Boise State University Department of Electrical and Computer Engineering ECE470 Electric Machines

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1 Boie State Univeity Depatment of Electical and Compute Engineeing ECE470 Electic Machine Deivation of the Pe-Phae Steady-State Equivalent Cicuit of a hee-phae Induction Machine Nomenclatue θ: oto haft angle meaued fom the tato phae-a axi to the oto phae-a axi α: Spatial angle along the tato peiphey meaued fom the axi of the tato phae-a winding β: Spatial angle along the oto peiphey meaued fom the axi of the oto phae-a winding f : Fequency of the tato cuent and voltage (f = 60 Hz) ω : Synchonou (electical) peed (ω = 2πf ) n : Synchonou peed in pm (n = ω (60/2π) = 3600 pm) ω: oto haft peed in ad/ n: Mechanical peed in pm : lip peed ( = (ω ω)/ω = (n n)/n ) f : Fequency of the oto cuent (f = f ) ω : Angula peed of oto cuent (ω = ω ) : adiu of oto cylinde l: Length of oto cylinde µ o : Pemeability of fee pace (µ o = 4π 10 7 ) l, l : Stato, oto leakage inductance L m, L m : Stato, oto magnetizing inductance M, M : Maximum tato-tato, oto-oto mutual inductance M : Maximum tato-oto mutual inductance, : Pe-phae tato, oto eitance g, P g : Effective eluctance, pemeance of ai gap 1

2 1 Flux-Cuent elationhip baxi baxi aaxi θ aaxi caxi caxi Figue 1: Schematic epeentation of a wo-pole, hee-phae Induction Machine Conide a thee-phae, two-pole, wound-oto induction machine with a mooth ai gap a hown in Figue 1. hi machine ha thee tato winding labeled (a-a, b-b, c-c ) and thee oto winding labeled (a-a, b-b, c-c ). All winding ae aumed to be inuoidally ditibuted along the tato and peipheie o that the magnetic field intenitie ceated by the coeponding tato and oto cuent in the ai gap ae given by: H a = N i a 2g H b = N i b 2g H c = N i c 2g H a = N i a 2g H b = N i b 2g H c = N i c 2g 4 π co α (1) 4 π co(α 2π 3 ) (2) 4 π co(α 2π 3 ) (3) 4 π co β = N i a 4 co(α θ) 2g π (4) 4 π co(β 2π 3 ) = N i a 4 2g π co(α θ 2π 3 ) (5) 4 π co(β 2π 3 ) = N i a 4 2g π co(α θ 2π 3 ) (6) 2

3 he flux linkage with coil a-a, b-b, c-c, a-a, b-b, and c-c ae given by: π/2 λ a = N µ o (H a H b H c H a H b H c )l dα l i a (7) π/2 π/6 λ b = N µ o (H a H b H c H a H b H c )l dα l i b (8) 7π/6 5π/6 λ a = N µ o (H a H b H c H a H b H c )l dα l i c (9) 11π/6 π/2 λ a = N µ o (H a H b H c H a H b H c )l dβ l i a (10) π/2 π/6 λ b = N µ o (H a H b H c H a H b H c )l dβ l i b (11) 7π/6 5π/6 λ c = N µ o (H a H b H c H a H b H c )l dβ l i c (12) 11π/6 whee l and l ae tato and oto leakage inductance, epectively. Integating thee equation eult in linea flux-cuent elationhip in the fom: λ a l L m M co 120 o M co 240 o i a λ b = M co 240 o l L m M co 120 o i b (13) λ c M co 120 o M co 240 o l L m i c λ a λ b λ c = M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ l L m M co 120 o M co 240 o M co 240 o l L m M co 120 o M co 120 o M co 240 o l L m i a i b i c i a i b i c i a i b i c (14) (15) (16) whee L m N 2 = M N 2 = L m N 2 = M N 2 = M = 4µ ol = 1 = P g (17) N N πg g and whee g and P g ae, epectively, the effective eluctance and pemeance of the ai gap. Execie 1: eify the mutual inductance by inpection of the magnetic axe. he magnetic coenegy of the coupling field i computed a: W m = W m(i a, i b, i c, i a, i b, i c, θ) = 1 2 λ ai a 1 2 λ bi b 1 2 λ ci c 1 2 λ ai a 1 2 λ bi b 1 2 λ ci c = 1 2 (l L m )(i 2 a i 2 b i 2 c) 1 2 (l L m )(i 2 a i 2 b i 2 c) M co θ(i a i a i b i b i c i c ) M co(θ 120 o )(i a i b i b i c i c i a ) M co(θ 120 o )(i a i b i b i c i c i a ) M (i a i b i b i c i c i a ) M (i a i b i b i c i c i a ) 3

