19 files changed, 613 insertions, 0 deletions
diff --git a/2090/CH3/EX3.1/Chapter3_Example1.sce b/2090/CH3/EX3.1/Chapter3_Example1.sce
new file mode 100755
index 000000000..55e6184de
--- /dev/null
+++ b/2090/CH3/EX3.1/Chapter3_Example1.sce
@@ -0,0 +1,34 @@
+clc
+clear
+//Input data
+E=20;//Methanol burned with excess air in percentage 
+p=1;//Pressure of air in bar
+t=27;//Temperature of air in degree centigrade
+O=32;//The molecular weight of oxygen
+N=28;//The molecular weight of nitrogen
+R=8314;//Universal gas constant in Nm/kmolK
+C=32;//Molecular weight of methanol
+CO=44;//Molecular weight of the carbondioxide 
+H=18;//Molecular weight of the water
+
+//Calculations
+S=[(1.8*O)+(6.768*N)]/C;//Stoichiometric air/fuel ratio
+A=[(1.8*O)+(6.768*N)]/C;//Actual air/fuel ratio
+M=1.8+6.768;//1 kmole of fuel reacts with air in kmole
+V=(M*R*(t+273))/(p*10^5);//Volume of air in m^3/kmole fuel
+T=(1+1.8+6.768);//The total number of moles in the reactants when excess air is supplied in moles
+Cm=(1/T);//Mole fraction of the methanol
+Om=(1.8/T);//Mole fraction of the oxygen
+Nm=(6.768/T);//Mole fraction of the nitrogen
+Mr=(Cm*C)+(Om*O)+(Nm*N);//Molecular weight of reactants 
+Tp=(1+2+6.768+0.3);//Total number of moles in the products in moles
+COm=(1/Tp);//Mole fraction of the carbondioxide
+Hp=(2/Tp);//Mole fraction of the water 
+Np=(6.768/Tp);//Mole fraction of the nitrogen
+Op=(0.3/Tp);//Mole fraction of the oxygen
+Mp=(COm*CO)+(Hp*H)+(Np*N)+(Op*O);//Molecular weight of products
+Pp=(Hp*p);//Partial pressure of water vapour in bar
+D=60;//The dew point is the saturation temp corresponding to partial pressure in degree centigrade
+
+//Output
+printf(' (a) The volume of air supplied per kmole of fuel = %3.1f m^3/kmole fuel \n (b) The molecular weight of the reactants = %3.2f \n The molecular weight of the products = %3.2f \n (c) The dew point of the products = %3.0f degree centigrade ',V,Mr,Mp,D)
diff --git a/2090/CH3/EX3.10/Chapter3_Example10.sce b/2090/CH3/EX3.10/Chapter3_Example10.sce
new file mode 100755
index 000000000..badbfe764
--- /dev/null
+++ b/2090/CH3/EX3.10/Chapter3_Example10.sce
@@ -0,0 +1,18 @@
+clc
+clear
+//Input data
+T=1500;//The given temperature in K
+
+//Calculations
+hfco=-393.52;//The enthalpy of formation for carbondioxide in MJ/kmol
+hf1=61.714;//The change in enthalpy for actual state and reference state in MJ/kmol
+HP=hfco+hf1;//The total enthalpy in the products side in MJ/kmol
+hfc=-110.52;//The enthalpy of formation for carbonmonoxide in MJ/kmol
+hf2=38.848;//The change in enthalpy of CO for actual and reference state in MJ/kmol
+hfo=0;//The enthalpy of formation for oxygen gas
+hf3=40.61;//The change in enthalpy of oxygen for different states in MJ/kmol
+HR=[hfc+hf2]+[0.5*(hfo+hf3)];//The total enthalpy in the reactants side in MJ/kmol
+H=HP-HR;//The enthalpy of combustion in MJ/kmol 
+
+//Output
+printf(' The enthalpy of combustion is  %3.3f MJ/kmol CO ',H)
diff --git a/2090/CH3/EX3.11/Chapter3_Example11.sce b/2090/CH3/EX3.11/Chapter3_Example11.sce
new file mode 100755
index 000000000..ae033086d
--- /dev/null
+++ b/2090/CH3/EX3.11/Chapter3_Example11.sce
@@ -0,0 +1,32 @@
+clc
+clear
+//Input data
+E=30;//The amount of excess air in percentage
+tp=400;//The temperature at which propane enters in K
+ta=300;//The temperature at which air enters in K
+T=900;//The temperature at which products leave in K
+m=83.7;//The average molar specific heat of propane at constant pressure in kJ/kmolK
+Mp=44;//The molecular weight of propane
+
+//Calculations
+hfc=-393.52;//The enthalpy of formation for carbondioxide in MJ/kmol
+hf1=28.041;//The change in enthalpy of CO2 for actual and reference state in MJ/kmol
+hfh=-241.82;//The enthalpy of formation for water in MJ/kmol
+hf2=21.924;//The change in enthalpy of water for actual and reference state in MJ/kmol
+hfn=0;//The enthalpy of nitrogen gas 
+hf3=18.221;//The change in enthalpy of nitrgen for actual and reference state in MJ/kmol
+hfo=0;//The enthalpy of oxygen gas 
+hf4=19.246;//The change in enthalpy of oxygen for actual and reference state in MJ/kmol
+HP=[3*(hfc+hf1)]+[4*(hfh+hf2)]+[24.44*(hfn+hf3)]+[1.5*(hfo+hf4)];//The total enthalpy in the products side in MJ/kmol
+hfp=-103.85;//The enthalpy of formation for propane in MJ/kmol
+R=0.0837;//Universal gas constant 
+hfo1=0;//The enthalpy of oxygen gas 
+hf11=0.054;//The change in enthalpy of oxygen gas for actual and reference state in MJ/kmol
+hfn1=0;//The enthalpy of nitrogen gas
+hfn22=0.054;//The change in enthalpy of nitrogen for actual and reference state in MJ/kmol
