From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 905/CH6/EX6.1/6_1.sce   |  83 ++++++++++++++++++++++++++++++
 905/CH6/EX6.10/6_10.sce |  97 +++++++++++++++++++++++++++++++++++
 905/CH6/EX6.11/6_11.sce |  42 +++++++++++++++
 905/CH6/EX6.12/6_12.sce |  71 +++++++++++++++++++++++++
 905/CH6/EX6.13/6_13.sce |  65 +++++++++++++++++++++++
 905/CH6/EX6.14/6_14.sce |  65 +++++++++++++++++++++++
 905/CH6/EX6.2/6_2.sce   |  53 +++++++++++++++++++
 905/CH6/EX6.3/6_3.sce   |  44 ++++++++++++++++
 905/CH6/EX6.4/6_4.sce   | 134 ++++++++++++++++++++++++++++++++++++++++++++++++
 905/CH6/EX6.7/6_7.sce   |  69 +++++++++++++++++++++++++
 10 files changed, 723 insertions(+)
 create mode 100755 905/CH6/EX6.1/6_1.sce
 create mode 100755 905/CH6/EX6.10/6_10.sce
 create mode 100755 905/CH6/EX6.11/6_11.sce
 create mode 100755 905/CH6/EX6.12/6_12.sce
 create mode 100755 905/CH6/EX6.13/6_13.sce
 create mode 100755 905/CH6/EX6.14/6_14.sce
 create mode 100755 905/CH6/EX6.2/6_2.sce
 create mode 100755 905/CH6/EX6.3/6_3.sce
 create mode 100755 905/CH6/EX6.4/6_4.sce
 create mode 100755 905/CH6/EX6.7/6_7.sce

(limited to '905/CH6')

diff --git a/905/CH6/EX6.1/6_1.sce b/905/CH6/EX6.1/6_1.sce
new file mode 100755
index 000000000..8d33946a8
--- /dev/null
+++ b/905/CH6/EX6.1/6_1.sce
@@ -0,0 +1,83 @@
+clear;
+clc;
+
+// Illustration 6.1
+// Page: 324
+
+printf('Illustration 6.1 -  Page: 324\n\n');
+
+// solution
+//*****Data*****//
+//   n-heptane - a    n-octane - b
+T1 = 303; // [K]
+P = 1; // [bar]
+D = 0.6;
+W = 0.4;
+zf = 0.5;
+ 
+// Parameters for componenr 'A'
+Tc_a = 540.3; // [K]
+Pc_a = 27.4; // [bar]
+A_a = -7.675;
+B_a = 1.371;
+C_a =-3.536;
+D_a = -3.202;
+
+// Parameters for component 'B'
+Tc_b = 568.8; // [K]
+Pc_b = 24.9; // [bar]
+A_b = -7.912;
+B_b = 1.380;
+C_b = -3.804;
+D_b = -4.501;
+
+// Using equation 6.5
+// x_a = 1-(T/Tc_a);
+// P_a = Pc_a*exp((A_a*x_a+B_a*x_a^1.5+C_a*x_a^3+D_a*x_a^6)/(1-x_a)); // [bar]
+
+// x_b = 1-(T/Tc_b);
+// P_b = Pc_b*exp((A_b*x_b+B_b*x_b^1.5+C_b*x_b^3+D_b*x_b^6)/(1-x_b)); // [bar]
+
+// m_a = P_a/P;
+// m_b = P_b/P;
+
+// Solution of simultaneous equation
+function[f]=F(e)
+    f(1) = e(2) - (e(3)*Pc_a*exp(((A_a*(1-(e(1)/Tc_a))+B_a*(1-(e(1)/Tc_a))^1.5+C_a*(1-(e(1)/Tc_a))^3+D_a*(1-(e(1)/Tc_a))^6))/(1-(1-(e(1)/Tc_a)))))/P;
+    f(2) = 1-e(2) - ((1-e(3))*Pc_b*exp((A_b*(1-(e(1)/Tc_b))+B_b*(1-(e(1)/Tc_b))^1.5+C_b*(1-(e(1)/Tc_b))^3+D_b*(1-(e(1)/Tc_b))^6)/(1-(1-(e(1)/Tc_b)))))/P;
