overview lectures

DISTRIBUTION DEVICE COORDINATION

Mike Diedesch & Jon Harms – Avista Utilities

Presented March 14, 2011 At the 28th Annual

HANDS‐ON Relay School Washington State University

Pullman, Washington

TABLE OF CONTENTS

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TABLE OF CONTENTS Table of Contents ………………………………………………………………..……………… 2 System Overview ………………………………………………………………………………… 3 Symmetrical Components …………………………………………………………………… 4 Transformer Protection Devices ……………………………………………………..…... 13 Relay Setting Criteria …………………………………………………………………………… 15 Figures & Tables …………………………………………………………………………………. 17 Feeder and Transformer Protection using Electromechanical Relays ……… 32 Feeder and Transformer Protection using Microprocessor Relays ……….… 42 Problem – Moscow Example ……………………………………………………………….. 54

SYSTEM OVERVIEW

3

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR

1 PHASE#4 ACSR #4 ACSR

#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

1 MILE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

Figure 1. System Overview of a Typical 115/13.8 kV Substation and a 13.8 kV Distribution Feeder. – Used to work problem at end of paper.

SYMMETRICAL COMPONENTS

4

SYMMETRICAL COMPONENTS Three Phase 13.8 kV Fault

A

B

C

a

b

c

RR

IA = 619 ∠-88° Ia = 5158 ∠-118° IB = 619 ∠ 152° Ib = 5158 ∠ 122° IC = 619 ∠ 32° Ic = 5158 ∠ 2°

Figure 2. Current distribution through a Delta-Wye high lead low transformer bank for a three phase 13.8 kV fault.

Figure 3. 13.8 kV phase currents for a 13.8 kV three-phase bus fault at Moscow. The formula for calculating a three phase fault magnitude is: Ia = Ib = Ic = Vn ÷ Z1n (Z1 or Z+) in Ω where:

• Vn (Va, Vb or Vc) is the line to neutral voltage at the fault. You only have positive sequence voltage and current since the system is balanced. That is; Ia = I+. The current angles are referenced to the system, which in this case is the 115 kV (see Figure 4).


5

• Z1 is the line to neutral positive (balanced) sequence impedance (series resistance and reactance) of the system (all generators, transformers and lines) to the point of fault. Here it is made up of the entire Avista system Z1 + the Z1 of the Moscow xfmr.

• The current will lag (with an ABC CCW rotation) the voltage by the Z1 system inductive impedance angle of the system (here, that’s 88°). NOTE: Since the voltage is zero at the fault, the angle as given by ASPEN does not represent the normal voltage angle.

• A three phase fault is similar to balanced load where the three currents are equal and 120° apart. Of course for a load, typically the current angle would be very close to the voltage angle.

Figure 4. 115 kV phase currents and voltages for the 13.8 kV bus three-phase fault at Moscow. • The current magnitudes are equal to the 13.8 kV currents divided by the transformer ratio of

115/13.8 = 8.33 or 5158/8.33 = 619 amps. This is the line current flowing in the line connected to the transformer and not what is flowing in the transformer windings. This is the current that a relay would see from a CT on the transformer high side bushing or a fuse would see.

• Note the current angles lead the 13.8 kV current angles by 30°. That’s because this transformer is connected high lead low (where the high side voltages lead the low side voltages for an ABC CCW rotation).

• Also note that Va is at 0° because ASPEN made the 115 kV system the reference for the system and the voltage magnitudes are about 92% of normal. Normal is 115 kV÷√3 = 66.4 kV and 61.1÷66.4 = 0.92.


6

Single Line to Ground 13.8 kV Fault A

B

C

a

b

c

RR

IA = 370 ∠-118° Ia = 5346 ∠-118° IB = 0 ∠ 0° Ib = 0 ∠ 0° IC = 370 ∠ 62° Ic = 0 ∠ 0°

Figure 5. Current distribution through a Delta-Wye high lead low transformer bank for a phase A single line to ground 13.8 kV fault.

Figure 6. 13.8 kV phase voltages and currents for a 13.8 kV single line to ground (SLG) bus fault on phase A at Moscow. The formula for calculating a SLG fault magnitude on Phase A is: Ia = 3Va/(Z1 + Z2 + Z0) where: • Va is again the line to neutral voltage at the fault. • Z1 is the line to neutral positive sequence impedance of the system. • Z2 is the line to neutral negative sequence impedance of the system (which is normally = to

Z1). • Z0 is the line to neutral zero sequence impedance of the system. This is generally a

different value from Z1 or Z2 because it includes the impedance of any neutral wires and the


7

ground plane. Here, since the transformer is connected delta on the 115 kV, we can only use the transformer Z0 (and not the Avista system Z0) because there is no ground connection from the transformer to the 115 kV system.

• Because the Z0 of the transformer is less than the total Z1 (system + transformer), the SLG fault is slightly higher than the three-phase fault (5346 A vs. 5158 A).

• The current will lag the voltage by the Z1, Z2 & Z0 system inductive impedance angle of the system (again 88°). NOTE: Since the phase B & C voltages are not zero, their angles are correct.

• 3I0 is the sum of the 3 phase currents and since Ib & Ic = 0, then 3I0 = Ia. This means the phase and ground overcurrent relays on the feeder breaker see the same amount of current.

Figure 7. 13.8 kV sequence voltages and currents (positive, negative and zero) for a 13.8 kV SLG bus fault on phase A at Moscow. • Note that V- & V0 are = ½ of V+ and in the opposite direction so that Va = 0 at the point of

fault. • Note that I+ (I1) & I- (I2) are at the same angle as 3I0 (which is the same as Ia) and Ia = 3I0 =

I+ + I- + I0 and I+ = I- = I0.


8

Figure 8. 115 kV phase voltages and currents for a 13.8 kV SLG fault on phase A at Moscow. Note Ia still leads the 13.8 kV Ia by 30° (-118°) the same as the three phase fault. Also note that Ia is much less than the Ia for a three-phase fault. That’s because I0 can’t flow on the 115 kV system for a SLG on the 13.8 kV (I0 circulates in the delta). The calculation is: 5346/(8.33*√3) = 370 amps. In other words the high side phase current is the √3 less than it was for the 3Ø fault. 5346 = the 13.8 kV A phase current. 8.33 is the transformer ratio.

Figure 9. 115 kV sequence voltages and currents for a 13.8 kV SLG fault on phase A at Moscow.

Note that the I+ & I- are 60° apart. That’s because I+ rotates a positive 30° and I- rotates a minus 30° for a HLL connection. Add them together and they equal Ia (this is where you get the √3 reduced magnitude from a 3Ø fault). Note that 3I0 = 0 since the transformer connection is delta – wye.


9

Phase-to-Phase 13.8 kV Fault A

B

C

a

b

c

RR

IA = 309 ∠-28° Ia = 0 ∠ 0° IB = 619 ∠ 152° Ib = 4467 ∠ 152° IC = 309 ∠-28° Ic = 4467 ∠-28°

Figure 10. Current distribution through a Delta-Wye high lead low transformer bank for a phase-to-phase 13.8 kV fault.

Figure 11. 13.8 kV phase voltages and currents for a 13.8 kV line-to-line bus fault on phases B & C at Moscow. • Note that Ib & Ic = 4467 A which is 4467/5158 = 86.6% of the current for a three phase

fault. That’s because the voltage used is VØ-Ø which is √3 times Vn but the impedance used is Z1 + Z2 or 2 times Z1 so the value is 1.732/2 = 86.6%. The formula for the B to C phase fault is: I = (Vb-Vc)/(Z1+Z2).

• Note that Ib & Ic are in the opposite direction. That is; the current flows out on B phase and in on C phase while Vb & Vc are in the same direction as Ib.

• Note that Vb & Vc are ½ VA and in the opposite direction.


10

Figure 12. 13.8 kV sequence voltages and currents for a 13.8 kV line-to-line bus fault on phases B & C at Moscow. • Note that I+ & I- are equal and in opposite directions so they will sum to 0. That’s because

this is given in terms of Ia, which is 0. In order to obtain the Ib & Ic currents you have to rotate Ia+ & Ia- by: (Ib = Ia+ rotated by 240° plus Ia- rotated by 120° = 2579∠122° + 2579∠182° = 4467∠152°) and (Ic = Ia+ rotated by 120° plus Ia- rotated by 240° = 2579∠2° + 2579∠302° = 4467∠-28°).

• The above shows that I2 = the phase current/√3 (4467/√3 = 2579). • Note that V+ & V- are in phase so when they are added together to get Va, the Va magnitude

is normal. To get Vb & Vc again you rotate the sequence voltages similar to the current above.


11

Figure 13. 115 kV phase voltages and currents for a 13.8 kV line-to-line bus fault on phases B & C at Moscow. • Note that Ib = 619 amps which is the same value as it saw for a 3Ø fault. That’s because of

the way the positive and negative sequence currents added after the 30° phase shift. • Also note that Ia & Ic added together = Ib and in the opposite direction to get the current in

and out of the delta winding.

Figure 14. 115 kV sequence voltages and currents for a 13.8 kV line-to-line bus fault on phases B & C at Moscow.

• Again the sequence values are given for phase A and you have to rotate them to get phases B & C. • Note that the I+ & I- are 120° apart (were 180° apart at 13.8 kV). That’s because I+ rotates a

positive 30° and I- rotates a minus 30° for a HLL connection. Add them together and they equal Ia.


12

TRANSFORMER PROTECTION DEVICES

13


TRANSFORMER PROTECTION USING 115 KV FUSES For the smaller 115/13 kV substation transformers up through 7.5 MVA, we generally use 115 kV fuses for protection simply because of the cost. Some advantages to this are: • Low cost. • Low maintenance. • Does not require a panel house or substation battery. • Some fuses that Avista has used are GE, Southern States and S&C. There are also several disadvantages to using fuses however, which are: • The interrupting rating can be as low as 1,200 amps (for some older models) and only

go up to 10,000 amps at 115 kV. By contrast a modern circuit switcher can have an interrupting rating of 25,000 amps and a breaker can have even higher (we typically use 40,000 amps interrupting).

• The fuses we generally use are rated to blow within 5 minutes at twice their nameplate rating. Thus, a 65-amp fuse will blow at 130 amps. This compromises the amount of overload we can carry in an emergency and still provide good sensitivity for faults.

• The fuse time current characteristic (TCC) is fixed (although you can buy a standard, slow or very slow speed ratio which are different inverse curves).

• The sensitivity to detect faults (especially low side Single Line to Ground faults) is not very good as compared to using a circuit switcher or breaker and low side neutral overcurrent relaying. This is because we use DELTA/WYE connected transformers so the phase current on the 115 kV is reduced by the √3 as opposed to a three-phase fault. For example a typical 13.8 kV SLG fault through a 7.5 MVA transformer is 4,000 amps at 13.8 kV and 4000/(8.3*√3) = 277 amps at 115 kV. Since we use 65 amp fuses for this size transformer, our margin to detect this fault is only 277/(2*65) = 2.13:1. By contrast a low set feeder ground relay could probably be set between 480 to 720 amps so our margin to detect this fault would be around 8.3 to 5.5:1 (4000/480 or 720).

• Some fuses can be damaged and then blow later at some high load point. • When only one 115 kV fuse blows, it subjects the customer to low distribution

voltages. For example the phase to neutral distribution voltages on two phases on the 13.8 kV become 50% of normal.

• No indication of faulted zone (transformer, bus or feeder). TRANSFORMER PROTECTION USING A CIRCUIT SWITCHER Some advantages to this over fuses are: • Higher interrupting. • Relays can be set to operate faster and with better sensitivity than fuses. • Three phase operation. • Provide better coordination with downstream devices.


14

Some disadvantages would be: • Higher cost. • Higher maintenance. • Requires a substation battery, panel house and relaying. • Transformer requires CT’s. TRANSFORMER PROTECTION USING A BREAKER This is very similar to using a circuit switcher with a couple of advantages such as: • Higher interrupting. • Somewhat faster tripping than a circuit switcher (3 cycles vs. 6 – 8 cycles). • Possibly less maintenance than a circuit switcher. • The CT’s would be located on the breaker so it would interrupt faults on the bus

section up to the transformer plus the transformer high side bushings.

RELAY SETTING CRITERIA

15

RELAY SETTING CRITERIA The distribution feeder relay settings have to meet the following criteria (as basically does any other protective device): • Protect the feeder conductor from thermal damage due to faults (not overloads). • Detect as low a fault current as possible (our general rule is 0.5 times the calculated

fault at the next device or end of feeder). • For all relays other than the phase overcurrent, this means to set the pickup as

sensitive as we can and the time lever as low as we can and still coordinate with the highest set downstream device.

