class: center, middle # EE-362 ELECTROMECHANICAL ENERGY CONVERSION-II # Recitation Hour ## Ozan Keysan [ozan.keysan.me](http://ozan.keysan.me) Office: C-113
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Tel: 210 7586 --- # Questions From: ## Yıldırım Üçtug ## Solved Problems, Feb 2004 --- # CH1 - Q2: ### Consider a 3-phase cylindrical rotor type AC machine. -- ## Rotor: 2-pole, single-layer, full-pitched winding in 12 slots. Each coil side has 10 turns. -- ## Stator: 2-pole, double-layer 10/9 full-pitched winding in 18 slots. Each coil side has 15 turns. -- ## a) Draw the rotor and stator winding diagram. --- ## The air-gap radius is 0.1 m, air-gap distance is 0.002 m, axial length 0.3 m ### Rotor is excited with: #### \\(I\_a(t)= 10 cos (100 \pi t)\\) ####\\(I\_b(t)= 10 cos (100 \pi t - 2 \pi /3) \\) ####\\(I\_c(t)= 10 cos (100 \pi t - 4 \pi /3)\\) ## b) Draw the MMF distribution produced by the rotor at t=0.15 s. --- ## The air-gap radius is 0.1 m, air-gap distance is 0.002 m, axial length 0.3 m ### Rotor is excited with: #### \\(I\_a(t)= 10 cos (100 \pi t)\\) ####\\(I\_b(t)= 10 cos (100 \pi t - 2 \pi /3) \\) ####\\(I\_c(t)= 10 cos (100 \pi t - 4 \pi /3)\\) ## c) Calculate the generated first harmonic open-circuit rms stator phase voltage when the rotor speed is: ### i) Zero, ii) 3000 rpm, iii) -3000 rpm --- # CH2 - Q4: -- ### The following observations are made on a 3-phase, 50-Hz, Y-connected, 380 V rms line-to-line induction motor with negligible stator winding resistance. -- ### i) At no-load the motor is rotating at 745 rpm. -- ## a) What is the number of poles of the machine? --- # CH2 - Q4: ### The following observations are made on a 3-phase, 50-Hz, Y-connected, 380 V rms line-to-line induction motor with negligible stator winding resistance. ### i) At no-load the motor is rotating at 745 rpm. ### ii) The stator can not be started with a heavy load, but it only starts when the load torque is reduced to 236 Nm ### iii) When the motor is running at rated speed, the load is gradually increased and it is found that the rotor speed reduced down to 600 rpm, but after it decelerates and stops. --- # CH2 - Q4: ### The following observations are made on a 3-phase, 50-Hz, Y-connected, 380 V rms line-to-line induction motor with negligible stator winding resistance. ### i) At no-load the motor is rotating at 745 rpm. ### ii) Starting Torque = 236 Nm ### iii) Maximum torque at 600 rpm. -- ## b-c) Calculate the referred rotor resistance and total leakage reactance (stator+rotor)? --- # CH2 - Q4: ### The following observations are made on a 3-phase, 50-Hz, Y-connected, 380 V rms line-to-line induction motor with negligible stator winding resistance. ### i) At no-load the motor is rotating at 745 rpm. ### ii) Starting Torque = 236 Nm ### iii) Maximum torque at 600 rpm. ### d) If the rotational power is 1518 W, find the air-gap power, rotor copper loss, internal mechanical power and the net output power when the rotor is rotating at 712.5 rpm --- # CH2 - Q4: ### The following observations are made on a 3-phase, 50-Hz, Y-connected, 380 V rms line-to-line induction motor with negligible stator winding resistance. ### i) At no-load the motor is rotating at 745 rpm. ### ii) Starting Torque = 236 Nm ### iii) Maximum torque at 600 rpm. ### e) Calculate the efficiency at 712.5 rpm --- # Solutions ## Don't try to write it up everyting! ## You can [download the solutions](https://www.dropbox.com/s/5z57zzdlzeba0z9/362_recitation.pdf?dl=1) --- # ODTUClass ## A 10 hp 3-phase, 60 Hz, 6-pole squirrel-cage induction motor is used to drive a load at a slip of 3%. The friction and windage torque is measured to be 6.1 Nm at this slip. The torque requirement of the load varies as the cube of its speed and is 67.4 Nm at 1200 rpm. -- ## Compute the electromagnetical torque (Te), the power transferred across the air-gap (Pg), adn the rotor copper loss (Pcu2). --- ## You can download this presentation from: [keysan.me/ee362](http://keysan.me/ee362) --- # Example ### A 480 V, 50 hp induction motor is drawing 60 A at 0.85 pf lagging. -- The stator copper losses are 2kW and the rotor copper losses are 700W. -- The friction and windage losses are 600 W, the core losses are 1800 W and the stray losses are negligible, find: -- - ### The air gap power. -- - ### The converted power. -- - ### The output power. -- - ### The efficiency of the motor -- - ### Torque of the motor -- - ### Rotational speed of the rotor ---