Chapter 4 Distillation 4. Vapor Liquid Equilibrium Relations The equilibrium in vapor-liquid equilibrium is restricted b the phase rule. For a binar miture in two phases, the degree of freedom F is given b F m π + 2 2 2 + 2 2 (4.-) As an eample we will use the benzene-toluene, vapor-liquid sstem. 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 sstem. If the pressure is fied at 200 kpa, onl one more variable can be set. If we set the temperature, the mole fraction of benzene (component ) can be determined from the following relations: sat PP and sat P2 P (4.-2a, b) sat sat In these epressions, P 200 kpa, P and P 2 are the vapor pressures of benzene and toluene at the specified temperature, respectivel. For ideal solution such as benzene-toluene sstem, the bubble and dew point temperatures of the miture will be between the saturation temperatures of benzene and toluene. At 200 kpa, the saturation temperatures of benzene and toluene are 377.3 K and 409.33 K, respectivel. If we set the miture temperature at 400 K, P sat 352.60 kpa and P sat 2 57.8406 kpa. Equations (4.-2a, b) become 352.6 200 and 57.84 200 (4.-3a, b) Solving these two equations, we obtain 0.270 and 0.3820. The vapor-liquid equilibrium relations for benzene ()-toluene (2) at a total pressure of 200 kpa are given as a boiling-point T diagram shown in Figure 4.-. 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.-, if we start with a cold liquid miture of 0.27 and heat the miture, it will start to boil at 400 K, and the composition of the first vapor in equilibrium is 0.382. As we continue boiling, the composition in the remaining liquid miture will decrease since is richer in benzene. If we start with a hot gas miture of 0.382 and cool the miture, it will start to condense at 400 K, and the composition of the first liquid drop in equilibrium is 0.27. As we continue condensing, the composition in the remaining gas miture will increase since is richer in toluene. The T diagram for an ideal miture can be calculated from the pure vapor-pressure data as illustrated in Eample 4.-. 4-
Figure 4.- Calculated T diagram of benzene and toluene at 200 kpa. Eample 4.- ---------------------------------------------------------------------------------- Construct a T diagram for a miture of benzene and toluene at 200 kpa. Benzene and toluene mitures ma be considered as ideal. Data: Vapor pressure, P sat, data: ln P sat A B/(T + C), where P sat is in kpa and T is in K. Compound A B C Benzene () 4.603 2948.78 44.5633 Toluene (2) 4.255 3242.38 47.806 Solution ------------------------------------------------------------------------------------------ The temperature for the T diagram should be between the boiling points of benzene and toluene given b T boil 2948.78 4.603 log(200) + 44.5633 377.3 K T 2 boil 3242.38 + 47.806 409.33 K 4.255 log(200) 4-2
The simplest procedure is to choose a temperature T between 377.3 K and 409.33 K, evaluate the vapor pressures, and solve for and from the following equations: i i P sat i (E-) P At 400 K, the vapor pressure of benzene and toluene are given b P sat ep(4.603 2948.78/(400 44.5633)) 352.60 kpa P sat 2 ep(4.255 3242.38/(400 47.806)) 57.8406 kpa Therefore 352.6 200 K.7608 (E-) 57.84 200 K 2 ( ) 0.7892 ( ) (E-) Substituting Eq. (E-2) into (E-) ields Therefore K K 2 ( ) K2 K K 2 0.7892.7608 0.7892 0.270 K.7608 0.3820 The Matlab program listed in Table E4.- plots the T diagram shown in Figure 4.-. Table E4.--------------------------------------------------------------------------- % Eample 4.-, T diagram for benzene-toluene miture at 200 kpa % P200; % kpa A[4.603 4.255]; B[2948.78 3242.38]; C[-44.5633-47.806]; % Boling point at 200 kpa TbB./(A-log(P))-C; fprintf('boiling point of Benzene at P %g, Tb %6.2f C\n',P,Tb()) fprintf('boiling point of Toluene at P %g, Tb %6.2f C\n',P,Tb(2)) Tlinspace(Tb(),Tb(2),50); Kep(A()-B()./(T+C()))/P; K2ep(A(2)-B(2)./(T+C(2)))/P; (-K2)./(K-K2); K.*; minround(tb()-);maround(tb(2)+); 4-3
