EE_387_200809SP_homework_08_solution

EE_387_200809SP_homework_08_solution - sa 1. r = r t isa =...

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1 Homework 08 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 2c o s ( ) 2s i n ( ) o s ( ( ) ) i n ( ( ) ) sa s e is sb s e is ra r e r ir rb r e r ir iI t iIt t t ωϕ ω =+ ′′ =− + + t r r θ = 0 0 . ) 0 ( = = r r 3 (16.67 10 ) (2 58 rad/s)(1/ 60 s)(180/ / rad) 348 r θπ π ×= ° 3 (33.33 10 ) (2 58 rad/s)(2/ 60 s)(180/ / rad) 696 336 r ° 1. sa a s sb b s ra rb rb ra r Φ K s Φ K sa a s sb b s ra rb ra r Φ K s Φ K sa a s sb b s ra rb ra r Φ K s Φ K rb rb
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2 Homework 08 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 pos( (0)) 29.2 si s ϕ Φ= =− ° G 3 pos( (16.67 10 )) (2 60)(1/ 60) 2 29.2 s is is is π πϕ Φ× = + =+ = ° G 3 pos( (33.33 10 )) (2 60)(2/ 60) 4 29.2 s is is is = + = ° G Stator windings are “standard” and currents are positive-sequence currents Î stator flux vector is in the direction of the instantaneous phase angle: 1. (cont.) pos( ( )) re i s tt ω + G Rotor windings are “standard” and currents are positive-sequence currents Î rotor flux vector is in the direction of the instantaneous phase angle: 33 pos( (0)) pos( (16.67 10 )) pos( (16.67 10 )) 174.7 rr r i r −− Φ × = Φ × = = ° GG G pos( ( )) (( ) ) r i r r ei r t t ωϕω ωϕ Φ=−+ + G
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3 Homework 08 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 2. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 v sa 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 sb 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 i 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 ra 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -200 0 200 rb t
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4 Homework 08 Solution © Jeffrey Mayer 2005-2009. All rights reserved.
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This note was uploaded on 10/08/2009 for the course EE 387 taught by Professor Jefferymayer during the Spring '08 term at Pennsylvania State University, University Park.

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EE_387_200809SP_homework_08_solution - sa 1. r = r t isa =...

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