13 files changed, 329 insertions, 0 deletions
diff --git a/2090/CH4/EX4.1/Chapter4_Example1.sce b/2090/CH4/EX4.1/Chapter4_Example1.sce
new file mode 100755
index 000000000..53b49c86b
--- /dev/null
+++ b/2090/CH4/EX4.1/Chapter4_Example1.sce
@@ -0,0 +1,12 @@
+clc
+clear
+//Input data
+r=8.5;//The compression ratio 
+sv=1.4;//The specific heat at constant volume in percent
+
+//Calculations
+n=1-(1/r)^(sv-1);//The efficiency of the otto cycle 
+ef=[((1-n)/n)*(sv-1)*(log(r))*(sv/100)]*100;//The percentage change in efficiency of an otto cycle and is negative
+
+//Output
+printf('The efficiency decreases by %3.3f percent ',ef)
diff --git a/2090/CH4/EX4.10/Chapter4_Example10.sce b/2090/CH4/EX4.10/Chapter4_Example10.sce
new file mode 100755
index 000000000..371862643
--- /dev/null
+++ b/2090/CH4/EX4.10/Chapter4_Example10.sce
@@ -0,0 +1,23 @@
+clc
+clear
+//Input data
+r=7.8;//Compression ratio 
+p=1;//The pressure at the start of compression in atm
+T1=335;//The temperature at the start of compression in K
+
+//Calculations
+W1=100;//Isentropic compression function for T1 in J/kg air K 
+W2=W1-(292*log(1/r));//Isentropic compression function in J/kg air K 
+T2=645;//The temperature corresponding to isentropic compression function in J/kg air K 
+V1=(292*T1)/(p*10^5);//Volume at initial in m^3/kg air 
+p2=p*(T2/T1)*r;//The pressure at the end of compression stroke in atm 
+V2=V1/r;//The volume per unit mass of air at the end of the compression stroke in m^3/kg air 
+U1=35;//Internal energy corresponding to temp T1 in kJ/kg air 
+U2=310;//Internal energy corresponding to temp T2 in kJ/kg air 
+W=U2-U1;//Work input during compression in kJ/kg air 
+E1=120;//Isentropic compression function at T1 
+E2=910;//Isentropic compression function at T2 
+p21=[exp((E2-E1)/292)];//The pressure at the end of compression stroke in atm 
+
+//Output
+printf('(a)At the end of the compression stroke, \n  The temperature is %3.0f K \n The pressure is %3.0f atm \n The volume per unit mass of air is %3.3f m^3/kg air \n The pressure is %3.0f atm \n (b)The work input during compression is %3.0f kJ/kg air ',T2,p2,V2,p21,W)
diff --git a/2090/CH4/EX4.11/Chapter4_Example11.sce b/2090/CH4/EX4.11/Chapter4_Example11.sce
new file mode 100755
index 000000000..6eff40792
--- /dev/null
+++ b/2090/CH4/EX4.11/Chapter4_Example11.sce
@@ -0,0 +1,21 @@
+clc
+clear
+//Input data
+p=65;//The pressure in the cylinder in bar
+r=10;//The compression ratio 
+V3=0.1;//The volume per unit mass of air at the start of expansion in m^3/kg air 
+p3=p*100;//The pressure in the cylinder after the completion of combustion in kN/m^2 
+
+//Calculations 
+T3=2240;//The temperature from the chart corresponding to p3,V3 in K
+u3=-1040;//The energy from the chart in kJ/kg air 
+s3=8.87;//The entropy from the chart in kJ/kg air K
+s4=s3;//Since the process is isentropic 
+V4=r*V3;//The volume per unit mass of air at the end of expansion stroke in m^3/kg air 
+T4=1280;//The temperature from the chart corresponding to p4,V4 in K 
+u4=-2220;//The energy from the chart in kJ/kg air 
+p4=4.25;//The pressure from the chart in bar 