% Solve for and at T 400 K T400; Kep(A()-B()./(T+C()))/P; K2ep(A(2)-B(2)./(T+C(2)))/P; (-K2)/(K-K2); K*; plot(,t,,t,':',[0 ],[T T],'--',[ ],[min T],'--',[ ],[min T],'--') ais([0 min ma]); grid on label(',');label('t(k)') legend('saturated liquid','saturated vapor') ----------------------------------------------------------------------------------------- When deviations from Raoult s law are large enough, the T and T curves can go through a maimum or a minimum. The etreme point (either minimum or maimum) is called azeotrope where the liquid mole fraction is equal to the vapor mole fraction for each species: i i (4.-) A sstem that ehibits a maimum in pressure (positive deviations from Raoult s law) will ehibits a minimum in temperature called minimum boiling azeotrope as shown the top part of Figure 4.- for a miture of chloroform and heane. The P diagram is plotted at 38 K and the T diagram is plotted at atm. This is the case when the like interaction is stronger than the unlike interaction between the molecules. The miture will require less energ 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 sstem will ehibit a minimum in pressure or a maimum in temperature called maimum boiling azeotrope. A miture of acetone and chloroform shows this behavior in the bottom part of Figure 4.-. The P diagram is plotted at 328 K and the T diagram is plotted at atm. The data for vapor pressure and Wilson model are from the Thermosolver program b Koretsk. This program can also plot P and T diagrams for different mitures. Table 4.- lists the Matlab program used to produce Figure 4.-2. 4-4
Chloroform-Heane Chloroform-Heane 0.55 342 340 P(atm) 0.5 T(K) 338 0.45 336 0.4 0 0.2 0.4 0.6 0.8, 334 0 0.2 0.4 0.6 0.8, P(atm) 0.95 0.9 0.85 0.8 Acetone-Chloroform T(K) 340 338 336 334 Acetone-Chloroform 0.75 0.7 0.65 0 0.2 0.4 0.6 0.8, 332 330 0 0.2 0.4 0.6 0.8, Figure 4.-2 Top: minimum boiling azeotrope for chloroform and n-heane sstem. Bottom: maimum boiling azeotrope for acetone and chloroform sstem Table 4.-2 ----------------------------------------------------------------------------------- % Figure 4.-2: Construct a T diagram for chloroform () and heane (2) miture % at a total pressure (atm) of P ; % Use the Wilson equation with parameters G2.2042; G2 0.39799; % Vapor pressure data: P(atm), T(K) psat 'ep(9.33984-2696.79/(t-46.4))'; p2sat 'ep(9.20324-2697.55/(t-48.78))'; % Estimate boiling points Tb 2696.79/(9.33984-log(P))+46.4; Tb2 2697.55/(9.20324-log(P))+48.78; [0.002.004.006.008.0.05]; 2linspace(.02,.92,46); 3[.93.94.95.96.97.98.985.990.995 ]; p[ 2 3];nplength(p); pp;tpp; Tp()Tb2;Tp(np)Tb; dt.0; TTb2; 4-5
for i2:np p(i);2-; % Evaluate activit coefficients tem + 2*G2; tem2 2 + *G2; gam ep(-log(tem)+2*(g2/tem-g2/tem2)); gam2 ep(-log(tem2)+*(g2/tem2-g2/tem)); for k:20 ft*gam*eval(psat)+2*gam2*eval(p2sat)-p; TT+dT; ft2*gam*eval(psat)+2*gam2*eval(p2sat)-p; dft(ft2-ft)/dt; etft/dft; TT-dT-eT; if abs(et)<.00, break,end end Tp(i)T; p(i)*gam*eval(psat)/p; fprintf('t(k) %8.2f, %8.4f; %8.4f, iteration %g\n',t,,p(i),k) end subplot(2,2,2); plot(p,tp,p,tp,':') ais([0 333 343]) label(',');label('t(k)');title('chloroform-heane') grid on legend('','') % % Construct a P diagramfor chloroform () and heane (2) miture % at a temperature (K) of T38; pvapeval(psat);p2vapeval(p2sat); p2-p; tem p + p2*g2; tem2 p2 + p*g2; gam ep(-log(tem)+p2.*(g2./tem-g2./tem2)); gam2 ep(-log(tem2)+p.*(g2./tem2-g2./tem)); pp.*gam*pvap;p2p2.*gam2*p2vap; Ppp+p2;pp./Pp; subplot(2,2,); plot(p,pp,p,pp,':') ais([0 0.4 0.6]) label(',');label('p(atm)');title('chloroform-heane') grid on legend('','',2) % % Figure 4.-2: Construct a T diagramfor acetone () and chloroform (2) miture % at a total pressure (atm) of P ; % Use the Wilson equation with parameters G2.324; G2.734; % Vapor pressure data: P(atm), T(K) psat 'ep(0.079-2940.46/(t-35.93))'; p2sat 'ep(9.33984-2696.79/(t-46.4))'; % Estimate boiling points 4-6
