Power System Protection by azlan4477

PREPARED BY: DR. ZULKIFFLI BIN ABDUL HAMID

Introduction  When faults occur, excessive currents will flow in the power system.  These fault currents can be several orders of magnitude larger than normal operating currents  If not removed quickly, they may cause:  Insulation damage  Conductor melting  Mechanical damage to windings and busbars  Fire and explosion

 It is therefore important to have power system protection in order to:  Remove the faulted equipment from the system  Maintain the unfaulted system in order to continue service

System Protection Components  Protection systems have three basic components:  Instrument transformer  Relays  Circuit-breakers

 Example:  Instrument transformer - current transformer (CT) & voltage transformer

(VT)  Relays – overcurrent relay (OC), directional relay (D), impedance relay & differential relay.

System Protection Components I I'

Basic protection system (per-phase)

System Protection Components  The function of the current transformer (CT):  To reproduce secondary current which will activate the relays.  It converts (step-down) primary currents I (in kA range) to secondary

currents I’ ( in 0 to 5A range) for convenience of measurement.  By having low magnitude of current: Safety: Provide electrical isolation from the power system so that personnel working with relays will work in a safer environment. Economy: Lower level inputs enable relays to be smaller, simpler & less expensive. Accuracy: Accurately reproduce power system currents & voltages over wide operating ranges

System Protection Components  The function of the overcurrent (OC) relay:  To discriminate (differentiate) between normal operation and fault condition.  Receive signal (or current) from the secondary winding of the current transformer.  Has an operating coil, which is connected to the CT secondary winding & a set of contacts.  When I’ exceeds a specified “pick up” value IP (i.e. I’ > IP) the operating coil causes the relay contacts to close.  When the relay contacts close, the breaker trip coil is energized, which then causes the circuit breaker to open.  The function of the circuit breaker (CB):  To separate between the faulty and non-faulty parts in power systems.  It opens when there is a fault happens.

System Protection Components Instrument transformer

Step-down the current (CT) or voltage (VT) to be sent to relay

Relay

Circuit breaker

Receive the current (CT) or voltage (VT).

Its contacts remains close if no fault happens.

Identify whether fault happens .

When fault happens, its contacts open and separate the faulted zone from the remaining system.

If yes, it tells the circuit breaker to open.

Instrument Transformer  There are two basic types of instrument transformers:  Voltage transformers (VT) formerly called potential transformer (PT)  Current transformers (CT)

 The VT reduces the primary voltage and the CT reduces the primary current to much lower, standardized levels suitable for operation of relays.

VT CT

: step down from primary voltage to secondary voltage (V to V’ ) : step down from primary current to secondary current (I to I’ )

Instrument Transformer

VT and CT schematic diagram

Instrument Transformer – voltage transformer (VT)  VT is usually modelled as an ideal transformer. The secondary voltage is found as:

 Where, V’ is the secondary voltage, V is the primary voltage and n is the VT turn ratio. V’ is in phase with V.  Standard VT ratios, n are given below:

Instrument Transformer – voltage transformer (VT)  For example, if the primary voltage of VT is 200 kV and its turn ratio is 300:1, then the secondary voltage will:

V ’=V/n = 200 kV / (300/1) = 0.67 kV

Instrument Transformer – current transformer (CT)  Approximate equivalent circuit of a CT is shown below:

 Where:

Z’ = secondary leakage impedance Xe = saturable or excitation reactance ZB = impedance of terminating device (relay). Sometimes called as burden

Instrument Transformer â&#x20AC;&#x201C; current transformer (CT) ď&#x201A;&#x2014; Standard CT ratios are given below:

Instrument Transformer – current transformer (CT)  The total impedance ZB of the terminating device is called the burden.  Associated with the CT equivalent circuit is an excitation curve that determines the relationship between the secondary voltages E’ and excitation current Ie.  Using the CT equivalent circuit and excitation curves, the following procedure can be used to determine CT performance. Step 1: Assume a CT secondary output current I’ Step 2: Compute E’ = (Z’ + ZB) I’ Step 3: Using E’, find Ie from the excitation curve Step 4: Compute I = n(I’ + Ie) Step 5: Repeat steps 1 to 4 for different values of I’, then plot I’ versus I  The CT error is given by:

Excitation curves for a multi ratio bushing CT with a C100 ANSI accuracy classification

