7/27/2019 Negative Sequence Relays

1/10

7/27/2019 Negative Sequence Relays

2/10

NEGATIVE SEQUENCE RELAY

The negative relays are also called phase unbalancerelays because these relays provide protectionagainst negative sequence component of unbalancedcurrents existing due to unbalanced loads or phase-phase faults.

The unbalanced currents are dangerous from

generators and motors point of view as these currentscan cause overheating. Negative sequence relays aregenerally used to give protection to generators andmotors against unbalanced currents.

7/27/2019 Negative Sequence Relays

3/10

DIAGRAM:

http://4.bp.blogspot.com/-zLSwVKC50i4/ThwtvhNx08I/AAAAAAAAA18/kFpSEGFWZC8/s1600/ball1159.jpeg

7/27/2019 Negative Sequence Relays

4/10

CONSTRUCTION: It consists of a resistance bridge network.

The magnitudes of the impedances of all the branchesof the network are equal.

The impedances Z1 and Z3 are purely resistive while theimpedances Z2 and Z4 are the combinations ofresistance and reactance.

The currents in the branches Z2 and Z4 lag by 60o from

the currents in the branches Z1 and Z3. The vertical branch B-D consists of inverse time

characteristics relay. The relay has negligibleimpedance.

7/27/2019 Negative Sequence Relays

5/10

PHASOR DIAGRAM:

http://1.bp.blogspot.com/-hjvGS4XeBXw/ThwvamB3wkI/AAAAAAAAA2A/xfwl8sYfmn4/s1600/ball1160.jpeg

7/27/2019 Negative Sequence Relays

6/10

The current IRgets divided into two equal parts I1 and I2. AndI2 lags I1 by 60o.

1+ 2= rs

Let I1 = I2 = I

The perpendicular is drawn from point A on the diagonalmeeting it at point B. This bisects the diagonal.... OB = IR/2

Now in triangle OAB,

cos 30 = OB/OA... 3/2 = (IR/2)/I

... I = IR/3 = I1 = I2 ............(1)

7/27/2019 Negative Sequence Relays

7/10

Now I1 leads IRby 30owhile I2 lags IRby 30

o.

Similarly the current IB gets divided into two equal partsI3 and I4. The current I3 lags I4 by 60

o. From equation (1) wecan write,

IB /3 = I3 = I4 ...............(2)

The current I4 leads by IBwhile current I3 lags IB by 30o.The current entering the relay at the junction point B in

the Fig. 1 is the vector sum of , and .

Irelay =1 +3 +Y

= IY+ (IR/3) (leads IRby 30o) + IB/3(lags IB by30o)

when the load is balanced and nonegative sequence currents exist.

7/27/2019 Negative Sequence Relays

8/10

1+ 3 = -Y... 1+ 3+ Y= 0

Thus the current entering the relay at point B is zero.Similarly the resultant current at junction D is also

zero. Thus the relay is inoperative for a balancedsystem.

http://4.bp.blogspot.com/-RuB795otd9g/Thwv3bOCH9I/AAAAAAAAA2E/ofBbfC2tTr8/s1600/ball1161.jpeg

7/27/2019 Negative Sequence Relays

9/10

UNDER FAULTY CONDITION:Now consider that there is unbalanced load on

generator or motor due to which negative sequencecurrents exist.

The component I1 and I3 are equal and opposite toeach other at the junction point B. Hence I1 andI3 cancel each other. Now the relay coil carries thecurrent I

Yand when this current is more than a

predetermined value, the relay trips closing thecontacts of trip circuit which opens the circuit breaker.

7/27/2019 Negative Sequence Relays

10/10

ZERO SEQUENCE CURRENT: Under zero sequence currents the total current of

twice the zero sequence current flows through therelay. Hence the relay operates to open the circuitbreaker.

To make the relay sensitive to only negativesequence currents by making it inoperative under theinfluence of zero sequence currents is possible byconnecting the current transformers in delta .Underdelta connection of current transformers, no zerosequence current can flow in the network.