assignment3

assignment3 - ) ( (2) Eq.(2) can be split into two as...

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1 ASSIGNMENT 3 Trigonometric Identities cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-cosAsinB Question 1 The inductance of a single-phase reluctance machine is: ] 2 cos [ d B A a L L L θ + = The single-phase winding is excited by a current t I i a ω cos = Let the rotor turn at synchronous speed so that δ + = t d (i) What is the electromechanical torque e T ? (ii) What is the time invariant component ) ( f T ave e = ? Sketch ) ( f T ave e = for o 180 . . . 180 + ° . (iii) What are the even harmonic pulsating components of torque, even e T ? (iv) The reluctance motor has a moment of inertia J and drives a load L T . The mechanical equation of motion is: L e m T T dt d J = (1) Let the rotor speed be the sum of a term at synchronous speed and perturbation terms m Δ due to the even harmonic pulsating torque components. Substituting in (1) L even e ave e m T T T dt d J + = Δ +
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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.

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assignment3 - ) ( (2) Eq.(2) can be split into two as...

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