Protection and Switchgear 5-23 Generator Protection 5.13 Negative Sequence Relays The negative relays are also called phase unbalance relays because these relays provide protection against negative sequence component of unbalanced currents existing due to unbalanced loads or phase-phase faults, The unbalanced currents are dangerous from generators and motors paint of view as these currents can cause overheating. Negative sequence relays are gene gene ly used to give protection to tors and motors against unbalanced currents. A negative sequence relay has a filter circuit which is operative only for negative sequence components. Low order of over current also can cause dangerous situations hence a negative sequence relay has low current settings. The earth relay provides protection for phase to earth fault but mot for phase to phase fault. A negative sequence relay provides protection against phase to phase faults which are responsible to produce negative sequence components. The Fig, 5.16 shows the schematic arrangement of negative phase sequence relay Fig. 5.16 Negative phase sequence relayProtection and Switchgear 5-24 Generator Protection Basically it consists of a resistance bridge network. The magnitudes of the impedances of all the branches of the network are equal. The impedances Z, and Zs and reactance. The currents in the branches Z; and Z, lag by 60° from the currents in the branches Z; and Z;. The vertical branch B-D consists of inverse time characteristics. relay, The relay has negligible impedance. The current I, gets divided into two equal parts [, and Is. And I, lags I, by 60° The phasor diagram is shown in the Fig. 5.17 Tat; = ip Let The perpendicular is drawn from point A on the diagonal meeting it at point B, as shown in the Fig. 5.17. This bisects the Fig. 5.17 diagonal, OB = Now in triangle OAB, OB cos 8 = OK BL 2 I b= Beheh Now I, leads Ip by 30° while I; lags Ip by 30° Similarly the current Ip gets divided into two equal parts 1, and 1, The current ly lags 1, by 60°. From equation (1) we can write, 1 ma hele @ 3 The current 1, leads Iz by 30° while current I; lags Ip by 30° The current entering the relay at the junction point B in the Fig. 5.16 is the vector sum of Ij, 1) and TyProtection and Switchgear 5-25 Generator Protection I 1 = Ty + -& (leads I, by 309) + —E (lags I, by ly a" 304 yg (aes Ie 304 The vector sum is shown in the Fig. 5.18 when the load is balanced and no negative sequence currents exist, hes ) C. T. secondary currents Fig. 5.18 It can be seen from the Fig. 5.18 that, ith itty hy =0 Thus the current entering the relay at point B is vero. Similarly the resultant current at junction D is also zero. Thus the relay is inoperative for a balanced system. Now consider that there is unbalanced load on generator or motor due to which negative sequence currents exist, The phase sequence of C.T. secondary currents is as shown in the Fig. 5.19 (a). The vector diagram of I, [, and ly is shown in the Fig. 5.19 (b) under this condition. The components I, and 1; are equal and opposite to each other at the junction point B. Hence |, and 1; cancel each other. Now the relay coil carries the current ly and when this current is more than a predetermined value, the relay trips closing the contacts of trip circuit which opens the circuit breaker.Protection and Switchgear 5-26 Generator Protection (a) €.7. secondary currents (b) Vector sum Fig. 5.19 Negative sequence currents Zero Sequence Currents : The zero sequence components of secondary currents are shown in the Fig. 5.20 (a). We know that, Fa by be le ly / { ty ly fa ty fa) f(b) fe) (@) Fig. 5.20 Zero sequence currents ip = Rely iy = Hel These sums are shown in the Fig. 5.20 (b) and (c), It can be seen from the Fig. 5.20 (d) that, Ty in phase with ly The total current through relay is +1; +17. Thus under zero sequence currents the total current of twice the zero sequence current flows through the relay, Hence the rates to open the circuit breakerProtection and Switchgear $-27 Generator Protection To make the relay sensitive to only negative sequence currents by making it inoperative under the influence of zero sequence currents is possible by connecting the current transformers in delta as shown in the F 5.21, Under del nsf no zero sequence current can flow in the connection of curre network, Fig. 5.21 Delta connection of C.T.s 5.13.1 Induction Type Negative Sequence Relay Another commonly used negative sequence relay is induction type. Its