iL Sed ssh ina 1! ~LitounpghousE burr aa 7 (9883; SL! cis)Example 22.1. A star-conmected, 3-phase, 10-MVA, 6:6 kV alternator has a per phase reac- tance of 10%. It is protected by Merz-Price circulating-current principle which is set to operate for fault currents not less than 175.4. Calculate the value of earthing resistance to be provided in order to ensure that only 10% of the alternator winding remains unprotected. Solution. Let r ohms be the earthing resistance required to leave 10% of the winding unprotected (portion NA). The whole arrangement is shown in the simplified diagram of Fig. 22.9. 3 ‘Voltage per phate, Vy oe =3810V all 7 = —10x108 V3 X 6-6 X10 Let the reactance per phase be x ohms. . to = LxHx8I5 s 100 or x = 04362 Reactance of 10% winding = 0-436 x 0-1 = 0.0436 2 EME. induced in 10% winding = Vj4,x 0-1 =3810x0-1=381V Impedance offered to fault by 10% winding is Z, = J{(0-0436) +77 Earth-fault current due to 10% winding 381_ 381 2, ffo-o1s6F a? ‘When this fault current becomes 175 A, the relay will trip. s 17s = —— 381 __ y(0-0436) +7? 381) or (0-0436 +9? = 2) or (0.04367 +7? = 4-715 or r=2a7Q‘Example 22.2. 4 star-connected, 3-phase, 10 MVA, 6-6 kV alternator is protected by Merz- Price cireulating-current principle using 1000/5 amperes current transformers. The star point of the alternator is earthed through a resistance of 7-5. If the minimum operating current for the relay is 05 A, calculate the percentage of each phase of the stator winding which is unprotected against earth faults when the machine is operating at normal voltage. Solution, Let x % of the winding be unprotected. Farthing resistance, Q 6-6 x 10°/¥3 = 3810 V operate the relay 1000 50.5 — 100.8 EME. induced in.x% winding = V,,, x (x/100) = 3810 x (2/100) = 38-1 x volts Earth fault current which x% winding will cause 38:1x _ 38:1x ‘Voltage per phase, Van r 75 eres ‘This current must be equal to 100 A. a: or Unprotected winding 100% 7-5 _ e994 Hence 19-69% of alternator winding is left unprotected. Example 22.3, A 10 MVA, 6-6 k¥; 3-phase siar-connected alternator is protected by Mer2-Price circulating current system. If the ratio of the current transformers is 1000/5, the minimum operating current for the relay is 0-75 4 and the neutral point earthing resistance is 6 Q, calculate : © the percentage of each of the siator windings which is unprotected against earth faults ‘when the machine is operating at normal voltage. () the minimum resistance to provide protection for 90% of the stator winding. Solution. Fig. 22.10 shows the circuit diagram. Current . Current transformers pare transtormers — 7 TOSS \ ~\ R C OOOO } — 7865} fos Y CSSTTS of B a + Main r=62 s/t Switch Pilot + wes( Letx% of the winding be unprotected. Barthingresistance, r= 6 Voliage perphase, Tig = 66% 108/3 = 3810 volts Minimnam fait curent which will operate the relay 1000 9.75 = 150 EMF. induced in x% of stator winding = Py X (8/100) ~ 3810 X (x/100) ~ 38-1 x volts Barth fault current which x% winding will cause B8-1x _ 38-1x + 2 amy ‘This must be equal to 150A. 38-1 ” 150 3 or x = 23-696 (i Let rohms be the minimum earthing resistance required to provide protection for 90% of stator winding. Then 10% winding would be unprotected i.e. x= 10%. 38-1x 38-Lx _ 38-110 150 150 Example 22.4, A star-connected, 3-phase, 10 MVA, 6-6 kV alternator is protected by circulat- ‘ing current protection, the star point bemg earthed via a resistance r. Estimate the value of earthing resistor if 85% of the stator winding is protected against earth faults. Assume an earth fault setting 0f 20%. Neglect the impedance of the alternator winding. Solution, Since 85% winding is to be protected, 15% would be unprotected. Let r ohms be the earthing resistance required to leave 15% of the winding unprotected. 6 Fullsload current = 1010" Lgrg 4 V3 X66 X10" ‘Minimum fault current which will operate the relay. 20% of full-load current 20 = 722x876 = ico 154 ‘Voltage induced in 15% of winding 15 66x10" o 130 2540 '30V3 volts - jo Earth fault current which 15% winding will cause — 330V3 r This current must be equal to 175 A. irs = 33048 r or 7 = B08 L270 175