Instrument Transformer – current transformer (CT)  For example, if the primary current of CT is 500 A and its turn ratio is 900:5, then the secondary current will (assume Ie = 0):

I ’=I/n = 500 A / (900/5) = 2.78 A

Relay  In this chapter, there are four types of relay to be covered:  Overcurrent relay  Directional relay  Impedance relay  Differential relay

Part 1 Overcurrent relay (OC)

Overcurrent relay  There are two types of overcurrent relays:  Instantaneous overcurrent relays  Time-delay overcurrent relays

Overcurrent relay - Instantaneous  Instantaneous overcurrent relays respond to the magnitude of their input current, I’ as shown by the trip and block regions.  If the magnitude of I’ exceeds a specified adjustable current magnitude IP called the pickup current, (i.e. I’ > IP), the relay contacts close instantaneously to energize the circuit breaker trip coil.  If I’ < IP, then the relay contacts remain open, blocking the trip coil.

I’ > IP : I’ < IP :

Relay trips, relay contacts close, circuit breaker open Relay not trip, relay contacts stay open, circuit breaker remain close

Overcurrent relay - Instantaneous

Trip

Block

Instantaneous overcurrent relay - block and trip regions

Overcurrent relay - Time-delay  Time-delay overcurrent relays also respond to the magnitude of their input current, I’ but with an intentional time delay.  The time delay depends on the magnitude of the relay input current.  For I’ > Ip :

The relay will trip. The time delay for the relay to trip will be small if I’ is large. Otherwise, the time delay will be large if I’ is small.  If I’ < IP, the relay remains in the blocking position.  These relays have two settings:  Current tap setting: The pickup current, IP in amperes  Time dial setting: The adjustable amount of time delay

Overcurrent relay - Time-delay td 1 < td 2 < td 3

(time delay)

Increase Iâ&#x20AC;&#x2122;

Trip

Block

Time-delay overcurrent relay block and trip regions

Protection using overcurrent relay  Many radial systems are protected by time-delay overcurrent relays.  These relays are coordinated to operate in sequence so that they interrupt minimum load during faults.  The breaker closest to the fault opens, while other upstream breakers with larger time delays remain closed.

Single line diagram of a 34.5kV radial system

Protection using overcurrent relay  Consider a fault at P1 to the right of breaker B3:  For this fault we want breaker B3 to open while B2 and B1 remain closed.  Under these conditions, only load L3 is interrupted.  We could select a longer time delay for the relay at B2 (as a back-up), so that

B3 operates first.  Thus, B3 provides primary protection for any fault to the right of B3.  Only if B3 fails to open will B2 open, after time delay, thus providing backup protection.

I fault

Protection using overcurrent relay  Similarly, consider a fault at P2 between B2 and B3.  We want B2 to open while B1 remains closed.  Under these conditions, loads L2 and L3 are interrupted.  Since the fault is closer to the source, the fault current will be larger than

the previous fault considered.  We could select a longer time delay for the relay at B1 (as a backup), so that B2 opens first.  Thus, B2 provides primary protection for faults between B2 and B3.  Similarly, B1 provides primary protection for faults between B1 and B2.

I fault

Part 2 Directional relay

Directional Relays ď&#x201A;&#x2014; Overcurrent relays can also be used with the use of directional element. This type of relay is called directional overcurrent relay or simply directional relay. ď&#x201A;&#x2014; Directional relays are designed to operate for fault currents in only one direction, i.e. in forward direction.

With the inclusion of directional capability, the overcurrent relay becomes directional overcurrent relay.

Directional Relays

Directional relay in series with overcurrent relay

Directional Relays ď&#x201A;&#x2014; Consider the directional relay D in previous figure, which is required to operate only for faults to the right of the CT. ď&#x201A;&#x2014; Since the line impedance is mostly reactive, a fault at P1 to the right of the CT will have a fault current I from bus 1 to bus 2 that lags the bus voltage V by an angle of almost 90 degrees. This fault current is said to be in the forward direction. ď&#x201A;&#x2014; On the other hand, a fault at P2, to the left of the CT, will have a fault current I that leads V by almost 90 degrees. This fault current is to said to be in the reverse direction.

