From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 593/CH4/EX4.1/ex4_1.sce | 18 ++++++++++++++++++
 593/CH4/EX4.2/ex4_2.sce | 43 +++++++++++++++++++++++++++++++++++++++++++
 593/CH4/EX4.3/ex4_3.sce | 25 +++++++++++++++++++++++++
 3 files changed, 86 insertions(+)
 create mode 100755 593/CH4/EX4.1/ex4_1.sce
 create mode 100755 593/CH4/EX4.2/ex4_2.sce
 create mode 100755 593/CH4/EX4.3/ex4_3.sce

(limited to '593/CH4')

diff --git a/593/CH4/EX4.1/ex4_1.sce b/593/CH4/EX4.1/ex4_1.sce
new file mode 100755
index 000000000..19eecc47d
--- /dev/null
+++ b/593/CH4/EX4.1/ex4_1.sce
@@ -0,0 +1,18 @@
+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
new file mode 100755
index 000000000..e1857709f
--- /dev/null
+++ b/593/CH4/EX4.2/ex4_2.sce
@@ -0,0 +1,43 @@
+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
new file mode 100755
index 000000000..da95ca411
--- /dev/null
+++ 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);
-- 
cgit