initial commit / add all books

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diff --git a/593/CH4/EX4.1/ex4_1.sce b/593/CH4/EX4.1/ex4_1.sce
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+clear;

+//clc();

+

+// Example 4.1

+// Page: 67

+printf("Example-4.1  Page no.-67\n\n");

+

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

+T = 671.7;//[R] Equilibrium temperature

+m_steam = 1;//[lbm] Condensing amount of the steam

+// Using values from the steam table [1], we find that

+delta_h_condensation = -970.3//[Btu/lbm] Enthalpy change of the steam

+delta_s_condensation = -1.4446;//[Btu/(lbm*R)] Entropy change of the steam

+

+// Gibb's free energy change of the steam is

+delta_g_condensation = delta_h_condensation - T*delta_s_condensation;//[Btu/lbm]

+

+printf("Gibb''s free energy change of the steam is %0.1f Btu/lbm",delta_g_condensation);

diff --git a/593/CH4/EX4.2/ex4_2.sce b/593/CH4/EX4.2/ex4_2.sce
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index 000000000..e1857709f
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+++ b/593/CH4/EX4.2/ex4_2.sce
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+clear;

+//clc();

+

+// Example 4.2

+// Page: 77

+printf("Example-4.2  Page no.-77\n\n");

+

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

+

+// let we denote graphite by 'g' and diamond by 'd' 

+// Gibb's free energies of graphite and diamond are given by

+g_g = 0.00;//[kJ/mol] 

+g_d = 2.90;//[kJ/mol]

+

+// Specific volumes of graphite and diamond are given by

+v_g = 5.31*10^(-1);//[kJ/(mol*kbar)]

+v_d = 3.42*10^(-1);//[kJ/(mol*kbar)]

+

+// Now from the equation 4.32 ( page 74) given in the book, we have

+// (dg/dP) = v , at constant temperature

+// where 'v' is specific volume

+// let us denote (dg/dP) by 'D' ,so

+D_g = v_g;//[J/(mol*Pa)] For graphite

+D_d = v_d;//[J/(mol*Pa)] For diamond

+

+// Now we can take our plot from P = 0( =1 ), however, total pressure is 1 atm. 

+// If we consider specific volumes of the given species to be constant with changing the pressure then g-P curve will be a straight line

+// So the equation of the line for graphite is 

+// g = D_g*P + g_g

+// and that for diamond

+// g = D_d*P + g_d

+

+P = [0:1:30]';

+

+plot2d(P,[ D_d*P+g_d D_g*P+g_g ],style=[color("darkgreen"),color("red")]);

+

+xlabel("Pressure, P, kbar");

+ylabel("Gibb''s free energy per mol, g, kJ/mol");

+

+printf(" Gibb''s free energy-pressure diagram for graphite-diamond system at 25 degC is as shown in the graphic window. ");

+hl=legend(['Diamond, slope = 0.342 (kJ/mol)/kbar';'Graphite, slope = 0.532 (kJ/mol)/kbar']);

+

+

diff --git a/593/CH4/EX4.3/ex4_3.sce b/593/CH4/EX4.3/ex4_3.sce
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index 000000000..da95ca411
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+++ b/593/CH4/EX4.3/ex4_3.sce
@@ -0,0 +1,25 @@
+clear;

+//clc();

+

+// Example 4.3

+// Page: 80

+printf("Example-4.3  Page no.-80\n\n");

+

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

+// We have the system which consists of isobutane and normal butane and isomerisaation is taking place between them 

+// The equilibrium constant for this reaction is given by

+// K = (mole fraction of isobutane)/(mole fraction of n-butane) = x_iso/x_normal

+

+// For this reaction, at 25C, 

+K = 4.52;

+

+// and

+// x_iso + x_normal = 1

+// so

+// K = x_iso/(1-x_iso)

+

+// solving for x_iso

+deff('[y]=f(x_iso)','y = x_iso/(1-x_iso)-K');

+x_iso = fsolve(0,f);

+

+printf(" Mole fraction of isobutane isomer in equilibrium is %0.2f",x_iso);