ANNEXURE I
एस.आर.पी.सी40वें प्रोटे क्शनउप–समितिकीबैठककेमऱएएजेंडे
Additional Agenda Points for 40thmeeting of Protection Sub-Committee of
SRPC
1. ग्रिडडडस्टबबन्स / ग्रिडइन्सडन्टकावववरण - Details of Grid incidents:
Sr.
No.
क्र.
सं.
घटनाकावववरण / Details
of Event
िारीख/ Date
&
सिय /Time
Tripping of Unit-1 at
Kudankulam
14-01-2014
at 19:20
hrs
2
Tripping of 400kV
Cudappah-Chitoor line
14-01-2014
at 10:03hrs
3
Tripping of 220kV
Sabarigiri-Theni line
15-01-2015
at hrs
1
टटप्पणी
/Remarks
कारण / 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
LBB protection got operated at
Sabarigiri end.
2. Failure or non-operation of Auto reclose during transient fault
Sr.
No.
1
Element
Date &
Time
400kV Raichur-Munirabad line
18-01-2015 at
1:39 hrs
Reason
Line tripped on R-phase to earth
fault. Line did not auto reclose at
Raichur
श्रेणी/
Category
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ANNEXURE
1. Agenda Items from the Record Notes of the Special Meeting regarding
the issues pertaining to Karnataka System and BMRCL- 16-01-2015
Page 1 of 2
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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
ANNEXURE III
ANNEXURE III
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ANNEXURE III
2
ANNEXURE III
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ANNEXURE III
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ANNEXURE-IV
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 = 100
220KV (Star)/ 11KV(Delta)
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ANNEXURE-IV
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
F1
0.14
0.15
1 p.u.
1 p.u.
N1
The above diagram can be simplified using Thevenin's theorem:
Equivalent resistance = 0.02*(0.14+0.15)/(0.02+0.14+0.15) = 0.0187
The positive sequence network becomes:
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ANNEXURE-IV
0.0187
F1
1 p.u
N1
Similarly negative sequence networks becomes:
F2
0.0187
N2
ZERO SEQUENCE NETWORK
0.02
0.14
Fo
0.05
3R
No
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ANNEXURE-IV
For zero sequence network, equivalent resistance = 0.02*0.14/(0.02+0.14) = 0.0175
Zero sequence network becomes:
0.0175
.
No
The total circuit is shown as below:
0.0187
F1
1p.u.
N1
0.0187
F2
N2
0.0175
F0
No
I1 = I2 = I0 = 1/ (0.0187+0.0187+0.0175) = 18.215
Out of this sequence currents, one portion flows through the GT neutral as shown below:
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ANNEXURE-IV
18.215 p.u.
Fo
0.02
0.14
No
The per phase sequence current flowing through GT neutral = 18.215 * 0.02/(0.02+0.14) = 2.277
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
SS
G
R
Positive, negative and zero sequence networks are shown below:
POSTIVE SEQUENCE NETWORK
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ANNEXURE-IV
0.14
0.02
F1
1 p.u
0.15
1 p.u
N
NEGATIVE SEQUENCE NETWORK
0.02
0.14
F2
0.15
N2
ZERO SEQUENCE NETWORK
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ANNEXURE-IV
0.02
0.14
Fo
0.05
3R
No
And its simplified forms are as given below:
POSITIVE SEQUENCE NETWOK
0.02
1 p.u
NEGATIVE SEQUENCE NETWORK
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ANNEXURE-IV
F2
0.02
N2
ZERO SEQUENCE NETWORK
F0
0.0175
N0
TOTAL CIRCUIT
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ANNEXURE-IV
0.02
1p.u.
0.02
0.0175
Per phase sequence current = 1 / (0.02+0.02+0.0175) = 17.39 p.u.
The total zero sequence current through GT neutral = 3*17.39*0.02/(0.02+0.14) = 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):
T
G
SS
F
9
R
ANNEXURE-IV
Its positive, negative and zero sequence networks are as shown below:
POSITIVE SEQUENCE NETWORK
F1
0.14
0.15
1 p.u.
N1
NEGATIVE SEQUENCE NETWORK
F2
0.15
0.14
N2
ZERO SEQUENCE NETWORKS
Fo
0.05
0.14
3R
No
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ANNEXURE-IV
which can be simplified and connected to get total network as shown below:
0.29
F1
1p.u.
N1
0.29
F2
N2
0.14
F0
No
Therefore, per phase sequence current = 1/(0.29+0.29+0.14) = 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 = 6.831 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.
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ANNEXURE-IV
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.
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