+    f(3) = (-W/D) - ((e(2)-zf)/(e(3)-zf));
+    funcprot(0);
+endfunction
+
+// Initial guess
+e = [400 0.6 0.4];
+y = fsolve(e,F);
+T = y(1); // [K] 
+Yd = y(2);
+Xw = y(3);
+
+printf("The composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage are %f, %f and %f K respectively\n\n",Yd,Xw,T);
+
+// For the capculation of the amount of heat to be added per mole of feed
+T0 = 298; // [K]
+lambdaA = 36.5; // [Latent heats of vaporization at To = 298 K ,kJ/mole]
+lambdaB = 41.4; // [Latent heats of vaporization at To = 298 K ,kJ/mole]
+CpA = 0.187; // [kJ/mole.K]
+CpB = 0.247; // [kJ/mole.K]
+CLA1 = 0.218; // [ 298-303 K,  kJ/mole.K]
+CLB1 = 0.253; // [ 298-303 K,  kJ/mole.K]
+CLA2 = 0.241; // [ 298-386 K,  kJ/mole.K]
+CLB2 = 0.268; // [ 298-386 K,  kJ/mole.K]
+// Bubble point calculated when 'D' approaches 0 and Dew point calculated when 'D' approaches 1
+Tbp = 382.2 // [Bubble point of the mixture, K]
+Tdp = 387.9 // [Dew point of mixture, K]
+
+HF = (T1-T0)*(Xw*CLA1+CLB1*(1-Xw)); // [kJ/mole]
+HW = (Tbp-T0)*(Xw*CLA2+CLB2*(1-Xw)); // [kJ/mole]
+HG = (Tdp-T0)*(Yd*CpA+(1-Yd)*CpB) + Yd*lambdaA +(1-Yd)*lambdaB; // [kJ/mole]
+
+f =1 // [feed]
+// Using equation 6.4
+deff('[y] = f14(Q)','y = W/D + (HG-(HF+Q/f))/(HW -(HF+Q/f))');
+Q = fsolve(40,f14);
+printf("The amount of heat to be added per mole of feed is %f kJ/mole\n\n",Q);
\ No newline at end of file
diff --git a/905/CH6/EX6.10/6_10.sce b/905/CH6/EX6.10/6_10.sce
new file mode 100755
index 000000000..39503f7cf
--- /dev/null
+++ b/905/CH6/EX6.10/6_10.sce
@@ -0,0 +1,97 @@
+clear;
+clc;
+
+// Illustration 6.10
+// Page: 371
+
+printf('Illustration 6.10 -  Page: 371\n\n');
+
+// solution
+//*****Data*****//
+// A-toluene   B-1,2,3-trimethyl benzene   C-benzene
+// Solution of above three are ideal 
+// Feed
+za = 0.40;
+zb = 0.30;
+zc = 0.30;
+// Bottom 
+FRAd = 0.95; // [recovery of toluene in distillate]
+FRBw = 0.95; // [recovery of 1,2,3-trimethyl benzene in the bottom]
+P = 1; // [atm]
+
+// First estimate of distillate composition 
+xc = 40/70;
+xa = 30/70;
+xb = 0;
+// The bubble point temperature for this solution is 
+Tb = 390; // [K]
+// The corresponding parameters for benzene, toluene and 1,2,3-trimethyl benzene
+// For toluene
+Tc_a = 568.8; // [K]
+Pc_a = 24.9; // [bar]
+A_a = -7.912;
