures. Table 4.1-1 lists the Matlab program used to produce Figure 4.1-2.
4-4
Chloroform-Hexane
Chloroform-Hexane
x
y
0.55
340
T(K)
P(atm)
x
y
342
0.5
338
336
0.45
334
0.4
0
0.2
0.4
0.6
0.8
1
0
0.2
0.4
x,y
0.6
0.8
1
x,y
Acetone-Chloroform
Acetone-Chloroform
1
340
x
y
0.95
x
y
338
336
0.85
T(K)
P(atm)
0.9
0.8
0.75
332
0.7
0.65
334
0
330
0.2
0.4
0.6
0.8
1
0
x,y
0.2
0.4
0.6
0.8
1
x,y
Figure 4.1-2 Top: minimum boiling azeotrope for chloroform and n-Hexane system.
Bottom: maximum boiling azeotrope for acetone and chloroform system
Table 4.1-2 ----------------------------------------------------------------------------------% Figure 4.1-2: Construct a Txy diagram for chloroform (1) and hexane (2) mixture
% at a total pressure (atm) of
P = 1;
% Use the Wilson equation with parameters
G12 = 1.2042; G21 = 0.39799;
% Vapor pressure data: P(atm), T(K)
p1sat = 'exp(9.33984-2696.79/(T-46.14))';
p2sat = 'exp(9.20324-2697.55/(T-48.78))';
% Estimate boiling points
Tb1 = 2696.79/(9.33984-log(P))+46.14;
Tb2 = 2697.55/(9.20324-log(P))+48.78;
x1=[0 .002 .004 .006 .008 .01 .015];
x2=linspace(.02,.92,46);
x3=[.93 .94 .95 .96 .97 .98 .985 .990 .995 1];
xp=[x1 x2 x3];np=length(xp);
yp=xp;Tp=xp;
Tp(1)=Tb2;Tp(np)=Tb1;
dT = .01;
T=Tb2;
4-5
for i=2:np
x1=xp(i);x2=1-x1;
% Evaluate activity coefficients
tem1 = x1 + x2*G12; tem2 = x2 + x1*G21;
gam1 = exp(-log(tem1)+x2*(G12/tem1-G21/tem2));
gam2 = exp(-log(tem2)+x1*(G21/tem2-G12/tem1));
for k=1:20
fT=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P;
T=T+dT;
fT2=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P;
dfT=(fT2-fT)/dT; eT=fT/dfT;
T=T-dT-eT;
if abs(eT)<.001, break ,end
end
Tp(i)=T;
yp(i)=x1*gam1*eval(p1sat)/P;
fprintf('T(K) = %8.2f , x=%8.4f;y = %8.4f, iteration = %g\n',T,x1,yp(i),k)
end
subplot(2,2,2); plot(xp,Tp,yp,Tp,':')
axis([0 1 333 343])
xlabel('x,y');ylabel('T(K)');title('Chloroform-Hexane')
grid on
legend('x','y')
%
% Construct a Pxy diagramfor chloroform (1) and hexane (2) mixture
% at a temperature (K) of
T=318;
p1vap=eval(p1sat);p2vap=eval(p2sat);
xp2=1-xp;
tem1 = xp + xp2*G12; tem2 = xp2 + xp*G21;
gam1 = exp(-log(tem1)+xp2.*(G12./tem1-G21./tem2));
gam2 = exp(-log(tem2)+xp.*(G21./tem2-G12./tem1));
p1=xp.*gam1*p1vap;p2=xp2.*gam2*p2vap;
Pp=p1+p2;yp=p1./Pp;
subplot(2,2,1); plot(xp,Pp,yp,Pp,':')
axis([0 1 0.4 0.6])
xlabel('x,y');ylabel('P(atm)');title('Chloroform-Hexane')
grid on
legend('x','y',2)
%
% Figure 4.1-2: Construct a Txy diagramfor acetone (1) and chloroform (2) mixture
% at a total pressure (atm) of
P = 1;
% Use the Wilson equation with parameters
G12 = 1.324; G21 = 1.7314;
% Vapor pressure data: P(atm), T(K)
p1sat = 'exp(10.0179-2940.46/(T-35.93))';
p2sat = 'exp(9.33984-2696.79/(T-46.14))';
% Estimate boiling points
4-6
Tb1 = 2940.46/(10.0179-log(P))+35.93;
Tb2 = 2696.79/(9.33984-log(P))+46.14;
x1=[0 .002 .004 .006 .008 .01 .015];
x2=linspace(.02,.92,46);
x3=[.93 .94 .95 .96 .97 .98 .985 .990 .995 1];
xp=[x1 x2 x3];np=length(xp);
yp=xp;Tp=xp;
Tp(1)=Tb2;Tp(np)=Tb1;
dT = .01;
T=Tb2;
for i=2:np
x1=xp(i);x2=1-x1;
% Evaluate activity coefficients
tem1 = x1 + x2*G12; tem2 = x2 + x1*G21;
gam1 = exp(-log(tem1)+x2*(G12/tem1-G21/tem2));
gam2 = exp(-log(tem2)+x1*(G21/tem2-G12/tem1));
for k=1:20
fT=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P;
T=T+dT;
