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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+//variable initialisation
+Vm=400 //input of motor in volt
+F=50 //supply frequency in hrtz
+P=4 //Number Of Poles
+R1=1 //resistance of stator in ohm
+R2=0.4 //resistance of rotor in ohm
+X1=1 //reactance of Motor in ohm
+X2=1 //reactance of Motor in ohm
+Xm=50 //reactance of Motor in ohm
+Vc=231 //Constant Voltage Source In Volt
+I1=28//Current from Constant Current Source in Amp
+//Solution
+Xe=(X1*Xm)/(X2+Xm)
+Sm=R2/(Xe+X2)
+Sm1=R2/(X2+Xm)
+Ve=(Vc*Xm)/(X1+Xm)
+Ws=(4*%pi*F)/(P)
+Test=(3/Ws)*(((Ve)^2)/(R2^2+(X2+Xe)^2))*R2
+Tem=(3/Ws)*(((Ve)^2)/(2*(X2+Xe)))
+Test1=(3/Ws)*(((I1*Xm)^2)/(R2^2+(X2+Xm)^2))*R2
+Tem1=(3/Ws)*(((I1*Xm)^2)/(2*(X2+Xm)))
+Im=I1*((R2/Sm1)+(%i*X2))/((R2/Sm1)+%i*(X2+Xm))
+V1=sqrt(3)*abs(Im)*Xm
+printf('\n\n The Slip for maximum torque for Voltage source=%0.1f\n\n',Sm)
+printf('\n\n The Slip for maximum torque for current source=%0.1f\n\n',Sm1)
+printf('\n\n The Starting Torque for Voltage source=%0.1f N-m\n\n',Test)
+printf('\n\n The Maximum Torque for Voltage Source=%0.1f N-m\n\n',Tem)
+printf('\n\n The Starting Torque for Current Source=%0.1f N-m\n\n',Test1)
+printf('\n\n The Maximum Torque for Current Source=%0.1f N-m\n\n',Tem1)
+printf('\n\n The Supply voltage required=%0.1f Volt\n\n',V1)
+//The answers vary due to round off error