The induction machine Solution

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EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 1 Class 6 - solution The Induction Machine Exercise 1 1. The velocity of the rotating magnetic field in [ rot/min ] is /g√840 /g8868/g340460/g√858 /g√868 2 /g3404 1000 /g4670/g√870/g√867/g√878//g√865/g√86√/g√866/g467√ The fractional slip in normal operating conditions is /g√87√ /g3404/g√858 /g8870 /g√858/g3404 0.036 Therefore, the velocity of the rotor in [rot/min ] is /g√840 /g3404 /g√840 /g8868/g46661/g3398/g√87√/g4667/g3404 964 /g4670/g√870/g√867/g√878//g√865/g√86√/g√866/g467√ while in [rad/s ] Ω /g3040/g34042/g8084/g√840 60 /g3404 100.95 /g4670/g√870/g√853/g√856//g√87√/g467√ Thus, the mechanical power developed by the rotor at the shaft is /g√848 /g3040/g3404 Ω/g3040/g√889/g3040/g3404 20.493 /g4670/g√863/g√849/g467√ 2. Neglecting the mechanical losses in the machine, the power balance in the rotor windings is: /g√848 /g3034/g3088/g3043 /g3404 /g√848/g8870/g30√√/g3397/g√848/g3040 or, considering the physical meaning of the slip: /g√848 /g8870/g30√√ /g√87√/g3404 /g√848/g8870/g30√√/g3397/g√848/g3040 from where we obtain the Joule power losses in the rotor phases /g√848 /g8870/g30√√/g3404/g√87√ 1/g3398/g√87√/g√848/g3040/g3404 0.7653 /g4670/g√863/g√849/g467√ EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 2 3. Neglecting the voltage drops in the stator phases and considering the phasor of the supply voltage V as phase origin, the emf induced in the stator is: /g√83√ /g8869/g3364/g3364/g3364/g3404/g√848 √3 /g3404 219.3931 /g4670/g√848/g467√ Hence, the emf induced in the rotor at unitary slip is: /g√83√ /g8870/g8869 /g3364/g3364/g3364/g3364/g3364/g3404/g√83√/g8869 /g√863/g3404 182.828 /g4670/g√848/g467√ Or, at slip s: /g√83√/g8870/g3404…