Chapter 3. Generator and Transformer Models; The Per-Unit System

Transcription

1 3.1 Introduction Chpter 3. Genertor nd Trnforer Model; The Per-Unit Syte 1. Ue of the "per phe" bi to repreent three-phe blnced yte.. Ue the π odel to decribe trniion line (ee Chpter 5). 3. Siple odel for genertor nd trnforer to tudy tedy tte blnced opertion. 4. One-line digr to repreent three phe yte. 5. The per-unit yte nd the ipednce digr on coon MA be. 3. Synchronou Genertor Note the coil ', bb', nd cc' re 10 prt. Axi of -coil i the x-xi hown. F r Thee re "concentrted" full pitch winding. In δr rel chine the F r n winding re ditributed ong ny lot, nd often c' b' re not full pitch. ω t We ue the r ψ r' winding produce inuoidl f ψ round the rotor b F c periphery (ngle round the irgp with repect to winding xi). ' Aue the rotor i excited with DC current I f producing flux φ which Rotor poition hown t ωt = π/ rotte with the rotor t peed ω. At tie t the rotor would hve oved n ngle ω t. Nφ t ω t = 0. The The flux linkge for n N-turn -winding will be xiu ( ) flux linkge with the -winding will go to 0 t ωt = π /. The flux linkge will go to Nφ t ωt = π, nd bck to 0 t ωt = 3 π /. When ωt = π the flux linkge will ( ) be bck to ( Nφ ). Thu whole cycle Thi repet every revolution of the rotor.

2 The flux linkge with winding "" i thu given by: λ = Nφcoωt Uing Frdy' lw, the voltge induced in phe "" i given by: dλ e = = ω N φ in ω t dt = E inωt x π = Ex co ωt Where Ex = ωnφ = π fnφ, ince π / 4.44, the r vlue of the generted voltge i given by: E = 4.44 fnφ c' r F r δr ωt ψ F r r' The frequency i function of peed nd the nuber of pole, thu: P n f = where n i 60 the peed in rp (the ynchronou peed) nd P i the nuber of pole (lwy n even nuber). b ψ F c If the phe "" i connected to lod, then current i will flow. Depending on the lod, thi current will hve ' Rotor poition hown t ω t = π/ phe ngle, y ψ (ee figure) lgging the generted voltge e which i long the x- xi. Agin, thi i hown in the figure the line "n". N.B. The line "n" i ttched to nd rotte with the rotor. Be creful lo tht the "Flux" vector re ptil while the voltge nd current vector re phor (in tie). The e i true for phe "b" nd "c" but they will "lg" the voltge in phe "" by 10 nd 40 degree repectively. b' n

3 Since e in ( ωt), then we hve: x x x ( ω ψ) i = I in t ( ω ψ ) ( ω ψ ) i = I in t 10 b i = I in t 40 Since f i proportionl to the current, we then hve: c ( ω ψ) F = F in t ( ω ψ ) ( ω ψ ) F = F in t 10 b F = F in t 40 c We now tke coponent of thee phor long the line "n" nd in qudrture with it. Along "n" we hve: ( ω ψ) ( ω ψ) ( ω ψ ) ( ω ψ ) ( ω ψ ) ( ω ψ ) F1 = Fin t co t + F in t 10 co t 10 + F in t 40 co t 40 Uing the identity inα coα in( α) F = 1 the bove eqution becoe: ( ω ψ) ( ω ψ ) ( ω ψ ) F1 = in t + in t 10 + in t 40 It i noted tht thi expreion i the u of three blnced phor hence i equl to zero. Next we conider the coponent of the f perpendiculr to "n": ( ω ψ ) ( ω ψ ) ( ω ψ ) ( ω ψ ) ( ω ψ ) ( ω ψ ) F = Fin t in t + F in t 10 in t 10 + F in t 40 in t 40 Uing the identity in α ( 1 coα) F = we hve: 1 { ( ω ψ) ( ω ψ ) ( ω ψ )} F = 3 co t + co t 10 + co t 40 The three coine ter dd to zero (blnced phor) thu we hve: 3

4 F S = 3 F Thu the reult of hving three pulting ingle phe fluxe produce (when pplied yetriclly) contnt flux perpendiculr to line "n" nd rotte t the e peed n the rotor. Thi flux i clled the "rture rection" to the field of the rotor F r. the vriou field re hown for one phe (y phe "") in the digr below: Note tht F i perpendiculr to line "n" nd rotte with it t the e peed. The fct tht F (which i proportionl to I ) i perpendiculr to line "n" indicte tht tht we cn theorize tht the rture current I i producing rective voltge drop prllel to line "n" due to inductnce in the chine. Run the cond "rotfield" in Mtlb window to ee deo of the rotting field. F r Fr n E r E r E F R I jx l I jx r I I Figure 3. Pge 53 θ i the ngle between nd I it i the power fctor ngle δ i the ngle between E nd it i the power ngle Firt the rotor field F r produce the no-lod generted voltge E (t zero rture current). Note tht E lg F r by 90 degree. E i clled the "excittion" voltge. It i directly proportionl to the field current. The voltge/current phor for phe "" re hown bove lgging the flux digr by 90 deg. Note the bove digr i "hybrid" cobining ptil nd teporl vector. 4

