initial commit / add all books

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diff --git a/389/CH8/EX8.1/Example8_1.sce b/389/CH8/EX8.1/Example8_1.sce
new file mode 100755
index 000000000..b6f720fe4
--- /dev/null
+++ b/389/CH8/EX8.1/Example8_1.sce
@@ -0,0 +1,42 @@
+clear;

+clc;

+

+// Illustration 8.1

+// Page: 278

+

+printf('Illustration 8.1 -  Page: 278\n\n');

+

+// solution

+

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

+P_star = 2*10^(5);// [N/square m]

+X_methane = 0.6;

+X_ethane = 0.2;

+X_propane = 0.08;

+X_nbutane = 0.06;

+X_npentane = 0.06;

+//******//

+

+MoleFraction = [0.6 0.2 0.08 0.06 0.06]

+Heading = ["Component" "Equilibrium Partial Pressure" "Vapour Pressue    " "Mole Fraction"];

+Component = ["Methane" "Ethane   " "Propane" "n-Butane" "n-Pentane"];

+VapPressure = [0 42.05 8.96 2.36 0.66];// [N/square m]

+Sum = 0;

+for i = 1:4

+    printf("%s \t",Heading(i));

+end

+printf("\n");

+for i = 1:5

+    printf("%s \t ",Component(i));

+    printf("%e \t \t \t",(MoleFraction(i)*P_star));

+    printf("%e \t \t",(VapPressure(i)*10^(5)));

+    if  (VapPressure(i) =  = 0)

+        printf("\t \n");

+        Sum = Sum+0;

+    else

+         printf("%f \n",(MoleFraction(i)*P_star)/(VapPressure(i)*10^(5)));

+         Sum = Sum+(MoleFraction(i)*P_star)/(VapPressure(i)*10^(5));

+

+end

+end

+printf("Mole Fraction Of solvent Oil is %f",1-Sum);
\ No newline at end of file
diff --git a/389/CH8/EX8.2/Example8_2.sce b/389/CH8/EX8.2/Example8_2.sce
new file mode 100755
index 000000000..95e54575f
--- /dev/null
+++ b/389/CH8/EX8.2/Example8_2.sce
@@ -0,0 +1,89 @@
+clear;

+clc;

+

+// Illustration 8.2

+// Page: 286

+

+printf('Illustration 8.2 - Page: 286\n\n');

+

+// solution

+

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

+// Absorber:

+G = 0.250;// [cubic m/s]

+Temp1 = 273+26;// [K]

+Pt = 1.07*10^(5);// [N/square m]

+y1 = 0.02;

+x2 = 0.005;

+//******//

+

+G1 = G*(273/Temp1)*(Pt/(1.0133*10^(5)))*(1/22.41);// [kmol/s]

+Y1 = y1/(1-y1);// [kmol benzene/kmol dry gas]

+Gs = G1*(1-y1);// [kmol dry gas/s]

+// For 95% removal of benzene:

+Y2 = Y1*0.05;

+X2 = x2/(1-x2);// [kmol benzene/kmol oil]

+// Vapour pressure of benzene:

+

+P_star = 13330;// [N/square m]

+X_star = zeros(20);

+Y_star = zeros(20);

+j = 0;

+for i = 0.01:0.01:0.20

+    j = j+1;

+    x = i;

+    X_star(j) = i;

+    deff('[Y] = f27(y)','Y = (y/(1+y))-(P_star/Pt)*(x/(1+x))');

+    Y_star(j) = fsolve(0,f27);

+end

+// For min flow rate:

+X1 = 0.176;// [kmolbenzene/kmol oil]

+DataMinFlow = [X2 Y2;X1 Y1];

+scf(6);

+plot(X_star,Y_star,DataMinFlow(:,1),DataMinFlow(:,2));

+minLs = (Gs*(Y1-Y2)/(X1-X2));// [kmol/s]

+// For 1.5 times the minimum:

+Ls = 1.5*minLs;// [kmol/s]

+X1_prime = (Gs*(Y1-Y2)/Ls)+X2;// [kmol benzene/kmol oil]

+DataOperLine = [X2 Y2;X1_prime Y1];

+plot(X_star,Y_star,DataMinFlow(:,1),DataMinFlow(:,2),DataOperLine(:,1),DataOperLine(:,2));

+xgrid();

+xlabel("moles of benzene / mole wash oil");

+ylabel("moles benzene / mole dry gas");

+legend("Equlibrium Line","Min Flow Rate Line","Operating Line");

+title("Absorption")

+printf("The Oil circulation rate is %e kmol/s\n",Ls);

+

+// Stripping

+Temp2 = 122+273;// [K]

+// Vapour pressure at 122 OC

+P_star = 319.9;// [kN/square m]

+Pt = 101.33;// [kN/square m]

+X_star = zeros(7);

+Y_star = zeros(7);

+j = 0;

+for i = 0:0.1:0.6

+    j = j+1;

+    x = i;

+    X_star(j) = i;

+    deff('[Y] = f28(y)','Y = (y/(1+y))-(P_star/Pt)*(x/(1+x))');

+    Y_star(j) = fsolve(0,f28);

+end

+X1 = X2;// [kmol benzene/kmol oil]

+X2 = X1_prime;// [kmol benzene/kmol oil]

