11 files changed, 268 insertions, 0 deletions
diff --git a/647/CH4/EX4.1/Example4_1.sce b/647/CH4/EX4.1/Example4_1.sce
new file mode 100755
index 000000000..9329c044f
--- /dev/null
+++ b/647/CH4/EX4.1/Example4_1.sce
@@ -0,0 +1,21 @@
+clear;
+clc;
+
+// Example: 4.1
+// Page: 118
+
+printf("Example: 4.1 - Page: 118\n\n");
+
+// Solution
+
+//*****Data*****//
+Qp = -327;// [kcal]
+T = 27 + 273;// [K]
+R = 2*10^(-3);// [kcal/K mol]
+//*************//
+
+// The reaction involved is:
+// C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)
+deltan = 2 - 3;
+Qv = Qp - deltan*R*T;// [kcal]
+printf("Value of Qv is %.2f kcal\n",Qv);
+\ No newline at end of file
diff --git a/647/CH4/EX4.10/Example4_10.sce b/647/CH4/EX4.10/Example4_10.sce
new file mode 100755
index 000000000..d8d3ed8f5
--- /dev/null
+++ b/647/CH4/EX4.10/Example4_10.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+
+// Example: 4.10
+// Page: 129
+
+printf("Example: 4.10 - Page: 129\n\n");
+
+// Solution
+
+//*****Data*****//
+T2 = 1273;// [K]
+T1 = 300;// [K]
+deltaH_300 = -11030;// [cal/mol]
+//*************//
+
+// The chemical reaction involved is:
+// N2 + 3H2 = 2NH3
+// (1/2)N2 + (3/2)H2 = NH3
+// deltaH_1273 = deltaH_300 + integrate('Cp_NH3(T) - (1/2)*Cp_N2(T) - (1/2)*Cp_H2(T)','T',1273,300);
+deltaH_1273 = deltaH_300 + integrate('(6.2 + 7.8*10^(-3)*T - 7.2*10^(-6)*T^2) - (1/2)*(6.45 + 1.4*10^(-3)*T) - (1/2)*(6.94 - 0.2*10^(-3)*T)','T',1273,300);// [cal]
+printf("Heat of formation at 1273 K is %d cal",deltaH_1273);
+\ No newline at end of file
diff --git a/647/CH4/EX4.11/Example4_11.sce b/647/CH4/EX4.11/Example4_11.sce
new file mode 100755
index 000000000..820cdc6c3
--- /dev/null
+++ b/647/CH4/EX4.11/Example4_11.sce
@@ -0,0 +1,24 @@
+clear;
+clc;
+
+// Example: 4.11
+// Page: 130
+
+printf("Example: 4.11 - Page: 130\n\n");
+
+// Solution
+
+//*****Data*****//
+CO2 = 13.4;// [percent by volume]
+N2 = 80.5;// [percent by volume]
+O2 = 6.1;// [percent by volume]
+//*************//
+
+// Basis : 100 cubic m of flue gas.
+Vol_N2_flue = N2;// [Volume of Nitrogen in flue gas, cubic m]
+Vol_O2_flue = O2;// [Volume of O2 in flue gas, cubic m]
+Vol_Air = N2/0.79;// [Volume of air supplied, cubic m]
+Vol_O2 = Vol_Air*0.21;// [Volume of O2 in air supply, cubic m]
+Vol_O2_cumbustion = Vol_O2 - Vol_O2_flue;// [Volume of O2 used up in cumbustion of the fuel, cubic m]
+Excess_Air = Vol_O2_flue/Vol_O2_cumbustion * 100;// [percent of excess air supplied]
+printf("Percent of excess air supplied is %.1f %%",Excess_Air);
+\ No newline at end of file
diff --git a/647/CH4/EX4.2/Example4_2.sce b/647/CH4/EX4.2/Example4_2.sce
new file mode 100755
index 000000000..f757a2a6f
--- /dev/null
+++ b/647/CH4/EX4.2/Example4_2.sce
@@ -0,0 +1,21 @@
+clear;
+clc;
+
+// Example: 4.2
+// Page: 119
+
+printf("Example: 4.2 - Page: 119\n\n");
+
+// Solution
+
+//*****Data*****//
+// Mg + (1/2)O2 = MgO ...............(1)
+deltaH1 = -610.01;// [kcal]
+// 2Fe + (3/2)O2 = Fe2O3 ............(2)
+deltaH2 = -810.14;// [kcal]
+//*************//
+
+// 3Mg + Fe2O3 = 3MgO + 2Fe .........(3)
+// Multiplying (1) by 3 and substracting from (2), we get (3):
+deltaH = 3*deltaH1 - deltaH2;// [kcal]
+printf("Heat produced in the reaction is %.1f kcal\n",deltaH);
+\ No newline at end of file
diff --git a/647/CH4/EX4.3/Example4_3.sce b/647/CH4/EX4.3/Example4_3.sce
new file mode 100755
index 000000000..1809a6117
--- /dev/null
+++ b/647/CH4/EX4.3/Example4_3.sce
@@ -0,0 +1,23 @@
+clear;
+clc;
+
+// Example: 4.3
+// Page: 121
+
+printf("Example: 4.3 - Page: 121\n\n");
+
+// Solution
+
+//*****Data*****//
+// 2H2(g) + O2(g) ---------------> 2H2O .....................(1)
+deltaH1 = -241.8*2;// [kJ/gmol H2]
+// C(graphite) + O2(g) =---------> CO2(g) ...................(2)
+deltaH2 = -393.51;// [kJ/gmol C]
+// CH4(g) + 2O2(g) ---------------> CO2(g) + 2H2O(l) ........(3)
+deltaH3 = -802.36;// [kJ/mol CH4]
