Overcurrent Protection

1. ECNG 3015
Industrial and Commercial
Electrical Systems
Lecturer
Prof Chandrabhan Sharma
#4
Overcurrent Relays

2. 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.

3. 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

4. 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

5. 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

6. 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

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

9. Typical time/current characteristics of 3 seconds standard I.D.M.T.
– Available p.s. = 2, 4, 6, 8

10. 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.

11. 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

12. 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

13. ∴ 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

14. 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

15. 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.

16. EARTH FAULT CONNECTIONS
Combined with Overcurrent Relays

17. For Economy, can use 2 x Overcurrent Relays

18. 4-Wire Systems
Earth Fault relay setting must be greater than “Normal” IN

19. 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.

20. 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!

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

22. 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:

23. 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).

24. 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

25. OVERALL DIFFERENTIAL PROTECTION OF STAR/DELTA
POWER TRANSFORMER

26. Recall:
Three-phase connections (Y/d)

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