• Coordinate with all downstream devices whether fuses or reclosers by the desired Coordinating Time Interval (CTI) – see Table 6. This is the minimum amount of time we want between the operation of the devices.

• Carry normal maximum load (phase overcurrent only). • Pickup the feeder in a cold load condition (≅ 2 times maximum normal load) or

pickup ½ of the next feeder load (≅ 1.0 +2*1/2 = 2 times normal load, phase overcurrent only).

• Most overhead feeders also use reclosing capability to automatically re-energize the feeder for temporary faults. Most distribution reclosing relays have the capability of providing up to three or four recloses (Avista generally uses either one or two). The reclosing relay also provides a reset time generally adjustable from about 10 seconds to three minutes. This means if we run through the reclosing sequence and trip again within the reset time, the reclosing relay will lockout and the breaker will have to be closed by manual means.

• Lockout only for faults within the protected zone. That is; won’t lockout for faults beyond fuses, line reclosers etc.

Distribution Fuse Protection/Saving Scheme: A lot of overhead residential or rural feeders will also use a ‘fuse protection scheme’. This is done by using both an instantaneous/fast OC trip and an inverse time delayed OC curve similar to a fuse curve (See Figures 18 & 19). The time delayed OC is set to coordinate with the maximum device out on the feeder. This means that if a fault occurs on a fused lateral, you want the following to happen: • Trip and clear the fault at the station (or line recloser) by the instantaneous trip before

the fuse is damaged for a lateral fault. • Reclose the breaker. That way if the fault were temporary the feeder is completely

re-energized and back to normal. • During the reclose the reclosing relay has to block the instantaneous trip from

tripping again. That way, if the fault still exists you force the time delay trip and the fuse will blow before you trip the feeder again thus isolating the fault and re-energizing most of the customers.

• Of course if the fault were on the main trunk the breaker will trip to lockout. • There are variations on this such as using more than one instantaneous trip or more

than one time delay trip, varying the different reclose times etc. For example Avista often uses one instantaneous trip and two time delayed trips with a fast (1/2 second) and a 15 second reclose.

RELAY SETTING CRITERIA

16

Industrial, URD or Network Feeders • Generally these would not use fused laterals so would not use a fuse protection

scheme. Depending on several factors they also may not use reclosing.

Distribution Transformers: The distribution transformer relay settings have to (should) meet the following criteria (there will be several compromises here): NOTE: These are for an outdoor bus arrangement and not switchgear which also uses a bus breaker between the transformer relaying and the feeder relaying. This also doesn’t include the transformer sudden pressure relay. • Protect the feeder conductor from thermal damage in case the feeder breaker can’t

trip. This normally can’t be done completely since the transformer relaying has to be set higher than the feeder relaying.

• Protect the transformer from thermal damage. • Detect as low a fault current as possible. It would be desirable to be just as sensitive

as the feeder relaying in case the feeder breaker can’t trip but this can’t always be done.

• For all relays other than the phase overcurrent, this means to set the pickup as sensitive as we can and the time lever as low as we can and still coordinate with the highest downstream device by the desired CTI (Table 6).

• Coordinate with all downstream devices, which is the feeder breaker relaying. • Carry normal maximum load (phase overcurrent only). • After an outage, pickup the station cold load (≅ 2 times maximum normal, phase

overcurrent only). • Differential relay operate for internal transformer faults only and account for CT and

transformer ratio mismatch.

FIGURES & TABLES

17

FIGURES AND TABLES

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor ACSR Conductor Damage Curves No.Comment Date 12/13/05

1

1. M15-515 Phase INST INST TD=1.000CTR=160 Pickup=7.A No inst. TP@5=0.048s

A

A. Conductor damage curve. k=0.08620 A=355107.0 cmilsConductor ACSR336.4 ACSR

B

B. Conductor damage curve. k=0.08620 A=167800.0 cmilsConductor ACSR AWG Size 4/04/0 ACSR

C

C. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 2/02/0 ACSR

D

D. Conductor damage curve. k=0.08620 A=83690.0 cmilsConductor ACSR AWG Size 1/01/0 ACSR

E

E. Conductor damage curve. k=0.08620 A=52630.0 cmilsConductor ACSR AWG Size 2#2 ACSR

FF. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4#4 ACSR

Figure 15. ACSR Conductor damage curves from ASPEN.

FIGURES & TABLES

18

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Copper Conductor Damage Curves No.Comment Date 12/13/05

1


A

A. Conductor damage curve. k=0.14040 A=105500.0 cmilsConductor Copper (bare) AWG Size 2/02/0 Copper

B

B. Conductor damage curve. k=0.14040 A=83690.0 cmilsConductor Copper (bare) AWG Size 1/01/0 Copper

C

C. Conductor damage curve. k=0.14040 A=52630.0 cmilsConductor Copper (bare) AWG Size 2#2 Copper

D

D. Conductor damage curve. k=0.14040 A=33100.0 cmilsConductor Copper (bare) AWG Size 4#4 Copper

EE. Conductor damage curve. k=0.14040 A=20820.0 cmilsConductor Copper (bare) AWG Size 6#6 Copper

Figure 16. Copper Conductor damage curves from ASPEN.

FIGURES & TABLES

19

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Transformer Damage Curves No.Comment Date 12/13/05

1


A

A. Transf. damage curve. 12.00 MVA. Category 3Base I=502.04 A. Z= 9.0 percent.MoscowCity#2 13.8kV - MOSCOWSUB115 115.kV T

B

B. Transf. damage curve. 18.00 MVA. Category 3Base I=753.00 A. Z= 10.0 percent.Mead 18/24/30 MVA 13.8 kV Transformer Damage Curve

Figure 17. 12/16/20 and 18/24/30 Transformer Damage Curves.

FIGURES & TABLES

20

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Commonly Used Fuse Curves No.Comment Date 12/13/05

1


2

2. M15 515 kEARNEY 200T Kearney 200TMinimum melt.

3

3. Moscow 515 Kear 140T Kearney 140TMinimum melt.

4

4. Moscow 515 S&C 100T S&C Link100TMinimum melt.

5

5. M15 515 S&C 80T 170-62-80Minimum melt.

6

6. M15 515 S&C 65T S&C Link 65TMinimum melt.

7


8


Figure 18. Commonly Used Distribution Fuses.

FIGURES & TABLES

21

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor VARIOUS SEL OVERCURRENT RELAY CURVES No.Comment Date 12/21/05

1

1. SEL EXT INV CURVE SEL-EI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.404s

2

2. SEL VERY INV CURVE SEL-VI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.3865s

3

3. SEL INV CURVE SEL-I TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.6419s

4

4. SEL MOD INV CURVE SEL-MI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.4875s

5

5. SEL SHORT TIME INV SEL-STI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.1608s

Figure 19. SEL Various Overcurrent Relay Curves. These are basically the same as various E-M relays.

FIGURES & TABLES

22

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Comparing 140T VS #4 ACSR Damage No.Comment 140T won't Protect #4 ACSR Below About 550 Amps Date 12/13/05

1


2

2. Moscow 515 Kear 140T Kearney 140TMinimum melt.

A

A. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4#4 ACSR

Figure 20. Comparing a 140T fuse Vs. #4 ACSR Damage curve. The 140T won’t protect the conductor below about 550 amps where the curves cross.

FIGURES & TABLES

23

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Comparing a 115 kV Fuse with Feeder Relays No.Comment Coordinating for a 3 Phase Fault. 100A Fuse sees 619 amps Date 1/11/06

1 1. M15-A-172 100A Fuse 176-22-100Minimum melt.I= 619.0A T= 2.15s H=8.33

2

2. M15-515 Phase Time CO-11 TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.3766sI= 5158.1A (32.2 sec A) T= 0.33s

FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 3LG

Figure 21. Comparing a 100 Amp S&C 115 kV fuse with the Moscow feeder phase relay. The 100-amp fuse blows at 200 amps and sees 619. Margin = 619/200 = 3.1:1.

FIGURES & TABLES

24

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Comparing a 115 kV Fuse with Feeder Relays No.Comment Coordinating for a SLG Fault. 100A Fuse sees 370 amps Date 1/11/06

1 1. M15-A-172 100A Fuse 176-22-100Minimum melt.I= 370.4A T= 7.28s H=8.33

2

2. M15 515 GND TIME CO-11 TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0192sI= 5346.5A (33.4 sec A) T= 0.29s

FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 1LG Type=A

Figure 22. Comparing a 100 Amp S&C 115 kV fuse with the Moscow feeder ground relay. The 100-amp fuse blows at 200 amps and sees 370. Margin = 370/200 = 1.85:1.

FIGURES & TABLES

25

Figure 23. Transformer inrush UNFILTERED current. Inrush is the current seen when energizing a transformer. The maximum peak current is about 1800 amps. Note Ia is the sum of Ib + Ic.

FIGURES & TABLES

26

Figure 24. Transformer inrush FILTERED current (filtered by SEL digital filters to show basically only 60 HZ). Maximum peak is about 700 amps.

FIGURES & TABLES

27

Universal Distribution Fuse Links Avista generally uses S&C type T universal fuse links used with open type cutouts for most of its applications. That’s because the S&C fuse link uses silver in the fuse link instead of a tin alloy. A silver link has both a total clear and a minimum melt curve associated with it’s melting characteristic and is not damageable. That basically means if the fuse hasn’t blown, it’s still all right. However, a tin alloy link is damageable with a damage curve that is about 75% of its minimum melt curve. This means it’s much harder to coordinate two sets of fuses with one another and it’s also harder to protect the fuse by tripping the breaker or recloser on an instantaneous trip than the S&C. By using the silver fuse link, we can also use every size made below. That is; fuses in series can be one size apart and still coordinate. With the tin alloy we could only use about ½ the ratings because the fuses one size apart wouldn’t coordinate. We said that the S&C fuse could be protected from a temporary fault by a fast clearing time of 3.5 cycles if the minimum melt curve was at 0.1 seconds. NOTE: The S&C silver link is only made up through 100 amps while the 140 and 200 amp fuses are a tin alloy. There are other types of fuses made but these are not shown. Table 1. Shows the maximum fault current for which S&C type T fuses can still be protected by a recloser/breaker instantaneous trip for temporary faults (minimum melt curve at 0.1 seconds): 6T – 120 amps 8T – 160 amps 10T – 225 amps 12T – 300 amps 15T – 390 amps 20T – 500 amps 25T – 640 amps 30T – 800 amps 40T – 1040 amps 50T – 1300 amps 65T – 1650 amps 80T – 2050 amps 100T – 2650 amps 140T – 3500 amps 200T – 5500 amps

FIGURES & TABLES

28

Table 2. Shows the maximum fault current for which S&C type T fuses can coordinate with one another. NOTE: These values were taken from the S&C data bulletin 350-170 of March 28, 1988 based on no preloading and then preloading of the source side fuse link. Preloading is defined as the source side fuse carrying load amps equal to it’s rating prior to the fault. This means there was prior heating of that fuse so it doesn’t take as long to blow for a given fault. Source Side Load Side Maximum Maximum Fuse Fuse Coordinating Coordinating Current Current No preload With preload

200T 140T 8,300 4,650 140T 100T 5,800 3,800 100T 80T 3,600 1,900 80T 65T 2,700 1,400 65T 50T 2,200 Too close 50T 40T 1,550 640 40T 30T 1,400 820 30T 25T 1,100 570 25T 20T 840 360 20T 15T 630 315 15T 12T 540 295 12T 10T 410 210 10T 8T 320 175 8T 6T 235 145

Table 3. Typical continuous and 8 hour emergency rating of the S&C T rated silver fuse links plus the 140T and 200T. Fuse Rating Continuous 8 Hour emergency 6T 7.8 8.8 8T 10 12 10T 13 15 12T 16 18 15T 22 25 20T 27 31 25T 36 41 30T 42 49 40T 52 59 50T 63 72 65T 88 100 80T 105 115 100T 120 135 140T 210 225 200T 295 320

