Transformer Curve Shift

[D!NNy]

	Hello, I am copying this data from SKM website: someone please help me to understand why there is curve shift for only a single phase to ground fault of Delta Wye Grounded configuration which is 0.58% where as line line fault current is 0.5pu on the primary side. I understand the explanation provided below that primary relay is seeing fault current of 1.0pu on at least one of the phases which a relay can trip the circuit breaker (assuming that there is no single phase tripping) except in the case of single phase to ground fault shown My question if there is a line line (Phase A and C as shown in the table) fault on secondary side of a 12kV to 480V 3X100kVA pole mounted transformer Delta Y grounded configuration, protected by the fuses on the primary side (individually operated). in this case fuses on the phase A and C has to be sized for 0.5pu transformer curve shift and phase C as 0.58pu. OR curve shift to applied 0.5pu for all phases and coordinate the fuses properly. Am i doing correct?? Thanks for the suggestions Have a great day.
	Through-Fault Current Protection Engineers must consider all possible fault scenarios on the secondary-side of the transformer in order to evaluate the level of through-fault current protection provided by a primary-side fuse, relay or circuit breaker. On a three-phase power system this would include; three-phase fault, single-line-to-ground fault, line-to-line fault and line-to-line-to-ground fault. It is common practice to view current distributions in per unit on the three-phase fault current base seen by the primary-side protective device. Three-Phase (3?) Secondary Fault The primary-side relays would each see 1.0 A p.u. fault current on a 3-? primary fault current base, figure 1.
	Single-Line-to-Ground (SLG) Secondary Fault The primary-side phase A and C relays would see 0.58 A p.u. fault current on a 3-? primary fault current base, figure 2.
	Line-to-Line (LL) Secondary Fault The primary-side phase A and C relays would see 0.50 A p.u. fault current and the phase B relay would see 1.0 A p.u. fault current on a 3-? primary fault current base, figure 3.
	Line-to-Line-Ground (LLG) Secondary Fault The primary-side phase A and C relays would see 0.58 A p.u. fault current and the phase B relay would see 1.0 A p.u. fault current on a 3-? primary fault current base, figure 4.
	Transformer secondaries were faulted and primary currents were calculated for each fault type and summarized in Table 2 for winding configurations listed in Table 1.
	Table 2 Fault Current Summary Table
	1. Shifting of the damage curve is dependent upon the amp rating of the NGR relative to the transformer primary-side FLA. Shifting is To Be Determined by the engineer on a case by case basis. For a sample case study see Example 3. Example 1 Plot the damage curve and set a primary-side 50/51 relay to protect a 1000kVA, 65?C, 4160-480/277V, ?-YG, oil-filled, substation transformer with an impedance of 6.0%. Consider secondary 3-? faults only when setting the primary-side 50/51 relay to protect the transformer during through-faults. Solution The protection is met based on the criteria given, see Figure 5. Now examine the protection provided by the 50/51 relay shown in figure 1 for faults of all types using Table 1. For a secondary LL fault the primary relays in phases A and C would only see 50% of the fault current they would otherwise see for a 3-? fault. This is not a problem though because the B-phase relay would see 100% current, and the transformer would be protected. The LLG fault is similar to the LL fault accept that the relays in phases A and C would see 58% current. But again the B-phase relay would see 100% current, so the transformer would be protected. However for the SLG fault case, the A and C-phase relays would see 58% current with no current flowing in the B-phase! In order to evaluate the through-fault current protection afforded by the relays for a secondary SLG fault, the damage curve of Figure 5 would have to be shifted 58% to the left. It is common practice to show both damage curves on the TCC. After shifting it becomes apparent that the original relay settings are inadequate, see Figure 6. Example 2 Repeat Example 1 but now consider secondary faults of all types when setting the primary-side 50/51 relay to protect the transformer during through-faults. Solution The protection is met based on the criteria given, see Figure 7.
	Fig. 5 Example 1 Though-Fault Current Protection (3-? Fault Only)

 

[kingpb]

If you have a delta-wye solidly grounded system, the transformer curve shift applies, and your relay settings should be set accordingly.

The explanation is telling you that the SLG scenario produces a situation where the fault current would be less, and therefore the relay, if set to high would not see the current and therefore not trip and protect the transformer. Don't be confused by the A, C designations. That is just for example, it could easily apply to any combination of faulted phases.

The example being used is with a 50/51 relay. A fuse curve shape is going to be much different and you need to plot it. You also need to make sure the switch that is being used with the fuses is not going to operate a single phase, i.e. only open the faulted phase(s). Any blown fuse should open all three phases.

 

[D!NNy]

If you have a delta-wye solidly grounded system, the transformer curve shift applies, and your relay settings should be set accordingly.

The explanation is telling you that the SLG scenario produces a situation where the fault current would be less, and therefore the relay, if set to high would not see the current and therefore not trip and protect the transformer. Don't be confused by the A, C designations. That is just for example, it could easily apply to any combination of faulted phases.

The example being used is with a 50/51 relay. A fuse curve shape is going to be much different and you need to plot it. You also need to make sure the switch that is being used with the fuses is not going to operate a single phase, i.e. only open the faulted phase(s). Any blown fuse should open all three phases.

But pole mount cutout fuses are operated individually, if the fault currents on the phases are of different amplitudes means they will have different clearing time. one might clear faster than other.

if there is single line to ground fault on one of phases (B) of secondary side of transformer then primary side fault current on phases A and C is see 0.58pu on the primary, we size the fuse based on that ........no we have LL fault on phases A and C on the secondary side of the transformer then fuses on the primary side of the transformer will see 0.5pu on phases A and C which is lower than 0.58pu fault current not necessarily protecting the transformer .........based on this do i have to apply the curve shift 0.5pu instead of 0.58pu?

Thanks

 

[kingpb]

You are trying to apply 3 phase relay protection to a bank of single phase distribution transformers that would be protected by individual fuses.

The type of fuses used have different operating characteristics. But regardless, I don't think there is any adjustment to the transformer curve needed other than the 0.58.