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diff --git a/2090/CH2/EX2.1/Chapter2_example1.sce b/2090/CH2/EX2.1/Chapter2_example1.sce
new file mode 100755
index 000000000..5dd63f29c
--- /dev/null
+++ b/2090/CH2/EX2.1/Chapter2_example1.sce
@@ -0,0 +1,38 @@
+clc

+clear

+//Input data

+d=20;//Cylinder bore diameter in cm

+L=25;//Stroke length in cm

+Vc=1570;//The clearance volume in cm^3

+P1=1;//Pressure at the beginning of the compression in bar

+T1=300;//Temperature at the beginning of the compression in K

+T3=1673;//The maximum temperature of the cycle in K

+pi=3.141;//Mathematical constant value of pi

+Cv=0.718;//specific heat at constant volume for air in kJ/kgK

+R=0.287;//Real gas constant in kJ/kgK

+g=1.4;//Isentropic index

+c=500;//Number of cycles per minute

+

+//Calculations

+Vs=(pi/4)*d^2*L;//Swept volume in cm^3

+V1=Vs+Vc;//According to pv diagram Total volume i.e sum of swept and clearance volume in cm^3

+V2=Vc;//Volume according to pv diagram in cm^3

+r=V1/V2;//Compression ratio

+T2=T1*r^(g-1);//In isentropic process, Temperature at point 2 in degree centigrade

+P2=P1*r^g;//In isentropic process, Pressure at point 2 in bar

+P3=P2*(T3/T2);//In constant volume, process Pressure at point 3 in bar

+T4=T3*(1/r)^(g-1);//In isentropic process, Temperature at point 4 in degree centigrade

+P4=P3*(1/r)^(g);//In isentropic process, Pressure at point 4 in bar

+no=(1-(1/r)^(g-1))*100;//Air standard efficiency of otto cycle

+Q1=Cv*(T3-T2);//Heat supplied in kJ/kg

+Q2=Cv*(T4-T1);//Heat rejected in kJ/kg

+W=Q1-Q2;//Work done per unit mass in kJ/kg

+m=[(P1*10^5*V1*10^-6)/(R*T1)]/1000;//The amount of mass in kg

+W1=W*m;//Work done in kJ

+pm=[(W1*10^3)/(Vs*10^-6)]/10^5;//Mean effective pressure in N/m^2

+P=W1*(c/60);//Power developed in kW

+

+//Output

+printf('Temperature at point 2 = %3.1f K \n Pressure at point 2 = %3.3f bar \n Pressure at point 3 = %3.2f bar \n Temperature at point 4 = %3.0f K \n Pressure at point 4 = %3.3f bar \n Air standard efficiency of otto cycle = %3.4f percent \n Work done = %3.2f kJ \n Mean effective pressure = %3.3f bar \n Power developed = %3.1f kW ',T2,P2,P3,T4,P4,no,W1,pm,P)

+

+

diff --git a/2090/CH2/EX2.10/Chapter2_example10.sce b/2090/CH2/EX2.10/Chapter2_example10.sce
new file mode 100755
index 000000000..71ee14072
--- /dev/null
+++ b/2090/CH2/EX2.10/Chapter2_example10.sce
@@ -0,0 +1,26 @@
+clc

+clear

+//Input data

+p1=1//Inlet pressure in bar

+T1=27+273//Temperature in K

+p2=4//pressure at point 2 in bar

+p3=16//Maximum pressure in bar

+Cv=0.573//specific heat at constant volume for gas in kJ/kgK

+Cp=0.761//specific heat at constant pressure for gas in kJ/kgK

+

+//Calculations

+g=(Cp/Cv)//Adiabatic index

+T2=(T1*(p2/p1)^((g-1)/g))//Temperature in K

+T3=(p3/p2)*T2//Temperature in K

+T4=T3*(p1/p3)^((g-1)/g)//Temperature in K

+Q1=Cv*(T3-T2)//Heat supplied in kJ/kg

+Q2=Cp*(T4-T1)//Heat rejected in kJ/kg

+W=Q1-Q2//Workdone in kJ/kg

+n=(W/Q1)*100//Efficiency of cycle 

+r=(p2/p1)^(1/g)//Compression ratio

+R=(Cp-Cv)//Universal gas constant in kJ/kg.K

+Vs=(R*1000*T1*(r-1))/(p1*10^5*r)//Swept volume in m^3/kg

+pm=(W/(Vs*100))//Mean effective pressure in bar

+

+//Output

+printf('(a) The work done per kg of gas is %3.1f kJ/kg \n (b) The efficiency of the cycle is %3.1f percent \n (c) Mean effective pressure is %3.2f bar',W,n,pm)

diff --git a/2090/CH2/EX2.2/Chapter2_example2.sce b/2090/CH2/EX2.2/Chapter2_example2.sce
new file mode 100755
index 000000000..254d00813
--- /dev/null
+++ b/2090/CH2/EX2.2/Chapter2_example2.sce
@@ -0,0 +1,22 @@
+clc

