Overcurrent and Directional Control

1. ECNG 6503 -2
Advanced Power Systems Practice
Lecturer
Prof Chandrabhan Sharma
University of the West Indies
Trinidad and Tobago

2. GRADING PROCEDURE
Grading Relay/Fuse
A. Fuse Behind Relay
Suggest:
1.2 tR + 0.1 + 0.1 = 0.6 tF
Allowance Circuit Safety Allowance for Relay
Breaker Margin Fuse Error (Fast) & CT Error (Slow)
Then tF = 2tR + 0.33 Seconds

3. B. Relay Behind Fuse

4. CO-ORDINATION OF I.D.M.T.
RELAYS

5. In General:-
1 . Discriminate at Max. Fault Level
- Relay will then discriminate at all
other current levels
2. The setting of the upstream relay
should be greater than the
downstream relay

6. PLUG SETTING MULTIPLIER (PSM)
- On Induction Disc relays current setting is made by
inserting a plug into a plug bridge
- Hence “Plug Setting”
- If relay setting and CT ratio are known, can find fault
current as a multiple of current setting – PSM
- Relay characteristics give operating times at
multiples of current setting (PSM)
- Therefore the characteristics can be applied to any
relay regardless of current setting and nominal
rating.

7. EXAMPLE
CT Ratio = 100/1 A
Fault Current = 1000 A
Relay current settings are made in % of CT ratings
Is = 100% of 100 A
Is = 100 A primary
PSM = If/Is = 1000/100 = 10
So read off operating time at 10 x current setting

8. TIME MULTIPLIER SETTING (TMS)
- Not a time setting in seconds
- Multiplying factor which is applied to the basic
relay operating time characteristic
For Grading:
Required operating time = TMS x operating time at TMS=1

10. IDMT RELAYS
Grade „B‟ with „A‟ at IFMAX
Both Relays Normal IDMT (3/10)
Relay A
Current Setting = 5 AMP = 100 AMP (Pr)
IFMAX = 1400 AMP = 14 x Setting
PSM = 14
Relay operating time at 14 x Setting and TMS of 0.05 is 0.13 seconds

11. Relay B
Current Setting = 5 AMP = 200 AMP (Pr)
IFMAX =1400 AMP = 7 x Setting
Relay operating time at 7 x setting and TMS of 1.0 is 3.6 sec.
Required operating time = 0.13 + 0.4 = 0.53 seconds
Therefore required TMS = 0.53/3.6 = 0.147
Use TMS = 0.15 for relay B

12. Available p.s. = 2, 4, 6, 8

13. OVERCURRENT RELAY
CO-ORDINATION
Given: Tap Settings available are: 2, 4, 6, 8
Take discrimination time = 0.5 sec.

14. Step 1: Start grading from extreme point i.e. load point
Select the lowest TMS = 0.1
∴Select P.S. = 4
For fault at D bus:
From IDMT curve for PSM 12.5 and TMS 1.0
Operating time = 2.7 sec.
∴Actual operating time = (2.7)(0.1) = 0.27 sec.

15. Step 2:
Relay C:
For fault at D at 2000A
Relay C should take (0.27 + 0.5) =0.77 sec.
∴Set P.S. for C at 2
For fault at D:
Operating time from characteristic (TMS =1.0, PSMC 16.67)=2.45 s
But relay must operate in 0.77 s

16. For Fault at C:
∴From characteristic, op. time = 2.2 s
∴ Actual op. time = (2.2)(.314) = 0.69 s
Step 3: Relay B.
For Fault of 3000 A (at C)
Operate TimeB = 0.69 + 0.5 = 1.19 s
IL(B)= 100A
∴ Select P.S. = 2

17. ∴ From characteristic → op. time = 2.5 s
But breaker should operate in 1.19 s
For fault at B:
∴ Operating time =2.2 s
Actual Operating time = 2.2 x 0.476 = 1.05 s
For Relay at A: IL = 100
∴ P.S. =2

18. For fault at B(5000A), relay A should back up = 1.05 + 0.5 = 1.55 s
∴ Time to operate from characteristics = 2.32 s

19. CONSIDERATIONS FOR DELTA-STAR
TRANSFORMERS

20. For DL fault on Y side

21. CONSIDERATIONS FOR DELTA-STAR
TRANSFORMERS
- For a ph-ph fault on star winding there is a 2:1:1
current distribution on the delta side.
- Should use an overcurrent relay in each phase.
- Must ensure correct grading margin between star
side relay at 0.866 x 3 ph current level and delta
side relay at 3-ph current.

