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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+//variable initialization
+Pout=30 //output of induction motor in kilowatt
+Vm=400 //input of motor in volt
+F0=50 //supply frequency in hrtz
+F1=40 // frequency in hrtz
+P=4 //number of poles
+Rs=0.33 //resistance of stator in ohm
+Rr=0.22 //resistance of rotor in ohm
+Xs=0.9 //reactance of stator in ohm
+Xr=0.9 //reactance of rotor in ohm
+
+//Solution
+Vs=Vm/sqrt(3)
+Sm=Rr/(sqrt((Rs^2)+((Xs+Xr)^2)))
+Ir=Vs/sqrt(((Rs+(Rr/Sm))^2)+((Xs+Xr)^2))
+cos_P=cosd(atand((Xs+Xr)/(Rs+(Rr/Sm))))
+Pi=sqrt(3)*Vm*Ir*cos_P
+P0=3*(Ir^2)*Rr*((1/Sm)-1)
+n=(P0/Pi)*100
+
+K=F1/F0//for frequency of 40 Hz
+Xs2=K*Xs
+Xr2=K*Xr
+Sm2=Rr/(sqrt((Rs^2)+((Xs2+Xr2)^2)))
+Vs2=K*Vs
+Ir2=Vs2/sqrt(((Rs+(Rr/Sm2))^2)+((Xs2+Xr2)^2))
+cos_p2=cosd(atand((Xs2+Xr2)/(Rs+(Rr/Sm2))))
+Pi2=3*Vs2*Ir2*cos_p2
+P02=3*(Ir2^2)*Rr*((1/Sm2)-1)
+n2=(P02/Pi2)*100
+printf('\n\n The Efficiency at breakdown torque with 50Hz=%0.1f\n\n',n)
+printf('\n\n The Efficiency at breakdown torque with 40Hz=%0.1f\n\n',n2)
+//The answers vary due to round off error