diff options
Diffstat (limited to '1808/CH2')
26 files changed, 606 insertions, 0 deletions
diff --git a/1808/CH2/EX2.1/Chapter2_Example1.sce b/1808/CH2/EX2.1/Chapter2_Example1.sce new file mode 100644 index 000000000..3e7bbb96e --- /dev/null +++ b/1808/CH2/EX2.1/Chapter2_Example1.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+//GASES IN THE ORDER CO2,CO,O2,N2
+p=[16.1,0.9,7.7,75.3];//percentage of gas
+m=[44,28,32,28];//molecular weight of gas
+
+//CALCULATIONS
+x1=p(1)/m(1);//individual moles per 100 kg of CO2 mixture
+x2=p(2)/m(2);//individual moles per 100 kg of CO mixture
+x3=p(3)/m(3);//individual moles per 100 kg of O2 mixture
+x4=p(4)/m(4);//individual moles per 100 kg of N2 mixture
+x=x1+x2+x3+x4;//Total moles per 100 kg mixture
+v1=(x1/x)*100;//percentage of gases on volume basis
+v2=(x2/x)*100;//percentage of gases on volume basis
+v3=(x3/x)*100;//percentage of gases on volume basis
+v4=(x4/x)*100;//percentage of gases on volume basis
+v=v1+v2+v3+v4;//total percentage of gases on volume basis
+
+//OUTPUT
+printf('(i)percentage of gases on volume basis is %3.3f ',v)
+
+
diff --git a/1808/CH2/EX2.10/Chapter2_Example10.sce b/1808/CH2/EX2.10/Chapter2_Example10.sce new file mode 100644 index 000000000..77b1fe9c0 --- /dev/null +++ b/1808/CH2/EX2.10/Chapter2_Example10.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//0.8062 CH4+0.0541 C2H6 +0.0187C3H8+0.0160C4H10+0.1050N2+a (O2+3.76 N2)=b (0.078 CO2+0.002 CO +0.07 O2 +0.85N2)+ c H2O ;//Combustion equation for 1 kmol of fuel mixture
+//b*(0.078+0.002)=0.8062+2*(0.0541)+3*(0.0160);//by carbon balance
+c=1.93;//Carbon balance
+a=2.892;//Oxygen balance
+
+//CALCULATIONS
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 12.93 (0.078 CO2 )+0.002(0+0.07 O2 +0.85 N2)+1.93 H2O ;//Balanced chemical equation
+xm=a*4.76/1;//Air fuel ratio on molar basis
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 1.0345 CO2+1.93 H2O+7.625 N2 ;//Balanced chemical equation
+xth=2*4.76;//Theoretical air fuel ratio
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)The air fuel ratio on molar basis %3.2f kmol of air/kmol of fuel \n (b)Percentage of theoretical air %3.1f percentage ',xm,nth)
diff --git a/1808/CH2/EX2.11/Chapter2_Example11.sce b/1808/CH2/EX2.11/Chapter2_Example11.sce new file mode 100644 index 000000000..c36e7935a --- /dev/null +++ b/1808/CH2/EX2.11/Chapter2_Example11.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+t1=25;//Temperature
+p1=1;//atmospheric pressure
+//H2+(1/2)O2=H20
+//Qcv+Reactants=Products
+
+//CALCULTIONS
+Qcv=-285838;//Enthalpy in kJ/kmol from the table
+
+//OUTPUT
+printf('(i)Enthalpy of formation of H2O is %3.1f kJ/kmol',Qcv)
+
+
+
diff --git a/1808/CH2/EX2.12/Chapter2_Example12.sce b/1808/CH2/EX2.12/Chapter2_Example12.sce new file mode 100644 index 000000000..0747edef8 --- /dev/null +++ b/1808/CH2/EX2.12/Chapter2_Example12.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//H2+(1/2)O2=H20
+//Qcv+Reactants=Products
+
+//CALCULTIONS
+hf=-241827;//From the tables of enthalpy
+dh=9515;//enthalpy change
+Qcv=hf+dh;//Enthalpy of H2O on kmol basis
+
+//OUTPUT
+printf('(i)Enthalpy of H2O on kmol basis %3.1f kJ/kmol',Qcv)
+
+
+