+//*************//
+
+// For standard heat of formation of methane, (a) + (b) - (c)
+// C + 2H2 ------------------------> CH4
+deltaHf = deltaH1 + deltaH2 - deltaH3;// [kJ/gmol]
+printf("The standard heat of formation of methane is %.2f kJ/gmol\n",deltaHf);
+\ No newline at end of file
diff --git a/647/CH4/EX4.4/Example4_4.sce b/647/CH4/EX4.4/Example4_4.sce
new file mode 100755
index 000000000..088d98ae9
--- /dev/null
+++ b/647/CH4/EX4.4/Example4_4.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+
+// Example: 4.4
+// Page: 122
+
+printf("Example: 4.4 - Page: 122\n\n");
+
+// Solution
+
+//*****Data*****//
+deltaH_C6H12O6 = -1273;// [kcal]
+deltaH_C2H5OH = -277.6;// [kcal]
+deltaH_CO2 = -393.5;// [kcal]
+deltaH_H2O = -285.8;// [kcal]
+//**************//
+
+// C6H12O6(s) = 2C2H5OH(l) + 2CO2(g) ..........................(A)
+deltaH_A = 2*deltaH_C2H5OH + 2*deltaH_CO2 - deltaH_C6H12O6;// [kJ]
+// C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l) ...................(B)
+deltaH_B = 6*deltaH_CO2 + 6*deltaH_H2O - deltaH_C6H12O6;// [kJ]
+printf("Energy supplied by reaction A is %.1f kJ\n",deltaH_A);
+printf("Energy supplied by reaction B is %.1f kJ\n",deltaH_B);
+if deltaH_A < deltaH_B
+    printf("Reaction A supplies more energy to the organism\n");
+else
+     printf("Reaction B supplies more energy to the organism\n");
+end
+\ No newline at end of file
diff --git a/647/CH4/EX4.5/Example4_5.sce b/647/CH4/EX4.5/Example4_5.sce
new file mode 100755
index 000000000..ad2f618be
--- /dev/null
+++ b/647/CH4/EX4.5/Example4_5.sce
@@ -0,0 +1,25 @@
+clear;
+clc;
+
+// Example: 4.5
+// Page: 122
+
+printf("Example: 4.5 - Page: 122\n\n");
+
+// Solution
+
+//*****Data*****//
+// Zn + S = ZnS ....................................................(A)
+deltaH_A = -44;// [kcal/kmol]
+// ZnS + 3O2 = 2ZnO + 2SO2 .........................................(B)
+deltaH_B = -221.88;// [kcal/kmol]
+// 2SO2 + O2 = 2SO3 ................................................(C)
+deltaH_C = -46.88;// [kcal/kmol]
+// ZnO + SO3 = ZnSO4 ...............................................(D)
+deltaH_D = -55.10;// [kcal/kmol]
+//***************//
+
+// Multiplying (A) by 2 & (D) by (2) and adding (A), (B), (C) & (D)
+// Zn + S + 2O2 = ZnSO4
+deltaH = 2*deltaH_A + deltaH_B + deltaH_C + 2*deltaH_D;// [kcal/kmol for 2 kmol of ZnSO4]
+printf("Heat of formation of ZnSO4 is %.2f kcal/kmol\n",deltaH/2);
+\ No newline at end of file
diff --git a/647/CH4/EX4.6/Example4_6.sce b/647/CH4/EX4.6/Example4_6.sce
new file mode 100755
index 000000000..11dbbb0ea
--- /dev/null
+++ b/647/CH4/EX4.6/Example4_6.sce
@@ -0,0 +1,25 @@
+clear;
+clc;
+
+// Example: 4.6
+// Page: 124
+
+printf("Example: 4.6 - Page: 124\n\n");
+
+// Solution
+
+//*****Data*****//
+// HC : Heat of Combustion
+HC_NH3 = -90.6;// [kcal]
+HC_H2 = -68.3;// [kcal]
+//*************//
+
+// Heat of combustion of NH3:
+// 2NH3 + 3O = N2 + 3H2O ............................ (A)
+// Heat of combustion of H2:
+// H2 + O = H2O ..................................... (B)
+// Multiplying (B) by 3 & substracting from (A), we get:
+// 2NH3 = N2 + 3H2 .................................. (C)
+// Hf : Heat of Formation
+Hf_NH3 = -(2*HC_NH3 - 3*HC_H2)/2;// [kcal]
+printf("Standard Heat of formation of NH3 is %.1f kcal",Hf_NH3);
+\ No newline at end of file
diff --git a/647/CH4/EX4.7/Example4_7.sce b/647/CH4/EX4.7/Example4_7.sce
new file mode 100755
index 000000000..fbfbde81c
--- /dev/null
+++ b/647/CH4/EX4.7/Example4_7.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+
+// Example: 4.7
+// Page: 125
+
+printf("Example: 4.7 - Page: 125\n\n");
+
+// Solution
+
+//*****Data*****//
+// HC : Heat of Combustion
+HC_C2H2 = -310600; // [cal]
+//**************//
+
+// C2H2 + (5/2)O2 = 2CO2 + H2O
+Q = -HC_C2H2;// [cal]
+// The gases present in the flame zone after combustion are carbon dioxide, water vapor and the unreacted nitrogen of the air.
