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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)