Distillation 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   + 2 = 2  2 + 2 = 2 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 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: 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 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. 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

Set 3, problem 7: The relative volatility, , of benzene to toluene at 1 atm is 2.5. Construct x-y and T- x-y diagrams for this system at 1 atm. On the same diagrams, plot the data obtained using the following vapor pressure data and ideal solution: For a constant relative volatility To calculate y-x and T-x-y curves from vapor pressure data, using Raoult's and Dalton's laws, as well as the sum of the mole fractions in the phases in equilibrium.

Equations (1) to (4) can be reduced to the following equations for the mole fractions of benzene (A) in terms of the K-values: Linear curve fitting vapor pressure with Pv = exp(A – B/T{K})

% Set 3 problem 7 P=[20 40 60 100 200 400 760]; Tb=[2.6 7.6 15.4 26.1 42.2 60.6 80.1]+273.15; Tt=[18.4 31.8 40.3 51.9 69.5 89.5 110.6]+273.15; y=log(P); xb=1.0./Tb;xt=1.0./Tt; c=polyfit(xb,y,1); Ab=c(2);Bb=c(1); c=polyfit(xt,y,1); At=c(2);Bt=c(1); % xy plot from relative volatiliy x=0:0.02:1; y=2.5*x./(1+1.5*x); % xy data

% xy data dT=(Tbt-Tbb)/14; T=Tbb:dT:Tbt; Tk=T+273.15; Pvb=exp(Ab+Bb./Tk);Pvt=exp(At+Bt./Tk); Kb=Pvb/760;Kt=Pvt/760; xb=(1-Kt)./(Kb-Kt);yb=Kb.*xb; % Plot xy diagram figure(1) plot(x,y,xb,yb,'o') axis([0 1 0 1]) grid xlabel('x');ylabel('y') legend('xy from relative volatility','xy from data')

dT=(Tbt-75)/25; T=75:dT:Tbt; Tk=T+273.15; Pvb=exp(Ab+Bb./Tk);Pvt=exp(At+Bt./Tk); Kb=Pvb/760;Kt=Pvt/760; xb=(1-Kt)./(Kb-Kt);yb=Kb.*xb; % Plot Txy diagram figure(2) plot(xb,T,yb,T) axis([0 1 75 111]) grid ylabel('T(C)');xlabel('x or y')

Better results with non linear curve fitting Better results with non linear curve fitting .Fit the vapor pressure data with Antoine equation Pv = exp[A – B/(T{C} + C)] function y=Bvapor(c) Ab=c(1);Bb=c(2);Cb=c(3); P=[20 40 60 100 200 400 760]; Tb=[2.6 7.6 15.4 26.1 42.2 60.6 80.1]; Pc=exp(Ab-Bb./(Tb+Cb)); y=sum((P-Pc).^2); function y=Tvapor(c) At=c(1);Bt=c(2);Ct=c(3); P=[20 40 60 100 200 400 760]; Tt=[18.4 31.8 40.3 51.9 69.5 89.5 110.6]; Pc=exp(At-Bt./(Tt+Ct)); y=sum((P-Pc).^2);

% Set 3 problem 7 c=fminsearch('Bvapor',[15 2500 200]) Ab=c(1);Bb=c(2);Cb=c(3); c=fminsearch('Tvapor',[16 3500 250]) At=c(1);Bt=c(2);Ct=c(3); Tbb=80.1;Tbt=110.6; % xy plot from relative volatiliy x=0:0.02:1; y=2.5*x./(1+1.5*x); % xy data dT=(Tbt-Tbb)/14; T=Tbb:dT:Tbt; Pvb=exp(Ab-Bb./(T+Cb));Pvt=exp(At-Bt./(T+Ct));

A liquid containing 70 mol% benzene and 30 mol% toluene is heated in a container at 1 atm until 25 mol% of the original liquid is evaporated. T = 88oC y = 0.84 x = 0.65

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 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 for a mixture of chloroform and hexane.

