51 files changed, 1276 insertions, 0 deletions
diff --git a/926/CH2/EX2.1/Chapter2_Example1.sce b/926/CH2/EX2.1/Chapter2_Example1.sce
new file mode 100644
index 000000000..47c5fb729
--- /dev/null
+++ b/926/CH2/EX2.1/Chapter2_Example1.sce
@@ -0,0 +1,20 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 1, Page 30
+//Title: Calculating volume of gas at standard conditions from given weight
+//=============================================================================
+clear
+clc
+
+//INPUT
+w = 25; //Weight of liquid chlorine in lb
+mw = 70.92; //molecular weight of chlorine gas in lb/lb mol
+
+//CALCULATION
+n = w/mw; //To find no of moles of chlorine gas in lb mol
+v = n*359; //To compute volume of chlorine gas in cu ft at standard conditions
+
+//OUTPUT
+mprintf('\n The volume of chlorine gas that will be occupied at standard conditions = %4.1f cu ft',v);
+
+//======================END OF PROGRAM=========================================
diff --git a/926/CH2/EX2.2/Chapter2_Example2.sce b/926/CH2/EX2.2/Chapter2_Example2.sce
new file mode 100644
index 000000000..12db85149
--- /dev/null
+++ b/926/CH2/EX2.2/Chapter2_Example2.sce
@@ -0,0 +1,20 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 2, Page 31
+//Title: Calculating mass from given volume at standard conditions
+//=============================================================================
+clear
+clc
+
+//INPUT
+v = 500; //Volume of gaseous propane in liters
+mw = 44.06; //Molecular weight of propane in g/g mole
+
+//CAlCULATION
+n = v/22.4; //To find the no of moles of propane in g mole
+m = n*mw; //To calculate the weight of propane in grams
+
+//OUTPUT
+mprintf('\n Weight of liquid propane formed after liquification = %3.0f grams',m);
+
+//==============================END OF PROGRAM=================================
diff --git a/926/CH2/EX2.3/Chapter2_Example3.sce b/926/CH2/EX2.3/Chapter2_Example3.sce
new file mode 100644
index 000000000..d1232a7a4
--- /dev/null
+++ b/926/CH2/EX2.3/Chapter2_Example3.sce
@@ -0,0 +1,34 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 3, Page 31
+//Title: Calculation of weight and volume from reaction stoichiometry
+//=============================================================================
+clear
+clc
+
+//INPUT
+w1 = 100; //Weight of hydrogen in lb(Basis of calculation)
+MW = [55.84,18.02,231.5,2.016]; //Atomic weight of iron, Steam, ferric oxide, hydrogen in lb/lb-mole
+//From the reaction stoichiometry
+stoic = [3,4,1,4]; //Stoichiometric coefficient of Fe, H2O, Fe3O4, H2
+
+//CALCULATION
+//part(a)
+n1 = w1/MW(4); //Moles of hydrogen produced in lb mole
+n2 = n1*stoic(1)/stoic(4); //Atoms of iron required in lb atom
+w2 = n2*MW(1); //Weight of iron required in lb
+n3 = n1*stoic(2)/stoic(4); //Moles of Steam required in lb mole
+w3 = n3*MW(2); //Weight of steam required in lb
+n4 = n1*stoic(3)/stoic(4); //Moles of ferric oxide required in lb mole
+w4 = n4*MW(3); //Weight of ferric oxide required in lb
+M1 = w2+w3; //Total input in lb
+M2 = w1+w4; //Total output in lb
+//part(b)
+v = n1*359; //volume of hydrogen at standard conditions in cu ft
+
+//OUTPUT
+mprintf('\n (a) To produce %3.0f lb of hydrogen, the weight of iron and steam required is %4.0f lb and %3.0f lb respectively',w1,w2,w3);
+mprintf('\n     The weight of ferric oxide formed = %4.0f lb',w4);
+mprintf('\n (b) Volume occupied by hydrogen at standard conditions = %5.0f cu ft',v);
+
+//============================END OF PROGRAM===================================
diff --git a/926/CH2/EX2.4/Chapter2_Example4.sce b/926/CH2/EX2.4/Chapter2_Example4.sce
new file mode 100644
index 000000000..db1bb49a9
--- /dev/null
+++ b/926/CH2/EX2.4/Chapter2_Example4.sce
@@ -0,0 +1,25 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, illustration 4, Page 36
+//Title: Expressing weight percent into mole percent
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 100; //Weight of solution in grams(Basis of calculation)
+w1 = 40; //Weight of sodium carbonate present in solution in grams
+MW = [106,18.02]; //Molecular weight of sodium carbonate and water respectively in g/g-mole
+
+//CALCULATION
+n1 = w1/MW(1); //To find the no of moles of sodium carbonate in g mole
+n2 = (W-w1)/MW(2); //To find the no of moles of water in g mole
+N = n1+n2; //Calculation of total no of moles in g mole
+x1 = n1*100/N; //Mole % of sodium carbonate
+x2 = n2*100/N; //Mole % of water
+
+//OUTPUT
+mprintf('\n mole percent of Na2CO3 = %4.2f',x1);
+mprintf('\n mole percent of H2O = %3.1f',x2);
+
+//================================END OF PROGRAM===============================
+\ No newline at end of file
diff --git a/926/CH2/EX2.5/Chapter2_Example5.sce b/926/CH2/EX2.5/Chapter2_Example5.sce
new file mode 100644
index 000000000..fa61b39bf
--- /dev/null
+++ b/926/CH2/EX2.5/Chapter2_Example5.sce
@@ -0,0 +1,25 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 5, Page 37
+//Title: Expressing mole percent into weight percent
+//=============================================================================
+clear
+clc
+
+//INPUT
+N = 100; //Total no of moles of solution in g mole(Basis of calculation)
+n1 = 25; //No of moles of napthalene present in solution in g mole
+MW = [128.1,78.1]; //Molecular weight of napthalene and benzene in g/g mole
+
+//CALCULATION
+w1 = n1*MW(1); //Weight of napthalene present in the solution in grams
+w2 = (N-n1)*MW(2); //Weight of benzene present in the solution in grams
+W = w1+w2; //Total weight of solution in grams
+m1 = w1*100/W; //Weight % of napthalene in solution
+m2 = w2*100/W; //Weight % of benzene in solution
+
+//OUTPUT
+mprintf('\n Weight percent of napthalene = %3.1f',m1);
+mprintf('\n Weight percent of benzene = %3.1f',m2);
+
+//===========================END OF PROGRAM====================================
+\ No newline at end of file
diff --git a/926/CH2/EX2.6/Chapter2_Example6.sce b/926/CH2/EX2.6/Chapter2_Example6.sce
new file mode 100644
index 000000000..f549541f3
--- /dev/null
+++ b/926/CH2/EX2.6/Chapter2_Example6.sce
@@ -0,0 +1,37 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 6, Page 37
+//Title: Calculation of composition, average molecular weight and density
+//=============================================================================
+clear
+clc
+
+//INPUT
+N = 100; //Total no of moles of natural gas in lb mole(Basis of calculation)
+v = [83.5,12.5,4]; //Volumetric percent of methane, ethane and nitrogen respectively
+MW = [16.03,30.05,28.02]; //Molecular weight of methane, ethane and nitrogen respectively
+
+//CALCULATION
+//part(a)
+x = v; //mol percent of methane, ethane and nitrogen respectively
+//part(b)
+w1 = v(1)*MW(1); //Weight of methane in lb
+w2 = v(2)*MW(2); //Weight of ethane in lb
+w3 = v(3)*MW(3); //Weight of nitrogen in lb
+W = w1+w2+w3; //Total weight of the mixture in lb
+m1 = w1*100/W; //weight percent of methane
+m2 = w2*100/W; //weight percent of ethane
+m3 = w3*100/W; //weight percent of nitrogen
+//part(c)
+AVG_MW = W/100; //Average molecular weight in lb/lb mole
+//part(d)
+v = N*359; //Volume of natural gas at standard conditions in cu ft
+rho = W/v; //Density of natural gas at standard conditions in lb/cu ft
+
+//OUTPUT
+mprintf('\n(a) The mole percent of methane, ethane and nitrogen are %3.1f, %3.1f and %2.0f respectively',x(1),x(2),x(3));
+mprintf('\n(b) The weight percent of methane ethane and nitrogen are %3.1f, %3.1f and %2.1f respectively',m1,m2,m3);
+mprintf('\n(c) The average molecular weight of natural gas is %4.2f lb/lb mole',AVG_MW);
+mprintf('\n(d) The density at standard conditions is %5.4f lb/cu ft',rho);
+
+//=============================END OF PROGRAM==================================
+\ No newline at end of file
diff --git a/926/CH2/EX2.7/Chapter2_Example7.sce b/926/CH2/EX2.7/Chapter2_Example7.sce
new file mode 100644
index 000000000..2f9179c2f
--- /dev/null
+++ b/926/CH2/EX2.7/Chapter2_Example7.sce