+Y1 = 0;// [kmol benzene/kmol steam]

+// For min. steam rate:

+Y2 = 0.45;

+DataMinFlow = [X2 Y2;X1 Y1];

+minGs = Ls*(X2-X1)/(Y2-Y1);// [kmol steam/s]

+slopeOperat = 1.5*(Y2-Y1)/(X2-X1);

+deff('[y] = f29(x)','y = slopeOperat*(x-X1)+Y1');

+x = 0:0.01:0.14;

+scf(7);

+plot(Y_star,X_star,DataMinFlow(:,1),DataMinFlow(:,2),x,f29);

+xgrid();

+xlabel("moles of benzene / mole wash oil");

+ylabel("moles benzene / mole dry gas");

+legend("Equlibrium Line","Min Flow Rate Line","Operating Line");

+title("Stripping");

+printf("The Steam circulation rate is %e kmol/s\n",1.5*minGs);

diff --git a/389/CH8/EX8.3/Example8_3.sce b/389/CH8/EX8.3/Example8_3.sce
new file mode 100755
index 000000000..b8cd0fcfe
--- /dev/null
+++ b/389/CH8/EX8.3/Example8_3.sce
@@ -0,0 +1,60 @@
+clear;

+clc;

+

+// Illustration 8.3

+// Page: 292

+

+printf('Illustration 8.3 - Page: 292\n\n');

+

+// solution

+

+// Since tower is a tray device:

+// Following changes in notation is made:

+// L1 to LNp

+// L2 to L0

+// X1 to XNp

+// X2 to X0

+// G1 to GNpPlus1

+// G2 to G1

+// Y1 to YNpPlus1

+// Y2 to Y1

+// x1 to xNp

+// x2 to x0

+// y1 to yNpPlus1

+// y2 to y1

+// From Illustration 8.2:

+yNpPlus1 = 0.02;

+Y1 = 0.00102;

+y1 = Y1/(1+Y1);

+GNpPlus1 = 0.01075;// [kmol/s]

+x0 = 0.005;

+m = 0.125;// [m = y_star/x]

+Ls = 1.787*10^(-3);// [kmol/s]

+Gs = 0.01051;// [kmol/s]

+XNp = 0.1190;

+LNp = Ls*(1+XNp);// [kmol/s]

+ANp = LNp/(m*GNpPlus1);

+X0 = x0/(1-x0);

+L0 = Ls*(1+X0);// [kmol/s]

+G1 = Gs*(1+Y1);// [kmol/s]

+A1 = L0/(m*G1);

+A = (ANp*A1)^0.5;

+// From Eqn. 5.55:

+Np = (log((yNpPlus1-(m*x0))/(y1-(m*x0))*(1-(1/A))+(1/A)))/log(A);

+printf("Absorber\n");

+printf("From Analytical Method, no. of theoretical trays required is %f \n",Np);

+// From Fig. 8.13 (Pg292):

+Np = 7.6;

+printf("From Graphical Method, no. of theoretical trays required is %f \n",Np);

+

+// Stripper

+SNp = 1/ANp;

+S1 = 1/A1;

+// Due to relative nonconstancy of the stripping factor,graphical method should be used.

+printf("Stripper\n");

+// From Fig. 8.11 (Pg 289):

+Np = 6.7;

+printf("From Graphical Method, no. of theoretical trays required is %f \n",Np);

+// From Fig. 5.16 (Pg 129):

+Np = 6.0;

+printf("From Fig. 5.16, no. of theoretical trays required is %f \n",Np);
\ No newline at end of file
diff --git a/389/CH8/EX8.4/Example8_4.sce b/389/CH8/EX8.4/Example8_4.sce
new file mode 100755
index 000000000..3d256f295
--- /dev/null
+++ b/389/CH8/EX8.4/Example8_4.sce
@@ -0,0 +1,88 @@
+clear;

+clc;

+

+// Illustration 8.4

+// Page: 295

+

+printf('Illustration 8.4 - Page: 295\n\n');

+

+// solution

+

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

+// a = CH4 b = C5H12

+Tempg = 27;// [OC]

+Tempo = 0;// [base temp,OC]

+Templ = 35;// [OC]

+xa = 0.75;// [mole fraction of CH4 in gas]

+xb = 0.25;// [mole fraction of C5H12 in gas]

+M_Paraffin = 200;// [kg/kmol]

+hb = 1.884;// [kJ/kg K]

+//********//

+

+Ha = 35.59;// [kJ/kmol K]

+Hbv = 119.75;// [kJ/kmol K]

+Hbl = 117.53;// [kJ/kmol K]

+Lb = 27820;// [kJ/kmol]

+// M = [Temp (OC) m]

+M = [20 0.575;25 0.69;30 0.81;35 0.95;40 1.10;43 1.25];

+// Basis: Unit time

+GNpPlus1 = 1;// [kmol]

+yNpPlus1 = 0.25;// [kmol]

+HgNpPlus1 = ((1-yNpPlus1)*Ha*(Tempg-Tempo))+(yNpPlus1*(Hbv*(Tempg-Tempo)+Lb));// [kJ/kmol]

+L0 = 2;// [kmol]

+x0 = 0;// [kmol]

+HL0 = ((1-x0)*hb*M_Paraffin*(Templ-Tempo))+(x0*hb*(Templ-Tempo));// [kJ/kmol]