+W=-(u4-u3);//Work of expansion in kJ/kg air 
+
+//Output
+printf('(a)At the end of expansion stroke, \n The pressure is %3.2f bar \n The temperature is %3.0f K \n The volume is %3.1f m^3/kg air \n (b)The work during the expansion stroke is %3.0f kJ/kg air ',p4,T4,V4,W)
diff --git a/2090/CH4/EX4.13/Chapter4_Example13.sce b/2090/CH4/EX4.13/Chapter4_Example13.sce
new file mode 100755
index 000000000..748c688c6
--- /dev/null
+++ b/2090/CH4/EX4.13/Chapter4_Example13.sce
@@ -0,0 +1,26 @@
+clc
+clear
+//Input data
+Tu=645;//The temperature at the end of compression process in K
+usu=310;//The internal energy at the end of compression process in kJ/kg air 
+pu=(15.4*1.013);//The pressure at the end of the compression process in bar 
+Vu=0.124;//The volume at the end of the compression process in m^3/kg air 
+e=1;//Equivalence ratio 
+f=0.065;//Burned gas fraction 
+
+//Calculations 
+ufu=-118.5-(2963*f);//Internal energy of formation in kJ/kg air 
+ub=usu-ufu;//The internal energy for constant volume adiabatic combustion in kJ/kg air 
+Vb=Vu;//The volume for constant volume adiabatic combustion in kJ/kg air 
+Tb=2820;//The temperature for constant volume adiabatic combustion corresponding to ub,Vb on the burnt gas chart in K 
+pb=6500;//The pressure for constant volume adiabatic combustion corresponding to ub,Vb on the burnt gas chart in kN/m^2 
+hfu=-129.9-(2958*f);//The enthalpy of formation in kJ/kg air 
+hsu=440;//The enthalpy from chart corresponding to temp Tu in kJ/kg air 
+hb=hsu+hfu;//The enthalpy for constant pressure adiabatic combustion in kJ/kg air 
+pb1=1560;//The pressure for constant pressure adiabatic combustion in kN/m^2 
+ub1=-700;//Trail and error along the pb internal energy in kJ/kg air 
+vb1=(118-ub1)/pb;//The volume in m^3/kg air 
+Tb1=2420;//The temperature for constant pressure adiabatic combustion corresponding to ub,Vb on the burnt gas chart in K 
+
+//Output
+printf('(a)For constant volume adiabatic combustion,\n The temperature is %3.0f K \n The pressure is %3.0f kN/m^2 \n (b)For constant pressure adiabatic combustion, \n The temperature is %3.0f K \n The pressure is %3.0f kN/m^2',Tb,pb,Tb1,pb1)
diff --git a/2090/CH4/EX4.14/Chapter4_Example14.sce b/2090/CH4/EX4.14/Chapter4_Example14.sce
new file mode 100755
index 000000000..1e45f27c9
--- /dev/null
+++ b/2090/CH4/EX4.14/Chapter4_Example14.sce
@@ -0,0 +1,38 @@
+clc
+clear
+//Input data
+r=8;//The compression ratio 
+T1=350;//The given temperature at the start of compression in K
+p=1;//The given pressure at the start of compression in bar
+f=0.08;//The exhaust residual fraction 
+cv=44000;//The calorific value in kJ/kg
+
+//Calculations
+W1=150;//Isentropic compression functions for corresponding temp T1 in J/kg air K
+W2=W1-(292*log(1/r));//Isentropic compression function in J/kg air K 
+T2=682;//The temperature corresponding to isentropic compression function in K 
+V1=(292*T1)/(p*10^5);//The initial volume in m^3/kg air 
+p2=p*(T2/T1)*r;//The pressure at point 2 in atm
+V2=V1/r;//The volume at point 2 in m^3/kg air 
+us2=350;//The internal energy corresponding to temp T2 in K
+us1=40;//The internal energy corresponding to temp T1 in K 