diff --git a/1808/CH2/EX2.13/Chapter2_Example13.sce b/1808/CH2/EX2.13/Chapter2_Example13.sce new file mode 100644 index 000000000..cdfe7f236 --- /dev/null +++ b/1808/CH2/EX2.13/Chapter2_Example13.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+p=50;//Power output in kW
+m=0.05;//mass flow rate in kg/s
+//C8H18 +2*12.5 O2 +2*12.5*3.76 N2= 8 CO2 +9 H2O +12.5 O2 +99N2 ;//Rate of heat transfer from the engine in kJ/kmol
+//Qcv+x1=x2+Wcv ;//rate of heat transfer
+x1=-249952;//inlet heat transfer in kJ/kmol
+x2=-2415445.5;//exit heat transfer in kJ/kmol
+
+//CALCULATIONS
+Wcv=(p/m)*114.28;//work done in J/kmol of fuel
+Qcv=x2+Wcv-x1;//Heat transfer rate from the engine in kJ/kmol
+
+//OUTPUT
+printf('(i)Heat transfer rate from the engine is %3.1f kJ/kmol',Qcv)
+
+
+
+
+
diff --git a/1808/CH2/EX2.14/Chapter2_Example14.sce b/1808/CH2/EX2.14/Chapter2_Example14.sce new file mode 100644 index 000000000..f52371727 --- /dev/null +++ b/1808/CH2/EX2.14/Chapter2_Example14.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//CH4+2O2=CO2+H2O ;//Combustion equation
+//Q=Up-Ur ;//Energy balance for the closed system
+hfco2=-393520;//enthalpy of CO2 From the table
+dhco2=28041;//change in enthalpy in KJ/kmol
+hfh2o=-241820;//enthalpy of H2O From the table
+dhh2o=21924;//change in enthalpy in KJ/kmol
+hfch4=-74850;//enthalpy of CH4 From the table
+t1=298;//initial temperature in K
+t2=900;//final temperature in K
+p1=1;//Pressure in atm
+R=8.314;//gas constant
+
+//CALCULATIONS
+Q=(hfco2+dhco2)+2*(hfh2o+dhh2o)-(hfch4)+3*R*(t1-t2);//Amount of heat transfer in kJ/kmol
+p2=p1*(t2/t1);//Final pressure in atmosphere
+
+//OUTPUT
+printf('(i)Amount of heat transfer is %3.2f kJ/kmol \n (ii)Final pressure is %3.2f atmosphere',Q,p2 )
+
+
+
diff --git a/1808/CH2/EX2.15/Chapter2_Example15.sce b/1808/CH2/EX2.15/Chapter2_Example15.sce new file mode 100644 index 000000000..72aa0995a --- /dev/null +++ b/1808/CH2/EX2.15/Chapter2_Example15.sce @@ -0,0 +1,43 @@ +clc
+clear
+//INPUT DATA
+//CH4 + 2O2 + 7.52N2=CO2 + 2H2O + 7.52N2 ;//Combustion equation with liquid water in the products
+t1=278;//atmospheric temperature
+t2=1000;//products temperature
+p1=1;//atmospheric pressure
+hfco2=-393520;//Acc.to tables with liquid water in the products enthalpy of CO2
+hfh2o=-285830;//Acc.to tables with liquid water in the products enthalpy of H2O
+hfch4=-74850;//Acc.to tables with liquid water in the products enthalpy of CH4
+
+hfco21=-393520;///Acc.to tables with water vapour in the products enthalpy of CO2
+hfh2o1=-241820;//Acc.to tables with water vapour in the products enthalpy of H2O
+hfch41=-74850;//Acc.to tables with water vapour in the products enthalpy of CH4
+
+h21co2=33368;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of CO2
+h21h2o=25978;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of H2O
+h21n2=21468;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of N2
+
+//CALCULATIONS
+hrp=1*hfco2+2*hfh2o-hfch4;//enthalpy of reactants and products in kJ/kmol
+hrpCH4=hrp/16.04;//Enthalpy of combustion of gaseous methane with liquid water in the products in kJ/kg
+
+hrp1=1*hfco21+2*hfh2o1-hfch41;//enthalpy of reactants and products in kJ/kmol
+hrpCH41=hrp1/16.04;//Enthalpy of combustion of gaseous methane with water vapour in the products in kJ/kg
+
+hrp2=1*(hfco21)+(h21co2)+2*(h21h2o)+2*(hfh2o)+7.52*(h21n2)-1*(hfch4);//enthalpy of reactants and products in kJ/kmol