+C5H12_absorbed = 0.98*xb;// [kmol]

+C5H12_remained = xb-C5H12_absorbed;

+G1 = xa+C5H12_remained;// [kmol]

+y1 = C5H12_remained/G1;// [kmol]

+LNp = L0+C5H12_absorbed;// [kmol]

+xNp = C5H12_absorbed/LNp;// [kmol]

+// Assume:

+Temp1 = 35.6;// [OC]

+Hg1 = ((1-y1)*Ha*(Temp1-Tempo))+(y1*(Hbv*(Temp1-Tempo)+Lb));// [kJ/kmol]

+

+// Eqn. 8.11:

+Qt = 0;

+deff('[y] = f30(HlNp)','y = ((L0*HL0)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(G1*Hg1)+Qt)');

+HlNp = fsolve(2,f30);

+

+deff('[y] = f31(TempNp)','y = HlNp-(((1-x0)*hb*M_Paraffin*(TempNp-Tempo))+(x0*hb*(TempNp-Tempo)))');

+TempNp = fsolve(35.6,f31);

+// At Temp = TempNp:

+mNp = 1.21;

+yNp = mNp*xNp;// [kmol]

+GNp = G1/(1-yNp);// [kmol]

+HgNp = ((1-yNp)*Ha*(TempNp-Tempo))+(yNp*(Hbv*(TempNp-Tempo)+Lb));// [kJ/kmol]

+// Eqn. 8.13 with n = Np-1

+deff('[y] = f32(LNpMinus1)','y = LNpMinus1+GNpPlus1-(LNp+GNp)');

+LNpMinus1 = fsolve(2,f32);// [kmol]

+

+// Eqn. 8.14 with n = Np-1

+deff('[y] = f33(xNpMinus1)','y = ((LNpMinus1*xNpMinus1)+(GNpPlus1*yNpPlus1))-((LNp*xNp)+(GNp*yNp))');

+xNpMinus1 = fsolve(0,f33);// [kmol]

+

+// Eqn. 8.15 with n = Np-1

+deff('[y] = f34(HlNpMinus1)','y = ((LNpMinus1*HlNpMinus1)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(GNp*HgNp))');

+HlNpMinus1 = fsolve(0,f34);// [kJ/kmol]

+deff('[y] = f35(TempNpMinus1)','y = HlNpMinus1-(((1-xNpMinus1)*hb*M_Paraffin*(TempNpMinus1-Tempo))+(xNpMinus1*hb*(TempNpMinus1-Tempo)))');

+TempNpMinus1 = fsolve(42,f35);// [OC]

+

+// Thecomputation are continued upward through the tower in this manner until the gas composition falls atleast to 0.00662.

+// Results = [Tray No.(n) Tn(OC) xn yn]

+Results = [4.0 42.3 0.1091 0.1320;3 39.0 0.0521 0.0568;2 36.8 0.0184 0.01875;1 35.5 0.00463 0.00450];

+scf(8);

+plot(Results(:,1),Results(:,4));

+xgrid();

+xlabel('Tray Number');

+ylabel('mole fraction of C5H12 in gas');

+

+scf(9);

+plot(Results(:,1),Results(:,2));

+xgrid();

+xlabel('Tray Number');

+ylabel('Temparature(OC)');

+

+// For the cquired y1

+Np = 3.75;

+printf("The No. of trays will be %f",Np);
\ No newline at end of file
diff --git a/389/CH8/EX8.5/Example8_5.sce b/389/CH8/EX8.5/Example8_5.sce
new file mode 100755
index 000000000..9dd6fdff6
--- /dev/null
+++ b/389/CH8/EX8.5/Example8_5.sce
@@ -0,0 +1,129 @@
+clear;

+clc;

+

+// Illustration 8.5

+// Page: 299

+

+printf('Illustration 8.5 - Page: 299\n\n');

+

+// solution

+

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

+// a = NH3 b = H2 c = N2 w = water

+P = 2;// [bars]

+Temp = 30;// [OC]

+L = 6.38;// [kg/s]

+W = 0.53;// [weir length,m]

+pitch = 12.5/1000;// [m]

+D = 0.75;// [Tower diameter,m]

+hW = 0.060;// [weir height,m]

+t = 0.5;// [tray spacing,m]

+//*******//

+

+// From Geometry of Tray Arrangement:

+At = 0.4418;// [Tower Cross section,square m]

+Ad = 0.0403;// [Downspout Cross section,square m]

+An = At-Ad;// [square m]

+Ao = 0.0393;// [perforation area,square m]

+Z = 0.5307;// [distance between downspouts,square m]

+z = (D+W)/2;// [average flow width,m]

+h1 = 0.04;// [weir crest,m]

+// From Eqn. 6.34

+Weff = W*(sqrt(((D/W)^2)-((((D/W)^2-1)^0.5)+((2*h1/D)*(D/W)))^2));// [m]

+q = Weff*(1.839*h1^(3/2));//[cubic m/s]

+// This is a recommended rate because it produces the liquid depth on the tray to 10 cm.

