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diff --git a/389/CH7/EX7.1/Example7_1.sce b/389/CH7/EX7.1/Example7_1.sce
new file mode 100755
index 000000000..43373c23a
--- /dev/null
+++ b/389/CH7/EX7.1/Example7_1.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+

+// Illustration 7.1

+// Page: 222

+

+printf('Illustration 7.1 - Page: 222\n\n');

+

+// Solution

+

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

+Temp1 = 273+26.1;// [K]

+P1 = 100;// [mm Hg]

+Temp2 = 273+60.6;// [K]

+P2 = 400;// [mm Hg]

+P = 200;// [mm Hg]

+//*****//

+

+deff('[y] = f12(T)','y = ((1/Temp1)-(1/T))/((1/Temp1)-(1/Temp2))-((log(P1)-log(P))/(log(P1)-log(P2)))');

+T = fsolve(37,f12);// [K]

+printf("At %f 0C, the vapour pressure of benzene is 200 mm Hg\n",T-273);
\ No newline at end of file
diff --git a/389/CH7/EX7.10/Example7_10.sce b/389/CH7/EX7.10/Example7_10.sce
new file mode 100755
index 000000000..1fd1a55c1
--- /dev/null
+++ b/389/CH7/EX7.10/Example7_10.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+

+// Illustration 7.10

+// Page:241

+

+printf('Illustration 7.10 - Page:241\n\n');

+

+// solution

+

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

+Tg = 60;// [OC]

+Y_prime = 0.050;// [kg toulene/kg air]

+//*****//

+

+// Wet Bulb temparature

+Dab = 0.92*10^(-5);// [square m/s]

+density_air = 1.060;// [kg/cubic cm];

+viscocity_G = 1.95*10^(-5);// [kg/m.s]

+Sc = viscocity_G/(density_air*Dab);

+// From Eqn. 7.28

+hG_by_kY = 1223*(Sc^0.567);// [J/kg.K]

+// Soln. of Eqn. 7.26 by trial & error method:

+// (Tg-Tw) = (Yas_prime-Y_prime)*(lambda_w/hG_by_kY)

+Tw = 31.8;// [OC]

+printf("Wet Bulb Temparature:%f OC\n",Tw);

+

+// Adiabatic Saturation Temparature

+C_air = 1005;// [J/kg.K]

+C_toulene = 1256;// [J/kg.K]

+Cs = C_air+(C_toulene*Y_prime);// [J/kg.K]

+// Soln. of Eqn. 7.21 by trial & error method:

+// (Tg-Tas) = (Yas_prime-Y_prime)*(lambda_as/Cs)

+Tas = 25.7;// [OC]

+printf("Adiabatic Saturation Temparature: %f OC\n",Tas);
\ No newline at end of file
diff --git a/389/CH7/EX7.11/Example7_11.sce b/389/CH7/EX7.11/Example7_11.sce
new file mode 100755
index 000000000..bdb7bba7f
--- /dev/null
+++ b/389/CH7/EX7.11/Example7_11.sce
@@ -0,0 +1,79 @@
+clear;

+clc;

+

+// Illustration 7.11

+// Page: 249

+

+printf('Illustration 7.11 - Page: 249\n\n');

+

+// solution

+

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

+L_min = 2.27;// [kg/square m.s]

+G_min = 2;// [kg/square m.s]

+L2_prime = 15;// [kg/s]

+Q = 270;// [W]

+Templ2 = 45;// [OC]

+Tempg1 = 30;// [OC]

+Tempw1 = 24;// [OC]

+Kya = 0.90;// [kg/cubic m.s]

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

+

+H1_prime = 72;// [kJ/kg dry air]

+Y1_prime = 0.0160;// [kg water/kg dry air]

+Templ1 = 29;// [OC]

+Cal = 4.187;// [kJ/kg]

+// Equilibrium Data:

+// Data  = [Temp.(OC),H_star(kJ/kg)]

+Data_star = [29 100;32.5 114;35 129.8;37.5 147;40 166.8;42.5 191;45 216];

+// The operating line for least slope:

+H2_star = 209.5;// [kJ/kg]

+Data_minSlope = [Templ1 H1_prime;Templ2 H2_star];

+deff('[y] = f14(Gmin)','y = ((L2_prime*Cal)/Gmin)-((H2_star-H1_prime)/(Templ2-Templ1))');

+Gmin = fsolve(2,f14);// [kg/s]

+Gs = 1.5*Gmin;// [kg/s]

