initial commit / add all books

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diff --git a/389/CH6/EX6.1/Example6_1.sce b/389/CH6/EX6.1/Example6_1.sce
new file mode 100755
index 000000000..515045c6b
--- /dev/null
+++ b/389/CH6/EX6.1/Example6_1.sce
@@ -0,0 +1,55 @@
+clear;

+clc;

+

+// Illustration 6.1

+// Page: 145

+

+printf('Illustration 6.1 - Page: 145\n\n');

+

+// solution

+

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

+// w = Gas flow rate per orifice

+w = 0.055/50;// [kg/s]

+L = 8*10^(-4);// [liquid flow rate, cubic m/s]

+d = 0.003;// [diameter of the orifice,m]

+viscocity_gas = 1.8*10^(-5);// [kg/m.s]

+//******//

+

+Re = 4*w/(%pi*d*viscocity_gas);

+Dp = 0.0071*Re^(-0.05);// [m]

+h = 3;// [height of vessel,m]

+P_atm = 101.3;// [kN/square m]

+Density_water = 1000;// [kg/cubic m]

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

+Temp = 273+25;// [K]

+P_orifice = P_atm+(h*Density_water*g/1000);// [kN/square m]

+P_avg = P_atm+((h/2)*Density_water*g/1000);// [kN/square m]

+Density_gas = (29/22.41)*(273/Temp)*(P_avg/P_atm);// [kg/cubic m]

+D = 1;// [dia of vessel,m]

+Area = (%pi*D^2)/4;// [square m]

+Vg = 0.055/(Area*Density_gas);// [m/s]

+Vl = L/Area;// [m/s]

+sigma = 0.072;// [N/m]

+// From fig. 6.2 (Pg 143)

+abscissa = 0.0516;// [m/s]

+Vg_by_Vs = 0.11;

+Vs = Vg/Vg_by_Vs;// [m/s]

+deff('[y] = f6(shi_g)','y = Vs-(Vg/shi_g)+(Vl/(1-shi_g))');

+shi_g = fsolve(0.5,f6);

+dp = ((Dp^3)*(P_orifice/P_avg))^(1/3);// [bubble diameter,m]

+// From eqn. 6.9

+a = 6*shi_g/dp;// [specific interfacial area,square m]

+printf("The Specific Interfacial Area is %f square m/cubic m\n",a);

+

+// For diffsion of Cl2 in H20

+Dl = 1.44*10^(-9);// [square m/s]

+viscocity_water = 8.937*10^(-4);// [kg/m.s]

+Reg = dp*Vs*Density_water/viscocity_water;

+Scl = viscocity_water/(Density_water*Dl);

+// From Eqn.6.11

+Shl = 2+(0.0187*(Reg^0.779)*(Scl^0.546)*(dp*(g^(1/3))/(Dl^(2/3)))^0.116);

+// For dilute soln. of Cl2 in H20

+c = 1000/18.02;// [kmol/cubic m]

+Fl = (c*Dl*Shl)/dp;// [kmol/square m.s]

+printf("Mass Transfer coeffecient is %f kmol/square m.s\n",Fl);
\ No newline at end of file
diff --git a/389/CH6/EX6.2/Example6_2.sce b/389/CH6/EX6.2/Example6_2.sce
new file mode 100755
index 000000000..70e02ecd3
--- /dev/null
+++ b/389/CH6/EX6.2/Example6_2.sce
@@ -0,0 +1,86 @@
+clear;

+clc;

+

+// Illustration 6.2

+// Page: 157

+

+printf('Illustration 6.2 - Page: 157\n\n');

+

+// solution

+

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

+// a = N2 b = H2O

+L = 9.5*10^(-4);// [cubic m/s]

+G = 0.061;// [kg/s]

+Temp = 273+25;// [K]

+//*****//

+

+printf("Construction Arrangement\n");