@@ -0,0 +1,54 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-2, Illustration 7, Page 39
+//Title: Calculation of composition and molality
+//=============================================================================
+clear
+clc
+
+//INPUT
+V = 1000; //Total volume of solution in cc
+rho = 1.148; //Density of solution in g/cc
+w1 = 230; //Weight of NaCl in solution in g
+MW = [58.5,18.020]; //Molecular weight of NaCl and water respectively in g/g mole
+rho_water = .998; //Density of water at given temperature in g/cc
+
+//CALCULATIONS
+W = V*rho; //Total weight of solution in g
+w2 = W-w1; //Weight of water in solution in g
+n1 = w1/MW(1); //To calculate no of moles of NaCl in g mole
+n2 = w2/MW(2); //To calculate no of moles of water in g mole
+N= n1+n2; //Total no of moles in g mole
+//Part(a)
+m1 = w1*100/W; //To compute weight percent of NaCl
+m2 = w2*100/W; //To compute weight percent of Water
+//part(b)
+v1 = w2/rho_water; //Volume of pure water in cc
+V1 = v1*100/V; //To compute volume percent of water
+//part(c)
+x1 = n1*100/N; //To calculate mole % of NaCl
+x2 = n2*100/N; //To calculate mole % of water
+//part(d)
+a1 = n1; //To calculate no of atoms of sodium in g atom
+a2 = n1; //To calculate no of atoms of chlorine in g atom
+a3 = 2*n2; //To calculate no of atoms of hydrogen in g atom
+a4 = n2; //To calculate no of atoms of oxygen in g atom
+A = a1+a2+a3+a4; //To calculate total no of atoms
+A1 = a1*100/A; //To calculate atomic percent of sodium
+A2 = a2*100/A; //To calculate atomic percent of chlorine
+A3 = a3*100/A; //To calculate atomic percent of hydrogen
+A4 = a4*100/A; //To calculate atomic percent of oxygen
+//part(e)
+m = n1*V/w2; //Molality of solution in lb mole NaCl/1000 lb H2O
+//part(f)
+M = w1/w2; //lb of NaCl per lb H20
+
+//OUTPUT
+mprintf('\n(a) Weight percent of NaCl and water are %3.0f and %3.0f respectively',m1,m2);
+mprintf('\n(b) Volumetric percent of water is %3.0f',V1);
+mprintf('\n(c) Mole percent of NaCl and water are %3.2f and %3.1f respectively',x1,x2);
+mprintf('\n(d) Atomic percent of sodium,chlorine,hydrogen and oxygen are %3.2f, %3.2f, %4.1f and %3.1f respectively',A1,A2,A3,A4);
+mprintf('\n(e) Molality of the solution is %3.2f lb mole of NaCl/1000 lb H2O',m);
+mprintf('\n(f) lb of NaCl per lb of water is %4.3f',M);
+
+//=================================END OF PROGRAM==============================
+\ No newline at end of file
diff --git a/926/CH3/EX3.1/Chapter3_Example1.sce b/926/CH3/EX3.1/Chapter3_Example1.sce
new file mode 100644
index 000000000..95e4356ef
--- /dev/null
+++ b/926/CH3/EX3.1/Chapter3_Example1.sce
@@ -0,0 +1,27 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 1, Page 54
+//Title: Calculation of volume
+//=============================================================================
+clear
+clc
+
+//INPUT
+W1 = 30; //Weight of chlorine in lb
+MW = 71; //Molecular weight of chlorine in lb/lb mole
+P1 = 760; //Pressure at standard conditions in mm Hg
+T1 = 492; //Temperature at standard conditions in degree R
+P2 = 743; //Given pressure in mm Hg
+T = 70; //Given temperature in degree F
+
+//CALCULATIONS
+n = W1/MW; //No of moles of chlorine in lb mole
+V1= n*359; //Volume of chlorine at standard conditions in cu ft
+T2 = 530; //Given temperature in degree R
+
+V2 = V1*(P1/P2)*(T2/T1); //Volume of chlorine at given conditions in cu ft
+
+//OUTPUT
+mprintf(' \n Volume occupied by %2.0f lb of chlorine at given temperature and pressure is %3.0f cu ft',W1,V2);
+
+//=================================END OF PROGRAM==============================
+\ No newline at end of file
diff --git a/926/CH3/EX3.10/Chapter3_Example10.sce b/926/CH3/EX3.10/Chapter3_Example10.sce
new file mode 100644
index 000000000..e1e4e0bae
--- /dev/null
+++ b/926/CH3/EX3.10/Chapter3_Example10.sce
@@ -0,0 +1,36 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 10, Page 62
+//Title: Calculation of volume change with change in composition
+//=============================================================================
+clear
+clc
+
+//INPUT
+v = 100; //Volume of gas entering in cu ft
+v1 = 75; //Volume of air before absorption in cu ft
+v2 = 25; //Volume of HCl before absorption in cu ft
+a = 98; //Percent of HCl removed by absorption
+P = [743,738]; //Pressure of gas before and after entering the system in mm Hg
+T = [120,80]; //Temperature of gas before and after entering the system in degree F
+T3 = 492; //Temperature at standard condition in degree R
+P3 = 760; //Pressure at standard conditions in mm Hg
+MW = 36.5; //Molecular weight of HCl in lb/lb mole
+
+//CALCULATIONS
+v3 = v2*(a/100); //Pure component volume of HCl absorbed in cu ft
+v4 =v2-v3; //Volume of HCl remaining in cu ft
+v5 = v1+v4; //Vulome of gas remaining in cu ft
+T1 = T(1)+460; //Temperature of gas entering in degree R
+T2 = T(2)+460; //Temperature of gas leaving in degree R
+v6 = v5*(P(1)/P(2))*(T2/T1); //Volume of gas leaving in cu ft
+x = v4*100/v5; //Percentage composition by volume of HCl
+y = 100-x; //Percentage composition by volume of air
+v7 = v3*(P(1)/P3)*(T3/T1); //Volume at standard condition of HCl absorbed in cu ft
+n = v7/359; //No of moles of HCl absorbed in lb mole
+w = n*MW; //Weight of HCl absorbed in lb
+
+//OUTPUT
+mprintf('\n (a) Percentage composition by volume of gases leaving the absorption apparatus are %3.2f percent HCl and %4.2f percent air \n (b) Weight of HCl removed per %3.0f cu ft of gas is %3.2f lb',x,y,v,w);
+
+//==============================END OF PROGRAM=================================
diff --git a/926/CH3/EX3.11/Chapter3_Example11.sce b/926/CH3/EX3.11/Chapter3_Example11.sce
new file mode 100644
index 000000000..38f96d9c7
--- /dev/null
+++ b/926/CH3/EX3.11/Chapter3_Example11.sce
@@ -0,0 +1,33 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 11, Page 63
+//Title: Calculation of volume change with change in composition
+//=============================================================================
+clear
+clc
+
+//INPUT
+v = 100; //Volume of gas entering in cu ft
+p = [59,0.5]; //Partial pressures of chlorine before entering and after leaving the absorption apparatus in mm Hg
+P = [740,743]; //Pressure of entering and leaving the apparatus in mm Hg
+T = [75,80]; //Temperature of the gas entering and leaving the apparatus in degree F
+T3 = 492; //Temperature at standard conditions in degree R
+P3 = 760; //Pressure at standard conditions in degree R
+MW = 71; //Molecular weight of chlorine in lb/lb mole
+
+//CALCULATIONS
+p1 = P(1)-p(1); //Partial pressure of inert gas entering in mm Hg
+p2 = P(2)-p(2); //Partial pressure of inert gas leaving in mm Hg
+T1 = T(1)+460; //Temperature of gas entering in degree R
+T2 = T(2)+460; //Temperature of gas leaving the apparatus in degree R
+v1 = v*(p1/p2)*(T2/T1); //Volume of inert gas leaving in cu ft
+v2 = v*(p(1)/P3)*(T3/T1); //Volume at standard conditions of chlorine entering in cu ft
+v3 = v*(p(2)/P3)*(T3/T2); //Volume at standard conditions of chlorine leaving in cu ft
+V = v2-v3; //Volume at standard conditions of chlorine absorbed in cu ft
+n = V/359; //No of moles of chlorine absorbed in lb mole
+W = n*MW; //Weight of chlorine absorbed in lb
+
+//OUTPUT
+mprintf('\n (a) Volume of gases leaving the apparatus per %3.0f cu ft entering is %3.1f cu ft \n (b) Weight of chlorine absorbed per %3.0f cu ft of gas entering is %3.2f lb',v,v1,v,W);
+
+//========================END OF PROGRAM=======================================
diff --git a/926/CH3/EX3.12/Chapter3_Example12.sce b/926/CH3/EX3.12/Chapter3_Example12.sce
new file mode 100644
index 000000000..eeae855a4
--- /dev/null
+++ b/926/CH3/EX3.12/Chapter3_Example12.sce
@@ -0,0 +1,71 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 12, Page 65
+//Title: Calculation of percentage composition by volume
+//=============================================================================