construction is similar to that of induction type over current relay. The schematic diagram of this type of relay is shown in the Fig. 5.22 Fig. 5.22 Induction type negative sequence relay The central limb of upper magnet carries the primary which has a centre tap. Due to this, the primary winding has three terminal 1, 2, and 3. The section 1 energized from the secondary of an auxiliary transformer to R-phase. The section 2 is directly energized from the Y-phase current The auxiliary transformer is a special device having an air gap in its magnetic circuit: With the help of this, the phase angle between its primary and secondary canProtection and Switchgear 5-28 Generator Protection be easily adjusted. In practice it is adjusted such that output current lags by 120° rather than usual 180° feom the input. So, Ix = Input current of auxiliary transformer Ij = Output current of auxiliary transformer and lags In by 120° Hene the relay primary carries the current which is phase difference of Ip; and 1, Positive Sequence Currents ‘The CT. secondary currents are shown in the Fig, 5.23 (a). The Fig. 5.23 (b) shows the position of vector fy; lagging Ix by 120° The Fig. 5.23 (c) shows the vector sum of Ip, and = ly. The phase difference of Ty, and ly is the vector sum of Igy and - ly. It can seen from the Fig, 5.23 (c) that the resultant is zero. Thus the relay primary current is zero and relay is inoperative for positive sequence currents. Is (a) 6.1. secondary currents (b} Current Ip, (0) Relay current Fig, 5.23 Positive sequence currents.Protection and Switchgear 5-29 Generator Protection @ shown in the egative sequence currents : The CT. secondary currents Fig. 5.24 {a). The Fig. 524 (b) shows the position of Fig. 5.24 (c) shows the vector difference of Igy and ly whi lagging I, by 120° The h is the relay current Under negative sequence currents, the vector difference of Igy and ly results into a current | as shown in the Fig. 5.24 (c). This current I flows through the primary coil of the relay (a) C.T. secondary currents (b) Current lay (¢) Relay current Fig, 5.24 Negative sequence currents Under the influence of current L the relay operates. The dise ro trip contacts and it opens the circuit breaker. tes to close the This relay is inoperative for zero phase sequence currents, But the relay can be made operative for the flow of zero sequence currents also by providing an additional winding on the central limb of the upper magnet of the relay. This winding is connected in the residual circuit of the three line C-T.s. This relay is called induction type negative and zero sequence felay.Protection and Switchgear 5-30 Generator Protection The schematic arrangement of induction type negative and zero sequence relay is shawn in the Fig, 5.25. Fig. 5.25 Induction type negative and zero sequence reli 5.14 Protection Against Unbalanced Loading When the load on the generator becomes unbalanced, negative phase sequence currents flow, The negative sequence components produce a rotating magnetic field which rotates at synchronous speed in a direction opposite to the direction of rotor ficld. Hence effectively the relative speed between the two is double the synchronous speed. Thus double frequency currents are induced. in the rotor, These currents cause severe heating of the rotor and can cause damage to the roter. The unbalanced stator currents also cause severe vibrations and heating of stator. Hence it is necessary to provide the negative sequence protection to the generators against the unbalanced load conditions The negative phase sequence filter alongwith the overcurrent relay provides the necessary protection against the unbalanced loads The relative asymmetry of a three phase generator is given by the ratio of negative sequence current to the rated current. Mathematically it can be expressed as, %S = «100 where % S = percentage asymmetry J, = Negative sequence current 1 = Rated currentProtection and Switchgear $-31 Generator Protection The negative sequence protection scheme is shown in the Fig. 5.26. Generwtor stator winding Roiny Fig. 5.26 Negative sequence protection A negative phase sequence filter is connected to the secondaries of the current transformers. A negative phase sequence filter consists of resistors and inductors. ‘These are so artanged that under normal operating conditions, the relay is inoperative. The filter cireuit is stable for the symmetrical overloads up to about three times the rated full load, When