+Density_L = 996;// [kg/s]

+Mw = 18.02;// [kg/kmol]

+L1 = 6.38/Mw;// [kmol/s]

+Ma = 17.03;// [kg/kmol]

+Mb = 28.02;// [kg/kmol]

+Mc = 2.02;// [kg/kmol]

+MavG = (0.03*Ma)+(0.97*(1/4)*Mb)+(0.97*(3/4)*Mc);// [kg/kmol]

+Density_G = (MavG/22.41)*(P/0.986)*(273/(273+Temp));// [kg/cubic m]

+G = 0.893;// [kg/s]

+sigma = 68*10^(-3);// [N/m]

+abcissa = (L/G)*(Density_G/Density_L)^0.5;

+// From Table 6.2 (Pg169):

+alpha = 0.04893;

+beeta = 0.0302;

+// From Eqn. 6.30

+Cf = ((alpha*log10(1/abcissa))+beeta)*(sigma/0.02)^0.2;

+// From Eqn. 6.29

+Vf = Cf*((Density_L-Density_G)/Density_G)^(1/2);// [m/s]

+// 80% of flooding value:

+V = 0.8*Vf;// [m/s]

+G = 0.8*G;// [kg/s]

+G1 = G/MavG;// [kmol/s]

+Vo = V*An/Ao;// [m/s]

+l = 0.002;// [m]

+Do = 0.00475;// [m]

+// From Eqn. 6.37

+Co = 1.09*(Do/l)^0.25;

+viscosity_G = 1.13*10^(-5);// [kg/m.s]

+Reo = Do*Vo*Density_G/viscosity_G;

+// At Reynold's No. = Reo

+fr = 0.0082;

+g = 9.81;// [m/s^2]

+// From Eqn. 6.36

+deff('[y] = f36(hD)','y = (2*hD*g*Density_L/(Vo^2*Density_G))-(Co*(0.40*(1.25-(Ao/An))+(4*l*fr/Do)+(1-(Ao/An))^2))');

+hD = fsolve(1,f36);

+// From Eqn. 6.31;

+Aa = (Ao/0.907)*(pitch/Do)^2;// [square m]

+Va = V*An/Aa;// [m/s]

+// From Eqn. 6.38

+hL = 6.10*10^(-3)+(0.725*hW)-(0.238*hW*Va*(Density_G)^0.5)+(1.225*q/z);// [m]

+// From Eqn. 6.42

+hR = 6*sigma/(Density_L*Do*g);// m

+// From Eqn. 6.35

+hG = hD+hL+hR;// [m]

+Al = 0.025*W;// [square m]

+Ada = min(Al,Ad);

+// From Eqn. 6.43

+h2 = (3/(2*g))*(q/Ada)^2;// [m]

+// From Eqn.6.44

+h3 = hG+h2;

+// since hW+h1+h3 is essentially equal to t/2, flooding will not occur

+abcissa = (L/G)*(Density_G/Density_L)^0.5;

+V_by_Vf = V/Vf;

+// From Fig.6.17, V/Vf = 0.8 & abcissa = 0.239

+E = 0.009;

+

+// At the prevailing conditions:

+Dg = 2.296*10^(-5);// [square m/s]

+viscosity_G = 1.122*10^(-5);// [kg/m.s]

+ScG = viscosity_G/(Density_G*Dg)

+Dl = 2.421*10^(-9);// [square m/s]

+

+// From Henry's Law:

+m = 0.850;

+A = L1/(m*G1);

+

+// From Eqn. 6.61:

+NtG = (0.776+(4.57*hW)-(0.238*Va*Density_G^0.5)+(104.6*q/Z))/(ScG^0.5);

+// From Eqn. 6.64:

+thetha_L = hL*z*Z/q;// [s]

+// From Eqn. 6.62:

+NtL = 40000*(Dl^0.5)*((0.213*Va*Density_G^0.5)+0.15)*thetha_L;

+// From Eqn. 6.52:

+NtoG = 1/((1/NtG)+(1/(A*NtL)));

+// From Eqn. 6.51:

+EoG = 1-exp(-NtoG);

+// From Eqn. 6.63:

+DE = ((3.93*10^(-3))+(0.0171*Va)+(3.67*q/Z)+(0.1800*hW))^2;// [square m/s]

+// From Eqn. 6.59:

+Pe = Z^2/(DE*thetha_L);

+// From Eqn. 6.58:

+eta = (Pe/2)*((1+(4*m*G1*EoG/(L1*Pe)))^0.5-1);

+// From Eqn. 6.57:

+EMG = EoG*(((1-exp(-(eta+Pe)))/((eta+Pe)*(1+(eta+Pe)/eta)))+((exp(eta)-1)/(eta*(1+(eta/(eta+Pe))))));

+// From Eqn. 6.60:

+EMGE = EMG/((1+(EMG*(E/(1-E)))));

+// From Eqn. 8.16:

+EO = log(1+EMGE*((1/A)-1))/log(1/A);

+Np = 14*EO;

+yNpPlus1 = 0.03;

+x0 = 0;

+// From Eqn. 5.54(a):

+deff('[y] = f37(y1)','y = ((yNpPlus1-y1)/(yNpPlus1-m*x0))-(((A^(Np+1))-A)/((A^(Np+1))-1))');

+y1 = fsolve(0.03,f37);

+printf("Mole Fraction Of NH3 in effluent is %e",y1);
\ No newline at end of file
diff --git a/389/CH8/EX8.6/Example8_6.sce b/389/CH8/EX8.6/Example8_6.sce
new file mode 100755
index 000000000..342322cb3
--- /dev/null
+++ b/389/CH8/EX8.6/Example8_6.sce
@@ -0,0 +1,100 @@
+clear;