+printf("Use 4 vertical wall baffles, 100 mm wide at 90 degree intervals.\n");

+printf("Use a 305 mm dameter, a six bladed disk flat blade turbine impeller, arranged axially, 300 mm from the bottom of vessel\n");

+printf("The sparger underneath the impeller will be in the form of a 240 mm dameter ring made of 12.7 mm tubing drilled in the top with 3.18 mm dia holes\n");

+Di = 0.305;// [m]

+Do = 0.00316;// [m]

+viscocity_a = 1.8*10^(-5);// [kg/m.s]

+Re_g = 35000;

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

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

+// w = Gas flow rate per orifice

+w = Re_g*%pi*Do*viscocity_a/4;// [kg/s]

+N_holes = G/w;

+Interval = %pi*240/round(N_holes);

+printf("The number of holes is %d at approx %d mm interval around the sparger ring\n",round(N_holes),round(Interval));

+

+viscocity_b = 8.9*10^(-4);// [kg/m.s]

+Sigma = 0.072;// [N/m]

+Density_b = 1000;// [kg/cubic m]

+D = 1;// [dia of vessel,m]

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

+// From Eqn. 6.18

+deff('[y] = f7(N)','y = (N*Di/(Sigma*g/Density_b)^0.25)-1.22-(1.25*D/Di)');

+N_min = fsolve(2,f7);// [r/s]

+N = 5;// [r/s]

+Re_l = ((Di^2)*N*Density_b/viscocity_b);

+// From fig 6.5 (Pg 152)

+Po = 5;

+P = Po*Density_b*(N^3)*(Di^5);

+h = 0.7;// [m]

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

+P_gas = P_atm+(h*Density_b*g/1000);// [kN/square m]

+Qg = (G/Ma)*22.41*(Temp/273)*(P_atm/P_gas);// [cubic m/s]

+// From Fig.6.7 (Pg 155)

+abcissa = Qg/(N*(Di^3));

+// abcissa is off scale

+Pg_by_P = 0.43;

+Pg = 0.43*P;// [W]

+Vg = Qg/(%pi*(D^2)/4);// [superficial gas velocity,m/s]

+check_value = (Re_l^0.7)*((N*Di/Vg)^0.3);

+vl = %pi*(D^2)/4;// [cubic m]

+// Since value<30000

+// From Eqn. 6.21, Eqn.6.23 & Eqn. 6.24

+K = 2.25;

+m = 0.4;

+Vt = 0.250;// [m/s]

+shi = 1;

+err = 1;

+while (err>10^(-3))

+    a = 1.44*((Pg/vl)^0.4)*((Density_b/(Sigma^3))^0.2)*((Vg/Vt)^0.5);// [square m/cubic m]

+    shin = (0.24*K*((viscocity_a/viscocity_b)^0.25)*((Vg/Vt)^0.5))^(1/(1-m));

+    Dp = K*((vl/Pg)^0.4)*((Sigma^3/Density_b)^0.2)*(shin^m)*((viscocity_a/viscocity_b)^0.25);// [m]

+    err = abs(shi-shin);

+    Vt = Vt-0.002;// [m/s]

+    shi = shin;

+end

+

+// For N2 in H2

+Dl = 1.9*10^(-9);// [square m/s]

+Ra = 1.514*10^(6);

+// By Eqn. 6.25

+Shl = 2.0+(0.31*(Ra^(1/3)));

+// For dilute soln.