+hrpCH42=hrp2/16.04;//Enthalpy of combustion of gaseous methane at 1000 K ,1atm with water vapour in the products in kJ/kg
+
+dhco2=(42769-9364);//From tables both reactants and products enthalpy
+dhh2o=(35882-9904);//From tables both reactants and products enthalpy
+dho2=(31389-8682);//From tables both reactants and products enthalpy
+dhch4=38189;//From tables both reactants and products enthalpy
+
+hrp3=1*(hfco2+dhco2)+2*(hfh2o1+h21h2o)-(hfch41+dhch4)-2*(dho2);//enthalpy of reactants and products in kJ/kmol
+hrpCH43=hrp3/16.04;//Enthalpy of combustion of gaseous methane at 1000 K ,1atm with water vapour and liqid water in the products in kJ/kg
+
+//OUTPUT
+printf('(i)Enthalpy of combustion of gaseous methane with liquid water in the \n products %3.2f kJ/kg of fuel\n(ii)Enthalpy of combustion of gaseous methane with water \n vapour in the products %3.2f kJ/kg of fuel\n ',hrpCH4,hrpCH41)
+printf('(iii)Enthalpy of combustion of gaseous methane at 1000 K ,1atm \n with water vapour in the products is %3.3f kJ/kg of fuel\n(iv)Enthalpy of combustion of gaseous methane at 1000 K ,1atm \n with water vapour and liqid water is the products is %3.2f kJ/kg of fuel',hrpCH42,hrpCH43)
+
+
+
diff --git a/1808/CH2/EX2.16/Chapter2_Example16.sce b/1808/CH2/EX2.16/Chapter2_Example16.sce new file mode 100644 index 000000000..91385906d --- /dev/null +++ b/1808/CH2/EX2.16/Chapter2_Example16.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+//C2H2+3O2=2CO2+2H2O ;//Chemical equation
+t1=298;//initial temperature in K
+t2=800;//Final temperature in K
+R=0.287;//gas constant in kJ/kgK
+dhCO2=22815;//From tables enthalpy of CO2 kJ/kmol
+dhH2O=17991;//From tables enthalpy of H2O kJ/kmol
+hfCO2=-393520;//From tables enthalpy of CO2 kJ/kmol
+hfH2O=-241827;//From tables enthalpy of H2O kJ/kmol
+hfC2H4=52283;//From tables enthalpy of C2H4 kJ/kmol
+
+//CALCULATIONS
+Q=2*(hfCO2+dhCO2)+2*(hfH2O+dhH2O)-1*(hfC2H4)-4*R*(t2-t1);//amount of heat transfer from the reactants
+
+//OUTPUT
+printf('(i)The amount of heat transfer from the reactants is %3.1f kJ/kmol of fuel ',Q)
+
diff --git a/1808/CH2/EX2.17/Chapter2_Example17.sce b/1808/CH2/EX2.17/Chapter2_Example17.sce new file mode 100644 index 000000000..ce9eb8720 --- /dev/null +++ b/1808/CH2/EX2.17/Chapter2_Example17.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+hc=4;//cylinders
+mf=2;//mass flow rate in g/s
+vd=2.8;//capacity in litres
+N=1500;//speed in rpm
+ta=303;//temperature in K
+Pa=101.325;//atmospheric pressure in kPa
+R=0.287;//gas constant in kJ/kgK
+xs =15.14;//air fuel ratio
+//C8H18+12.5(O2+3.773N2)=8CO2+9H2O+47.16 N2;//Chemical equation
+
+//CALCULATIONS
+ma1=xs*mf;//mass of air in g/s
+ma=ma1/50;//mass of air in g/cylinder/cycle
+ro=Pa/(R*ta);//density in kg/m^3
+nv=ma/(ro*vd/4);//Volumetric efficiency
+
+//OUTPUT
+printf('Volumetric effiiciency is %3.4f ',nv)
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.18/Chapter2_Example18.sce b/1808/CH2/EX2.18/Chapter2_Example18.sce new file mode 100644 index 000000000..2ee095647 --- /dev/null +++ b/1808/CH2/EX2.18/Chapter2_Example18.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//C4H10+6.5(O2+3.773N2) = 4CO2+5H2O+24.5245N2 ;//Chemicl equation
+x=0.9;//equivalent ratio