+HR=[1*(hfp+(R*(tp-ta)))]+[6.5*(hfo1+hf11)]+[24.44*(hfn1+hfn22)];//The total enthalpy on the reactants side in MJ/kmol
+Q=HP-HR;//The amount of heat liberated in MJ/kmol
+Q1=[-Q/Mp];//The amount of heat liberated in MJ/kg
+
+//Output
+printf(' The amount of heat transfer per kg of fuel is  %3.0f MJ/kg',Q1)
diff --git a/2090/CH3/EX3.12/Chapter3_Example12.sce b/2090/CH3/EX3.12/Chapter3_Example12.sce
new file mode 100755
index 000000000..ed7684ce8
--- /dev/null
+++ b/2090/CH3/EX3.12/Chapter3_Example12.sce
@@ -0,0 +1,29 @@
+clc
+clear
+//Input data
+Ta=150;//The presence of Theoretical air
+
+//Calculations
+hfc=-393.52;//The enthalpy of formation for carbondioxide in MJ/kmol
+hfh=-285.8;//The enthalpy of formation for water in MJ/kmol
+hfon=0;//The enthalpy of formation for oxygen and nitrogen gas 
+hfch=-74.87;//The enthalpy of formation for methane in MJ/kmol
+HP=[hfc]+[2*hfh];//The total enthalpy on the products side in MJ/kmol
+HR=1*hfch;//The total enthalpy on the reactants side in MJ/kmol
+H=HP-HR;//The total change in enthalpy of reactants and products in MJ/kmol
+np=2;//Number of moles of product
+nr=4;//Number of moles of reactant
+n=np-nr;//The difference in moles
+R=8.314*10^-3;//Universal gas constant 
+t=298;//The temperature in K
+U1=H-[n*R*t];//The standard internal energy in MJ/kmol
+hfh1=-241.82;//The enthalpy of formation for water in MJ/kmol
+HP1=[1*hfc]+[2*hfh1];//The total enthalpy on the products side in MJ/kmol
+H1=HP1-HR;//The change in enthalpy for reactants and products in MJ/kmol
+np1=4;//Number of moles of product
+nr1=4;//Number of moles of reactant
+n1=np1-nr1;//The difference in moles
+U2=H1-[n1*R*t];//The standard internal energy in MJ/kmol
+
+//Output
+printf(' (a)The water as liquid , \n    The standard enthalpy of combustion is  %3.2f MJ/kmol \n    The standard internal energy of combustion is  %3.2f MJ/kmol \n (b)The water as a gas , \n    The standard enthalpy of combustion is  %3.2f MJ/kmol \n    The standard internal energy of combustion is  %3.2f MJ/kmol ',H,U1,H1,U2)
diff --git a/2090/CH3/EX3.13/Chapter3_Example13.sce b/2090/CH3/EX3.13/Chapter3_Example13.sce
new file mode 100755
index 000000000..46ad9af9a
--- /dev/null
+++ b/2090/CH3/EX3.13/Chapter3_Example13.sce
@@ -0,0 +1,28 @@
+clc
+clear
+//Input data
+cv=44000;//The lower calorific value of liquid fuel in kJ/kg
+C=84;//The carbon content present in the fuel in percentage
+H=16;//The hydrogen content present in the fuel in percentage
+t=25;//The temperature in degree centigrade
+hfg=2442;//The enthalpy of vaporization for water in kJ/kg
+c=12;//Molecular weight of carbon 
+h=2;//Molecular weight of hydrogen
+co2=44;//Molecular weight of carbondioxide
+h2o=18;//Molecular weight of water 
+o2=32;//Molecular weight of oxygen
+R=8.314;//Universal gas constant in J/molK
+
+//Calculations
+CO2=[0.84*(co2/c)];//The amount of carbondioxide present per kg of fuel in kg
+H2O=[0.16*(h2o/h)];//The amount of water present per kg of fuel in kg
+cvd=H2O*hfg;//The difference in the higher and lower calorific value in kJ/kg fuel 
+HHV=cv+cvd;//The higher calorific value of the liquid fuel in kJ/kg fuel
+np=3.08/co2;//number of moles of product in kmol/kg fuel
+nr=3.52/o2;//The number of moles of reactant in kmol/kg fuel
+n=np-nr;//The difference in the moles
+HHVv=HHV+[n*R*(t+273)];//The higher calorific value at constant volume in kJ/kg fuel
+LHVv=cv+[n*R*(t+273)];//The lower calorific value at constant volume in kJ/kg fuel
+
+//Output
+printf(' The higher calorific value at constant pressure = %3.0f kJ/kg fuel \n The higher calorific value at constant volume = %3.0f kJ/kg fuel \n The lower calorific value at constant volume = %3.0f kJ/kg fuel',HHV,HHVv,LHVv)
diff --git a/2090/CH3/EX3.14/Chapter3_Example14.sce b/2090/CH3/EX3.14/Chapter3_Example14.sce
new file mode 100755
index 000000000..89e42aba6
--- /dev/null
+++ b/2090/CH3/EX3.14/Chapter3_Example14.sce
@@ -0,0 +1,31 @@
+clc
+clear
+//Input data
+E=100;//The amount of excess air in percent
+T=298;//The temperature of reactants in K
+nc=1;//Number of moles of propane
+
+//Calculations
+hfch=-103.85;//Enthalpy of formation for propane in MJ/kmol fuel
+HR=nc*hfch;//Total enthalpy on the reactants side in MJ/kmol fuel
+hfc=-393.52;//Enthalpy of formation for carbondioxide in MJ/kmol fuel
+hfh=-241.82;//Enthalpy of formation for water in MJ/kmol fuel
+hfon=0;//Enthalpy of formation for both oxygen and nitrogen gas
+x=HR-[(3*hfc)+(4*hfh)+(5*hfon)+(37.6*hfon)];//For adiabatic combustion enthalpy obtained for equating reactants and products in MJ/kmol fuel