Directional Relays  The directional relay has two inputs, namely the secondary voltage of VT and secondary current of CT:

V ' ∠0° & I ' ∠φ

The condition to trip:

− 180° < (φ − φ1 ) < 0° [Trip ]

Block

Otherwise

Trip

[ Block ]

φ : Angle of current, I’ φ1 : Boundary between trip & block

I' Directional relay block and trip regions in the complex plane

regions (typically 2 to 8 degrees)

Directional Relays - torque  The torque that will cause a directional relay to trip; (i.e. to close its contacts) can be calculated as follows:

τ=

1 V ' I ' sin(φ1 − φ ) 2πf

= kV ' I ' sin(φ1 − φ ) = kV ' I ' sin(θV − θ I )  Where, k = 1/ω

Directional Relays  From torque equation:

τ = kV ' I ' sin(φ1 − φ )  For faults in the forward direction:  The current lags the voltage, the angle (φ1 – φ) is close to +90 degrees.

 This results in maximum positive torque on the rotating disc of relay, which

would cause the relay contacts to close (i.e. trips).

 For faults in the reverse direction:  The current leads the voltage, and (φ1 – φ) is to close –90 degrees.

 This results in maximum negative torque on the rotating disc of relay,

which would cause the relay contacts to open (i.e. block).

Directional Relays - operation  Fault happens in forward direction of directional relay:

I lags V  (φ1 – φ) = +90 degrees  τ = +maximum  relay contacts close  directional relay trips

 Fault happens in reverse direction of directional relay:

I leads V  (φ1 – φ) = –90 degrees  τ = –maximum  relay contacts open  directional relay not trip

Directional Relays So, for a protection system that has both overcurrent relay and directional relay, the circuit breaker will open if both conditions as follows are satisfied: Overcurrent relay (OC): Iâ&#x20AC;&#x2122; > Ip Directional relay (D): the fault happens in forward direction (I lags V)

Example

Protection of Two Source System with Directional Relays ď&#x201A;&#x2014; It becomes difficult and in some cases, impossible to coordinate overcurrent relays when there are two or more sources at different locations. ď&#x201A;&#x2014; Consider the system with two sources as follows, where B12, B21, B23 and B32 are circuit breakers with directional relays, while B1 and B3 are circuit breakers with overcurrent relays.

System with two sources

Protection using Directional Relays – line fault  Suppose there is a fault at P1.  The fault happens in forward direction of B23, B32 and B12.  We want B23 and B32 to clear the fault as they are close to the fault location.  B12 could functions as a backup by setting its time delay longer than B23. If

B23 fails to open, B12 will replace its role.  B3 could also function as a backup if B32 fails to open.

Protection using Directional Relays – line fault  Now consider a fault at P2 instead.  The fault happens in forward direction of B12, B21 and B32.  We want B12 and B21 to clear the fault as they are close to the fault location.  B32 could functions as a backup by setting its time delay longer than B21. If

B21 fails to open, B32 will replace its role.  B1 could also function as a backup if B12 fails to open.

Protection using Directional Relays – bus fault  If the fault happens at:  Bus 1: B1 and B21 will operate to clear the fault.  Bus 2: B12 and B32 will operate to clear the fault.  Bus 3: B3 and B23 will operate to clear the fault.

Zones of Protection  A fundamental concept in protection is the division of a system into protective zones.  If a fault occurs anywhere within zone, action will be taken to isolate that zone from the rest of the system.  Division of protective zones can be performed using three steps as follows:  Step 1: define the zones for line components such as transmission lines &

transformers from one bus to another bus (each zone will have two breakers)  Step 2: define the zones for bus (any breakers connected to the bus are in the same zone).  Step 3: define the zones for terminal components such as generators, motors and loads (each zone will have one breaker).

Zones of Protection  Protective zones have the following characteristic:  Zones are overlapped.  Circuit breakers are located in the overlap regions.  For a fault happens in a zone, all circuit breakers in that zone open to clear

the fault.  For a fault happens in an overlap region between two zones, all circuit breakers in both zones open to clear the fault.  Neighbouring zones are overlapped in order:  To avoid the possibility of unprotected areas  Without overlap, the small area between two neighbouring zones would

not be located in any zone and thus would not be protected.

Zones of Protection Zone for generator

Zone for bus

Zone for motor Zone for line

Power system protective zones

Zone for transformer

Zones of Protection  From the previous power system:

Zones for terminal component: Zone 1, 3 and 9. Zone for bus: zone 4 and 7. Zone for line component: zone 2, 5, 6, 8 and 10.  If a fault occurs at P1, then the two breakers in zone 5 should open.  If a fault occurs at P2 within the overlap region of zone 4 and 5, then all five breakers in zones 4 and 5 should open.  Clearly, if a fault occurs within an overlap region,  Two zones will be isolated and a larger part of the system will be lost from

service  To minimize this possibility, overlap regions are kept as small as possible.

