How Directional overcurrent relay works?

Answer 1

I don't know how you got this posted in the Dodge forum but I will explain.

The directional overcurrent relay is a relay that will provide overcurrent protection in a directional manner. I know this sounds simplistic but let me give you a scenario.

A large industrial company has its own electrical generation provided by a few generators. One might think that the company would have no outside connection to a public utility because they generate their own electricity.

However this is not the case as having a tie to the utility affords a few advantages. Some advantages are that synchronization of the industrial companies generators can easily be maintained at 60Hz. Additional inrush current can be easily provided by the utility, whereas should the industrial company only rely on their generators a larger voltage swing might occur when a large motor is started. The industrial customer can also be provided with backup power levels in case of some failure with their own generation.

So you do want to provide a tie to the utility. You must protect the tie against overcurrents. However, if something were to happen to cause the utility power to fail you certainly do NOT want to try to power all the utilities other customers.

Thats where the directional overcurrent relay comes into play. It will allow power to flow and protect a circuit as long as power is coming into a plant by a tie line. However should power try to flow out of the utility tie the directional overcurrent relay will trip.

A mechanical directional overcurrent relay is actually a combination of a directional relay and an overcurrent relay. The directional portion is closely resembles a watt-hour meter. A potential transformer is required to provide a reference and if current is flowing one direction then a positive torque is placed on a mechanical disk. If current is flowing in the other direction then a negative torque is placed on the disk.

Should the CT and PT connections be made such that positive torque is placed on the disk when current is flowing out of the industrial customer and to the utility then the disk will rotate and cause a contact to close. The closing contact will operate a breaker trip coil. However if current is flowing from the utility to the industrial customer then negative torque will be placed on the disk and it will be stopped by mechanical stops and the breaker will continue to remain closed.

Of course you can use normally open or closed contacts to make the relays operate when and how you want.

Other applications are as a reverse power relay for a generator and for line protection in a grid type system.

Answer 2

Directional relays have protection zones that include all of the power system situated in only one direction from the relay location. (This is in contrast to magnitude relays which are not directional, i.e., they trip based simply on the magnitude of the relay.)

Consider the one-line diagram in Fig. 1.

Fig. 1

If the relays R1 and R2 in Fig. 1 are directional relays, then

- R1 "looks" to the left but not to the right, and

- R2 "looks" to the right but not to the left.

In order to understand how the directional relay works, first, consider that R2 measures the phasors V2 and I23. Now define the following parameters associated with Fig. 1:

· L23: length of circuit 2-3.

· x: distance from R2 to a fault on circuit 2-3.

· λx=x/L23: the fraction of the circuit length between the relay R2 and the fault at point x.

· I23: the current in circuit 2-3 resulting from the fault x on circuit 2-3 (a phasor).

· V2­: the bus 2 voltage (a phasor).

· Z23: total series impedance of circuit 2-3.

If a fault occurs on circuit 2-3, at point x, then the fraction of total circuit length is λx. If the circuit has uniform impedance per unit length, then the impedance between the relay R2 and the fault point is λxZ23, and with the bus 2 voltage being V2, the current flowing into circuit 2-3 from bus 2 is:

(1)

But recall that for transmission lines, it is generally the case that R<

(2)

In that case, eq. (1) becomes:

(3)

Recognizing that 1/j=-90°, eq. (3) becomes:

(4)

Therefore I23 lags V2 by 90°.

Now if the fault occurs on circuit 1-2, at point y in Fig. 1, we can repeat the same analysis as eqs. (1)-(4), except for point y, where we use λy=y/L12, The result will be

(5)

But R2 measures I23, not I21. Reference to Fig. 1 results in the conclusion that

(6)

Therefore, in this case, I23 leads V2 by 90°.

From this simple analysis, we can establish a logic for the directional relay R1.

Define θ23 as the angle of the phasor I23­, i.e.,

(7)

Then if we

· trip when current exceeds pickup & θ23=-90°and

· block if θ23=+90°,

the relay will be directional.