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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+pd=16;//delivery pressure in bar
+n=1.25;//index of compression
+m=10;//mass flow rate
+R=0.287;//gas constant
+t1=288;//temperature in K
+cp=1.005;//specific pressure
+k1=0.04;//clearance retio
+k2=0.06;//clearance ratio
+N=400;//speed in rpm
+g=1.4;//constant
+
+
+//CALCULATIONS
+p2=sqrt(p1*pd);//intermediate pressure in bar
+ipm=2*(n/(n-1))*(m/60)*R*t1*((p2/p1)^((n-1)/n)-1);//power required in kW
+pi=(m*R*t1/60)*log(pd/p1);//isothermal power
+niso=(pi/ipm)*100;//isothermal efficiency in percentage
+va=m*R*t1/(p1*10^2);//free air delivered in m^3/min
+t2=t1*(p2/p1)^((n-1)/n);//temperature in K
+Qr=(m/60)*cp*(t2-t1);//heat rejected in intercooler in kW
+nvl=1+k1-k1*((p2/p1)^(1/(n*2)));//volumetric efficiency in percentage
+vsl=va/(N*nvl);//swept volume in m^3
+nv2=(1+k2-(k2*((pd/p1)^(1/(n*2)))))*100;//volumetric efficiency in percentage
+vsh=va/(2*((pd/p1)*N*nv2)^(1/2));//swept volume
+Ql=(g-n)*m*R*(t2-t1)/((g-1)*(n-1));//heat transferred in LP
+Qh=(g-n)*m*R*(t2-t1)/((g-1)*(n-1));//heat transferred in HP
+t6=t1*(pd/p1)^((n-1)/n);//temperature in K
+Qi=(m/60)*cp*(t2-t1);//Heat trnsferred in intercooler
+
+//OUTPUT
+printf('(i)The power required is %3.3f kW \n (ii)The isothermal efficiency is %3.3f percentage \n (iii)The free air delivered is %3.4f m^3/min \n (iv)The heat rejected in intercooler is %3.3f kW \n (v)swept volume is %3.5f m^3 \n swept volume is %3.5f m^3 \n (vi)net heat transferred in intercooler is %3.3f kW',ipm,niso,va,Qr,vsl,vsh,Qi)
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