FIGURES & TABLES

29

Table 4. Conductor current ratings for various sizes of ACSR conductor at 25°C ambient taken from the Westinghouse Transmission & Distribution book. Conductor Rating 556 730 336.4 530 4/0 340 2/0 270 1/0 230 #2 180 #4 140 Table 5. Conductor current ratings for various sizes of copper conductor at 25°C ambient taken from the Westinghouse Transmission & Distribution book. Conductor Rating 2/0 360 1/0 310 #2 230 #4 170 #6 120 Table 6. Typical Coordinating Time Intervals (CTI) that Avista generally uses between protective devices. Other utilities may use different times. DEVICES: CTI (Seconds) Relay – Fuse Total Clear 0.2 Relay – Series Trip Recloser 0.4 Relay – Relayed line Recloser 0.3 Low Side Xfmr Relay – Feeder Relay 0.4 High Side Xfmr Relay – Feeder Relay 0.4 Xfmr Fuse Min Melt – Feeder Relay 0.4 Table 7. Typical minimum conductor that can be protected by the Avista feeder settings and S&C type T fuses. These are what Avista uses. Other companies may use different values. The relay phase pickup for a given feeder rating is typically about twice the feeder rating and we use Extremely Inverse curves. Feeder Setting or Fuse MIN Conductor Size 500 AMP fdr Setting #2CU or 1/0 ACSR 300 AMP fdr setting #4CU or #2ACSR 200T #2CU 140T #4CU or #2ACSR 100T #6CU or #4ACSR 65T #8CU

FIGURES & TABLES

30

Table 8. IEEE numbers for various relays, breakers etc, that could be used in a distribution substation. Avista sometimes adds letters to these such as F for feeders, T for transformers, B for bus and BF for breaker failure. 2 – Time delay relay. 27 – Undervoltage relay. 43 – Manual transfer or selective device. We use these for cutting in and out instantaneous overcurrent relays, reclosing relays etc. 50 (or 50P) – Instantaneous overcurrent phase relay. 50N (or 50G) – Instantaneous overcurrent ground (or neutral) relay. 50Q – Instantaneous Negative Sequence overcurrent relay. 51 (or 51P) – Time delay overcurrent phase relay. 51N (or 51G) – Time delay overcurrent ground (or neutral) relay. 51Q – Time delay Negative Sequence overcurrent relay. 52 – AC circuit breaker. 52/a – Circuit breaker auxiliary switch closed when the breaker is closed. 52/b – Circuit breaker auxiliary switch closed when the breaker is open. 59 – Overvoltage relay. 62 – Time Delay relay 63 – Sudden pressure relay. 79 – AC Reclosing relay. 81 – Frequency relay. 86 – Lock out relay which has several contacts. Avista uses 86T for a transformer lockout, 86B for a bus lockout etc. 87 – Differential relay. 94 – Auxiliary tripping relay. Table 9. Definitions and glossary of terms used in this paper. • 5 amps per MVA = amps per phase at 1 MVA 3Ø power at 115 kV = 1000000/(3*[115000÷√3]) =

5.02 amps. • 41.8 amps per MVA = amps per phase at 1 MVA 3Ø power at 13.8 kV = 1000000/[3*(13800÷√3)]

= 41.84 amps. • Z1 is the line to neutral positive (balanced) sequence impedance (resistance and reactance) of the

system (all generators, transformers and lines) to the point of interest. • Z2 is the line to neutral negative sequence impedance of the system (which is normally = to Z1). • Z0 is the line to neutral zero sequence impedance of the system. This is generally a different value

from Z1 or Z2 because it includes the impedance of any neutral wires and the ground plane. At the 13.8 kV bus, since the transformer is connected delta on the 115 kV, we can only use the transformer Z0 (and not the Avista system Z0) because there is no ground connection from the transformer to the 115 kV system.

• Va ( or Van) is the line to neutral voltage for Phase A. Vb & Vc are the line to neutral voltages for those phases. The Ø-Ø voltage = √3*Vn.

• V1 (or V+) is the positive sequence voltage for whichever phase you are referring to. • V2 (or V-) is the negative sequence voltage for whichever phase you are referring to. • V0 (or V0) is the zero sequence voltage for whichever phase you are referring to. • Ia is the single phase current for Phase A, Ib for phase B and Ic for phase C. • I1 (or I+) is the positive sequence current for whichever phase you are referring to.

FIGURES & TABLES

31

• I2 (or I-) is the negative sequence current for whichever phase you are referring to. • I0 (or I0) is the zero sequence current for whichever phase you are referring to. • P is for phase, N is for neutral & Q is for negative sequence. • 3I0 is the total ground or neutral current for ground faults. This is the current that flows up the

transformer neutral. • CCW is counterclockwise rotation of the system. That is; phase A leads phase B by 120° and leads

phase C by 240°. • HLL is a high lead low transformer connection. That is; the phase A voltage to the high side

windings leads the low side voltage by 30° (other angles are possible). • LLH is a low lead high transformer connection. That is; the phase A voltage to the high side

windings lags the low side voltage by 30° (other angles are possible). • SLG is single line to ground fault. • The letter ‘a’ is an operator that when applied to voltages or currents rotates them by plus (or

CCW)120°. The letter ‘a2’ shifts them by 240°. EXP: using phase A as a reference we would have Vb = a2Va1 + aVa2 + V0 or said another way Vb = Va1∠240° + Va2∠120° + V0. Vc = aVa1 + a2Va2 + V0.

• TCC is Time Current Characteristic. This is generally done on a log-log graph like is shown in this paper. The vertical axis is the time and the horizontal is the current magnitude.

• EI = EXT INV = Extremely Inverse Time Current Curve. • VI = VERY INV = Very Inverse Time Current Curve. • INV = Inverse Time Current Curve. • MI = MOD INV = Moderately Inverse Time Current Curve. • STI = Short Time Inverse Time Current Curve. • TL = TD = time lever (or dial) used with inverse time overcurrent curves. This shifts the curve up

and down on the TCC graph. However, it does not change the shape of the curve. • PU = Pickup value of the unit. This moves the curve right and left on the TCC graph. • CT = Current Transformer. • BCT = Bushing Current Transformer. • PT = VT = Potential (or voltage) transformer. • E-M = Electromechanical relay. • MP = Microprocessor relay. • TC = Torque Control. This controls whether or not an element will be able to operate. We used this

to control certain instantaneous elements in a transformer MP relay from a feeder MP relay. The term torque control is from the E-M relaying.

• DC Offset – When a fault occurs at the voltage peak, the current can be offset from the zero axis by as much as 1.6 times because of the ratio of reactance to resistance of the system (the higher the ratio the greater the offset). This is called Asymmetrical current. The offset then decays over a few cycles so that the positive and negative current magnitudes are basically equal. This is called Symmetrical current.

• CTI – Coordinating Time Interval. This is the minimum operating time desired between two protective devices.

FEEDER AND TRANSFORMER PROTECTION WITH E-M RELAYS

32

FEEDER AND TRANSFORMER PROTECTION USING ELECTROMECHANICAL RELAYING

Feeder - The normal electromechanical feeder relaying consists of 3 phase and 1 ground overcurrent relay. Each relay has both an instantaneous and time delay unit. Typical settings for a normal 500 amp feeder having to coordinate with 140T line fuses would be as follows: 1. Phase overcurrent:

• Set 51P pickup ≅ 2*500 = 1000 amps to pickup cold load or ½ of the next feeder cold load. Since we use 800/5 CT ratios, the setting would be tap 6 = 960 amps primary.

• 51P Time Lever(TL) – This would be set to provide ≅ 0.2 seconds coordination (see Table 6) between the relay and the maximum fuse size which is usually a 140T. This is the coordination time that Avista feels comfortable with basically because the fuse total clear curve is a maximum value and all deviations should be negative. However, other utilities may use a different number. This would be done assuming a fused lateral across the street from the sub and would be for either a 3Ø or SLG fault (I.E. maximum feeder fault). A typical curve would be ≅ TL 1.75 - 2.5. This would vary depending on the 13.8 kV fault duty.

• 50P - The instantaneous unit would be set similar to the 51P. However, since it will pickup whenever the feeder is closed in due to inrush and cold load (which can be as high as 5 times the normal load), it has to be set higher than the 51P so it can drop back out before the block of the instantaneous unit by the reclosing relay can reset which may be as low as 3 seconds from the lockout position. A typical setting would be 7.0 amps = 1120 amps primary (116.7% of the 51P).

2. Ground overcurrent: • 51N pickup – Since this doesn’t operate on load, it can be set down to merely

coordinate with the 140T across the street with 0.2 seconds coordination time. A typical minimum setting would be tap 3 = 480 amps PU. See Figure 28.

• 51N TL – This would be set for the same criteria as the phase TL. A typical setting would be 4.0 – 5.0.

• 50N – This is set the same PU as the time or 3.0 amps sec = 480 amps primary. 3. Reclosing (79) – We will use either one fast or one fast and one time delay reclose to

lockout. We will normally block the instantaneous unit from tripping after the first trip to provide for a fuse protecting scheme. The reset time Avista normally uses is from 90 to 180 seconds. The time to reset from the lockout position is 3 to 6 seconds for an E-M reclosing relay.


33

Transformer - The normal transformer relaying will consist of 3 high side phase and 1 low side ground overcurrent relays plus a sudden pressure. We could also use a high side ground but generally haven’t because it won’t detect ground faults through the transformer. At switching stations we also used a low side BØ OC relay for breaker failure purposes and we fused the DC control circuit of the high side phase overcurrent relays separately from the other relaying. This BØ relay has to be set similar to the high side except it’s worst coordination case is a 3Ø fault. We would then use a breaker failure overcurrent relay to operate a bus lockout relay to clear the fault. Typical settings for both a 12/16/20 MVA and an 18/24/30 MVA transformer coordinating with a 500 amp feeder would be: 1. Phase overcurrent (on 115 kV transformer BCT’s):

• 51P pickup – This has to be able to pick up the entire transformer load cold load or ≅ 2*normal. Our typical rule of thumb for this is 2.4*highest MVA rating. For a 12/16/20 (5 amps per MVA) this is 2.4*20*5 = 240 amps (1920 amps at 13.8 kV). For an 18/24/30 this is 2.4*30*5 = 360 amps (2880 amps at 13.8 kV). The pickup is so high that we can’t be as sensitive as the feeder phase relay so can’t reach to the same fault points on the feeder. Thus the ability to detect multi phase faults and protect the feeder conductor are compromised.

• 51P time lever – This relay has to coordinate with the feeder phase relays for a maximum feeder fault with 0.4 sec coordination time (Table 6). Avista uses 0.4 sec coordination time because there is more uncertainty about the actual ratio of amps seen on the high side vs. the low side. The worst case for this is a Ø-Ø fault since the feeder phase relays see 86.6% of what they saw for a 3Ø fault but yet one of the high side relays see the same as for a 3Ø fault because of the delta/wye connection (see figure 13). A typical TL would be ≅ 1.2 – 2.0.

• 50P – This is a direct trip element and is set to not detect a maximum fault on the 13.8 kV bus. Since the inst unit is a hinged armature device, it responds to current ‘DC Offset’ so the setting we have used is √3*MAX 13.8 kV fault. A typical setting is 900 to 1500 amps primary.

2. Ground overcurrent (on 13.8 kV transformer BCT’s): • 51N pickup – This relay will be set to coordinate with the feeder phase relays in

case the feeder ground relay is out of service. The minimum pickup to coordinate with a 500 amp feeder phase relay is 960 amps. We again compromise our ability to detect feeder faults since the feeder ground is twice as sensitive as the transformer ground.

• 51N time lever – This would normally be ≅ 4.0 – 6.0. This could be set to coordinate with 0.3 seconds margin since we are on the same voltage. However, a little more conservative setting would still use the 0.4 seconds because the transformer unbalance can be greater than any one feeder so it might see slightly more current for a fault. However, the phase OC relay also sees at least some load that the ground relay doesn’t see so the 0.3 seconds may be acceptable. Avista prefers the 0.4 seconds for all elements.

• 50N – We can’t use this element because we can’t coordinate it with the feeder relays.


34

13kV BUS

115 kV SYSTEM

5051

63

1200/5A MRCONN 600/5

86T

51N

2000/5A MRCONN 1200/5

5051

50N51N

1200/5A MRCONN 800/5

600ASPTT

REG

560ABKR515

79

TRIPCLOSE

E/M

E/M

E/M

E/M

3-E/M

MA-172 CS

MOSCOW

XFMR 12/16/20 MVA

115/13.8kVDELTA/WYEHIGH LEAD LOW

Figure 25. Moscow 115/13.8 kV Partial One Line.


