This document provides an overview of advanced power system protection topics including differential protection, busbar protection, linear couplers, and pilot wire protection. It discusses the principles and applications of differential protection including Merz-Price and balanced voltage schemes. It also covers special considerations for differential protection such as phase shift, tap changing transformers, and inrush current. The document then summarizes busbar protection and how it uses a pure earth fault system to measure fault current flowing from the switchgear to earth. Finally, it examines linear couplers and how they are used in differential protection systems as well as the performance of pilot wire protection schemes that can employ either balanced voltage or circulating current principles.

1. ECNG 6503 -3
Advanced Power Systems Practice
Lecturer
Prof Chandrabhan Sharma
University of the West Indies
Trinidad and Tobago

2. Topics:
1 Differential Protection
2 Busbar Protection
3 Linear Couplers
4 Pilot wire Protection

3. DIFFERENTIAL PROTECTION
(Merz-Price Systems)
(Unit Protection)
3 Categories of Protection:
• Input and Output Terminals are close
→ transformer
• Where ends are Geographically Separate
→ overhead/underground cables
→ pilots at supply frequency used
• Very long lines P.L.C.
→ end → end signalling achieved
by modulated carrier waveform.
Main difference is in way signals derived & transmitted.

4. Why needed?:-
Overcomes application difficulties of simple O.C. Relays
when applied to complex networks e.g. Co-ordination
problems and excessive fault clearance times.
Principle:-
Measurement of current at each end of feeder and
transmission of information between each end of feeder.
Protection should operate for fault inside protected Zone only
→Stable for others.
i.e. Instantaneous Operation Possible!

5. MERZ-PRICE SYSTEM
Circulating Current Mode
- Current confined to series path. No currents in relay
- Both breakers tripped..

6. BALANCED VOLTAGE SCHEME
When C.T. secondaries are short, can be applied to transformer,
busbars, machines etc.
Where boundaries are at distance, communication channel required.

7. SIMPLE LOCAL DIFFERENTIAL PROTECTION

8. UNIT PROTECTION INVOLVING DISTANCE BETWEEN
CIRCUIT BREAKERS

9. EARLY MERZ-PRICE BALANCED VOLTAGE
SYSTEMS FOR FEEDERS
Two Problems:
1. Maloperation due to unequal open circuit secondary voltages
of the two transformers for thru fault currents.
2. High Output voltages of CT’s cause capacitance currents to
flow through relay since capacitive currents are proportional to
pilot length, relay insensitive for all but very short lines.

10. ACCURACY OF C.T.s
- Accuracy of relaying signals (balanced) for faults
external to protected zone.
- Depends on characteristics of C.T. and burdens
- Unequal magnetic saturation in 2 C.T.s due to unequal
lead burdens
- Need to define range of unbalance conditions for
which protection is stable and inoperative
- Use of complex ratio of IA/N and IB/N

11. Ideally, under conditions of exact balance, the point (1.0 +j 0.0)
would be the conditions for through faults in relay signal ratio plane
NOTE: All unbalance defined w.r.t. (1.0 +j 0.0)
Zone of uncertainty is area enclosed by circle.

12. STABILITY MARGINS
Relay Setting corresponds to minimum value of current for
which relay operates.
OR
Related to maximum out-of-balance current for which the
protection (relay) is required to remain stable.
There are two methods to achieve stability margins:
1. Low-impedance relay and a current setting that would give
the required thru-fault stability
2. High-impedance relay having an inherent low current setting.

13. By insertion of stabilising resistor (Rs) the stability margin may be
obtained.

14. SATURATION OF ONE C.T. (Equivalent cct)

15. Where: R1, R2 → lead resistances
RCA, RCB → core less
For Transformer B saturated than:
Tx. B does not contribute to relay current
If RS ›› R2 + RB
Neglecting leakage reactance XB, then
∴RS is chosen s.t. the largest value of IA will not cause relay
operation.