+clc;

+

+// Illustration 8.6

+// Page: 304

+

+printf('Illustration 8.6 - Page: 304\n\n');

+

+// solution

+

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

+// Gas:

+// In:

+y_prime1 = 0.02;

+Y_prime1 = 0.0204;// [mol/mol dry gas]

+// Out:

+y_prime2 = 0.00102;

+Y_prime2 = 0.00102;// [mol/mol dry gas]

+// Non absorbed gas:

+MavG = 11;// [kg/kmol]

+G = 0.01051;// [kmol/s nonbenzene]

+Gm = 0.01075;// [kmol/s]

+T = 26;// [OC]

+viscosity_G = 10^(-5);// [kg/m.s]

+DaG = 1.30*10^(-5);// [square m/s]

+

+// Liquid:

+// In:

+x_prime2 = 0.005;

+X_prime2 = 0.00503;// [mol benzene/mol oil]

+// Out:

+x_prime1 = 0.1063;

+X_prime1 = 0.1190;// [mol benzene/mol oil]

+// Benzene free oil:

+MavL = 260;// [kg/kmol]

+viscosity_L = 2*10^(-3);// [kg/kmol]

+Density_L = 840;// [kg/cubic cm]

+L = 1.787*10^(-3);// [kmol/s]

+DaL = 4.77*10^(-10);// [square m/s]

+sigma = 0.03;// [N/square m]

+m = 0.1250;

+//*******//

+

+A = 0.47^2*%pi/4;// [square m]

+// At the bottom:

+L_prime1 = ((L*MavL)+(X_prime1*L*78))/A;// [kg/square m.s]

+// At the top

+L_prime2 = ((L*MavL)+(X_prime2*L*78))/A;// [kg/square m.s]

+L_primeav = (L_prime1+L_prime2)/2;// [kg/square m.s]

+// At the bottom

+G_prime1 = ((G*MavG)+(Y_prime1*G*78))/A;// [kg/square m.s]

+// At the top

+G_prime2 = ((G*MavG)+(Y_prime2*G*78))/A;// [kg/square m.s]

+G_primeav = (G_prime1+G_prime2)/2;// [kg/square m.s]

+

+// From Illustration 6.6:

+Fga = 0.0719;// [kmol/cubic cm.s]

+Fla = 0.01377;// [kmol/cubic cm.s]

+// Operating Line:

+X_prime = [0.00503 0.02 0.04 0.06 0.08 0.10 0.1190];

+x_prime = zeros(7);

+Y_prime = zeros(7);

+y_prime = zeros(7);

+for i = 1:7

+    x_prime(i) = X_prime(i)/(1+X_prime(i));

+    deff('[y] = f38(Y_prime)','y = (G*(Y_prime1-Y_prime))-(L*(X_prime1-X_prime(i)))');

+    Y_prime(i) = fsolve(Y_prime1,f38);

+    y_prime(i) = Y_prime(i)/(1+Y_prime(i));

+end

+deff("[y] = f39(x)","y = m*x")

+x = [0:0.01:0.14];

+

+// Interface compositions are determined graphically and according to Eqn. 8.21:

+yi = [0.000784 0.00285 0.00562 0.00830 0.01090 0.01337 0.01580];

+ylog = zeros(7);

+y_by_yDiffyi = zeros(7);

+for i = 1:7

+    ylog(i) = log10(yi(i));

+    y_by_yDiffyi(i) = y_prime(i)/(y_prime(i)-yi(i));

+end

+scf(10);

+plot(x_prime,y_prime,x,f39,x_prime,yi);

+legend("Operating Line","Equilibrium Line","Interface Composition");

+xgrid();

+xlabel("mole fraction of benzene in liquid");

+ylabel("mole fraction of benzene in gas");

+scf(11);

+plot(ylog,y_by_yDiffyi);

+xgrid();

+xlabel("log y");

+ylabel("y/(y-yi)");

+title("Graphical Integration Curve");

+// Area under the curve:

+Ac = 6.556;

+// Eqn. 8.28:

+NtG = (2.303*Ac)+1.152*(log10((1-y_prime2)/(1-y_prime1)));

+Gav = (Gm+(G/(1-Y_prime2)))/(2*A);// [kmol/square m.s]

+HtG = Gav/Fga;// [m]

+Z = HtG*NtG;// [m]

+printf("The depth of packing recquired is %f m",Z);
\ No newline at end of file
diff --git a/389/CH8/EX8.7/Example8_7.sce b/389/CH8/EX8.7/Example8_7.sce
new file mode 100755
index 000000000..98068010c
--- /dev/null
+++ b/389/CH8/EX8.7/Example8_7.sce
@@ -0,0 +1,92 @@
+clear;

+clc;

+

+// Illustration 8.7

+// Page: 312

+

+printf('Illustration 8.7 - Page: 312\n\n');

+

+// solution

+

+// Fom Illustration 8.6:

+y1 = 0.02;

+y2 = 0.00102;

+m = 0.125;

+x2 = 0.005;

+x1 = 0.1063;

+

+// Number of transfer units:

+// Method a:

+y1_star = m*x1;