+// For the Operating Line:

+y = deff('[y] = f15(H2)','y = ((H2-H1_prime)/(Templ2-Templ1))-((L2_prime*Cal)/Gs)');

+H2 = fsolve(2,f15);// [kJ/kg dry air]

+Data_opline = [Templ1 H1_prime;Templ2 H2];

+scf(4);

+plot(Data_star(:,1),Data_star(:,2),Data_minSlope(:,1),Data_minSlope(:,2),Data_opline(:,1),Data_opline(:,2));

+xgrid();

+legend('Equilibrium line','Minimum Flow Rate Line','Operating Line');

+xlabel("Liquid Temperature, 0C");

+ylabel("Enthalphy Of Air Water vapour, kJ / kg dry air");

+// Tower cross section Area:

+Al = L2_prime/L_min;// [square m]

+Ag = Gs/G_min;// [square m]

+A = min(Al,Ag);// [square m]

+// Data from operating line:

+// Data1 = [Temp.(OC),H_prime(kJ/kg)]

+Data1 = [29 72;32.5 92;35 106.5;37.5 121;40 135.5;42.5 149.5;45 163.5];

+// Driving Force:

+Data2 = zeros(7,2);

+// Data2 = [Temp[OC],driving Force]

+for i = 1:7

+    Data2(i,1) = Data1(i,1);

+    Data2(i,2) = 10^2/(Data_star(i,2)-Data1(i,2));

+end

+// The data for operating line as abcissa is plotted against driving force;

+Area = 3.25;

+// From Eqn. 7.54

+deff('[y] = f16(Z)','y = Area-(Kya*Z/G_min)');

+Z = fsolve(2,f16);

+printf("The height of tower is %f m\n",Z);

+NtoG = 3.25;

+HtoG = G_min/Kya;// [m]

+

+// Make up water

+// Assuming the outlet air is essentially saturated:

+Y2_prime = 0.0475;// [kg water/kg dry air]

+E = G_min*(A)*(Y2_prime-Y1_prime);// [kg/s]

+// Windage loss estimated as 0.2 percent

+W = 0.002*L2_prime;// [kg/s]

+ppm_blowdown = 2000;// [ppm]

+ppm_makeup = 500;// [ppm]

+// Since the weight fraction are proportional to the corresponding ppm values:

+B = (E*ppm_makeup/(ppm_blowdown-ppm_makeup))-W;// [kg/s]

+M = B+E+W;// [kg/s]

+printf("The makeup water is estimated to be %f kg/s\n",M);
\ No newline at end of file
diff --git a/389/CH7/EX7.12/Example7_12.sce b/389/CH7/EX7.12/Example7_12.sce
new file mode 100755
index 000000000..ca25c5bb8
--- /dev/null
+++ b/389/CH7/EX7.12/Example7_12.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+

+// Illustration 7.12

+// Page: 252

+

+printf('Illustration 7.12 - Page: 252\n\n')

+// solution

+

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

+Tempg1 = 32;// [OC]

+Tempw1 = 28;// [OC]

+//******//

+

+H1 = 90;// [kJ/kg]

+H1_prime = 72;// [kJ/kg dry air]

+H2_prime = 163.6;// [kJ/kg dry air]

+deff('y = f17(H2)','y = (H2-H1)-(H2_prime-H1_prime)');

+H2 = fsolve(2,f17);// [kJ/kg dry air]

+// Slope of Operating Line same as Operating Line as Illustration 7.11

+slopeOperat = (163.5-72)/(45-29);

+deff("[y] = f18(Temp)","y = slopeOperat*(Temp-Tempg1)+H1");

+Temp = 30:0.01:45;

+// Equilibrium Data:

+// Data  = [Temp.(OC),H_star(kJ/kg)]

+Data_star = [29 100;32.5 114;35 129.8;37.5 147;40 166.8;42.5 191;45 216];

+scf(5);

+plot(Data_star(:,1),Data_star(:,2),Temp,f18);

+xgrid();

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

+xlabel("Liquid Temperature, C");

+ylabel("Enthalphy Of Air Water vapour, kJ/kg dry air");

+// The Value for NtoG & HtoG will be same as in Illustration 7.11

+NtoG = 3.25;

+HtoG = 2.22;// [m]

+// By hit & trial method:

+Temp = 37.1;// [OC]