+
+//CALCULATIONS
+xs=(12*4+1*10)/(31.0245*28.962);//air fuel ratio
+xa=x*xs;//exhaust gas composition
+
+//OUTPUT
+printf('Exhaust gas compositin is %3.4f \n',xa)
+printf('C4H10+1.11*6.5( O2+3.773 N2) = 4 CO2+5 H2O+0.7150 O2+27.22 N2' )
+
+
+
diff --git a/1808/CH2/EX2.19/Chapter2_Example19.sce b/1808/CH2/EX2.19/Chapter2_Example19.sce new file mode 100644 index 000000000..553b942f3 --- /dev/null +++ b/1808/CH2/EX2.19/Chapter2_Example19.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+//CH4 + 2 O2 = CO2+2 H2O ;//Chemical equation
+hfco2=-393.52;//enthalpy of CO2 in MJ/kmol of CH4
+hfh2o=-241.83;//enthalpy of H2O in MJ/kmol of CH4
+hfch4=-77.87;//enthalpy of CH4 in MJ/kmol of CH4
+hfo2=0;//enthalpy of O2 in MJ/kmol of CH4
+
+//CALCULATIONS
+
+Qp=hfco2+2*hfh2o-(hfch4+2*hfo2);//Lower heating value in MJ/kmol
+Qp1=-Qp/(21*1-1*4);//Lower heating value in MJ/kg
+dU=Qp1;//lower heating values
+
+//OUTPUT
+printf('np=nr \n')
+printf('Lower heating values are \n Qp %3.1f MJ/kg of CH4 \n du %3.1f MJ/kg of CH4 \n Qp1=dU \n ',Qp1,dU)
+
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.2/Chapter2_Example2.sce b/1808/CH2/EX2.2/Chapter2_Example2.sce new file mode 100644 index 000000000..74b8d1ef9 --- /dev/null +++ b/1808/CH2/EX2.2/Chapter2_Example2.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//GASES IN THE ORDER CO2,CO,O2,N2
+p=[10,8,1.5,80.5];//percentage of gas
+m=[44,32,28,28];//molecular weight of gas
+
+//CALCULATIONS
+x1=m(1)*(p(1)/100);//individual moles per 100 kg of CO2 mixture
+x2=m(2)*(p(2)/100);//individual moles per 100 kg of CO mixture
+x3=m(3)*(p(3)/100);//individual moles per 100 kg of O2 mixture
+x4=m(4)*(p(4)/100);//individual moles per 100 kg of N2 mixture
+x=x1+x2+x3+x4;//Total moles per 100 kg mixture
+v1=(x1/x)*100;//percentage of gases on volume basis
+v2=(x2/x);//percentage of gases on volume basis
+
+
+
+//OUTPUT
+printf('(a)percentage of each gas by mass is %3.2f \n (b)mass of oxygen per kg of dry flue gases is %3.3f kg',v1,v2)
+
+
+
+
diff --git a/1808/CH2/EX2.20/Chapter2_Example20.sce b/1808/CH2/EX2.20/Chapter2_Example20.sce new file mode 100644 index 000000000..1abfcd59f --- /dev/null +++ b/1808/CH2/EX2.20/Chapter2_Example20.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+mf=0.4;//flue flow rate in g/s
+ma=5.6;//air flow rate in g/s
+hbc=1.8;//gasoline with H/C ratio
+//a CH1.87+b O2+c N2=13 CO2 +2.8 CO +0.933 H2+ d H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+c=83.267;//Composition of N2
+b=22.069;//Composition of O2
+a=15.8;//Composition of CH1.87
+d=13.84;//Composition of H2O
+
+//CALCULATIONS
+xa=mf/ma;//air fuel ratio
+//CH1.87+1.4675(O2+3.773N2)=CO2+0.935 H2O+ 5.536 N2 ;//CHEMICAL COMPOSITION
+xs=(12*1+1*hbc)/(202+86);//air fuel ratio
+x=(xa/xs);//Equivlent ratio from the fuel and air flow rate
+
+//15.8 CH1.87+22.069 O2+83.267 N2=13 CO2 +2.8 CO +0.937 H2+ 13.84 H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+//CH1.87+1.397 O2+5.27 N2=0.823 CO2 +0.177 CO +0.059 H2+ 0.876 H2O+ 5.27 N2 ;//CALCULATED EXHAUST GAS COMPOSITION
+xa1=(d)/(1.397*32+5.27*28);//air fuel ratio
+x1=xa1/xs;//Equivalent ratio of calculated exhaust gas composition
+
+//OUTPUT
+printf('(i)The Equivlent ratio from the fuel and air flow rate is %3.3f \n (ii)TheEquivalent ratio of calculated exhaust gas composition is %3.3f',x,x1)