16. For Faults within Zone:
Let IR0 be the relay minimum current
ImA, ImB is magnetising current of transformer A and B
∵ Minimum fault current referred to primary is
Ifs = N(IRO + IMA + IMB)
i.e. for IRO = 0.8 A
ImA = ImB = 0.1 A
N = 500:1
∵ Ifs = 500(0.8 + 0.1 + 0.1)
= 500 A (min. fault current at which protection operates)

17. Also: if protection stable under thru fault condition.
∵ for fault if 16kA, with one transformer saturated
→ RB=0.3 Ω; R1= R2=2.2 Ω

18. THROUGH FAULT
FIG 6. SINGLE-PHASE DIFFERENTIAL LOOP

19. FIG 7. BIASED DIFFERENTIAL RELAY

20. BIASED DIFFERENTIAL RELAY
CT2 provides phasor addition of currents circulating
Let SO= differential current required to operate relay
and
SB = bias
then % bias = B % = SO/SB x 100

21. WITH TRANSFORMER B SATURATED
FIG 8(a) EQUIVALENT CIRCUIT OF BIASED DIFFERENTIAL RELAY
(Transformer B saturated)

22. Neglecting impedance of biased cct and leakage reactance of
saturated transformer.

23. N.B.
A specified value of fractional (%) bias is set by obtaining/ proportioning the
number of turns on the operate and bias windings.

24. % WINDING PROTECTED

25. CIRCULATING CURRENT PROTECTION
For generated voltage V ∴ x = 1- Is/(V/R)
and for a fault position x
Let setting current of
relay be Is
∴ x = 1 - Is/IR
Where x = % winding protected
and (1-x) = % winding not protected

26. SPECIAL CONSIDERATIONS
• Phase Shift between primary and secondary
• Winding turn ratios and effect of tap changers
• Different Zero – Phase – Sequence circuits depending on
connection
• Magnetising Inrush
Derived signals must take these three concerns into
account when Differential Relays are applied.

27. ANALYSIS OF THE MERZ-PRICE UNIT PROTECTION
Relay operates when IA≠ IB
but IC = IA- IB ……… (1)
∴ Relay operates when
Let Nr = kNo where o<k<1 ………. (3)
Sub into (2) equations (1) and (3)
∴ From which for threshold we have:

28. Question:
A 1 100kVA 2400/240V transformer is to be differentially protected.
Choose the C.T. ratios. Determine the ratio Nr/No if the relay is to tolerate
a mismatch in current of up to 20% of primary current (bias).

29. Solution:
Rated Primary Current of T/F : = 100/2.4 = 41.7 A
∴ Rated secondary = 41.7 x 10 = 417 A
For C.T. Secondary = 1A
Primary C.T. ratio = 500:5
Secondary C.T. ratio = 5000:5
For 20% mismatch (bias) → Reduction in Sensitivity
→ IA = 0.8 IB (At threshold)
or 2 - k = 1.6 + 0.8k k = 0.222

30. OVERALL DIFFERENTIAL PROTECTION OF STAR/DELTA
POWER TRANSFORMER

31. Must Consider: * Magnetising inrush
• Phase shift between primary and secondary
• Turns ratio
• Zero sequence cct.
These can be compensated for by C.T. Connection
For the above:

32. TAP CHANGING

33. This can be represented by Figure (b)
Given a transformer with max. off-nominal position 20%
And fault current = 10 (rated)
Iob = 2.5 (Rating)
= normal

35. MAGNETISING INRUSH
• Harmonic Restraint → 2nd Harmonic to restraint winding
• Application of time delay (usually 0.2s)
• Desensitizing relay
• Mismatch of C.T.’s

36. Busbar Protection

37. Busbar Protection
• Purely Earth Fault System.
• Simply a measurement of fault current flowing from switchgear to
earth.
• No other earth connection should be present.
• Switchgear must be isolated as a whole i.e. standing on
concrete (grout).