23. EARTH FAULT PROTECTION
- Earth fault current may be limited.
- Sensitivity and speed requirements may not be
met by overcurrrent relays.
- Use separate earth fault relays.
- Connect to measure residual (zero sequence)
currents.
- Therefore can be set to values less than full load
current.

24. EARTH FAULT CONNECTIONS
Combined with Overcurrent Relays

25. For Economy, can use 2 x Overcurrent Relays

26. 4-Wire Systems
Earth Fault relay setting must be greater than “Normal” IN

27. Use CT in Neutral
Earth Fault setting is now independent of IN BUT must use
3 x Overcurrent Relays

28. EARTH-FAULT RELAYS SETTING
RANGE

29. EARTH-FAULT RELAY BURDEN
Due to Low current setting, more coil turns are required on
electromechanical relays to give operation
e.g.
At setting current overcurrent and earth fault relays have similar VA Burden
Earth Fault Relay ← CDG 11→Overcurent Relay
Setting = 0.2 AMP Setting = 1.0 AMP
Burden = 3 VA Burden = 3 VA
Impedance = 75 OHMS Impedance = 3 OHMS
i.e. ZE/F = 25 x ZO/C (AT SETTING)

30. N.B.
As setting is lowered∵# turns must increase so as to maintain
ampere-turns at operate level.
As number of turns increase → wire (length) increases →
increase in resistance.
Rule of thumb → above 20 x setting current
relay magnetic circuit goes into saturation.
When this happens, (saturation), relay burden remains constant
i.e. I2Ƶ = constant (at saturation).

31. EARTH-FAULT RELAYS – EFFECTIVE
SETTING
CONSIDER

32. Voltage developed across earth fault relay is applied to “B” and “C”
phase CT‟s which will therefore take appropriate magnetising current.
*
Overall Effective Setting (Or Primary Operating Current)
IEFF = CT RATIO (IS + 3IMAG) *
* will be supplied by „A‟
Since fuses cannot discriminate between phase ; earth faults,
grading of EF relays with fuses is not possible!

33. EFFECTIVE SETTING
Given a 3 VA relay, 20% ( i.e. 1 A bias)
∴ at setting, Ƶrelay = VA/I2 = 3/(0.2)2 = 75 Ω
Same relay but 100% setting
Ƶr = 3/12 = 3 Ω
∴ 20% setting has Ƶr (20%) = Ƶr (100%) (x 25)
But burden at 20% setting and rated I
I2Ƶ = (1)2(75) = 75 VA (assuming no saturation)
* Lower due to magnetic saturation which reduces Ƶr

34. EARTH-FAULT EFFECTIVE SETTING
(CDG11 3VA)

35. SENSITIVE EARTH-FAULT
PROTECTION
Used when Earth Fault Level is severely limited.
e.g. High Resistance Earthed Systems
As very low settings are required:-
Essential to use low burden relay (usually use static relay)
Also can use core balance CT (CBCT) for better sensitivity
Turns Ratio is not related to Full Load Current.

36. CORE BALANCE CT WITH CABLE
Circuit

37. CORE – BALANCE C.T. WITH CABLE

38. NEED FOR DIRECTIONAL CONTROL
Generally required if current can flow in both directions through
a relay location
e.g. - Parallel feeder circuits
- Ring main circuits
Relays operate for current flow in direction indicated
(Typical operating times shown)

39. PARALLEL FEEDERS
Non-Directional Relays:-

40. PARALLEL FEEDERS
Consider Fault on one feeder:-
Relays „C‟ and „D‟ see the same fault current (I2). Since „C‟
and „D‟ have similar settings both feeders will trip.

41. Solution !!: - Directional Control at „C‟ and „D‟
Relay „D‟ does not operate due to current flow in „Wrong‟
Direction.