Example ď&#x201A;&#x2014; Draw the protective zones for the power system shown below. Which circuit breakers should open for a fault at P1 and P2?

Solution

Fault at P1: Breakers B24 and B42 will open Fault at P2: All breakers in zone 4 and 5 will open (B21, B23, B24 and B42)

Part 3 Impedance relay

Impedance Relays  Commonly used to protect buses and transmission lines.  These relays operate based on voltage-to-current ratio, which in turn produces impedance.  It is also called a distance relay because the impedance is proportional to the distance in transmission lines.  The reach of an impedance relay denotes how far down the line the relay detects fault.

For example, an 80% reach means that the relay will detect any (solid three phase) fault between the relay and 80% of the line length.

Impedance Relays  Impedance relay block and trip regions are shown below  The relay trips for |Z| < |Zr|  Where Z is the voltage-to-current ratio at the relay location  Zr is an adjustable relay setting.

Im(Z )

Block

Trip Re(Z ) Impedance relay block and trip region

Impedance Relays ď&#x201A;&#x2014; Consider an impedance relay for breaker B12, for which Z = V1 / I12

Note: Zone 1, 2 and 3 are the protective zones covered by breaker B12.

Impedance Relays  During normal operation,  Load currents are usually much smaller than fault currents, and the ratio Z

= V / I has a large magnitude.  Therefore Z will lie outside the circle (i.e. in block region), and the relay will not trip during normal operation.  During a three-phase fault at P1, however,  Z appears to relay B12 to be the line impedance from the relay to the fault

location.  If |Z| < |Zr| (i.e. in trip region), then the B12 relay will trip.

Impedance Relays  Also, during a three-phase fault at P3,  Z appears to relay B12 to be the negative of the line impedance from the

relay to the fault  As long as |Z| < |Zr| (i.e. in trip region), the B12 relay will trip regardless of fault location (forward or reverse direction).  Thus, the impedance relay is not directional!

Impedance Relays  Two ways to include directional capability with an impedance relay are (i.e. to make it operate in forward direction only) :  Directional restraint: by including a directional relay in series with an

impedance relay  Mho relay: by offsetting the centre of the impedance circle from the origin.

 If either of these relays is used at B12 in the previous power system, a fault at P1 will result in a trip decision, but a fault at P3 will result in a block decision.

With the inclusion of directional capability, the impedance relay becomes directional impedance relay.

Impedance Relays Zr

Zr Trip

Impedance relay with directional restraint

Block

Trip

Block

Modified impedance relay (mho relay)

Impedance Relays  It is common practice to use three directional impedance relays per phase, with increasing:  Reaches (i.e. Zr)  Time delays.

 Thus, there will be three protective zones covered by the relays per-phase. Zr3

Zr2 Z r1

Impedance relay with directional restraint (per-phase)

Zr3

Zr2 Z r1

Mho relay (per-phase)

Impedance Relays ď&#x201A;&#x2014; So in practice, each directional impedance relay will have 3 protective zones, as shown below:

Note: Zone 1, 2 and 3 are the protective zones covered by breaker B12.

Impedance Relays  For example, consider B12 with three relays per phase for the previous power system.

Zone 1 (relay 1):  Provide primary protection for line 1-2.  Relay is typically set for an 80% reach of line 1-2. Any fault happens between the relay and 80% of the line length will be cleared by this relay.  Instantaneous operation (no delay) Zone 2 (relay 2):  Provide backup protection for line 1-2 as well as remote backup for line 2-3 or 2-4.  Relay is set for about 120% reach of line 1-2, extending beyond bus 2.  Typical time delay of 0.2 to 0.3 seconds. Zone 3 (relay 3):  Provide remote backup for neighbouring lines.  Relay is set for 100% reach of line 1-2 plus 120% of either line 2-3 or 2-4 whichever is longer. It extends beyond bus 3 and 4.  Typical time delay one second. 58

Impedance Relays  In order to determine the tripping status of impedance relay, the secondary impedance of the relay’s transformer, Z’ must be determined.  Secondary impedance:

V ' (Vφ / nVT ) Vφ  nCT =  Z'= = I ' ( I φ / nCT ) I φ  nVT

 n  = Z  CT   nVT

  

 Where:

V’ = secondary voltage of VT I’ = secondary current of CT Vφ = phase voltage Iφ = phase current (assumed to be the same as line current) Z = primary impedance nVT and nCT = VT and CT turn ratios respectively

Impedance Relays  After that, the setting impedance, Zr is calculated based on the percentage of reach defined for each zone.  The setting impedance of a zone:

 nCT Z r = rZ line   nVT

  

 Where, r is the reach expressed in fraction and Zline is the line impedance.  By comparing |Z’| < |Zr|, we can determine whether the relay trips or not.