+B_a = 1.380;
+C_a =-3.804;
+D_a = -4.501;
+// For 1,2,3-trimethyl benzene
+Tc_b = 664.5; // [K]
+Pc_b = 34.5; // [bar]
+A_b = -8.442;
+B_b = 2.922;
+C_b =-5.667;
+D_b = -2.281;
+// For benzene
+Tc_c = 540.3; // [K]
+Pc_c = 27.4; // [bar]
+A_c = -7.675;
+B_c = 1.371;
+C_c =-3.536;
+D_c = -3.202;
+
+
+// At the estimated reboiler temperature of 449.3 K
+Tr = 449.3; // [K]
+// P = [Toluene;1,2,3-trimethyl benzene;Benzene]
+P1 = zeros(3,6);
+// P = [Tc Pc A B C D]
+P1 = [568.8 24.9 -7.912 1.380 -3.804 -4.501;664.5 34.5 -8.442 2.922 -5.667 2.281;540.3 27.4 -7.675 1.371 -3.536 -3.202;];
+
+for i=1:3
+    P1(i) = P1(i,2)*exp((P1(i,3)*(1-Tr/P1(i,1))+P1(i,4)*(1-Tr/P1(i,1))^1.5+P1(i,5)*(1-Tr/P1(i,1))^3+P1(i,6)*(1-Tr/P1(i,1))^6)/(1-(1-Tr/P1(i,1))));
+end
+PA1 = P1(1); // [bar]
+PB1 = P1(2); // [bar]
+PC1 = P1(3); // [bar]
+alphaAB1 = PA1/PB1;
+alphaCB1 = PC1/PB1;
+
+// At the estimated distillate temperature of 390 K
+Td = 390; // [K]
+// P = [Toluene;1,2,3-trimethyl benzene;Benzene]
+P2 = zeros(3,6);
+// P = [Tc Pc A B C D]
+P2 = [568.8 24.9 -7.912 1.380 -3.804 -4.501;664.5 34.5 -8.442 2.922 -5.667 2.281;540.3 27.4 -7.675 1.371 -3.536 -3.202;];
+
+for i=1:3
+    P2(i) = P2(i,2)*exp((P2(i,3)*(1-Td/P2(i,1))+P2(i,4)*(1-Td/P2(i,1))^1.5+P2(i,5)*(1-Td/P2(i,1))^3+P2(i,6)*(1-Td/P2(i,1))^6)/(1-(1-Td/P2(i,1))));
+end
+
+PA2 = P2(1); // [bar]
+PB2 = P2(2); // [bar]
+PC2 = P2(3); // [bar]
+alphaAB2 = PA2/PB2;
+alphaCB2 = PC2/PB2;
+
+// The geometric-average relative volatilities are
+alphaAB_avg = sqrt(alphaAB1*alphaAB2);
+alphaCB_avg = sqrt(alphaCB1*alphaCB2);
+
+// From equation 6.66
+Nmin = log(FRAd*FRBw/((1-FRAd)*(1-FRBw)))/log(alphaAB_avg);
+
+// From equation 6.67
+FRCd = alphaCB_avg^Nmin/((FRBw/(1-FRBw))+alphaCB_avg^Nmin); // [fractional recovery of benzene in the distillate]
+
+printf("The number of equilibrium stages required at total reflux is %f.\n",Nmin);
+printf("The recovery fraction of benzene in the distillate is %f.\n\n",FRCd);
+printf('Thus, the assumption that virtually all of the LNK will be recovered in the distillate is justified.');
\ No newline at end of file
diff --git a/905/CH6/EX6.11/6_11.sce b/905/CH6/EX6.11/6_11.sce
new file mode 100755
index 000000000..f045cde12
--- /dev/null
+++ b/905/CH6/EX6.11/6_11.sce
@@ -0,0 +1,42 @@
+clear;
+clc;
+
+// Illustration 6.11
+// Page: 376
+
+printf('Illustration 6.11 -  Page: 376\n\n');