unbalanced load occurs, the negative phase sequence filter circuit produces an output proportional to the negative phase sequence components. This is directed through the relay coil. Hence the relay operates to open the circuit breaker to isolate the generator. Examples with Solutions imp Example 5.3: The neutral point of a 11 kV alternator is earthed through a resistance of 12 $2, the relay ts set to operate when there is out of balance current of 0.8 A. The CTs have a ratio of 2005. What percentage of the winding is pratected against carth faults. What must be the minimum value of eartiung resistance required to give 90% of protection to encis piuase ?Protection and Switchgear 5-32 Generator Protection Solution ; The given values are, Vo= kV OR 2 — CT. ratio = 2000/5 i, = relay current = 03 A L, = minimum operating line current (CT. primary) 2000. . 0.8» 2000 tg x20 9.80200 = 220A V = line to neutral volta; = My ar) = = 6350.8529 V . Ri, _ 12320 Winding unprotected *100 = Bea epg = 046% % Winding protected = 100 ~ 6046 = 39.53 % Thus with R= 12.Q only 39,53 % winding is protected. It is necessary to give 90% protection. % Winding unprotected = 100 - 90 = 10% Rely 10° 100 Rx 320 10 = 50.6530 "1? R = 1.98460 This is the minimum value of resistance to give 90% protection to the la B Pf machine, am Example $4: A 50 MVA, 3 phase, 33 LV synchronous generator is protected by the Merz-Price protection using 1000/5 ratio CTs. It is provided with restricted earth fault protection with the earthing resistance of 7.5. . Caleulate the percentage of winding unprotected in eack phase against earth faults if the minimum operating current of the relay is 0.5 A Solution : The given values are, Vi = RV C.T. ratio = 1000/5 R=750Protection and Switehgear 5-33 Genorator Protection i, = O5 A = relay current minimum operating current (primary) Vi, _ 33*10° 3 3 = 1905255 V RI % Winding unprotected = —[# x 100 Example 5.5: A 13.2 kV, 3 phase, 100 MW at 0.8 pf lag, alternator has reactance of 0.2 pus. If it is equipped with a circilating current differential protection set to operate at least at 500 A fault current, determine the magnitude of the neutral grounding resistance that leaves the 10% of the winding unprotected. Solution : The given values are, Vi = 12KV cos#=08 P=100MW X=02 pu Now P= V3V I, cos} 100«10* = ¥3x 13.2% 10" xf, « 08 I, = 5467.33 A = [= full load current The p.u, reactance is given by, go Xe I here X w cemckance per phat og = _SH67.33% where V « St (13.280 ) v3 v3 X = 0.2787 0 per phase % of unprotected winding = 10% Reactance of unprotected winding = {10 «0.2787Protection and Switchgear 5-34 Generator Protection = 002787: Voltage induced in 10% of unprotected winding, 10 _ 10, 13.2108 = Too" - 00" Bi = 762.1023 V Let this voltage be v = 762.1023 V ze J where Z = Impedance offered to the fault r = Resistance in neutral x = Reactance of 10% of winding Now z-t where v= Voltage induced in 10% winding = 762.1023 V i = Fault current = 500 A peavey _ 762.1023 300 = 1.5242 P+ (0.02787F = 2.3232 2.9224 a rs 15240 This is the required resistance in neutral earthing. ‘wb Example 5.6 : Ar alternator stator winding protected by a percentage differential relay is shown in the Fig, 5.27. The relay has 15% slope of characteristics (1, — Iy) against (1, + 1y2). The high resistance ground fault has occursed newr the grounded neutral end of the generator winding while the generator is carrying load. The currents flowing at each end of the generator winding are also shown. Assuming C.T. ratio to be 50/5 A, wit the relay operate to trip the circuit breaker ?Protection and Switchgear 5-35 Generator Protection 300+j0A 340+ )0A (ay ©) Fig. 5.27 Solution : From the given current at two ends, let us calculate C.T. secondary s at two ends, i, (300 +] 0) xg =3.4 344 (340 +40) x2 and iy 07 The directions of currents are shown in the Fig. 5.28. Fig. 5.28 The current flowing through the relay coil is iy ~ iz. = 3-34=-04A While = 3S toa2a From the characteristics of 15 % slope, corresponding, to +! = the out of balance current required is,Protection and Switchgear 5-36 Generator Protection ‘This is shown in the 15 Siope ~ ig. 5.29. ‘Thus iy ~ ip must be ly-ip more than 0.48 A ie. O48 above the line to operate the relay but actual point is located below the line in negative torque region. Hence the relay will not operate. Example 5.7: A 6.6 KV, star connected alternator has a transient reactance of 2.0 per phase and negligible winding resistance. It