+hfn=x/37.6;//trail to get the change in enthalpy of nitrogen in MJ/kmol
+T1=1500;//Assuming the products temperature for fist trail in K
+hfc1=61.714;//The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel
+hfh1=48.095;//The change in enthalpy for water for trail temp in MJ/kmol fuel
+hfo1=40.61;//The change in enthalpy for oxygen for trail temp in MJ/kmol fuel
+hfn1=38.405;//The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel
+HP1=(HR-x)+(3*hfc1)+(4*hfh1)+(5*hfo1)+(37.6*hfn1);//Total enthalpy of products for first trail in MJ/kmol fuel
+T2=1600;//Assuming the products temperature for second trail in K
+hfc2=67.58;//The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel
+hfh2=52.844;//The change in enthalpy for water for trail temp in MJ/kmol fuel
+hfo2=44.279;//The change in enthalpy for oxygen for trail temp in MJ/kmol fuel
+hfn2=41.903;//The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel
+HP2=(HR-x)+(3*hfc2)+(4*hfh2)+(5*hfo2)+(37.6*hfn2);//Total enthalpy of products for second trail in MJ/kmol fuel
+Te=[[(HR-HP1)/(HP2-HP1)]*(T2-T1)]+T1;//The eatimated adiabatic flame temperature in K
+
+//Output
+printf(' The adiabatic flame temperature for steady-flow process is  %3.1f K',Te)
diff --git a/2090/CH3/EX3.15/Chapter3_Example15.sce b/2090/CH3/EX3.15/Chapter3_Example15.sce
new file mode 100755
index 000000000..6b1117fca
--- /dev/null
+++ b/2090/CH3/EX3.15/Chapter3_Example15.sce
@@ -0,0 +1,60 @@
+clc
+clear
+//Input data
+T=600;//The initial temperature of air in K
+p=1;//The initial pressure of air in atm
+R=8.314;//Universal gas constant in J/molK
+Tr=298;//The temperature of reactants in K
+a=4.503;//Given Constant 
+b=-8.965*10^-3;//Given Constant
+c=37.38*10^-6;//Given Constant
+d=-36.49*10^-9;//Given Constant
+e=12.22*10^-12;//Given Constant
+
+//Calculations
+hfc=-393.52;//Enthalpy of formation for carbondioxide in MJ/kmol fuel
+hfh=-241.82;//Enthalpy of formation for water in MJ/kmol fuel
+hfn=0;//Enthalpy of formation for nitrogen gas
+HP=[1*hfc]+[2*hfh]+[7.52*hfn];//The enthalpy on the products side in MJ/kmol fuel
+hch=[R*[(a*(T-Tr))+((b/2)*(T^2-Tr^2))+((c/3)*(T^3-Tr^3))+((d/4)*(T^4-Tr^4))+((e/5)*(T^5-Tr^5))]]/1000;//The change in enthalpy of the methane in MJ/kmol
+hfc1=-74.87;//The enthalpy of formation for methane in MJ/kmol fuel 
+hfh1=9.247;//The change in enthalpy of the water in MJ/kmol
+hfn1=8.891;//The change in enthalpy of nitrogen in MJ/kmol
+HR=[(hfc1+hch)+(2*hfh1)+(7.52*hfn1)];//The enthalpy on the reactants side in MJ/kmol
+x=HR-HP;//The enthalpy for the remaining gases in the product side in MJ/kmol
+hfn2=x/7.52;//The guess enthalpy for the nitrogen gas in MJ/kmol
+Tc=3700;//The corresponding temperature for the enthalpy of guess nitrogen in K
+T1=2800;//The temperature assumed for the first trail in K
+hco1=140.444;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hh1=115.294;//The change in enthalpy for the assume temp for water in MJ/kmol
+hn1=85.345;//The change in enthalpy for the assume temp for nitrogen in MJ/kmol
+HP1=hco1+(2*hh1)+(7.52*hn1)+(HR-x);//The total enthalpy on the products side for first trail in MJ/kmol fuel
+T2=2500;//The temperature assumed for the second trail in K
+hco2=121.926;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hh2=98.964;//The change in enthalpy for the assume temp for water in MJ/kmol
+hn2=74.312;//;//The change in enthalpy for the assume temp for nitrogen in MJ/kmol
+HP2=hco2+(2*hh2)+(7.52*hn2)+(HR-x);//The total enthalpy on the products side for the second trail in MJ/kmol
+T3=2600;//The temperature fo the third trail in K
+hco3=128.085;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hh3=104.37;//The change in enthalpy for the assume temp for water in MJ/kmol
+hn3=77.973;//The change in enthalpy for the assume temp for nitrogen in MJ/kmol
+HP3=hco3+(2*hh3)+(7.52*hn3)+(HR-x);//The total enthalpy on the products side for the third trail in MJ/kmol
+Ta1=[[(HR-HP2)/(HP3-HP2)]*(T3-T2)]+T2;//The adiabatic temperature for constant pressure process in K
+UR1=HR-(10.52*R*10^-3*T);//The internal energy of reactant in MJ/kmol fuel
+Tc1=3000;//Assume temperature for first trail in K
+hcoa1=146.645;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hha1=120.813;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hna1=89.036;//The change in enthalpy for the assume temp for nitrogen in MJ/kmol
+UP1=hcoa1+(2*hha1)+(7.52*hna1)+(HR-x)-(0.08746*Tc1);//The internal energy of products in MJ/kmol fuel
+Tc2=3200;//Assume temperature for the second trail in K
+hcoa2=165.331;//;//The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