+
+// solution
+//*****Data*****//
+// 1-toluene   2-1,2,3--trimethylbenzene   3-benzene
+// Basis: 100 kmol/h of feed
+F = 100; // [kmole/h]
+// Since feed is saturated, therefore
+q = 0;
+// From example 6.10
+x1d = 0.3;
+x2d = 0.3;
+x3d = 0.4;
+a12 = 3.91;
+a32 = 7.77;
+a22 = 1; 
+// Equ 6.78 gives
+deff('[y] = f14(Q)','y = 1- a12*x1d/(a12-Q)-a22*x2d/(a22-Q)-a32*x3d/(a32-Q)');
+Q = fsolve(2,f14);
+
+// From the problem statement
+// d1 = D*x1d    d2 = D*x2d
+d1 = F*x1d*0.95; // [kmol/h]
+d2 = F*x2d*0.05; // [kmol/h]
+d3 = F*x3d*0.997; // [kmol/h]
+
+// Summing the three distillate, d1,d2 and d3
+D = d1+d2+d3; // [kmole/h]
+
+Vmin = a12*d1/(a12-Q)+a22*d2/(a22-Q)+a32*d3/(a32-Q);
+
+// From the mass balance 
+Lmin = Vmin-D; // [kmol/h]
+// Minimum reflux ratio
+Rmin = Lmin/D;
+printf("The minimum reflux ratio is %f\n\n",Rmin);
\ No newline at end of file
diff --git a/905/CH6/EX6.12/6_12.sce b/905/CH6/EX6.12/6_12.sce
new file mode 100755
index 000000000..48339d962
--- /dev/null
+++ b/905/CH6/EX6.12/6_12.sce
@@ -0,0 +1,71 @@
+clear;
+clc;
+
+// Illustration 6.12
+// Page: 377
+printf('Illustration 6.12 -  Page: 377\n\n');
+
+// solution
+//*****Data*****//
+// Componenets  A-propane  B-pentane  C-methane  D-ethane  E-butane  F-hexane
+// x-mole fraction  a-relative volatility
+xA = 0.25;  aA = 4.08;
+xB = 0.11;  aB = 1.00;
+xC = 0.26;  aC = 39.47;
+xD = 0.09;  aD = 10.00;
+xE = 0.17;  aE = 2.11;
+xF = 0.12;  aF = 0.50;
+// Since propane and pentane are light and heavy key respectively
+// Methane and ethane are LNK, hexane is a HNK, while butane is a “sandwich component,” meaning that it has a volatility intermediate between the keys.
+
+FRlkd = 0.98;
+FRhkd = 0.01;
+// For methane
+D_CR = (aC-1)/(aA-1)*FRlkd + (aA-aC)/(aA-1)*FRhkd;
+// For ethane
+D_DR = (aD-1)/(aA-1)*FRlkd + (aA-aD)/(aA-1)*FRhkd;
+// For butane
+D_ER = (aE-1)/(aA-1)*FRlkd + (aA-aE)/(aA-1)*FRhkd;
+// For hexane
+D_FR = (aF-1)/(aA-1)*FRlkd + (aA-aF)/(aA-1)*FRhkd;
+// Since the feed is 66% vaporized
+q = 1-0.66;
+
+// Now equation 6.82 is solved for two values of Q
+deff('[y] = f14(Q1)','y = 0.66 - aA*xA/(aA-Q1)-aB*xB/(aB-Q1)-aC*xC/(aC-Q1)-aD*xD/(aD-Q1)-aE*xE/(aE-Q1)-aF*xF/(aF-Q1)');
+Q1 = fsolve(1.2,f14);
+
+deff('[y] = f15(Q2)','y = 0.66 - aA*xA/(aA-Q2)-aB*xB/(aB-Q2)-aC*xC/(aC-Q2)-aD*xD/(aD-Q2)-aE*xE/(aE-Q2)-aF*xF/(aF-Q2)');
+Q2 = fsolve(2.5,f15);