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.21/Chapter2_Example21.sce b/1808/CH2/EX2.21/Chapter2_Example21.sce new file mode 100644 index 000000000..8cb5a22c3 --- /dev/null +++ b/1808/CH2/EX2.21/Chapter2_Example21.sce @@ -0,0 +1,13 @@ +clc
+clear
+//INPUT DATA
+//H2+ 0.5(O2+3.773 N2)= H2O+1.887 N2 ;//CHEMICAL EQUATION
+//22.3 H2 +18.606 O2 +70.2 N2 = 22.3 H2O+ 7.44 O2+70.2 N2 ;//EXHAUST GASES CHEMICAL EQUATION
+
+//CALCULATIONS
+xs=2*1/(0.5*32+0.5*3.773*28);//air fuel ratio From the combustion equation
+xa=(1*2)/(0.8343*32+3.148*28);//air fuel ratio From the combustion equation
+x=xa/xs;//Equivalent ratio
+
+//OUTPUT
+printf('Equivalent ratio is %3.3f',x)
diff --git a/1808/CH2/EX2.22/Chapter2_Example22.sce b/1808/CH2/EX2.22/Chapter2_Example22.sce new file mode 100644 index 000000000..dd86f0d71 --- /dev/null +++ b/1808/CH2/EX2.22/Chapter2_Example22.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//C8H18+12.5(O2+3.773N2)=8 CO2 +9 H2O +47.16 N2 ;//FUEL COMPOSITION
+n=60.66;//number of moles of air
+
+//CALCULATIONS
+n1=8+9+47.16;//number of moles of air and product
+xs= 15.14/1;//air fuel ratio
+xs1=1/xs;//fuel air ratio
+Mr=(1/n)*(114.15+59.66*28.96);//Molecular weights of reactants
+Mp=(1/n1)*(8*44+9*18+47.16*28);//Molecular weights of products
+
+//OUTPUT
+printf('(i)number of moles of air and product %3.2f \n (ii)(A/F)s %3.2f \n (F/A)s %3.2f \n (iii)Molecular weights of reactants %3.2f \n Molecular weights of products %3.2f',n1,xs,xs1,Mr,Mp)
+
+
diff --git a/1808/CH2/EX2.23/Chapter2_Example23.sce b/1808/CH2/EX2.23/Chapter2_Example23.sce new file mode 100644 index 000000000..bbb074647 --- /dev/null +++ b/1808/CH2/EX2.23/Chapter2_Example23.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//CH4+2O2=CO2+2H2O ;//STOICHIOMETRIC REACTION
+//CASE I
+//H2O in the products is liquid
+//CASE II
+//H2O in the products is gas
+Hr=-74.87;//enthalpy of reactants
+Hp1=-964.2;//enthalpy of products
+Hp2=-876.18;//enthalpy of products
+R=8.314*10^-3;//gas constant
+t=298;//initial temperature in K
+
+//CLCULATIONS
+dH1=Hp1-Hr;//Enthalpy increase in MJ/kmol
+dH2=Hp2-Hr
+dU1=dH2-((1-3)*R*t);//internal energy in MJ/kmol
+dU=Hp2;//internal energy in MJ/kmol
+
+//OUTPUT
+printf('Enthalpy increase is %3.2f MJ/kmol of CH4 \n internal energy increase is %3.2f MJ/kmol of CH4 \n',dH2,dU1)
+printf('H2O in the products and internal energy increase are same \n')
+printf('np=nr')
diff --git a/1808/CH2/EX2.24/Chapter2_Example24.sce b/1808/CH2/EX2.24/Chapter2_Example24.sce new file mode 100644 index 000000000..e93581ce1 --- /dev/null +++ b/1808/CH2/EX2.24/Chapter2_Example24.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//CH2+(3/2) (O2+3.773 N2)= CO2+H2O+5.66N2 ;//STOICHIOMETRIC EQUATION
+dU=-43.2;//Internal energy in MJ/kg
+
+//CALCULATIONS
+dH=dU+(7.66-7.16)*8.3143*10^-3*298/14;//ENTHALPY CHANGE
+Hp=-((1*-393.52)+(-241.8))/(221.4);//enthalpy of products per kg of mixture
+Hr=Hp-((-43.1*14)/(221.4));//Enthalpy of reactants per kg of mixture
+
+//OUTPUT
+printf('enthalpy of products per kg of mixture %3.2f MJ/kg \n enthalpy of reactants per kg of mixture %3.2f MJ/kg',Hp,Hr)
+
+
+