+Wc=us2-us1;//Adiabatic compression work in kJ/kg air 
+ufu=-118.5-(2963*f);//The internal energy of formation in kJ/kg air 
+u3=us2+ufu;//The internal energy at point 3 in kJ/kg air 
+V3=V2;//The volume at point 3 in m^3/kg air 
+T3=2825;//The temperature at point 3 corresponding to u3,V3 on the burned gas chart in K
+p3=7100;//The pressure at point 3 in kN/m^2 
+s3=9.33;//Entropy at point 3 in kJ/kg air K 
+s4=s3;//Entropy is same in kJ/kg air K 
+V4=V1;//The volume at point 4 in m^3/kg air 
+u4=-1540;//The internal energy at point 4 corresponding to V4,s4 in kJ/kg air 
+p4=570;//The pressure at point 4 in kN/m^2 
+T4=1840;//The temperature at point 4 in K 
+We=u3-u4;//The expansion work in kJ/kg air 
+Wn=We-Wc;//The net work output in kJ/kg air 
+nth=[(Wn)/((1-f)*0.0662*cv)]*100;//The indicated thermal efficiency in percent 
+imep=((Wn*1000)/(V1-V2))/10^5;//The indicated mean effective pressure in bar 
+nv=[((1-f)*287*298)/(1.013*10^5*(1-0.125))]*100;//The volumetric efficiency in percent 
+
+//Output
+printf('(a)At point 2, \n The temperature is %3.0f K \n The pressure is %3.1f atm \n At point 3, \n The temperature is %3.0f K \n The pressure is %3.0f kN/m^2 \n At point 4, \n The temperature is %3.0f K \n The pressure is %3.0f kN/m^2 \n (b)The indicated thermal efficiency is %3.1f percent \n (c)The indicated mean effective pressure is %3.0f bar \n (d)The volumetric efficiency is %3.1f percent',T2,p2,T3,p3,T4,p4,nth,imep,nv)
diff --git a/2090/CH4/EX4.2/Chapter4_Example2.sce b/2090/CH4/EX4.2/Chapter4_Example2.sce
new file mode 100755
index 000000000..4da6abaf4
--- /dev/null
+++ b/2090/CH4/EX4.2/Chapter4_Example2.sce
@@ -0,0 +1,19 @@
+clc
+clear
+//Input data
+r=18;//The compression ratio 
+l=6;//The cut off taking place corresponding of the stroke in percent
+sc=2;//The specific heat at constant volume increases in percent
+cv=0.717;//The specific heat at constant volume in kJ/kgK
+R=0.287;//Gas constant in kJ/kgK
+
+//Calculations
+Vs=(r-1);//The ratio of swept volume and volume 2
+B=((l/100)*Vs)+1;//The cut off ratio 
+cp=cv+R;//The specific heat at constant pressure in kJ/kgK
+R1=cp/cv;//The ratio of specific heats 
+n=1-[[[[(1/r)^(R1-1)]*(B^R1-1)]/(R1*(B-1))]];//The efficiency of the diesel cycle 
+dn=[((1-n)/n)*[(R1-1)*((log(r))-(((B^R1)*log(B))/(B^R1-1))+(1/B))]*(sc/100)]*100;//The efficiency decrease in percent
+
+//Output
+printf('The efficiency decreases by %3.3f percent ',dn)
diff --git a/2090/CH4/EX4.3/Chapter4_Example3.sce b/2090/CH4/EX4.3/Chapter4_Example3.sce
new file mode 100755
index 000000000..6a0a69b74
--- /dev/null
+++ b/2090/CH4/EX4.3/Chapter4_Example3.sce
@@ -0,0 +1,25 @@
+clc
+clear
+//Input data
+r=8;//The compression ratio
+af=15;//Air/fuel ratio
+p1=1;//The pressure at the beginning of a compression stroke in bar
+t=60;//The temperature at the beginning of a compression stroke in degree centigrade
+cv=44000;//The calorific value of the fuel in kJ/kg
+n=1.32;//The index of the compression 
+Cv=0.717;//specific heat at constant volume in kJ/kgK
+
+//Calculations
+T1=t+273;//The temperature at the beginning of a compression stroke in K
