8 THREE PHASE A.C. CIRCUITS

Transcription

1 8 THREE PHSE.. IRUITS The signls in hpter 7 were sinusoidl lternting voltges nd urrents of the so-lled single se type. n emf of suh type n e esily generted y rotting single loop of ondutor (or single winding), hene single se, in mgneti field. In prtie single se lternting emf is not suffiient nd stisftory for mny prtil pplitions, suh s.. motors. For this reson multise.. emfs (produed y multiple windings) re generted nd utilised in mny eletril systems. The three se system is y fr the most ommon multise system used for genertion, trnsmission, nd hevy power utilistion of.. eletri energy euse of its eonomi nd operting dvntges. One type of n idel three se genertor, ontins three identil windings displed y 120 from eh other in spe nd rotting within onstnt mgneti field. Suh soure genertes three sinusoidl voltges of equl mplitudes displed form eh other y 120 in time. The three se sheme of power trnsmission offers the dvntges of using the. mode, onstnt power flow, nd high power trnsfer pility. In generl, three se systems hve numer of dvntges over single se systems, like: More effiient power trnsmission. More ost effetive (the totl numer of ondutor mteril less thn in the single se). Motors nd ontrol ger re simpler, more effiient nd less ostly. Industry mkes use of 6, 12 nd 24 se systems tht n e produed form three se supply Single Phse.. Genertors: the priniples single se lternting sinusoidl voltge n e generted y rotting with onstnt ngulr veloity ω single winding in onstnt mgneti field. s rottes ntilokwise in the mgneti field its sides nd ut the mgneti flux nd emf is indued in them y motionl indution. The indued emf t oth sides of the winding t time t is e () t = 2lv 2lrω sinϑ = 2lrω sinω t (8.1) h = the motion referred from the horizontl position t time t =0. e S ω i() t l v ω = ω v h v v h r ϑ ross setion perpendiulr to the xis of rottion r winding 8.2 Three Phse lned.. Genertors three se supply uses the sme priniple s the single se, where three oils ngulrly displed etween eh other round the sme xis of rottion re used, insted of one oil. Eh oil is lled se winding. The three ses re usully given the nmes red se (R se), or se yellow se (Y se), or se nd lue se ( se), or se lned or symmetril three se supply onsists of three identil oils displed y 120 in spe from eh other. Therefore, lned or symmetril three se supply hs sinusoidl lternting se voltges (or urrents), whih re equl in mgnitude nd displed in time from one nother y se of 120. It is ssumed tht the lod t eh se is lso lned (see lter), or well uffered. If either of the ove onditions do not pply, the supply is sid to e imlned

2 n elementry three se, two pole genertor is shown ellow. It hs three identil sttor oils (,, nd ), of one or more turns, displed y 120 in spe from eh other nd The rotor rries field winding exited y the d.. supply through rushes nd slip rings nd is driven t n ngulr veloity ω in suh wy tht the flux is distriuted sinusoidlly over the poles. Here the se windings re stti within rotting mgneti field, ut the priniple of indution is the sme s in the single se genertor. Field winding Field system or rotor Phse Rottion S rmture or sttor rmture oil sides The voltge signls from lned emf genertor re given y U e = sin( ωt) e = sin( ωt 120 ) (8.2) e = sin( ωt 240 ) W Using the sptil geometry of the oils in the rmture, the sor digrm in polr form of the ove signls is the following ote tht E + E + E =0 or e () t + e () t + e () t = 0 (8.3) E E 120 E Winding onnetions for.. lned Genertors The se windings my e onneted in either wye or delt onnetions. oltge sor digrm in the WYE onnetion I I lned WYE Supply onnetion wye or str onnetion (represented s Y) results y onneting together ll three primed or unprimed terminls to form ommon terminl known s the neutrl of the wye. If neutrl ondutor is used, the system is known s four wire, three se system, otherwise it is three wire, three se system. e n e n n e n, or I I I = I = 3 (8.4) where n stnds for se quntities, nd, nd - stnds for line, nd line-to-line quntities respetively. = 0 = 90 = 3 90 = 120 = 30 = 30 3 = 240 = 150 = (8.5)

