initial commit / add all books

10 files changed, 723 insertions, 0 deletions

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