+p2=p1*(r)^n;//The pressure at the end of a compression stroke in bar
+T2=T1*r^(n-1);//The temperature at the end of a compression stroke in K
+f=(1/(af+1));//The amount of fuel present in 1 kg of mixture in kg
+a=(af/(af+1));//The amount of air present in 1 kg of mixture in kg
+q23=cv/(af+1);//The heat transfer during process 2-3 per kg of mixture in kJ/kg
+T3=[[-10430+[(10430)^2+(4*494.8*10^5)]^(1/2)]/2];//The temperature at point 3 in K
+p3=(T3/T1)*(r)*p1;//The pressure at point 3 in bar
+T31=(q23/Cv)+T2;//The pressure at point 3 in K
+p31=(T31/T1)*r*p1;//The pressure at point 3 in bar
+
+//Output
+printf('(a) The Maximum temperature in the cylinder T3 = %3.0f K \n The Maximum pressure in the cylinder P3 = %3.0f bar \n (b)With constant value of Cv \n The Maximum temperature in the cylinder T3 = %3.0f K \n The Maximum pressure in the cylinder P3 = %3.1f bar ',T3,p3,T31,p31)
diff --git a/2090/CH4/EX4.4/Chapter4_Example4.sce b/2090/CH4/EX4.4/Chapter4_Example4.sce
new file mode 100755
index 000000000..9163d5ca9
--- /dev/null
+++ b/2090/CH4/EX4.4/Chapter4_Example4.sce
@@ -0,0 +1,23 @@
+
+clc
+clear
+//Input data
+r=21;//The compression ratio 
+af=29;//Air/fuel ratio
+T=1000;//The temperature at the end of compression in K
+cv=42000;//The calorific value of the in kJ/kg
+R=0.287;//Gas constant in kJ/kgK
+
+//Calculations 
+q23=cv/(af+1);//Heat transfer during the process 2-3 per kg of mixture in kJ
+T3=[-0.997+[((0.997)^2)+(4*2411*14*10^-6)]^(1/2)]/(28*10^-6);//The temperature during the process 2-3 in K
+function y=f1(x),y=(0.997+(28*10^-6)*x),endfunction
+I=intg(T,T3,list(f1));//Integrating the above function
+abs(I)
+V3=(T3/T);//The ratio of volumes at 2 and 3 points 
+Vs=(r-1);//Swept volume in terms of V2 
+V=V3-1;//The difference in the volume at 2 and 3 points
+pc=(V/Vs)*100;//The percentage stroke during which the combustion is completed in percent
+
+//Output
+printf('The percentage of stroke at which combustion is complete = %3.3f percent ',pc)
diff --git a/2090/CH4/EX4.5/Chapter4_Example5.sce b/2090/CH4/EX4.5/Chapter4_Example5.sce
new file mode 100755
index 000000000..d5f31ce97
--- /dev/null
+++ b/2090/CH4/EX4.5/Chapter4_Example5.sce
@@ -0,0 +1,58 @@
+
+clc
+clear
+//Input data
+r=16;//The compression ratio 
+l=6;//The cut-off of the stroke in percent
+p3=70;//The maximum pressure obtained in bar
+p1=1;//The pressure at the beginning of compression in bar
+T1=(100+273);//The temperature at the beginning of compression in K
+R=0.287;//Gas constant in kJ/kgK
+g=1.4;//Assume the isentropic index 
+
+//Calculations
+T2=T1*(r)^(g-1);//The temperature at point 2 in K
+function y=f1(x),y=(0.716+(125*10^-6)*x),endfunction
+I=intg(373,1131,list(f1));//Integrating the above function
+abs(I)
+Cv=(1/(1131-373))*I;//The specific heat at constant volume for the temp range T2 and T3 in kJ/kgK
+Cp=Cv+R;//The specific heat at constant pressure in kJ/kgK
+g1=Cp/Cv;//The ratio of specific heats 
+T21=T1*(r)^(g1-1);//The temperature at point 2 in K
+function y=f1(x),y=(0.716+(125*10^-6)*x),endfunction
+I1=intg(373,995,list(f1));//Integrating the above function
+abs(I1)
