11 files changed, 297 insertions, 0 deletions
diff --git a/647/CH3/EX3.1/Example3_1.sce b/647/CH3/EX3.1/Example3_1.sce
new file mode 100755
index 000000000..4b7c21a9a
--- /dev/null
+++ b/647/CH3/EX3.1/Example3_1.sce
@@ -0,0 +1,19 @@
+clear;
+clc;
+
+// Example: 3.1
+// Page: 88
+
+printf("Example: 3.1 - Page: 88\n\n");
+
+// Solution
+
+//*****Data*****//
+P = 2*10^5;// [Pa]
+T = 273 + 37;// [K]
+R = 8.314;// [J/mol K]
+//****************//
+
+// Since the gas behaves ideally:
+V = R*T/P;// [cubic metres]
+printf("Molar Volume of the gas is %.2e cubic metres",V);
+\ No newline at end of file
diff --git a/647/CH3/EX3.10/Example3_10.sce b/647/CH3/EX3.10/Example3_10.sce
new file mode 100755
index 000000000..7f8a27f88
--- /dev/null
+++ b/647/CH3/EX3.10/Example3_10.sce
@@ -0,0 +1,42 @@
+clear;
+clc;
+
+// Example: 3.10
+// Page: 103
+
+printf("Example: 3.10 - Page: 103\n\n");
+
+// Solution
+
+//*****Data*****//
+beeta = 1.487*10^(-3);// [1/OC]
+alpha = 62*10^(-6);// [1/bar]
+V1 = 1.287;// [cubic cm /g]
+//************//
+
+// Solution (a)
+// The value of derivative (dP/dT) at constant V:
+// dV/V = beeta*dT - alpha*dP
+// dV = 0
+// dP/dT = beeta/alpha
+// Value = dP/dT
+Value = beeta/alpha;// [bar/OC]
+printf("Value of derivative is %.2f bar/OC\n",Value);
+
+// Solution (b)
+P1 = 1;// [bar]
+T1 = 20;// [OC]
+T2 = 30;// [OC]
+// Applying the same equation:
+P2 = P1 +(beeta/alpha)*(T2 - T1);// [bar]
+printf("The pressure generated by heating at constant Volume is %.2f Pa\n",P2);
+
+// Solution (c)
+T2 = 0;// [OC]
+T1 = 20;// [OC]
+P2 = 10;// [bar]
+P1 = 1;// [bar]
+// The change in Volume can be obtained as:
+V2 = V1*exp((beeta*(T2 - T1)) - alpha*(P2 - P1));// [cubic cm/g]
+deltaV = V2 - V1;// [cubic cm/g]
+printf("The change in Volume is %.3f cubic cm/g\n",deltaV);
+\ No newline at end of file
diff --git a/647/CH3/EX3.11/Example3_11.sce b/647/CH3/EX3.11/Example3_11.sce
new file mode 100755
index 000000000..6ac349201
--- /dev/null
+++ b/647/CH3/EX3.11/Example3_11.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+
+// Example: 3.11
+// Page: 107
+
+printf("Example: 3.11 - Page: 107\n\n");
+
+// Solution
+
+//*****Data*****//
+Tc = 513.9;// [K]
+Pc = 61.48*10^5;// [Pa]
+//************//
+
+Tr = 0.7;
+T = Tr*Tc - 273.15;// [OC]
+P_sat = 10^(8.112 - (1592.864/(T + 226.184)));// [mm Hg]
+P_sat = P_sat*101325/760;// [Pa]
+Pr_sat = P_sat/Pc;// [Pa]
+omega = -1 - log10(Pr_sat);// [Acentric factor]
+printf("Acentric factor is %.4f",omega);
+\ No newline at end of file
diff --git a/647/CH3/EX3.2/Example3_2.sce b/647/CH3/EX3.2/Example3_2.sce
new file mode 100755
index 000000000..80b412c52
--- /dev/null
+++ b/647/CH3/EX3.2/Example3_2.sce
@@ -0,0 +1,19 @@
+clear;
+clc;
+
+// Example: 3.2
+// Page: 89
+
+printf("Example: 3.2 - Page: 89\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 8;// [cubic m]
+P1 = 300;// [kPa]
+V2 = 2;// [cubic m]
+//**************//
+
+// Apptying  the ideal gas Eqn. & since the Temperature remains constant:
+P2 = P1*V1/V2;// [kPa]
+printf("The pressure of air after compression is %d kPa\n",P2);
+\ No newline at end of file