4 he developed electomagnetic toque i: e = e (i a, i b, i c, i a, i b, i c, θ) = W m θ e = M in θ(i a i a i b i b i c i c ) M in(θ 120 o )(i a i b i b i c i c i a ) M in(θ 120 o )(i a i c i b i a i c i b ) (18) 2 Model of a hee-phae Wound-oto Induction Moto We will aume moto opeation in the following analyi, that i, the induction machine convet electical enegy to mechanical enegy. (Geneato opeation, paticulaly in wind tubine ytem, i poible and involve the conveion of mechanical enegy into electical enegy). he mathematical model fo thi machine i compoed of ix diffeential equation given by Kichhoff voltage law fo the ix tato and oto winding, and two diffeential equation fo the oto haft given by Newton econd law fo otating bodie: v a = i a dλ a v a = i b dλ b v a = i c dλ c v a = ext i a = i a dλ a v b = ext i b = i b dλ b v c = ext i c = i c dλ c dθ = ω J dω = e m In addition, the following contitutive equation ae pat of the mathematical model: λ a l L m M co 120 o M co 240 o i a λ b = M co 240 o l L m M co 120 o i b (27) λ c M co 120 o M co 240 o l L m i c λ a λ b λ c = M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ M co(θ 120 o ) M co(θ 120 o ) M co(θ 120 o ) M co θ l L m M co 120 o M co 240 o M co 240 o l L m M co 120 o M co 120 o M co 240 o l L m i a i b i c i a i b i c i a i b i c (19) (20) (21) (22) (23) (24) (25) (26) (28) (29) (30) 4

5 and e = M in θ(i a i a i b i b i c i c ) M in(θ 120 o )(i a i b i b i c i c i a ) M in(θ 120 o )(i a i c i b i a i c i b ) (31) In the above model, the oto winding ae nomally hot-cicuited though balanced extenal eito. hee eito, a will be een late, will allow the haping of the toque-peed chaacteitic fo diffeent application. he input in the above model ae the thee tato voltage v a (t), v b (t), v c (t), and the mechanical load toque m (t). Aume balanced input voltage and a contant load toque: v a (t) = 2 co(ω t θ v ) (32) v b (t) = 2 co(ω t θ v 120 o ) (33) v c (t) = 2 co(ω t θ v 120 o ) (34) m (t) = m o find the teady-tate cuent i a (t), i b (t), i c (t), i a (t), i b (t), i c (t), and the teady-tate haft peed ω m, we could imulate the above model tating fom abitay initial condition until teady tate i eached. Intead, we will gue gue at the geneal fom of thee olution and ty to obtain analytical olution. If thee geneal equation atify the diffeential equation and the contitutive equation, then, by unicity of thee olution, they will be the ame a the imulated epone in teady tate. heefoe, let u aume that the geneal olution have the following fom: i a (t) = 2I co(ω t θ i ) (36) i b (t) = 2I co(ω t θ i 120 o ) (37) i c (t) = 2I co(ω t θ i 120 o ) (38) i a (t) = 2I co(ω t θ i ) (39) i b (t) = 2I co(ω t θ i 120 o ) (40) i c (t) = 2I co(ω t θ i 120 o ) (41) ω(t) = ω θ(t) = ωt θ o whee ω = ω, f = f, and i the lip peed: = ω ω ω = ω = (1 )ω (44) hu, thee ae ix unknown vaiable to be detemined: I, θ i, I, θ i, ω o, and θ o. In othe wod, we need to come up with ix independent equation that will allow u to olve fo thee ix unknown, povided the eight diffeential equation ae all atified in teady tate. Execie 2: eify that the mechanical equation can be be atified in teady tate by a contant electomagnetic toque equal to the mechanical load toque, that i: J dω = 0 = e m = e = m = contant (45) (35) (42) (43) 5