+clear
+clc
+
+//INPUT
+n = 1; //Total no moles of NH3 in lb mole
+v = 100; //Volume of NH3 entering in cu ft
+a = [1,2,1,1]; //Stoichiometric coefficients of NH3, O2, HNO3 and H2O in overall reaction 
+b = [4,5,6,4];//Stoichiometric coefficients of NH3, O2, HNO3 and H2O in reaction 1
+a1 = .21; //lb moles of O2 in 1 lb mole of air
+b1 = .79; //lb moles of N2 in 1 lb mole of air
+a2 = .2; //Amount of excess O2
+T = [20,700]; //Temperature at which gases enters the process and leave the catalyzer in degree C
+P = [755,743]; //Pressure at which gases enters the process and leaves the catalyzer in mm Hg
+T2 = 273; //Temperature at standard conditions in K
+P2 = 760; //Pressure at standard conditions in mm Hg
+V = 359; //Volume at standard conditons in cu ft
+N = .85; //lb moles of NH3 oxidised in catalyzer
+c = .9; //Nitric oxide entering the tower oxidised to Nitric acid
+MW = 63; //Molecular weight of HNO3 in lb/lb mole
+
+//CALCULATIONS
+//part(a)
+T1 = T(1)+273;
+T3 = T(2)+273;
+n1 = a(2)*n; //O2 required in lb moles
+n2 = n1*(n+a2); //O2 supplied in lb moles
+n3 = n2/a1; //Air supplied in lb moles
+v1 = V*(T1/T2)*(P2/P(1)); //Volume of NH3 in cu ft
+v2 = n3*v1; //Volume of air supplied
+v3 = v2*v/v1; //Volume of air per 100 ft of NH3 in cu ft
+//part(b)
+n4 = b1*n3; //N2 present in air in lb moles
+n5 = n3+n; //Total lb moles of gas entering the catalyzer
+x1 = n*100/n5; //Composition of NH3 by volume %
+x2 = n2*100/n5; //Composition of O2 by volume %
+x3 = n4*100/n5; //Composition of N2 by volume %
+//Part(c)
+n6 = n - N; //lb moles of NH3 leaving catalyzer
+n7 = b(2)*N/b(1); //lb moles of O2 consumed in catalyzer
+n8 = n2 - n7; //lb moles of O2 leaving catalyzer
+n9 = b(4)*N/b(1); //lb moles of NO formed in catalyzer
+n10 = b(3)*n9/b(4); //lb moles of H2O formed in catalyzer
+N1 = n4+n6+n8+n9+n10; //lb moles of total quantity of gas leaving catalyzer
+y1 = n9*100/N1; //Composition of NO by volume %
+y2 = n10*100/N1; //Composition of H2O by volume %
+y3 = n6*100/N1; //Composition of NH3 by volume %
+y4 = n8*100/N1; //Composition of O2 by volume %
+y5 = n4*100/N1; //Composition of N2 by volume %
+//part(d)
+N2 = n*v/v1; //lb moles of NH3 per 100 cu ft
+N3 = N1*N2; //lb moles of gas leaving catalyzer
+v4 = N3*V; //Volume at standard conditions of gas leaving catalyzer in cu ft
+v5 = v4*(P2/P(2))*(T3/T2); //Volume of gas laeving catalyzer per 100 cu ft NH3 entering in cu ft
+//part(e)
+N4 = N2*n9; //lb moles of NO produced in catalyzer
+N5 = N4*c; //lb moles of NO oxidised in tower
+W = N5*MW; //Weight of HNO3 formed in lb
+
+//OUTPUT
+mprintf('\n (a) Volume of air per %3.0f cu ft NH3 entering is %4.0f cu ft',v,v3);
+mprintf('\n (b) Percentage composition by volume of gases entering catalyzer:- \n    NH3 = %2.1f \n    O2 = %3.1f \n    N2 = %3.1f',x1,x2,x3);
+mprintf('\n (c) Percentage composition by volume of gases leaving catalyzer:- \n    NO = %2.1f \n    H2O = %3.1f \n    NH3 = %2.1f \n    O2 = %3.1f \n    N2 = %3.1f',y1,y2,y3,y4,y5);
+mprintf('\n (d) Volume of gases leaving catalyzer per %3.0f cu ft of NH3 entering is %4.0f cu ft',v,v5);
+mprintf('\n (e) Weight of HNO3 produced per %3.0f cu ft of NH3 entering is %3.1f lb',v,W);
+
+//========================END OF PROGRAM======================================
+\ No newline at end of file
diff --git a/926/CH3/EX3.2/Chapter3_Example2.sce b/926/CH3/EX3.2/Chapter3_Example2.sce
new file mode 100644
index 000000000..ee3759acb
--- /dev/null
+++ b/926/CH3/EX3.2/Chapter3_Example2.sce
@@ -0,0 +1,26 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 2, Page 54
+//Title: Calculation of weight
+//=============================================================================
+clear
+clc
+
+//INPUT
+V2 = 100; //Volume of water vapor in cu ft
+MW = 18; //Molecular weight of water vapor in lb/lb mole
+P1 = 760; //Pressure at standard conditions in mm Hg
+T1 = 273; //Temperature at standard conditions in K
+P2 = 15.5; //Given pressure in mm Hg
+T = 23; //Given temperature in degree C
+
+//CALCULATIONS
+T2 = T+273; //Given temperature in K
+V1 = V2*(P2/P1)*(T1/T2); //Volume of water vapor at standard conditions in Cu ft
+n = V1/359; //Moles of water vapor in lb mole
+W = n*MW; //Weight of water vapor in lb
+
+//OUTPUT
+mprintf('\n Weight of %3.0f cu ft of water vapor at given conditions is %5.4f lb',V2,W);
+
+//=================================END OF PROGRAM==============================
+\ No newline at end of file
diff --git a/926/CH3/EX3.3/Chapter3_Example3.sce b/926/CH3/EX3.3/Chapter3_Example3.sce
new file mode 100644
index 000000000..fea140094
--- /dev/null
+++ b/926/CH3/EX3.3/Chapter3_Example3.sce
@@ -0,0 +1,26 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 3, Page 54
+//Title: Calculation of pressure
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 10; //Weight of CO2 in lb
+MW = 44; //Molecular weight of CO2 in lb/lb mole
+P1 = 14.7; //Pressure at standard condition in psi
+T1 = 273; //Temperature at standard conditions in K
+V2 = 20; //Volume of CO2 after compressing in cu ft
+T = 30; //Given temperature in degree C
+
+//CALCULATIONS
+n = W/MW; //No of noles of CO2 in lb mole
+V1 = n*359; //Volume of CO2 at standard conditions in cu ft
+T2 = T+273; //Given temperature in K
+P2 = P1*(V1/V2)*(T2/T1); //Pressure at given conditions in psi
+
+//OUTPUT
+mprintf('\n Pressure required to compress %2.0f lb of CO2 to a volume of %2.0f cu ft at given conditions is %3.1f psi',W,V2,P2);
+
+//=================================END OF PROGRAM==============================
+\ No newline at end of file
diff --git a/926/CH3/EX3.4/Chapter3_Example4.sce b/926/CH3/EX3.4/Chapter3_Example4.sce
new file mode 100644
index 000000000..426502ab9
--- /dev/null
+++ b/926/CH3/EX3.4/Chapter3_Example4.sce
@@ -0,0 +1,26 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 4, Page 55
+//Title: Calculation of temperature
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 10; //Weight of nitrogen in lb
+MW = 28; //Molecular weight of nitrogen in lb/lb mole
+P1 = 14.7; //Pressure at standard conditions in psi
+T1 = 273; //Temperature at standard conditions in K
+P2 = 150; //Maximum pressure in psi
+V2 = 30; //Given volume of nitrogen in cu ft
+
+//CALCULATIONS
+n = W/MW; //No of moles of Nitrogen at standard conditions in lb mole
+V1 = n*359; //Volume of nitrogen at standard conditions in cu ft
+T2 = T1*(P2/P1)*(V2/V1); //Temperature at given conditions in K
+T = T2-273; //Temperature at given conditions in  degree C
+
+//OUTPUT
+mprintf('\n The maximum temperature to which %2.0f lb of nitrogen can be heated is %4.0f K or %4.0f degree C',W,T2,T);
+
+//========================= END OF PROGRAM ====================================
+\ No newline at end of file
diff --git a/926/CH3/EX3.5/Chapter3_Example5.sce b/926/CH3/EX3.5/Chapter3_Example5.sce
new file mode 100644
index 000000000..cda2af729
--- /dev/null
+++ b/926/CH3/EX3.5/Chapter3_Example5.sce
@@ -0,0 +1,29 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 5, Page 55
+//Title: Calculation of percentage dissociation of gas
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 17.2; //Weight of N2O4 in grams
+V1 = 11450; //Volume occupied by N2O4 in cc
+P1 = 720; //Given pressure in  mm hg
+T = 100; //Given temperature in degree C
+MW = 92; //Molecular weight of N2O4 in g/g mole
+V2 = 22400; //Volume at standard conditions in cc
+P2 = 760; //Pressure at standard conditions in mm Hg
+T2 = 273; //Temperature at standard conditions in K
+
+//CALCULATIONS
+T1 = T+273; //Given temperature in K
+n1 = W/MW; //No of moles of N2O4 initially present in g mole
+n2 = (V1/V2)*(T2/T1)*(P1/P2); //Total g moles present after dissociation
+x = n2-n1; //g moles of N2O4 dissociated
+X = x*100/n1; //Percentage dissociation
+
+//OUTPUT
+mprintf('\n Percentage dissociation of N2O4 to NO2 is %2.0f percent',X);
+
