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diff --git a/2090/CH10/EX10.1/Chapter10_example1.sce b/2090/CH10/EX10.1/Chapter10_example1.sce
new file mode 100755
index 000000000..891ce3361
--- /dev/null
+++ b/2090/CH10/EX10.1/Chapter10_example1.sce
@@ -0,0 +1,31 @@
+clc

+clear

+//Input data

+bsfc=0.3;//The brake specific fuel consumption in kg/kWh

+bp=250;//The brake power in kW

+N=1500;//Number of cycles per min in rpm

+CA=15;//Crank angle in degrees

+pi1=30;//The pressure of air in the cylinder at the beginning of the injection in bar

+pi2=60;//The pressure of air in the cylinder at the end of the injection in bar

+pf1=220;//The fuel injection pressure at the beginning in bar

+pf2=550;//The fuel injection pressure at the end in bar

+Cd=0.65;//The coefficient of discharge for the injector 

+df=850;//The density of the fuel in kg/m^3

+p1=1.013;//The atmospheric pressure in bar

+n=4;//The number of orifices used in the nozzle

+x=6;//Number of cylinders

+pi=3.141;//The mathematical constant of pi

+

+//Calculations

+mf=bsfc*bp/60;//The mass flow rate of fuel in kg/min

+F=(mf/(N/2))*(1/x);//Fuel injected per cycle per cylinder in kg

+s=(CA/360)/(N/60);//Duration of injection in s

+mf1=F/s;//Mass of fuel injected per second

+p1=pf1-pi1;//Pressure difference at the beginning in bar

+p2=pf2-pi2;//Pressure difference at the end in bar

+pa=(p1+p2)/2;//Average pressure difference in bar

+Af=[mf1/(Cd*(2*df*pa*10^5)^(1/2))]*10^6;//Area of cross section of the nozzle in mm^2

+do=[(Af/n)*(4/pi)]^(1/2);//The diameter of the orifice in mm

+

+//Output

+printf('The nozzle area required per injection = %3.3f mm^2 \n The diameter of the orifice = %3.2f mm ',Af,do)

diff --git a/2090/CH10/EX10.2/Chapter10_example2.sce b/2090/CH10/EX10.2/Chapter10_example2.sce
new file mode 100755
index 000000000..9cbb644c6
--- /dev/null
+++ b/2090/CH10/EX10.2/Chapter10_example2.sce
@@ -0,0 +1,25 @@
+clc

+clear

+//Input data

+bp=30;//The brake power of the engine in kW

+N=3000;//The engine speed in rpm 

+bsfc=0.28;//The brake specific fuel consumption in kg/kWh 

+API=35;//The API

+p2=160;//The pressure at which fuel is injected in bar

+CA=28;//The crank angle in degrees

+p1=35;//The pressure in the combustion chamber in bar

+Cv=0.92;//The coefficient of velocity 

+pi=3.141;//The mathematical constant of pi

+

+//Calculations

+S=141.5/(131.5+API);//Specific gravity

+df=S*1000;//The density of the fuel in kg/m^3

+D=(CA/360)/(N/60);//Duration of injection in s

+F=(bsfc*bp)/((N/2)*60);//Fuel consumption per cycle in kg

+mf=F/D;//Mass flow rate of fuel in kg/s

+Cf=Cv*((2*(p2-p1)*10^5)/df)^(1/2);//Velocity of injection of the fuel in m/s

+Af=[mf/(df*Cf)]*10^6;//Area of the fuel orifice in mm^2

+d=(4*Af/pi)^(1/2);//The diameter of the orifice in mm

+

+//Output

+printf('The velocity of injection of the fuel = %3.1f m/s \n The diameter of the fuel orifice = %3.3f mm ',Cf,d)

diff --git a/2090/CH10/EX10.3/Chapter10_example3.sce b/2090/CH10/EX10.3/Chapter10_example3.sce
new file mode 100755
index 000000000..c513826a0
--- /dev/null
+++ b/2090/CH10/EX10.3/Chapter10_example3.sce
@@ -0,0 +1,18 @@
+clc