+printf("The Temperature to which water is to be cooled is %f OC\n",Temp);
\ No newline at end of file
diff --git a/389/CH7/EX7.13/Example7_13.sce b/389/CH7/EX7.13/Example7_13.sce
new file mode 100755
index 000000000..6e967205b
--- /dev/null
+++ b/389/CH7/EX7.13/Example7_13.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+

+// Illustration 7.13

+// Page: 254

+

+printf('Illustration 7.13\n\n');

+

+// solution

+

+// Given

+Tempg1=65;// [OC]

+Y1_prime=0.0170;// [kg water/kg dry air]

+// Using adiabatic satursion line on Fig. 7.5 (Pg 232)

+Tempas=32;// [OC]

+Yas_prime=0.0309;// [kg water/kg dry air]

+Tempg2=45;// [OC]

+Z=2;// [m]

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

+

+Y2_prime=0.0265;// [kg water/kg dry air]

+deff('[y]=f19(Kya_by_Gs)','y=log((Yas_prime-Y1_prime)/(Yas_prime-Y2_prime))-(Kya_by_Gs*Z)');

+Kya_by_Gs=fsolve(1,f19);// [1/m]

+

+// For the extended chamber:

+Z=4;// [m]

+deff('[y]=f20(Y2_prime)','y=log((Yas_prime-Y1_prime)/(Yas_prime-Y2_prime))-(Kya_by_Gs*Z)');

+Y2_prime=fsolve(0.029,f20);//[kg water/kg dry air] 

+// With the same adiabatic curve:

+Tempg2=34;// [OC]

+printf("The Outlet Conditions are:\n");

+printf("Absolute Humidity is %f kg water/kg dry air\n",Y2_prime);

+printf("Dry Bulb Temperature is %f OC\n",Tempg2);
\ No newline at end of file
diff --git a/389/CH7/EX7.14/Example7_14.sce b/389/CH7/EX7.14/Example7_14.sce
new file mode 100755
index 000000000..f3bfd7c51
--- /dev/null
+++ b/389/CH7/EX7.14/Example7_14.sce
@@ -0,0 +1,113 @@
+clear;

+clc;

+

+// Illustration 7.14

+// Page: 256

+

+printf('Illustration 7.14 - Page: 256\n\n');

+

+// solution

+

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

+// a = N2 b = CO

+// Entering gas

+Y1_prime = 0;// [kg water/kg dry air]

+Pt = 1;// [atm]

+Tempg1 = 315;// [OC]

+G_prime = 5;// [square m/s]

+

+// Temp of the tower:

+Templ2 = 18;// [OC]

+Density_L2 = 1000; //[kg/square m]

+viscocity_L2 = 1.056*10^(-3);// [kg/m.s]

+Tempg2 = 27;// [OC]

+

+Mb = 28;// [kg/kmol]

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

+Density_G1 = (Mb/22.41)*(273/(Tempg1+273));// [kg/square m]

+G1 = G_prime*(Density_G1);// [kg/s]

+

+// Since the outlet gas is nearly saturated:

+Y_prime = 0.024;// [kg water/kg dry air]

+Y2_prime = 0.022;// [kg water/kg dry air, assumed]

+G2 = G1*(1+Y2_prime);// [kg/s]

+Mav = (1+Y2_prime)/((1/Mb)+(Y2_prime/Ma));// [kg/kmol]

+Density_G2 = (Mav/22.4)*(273/(Templ2+273));// [kg/square m]

+L2_by_G2 = 2;

+abcissa = L2_by_G2*(Density_G2/(Density_L2-Density_G2))^(1/2);

+// From Fig. 6.34:

+// For a gas pressure drop of 400 N/square m/m

+ordinate = 0.073;

+// From Table 6.3:

+Cf = 65;

+J = 1;

+deff('[y] = f21(G2_prime)','y = ((G2_prime^2)*Cf*(viscocity_L2^0.1)*J/(Density_G2*(Density_L2-Density_G2)))-ordinate');

+// Tentative data:

+G2_prime = fsolve(2,f21);// [kg/square m.s]

+Area = G1/G2_prime;// [square m]

+dia = sqrt(4*Area/%pi);// [m]

+

+// Final data:

+dia = 1.50;// [m]

+Area = %pi*dia^2/4;// [square m]

+Gs_prime = G1/Area;// [kg/square m.s]

+G2_prime = G2/Area;// [kg/square m.s]

+L2_prime = L2_by_G2*G2_prime;// [kg/square m.s]

+// From Eqn. 7.29:

+deff('[y] = f22(L1_prime)','y = (L2_prime-L1_prime)-(Gs_prime*(Y2_prime-Y1_prime))');

+L1_prime = fsolve(2,f22);

+Cb = 1089;// [J/kg.K]

+Ca = 1884;// [J/kg.K]

+Cs1 = Cb+(Y1_prime*Ca);// [J/(kg dry air).K]

+Cs2 = Cb+(Y2_prime*Ca);// [J/(kg dry air).K]

+Tempo = Templ2;// [base temp.,K]

+lambda = 2.46*10^6;// [J/kg]

+CaL = 4187;// [J/kg K]

+// From Eqn. 7.31:

+deff('[y] = f23(Templ1)','y = ((L2_prime*CaL*(Templ2-Tempo))+(Gs_prime*Cs1*(Tempg1-Tempo)))-((L1_prime*CaL*(Templ1-Tempo))+(Gs_prime*(Cs2*(Tempg2-Tempo))+(Y2_prime*lambda)))');

+Templ1 = fsolve(2,f23);

+// At Templ1 = 49.2 OC

+viscocity_L = 0.557*10^(-3);// [kg/m.s]

+Density_L = 989;// [kg/square m]

+K = 0.64;// [w/m.K]

+Prl = CaL*viscocity_L/K;

+

+// For Entering Gas:

+viscocity_G1 = 0.0288*10^(-3);// [kg*/m.s]

+Dab = 0.8089*10^(-4);// [square m/s]

+ScG = viscocity_G1/(Density_G1*Dab);

+PrG = 0.74;

+

+// From Illustration 6.7:

+a = 53.1;// [square m/square m]

+Fga = 0.0736;// [kmol/square m]

+Hga = 4440;// [W/square m.K]

+Hla = 350500;// [W/square m.K]

+// At the bottom, by several trial:

+Tempi = 50.3;// [OC]

+pai = 93.9/760;// [atm]

+paG = 0;// [atm]

+// By Eqn. 7.64:

+dY_prime_by_dZ = -(Ma*Fga/Gs_prime)*log((1-(pai/Pt))/(1-(paG/Pt)));// [(kg H2O/kg dry gas)/m]

+Hg_primea = -(Gs_prime*Ca*dY_prime_by_dZ)/(1-exp((Gs_prime*Ca*dY_prime_by_dZ)/(Hga)));// [W/square m.K]

+dTempg_by_dZ = -(Hg_primea*(Tempg1-Tempi)/(Gs_prime*Cs1));// [OC/m]

+Tempi = (Templ1)+((Gs_prime*(Cs1*dTempg_by_dZ)+((Ca*(Tempg1))-(CaL*(Templ1))+(((CaL-Ca)*(Tempo))+lambda))*dY_prime_by_dZ)/((Gs_prime*CaL*dY_prime_by_dZ)-Hla));// [OC]

+// Assume:

+delta_Tempg = -30;// [OC]

+delta_Z = delta_Tempg/(dTempg_by_dZ);// [m]

+Tempg = Tempg1+delta_Tempg;// [OC]

+Y_prime = Y1_prime+(dY_prime_by_dZ)*delta_Z;// [kg H2O/kg dry gas]

+paG = Y_prime/(Y_prime+(Ma/Mb));// [atm]

+Cs = Cb+Ca*(Y_prime);// [J/(kg dry air).K]

+// Water balance, From Eqn. 7.29:

+deff('[y] = f24(L_prime)','y = (L2_prime-L_prime)-(Gs_prime*(Y_prime-Y1_prime))');

+L_prime = fsolve(2,f24);// [kg/square m.s]

+

+deff('[y] = f25(Templ)','y = ((L_prime*CaL*(Templ-Tempo))+(Gs_prime*Cs1*(Tempg1-Tempo)))-((L1_prime*CaL*(Templ1-Tempo))+(Gs_prime*(Cs*(Tempg-Tempo))+(Y_prime*lambda)))');

+Templ = fsolve(2,f25);

+// This process is repeated several times until gas temp falls to Tempg2

+// The value of Y2_prime was calculated to be 0.0222 which is sufficiently close to the assumed value.

