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+clc
+clear
+//Input data
+n=6;//Number of cylinders
+Do=0.03;//Orifice diameter in m
+Cd=0.6;//Coefficient of discharge 
+H=0.14;//Pressure drop across the orifice
+d=0.1;//The diameter of the bore in m
+L=0.11;//The length of the stroke in m
+W=540;//Brake load in N
+N=2500;//Engine speed in rpm
+ch=83/17;//C/H ratio by mass
+p=1;//Ambient pressure in bar
+t=18;//Time taken for fuel consumption in s
+f=100;//The amount of fuel consumption in cc
+T=300;//Ambient air temperature in K
+df=780;//The density of the fuel in kg/m^3
+R=287;//Real gas constant in J/kgK
+g=9.81;//Gravitational force constant in m/s^2
+pi=3.141;//Mathematical constant
+dhg=13600;//Density of Hg in kg/m^3
+
+//Calculations
+da=(p*10^5)/(R*T);//The density of air in kg/m^3
+Va=(Cd*(pi/4)*Do^2*[2*g*H*(dhg/da)]^(1/2));//Volume flow rate of air in m^3/s
+Vs=(pi/4)*d^2*L*(N/(2*60))*n;//Swept volume per second in m^3/s
+nv=(Va/Vs)*100;//Volumetric efficiency in percent
+bp=(W*N)/(20000);//The brake power in kW
+bmep=[(bp*1000)/(L*(pi/4)*d^2*(N/(2*60))*n)]/10^5;//The brake mean effective pressure in bar
+T=(60*bp*1000)/(2*pi*N);//Torque in Nm
+mf=(f/18)*(780/1000)*(1/1000)*3600;//Mass flow rate of fuel in kg/h
+bsfc=mf/bp;//The brake specific fuel consumption in kg/kWh
+so=(0.83*(32/12))+(0.17*(8/1));//Stoichiometric oxygen required per kg of fuel in kg/kg fuel
+sa=so/bsfc;//Stoichiometric air required in kg/kg fuel
+maa=Va*da;//Actual mass flow rate of air in kg/s
+af=(maa*3600)/mf;//Actual air fuel ratio 
+pea=[(af-sa)/sa]*100;//Percentage of excess air in percent
+
+//Output
+printf('The volumetric efficiency = %3.1f percent \n The brake mean effective pressure = %3.2f bar \n The brake power = %3.1f kW \n The Torque = %3.1f Nm \n The brake specific fuel consumption = %3.3f kg/kWh \n The percentage of excess air = %3.1f percent ',nv,bmep,bp,T,bsfc,pea)
+
+
+
+