+clear

+//Input data

+CV=42000;//The calorific value of the fuel in kJ/kg

+pa=5//Percentage of compression

+Pa=1.2;//Pressure in the cylinder at 5% compression stroke

+pb=75//Percentage of compression

+Pb=4.8;//Pressure in the cylinder at 75% compression stroke

+g=1.3;//polytropic index

+g1=1.4//Isentropic index

+n=0.6;//Air standard efficiency

+

+

+//Calculations

+V=(Pb/Pa)^(1/1.3);//Ratio of volumes

+r=(V*(pb/100)-(pa/100))/((1-(pa/100))-(V*(1-(pb/100))))//Compression ratio

+n1=((1-(1/r)^(g1-1)))*100//Relative efficiency

+nthj=n*(n1/100)//Indicated thermal efficiency

+x=(1/(CV*nthj))*3600//Specific fuel consumption in kg/kW.h

+

+//Output

+printf('The compression ratio of the engine is %3.1f \n The specific fuel consumption is %3.3f kg/kW.h',r,x)

diff --git a/2090/CH2/EX2.4/Chapter2_example4.sce b/2090/CH2/EX2.4/Chapter2_example4.sce
new file mode 100755
index 000000000..ff82428bb
--- /dev/null
+++ b/2090/CH2/EX2.4/Chapter2_example4.sce
@@ -0,0 +1,38 @@
+clc

+clear

+//Input data

+d=0.2;//The diameter of the cylinder bore in m

+L=0.3;//The length of the stroke in m

+P1=1;//The pressure at the beginning of the compression in bar

+T1=300;//The temperature at the beginning of the compression in K

+r=16;//Compression ratio

+V=0.08;//Cutt off takes place at 8& of the stroke

+pi=3.141;//Mathematical constant value of pi

+R=0.287;//Real gas constant in kJ/kgK

+g=1.4;//Isentropic index

+Cp=1.005;//Specific heat at constant prassure in kJ/kgK

+Cv=0.718;//specific heat at constant volume for air in kJ/kgK

+

+//Calculations

+Vs=(pi/4)*d^2*L;//Swept volume in m^3

+Vc=Vs/(r-1);//Clearance volume in m^3

+V2=Vc;//Volume at point 2 in m^3

+V1=Vs+Vc;//Volume at point 1 in m^3

+m=(P1*10^5*V1)/(R*T1);//The amount of mass in kg

+P2=P1*(r^g);//Pressure at point 2 in bar

+P3=P2;//Pressure at point 3 in bar

+T2=T1*r^(g-1);//Temperature at point 2 in K

+V3=(V*Vs)+V2;//Volume at point 3 in m^3

+C=V3/V2;//Cut off ratio

+T3=C*T2;//Temperature at point 3 in K

+P4=P3*(C/r)^g;//Pressure at the point 4 in bar

+T4=T3*(C/r)^(g-1);//Temperature at point 4 in K

+V4=V1;//Volume at point 4 in m^3

+Q1=[m*Cp*(T3-T2)]/1000;//Heat supplied in kJ

+Q2=[m*Cv*(T4-T1)]/1000;//Heat rejected in kJ

+W=[Q1-Q2];//Work done per cycle in kJ

+na=(W/Q1)*100;//Air standard efficiency

+Pm=[W*1000/Vs]/10^5;//Mean effective pressure in bar

+

+//Output

+printf('(a) Volume at point 2 = %3.6f m^3 \n Volume at point 1 = %3.6f m^3 \n Pressure at point 2 = %3.1f bar \n Temperature at point 2 = %3.1f K \n Temperature at point 3 = %3.0f K \n Pressure at point 4 = %3.3f bar \n Temperature at point 4 = %3.1f K \n Volume at point 4 = %3.6f m^3 \n (b) cut off ratio = %3.2f \n (c) Work done per cycle = %3.3f kJ \n (d) air standard efficiency = %3.2f percent \n (e)Mean effective pressure = %3.2f bar ',V2,V1,P2,T2,T3,P4,T4,V4,C,W,na,Pm)

diff --git a/2090/CH2/EX2.5/Chapter2_example5.sce b/2090/CH2/EX2.5/Chapter2_example5.sce
new file mode 100755
index 000000000..483d22229
--- /dev/null
+++ b/2090/CH2/EX2.5/Chapter2_example5.sce
@@ -0,0 +1,23 @@
+clc