+hha2=137.553;//The change in enthalpy for the assume temp for water in MJ/kmol
+hna2=100.161;//The change in enthalpy for the assume temp for nitrogen in MJ/kmol
+UP2=hcoa2+(2*hha2)+(7.52*hna2)+(HR-x)-(0.08746*Tc2);//The internal energy of products in MJ/kmol fuel
+Tu=[[(UR1-UP1)/(UP2-UP1)]*(Tc2-Tc1)]+Tc1;//The adiabatic flame temperature at constant pressure process in K
+
+//Output
+printf('The adiabatic flame temperature at  \n    (a)Constant pressure process is %3.0f K \n    (b)Constant volume process is %3.1f K',Ta1,Tu)
+
+
+
diff --git a/2090/CH3/EX3.16/Chapter3_Example16.sce b/2090/CH3/EX3.16/Chapter3_Example16.sce
new file mode 100755
index 000000000..5f611e0e0
--- /dev/null
+++ b/2090/CH3/EX3.16/Chapter3_Example16.sce
@@ -0,0 +1,68 @@
+clc
+clear
+//Input data
+T=600;//Temperature at constant pressure process in K
+p=1;//The pressure in atm
+E=50;//The amount of excess air in percent
+L=20;//The amount of less air in percent
+cp=52.234;//Specific constant for methane in kJ/kmolK
+T1=298;//Assume the normal temperature in K
+
+//Calculations
+hfch=-74.87;//The enthalpy of formation for carbondioxide in MJ
+hch=cp*(T-T1)*10^-3;//The change in enthalpy of carbondioxide in MJ
+ho=9.247;//The change in enthalpy of oxygen in MJ
+hn=8.891;//The change in enthalpy of nitrogen in MJ
+HR=hfch+hch+(3*ho)+(11.28*hn);//The total enthalpy on the reactants side in MJ
+hfc1=-393.52;//The enthalpy of formation of carbondioxide in MJ
+hfh1=-241.82;//The enthalpy of formation of water in MJ
+HP=hfc1+(2*hfh1);//The enthalpy of products side in MJ
+x=HR-HP;//The change in enthalpy for the remaining in MJ
+hn2=x/11.28;//The enthalpy of nitrogen assumed to be in MJ/kmol
+Tc=2800;//The corresponding temperature in K
+T1=2000;//The temperature for first trail in K
+hfc11=91.45;//The enthalpy for the assume temp for carbondioxide in MJ
+hfh11=72.689;//The change in enthalpy for the assume temp for water in MJ
+hfn11=56.141;//The change in enthalpy for the assume temp for nitrogen in MJ
+hfo11=59.199;//;//The change in enthalpy for the assume temp for oxygen in MJ
+HP1=hfc11+(2*hfh11)+(11.28*hfn11)+(hfo11)+(HR-x);//The total enthalpy on the products side for first trail in MJ
+T2=2100;//The temperature for second trail in K
+hfc22=97.5;//The enthalpy for the assume temp for carbondioxide in MJ
+hfh22=77.831;//The change in enthalpy for the assume temp for water in MJ
+hfn22=59.748;//The change in enthalpy for the assume temp for nitrogen in MJ
+hfo22=62.986;//;//The change in enthalpy for the assume temp for oxygen in MJ
+HP2=hfc22+(2*hfh22)+(11.28*hfn22)+(hfo22)+(HR-x);//The total enthalpy on the products side for second trail in MJ
+Ta1=[[(HR-HP1)/(HP2-HP1)]*(T2-T1)]+T1;//The adiabatic temperature for constant pressure process in K
+X=2*[2-1.6];//By balance oxygen
+hfchr=-74.87;//The enthalpy of formation for methane in MJ
+hor=9.247;//The change in enthalpy for oxygen in MJ
+hnr=8.891;//The change in enthalpy for nitrogen in MJ
+HRr=hfchr+hch+(1.6*hor)+(6.01*hnr);//The total enthalpy on reactants side in MJ
+hfcop=-110.52;//The formation of enthalpy for carbonmoxide in MJ
+hfcp=-393.52;//The formation of enthalpy for carbondioxide in MJ
+hfhp=-241.82;//The formation of enthalpy for water in MJ
+HPp=(0.8*hfcop)+(0.2*hfcp)+(2*hfhp);//The enthalpy on product side in MJ
+Tp1=2000;//The temperature for first trail in K
+hco11=56.739;//The change in enthalpy for CO in MJ
+hco211=91.45;//The change in enthalpy for CO2 in MJ
+hh11=72.689;//The change in enthalpy for water in MJ
+hn11=56.141;//The change in enthalpy for nitrogen in MJ
+HPp1=(0.8*hco11)+(0.2*hco211)+(2*hh11)+(6.016*hn11)-HPp;///The enthalpy on the products side for trail temp in MJ
+Tp2=2400;//The temperature for second trail in K
+hco22=71.34;//The change in enthalpy for CO in MJ
+hco222=115.788;//The change in enthalpy for CO2 in MJ
+hh22=93.604;//The change in enthalpy for water in MJ
+hn22=70.651;//The change in enthalpy for nitrogen in MJ
+HPp2=(0.8*hco22)+(0.2*hco222)+(2*hh22)+(6.016*hn22)+HPp;///The enthalpy on the products side for trail temp in MJ
+Tp3=2300;//The temperature for first trail in K
+hco33=67.676;//The change in enthalpy for CO in MJ
+hco233=109.671;//The change in enthalpy for CO2 in MJ
+hh33=88.295;//The change in enthalpy for water in MJ
+hn33=67.007;//The change in enthalpy for nitrogen in MJ
+HPp3=(0.8*hco33)+(0.2*hco233)+(2*hh33)+(6.016*hn33)+HPp;///The enthalpy on the products side for trail temp in MJ
+Ta2=[[(HRr-HPp3)/(HPp2-HPp3)]*(Tp2-Tp3)]+Tp3;//The adiabatic temperature for constant pressure process in K