+// Z = sum of all delta_Z

+Z = 1.54;// [m]

+printf("The diameter of tower is %f m\n",dia);

+printf("The packed height is %f m\n",Z);
\ No newline at end of file
diff --git a/389/CH7/EX7.15/Example7_15.sce b/389/CH7/EX7.15/Example7_15.sce
new file mode 100755
index 000000000..bbfab304a
--- /dev/null
+++ b/389/CH7/EX7.15/Example7_15.sce
@@ -0,0 +1,107 @@
+clear;

+clc;

+

+// Illustration 7.15

+// Page: 267

+

+printf('Illustration 7.15 - Page: 267\n\n');

+

+// solution

+

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

+w = 0.75;// [m]

+OD = 19.05/1000;// [m]

+l = 3.75;// [m]

+n = 20;

+t = 1.65/1000;// [m]

+Ws = 2.3;// [kg/s]

+Wal = 10;// [kg/s]

+Wt = 4;// [kg/s]

+Density = 800;// [kg/cubic m]

+viscocity = 0.005;// [kg/m.s]

+K = 0.1436;// [W/m.K]

+Ct = 2010;// [J/kg.K]

+Cal = 4187;// [J/kg.K]

+Y1_prime = 0.01;// [kg H2O/kg dry air]

+Y2_prime = 0.06;// [kg H2O/kg dry air]

+TempT = 95;// [OC]

+//*****//

+

+Free_area = (w-(n*OD))*l;// [square m]

+Gs_min = 2.3/Free_area;// [kg/square m.s]

+Yav_prime = (Y1_prime+Y2_prime)/2;// [kg H2O/kg dry air]

+// From Eqn. 7.86:

+ky = 0.0493*(Gs_min*(1+Yav_prime))^0.905;// [kg/square m.s.delta_Y_prime]

+// From Fig. 7.5:

+H1_prime = 56000;// [J/kg]

+Ao = 400*%pi*OD*l;// [square m]

+// Cooling water is distributed over 40 tubes & since tubes are staggered

+geta = Wal/(40*2*l);// [kg/m.s]

+geta_by_OD = geta/OD;// [kg/square m.s]

+// Assume:

+TempL = 28;// [OC]

+// From Eqn. 7.84:

+hL_prime = (982+(15.58*TempL))*(geta_by_OD^(1/3));// [W/square m.K]

+// From Eqn. 7.85:

+hL_dprime = 11360;// [W/square m.K]

+// From Fig. 7.5 (Pg 232)

+m = 5000;// [J/kg.K]

+Ky = 1/((1/ky)+(m/hL_dprime));

+ID = (OD-(2*t));// [m]

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

+Gt_prime = Wt/(n*Ai);// [kg/square m.s]

+Re = ID*Gt_prime/viscocity;

+Pr = Ct*viscocity/K;

+// From a standard correlation:

+hT = 364;// [W/square m.K]

+Dav = (ID+OD)/2;// [m]

+Zm = (OD-ID)/2;// [m]

+Km = 112.5;// [W/m.K]

+// From Eqn. 7.67:

+Uo = 1/((OD/(ID*hT))+((OD/Dav)*(Zm/Km))+(1/hL_prime));// [W/square m.K]

+// From Eqn. 7.75:

+alpha1 = -(((Uo*Ao)/(Wt*Ct))+((Uo*Ao)/(Wal*Cal)));

+alpha2 = m*Uo*Ao/(Wt*Ct);

+// From Eqn. 7.76:

+beeta1 = Ky*Ao/(Wal*Cal);

+beeta2 = -((m*Ky*Ao/(Wal*Cal))-(Ky*Ao/Ws));

+y = deff('[y] = f26(r)','y = (r^2)+((alpha1+beeta2)*r)+((alpha1*beeta2)-(alpha2*beeta1))');

+r1 = fsolve(10,f26);

+r2 = fsolve(0,f26);

+beeta2 = 1.402;

+// From Eqn. 7.83:

+// N1-(M1*(r1+alpha1)/beeta1) = 0............................................(1)

+// N2-(M2*(r2+alpha2)/beeta2) = 0............................................(2)

+// From Eqn. 7.77:

+// At the top:

+x1 = 1;

+// TempL2+(M1*exp(r1*x1))+(M2*exp(-(r2*x1))) = TempL.........................(3)

+// From Eqn. 7.78:

+// At the bottom:

+x2 = 0;

+// H1_star-N1-N2 = H1_prime..................................................(4)

+// From Eqn. 7.80:

+// ((M1/r1)*(exp(r1)-1))+((M2*r2)*(exp(r2)-1)) = (Tempt-TempL)...............(5)

+// From Eqn. 7.81:

+// ((N1/r1)*(exp(r1)-1))+((N2*r2)*(exp(r2)-1)) = (H1_star-H1_prime)..........(6)