+clear

+//Input data

+d=0.8*10^-3;//The diameter of an orifice in m

+A=1.65*10^-6;//The cross sectional area in m^2

+Cd=0.9;//The discharge coefficient of the orifice 

+Cp=0.85;//The coefficient of the passage

+p1=170;//The injection pressure in bar

+p2=25;//The compression pressure of the discharge in bar

+df=850;//The density of the fuel in kg/m^3

+

+//Calculations

+Q=[(145/(22.931*10^9))^(1/2)]*10^6;//Adding two equations and solving then the discharge in cm^3/s

+p=170-(2.161*10^9*(Q/10^6)^2);//Pressure immediately formed before the orifice in bar

+Cf=Cd*((2*(p-p2)*10^5)/df)^(1/2);//The velocity of fuel flow through the orifice in m/s

+

+//Output

+printf('The discharge of fuel through the injector = %3.1f cm^2/s \n The jet velocity through the orifice = %3.1f m/s ',Q,Cf)

diff --git a/2090/CH10/EX10.4/Chapter10_example4.sce b/2090/CH10/EX10.4/Chapter10_example4.sce
new file mode 100755
index 000000000..8b93d9cdc
--- /dev/null
+++ b/2090/CH10/EX10.4/Chapter10_example4.sce
@@ -0,0 +1,20 @@
+clc

+clear

+//Input data

+s=20;//Spray penetration in cm

+t1=15.7;//The spray penetration of 20 cm in ms

+pi1=150;//The injection pressure in bar

+pi2=450;//The injection pressure to be used in bar

+p2=15;//The combustion chamber pressure in bar

+d1=0.34;//The diameter of the orifice in mm

+s1=20;//The penetration for an orifice in cm

+d2=0.17;//If the diameter of the orifice in cm

+t11=12;//The spray penetration in ms

+

+//Calculations

+t2=(t1*(pi1-p2)^(1/2))/(pi2-p2)^(1/2);//The time required for the spray to penetrate in ms

+s2=d2*(s1/d1);//The penetration of the orifice in cm

+t21=t11*(d2/d1);//The time required for the spray to penetrate in ms

+

+//Output

+printf('(a) The time required for the spray to penetrate = %3.2f ms \n (b) The spray penetration of the orifice = %3.0f cm \n The time required for the spray to penetrate = %3.0f ms ',t2,s2,t21)

diff --git a/2090/CH10/EX10.5/Chapter10_example5.sce b/2090/CH10/EX10.5/Chapter10_example5.sce
new file mode 100755
index 000000000..bb758028b
--- /dev/null
+++ b/2090/CH10/EX10.5/Chapter10_example5.sce
@@ -0,0 +1,22 @@
+clc

+clear

+//Input data

+v=6.5;//The volume of fuel in the barrel in cc

+d=0.3;//The dimeter of fuel pipe line in cm

+l=65;//The length of the fuel pipe line in cm 

+vi=2.5;//The volume of fuel in the injection valve in cc

+K=78.5*10^-6;//The coefficient of compressibility of the oil per bar

+p1=1;//The atmospheric pressure in bar

+p2=180;//The pressure due to pump in bar

+v3=0.1;//The pump displacement necessary for the fuel in cc

+e=0.75;//The effective stroke of the plunger in cm

+pi=3.141;//Mathematical constant of pi

+

+//Calculations

+V1=v+((pi*d^2)/4)*l+vi;//The total initial volume in cc

+V=K*V1*(p2-p1);//Change in volume due to compression in cc

+T=(V)+v3;//Total displacement of the plunger in cc

+L=T*(4/pi)*(1/(e^2));//Effective stroke of the plunger in cm

+

+//Output

+printf('(a) The total displacement of the plunger = %3.3f cc \n (b) The effective stroke of the plunger = %3.3f cm',T,L)

diff --git a/2090/CH10/EX10.6/Chapter10_example6.sce b/2090/CH10/EX10.6/Chapter10_example6.sce
new file mode 100755
index 000000000..5c575db5f
--- /dev/null
+++ b/2090/CH10/EX10.6/Chapter10_example6.sce
@@ -0,0 +1,29 @@
+clc