+hccc=-283.022;//The only combustible substance is CO in MJ/kmol
+Q=-0.8*hccc;//The thermal energy loss in MJ/kmol fuel
+
+//Output
+printf(' The adiabatic flame temperature having \n    (a)50 percent excess air is %3.1f K \n    (b)20 percent less air is %3.1f K \n The loss of thermal energy due to incomplete combustion is %3.1f MJ/kmol fuel',Ta1,Ta2,Q)
diff --git a/2090/CH3/EX3.18/Chapter3_Example18.sce b/2090/CH3/EX3.18/Chapter3_Example18.sce
new file mode 100755
index 000000000..e98a9bd37
--- /dev/null
+++ b/2090/CH3/EX3.18/Chapter3_Example18.sce
@@ -0,0 +1,27 @@
+clc
+clear
+//Input data
+T1=3000;//Given temperature in K
+T2=4000;//Given temperature in K
+p=1;//The pressure in atm
+KP1=1.117;//Natural logarithm of equilibrium constant at 3000 K 
+KP2=-1.593;//Natural logarithm of equilibrium constant at 4000 K
+
+//Calculations
+Kp1=exp(KP1);//The value of equilibrium constant at 3000 K
+Kp2=exp(KP2);//The value of equilibrium constant at 4000 K
+a1=0.4;//The dissociation of 1 mole of CO2 for the first trail
+a2=0.5;//The dissociation of 1 mole of CO2 for the second trail 
+K1=3.674;//The value of equilibrium constant for the first trail 
+K2=2.236;//The value of equilibrium constant for the second trail
+a12=[[(K1-Kp1)/(K1-K2)]*(a2-a1)]+a1;//The approximate dissociation of 1 mole of CO2
+A12=a12*100;//The amount of CO2 will dissociate in percent
+a3=0.9;//The dissociation of 1 mole of CO2 for the first trail
+a4=0.89;//The dissociation of 1 mole of CO2 for the second trail
+K3=0.1995;//The value of equilibrium constant for the first trail 
+K4=0.2227;//The value of equilibrium constant for the second trail 
+a23=[[(Kp2-K4)/(K3-K4)]*(a3-a4)]+a4;//The approximate dissociation of 1 mole of CO2
+A23=a23*100;//The amount of CO2 will dissociate in percent
+
+//output
+printf('The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \n    (a) at 3000 K and 1 atm pressure = %3.1f percent \n    (b) at 4000 K and 1 atm pressure = %3.2f percent ',A12,A23)
diff --git a/2090/CH3/EX3.19/Chapter3_Example19.sce b/2090/CH3/EX3.19/Chapter3_Example19.sce
new file mode 100755
index 000000000..09ae29d7f
--- /dev/null
+++ b/2090/CH3/EX3.19/Chapter3_Example19.sce
@@ -0,0 +1,21 @@
+clc
+clear
+//Input data
+p=1;//Initial pressure in atm
+T=300;//Initial temperature in K
+Tc=2400;//To calculate the molefraction of the products at this temperature in K
+KP1=3.866;//Natural logarithm of equilibrium constant at 2400 K for the equation 
+
+//Calculations
+K1=exp(KP1);//The value of equilibrium constant at 2400 K
+nr=1+0.5;//The number of moles of reactants
+Pp=(p*Tc)/(nr*T);//Pressure exercted on the products side per mole in atm/mole
+a=0.098;//The dissociation of 1 mole of CO2
+np=(a+2)/2;//The number of moles of products
+xco=[2*(1-a)]/(2+a);//Mole fraction of CO2
+xc=[2*a]/(2+a);//Mole fraction of CO
+xo=a/(2+a);//Mole fraction of O2
+PP=5.333*np;//Pressure of the product in bar
+
+//output
+printf('Mole fraction of the carbondioxide is %3.4f \n Mole fraction of the carbonmonoxide is %3.4f \n Mole fraction of oxygen is %3.4f \n Pressure of the product is %3.3f bar',xco,xc,xo,PP)
diff --git a/2090/CH3/EX3.2/Chapter3_Example2.sce b/2090/CH3/EX3.2/Chapter3_Example2.sce
new file mode 100755
index 000000000..3da118877
--- /dev/null
+++ b/2090/CH3/EX3.2/Chapter3_Example2.sce
@@ -0,0 +1,37 @@
+clc
+clear
+//Input data
+C1=40;//The content of C7H16 in the fuel in percentage
+C2=60;//The content of C8H18 in the fuel in percentage
+d=0.12;//The diameter of the bore in m
+l=0.145;//The length of the bore in m
+r=8.5;//Compression ratio 
+p=1.1;//Pressure at exhaust stroke in bar
+T=720;//The temperature at the exhaust stroke in K
+pi=3.141;//Mathematical constant pi
+O=32;//The molecular weight of oxygen
+N=28;//The molecular weight of nitrogen
+C3=100;//Molecular weight of C7H16
+C4=114;//The molecular weight of C8H18
+R=8314;//Universal gas constant in Nm/kmolK
+CO2=44;//Molecular weight of the carbondioxide 
+C5=28;//Molecular weight of the carbonmonoxide
+H=18;//Molecular weight of the water
+
+//Calculations
+N2=100-(12+1.5+2.5);//Percentage of nitrogen in the dry products of combustion
+Y=84/3.76;//The number of moles oxygen is supplied
+X=13.5/7.6;//Moles of carbon
+Z=(22.34-15.25)*2;//The number of moles of hydrogen
+Hl=(6.4+10.8)/2;//Number of moles of hydrogen on L.H.S
+Hr=7.98;//Number of moles of hydrogen on R.H.S
+Hd=Hl-Hr;//Difference of hydrogen moles
+A=[[12.58*(O+(3.76*N))]/[((C1/100)*C3)+((C2/100)*C4)]];//The Air/fuel ratio
+Vs=(pi/4)*d^2*l;//Swept volume of the cylinder in m^3
+Vc=Vs/(r-1);//Clearance volume in m^3
