एस.आर.प.स 40व प र ट क शनउप सम त क ब ठकक मऱएएज ड Additional Agenda Points for 40 th meeting of Protection Sub-Committee of SRPC

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2 ANNEXURE I एस.आर.प.स 40व प र ट क शनउप सम त क ब ठकक मऱएएज ड Additional Agenda Points for 40 th meeting of Protection Sub-Committee of SRPC 1. ग र डडडस टबबन स / ग र डइन सडन टक वववरण - Details of Grid incidents: Sr. No. क र. स. 1 2 घटन क वववरण / Details of Event Tripping of Unit-1 at Kudankulam Tripping of 400kV Cudappah-Chitoor line र ख/ Date & स य /Time at 19:20 hrs at 10:03hrs क रण / Reason Y-phase Differential protection of Generating transformer of Unit-1 got operated which resulted in tripping of Unit-1 Line tripped at Cudappah end only टटप पण /Remarks श र ण / Category -- 3 Tripping of 220kV Sabarigiri-Theni line at hrs LBB protection got operated at Sabarigiri end. 2. Failure or non-operation of Auto reclose during transient fault Sr. No. Element Date & Time Reason 1 400kV Raichur-Munirabad line at 1:39 hrs Line tripped on R-phase to earth fault. Line did not auto reclose at Raichur

3 ANNEXURE II 1. Agenda Items from the Record Notes of the Special Meeting regarding the issues pertaining to Karnataka System and BMRCL Page 1 of 2

4 ANNEXURE II 2. PGCIL SR-1 Agenda Items Sub: Over Voltage grading of newly commissioned lines in SR This is in reference to the minutes of the 39th PCSC meeting forwarded. As recorded in the minutes, the Grading of Over Voltage settings need to be revised and recommended after the addition of newly added elements in Southern Region. Presently the settings recommended by PGCIL (SR-1) CC Engg group forwarded at the time of commissioning is implemented and put into service. Subsequent to the receipt of the settings recommended by SRLDC / SRPC / PCC groups, the same shall be adopted on all the feeders. Page 2 of 2

5 ANNEXURE III

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10 Suggestions for providing settings for High set of Earth fault relay on HV side of Generator Transformer (with Star connection on HV side and Delta connection on LV side): Jayarajan.C.N, Executive Engineer, Protection and Monitoring Division, Kerala State Electricity Board, Kerala, India It may be noted that the high set unit of Earth fault relay on HV side of Generator Transformer may un necessarily operate for external fault even though the primary relay corresponding to that fault had operated if the relay relay has not been set properly. For explaining this let us take a bus fault for analysis. A simple diagram is given below. Hope this will suffice for calculations. See the figure above. Here S represents the Fault MVA source, T is the Generator Transformer and G is the Generator. And let us take a line to ground fault occurs on one phase of 220KV bus. Now, we can assume the following values: Fault MVA source S: Here, 5000MVA has been taken as maximum fault level for calculation. MVA = 5000, Positive Sequence impedance (p.u) Xs1 = 1 Negative Sequence impedance(p.u) Xs2 =1 Zero sequence impedance (p.u) Xs0= 1 Per Unit values are based on 5000 MVA. Resistance values are neglected. Generator Transformer T: MVA = KV (Star)/ 11KV(Delta) 1

11 p.u reactance = 0.15 Solidly earthed. Generator G: MVA = 100 Positive Sequence Impedance(p.u.) Xg1 = 0.15 Negative Sequence Impedance(p.u) Xg2 = 0.15 Zero sequence Impedance (p.u) Xg0 = 0.05 Resistance earthed with resistance value = R. It may be noted that zero sequence impedance and resistance R will not come into calculations as the same will become clear from following diagrams. Base MVA can be taken as 100MVA, with this p.u. impedances of Fault MVA source can be rewritten as: Positive Sequence impedance (p.u) Xs1 = 1*100/5000 = 0.02 Negative Sequence impedance(p.u) Xs2 =1*100/5000 = 0.02 Zero sequence impedance (p.u) Xs0= 1*100/5000 = 0.02 Positive Sequence Network: POSTIVE SEQUENCE NETWORK 0.02 F p.u p.u. N1 The above diagram can be simplified using Thevenin's theorem: Equivalent resistance = 0.02*( )/( ) = The positive sequence network becomes: 2

