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Chapter 4 Distillation

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Chapter 4 Distillation 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: y Psat 1− y Psat 1 = 1 and 1 = 2 (4.1-2a, b) − x1 P 1 x1 P sat sat In these expressions, P = 200 kPa, P1 and P2 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, sat sat P1 = 352.160 kPa and P2 = 157.8406 kPa. Equations (4.1-2a, b) become y 352.16 1− y 157.841 1 = and 1 = (4.1-3a, b) − x1 200 1 x1 200 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 bubble- point 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 A B C Benzene (1) 14.1603 2948.78 − 44.5633 Toluene (2) 14.2515 3242.38 − 47.1806 Solution ------------------------------------------------------------------------------------------ The temperature for the Txy diagram should be between the boiling points of benzene and toluene given by boil 2948 78. T1 = + 44.5633 = 377.31 K 14 .1603 − log( 200 ) boil 3242 38. T2 = + 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 Pi = (E-1) xi P At 400 K, the vapor pressure of benzene and toluene are given by sat P1 = exp(14.1603 − 2948.78/(400 − 44.5633)) = 352.160 kPa sat P2 = exp(14.2515 − 3242.38/(400 − 47.1806)) = 157.8406 kPa Therefore y1 352.16 = ⇒ y1 = K1 x1 = 1.7608 x1 (E-1) x1 200 1 − y 157.841 1 = ⇒ 1 − y = K (1 − x ) = 0.7892 (1 − x ) (E-1) − 1 2 1 1 1 x1 200 Substituting Eq. (E-2) into (E-1) yields 1 − K1 x1 = K2 (1 − x1) Therefore 1 − K 1 − .0 7892 x = 2 = = 0.2170 1 − − K1 K2 1.7608 0.7892 y1 = K1 x1 = 1.7608 x1 = 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 342 x y y 0.55 340 0.5 338 T(K) P(atm) 336 0.45 334 0.4 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x,y x,y Acetone-Chloroform Acetone-Chloroform 1 340 x x 0.95 y 338 y 0.9 336 0.85 T(K) 334 P(atm) 0.8 0.75 332 0.7 330 0.65 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x,y 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));
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