+M=[(6.757*CO2)+(0.8446*C5)+(1.408*O)+(47.3*N)+(8.6*H)]/[6.757+0.8446+1.408+47.3+8.6];//Molecular weight of the product
+R1=R/M;//Gas constant in J/kgK
+m=[(p*10^5)*Vc]/[R1*T];//Mass of the exhaust gases in the clearance space in kg
+
+//Output 
+printf('(a)The air/fuel ratio =%3.2f \n (b)The mass of the exhaust gases in the clearance space =%3.7f kg ',A,m)
diff --git a/2090/CH3/EX3.20/Chapter3_Example20.sce b/2090/CH3/EX3.20/Chapter3_Example20.sce
new file mode 100755
index 000000000..a859826b3
--- /dev/null
+++ b/2090/CH3/EX3.20/Chapter3_Example20.sce
@@ -0,0 +1,34 @@
+clc
+clear
+//Input data
+t=25;//The temperature of air in degree centigrade
+p=1;//The pressure of air in atm
+T1=2200;//Given first temperature in K
+T2=2400;//Given second temperature in K
+h1=59.86;//The change in enthalpy of hydrogen at 2200 K in MJ/kmol
+h2=66.915;//The change in enthalpy of hydrogen at 2400 K in MJ/kmol
+T=298;//The temperature of air in K
+
+//Calculations
+HR=0;//The total enthalpy on the reactants side since all the reactants are elements
+Kp1=-6.774;//Natural logarithm of equilibrium constant at 2200 K for the equation 
+K1=exp(Kp1);//The value of equilibrium constant at 2200 K
+a1=0.02;//By trail and error method the degree of dissociation of H2O
+hfh=-241.82;//The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol
+hfh1=83.036;//The change in enthalpy of water at 2200 K in MJ/kmol
+hfd1=59.86;//The change in enthalpy of hydrogen at 2200 K in MJ/kmol
+hfo1=66.802;//The change in enthalpy of oxygen at 2200 K in MJ/kmol
+hfn1=63.371;//The change in enthalpy of nitrogen at 2200 K in MJ/kmol
+HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1);//The enthalpy on the products side in MJ/kmol 
+a2=0.04;//By trail and error method the degree of dissociation of H2O at 2400 K
+hfh2=93.604;//The change in enthalpy of water at 2400 K in MJ/kmol
+hfd2=66.915;//The change in enthalpy of hydrogen at 2400 K in MJ/kmol
+hfo2=74.492;//The change in enthalpy of oxygen at 2400 K in MJ/kmol
+hfn2=70.651;//The change in enthalpy of nitrogen at 2400 K in MJ/kmol
+HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2);//The enthalpy on the products side in MJ/kmol 
+H1=HP1-HR;//The total change in enthalpy at 2200 K in MJ/kmol
+H2=HP2-HR;//The total change in enthalpy at 2400 K in MJ/kmol
+Tl=[[[T2-T1]/[HP2-HP1]]*[HR-HP1]]+T1;//The required temperature in K
+
+//Output
+printf('The adiabatic flame temperature taking dissociation into account is %3.0f K',Tl)
diff --git a/2090/CH3/EX3.3/Chapter3_Example3.sce b/2090/CH3/EX3.3/Chapter3_Example3.sce
new file mode 100755
index 000000000..3a701bdda
--- /dev/null
+++ b/2090/CH3/EX3.3/Chapter3_Example3.sce
@@ -0,0 +1,44 @@
+clc
+clear
+//Input data
+C=0.86;//The amount of carbon content in the 1kg of fuel by weight in kg
+H=0.05;//The amount of hydrogen content in the 1kg of fuel by weight in kg
+O=0.02;//The amount of oxygen content in the 1kg of fuel by weight in kg
+S=0.005;//The amount of sulphur content in the 1kg of fuel by weight in kg
+N=0.065;//The amount of nitrogen content in the 1kg of fuel by weight in kg
+E=25;//The amount of excess air supplied in percentage
+o=32;//Molecular weight of the oxygen
+co=44;//Molecular weight of the carbondioxide 
+c=12;//Molecular weight of the carbon
+s=32;//Molecular weight of the sulphur
+so=64;//Molecular weight of sulphur dioxide
+n=28;//Molecular weight of the nitrogen
+
+//Calculations
+o1=(o/c)*C;//The amount of oxygen required for 0.86 kg of carbon in kg
+coa=(co/c)*C;//The amount of carbondioxide produced for 0.86 kg of carbon in kg
+o2=(o/4)*H;//The amount of oxygen required for 0.05 kg of hydrogen in kg
+h2=(36/4)*H;//The amount of water produced for 0.05 kg of hydrogen in kg
+o3=(o/s)*S;//The amount of oxygen required for 0.005 kg of sulphur in kg
+s1=(so/s)*S;//The amount of sulphur dioxide produced for 0.005 kg of sulphur in kg
+To=o1+o2+o3;//Total oxygen required for the complete combustion of fuel in kg
+Tt=To-O;//The amount of oxygen required per kg of fuel for complete combustion theoretically in kg
+As=(Tt*100)/23;//Stoichiometric air/fuel ratio
+as=As*(1+(E/100));//The actual quantity of air supplied per kg of fuel in kg
+o2a=0.23*(E/100)*As;//The oxygen in the excess air in kg
+n2a=0.77*(1+(E/100))*As;//The nitrogen in the air in kg
+n2e=n2a+N;//Total nitrogen in the exhaust in kg
+Tw=coa+n2e+o2a;//Total weight in kg
+pco=(coa/Tw)*100;//Percentage composition of carbondioxide
+pn=(n2e/Tw)*100;//Percentage composition of nitrogen
+po=(o2a/Tw)*100;//Percentage composition of oxygen
+mco=(coa/co);//Moles of carbondioxide