+y2_star = m*x2;

+yDiffy_star1 = y1-y1_star;

+yDiffy_star2 = y2-y2_star;

+yDiffy_starm = (yDiffy_star1-yDiffy_star2)/log(yDiffy_star1/yDiffy_star2);

+// From Eqn. 8.48:

+NtoG = (y1-y2)/yDiffy_starm;

+printf("NtoG according to Eqn. 8.48: %f\n",NtoG);

+

+// Mehod b:

+// From Illustration 8.3:

+A = 1.424;

+NtoG = (log((((y1-(m*x2))/(y2-(m*x2)))*(1-(1/A)))+(1/A)))/(1-(1/A));

+printf("NtoG according to Eqn. 8.50: %f\n",NtoG);

+

+// Method c:

+// Operating Line:

+// From Illustration 8.3:

+X_prime = [0.00503 0.02 0.04 0.06 0.08 0.10 0.1190];

+x_prime = [0.00502 0.01961 0.0385 0.0566 0.0741 0.0909 0.1063]

+Y_prime = [0.00102 0.00357 0.00697 0.01036 0.01376 0.01714 0.0204];

+y_prime = [0.00102 0.00356 0.00692 0.01025 0.01356 0.01685 0.0200];

+deff("[y] = f2(x)","y = m*x")

+x = [0:0.01:0.14];

+scf(12);

+plot(x_prime,y_prime,x,f2);

+legend("Operating Line","Equilibrium Line",);

+xgrid();

+xlabel("mole fraction of benzene in liquid");

+ylabel("mole fraction of benzene in gas");

+// From graph:

+NtoG = 8.7;

+printf("NtoG from graph: %f\n",NtoG);

+

+// Method d:

+// from Fig 8.10:

+Y_star = [0.000625 0.00245 0.00483 0.00712 0.00935 0.01149 0.01347];

+ordinate = zeros(7);

+for i = 1:7

+    ordinate(i) = 1/(Y_prime(i)-Y_star(i));

+end

+scf(13);

+plot(Y_prime,ordinate);

+xgrid();

+xlabel("Y");

+ylabel("1/(Y-Y*)");

+title("Graphical Integration");

+// Area under the curve:

+Ac = 8.63;

+// From Eqn. 8.36:

+NtoG = Ac+(1/2)*log((1+y2)/(1+y1));

+printf("NtoG from graphical integration: %f\n",NtoG);

+

+// Height of transfer units:

+NtoG = 9.16;

+// From Illustration 6.6:

+Fga = 0.0719;// [kmol/cubic m.s]

+Fla = 0.01377;// [kmol/cubic m.s]

+Gav = 0.0609;// [kmol/square m.s]

+L = 1.787*10^(-3);// [kmol/s]

+X1 = x1/(1-x1);

+X2 = x2/(1-x2);

+Area = 0.1746;// [square m]

+Lav = L*((1+X1)+(1+X2))/(2*Area);

+// From Eqn. 8.24:

+Htg = Gav/Fga;// [m]

+// From Eqn. 8.31:

+Htl = Lav/Fla;// [m]

+// since Solutions are dilute:

+HtoG = Htg+Htl/A;// [m]

+printf("HtoG: %f m\n",HtoG);

+Z = HtoG*NtoG;// [m]

+printf("The depth of packing recquired is %f m",Z);
\ No newline at end of file
diff --git a/389/CH8/EX8.8/Example8_8.sce b/389/CH8/EX8.8/Example8_8.sce
new file mode 100755
index 000000000..bc98fbb0b
--- /dev/null
+++ b/389/CH8/EX8.8/Example8_8.sce
@@ -0,0 +1,187 @@
+clear;

+clc;

+

+// Illustration 8.8

+// Page: 317

+

+printf('Illustration 8.8 - Page: 317\n\n');

+

+// Solution

+

+//***Data***

+// a:NH3 b:air c:H2O

+ya = 0.416;// [mole fraction]

+yb = 0.584;// [mole fraction]

+G1 = 0.0339;// [kmol/square m.s]

+L1 = 0.271;// [kmol/square m.s]

+TempG1 = 20;// [OC]

+//********//

+

+// At 20 OC

+Ca = 36390;// [J/kmol]

+Cb = 29100;// [J/kmol]

+Cc = 33960;// [J/kmol]

+lambda_c = 44.24*10^6;// [J/kmol]

+// Enthalpy base = NH3 gas, H2O liquid, air at 1 std atm.

+Tempo = 20;// [OC]

+lambda_Ao = 0;// [J/kmol]

+lambda_Co = 44.24*10^6;// [J/kmol]

+

+// Gas in:

+Gb = G1*yb;// [kmol air/square m.s]

+Ya1 = ya/(1-ya);// [kmol NH3/kmol air]

+yc1 = 0;// [mole fraction]

+Yc1 = yc1/(1-yc1);// [kmol air/kmol NH]

+// By Eqn 8.58:

+Hg1 = (Cb*(TempG1-Tempo))+(Ya1*(Ca*(TempG1-Tempo))+lambda_Ao)+(Yc1*(Cc*(TempG1-Tempo)+lambda_Co));// [J/kmol air]

+

+// Liquid in:

+xa1 = 0;// [mole fraction]

+xc1 = 1;// [mole fraction]