12 F1 1 p.u N1 Similarly negative sequence networks becomes: F2 N2 ZERO SEQUENCE NETWORK 0.02 Fo R No 3

13 For zero sequence network, equivalent resistance = 0.02*0.14/( ) = Zero sequence network becomes: No The total circuit is shown as below: F1 1p.u. N F2 N F0 No I1 = I2 = I0 = 1/ ( ) = Out of this sequence currents, one portion flows through the GT neutral as shown below: 4

14 p.u. Fo No The per phase sequence current flowing through GT neutral = * 0.02/( ) = p.u. Total zero sequence current flowing through the transformer = 3*2.277 = 6.831p.u...(1) Now we can imagine that the generator is out of service but GT is in service as shown in the figure below: 220 kv T S G R Positive, negative and zero sequence networks are shown below: POSTIVE SEQUENCE NETWORK 5

15 0.02 F p.u p.u N NEGATIVE SEQUENCE NETWORK 0.02 F N2 ZERO SEQUENCE NETWORK 6

16 0.02 Fo R No And its simplified forms are as given below: POSITIVE SEQUENCE NETWOK p.u NEGATIVE SEQUENCE NETWORK 7

17 0.02 F2 N2 ZERO SEQUENCE NETWORK F0 N0 TOTAL CIRCUIT 8

18 0.02 1p.u Per phase sequence current = 1 / ( ) = p.u. The total zero sequence current through GT neutral = 3*17.39*0.02/( ) = 6.52 p.u....(2) It may be noted that the current flowing through GT neutral is almost same even though the generator is in service or not. Now let us see what happens when there is no source and 220KV bus is charged from generator (dead bus charging): S T G F R 9

19 Its positive, negative and zero sequence networks are as shown below: POSITIVE SEQUENCE NETWORK F1 1 p.u. N1 NEGATIVE SEQUENCE NETWORK F2 N2 ZERO SEQUENCE NETWORKS Fo R No 10

20 which can be simplified and connected to get total network as shown below: 0.29 F1 1p.u. N F2 N F0 No Therefore, per phase sequence current = 1/( ) = 1.39 p.u. Total zero sequence current through GT neutral = 3*1.39 = 4.16 p.u...(3) Conclusion: 1. While comparing the total fault current contributions through Generator Transformer neutral vide equations (1), (2) and (3), it can be seen that the condition (1), i.e., the situation where system, GT and generator are in service, give maximum current. Maximum fault current through GT neutral = p.u It may be noted that it is very difficult to get exact sequence network picture. Besides, it offset due to X/R due to actual system conditions to be considered while providing setting. Therefore, minimum setting shall be = 1.3*6.831 p.u = 8.88 p.u. = 9 p.u Please note that the above value may sometimes be changed to higher value for stability if instantaneous delay is provided for the highest relay operation, if necessary. However this value can be selected which may suffice for most of applications with 5 cycles delay. 11

21 2. In the absence of exact system configuration for analysis, settings have to be provided carefully. For avoiding maloperations due to DC offset in CTs, X/R in systems etc. 100 ms (5 cycles) may be given as delay. This delay is acceptable for the following reasons: a. The protection area of high set unit of earth fault relay is between CTs on HV side of GT and (including) transformer HV winding. b. This area is normally covered by Restricted Earth Fault protection and Transformer differential protection which are more sensitive than high set unit earth fault relay. c. For the above reasons in (a) and (b), the function of high set unit of earth fault relay in HV side of Generator transformer is to act as back up of HV side REF and transformer differential relay for faults on HV side of transformer. 3. Please note that this high set protection is effective only when the earth fault relay is located in the residual connection of HV side phase CTs and not on the neutral CT of the transformer. 12