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+clear;

+clc;

+

+// Illustration 6.4

+// Page: 342

+

+printf('Illustration 6.4 -  Page: 342\n\n');

+

+// solution

+//*****Data*****//

+T = 298; // [K]

+Fa = 200; // [feed, kmole/hr]

+zf = 0.6;

+yd = 0.95; xd = yd;

+xw = 0.05;

+q = 0.5; // [Lf/F]

+//*****//

+

+printf('Illustration 6.4(a) -  Page: 342\n\n');

+// Solution (a)

+

+// Solution of simultaneous equation 

+function[f]=F(e)

+    f(1) = Fa - e(1)-e(2);

+    f(2) = zf*Fa - yd*e(1) - xw*e(2);

+    funcprot(0);

+endfunction

+

+// Initial guess

+e = [120 70];

+y = fsolve(e,F);

+D = y(1);

+W = y(2);

+printf("Quantity of liquid and vapor products are %f kmole/h and %f kmole/h respectively\n\n",D,W);

+

+

+printf('Illustration 6.4(b) -  Page: 342\n\n');

+// Solution(b)

+// VLE data is generated in the same manner as generated in Example 6.1 by applying Raoult's law

+// VLE_data = [T,x,y]

+VLE_data = [379.4 0.1 0.21;375.5 0.2 0.37;371.7 0.3 0.51;368.4 0.4 0.64;365.1 0.5 0.71;362.6 0.6 0.79;359.8 0.7 0.86;357.7 0.8 0.91;355.3 0.9 0.96];

+// From figure 6.14

+// The minimum number of equilibrium stages is stepped off between the equilibrium curve and the 45 degree Iine, starting from the top, giving

+Nmin = 6.7;

+printf("The minimum number of theoretical stages is %f\n\n",Nmin);

+

+printf('Illustration 6.4(c) -  Page: 342\n\n');

+// Solution(c)

+// Slope of q-line = Lf/F/(1-(Lf/F))

+s = q/(1-q);

+// For minimum reflux ratio

+// From figure 6.12 y-intercept is

+i = 0.457;

+// Therefore Rmin is

+Rmin = xd/i -1;

+printf("The minimum reflux ratio is %f mole reflux/mole distillate\n\n",Rmin);

+

+printf('Illustration 6.4(d) -  Page: 343\n\n');

+// Solution(d)

+R = 1.3*Rmin;

+// The y-intercept of the rectifying-section operating line is

+ia = xd/(R+1);

+// The operating line for the stripping section is drawn to pass through the point x = y = xw = 0.05 on the 45" line and the point of intersection of the q-line   // and the rectifying-section operating line.

+// Therefore from figure 6.15 

+Nact = 13;

+// But it include boiler

+Nact1 = Nact-1;

+printf("The number of equilibrium stages for the reflux ratio specified is %f\n",Nact1);

+// For the optimal feed-stage location, the transition from one operating line to the other occurs at the first opportunity

+// after passing the operating-line intersection 

+// Therefore from figure 6.15 shows that 

+printf("The optimal location of the feed stage for the reflux ratio specified is sixth from the top\n\n");

+

+printf('Illustration 6.4(e) -  Page: 344\n\n');

+// Solution(e)

+L = R*D; // [kmole/h]

+V = L+D; // [kmole/h]

+// From equation 6.27

+Lst = L+q*Fa; // [kmole/h]

+// From equation 6.28

+Vst = V+(q-1)*Fa; // [kmole/h]

+

+// For 50% vaporization of the feed ( zf = 0.60), from calculations similar to those illustrated in Example 6.1, the separator temperature and the equilibrium     // compositions are

+Tf = 365.5; // [K]

+yf = 0.707;

+xf = 0.493;

+

+// Latent heat vaporisation data at temperature T = 298 K

+lambdaA = 33.9; // [kJ/mole]

+lambdaB = 38; // [kJ/mole]

+// Heat capacities of liquids (298-366 K)

+Cla = 0.147; // [kJ/mole.K]

+Clb = 0.174; // [kJ/mole.K]

+// Heat capacities of gases, average in the range 298 to 366 K

+Cpa = 0.094; // [kJ/mole.K]

+Cpb = 0.118; // [kJ/mole.K]

+// Substituting in equation 6.6 gives

+Hf = 0;

+Hlf = (Tf-T)*(xf*Cla+(1-xf)*Clb); // [kJ/mole of liquid feed]

+// From equation 6.7

+Hvf = (Tf-T)*(yf*Cpa+(1-yf)*Cpb) + yf*lambdaA + (1-yf)*lambdaB; // [kJ/mole of vapor feed]

+

+Lf = Fa*q; // [kmole/h]

+Vf = Fa*(1-q); // [kmole/h]

+// From equation 6.3

+Qf = (Hvf*Vf +Hlf*Lf-Fa*Hf)*1000/3600; // [kW]

+

+

+Tlo = 354.3; // [Bubble point temperature, K]

+T1 = 355.8; // [Dew point temperature, K]

+y1 = 0.95; // [composition of saturated vapor at dew point]

+x0 = 0.95; // [composition of saturated liquid at bubble point]

+Hv1 = (T1-T)*(y1*Cpa+(1-y1)*Cpb) + y1*lambdaA + (1-y1)*lambdaB; // [kJ/mole of vapor feed]

+Hlo = (Tlo-T)*(x0*Cla+(1-x0)*Clb); // [kJ/mole of liquid feed]

+

+// An energy balance around condenser

+Qc = V*(Hv1-Hlo)*1000/3600; // [kW]

+

+// A flash-vaporization calculation is done in which the fraction vaporized is known (53.8/75.4 = 0.714) and the concentration

+// of the liquid residue is fixed at xw = 0.05

+// The calculations yield

+Tr = 381.6; // [K]

+x12 = 0.093;

+y13 = 0.111;

+T12 = 379.7; // [Bubble point of the liquid entering in the reboiler, K]

+

+Hl12 = (T12-T)*(x12*Cla+(1-x12)*Clb); // [kJ/mole of liquid feed]

+Hv13 = (Tr-T)*(y13*Cpa+(1-y13)*Cpb) + y13*lambdaA + (1-y13)*lambdaB; // [kJ/mole of vapor feed]

+

+Hlw = (Tr-T)*(xw*Cla+(1-xw)*Clb); // [kJ/mole of liquid feed]

+

+// An energy balance around the reboiler 

+Qr = (Vst*Hv13+W*Hlw-Lst*Hl12)*1000/3600; // [kW]

+printf("The thermal load of the condenser, reboiler, and feed preheater are %f kW,  %f kW and %f kW respectively\n\n",Qc,Qr,Qf);
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