3 urrent sor digrm in the DET onnetion lned DET Supply onnetion lthough lmost ll.. genertors hve their windings onneted in wye onnetion, the ft tht the sum of the three emfs is equl to zero in the lned se (see Eqn. 8.3), mkes delt onnetion possile. delt or mesh onnetion (represented s ) n e hieved y onneting in series the three windings of the three ses, i.e. terminls to, to, nd to. I I, or I = I 90 I F I = I I H K I I I Sine the supply is lned, the urrent sors I, I, nd I re displed y equl ses from the orresponding voltge sors,, nd ( 90 in the idel se of non-resistor se windings). Hene the urrent sor digrm eomes s the following. I I I I, or = I I I = 3I (8.6) I Exmple 8.1 I = I I I = 3I I I I I I = I I I I I = 0 = 90 = 3 90 I I = I I I I I = 120 = 30 = 30 3 I I = I I I I I = 240 = 150 = (8.7) str onneted set of unlned three se voltges is s follows: = = = lulte the mgnitudes of the line voltges. Solution j = = j = = = = 513. j141 The line voltges re = = j1628. = = j1752. = = 166. j338 So the mgnitudes re = =2075. = =3384. = =

4 8.3.1 lned od WYE onnetion 8.3 Three Phse lned ods I three se lod is sid to e lned if the lod impednes in eh of the three ses re the sme (oth in mgnitude nd in se). Y If oth the supply nd the lod re lned, the supply lod system is lled lned system. I I Y I Y Three se lods n e onneted in either wye or delt onnetions. = = = = = = = = 3 I = I = I = I = I I = I + I + I = 0 (8.8) lned od DET onnetion Y nd Y trnsformtions I I wye onneted network n e onverted to delt onneted network nd vie vers. These trnsformtions re useful for the nlysis of mny network prolems nd they re given elow. I I I = = = = I y equivlene it is ment tht in oth onnetions, the orresponding urrents entering eh node nd the orresponding voltge differenes etween nodes re equl in oth representtions. For the lned se, eh Y impedne is one third of eh impedne wheres for eh impedne is three times eh Y impedne. I = I = I = I = I (8.9) I = I = I = I = 3I

5 8.3.3 Single ine Equivlent iruits Due to the symmetry in the lned three se system, the network n e solved y solving only for the single line se. Y trnsformtion is needed, in the se of onnetion. I Y Trnsformtion Y Trnsformtion 0 Y = + + = + + = + + For the lned se + + = + + = + + = (8.10) I 0 Y = = = = 1 Y 3 = = = = 3 Y (8.11) Power in lned Three Phse etworks The power delivered y three se supply is the sum of power delivered y eh of the three ses. In the lned se it is three times the power in ny single se. So the totl verge power (in Wtts) is where I is the se voltge is the se urrent P =3 I osϕ (8.12) osϕ is the lod power ftor ϕ is the se differene etween nd I (positive for leding ). The totl retive power (in Rs) is given y Q= 3 I sinϕ = 3 I sinϕ (8.14) nd the totl omplex power is S = P+ jq (8.15) The totl pprent power (in s) is then S= S = P2+ Q2= 3 I = 3 I (8.16) In terms of line quntities Eqn. (8.12) eomes P= 3 I osϕ (8.13)

6 Exmple 8.2 lned 433, wye onneted, three se, four wire supply is onneted to the following four wire, wye onneted lod: se: 10 Ω resistor se: n 8 Ω resistor in series with 2 Ω indutive retne se: 4 Ω resistor in series with 5 Ω pitive retne. Determine the urrent in eh se of the lod nd in the neutrl wire. Solution onventionlly, it is ssumed tht 433 is the rms vlue of the line to line voltge of the supply system. Therefore, the mgnitude of the se voltge is = / 3= 433/ 3= 250 ssume tht the se sequene is nd tke the se s referene. So = 0, = 120, = 240 e n e n e n =10 Ω, So I = / = = j = = Ω, So I = / = = j = 4 5 = Ω, So I = / = = I I I = I + I + I = 25+ ( j) + ( j) = j = I I Exmple 8.3 three se, motor provides full lod mehnil output of 10 kw when its power ftor is 0.8 (lgging) nd its effiieny is 90 per ent. lulte t full lod (1) the power onsumed y the motor, (2) its line urrent, nd (3) the pprent nd retive power onsumed. Solution (1) The input power to the motor is (3) Sine the power ftor is 0.8, then ϕ = os 1 ( 08. ) = lgging. So pprent power: S= 3 I = = 139. k Retive power: S= 3 I sin ϕ = sin( ) = 833. kr P = output power / effiieny = 10/ 09. = 111. kw (2) The input power is equl to the verge power supplied, so P= 3 I I = P = osϕ. = 3 os ϕ