Week10tue - ECE 410/510 Electric Vehicles Part I Electric Drives and Controllers Week 10 Tuesday March 8 1 Maseeh College of Engineering and

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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
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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
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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 e = p 2 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 = rl N s p ˆ B ms , since = p 2 syn so ˆ E m = 2 rlN s syn ˆ B ms For a PMAC drive ˆ E m = ˆ E f ˆ B ms = ˆ B r syn = m ˆ E f = 2 rlN s ˆ B r m since k E = ˆ E f m = 0.5 V / rad ( mech ) / s ˆ E f = k E 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 m (0) = 90 p / 2 = 45 , recall i s ( t ) = p 2 m ( t ) ± π 2 , i s ( t ) = p 2 m t = e t = 1047.2 × t rad i a ( t ) = ˆ I a cos( i s ( t )) = 6.67cos( is ( t )) I a = 6.67 0 A V a = E fa + j 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
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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
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This note was uploaded on 12/08/2011 for the course ECE 410 taught by Professor Maseeh during the Spring '11 term at Portland State.

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Week10tue - ECE 410/510 Electric Vehicles Part I Electric Drives and Controllers Week 10 Tuesday March 8 1 Maseeh College of Engineering and

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