This document discusses overcurrent relays and differential protection. It explains plug setting multipliers and time multiplier settings which are used to determine operating times from relay characteristics. It provides an example of coordinating overcurrent relays on a system. The document also discusses earth fault protection, differential protection using Merz-Price systems, and the differential protection of a star/delta power transformer accounting for inrush current and phase shifts.
Power System Analysis was a core subject for Electrical & Electronics Engineering, Based On Anna University Syllabus. The Whole Subject was there in this document.
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Power System Analysis was a core subject for Electrical & Electronics Engineering, Based On Anna University Syllabus. The Whole Subject was there in this document.
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As a final year project in PESIT Bangalore
Amplitude and phase comparators
Over current relays
Directional relays
Distance relays
Differential relay.
Static Relays: Comparison with electromagnetic relay
Classification and their description
Over current relays
Directional relay
Distance relays
Differential relay
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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
7.
8. OVERCURRENT RELAY
CO-ORDINATION
Given: Tap Settings available are: 2, 4, 6, 8
Take discrimination time = 0.5 sec.
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.
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
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