+// Since (5/2) mole of oxygen were required for combustion, nitrogen required would be 10 mol.
+// Hence the composition of the resultant gas would be 2 mol CO2, 1  ol H2 & 10 mol N2.
+// Q = integrate('Cp(T)','T',T,298);
+// On integrating we get:
+// Q = 84.52*(T - 298) + 18.3*10^(-3)*(T^2 - 298^2)
+deff('[y] = f(T)','y = Q - 84.52*(T - 298) - 18.3*10^(-3)*(T^2 - 298^2)');
+T = fsolve(7,f);// [K]
+printf("The maximum attainable temperature is %.1f K",T);
+\ No newline at end of file
diff --git a/647/CH4/EX4.8/Example4_8.sce b/647/CH4/EX4.8/Example4_8.sce
new file mode 100755
index 000000000..1c8a60d43
--- /dev/null
+++ b/647/CH4/EX4.8/Example4_8.sce
@@ -0,0 +1,34 @@
+clear;
+clc;
+
+// Example: 4.8
+// Page: 126
+
+printf("Example: 4.8 - Page: 126\n\n");
+
+// Solution
+
+//*****Data*****//
+Cp_CO2 = 54.56;// [kJ/mol K]
+Cp_O2 = 35.20;// [kJ/mol K]
+Cp_steam = 43.38;// [kJ/mol K]
+Cp_N2 = 33.32;// [kJ/mol K]
+// 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(g)
+deltaH_273 = -1560000;// [kJ/kmol]
+//************//
+
+// Since the air is 25% in excess of the amount required,the combustion may be written as:
+// C2H6(g) + (7/2)O2(g) = 2CO2(g) + 3H2O(g)
+// 25% excess air is supplied.
+// Since the air contains N2 = 79% and O2 = 21%
+// C2H6(g) + 3.5O2(g) + 0.25*3.5O2(g) + (4.375*(79/21))N2 = 2CO2 + 3H2O + 0.875O2 + 16.46N2 .................. (A)
+// Considering the reaction (A),
+// Amount of O2:
+O2 = 3.5 + 3.5*0.25;// [mol]
+// Amount of N2 required:
+N2 = 4.375*(79/21);// [mol]
+// Let the initial temperature of ethane and air be 0 OC and the temperature of products of combustion be T OC
+// Since heat librated by combustion = heat accumulated by combustion products
+Q = -deltaH_273;// [kJ/mol K]
+T = Q/(2*Cp_CO2 + 3*Cp_steam + 0.875*Cp_O2 + N2*Cp_N2);// [OC]
+printf("The theoretical temperature of combustion is %d degree Celsius",T);
+\ No newline at end of file
diff --git a/647/CH4/EX4.9/Example4_9.sce b/647/CH4/EX4.9/Example4_9.sce
new file mode 100755
index 000000000..c6c94db83
--- /dev/null
+++ b/647/CH4/EX4.9/Example4_9.sce
@@ -0,0 +1,19 @@
+clear;
+clc;
+
+// Example: 4.9
+// Page: 129
+
+printf("Example: 4.9 - Page: 129\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 273;// [K]
+T2 = 253;// [K]
+deltaH_273 = 1440;// [cal/mol]
+Cp = 8.7;// [cal/mol]
+//**************//
+
+deltaH_253 = deltaH_273 + Cp*(T2 - T1);// [cal/mol]
+printf("Laten heat of ice at -20 OC is %d cal/mol\n",deltaH_253);
+\ No newline at end of file