38. • Resistance earth, connection of frame is as shown
•Ƶ electrode < 1Ω ; Ƶframe ≃ 10Ω
I1 ≃ 10% [I1+ I2] = 0.1 IF ∴ Setting is about ≃ Is=30% IF(normally)
• Foundations bolts should not touch steel reinforcement
• Grouting gives adequate insulation

39. BUSBAR DIFFERENTIAL SCHEME
3
• Two current balance loops.
• Plus check loop → safeguard against incorrect tripping due
to faults in differential system.
• Provides for both phase and earth faults.

40. Summation Schemes By using different C.T. ratios, can differentiate between faults.
Phase and earth fault
circulating current
scheme using single
element relay and
current transformers
with different ratios in
each phase

41. LINEAR COUPLERS - (L.C.)
- Use of magnetic C.T. lead to errors in
reproduction when fault at/near saturation
- D.C. saturation even greater problem
Solution:
- Linear Couplers (no iron core)
- Multi-restraint, variable percentage differential
relay
- High impedance, voltage operated differential
relay with series resonant cct. To limit
sensitivity to dc.

42. L.C. DIFFERENTIAL SYSTEM
Of the 3 systems used (before) L.C. are:
- Fastest (operation)
- Easiest to apply, set and maintain.
- L.C. are air core reactors wound on non-
magnetic toroidal cores
For L.C.
Esec = M Ipri (volts)
Where Esec – induced secondary voltage
M - mutual reactance of L.C.
Ipri - primary current in cct.
** Since L.C. has negligible response to DC, only s.s.
conditions need be considered.

43. OPERATION (Busbar Protection)

44. ∴ From the cct. given before
Where: IR = current in L.C. secondary and relay
Esec = voltage induced L.C. secondary and relay
Ipri = primary current in each cct.
M = mutual reactance
ƵC = self-impedance of L.C. secondary
ƵR = relay impedance
N.B. For max. energy transfer, relay tap ≃ total self-impedance of all LC’s
in cct.

45. Question:
A section of a 230kV bus has 6 circuits (i.e. 3 Tx lines, 2 T/Fs to the 115 kV
system and 1 bus tie). The max. external fault is 23,800 A r.m.s and the
min. internal fault is 1800 A r.m.s. Six linear couplers are to be used in a
differential system to protect the bus. The recommended application range
between max. external fault and min. internal fault is 25:1. The mutual
reactance of each linear coupler is 0.005Ω. Determine the setting of the
relay given its recommended range is 15 mA ∼ 28 mA. The impedances of
each linear coupler cct are:
Transformer #1 = (9.8 + j10.75) Ω
Transformer #2 = (3.9 + j6.6) Ω
Line A = (4.3 + j7.2) Ω ; Line B = (4.3 + j7.5) Ω
Line C = (3.0 + j3.6) Ω ; bus tie = (10.7+ j12.7) Ω

47. PERFORMANCE OF PILOT WIRE
PROTECTION SCHEMES
As any Differential Scheme, pilot wire may employ either of the
following principles:-
(a) Balanced Voltage
(b) Circulating Current
(a) Balanced Voltage (longitudinal differential protection)
A Voltage Driven Scheme

48. Also Vs= Vin ; Is = Iin + I1 = Yc Vin + Iin

49. Similarly for Vo = VR, Io = IR + I2 = (Yc VR+ IR)
Sub (1) & (3) into (2)

50. (b) Circulating Current (transverse differential protection)
(Current Driven Scheme)
Note the interchange of the op. and res. windings.
Applications:- (i) Parallel lines
(ii)Split winding generators
i.e. areas where current usually equal.

51. BALANCED-VOLTAGE PILOT WIRE
PROTECTION (FEEDER)
The operating and restraint ccts. take the form of an amplitude
comparator.
For operation:
Effective operating Ampere turns > Effective Restraint Ampere turns
For threshold conditions (Receiving End)
k1IR > k2VR Yr
where Yr = admittance of restraint cct.
(neglecting impedance of operating cct.)