+// From Eqn. 7.91 & Eqn. 7.92:

+// Uo*Ao*(TempT-TempL)=Ky*Ao*(H1_star-H1_prime)..............................(7)

+

+// Elimination of M's & N's by solving Eqn. (1) to (4) and (7) simultaneously:

+// and from Fig. 7.5 (Pg 232):

+TempL1=28;// [OC]

+H1_star=(Uo*Ao*(TempT-TempL)/(Ky*Ao))+H1_prime;// [J/kmol]

+// Solving (1) to (4) simultaneously:

+a = [1 -(r1+alpha1)/beeta1 0 0;0 0 1 -(r2+alpha1)/beeta1;0 exp(r1*x1) 0 exp(r2*x1);1 0 1 0];

+b = [0;0;TempT-TempL1;H1_star-H1_prime];

+soln = a\b;

+N1 = soln(1);

+M1 = soln(2);

+N2 = soln(3);

+M2 = soln(4);

+// By Eqn. 5

+delta_Temp = ((M1/r1)*(exp(r1)-1))+((M2*r2)*(exp(r2)-1));// [OC]

+Q = Uo*delta_Temp*Ao;

+TempT1 = TempT-(Q/(Wt*Ct));// [OC]

+H2_prime = Q/(Ws)+H1_prime;// [J/kg]

+printf("Temparature to which oil was cooled: %f OC\n",TempT1);
\ No newline at end of file
diff --git a/389/CH7/EX7.2/Example7_2.sce b/389/CH7/EX7.2/Example7_2.sce
new file mode 100755
index 000000000..0cc483112
--- /dev/null
+++ b/389/CH7/EX7.2/Example7_2.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+

+// Illustration 7.2:

+// Page: 223

+

+printf('Illustration 7.2 - Page: 223\n\n');

+printf('Illustration 7.2 (b)\n\n');

+

+// Solution (b)

+

+// At 100 OC,

+PH2O = 760;// [Vapour pressure of water, mm of Hg]

+// From Fig. 7.2 (Pg 224)

+// At this value,

+PC6H6 = 1400;// [Vapour pressure of benzene, mm of Hg]

+printf("Vapour Pressure of benzene at 100 OC is %d mm of Hg\n\n", PC6H6);

+

+printf('Illustration 7.2 (c)\n\n');

+

+// Solution (c)

+

+// Reference: H20

+// At 25 OC

+m = 0.775;

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

+lambdar = 2443000;// [N/m.kg]

+M = 78.05;// [kg/kmol]

+// From Eqn. 7.6:

+lambda = m*lambdar*Mr/M;// [N/m.kg]

+printf("Latent Heat of Vaporization at 25 OC is %f kN/m.kg\n",lambda/1000);
\ No newline at end of file
diff --git a/389/CH7/EX7.3/Example7_3.sce b/389/CH7/EX7.3/Example7_3.sce
new file mode 100755
index 000000000..24d34674e
--- /dev/null
+++ b/389/CH7/EX7.3/Example7_3.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+

+// Illustration 7.3

+// Page: 226

+

+printf('Illustration 7.3 - Page: 226\n\n');

+

+// solution

+

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

+m = 10;// [kg]

+Cvap = 1.256;// [kJ/kg.K]

+Cliq = 1.507;// [kJ/kg.K]

+Temp1 = 100;// [OC]

+Temp4 = 10;// [OC]

+//******//

+

+// Using Fig 7.2 (Pg 224):

+Temp2 = 25;// [OC]

+// Using the notation of Fig. 7.3:

+H1_diff_H2 = Cvap*(Temp1-Temp2);// [kJ/kg]

+// From Illustration 7.2:

+H2_diff_H3 = 434;// [Latent Heat of Vaporisation, kJ/kg]

+H3_diff_H4 = Cliq*(Temp2-Temp4);// [kJ/kg]

+H1_diff_H4 = H1_diff_H2+H2_diff_H3+H3_diff_H4;// [kJ/kg]

+H = m*H1_diff_H4;// [kJ]

+printf("Heat evolved for 10 kg Benzene is %f kJ\n",H);
\ No newline at end of file
diff --git a/389/CH7/EX7.4/Example7_4.sce b/389/CH7/EX7.4/Example7_4.sce
new file mode 100755
index 000000000..cb544325e
--- /dev/null
+++ b/389/CH7/EX7.4/Example7_4.sce
@@ -0,0 +1,43 @@
+clear;

