Question #189583

A 6 pole generator has a lap wound armature with 40 slots with 20 conducters per slot. The Flux per pole is 25mWb. The terminal voltage is 288v and the voltage regulation of the generator is 4%. Determine the speed of the generator


Expert's answer

Eg=PϕNZ60AE_g=\frac{P\phi NZ}{60A}


Eg=V+0.04V=288+0.04⋅288=299.52 (V)E_g=V+0.04V=288+0.04\cdot288=299.52\ (V)


N=60AEgPϕZ=60⋅6⋅299.526⋅0.025⋅(40⋅20)=682 (rpm)N=\frac{60AE_g}{P\phi Z}=\frac{60\cdot6\cdot 299.52}{6\cdot0.025\cdot (40\cdot20)}=682\ (rpm) . Answer

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