is protected by circulating current Merz-Price protection, The alternator neutral is earthed through the resistance of 7.5 Q. The relays are set fa operate when there is out of balance current of 1 A in secondary of SO0/S A current transformers. How much % of winding is protected against earth fault? Solution : The given values, Vi, = 66kV¥ X=20/phase r=752 CT. = 500/5 Let the x % of winding is unprotected. Reactance of unprotected windi iw0* 25002x2 Vy _ 6.6%10° vy = b= = 3810511 V .. full voltay 3 ee ¥ = Voltage across unprotected winding x = jgg* 3810-511 = 38.2051 x V r= 750 Z = Impedance offered to the fault = r+j 02%) = 75 +j @02x) Oo [Z| = y7.5)° +(0.02x)* fault current = out of balance secondary current « C.T. ratioProtection and Switchgear 5-37 Generator Protection 500 = Ix = 100A ¥ IZ] 7 I> oe? apa? 38.10511x (75? +O.02x)? = = (757 + (0.02 x)? = 0.1452 <7 56.25 + 4x10? = 0.14527 0.1448 P= 56.25 x? = 388.4668 x = 19.7% This is % of winding unprotected. 4% of winding protected = 100 - 19.7 = 80.29 % ‘mm Example 5.8: A synchronous generator rated at 20 KV protected by circulating current system having neutral grounded through a resistance of 15 0. The differential protection relay is set to opertte then there is an out of balance current of 3 A. The CTs have ratio of 1000/5 A, Determine, i} The % of winding remains unprotected ti) Value of earth resistance to achieve 75% protection of winding Solution : The given values are, Vi = 20kV, 3A, R= 159, CT. ratio = 1000/5 i I, = minimum line operating current (C.T. primary) = j, «1000 , 3*1000 = bang = 600A y = Ye. 20410" v3 = s7¥ % X = % of winding unprotectedProtection and Switchgear 5-38 Generator Protection = 794% ii) We want 75% protection. % X = 100~75 = 25% Remaining conditions are same except R. RI 2 = 100 V R600 28 = yey «100 25% 11547 RS Spx 100 = 4811 Tl jamb Example 5.9: The Fig. 530 shows the percentage differential relay used for the protection of an alternator winding, The relay has minimum pickup current of 0.25 A and has a % slope of 10%. A high resistance ground fault occurs wear the grounded neutral end of the generator winding with the current distribution as shown in the Fig, 5.30. Assinte a C.T. ratio of 400 : 5, determine if relay will operate. s is the required earth resistance. 3609 )0 A e840 +j0 A RF Foun Fig. 5.30 Solution : For the given current at the two ends, the CT. secondary currents can be obtained using C.7. ratio as, (380 + j 0) x5 = 475A 5 (340 + j 0) qo PA 4.75 -4.25=05A4 _ 4.7544.25 2 = 454 So i: — iy current flows through operating coil while 45 A flows through the restraining coilitchgear 5-39 Generator Protection With the minimum pickup current of 0.25 A, and slope 10° as shown in the Fig. 5.31. the operating characteristics ve torque Minimum pickup current ‘The equation of the characteristics mxte At origin, x = 0 but y = 0.25 y= c= 025 Hence characte: ic equation is, where m = slope = 0.1 For 4.5 we get, (O.1x4.5)+0.25 = O74 Thus to operate the relay, i, - i, must be greater than 0.7 A when 2742 is 4.5 A. But actually i, - iz = 0.5 A as shown. It is located below the line in negative torque region hence relay will not operate, iim Example 8.10: An alternator rated 10 KV protected ly balanced circulating current system has its neutral grounded Hiraugh a resistance of 10 ohms. The protective relay is set fo operate when there is an out of balance current of 1.8 amperes in the pilot wires which are connected fo the secondary of current transformers with ratio 1000/5.Protection and Switchgear 5-40 Generator Protection Determine : i) The percentage winding which remains unprotected. ii) The minimum value of the earthing resistance required to protect 80% of the winding, (V.T.U August-2002) Solution : V, = 10 kV, R = 10.0, C-T. ratio = 1000/5, j, = 1.8 A 1, = minimum operating line current (C.