A circuit of 500 V for single-phase asynchronous motor has synchronous impedance of 3.2 Ω. The armature resistance is 0.3 Ω. Determine the induced emf if the machine develops 31.3 kW and the mechanical losses equals 4 kW
E=(Vcosϕ+IR)2+(Vsinϕ+IX)2E=(5003+IR)2+(IX)2E= \sqrt{(V \cos \phi+IR)^2+(V \sin \phi+IX)^2}\\ E= \sqrt{(\frac{500}{\sqrt3}+IR)^2+(IX)^2}\\E=(Vcosϕ+IR)2+(Vsinϕ+IX)2E=(3500+IR)2+(IX)2
Input power
4kW+31.3kW35.3kW4 kW +31.3 kW\\ 35.3kW4kW+31.3kW35.3kW
35.3kW=3VIcosϕI=35.3∗1033∗500∗40.76A ⟹ Ephase=327.94V35.3kW= \sqrt3 VI \cos \phi\\ I= \frac{35.3*10^3}{\sqrt3 *500}*40.76A\\ \implies E_{phase}= 327.94 V35.3kW=3VIcosϕI=3∗50035.3∗103∗40.76A⟹Ephase=327.94V
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