+//=========================END OF PROGRAM======================================
+\ No newline at end of file
diff --git a/926/CH3/EX3.6/Chapter3_Example6.sce b/926/CH3/EX3.6/Chapter3_Example6.sce
new file mode 100644
index 000000000..a079b3f4b
--- /dev/null
+++ b/926/CH3/EX3.6/Chapter3_Example6.sce
@@ -0,0 +1,25 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 6, Page 59
+//Title: Calculation of Average molecular weight
+//=============================================================================
+clear
+clc
+
+//INPUT
+N = 1; //Total no moles of mixture
+V = [13.1,7.7,79.2]; //Composition of Carbon dioxide, Oxygen and nitrogen respectively by volume %
+MW = [44,32,28]; //Molecular weight of CO2, O2 and N2 respectively in g/g mole
+
+//CALCULATION
+n = V/100; // No of moles of CO2, O2, n2 in g mole
+W1 = n(1)*MW(1); //Weight of CO2 in grams
+W2 = n(2)*MW(2); //Weight of O2 in grams
+W3 = n(3)*MW(3); //Weight of N2 in grams
+W = W1+W2+W3; //Total weight of flue gas in grams
+AV_MW = W/N; //Average molecular weight of flue gas in g/g mole
+
+//OUTPUT
+mprintf('\n The average molecular weight of flue gas is %4.2f g/g mole',AV_MW);
+
+//========================END OF PROGRAM======================================
+\ No newline at end of file
diff --git a/926/CH3/EX3.7/Chapter3_Example7.sce b/926/CH3/EX3.7/Chapter3_Example7.sce
new file mode 100644
index 000000000..80562526a
--- /dev/null
+++ b/926/CH3/EX3.7/Chapter3_Example7.sce
@@ -0,0 +1,30 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 7, Page 59
+//Title: Calculation of density
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 1; //Total weight of mixture in lb
+w = [.111,.889]; //Weight of hydrogen and oxygen respectively in lb
+MW = [2,32]; //Molecular weight of hydrogen and oxygen respectively in lb/lb mole
+T = 30; //Given temperature in degree C
+T2 = 273; //Temperature at standard conditions in K
+P1 = 29; //Given prssure in in. Hg
+P2 = 29.92; //Pressure at standard conditions in in. Hg
+
+//CALCULATIONS
+n1 = w(1)/MW(1); //No of moles of H2 in lb mole
+n2 = w(2)/MW(2); //No of moles of O2 in lb mole
+N = n1+n2; //Total molal quantity in lb mole
+T1 = T+273; //given temperature in K
+V2 = N*359; //Volume at standard condition in cu ft
+V1= V2*(P2/P1)*(T1/T2); //Volume of the mixture at given conditions in cu ft
+rho = W/V1; //Density of the mixture at given conditions in lb per cu ft
+
+//OUTPUT
+mprintf('\n The density of the mixture at %2.0f in. Hg and %2.0f degree C is %5.4f lb per cu ft',P1,T,rho);
+
+//====================END OF PROGRAM===========================================
diff --git a/926/CH3/EX3.8/Chapter3_Example8.sce b/926/CH3/EX3.8/Chapter3_Example8.sce
new file mode 100644
index 000000000..110a96fb7
--- /dev/null
+++ b/926/CH3/EX3.8/Chapter3_Example8.sce
@@ -0,0 +1,30 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 8, Page 61
+//Title: Calculation of density of air
+//=============================================================================
+clear
+clc
+
+//INPUT
+N = 1; //Total no moles of air in g mole
+n = [.21,.79]; //No of moles of O2 and N2 respectively in g mole
+MW = [32,28]; //Moleculaw weight of O2 and N2 respectively in g/g mole
+T = 70; //Given temperature in dergee F
+P1 = 741; //Given pressure in mm Hg
+T2 = 492;  //Temperature at standard conditions in degree R
+P2 = 760; //Pressure at standard conditions in mm Hg
+V2 = 22.41; //Volume at standard conditions in liters
+
+//CAlCULATIONS
+w1 = n(1)*MW(1); //Weight of O2 in grams
+w2 = n(2)*MW(2); //Weight of N2 in grams
+W = w1+w2; //Total weight of air in grams
+T1 = T+460; //Given temperature in dergree R
+V1 = V2*(P2/P1)*(T1/T2); //Volume of air at given conditions in liters
+rho = W/V1; //Dendity of air at given conditions in grams per liter
+
+//OUTPUT
+mprintf('\n The density of air at %3.0f mm Hg and %2.0f degree F is %4.3f grams per liter', P1,T,rho);
+
+//==========================END OF PROGRAM=====================================
diff --git a/926/CH3/EX3.9/Chapter3_Example9.sce b/926/CH3/EX3.9/Chapter3_Example9.sce
new file mode 100644
index 000000000..216f6ce27
--- /dev/null
+++ b/926/CH3/EX3.9/Chapter3_Example9.sce
@@ -0,0 +1,39 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 9, Page 61
+//Title: Calculation of volume change with change in composition
+//=============================================================================
+clear
+clc
+
+//INPUT
+Nt = 1; //Total no of moles of combustion gas in g mole
+N = [.792,.072,.136]; //No of moles of N2, O2 and CO2 respectively before evaporation in g mole
+T1 = 200; //Temperature of gases passed into evaporator in degree C
+P1 = 743; //Pressure of gases passed into evaporator in mm Hg
+n = [.483,.044,.083,.39]; //No of moles of N2, O2, CO2 and water respectively after evaporation in g mole
+T2 = 85; //Temperature of gases after evaporation in degree C
+P2 = 740; //Pressure of gases after evaporation in mm Hg
+MW = [28,32,44,18]; //Molecular weight of N2, O2, CO2 and water respectively in g/ g mole
+R = 82.1; //Ideal gas constant in cc-atm per K
+v = 100; //Volume of gas entering in cu ft
+
+//CALCULATIONS
+P = P1/760; //Pressure of gases passed to evaporator in atm
+T3 = T1+273; //Temperature of gases passed to evaporator in K
+v1 = (N(1)+N(2)+N(3))*R*T3/P; //Total volume of gases passed into evaporator in cc
+v2 = v1*3.2808^3*(10^-6); //Total volume of gases passed into evaporator in cu ft
+n1 = Nt/(n(1)+n(2)+n(3)); //No of moles of gases leaving evaporator in g mole
+n2 = n1 -Nt; //No of moles of water leaving the evaporator in g mole
+p = P2/760; //Pressure of gases leaving evaporator in atm
+T4 = T2+273; //Temperature of gases laving evaporator in K
+V1 = (N(1)+N(2)+N(3)+n2)*R*T4/p; //Volume of gases leaving in cc
+V2 = V1*3.2808^3*10^-6; //Volume of gases leaving evaporator in cu ft
+V = (V2/v2)*v; //Volume of gas leaving in cu ft per 100 cu ft entering
+w = n2*MW(4)*2.2046*10^-3; //weight of water leaving evaporator in lb
+W = w*(v/v2); //Weight of water evaporated in lb per 100 ft of gas entering 
+
+//OUTPUT
+mprintf('\n Weight of water evaporated per %3.0f cu ft of gas entering is %3.2f lb',v,W);
+
+//===========================END OF PROGRAM====================================
+\ No newline at end of file
diff --git a/926/CH5/EX5.1/Chapter5_Example1.sce b/926/CH5/EX5.1/Chapter5_Example1.sce
new file mode 100644
index 000000000..d2ca16bb0
--- /dev/null
+++ b/926/CH5/EX5.1/Chapter5_Example1.sce
@@ -0,0 +1,52 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 1, Page 111
+//Title: Calculation of composition of a saturated mixture
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [442,745 760]; //Vapor pressure of ethyl ether, working pressure and standard pressure in mm Hg
+T1 = [20 0]; //Temperature of system and standard temperature in degree C
+V = 1; //cu ft of mixture(Basis of calculation for part(a) )
+MW = [74 28]; //Molecular weight of ethyl ether and nitrogen in lb/lb-mole
+
+//CALCULATIONS
+//Part(a)
+V1 = V*(P(1)/P(2)); //Pure component volume of vapor in cu ft
+v1 = V1*100; //Composition by volume of Ether vapor
+v2 = (1-V1)*100; //Composition by volume of Nitrogen
+//Part(b)
+W1 = V1*MW(1); //lb of ether vapor present
+W2 = (1-V1)*MW(2); //lb of Nitrogen present 
+W = W1+W2; //Total weight of mixture in lb
+w1 = W1*100/W; //Composition by weight of Ether vapor
+w2 = W2*100/W; //Composition by weight of Nitrogen
+//Part(c)
+T = T1+273; //Converting temperature in K
+V2 = 359*(P(3)/P(2))*(T(1)/T(2)); //cu ft of mixture
+w3 = W1/V2; //lb of vapor per cu ft of mixture
+//Part(d)
+w4 = W1/W2; //lb of vapor per lb lb of nitrogen
+//Part(e)
+v3 = V1/(1-V1); //lb mole of vapor per lb moles of Nitrogen
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Composition by Volume \n     Ether Vapor %3.1f %% \n     Nitrogen %3.1f %% ',v1,v2);