6 Solution: e = M in θ(i a i a i b i b i c i c ) M in(θ 120 o )(i a i b i b i c i c i a ) M in(θ 120 o )(i a i c i b i a i c i b ) = M i a [i a in θ i b in(θ 120 o ) i c in(θ 120 o )] M i b [i a in(θ 120 o ) i b in θ i c in(θ 120 o )] M i c [i a in(θ 120 o ) i b in(θ 120 o ) i c in θ] = 9 2 M I I in(θ i θ o θ i ) = contant = m (46) Now, conide the tato equation fo phae a: v a (t) = i a (t) dλ a 0 = ( ext )i a (t) dλ a Since v a (t) i a inuoidal voltage at f = 60 Hz, and i a (t) i alo aumed inuoidal at 60 Hz, we need to veify that the tato flux linkage λ a (t) ae alo inuoidal at 60 Hz. Execie 3: eify that the tato flux linkage λ a (t) ae inuoidal at 60 Hz in teady tate, that i: Solution: λ a (t) = 2Λ co(ω t ϕ ) (49) Since i a i b i c = 0 and i a i b i c = 0 in teady tate, λ a = (l L m )i a M (i b i c ) M [i a co θ i b co(θ 120 o ) i c co(θ 120 o )] = (l L m M )i a M (i a co θ i b co(θ 120 o ) i c co(θ 120 o ) = (l L m M ) 2I co(ω t θ i ) 3 2 M 2I co[(ω ω )t θ o θ i ] = (l 3 2 L m) 2I co(ω t θ i ) 3 2 M 2I co(ω t θ i θ o ) = 2Λ co(ω t ϕ ) (50) whee we ued the fact that ω ω = (1 )ω ω = ω and M = L m /2. he coeponding phao fo λ a i Λ = (l 3 2 L m)i θ i 3 2 M I (θ i θ o ) (51) Similaly, conide the oto equation fo phae a: 0 = ( ext )i a (t) dλ a Since i a (t) i aumed inuoidal at lip fequency f = f, we need to veify that the flux linkage λ a (t) ae alo inuoidal at lip fequency. (47) (48) (52) 6

7 Execie 4: eify that the oto flux linkage λ a (t) ae inuoidal at lip fequency f = f in teady tate, that i: Solution: λ a (t) = 2Λ co(ω t ϕ ) (53) Since i a i b i c = 0 and i a i b i c = 0 in teady tate, λ a = (l L m )i a M (i b i c ) M [i a co θ i b co(θ 120 o ) i c co(θ 120 o )] = (l L m M )i a M (i a co θ i b co(θ 120 o ) i c co(θ 120 o ) = (l L m M ) 2I co(ω t θ i ) 3 2 M 2I co[(ω ω)t θ i θ o ] = (l 3 2 L m) 2I co(ω t θ i ) 3 2 M 2I co(ω t θ i θ o ) = 2Λ co(ω t ϕ ) (54) whee we ued the fact that ω ω = ω and M = L m /2. he coeponding phao fo λ i ˆΛ = (l 3 2 L m)i θ i 3 2 M I (θ i θ o ) (55) Note that ˆΛ i a phao coeponding to a inuoidal wavefom with fequency f = f, not f. he tato and oto equation yield v a = i a dλ a 0 = ( ext )i a dλ a (56) (57) θ = I θ i jω Λ (58) 0 = I θ i jω ˆΛ (59) Uing the expeion fo Λ and ˆΛ, thee two equation can be manipulated to yield θ = I θ i jω (l 3 2 L m)i θ i jω 3 2 M I (θ i θ o ) (60) 0 = I θ i jω 3 2 M I (θ i θ o ) jω (l 3 2 M )I θ i (61) Multiplying thi lat equation by e jθ o and uing ω = ω yield 0 = ext 3 I (θ i θ o ) jω 2 M I θ i jω (l 3 2 M )I (θ i θ o ) (62) Defining 60-Hz phao Ṽ = θ v, Ĩ = I θ i, and Ĩ = I (θ i θ o ), two phao equation can be obtained a Ṽ = Ĩ jω (l 3 2 L m)ĩ jω 3 2 M Ĩ (63) 0 = ext 3 Ĩ jω 2 M Ĩ jω (l 3 2 L m)ĩ (64) 7