35

52/a

TC

R

86T

101

86T

86T

ICS

TOC

ICS

IIT

ICS

TOC

ICS

IIT

ICS

TOC

ICS

IIT

ICS

TOC

ICS

AØ 50/51 BØ 50/51 CØ 50/51 GND 51N

63

MOSCOW 86T AND CS A-172 TRIP CKTS

Figure 26. Moscow 86T and Circuit Switcher Partial Schematic.


36

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

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300400500700

1000

.01

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1

.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor M15 A-172 CS with fdr 515 with Kearney 140T Fuse No.Comment Three Phase Fault Date 12/12/05

1

1. Moscow 515 Kear 140T Kearney 140TMinimum melt.I= 5158.1A T= 0.10s

2


3

3. M15 A-172 Phase CO- 9 TD=1.500CTR=120 Pickup=2.A Inst=1200A TP@5=0.3605sI= 619.0A (5.2 sec A) T= 1.03s H=8.33


Figure 27. Coordinating E-M Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using E-M Relays with the 500 Amp Feeder for a Three Phase Fault.


37

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

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300400500700

1000

.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS with fdr 515 with 140T fuses No.Comment Single Line to Ground Fault Date 12/12/05

1


2


3


4 4. M15 A-172 GND TIME CO-11 TD=4.000CTR=240 Pickup=4.A No inst. TP@5=1.0192sI= 5346.5A (22.3 sec A) T= 0.83s


Figure 28. Coordinating E-M Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using E-M Relays with the 500 Amp Feeder for a Single Line to Ground Fault.


38

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

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300400500700

1000

.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS A-172 with fdr 515 with 140T Fuse No.Comment Line to Line Fault Date

1


2


3


FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L LL Type=B-C

Figure 29. Coordinating E-M Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using E-M Relays with the 500 Amp Feeder for a Line to Line Fault.


39

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

345

7

10

20

304050

70

100

200

300400500

700

1000

2

345

7

10

20

304050

70

100

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300400500

700

1000

.01

.02

.03

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLH

For M15 515 with P584 with 100T No.

Comment Three Phase Fault Example Date 12/5/05

1

1. M15 P584 100T Fuse S&C Link100TTotal clear.I= 3452.9A T= 0.08s

2

2. M15 P584 PHASE TIME CO-11 TD=4.100CTR=100 Pickup=3.A No inst. TP@5=1.0492sI= 3452.9A (34.5 sec A) T= 0.28s

3


FAULT DESCRIPTION:Bus Fault on: 0 511-P584 13.8 kV 3LG

Figure 30. Coordinating E-M Relays for a Line Recloser Using a 100T fuse and Coordinating a 500 Amp Feeder Using E-M Relays with the Line Recloser for a Three Phase Fault.


40

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

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.7

1

.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor M15 515 with P584 with 100T fuse No.Comment Single Line to Ground Fault Date 1/11/06

1


2


3

3. M15 P584 GND TIME CO-11 TD=3.500CTR=100 Pickup=3.A No inst. TP@5=0.8938sI= 2761.9A (27.6 sec A) T= 0.33s

4 4. M15 515 GND TIME CO-11 TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0192sI= 2762.0A (17.3 sec A) T= 0.79s

FAULT DESCRIPTION:Bus Fault on: 0 511-P584 13.8 kV 1LG Type=A

Figure 31. Coordinating E-M Relays for a Line Recloser Using a 100T fuse and Coordinating a 500 Amp Feeder Using E-M Relays with the Line Recloser for a Single Line to Ground Fault.


41

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

345

7

10

20

304050

70

100

200

300400500

700

1000

2

345

7

10

20

304050

70

100

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300400500

700

1000

.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLH

For M15 515 with P584 with 100T No.

Comment Line to Line Fault Example Date 12/5/05

1


2


3


FAULT DESCRIPTION:Bus Fault on: 0 511-P584 13.8 kV LL Type=B-C

Figure 32. Coordinating E-M Relays for a Line Recloser Using a 100T fuse and Coordinating a 500 Amp Feeder Using E-M Relays with the Line Recloser for a Line to Line Fault.

FEEDER AND TRANSFORMER PROTECTION WITH MP RELAYS

42

FEEDER AND TRANSFORMER PROTECTION USING MICROPROCESSOR RELAYS

The following is at a new station where we can use microprocessor relaying on all of the feeders plus the transformer. We will use SEL 351 and 587 relays for our example since that is Avista’s present standard. Of course there are some general advantages to using MP relaying over E-M. Some of these are: 1. The taps are selectable in very small increments. 2. There are several independent units contained in the relay (up to 6 instantaneous and

2 time delay). 3. Internal logic programming capability. 4. Less CT burden (CT’s can saturate with high burdens and high fault values and won’t

produce a good secondary current to the relay). 5. Event reporting. 6. Remote Communication capability. Feeder – The normal relaying will use the SEL 351 phase, negative sequence and ground elements. We use one phase and ground instantaneous element. NOTE: We don’t use the instantaneous negative sequence because of the possibility of false trips due to motor contributions for external faults. The time delay pickup units for phase, negative sequence and ground are used to operate an output contact to torque control some low set fast elements on the 50/51 transformer relay. The relay is also programmed to provide a breaker failure output in case the breaker fails to open. Again, for a 500 amp feeder we have: 1. Phase overcurrent:

• 51P pickup - Set the pickup at 960 amps for the same reasons as the E-M relays. • 51P TL – Set the same as the E-M relays for the same reasons. • 50P pickup – Set similar to the E-M 50PF pickup. NOTE: This unit has a higher

dropout characteristic than the E-M relays so it could probably be set closer to the 51P pickup.

2. Ground overcurrent: • 51G pickup – This is actually the residual element in the 351 which is obtained by

adding the 3 phase currents. There is also a neutral element (single coil) that is not used. The pickup is the same as the E-M relay or 480 amps.

• 51G TL – This is the same as the E-M relay. • 50G – This is the same as the E-M relay = 480 amps.

3. Negative Sequence: • 51Q pickup – This element can be set for similar criteria to the ground element

since it doesn’t respond to load. Similar to the ground which responds to 3I0, this unit responds to 3I2. That is; it should be set as sensitive as possible with the lowest time lever and still coordinate with the 140T with 0.2 seconds margin. The worst coordination case for this element is for a Ø-Ø fault (see figures 8 & 11 for a comparison). For this fault if the fuse sees 6000 amps Ø flow, the negative sequence current is 3464 (6000÷√3) and 3I2 = 10,392 amps. The similarity to the ground pickup is 480*√3 = 831. Thus we wind up with ≅ 830 amps minimum pickup to coordinate with the 140T. The advantage to this element is that it is


43

more sensitive than the phase element for the Ø-Ø fault. The difference in sensitivity is: 960*√3÷830 = 2.0.

• 51Q TL – The criteria is the same as for the ground unit. • 50Q pickup – We don’t use this element because of the possibility of false trips.

4. Reclosing – This is the same as the E-M reclosing. The major difference is that we can choose the time from lockout to reset to allow more time for the cold load to subside and drop out the inst unit. We have been choosing 30 seconds.

Transformer - The normal transformer relaying will consist of a SEL 351 connected to the 115 kV BCT’s and the neutral element (single coil) connected to the 13.8 kV BCT’s (or possibly a neutral CT). In addition to the normal tripping elements, the SEL 351 will also use ‘fast’ (≅ 4 cycles) elements that are torque controlled by the feeder time pickup elements. That is; if an overcurrent element picks up on a feeder, it will block the 351 torque controlled element. Of course this can’t be used at a station that has any feeders that use E-M relays. NOTE: One of the major differences between using all microprocessor relays and E-M relays is that with the microprocessor relays, you can coordinate like elements with like elements only. For example, with the E-M we coordinate the transformer ground with the feeder phase because the feeder ground could be out of service. With microprocessors, the entire relay would be out of service and not just one element. Therefore, we can coordinate the transformer ground with just the feeder ground and obtain a more sensitive setting. Typical settings for both a 12/16/20 MVA and an 18/24/30 MVA transformer coordinating with the 500-amp feeder above would be: 1. SEL 351 Phase overcurrent (on 115 kV transformer BCT’s):

• 51P pickup – This is identical to the E-M relay setting or 240 amps for a 20 MVA and 360 amps for a 30 MVA.

• 51P TL – This has to coordinate with the feeder phase for a maximum feeder fault with 0.4 seconds coordination. In this case we could consider this to be a 3Ø fault since we will have negative sequence relays on the feeder to take care of the Ø-Ø fault.

• 50P1 pickup – This is the direct trip phase element and is set to not detect a 13.8 kV fault similar to the E-M instantaneous setting. However, this does not respond to DC offset so our setting is ≅ 1.3*MAX 13.8 kV fault.

• 50P2 pickup – This is the torque controlled element. We have set this to account for inrush and still be as sensitive as possible. • Inrush - For this unit we need to consider transformer inrush current since it

can operate ‘fast’ (≅ 4 cycles time delay). A rule of thumb we have used is that a typical transformer inrush RMS current can be as high as ≅ 8 times the transformer base rating (old E-M differential relays had the unrestrained overcurrent trip fixed at ten times the TAP of the restrained trip and the TAP would be set at ≅ transformer full load). Note that the SEL 351 only responds to the 60 HZ fundamental and that this fundamental portion of inrush current is ≅ 60% of the total. So to calculate a setting for a SEL 351, we could use the 8 times rule of thumb along with the 60% value. For a 12/16/20 MVA transformer, the calculation would be 8*12*5*0.6 = 288 amps. For an


44

18/24/30 it would be 8*18*5*0.6 = 432 amps. Generally, we would set this unit at about 1.5 times the time unit to be safe. So for a 12/16/20 we would set 1.5*240 = 360 amps and for an 18/24/30 we would set 1.5*360 = 540 amps. NOTE: For comparison I have included some inrush characteristics from a SEL 351 that energized 6 – 2.25 MVA (13.5 MVA total) 480/13.8 kV transformers from the 13.8 kV side (see Figures 23 & 24). These transformers were connected delta (480V)/wye(13.8 kV). The A phase was the highest current. The filtered current (that the elements respond to) to the relay showed a maximum AØ of 944 amps and this died down to 250 amps in 26 cycles. See Figure 24. The highest current was 944÷41.8 = 22.6 MVA which is 22.6÷13.5 = 1.67 times the transformer total rating. Using these values for a 12 MVA we would have 1.67*12*5 = 100 amps. The unfiltered current to the relay showed a maximum A-phase current of 2016 amps, which died down to 504 amps in 26 cycles. See Figure 23. This maximum is 2016÷41.8 = 48.2 MVA = 48.2÷13.5 = 3.6 times the total transformer rating.

2. SEL 351 ground (residual) overcurrent (on 115 kV BCT’s) NOTE: This is calculated from the phase quantities:

• 51G pickup – We can set this very low since it doesn’t have to coordinate with anything. We have been setting ≅ 120 amps primary.

• 51G TL – Also can set low. Have been setting at TL 1.0. • 50G pickup – This can also be set low. We have been choosing a setting equal to

the 50P2 TC setting simply because we don’t feel we need more sensitivity but it really can be set as low as the time unit since the inrush to the delta doesn’t contain any 3I0 current. A typical setting is 120 to 360 amps.

3. SEL 351 neutral overcurrent (on 13.8 kV BCT’s): • 51N pickup – We can set as low as possible and still coordinate with the feeder

ground settings (not phase). Since the transformer can have more unbalance than the feeder we have been setting ≅ 1.33 times the feeder ground pickup or 640 amps when coordinate with 500 amp feeders.

• 51N TL – We coordinate with the feeder ground with 0.4 seconds margin. • 50N1 – We can’t set a direct trip element so we use this as a torque controlled

element only. Again we need to set above the feeder ground setting and to be a little conservative we have been setting ≅ 720 amps with 4 cycles time delay.

4. SEL 351 negative sequence (on 115 kV BCT’s): • 51Q pickup – We can set this as low as possible and still coordinate with the

feeder negative sequence relay. The worst case is for a Ø-Ø fault. If we simply take the transformer ratio of 8.33:1 and the 830 amp feeder setting, we could set this down to 830÷8.33 = 100 amps. However, because of the transformer, we are a little more conservative and have used about 200 amps or 2.0 times. For comparison with the phase overcurrent sensitivity for 13.8 kV Ø-Ø fault we have the following example: At Moscow (if we had MP relays) on a 12/16/20 for a 13.8 kV Ø-Ø fault the 115 kV sees 619 amps Ø flow and 309 amps I2. See Figures 13 & 14. The margin for the phase OC PU to see the fault is 619÷240 = 2.58. The margin for the negative sequence OC PU would be: 3*309*200 = 4.64


45

or 1.8 times greater than the phase OC. NOTE: It would be 2 times like the feeder but one of the phases sees the same current as for a 3Ø fault.

