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Unformatted text preview: ) ( (2) Eq.(2) can be split into two as follows: L ave e T T dt d J = (3) even e m T dt d J = (4) The steady-state operating speed is constant so that = dt d and (3) becomes: L ave e T T = (5) (v) What is the maximum load torque max L T which the reluctance motor can drive? 2 (vi) Using your sketch of ) ( f T ave e = of (ii), illustrate how changes as L T is increased from L T =0 to max 25 . L L T T = , max 5 . L L T T = , max 75 . L L T T = and max L L T T = . (vii) By integrating (4), solve for m . Show that the magnitudes of the perturbation speed components are small when the moment of inertia J and the frequency are large. (You can neglect the constant of integration.)...
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This note was uploaded on 02/23/2010 for the course ECSE 462 taught by Professor Bon-tekooi during the Spring '09 term at McGill.
- Spring '09