Minimum boiling azeotrope: 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. Maximum boiling azeotrope: 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

A liquid containing 60 mol% toluene and 40 mol% benzene is continuously distilled in a single equilibrium stage at 1 atm. What percent of benzene in the feed leaves as vapor if 90% of the toluene entering in the feed leaves as liquid? Assume a relativity of 2.3 and obtain the solution graphically. Let F = 1, y = mole fraction of benzene in V x = mole fraction of benzene in L Benzene balance: xF = fy + (1 – f)x Equilibrium relation: y=2.3*x/(1+1.3*x) Constraint: 0.9*(1- xF) = (1-f)*(1-x) 3 unknowns: x, y, f

Benzene balance: xF = fy + (1 – f)x Equilibrium relation: y=2.3*x/(1+1.3*x) Constraint: 0.9*(1- xF) = (1-f)*(1-x) Procedure: Assume x, calculate y=2.3*x/(1+1.3*x), Calculate f from material balance xF = fy + (1 – f)x check if 0.9*(1- xF) = (1-f)*(1-x)

xf=0.4; x=input('x = '); y=2.3*x/(1+1.3*x); f=(xf-x)/(y-x); CtoinL=(1-f)*(1-x); RtoinL=0.9*(1-xf); fprintf('Required toluene in L = %g, Calculated toluene in L = %g\n',RtoinL,CtoinL) PbinV=y*f*100/xf; fprintf('Percent of benzene in the feed leaves as vapor = %g\n',PbinV) s4p5a x = .37 Required toluene in L = 0.54, Calculated toluene in L = 0.53763 Percent of benzene in the feed leaves as vapor = 21.0623

Equilibrium relation: y=2.3*x/(1+1.3*x) f = 1-0.54./(1-x) Constraint: 0.9*(1- xF) = (1-f)*(1-x) Benzene balance: xF = fy + (1 – f)x y = (0.4-(1-f).*x)./f x=0.1:0.02:0.42; xe=0:0.02:1; ye=2.3*xe./(1+1.3*xe); f=1-0.54./(1-x); ym=(0.4-(1-f).*x)./f; plot(xe,ye,x,ym) grid on xlabel('x');ylabel('y')

Save these codes as flash.m Solving set of nonlinear equations using Matlab fminsearch function Benzene balance: xF = fy + (1 – f)x; xF = 0.4 Equilibrium relation: y=2.3*x/(1+1.3*x) Constraint: 0.9*(1- xF) = (1-f)*(1-x) function y=flash(p) x=p(1);y=p(2);f=p(3) f1=y-2.3*x/(1+1.3*x); f2=0.4-f*y-(1-f)*x; f3=0.54-(1-f)*(1-x); y=f1*f1+f2*f2+f3*f3; Save these codes as flash.m At the command window type: z=fminsearch('flash',[0.3 0.5 0.2]) z = 3.7106e-001 5.7573e-001 1.4144e-001

Single-Stage Equilibrium Contact for Vapor-Liquid System 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.

A. A steady-state open system is an example of a system that is at equilibrium. B. Cp (specific heat capacity in J/kg.K) is an intensive property. A. A and B are true B. Only A is true C. Only B is true D. A and B are false

A. If G is known as a function of T and P, the other properties (H, S, U, V, A) can easily be determined. B. If H is known as a function of T and P, the other properties (G, S, U, V, A) can easily be determined. . A. A and B are true B. Only A is true C. Only B is true D. A and B are false

A. B. A. A and B are true B. Only A is true C. Only B is true D. A and B are false

A. Energy balance is not required for isothermal system B. A real system behavior is between isothermal and adiabatic A. A and B are true B. Only A is true C. Only B is true D. A and B are false

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. 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 Derive an equation relating the amount of liquid remaining in the still to the mole fraction of benzene in this liquid.

System: Moles L of liquid left in the still with composition x = 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 The composition in the still pot changes with time with x as the mole fraction of benzene in the liquid phase.

Distillation with Reflux

McCabe-Thiele Method The McCabe Thiele method is a graphical procedure to solve alternately the equilibrium relation and the material balance for binary distillation. Total material balance for tray n gives   Vn+1 + Ln-1 = Vn + Ln A component balance on the more volatile component, A, gives yn+1Vn+1 + xn-1Ln-1 = ynVn + xnLn

The McCabe-Thiele method is valid when we have constant molal overflow in both the rectifying and stripping sections of the distillation column. Hence V1 = V2 = V3 = … = Vn = Vn+1 = constant L0 = L1 = L2 = … = Ln-1 = Ln = constant Vm-1 = Vm = Vm+1 = constant Lm-1 = Lm = Lm+1 = constant Even for CMO, Vn  Vm and Ln = Lm. The constant molal overflow is justified if the heat of mixing is negligible for the two components in the mixture (ideal solution) and the molar heats of vaporization for the two components are almost the same.

