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diff --git a/905/CH7/EX7.6/7_6.sce b/905/CH7/EX7.6/7_6.sce
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+clear;
+clc;
+
+// Illustration 7.6
+// Page: 449
+
+printf('Illustration 7.6 -  Page: 449\n\n');
+
+// solution
+//*****Data*****//
+// C-styrene    A-ethylbenzene    B-diethylene glycol
+F = 1000; // [kg/h]
+XF = 0.6; // [wt fraction of styrene]
+XPE = 0.9;
+XN = 0.1;
+// All above fractions are on solvent basis
+// Equilibrium Data for Ethylbenzene (A)-Diethylene Glycol (B)-Styrene (C) at 298 K
+// Data_eqm = [X Y];
+// X - kg C/kg (A+C) in raffinate solution
+// Y - kg C/kg (A+C) in extract solution
+Data_eqm = [0 0;0.087 0.1429;0.1883 0.273;0.288 0.386;0.384 0.48;0.458 0.557;0.464 0.565;0.561 0.655;0.573 0.674;0.781 0.863;0.9 0.95;1 1];
+//*****//
+
+printf('Illustration 7.6(a) -  Page: 449\n\n');
+// Solution(a)
+
+// Minimum theoretical stages are determined on the XY equilibrium distribution diagram, stepping them off from the diagonal line to the equilibrium curve, beginning at XPE = 0.9 and ending at XN = 0.1
+
+Data_opl = [0 0;0.09 0.09;0.18 0.18;0.27 0.27;0.36 0.36;0.45 0.45;0.54 0.54;0.63 0.63;0.72 0.72;0.81 0.81;0.90 0.90;1 1;];
+
+scf(1);
+plot(Data_eqm(:,1),Data_eqm(:,2),Data_opl(:,1),Data_opl(:,2));
+xgrid();
+legend('Equilibrium line','Operating line');
+xlabel("X,kg C/kg (A+C) in raffinate solution");
+ylabel("Y,kg C/kg (A+C) in extract solution");
+
+// Figure 7.20
+Nmin = 9; // [number of ideal stages]
+
+printf("The minimum number of theoretical stages are %f.\n\n",Nmin);
+
+printf('Illustration 7.6(b) -  Page: 450\n\n');
+// Solution(b)
+
+// Since the equilibrium-distribution curve is everywhere concave downward// ,the tie line which when extended passes through F provides the minimum
+// reflux ratio
+// From figure 7.19
+NdeltaEm = 11.04;
+NE1 = 3.1;
+// From equation 7.30
+// Y = R_O/P_E, external reflux ratio
+Ymin = (NdeltaEm-NE1)/NE1; // [kg reflux/kg extract product]
+
+printf("The minimum extract reflux ratio is %f kg reflux/kg extract product.\n\n",Ymin);
+
+printf('Illustration 7.6(c) -  Page: 450\n\n');
+// Solution(c)
+
+Y = 1.5*Ymin; // [kg reflux/kg extract product]
+// From equation 7.30
+NdeltaE = Y*NE1+NE1;
+// From figure 7.19
+NdeltaR = -24.90;
+// From figure 7.21
+N = 17.5; // [number of equilibrium stages]
+
+// From figure 7.19
+// For XN = 0.1 NRN = 0.0083
+NRN = 0.0083;
+// Basis: 1 hour
+
+// e = [P_E R_N] 
+// Solution of simultaneous equation
+function[f]=G(e)
+    f(1) = F - e(1) - e(2);
+    f(2) = F*XF-e(1)*XPE-e(2)*XN;
+    funcprot(0);
+endfunction
+// Initial guess:
+e = [600 300];
+y = fsolve(e,G);
+P_E = y(1); // [kg/h]
+R_N = y(2); // [kg/h]
+
+R_O = Y*P_E; // [kg/h]
+E_1 = R_O+P_E; // [kg/h]
+
+B_E = E_1*NE1; // [kg/h]
+E1 = B_E+E_1; // [kg/h]
+RN = R_N*(1+NRN); // [kg/h]
+S = B_E+R_N*NRN; // [kg/h]
+
+printf("The number of theoretical stages are %f.\n",N);
+printf('The important flow quantities at an extract reflux ratio of 1.5 times the minimum value are\n\n');
+printf(" PE = %f kg/h\n RN = %f kg/h\n RO = %f kg/h\n E1 = %f kg/h\n BE = %f kg/h\n E1 = %f kg/h\n RN = %f kg/h\n S = %f kg/h\n",P_E,R_N,R_O,E_1,B_E,E1,RN,S);
+\ No newline at end of file