42. PARALLEL FEEDERS
SETTING PHILOSOPHY FOR DIRECTIONAL RELAYS
- Load current always flows in „Non-operate‟ direction.
- Any current flow in „Operate‟ direction is indicative of a fault condition.
Thus relays „C‟ and „D‟ may be set
- Sensitive
- Fast Operating Time

43. Usually, relays are set:-
- 50% Full Load Current (Note Thermal Rating)
- Minimum T.M.S. (0.1)
Grading Procedure :
1. Grade A (and B) with E assuming one feeder in service.
2. Grade A with D (and B with C) assuming both feeders in
service.

44. RING MAIN CIRCUIT
With Ring closed:
Both load and fault current may flow in either direction along
feeder circuits thus, directional relays are required.
Note: Directional relays “Look into” feeder.
Need to establish principle for Relay Co-ordination.

45. RING MAIN CIRCUIT
GRADING PROCEDURE FOR IDMT RELAYS
- Grading Margin is established at highest current level seen
by both relays
- Highest Fault Level occurs with ring closed with each
branch contributing a proportion of the total fault current
- Highest „Branch” Current occurs with ring open

46. Case A.
Case B.
Clearly, case B gives highest branch current.

47. RING MAIN CIRCUIT
Procedure:
1. Open Ring at A
Grade : A‟ – E‟ – D‟ – C‟ – B‟
2. Open Ring at A‟
Grade A – B – C – D – E
Typical operating times shown
Note: Relays B, C, D’, E’ may be non directional

48. NON-DIRECTIONAL RELAYS ON RING
CIRCUITS
General Rule:
• Relays at Source Substation
• Relays with the higher time setting at load
substations where relays have different operating
times

49. ODD NUMBER OF CIRCUITS
EVEN NUMBER OF CIRCUITS (Substation ‘X’ added)

50. PARALLEL FEEDER – APPLICATION NOTE
Grade B with C at IF(1)
Grade B with D at IF(2) (IN PRACTICE)
Grade A with B at IF – But check that sufficient margin exists for bus fault at Q when relay
‘A’ sees total fault current IF(2) but relay ‘B’ sees only IF(2) /2

51. LINE WITH TWO FEEDERS
BREAK INTO TWO

52. GRADED SYSTEM
RING MAIN (ONE SOURCE)

53. RING MAIN (WITH ONE SOURCE AND INTERTIE)
With inter-tie set „X‟ and „Y‟ at lowest value of 0.1 & let them be
non-directional relays. Therefore, for fault at F1 the inter-tie
would trip first then the faulted line, allowing the rest of the
system Grading to operate.

54. RING SYSTEMS WITH TWO SOURCES
Discrimination between all relays not possible due to different
requirements under different ring operating conditions
For F1:- B‟ must operate & A
NOT COMPATIBLE
For F2:- D must operate & A‟

55. RING SYSTEMS WITH TWO SOURCES
Option 1
Trip least important source instantaneously then treat as normal ring main.
Option 2 (onerous)
Fit pilot wire protection to circuit A - B and consider as common source
busbar. Then grade rest of system as two in feeds.

56. ESTABLISHING DIRECTION
The direction of alternating current may only be determined with
respect to a common reference. Recall that direction implies phase
comparison. The most convenient reference quantity is the
system voltage. In relaying terms, the reference is called the
polarising signal.
Which voltage to use?
e.g. For „A‟ phase relay:-
* VA not usually used i.e. for fault on „A‟ phase, VA = 0

57. RELAY CONNECTION
The angle by which the current applied to the relay is displaced from
the voltage* applied to the relay at system unity power factor.
Example 90˚Connection: IA & VB-C
MAXIMUM TORQUE ANGLE
The Theoretical angle by which the current applied to the relay must
be displaced from the voltage* applied to the relay to produce
maximum torque.
Example 45˚
* This is also usually the polarising quantity

58. ELECTROMECHANICAL RELAY
- Will develop torque by inter-action of two fluxes.
- Maximum torque when fluxes are 90˚ apart.
T ∝ 1 2 sin
General Phasor Diagram :-
Maximum Torque Line.(Direction of I for Max Torque)
= maximum torque angle
If I (or I) is displaced by
from V, then maximum
torque is obtained
V : APPLIED VOLTAGE (POLARISING SIGNAL)
I : APPLIED CURRENT
IV : CURRENT IN VOLTAGE COIL
V : FLUX PRODUCED IN VOLTAGE COIL
I : FLUX PRODUCED IN CURRENT COIL

59. DIRECTIONAL RELAY
Applied Voltage: VA
Applied Current : IA
Assuming Voltage Coil Angle ≏ 90˚
Question:-
- Is this connection suitable for a typical Power System?
(Consider relative phase of VA & IA under fault conditions)
(Consider “Close up” Fault, VA 0)

60. 90˚ Connection - 45˚ M.T.A.