diff --git a/647/CH3/EX3.3/Example3_3.sce b/647/CH3/EX3.3/Example3_3.sce
new file mode 100755
index 000000000..f7d8a16ca
--- /dev/null
+++ b/647/CH3/EX3.3/Example3_3.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+
+// Example: 3.3
+// Page: 89
+
+printf("Example: 3.3 - Page: 89\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 6;// [cubic m]
+P1 = 500;// [kPa]
+R = 0.287;// [kJ/kg K]
+//*************//
+
+// Applying the charectarstic equation to the gas initially:
+// P1*V1 = m1*R*T1.......................................(i)
+// Applying the charectarstic equation to the gas which was left in the vessel after one-fifth of the gas has been removed:
+// P2*V2 = m2*R*T2.......................................(ii)
+// V2 = V1;
+// T2 = T1;
+// m2 = (4/5)*m1;
+// Eqn (ii) becomes:
+// P2*V1 = (4/5)*m1*R*T1..................................(iii)
+// Dividing eqn (i) by eqn (iii), we get:
+P2 = (4/5)*P1;// [kPa]
+printf("The pressure of the remaining air is %d kPa\n",P2);
+\ No newline at end of file
diff --git a/647/CH3/EX3.4/Example3_4.sce b/647/CH3/EX3.4/Example3_4.sce
new file mode 100755
index 000000000..b2dff7a14
--- /dev/null
+++ b/647/CH3/EX3.4/Example3_4.sce
@@ -0,0 +1,33 @@
+clear;
+clc;
+
+// Example: 3.4
+// Page: 90
+
+printf("Example: 3.4 - Page: 90\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 450 + 273;// [K]
+P1 = 3;// [bar]
+//***************//
+
+// Soluton(a)
+// From Fig. 3.7, (Page 90)
+// Since the weight remains the same, therefore, the final pressure is equal to the initial pressure.
+// Therefore it is a constant pressure process.
+P2 = P1;// [bar]
+// Volumetric Ratio:
+V2_by_V1 = 2.5/(2.5 + 2.5);
+// Applying ideal gas law & P1 = P2
+T2 = T1*V2_by_V1;// [K]
+printf("Final Temperature of the air when the piston reaches stop is %.1f K\n",T2);
+
+// Solution (b)
+// When the piston rests ot the stops, the pressure exerted by the weight, air & the atmosphere will be different. But there will beno further decrease in volume.
+// This is a constant volume process.
+T3 = 273 + 30;// [K]
+// Applying ideal gas law & V2 = V3
+P3 = T3*P2/T2;// [bar]
+printf("Pressure of air inside the cylinder is %.2f bar\n",P3);
+\ No newline at end of file
diff --git a/647/CH3/EX3.5/Example3_5.sce b/647/CH3/EX3.5/Example3_5.sce
new file mode 100755
index 000000000..cf403fe21
--- /dev/null
+++ b/647/CH3/EX3.5/Example3_5.sce
@@ -0,0 +1,25 @@
+clear;
+clc;
+
+// Example: 3.5
+// Page: 95
+
+printf("Example: 3.5 - Page: 95\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1.373;// [kg]
+P = 1.95*10^(6);// [Pa]
+V = 0.1;// [cubic m]
+a = 422.546*10^(-3);// [cubic m/square mol]
+b = 37*10^(-6);// [cubic m/mol]
+M = 17*10^(-3);// [kg/mol]
+R = 8.314;// [J/mol K]
+//****************//
+
+n = m/M;// [moles]
+Vm = V/n;// [molar volume, cubic m]
+// Applying Van der Waals equation of state:
+T = (P + (a/Vm^2))*((Vm - b)/R);// [K]
+printf("The temperature at which ammonia exists in the cylinder is %.1f K\n",T)
+\ No newline at end of file
diff --git a/647/CH3/EX3.6/Example3_6.sce b/647/CH3/EX3.6/Example3_6.sce
new file mode 100755
index 000000000..45ac1647d
--- /dev/null
+++ b/647/CH3/EX3.6/Example3_6.sce
