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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+pmi=6;//Mean effective pressure in bar
+L=0.45;//Stroke in m
+d=0.3;//Rope diameter in m
+N=12000;//Total revolutions made
+nc=1;//number of cylinders
+n=2;//for four cylinders
+D=1.8;//Brake drum diameter in m
+x=0.02136;//difference of W and S
+cpw=4.18;//specific pressure of water
+cpe=1.005;//specific pressure of air
+cv=45000;//calorific value
+two=60;//outlet water temperature
+twi=15;//inlet water temperature
+te=300;//exhaust gas temperature in Degree C
+ta=20;//room temperature in Degree C
+mf=7.6;//mass flow rate in kg/h
+mw=550;//water flo rate in kg/h
+me=367.6;//total flow rate in kg/h
+
+
+//CALCULATIONS
+IP=(pmi*102*L*(3.14*(d^2)/4)*N*nc)/(60*60*n);//Indicated power in kW
+BP=((x)*3.14*(D+d)*N)/60;//Brake power in kW
+nit=(IP/(mf*cv/3600))*100;//Indicetad thermal efficiency in percentage
+nm=(BP/IP)*100;//mechanical efficiency in percentage
+Qs=mf*cv/60;//heat supplied in kJ/min
+a11=(BP/Qs)*100;//% of heat equivalent to BP
+Qw=(mw*cpw*(two-twi))/60;//Heat loss to cooling water in kW
+b11=(Qw/Qs)*100;//% of heat lost to cooling water
+Qe=(me*cpe*(te-ta))/60;//Heat loss to exhaust gases in kW
+c11=(Qe/Qs)*100;//% of heat lost to exhaust gases
+Qu=(Qs-(BP*60+Qw+Qe));//Enthalpy of unaccount in kW
+d11=(Qu/Qs)*100;//unaccounted heat in percentage
+
+//OUTPUT
+printf('(i)Indicated power is %3.2f kW \n brake power is %3.2f kW \n (ii)Indicated thermal efficiency is %3.2f percentage \n (iii)Mechanical efficiency is %3.2f percentage \n (iv)HEAT BALANCE SHEET \n (I)Heat supplied %3.i kJ/min \n (II)Heat utilised in the system is %3.2f kW',IP,BP,nit,nm,Qs,Qu)