+mprintf('\n (b) Composition by Weight \n     Ether Vapor %3.1f %% \n     Nitrogen %3.1f %% ',w1,w2);
+mprintf('\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mprintf('\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mprintf('\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+
+// File output
+fd= mopen('.\Chapter5_Example1_Output.txt','w');
+mfprintf(fd,'\n (a) Composition by Volume \n     Ether Vapor %3.1f %% \n     Nitrogen %3.1f %%',v1,v2);
+mfprintf(fd,'\n (b) Composition by Weight \n     Ether Vapor %3.1f %% \n     Nitrogen %3.1f %%',w1,w2);
+mfprintf(fd,'\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mfprintf(fd,'\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mfprintf(fd,'\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.1/Chapter5_Example1_Output.txt b/926/CH5/EX5.1/Chapter5_Example1_Output.txt
new file mode 100644
index 000000000..685f90ae6
--- /dev/null
+++ b/926/CH5/EX5.1/Chapter5_Example1_Output.txt
@@ -0,0 +1,10 @@
+
+ (a) Composition by Volume 
+     Ether Vapor 59.3 % 
+     Nitrogen 40.7 %
+ (b) Composition by Weight 
+     Ether Vapor 79.4 % 
+     Nitrogen 20.6 %
+ (c) Weight of ether per cu ft of mixture is 0.112 lb 
+ (d) Weight of vapor per lb Nitrogen is 3.86 lb 
+ (e) Moles of vapor per mole of nitrogen is 1.459
+\ No newline at end of file
diff --git a/926/CH5/EX5.11/Chapter5_Example11.sce b/926/CH5/EX5.11/Chapter5_Example11.sce
new file mode 100644
index 000000000..2c2a839d3
--- /dev/null
+++ b/926/CH5/EX5.11/Chapter5_Example11.sce
@@ -0,0 +1,35 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 11, Page 127
+//Title: Calculation of dry bulb and wet bulb temperature
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [190 90]; //Dry bulb and wet bulb temperature of air entering dryer in degree F
+n = 0.028; //lb-mole of water evaporated per lb mole of dry air entering
+
+//DATA FROM GRAPH
+mH = 0.011; //Molal humidity corresponding to DBT of 190 degree F and WBT of 90 degree F from Fig 19, Pg 120
+DBT = 116; //Dry bulb temperature in degree F corresponding to molal humidity of 0.039 from Fig 19, Pg 120
+PS = 35; //Percentage saturation corresponding to molal humidity of 0.039 and DBT of 116 degree F from Fig 19, Pg 120
+
+//CALCULATIONS
+mH1 = n+mH; //Molal humidity of leaving air
+WBT = T(2); //Wet bulb temperature of leaving air in degree F
+
+//OUTPPUT
+// Console output
+mprintf('\n Dry bulb temperature of air leaving the drier = %3.0f degree F',WBT);
+mprintf('\n wet bulb temperature of air leaving the drier = %2.0f degree F',WBT);
+mprintf('\n Percentage saturation of air leaving the drier is %2.0f percent',PS);
+
+// File output
+fd= mopen('.\Chapter5_Example11_Output.txt','w');
+mfprintf(fd,'\n Dry bulb temperature of air leaving the drier = %3.0f degree F',WBT);
+mfprintf(fd,'\n wet bulb temperature of air leaving the drier = %2.0f degree F',WBT);
+mfprintf(fd,'\n Percentage saturation of air leaving the drier is %2.0f percent',PS);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.11/Chapter5_Example11_Output.txt b/926/CH5/EX5.11/Chapter5_Example11_Output.txt
new file mode 100644
index 000000000..8866d0175
--- /dev/null
+++ b/926/CH5/EX5.11/Chapter5_Example11_Output.txt
@@ -0,0 +1,4 @@
+
+ Dry bulb temperature of air leaving the drier =  90 degree F
+ wet bulb temperature of air leaving the drier = 90 degree F
+ Percentage saturation of air leaving the drier is 35 percent
+\ No newline at end of file
diff --git a/926/CH5/EX5.12/Chapter5_Example12.sce b/926/CH5/EX5.12/Chapter5_Example12.sce
new file mode 100644
index 000000000..669dd7fa4
--- /dev/null
+++ b/926/CH5/EX5.12/Chapter5_Example12.sce
@@ -0,0 +1,26 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 12, Page 127
+//Title: Calculation of molal humidity
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = 1; //Pressure of entering gas in atm
+DBT = 120; //Temperature of entering gas in degree F
+
+//DATA FROM GRAPH
+WBT = 71; //Wet bulb temperature in degree F corresponding to DBT of 120 degree F from Fig 20 Page 122
+mH = 0.027; //Molal humidity corresponding to DBT of 120 degree F from Fig 20 Page 122
+
+//OUTPUT
+// Console output
+mprintf('\n The temperature and molal humidity of saturated carbon dioxide leaving the chamber is %2.0f degree F and %4.3f respectively',WBT,mH);
+
+// File output
+fd= mopen('.\Chapter5_Example12_Output.txt','w');
+mfprintf(fd,'\n The temperature and molal humidity of saturated carbon dioxide leaving the chamber is %2.0f degree F and %4.3f respectively',WBT,mH);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH5/EX5.12/Chapter5_Example12_Output.txt b/926/CH5/EX5.12/Chapter5_Example12_Output.txt
new file mode 100644
index 000000000..ead55fe14
--- /dev/null
+++ b/926/CH5/EX5.12/Chapter5_Example12_Output.txt
@@ -0,0 +1,2 @@
+
+ The temperature and molal humidity of saturated carbon dioxide leaving the chamber is 71 degree F and 0.027 respectively
+\ No newline at end of file
diff --git a/926/CH5/EX5.2/Chapter5_Example2.sce b/926/CH5/EX5.2/Chapter5_Example2.sce
new file mode 100644
index 000000000..426e5302a
--- /dev/null
+++ b/926/CH5/EX5.2/Chapter5_Example2.sce
@@ -0,0 +1,34 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 2, Page 113
+//Title: Calculation of relative saturation and percentage saturation
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [745 184.8] //Working pressure and vapor pressure of acetone at 20 degree C in mm Hg
+v = 14.8 //Composition by volume of acetone
+
+//CALCULATIONS
+P1 = P(1)*v/100; //Partial pressure of acetone in mm Hg
+RS = P1*100/P(2); //Relative saturation of mixture
+n1 = v/100; //lb mole of acetone
+n2 = 1-n1; //lb mole of nitrogen
+n3 = n1/n2; //lb moles of acetone per lb moles of nitrogen
+V1 = P(2)*100/P(1); //Percentage by volume of acetone
+n4 = V1/100; //lb moles of acetone
+n5 = 1-n4; //lb moles of nitrogen
+n6 = n4/n5; //Moles of acetone per moles of nitrogen
+PS = n3*100/n6; //Percentage saturation
+
+//OUTPUT
+//Console output
+mprintf('\n Relative saturation of the mixture at given conditions = %3.1f %% \n Percentage saturation = %3.1f %%',RS,PS);
+
+// File output
+fd= mopen('.\Chapter5_Example2_Output.txt','w');
+mfprintf(fd,'\n Relative saturation of the mixture at given conditions = %3.1f %% \n Percentage saturation = %3.1f %%',RS,PS);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.2/Chapter5_Example2_Output.txt b/926/CH5/EX5.2/Chapter5_Example2_Output.txt
new file mode 100644
index 000000000..8cabf5efd
--- /dev/null
+++ b/926/CH5/EX5.2/Chapter5_Example2_Output.txt
@@ -0,0 +1,3 @@
+
+ Relative saturation of the mixture at given conditions = 59.7 % 
+ Percentage saturation = 52.7 %
+\ No newline at end of file
diff --git a/926/CH5/EX5.4/Chapter5_Example4.sce b/926/CH5/EX5.4/Chapter5_Example4.sce
new file mode 100644
index 000000000..67097606e
--- /dev/null
+++ b/926/CH5/EX5.4/Chapter5_Example4.sce
@@ -0,0 +1,36 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 4, Page 114
+//Title: Calculation of dew point
+//=============================================================================
+clear
+clc
+
+//INPUT
+v1 = 10.1; //Composition by volume of benzene
+P = [750 750 700]; //Various pressurea in mm Hg at which dew point is to be computed
+
+//CALCULATIONS
+P1 = P(1)*v1/100; //Partial pressure of benzene in mm Hg at 25 degree C and 750 mm Hg
+T1 = 20; //Temperature in degree C corresponding to pressure P1 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+P2 = P(2)*v1/100; //Partial pressure of benzene in mm Hg at 30 degree C and 750 mm Hg
+T2 = 20; //Temperature in degree C corresponding to pressure P2 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+P3 = P(3)*v1/100; //Partial pressure of benzene in mm Hg at 30 degree C and 700 mm Hg
+T3 = 18.7; //Temperature in degree C corresponding to pressure P3 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+
+//OUTPUT
+// Console output