+
+// Basis: 100 mole of feed
+F = 100; // [mole]
+// Let d1 = Dxad, d2 = Dxbd, d3 = Dxcd, and so on
+d1 = F*xA*FRlkd; // [moles of propane]
+d2 = F*xB*FRhkd; // [moles of pentane]
+d3 = F*xC; // [moles of methane]
+d4 = F*xD; // [moles of ethane]
+d6 = F*xF*0; // [moles of hexane]
+// And d5 is unknown
+// Applying equation 6,78 for each value of Q
+
+// Solution of simultaneous equation 
+function[f]=H(e)
+    f(1) = e(1) - aA*d1/(aA-Q1)-aB*d2/(aB-Q1)-aC*d3/(aC-Q1)-aD*d4/(aD-Q1)-aE*e(2)/(aE-Q1)-aF*d6/(aF-Q1);
+    f(2) = e(1) - aA*d1/(aA-Q2)-aB*d2/(aB-Q2)-aC*d3/(aC-Q2)-aD*d4/(aD-Q2)-aE*e(2)/(aE-Q2)-aF*d6/(aF-Q2);
+    funcprot(0);
+endfunction
+
+// Initial guess
+e = [90 5];
+y = fsolve(e,H);
+Vmin = y(1); // [mole]
+d5 = y(2); // [d5 = Dxed, mole]
+
+// From equ 6.84
+D = d1+d2+d3+d4+d5+d6; // [mole]
+// From mass balance 
+Lmin = Vmin-D; // [mole]
+// For minimum reflux ratio
+Rmin = Lmin/D;
+printf("The minimum reflux ratio is %f\n\n",Rmin);
\ No newline at end of file
diff --git a/905/CH6/EX6.13/6_13.sce b/905/CH6/EX6.13/6_13.sce
new file mode 100755
index 000000000..5e05bd264
--- /dev/null
+++ b/905/CH6/EX6.13/6_13.sce
@@ -0,0 +1,65 @@
+clear;
+clc;
+
+// Illustration 6.13
+// Page: 380
+printf('Illustration 6.13 -  Page: 380\n\n');
+
+// solution
+//*****Data*****//
+// A-benzene  B-toluene  C-1,2,3-trimethylbenzene
+// From example 6.10
+Nmin = 4.32; // [stages]
+// From example 6.11
+Rmin = 0.717; // [minimum reflux ratio]
+// For R = 1
+R = 1;
+X = (R-Rmin)/(R+1);
+// From equ 6.88
+Y = 1-exp((1+54.4*X)/(11+117.2*X)*(X-1)/sqrt(X));
+// Fro equ 6.86
+N = (Y+Nmin)/(1-Y);
+// From example 6.10 99.7% of the LNK (benzene) is recovered in the distillate// , 95% of the light key is in the distillate, and 95% of the heavy key is in// the bottoms
+
+// For a basis of 100 mol of feed, the material balances for three components // are
+// For distillate
+nAd = 39.88; // [LNK, moles of benzene]
+nBd = 28.5; // [LK, moles of toluene]
+nCd = 1.50; // [HK, moles of 1,2,3-trimethylbenzene]
+nTd = nAd+nBd+nCd; // [total number of moles]
+xAd = nAd/nTd;
+xBd = nBd/(nTd);
+xCd = nCd/(nTd);
+
+// For bottoms
+nAb = 0.12;
+nBb = 1.50;
+nCb = 28.50;
+nTb = nAb+nBb+nCb;
+xAb = nAb/nTb;
+xBb = nBb/nTb;
+xCb = nCb/nTb;
+
+D = nTd;
+W = nTb;
+// From problem statement
+Zlk = 0.3;
+Zhk = Zlk;
+// Substituting in equation 6.89
+// T = Nr/Ns