Impedance Relays ď&#x201A;&#x2014; Relay connections for a three-zone impedance relay with directional restraint are shown below:

Relay connections for a three zone directional impedance relay (per-phase)

Example  Table below gives positive sequence line impedances as well as CT and VT ratios at B12 for the 345kV system shown in the previous diagram.  Determine the settings Zr1, Zr2 and Zr3 for the B12 three-zone, directional

impedance relays. Consider only solid, three-phase faults.  Maximum current for line 1-2 during emergency loading conditions is 1500 A at a power factor of 0.95 lagging. Verify that B12 does not trip during this condition.

Impedance Relays ď&#x201A;&#x2014; So in practice, each directional impedance relay will have 3 protective zones, as shown below:

Note: Zone 1, 2 and 3 are the protective zones covered by breaker B12.

Solution  The ratio of CT to VT:

nCT 1500 / 5 1 = = nVT 3000 / 1 10  Zone 1: 80% reach of line 1-2

 nCT Z r1 = (80%) Z1− 2   nVT

 1  = 0.8(8 + j 50)  = 4.05∠80.9° Ω  10  

Solution  Zone 2: 120% reach of line 1-2

 nCT Z r 2 = (120%) Z1− 2   nVT

 1  = 1.2(8 + j 50)  = 6.08∠80.9° Ω  10  

 Zone 3: 100% of line 1-2 plus 120% of either line 2-3 or 2-4 whichever is longer

 nCT Z r 3 = (100%) Z1− 2   nVT

  nCT  + (120%) Z 2 −3    nVT

  

1 1 = 1.0(8 + j 50)  + 1.2(8 + j 50)  = 11.14∠80.9° Ω  10   10  65

Solution  At emergency: V = 345 kV and I = 1500A with pf = 0.95 lagging  Phase voltage and current:

Vφ =

345 = 199.2∠0° kV 3

I φ = 1500∠ − cos −1 0.95 = 1500∠ − 18.19°A  Secondary impedance:

Z'=

Vφ  nCT  I φ  nVT

 199.2∠0° k  1   =   = 13.28∠18.19° Ω  1500∠ − 18.19°  10 

 Since |Z’| > |Zr1|, |Zr2| and |Zr3|, the impedance is in block region. So, no relay will trip.

Part 4 Differential relay 67

Differential Relays  Commonly used to protect generators, buses, power transformers and

transmission lines.

 These relays operate based on the difference in current entering and

leaving a component.

 The entering current is noted as input current, I1’ while the leaving

current is noted as output current, I2’.

 The condition for differential relay to trip:

I1 '−I 2 ' ≠ 0

⇒ relay will trip

I1 '−I 2 ' ≈ 0

⇒ relay will not trip

Differential Relays ď&#x201A;&#x2014; Figure below illustrates the basic method of differential relaying for

generator protection.

Per-phase differential relaying for generator protection

Differential Relays  When the relay in any one phase operates, all three phases of the main

circuit breaker will open, as well as the generator neutral and field breakers.

 The tripping of differential relay is as follows:  When no fault happens in the generator windings (stator windings):  the current in the relay operating coil is zero, i.e. I1’ - I2’ = 0  relay does not operate.

 When there is fault happens in the generator windings (stator

windings):  the current in the relay operating coil is not zero, i.e. I1’ - I2’ ≠ 0  the relay operates.

Differential Relays - structure ď&#x201A;&#x2014; An electromechanical differential relay called a balance beam relay is

shown below:

Balance beam differential relay

Differential Relays - structure  The relay contacts close if the downward force on the right side exceeds

the downward force on the left side.

