2. Material used:  Tin  Lead  Zinc  Silver  Antimony  Copper  Aluminum

3. Where, I = current through the fuse wire ℓ= resistivity of the fuse material d = diameter of the fuse wire L = length of the fuse wire K 1 = parameter constant

4. CB short circuit current carrying capacity >Operating current of fuse >Max over-load current Time in sec Current in Amp

5. Current Time I chopping T 1 = Melting Time T 2 = Arcing Time T = Total operating time Hence, T = T 1 + T 2 T1T1 T2T2 T = T 1 + T 2 Fault current without fuse

6. LoadLoad Fuse Source Series trip When, Over-load current or fault current > Fuse rating Fuse will be burnt and load will be OFF

7. LoadLoad Fuse Shunt trip When, CT secondary current> Fuse rating Fuse will be burnt and CT secondary current will flow through trip coil and CB will trip. TC

8. Fuses are classified on the basis of  Continuous current carrying capacity  Overload capacity  Response characteristics  Voltage rating

10. Use of fuses  Motor  Air-conditioner  Laptop  Cell-phone  Printers

11. LoadLoad TC Relay Battery CB  CT Secondary current exceeds relay setting current  Relay NO contact will make according to relay characteristics.  Trip coil will get energized  Circuit breaker will trip.

12.  O/C relay is more accurate and versatile than Fuse protection  O/C relay is more costly and more complicated than fuse.  Time gradation is difficult in case of fuse protection  Once fuse melts protection cannot be re-commissioned until it is replaced.  Unlike relay, fuse characteristics is temperature dependent.

13. B R N 1. Two-overcurrent & one earth fault relay

14. 2. Three-overcurrent & one-earth fault relay B Y R N

15.  Overcurrent relay current setting = 125% of nominal load current  Under 3-ph balanced load condition, Earth Fault relay will sense zero current. Thus this relay has to be very sensitive. Earth fault current setting = 20% of rated current

16. B R N 1. Two-overcurrent & one earth fault relay 2. Three-overcurrent & one-earth fault relay B Y R N E/F relay will operate in both the cases

17. B Y R N 1. Two-overcurrent & one earth fault relay 2. Three-overcurrent & one-earth fault relay B R N Only R-ph O/C relay will operate Both R & Y-ph O/C relay will operate

18. Relay only operates when PSM > 1 Let t = operating time from relay characteristics Hence, Relay operating time = t x TMS Where TMS = time multiplier setting

21. Now-a-days numerical relay has inbuilt three overcurrent and one earth fault element. B Y R N Numerical relay

22. Protection healthy I input Calculate I rms Compute PSM =I input / I rms If PSM>1 Start timer T 1 If Type = DT DT delay t op If Type = SI Compute t op If Type = VI Compute t op If Type = EI Compute t op If T op >T 1 Alarm Trip yes no

23.  Less wear and tear  Self supervision facility is available  Setting range is comparatively wide  Choice of operating characteristics is possible  Relay resetting time is adjustable

24.  Waveform record and post-fault analysis  Remote communication and time synchronization  Over-voltage and under-voltage protection  Over-frequency and under-frequency protection  Trip circuit supervision  CT supervision  VT supervision

26. Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 Assume out-going feeders are radial. For electromechanical relay let T1 = Operating time of the incomer relay T2 = Resetting time of the incomer relay t1 = Operating time of the feeder relay t2 = Resetting time of the feeder relay And T1 > t1 For numerical relay T2, t2 = 0 51

27. Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 At t = 0 51 Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 51 At t = t1 For Electro-mechanical relay

28. Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 At t = t1 + Δt 51 Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 51 If T1 < (2.t1-Δt) For Electro-mechanical relay

29. Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 At t = 0 51 Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 51 At t = t1 For Numerical relay

30. Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 At t = t1 + Δt 51 Incomer Out-going Feeder-1 Out-going Feeder-2 Out-going Feeder-3 51 At t = 2.t1 + Δt For Numerical relay

31. For Electro-mechanical relay time Angular displacement of disc T1 T2 t1 t2 Incomer Feeder-1 Feeder-2 ΔtΔt (T1+ 2Δt) (2t1+ Δt) time Angular displacement of disc T1 t1 ΔtΔt T1

32. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 Z p.u. =0.2 75MVA Tr.-1 Z p.u. =0.2 20MVA Tr.-1 Z p.u. =0.6 Out-going feeder B/S I 11,T 11 I 12,T 12 I 13,T 13 I 21,T 21 I 22,T 22 I 23,T 23 I 21,T 21 I 23,T 23 I 31,T 31 I 32,T 32 I 33,T 33 Current setting= I **, TMS=T ** Let base MVA=100