• 51Q TL – Set for 0.4 seconds coordination. • 50Q1 pickup and time – This is a direct trip element and is set not to detect a 13.8

kV fault. A typical setting would be ≅ 1200 – 1800 amps or 1.3*MAX 13.8 kV fault. NOTE: We time delay this by 2 cycles (along with the high setting) to avoid false trips.

• 50Q2 pickup and time – This is a torque controlled element and can be set low with 4 cycles time delay. Since the unit responds to inrush and can see as much as the phase instantaneous units, we need to set similar to the 50P2 unit (360 & 540). NOTE: On the transformer inrush tests we observed a maximum I2 filtered current of 331 so 3I2 = 993 (the max Ø was 944) so in terms of transformer rating this is 1.75 times the total rating. This says that we really should be using the same pickup values as the Ø.

• SEL 587 – This relay is connected to both the high and low side transformer BCT’s. Basically a normal E-M transformer differential relay contains a harmonic restrained differential unit which uses a 2nd harmonic restraint to prevent a trip during transformer inrush and a variable percentage trip characteristic to prevent a trip for external faults in the presence of CT saturation. That is; the higher the current through the restraint coils, the greater percentage of that current needs to flow in the operate coil to get a trip. The relay also has taps for the two windings to correct the high and low current magnitudes so similar currents flow in each restraint winding for external faults. There are generally about 8 taps between 2.9 and 8.7 amps. There is also an unrestrained differential unit set around 10 times the upper tap value which was generally chosen to be the same as the transformer full load current. Normally the unrestrained unit would operate slightly faster for high magnitude transformer faults. Since the distribution transformer is connected delta – wye the transformer CT’s have to set wye – delta to compensate for the phase shift. See Figure 33.

• Some of the advantages and ways we use the MP SEL 587 relay over the E-M relays are: • In addition to similar differential units as above, it contains independent

overcurrent units for both windings similar to a MP overcurrent relay. Thus it can also operate for external faults and completely backup the transformer overcurrent relay.

• Different differential slopes can be set in the relay, which provides greater flexibility.

• You can use more than one harmonic to restrain the differential. • The transformer phase angle can be set in the relay, which allows you to set all

CT’s in wye. • The unrestrained OC element can be set independently from the differential tap

value. We use this feature to set this based on the transformer BASE rating and not the FOA rating to avoid tripping on transformer inrush. That is; on a 12/16/20 we use the 12 MVA rating and not the 20 MVA rating because the inrush characteristic is a function of the base rating.


46

• Again you can torque control the instantaneous elements like we did for the 351 but we didn’t feel this was needed. Instead we used direct trip inst elements on the high side winding while the low side winding can’t use inst elements.


47

87R1

87OP

87R2

87R1

87OP

87R2

PRI I

PRI I

INTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN OPPOSITE DIRECTIONS, ADD AND THEN FLOW THROUGH THE OPERATE COIL ANDBACK TO THE RESPECTIVE CT'S

SEC I

SEC I

PRI I

PRI I

SEC I

SEC I

EXTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN THE SAME DIRECTION AND THENCIRCULATE BACK THROUGH THE CT'S. THEY DO NOTFLOW THROUGH THE OPERATE COIL

CURRENT FLOW THROUGH AN E/M 87 DIFFERENTIAL RELAY FOR AN INTERNAL AND EXTERNAL FAULT

BCT'S

BCT'S

BCT'S

BCT'S

CT POLARITY MARK

CT POLARITY MARK

Figure 33. Current Flow Through an E-M Differential Relay for an External and an Internal Fault.


48

13kV BUS

115 kV SYSTEM

5051

63

1200/5A MRCONN 600/5

86T

2000/5A MRCONN 2000/5

5051

50G51G

1200/5A MRCONN 800/5

600ASPTT

REG

1200ABKR12F1

79

TRIPCLOSE

MA-699 CS

MEAD

XFMR 18/24/30 MVA

115/13.8kVDELTA/WYELOW LEAD HIGH

50G51G

50N51N

5051 87 50

51

50N51N

50N51N

351 587

50BF

62BF

94TX1

86TX1

351

TORQUE CONTROL

Figure 34. Mead 115/13.8 kV Partial One Line.


49

W

86TX1 AR

50/51T1 351OUT101

86T1

FDR BF CKTS

50/51F1 12F1 351OUT104

50/51F1 12F1 351OUT103

AR

94T1

AR

86TX1

FAST TRIP BLOCKING

50/51T1 351

50/51T1 351 P.S.

A-699 CS 52/a

IN101

50/51T1-351 POWER SUPPLY

86T1

TIR

86T1

S+

R+

BFTGT

TIR

101 T 86T1

50/51T1 351OUT102R

TC

52/a

A-699 CKT SWTR TRIP CKT ALARMS to SCADA

50/51T1 351OUT105MAJOR 50/51T1

351OUT108ALARM

SCADA

50/51T1 351OUT107MINOR IN

10350/51T1 351 TCM

IN102

94T1 AR

50/51T1 351

FDR FASTTRIP BLK

87T1587OUT2

86T1

87T1587OUT1

IN104

50/51T1 351 BFI

50/51T1 351OUT106

BKR FLR

87T1 587OUT108ALARM

87T1587

87T1 587 P.S.

A-699 CS 52/a

IN1

87T1-587 POWER SUPPLY

CKT SWTCHR CLOSE CIRCUIT

84X IN106

.

.

.

50/51T1 351

63XP

86T1 (IfUsed)

Figure 35. Mead circuit switcher and lockout circuit partial schematic.


50

Typical Distribution Feeder Schematic with SEL351 - 50/51F1

52/a

TC

R IN104

W IN103

50/51F1 351OUT 101

INST Ø& GNDTRIP SWDBL POLE 101

T201 T

50/51F1 351OUT106 SH0

50/51F 351TCM 50/51F1

351INST TRIPENABLED

43I 43I

101 C

201 C OUT

102CLOSE

DBL-POLE

101SC 50/51F1

351 NON-RECLOSETO CLOSE

CKT

50/51F1 - AC RECLOSE CIRCUIT

351PS

52/a

50/51F1 351

IN101

ALARMS to SCADA

50/51F1 351OUT107MINORALARMS/ TEST

50/51F1 351OUT105MAJORALARMS

50/51F1 - 351 DC POWER SUPPLY

50/51F1 351

50/51F1 351OUT108 SELALARM

IN102

43RN.O.

43RN.C.

43H

43H43H

43HHOT LINE

HOLDSTATUS

43H

SCADA

Figure 36. Mead Feeder Partial Schematic.


51

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Coordinating Mead A-699 with 12F1 with S&C 140T No.Comment Three Phase Fault Date 12/12/05

1 1. MEA S&C 140T S&C Link140TTotal clear.I= 7205.9A T= 0.06s

2

2. MEA 12F1 PHASE TIME SEL-EI TD=2.000CTR=160 Pickup=6.A No inst. TP@5=0.5429sI= 7205.9A (45.0 sec A) T= 0.28s

3

3. MEA A-699 PHASE SEL-VI TD=0.750CTR=120 Pickup=3.A Inst=1128A TP@5=0.1935sI= 864.6A (7.2 sec A) T= 0.68s H=8.33

FAULT DESCRIPTION:Close-In Fault on: 0 MEAD BUS 13.8kV - 0 MEA FDR BUS 13.8kV 2L 3LG

Figure 37. Coordinating MP Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using MP Relays with the 500 Amp Feeder for a Three Phase Fault.


52

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Coordinating Mead A-699 with 12F1 with S&C 140T No.Comment Single Line to Ground Fault Date 12/12/05


2

2. MEA 12F1 GND TIME SEL-EI TD=4.400CTR=160 Pickup=3.A No inst. TP@5=1.1944sI= 7399.5A (46.2 sec A) T= 0.26s

3

3. MEA A-699 GND TIME SEL-EI TD=8.600CTR=400 Pickup=1.6A No inst. TP@5=2.3345sI= 7399.5A (18.5 sec A) T= 0.67s

FAULT DESCRIPTION:Close-In Fault on: 0 MEAD BUS 13.8kV - 0 MEA FDR BUS 13.8kV 2L 1LG Type=A

Figure 38. Coordinating MP Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using MP Relays with the 500 Amp Feeder for a SLG Fault.


53

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Coordinating Mead A-699 with 12F1 with S&C 140T No.Comment Phase to Phase Fault Date 12/12/05


2

2. MEA 12F1 NEG SEQ TIM SEL-EI TD=4.000CTR=160 Pickup=5.2A No inst. TP@5=1.0858sI=10910.4A (68.2 sec A) T= 0.27s

3

3. MEA A-699 NEG SEQ SEL-VI TD=4.500CTR=120 Pickup=1.25A Inst=1692A TP@5=1.1609sI= 1309.3A (10.9 sec A) T= 0.67s H=8.33

4

4. MEA A-699 PHASE SEL-VI TD=0.750CTR=120 Pickup=3.A Inst=1128A TP@5=0.1935sI= 864.6A (7.2 sec A) T= 0.68s H=8.33

FAULT DESCRIPTION:Close-In Fault on: 0 MEAD BUS 13.8kV - 0 MEA FDR BUS 13.8kV 2L LL Type=B-C

Figure 39. Coordinating MP Relays for a 500 Amp Feeder Using a 140T fuse and Coordinating a CS Using MP Relays with the 500 Amp Feeder for a Ø-Ø Fault.

MOSCOW FEEDER 515 PROTECTION EXAMPLE

54

PROBLEM – MOSCOW EXAMPLE

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

1 MILE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

Figure 40. Moscow feeder 515 (same as Figure 1)


55

Generally before we start the protection design on a feeder we know some things about the feeder such as: • The general feeder rating ( 300 or 500 amps). If it’s a 500 amp feeder we know the

phase pickup will be around 960 amps. In this case the Moscow feeder #515 is a 500 amp feeder and it has existed for some time.

• We know what the bus fault duty is. In this case the 3Ø fault is 5,188 amps and the SLG is 5,346. This tells us that the highest fuse size will probably be a 140T so the ground pickup will be about 480 amps or higher. NOTE: In rare cases we may have to use a 200 amp fuse or larger but we don’t like to use them if we don’t have to because we have to raise the ground pickup to coordinate with it and this reduces our sensitivity to detect faults.

• This feeder already was using a line recloser at the location shown and we found out from the Field Engineer that they wanted to keep a recloser there. In fact the reason we were looking at this feeder was because we were replacing the older recloser with a newer relayed recloser.

• We then have to run feeder fault studies (normally done by a Distribution Engineer) and look at the feeder configuration so we know what fusing and relay settings to use.

• We then generally start near the end of the feeder and work back towards the substation. A lot of times though we have to go back and forth because our first choice of a fuse size may not work correctly when we figure out the protection towards the substation and we have to make compromises.

The following is an example of how we arrive at those answers. NOTE: The actual Moscow feeder 515 does not look exactly like this even though the substation and line recloser settings we will arrive at are the actual settings. • POINT 8: This is a customers load and we are using 3-250 KVA transformers to

serve the load. The full load of this size of bank is 31.4 amps. The Avista transformer-fusing standard says to use a 65T on this transformer so that’s what we’ll choose.

• POINT 7: This is the end of a 3Ø lateral and the fault duty is: 3Ø = 322 amps and

SLG = 271 amps. • POINT 6A: The fault duty at PT 6 is: 3Ø = 1,907 amps and SLG is 1,492. It is

feeding a short single phase #4 ACSR lateral with not much load on it (that’s why we didn’t include the fault current at the end of the lateral).

o Temporary Faults - From Table 1 we see that a 65T fuse can be protected up to 1,650 amps for temporary faults and the SLG is 1,492 amps (SLG because it’s a single phase lateral).

o Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a 100T or smaller fuse.

o Fault Detection - We aren’t worried about detecting the fault at the end of the feeder or coordinating with other fuses.

o Loading – Assume the load is small so don’t need to worry. o So what fuse size should we use?


