6.641 Electromagnetic Fields, Forces, and Motion

Transcription

1 MIT OpenCourseWare Electromagnetic Fields, Forces, and Motion Spring 9 For information about citing these materials or our Terms of Use, visit:

2 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 13: Magnetoquasistatic Forces I. MQS Energy Method of Forces A. Circuit Approach dλ d di dt + i v= = dt dt L ( ) i =L ( ) p= vi =L ( ) di + i i dt dl () dt d 1 dl () =L ( ) i + i dt dt dl () dt d 1 = L ( ) i + i dt dt 1 dl () dl () d 1 d = ( )i + 1 L i dt d dt dw m d 1 vi= + f m ( ) W = L i, f = i dt dt d 1 dl () λ =L ( ) i f = 1 i d dl () Prof. Markus Zahn Page 1 of 14

3 1 λ dl () = L ( ) d = 1 λ d 1 d L ( ) B. Energy Method dλ dw m d vi = i = + f dw m = id λ f d dt dt dt Wm W m, i = f = λ λ =constant =constant W = f d i d λ m + λ= = constant i= λ L ( ) λ λ W = dλ = L m L ( ) ( ) =cons tan t f = W m λ=cons tan t = 1 λ d d 1 L ( ) Prof. Markus Zahn Page of 14

4 1 1 dl () = λ L ( ) d 1 dl () = i d II. Force on a Wire over a Perfectly Conducting Plane Depth D Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. µ D [See Haus & Melcher p. 343, L ( ) = ln + 1 π R R take ½ of Eq. (1) which is the inductance between cylinders] A. Energy Method 1 dl ( ) µ i D 1 f = i = d 4π R ( ) 1 R Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. Prof. Markus Zahn Page 3 of 14

5 B. Method of Images Approach with Lorentz Force µ i µ i D µ i D f = id = = π ( a ) 4πa 4π R C. Demonstration: Steady State Magnetic Leviation Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. Prof. Markus Zahn Page 4 of 14

6 III. One Turn Loop H = I, Φ = µ H xl, L x = Φ z D µ xl = z ( ) I D µ xl = I D A. Energy Method 1 dl (x) 1 µ l f = I = I x dx D B. Lorentz Force Law f = J B dv V Prof. Markus Zahn Page 5 of 14

7 Model surface current K = I as volume current of small thickness δ y D J = y I Dδ z H = J H I = J y = Hz = x Dδ Dδ I (x ( + δ )) f = J µ H dx dy dz x y z V + δ ( )) l D = I µ I Dδ (x + δ x= D δ dx +δ µ I l x = + δ x Dδ x = ( ) µ I l (+δ) = ( +δ ) + +δ ) Dδ ( µ I l 1 = ( + δ ) + + δ Dδ µ I l 1 = Dδ δ µ I l = + 1 f = 1 K B D S ds ½ comes from integrating uniform volume current over small thickness δ General formula: f = K Bav ds S For our case: B = B + B metal air = 1 av B air Prof. Markus Zahn Page 6 of 14

8 IV. Lifting of Magnetic Fluid A. Energy Method Approach Ni H= s Φ =H µ + µ (l ) d = Nd µ+µ (l ) i s λ =NΦ = Nd µ+µ (l ) i s λ Nd L ( ) = = i s µ + µ ( l ) f = 1 dl i = 1 N i d s d (µ µ ) f = ρ ghds = 1N i d m (µ µ ) s 1 N i d h= (µ µ ) s ρ m g d Prof. Markus Zahn Page 7 of 14

9 B. Magnetization force F = µ (M i ) H X x H x Hx = µ M x x + M y y + M z H x z = z H H= J = x = y H y x H H y x F X = µ M x + M y x x B = µh = µ (H + M) M = µ µ 1 H µ H x µ H y F = µ 1 H + 1 H y x x µ x µ x = µ µ 1 1 ( µ x H x + H y ) f = F dx x x dy dz = (µ µ ) h s d (H x x x = y = z = + H y ) dxdydz (µ µ = ds H ) ( x + H y ) h x = (µ µ = ) d s = 1 (µ µ ) d Ni s N i s Prof. Markus Zahn Page 8 of 14