61. 90˚ Connection - 30˚ M.T.A.

62. 30˚ Connection - 0˚ M.T.A.

63. SELECTION OF M.T.A
OVERCURRENT RELAYS
90˚ Connection 30˚ M.T.A. – Plain Feeder, Zero Sequence
(Lead) Source, “Behind Relay”
90˚ Connection 45˚ M.T.A. – Transformer Feeder, Zero Sequence
(Lead) Sequence Source, in front of relay

64. DIRECTIONAL RELAY- C.T.
CONNECTIONS
Clearly, correct polarity of current coil and voltage coil is essential to
ensure correct application.
- Refer to manufacturers diagrams
Philosophy of 2 x Overcurrent / 1 x Earthfault may be retained if 90˚
Connection is used for overcurrent elements.
i.e.

65. Must use 3 x Overcurrent elements if 30˚ Connection is
used
i.e.

66. DIRECTIONAL EARTHFAULT
Requirements are similar to directional overcurrent.
Viz. Need operating Signal
and Polarising Signal
Operating Signal
- Obtained from residual connection of line C.T.‟s
i.e. IOP = 3IO
Polarising Signal
The use of either Phase-Neutral or Phase-Phase voltage as
the “Reference” becomes inappropriate for comparison
with Residual current
Most appropriate polarising signal is RESIDUAL VOLTAGE.

67. RESIDUAL VOLTAGE
May be obtained from „Broken‟ Delta V.T. Secondary.
Notes:
1. V.T. Primary must be earthed
2. V.T. must be of „5Limb‟ Construction (Or single 3 x Single ph. units).

68. RESIDUAL VOLTAGE
SOLIDLY EARTHED SYSTEM
Residual Voltage at R (Relaying Point) is dependent upon ƵS/ƵL RATIO

69. DIRECTIONAL CONTROL
ELECTROMAGNETIC RELAY

70. STATIC RELAY

71. CURRENT POLARISING
A solidly earthed, high fault level(low source impedance) system
may result in a small value of residual voltage at the relaying point.
If residual voltage is too low to provide a reliable polarising signal
then a current polarising signal may be used as an alternative.
The current polarising signal may be derived from a C.T. located in a
suitable system neutral to earth connection.

72. MAXIMUM TORQUE ANGLE
0˚- Resistance Earthed systems
45˚(I LAGS V) – Distribution Systems (Solidly Earthed)
60˚(I LAGS V) – Transmission Systems (Solidly earthed)
ZERO SEQUENCE NETWORK:
VO = (-) IO (ƵSO + 3R)

73. CURRENT POLARISATION
INCORRECT

74. CURRENT POLARISATION
CORRECT

75. CURRENT POLARISATION
CORRECT
If (ƵLO + ƵSO) is positive

76. CURRENT POLARISATION

77. RESISTANCE EARTHED SYSTEM

78. AUTO TRANSFORMER
Neutral connection is suitable for current polarising if earthfault
current flows “UP” the neutral for faults on HV side.

79. Check:
For Correct Application
(Note: There is also a possibility that Neutral
Current may be zero)
Alternative: use C.T. in one leg of Δ winding

80. TYPE OF FEEDER
90⁰ - 30⁰ Connection
FIGURE 5a
OPERATING AND NON-OPERATING ZONES OF QUADRATURE
CONNECTED RELAYS FOR DIFFERENT SYSTEM FAULT CONDITIONS.

81. TYPE OF FEEDER
90⁰ - 45⁰ Connection
FIGURE 5b
OPERATING AND NON-OPERATING ZONES OF QUADRATURE
CONNECTED RELAYS FOR DIFFERENT SYSTEM FAULT CONDITIONS.

82. TYPE OF FEEDER
90⁰ - 45⁰ Connection
FIGURE 5c
OPERATING AND NON-OPERATING ZONES OF QUADRATURE
CONNECTED RELAYS FOR DIFFERENT SYSTEM FAULT CONDITIONS.