+T = (Zhk/Zlk*W/D*(xBb/xCd)^2)^0.206;
+
+// Solution of simultaneous equation 
+function[f]=H(e)
+    f(1) = e(1)-e(2)*T;
+    f(2) = e(1)+e(2)-N;
+        funcprot(0);
+endfunction
+
+// Initial guess
+e = [5 4];
+y = fsolve(e,H);
+Nr = y(1); // [number of stages in rectifying section]
+Ns = y(2); // [number of stages in stripping section]
+disp(Ns,Nr);
+printf('Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed.');
diff --git a/905/CH6/EX6.14/6_14.sce b/905/CH6/EX6.14/6_14.sce
new file mode 100755
index 000000000..3b8d49507
--- /dev/null
+++ b/905/CH6/EX6.14/6_14.sce
@@ -0,0 +1,65 @@
+clear;
+clc;
+
+// Illustration 6.14
+// Page: 387
+printf('Illustration 6.14 -  Page: 387\n\n');
+
+// solution
+//*****Data*****//
+// a-acetone  b-methanol  c-water
+yna = 0.2971; yn1a = 0.17; ynIa = 0.3521; mnIa = 2.759; xna = 0.1459;
+ynb = 0.4631; yn1b = 0.429; ynIb = 0.4677; mnIb = 1.225; xnb = 0.3865;
+ync = 0.2398; yn1c = 0.4010; ynIc = 0.1802; mnIc = 0.3673; xnc = 0.4676;
+
+Fabv = 4.927; // [mol/square m.s]
+Facv = 6.066; // [mol/square m.s]
+Fbcv = 7.048; // [mol/square m.s]
+aI = 50; // [square m]
+Vn1 = 188; // [mol/s]
+Vn = 194.8; // [mol/s]
+//*****//
+printf('Illustration 6.14(a) -  Page: 387\n\n');
+// Solution(a)
+
+ya = (yna+ynIa)/2;
+yb = (ynb+ynIb)/2;
+yc = (ync+ynIc)/2;
+
+Rav = ya/Facv+yb/Fabv+yc/Facv;
+Rbv = yb/Fbcv+ya/Fabv+yc/Fbcv;
+
+Rabv = -ya*(1/Fabv-1/Facv);
+Rbav = -yb*(1/Fabv-1/Fbcv);
+// Thus in matrix form
+Rv = [Rav Rabv;Rbav Rbv];
+kv = inv(Rv); // [inverse of Rv]
+// From equ 6.99
+b = [yna-ynIa;ynb-ynIb];
+J = kv*b;
+
+// From equ 6.98
+Jc = -sum(J); // [mol/square m.s]
+
+printf("The molar diffusional rates of acetone, methanol and water are %f mol/square m.s, %f mol/square m.s and %f mol/square m.s respectively.\n\n",J(1,1),J(2,1),Jc);
+
+printf('Illustration 6.14(b) -  Page: 388\n\n');
+// Solution(b)
+Ntv = Vn1-Vn; // [mol/s]
+
+// From equation 6.94
+Nta = aI*J(1,1)+ya*Ntv;
+Ntb = aI*J(2,1)+yb*Ntv;
+Ntc = aI*Jc+yc*Ntv;
+printf("The mass transfer rates of acetone, methanol and water are %f mol/s ,%f mol/s and %f mol/s respectively.\n\n",Nta,Ntb,Ntc);
+
+printf('Illustration 6.14(c) -  Page: 389\n\n');
+// Solution(c)
+
+// Approximate values of Murphree vapor tray efficiency are obtained from   // equation 6.105
+
+EMG_a = (yna-yn1a)/(mnIa*xna-yn1a);
+EMG_b = (ynb-yn1b)/(mnIb*xnb-yn1b);