+mprintf('\n Dew point of benzene vapor and air mixture at \n (a) 25 degree C and 750 mm Hg = %2.0f degee C ',T1);
+mprintf('\n (b) 30 degree C and 750 mm Hg = %2.0f degree C ',T3);
+mprintf('\n (c) 30 degree C and 700 mm Hg = %3.1f degree C',T3);
+mprintf('\n Above results shows that the dew point does not depend on temperature but vary with the total pressure');
+
+// File output
+fd= mopen('.\Chapter5_Example4_Output.txt','w');
+mfprintf(fd,'\n Dew point of benzene vapor and air mixture at \n (a) 25 degree C and 750 mm Hg = %2.0f degee C ',T1);
+mfprintf(fd,'\n (b) 30 degree C and 750 mm Hg = %2.0f degree C ',T3);
+mfprintf(fd,'\n (c) 30 degree C and 700 mm Hg = %3.1f degree C',T3);
+mfprintf(fd,'\n Above results shows that the dew point does not depend on temperature but vary with the total pressure');
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
diff --git a/926/CH5/EX5.4/Chapter5_Example4_Output.txt b/926/CH5/EX5.4/Chapter5_Example4_Output.txt
new file mode 100644
index 000000000..a57b61d69
--- /dev/null
+++ b/926/CH5/EX5.4/Chapter5_Example4_Output.txt
@@ -0,0 +1,6 @@
+
+ Dew point of benzene vapor and air mixture at 
+ (a) 25 degree C and 750 mm Hg = 20 degee C 
+ (b) 30 degree C and 750 mm Hg = 19 degree C 
+ (c) 30 degree C and 700 mm Hg = 18.7 degree C
+ Above results shows that the dew point does not depend on temperature but vary with the total pressure
+\ No newline at end of file
diff --git a/926/CH5/EX5.5/Chapter5_Example5.sce b/926/CH5/EX5.5/Chapter5_Example5.sce
new file mode 100644
index 000000000..29cb289c8
--- /dev/null
+++ b/926/CH5/EX5.5/Chapter5_Example5.sce
@@ -0,0 +1,54 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 5, Page 116
+//Title: Calculation of different concentrations in vaporization process
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [750 760]; //Barometric pressure and standard pressure in mm Hg
+PS = [116 185]; //Vapor pressure of acetone at 10 degree C and 20 degree C in mm Hg
+T = [30 25 0]; //Temperature of nitrogen entering evaporator and standard temperature in degree C
+n = 1; //lb mole of nitrogen (Basis of calculation in part(c),part(d) and part(e))
+MW = 58; //Molecualar weight of acetone in lb/lb mole
+
+//CALCULATIONS
+//Part(a)
+//Entering gases
+PP1 = PS(1); //Partial pressure of acetone in mm Hg
+PP2 = P(1)-PP1; //Partial pressure of nitrogen in mm Hg
+n1 = PP1/PP2; //Moles of acetone per mole of nitrogen
+//leaving gases
+PP3 = PS(2); //Partial pressure of acetone in mm Hg
+PP4 = P(1)-PS(2); //Partial pressure of nitrogen in mm Hg
+n2 = PP3/PP4; //Moles of acetone per mole of nitrogen
+//Part(b)
+n3 = n2-n1; //lb mole of acetone evaporated
+//Part(c)
+Tkelvin = T+273; //Converting temperature fron degree C to kelvin
+n4 = n+n1; //lb mole of total gas entering the process
+v1 = n4*359*(P(2)/P(1))*(Tkelvin(1)/Tkelvin(3)); //Volume of gas entering in cu ft
+m1 = MW*n3; //Weight of acetone evaporated
+m2 = m1*1000/v1; //lb of acetone evaporated per 1000 cu ft of gas entering
+//Part(d)
+n5 = n+n2; //lb mole of total gas leaving the process
+v2 = n5*359*(P(2)/P(1))*(Tkelvin(2)/Tkelvin(3)); //Volume of gas leaving in cu ft
+v3 = v2*1000/v1; //cu ft of gas leaving per 1000 cu ft of gas entering the process
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Vapor concentration of gases entering and leaving = %4.3f and %4.3f respectively',n1,n2);
+mprintf('\n (b) Moles of acetone evaporated = %4.3f lb moles',n3);
+mprintf('\n (c) Weight of acetone evaporated per 1000 cu ft of gas entering = %3.1f lb',m2);
+mprintf('\n (d) Volume of gas leaving per 1000 cu ft of gas entering = %4.0f cu ft',v3);
+
+// File output
+fd= mopen('.\Chapter5_Example5_Output.txt','w');
+mfprintf(fd,'\n (a) Vapor concentration of gases entering and leaving = %4.3f and %4.3f respectively',n1,n2);
+mfprintf(fd,'\n (b) Moles of acetone evaporated = %4.3f lb moles',n3);
+mfprintf(fd,'\n (c) Weight of acetone evaporated per 1000 cu ft of gas entering = %3.1f lb',m2);
+mfprintf(fd,'\n (d) Volume of gas leaving per 1000 cu ft of gas entering = %4.0f cu ft',v3);
+mclose(fd);
+
+//===========================END OF PROGRAM====================================
diff --git a/926/CH5/EX5.5/Chapter5_Example5_Output.txt b/926/CH5/EX5.5/Chapter5_Example5_Output.txt
new file mode 100644
index 000000000..a1b23ddc3
--- /dev/null
+++ b/926/CH5/EX5.5/Chapter5_Example5_Output.txt
@@ -0,0 +1,5 @@
+
+ (a) Vapor concentration of gases entering and leaving = 0.183 and 0.327 respectively
+ (b) Moles of acetone evaporated = 0.144 lb moles
+ (c) Weight of acetone evaporated per 1000 cu ft of gas entering = 17.5 lb
+ (d) Volume of gas leaving per 1000 cu ft of gas entering = 1104 cu ft
+\ No newline at end of file
diff --git a/926/CH5/EX5.6/Chapter5_Example6.sce b/926/CH5/EX5.6/Chapter5_Example6.sce
new file mode 100644
index 000000000..e32eab562
--- /dev/null
+++ b/926/CH5/EX5.6/Chapter5_Example6.sce
@@ -0,0 +1,56 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 6, Page 118
+//Title: Calculation of different concentrations in vaporization process
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [750 760]; //Given pressure and standard pressure in mm Hg
+RH = 0.8; //Relative humidity
+PS = [17.5 9.2]; //Vapor pressure of water in mm Hg at 20 degree C and 10 degree C
+P_new = 35; //Increased pressure by condensing out some water in psi
+V = 1000; //Volume of original wet water in cu ft (basis for calculation in part (c))
+T = [20 0]; //Given temperature and standard temperature in degree C
+T_new = 10; //Reduced temperature by condensing out some water in degree C
+MW = 18; //Molecular weight of water in lb/lb-mole
+
+//CALCULATIONS
+//Part(a)
+PP1 = RH*PS(1); //initial partial pressure of water in mm Hg
+MH1 = PP1/(P(1)-PP1); //Initial molal humidity
+//Part(b)
+PP2 = PS(2); //Final partial pressure of water in mm Hg
+P_new1 = 35*(P(2)/14.7); //Final total pressure in mm Hg
+MH2 = PP2/(P_new1-PP2); //Final molal humidity
+//Part(c)
+T1 = T+273; //Given temperature and standard temperature in K
+PP3 = P(1)-PP1; //Partial pressure of dry air in mm Hg
+PV = V*(PP3/P(2))*(T1(2)/T1(1)); //partial volume of dry air at standard condition in cu ft
+n1 = PV/359; //Moles of dry air in lb-moles
+n2 = n1*MH1; //lb-mole of water originally present
+n3 = n1*MH2; //lb-mole of water finally present
+n4 = n2-n3; //lb-mole of water condensed
+w1 = n4*MW; //lb of water condensed
+//Part(d)
+T_new1 = T_new+273; //Final temperature in K
+n5 = n1+n3; //Total wet air finally present in lb-moles
+v1 = n5*359*(P(2)/P_new1)*(T_new1/T1(2)); //Final volume of wet air in cu ft
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Initial molal humidity of air = %5.4f',MH1);
+mprintf('\n (b) Final molal humidity of air = %5.4f',MH2); 
+mprintf('\n (c) Amount of water condensed = %4.3f lb',w1);
+mprintf('\n (d) Final volume of wet air = %3.0f cu ft',v1);
+
+// File output
+fd= mopen('.\Chapter5_Example6_Output.txt','w');
+mfprintf(fd,'\n (a) Initial molal humidity of air = %5.4f',MH1);
+mfprintf(fd,'\n (b) Final molal humidity of air = %5.4f',MH2); 
+mfprintf(fd,'\n (c) Amount of water condensed = %4.3f lb',w1);
+mfprintf(fd,'\n (d) Final volume of wet air = %3.0f cu ft',v1);
+mclose(fd);
+
+//===============================END OF PROGRAM=================================
diff --git a/926/CH5/EX5.6/Chapter5_Example6_Output.txt b/926/CH5/EX5.6/Chapter5_Example6_Output.txt
new file mode 100644
index 000000000..10f9c5210
--- /dev/null
+++ b/926/CH5/EX5.6/Chapter5_Example6_Output.txt
@@ -0,0 +1,5 @@
+
+ (a) Initial molal humidity of air = 0.0190
+ (b) Final molal humidity of air = 0.0051
+ (c) Amount of water condensed = 0.629 lb
+ (d) Final volume of wet air = 395 cu ft
+\ No newline at end of file
diff --git a/926/CH5/EX5.7/Chapter5_Example7.sce b/926/CH5/EX5.7/Chapter5_Example7.sce
new file mode 100644
index 000000000..dcc8a2c22
--- /dev/null