56

• POINT 6B: Here we have a 3Ø lateral feeding to a 65T at the end so what fuse size should we use?

o Temporary Faults - From Table 1 we see that an 80T fuse can be protected up to 2,050 amps for temporary faults and the 3Ø fault is 1,907 amps so we could choose an 80T or higher from that standpoint.

o Coordination - From Table 2 we see that an 80T will coordinate with a 65T at 2,700 amps with no preloading or 1,400 amps with preloading and the maximum fault at the 65T is 558 amps. So we could choose an 80T or higher.


o Fault Detection - Under the Relay Setting Criteria we want to detect the minimum end fault with a 2:1 margin. The SLG is 463 so the maximum fuse we could use would be: 463/2 (2:1 margin) = 231 amps so the largest fuse we could use is 231/2 (blows at twice the rating) = 115 so we could use a 100T or smaller fuse for this criteria.

o Loading – Assume the load at the end is all there is so no need to worry. o So what fuse size should we use?

• POINT 6C: Here we have a 3Ø lateral feeding to point 7 with a SLG of 271 amps. o Temporary Faults - From Table 1 we see that an 80T fuse can be protected up

to 2,050 amps for temporary faults and the 3Ø fault is 1,907 amps so we could choose an 80T or higher from that standpoint.


o Fault Detection - Under the Relay Setting Criteria we want to detect the end fault with a 2:1 margin. The SLG is 271 so the maximum fuse we could use would be: 271/2 (2:1 margin) = 135 amps so the largest fuse we could use is 135/2 (blows at twice the rating) = 67 so we could use a 65T or smaller fuse for this criteria. NOTE: Protecting the fuse for temporary faults is not as important as detecting the fault with adequate margin so we would sacrifice the fuse protection in order to gain sensitivity to detect the fault.


• POINT 5: This is the point of the line recloser. Here we have a 3Ø line of #4 ACSR

feeding to point 6. o Temporary Faults - The fault duties at the recloser are: 3Ø = 3,453 amps and

SLG = 2,762. We will assume we have a fused lateral just beyond the line recloser and we want to protect it for temporary faults. From Table 1 we see that we can protect a 100T fuse up to 2,650 and a 140T up to 3,500 amps. Therefore, from this standpoint we would want to be able to use a 140T fuse beyond the recloser.

o Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a 100T or smaller fuse. That means that we won’t be able to set the line recloser to coordinate with a 140T and still protect the conductor. Therefore, we have to compromise and we will sacrifice the fuse protection for


57

temporary faults in order to gain protection for the conductor. NOTE: This normally isn’t too bad because most faults are SLG and are somewhat less than the calculated value because of fault resistance and distance to the fault out on the lateral.

o Coordination – We will try and coordinate with a 100T fuse and still protect the #4 ACSR conductor. Once we determine the overcurrent pickup values, we will choose a Time Lever to provide 0.2 seconds (see Table 6) coordination with the fuse and then will recheck to determine if we are satisfied with the conductor protection.

o Fault Detection & Loading - The loading information we obtained from the Distribution Group is 84 amps normal and cold load ≅ 2 times = 168 amps. Therefore, we can set the phase PU down to ≅ 200 amps which would carry the load and pick it up cold load. However, #4 ACSR can carry around 140 amps (see Table 4) so we would probably want to set about 300 amps to allow for load growth. Also the minimum pickup we can use and still coordinate with a 100T is 300 amps (see Figure 32). NOTE: We do not set the phase overcurrent to protect the conductor from normal or emergency loading.

The Ø-Ø fault at point 6 is 0.866*1907 = 1,651 so our margin to detect that fault would be 1651/300 = 5.5:1 so no problem with the 300 amps Ø PU from that standpoint (used Ø-Ø because it’s the minimum multi phase fault).

o SLG Fault Detection – The SLG at point 6 is 1,492 so we could set the ground up to 1492/2 = 745 amps and still detect the fault. However, our criteria says to set as low as possible and still coordinate with the largest downstream device and from above we’re trying to use a 100T. Again this is 300 amps so in this case the ground PU will be set the same as the phase.

o Instantaneous Units – The phase instantaneous will be set at 350 A and the ground instantaneous unit will be set at 300 A. Note that they can’t reach all the way to point 7 so they can’t protect the point 6C fuse from temporary faults occurring near the end of the lateral.

• POINT 4: The fault duty at PT 4 is: 3Ø = 1,210 amps and SLG is 877. It is feeding a

short single-phase #4 ACSR lateral with not much load on it (that’s why we didn’t include the fault current at the end of the lateral).

o Temporary Faults - From Table 1 we see that a 40T fuse can be protected up to 1,040 amps for temporary faults and the SLG is 877 amps (single phase lateral).


o Fault Detection - We aren’t worried about detecting the fault at the end of the feeder or coordinating with other fuses.



58

• POINT 3A: The fault duty at point 3 is: 3Ø = 3,699 amps and the SLG is 3,060. Here we have a 2/0 ACSR 3Ø lateral feeding to the next fuse with a SLG of 877 amps.

o Temporary Faults - From Table 1 we see that a 140T fuse can be protected up to 3,500 amps for temporary faults and the 3Ø fault is 3,699 amps. This appears to indicate we would want to use a 200T fuse. However, this is a very large fuse and raises the substation ground relay settings a lot (480 to 720 amps) so we don’t like to use it. Therefore, we will sacrifice the fuse protection in order to gain more sensitive settings at the substation recloser so will tentatively choose a 140T and check a couple of other things.

o Coordination - From Table 2 we see that a 140T will coordinate with a 100T at 5,800 amps with no preloading or 3,800 amps with preloading. Thus if we added another lateral just beyond point 3A and used a 100T we could coordinate with it with the 140T. Note however that the 100T can be protected up to 2,650 amps so the protection for temporary faults would be further compromised on that branch.

o Conductor Protection - From Table 7 we see that 1/0 ACSR (or higher) can be protected with the 500 amp feeder settings at the substation so we won’t worry about the conductor protection.

o Fault Detection - Under the Relay Setting Criteria we want to detect the end fault with a 2:1 margin. The SLG is 877 so the largest fuse we could use would be: 877/2 (2:1 margin) = 438 amps so the largest fuse we could use is 438/2 (blows at twice the rating) = 219 so we could use a 200T or smaller fuse for this criteria.

o Loading – 2/0 ACSR can carry about 270 amps (see Table 4). Therefore, if this lateral starts to load up to it’s rating we wouldn’t be able to carry the load with a 140T or even a 200T fuse. The substation settings could protect the conductor but wouldn’t be able to detect a fault at point 4 with adequate margin. We would be forced to add another line recloser or maybe move the fuse out on the lateral until it could carry the load and then check to see if the substation could detect the faults at that point.

o So what fuse size should we use? • POINT 3B: Here we have a #2 ACSR 3Ø short lateral with only a small amount of

load. o Temporary Faults – Same story as for point 3A. Would choose a 140T fuse. o Coordination – There is nothing to coordinate with so no problem. o Conductor Protection - From Table 7 we see that #2 ACSR can be protected

with a 140T. o Fault Detection – It’s a short lateral so no problem. o Loading – No problem. o So what fuse size should we use?

• POINT 3C: The fault duty at point 3 is: 3Ø = 3,699 amps and the SLG is 3,060.

Here we have a 1 mile #4 ACSR 3Ø lateral (main lateral) feeding to the line recloser with a SLG of 3,060 amps.


59

o Temporary Faults - From Table 1 we see that a 140T fuse can be protected up to 3,500 amps for temporary faults and the 3Ø fault is 3,699 amps. This again appears to indicate we would want to use a 200T fuse.

o Coordination – We set the line recloser up to coordinate with a 100T fuse so we wouldn’t be able to coordinate with the recloser by using a 140T so this would indicate we need a 200T.

o Conductor Protection - From Table 7 we see that #4 ACSR requires a 100T fuse or smaller to be protected.

o The above coordination and conductor protection dilemma means that we can’t solve this problem by fusing this lateral. We also can’t remove the fuse because the substation settings can’t protect the conductor either.

o What are some possible solutions? • POINT 2A: Here we have a #4 ACSR 3Ø short lateral with only a small amount of

load. The fault duty at point 2 is: 3Ø = 5,158 amps and SLG = 5,346 (same as at the 13.8 kV bus).

o Temporary Faults – Basically same story as for point 3A. Would choose a 140T fuse because don’t want to use a 200T.

o Coordination – There is nothing to coordinate with so no problem. o Conductor Protection - From Table 7 we see that #4 ACSR can be protected

with a 100T fuse. Since the conductor protection is more important than fuse protection we have to lower this down to a 100T even though our fuse protection is greatly compromised.

o Fault Detection – It’s a short lateral so no problem. o Loading – No problem. o So what fuse size should we use?

• POINT 2B: Here we have a #2 ACSR 3Ø short lateral with only a small amount of

load. o Temporary Faults – Basically same story as for point 3A. Would choose a

140T fuse because don’t want to use a 200T. o Coordination – There is nothing to coordinate with so no problem. o Conductor Protection - From Table 7 we see that #2 ACSR can be protected

with a 140T fuse. o Fault Detection – It’s a short lateral so no problem. o Loading – No problem. o So what fuse size should we use?

• POINT 1: Substation recloser. We already know this is a 500 amp feeder so we have

to set to carry that load (even if the feeder is not presently loaded that amount). NOTE: Per the Distribution group the load is 325 amps and cold load 650.

o Temporary Faults – We have decided that a 140T fuse is the maximum fuse we will use on the feeder even though it can’t be protected at the maximum fault duties.


60

o Conductor Protection – The main trunk is 556 ACSR and from Table 7 a 500 amp feeder setting can protect 1/0 ACSR or higher so no problem since all laterals are fused.

o Coordination – We will coordinate with a 140T fuse at the maximum SLG fault duty of 5,346 amps with 0.2 seconds coordination time per Table 4. We will also coordinate with the line recloser settings with 0.3 seconds coordination time again per Table 4.

o Fault Detection & Loading – We will set the phase pickup to carry 500 amps normally plus pick up cold load so will set at 960 amps.

o The Ø-Ø fault at point 5 is 0.866*3453 = 2990 so our margin to detect that fault would be 2990/960 = 3.1:1 so no problem.

o SLG Fault Detection – Again per our criteria we will set as low as possible and still coordinate with the maximum downstream devices with the same coordination times as above (see 825).

o The SLG at point 5 is 2,762 so our margin to detect that fault is: 2762/480 = 5.7:1 so no problem.

o Instantaneous Units – The phase instantaneous will be set at 1,120A and the ground instantaneous unit will be set at 480 A.

1

DISTRIBUTION PROTECTION OVERVIEW

Mike Diedesch & Jon Harms – Avista Utilities

Presented March 14, 2011

At the 28th AnnualHANDS-ON Relay School

Washington State UniversityPullman, Washington

2

Table of Contents

System Overview ………………………………………………………....… 3

IEEE Device Designations …………………………………………….…….. 7

Symmetrical Components ……………………………………………..…….. 8

Distribution (3-30MVA ) Transformer Protection ………………….…..…... 14

Overcurrent Relay Curves …………………………………………..……. 22

Distribution Fuse Protection ………………………………………….……. 24

Conductors ………………..………..………………………………………. 28

Electromechanical Relays used on Distribution Feeders…………….…….… 31

Coordinating Time Intervals …………………………………………….… 36

Transformer Relay Settings using Electromechanical Relays ………….…… 48

Transformer & Feeder Protection using Microprocessor Relays ………….… 52

Transformer Differential Protection ……………………………………….… 61

Moscow 13.8kV Feeder Coordination Example ……………………………. 64

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

3

System Overview – Inside the Substation Fence

4

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

System Overview – Outside the Substation Fence

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

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TIME-CURRENT CURVES @ Voltage 13.8 kVHORS 2010 By JDH

For Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base.olr No.

Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A Date 11-25-2008

TRANSFORMER PROTECTION

STATION VCB 252 PROTECTION

MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR

MAXIMUM MIDLINE FUSE

Fault I=5665.9 A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096sIa= 679.8A (5.7 sec A) T= 0.78s H=8.33

2

2. IDR 252 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072sIa= 5665.9A (35.4 sec A) T= 0.30s

3. IDR 252 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s3Io= 0.0A (0.0 sec A) T=9999s

4

4. 252 140T FUSE stn S&C Link140TTotal clear.Ia= 5665.9A T= 0.09s

5

5f

5. Phase unit of recloser MID LINE OCR Fast: ME-341-B Mult=0.2 Slow: ME-305-A Add=1000. Ia= 5665.9A T(Fast)= 0.03s

6

6. T FUSE S&C Link 50TMinimum melt.Ia= 5665.9A T= 0.01s

A

A. Conductor damage curve. k=0.06710 A=556000.0 cmilsConductor AACFEEDER 252 SMALLEST CONDUCTOR TO PROTECT

B

B. Transf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.2 percent.IDAHO RD 12/16/20 MVA XFMR

FAULT DESCRIPTION:Bus Fault on: 0 IDR 252 13.8 kV 3LG

5

System Overview – Transformer Damage Curve, Time-Current Relay & Fuse Coordination Curves

6

System Overview 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

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TIME-CURRENT CURVES @ Voltage 13.8 kVINT 12F2 & MID 466R By JDHFor Indian Trail feeder 12F2 and 12F2 Midline 466R in INT 2007base.olr No.Comment AT SUB: 3LG= 5719A, LL=4951A, SLG= 5880A Date 1-29-08


STATION VCB 12F2 PROTECTION

EAST BRANCH - FEEDER MAXIMUM FUSE

FAULT DUTY AT 140T - WHEN FED FROM NORTH (N.O. PT)

12F2 MIDLINE 466R PROTECTION

WEST OR NORTH BRANCH - FEEDER MAXIMUM FUSE

FAULT DUTY AT 150E; 3LG=2318, LL=2008, SLG=1684

3LG=1932, LL=1673, SLG=1388

FAULT DUTY AT 140T - WHEN FED FROM WEST MID 466R 3LG=2093, LL=1813, SLG=1520

9&15

11&4

1

1. INT A-742 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s H=8.33

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2. INT A-742 51G 351 SEL-VI TD=1.000CTR=600/5 Pickup=1.A No inst. TP@5=0.258s H=8.33

3

3. INT A-742 51N 351 SEL-EI TD=5.600CTR=1200/5 Pickup=3.A No inst. TP@5=1.5201s

4

4. INT A-742 51Q 351 SEL-EI TD=4.200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s H=8.33

5

5. INT F2 150E FUSE SMU-20_150ETotal clear.

6

6. INT 12F2 50P 351S INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

7

7. INT 12F2 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s

8

8. INT 12F2 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

9

9. INT 12F2 51G 351S SEL-EI TD=3.900CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s

10

10. INT 12F2 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11

11. INT A-742 51Q 587 W2 SEL-EI TD=4.300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

12

12. F2 MID 466R 50P 351R INST TD=1.000CTR=500/1 Pickup=1.68A No inst. TP@5=0.048s

13

13. F2 MID 466R 51P 351R SEL-EI TD=2.300CTR=500/1 Pickup=1.44A No inst. TP@5=0.6243s

14

14. F2 MID 466R 50G 351R INST TD=1.000CTR=500/1 Pickup=0.89A No inst. TP@5=0.048s

15

15. F2 MID 466R 51G 351R SEL-EI TD=4.200CTR=500/1 Pickup=0.89A No inst. TP@5=1.1401s

16

16. F2 MID 466R 51Q 351R SEL-EI TD=3.800CTR=500/1 Pickup=1.54A No inst. TP@5=1.0315s

17

17. INT F2 140T FUSE stn S&C Link140TTotal clear.

A

A. Transf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.4 percent.INDIAN TRAIL 12/16/20 MVA XFMR

B

B. Conductor damage curve. k=0.14040 A=52630.0 cmilsConductor Copper (bare) AWG Size 2NORTH BRANCH SMALLEST TRUNK CONDUCTOR

C

C. Conductor damage curve. k=0.04704 A=350000.0 cmilsCable XLPE 90C/250CEAST BRANCH SMALLEST TRUNK CONDUCTOR - 350 CN15

D

D. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 2/0WEST BRANCH SMALLEST TRUNK CONDUCTOR

An example of a completed 13.8kV Feeder Coordination Study with 20 Time-Current Curves representing:

Instantaneous & Time-Delay Curves for the: - Transformer high side protection,- Transformer low side protection,- Station feeder breaker protection,- Midline feeder breaker protection.

Fuses

Transformer Damage Curve

Conductor Damage Curves

7

2 – Time delay relay.27 – Undervoltage relay.43 – Manual transfer or selective device.

We use these for cutting in and out instantaneous overcurrent relays, reclosing relays etc.

50 (or 50P) – Instantaneous overcurrent phase relay.

50N (or 50G) – Instantaneous overcurrentground (or neutral) relay.

50Q – Instantaneous Negative Sequence overcurrent relay.

51 (or 51P) – Time delay overcurrent phase relay.

51N (or 51G) – Time delay overcurrent ground (or neutral) relay.

51Q – Time delay Negative Sequence overcurrent relay.

52 – AC circuit breaker.

IEEE Device Designations

Avista sometimes adds letters to these such as F for feeders, T for transformers, B for bus and BF for breaker failure.

52/a – Circuit breaker auxiliary switch closedwhen the breaker is closed.

52/b – Circuit breaker auxiliary switch closed when the breaker is open.

59 – Overvoltage relay.62 – Time Delay relay63 – Sudden pressure relay.79 – AC Reclosing relay.81 – Frequency relay.86 – Lock out relay which has several contacts.

Avista uses 86T for a transformer lockout, 86B for a bus lockout etc.

87 – Differential relay.94 – Auxiliary tripping relay.

8

Symmetrical Components

Symmetrical Components for Power Systems Engineering, J. Lewis Blackburn,

There are three sets of independent components in a three-phase system: positive, negative and zero for both current and voltage. Positive sequence voltages (Figure 1) are supplied by generators within the system and are always present. A second set of balanced phasors are also equal in magnitude and displaced 120 degrees apart, but display a counter-clockwise rotation sequence of A-C-B (Figure 2), which represents a negative sequence. The final set of balanced phasors is equal in magnitude and in phase with each other, however since there is no rotation sequence (Figure 3) this is known as a zero sequence.

9

Symmetrical Components

Three phase (3LG) fault - Positive sequence currents – for setting phase elements in relays. Single phase (1LG) fault - Zero sequence currents – for setting ground elements in relays. Phase -Phase (L-L) fault - Negative sequence currents – for setting negative sequence elements in relays.

Symmetrical Components Notation:Positive Sequence current = I + = I1Negative Sequence current = I - = I2Zero Sequence current = I 0 = I0

Phase Current notation:IA – High side AmpsIa – Low side Amps

8.33= 115/13.8 = transformer Voltage ratioPhasor diagram from Fault-study Software.

Examples of three 13.8kV faults showing the current distribution through a Delta-Wye high-lead-low transformer bank :

10

Symmetrical Components – Positive Sequence, 3LG 13.8 kV FaultYou have only positive sequence voltage and current since the system is balanced.

A

B

C

a

b

c

RR

IA = 619 ∠-88° Ia = 5158 ∠-118°IB = 619 ∠ 152° Ib = 5158 ∠ 122°IC = 619 ∠ 32° Ic = 5158 ∠ 2°Phase current = Sequence currentThat is; Ia = I+.

Phase currents and voltages for the 115kV side.

IA = Ia / 8.33 = 5158A / 8.33IA = 619A

11

Symmetrical Components – Zero Sequence, 1LG 13.8 kV Fault

115kV side sequence currents and voltages

A

B

C

a

b

c

RR

3I0 is the sum of the 3 phase currents and since Ib & Ic= 0, then 3I0 = Ia. This means the phase and ground overcurrent relays on the feeder breaker see the same amount of current.

5346/(8.33*√3) = 370 amps. So the high side phase current is the √3 less as compared to the 3Ø fault.

Digital relays ground elements set using 3I0. 13.8kV side sequence currents and voltages

IA = 370 ∠-118° Ia=3I0=5346 ∠-118°IB = 0 ∠ 0° Ib = 0 ∠ 0°IC = 370 ∠ 62° Ic = 0 ∠ 0°Ia = 3Va/(Z1 + Z2 + Z0)

12

Symmetrical Components – Negative Sequence, L-L 13.8 kV Fault

115kV side sequence currents and voltages

13.8kV side sequence currents and voltages

A

B

C

a

b

c

RR

IA = 309 ∠-28° Ia = 0 ∠ 0°IB = 619 ∠ 152° Ib = 4467 ∠ 152°IC = 309 ∠-28° Ic = 4467 ∠-28°

3LG 13.8kV fault = 5158A,Ib=Ic=4467 A, 4467/5158 = 86.6%=√3/2IB3LG = IBLL = 619AIA & IC = IB or Ia & Ic = IbI2 = the phase current/√3 = 4467/√3 = 2579

Digital relays 50Q/51Q elements set using 3I2.3I2 = √3 x Ib = 7737,

13

Symmetrical Components - Summary of 13.8 kV Faults

IA = 309 Ia = 0IB = 619 Ib = 4467IC = 309 Ic = 4467

5158 x √3/2 = 4467 Ib x √3 = 7737 = 3I2 4467/(8.33x √3) = 309309 x 2 = 619

3LG

L-L

SLG

IA = 619 Ia = 5158IB = 619 Ib = 5158IC = 619 Ic = 5158

IA = 370 Ia=5346IB = 0 Ib = 0IC = 370 Ic = 0

5158 / 8.33 = 619 Ia = 5158 = I1

5346/(8.33x√3) = 370 Ia = 5346 = 3I0

If you have a Delta-Wye transformer bank, and you know the voltage ratio and secondary phase current values for 13.8kV 3LG (5158) and SLG (5346) faults, you can find the rest:

14

Transformer Protection - Damage Curve – ANSI/IEEE C57.109-1985

The main damage curve line shows only the thermal effect from transformer through-fault currents. It is graphed from data entered below (MVA, Base Amps, %Z ): 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

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T IME-CURRENT CURVES @ Voltage By

For No.

Comment Date

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1. M15CTR=

A

A. T ransf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.2 percent.

The dog leg on the curve is added to allow for additional thermal and mechanical damage from (typically more than 5) through-faults over the life of a transformer serving overhead feeders.

Dog leg curve - 10 times base current at 2 seconds.

Main curve - 25 times base current at 2 seconds.

Time at 50% of the maximum per-unit through fault current = 8 seconds.

Avista has Category III size (5-30MVA) Distribution Transformers in service per the above standard.

15

Transformer Protection using 115 kV Fuses

16

Transformer Protection using 115 kV Fuses

Used at smaller substations up to 7.5 MVA transformer due to low cost of protection.Other advantages are:- Low maintenance- Panel house & station battery not required

There are also several disadvantages to using fuses however which are:• Low interrupting rating from 1,200A (for some older models) up to 10,000A at 115 kV. By contrast a circuit switcher can have a rating of 25KAIC and our breakers have normally 40KAIC. • The fuses we generally use are rated to blow within 5 minutes at twice their nameplate rating. Thus, a 65 amp fuse will blow at 130 amps. This compromises the amount of overload we can carry in an emergency and still provide good sensitivity for faults.

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

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TIME-CURRENT CURVES @ Voltage 13.8 KVROK 451R2 By JDH

For ROK FDR 451, BUS FAULTS: 3LG=3580A, SLG=3782A, L-L=3100A No.

Comment ASPEN FILE: ROK NEW XFMR (2005 BASE).OLR Date 2-13-07

STATION TRANSFORMER PROTECTION

FEEDER VCR PROTECTION

MAXIMUM FEEDER FUSE

1

1. LAT RP4211 50N CO-11 INST TD=1.000CTR=500/5 Pickup=2.A No inst. TP@5=0.048s

2

2. SMD-2B 65E VERY SLOW 176-19-065Minimum melt. H=8.33

3

3. LAT RP4211 50P CO-11 INST TD=1.000CTR=500/5 Pickup=3.5A No inst. TP@5=0.048s

4

4. LAT RP4211 51P CO-11 CO-11 TD=2.000CTR=500/5 Pickup=3.A No inst. TP@5=0.5043s

5

5. LAT RP4211 51N CO-11 CO-11 TD=4.000CTR=500/5 Pickup=2.A No inst. TP@5=1.0192s

6

6. LAT RP4211 FUSE S&C Link 65TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 7.3 percent.Rockford 13.8kV - ROCKFORD115 115.kV 1 T

17

• The fuse time current characteristic (TCC) is fixed (although you can buy a standard, slow or very slow speed ratio which are different inverse curves).

• The sensitivity to detect lo-side SLG faults isn’t as good as using a relay on a circuit switcher or breaker. This is because we use DELTA/WYE connected transformers so the phase current on the 115 kV is reduced by the √3 as opposed to a three phase fault.