8 3 Equivalent Cicuit of a hee-phae Induction Moto Equivalent cicuit epeentation fo the induction machine have been developed that make the peviou equation eay to emembe and imple to olve. A wod of caution i in ode hee: he peviou phao equation wee deived uing the tato and oto cicuit. It hould be kept in mind that thee cicuit have diffeent inuoidal fequencie! he actual oto phao Î coepond to a inuoidal wavefom with fequency f = f. hi phao i defined a: Î = I θ i (65) he newly-defined oto phao Ĩ i a fictitiou 60-Hz phao efeed to the tato and defined a: Ĩ = I (θ i θ o ) (66) hu, the two phao Î and Ĩ have the ame m magnitude, but thei phae ae diffeent. ecall that: L m N 2 = L m N 2 = M N N (67) efeing all oto vaiable to the tato by mean of an ideal tanfome with tun atio: a = N N and defining (68) ext = l 3 2 L m = ( N N ( N N Ĩ = N Ĩ N N M = L m N ) 2 ( ext ) (69) ) 2 (l 3 2 L m) = ( N N ) 2 l L m (70) the peviou equation on page 7 become: [ Ṽ = jω (l 3 ] 2 L 3 m) Ĩ jω 2 LmĨ (73) [ 3 0 = jω 2 L mĩ ext jω (l 3 ] 2 L m) Ĩ (74) hee two equation can be combined to give the following cicuit (71) (72) 8

9 I jx l N :N jx l I j1.5x m 1 (a) I jx l jx l I j1.5x m 1 (b) Figue 2: Equivalent Cicuit epeentation of a hee-phae Induction Machine. (a) Uing an Ideal wo-winding anfome Between the Stato and oto Side. (b) Afte efeing all oto Quantitie to the Stato Side. 9

10 Pefomance Analyi of an Induction Machine I jx l jx l I I jx l jx l I jx M jx M (a) (b) Figue 3: (a) Exact Equivalent Cicuit (b) Appoximate Equivalent Cicuit Note: X M = { X m = ω L m fo a two-phae induction machine (3/2)X m = (3/2)ω L m fo a thee-phae induction machine ω = 2πf = 2π60 elec. ad/ ( ( 2 2 ω m = ω = 2π60 mech. ad/ p) p) mech. ad n = ω m 1 ( 2 n = 3600 pm p) 2π mech. ad 1 ev. 1 mn 60 = ω m ω m = n n ω m n ω m = Actual oto haft peed in mech. ad/ n = Actual oto haft peed in mech. pm = ( 2 2πf p) 60 2π = 120f p Poblem: Fo a given lip, compute the efficiency η of a thee-phae induction machine uing: (i) he exact equivalent cicuit; (ii) he appoximate equivalent cicuit. pm Uing the exact equivalent cicuit: ( ) Z in = ( jx l ) (jx M ) jx l = Z in ϕ = in jx in Ĩ = Ṽ Z in = Z in jx M ϕ Ĩ = / j(x l XM)Ĩ S in,3ϕ = 3ṼĨ = 3 Ṽ Ĩ co ϕ j3 Ṽ Ĩ in ϕ P in,3phi = 3 Ṽ Ĩ co ϕ Q in,3phi = 3 Ṽ Ĩ in ϕ 10