Tb 2940.46/(0.079-log(P))+35.93; Tb2 2696.79/(9.33984-log(P))+46.4; [0.002.004.006.008.0.05]; 2linspace(.02,.92,46); 3[.93.94.95.96.97.98.985.990.995 ]; p[ 2 3];nplength(p); pp;tpp; Tp()Tb2;Tp(np)Tb; dt.0; TTb2; for i2:np p(i);2-; % Evaluate activit coefficients tem + 2*G2; tem2 2 + *G2; gam ep(-log(tem)+2*(g2/tem-g2/tem2)); gam2 ep(-log(tem2)+*(g2/tem2-g2/tem)); for k:20 ft*gam*eval(psat)+2*gam2*eval(p2sat)-p; TT+dT; ft2*gam*eval(psat)+2*gam2*eval(p2sat)-p; dft(ft2-ft)/dt; etft/dft; TT-dT-eT; if abs(et)<.00, break,end end Tp(i)T; p(i)*gam*eval(psat)/p; fprintf('t(k) %8.2f, %8.4f; %8.4f, iteration %g\n',t,,p(i),k) end subplot(2,2,4); plot(p,tp,p,tp,':') ais([0 329 340]) label(',');label('t(k)');title('acetone-chloroform') grid on legend('','') % % Construct a P diagramfor acetone () and chloroform (2) miture % at a temperature (K) of T328; pvapeval(psat);p2vapeval(p2sat); p2-p; tem p + p2*g2; tem2 p2 + p*g2; gam ep(-log(tem)+p2.*(g2./tem-g2./tem2)); gam2 ep(-log(tem2)+p.*(g2./tem2-g2./tem)); pp.*gam*pvap;p2p2.*gam2*p2vap; Ppp+p2;pp./Pp; subplot(2,2,3); plot(p,pp,p,pp,':') ais([0 0.65.0]) label(',');label('p(atm)');title('acetone-chloroform') grid on legend('','',2) 4-7
4.2 Single-Stage Equilibrium Contact for Vapor-Liquid Sstem Consider an equilibrium stage where a vapor stream V 2 is in contact with a liquid stream L 0. The two streams V and L leaving the equilibrium stage is in equilibrium. The compositions in streams V and L must be solved from the material balance, the energ balance, and the equilibrium relations. For a binar miture of A and B, if sensible heat effects are small and the latent heats of both compounds are the same, then when mole of A condenses, mole of B must vaporize. Hence, the total molar flow V will equal the total molar flow V 2. Similarl, the total molar flow L will equal the total molar flow L 0. This situation is called constant molal overflow (CMO). When CMO is valid, the compositions in streams V and L can be solved from onl the material balance and the equilibrium relations. The energ balance is not required since it is satisfied when the material balance is satisfied. L 0 V Equilibrium stage L V 2 Figure 4.- Schematic of an equilibrium stage. Eample 4.2- ---------------------------------------------------------------------------------- A vapor at the dew point and 200 kpa containing a mole fraction of 0.40 benzene () and 0.60 toluene (2) and 00 kmol total is brought into contact with 0 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 eit streams. Data: Vapor pressure, P sat, data: ln P sat A B/(T + C), where P sat is in kpa and T is in K. Compound A B C Benzene () 4.603 2948.78 44.5633 Toluene (2) 4.255 3242.38 47.806 Solution ------------------------------------------------------------------------------------------ For CMO, L L 0 0 kmol, V 2 V 00 kmol. Making a balance on benzene gives L 0 0 + V 2 2 L + V 0(0.30) + 00(0.40) 0 + 00 + 0 7.3 0.73. (E-) Since the two streams V and L are in equilibrium, we have 4-8
sat P 200 200 ep(4.603 2948.78/(T 44.5633)) (E-2) sat P2 200 200( ) ( )ep(4.255 3242.38/(T 47.806)) (E-3) The three equations (E-,2,3) can be solved for T,, and either b graphical or numerical method. A) Graphical method We can choose a range of temperatures between the boiling point of pure benzene (377.3 K) and pure toluene (409.33 K) and solve equations (E-,2) for, and then plot versus to obtain the equilibrium curve. We then plot the material balance equation (E-) to obtain the operating line. The intersection of the equilibrium curve and the operating line provides the composition and of the streams eiting 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 b Therefore P sat ep(4.603 2948.78/(400 44.5633)) 352.60 kpa P 2 sat ep(4.255 3242.38/(400 47.806)) 57.8406 kpa 352.6 200 K.7608 (E-2) 57.84 200 K 2 ( ) 0.7892 ( ) (E-3) Substituting Eq. (E-3) into (E-2) ields Therefore K K 2 ( ) K2 K K 2 0.7892.7608 0.7892 0.270 K.7608 0.3820 Figures E4.2-a, 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-b with 0.265 and 0.4425. The Matlab program is listed in Table E4.2-. 4-9
Figure E4.2-a Equilibrium diagram for benzene-toluene miture at 200 kpa Figure E4.2-b Equilibrium diagram for benzene-toluene miture at 200 kpa Table E4.2- Matlab program for Figure E4.2-a,b ----------------------------------------- % Eample 4.2-, Composition of benzene-toluene miture at 200 kpa leaving an equilibrium stage % P200; % kpa A[4.603 4.255]; B[2948.78 3242.38]; C[-44.5633-47.806]; % Boling point at 200 kpa TbB./(A-log(P))-C; 4-0