+clear

+//Input data

+Pm=7;//The mean effective pressure of a diesel cycle in bar

+r=12;//The compression ratio

+P1=1;//Initial pressure in bar

+g=1.4;//Isentropic index

+

+//Calculations

+function[f] = F(x);//function definition

+    f = 45.4*x- 12*x^1.4 -64.2;

+endfunction

+x = 1;//Initial guss

+function[z] = D(x)//Derivative

+    z= 3*x^2 - 3;

+endfunction

+y = fsolve(x,F, D);

+C=y;//The cut off ratio 

+na=[1-(1/(r^(g-1)))*[((C^g)-1)/(g*(C-1))]]*100;//Air standard efficiency

+

+//Output 

+printf('The cut off ratio = %3.1f \n  The air standard efficiency = %3.2f percent ',C,na)

+

diff --git a/2090/CH2/EX2.6/Chapter2_example6.sce b/2090/CH2/EX2.6/Chapter2_example6.sce
new file mode 100755
index 000000000..987b51114
--- /dev/null
+++ b/2090/CH2/EX2.6/Chapter2_example6.sce
@@ -0,0 +1,18 @@
+clc

+clear

+//Input data

+m=30;//The air fuel ratio by mass

+T1=300;//The temperature of air at the beginning of the compression in K

+r=16;//The compression ratio

+CV=42000;//The calorific value of the fuel in kJ/kg

+g=1.4;//Isentropic index

+Cp=1.005;//Specific heat at constant prassure in kJ/kgK

+

+//Calculations

+T2=T1*(r^(g-1));//Temperature at point 2 in K

+T3=[(1/m)*(CV/Cp)]+T2;//Temperature at point 3 in K

+C=T3/T2;//The cut off ratio

+n=(1-[(1/r^(g-1))*[((C^g)-1)/(g*(C-1))]])*100;//The ideal efficiency of the engine based on the air standard cycle

+

+//Output

+printf(' The ideal efficiency of the engine based on the air standard cycle = %3.1f percent ',n)

diff --git a/2090/CH2/EX2.7/Chapter2_example7.sce b/2090/CH2/EX2.7/Chapter2_example7.sce
new file mode 100755
index 000000000..b3c9dde24
--- /dev/null
+++ b/2090/CH2/EX2.7/Chapter2_example7.sce
@@ -0,0 +1,16 @@
+clc

+clear

+//Input data

+p1=1//Inlet pressure in bar

+p2=32.425//Pressure at the end of isentropic compression in bar

+r=6//Ratio of expansion

+r1=1.4//Isentropic index

+

+//Calculations

+rc=(p2/p1)^(1/r1)//Compression ratio

+b=(rc/r)//cut-off ratio

+n=(1-((b^r1-1)/(rc^(r1-1)*r1*(b-1))))*100//Air-standard efficiency

+pm=((p1*rc^r1*n/100*r1*(b-1))/((r1-1)*(rc-1)))//Mean effective pressure in bar

+

+//Output

+printf('Air-standard efficiency is %3.2f percent \n Mean effective pressure is %3.3f bar',n,pm)

diff --git a/2090/CH2/EX2.8/Chapter2_example8.sce b/2090/CH2/EX2.8/Chapter2_example8.sce
new file mode 100755
index 000000000..1fe96216a
--- /dev/null
+++ b/2090/CH2/EX2.8/Chapter2_example8.sce
@@ -0,0 +1,30 @@
+clc

+clear

+//Input data

+rc=15//Compression ratio

+p1=1//Pressure at which compression begins in bar

+T1=27+273//Temperature in K

+pm=60//Maximum pressure in bar

+h=2//Heat transfered to air at constant volume is twice that at constant pressure