+mn=(n2e/n);//Moles of nitrogen
+mo=(o2a/o);//Moles of oxygen
+Tm=mco+mn+mo;//Total moles
+vco=(mco/Tm)*100;//Volumetric analysis of carbondioxide in percentage
+vn=(mn/Tm)*100;//Volumetric analysis of nitrogen in percentage
+vo=(mo/Tm)*100;//Volumetric analysis of oxygen in percentage
+
+//Output
+printf(' (a)Stoichiometric air/fuel ratio = %3.2f \n (b)The percentage of dry products of combustion by weight : \n   CO2 = %3.2f percent \n   N2 = %3.2f percent \n   O2 = %3.2f percent \n (c)The percentage of dry products of combustion by volume : \n   CO2 = %3.2f percent \n   N2 = %3.2f percent \n   O2= %3.2f percent ',As,pco,pn,po,vco,vn,vo)
diff --git a/2090/CH3/EX3.4/Chapter3_Example4.sce b/2090/CH3/EX3.4/Chapter3_Example4.sce
new file mode 100755
index 000000000..643f00dae
--- /dev/null
+++ b/2090/CH3/EX3.4/Chapter3_Example4.sce
@@ -0,0 +1,28 @@
+clc
+clear
+//Input data
+CO=12;//The composition of carbondioxide of combustion by volume in percentage 
+C=0.5;//The composition of carbonmoxide of combustion by volume in percentage 
+O=4;//The composition of oxygen of combustion by volume in percentage 
+N=83.5;//The composition of nitrogen of combustion by volume in percentage 
+o=32;//Molecular weight of the oxygen
+co=44;//Molecular weight of the carbondioxide 
+c=12;//Molecular weight of the carbon
+s=32;//Molecular weight of the sulphur
+so=64;//Molecular weight of sulphur dioxide
+n1=28;//Molecular weight of the nitrogen
+h=2;//Molecular weight of the hydrogen
+
+//Calculations
+m=12+0.5;//Balancing carbon
+x=N/3.76;//Balancing nitrogen
+z=[x-(CO+(C/2)+O)]*2;//Balancing oxygen
+n=z*h;//Balancing hydrogen
+Af=[(x*o)+(N*n1)]/[(m*c)+(n)];//Air/fuel ratio
+As=[(18.46*o)+(69.41*n1)]/173.84;//Stoichiometric air/fuel ratio
+Ta=(Af/As)*100;//Percent theoretical air
+mc=[(m*c)/173.84]*100;//Composition of carbon on mass basis in percent
+mh=(n/173.84)*100;//Composition of hydrogen on mass basis in percent
+
+//Output
+printf(' (a)The air/fuel ratio = %3.2f \n (b)The percent theoretical air = %3.1f percent \n (c)The percentage composition of fuel on a mass basis : \n   C = %3.1f percent \n   H = %3.1f percent ',Af,Ta,mc,mh)
diff --git a/2090/CH3/EX3.5/Chapter3_Example5.sce b/2090/CH3/EX3.5/Chapter3_Example5.sce
new file mode 100755
index 000000000..a89a6c555
--- /dev/null
+++ b/2090/CH3/EX3.5/Chapter3_Example5.sce
@@ -0,0 +1,25 @@
+clc
+clear
+//Input data
+C=86;//The composition of carbon in the fuel by weight in percentage
+H=14;//The composition of hydrogen in the fuel by weight in percentage
+e=1.25;//Equivalent ratio
+o=32;//Molecular weight of the oxygen
+co=44;//Molecular weight of the carbondioxide 
+c=12;//Molecular weight of the carbon
+s=32;//Molecular weight of the sulphur
+so=64;//Molecular weight of sulphur dioxide
+n=28;//Molecular weight of the nitrogen
+h2=2;//Molecular weight of the hydrogen
+Fc=0.86;//Fraction of C
+
+//Calculations
+Ra=1/Fc;//Relative air/fuel ratio
+x=2*[1+(0.9765/2)-(1.488*0.8)];//By oxygen balance
+Tm=0.5957+0.4043+4.476;//Total number of moles of dry exhaust gas
+vc=(0.5957/Tm)*100;//Volumetric analysis of carbonmonoxide of combustion in percentage
+vco=(0.4043/Tm)*100;//Volumetric analysis of carbondioxide of combustion in percentage
+vn=(4.476/Tm)*100;//Volumetric analysis of nitrogen of combustion in percentage
+
+//Calculations
+printf(' The percentage analysis of dry exhaust gas by volume : \n    CO = %3.2f percent \n    CO2 = %3.2f percent \n    N2 = %3.2f percent ',vc,vco,vn)
diff --git a/2090/CH3/EX3.6/Chapter3_Example6.sce b/2090/CH3/EX3.6/Chapter3_Example6.sce
new file mode 100755
index 000000000..0a6144c0f
--- /dev/null
+++ b/2090/CH3/EX3.6/Chapter3_Example6.sce
@@ -0,0 +1,16 @@
+clc
+clear
+//Input data
+t=25;//The temperature of both reactants and products in degree centigrade
+p=1;//The pressure of both reactants and products in bar
+
+//Calculations 
+h=0;//Enthalpy of all elements at given temp and pressure 
+hf1=-103.85;//The enthalpy of the compound C3H8 in the reactants side at given temp and pressure in MJ/kmol
+hf2=-393.52;//The enthalpy of carbondioxide for the given temp and pressure in MJ/kmol
+hf3=-285.8;//The enthalpy of the water for the given temp and pressure in MJ/kmol
+hf4=[3*hf2]+[4*hf3];//Total enthalpy in the products side in MJ/kmol
+Q=hf4-hf1;//The heat transfer per mole of fuel in MJ/kmol fuel
+
+//Output
+printf(' The heat transfer per mole of fuel = %3.2f kJ/mol fuel',Q)
diff --git a/2090/CH3/EX3.7/Chapter3_Example7.sce b/2090/CH3/EX3.7/Chapter3_Example7.sce
new file mode 100755
index 000000000..2f072825a
--- /dev/null