fprintf('boiling point of Benzene at P %g, Tb %6.2f C\n',P,Tb()) fprintf('boiling point of Toluene at P %g, Tb %6.2f C\n',P,Tb(2)) % Solve for equilibrium curve Tlinspace(Tb(),Tb(2),50); Kep(A()-B()./(T+C()))/P; K2ep(A(2)-B(2)./(T+C(2)))/P; (-K2)./(K-K2); K.*; % Solve for operating line p[0.2 0.6]; p 0.73 -.*p; plot(,,':',p,p) ais([0 0 ]); grid on label(', mole fraction of benzene in liquid phase');label('') legend('equilibrium curve','operating line',2) ------------------------------------------------------------------ B) Numerical method. We need to solve the following equations: 0.73. 200 ep(4.603 2948.78/(T 44.5633)) (E-) (E-2) 200( ) ( )ep(4.255 3242.38/(T 47.806)) (E-3) Substituting (E-) into (E-2) and (E-3) we obtain 46 220 ep(4.603 2948.78/(T 44.5633)) 0 (E-2a) 54 + 220 ( )ep(4.255 3242.38/(T 47.806)) 0 (E-3a) Equations (E-2a, and E-3a) can be solved b the following Matlab statement >> pfminsearch('estage',[.5 400]) p 0.264 398.3056 The function estage is written as follow function estage(p) p(); Tp(2); f46-220*-*ep(4.603-2948.78/(t-44.5633)); f254+220*-(-)*ep(4.255-3242.38/(t-47.806)); f*f+f2*f2; Therefore 0.264 and 0.73. 0.73.(0.264) 0.4425 4-
4.3 Simple Batch or Differential Distillation In a simple batch (or, more precisel, semi-batch) distillation unit, liquid is first charged to a heated kettle. The liquid charge is boiled slowl and the vapors are withdrawn as rapidl as the 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. Initiall, a charge of L i moles of a binar miture with a mole fraction i of the more volatile component A is placed in the still. At an given time, L is the moles of liquid left in the still with composition. V is the instant vapor flow rate leaving the still with the mole fraction of the more volatile component A. The vapor leaving the still ma be considered to be in equilibrium with the remaining liquid. Eample 4.3-. ---------------------------------------------------------------------------------- A semi-batch distillation unit is charged with 00 mol of a 60 mol% benzene-40 mol% toluene miture. At an given instant, the benzene mole fraction in the vapor flow rate,, and the benzene mole fraction in the remaining liquid,, are related b the equilibrium relation 2.6 +.6 Derive an equation relating the amount of liquid remaining in the still to the mole fraction of benzene in this liquid. Solution ------------------------------------------------------------------------------------------ Step #: Define the sstem.. Sstem: Moles L of liquid left in the still with composition Let V ɺ moles vapor rate to condenser mole fraction of the more volatile component in the vapor L moles of liquid in the still mole fraction of the more volatile component in the liquid L The composition in the still pot changes with time with as the mole fraction of benzene in the liquid phase. 4-2
Step #2: Find equation that relates L to. At an given instant, the mole of benzene in the still is L. 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: Appl the benzene balance. d ( L) V ɺ (E-) dt Total balance is required since there are more than one unknown in Eq. (E-) dl V ɺ dt The product rule of derivative can be applied to the LHS of Eq. (E-) d dl L + V ɺ dt dt (E-2) V ɺ and can be eliminated from Eq.(E-2) b using the total balance and the equilibrium relation. d dl L + dt dt 2.6 dl +.6 dt d L dt 2.6 dl dl +.6 dt dt 2.6 +.6 dl dt dl L d 2.6 +. 6 d 2.6.6 +.6 2 +.6 d.6( ) The epression +.6.6( ) can be epanded b partial fraction +.6.6( ) a.6 + b +.6 a( ) + b(.6) 0 > a > 2.6.6b > b 2.6/.6 4-3
+.6.6( ) 2.6 +.6.6( ) dl L 2.6 d + d.6( ).6 Step #4: Specif the boundar conditions for the differential equation. At L 00, 0.6 Step #5: Solve the resulting equation and verif the solution. L 00 dl L d 0.6. 6 2.6d +.6.6( 0 ) ln L 2.6 ln ln( ) 00.6.6 0.6 L ln 0.625 ln.625 ln 00. 6. 4 ln.6 0.625.4. 625 L 00.6 0.625.4. 625 A plot of L versus ( ) shows that the liquid remaining in the still becomes progressivel leaner in benzene; when L 0, ( ) (pure toluene). 4-4