@@ -0,0 +1,52 @@
+clear;
+clc;
+
+// Example: 3.6
+// Page: 96
+
+printf("Example: 3.6 - Page: 96\n\n");
+
+// Solution
+
+//*****Data*****//
+P = 15*10^5;// [Pa]
+T = 773;// [K]
+R = 8.314;// [J/mol K]
+//**************//
+
+// Solution (a)
+printf("Ideal Equation of State\n")
+// Applying ideal Eqn. of State:
+Vm = R*T/P;// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (b)
+printf("Van der Wall Equation of State\n");
+a = 0.2303;// [Nm^4/square mol]
+b = 4.3073*10^(-5);// [cubic m/mol]
+deff('[y] = f1(Vm)','y = P - (R*T/(Vm-b)) + (a/Vm^2)');
+Vm = fsolve(Vm,f1);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+//Solution (c)
+printf("Virial Equation of State\n");
+// Z = 1 + B/V
+// (P*V/(R*T)) = (1 + B/V)
+// V^2 - V*R*T/P - B*R*T/P = 0
+B = 1.3697*10^(-5);// [cubic m/mol]
+deff('[y] = f2(Vm)','y = Vm^2 - (Vm*R*T/P) - (B*R*T/P)');
+Vm = fsolve(7,f2);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (d)
+printf("Redlich Kwong Equation of State\n");
+Tc = 190.6;// [K]
+Pc = 45.99*10^5;// [Pa]
+a = 0.4278*R^2*Tc^2.5/Pc;// [N/m^4 square mol]
+b = 0.0867*R*Tc/Pc;// [cubic m/mol]
+deff('[y] = f3(Vm)','y = P - (R*T/(Vm - b)) + (a/((T^0.5)*Vm*(Vm+b)))');
+Vm = fsolve(Vm,f3);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+\ No newline at end of file
diff --git a/647/CH3/EX3.7/Example3_7.sce b/647/CH3/EX3.7/Example3_7.sce
new file mode 100755
index 000000000..a1e824d8e
--- /dev/null
+++ b/647/CH3/EX3.7/Example3_7.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 3.7
+// Page: 101
+
+printf("Example: 3.7 - Page: 101\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 3.7 on page number 101 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 3.7 on page 101 of the book.")
+\ No newline at end of file
diff --git a/647/CH3/EX3.8/Example3_8.sce b/647/CH3/EX3.8/Example3_8.sce
new file mode 100755
index 000000000..220e5c9f5
--- /dev/null
+++ b/647/CH3/EX3.8/Example3_8.sce
@@ -0,0 +1,31 @@
+clear;
+clc;
+
+// Example: 3.8
+// Page: 101
+
+printf("Example: 3.8 - Page: 101\n\n");
+
+// Solution
+
+//*****Data*****//
+T = 500;// [K]
+P = 8*10^6;// [Pa]
+R = 8.314;// [J/mol K]
+//*************//
+
+// Solution (a)
+// By ideal gas equation of state:
+printf("Ideal Equation of State\n")
+Vm = R*T/P;// [cubic m/mol]
+printf("Molar Volume of gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (b)
+// By Virial Equation of State:
+printf("Virial Equation of State\n");
+B = -0.265*10^(-3);// [cubic m/mol]
+C = 0.3025*10^(-7);// [m^6/square mol]
+deff('[y] = f(Vm)','y = (P*Vm/(R*T)) - 1 -(B/Vm) - (C/Vm^2)');
+Vm = fsolve(Vm,f);
+printf("Molar Volume of gas is %.2e cubic m/mol\n",Vm);
+\ No newline at end of file
diff --git a/647/CH3/EX3.9/Example3_9.sce b/647/CH3/EX3.9/Example3_9.sce
new file mode 100755
index 000000000..8549376d6
--- /dev/null
+++ b/647/CH3/EX3.9/Example3_9.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 3.9
+// Page: 102
+
+printf("Example: 3.9 - Page: 102\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 3.9 on page number 102 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 3.9 on page 102 of the book.")
+\ No newline at end of file