Thursday 1 December 2011

Restricted Earth fault Protection in Transformers & Generators


Restricted Earth fault Protection in

Transformers & Generators

1.0            General :
Transformers and generators are voltage sources. They are traditionally protected by an Over current + Earth fault relay , normally mounted in the breaker panel. This is shown in fig. 1. It should be noted that this protection alone is not adequate. 
When an earth fault occurs within the zone defined as A in the fig-1, or within the machine, the fault current will circulate within the zone or within the machine. The fault current will not flow through the CTs connected to the O/C +E/F relays near the breaker. This will cause a no trip situation when there is a fault in the zone A (Internal fault).

Consequently, a separate scheme is required to detect internal earth faults in zone A. This scheme is called Restricted Earth Fault (REF) scheme.
 
                                                                        


It should be noted that the fault currents in Zone A is limited by the  impedance of the equipments in the zone – for transformers and generators it is very low – the fault currents can rise very fast and damage the equipment. Consequently REF protection is of utmost importance for generators and transformers.

2.0            Why is this called Restricted E/F ?
The name Restricted is derived – since the objective of the protection is to detect the earth fault in the specific zone , restricted to the zone, starting from the breaker to the machine terminals. In case of a generator, the machine terminal is the neutral point. In case of a transformer , the machine terminal becomes the star point of either the primary or secondary winding or both. In case of delta winding, it is the winding itself.                                                                                                                  

3.0            How can we detect internal earth faults ?
It is well established that the sum of currents at the beginning of the zone A should be equal to the sum of currents exiting the zone.    Two sets of CTs are
used to derive these sum of currents at the inlet and exit. A fault in the zone will result in a difference in current .
………..2
-2-

An over current relay is used to measure the  difference in the sum of these currents.   

Please refer to fig.2 where a typical REF scheme shown . The REF relay is connected between P & S . It will pick up if there is enough voltage across P-S to drive the pick up current through the relay.
       

There are two current sources in the CT secondary circuit PQRSTU :
a)       the current produces by CT  X ( loop RSPQ) 
b)      the current produced by CT  Y ( loop UPST) 
Under normal conditions, vectorial sum of the currents of these two CTs will be equal and opposite in the branch P-S (ie) the resultant current in the branch PS will be zero. In this situation , the currents of CT X and CT Y will circulate in the loop PQRSTU. The voltage across P-S will be zero.

When there is an earth fault within the zone A, the currents of  X & Y CTs will not be equal – a small difference of current will flow in the branch P-S. This current will result in a voltage across P-S – since the current flows through the relay impedance. If this voltage is adequate to operate the relay , the relay will pick up – thus detecting a fault within the zone.
…………..3
-3-

4.0            What if a fault occurs outside the zone ?
For a fault outside the zone, the currents through CTs X & Y will be the same – the resultant current through P-S will still be zero and the relay will not not pick up. In this case the fault has to be cleared by another O/C + E/F relay connected near the breaker.

5.0            Why is a Stabilising resistor required in REF scheme ?
We have said in the sections 3 & 4 that the REF relay will detect a fault within a zone restricted between two CTs and it will not detect a fault out side the zone defined by the CTs. This is true, only for ideal conditions- where the two CTs are perfectly matched. Following mis matches will occur under practical situations in the field :

a)                 the CT secondary impedances may not be equal
b)                the lead wires connecting the CT secondaries to the relay may not have equal resistances
c)                 the CTs may have different ratio error and phase angle error – due to this , the secondary cuurents will not be equal even if the primary currents are same..
d)                the CTs may have different saturation characteristics – this will cause small difference in the secondary currents for a same primary current

The cumulative effect all the above, can make the relay trip even when there is a full load current flowing in the primaries – though the primary side currents are same, the secondary side currents need not be the same – a voltage sufficient to trip the REF relay may develop across P-S and hence the relay will trip.

To make the relay insensitive to this voltage produced by CT mismatch, a resistor is added in series with the relay.  Once this resistor is added, the relay will need a voltage which is higher than  the voltage produced by the CT mismatch. This resistor is called stabilizing resistor – this is an important component in REF scheme –since this ensures stability in the scheme by avoiding spurious tripping.

6.0            What is knee point voltage ?
It should be noted that actual input crcuit to trip mechanism (consisting of the
relay + stabilizing resistor) has become a high impedance circuit.   If the relay
has  to  trip,  the CT  secondaries   should  produce  sufficiently  high  enough
voltage to activate the relay,  after allowing for the drop across the stabilizing
resistor.
……..4
-4-

To ensure the CTs produce enough voltage,  an additional specification – the
Knee poit voltage - is included for the CTs used for REF protection .

