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+clc
+clear
+//INPUT DATA
+a=450;//Area of indicator diagram mm^2
+S=9.806;//Spring number
+l=50*1.2;//Length of diagram
+d=0.15;//bore in m
+L=0.25;//stroke in m
+N=400;//engine speed in rpm
+nc=1;//number of cylinders
+n=2;//for single cylinder
+mf=3;//mass flow rate in kg/h
+cv=44200;//calorific value
+dTc=42;//rise of temperature for cooling water in Degree C
+cpw=4.18;//specific pressure
+mw=4;//mass of water
+T=225;//Brake torque in Nm
+
+//CALCULATIONS
+pmi=a*S/l;//mean effective pressure in N/cm^2
+IP=((pmi/10)*L*(3.14*(d^2)/4)*N*nc)/(60*n);//Indicated power in kW
+BP=(2*3.14*N*T)/60000;//brake power in kW
+nm=(BP/IP);//Meahanical efficiency in percentage
+nbt=(BP*3600/(mf*cv))*100;//Brake thermal efficiency in percentage
+bsfc=mf/BP;//Brake specific fuel consumption in kg/kWh
+Qs=mf*cv/3600;//Heat supplied in kW
+a11=(BP/Qs)*100;//% of heat equivalent to BP
+Qw=(mw*cpw*(dTc))/60;//Heat lost to cooling water in kW
+b11=(Qw/Qs)*100;//% of heat lost to cooling water
+Qe=(Qs-(BP+Qw));//heat utilised in the system
+c11=(Qe/Qs)*100;//% of heat lost to exhaust gases and radiation
+
+//OUTPUT
+printf('(i)Mechanical efficiency is %3.2f percentage \n(ii)Brake thermal efficiency is %3.2f percentage \n (iii)Brake specific fuel consumption is %3.3f kg/kW.hr \n(iv)\n(I)heat supplied is %3.3f kW \n (II)Heat utilised in the system %3.2f \n percentage',nm,nbt,bsfc,Qs,Qe)