12 files changed, 398 insertions, 0 deletions
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