Knee point voltage (KPV)is defined as the point on the magnetizing curve (of
the material used for the CT core) where the core will need 50% increase in
the magnetizing force (ampere turns) to cause a 10% increse in the flux
density.(voltage build up across secondary). In effect, KPV defines the end of
the linear portion of the BH curve .  Higher the KPV, larger is the linear zone
and better will be secondary output for higher fault currents. Higher the KPV,
better are  the chances of a high impedance relay trip.

7.0            How can we calculate the value of the stabilizing resistor and KPV ?
The value of stabilizing resistor and KPV will depend on the following parameters which are unique to a given feeder:

a)                 Impedance of the CT secondaries
b)                Lead wire resistance between the CT secondaries and the REF relay
c)                 The impedance of the REF relay -  this can vary with respect to pick up setting .  We have to consider the relay impedance at the pick up setting being contemplated for the feeder.
d)                The maximum fault current which can occur on the CT secondary side – CT should not saturate under this maximum fault conditions. If the CT saturates, it will offer an alternate path for the resultant current – and the REF relay may not trip.

A detailed method is provided in annexure 1, for calculation of stabilizing resistor and KPV

8.0            What are L&T solutions for REF protections ?
L&T manufactures a high impedance Over curret relay – which is ideally suited for the REF protection of generators and transformers. Typical schemes are shown in fig. 3 &4.

L&T offers a unique feature in the REF relay SC14S – in additon to instantaneous trip, user can select a definite time delay of either 100 millisec or 200 millisec. In case of small transformers & generators ( upto 5 MVA), the feeder trips during breaker closing. This mainly is due to the large inrush current causing momentary difference in CT secondary currents due to mismatch in saturation characteristics. A 100 millisec time delay will help in this case.

…………..5
-5-


Fig.3 shows a typical scheme for REF protection for generators. The scheme envisages the following ;

a) 3 nos. phase CTs
b) 1 no. neutral CT
c) 1 no.  Relay SC14S
d) 1 no. Stabilising resistor
 


…………6
-6-

Fig.4 shows a typical scheme for REF protection for transformers. It should be noted that transformers will need two REF schemes – one on the primary side and the other on the secondary side.

For the transformer primary side , which is usually delta connected, following are envisaged in the REF scheme :

a) 3 nos. phase CTs
b) 1 no.  Relay SC14S
d) 1 no. Stabilising resistor

For the transformer secondary side, which is usually star connected, following are envisaged in the REF scheme :

a) 3 nos. phase CTs
b) 1 no. neutral CT
b) 1 no.  Relay SC14S
d) 1 no. Stabilising resistor



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NEGATIVE SEQUENCE PROTECTION FOR AC MOTORS


NEGATIVE SEQUENCE PROTECTION FOR AC MOTORS

1.0              Need for Negative Sequence Protection

Primary cause of motor failure is excessive heating, which if sustained over long time periods will result in motor burn out.  Over heating also reduces the life of motor. If a motor is continuously over heated by just 10 degrees, its life can get reduced by almost 50%.

Over heating normally occurs due to over current, which in turn may be due to over loads or locked rotor condition or low voltage or phase failure or repeat starts or phase unbalance.

Bimetallic relays are most economical solution for heating due to over loads. However they suffer from inherent deficiencies like poor accuracy, rigid inverse time characteristics, poor repeatability etc. They are totally insensitive to current unbalance, which is one of the major contributors to over heating in motors.

Though the three-phase motor is supposed to be a balanced load, current unbalance occurs frequently in motor feeders due to following:

a)         voltage unbalance in the feeder supply
b)                  phase reversal
c)                  single phasing

Current unbalance in a motor is best represented by the presence of excessive negative sequence component in the motor current. Consequently it is necessary to protect motors against negative sequence .

2.0       Supply Unbalance & Sequence components :

When the power supply to the motor is unbalanced, the unbalanced voltage and the resulting unbalanced currents in the three phases can be resolved into three balanced components as follows :

a)                  Positive Sequence component : This component is in the same phasesequence as that of the motor current. All its three phases are perfectly balanced – they are equal in magnitude and are displaced by 120 degrees. The positive sequence component represents the amount of balance in the power supply  and consequently  is instrumental in delivering useful power.

b)                  Negative Sequence component : This component has a phase sequence opposite to that of the motor current hence the name negative sequence. It represents the amount of unbalance in the feeder . All its three phases are perfectly balanced – they are equal in magnitude and are displaced by 120 degrees. This component does not produce useful power – however by being present it contributes to the losses and causes temperature rise.

      ………2
-2-

c)                  Zero Sequence component : This , if present, represents extent of earth fault in the feeder. All its three phases are in the same direction.

3.0       Effect of Unbalance in Motors :

3.1              When the supply voltage is unbalanced, the positive sequence component reduces and results in low power delivery.