5 Second, ue the rture crrie current to lod, now the rture rection flux F i produced. Thi i perpendiculr to line "n". The two fluxe (due to rotor nd rture) cobine together to for the "reultnt" flux F r. The reultnt flux induce the generted on-lod ef E r. The rture f F induce the voltge E r known the rture rection voltge. In ll ce, ech f produce voltge lgging the f by 90 degree. Note tht the voltge E r led F (hence I ) by 90 degree. Thu we cn theorize n inductor odel for thi reltionhip with rectnce X r i.e. Er = jxri. X r i known the rectnce of rture rection. Thu we now hve the circuit eqution: E= Er + jxri The terinl voltge i found by conidering the rture reitnce nd lekge rectnce thu: Which cn be iplified to: ( ) E= + R + j X + X I l r [ ] E = + R + jx I where X = Xl + Xr i known the ynchronou rectnce. The ngle between nd I i θ the power fctor ngle. The power ngle i the ngle between E nd clled δ. The circuit odel of the chine i hown below: The reitnce R i uch ller thn the ynchronou rectnce X nd i often neglected. Thi i hown in the iplified one line digr below where the chine i hown connected to n infinite E jx I I + E - R jx + - Lod bu. An infinite bu h voltge (uully ued t n ngle zero) which doe not chnge no tter how the genertor i controlled or excited. A figure of erit of genertor i the percent voltge regultion R defined : nl rted E rted R = 100= 100 rted rted 5

6 The ller the R the better. In rel genertor, the regultion i few percent. 3.3 Stedy-Stte Chrcteritic, Cylindricl Rotor Power fctor control. Mot genertor re connected to lrge power grid. Thi i n infinite bu (it voltge, ngle nd frequency re contnt). Auing the genertor h ll lekge rectnce nd ll rture reitnce, then it odel i hown below: We hve two wy of clculting the power (per phe): E jx I 1. P1 φ = I coθ E. P1 φ = in δ X Equting thee two eqution we hve E inδ = X I coθ. Auing the power i contnt, then fro eqution 1 we hve: I co θ = cont. Thi locu i hown the verticl dhed line in the figure below. Fro eqution, uing the power i contnt we hve: E in δ = cont. Thi locu i the horizontl dhed line hown. I 3 E 3 E E 1 Note tht for thi lod, the iniu rture δ 1 current i I when the power fctor of the q I 1 genertor i unity. Note tht E i directly bove I the voltge. Thu I nd re in phe, 1 producing unity power fctor. If the excittion i increed, the ef of the genertor i increed to oe vlue y E 1 hown. Clerly the voltge of the genertor led the current hence it i like n inductor conuing vr (lgging power fctor). On the other hnd, if the excittion i reduced below tht for unity power fctor, the ef of the genertor i ller, y E 3 which lg I 3. Now the power fctor of the genertor i "leding", i.e. it i like cpcitor (current leding voltge). A plot of the vrition of I function of field current i known -curve. A fily of uch curve, one t ech lod i known the -curve of the genertor. The iniu point on ech curve correpond to unity power fctor. To the right (higher excittion) i the lgging power fctor re nd to the left i the leding power fctor zone. 6

7 Power ngle chrcteritic. Conider the chine connected to n infinite bu nd neglecting rture reitnce, the chine odel i hown below: S = * 3φ 3I But we hve I E δ 0 = thu for coplex power: γ E S3 = 3 3 φ γ δ γ thu the rel nd iginry power re given by: E jx I E P3 = 3 co( ) 3 co φ γ δ γ E Q3 = 3 in( ) 3 in φ γ δ γ nd if we neglect the rture reitnce we hve: (Note tht γ = 90 X ) nd E P3 φ = 3 inδ nd Q3 φ = 3 ( E coδ ). Thu the rel power i function of X X the power ngle δ nd vrie inδ. The xiu power i given by: E Px(3 φ ) = 3. X The ngle δ trt t zero when the chine doe not deliver ny power. A the chine i loded, the power delivered incree nd the ngle δ incree. A the ngle δ reche 90, xiu power i reched. If prie over power i increed beyond thi point, the chine will peed up nd looe ynchroni. Thi en the tie between the chine nd the power grid cn hndle only the xiu power hown bove. In rel opertion, the power ngle i very ll of the order of 10 degree nd le. The power ngle δ cn "wing" by lrger ount under trnient condition. The eqution for Q how tht Q3 φ 3 ( E ). Thu if E > the genertor X deliver rective power nd the genertor i id to be overexcited. If E <, the genertor i underexcited nd the rective power delivered to the bu i negtive, thu the genertor i receiving rective power fro the bu. Genertor re uully operted in the overexcited ode ince they need to upply the yte with rective power: ot our doetic lod re rective. Exple 3.1 7