+EMG_c = (ync-yn1c)/(mnIc*xnc-yn1c);
+
+printf("The Murphree vapor tray efficiencies for acetone, methanol and water are %f, %f and %f respectively.\n\n",EMG_a,EMG_b,EMG_c);
\ No newline at end of file
diff --git a/905/CH6/EX6.2/6_2.sce b/905/CH6/EX6.2/6_2.sce
new file mode 100755
index 000000000..9b730e030
--- /dev/null
+++ b/905/CH6/EX6.2/6_2.sce
@@ -0,0 +1,53 @@
+clear;
+clc;
+
+// Illustration 6.2
+// Page: 326
+
+printf('Illustration 6.2 -  Page: 326\n\n');
+
+// solution
+//*****Data*****//
+// a-benzene   b-toluene   c-orthoxylene
+T = 373; // [K]
+P = 101.3; // [kPa]
+Pa = 182.7; // [kPa]
+Pb = 73.3; // [kPa]
+Pc= 26.7; // [kPa]
+Zfa = 0.5;
+Zfb = 0.25;
+Zfc = 0.25;
+//*****//
+// Therefore
+ma = Pa/P;
+mb = Pb/P;
+mc = Pc/P;
+// Let Feed is 1 kmole
+// Therefore D+W = 1
+
+// Solution of simultaneous equation
+function[f]=F(e)
+    f(1) = e(1)+e(2)-1;
+    f(2) = e(2)/e(1) + (e(3)-Zfa)/(e(4)-Zfa);
+    f(3) = e(3)-ma*e(4);
+    f(4) = e(5)-mb*e(6);
+    f(5) = 1-e(3)-e(5) -mc*(1-e(4)-e(6));
+    f(6) = e(2)/e(1) + (e(5)-Zfb)/(e(6)-Zfb);
+    funcprot(0);
+endfunction
+
+// Initial guess
+e = [0.326 0.674 0.719 0.408 0.198 0.272];
+y = fsolve(e,F);
+D = y(1);
+W = y(2);
+Yad = y(3);
+Xaw = y(4);
+Ybd = y(5);
+Xbw = y(6);
+Ycd = 1-Yad-Ybd;
+Xcw = 1-Xaw-Xbw;
+
+printf("The amounts of liquid and vapor products are %f and %f respectively\n\n",D,W);
+printf("The vapor compositions of components A, B and C are %f, %f and %f respectively\n\n",Yad,Ybd,Ycd);
+printf("The liquid composition of components A, B and C are %f, %f and %f respectively\n\n",Xaw,Xbw,Xcw); 
\ No newline at end of file
diff --git a/905/CH6/EX6.3/6_3.sce b/905/CH6/EX6.3/6_3.sce
new file mode 100755
index 000000000..f0c585fe9
--- /dev/null
+++ b/905/CH6/EX6.3/6_3.sce
@@ -0,0 +1,44 @@
+clear;
+clc;
+
+// Illustration 6.3
+// Page: 328
+
+printf('Illustration 6.3 -  Page: 328\n\n');
+
+// solution
+//*****Data*****//
+//   n-heptane - a    n-octane - b
+P = 1; // [bar]
+
+// Basis:
+F = 100; //  [mole]
+// Therefore
+D = 60; // [mole]
+W = 40; // [mole]
+xf = 0.5;
+// Substituting in equation 6.11 yields
+// log(F/W) = Integration of dx/(y_star-x) from xw to 0.50
+
+// The equilibrium-distribution data for this system can be generated by calculating the liquid composition (x = xw) at the dew point (D = l.O).for different feed  // compositions (y_star = z).