diff --git a/1808/CH2/EX2.25/Chapter2_Example25.sce b/1808/CH2/EX2.25/Chapter2_Example25.sce new file mode 100644 index 000000000..1ed69e3fe --- /dev/null +++ b/1808/CH2/EX2.25/Chapter2_Example25.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//0.506 H2 +0.1 CO+0.26 CH4 +0.04 C4H8 +0.004 O2 +0.03 CO2 +0.06N2 +0.21*8 02+0.79 8 N2= a CO2 +b H20 +c O2 +d N2 ;//COMBUSTION EQUATION
+a=0.55;//CARBON BALANCE
+b=1.186;//HYDROGEN BALANCE
+c=0.621;//OXYGEN BALANCE
+d=6.38;//NITROGEN BALANCE
+
+//CALCULATIONS
+n=a+c+d;//Total moles of dry products
+nCO2=(a/n)*100;//Analysis of products by volume
+nO2=(c/n)*100;//Analysis of products by volume
+nN2=(d/n)*100;//Analysis of products by volume
+
+//OUTPUT
+printf('Analysis of products by volume \n(i)CO2 %3.2f \n (ii)O2 %3.2f \n (iii)N2 %3.2f ',nCO2,nO2,nN2)
diff --git a/1808/CH2/EX2.26/Chapter2_Example26.sce b/1808/CH2/EX2.26/Chapter2_Example26.sce new file mode 100644 index 000000000..e481ac915 --- /dev/null +++ b/1808/CH2/EX2.26/Chapter2_Example26.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+a=7;//Composition
+b=9;//Composition
+//C7H9O0.2813 ;//GIVEN FUEL
+//C7H9O0.2813N0.107 + m O2 + m (79/21) N2 = x CO2 + y H2O + z N2 ;//COMBUSTION EQUATION
+
+//CALCULATIONS
+x=7;//By balancing method composition of CO2
+y=4.5;//By balancing method composition of H2O
+m=9.11;//By balancing method composition of O2
+z=35.4;//By balancing method composition of N2
+
+//C7H9O0.2813N0.107 + 9.11 O2 + 9.11*(79/21) N2 = 7 CO2 + 4.5 H2O + 35.4 N2 ;//COMBUSTION EQUATION CALCULATED
+xs=m*32+m*(79/21)*28/(100);//air fuel ratio
+
+//(ii)Percentage composition of dry flue gases increased by 20% excess air
+//C7H9O0.2813N0.107 + (1.2*9.11) O2 + (1.2*9.11)*(79/21) N2 = 7 CO2 + 4.5 H2O + 1.2*35.4 N2 + (0.2*9.11) O2 ;//COMBUSTION EQUATION CALCULATED
+n=7+4.5+0.2*9.11+1.2*35.4;//Total number of moles of dry flue gases by volume
+nCO2=(7/n)*100;//Percentage composition of dry flue gases by volume
+nO2=(1.822/n)*100;//Percentage composition of dry flue gases by volume
+nN2=(42.48/n)*100;//Percentage composition of dry flue gases by volume
+
+//OUTPUT
+printf('Percentage composition of dry flue gases by volume \n(i)CO2 %3.2f \n (ii)O2 %3.2f \n (iii)N2 %3.2f ',nCO2,nO2,nN2)
+
diff --git a/1808/CH2/EX2.3/Chapter2_Example3.sce b/1808/CH2/EX2.3/Chapter2_Example3.sce new file mode 100644 index 000000000..1d53e20a5 --- /dev/null +++ b/1808/CH2/EX2.3/Chapter2_Example3.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+//C8H18 + XO2 = a Co2 + b H2O// Stoichiometric equation for combustion of octane
+
+//CALCULATIONS
+
+//Carbon balance
+a=8//8C=aC
+
+//Hydrogen balance
+b=9//18H=2*b
+
+//Oxygen balance
+X=(16+9)/2//2XO=2aO+9
+
+//Combustion equation
+Y=3.76*X//moles of nitrogen in the air for Y moles of Oxygen
+N=(X+Y)/1//moles of air to one mole of fuel
+M=(N)*(29/114)//mass of air required for 1 Kg of Fuel
+
+//OUTPUT
+printf("(i)Theoretical air fuel ratio for combustion of octane is %3.2f kg of air/kg",M)
diff --git a/1808/CH2/EX2.4/Chapter2_Example4.sce b/1808/CH2/EX2.4/Chapter2_Example4.sce new file mode 100644 index 000000000..cad029031 --- /dev/null +++ b/1808/CH2/EX2.4/Chapter2_Example4.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+//C8H18 + 12.5 O2 + (12.5*3.76) N2 = 8 Co2 + 9 H2O + 47 N2// Stoichiometric equation for combustion of octane with 100 percent of air
