POLYPHASE CIRCUITS. Introduction:

Transcription

1 POLYPHASE CIRCUITS Introduction: Three-phse systems re commonly used in genertion, trnsmission nd distribution of electric power. Power in three-phse system is constnt rther thn pulsting nd three-phse motors strt nd run much better thn single-phse motors. A three-phse system is genertor-lod pir in which the genertor produces three sinusoidl voltges of equl mplitude nd frequency but differing in phse by 0 from ech other. As shown below: () Schemtic lyout of genertor of - phse (b) Wveform of induced emf in R,Y,B coils R = m sin ωt Y = m sin (ωt-0 o ) B = m sin (ωt-40 o ) R = m 0 Y = m -0 o B = m 0 o OR (c) phsor Digrm Algebric or ector sum t ny time of ll emfs = R + Y + B = 0 Phse sequence: Order in which emfs or induced current ttin their pek vlue, i.e. here it is R Y B or sometimes lso specified s -b-c

2 Fig. A three-phse system is shown in Fig. In specil cse ll impednces re identicl = b = c = Such lod is clled blnced lod nd is described by equtions I I I b b c c. Using KCL, we hve I I I I n b c b c, where. 0 j j jsin 40 cos 40 jsin0 cos0 e e m m j40 j0 m c b Setting the bove result into (4), we obtin I n 0.

3 Since the current flowing through the fourth wire is zero, the wire cn be removed (see Fig.) I n b I b n c I c Fig. The system of connecting the voltge sources nd the lod brnches, s depicted in Fig., is clled the Y system or the str system. Point n is clled the neutrl point of the genertor nd point n is clled the neutrl point of the lod. Ech brnch of the genertor or lod is clled phse. The wires connecting the supply to the lod re clled the lines. In the Y-system shown in Fig. ech line current is equl to the corresponding phse current, wheres the line-to-line voltges ( or simply line voltges ) re not equl to the phse voltges. Methods of interconnection of three phse circuit. (i) Y- connection or Str connection (ii) Δ connection or Delt connection Y- connection:

4 Useful Definitions: (i) (ii) (iii) (iv) Phse voltge nd Phse current: oltge of ny one of phses w.r.t. neutrl or str point Line voltge nd Line current: oltge between ny two phse nd current in ny line Blnced system or lod: oltge in ll three phse re equl in mgnitude nd displced from ech other by equl ngle Unblnced system or lod: if one of two bove condition or both is not stisfied it is known s unblnced system of lod. Now we consider the Y-connected genertor sources (see Fig. ). n b b b c c c bc Fig. The phsors of the phse voltges cn be generlly written s follows me j j0. b e j40 c e We determine the line voltges b, bc, c ( see Fig.). Using KL, we obtin o o b b e jtn j j e j0. Thus, Holds nd similrly we obtin j b e 0o.

5 bc b j e 0o c c j e 0o. The phsor digrm showing the phse nd line voltges is shown in Fig.4. Fig.4 Thus, the line voltges b, bc, c form symmetricl set of phsors leding by 0 the set representing the phse voltges nd they re times greter. b bc c. Reltionship between line nd phse quntities in delt connection: (i) oltge: The voltge between two lines is L. As shown in fig. the phse coils re connected between lines, the mgnitude of phse voltge will be equl to mgnitude of line voltge L. ERY = EYB = EBR =L = Ph

6 (ii) Current: Blnced phse delt connected lod Phsor digrm Since the system is blnced the mgnitude of the phse currents i.e. IRY, IYB, IBR re equl but they re displyed 0 o from one nother s shown in figure. The currents re ssumed s lgging behind the respective phse voltges by n ngle of Φ. Similrly IR, IY nd IB re currents flowing through line R, Y nd B respectively. The vlues of line current cn be esily determined by pplying KCL t terminl R, Y nd B. ccording to KCL; IR = IRY - IBR = IRY + (-IBR) = (IRY + IBR + IRY * IBR cos 60 o ) But IRY = IYB = IBR = IPh Hence IR = IPh Thus IR = IY= IB= IPh Expression of Power: Apprent Power: For three phse power S = *Ph * IPh Here Ph=L nd IPh= (IL/ ) Hence S = * L * (IL/ ) = L IL Active power P: For three phse power P = *Ph * IPh* cos Φ = L IL cos Φ