+clear

+//Input data

+n=4;//Number of cylinders 

+N=2500;//The engine speed in rpm 

+P=90;//The power produced by the engine in kW

+bsfc=0.28;//The brake specific fuel consumption in kg/kWh

+v=3.5;//The volume of fuel in the barrel in cc

+vp=2.5;//Volume of fuel in the pipe line in cc

+vi=2;//The fuel inside the injector in cc

+p1=280;//The average injection pressure in bar

+p2=30;//The compression pressure of air during injection in bar

+df=850;//The density of the fuel in kg/m^3

+K=80*10^-6;//The coefficient of compressibility of fuel per bar

+pi=1;//The pressure with which fuel enter into the barrel in bar

+

+//Calculations

+F=(bsfc*P)/((N/2)*60);//Fuel consumption per cycle in kg

+F1=F/n;//Fuel consumption per cylinder in kg/cycle

+Vf=[F1/df]*10^6;//Volume of fuel injected per cylinder per cycle in cm^3

+V1=v+vp+vi;//Total initial volume in cc

+V=K*V1*(p1-pi);//Change in volume due to compression in cc

+Vp=Vf+V;//Volume displaced by plunger in cc

+W=[(1/2)*(p1-pi)*10^5*V*10^-6]+[(p1-p2)*10^5*Vf*10^-6];//Pump work per cycle in J

+P1=(W*N)/(2*60*1000);//Power lost per cylinder in kW

+P2=P1*4;//Total power lost for pumping the fuel in kW

+

+//Output 

+printf('The displacement volume of one plunger per cycle = %3.4f cc \n Total power lost for pumping the fuel = %3.3f kW',Vp,P2)

diff --git a/2090/CH10/EX10.7/Chapter10_example7.sce b/2090/CH10/EX10.7/Chapter10_example7.sce
new file mode 100755
index 000000000..6dfe74ce0
--- /dev/null
+++ b/2090/CH10/EX10.7/Chapter10_example7.sce
@@ -0,0 +1,29 @@
+clc

+clear

+//Input data

+v1=0.3;//Velocity of the pump plunger in m/s

+l=0.575;//The length of the fuel pipe in m

+A=1/20;//The cross sectional area of pipe to the plunger cylinder

+a=1/40;//The area of nozzle hole to the pipe 

+p1=27.6;//Initial pressure in the line in bar 

+p2=27.6;//The compression pressure of the engine

+K=17830*10^5;//The bulk modulus of fuel in N/m^2

+df=860;//The density of the fuel in kg/m^3

+pi=3.141;//Mathematical constant of pi

+

+//Calculations

+Vs=(K/df)^(1/2);//The velocity of pressure disturbances in m/s

+t=l/Vs;//Time taken by the disturbance to travel through pipe line in s

+Vp=(1/A)*v1;//The velocity of the fuel at the inlet of the pipe line in m/s

+p=[(K/Vs)*Vp]/10^5;//The change in pressure in bar

+pi=p+p1;//The pressure according to change in velocity in bar

+po=p1+p;//The change in total to the disturbance pressure in bar 

+vc=Vp-(a*((2*(po-p2))/df)^(1/2));//Change in the velocity in m/s

+pr=26.8;//By trail method the first reflected pressure wave from velocity in bar

+Vc=pr*(Vs/(K/10^5));//The change in velocity in m/s

+po1=p1+p+pr;//The pressure at the orifice end of the pipe in bar

+vo=a*((2*(po1-p2)*10^5)/df)^(1/2);//The velocity at the oriface end of the pipe in m/s

+

+//Output

+printf('(a)The velocity of the pressure disturbance = %3.0f m/s \n (b) The time taken by the disturbance to travel through the pipe line = %3.4f s \n (c) The velocity at the pump end of the pipe line as the plunger moves = %3.0f m/s \n The pressure at the pump end of the pipe line as the plunger moves =  %3.1f bar \n (d) The magnitude of the first reflected pressure = %3.2f bar \n The magnitude of the first reflected velocity wave = %3.2f m/s \n (e)The pressure at the oriface end of the pipe line after the first reflection = %3.1f bar \n The velocity at the oriface end of the pipe line after the first reflection = %3.2f m/s ',Vs,t,Vp,pi,pr,Vc,po1,vo)

+