 The electromagnetic force on the right, FR is given by:

FR = [ N 0 ( I1 '− I 2 ' )]2  Similarly, the electromagnetic force on the left, restraining coil is

proportional to

N  FL =  r ( I1 '+ I 2 ' )  2   The relay will trip if:

FR > FL 72

Differential Relays - structure  Hence:

N  [ N 0 ( I1 '− I 2 ' )]2 >  r ( I1 '+ I 2 ' )  2 

 Taking the square root:

N 0 ( I1 '− I 2 ' ) > ( I1 '− I 2 ' ) >

Nr ( I1 '+ I 2 ' ) 2

k ( I1 '+ I 2 ' ) 2

⇒ k=

Nr N0

2( I1 '− I 2 ' ) > k ( I1 '+ I 2 ' ) 73

Differential Relays  Expanding and simplifying the equation:

I1 ' (2 − k ) > I 2 ' (k + 2)  Finally, the condition for the relay to trip becomes:

I1 ' 2 + k > I2 ' 2 − k

( trip)

Differential Relays  Therefore, a differential relay will trip if one of the following conditions

is satisfied:

 2−k  I 2 ' < I1 '    2+k 

2+k  I 2 ' > I1 '    2−k 

( trip)

 The relay will not trip (block) if the following condition is satisfied:

2+k   2−k  I1 '    < I 2 ' < I1 '   2−k   2+k 

(block)

Differential Relays â&#x20AC;&#x201C; curve with k = 0.1 ď&#x201A;&#x2014; The previous inequalities are plotted to obtain the block and trip

regions of the differential relay for k = 0.1, as follows: Upper bound

I 2 ' = 1.11 I1 '

I 2 ' = 0.905 I1 '

Lower bound

Differential relay block and trip regions

Differential Relays – curve with k = 0.1  For upper bound, the percentage of mismatch between I1’ and I2’ is:

I 2 '− I1 ' 1.11 I1 '− I1 ' ×100% = ×100% = 11% I1 ' I1 '  For lower bound, the percentage of mismatch between I1’ and I2’ is:

I 2 '− I1 ' 0.905I1 '− I1 ' ×100% = ×100% = 9.5% I1 ' I1 '  Hence, it can be understood that the relay will block for:  up to 11% mismatch between I1’ and I2’  down to 9.5% mismatch between I1’ and I2’  In other words, the relay will trip if the mismatch between the currents is either below 9.5% or above 11%.

Differential Relays  Note that as k increases, the block region becomes larger (i.e. the relay

becomes less sensitive).

 Advantages of differential relay are:  provides primary zone protection without backup.  coordination with protection in adjacent zones is eliminated  permits high speed tripping.  precise settings are unnecessary.  the need to calculate fault current and voltages is avoided.

Differential Relays â&#x20AC;&#x201C; bus protection ď&#x201A;&#x2014; Differential relay for bus protection is illustrated by the following

single line diagram:

Single line diagram of differential bus protection

Differential Relays – bus protection  In practice, three differential relays are required, one for each phase.  Operation of any one relay would cause all of the three-phase circuit

breakers connected to the bus to open, thereby isolating the threephase bus from service.

 When there is no fault happens at the bus:  the current in the relay operating coil is zero, i.e. I1’ + I2’ - I3’ = 0  relay does not operate

 When there is a fault happens at the bus:  the current in the relay operating coil is not zero, i.e. I1’ + I2’ - I3’ ≠ 0  relay operates

Differential Relays – transformer protection  The protection method used for power transformers depends on the

transformer MVA rating.

 Fuses are often used to protect transformers with small MVA ratings,

whereas differential relays are commonly used to protect transformers with ratings larger than 10MVA.

Differential Relays â&#x20AC;&#x201C; transformer protection ď&#x201A;&#x2014; The differential protection method is illustrated as follows for a single-phase, two-winding transformer.

Differential protection of a single phase, two winding transformer (perphase)

Differential Relays – transformer protection  Denoting the turns ratio of the primary CT (CT1) and secondary CT (CT2) as follows:

1 a1 = n1

1 a2 = n2

 The secondary currents of CT1 and CT2 are:

 And the current in the relay operating coil is:

Differential Relays – transformer protection  When no fault happens, the primary and secondary currents of the twowinding transformer are related by:

 Where, N1 and N2 are the primary and secondary winding of two-winding transformer. Substitute (3) into (2), we get

Differential Relays – transformer protection  To prevent the relay from tripping at no fault, the differential relay current I’ must be zero. Hence, at no fault:

(At no fault)

Differential Relays – transformer protection  When there is a fault in the transformer, the differential relay current must be:

 The relay will only trip if the following operating conditions is satisfied.