8 A 50-MA, 30-k, three phe, 60 Hz ynchronou genertor h X = 9 Ω per phe nd negligible reitnce. The genertor i delivering rted power t 0.8 power fctor lgging t the rted terinl voltge to n infinite bu. () Deterine the excittion voltge per phe E nd the power ngle δ. (b) With the excittion held contnt t the vlue found in (), the driving torque i reduced until the genertor i delivering 5 MW. Deterine the current nd the power fctor. (c) If the genertor i operting t the excittion voltge of prt (), wht i the tedytte xiu power the chine cn deliver before loing ynchroni? Alo, find the rture current correponding to thi xiu power. The Mtlb progr follow: R=0; X = 9; = R + j*x; S = 40 + j*30; = 30/qrt(3); % MW nd k I1 = conj(s)*1000/(3*conj()); % Ap I1M = b(i1), I1ng=ngle(I1)*180/pi E1 = + *I1*0.001; % k E1M = b(e1), delt1 = ngle(e1)*180/pi % k, deg dip('(b)') P = 5; % MW delt = in(p*x/(3*b(e1)*)); deltd = delt*180/pi E = E1M*(co(delt) +j*in(delt)); % k I = (E - )*1000/; IM = b(i), Ing=ngle(I)*180/pi PF = co(ngle(i)) dip('(c)') Px = 3 *E1M*/X % kw E3 = E1M*(co(pi/) +j*in(pi/)); % k I3 = (E3 -)*1000/; % A I3M = b(i3), I3ng=ngle(I3)*180/pi PF = co(ngle(i3)) I1M = I1ng = E1M = delt1 = (b) deltd = IM = Ing = PF = (c) Px =

9 I3M = 3.489e+003 I3ng = PF = Exple 3. The genertor of exple 3.1 i delivering 40 MW t terinl voltge of 30 k. Copute the power ngle, rture current, nd power fctor when the field current i djuted for the following excittion: () The excittion voltge i decreed to 79. percent of the vlue found in exple 3.1. (b) The excittion voltge i decreed to 59.7 percent of the vlue found in exple 3.1. (c) Find the iniu excittion below which the genertor will loe ynchroni. The Mtlb progr follow: R=0; X = 9; = R + j*x; = 30/qrt(3); P = 40; E1M = 0.79*3.558; delt1 = in(p*x/(3*b(e1m)*)); delt1d = delt1*180/pi E1 = E1M*(co(delt1) +j*in(delt1)); I1 = (E1 - )*1000/; I1M = b(i1), I1ng=ngle(I1)*180/pi PF = co(ngle(/i1)) dip('(b)') EM = 0.595*3.558; delt = in(p*x/(3*em*)); deltd = delt*180/pi E = EM*(co(delt) +j*in(delt)); I = (E - )*1000/; IM = b(i), Ing=ngle(I)*180/pi PF = co(ngle(i)) dip('(c)') E3M = P *X/(3**1) E3=E3M*(co(pi/)+j*in(pi/)); I3 = (E3 -)*1000/; I3M = b(i3), I3ng=ngle(I3)*180/pi PF = co(ngle(i3)) delt1d = I1M = I1ng = PF = (b) deltd = IM =

10 Ing = PF = (c) E3M = 6.98 I3M =.078e+003 I3ng = PF = Note: in () the genertor i operting t unity power fctor. In both (b) nd (c) the genertor i under-erexcited nd i receiving rective power (power fctor of lod i leding, i.e. cpcitive lod nd the genertor i inductive.) In the rel world the lod i uully inductive (lgging power fctor) nd the genertor i over-excited, delivering rective power. Exple 3.3 For the genertor of exple 3.1, contruct the v-curve for the rted power of 40 MW with vrying field excittion fro 0.4 power fctor leding to 0.4 powr fctor lgging. Aue the open-circuit chrcteritic in the operting region i given by E = 000I f. The Mtlb progr follow: P = 40; % rel power, MW = 30/qrt(3)+ j*0; % phe voltge, k = j*9; % ynchronou ipednce ng = co(0.4); thet=ng:-0.01:-ng; % Angle fro 0.4 leding to 0.4 lgging pf P= P*one(1,length(thet)); % generte P rry of e ize I = P./(3*b()*co(thet)); % current gnitude ka I = I.*(co(thet) + j*in(thet)); % current phor E = +.*I; % excittion voltge phor E = b(e); % excittion voltge gnitude, k If = E*1000/000; % field current, A plot(if, I), grid xlbel('if, A'), ylbel('i, ka') text(3.4, 1, 'Leding pf'), text(13, 1, 'Lgging pf') text(9,.71, 'Upf') 10

11 I, ka Leding pf Lgging pf 0.8 Upf If, A Note tht the over-excited genertor (lod i inductive, genertor i cpcitive) i the norl wy to operte the genertor ince ot lod re inductive. If the lod becoe cpcitive, then the genertor i operted in the under-excited ode. 11