-T. primary) 1000 1000 = ign Eo = BK = 360A fs Vi _ 10x10° V = line to neutral voltage = —L = = 57735026 V eee ‘ is e199 = 10360. i) % Winding unprotected = “72100 = FEES x 100 = 62.3538 % ii) It is necessary to give 80 % protection. = % Winding unprotected = 100 - 80 = 20 % Rx 360 20 = 5795-5026 "100 R = 320750 _ Minimum earthing resistance required ‘me Example 5.11: An allernator stator winding protected by a percentage differential relay is shown in Fig. 5.324). The relay has 0.15 amp minimuent pick up and a 12% i, +i, 2 near the grounded neutral end of the generator winding while generator is carrying load. The currents flowing at each end of the generator winding are shown in Fig, 5.32, Assuming that the CT's have 400/5 amps ratio and no inaccuracies will the relay trip the generator CB under this fault condition. slope of characteristics liz = is) V, ( } A high resistance ground fault has occured Fig. 5.32(a)Protection and Switchge: 5-41 Generator Protection Positive torque “a— Slope 12% iy = Fig. 5.32(b) (AU-April-2004) Solution : From the given current at two ends, let us calculate C:T. secondary curents at two ends, = (300+ j0}x B75 A 400 iy = (300+ j0)x 45A ‘The relay coil current = i, - iy = 0.75 A p+ ni 2 = 41254 : From the characteristics of 12 % slope, corresponding to | a) » the out of Uz balance current required is, i) - i, = slope +f 4 = )=0.12 «4.125 = 0.495 A if } Thus iy — i, must be more thatn 0.495 A for relay to operate. And actually it is 075 A. Hence the relay will operate : Aijnig) 4 18-18 95 12 % slope O.75A 0.495AProtection and Switchgear Generator Protection Actual point is located in positive torque region. Slope of line through actual point is = fick). (izia) 2 18.18 % As this slope is more than 12% of the characteristics given, relay will operate. wap Example §.12 : A 3 phase, 10 MVA, 6.6 kV alternator supplies a load of 8 MVA at O8 pf. and is being protected through Merz-price circulating current system and its relays are so set that they do not operate until the out of balance current ocurs at 20% of full lood current. Calculate the value of earth resistance to be provided in order to ensure that only 10% of alternator winding remains unprotected. Assume alternator reactance of 10% Neglect resistance of the alternator. (AU-Dec.-2004) Solution : V, = 66 kV, Rating = 10 MVA, V = V_/J3 = 38105117 V Rating = ¥3 Vid, 2 1, = lx - ami a load current V3 x 6.6% 105 % Reactance = 7 x 100 X = Reactance per phase _ S74.7731X 10 = ssrosii7 “1 X = 043560 Reactance of unprotected winding = x 0.s356 = 0.09356 2 v = Voltage induced in unprotected winding 10 = jqp* ¥ = 38105117 V i = Fault current = 20% of I, = 174.954 A impedance offered to fault = 2.1780 But Zz A+ j 0.08356. |Z] = JR? +(0.04356)? (2.178)? = R? + (0.04356)? R= 21770 Earth resistance requiredProtection and Switchgear 5-43 Generator Protection ‘tm Example 5.13: Current transformers of current ratio of 1000/5 A are used for protection of a star connected 3 phase, 10 MVA, 6.6 RV alternator. if the relay is set to operates for @ minimum current of 0.5 A, Calestlate the percentage of each phase stator winding which is uprotected agains earth fault when the machine operates at normal woltage. Asstume that star point of alternator is earthed tirougit a resistance of 7.5.Q. {AU-Apri Solution : V, = 6.6 10° V, R= 752, CI. ratio = 1000/5, i, = 0.5 I, = ig x CIT. ratio = 0.5 xe = 100A Vv yee 100 = sere = 100 = 19.682% % winding unprotected = ti Example 5.14: A 500 kVA, 6.6 kV star connected alternator has a syncironous reaclance of 1.02. per phase and negligible resistance. The differential relay operates if the out of balance current through it exceeds 30% ofthe normal full load current of the alternator. The star point of the alternator is earthed through a resistance of 5 © . What percent of the stator twinding is left unprotected ? Show that the effect of the alternator reactance cm be neglected. {AU-Dec,-2005) Solution : The full load current, Va 500 «10° Le ate V3V, V3x66x10° Out of balance curent = 30% I, = 0.3 x 43.738 = 13.1214 A Let winding unprotected = x % 43.738 A Impedance of x% winding = Fog OH) Value of earthing resistance = 5 Total impedance at fault of the fault circuit. = [5 +j001 x) Voltage induced in x % of winding 6.6103 Voltage induced Out of balance current = ImpedanceProtection and Switchgear 5-44 ‘Generator Protection 13.1214 = (1) This is to be solved by trial and error method. Hence effect of alternator reactance can be neglected. 66x 13.1214 = —= V3 x = 17217 % ...Winding unprotetced The equation (1) can be solved as, 8 = +» Considering magnitude, 25+(0.01x)? = 8.4334 x? x L7217% This shows that the reactance can be neglected without any error. w Questions 1. Which are the narions types of frufts ahich can accur in a generator ? Explain: in brief 2. Why the pro 3, Wich are the various abnormal rinning conditions, wluich may exist in a generator ? What are their effects and how these effects cin Be minimized ? 4. Explain the basic differential protection scheme. What are its disadvantages ? tes Explain the basic percentage differential protection scheme. Draw its operating charac showing positive and negative torque regions. 