52. Assuming Linearity of current/voltage relationship
IR VR
∴ IR =VRYPR
Where
YPR = admittance presented by pilot cct to the receiving end relay
∴ k1 VRYPR > k2VRYr
or YPR > k Yr for k = k2/ k1
i.e.
Operating and restraint ccts for a 2-input comparator for which
tripping occurs when the admittance of the pilot cct. exceeds k
times the admittance of the restraint cct.

53. Criterion for threshold:
The centre of the stability circle should be the point 1 0 on the
Vs/VR plane. In practical pilots, transference of the admittance
characteristic to the IA/IB plane necessitates compensation to
present zero admittance for Vs/VR= 1 0

54. PRACTICAL PILOT WIRE SCHEMES
Comp - Compensation cct.
Yc - admittance of compensation cct.
Ideally the cct is tuned so that the overall compensated
pilot cct. (Yp+c) presents zero admittance to the relay for
condition Vs/VR=1 0.

55. EVALUATION OF REQUIRED COMPENSATION(Yc)
- Assume a symmetrical pilot with network parameters
A,B,C and D
- Let A’, B’, C’ and D’ be the parameters for the
compensated network
- Neglect impedance of series op. cct and given that Yc
is admittance of compensating network at either end.
where A’ = A + BYc ; B’ = B
C’ = AYc + C + Yc(BYc +D) ; D’ = BYc+D

58. Centre of circle = 1 0

59. Assuming a symmetrical pilot, the sending end characteristics
can be found from inversion of the receiving end.

60. CIRCULATING CURRENT FEEDER
PROTECTION SYSTEM
Dual of Balanced-voltage scheme
Note : ** Ƶc is in series with pilot instead of shunt Yc **
Neglecting impedance of restraint circuit, the threshold condition
is found by equating the operating and restraint quantities, i.e.
I1 > I2

61. CIRCULATING CURRENT FEEDER PROTECTION
SYSTEM

62. BRIDGE – Type Comparator Relay
(circulating current)

63. General Locus of the fault Performance of the
Differential Protection Systems
Where k1’ , k2’ , k3’ and k4’ may be expressed in terms of
distribution factors C1 and C0 for +ve and zero phase sequence,
the ratio zero to +ve phase sequence impedance and the
constants M, N, P of the summation device.

64. Determination of k1’ , k2’ , k3’ and k4’ for a S-L-G fault

65. Assumptions:
1. X/R ratio = constant
2. Complete transposition of line
3. Shunt capacitance neglected
4. Ƶ1 = Ƶ2
If Ƶ1 = Ƶ2 C1 = C2
IA0 = (1 - C0)(IA1 - IB1)
IB0 = - C0 (IA1 - IB1)
where Cn = distribution factor
IA2 = (1 - C1)(IA1 - IB1)
= Impedance ratio
IB2 = - C1 (IA1 - IB1)

66. Where k1 = M(1 - C0) + N + P (1 - C1)
k2 = - [M(1 - C0) + P (1 - C1)]
k3 = - [MC0 + PC1]
k4 = MC0 + N + PC1
Substitute into equation : IA1/ IB1 in terms of (EA / EB)
We can simplify the network as :

67. From analysis of the preceding network
EA1 = IA1C1Ƶ1 + (IA1 - IB1) C02Ƶ1
And EA1 - EB1 = IA1C1Ƶ1 + IB1(1 - C1)Ƶ1
From which we obtain :
Recall that EA1 = (EA/ 3) i.e. pre-fault line voltage

68. N.B. If resistance was not zero but = Rf then
IA/ IB1 IN TERMS OF (EA / EB)
Substitute equation(2) into (1) yields

69. Where
The poles of the set of orthogonal circles being (k1’ / k3’ )
and (k2’ / k4’ ) and all circles will pass through there poles.
Further Reading
Jones, D and Rushton J : “Generalized locus charts for the
evaluation of differential feeder protection performance”
Institute of Electrical Engineers 113 pp 1194 (1966)
GO TO EXAMPLE 7.2 pg. 259 of “Unit Protection” handout and
EXAMPLE 7.3 pg. 265 as well as EXAMPLE 7.4 pg. 273