+Hl1 = 0;// [J/kmol air]

+

+//Gas out:

+Ya2 = Ya1*(1-0.99);// [kmol NH3/kmol air]

+// Assume:

+TempG2 = 23.9;// [OC]

+yc2 = 0.0293;

+deff('[y] = f(Yc2)','y = yc2-(Yc2/(Yc2+Ya2+1))');

+Yc2 = fsolve(0.002,f);// [kmol H2O/kmol air]

+Hg2 = (Cb*(TempG2-Tempo))+(Ya2*(Ca*(TempG2-Tempo))+lambda_Ao)+(Yc2*(Cc*(TempG2-Tempo)+lambda_Co));// [J/kmol air]

+

+// Liquid out:

+Lc = L1-(Yc1*Gb);// [kmol/square m.s]

+La = Gb*(Ya1-Ya2);// [kmol/square m.s]

+L2 = La+Lc;// [kmol/square m.s]

+xa = La/L2;

+xc = Lc/L2;

+// At xa & tempo = 20 OC

+delta_Hs = -1709.6*1000;// [J/kmol soln]

+

+// Condition at the bottom of the tower:

+// Assume:

+TempL = 41.3;// {OC}

+// At(TempL+TempG1)/2:

+Cl = 75481;// [J/kmol]

+deff('[y] = f40(Cl)','y = Hl1+Hg1-((Gb*Hg2)+(L2*(Cl*(TempL-Tempo)+delta_Hs)))');

+Cl = fsolve(7,f40);// [J/kmol.K]

+

+// For the Gas:

+MavG = 24.02;// [kg/kmol]

+Density_G = 0.999;// [kg/cubic m]

+viscosity_G  =  1.517*10^(-5);// [kg/m.s]

+kG  =  0.0261;// [W/m.K]

+CpG  =  1336;// [J/kg.K]

+Dab  =  2.297*10^(-5);// [square m/s]

+Dac  =  3.084*10^(-5);// [square m/s]

+Dcb  =  2.488*10^(-5);// [square m/s]

+PrG  =  CpG*viscosity_G/kG;

+

+// For the liquid:

+MavL  =  17.97;// [kg/kmol]

+Density_L  =  953.1;// [kg/cubic m]

+viscosity_L  =  6.408*10^(-4);// [kg/m.s]

+Dal  =  3.317*10^(-9);// [square m/s]

+kl = 0.4777;// [W/m.K]

+ScL = viscosity_L/(Density_L*Dal);

+PrL = 5.72;

+sigma = 3*10^(-4);

+G_prime = G1*MavG;// [kg/square m.s]

+L_prime = L2*MavL;// [kg/square m.s]

+// From data of Chapter 6:

+Ds = 0.0472;// [m]

+a = 57.57;// [square m/cubic m]

+shiLt = 0.054;

+e = 0.75;

+// By Eqn. 6.71:

+eLo = e-shiLt;

+// By Eqn. 6.72:

+kL = (25.1*Dal/Ds)*(Ds*L_prime/viscosity_L)^0.45*ScL^0.5;// [m/s]

+c = Density_L/MavL;// [kmol/cubic m]

+Fl = kL*c;// [kmol/cubic m]

+// The heat mass transfer analogy of Eqn. 6.72:

+hL = (25.1*kl/Ds)*(Ds*L_prime/viscosity_L)^0.45*PrL^0.5;// [m/s]

+// The heat transfer analogy of Eqn. 6.69:

+hG = (1.195*G_prime*CpG/PrG^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [W/square m.K]

+// To obtain the mass transfer coeffecients:

+Ra = 1.4;

+Rc = 1-Ra;

+// From Eqn. 8.83:

+Dam = (Ra-ya)/(Ra*((yb/Dab)+((ya+yc1)/Dac))-(ya/Dac));// [square m/s]

+Dcm = (Rc-yc1)/(Rc*((yb/Dcb)+((ya+yc1)/Dac))-(yc1/Dac));// [square m/s]

+ScGa = viscosity_G/(Density_G*Dam);

+ScGc = viscosity_G/(Density_G*Dcm);

+// By Eqn. 6.69:

+FGa = (1.195*G1/ScGa^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [kmol/square m.K]

+FGc = (1.195*G1/ScGc^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [kmol/square m.K]

+Ra = Ra-0.1;

+// From Eqn. 8.80:

+scf(14);

+for i = 1:3

+    deff('[yai] = f41(xai)','yai = Ra-(Ra-ya)*((Ra-xa)/(Ra-xai))^(Fl/FGa)');

+    xai = xa:0.01:0.10;

+    plot(xai,f41)

+    Ra = Ra+0.1;

+end

+xgrid();

+xlabel("Mole fraction NH3 in the liquid, xa");

+ylabel("Mole fraction NH3 in the gas ya");

+title("Operating Line curves");

+Rc = Rc-0.1;

+// From Eqn. 8.81:

+scf(15);

+for i = 1:3

+    deff('[yci] = f42(xci)','yci = Rc-(Rc-yc1)*((Rc-xc)/(Rc-xci))^(Fl/FGc)');

+    xci = xc:-0.01:0.85;

+    plot(xci,f42)

+    Rc = Rc+0.1;

+end

+xgrid();

+xlabel("Mole fraction H2O in the liquid, xc");

+ylabel("Mole fraction H2O in the gas, yc");

+title("Operating line Curves");

+// Assume:

+Tempi = 42.7;// [OC]

+// The data of Fig. 8.2 (Pg 279) & Fig 8.4 (Pg 319) are used to draw the eqb curve of Fig 8.25 (Pg 320).

