clc clear //Input data t=20//Trial time in minutes NL=680//Net brake load in N mep=3//Mean effective pressure in bar N=360//Speed in rpm Fc=1.56//Fuel consumption in kg cw=160//Cooling water in kg Tw=32//Temperature of water at inlet in degree C Wo=57//Water outlet temperature in degree C a=30//Air in kg Ta=27//Room temperature in degree C Te=310//Exhaust gas temperature in degree C d=210//Bore in mm l=290//Stroke in mm bd=1//Brake diameter in m cv=44//Calorific value in MJ/kg st=1.3//Steam formed in kg per kg fuel in the exhaust cp=2.093//Specific heat of steam in exhaust in kJ/kg.K cpx=1.01//Specific heat of dry exhaust gases in kJ/kg.K cpw=4.187//Specific heat of water in kJ/kg.K //Calculations ip=(mep*100*(l/1000)*(3.14/4)*(d/1000)^2*N)/60//Indicated Power in kW bp=((2*3.14*N*(NL*(1/2)))/60)/1000//Brake power in kW nm=(bp/ip)*100//Mechanical efficiency in percent qs=(Fc*cv*10^3)//Heat supplied in kJ qip=(ip*t*60)//Heat equivalent of ip in kJ qcw=(cw*cpw*(Wo-Tw))//Heat carried away by cooling water in kJ tm=(Fc*a)//Toatl mass of exhaust gas in kg ms=(st*Fc)//Mass of steam formed in kg mde=(tm-ms)//Mass of dry exhaust gas in kg Ed=(mde*cpx*(Te-Ta))//Energy carried away by dry exhaust gases in kJ Es=(ms*((cpw*(100-Ta))+2257.9+(cp*(Te-100))))//Energy carried away by steam in kJ TE=(Ed+Es)//Total energy carried away by exhaust gases in kJ ue=(qs-(qip+qcw+TE))//Unaccounted energy in kJ pqip=(qip/qs)*100//Percentage pqcw=(qcw/qs)*100//Percentage pTE=(TE/qs)*100//Percentage pue=(ue/qs)*100//Percentage //Output printf('Indicated power is %3.2f kW \n Brake power is %3.3f kW \n\n ENERGY BALANCE SHEET \n (in kJ) (in percent)\n 1. Energy equivalent in ip %3.0f %3.2f \n 2. Energy carried away by cooling water %3.0f %3.2f \n 3. Energy carried away by exhaust gases %3.0f %3.2f \n 4. Unaccounted for energy loss %3.0f %3.2f \n Total %3.0f %3.2f',ip,bp,qip,pqip,qcw,pqcw,TE,pTE,ue,pue,qs,(pqip+pqcw+pTE+pue))