 Also, the value of k in the above equations can be selected to control the size of the block region of differential relay

Differential Relays – transformer protection  Hence for transformer protection using differential relay:

When there is no fault, the following conditions must be satisfied:

I 1  N1 / N 2   = 0 I ' = I1 '− I 2 ' = 1 − n1  n2 / n1   When there is a fault, the following conditions must be satisfied:

I ' = I1 '− I 2 ' =

I 1  N1 / N 2   ≠ 0 1 − n1  n2 / n1  87

Example ď&#x201A;&#x2014; A single-phase, two-winding, 10 MVA, 80/20 kV transformer has differential relay protection with two current transformers at primary and secondary sides. Determine the current in the two-winding transformer and in the CTs at rated condition. Hence, select suitable CT ratios based on the following table.

Example  Rated primary & secondary currents for two-winding transformer:

S 10 M I1 = = = 125 A V1 80k I2 =

S 10 M = = 500 A V2 20k

 Since:  I1 = 125 A, choose 150:5 tap setting for CT1  I2 = 500 A, choose 600:5 tap setting for CT2

CT ratio must be than rated current

Example ď&#x201A;&#x2014; So, the secondary currents of CT1 and CT2 are:

I1 125 I1 ' = = = 4.167 A n1 (150 / 5) I2 ' =

I2 500 = = 4.167 A n2 (600 / 5)

Common Problem in Differential Transformer Protection  A common problem in differential transformer protection is the mismatch of relay currents (I1’ and I2’) that occurs when standard CT ratios are used.  If the primary winding in the previous example has a 138 kV instead of 80 kV rating, then the rated primary current is:

I1 =

S 10 M = = 72.46 A V1 138k

 A CT ratio of 100:5 is selected, so the relay current at 138 kV side is:

72.46 I1 ' == = 3.62 A (100 / 5)  Hence, there is a mismatch between I1’ and I2’ (3.62 A and 4.17 A respectively). The mismatch is about 15%. 91

Common Problem in Differential Transformer Protection  Solutions to this problem can be solved through:  auxiliary CT connected to the standard CT  use tap settings on the relays

Example  A single phase, 30 MVA, 240 kV / 65 kV kV two-winding transformer is protected by differential relays with taps. Assume that the available relay tap settings are 5:5, 5:5.5, 5:6.6, 5:7.3, 5:8, 5:9 and 5:10, giving relay tap ratios of 1.00, 1.10, 1.32, 1.46, 1.60, 1.80, and 2.00  Determine currents in the transformer and in the CTs at rated conditions.  Select CT ratios and relay tap settings.  Hence, calculate the mismatch in the relay currents.

Example  Rated primary & secondary currents for two-winding transformer:

S 30 M = 125 A I1 = = V1 240k I2 =

S 30 M = = 461.54 A V2 65k

 Since:  I1 = 125 A, choose 150:5 tap setting for CT1  I2 = 461.54 A, choose 500:5 tap setting for CT2

Example  So, the secondary currents of CT1 and CT2 are:

I1 125 I1 ' = = = 4.167 A n1 (150 / 5) I2 ' =

I2 461.54 = = 4.615 A n2 (500 / 5)

 Relay tap setting can be determined from:

I 2 ' 4.615 = = 1.11 I1 ' 4.167  Hence, the closest tap ratio is 1.10. So choose relay tap setting of T1’:T2’ = 5:5.5.

Example  The percentage of mismatch is given by:

I 2 ' / T2 '− I1 ' / T1 ' × 100% I1 ' / T1 '

4.615 / 5.5 − 4.167 / 5 ×100% 4.167 / 5

= 0.683%  This is a good mismatch; since transformer differential relays typically have their block regions adjusted between 20% and 60%. So, 0.683% mismatch gives an ample safety margin in the event of CT and relay differences.

3-Phase Transformer Protection with Differential Relays  Figure 10.37 illustrates differential protection of a three-phase Y-Δ two-winding transformer.  Note that a Y-Δ transformer produces 30 degrees phase shifts in the line currents.  The CTs must be connected to compensate for the 30 degrees phase shifts, such that the CT secondary currents as seen by the relays are in phase.  The correct phase angle relationship is obtained by connecting CTs on the Y side of the transformer in Δ and CTs on the Δ side in Y.

3-Phase Transformer Protection with Differential Relays

Differential protection of a three-phase