Tham khảo tài liệu '(mcgraw-hill) (instructors manual) electric machinery fundamentals 4th edition episode 1 part 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | J J V __120Z0 V_ L A R1 jX1 Rf jXF Q j Q Q Q IL IA A b The stator copper losses are Pscl 3Ia2 R1 3 A 2 Q 1205 W c The air gap power is Pag 3I22 2 3IA Rf s R Note that 3Ia2Rf is equal to 3I22 since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit. R2 Pag 3I22 3Ia2Rf 3 A 2 Q kW d The power converted from electrical to mechanical form is Pconv 1 - s Pag 1 - kW kW e The induced torque in the motor is Pag kW ind m 2n rad sync 3600 r min Y - . N m 1 min 60 s f The output power of this motor is Pout Pconv - Pmech - core - Pmlsc kW - 250 W - 180 W - 0 W kW The output speed is nm 1 - s nsync 1 - 3600 r min 3420 r min Therefore the load torque is p . T OUT Tload mm 3420 r min kW 4 . N m 2 n rad 1 min 1 r 60 s g The overall efficiency is p __ p n T x 100 out P 1 IN 3V Ia cos kW x 100 n x100 3 120 V A h The motor speed in revolutions per minute is 3420 r min. The motor speed in radians per second is 2 n rad 1 min m 3420 r min 358 rad s m 1 r 60 s 7-8. For the motor in Problem 7-7 what is the slip at the pullout torque What is the pullout torque of this motor 175 7. z TH V TH I Xi j M Solution The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply and then using that with the rotor circuit model. jXM Ri jX1 _ j15 Q Q Q R1 j X1 XM Q j Q 15 Q Mm 1 Ml 1 ƠI m Q Q jX M V _ j15 Q R1 j X1 XM ộ Q j Q 15 Q 120Z0 V V The slip at pullout torque is V max R2 V max XTH X2 2 Q__ 7 Q 2 Q Q 2 The pullout torque of the motor is T .