+c = 1000/Mb;// [kmol/cubic m]

+Fl = Shl*c*Dl/Dp;// [kmol/square m.s]

+printf("The average gas-bubble diameter is %e m\n",Dp);

+printf("Gas Holdup:%f\n",shi);

+printf("Interfacial area:%e square m/cubic m \n",a);

+printf("Mass transfer coeffecient:%e kmol/square m.s\n",Fl);
\ No newline at end of file
diff --git a/389/CH6/EX6.3/Example6_3.sce b/389/CH6/EX6.3/Example6_3.sce
new file mode 100755
index 000000000..f18c23bdf
--- /dev/null
+++ b/389/CH6/EX6.3/Example6_3.sce
@@ -0,0 +1,169 @@
+clear;

+clc;

+

+// Illustration 6.3

+// Page: 174

+

+printf('Illustration 6.3 - Page: 174\n\n');

+

+// solution

+

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

+// a = methanol b = water

+G = 0.100;// [kmol/s]

+L = 0.25;// [kmol/s]

+Temp = 273+95;// [K]

+XaG = 0.18;// [mol % in gas phase]

+MaL = 0.15;// [mass % in liquid phase]

+//*****//

+

+Ma = 32;// [kg/kmol]

+Mb = 18;// [kg/kmol]

+Mavg_G = XaG*Ma+((1-XaG)*Mb);// [kg/kmol]

+Density_G = (Mavg_G/22.41)*(273/Temp);// [kg/cubic cm]

+Q = G*22.41*(Temp/273);// [cubic cm/s]

+Density_L = 961;// [kg/cubic cm]

+Mavg_L = 1/((MaL/Ma)+(1-MaL)/Mb);// [kg/kmol]

+q = L*Mavg_L/Density_L;

+

+// Perforations

+printf("Perforations\n");

+printf("Do = 4.5mm on an equilateral triangle pitch 12 mm between the hole centres, punched in sheet metal 2 mm thick\n");

+Do = 0.0045;// [m]

+pitch = 0.012;// [m]

+// By Eqn.6.31

+Ao_by_Aa = 0.907*(Do/pitch)^2;

+printf("The ratio of Hole Area By Active Area is:%f\n",Ao_by_Aa);

+printf("\n");

+

+// Tower Diameter

+printf("Tower Diameter\n");

+t = 0.50;// [tray spacing,m]

+printf("Tower Spacing:%f m\n",t);

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

+abcissa = (q/Q)*(Density_L/Density_G)^0.5;

+// From Table 6.2 (Pg 169)

+alpha = (0.0744*t)+0.01173;

+beeta = (0.0304*t)+0.015;

+if (abcissa<0.1) 

+    abcissa = 0.1;

+end

+sigma = 0.040;// [N/m]

+// 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]

+// Using 80% of flooding velocity

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

+An = Q/V;// [square m]

+// The tray area used by one downspout = 8.8%

+At = An/(1-0.088);// [square m]

+D = (4*At/%pi)^(1/2);// [m]

+// Take D = 1.25 m

+D = 1.25; //[m]

+At = %pi*(D^2)/4;// [corrected At, square m]

+W = 0.7*D;// [weir length,m]

+Ad = 0.088*At;// [square m]

+// For a design similar to Fig 6.14 (Pg 168)

+// A 40 mm wide supporting ring, beams between downspouts and a 50 mm wide disengaging & distributing zones these areas total 0.222 square m

+Aa = At-(2*Ad)-0.222;

+printf("Weir Length:%f\n",W);

+printf("Area for perforated sheet: %f square m\n",Aa);

+printf("\n");

+

+// Weir crest h1 & Weir height hw

+printf("Weir crest h1 & Weir height hw\n")

+h1 = 0.025;// [m]

+h1_by_D = h1/D;

+D_by_W = D/W;

+// From Eqn. 6.34

+Weff_by_W = sqrt(((D_by_W)^2)-((((D_by_W)^2-1)^0.5)+(2*h1_by_D*D_by_W))^2);

+// Set hw to 50 mm

+hw = 0.05;// [m]

+printf("Weir crest: %f m\n",h1);

+printf("Weir height: %f m\n",hw);

+printf("\n");

+

+// Dry Pressure Drop

+printf("Dry Pressure Drop\n");

+l = 0.002;// [m]

+// From Eqn. 6.37

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

+Ao = 0.1275*Aa;// [square m]