Equations for rectifying section Making a balance over the top part of the tower on component A gives yn+1Vn+1 = xnLn + xDD Since CMO is assumed, Vn+1 = V = constant and Ln = L = constant. yn+1 = xnL/V + xDD/V In terms of the reflux ratio R = L/D

Equations for stripping section Making a mole balance on component A gives ym+1V m+1 + xBB = xmL m

A. The two passing streams of an equilibrium stage usually are not in equilibrium. B. In an actual distillation column, the two streams leaving a stage are in equilibrium . A. A and B are true B. Only A is true C. Only B is true D. A and B are false

For binary distillation using McCabe Thiele diagram A. The operating line in the rectifying section has a slope greater or equal to one. B. The operating line in the stripping section has a slope greater or equal to one. A. A and B are true B. Only A is true C. Only B is true D. A and B are false

A. The total reboiler is counted as an equilibrium stage. B. The partial condenser is counted as an equilibrium stage. A. A and B are true B. Only A is true C. Only B is true D. A and B are false

Effect of feed conditions Let q be the ratio of the moles of saturated liquid in the feed to the total amount of feed then F = qF + (1  q)F The magnitude of q is related to the thermal condition of the feed. Making enthalpy and material balance around the feed plate

The feed-line or q-line equation At the intersection of the two operating lines y = yn+1 = ym+1 and x = xn = xm. Subtracting

[Eq. (1)] and (xBB + xDD) = xFF into into Eq. (1)

With binary distillation, the degree of freedom is then F = 2 + 6 = 8. For a design problem we can specify feed rate F, feed composition xF, feed quality q, distillate composition xD, distillate temperature (saturated liquid at the specified pressure), bottoms composition xB, external reflux ratio L0/D, and optimum feed plate. With these variables defined, D and B can be obtained from the external mass balances and QR and QC can be solved from the energy balances. Then the number of equilibrium stages can be solved from the stage-by-stage calculations or internal balances. A stage-by-stage calculation involves the successive solutions of the equilibrium relations and the material balance. The procedure is straight forward with no iteration.

A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed flow rate is 2,500 lbmol/hr and feed temperature is 30oC. The column is at 1 atm. A distillate that is 97 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. The external reflux ratio is L0/D = 3. Bottoms from the partial reboiler is 98 mole % n-hexane. Find D, B, QR, QC, and the number of equilibrium stages. 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 n-pentane (1) 13.9778 2554.6  36.2529 n-hexane (2) 14.0568 2825.42  42.7089 Heat of evaporation for n-pentane, C5 = 11,369 Btu/lbmol, CpL,C5 = 39.7 Btu/lbmoloF Heat of evaporation for n-hexane, C6 = 13,572 Btu/lbmol, CpL,C6 = 51.7 Btu/lbmoloF

D + B = 2,500 0.97D + 0.02B = (0.4)(2,500) = 1,000 B = 1,500 lbmoles/hr and D = 1000 lbmoles/hr. Condenser duty QC = V1(H1  hD)  V1Hevap = V1(0.97C5 + 0.03C6) V1 = L0 + D = (L0/D + 1)D = 4D = (4)(1000) = 4000 lbmoles/hr QC = (4,000)(0.97×11,369 + 0.03×13,572) = 4.574×107 Btu/hr

Since the distillate is almost pure pentane and the bottoms product is almost pure hexane, the boiling temperatures of pure pentane and hexane at 1 atm are used for the temperatures of the reflux stream and the reboiler, respectively. Hence TD  309 K and TB  342 K Since the boiling point of a 40 mole % n-pentane is 324.79 K, the feed enters the column at 30oC (or 303.15 K) is a subcooled liquid. Making an energy balance over the column, we obtain FhF + QR = DhD + QC + BhB Let the reference state be liquid at 30oC, then hF = 0.

Reboiler duty FhF + QR = DhD + QC + BhB Let the reference state be liquid at 30oC, then hF = 0. hD = CpL,C5(TD  TF) = (39.7)(309  303)(1.8) = 428.8 Btu/lbmol hB = CpL,C6(TB  TF) = (51.7)(342  303)(1.8) = 3,629.3 Btu/lbmol QR = DhD + QC + BhB = (1,500)( 3,629.3) + 4.574×107 + (1,000)(428.8) = 5.1613×107 Btu/hr

Operating line for the rectifying section R = 3 and xD = 0.97 The q-line is determined next. Since feed is subcooled liquid we have

In this equation HV  hL = latent heat = 0.4C5 + 0.6C6 = (0.4)(11,369) + (0.6)( 13,572) HV  hL = 12,691 Btu/lbmol CpL = 0.4CpL,C5 + 0.6CpL,C6 = (0.4)(39.7) + (0.6)(51.7) = 46.9 Btu/lbmoloF

The q-line At the intersection of the q-line and the enriching operating line y = yn+1 or 13.5x  5 = 0.75x + 0.2425 x = 0.41118, y = 0.55088 The stripping operating line can be plotted by connecting two points (x = xB, y = xB) and (x= xq = 0.41118, y = yq = 0.55088).