fT2=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P;
dfT=(fT2-fT)/dT; eT=fT/dfT;
T=T-dT-eT;
if abs(eT)<.001, break ,end
end
Tp(i)=T;
yp(i)=x1*gam1*eval(p1sat)/P;
fprintf('T(K) = %8.2f , x=%8.4f;y = %8.4f, iteration = %g\n',T,x1,yp(i),k)
end
subplot(2,2,4); plot(xp,Tp,yp,Tp,':')
axis([0 1 329 340])
xlabel('x,y');ylabel('T(K)');title('Acetone-Chloroform')
grid on
legend('x','y')
%
% Construct a Pxy diagramfor acetone (1) and chloroform (2) mixture
% at a temperature (K) of
T=328;
p1vap=eval(p1sat);p2vap=eval(p2sat);
xp2=1-xp;
tem1 = xp + xp2*G12; tem2 = xp2 + xp*G21;
gam1 = exp(-log(tem1)+xp2.*(G12./tem1-G21./tem2));
gam2 = exp(-log(tem2)+xp.*(G21./tem2-G12./tem1));
p1=xp.*gam1*p1vap;p2=xp2.*gam2*p2vap;
Pp=p1+p2;yp=p1./Pp;
subplot(2,2,3); plot(xp,Pp,yp,Pp,':')
axis([0 1 0.65 1.0])
xlabel('x,y');ylabel('P(atm)');title('Acetone-Chloroform')
grid on
legend('x','y',2)
4-7
4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System
Consider an equilibrium stage where a vapor stream V2 is in contact with a liquid stream L0.
The two streams V1 and L1 leaving the equilibrium stage is in equilibrium. The compositions
in streams V1 and L1 must be solved from the material balance, the energy balance, and the
equilibrium relations. For a binary mixture of A and B, if sensible heat effects are small and
the latent heats of both compounds are the same, then when 1 mole of A condenses, 1 mole
of B must vaporize. Hence, the total molar flow V1 will equal the total molar flow V2.
Similarly, the total molar flow L1 will equal the total molar flow L0. This situation is called
constant molal overflow (CMO). When CMO is valid, the compositions in streams V1 and L1
can be solved from only the material balance and the equilibrium relations. The energy
balance is not required since it is satisfied when the material balance is satisfied.
L0
V1
Equilibrium stage
L1
V2
Figure 4.1-1 Schematic of an equilibrium stage.
Example 4.2-1 ---------------------------------------------------------------------------------A vapor at the dew point and 200 kPa containing a mole fraction of 0.40 benzene (1) and
0.60 toluene (2) and 100 kmol total is brought into contact with 110 kmol of a liquid at the
boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams
are contacted in a single stage, and the outlet streams leave in equilibrium with each other.
Assume constant molar overflow, calculate the amounts and compositions of the exit
streams.
Data: Vapor pressure, Psat, data: ln Psat = A − B/(T + C), where Psat is in kPa and T is in K.
Compound
Benzene (1)
Toluene (2)
A
14.1603
14.2515
B
2948.78
3242.38
C
− 44.5633
− 47.1806
Solution -----------------------------------------------------------------------------------------For CMO, L1 = L0 = 110 kmol, V2 = V1 = 100 kmol. Making a balance on benzene gives
L0x0 + V2y2 = L1x1 + V1y1
110(0.30) + 100(0.40) = 110x1 + 100y1
11x1 + 10y1 = 7.3 ⇒ y1 = 0.73 − 1.1x1
Since the two streams V1 and L1 are in equilibrium, we have
4-8
(E-1)
y1 P1sat
=
⇒ 200y1 = x1exp(14.1603 − 2948.78/(T − 44.5633))
x1 200
(E-2)
1 − y1 P2sat
=
⇒ 200(1 − y1) = (1 − x1)exp(14.2515 − 3242.38/(T − 47.1806))
1 − x1 200
(E-3)
The three equations (E-1,2,3) can be solved for T, x1, and y1 either by graphical or numerical
method.
