Protection Systems & Devices (Relays)
A number of fault conditions can arise within a power transformer. These include:
Earth fault on a transformer winding.
Core faults due to insulation breakdown which sufficient eddy current to flow causing over heating.
Inter-turn faults occur due to winding flashover caused by line surges.
Phase to phase faults. This are rare in occurrence but will result in a substantial currents of magnitude similar to earth faults.
Tank faults due to loss of oil which produces abnormal temperature rises.
External abnormal conditions such as overloading, overvoltages due to transient core losses and corresponding temperature rise.
Fuses usually protect transformer with capacity less than 500 KVA in industry and 2500 KVA in residential areas. With ratings up to 5000 KVA in residential areas, instantaneous and time delay over current relays may be more desirable. For industrial loads greater than 1500 KVA and for transformers that are part of the bulk power system it is recommended to use differential protection on harmonic restraint percentage differential relays. Also, the higher the voltage, the more sophisticated and costly the protective device.
1. Differential Protection
The protection of transformers is usually performed by differential protection. The differential protection responds to the vector difference between two similar quantities. The C.T. connected on the transformer windings should be arranged so that the same current is flowing between the two sides.
A General Rule Is To Connect The CTs On Any Star Windings In Delta and In Any Delta Windings Connect CTs in Star
Two basic requirements that the relay connection must satisfy are:
The relay must not operate for loads or external faults.
The relay must operate for internal faults.
Fig. 1 represents differential protection of Delta-Star transformer
Fig. 2 shows a star-star transformer to which circulating current protection has been applied. Here it will be noted that the current transformers on both sides are connected in delta.
Fig. 3 (a) is included to show how had the current transformers been connected in star, operation of the protective relay would occur on a fault outside the protected zone which we wish to avoid while Fig 3 (b) shows how this can be avoided by connecting the current transformer secondaries in delta.
Problems arising in differential protection applied to transformers
Simple differential protection system is inadequate because the following difficulties arise:
Difference in length of pilot wires on either sides of relays. The difficulty is overcome by connecting adjustable resistors to pilot wires. These are adjusted on site to get equipotential points on pilot wires. Taps can be provided on operating coil and restraining coil of relay for adjusting the balance.
Difference in CT ratios due to error difference at high values of short circuit currents. Because of this difference relay operates for external faults. This difficulty is over come by using biased (percentage) differential relay. In such a relay a restraining coil is connected to pilot wires. The current flowing through the restraining coil can be taken as (I1 + I2 ) /2 . With increase in current the restraining torque increases too and the current due to the CT inaccuracy is not enough to casue the relay operation.
Tap changing alters the ratio of voltage ( and currents ) between HV side and LV side. Differential protection should be provided with bias (restrain) which exceeds the effect of variation in secondary current due to tap changing.
Magnetizing current inrush:
When power transformers are switched on, initially there is no induced e.m.f., the conditions is similar to switching an inductive circuit. Since the resistance of the coil is low, a large inrush of magnetizing current takes place. The magnitude of which depend on circuit conditions and the voltage at the instant of switching. Maximum values of 6 to 8 times the rated current can flow in the winding. Usually this high current decays after few cycles to the normal current but in some cases it may take 2 – 4 seconds.
Formerly, the relay was provided with time lag of 0.2 – 1 second. By this time, the inrush current would vanish and the relay does not trip unnecessary. However for many faults, the relay time lag might cause substantial damage to the transformer.
Next development was the use of kick of fuses to shunt the relay coils as shown in fig. 2. These fuses are of the time limit type that do not operate in the time of switching under sustained fault conditions, the fuses operate and the current then passes through the relay coil and trip the C.B. This also is a slow protection and may cause some problems. It also depends on the fuse.
The next development was to desensitizing the relay for a short period of 0.1 to1 sec during switching. After this time the shunt across the relay coil is removed. This method can lead to switching on a transformer for long period during faults. The latest method adopted is harmonic current restraint.
Since inrush current has very high contents of 2nd and 3rd harmonic currents, which may reach 65% and 25% of the fundamental respectively, the restraining differential relay senses only the fundamental component. Because the harmonic component of the short circuit current is negligible, this relay operates at faults but not sensitive to switching current. The operating coil in these relays will receive only the fundamental component of the differential current. The harmonics are usually separated and fed back into the restraining coil.
The overload fuses shown in fig.2 provide a form of back up protection. In the event of sustained through fault, damage may be caused to the transformer. One or more of the overload fuses will operate; leaving the relay to be fed from one of CTs and thus causing relay operation.