+clc;

+

+// Illustration 7.4

+// Page: 227

+

+printf('Illustration 7.4 - Page: 227\n\n');

+

+// solution

+

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

+// A = benzene vapour; B = Nitrogen Gas

+P = 800;// [mm Hg]

+Temp = 273+60;// [K]

+pA = 100;// [mm Hg]

+//******//

+

+pB = P-pA;// [mm Hg]

+MA = 78.05;// [kg/kmol]

+MB = 28.08;// [kg/kmol]

+

+// Mole Fraction

+printf("On the Basis of Mole Fraction\n");

+yAm = pA/P;

+yBm = pB/P;

+printf("Mole Fraction of Benzene is %f\n",yAm);

+printf("Mole Fraction of Nitrogen is %f\n",yBm);

+printf("\n");

+

+// Volume Fraction

+printf("On the Basis of Volume Fraction\n");

+// Volume fraction is same as mole Fraction

+yAv = yAm;

+yBv = yBm;

+printf("Volume Fraction of Benzene is %f\n",yAv);

+printf("Volume Fraction of Nitrogen is %f\n",yBv);

+printf("\n");

+

+// Absolute Humidity

+printf("On the basis of Absolute humidity\n")

+Y = pA/pB;// [mol benzene/mol nitrogen]

+Y_prime = Y*(MA/MB);// [kg benzene/kg nitrogen]

+printf("The concentration of benzene is %f kg benzene/kg nitrogen\n",Y_prime);
\ No newline at end of file
diff --git a/389/CH7/EX7.5/Example7_5.sce b/389/CH7/EX7.5/Example7_5.sce
new file mode 100755
index 000000000..cf90effd9
--- /dev/null
+++ b/389/CH7/EX7.5/Example7_5.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+

+// Illustration 7.5

+// Page: 228

+

+printf('Illustration 7.5 - Page: 228\n\n');

+

+printf('Illustration 7.5 (a)\n\n');

+// solution(a)

+

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

+// A = benzene vapour; B = Nitrogen Gas

+P = 1;// [atm]

+//*****//

+

+MA = 78.05;// [kg/kmol]

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

+// Since gas is saturated, from Fig. 7.2 (Pg 224):

+pA = 275/760;// [atm]

+Y = pA/(P-pA);// [kmol benzene/kmol nitrogen]

+Y_prime = Y*(MA/MB);// [kg benzene/kg nitrogen]

+printf("The concentration of benzene is %f kg benzene/kg nitrogen\n\n",Y_prime);

+

+printf('Illustration 7.5 (b)\n\n');

+// solution(b)

+

+// A = benzene vapour; B = CO2

+MA = 78.05;// [kg/kmol]

+MB = 44.01;// [kg/kmol]

+// Since gas is saturated, from Fig. 7.2:

+pA = 275/760;// [atm]

+Y = pA/(P-pA);// [kmol benzene/kmol CO2]

+Y_prime = Y*(MA/MB);// [kg benzene/kg CO2]

+printf("The concentration of benzene is %f kg benzene/kg CO2\n",Y_prime);
\ No newline at end of file
diff --git a/389/CH7/EX7.6/Example7_6.sce b/389/CH7/EX7.6/Example7_6.sce
new file mode 100755
index 000000000..009a721ef
--- /dev/null
+++ b/389/CH7/EX7.6/Example7_6.sce
@@ -0,0 +1,62 @@
+clear;

+clc;

+

+// Illustration 7.6

+// Page: 234

+

+printf('Illustration 7.6 - Page: 234\n\n');

+

+// solution

+

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

+// A = water vapour; B = air

+TempG = 55;// [OC]

+P = 1.0133*10^(5);// [N/square m]

+Y_prime = 0.030;// [kg water/kg dry air]

+//******//

+

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

+MB = 28.97;// [kg/kmol]

+

+// Percent Humidity

+// From psychrometric chart, at 55 OC

+Ys_prime = 0.115;// [kg water/kg dry air]

+percent_Humidity = (Y_prime/Ys_prime)*100;

+printf("The sample has percent Humidity =  %f %%\n",percent_Humidity);

+

+// Molal Absolute Humidity

+Y = Y_prime*(MB/MA);// [kmol water/kmol dry air]

+printf("Molal Absolute Humidity of the sample is %f kmol water/kmol dry air\n",Y);

+

+// Partial Pressure

+pA = Y*P/(1+Y);// [N/square m]