+Cv1=(1/(995-373))*I1;//The specific heat at constant volume for the temp range T2 and T3 in kJ/kgK
+Cp1=Cv1+R;//The specific heat at constant pressure in kJ/kgK
+g2=Cp1/Cv1;//The ratio of specific heats
+T22=T1*(r)^(g2-1);//The temperature at point 2 in K
+p2=(T22/T1)*r*p1;//The pressure at point 2 in bar
+T3=(p3/p2)*T22;//The temperature at point 3 in K
+V=[(l/100)*(r-1)]+1;//The ratio of volumes at 3-4 points
+T4=(V)*T3;//The temperature at point 4 in K
+p4=p3;//The pressure at point 4 in bar
+g3=1.3;//Assume isentropic index
+V5=r/V;//The ratio of volumes at 4-5 process 
+T5=T4*(1/V5)^(g3-1);//The temperature at point 5 in K
+Cv2=[[0.716*(T5-T4)]+[62.5*10^-6*(T5^2-T4^2)]]/(T5-T4);//The specific heat at constant volume for the temp range T5 and T4 in kJ/kgK
+Cp2=Cv2+R;//The specific heat at constant pressure in kJ/kgK
+g4=Cp2/Cv2;//The ratio of specific heats 
+T51=T4*(1/V5)^(g4-1);//The temperature at point 5 in K
+Cv3=[[0.716*(T51-T4)]+[62.5*10^-6*(T51^2-T4^2)]]/(T51-T4);//The specific heat at constant volume for the temp range T5 and T4 in kJ/kgK
+Cp3=Cv3+R;//The specific heat at constant pressure in kJ/kgK
+g5=Cp3/Cv3;//The isentropic index
+T52=T4*(1/V5)^(g5-1);//The temperature at point 5 in K
+p5=(T52/T1)*p1;//The pressure at point 5 in bar
+
+//Output
+printf('The pressure and temperature at all points of the cycle \n at point 2 , Temperature T2 = %3.0f K    and Pressure P2 = %3.2f bar \n at point 3 , Temperature T3 = %3.0f K    and Pressure P3 = %3.0f bar \n at point 4 , Temperature T4 = %3.0f K    and Pressure P4 = %3.0f bar \n at point 5 ,Temperature T5 = %3.0f K    and Pressure P5 = %3.2f bar ',T22,p2,T3,p3,T4,p4,T52,p5)
+
+
+
+
+
+
+
+
+
+
diff --git a/2090/CH4/EX4.6/Chapter4_Example6.sce b/2090/CH4/EX4.6/Chapter4_Example6.sce
new file mode 100755
index 000000000..7de8564e0
--- /dev/null
+++ b/2090/CH4/EX4.6/Chapter4_Example6.sce
@@ -0,0 +1,30 @@
+clc
+clear
+//Input data
+r=8;//Compression ratio
+lcv=44000;//The lower heating value of the fuel in kJ/kg
+af=15;//The air/fuel ratio
+Cv=0.71;//The specific heat at constant volume in kJ/kgK
+p=1;//The pressure at the beginning of the compression in bar
+t=60;//The temperature at the beginning of the compression in degree centigrade
+Mo=32;//Molecular weight of oxygen
+Mn=28.161;//Molecular weight of nitrogen
+Mh=18;//Molecular weight of water 
+n=1.3;//Polytropic index
+
+//Calculations
+T1=(t+273);//The temperature at the beginning of the compression in K
+sa=[12.5*[Mo+(3.76*Mn)]]/[(12*8)+(1*Mh)];//The stoichiometric air fuel ratio
+Y=af*[[(12*8)+(1*Mh)]/(Mo+(3.76*Mn))];//To balance the oxygen and nitrogen 
+x=(12.5-Y)*2;//By oxygen balance 
+nb=1+Y+(Y*3.76);//Number of moles before combustion 
+na=x+7.8+9+46.624;//Number of moles after combustion 
+Me=[(na-nb)/nb]*100;//The percentage molecular expansion in percent
+T2=T1*(r)^(n-1);//The temperature at point 2 in K
+T3=[lcv/(af+1)]*(1/Cv)+(T2);//The temperature at point 3 in K
+p3=r*(T3/T1)*p;//The pressure at point 3 in bar
+p31=p3*(na/nb);//The pressure at point 3 with molar expansion in bar
+
+//Output