+Vo = Q/Ao;// [m/sec]

+viscocity_G = 1.25*10^(-5);// [kg/m.s]

+Re = Do*Vo*Density_G/viscocity_G;

+// From "The Chemical Engineers Handbook," 5th Edition fig 5.26

+fr = 0.008;

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

+// From Eqn. 6.36

+deff('[y] = f(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,f);

+printf("Dry Pressure Drop:%f m\n",hd);

+printf("\n");

+

+// Hydraulic head hl

+printf("Hydraulic head hl");

+Va = Q/Aa;// [m/s]

+z = (D+W)/2;// [m]

+// 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]

+printf("Hydraulic head: %f m\n",hl);

+printf("\n");

+

+//Residual Pressure drop hr

+printf("Residual Pressure drop hr\n");

+// From Eqn. 6.42

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

+printf("Residual Pressure Drop:%e m\n",hr);

+printf("\n");

+

+// Total Gas pressure Drop hg

+printf("Total Gas pressure Drop hg\n")

+// From Eqn. 6.35

+hg = hd+hl+hr;// [m]

+printf("Total gas pressure Drop: %f m\n",hg);

+printf("\n");

+

+// Pressure loss at liquid entrance h2

+printf("Pressure loss at liquid entrance h2\n");

+// Al: Area for the liquid flow under the apron

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

+Ada = min(Al,Ad);

+// From Eqn. 6.43

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

+printf("Pressure loss at liquid entrance:%e m\n",h2);

+printf("\n");

+

+// Backup in Downspout h3

+printf("Backup in Downspout h3\n");

+// From Eqn.6.44

+h3 = hg+h2;

+printf("Backup in Downspout:%f m\n",h3);

+printf("\n");

+

+// Check on Flooding

+printf("Check on Flooding\n");

+if((hw+h1+h3)<(t/2))

+    printf("Choosen Tower spacing is satisfactory\n");

+else

+    printf("Choosen Tower spacing is  not satisfactory\n")

+end

+printf("\n");

+

+// Weeping Velocity

+printf("Weeping Velocity\n");

+printf("For W/D ratio %f weir is set at %f m from the center from the tower\n",W/D,0.3296*D);

+Z = 2*(0.3296*D);// [m]

+// From Eqn.6.46

+deff('[y] = f8(Vow)','y = (Vow*viscocity_G/(sigma))-(0.0229*((viscocity_G^2/(sigma*Density_G*Do))*(Density_L/Density_G))^0.379)*((l/Do)^0.293)*(2*Aa*Do/(sqrt(3)*(pitch^3)))^(2.8/((Z/Do)^0.724))');

+Vow = fsolve(0.1,f8);// [m/s]

+printf("The minimum gas velocity through the holes below which excessive weeping is likely: %f m/s\n",Vow);

+printf("\n");

+

+// Entrainment

+printf("Entrainment\n");

+V_by_Vf = V/Vf;

+// From Fig.6.17 (Pg 173), V/Vf = 0.8 & abcissa = 0.0622

+E = 0.05;

+printf("Entrainment:%f\n",E);
\ No newline at end of file
diff --git a/389/CH6/EX6.4/Example6_4.sce b/389/CH6/EX6.4/Example6_4.sce
new file mode 100755
index 000000000..e10305aca
--- /dev/null
+++ b/389/CH6/EX6.4/Example6_4.sce
@@ -0,0 +1,53 @@
+clear;

+clc;

+

+// Illustration 6.4

+// Page: 183

+

+printf('Illustration 6.4 - Page: 183\n\n');

+

+// solution

+

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

+//From Illustrtion 6.3:

+G = 0.100;// [kmol/s]

+Density_G = 0.679;// [kg/cubic m]

+q = 5*10^(-3);// [cubic m/s]

+Va = 3.827;// [m/s]

+z = 1.063;// [m]

+L = 0.25;// [kmol/s]