+printf("The Partial Pressure Of Water is %f N/square m\n",pA);

+

+// Relative Humidity

+pa = 118*133.3;// [vapour pressure of water at 55 OC,N/square m]

+relative_Humidity = (pA/pa)*100;

+printf("The sample has relative Humidity =  %f %%\n",relative_Humidity);

+

+// Dew Point

+// From psychrometric chart,

+dew_point = 31.5;// [OC]

+printf("Dew point Of the Sample is %f Oc\n",dew_point);

+

+// Humid Volume

+// At 55 OC

+vB = 0.93;// [specific volume of dry air,cubic m/kg]

+vsB = 1.10;// [specific volume of saturated air,cubic m/kg]

+vH = vB+((vsB-vB)*(percent_Humidity/100));// [cubic m/kg]

+printf("The Humid Volume of the Sample is %f cubic m/kg\n",vH);

+

+// Humid Heat

+CB = 1005;// [J/kg.K]

+CA = 1884;// [J/kg.K]

+Cs = CB+(Y_prime*CA);// [J/kg]

+printf("The Humid Heat is %f J/kg dry air.K\n",Cs);

+

+// Enthalpy

+HA = 56000;// [J/kg dry air]

+HsA = 352000;// [J/kg dry air]

+H_prime = HA+((HsA-HA)*(percent_Humidity/100));// [J/kg dry air]

+printf("The Enthalphy of the sample is %f J/kg dry air\n",H_prime);
\ No newline at end of file
diff --git a/389/CH7/EX7.7/Example7_7.sce b/389/CH7/EX7.7/Example7_7.sce
new file mode 100755
index 000000000..5128817c8
--- /dev/null
+++ b/389/CH7/EX7.7/Example7_7.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+

+// Illustration 7.7

+// Page: 236

+

+printf('Illustration 7.7 - Page: 236\n\n');

+

+// solution

+

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

+// A = water vapour; B = air

+V = 100;// [m^3]

+Tempi = 55;// [OC]

+Tempf = 110;// [OC]

+//*****//

+

+// From Illustration 7.6

+vH = 0.974;// [m^3/kg]

+Cs = 1061.5;// [J/kg]

+WB = V/vH;// [kg]

+Q = WB*Cs*(Tempf-Tempi);// [J]

+printf("Heat recquired is %e J\n",Q);
\ No newline at end of file
diff --git a/389/CH7/EX7.8/Example7_8.sce b/389/CH7/EX7.8/Example7_8.sce
new file mode 100755
index 000000000..c458eda0d
--- /dev/null
+++ b/389/CH7/EX7.8/Example7_8.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+

+// Illustration 7.8

+// Page: 237

+

+printf('Illustration 7.8 - Page: 237\n\n');

+

+// solution

+

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

+Y_prime1 = 0.030;// [kg water/kg dry air]

+Temp1 = 83;// [OC]

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

+

+// From the psychrometric chart, the condition at 90 OC

+Temp2 = 41.5;// [OC]

+Y_prime2 = 0.0485;// [kg water/kg dry air]

+printf("The Outlet Air condition are:\n");

+printf("Temp. =  %f OC\n",Temp2);

+printf("Absolute Humidity =  %f kg water/kg dry air\n",Y_prime2);
\ No newline at end of file
diff --git a/389/CH7/EX7.9/Example7_9.sce b/389/CH7/EX7.9/Example7_9.sce
new file mode 100755
index 000000000..098a6d1bf
--- /dev/null
+++ b/389/CH7/EX7.9/Example7_9.sce
@@ -0,0 +1,24 @@
+clear;

+clc;

+

+// Illustration 7.9

+// Page:240

+

+printf('Illustration 7.9 - Page:240\n\n');

+

+// solution

+

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

+Tempw = 35;// [OC]

+Tempg = 65;// [OC]

+//******//

+

+// From psychrometric chart

+lambda_w = 2419300;// [J/kg]

+Y_prime_w = 0.0365;// [kg H2O/kg dry air]

+// From fig 7.5(a)

+hG_by_kY = 950;// [J/kg]

+// From Eqn. 7.26

+deff('[y] = f13(Y_prime)','y = (Tempg-Tempw)-((lambda_w*(Y_prime_w-Y_prime))/hG_by_kY)');

+Y_prime = fsolve(2,f13);// [kg H2O/kg dry air]

+printf("Humidity of air is %f kg H2O/kg dry air\n",Y_prime);
\ No newline at end of file