+g=1.4;//Isentropic index

+Cv=0.718;//specific heat at constant volume for air in kJ/kgK

+Cp=1.005;//specific heat at constant pressure for air in kJ/kgK

+R=0.287;//Real gas constant in kJ/kgK

+

+//Calculations

+T2=(T1*rc^(g-1))//Temperature in K

+p2=(p1*rc^g)//Pressure in bar

+T3=(T2*(pm/p2))//Temperature in K

+T4=(Cv*(T3-T2))/(2*Cp)+T3//Temperature in K

+b=(T4/T3)//Cut-off ratio

+T5=(T4*(b/rc)^(g-1))//Temperature in K

+p5=(p1*(T5/T1))//Pressure in bar

+Q1=(Cv*(T3-T2))+(Cp*(T4-T3))//Heat supplied per unit mass in kJ/kg

+Q2=Cv*(T5-T1)//Heat rejected per unit mass in kJ/kg

+W=(Q1-Q2)//Workdone in kJ/kg

+n=(W/Q1)*100//Air standard efficiency

+Vs=((1*R*1000*T1)/(p1*10^5))*(1-1/rc)//Swept volume in m^3/kg

+pmean=((W*1000)/Vs)/10^5//Mean-effective pressure in bar

+

+//Output

+printf('(a) The pressures and temperatures at the cardinal points of the cycle are \n T2 = %3.0f K  p2 = %3.1f bar \n T3 = %3.0f K  p3 = %3.0f bar \n T4 = %3.0f K  p4 = %3.0f bar \n T5 = %3.0f K  p5 = %3.2f bar \n (b) The cycle efficiency is %3.0f percent \n (c) The mean effective pressure of the cycle is %3.2f bar',T2,p2,T3,pm,T4,pm,T5,p5,n,pmean)

diff --git a/2090/CH2/EX2.9/Chapter2_example9.sce b/2090/CH2/EX2.9/Chapter2_example9.sce
new file mode 100755
index 000000000..744058e53
--- /dev/null
+++ b/2090/CH2/EX2.9/Chapter2_example9.sce
@@ -0,0 +1,35 @@
+clc

+clear

+//Input data

+r=12;//Compression ratio

+B=1.615;//Cut off ratio

+p3=52.17;//Maximum pressure in bar

+p4=p3;//Maximum pressure in bar

+p1=1;//Initial pressure in bar

+T1=(62+273);//Initial temperature in K

+n=1.35;//Indices of compression and expansion

+g=1.4;//Adiabatic exponent

+mR=0.287;//Real gas constant in kJ/kgK

+Cv=0.718;//specific heat at constant volume for air in kJ/kgK

+Cp=1.005;//specific heat at constant pressure for air in kJ/kgK

+

+//Calculations

+T2=T1*r^(n-1);//The temperature at point 2 in K

+p2=p1*(r)^n;//The pressure at point 2 in bar

+T3=T2*(p3/p2);//The temperature at point 3 in K

+T4=T3*B;//The temperature at point 4 in K

+T5=T4*(B/r)^(n-1);//The temperature at point 5 in K

+Q12=[(g-n)/(g-1)]*mR*[(T1-T2)/(n-1)];//Heat transfer during the process 1-2 for unit mass in kJ/kg

+Q23=Cv*(T3-T2);//Heat transfer during the process 2-3 for unit mass in kJ/kg

+Q34=Cp*(T4-T3);//Heat transfer during the process 3-4 for unit mass in kJ/kg

+Q45=((g-n)/(g-1))*mR*((T4-T5)/(n-1));//Heat transfer during the process 4-5 for unit mass in kJ/kg

+Q51=Cv*(T1-T5);//Heat transfer during the process 5-1 for unit mass in kJ/kg

+Q1=Q23+Q34+Q45;//Heat supplied in kJ/kg

+Q2=-Q12+(-Q51);//Heat rejected in kJ/kg

+W=Q1-Q2;//Work done in kJ/kg

+E=[W/Q1]*100;//Efficiency in percentage

+Vs=[(mR*T1)/p1]*(r-1)/r;//Swept volume for unit mass in m^3/kg

+pm=[W*1000/Vs]/10^3;//Mean effective pressure in bar

+

+//Output

+printf(' (a)The temperature at cardinal points ,\n     T2 = %3.0f K\n     T3 = %3.0f K \n     T4 = %3.0f K \n     T5 = %3.0f K \n (b) The cycle efficiency = %3.1f percent \n (c) The mean effective pressure of the cycle = %3.3f bar ',T2,T3,T4,T5,E,pm)