A) Graphical method
We can choose a range of temperatures between the boiling point of pure benzene (377.31 K)
and pure toluene (409.33 K) and solve equations (E-1,2) for x1, and y1 then plot y1 versus x1
to obtain the equilibrium curve. We then plot the material balance equation (E-1) to obtain
the operating line. The intersection of the equilibrium curve and the operating line provides
the composition x1 and y1 of the streams exiting the equilibrium stage. The following is a
procedure to obtain a point on the equilibrium curve:
At 400 K, the vapor pressure of benzene and toluene are given by
P1sat = exp(14.1603 − 2948.78/(400 − 44.5633)) = 352.160 kPa
P2sat = exp(14.2515 − 3242.38/(400 − 47.1806)) = 157.8406 kPa
Therefore
y1 352.16
=
⇒ y1 = K1 x1 = 1.7608x1
x1
200
(E-2)
1 − y1 157.841
=
⇒ 1 − y1 = K2 (1 − x1) = 0.7892 (1 − x1)
1 − x1
200
(E-3)
Substituting Eq. (E-3) into (E-2) yields
1 − K1 x1 = K2 (1 − x1)
Therefore
x1 =
1 − K2
1 − 0.7892
=
= 0.2170
K1 − K 2
1.7608 − 0.7892
y1 = K1 x1 = 1.7608x1 = 0.3820
Figures E4.2-1a, and b show the intersection of the equilibrium curve and the operating line.
The intersection point can be zoom in as shown in Figure E4.2-1b with x1 = 0.2615 and y1 =
0.4425. The Matlab program is listed in Table E4.2-1.
4-9
Figure E4.2-1a Equilibrium diagram for benzene-toluene mixture at 200 kPa
Figure E4.2-1b Equilibrium diagram for benzene-toluene mixture at 200 kPa
Table E4.2-1 Matlab program for Figure E4.2-1a,b
----------------------------------------% Example 4.2-1, Composition of benzene-toluene mixture at 200 kPa leaving an
equilibrium stage
%
P=200; % kPa
A=[14.1603 14.2515]; B=[2948.78 3242.38]; C=[-44.5633 -47.1806];
% Boling point at 200 kPa
Tb=B./(A-log(P))-C;
4-10
fprintf('Boiling point of Benzene at P = %g, Tb = %6.2f C\n',P,Tb(1))
fprintf('Boiling point of Toluene at P = %g, Tb = %6.2f C\n',P,Tb(2))
% Solve for equilibrium curve
T=linspace(Tb(1),Tb(2),50);
K1=exp(A(1)-B(1)./(T+C(1)))/P;
K2=exp(A(2)-B(2)./(T+C(2)))/P;
x=(1-K2)./(K1-K2); y = K1.*x;
% Solve for operating line
xp=[0.2 0.6];
yp = 0.73 - 1.1*xp;
plot(x,y,':',xp,yp)
axis([0 1 0 1]);
grid on
xlabel('x, mole fraction of benzene in liquid phase');ylabel('y')
legend('Equilibrium curve','Operating line',2)
-----------------------------------------------------------------B) Numerical method. We need to solve the following equations:
y1 = 0.73 − 1.1x1
(E-1)
200y1 = x1exp(14.1603 − 2948.78/(T − 44.5633))
(E-2)
200(1 − y1) = (1 − x1)exp(14.2515 − 3242.38/(T − 47.1806))
(E-3)
Substituting (E-1) into (E-2) and (E-3) we obtain
146 − 220x1 − x1exp(14.1603 − 2948.78/(T − 44.5633)) = 0
(E-2a)
54 + 220x1 −(1 − x1)exp(14.2515 − 3242.38/(T − 47.1806)) = 0
(E-3a)
Equations (E-2a, and E-3a) can be solved by the following Matlab statement
>> p=fminsearch('estage',[.5 400])
p = 0.2614 398.3056
The function estage is written as follow
function y=estage(p)
x=p(1);
T=p(2);
f1=146-220*x-x*exp(14.1603-2948.78/(T-44.5633));
f2=54+220*x-(1-x)*exp(14.2515-3242.38/(T-47.1806));
y=f1*f1+f2*f2;
Therefore x1 = 0.2614 and y1 = 0.73 − 1.1x1 = 0.73 − 1.1(0.2614) = 0.4425
4-11
4.3 Simple Batch or Differential Distillation
In a simple batch (or, more precisely, semi-batch) distillation unit, liquid is first charged to a
heated kettle. The liquid charge is boiled slowly and the vapors are withdrawn as rapidly as
they form to a condenser, where the condensed vapor (distillate) is collected. The first
portion of vapor condensed will be richest in the more volatile component A. As vaporization
proceeds, the vaporized product becomes leaner in A. Initially, a charge of Li moles of a
binary mixture with a mole fraction xi of the more volatile component A is placed in the still.