33. Considering CB short circuit current rating, T 11 > T 12 > T 13 T 21 > T 22 > T 23 T 31 > T 32 > T 33

34. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 75MVA Tr.-1 20MVA Tr.-1 Out-going feeder B/S Case-1: Fault at 132kV bushing of 75MVA Tr. (T-op) 132kv incomer > (T-op) 132kV B/S > (T-op) 132 kV of 75MVA Tr,

35. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 75MVA Tr.-1 20MVA Tr.-1 Out-going feeder B/S Case-2: Fault at 33kV Bus bar (T-op) 132kv incomer > (T-op) 132kV B/S > (T-op) 132 kV of 75MVA Tr, And (T-op) 33kv of 75MVA Tr. > (T-op) 33kV B/S

36. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 75MVA Tr.-1 20MVA Tr.-1 Out-going feeder B/S Case-3: Fault at 33kV bushing of 20MVA Tr. (T-op) 33kV of 75MVA Tr. > (T-op) 33kV B/S > (T-op) 33kV of 20MVA Tr.

37. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 75MVA Tr.-1 20MVA Tr.-1 Out-going feeder B/S Case-4: Fault at 6kV Bus bar (T-op) 33kV of 75MVA Tr. > (T-op) 33kV B/S > (T-op) 33kV of 20MVA Tr And (T-op) 6kv of 20MVA Tr. > (T-op) 6kV B/S

38. 132kV Incomer-1 132kV Incomer-2 132kV Bus bar 33kV Bus bar 6kV Bus bar 75MVA Tr.-2 75MVA Tr.-1 20MVA Tr.-1 Out-going feeder B/S Case-5: Fault at 6kV outgoing feeder (T-op) 6kV of 20MVA Tr. > (T-op) 6kV B/S > (T-op) 6kV outgoing feeder

39. Normal direction(forward) of power flow 51 67 51= Non-directional O/C Relay 67= Directional O/C Relay Trip This relay operates only when fault occurs in a particular direction. For example, 1. Parallel interconnector: Sink end CB of the faulty interconnector will trip through Directional relay operation since it is sensing reverse current.

40. Normal direction(forward) of power flow 51 67 51= Non-directional O/C Relay 67= Directional O/C Relay Trip 2. Parallel Transformer feeder : LV side CB of the faulty transformer feeder will trip through directional overcurrent relay operation.

41. Normal direction(forward) of 30MW power flow Lock-out relay 3. Islanding scheme: By the directional relay operation  synchronization get lost  Lock-out relay operates  Two systems get isolated with part of the load is shed. Load 100MW 30MW 50MW 80MW 51 Trip 67 Trip System-1 System-2

42. It requires both voltage and current inputs. Under normal operating condition residual voltage and current will be zero. V a + V b + V c = 0 I a + I b + I c = 0 B Y R N Directional O/C & E/F Relay

43. During any fault a voltage phasor is taken as reference. The position of the fault current phasor w.r.t. the voltage phasor determines the direction of fault – Forward or reverse. V ref α Max. ‘+’ torque line Max. ‘-’ torque line Forward zone Reverse zone IfIf α = Max. Torque angle along which relay is max. sensitive

44. In general V ref is cross polarized. For example, Type of faultVoltage reference Neutral currentV0V0 R-Phase faultV YB Y-Phase to B-Phase faultV RN 3-Phase short circuit faultMemorized voltage (Numerical relay) Switch ON to 3-Phase Short circuit faultNon-directional behavior

45. Quadrature Connection: In case of numerical relay V ref is rotated through an angle (δ) w.r.t. which ±90 ⁰ is the forward direction. Type of faultsValue of the angle δ Phase – Phase fault -90 ⁰ Phase – Earth fault +90 ⁰

46. Phase-Phase fault: Let phase ‘Y’ gets shorted with phase ‘B’. Then phase reference will be V RN. For line side CT neutral (say) forward direction of operation is shown. Directional O/C & E/F Relay V RN δ = -90 ⁰ Forward zone I f =I Y = -I B IfIf V YN V BN -V BN V YB V Ref Reverse zone

47. Phase-Earth fault: Let phase ‘R’ gets shorted with earth. Then phase reference will be V YB. For bus side CT neutral forward direction of operation is shown. Directional O/C & E/F Relay V RN δ = +90 ⁰ Forward zone I f =I R IfIf V YN V BN -V BN V YB V Ref Reverse zone