+++ b/926/CH5/EX5.7/Chapter5_Example7.sce
@@ -0,0 +1,35 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 7, Page 119
+//Title: Calculation of condensation pressure
+//=============================================================================
+clear
+clc
+
+//INPUT
+VP = [229.2 89.1]; //Vapour pressure of acetone at 25 degree C and 5 degree C
+P = 750; //Pressure of given mixture in mm Hg
+w = 90; //Amount of acetone to be condensed;
+
+
+//CALCULATION
+n1 = VP(1)/P; //lb moles of acetone present
+n2 = 1-n1; //lb moles of flue gases present
+n3 = (1-(w/100))*n1; //lb moles of acetone in final mixture
+n4 = n2+n3; //lb moles of final mixture of gas
+PP = VP(2); //Partial pressure of acetone in final mixture
+m = n3*100/n4; //Mole percent of acetone in final mixture
+Pf = PP/(m/100); //Final pressure in mm Hg
+
+//OUTPUT
+// Console output
+mprintf('\n Final pressure after condensation = %.0f mm Hg',Pf);
+
+// File output
+fd= mopen('.\Chapter5_Example7_Output.txt','w');
+mfprintf(fd,'\n Final pressure after condensation = %.0f mm Hg',Pf);
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
+
+// Remark: Difference between the nswer given in the textbook (2110 mm Hg) and that computed using scilab code (2114 mm Hg) is due to round off error 
diff --git a/926/CH5/EX5.7/Chapter5_Example7_Output.txt b/926/CH5/EX5.7/Chapter5_Example7_Output.txt
new file mode 100644
index 000000000..5389ba5df
--- /dev/null
+++ b/926/CH5/EX5.7/Chapter5_Example7_Output.txt
@@ -0,0 +1,2 @@
+
+ Final pressure after condensation = 2114 mm Hg
+\ No newline at end of file
diff --git a/926/CH5/EX5.8/Chapter5_Example8.sce b/926/CH5/EX5.8/Chapter5_Example8.sce
new file mode 100644
index 000000000..e36a859ad
--- /dev/null
+++ b/926/CH5/EX5.8/Chapter5_Example8.sce
@@ -0,0 +1,43 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 8, Page 123
+//Title: Estimation of humidity and saturation
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [100 85]; //Initial and wet bulb temperature of entering air in degree F
+P = 1; //Pressure of entering air in atm
+n = 1; //lb-moles of moisture free air
+T1 = 492; //Temperature corresponsing to STP in Rankine
+T2 = [120 115.3]; //Initial and wet bulb temperature of air in degree F in part(b)
+MW = 18; //Molecular weight of water
+
+//DATA FROM GRAPH
+PS = [52 84]; //Percentage saturation at WBT of 85 and 115.3 degree C respectively obtained from Fig 20 Page 122
+MH = [0.037 0.11 ]; //Molal humidity at WBT of 85 and 115.3 degree C respectively obtained from Fig 20 Page 122
+DP = 80.5; //Dew point corresponding to WBT of 85 degree F obtained from Fig 20 Page 122
+
+
+//CALCULATION
+//Part(b)
+n1 = n+MH(1); //lb-moles of wet air entering
+T3 = T+460; //initial temperature of entering air in Rankine
+v = n1*359*(T3(1)/T1); //Volume of wet air entering in cu ft
+n2 = MH(2)-MH(1); //lb-moles of water evaporated
+w1 = n2*MW; //lb of water evaporated
+w2 = w1*1000/v; //lb of water evaporated per 1000 cu ft of entering wet air
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Molal humidity of the air = %4.3f \n     Percentage saturation = %2.0f %%\n     Dew point = %3.2f degree F',MH(1),PS(1),DP);
+mprintf('\n (b) Percentage saturation of the air leaving the evaporator = %2.0f %% \n     weight of the water evaporated = %2.1f lb',PS(2),w2);
+
+// File output
+fd= mopen('.\Chapter5_Example8_Output.txt','w');
+mfprintf(fd,'\n (a) Molal humidity of the air = %4.3f \n     Percentage saturation = %2.0f %%\n     Dew point = %3.2f degree F',MH(1),PS(1),DP);
+mfprintf(fd,'\n (b) Percentage saturation of the air leaving the evaporator = %2.0f %% \n     weight of the water evaporated = %2.1f lb',PS(2),w2);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH5/EX5.8/Chapter5_Example8_Output.txt b/926/CH5/EX5.8/Chapter5_Example8_Output.txt
new file mode 100644
index 000000000..02693e6fb
--- /dev/null
+++ b/926/CH5/EX5.8/Chapter5_Example8_Output.txt
@@ -0,0 +1,6 @@
+
+ (a) Molal humidity of the air = 0.037 
+     Percentage saturation = 52 %
+     Dew point = 80.50 degree F
+ (b) Percentage saturation of the air leaving the evaporator = 84 % 
+     weight of the water evaporated = 3.1 lb
+\ No newline at end of file
diff --git a/926/CH5/EX5.9/Chapter5_Example9.sce b/926/CH5/EX5.9/Chapter5_Example9.sce
new file mode 100644
index 000000000..b6ddb4c9d
--- /dev/null
+++ b/926/CH5/EX5.9/Chapter5_Example9.sce
@@ -0,0 +1,27 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 9, Page 124
+//Title: Estimation of dry bulb and wet bulb temperature
+//=============================================================================
+clear
+clc
+
+//INPUT
+m = [12.1 0.1 7.6 80.2]; //Composition of CO2, CO, O2, N2 in percentage
+
+//DATA FROM GRAPH
+//Part(a)
+T1 = 87; //Wet bulb temperature obtained from Fig 19, Page 120 & Fig 20, 122
+//Part(b)
+T2 = 140; //Dry bulb temperature obtained from Fig 19, Page 120 & Fig 20, 122
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Wet bulb temperature = %2.0f degree F \n (b) Dry bulb temperature = %3.0f degree F',T1,T2);
+
+// File output
+fd= mopen('.\Chapter5_Example9_Output.txt','w');
+mfprintf(fd,'\n (a) Wet bulb temperature = %2.0f degree F \n (b) Dry bulb temperature = %3.0f degree F',T1,T2);
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
diff --git a/926/CH5/EX5.9/Chapter5_Example9_Output.txt b/926/CH5/EX5.9/Chapter5_Example9_Output.txt
new file mode 100644
index 000000000..b72f90091
--- /dev/null
+++ b/926/CH5/EX5.9/Chapter5_Example9_Output.txt
@@ -0,0 +1,3 @@
+
+ (a) Wet bulb temperature = 87 degree F 
+ (b) Dry bulb temperature = 140 degree F
+\ No newline at end of file
diff --git a/926/CH8/EX8.2/Chapter8_Example2.sce b/926/CH8/EX8.2/Chapter8_Example2.sce
new file mode 100644
index 000000000..d62429dc2
--- /dev/null
+++ b/926/CH8/EX8.2/Chapter8_Example2.sce
@@ -0,0 +1,38 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 2, Page 275
+//Title: Calculation of molal heat of vapourization
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [1 8]; //Given pressure in atm
+Tb = [56.5 133.1]; //Boiling point of acetone at 1 atm and 8 atm pressure in degree C
+TC = 508.7; //Critical temperature of acetone in K
+PC = 46.6; //Critical pressure of acetone in atm
+Com_fac = [.966 .822]; //(ZG - ZL) value of acetone at 1 and 8 atm respectively from Table 26A, Page 275
+A = 3.0644; //Vapor pressure constant of acetone from Table 8, Page 95
+b = .180; //Vapor pressure constant of acetone from Table 8, Page 95
+R = 1.987; //gas constant in 
+
+//CALCULATION
+TB = Tb+273.15; //Boiling point of acetone at 1 atm and 8 atm pressure in K
+TR = TB/TC; //reduced temperature of acetone at 1 and 8 atm pressure respectively in degree C
+lamda = 2.303*Com_fac*R*TC*A;
+lamda1 = 2.303*Com_fac(1)*R*TC*(A+(40*TR(1)^2*(TR(1)-b)*exp(-20*(TR(1)-b)^2)));
+lamda2 = 2.303*Com_fac(2)*R*TC*(A+(40*TR(2)^2*(TR(2)-b)*exp(-20*(TR(2)-b)^2)));
+
+//OUTPUT
+// Console Output
+mprintf('\n Molal heat of vaporization of acetone at \n   1 atm pressure         8 atm pressure: \n   %4.0f cal/g-mole        %4.0f cal/g-mole   -------from equation (32), Page 275 \n   %4.0f cal/g-mole        %4.0f cal/g-mole   -------from equation (33), page 275',lamda1,lamda2,lamda(1),lamda(2));
+
+// File output
+fd= mopen('.\Chapter8_Example2_Output.txt','w');
+mfprintf(fd,'\n Molal heat of vaporization of acetone at \n   1 atm pressure         8 atm pressure: \n   %4.0f cal/g-mole        %4.0f cal/g-mole   -------from equation (32), Page 275 \n   %4.0f cal/g-mole        %4.0f cal/g-mole   -------from equation (33), Page 275',lamda1,lamda2,lamda(1),lamda(2));
+mclose(fd);
+
+//=================================END OF PROGRAM==============================
+
+//Remarks
+//Difference between the solution computed by scilab and that given in book is due to round off error. The solution computed by scilab matches when the same is computed manually.