+hL = 0.0106;// [m]

+hW = 0.05;// [m]

+Z = 0.824;// [m]

+E = 0.05;

+ya = 0.18;// [mole fraction methanol]

+

+// a:CH3OH b:H2O

+Ma = 32;// [kg/kmol]

+Mb = 18;// [kg/kmol]

+// From Chapter 2:

+ScG = 0.865;

+Dl = 5.94*10^(-9);// [square m/s]

+// From Eqn. 6.61:

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

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

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

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

+// For 15 mass% methanol:

+xa = (15/Ma)/((15/Ma)+(85/Mb));

+// From Fig 6.23 (Pg 184)

+mAC = -(NtL*L)/(NtG*G);// [Slope of AC line]

+meqb = 2.50;// [slope of equilibrium line]

+// From Eqn. 6.52:

+NtoG = 1/((1/NtG)+(meqb*G/L)*(1/NtL));

+// From Eqn. 6.51:

+EOG = 1-exp(-NtoG);

+// From Eqn. 6.59:

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

+// From Eqn. 6.58:

+eta = (Pe/2)*((1+(4*meqb*G*EOG/(L*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)));

+printf("Effeciency of Sieve trays: %f",EMGE);
\ No newline at end of file
diff --git a/389/CH6/EX6.5/Example6_5.sce b/389/CH6/EX6.5/Example6_5.sce
new file mode 100755
index 000000000..d10ecdec1
--- /dev/null
+++ b/389/CH6/EX6.5/Example6_5.sce
@@ -0,0 +1,65 @@
+clear;

+clc;

+

+// Illustration 6.5

+// Page: 200

+

+printf('Illustration 6.5 - Page: 200\n\n');

+

+// solution

+

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

+G = 0.80;// [cubic m/s]

+P = 10^2;// [kN/square m]

+XaG = 0.07;

+Temp = 273+30;// [K]

+L = 3.8;// [kg/s]

+Density_L = 1235;// [kg/cubic m]

+viscocity_L = 2.5*10^(-3);// [kg/m.s]

+//******//

+

+// a = SO2 b = air

+

+// Solution (a) 

+

+// Since the larger flow quantities are at the bottom for an absorber, the diameter will be choosen to accomodate the bottom condition

+Mavg_G = XaG*64+((1-XaG)*29);// [kg/kmol]

+G1 = G*(273/Temp)*(P/101.33)*(1/22.41);// [kmol/s]

+G2 = G1*Mavg_G;// [kg/s]

+Density_G = G2/G;// [kg/cubic m]

+// Assuming Complete absorption of SO2

+sulphur_removed = G1*XaG*64;// [kg/s]

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

+//From Fig. 6.24, using gas pressure drop of 400 (N/square m)/m

+ordinate = 0.061;

+// For 25 mm ceramic Intalox Saddle:

+Cf = 98;// [Table 6.3 Pg 196]

+J = 1;

+G_prime = (ordinate*Density_G*(Density_L-Density_G)/(Cf*viscocity_L^0.1*J))^0.5;// [kg/square m.s]

+A = G2/G_prime;// [square m]

+D = (4*A/%pi)^0.5;// [m]

+printf("The Tower Diameter is %f m\n",D);

+

+// Solution (b)

+

+// Let

+D = 1;// [m]

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

+// The pressure drop for 8 m of irrigated packing

+delta_p = 400*8;// [N/square m]

+// For dry packing

+G_prime = (G2-sulphur_removed)/A;// [kg/square m.s]

+P = P-(delta_p/1000);// [kN/square m]

+Density_G = (29/22.41)*(273/Temp)*(P/101.33);// [kg/cubic m]

+// From Table 6.3 (Pg 196)

+Cd = 241.5;

+// From Eqn. 6.68

+delta_p_by_z = Cd*G_prime^2/Density_G;// [N/square m for 1m of packing]