10 V. Magnetic Actuator Hi ds = H 1 (x + a) + H (a x) = N1i 1 + Ni C HA µ HA = µ HA H = A 1 H (a x) + (a + x) A =Ni 1 1 A 1 + N i (Ni 11 + Ni ) A 1 H = A a x + x A ( ) (a + ) 1 Prof. Markus Zahn Page 9 of 14

11 (Ni 11 + Ni ) A H 1 = A a x + a + x A 1 ( ) ( ) µ N A A (Ni + N i ) λ =N µ H A = A a x + a + x A λ =N µ H A = 1 ( ) ( ) µ N A A Ni + N i A ( a x) + ( a + x) A 1 ( ) λ =L x i + M x i 1 1 ( ) 1 ( ) λ =M x i ( ) + ( ) x i 1 L µ A A N µ A A N µ A A N N ; M( x ) = ( x + a + x) A A ( a x) + ( a + x ) A L = L 1 ( x ) = ; A a + ) ( x ) ( x) ( a + x A A a ) ( dw = i d λ +i d λ 1 1 f dx di ( 1 λ 1 + i λ w) = λ 1 di 1 + λ di + fdx w (co energy) = L x ( ) L ( x ) 1 dw ' = λ 1 di 1 + λ di + f dx w' f= + x i,i 1 constant Prof. Markus Zahn Page 1 of 14

12 dw ' = f dx + λ 1 di 1 + λ di 1 3 i1 = i = i = i 1=constant x=cons tan t x=cons tan t dw ' = L 1 ( x) i 1 di 1 + ( M( x) i 1 + L ( x) i ) di i= i 1=constant x=constant x=cons tan t 1 = L x i 1 ( ) 1 + M( x) i1i 1 + L x i ( ) w' 1 dl 1 dl dm f= + = i 1 + i + ii x i,i 1 dx 1 dx dx 1 VI. Synchronous Machine Prof. Markus Zahn Page 11 of 14

13 λ λ =L i + Mi cos θ as s as r =L i + Mi sin θ bs s bs r λ r = L r i r + M ( i as sin θ + i bs sin θ) dw = i d λ +i d λ + i d λ as as bs bs r r e T d θ ( as as i bs bs r r ) = dw' d w i λ λ i λ w' = i as λ as + ibs λ bs + ir λ r w co energy e dw' = λ as di as + λ bs di bs + λ r di r + T d θ Prof. Markus Zahn Page 1 of 14

14 w'= Td θ+ λ as di as + λ bs di bs + λ r di r i as = θ=cons tan t θ=cons tan t θ=cons tan t i bs = i bs = i as =cons tan t i as =constant ir = ir = i r = i bs =constant w ' = T θ+ L i di + L i di + L i + M i cos θ+i sin θ di s as as s bs bs r r ( as bs ) r i as = θ=constant i bs = i as =constant i = i =constant r bs w ' = 1 L i + 1 L i + 1 L i + Mi (i cos θ + i sin θ) s as s bs r r r as bs w' T e = + = Mi r ( i as θ i,i,i as bs r sin θ + i bs cos θ) Balanced phase currents i =I cos ωt, i =I sin ωt, i =I, θ = ω t + γ as s bs s r r m e T = MI I ( cos ωt sin θ + sin ωt cos θ) = MI I sin ( ω t θ r s r s <T e > ω = ω m ) =MI r I s sin ((ω ω m ) t γ) e T = MIr Is sinγ d θ dθ J = T e β dt dt Prof. Markus Zahn Page 13 of 14

15 θ = ω t + γ +γ ' t, γ ' t <<γ m ( ) ( ) MI I sin γ r s βω sin γ = MI r I s βω = Pullout when sin γ = 1 βω = MI r I s Hunting transients: sin (γ +γ' ) sin γ cos γ+ ' cos γ sin γ' sin γ +γ ' cos γ d γ ' J = MI I cos γ γ' βγ ' = (MI I cos γ dt r s r s + β ) γ ' d γ ' +ω γ '= ; ω = MI I cos γ +β J dt r s γ '= A 1 sin ω t + A cos ω t Stable if ω > ( ω real) Unstable if ω < ( ω imaginary) Prof. Markus Zahn Page 14 of 14