2. Frame leakage protection
3. Restricted earth fault protection ( differential protection )
Earth faults on secondary side are not reflected on primary side when the primary when the primary winding is delta connected or has unearthed star point. In such cases, an earth fault relay connected in residual circuit of 3 CTs on primary side operates on internal faults in primary windings only. Because earth faults on secondary side do not produce zero sequence currents on primary side, restricted earth fault protection may then be used for high speed tripping for faults on star connected earthed secondary winding of power transformers.
Figure 5 shows the connections of the earth fault relays connected in the residual circuit of the line CTs. Figure 6 shows the connection of the restricted earth fault protection relay in the secondary side and earth fault protection in the primary side.
If the fault F1 is beyond the transformer windings, I1 and I2 will flow so that the current in the earth fault relay is negligible. For earth fault within the transformer star connected windings, I2 flows and I1 is negligible. Hence I2 causes the relay to trip the circuit.
When fault occurs very near the neutral point of the transformer, the voltage available for driving the earth fault current is small and the fault current would be low. If the relay is adjusted to sense such small currents, it may operate under normal unbalance conditions. It is common to set the relay to pick up at about 15% of the rated current. Such setting leaves a portion of the windings unprotected. Therefore it is called unrestricted.
4. Bucholoz Protection: it is frequently used in transformers.
Example 1: Describe with the help of a neat diagram the connections of differential protection of a transformer. A 3-phase 33/6.6 kV star/delta connected transformer is protected by Differential system. The CT’s on LT side have a ratio of 300/5. Show that the CT’s on HT side will have a ratio 60 : 5/√3
Solution: CT’s on delta side are star connected. Hence the secondary phase currents are equal to currents in pilot wires. CT’s on star connected side are delta connected hence current in secondary is equal to current in pilot wires divided by √3.
Assume 300 A is flowing in the lines on LT side
√3 x 6.6 x 300 = √3 x 33 x I
I = 60 A ( current in HT lines )
which is primary current of CT on HT side.
Current in pilot wires: On the delta side of transformers the CT secondaries are star connected. Their secondary current is 5 Amp. Hence current fed is pilot wires from LT side is 5 Amperes. Same current is fed from CT connections on HT side which are delta connected.
Hence secondary current of CT’s on HT side is 5 / √3 Amp.
Hence CT ratio on HT side is 60 : 5/√3
Example 2: A 30 MVA, 11.5 kV/ kV, star-delta power transformer to be protected by differential protection. The high voltage side phase lags behind low voltage side by 30o. Formulate the complete differential protection for the transformer by selecting CT ratios, CT connections. The continuous current carrying capacity of restraining coils of the differential relay should not exceed 5 Amp. CT ratio is 3000/5 on 11.5 kV side. Determine CT ratio on 69 kV side.
Solution: Draw work sheet for connection of differential relays showing the main transformer, CT’s, operating and restraining coils of CT’s (Fig. 7). Connect the pilot wires with operating coils and restraining coils as described in the earlier section.
Calculate full load current for a 30 MVA, 11.5 start/69 delta power transformer.
On 11.5 kV side
Ip = 30000 = 1505 A
√3 x 11.5
CT ratio = 3000 / 5 = 600 ( given )
Is = 1505 = 2.51 A
since 11.5 kV side is star connected, CT secondaries will be delta connected. Hence current fed into pilot wires from 11.5 kV side CT secondaries is
√3 x 2.51 = 4.35 A
On 69 kV side
Ip = 30000 = 251 A
√3 x 69
Current in secondary CT’s = current in pilot wires. Since 69 kV side CT secondaries are connected in star = 4.35 A
hence CT ratio = 251 / 4.35 = 57.7
select CT ratio = 60
secondary current = 5 A
primary current = 60 x 5 = 300
ratio on 69 kV side = 300/5
Example 3: Consider a delta/star connected, 15 MVA, 33/11 kV transformer with differential protection applied, for the current transformer ratios shown in figure 8. Calculate the relay currents on full load. Find the minimum relay current setting to allow 125 percent overload.
The HV line current is given by
Ip = 15x106 = 262.43 A
√3 x 33x103
The LV line current is
Is = 15x106 = 787.30 A
√3 x 11x103
The CT current on the HV side is thus
ip = 262.43 ( 300/5) = 4.37 A
The CT current in the LV side is
is = 787.30 (5/2000) √3 = 3.41
Note that we multiply by √3 to obtain the values on the line side of the delta connected CT’s. The relay current at normal load is therefore
ir = ip – is = 4.37 – 3.41 = 0.9648 A
with 1.25 overload ratio, the relay setting should be
Ir = 1.25 (0.948) = 1.206 A
Plug Setting = 1.206/5 = 24.1 %