ece643_e1_au09_ans

# ece643_e1_au09_ans - rotor field Magnetic axis of a/2 for n...

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1. Note that because of the coil location, the two g2 gaps are in parallel. 10 * 4 A 250 N 10 * * 4 * * * * 2 4 core 7 0 0 2 0 2 2 0 1 1 2 1 2 A N L A g A g N L core core g core g g g ece 643 exam1 answers au/09 2.a. ideal magnetic structure = b. X q < X d for salient pole machine c. X q = X d for cylindrical rotor machine d.1. B rpm about the same as B r ferro 2 10 * 2 . 0 10 * 4 * 10 * * 4 * 250 10 * 2 . 0 2 2 1 4 7 2 2 2 2 1 g L g g g _ 2. |H c_pm |> |H cferro | 3. E af = V a + I a *jX q V a = 22*10 3 / 3 I a = 12,500 @ +cos 1 (0.9) E af = 12,272 @ +47.17 0 angle of I q = angle of E af ’= 47.17 0 Magnetic axis of 4. from Faraday’s law, e(t) = N [d (t)/dt], thus when (t) at a max, e(t) = 0 and when (t) at 0 then (t) is either a positive or negative maximum. [a] Therefore, e aa ’(t) = 0 when 0 = n for n = 0, 1, 2, …. [b] And e aa ’(t) is a maximum (either pos or neg) at n /2 for n = 1 3 5 [c] Positive magnetic axis of + + a b’ c’
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Unformatted text preview: rotor field Magnetic axis of a /2 for n 1, 3, 5, … [c] Positive magnetic axis of stator phases perpendicular to plane of coil – their orientation is NOT affected by the rotor for a cylindrical rotor machine. I aa’ = 0 at the instant of time indicated, whereas i cc’ = ‐ i bb’ . bb’ is directed at +120 and since I cc’ is negative, cc’ at + 60 , thus the resultant at +90 . + a’ b c + Plane of coil of rotor winding 5. b X a X c x b x a L(x) I x L W x x i W f fld fld fld fld 2 2 2 ' ' * * * 2 * 1 * * * ) ( * 2 1 )] , ( [ I I f ) ( 2 ) , (...
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## This note was uploaded on 12/07/2011 for the course ECE 600 taught by Professor Clymer,b during the Fall '08 term at Ohio State.

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