6. Draw and explain the Merz-Price protection of alternatar stator windings. State its advontages. 7. What i the role of auxiliary relay, én Mera-Price protection: ? 8. Explain the resteictod carth fault protection: of generators. 9. Derive the expression for the percentage of winding unprotected in the restricted earth fault protection. 10. Explain the operation of unrestricted earth fault protection scheme. 11. Draw anit explain balanced earth fanit protection scheme, 12. Is it possible that 100 % winding of generatar is protected against earth faults ? How ? 13. Suggest the scheme for interturn fut protection for Stator of alternator.Protection and Switchgear 5-45 Generator Protection 14, 15, 16. i”, 18, 20. 1. Explain the negative phase sequence protection for the generators. What are the methods to provide rotor earth fault protection ? How the protection against loss of excitation is provided in generators ? Why it is important ” A. generator is provided with restricted carth-fault protection. The ratings are 11 kV, 5000 RVA The percentage of winding protected against phase to ground fault # 80%, The relay setting such that i trips for 25% ont of balance. Calewlate the resistance to We added ir neucteat to ground connection tAns, 1.94.0) The neutral point of a 19,000 V alternator is earthed through a resistance of 10 olkms, the relay is set to operate when there is an out of balance current of 1A. The CT's have a ratio of 1000/5. What percentage of the winding i protected against fault to earth and what must be mininun ulus of earthing resistance to give 90% protection to each phase winding ? (Ans, : 625%, 2.88.0) A 3 pase, 2 pole, 11 RV, 10,000 AVA olternater hes neutral exrthed through a resistance of 7 kis. The machine has current balance protection which operates upon out of balance current ed 20% of full toad. Determine % of winding protected against earth fauit. (Ans. :88.4%) 5.34{a) shows percentage differentiat relay applied to the protection of an alternator winding, The relay es 10% slope of characteristics I~ 1) 05 (Ty + Ip. A height resistance ground fault occurred near the grounded neutral ead of the generator winding wile generator is anrrying load. As a consequence, the currents in amperes flowing at cach cra of the winding are shorn in Fig. 5.34(b). Assuming C.T. ratio of 400/5 amperes, will the relay operate to trip the breaker. (Ans. : Relay will not operate) 320+ 10 Fig. 5.34(b)Protection and Switchge: 5-46 Generator Protection a. 23 ATL RV, 3 phase alternator has full load rated current of 200 A. Renctance of armature winding is 5 percent, The differential protection system is set to opennte on earth finult currents of more than 200 A. Find the neutral earthing resistance, which gives earth fult protection to 90% of stator winding (Ans, : 3.145 0) A starsconnected S-phase 10 MVA, 66 kV alternator has a per phase reactance of 10%, It is protected by Merz-Price circalating-current principle which is set to operate for fault currents not less than 175 A, Calculate the caine of earthing resistance to be procided in onder to ensure that nly 10% of the alternator winding remains unprotected. (Ans, : 2.1710) A star connected, 3 phase, 10 MVA, 6,6 KV alternator és protected by Merz-Price circulating current principle using 1000/5 anaperes: current transformers. The star point of the alternator is varthed through a resistance of 7.5.0 . If the minimum operating current for the relay is 0.5 A, caleylate the percentage of each phase of the stator winding which is unprotected against earth {faults token the mactine is operating at normal voltage. (Ans. : 19.69%) A 6600 volt J-phase turbonalterniator has a niarininn continuous ating of 2.000 KW at 0.8 py and its, reactance is 12.5%. 1 is equipped with Mers-Price circuliting current protection which is set fo operate at fault currents not tess that 200 amperes. Find rohat value of the newtrel earthing resistance leaves 10% of the windings unprotected ? (Ans. = 1.89.9) ASO MVA, S-phase, 33 RV alternator is being protected by the use of circulating current balance scheme tsing 2000/5 ampere current transformer, The neutral of the generator is carted through 3 NGR of 7.5 ohms, If the pick wp current for the relay is just above 0.5 ampere, determine what percentage of the winding of each phase unprotected against earth then the machine eperties at nominal voltage. (Ans, : 7.88%) goa