At any given time, L is the moles of liquid left in the still with composition x. V is the instant
vapor flow rate leaving the still with the mole fraction y of the more volatile component A.
The vapor leaving the still may be considered to be in equilibrium with the remaining liquid.
Example 4.3-1. ---------------------------------------------------------------------------------A semi-batch distillation unit is charged with 100 mol of a 60 mol% benzene-40 mol%
toluene mixture. At any given instant, the benzene mole fraction in the vapor flow rate, y, and
the benzene mole fraction in the remaining liquid, x, are related by the equilibrium relation
y=
2 .6 x
1 + 1. 6 x
Derive an equation relating the amount of liquid remaining in the still to the mole fraction of
benzene in this liquid.
Solution -----------------------------------------------------------------------------------------Step #1: Define the system.
System: Moles L of liquid left in the still with composition x
.
Let Vɺ = moles vapor rate to condenser
y = mole fraction of the more volatile component in
the vapor
L = moles of liquid in the still
x = mole fraction of the more volatile component in
the liquid
L
The composition in the still pot changes with time with x as the mole fraction of benzene
in the liquid phase.
4-12
Step #2: Find equation that relates L to x.
At any given instant, the mole of benzene in the still is Lx. A benzene balance will
give an equation relating the amount of liquid remaining in the still to the mole fraction
of benzene in this liquid.
Step #3: Apply the benzene balance.
d ( Lx )
= − y Vɺ
dt
(E-1)
Total balance is required since there are more than one unknown in Eq. (E-1)
dL
= − Vɺ
dt
The product rule of derivative can be applied to the LHS of Eq. (E-1)
L
dx
dL
+x
= − y Vɺ
dt
dt
(E-2)
Vɺ and y can be eliminated from Eq.(E-2) by using the total balance and the
equilibrium relation.
L
dx
dL
2.6 x dL
+x
=
dt
dt
1 + 1.6 x dt
L
dx
2.6 x dL
dL  2.6 x
 dL
=
−x
=
− x
dt
dt
1 + 1.6 x dt
 1 + 1 .6 x
 dt
dL
=
L
dx
dx
1 + 1 .6 x
=
=
dx
2
2.6 x
2
.
6
x
−
x
−
1
.
6
x
1
.
6
x
(
1
−
x
)
−x
1 + 1 .6 x
1 + 1 .6 x
The expression
1 + 1 .6 x
can be expanded by partial fraction
1.6 x (1 − x )
a
1 + 1 .6 x
b
=
+
⇒ 1 + 1.6x = a(1 − x) + b(1.6x)
1.6 x (1 − x ) 1.6 x 1 − x
x = 0 => 1 = a
x = 1 => 2.6 = 1.6b => b = 2.6/1.6
4-13
1
1 + 1 .6 x
2 .6
=
+
1.6 x (1 − x ) 1.6 x 1.6(1 − x )
1
dL
2 .6
dx
=
dx +
L 1 .6 x
1.6(1 − x )
Step #4: Specify the boundary conditions for the differential equation.
At L = 100, x = 0.6
Step #5: Solve the resulting equation and verify the solution.
dL
∫100 L =
L
dx
∫0.6 1.6 x +
x
∫
x
0.6
2.6dx
1.6(1 − x )
x
L
2.6
 1

ln
=  ln x −
ln(1 − x )
100
1
.
6
1
.
6

 0. 6
 x  0.625  1 − x  −1.625 
L
x
1− x
ln
= 0.625 ln − 1.625 ln
= ln   


.4
100
.6
 .6   .4 

 x  0.625  1 − x  −1.625 
L = 100   


 .6   .4 

A plot of L versus (1 − x) shows that the liquid remaining in the still becomes
progressively leaner in benzene; when L → 0, (1 − x) → 1 (pure toluene).
4-14