diff --git a/926/CH8/EX8.2/Chapter8_Example2_Output.txt b/926/CH8/EX8.2/Chapter8_Example2_Output.txt
new file mode 100644
index 000000000..9f82819c2
--- /dev/null
+++ b/926/CH8/EX8.2/Chapter8_Example2_Output.txt
@@ -0,0 +1,5 @@
+
+ Molal heat of vaporization of acetone at 
+   1 atm pressure         8 atm pressure: 
+   7112 cal/g-mole        5878 cal/g-mole   -------from equation (32), Page 275 
+   6891 cal/g-mole        5864 cal/g-mole   -------from equation (33), Page 275
+\ No newline at end of file
diff --git a/926/CH8/EX8.3/Chapter8_Example3.sce b/926/CH8/EX8.3/Chapter8_Example3.sce
new file mode 100644
index 000000000..888c33113
--- /dev/null
+++ b/926/CH8/EX8.3/Chapter8_Example3.sce
@@ -0,0 +1,28 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 3, Page 278
+//Title: Calculation of heat of vaporization
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = 200; //Given temperature in degree F
+TC = 232.5; //Critical temperature of Freon 12 in degree F
+SRTR  = .516; //Heat of vaporization factor of Freon 12 from Table 27, Page 278
+
+//CALCULATION
+TR = (T+460)/(TC+460); //Reduced temperature of freon 12 at 200 degree F
+Lamda_1 = 7707; //Molal heat of vaporization of water at TR=0.951 in BTU per lb-mole obtained from Table page 279
+Lamda_2 = Lamda_1*SRTR; //Heat of vaporization of freon 12 at 200 degree F
+
+//OUTPUT
+// Console Output
+mprintf('\n Heat of vaporization of freon 12 at 200 degree F = %4.0f BTU per lb-mole',Lamda_2);
+
+// File Output
+fd= mopen('.\Chapter8_Example3_Output.txt','w');        // PRN: File name and path changed NAR 
+mfprintf(fd,'\n Heat of vaporization of freon 12 at 200 degree F = %4.0f BTU per lb-mole',Lamda_2);
+mclose(fd);
+
+//=================================END OF PROGRAM==============================
diff --git a/926/CH8/EX8.3/Chapter8_Example3_Output.txt b/926/CH8/EX8.3/Chapter8_Example3_Output.txt
new file mode 100644
index 000000000..38070c3ae
--- /dev/null
+++ b/926/CH8/EX8.3/Chapter8_Example3_Output.txt
@@ -0,0 +1,2 @@
+
+ Heat of vaporization of freon 12 at 200 degree F = 3977 BTU per lb-mole
+\ No newline at end of file
diff --git a/926/CH8/EX8.4/Chapter8_Example4.sce b/926/CH8/EX8.4/Chapter8_Example4.sce
new file mode 100644
index 000000000..787479ca4
--- /dev/null
+++ b/926/CH8/EX8.4/Chapter8_Example4.sce
@@ -0,0 +1,33 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 4, Page 281
+//Title: Calculation of heat of vaporization
+//=============================================================================
+clear
+clc
+
+//INPUT
+Tb = 78; //Normal boiling point of ethyl alcohol in degree C   
+Tc = 243; //Critical temperature of ethyl alcohol in degree C   
+T = 180; //Given temperature of ethyl alcohol in degree C   
+Lamda_1 = 204; //latent heat of vaporisation of ethyl alcohol at normal boiling point in cal per gram
+
+//CALCULATION
+TB = 273.15+Tb; //Normal boiling point of ethyl alcohol in K   
+TC = 273.15+Tc; //Critical temperature of ethyl alcohol in K
+T1 = 273.15+T; //Given temperature of ethyl alcohol in K 
+
+Tr1 = TB/TC; //Reduced temperature with reference to boiling point
+Tr2 = T1/TC; //Reduced temperature with reference to temperature at which heat of vaporization is to be estimated
+Lamda_2 = Lamda_1*((1-Tr2)/(1-Tr1))^0.38; //Heat of vaporization at given temperature in cal per gram
+
+//OUTPUT
+// Console Output
+mprintf('\n Heat of vaporization at a temperature of %3.0f degree C = %3.0f cal per gram',Tb,Lamda_2);
+
+// File Output
+fd= mopen('.\Chapter8_Example4_Output.txt','w');
+mfprintf(fd,'\n Heat of vaporization at a temperature of %3.0f degree C = %3.0f cal per gram',Tb,Lamda_2);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH8/EX8.4/Chapter8_Example4_Output.txt b/926/CH8/EX8.4/Chapter8_Example4_Output.txt
new file mode 100644
index 000000000..30bf68959
--- /dev/null
+++ b/926/CH8/EX8.4/Chapter8_Example4_Output.txt
@@ -0,0 +1,2 @@
+
+ Heat of vaporization at a temperature of 180 degree C = 141 cal per gram
+\ No newline at end of file
diff --git a/926/CH8/EX8.5/Chapter8_Example5.sce b/926/CH8/EX8.5/Chapter8_Example5.sce
new file mode 100644
index 000000000..7a105fc42
--- /dev/null
+++ b/926/CH8/EX8.5/Chapter8_Example5.sce
@@ -0,0 +1,35 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 5, Page 281
+//Title: Calculation of enthalpy of steam
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [350 32]; //Given and liquid state temperature of steam in degree F
+P = 50; //Given pressure in psi
+TS = 281; //Saturation temperature of steam at 50 psi in degree F obtained from Table 5 Page 83
+CP1 = 1.006; //Mean specific heat of water between 32-281 degree F in BTU per lb degree F
+CP2 = 9.2; //Mean heat capacity of water vapor between 32-281 degree F in BTU per lb-mole degree F
+lamda1 = 924; //Latent heat of vaporization of water at 281 degree F in BTU per lb
+MW = 18; //Molecular weight of water in lb/lb-mole
+
+//CALCULATION
+lamda2 = (TS-T(2))*CP1; //Enthalpy of liquid water at 281 degree F in BTU per lb
+lamda3 = (T(1)-TS)*CP2/MW; //Superheat of vapor in BTU per lb
+lamda = lamda1+lamda2+lamda3; //Total enyhalpy in BTU per lb
+
+//OUTPUT
+// Console Output
+mprintf('\n Total enthalpy of 1 lb steam = %5.1f BTU per lb',lamda);
+
+// File Output
+fd= mopen('.\Chapter8_Example5_Output.txt','w');
+mfprintf(fd,'\n Total enthalpy of 1 lb steam = %5.1f BTU per lb',lamda);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
+
+// Remark
+// Difference between the solution computed by scilab and that given in book is due to round off error. For instance, enthalpy computed by (281-32)*1.006 should be 250.494 and not 2501. as printed in the textbook.
diff --git a/926/CH8/EX8.5/Chapter8_Example5_Output.txt b/926/CH8/EX8.5/Chapter8_Example5_Output.txt
new file mode 100644
index 000000000..5bc70dc7f
--- /dev/null
+++ b/926/CH8/EX8.5/Chapter8_Example5_Output.txt
@@ -0,0 +1,2 @@
+
+ Total enthalpy of 1 lb steam = 1209.8 BTU per lb
+\ No newline at end of file
diff --git a/926/CH8/EX8.6/Chapter8_Example6.sce b/926/CH8/EX8.6/Chapter8_Example6.sce
new file mode 100644
index 000000000..5e344102c
--- /dev/null
+++ b/926/CH8/EX8.6/Chapter8_Example6.sce
@@ -0,0 +1,35 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 6, Page 282
+//Title: Calculation of total enthalpy
+//=============================================================================
+clear
+clc
+
+//INPUT
+AW = 65.4; //Atomic weight of zinc
+T = [1000 0 419 907]; //Given temperature, solid state temperature, melting point and boiling point of zinc in degree C
+CP = [0.105 0.109]; //Mean specific heat of solid from 0-419 degree C and liquid from 419-907 degree C in cal per gram degree C obtained from Fig 63, Page 260
+lamda1 = 1660; //Heat of fusion in cal per g-atom obtained from Table 24, Page 272
+CP1 = 4.97; //Molal heat capacity of zinc vapor at constant preesure in cal per g-mole
+
+//CALCULATION
+T1 = T+273; //Given temperature, solid state temperature, melting point and boiling point of zinc in K
+lamda2 = T1(4)*(8.75+4.571*log10(T1(4))); //Heat of vaporization at normal boiling point in cal per g-mole
+Lamda1 = CP(1)*(T(3)-T(2)); //Heat absorbed by solid in cal per gram
+Lamda2 = lamda1/AW; //Heat of fusion in cal per gram
+Lamda3 = CP(2)*(T(4)-T(3)); //Heat absorbed by liquid in cal per gram
+Lamda4 = lamda2/AW; //Heat of vaporization in cal per gram
+Lamda5 = CP1*(T(1)-T(4))/AW; //Heat absorbed by vapor in cal per gram
+Lamda = Lamda1+Lamda2+Lamda3+Lamda4+Lamda5; //Total enthalpy in cal per gram
+
+//OUTPUT
+// Console Output
+mprintf('\n Total enthalpy of zinc vapor at 1000 degree C = %3.0f cal per gram',Lamda);
+
+// File Output
+fd= mopen('.\Chapter8_Example6_Output.txt','w');
+mfprintf(fd,'\n Total enthalpy of zinc vapor at 1000 degree C = %3.0f cal per gram',Lamda);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH8/EX8.6/Chapter8_Example6_Output.txt b/926/CH8/EX8.6/Chapter8_Example6_Output.txt
new file mode 100644
index 000000000..395b32efe
--- /dev/null
+++ b/926/CH8/EX8.6/Chapter8_Example6_Output.txt
@@ -0,0 +1,2 @@
+
+ Total enthalpy of zinc vapor at 1000 degree C = 541 cal per gram
+\ No newline at end of file
diff --git a/926/CH8/EX8.7/Chapter8_Example7.sce b/926/CH8/EX8.7/Chapter8_Example7.sce
new file mode 100644
index 000000000..98f908a51
--- /dev/null
+++ b/926/CH8/EX8.7/Chapter8_Example7.sce
@@ -0,0 +1,35 @@
+//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 7, Page 283
+//Title: Calculation of total enthalpy
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [100 32]; //Given temperature and standard temperature of dry air in degree F
+MW = [18 29]; //Molecular weight of water and air respectively
+
+//DATA FROM GRAPH
+H = 0.0215; //humidity in terms of lb of water per lb of dry air at given conditions obtained from Fig 20, Page 122
+DP = 79; //Dew point corresponding to above humidity in degree F obtained from Fig 20, Page 122
+CP = [8.02 6.95]; //Molal heat capacities of water vapor between 79-100 degree F and 32-100 degree F in BTU per lb-mole degree F obtained from Fig 62, Page 259
+lamda = 1046; //Heat of vaporization at 79 degree F in BTU per lb obtained from Fig 19, Page 120
+
+//CALCULATION
+Lamda1 = (T(1)-T(2))*CP(2)/MW(2); //Sensible enthalpy of air in BTU per lb
+Lamda2 = (DP-T(2))*H; //Sensible enthalpy of liquid water in BTU per lb
+Lamda3 = lamda*H; //Latent heat of water in BTU per lb
+Lamda4 = (T(1)-DP)*H*CP(1)/MW(1); //Superheat of water vapor in BTU per lb
+Lamda = Lamda1+Lamda2+Lamda3+Lamda4; //Total enthalpy in BTU per lb of dry air
+
+//OUTPUT
+// Console Output
+mprintf('\n Total enthalpy of dry air is %2.0f BTU per lb',Lamda);
+
+// File Output
+fd= mopen('.\Chapter8_Example7_Output.txt','w');
+mfprintf(fd,'\n Total enthalpy of dry air is %2.0f BTU per lb',Lamda);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH8/EX8.7/Chapter8_Example7_Output.txt b/926/CH8/EX8.7/Chapter8_Example7_Output.txt
new file mode 100644
index 000000000..b8465afba
--- /dev/null
+++ b/926/CH8/EX8.7/Chapter8_Example7_Output.txt
@@ -0,0 +1,2 @@
+
+ Total enthalpy of dry air is 40 BTU per lb
+\ No newline at end of file