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# week10tue - ECE 410/510 Electric Vehicles Part I Electric...

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Maseeh College of Engineering and Computer Science 3/8/11 1 ECE 410/510 - Electric Vehicles Part I – Electric Drives and Controllers Week 10 – Tuesday March 8

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Electrical and Computer Engineering ECE 410 / 510 – Electric Vehicles I Hammerstrom/Butts Week 10 – March 8, 10 Tue: Finish batteries Thu: Review for final exam Lab 7 due Week 11 – FINAL EXAM, Tuesday March 15, 7:30-9:20, Same Room 3/8/11 2
Maseeh College of Engineering and Computer Science 3/8/11 3 Lab 5 Problem Solutions

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Electrical and Computer Engineering ECE 410 / 510 – Electric Vehicles I Hammerstrom/Butts Problem 9-6 In a 2-pole, three-phase machine with N s =100, calculate and at a time t if at that time the stator currents are as follows (a) i a =10A, i b =-5A, i c =-5A; (b) i a =-5A, i b =10A, i c =-5A; (c) i a =-5A, i b =-5A, i c =10A Just do (a): 3/8/11 4 i s F s F s ( t ) = ( N s / 2) i s ( t ) i s ( t ) = i a ( t ) + i b ( t ) 120 + i c ( t ) 240 i a = 10 A , i b = i c = 5 A i s = 10 5 120 5 240 = 15 0 F s = 750 0 A turns
Electrical and Computer Engineering ECE 410 / 510 – Electric Vehicles I Hammerstrom/Butts Problem 10-6 A PMAC machine has 4 poles and three phases k T and k E are both 0.5 (MKS units) The stator inductance is 15 mH (neglect winding resistance) The motor is supplying a torque of 5 Nm at a speed of 5000 rpm in a balanced sinusoidal steady state Draw a phasor diagram showing and along with their calculated values 3/8/11 5 V a I a p = 4, ω m = 523.6 rad / s ω e = p 2 ω m = 1047.2 rad ( elect ) / s k T = T em ˆ I s = π 2 rlN s ˆ B r = 0.5 N m / A It's independent of p so ˆ I s = T em k T ˆ E m = π rl N s p ω ˆ B ms , since ω = p 2 ω syn so ˆ E m = π 2 rlN s ω syn ˆ B ms For a PMAC drive ˆ E m = ˆ E f ˆ B ms = ˆ B r ω syn = ω m ˆ E f = π 2 rlN s ˆ B r ω m since k E = ˆ E f ω m = 0.5 V / rad ( mech ) / s ˆ E f = k E ω m

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Electrical and Computer Engineering ECE 410 / 510 – Electric Vehicles I Hammerstrom/Butts 3/8/11 6 ˆ E f = k E ω m = 0.5 × 523.6 = 261.8 V ˆ I S = T em k T = 5 k T = 10 A , ˆ I a = 2 3 ˆ I S = 6.67 A If we assume θ i s t = 0 = 0 , then θ m (0) = 90 p / 2 = 45 , recall θ i s ( t ) = p 2 θ m ( t ) ± π 2 , θ i s ( t ) = p 2 ω m t = ω e t = 1047.2 × t rad i a ( t ) = ˆ I a cos( θ i s ( t )) = 6.67cos( θ is ( t )) I a = 6.67 0 A V a = E fa + j ω e L s I a = 261.8 + j 1047.2 × (15 × 10 3 ) × 6.67 0 V a = 282.0 21.83 V Few did the analysis of why k T and k E gave the right answers, but I didn’t take points off for that
Electrical and Computer Engineering ECE 410 / 510 – Electric Vehicles I Hammerstrom/Butts Lab 5 Comments Several people used Matlab to plot results for both problems, I would have preferred phasor notation for problem 9-6, but the problem description didn’t require it, so no points were subtracted, in phasor notation you notice immediately that all three parts a), b), and c) are each 120 degrees apart Efficiency calculations were all over the map, the easiest is to use the ratio of the power in (VI) to power out (V2/R) Next year, we’ll make sure to put the boards at a station with a better oscilloscope!

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