7 Power mesurement: Three wttmeter method: This method is employed for mesurement of power in phse, 4 wire system. It cn lso be employed in phse wire delt connected lod, where power consumed by ech lod is required to be determined seprtely. Two wttmeter method: The connection digrm for the mesurement of power in three-phse circuit using two wttmeters is given in Fig. This is irrespective of the circuit connection str or delt. The circuit my be tken s unblnced one, blnced type being only specil cse. Plese note the connection of the two wttmeters. The current coils of the wttmeters, &, re in series with the two phses, R &B, with the pressure or voltge coils being connected cross Y R - nd Y B - respectively. Y is the third phse, in which no current coil is connected. Mesurement of power by Two wttmeter method:

8 One wttmeter method: The circuit digrm for mesuring power for blnced three-phse lod is shown in Fig. The only ssumption mde is tht, either the neutrl point on the lod or source side is vilble. The wttmeter mesures the power consumed for one phse only. The totl power is three times the bove reding, s the circuit is blnced. Mesurement of power by single wttmeter method: Comments on two wttmeter method: ) When the blnced lod is only resistive, i.e. power fctor is ( cos Φ = 0 ), the redings of the two wttmeters re equl nd positive. b) When power fctor is 0.5 lgging, one of the wttmeter W gives zero reding nd whole the power is mesured by wttmeter W. c) When power fctor is more thn 0.5 but less thn unity, both wttmeter reds positive, W reds more while W reds less thn W. d) When power fctor is less thn 0.5 but more thn zero, wttmeter W gives negtive wheres wttmeter W gives positive. e) When power fctor is zero, the redings of two wttmeters re equl in mgnitude but in opposite direction. Per-Phse Anlysis In blnced three-phse circuits the currents nd voltges in ech phse re equl in mgnitude nd displced from ech other by 0. This chrcteristic results in simplified procedure to nlyze blnced three-phse circuits. In this procedure it is necessry only to compute results in one phse nd subsequently predict results in the other phses by using the reltionship tht exists mong quntities in the other phses.

9 Exmple Three blnced three-phse lods re connected in prllel. Lod is Y-connected with n impednce of 50 + j50 / ; lod is -connected with n impednce of j600 / ; nd lod is ka t 0.6 pf leding. The lods re fed from distribution line with n impednce of + j4 /. The mgnitude of the line-to-neutrl voltge t the lod end of the line is 4.8 k. ) Clculte the totl complex power t the sending end of the line. b) Wht percent of the verge power t the sending end of the line is delivered to the lod? Solution The per-phse equivlent circuit is first constructed. For lod, Y-connected blnced lod, the perphse impednce is 50 + j50. For lod, -connected blnced lod, the per-phse impednce is the equivlent Y-connected lod which is / = 00 + j00. We will represent lod in terms of the complex power it bsorbs per-phse. This is given by S / j0.8 9,008 j5,44a The voltge cross the per-phse equivlents of these lods hs been specified s 4800 which is the lineto-neutrl voltge t the lod end. The per-phse equivlent circuit is shown below in Figure bellow: A Jj4 Iline + + I line-to-line n Jj50 Jj00 S / Nn Fig.: Per-Phse Equivlent Circuit

10 I ,008 j5,44 50 j50 00 j j j.7046a( rms) A( rms) j j5.8 In the bove step, the totl current lods. For lods &, we use the expression S I. I is obtined by summing the individul currents through the three I, nd for lod the current is determined using In the distribution line, P Q loss loss I I eff eff R X 8,58.04W ,4.4AR In ech lod, P 8.8 Q P Q 8.8 j9.6 j ,70W 50 8,40AR ( bs) j7.846 j ,507.0W 00 06,8.0AR ( bs) P 95,040 S Q totl ,5 57,04W 95, ,0AR j68,546 A ( lod end )

11 Check : S S totl sending j ,5 649, j68,546 8,58.04 j6,770.4a 6,5 j48,4.4a j68,546a % P 6,5 649, Exmple A blnced 0 volt (rms) three phse source is furnishing 6 ka t 0.8 pf lgging to two - connected prllel lods. One lod is purely resistive lod drwing kw. Determine the phse impednce of the second lod. Solution The totl complex power bsorbed by the lod is given by S j A Note it is specified in the problem tht cos 0.8 sin sin cos S 4980 j46. 58A Lod bsorbs S 000 j0a As result, Lod must bsorb S S 980 j46. 58A The power bsorbed by Lod per phse is S 980 j j5. 566A S / LL / Hence, 0.55 j j5.566

12 .55 j ************