ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE 17 FORCES OF ELECTRIC ORIGIN ENERGY APPROACH(1)

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1 ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE 17 FORCES OF ELECTRIC ORIGIN ENERGY APPROACH(1) Aknowledgent-These hndouts nd leture notes given in lss re bsed on teril fro Prof. Peter Suer s ECE 330 leture notes. Soe slides re tken fro Ali Bzi s presenttions Dislier- These hndouts only provide highlights nd should not be used to reple the ourse tetbook. 3/5/2018

2 MECHANICAL EQUATIONS Newton s lw sttes tht elertion fore in the positive diretion is equl to the lgebri su of ll the fores ting on the ss in the positive diretion 3/5/2018 2

3 MECHANICAL EQUATIONS In liner or trnsltionl syste with displeent, veloity, ss M, nd different fores (long ): d dt d dt e M f f spring f dper f et In rottionl syste with displeent θ,rottionl speed ω, oent of inerti J, nd different torques: d d e, J T T T T dt dt spring dper et 3/5/2018 3

4 MECHANICAL EQUATIONS Fores nd torques with supersript e re of eletril origin. In trnsltionl syste, power is d P f f dt In rottionl syste, power is d P T T dt 3/5/2018 4

5 ENERGY CONVERSION In n eletro-ehnil energy onversion syste, energy is trnsferred fro the eletril side to the ehnil side s follows: (ehnil to eletril trnsfer is bkwrds): 3/5/2018 5

6 ENERGY CONVERSION By differentiting the energy with respet to tie, power trnsfer n be shown to be: 3/5/2018 6

7 ENERGY CONVERSION Negleting the field losses, we get the siple reltion for the oupled syste (shown dotted) 3/5/2018 7

8 ENERGY CONVERSION The fore of eletril origin n be derived using the energy funtion W : Choosing λ nd s independent vribles:, f dw dt = vi 3/5/ f e d dt dw d e d = i f dt dt dt,, d, dw W W d = dt dt dt W i(, ) =, e W, (, ) =

9 In rottionl syste ENERGY CONVERSION dw d e d = i T dt dt dt Choosing λ nd θ s independent vribles: dw,,, W d W d = dt dt dt i (, ) = W,, f (, ) = e W, 3/5/2018 9

10 FORCES OF ELECTRICAL ORIGIN When the syste oves fro one operting point to nother dw t d d dt = ( i (, ) f (, ) ) dt dt t dt dt t b b e t Chnge vrible b b b e dw = i (, ) d f (, ) d W b b e W = i (, ) d f (, ) d b 3/5/

11 FORCES OF ELECTRICAL ORIGIN Integrte keeping λ onstnt t λ, then integrte λ keeping t b If λ = 0 nd W = 0 (f e = 0), then: W = b i (, ) d b W b b e W = i (, ) d f (, ) d b b 0 Letting λ b ny λ nd b ny, then: W (, ) = i (, ) d 0 3/5/

12 To opute ENERGY AND CO-ENERGY, we need to epress i = i(, ) W (, ) fro the flu linkge eqution. This would require solving for i fro = ( i, ). Prtiulrly in ulti-port systes, this ould be quite oplited. We n void this proble by defining quntity lled o-energy diretly oputble fro = ( i, ) e nd then using it to opute f. 3/5/

13 ENERGY AND CO-ENERGY Choosing λ nd s independent vribles, define: W i W Using i nd s independent vribles dw ( i, ) W ( i, ) di W ( i, ) d = dt i dt dt dw ( i, ) di d dw = i dt dt dt dt dw ( i, ) di d d = ( e d i i f ) dt dt dt dt dt 3/5/

14 ENERGY AND CO-ENERGY The left sides of these equtions re the se, opring ters: Coputtion of f W ( i, ) = i e ( i, ) = W W Method # 1 W ( i, ) ( i, ) W ( ( i, )) 3/5/

15 Method # 2 ENERGY AND CO-ENERGY dw di e d = f dt dt dt dw t d i d dt = ( f ) dt dt t dt dt t b b e t Chnge of vribles using i nd s independent vribles i b b e W W = ( i, ) di f ( i, ) d b i 3/5/

16 ENERGY AND CO-ENERGY Choose pth Integrte keeping i onstnt t i, then integrte λ keeping t b b b e W W = ( i, ) di f ( i, ) d Use i = 0 ( so f e = 0 ) nd ssue = 0 Let i b be ny i nd b be ny i b i b i 0 3/5/ W W ( i, ) = ( i, ) di ( onst.)

17 EXAMPLE : The en length of the gneti pth A : Cross-setionl re, the gneti iruit hs finite μ. Find the fore of eletri origin f e. 3/5/

18 EXAMPLE Apere s iruitl lw gives: H H H Ni 1 2 = Applying Guss s lw round the oving prt: H A H = Applying Guss s lw to the losed surfe round the upper fied surfe H A = H A H 3/5/ = H = H H A 0 1 1

19 Solve for H 1 : H 2 H = Ni The flu linkge of the oil: 1 H 1 = EXAMPLE 0 = N 3/5/ Ni 2 = N 0H 1 A 2 N 0Ni N i = A = A 0 A

20 The o-energy EXAMPLE is The fore of eletri origin: f e Note tht = ( i, ) di = is the relutne of the iron pth, nd is the relutne of the two ir gps in series. W f e i 0 = W 2 A N i N i 2 N i = = 2 A A 2 2 A A A A W A 2 A A 3/5/