7 files changed, 215 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