+++ b/2090/CH3/EX3.7/Chapter3_Example7.sce
@@ -0,0 +1,29 @@
+clc
+clear
+//Input data
+t=25;//The temperature of the air entering the diesel engine in degree centigrade 
+T=600;//The temperature at which the products are released in K
+Ta=200;//Theoretical air used in percentage 
+Q=-93;//Heat loss from the engine in MJ/kmol fuel
+f=1;//The fuel rate in kmol/h
+
+//Calculations 
+hfr=-290.97;//The enthalpy of C12H26 for the given conditions in the reactants side in MJ/kmol
+h1=-393.52;//Enthalpy of carbondioxide at formation state in MJ/kmol
+h11=12.916;//The change in enthalpy for the given temp of CO2 in MJ/kmol
+hfc=h1+h11;//The enthalpy of the carbondioxide in MJ/kmol
+h2=-241.82;//The enthalpy of water at formation state in MJ/kmol
+h22=10.498;//The change in enthalpy for the given temp of water in MJ/kmol
+hfh=h2+h22;//The enthalpy of the water in MJ/kmol
+h3=0;//Enthalpy of the oxygen gas 
+h33=9.247;//The change in enthalpy for the given temp of oxygen in MJ/kmol
+hfo=h3+h33;//The enthalpy of oxygen in MJ/kmol
+h4=0;//The enthalpy of the nitrogen gas
+h44=8.891;//The change in enthalpy of the nitrogen for the given temp in MJ/kmol
+hfn=h4+h44;//The enthalpy of nitrogen in MJ/kmol
+hfp=(12*hfc)+(13*hfh)+(18.5*hfo)+(139.12*hfn);//The total enthalpy on the products side in MJ/kmol
+W=Q+hfr-hfp;//The work in MJ/kmol fuel
+W1=(f*W*10^3)/3600;//The work in kW
+
+//Output
+printf('The work for a fuel rate of 1 kmol/h is %3.1f kW',W1)
diff --git a/2090/CH3/EX3.8/Chapter3_Example8.sce b/2090/CH3/EX3.8/Chapter3_Example8.sce
new file mode 100755
index 000000000..00f40c536
--- /dev/null
+++ b/2090/CH3/EX3.8/Chapter3_Example8.sce
@@ -0,0 +1,31 @@
+clc
+clear
+//Input data
+P=600;//Power of an engine in kW
+t=25;//Temperature at which fuel is used in degree centigrade
+Ta=150;//Theoretical air used in percentage
+T1=400;//The temperature at which air enters in K
+T2=700;//The temperature at which the products of combustion leave in K
+Q=-150;//The heat loss from the engine in kW
+C=12;//Molecular weight of carbon
+h=1;//Molecular weight of hydrogen
+
+//Calculations
+hfc=-259.28;//The enthalpy of the compound C8H18 for the given conditions in MJ/kmol fuel
+hfo1=3.029;//The enthalpy of the oxygen gas in MJ/kmol fuel
+hfn1=2.971;//The enthalpy of the nitrogen gas in MJ/kmol fuel
+HR=(hfc)+(1.5*12.5*hfo1)+(1.5*12.5*3.76*hfn1);//The total enthalpy on the reactants side in MJ/kmol fuel
+hfco=-393.52;//The enthalpy of carbondioxide for formation state in MJ/kmol fuel
+hfco1=17.761;//The change in enthalpy of the carbondioxide for temp difference in MJ/kmol fuel
+hfh=-241.82;//The enthalpy of water for formation state in MJ/kmol fuel
+hfh1=14.184;//The change in the enthalpy of the water for temp difference in MJ/kmol fuel
+hfo2=12.502;//The enthalpy of the oxygen gas in MJ/kmol fuel
+hfn2=11.937;//The enthalpy of the nitrogen gas in MJ/kmol fuel
+HP=(8*(hfco+hfco1))+(9*(hfh+hfh1))+(6.25*hfo2)+(70.5*hfn2);//The total enthalpy on the products side in MJ/kmol fuel
+H=HP-HR;//The total change in enthalpy of reactants and products in MJ/kmol fuel
+nf=([Q-P]*3600)/[H*10^3];//The fuel rate in kmol/s
+M=(8*C)+(18*h);//Molecular weight of fuel 
+mf=nf*M;//The fuel consumption in kg/h
+
+//Output
+printf(' The fuel consumption for complete combustion is %3.2f kg/h',mf)
diff --git a/2090/CH3/EX3.9/Chapter3_Example9.sce b/2090/CH3/EX3.9/Chapter3_Example9.sce
new file mode 100755
index 000000000..6430c9b52
--- /dev/null
+++ b/2090/CH3/EX3.9/Chapter3_Example9.sce
@@ -0,0 +1,21 @@
+clc
+clear
+//Input data
+t=25;//Temperature at which fuel is used for combustion in degree centigrade 
+p=1;//The pressure at which fuel is used in bar
+T=400;//The temperature of the products of combustion in K
+R=8.314*10^-3;//Universal gas constant
+
+//Calculations
+hfc=-103.85;//Enthalpy of the compound C3H8 in MJ/kmol fuel 
+HR=[1*(hfc-(R*(t+273)))]+[5*(-R*(t+273))];//The total enthalpy of the reactants in MJ/kmol fuel
+hfco=-393.52;//The enthalpy of the carbondioxide in MJ/kmol fuel
+hfco1=4.008;//The change in enthalpy of the carbondioxide for the given conditions in MJ/kmol fuel
+hfh=-241.82;//The enthalpy of the water in MJ/kmol fuel
+hfh1=3.452;//The change in enthalpy of the water for the given conditions in MJ/kmol fuel
+HP=[3*(hfco+hfco1-(R*T))]+[4*(hfh+hfh1-(R*T))];//The total enthalpy of the products in MJ/kmol fuel
+Q=HP-HR;//The total change in the enthalpy of reactans and products in MJ/kmol fuel
+Q1=-Q;//Heat liberated in kJ/mol propane
+
+//Output
+printf('The heat transfer per mole of propane = %3.1f kJ/mol propane',Q1)