+pressure_drop = delta_p+delta_p_by_z;// [N/square m]

+V = 7.5;// [m/s]

+head_loss = 1.5*V^2/2;// [N.m/kg]

+head_loss = head_loss*Density_G;// [N/square m]

+Power = (pressure_drop+head_loss)*(G2-sulphur_removed)/(Density_G*1000);// [kW]

+eta = 0.6;

+Power = Power/eta;// [kW]

+printf("The Power for the fan motor is %f kW\n",Power);
\ No newline at end of file
diff --git a/389/CH6/EX6.6/Example6_6.sce b/389/CH6/EX6.6/Example6_6.sce
new file mode 100755
index 000000000..0e38ff926
--- /dev/null
+++ b/389/CH6/EX6.6/Example6_6.sce
@@ -0,0 +1,72 @@
+clear;

+clc;

+

+// Illustration 6.6

+// Page: 204

+

+printf('Illustration 6.6 - Page: 204\n\n');

+

+// solution

+

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

+// Gas

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

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

+Pt = 107;// [kN/square m]

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

+Temp = 273+27;// [K]

+G_prime = 0.716;// [kg/square m.s]

+

+// Liquid:

+Mavg_L = 260;

+viscocity_L = 2*10^(-3);// [kg/m.s]

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

+sigma = 3*10^(-2);// [N/m]

+Dl = 4.71*10^(-10);// [square m/s]

+//******//

+

+//Gas:

+Density_G = (Mavg_G/22.41)*(Pt/101.33)*(273/Temp);// [kg/cubic m]

+ScG = viscocity_G/(Density_G*Dg);

+G = G_prime/Mavg_G;// [kmol/square m.s]

+

+// Liquid:

+L_prime = 2.71;// [kg/square m.s]

+ScL = viscocity_L/(Density_L*Dl);

+

+// Holdup:

+// From Table 6.5 (Pg 206), L_prime = 2.71 kg/square m.s

+Ds = 0.0472;// [m]

+beeta = 1.508*Ds^0.376;

+shiLsW = 5.014*10^(-5)/Ds^1.56;// [square m/cubic m]

+shiLtW = (2.32*10^(-6))*(737.5*L_prime)^beeta/(Ds^2);// [square m/cubic m]

+shiLoW = shiLtW-shiLsW;// [square m/cubic m]

+H = (1404*(L_prime^0.57)*(viscocity_L^0.13)/((Density_L^0.84)*((3.24*L_prime^0.413)-1)))*(sigma/0.073)^(0.2817-0.262*log10(L_prime));

+shiLo = shiLoW*H;// [square m/cubic m]

+shiLs = 4.23*10^(-3)*(viscocity_L^0.04)*(sigma^0.55)/((Ds^1.56)*(Density_L^0.37));// [square m/cubic m]

+shiLt = shiLo+shiLs;// [square m/cubic m]

+

+// Interfacial Area:

+// From Table 6.4 (Pg 205)

+m = 62.4;

+n = (0.0240*L_prime)-0.0996;

+p = -0.1355;

+aAW = m*((808*G_prime/(Density_G^0.5))^n)*(L_prime^p);// [square m/cubic m]

+// From Eqn. 6.73

+aA = aAW*shiLo/shiLoW;// [square m/cubic m]

+// From Table 6.3 (Pg 196)

+e = 0.75;

+// From Eqn. 6.71

+eLo = e-shiLt;

+// From Eqn. 6.70

+deff('[y] = f9(Fg)','y = ((Fg*ScG^(2/3))/G)-1.195*((Ds*G_prime)/(viscocity_G*(1-eLo)))^(-0.36)');

+Fg = fsolve(1,f9);// [kmol/square m.s]

+// From Eqn. 6.72:

+deff('[y] = f10(Kl)','y = (Kl*Ds/Dl)-(25.1*(Ds*L_prime/viscocity_L)^0.45)*ScL^0.5');