+y_star = [0.5 0.55 0.60 0.65 0.686 0.70 0.75];
+x = [0.317 0.361 0.409 0.460 0.5 0.516 0.577];
+for i = 1:7
+    f(i) = 1/(y_star(i)-x(i));
+end
+
+area = [0.317 5.464;0.361 5.291;0.409 5.236;0.460 5.263;0.5 5.376;0.516 5.435;0.577 7.78];
+// LHS of equation 6.11
+a = log(F/W);
+
+scf(4);
+plot(area(:,1),area(:,2));
+xgrid();
+legend('area under curve');
+xlabel("x");
+ylabel("1/(y_satr-x)");
+
+// When the area becomes equal to 0.916, integration is stopped; this occurs at 
+xw = 0.33; // [mole fraction of heptane in residue]
+yd =( F*xf-W*xw)/D; // [mole fraction of heptane]
+printf("The composition of the composited distillate and the residue are %f and %f respectively\n\n",yd,xw); 
\ No newline at end of file
diff --git a/905/CH6/EX6.4/6_4.sce b/905/CH6/EX6.4/6_4.sce
new file mode 100755
index 000000000..577962c84
--- /dev/null
+++ b/905/CH6/EX6.4/6_4.sce
@@ -0,0 +1,134 @@
+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);
\ No newline at end of file
diff --git a/905/CH6/EX6.7/6_7.sce b/905/CH6/EX6.7/6_7.sce
new file mode 100755
index 000000000..566c798f5
--- /dev/null
+++ b/905/CH6/EX6.7/6_7.sce
@@ -0,0 +1,69 @@
+clear;
+clc;
+
+// Illustration 6.7
+// Page: 358
+
+printf('Illustration 6.7 -  Page: 358\n\n');
+
+// solution
+//*****Data*****//
+// a-benzene   b-toluene
+xa = 0.46;
+xb = 0.54;
+Tb = 395; // [bottom temp., K]
+Tt = 360; // [top temp., K]
+alphab = 2.26;
+alphat = 2.52;
+D = 1.53; // [diameter of column, m]
+f = 0.81; // [flooding]
+deltaP = 700; // [average gas-pressure drop, Pa/tray]
+//*****//
+
+Tavg = (Tb+Tt)/2; // [K]
+alpha_avg = (alphab+alphat)/2;
+
+printf('Illustration 6.7(a) -  Page: 359\n\n');
+// Solution(a)
+
+// Constants for components 'a' and 'b'
+Aa = 4.612;
+Ba = 148.9;
+Ca = -0.0254;
+Da = 2.222*10^-5;
+ua = exp(Aa+Ba/Tavg+Ca*Tavg+Da*Tavg^2); // [cP]
+
+Ab = -5.878;
+Bb = 1287;
+Cb = 0.00458;
+Db = -0.450*10^-5;
+
+ub = exp(Ab+Bb/Tavg+Cb*Tavg+Db*Tavg^2); // [cP]
+
+// At the average column temperature 
+ul = ua^xa*ub^xb; // [cP]
+K = alpha_avg*ul;
+// From the O’Connell correlation
+Eo = 0.52782-0.27511*log10(K) + 0.044923*(log10(K))^2;
+printf("The overall tray efficiency using the O’Connell correlation is %f.\n\n",Eo);
+
+printf('Illustration 6.7(b) -  Page: 359\n');
+// Solution(b)
+
+Nideal = 20; // [number of ideal stages]
+Nreal = Nideal/(Eo); // [nnumber of real stages]
+disp(Nreal);
+// Since real stages cannot be fractional, therefore
+Nreal = 34; 
+// From Table 4.3 tray spacing 
+t = 0.6; // [m]
+// Adding 1 m over the top tray as an entrainment separator and 3 m beneath // the bottom tray for bottoms surge capacity, the total column height is
+Z = 4+Nreal*t; // [m]
+printf("The number of real trays and the total tower height are %f and %f m respectively.\n\n",Nreal,Z);
+
+printf('Illustration 6.7(c) -  Page: 359\n\n');
+// Solution(c)
+
+// Total gas pressure drop
+deltaPc = deltaP*Nreal/1000; // [kPa]
+printf("The total gas-pressure drop through the column is %f kPa.\n\n",deltaPc); 
\ No newline at end of file
-- 
cgit