Lecture4 - Lecture 4 RELUCTANCE MACHINES The inductances in...

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Lecture 4 RELUCTANCE MACHINES The inductances in the 2x2 inductance matrix are: )] ( 2 cos [ x d B A xx L L L θ + = )] ( 2 cos [ y d B A yy L L L + = )] 2 cos( ) (cos( [ y x d B y x A yx xy L L L L θθ + = = Single-Phase Reluctance Machine Consider the x-winding as the a-phase with x =0. Then, ] 2 cos [ d B A a L L L + = . If the a- phase current is i a , the magnetic co-energy is: W m 1 =0.5L a i a 2 . The torque is 2 5 . 0 a d a e i L T = which is (1) 2 ] 2 sin 2 [ 5 . 0 a d B e i L T = How does one make this machine produce a non-zero time invariant torque, so that as a motor it can drive a constant load torque T L ? Newton’s Law of Motion governing the rotational motion is: L e m T T dt d J = ω ( 2 ) where J is the moment of inertia of the shaft system and T L is the load torque. So far, most of the class notes are on analysis. Answering the above question requires synthesis, which is an art. Trial and error is applied most of the time. Often, we use results from known experience. For example, one can propose using the ac mains where cos( ) a iI t δ = + and assume that d t = . Substituting in (1) 2 0.5[ 2 sin(2 )][0.5 (1 cos(2 2 )] eB TL t I t ωδ =− + + Therefore 2 0.5 [sin(2 ) 0.5sin(4 2 ) 0.5sin 2 ] TI L t t + + + (3) Substituting (3) in (2)
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2 0.5 [sin(2 ) 0.5sin(4 2 ) 0.5sin 2 ] m BL d JI L t t dt T ω ωω δ =− + + + (4) Since d t θ = , the speed is already fixed at = m the synchronous speed of the supply frequency. However, one can expect a small perturbation speed Δ around the equilibrium , that is 0 mm m =+ Δ . Therefore 2 0 () 0.5 [sin(2 2 ) 0.5sin(4 2 ) 0.5sin 2 ] B L d L t t
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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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Lecture4 - Lecture 4 RELUCTANCE MACHINES The inductances in...

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