+// By interpolation of operating line curves with eqb line and the condition: xai+xci = 1;

+Ra = 1.38;

+Rc = 1-Ra;

+xai = 0.0786;

+yai = f41(xai);

+xci = 1-xai;

+yci = f42(xci);

+// From Eqn. 8.77:

+dYa_By_dZ = -(Ra*FGa*a/Gb)*log((Ra-yai)/(Ra-ya));// [kmol H2O/kmol air]

+// From Eqn. 8.78:

+dYc_By_dZ = -(Rc*FGc*a/Gb)*log((Rc-yci)/(Rc-yc1));// [kmol H2O/kmol air]

+// From Eqn. 8.82:

+hGa_prime = -(Gb*((Ca*dYa_By_dZ)+(Cc*dYc_By_dZ)))/(1-exp(Gb*((Ca*dYa_By_dZ)+(Cc*dYc_By_dZ))/(hG*a)));// [W/cubic m.K]

+// From Eqn. 8.79:

+dtG_By_dZ = -(hGa_prime*(TempG1-Tempi))/(Gb*(Cb+(Ya1*Ca)+(Yc1*Cc)));// [K/m]

+// When the curves of Fig. 8.2 (pg 279) & 8.24 (Pg 319) are interpolated for concentration xai and xci, the slopes are:

+mar = 0.771;

+mcr = 1.02;

+lambda_c = 43.33*10^6;// [J/kmol]

+// From Eqn. 8.3:

+Hai = Ca*(Tempi-Tempo)+lambda_Ao-(mar*lambda_c);// [J/kmol]

+Hci = Cc*(Tempi-Tempo)+lambda_Co-(mcr*lambda_c);// [J/kmol]

+// From Eqn. 8.76

+Tempi2 = TempL+(Gb/(hL*a))*(((Hai-Ca*(TempG1-Tempo)-lambda_Ao)*dYa_By_dZ)+((Hci-Cc*(TempG1-Tempo)-lambda_Co)*dYc_By_dZ)-((Cb+(Ya1*Ca)+(Yc1*Cc))*dtG_By_dZ));// [OC]

+// The value of Tempi obtained is sufficiently close to the value assumed earlier.

+

+deltaYa=-0.05;

+// An interval of deltaYa up the tower

+deltaZ = deltaYa/(dYa_By_dZ);// [m]

+deltaYc = (dYc_By_dZ*deltaZ);

+// At this level:

+Ya_next = Ya1+deltaYa;// [kmol/kmol air]

+Yc_next = Yc1+deltaYc;// [kmol H2O/kmol air]

+tG_next = TempG1+(dtG_By_dZ*deltaZ);// [OC]

+L_next = L1+Gb*(deltaYa+deltaYc);// [kmol/square m.s]

+xa_next = ((Gb*deltaYa)+(L1*xa))/L_next;// [mole fraction NH3]

+Hg_next = (Cb*(tG_next-Tempo))+(Ya_next*(Ca*(tG_next-Tempo))+lambda_Ao)+(Yc_next*(Cc*(tG_next-Tempo)+lambda_Co));// [J/kmol air]

+Hl_next = (L1*Hl1)+(Gb*(Hg_next-Hg2)/L_next);// [J/kmol]

+// The calculation are continued where the specified gas outlet composition are reached.

+// The packed depth is sum of all deltaZ

+Z = 1.58;// [m]

+printf("The packed depth is: %f m\n",Z);
\ No newline at end of file
diff --git a/389/CH8/EX8.9/Example8_9.sce b/389/CH8/EX8.9/Example8_9.sce
new file mode 100755
index 000000000..40345e67c
--- /dev/null
+++ b/389/CH8/EX8.9/Example8_9.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+

+// Illustration 8.9

+// Page: 327

+

+printf('Illustration 8.9 - Page: 327\n\n');

+

+// solution

+

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

+// C1=CH4 C2=C2H6 C3=n-C3H8 C4=C4H10

+Abs=0.15;// [Total absorption,kmol]

+T=25;// [OC]

+y1=0.7;// [mol fraction]

+y2=0.15;// [mol fraction]

+y3=0.10;// [mol fraction]

+y4=0.05;// [mol fraction]

+x1=0.01;// [mol fraction]

+x_involatile=0.99;// [mol fraction]

+L_by_G=3.5;// [mol liquid/mol entering gas]

+//******//

+

+LbyG_top=L_by_G/(1-y2);

+LbyG_bottom=(L_by_G+y2)/1;

+LbyG_av=(LbyG_top+LbyG_bottom)/2;

+// The number of eqb. trays is fixed by C3 absorption:

+// For C3 at 25 OC;

+m=4.10;

+A=LbyG_av/m;

+Frabs=0.7;// [Fractional absorption]

+X0=0;

+// From Eqn. 8.109:

+deff('[y]=f43(Np)','y=Frabs-((A^Np)-A)/((A^Np)-1)');

+Np=fsolve(2,f43);

+printf("Number of trays required is %f \n",Np);
\ No newline at end of file