11 P,lo = 3 Ĩ 2 P g = 3 Ĩ 2 = P,lo P e ( ) 1 P e = 3 Ĩ 2 = (1 )P g P,lo = 3 Ĩ 2 = P g In teady tate, J dω m = e m = e ( m,out ot ) = 0 = e = m = m,out m,ot ω m e = ω m = ω m m,out ω m ot = P e = P m = P m,out P ot η = P m,out P m,out P m,out 100% = 100% = 100% P in,3phi P m,out P loe P m,out P,lo P,lo P ot Note: P ot = P windage P fiction P coe. Uing the appoximate equivalent cicuit: Ṽ Ĩ = ( / j(x l X l ) = Ĩ = P,lo = 3 Ĩ 2 Ṽ ( /) 2 (X l X l )2 P g = 3 Ĩ 2 = P,lo P e ( ) 1 P e = 3 Ĩ 2 = (1 )P g P,lo = 3 Ĩ 2 = P g η = P m,out P in,3phi 100% = Note: P ot = P windage P fiction P coe. P m,out P m,out P loe 100% = P m,out P m,out P,lo P,lo P ot 100% 11

12 evolving Stato and oto Magnetic Field in the Ai Gap of a wo-phae Induction Machine i a = 2I co(ω t θ i ) i b = 2I in(ω t θ i ) H a = N ( ) i a 4 p 2g π co 2 α = N 4 ( ) p 2I co(ω t θ i ) co 2g π 2 α H b = N ( ) i b 4 p 2g π in 2 α = N 4 ( ) p 2I in(ω t θ i ) in 2g π 2 α H = H a H b = N 4 ( ) p 2I [co 2g π 2 α [ ] p 2 α ω t θ i H = N 2g 4 2I co π ( ) p co(ω t θ i ) in 2 α in(ω t θ i )] At a given time, the eultant tato magnetic field i maximum at an angle α on the tato peiphey uch that p 2 α ω t θ i = 0 = α = 2 p (ω t θ i ) = dα = 2 p ω = ω m hu, the tato magnetic field evolve at mechanical ynchonou peed (ω m o n ) with epect to the tato and at lip mechanical peed (ω m o n ) with epect to the oto. i a = 2I co(ω t θ i ) i b = 2I in(ω t θ i ) H a = N ( ) i a 4 p 2g π co 2 β = N 4 ( ) p 2I co(ω t θ i ) co 2g π 2 β H b = N ( ) i b 4 p 2g π in 2 β = N 4 ( ) p 2I in(ω t θ i ) in 2g π 2 β H = H a H b = N 4 ( ) p 2I [co 2g π 2 β [ ] p 2 β ω t θ i H = N 2g 4 2I co π ( ) p co(ω t θ i ) in 2 β in(ω t θ i )] At a given time, the eultant oto magnetic field i maximum at an angle β on the oto peiphey uch that p 2 β ω t θ i = 0 = β = 2 p (ω t θ i ) = dβ = 2 p ω = 2 p ω = ω m hu, the oto magnetic field evolve at lip mechanical peed (ω m o n ) with epect to the oto and at mechanical ynchonou peed (ω m o n ) with epect to the tato. 12

13 oque-slip Chaacteitic of a hee-phae Induction Machine Z th = th jx th I Z th = th jx th I th th (a) (b) Figue 4: hevenin Equivalent of the Induction Machine Equivalent iewed by the oto eitance. (a) Equivalent hevenin epeentation fo Finding Maximum Electomagnetic oque. (b) Equivalent hevenin epeentation fo Finding Maximum Electomagnetic Powe. Poblem Statement: Aume that the tato upply voltage magnitude Ṽ i fixed. Expe the electomagnetic toque e a a function of the vaiable lip. Solution: Z th = X l ( jx l ) (jx M ) = th jx th Ṽ th = th = jx M j(x l X M )Ṽ = th θ th X M Ṽ 2 (X l X M ) 2 θ th = 90 o tan 1 (X l X M ) Ṽ th Ĩ = / = jx Ĩ = th Ṽth ( /) 2 X 2 th e = P e ω m = (1 )P g (1 )ω m = P g ω m = 3 Ĩ 2 / ω m = 3 Ṽth 2 ω m [( /) 2 X 2 th ] 13

