Lecture12-2

Lecture12-2 - Lecture 12 (revised October 2009) The...

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Lecture 12 (revised October 2009) The revision is minor. The correction is in the –ve sign of j (in yellow) The manipulations follow three steps: Multiply the second and fourth row by – j. Thereafter in equation (6) the voltages are e 1 +je 2 , e 3 +je 4 , i 1 +ji 2 and i 3 +ji 4 as below: 12 34 () ( ) SS mR m R ej e R j M j M ij i e j nM Rj n M i ωω ++ + + ⎡⎤ = ⎢⎥ +− + + + ⎣⎦ A A (6) Just before equation (15), Earlier the following definitions have been made: S I i = + ± , R I i = + ± . End of Note on Revision Derivation of Equivalent Circuit from Application of γ - δ Frame From (20) in Lecture 10, the stator current vector in the d-q frame is + + = ) sin( ) cos( S S S Sq Sd t t I i i ϕ ω ( 1 ) In (8) of Lecture 11, the vector is transformed to the γ - δ frame as = S S S S S I i i sin cos ( 2 ) Equations (1) and (2) are related by the transformation matrix as: = S S Sq Sd i i t t t t i i cos sin sin cos ( 3 ) One can define a complex number S I i =+ ± and express
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] ~ Re[ t j S Sd S Sa e I i i i ω α = = = ( 4 ) In (4), one is reminded that the a -axis, the -axis and the d -axis of the stator remain in the same position. The γ - axis, however, rotates at the speed of the stator frequency . Electrical engineers are familiar with ac voltages and currents expressed as complex numbers, such as S I ~ , associated with a synchronously rotating unit phasor t j e . This lecture shows how the equivalent circuit of the induction machine is easily derived from (5) of Lecture 11.
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Lecture12-2 - Lecture 12 (revised October 2009) The...

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