+printf('The percentage molecular expansion is %3.0f percent \n (a) Without considering the molecular expansion \n The maximum temperature is %3.0f K \n The maximum pressure is %3.0f bar \n (b) With molecular expansion \n The maximum temperature is %3.0f K \n The maximum pressure is %3.1f bar ',Me,T3,p3,T3,p31)
+
diff --git a/2090/CH4/EX4.7/Chapter4_Example7.sce b/2090/CH4/EX4.7/Chapter4_Example7.sce
new file mode 100755
index 000000000..7f0cde4dd
--- /dev/null
+++ b/2090/CH4/EX4.7/Chapter4_Example7.sce
@@ -0,0 +1,16 @@
+clc
+clear
+//Input data
+f=0.03;//The residual fraction of an engine
+e=1.2;//The equivalence ratio
+
+//Calculations
+F=0.0795;//Fuel/air ratio for corresponding equivalence ratio 
+T=1+F;//Total mass in kg
+fa=1-f;//Fresh air in kg
+ff=F*(fa);//Fresh fuel in kg
+ra=f;//Air in residual in kg
+rf=ra*F;//Fuel in residual in kg
+
+//Output
+printf('Fresh air = %3.2f kg \n Fresh fuel = %3.6f kg \n Air in residual = %3.2f kg \n Fuel in residual = %3.6f kg ',fa,ff,ra,rf)
diff --git a/2090/CH4/EX4.8/Chapter4_Example8.sce b/2090/CH4/EX4.8/Chapter4_Example8.sce
new file mode 100755
index 000000000..0dea1651f
--- /dev/null
+++ b/2090/CH4/EX4.8/Chapter4_Example8.sce
@@ -0,0 +1,22 @@
+clc
+clear
+//Input data
+T=800;//The given temperature in K
+e=1;//The equivalence ratio 
+
+//Calculations
+hi=154.723;//Sensible Enthalpy for isooctane at 800 K in MJ/kmol 
+ho=15.841;//Sensible Enthalpy for oxygen at 800 K in MJ/kmol 
+hn=15.046;//Sensible Enthalpy for nitrogen at 800 K in MJ/kmol
+nc=0.00058;//Number of kmoles of C8H18 for equivalence ratio for 1 kg of air 
+no=0.00725;//Number of kmoles of oxygen for equivalence ratio for 1 kg of air 
+nn=0.0273;//Number of kmoles of nitrogen for equivalence ratio for 1 kg of air 
+Hs=(nc*hi)+(no*ho)+(nn*hn);//Total sensible enthalpy of reactants in MJ per kg of air
+Hs1=Hs*1000;//Total sensible enthalpy of reactants in kJ per kg of air 
+R=8.314;//Gas constant in kJ/kgK
+n=nc+no+nn;//Total number of kmoles for 1 kg of air
+Us=Hs-(n*R*10^-3*(T-298));//sensible internal energy of reactants in MJ per kg of air 
+Us1=Us*1000;//sensible internal energy of reactants in kJ per kg of air
+
+//Output
+printf('Total sensible enthalpy of reactants = %3.1f kJ/kg air \n Sensible internal energy of reactants = %3.1f kJ/kg air ',Hs1,Us1)
diff --git a/2090/CH4/EX4.9/Chapter4_Example9.sce b/2090/CH4/EX4.9/Chapter4_Example9.sce
new file mode 100755
index 000000000..8de0a00c5
--- /dev/null
+++ b/2090/CH4/EX4.9/Chapter4_Example9.sce
@@ -0,0 +1,16 @@
+clc
+clear
+//Input data
+T=500;//The given temperature in K
+e=1;//Equivalence ratio 
+
+//Calculations 
+Ai=0.0662;//The amount of isooctane for 1 kg of air in kg
+Ta=298;//Consider the ambient temperature in K 
+E=[[0.0662*[(0.44*log(T/Ta))+(3.67*10^-3*(T-Ta))]]+[(0.921*log(T/Ta))+(2.31*10^-4*(T-Ta))]]*1000;//Isentropic compression function in J/kg air 
+R=8.314;//Gas constant in kJ/kgK
+Ri=R/114;//Gas constant for isooctane in kJ/kgK
+W=[0.5874-(0.662*Ri*log(T/Ta))-(0.287*log(T/Ta))]*1000;//Gas constant for isooctane in kJ/kgK
+
+//Output
+printf('The isentropic compression functions at 500 K for the unburned, \n isooctsne-air mixture are %3.1f J/kg air and %3.1f J/kg air ',E,W)