14 Slip at Maximum Electomagnetic oque oque e, m e,max e2 e1 e1,tat m e2,tat > ,max 1,max lip Figue 5: Plot of the oque-slip Chaacteitic of an Induction Machine fo wo Diffeent alue of oto eitance. Note that ince e = P e = (1 )P g = P g ω m (1 )ω m ω m maximizing the electomagnetic powe e,max i equivalent to maximizing the ai-gap powe P g, not the electomagnetic powe P e. Fom the maximum powe tanfe theoem, the lip max at which maximum electomagnetic toque e,max o maximum ai-gap powe P g,max will occu i uch that max = 2 th X2 th Special Cae: Uing the appoximate equivalent cicuit, Ṽth = th = X th = X l X l max = e,max = 2 (X l X l )2 = max = 2 (X l X l )2 3 Ṽth 2 max ω m [( / max ) 2 Xth 2 ] = 3 Ṽ 2 2 (X l X l )2 ω m [( 2 2(X l X l )2 )] heefoe, the maximum electomagnetic toque (pull-out toque) e,max i independent of the oto eitance. If the oto teminal ae available a in a thee-phae wound-oto induction machine (a oppoed to a quiel-cage induction moto), then the toque-lip chaacteitic can be alteed by changing the lip at which maximum electomagnetic toque occu. 14

15 Slip at Maximum Electomagnetic Powe Fom the maximum powe tanfe theoem, the lip max at which maximum electomagnetic powe P e,max will occu i uch that ( (1 max max max ) = ( th ) 2 X 2 th Special Cae: Uing the appoximate equivalent cicuit, ( ) 1 1 = ( ) 2 (X l X l )2 max = ( ) 2 (X l X l )2 Ṽth Ĩ = ( / max) 2 Xth 2 ( 1 ) P e,max = 3 max Ĩ 2 max 15

16 Appendix: Maximum Powe anfe in DC and AC Cicuit Poblem #1: I L L L Find the maximum powe P L,max that can be deliveed to the vaiable load eitance L by the DC ouce voltage though the (known) eitance. Solution: ( ) P L = L I 2 2 L 2 = L = L ( L ) 2 dp L = 2 ( L ) 2 2 L ( L ) d L ( L ) 4 = (2 2 L ) 2 ( L ) 4 = 0 = L = P L,max = 2 (2 ) 2 = 2 4 Poblem #2: jx I L L L jx L Find the maximum powe P L,max that can be deliveed to the vaiable load impedance Z L = L jx L by the AC ouce voltage Ṽ though the (known) impedance Z = jx. Solution: P L = L I 2 L = L ( ) 2 = ( L ) 2 (X X L ) 2 L 2 ( L ) 2 (X X L ) 2 dp L = L 2 2(X X L ) dx L [( L ) 2 (X X L ) 2 ] 2 = 0 = X L = X 16

17 dp L = 2 [( L ) 2 (X X L ) 2 ] 2 L ( L ) d L [( L ) 2 (X X L ) 2 ] 2 = 0 = L = = Z L = L jx L = jx = Z P L,max = 2 (2 ) 2 = 2 4 Poblem #3: jx I L L L Find the maximum powe P L,max that can be deliveed to the vaiable load eitance Z L = L by the AC ouce voltage Ṽ though the (known) impedance Z = jx. Solution: 2 P L = L IL 2 = L ( L ) 2 X 2 = L 2 ( L ) 2 X 2 dp L = 2 [( L ) 2 X 2 ] 2 L( L ) d L [( L ) 2 X 2 = 2 [( 2 X2 ) 2 L ] ]2 [( L ) 2 X 2 = 0 ]2 = L = 2 X2 = ZL = L = 2 X2 = Z 2 2 P L,max = X2 2 2 = X2 ( 2 X2 )2 X 2 2( 2 X2 2 X2 ) P L,max = ( X2 ) <