+//C8H18 + 12.5 O2 + (12.5*3.76) N2 = 8 Co2 + 9 H2O + 47 N2// Stoichiometric equation for combustion of octane with 200 percent of air
+a=8;//Carbon balance
+b=9;//Hydrogen balance
+d=94;//Nitrogen balance
+c=12.5;//Oxygen balance
+
+//CALCULATIONS
+x=a+b+c+d;//Total moles of products
+x1=100*a/x;//Molal analysis of CO2
+x2=100*b/x;//Molal analysis of H20
+x3=100*c/x;//Molal analysis of O2
+x4=100*d/x;//Molal analysis of N2
+
+//OUTPUT
+printf('(i)Molal analysis of CO2 is %3.2f percentage \n (ii)Molal analysis of H2O is %3.2f percentage \n (iii)Molal analysis of O2 is %3.2f percentage \n (iv)Molal analysis of N2 is %3.2f percentage',x1,x2,x3,x4)
+
diff --git a/1808/CH2/EX2.5/Chapter2_Example5.sce b/1808/CH2/EX2.5/Chapter2_Example5.sce new file mode 100644 index 000000000..9cdb45e4b --- /dev/null +++ b/1808/CH2/EX2.5/Chapter2_Example5.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+//a.CH4 +b O2 + c N2=10 CO2 + 0.53 CO+2.37 O2 +d H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+c=87.1;//Nitrogen balance
+b=23.16;//(c/b)=3.76
+d=21.60;//d=2a Hydrogen balance
+a=10.54;//Carbon balance
+//10.54.CH4 +23.16 O2 + 87.1 N2=10 CO2 + 0.53 CO+2.37 O2 +21.06 H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+//CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 // Stoichiometric equation for combustion of Methane with 100 percent of air
+
+//CALCULATIONS
+N=(2.2+8.27)/1;//air fuel ratio on mole basis
+M=N*29/(12+4);//air fuel ratio on mass basis
+Nt=(2+7.52)*(29/16);//theoritical air fuel ratio
+nt=(M/Nt)*100;//Percentage theoritical air
+
+//OUTPUT
+printf('(a) CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 \n')
+printf('(b)The air fuel ratio is %3.2f kg of air/kg of fuel \n (c)Percentage theoritical air is %f ',M,nt)
+
diff --git a/1808/CH2/EX2.6/Chapter2_Example6.sce b/1808/CH2/EX2.6/Chapter2_Example6.sce new file mode 100644 index 000000000..cb1da6afd --- /dev/null +++ b/1808/CH2/EX2.6/Chapter2_Example6.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+//CaHb + d O2 + c N2 = 8 Co2 + 0.9 CO +8.8 O2+e H2O + 82.3 N2// Stoichiometric equation for combustion of dry products with 100 percent of air
+c=82.3;//Nitrogen balance
+d=21.9;//(c/d)=3.76
+e=9.3;//Oxygen balance
+a=8.9;//Carbon balance
+b=18.6;//Hydrogen balance
+mf=125.4;//Mass of fuel
+Ma=29;//mass of air
+
+//C8.9H18.6 + 21.9 O2 + 82.3 N2 = 8 Co2 + 0.9 CO +8.8 O2+9.3 H2O + 82.3 N2// Stoichiometric equation for combustion of dry products with 100 percent of air
+
+//CALCULATIONS
+xm=((c+d)*Ma)/mf;//Air fuel ratio on mass basis
+xc=(a*12/(mf))*100;//Carbon composition on mass basis
+xh=(b*1/(mf))*100;//Hydrogen composition on mass basis
+
+//C8.9H18.6 +13.5O2 +(13.5*3.76)N2 = 8.9CO2 +9.3H2O +50.8N2//Theoretical combustion equation on mass basis
+xth=(13.5+50.8)*Ma/(mf);//Air fuel ratio of theoretical air on mass basis
+nth=(xm/xth)*100;//Percentage of theoretical air om mass basis
+
+//OUTPUT
+printf('(a)Air fuel ratio is %3.1f kg of air/kg of fuel \n (b)Composition of fuel on mass basis is %3.1f percentage \n (c)Percentage of theoretical air om mass basis %3.i percentage',xm,xh,nth)
+
+
+
+