+Kl = fsolve(1,f10);// [(kmol/square m.s).(kmol/cubic m)]

+// Since the value of Kl is taken at low conc., it can be converted into Fl

+c = (Density_L/Mavg_L);// [kmol/cubic m]

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

+printf("The volumetric coeffecients are\n");

+printf("Based on Gas Phase %f kmol/cubic m.s\n",Fg*aA);

+printf("based on Liquid Phase %f kmol/cubic m.s\n",Fl*aA);
\ No newline at end of file
diff --git a/389/CH6/EX6.7/Example6_7.sce b/389/CH6/EX6.7/Example6_7.sce
new file mode 100755
index 000000000..e9e8ddfde
--- /dev/null
+++ b/389/CH6/EX6.7/Example6_7.sce
@@ -0,0 +1,64 @@
+clear;

+clc;

+

+// Illustration 6.7

+// Page: 207

+

+printf('Illustration 6.7 - Page: 207\n\n');

+

+// solution

+

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

+// Air

+G_prime = 1.10;// [kg/square m.s]

+viscocity_G = 1.8*10^(-5);// [kg/m.s]

+ScG = 0.6;// [for air water mixture]

+Temp1 = 273+20;// [K]

+

+// Water

+L_prime = 5.5;// [kg/square m.s]

+//*****//

+

+// Air:

+Ma = 29;// [kg/kmol]

+G = G_prime/Ma;// [kmol/square m.s]

+Density_G = (Ma/22.41)*(273/Temp1);

+Cpa = 1005;// [N.m/kg.K]

+PrG = 0.74;

+

+// Liquid:

+kth = 0.587;// [W/m.K]

+Cpb = 4187;// [N.m/kg.K]

+viscocity_L = 1.14*10^(-3);// [kg/m.s]

+

+// From Table 6.5 (Pg 206)

+Ds = 0.0725;// [m]

+beeta = 1.508*(Ds^0.376);

+shiLtW = (2.09*10^(-6))*(737.5*L_prime)^beeta/(Ds^2);// [square m/cubic m]

+shiLsW = 2.47*10^(-4)/(Ds^1.21);// [square m/cubic m]

+shiLoW = shiLtW-shiLsW;// [square m/cubic m]

+// From Table 6.4 (Pg 205)

+m = 34.03;

+n = 0;

+p = 0.362;

+aAW = m*(808*G_prime/Density_G^0.5)^(n)*L_prime^p;// [square m/cubic m]

+// From Eqn. 6.75

+aVW = 0.85*aAW*shiLtW/shiLoW;// [square m/cubic m]

+// From Table 6.3

+e = 0.74;

+eLo = e-shiLtW;

+// From Eqn. 6.70

+deff('[y] = f11(Fg)','y = ((Fg*ScG^(2/3))/G)-1.195*((Ds*G_prime)/(viscocity_G*(1-eLo)))^(-0.36)');

+Fg = fsolve(1,f11);// [kmol/square m.s]

+// Since the liquid is pure water. It has no mass trnsfer coeffecient.

+// For such process we need convective heat transfer coeffecient for both liquid & gas.

+// Asuming Jd = Jh

+// From Eqn. 6.70

+Jh = 1.195*((Ds*G_prime)/(viscocity_G*(1-eLo)))^(-0.36);

+Hg = Jh*Cpa*G_prime/(PrG^(2/3));// [W/square m.K]

+PrL = Cpb*viscocity_L/kth;

+// Heat transfer analog of Eqn. 6.72

+Hl = 25.1*(kth/Ds)*(Ds*L_prime/viscocity_L)^0.45*PrL^0.5;// [W/square m.K]

+printf("The volumetric coeffecients are\n");

+printf("Based on Gas Phase %f W/cubic m.K\n",Hg*aVW);

+printf("based on Liquid Phase %f W/cubic m.K\n",Hl*aVW);
\ No newline at end of file