An unbalance of only 5% ( R phase = 415 V , Y phase = 415 V, B phase = 436 V) can reduce the out put by 25%, even though the motor continues to draw the same current as at the time of balanced condition. This means that the motor current has to increase under unbalanced condition in order to deliver the same power. This increase will result in motor heating.

3.2              During supply unbalance, negative sequence currents flow through the stator windings. This results in induction of  negative sequence voltage in the rotor windings. Since the rotor is short circuited, this will result in abnormal current flow in the rotor and damage the rotor winding. A voltage unbalance of the order of 3% can increase the heating by nearly 20% in the rotor.

3.3              The negative sequence impedance of the motor is approximately same as the locked rotor impedance which in turn is approximately one sixth of normal motor impedance. Due to this even small voltage unbalance can produce large negative sequence current in the motor.

3.4              The frequency of the negative sequence current induced in the rotor will be equal to (supply frequency) x (2-slip) Hz. This is due to the fact that it is revolving in the opposite direction . This frequency will be typically around 99 Hz during normal motor operation . Due to skin effect, high frequency negative sequence currents encounter high rotor resistance. This inturn results in over heating. The increased resistance is typically 5 times the normal positive sequence resistance.

4.0       Reasons for Unbalance :

Voltage unbalance can be due to many reasons. These include unsymmetrical loading, blown fuses in capacitor banks and single phasing.

5.0              Single phasing :

Single phasing causes worst kind of unbalance – it produces equal amount of positive and negative sequence components. The prime causes for single phasing are :

a)                  broken or burn out of connecting leads to motor
b)                  blown fuses
c)                  faulty contact in switching element



……….3
-3-


To understand the effect  of single phasing in motors, consider the condition shown below .











It can be seen that the current in winding C will be larger than that in A and B and consequently it will burn out. Typical currents in the windings are shown below >
 

Motor current              Current in windings when single phasing has occured          
(% of full load)           (% of full load)
 

3 phase current           Lines L1 & L3                        Winding A&B             Winding C
 


            50                                80                               50                               108
            55                                 90                               58                               118
            60                                102                              62                               131
            65                                120                              70                               147
            70                                130                              79                               161
            75                                147                              87                               180
            80                                165                              95                               198
            85                                180                              102                              215
            90                                200                              111                              235
            95                                222                              120                              258
            100                              243                              129                              285
 

From the above table , we can see that in a motor running at 50% load, on single phasing in L2,

a)         the healthy lines L1 and L3 will carry 80% of full load
b)         the windings will carry 50% of full load
c)         the winding C carries more than twice full load (108%)

A thermal over load relay in this case ,  will  not trip since it is seeing only 80% current in healthy lines L1 & L3 while the winding C is already over stressed. This situation gets worse with increasing loads at the time of single phasing.

A negative sequence calculation would have helped in this case.

…………..4
-4-

6.0              Phase reversal :

Phase reversal in a motor can be very dangerous – the motor will run in the opposite direction resulting in severe damage to gear boxes, material flow problems and hazard to operating personnel.

Phase reversal in a motor feeder produces a negative sequence of nearly 100% and is well detected by a relay with negative sequence protection.

Considering the above details, it can be easily concluded that a motor protection relay will provide an effective protection to the motor only when it is equipped to measure the negative sequence component level in the motor feeder.

REF relay setting calculation



REF RELAY SETTING CALCULATION


Transformer Data :

  8 MVA , 6.6KV/ 415 ,  Impedance = 0.0835 p.u., Tap + 10% ( 440Kv – 360KV

CTR = 800 /1

Relay  : CAG 14 / P120

 

                                Fault MVA = MVA  / Transformer Impedance
                                        =  8 / 0.0835 = 95.8084

Fault Current  =    Fault MVA  x 1000 / ( 1.732 x  Voltage rating )
                      =   95.8084 x 1000 ( 1.732 x 6.6 )
                      =    8281.06  Amp primary
                          = 8281.06 / 800  = 10.4763  Amp Secondary

Fault Voltage developed across the relay   =  fault current  x loop resistance
 at the time of Maximum fault conditions

                                    V s       =    If ( Rct + 2 RL)     RL  = CT Lead Resistance
                                                 =   10.4763  x (  Rct + 2 RL)
                                                =    10.4763  x ( 4 + 2)  ( assumed RL = 1 ohms)
                                                =     62.8579 volts
The relay has to be set at   Is = 0.1 In for maximum sensitivity

The stabilizing resistor shall be set at value of resistance during fault minus the relay resistance

                                                =   62.85    -    1 VA
0.1                            (0.1) Square       
                                            = 628.5      -  100
                                                    R     =     528.5 ohms