diff --git a/1808/CH2/EX2.7/Chapter2_Example7.sce b/1808/CH2/EX2.7/Chapter2_Example7.sce new file mode 100644 index 000000000..1db5e6f6e --- /dev/null +++ b/1808/CH2/EX2.7/Chapter2_Example7.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+//aC4H10+ b O2+c N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+d H2O ;//Combustion equation
+c=82.9;//Nitrogen balance
+b=22.04;//(c/b)=3.76
+a=2.22;//Carbon balance
+d=11.1;//Hydrogen balance
+Ma=29;//mass of air
+
+//2.22C4H10+ 22.04 O2+82.9 N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+11.1 H2O ;//Combustion equation
+
+//CALCULATIONS
+//C4H10+ 9.92 O2+37.37 N2= 3.51 CO2+0.495 CO +3.69 O2+37.37 N2+5 H2O ;//Combustion equation dividing by 2.22 yields one mole of fuel
+xm=((9.92+37.37)*Ma)/(12*4+10);//air fuel ratio on mass basis
+
+//C4H10+ 6.5 O2+(6.5*3.76) N2= 4 CO2++24.44 N2+5 H2O ;//Theoretical Combustion equation
+xth=((6.5+24.44)*Ma)/(12*4+10);//Theoretical air fuel ratio
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)Percentage of theoretical air %3.2f percentage',nth)
+
+
+
+
diff --git a/1808/CH2/EX2.8/Chapter2_Example8.sce b/1808/CH2/EX2.8/Chapter2_Example8.sce new file mode 100644 index 000000000..bf0fddf60 --- /dev/null +++ b/1808/CH2/EX2.8/Chapter2_Example8.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+a=40;//percentage of H2 fuel
+b=10;//percentage of CO fuel
+c=3;//percentage of N2 fuel
+d=40;//percentage of CH4 fuel
+e=5;//percentage of CO2 fuel
+f=2;//percentage of O2 fuel
+N=4.76;//Amount of nitrogen require for complete combustion
+
+//CALCULATIONS
+x=a+b+c+d+e;//Total volumetric analysis of fuel
+//H2+ (1/2)O2= H2O
+X1=a/2;//Moles of oxygen required
+//CO+(1/2)O2=CO2
+X2=b/2;//Moles of oxygen required
+//CH4+2O2=CO2+2H2O
+X3=2*d;//Moles of oxygen required
+X4=-f;//Moles of oxygen required
+Y=X1+X2+X3+X4;//Moles of oxygen required for 100 moles of gas
+Z=Y*N;//Moles of air required for 100 moles of gas
+Z1=Z/x;//Theoritical air required for 1 mole of gas
+
+//OUTPUT
+printf('Theoritical air required for 1 mole of gas %3.1f mole of air',Z1)
+
+
+
+
+
diff --git a/1808/CH2/EX2.9/Chapter2_Example9.sce b/1808/CH2/EX2.9/Chapter2_Example9.sce new file mode 100644 index 000000000..2f82f7c96 --- /dev/null +++ b/1808/CH2/EX2.9/Chapter2_Example9.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+a=74;//Mass of constituent C
+b=4.3;//Mass of constituent H2
+c=2.7;//Mass of constituent S
+d=1.5;//Mass of constituent N2
+e=5.5;//Mass of constituent H2O
+f=5;//Mass of constituent O2
+g=7;//Mass of constituent ash
+
+a1=6.166;//Moles of constituent C
+b1=1.075;//Moles of constituent H2
+c1=0.084;//Moles of constituent S
+d1=0.053;//Moles of constituent N2
+e1=0.3055;//Moles of constituent H2O
+f1=0.156;//Moles of constituent O2
+g1=0;//Moles of constituent ash
+X1=26.955;//Moles of products N2
+
+
+//CALCULATIONS
+//C+O2=CO2
+x1=a1;//Moles of CO2 required
+//H2+(1/2)O2=H20
+x2=b1/2;//Moles of H2 required
+//S+O2=SO2
+x3=c1;//Moles of O2 required
+x4=d1;//Moles of O2 required
+x5=e1;//Moles of O2 required
+x5=f1;//Moles of O2 required
+x6=g1;//Moles of O2 required
+X=x1+x2+x3+x4+x5+x6;//total moles of products
+Y=a1+(b1+e1)+(2*X1)+(X)+c1;//Total moles of products required
+
+//OUTPUT
+printf('For 100 percentage excess air used,Total moles of products required is %3.3f',Y)
+
+
+
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