• Some fuses can be damaged and then blow later at some high load point.

• When only one 115 kV fuse blows, it subjects the customer to low distribution voltages. For example the phase to neutral distribution voltages on two phases on the 13.8 kV become 50% of normal.

• No indication of faulted zone (transformer, bus or feeder).

Transformer Protection using 115 kV Fuses - continued

A

B

C

a

b

c

1.0 PU

0.5 PU0.5 PU

18

Transformer Protection using a Circuit SwitcherShowing Avista’s present standard using Microprocessor relays.

13.8kV 115kV

19

Transformer Protection using a Circuit SwitcherShowing Avista’s old standard using Electromechanical relays.

20

Transformer Protection using a Circuit Switcher

Some advantages to this over fuses are:• Higher interrupting.• Relays can be set to operate faster

and with better sensitivity than fuses.

• Three phase operation.• Provide better coordination with

downstream devices.

Some disadvantages would be:• Higher cost.• Higher maintenance.• Requires a substation battery, panel

house and relaying.• Transformer requires CT’s.

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kVIDR FEEDER 251 By JDH





MAXIMUM FEEDER FUSE

HI-SIDE CT'S

LO-SIDE CT'S

MINIMUM FAULT TO DETECT:3LG=2460A, SLG=1833A, L-L=2130A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s H=8.33

2

2. IDR A-777 51G 351 SEL-VI TD=1.000CTR=600/5 Pickup=1.A No inst. TP@5=0.258s H=8.33

3

3. IDR A-777 51N 351 SEL-EI TD=5.700CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s

4

4. IDR A-777 51Q 351 SEL-EI TD=4.200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s H=8.33

5

5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

6

6. IDR 251 50P 351S INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

7

7. IDR 251 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s

8

8. IDR 251 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

9

9. IDR 251 51G 351S SEL-EI TD=3.900CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s

10

10. IDR 251 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11

11. 251 140T FUSE stn S&C Link140TTotal clear.

A

A. Conductor damage curve. k=0.08620 A=133100.0 cmilsConductor ACSR AWG Size 2/0FEEDER 251 SMALLEST CONDUCTOR TO PROTECT

B


21

Transformer Protection using a Breaker

This is very similar to using a circuit switcher with a couple of advantages such as:• Higher interrupting – 40kAIC for the one shown below.• Somewhat faster tripping than a circuit switcher (3 cycles vs. 6 – 8 cycles).• Possibly less maintenance than a circuit switcher.• The CT’s would be located on the breaker so it would interrupt faults on the bus section up to

the transformer plus the transformer high side bushings.

22

Figure 19. SEL Various Overcurrent Relay Curves.

Extremely Inverse – steepestVery InverseInverseModerately InverseShort Time Inverse – least steep

The five curves shown here have the same pickup settings , but different time dial settings.

These are basically the same as various E-M relays.

Avista uses mostly extremely inverse on feeders to match the fuse curves.

Overcurrent Relay Curves10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage By

For No.

Comment Date

1

1. EXTREMELY INVERSE SEL-EI TD=15.000CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s

2

2. INVERSE SEL-I TD=2.000CTR=800/5 Pickup=6.A No inst. TP@5=0.8558s

3

3. MODERATELY INVERSE SEL-2xx-MI TD=1.000CTR=800/5 Pickup=6.A No inst. TP@5=0.324s

4

4. SHORT TIME INVERSE SEL-STI TD=1.000CTR=800/5 Pickup=6.A No inst. TP@5=0.1072s

5 5. VERY INVERSE SEL-VI TD=6.000CTR=800/5 Pickup=6.A No inst. TP@5=1.5478s

23

Same Time Dial = 9Right curve picks up at 960 AmpsLeft curve picks up at 320Amps

Overcurrent Relay Curves - Example of Different Pickup & Time Dial Settings

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1


For No.

Comment Date

1

1. EXT INV 1 SEL-EI TD=15.000CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s

2 2. ~EXT INV 1 SEL-EI TD=12.000CTR=800/5 Pickup=6.A No inst. TP@5=3.2574s

3

3. ~EXT INV 2 SEL-EI TD=9.000CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s

4


5


10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1


For No.Comment Date

1

1. EXT INV 1 SEL-EI TD=9.000CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s

2


3


4


5


Same Pickup = 960 AmpsTop curve Time Dial = 15Bottom curve Time Dial = 2

24

Distribution Fusing – Fuse Protection/Saving Scheme

Fault on fused lateral on an overhead feeder:- Station or midline 50 element protects fuse. - Station or midline 51 element back up fuse.

During fault: • Trip and clear the fault at the station (or line recloser) by the instantaneous trip before the fuse is damaged for a lateral fault. • Reclose the breaker. That way if the fault were temporary the feeder is completely re-energized and back to normal. • During the reclose the reclosing relay has to block the instantaneous trip from tripping again. That way, if the fault still exists you force the time delay trip and the fuse will blow before you trip the feeder again thus isolating the fault and re-energizing most of the customers.• Of course if the fault were on the main trunk the breaker will trip to lockout.

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kVM15 515 51N to 140T By Protection

For Aspen File: HORS M15 EXP.olr No.

Comment Date 3/06

STATION FEEDER RELAYING

MAXIMUM FEEDER FUSE

1

1. M15 515 GND TIME CO-11 TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0192s

2

2. Moscow 515 Kear 140T Kearney 140TTotal clear.

3

3. M15 50N CO-11 INST TD=1.000CTR=160 Pickup=3.A No inst. TP@5=0.048s

25

Shows the maximum fault current for which S&C type T fuses can still be protected by a recloser/breaker instantaneous trip for temporary faults (minimum melt curve at 0.1 seconds):6T – 120 amps

8T – 160 amps10T – 225 amps12T – 300 amps15T – 390 amps20T – 500 amps25T – 640 amps30T – 800 amps40T – 1040 amps50T – 1300 amps65T – 1650 amps80T – 2050 amps100T – 2650 amps140T – 3500 amps200T – 5500 amps

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1


For No.

Comment Date

1

1. BR1 S&C 140T S&C Link140TMinimum melt.

2

2. ENDTAP S&C 40T S&C Link 40TMinimum melt.

3

3. M15 515 S&C 80T S&C Link 80TMinimum melt.

Distribution Fusing – Fuse Protection for Temporary Faults

26

NOTE: These values were taken from the S&C data bulletin 350-170 of March 28, 1988 based on no preloading and then preloading of the source side fuse link. Preloading is defined as the source side fuse carrying load amps equal to it’s rating prior to the fault. This means there was prior heating of that fuse so it doesn’t take as long to blow for a given fault.

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1


For No.

Comment Date

1

1. T FUSE S&C Link100TMinimum melt.

2


Distribution Fusing – Fuse to Fuse Coordination

27

Typical continuous and 8 hour emergency rating of the S&C T rated silver fuse links plus the 140T and 200T.

General Rule Fuse Blows at 2X Rating in 5 Minutes 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

45

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

45

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

T IME-CURRENT CURVES @ Voltage By

For No.

Comment Date

1

1. BR1 S&C 140T S&C Link140TMinimum melt.

Distribution Fusing – S&C T-Fuse Current Ratings

28

Copper Conductor Damage Curves ACSR Conductor Damage Curves

Conductor Damage Curves

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor ACSR Conductor Damage Curves No.Comment Date 12/13/05

1


A

A. Conductor damage curve. k=0.08620 A=355107.0 cmilsConductor ACSR336.4 ACSR

B

B. Conductor damage curve. k=0.08620 A=167800.0 cmilsConductor ACSR AWG Size 4/04/0 ACSR

C

C. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 2/02/0 ACSR

D

D. Conductor damage curve. k=0.08620 A=83690.0 cmilsConductor ACSR AWG Size 1/01/0 ACSR

E

E. Conductor damage curve. k=0.08620 A=52630.0 cmilsConductor ACSR AWG Size 2#2 ACSR

FF. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4#4 ACSR

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Copper Conductor Damage Curves No.Comment Date 12/13/05

1


A

A. Conductor damage curve. k=0.14040 A=105500.0 cmilsConductor Copper (bare) AWG Size 2/02/0 Copper

B

B. Conductor damage curve. k=0.14040 A=83690.0 cmilsConductor Copper (bare) AWG Size 1/01/0 Copper

C

C. Conductor damage curve. k=0.14040 A=52630.0 cmilsConductor Copper (bare) AWG Size 2#2 Copper

D

D. Conductor damage curve. k=0.14040 A=33100.0 cmilsConductor Copper (bare) AWG Size 4#4 Copper

EE. Conductor damage curve. k=0.14040 A=20820.0 cmilsConductor Copper (bare) AWG Size 6#6 Copper

29

Conductor Rating556 730

336.4 5304/0 3402/0 2701/0 230#2 180#4 140

TABLE 4 – ACSR Ampacity Ratings

Conductor Rating

2/0 3601/0 310#2 230#4 170#6 120

TABLE 5 – CopperAmpacity Ratings

Conductor AmpacitiesConductor at 25°C ambient taken from the Westinghouse Transmission & Distribution book.

30

Figure 20. Comparing a 140T fuse Vs. #4 ACSR Damage curve. The 140T won’t protect the conductor below about 550 amps where the curves cross.

Conductor Protection Graph

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV#4 ACSR & 140T By Protection

For Aspen File: HORS M15 EXP.olr No.

Comment Date 3/06

1

1. Moscow 515 Kear 140T Kearney 140TTotal clear.

A

A. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4

31

Electromechanical Relays used on Distribution Feeders

Avista’s standard relay package includes the following:

3 phase 50/51 CO-11 relays,

1 ground 50/51 CO-11 relay

1 reclosing relay.

Avista Distribution Feeder Standard until the mid-1990’s

32

Objectives:− Protect the feeder conductor− Detect as low a fault current as possible (PU = 50% EOL fault amps) − Other than 51P, set pickup and time dial as low as possible and still have minimum Coordinating Time Interval (CTI – table 6) to next device. CTI is minimum time between operation of adjacent devices. − Carry normal maximum load (phase overcurrent only).− Pickup the feeder in a cold load condition (≅ 2 times maximum normal load) or pickup ½ of the next feeder load·

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 KV By

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1


2

2. LAT421 50P CO-9 INST TD=1.000CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s

3

3. LAT421 51P CO-9 CO-9 TD=1.900CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s

4

4. LAT421 50N CO-9 INST TD=1.000CTR=600/5 Pickup=3.A No inst. TP@5=0.048s

5

5. LAT421 51N CO-9 CO-9 TD=3.000CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s

6

6. LAT421 FUSE S&C Link100TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 6.9 percent.Latah Jct 13.8kV - LATAHJCT115 115.kV T

B

B. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 1/0


33

FIGURE 16-1

13 kV BUS

MIN FAULT3LG = 2000 A1LG = 1000 A

FEEDER SETTINGS51P = 2000 / 2 = 1000 A51N = 1000 / 2 = 500 A


34

FIGURE 16-2

13 kV BUS

FAULT at MIDLINE3LG = 2000 A1LG = 1000 A

FEEDER SETTINGS51P = 2000 / 2 = 1000 A51N = 1000 / 2 = 500 A

MIN FAULT at END OF LINE3LG = 1000 A1LG = 500 A

MIDLINE SETTINGS51P = 1000 / 2 = 500 A51N = 500 / 2 = 250 A


35

13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A

SECTION LOAD = 500 A

FIGURE 16-3


36

FIGURE 16-4

13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A N.O.





37

• Most overhead feeders also use reclosing capability to automatically re-energize the feeder for temporary faults. Most distribution reclosing relays have the capability of providing up to three or four recloses.

-- Avista uses either one fast or one fast and one time delayed reclose to lockout.

• The reclosing relay also provides a reset time generally adjustable from about 10 seconds to three minutes. This means if we run through the reclosing sequence and trip again within the reset time, the reclosing relay will lockout and the breaker will have to be closed by manual means. The time to reset from the lockout position is 3 to 6 seconds for EM reclosing relays.

-- Avista uses reset times ranging from 90 to 180 seconds.

• Lockout only for faults within the protected zone. That is; won’t lockout for faults beyond fuses, line reclosers etc.

• Most distribution reclosing relays also have the capability of blocking instantaneous tripping.

-- Avista normally blocks the INST tripping after the first trip to provide for a Fuse Protecting Scheme.


Reclosing

38

13 kV BUS

RECLOSING SEQUENCE

Closed

Open0.5" 12" LOCKOUT

RESET = 120"(INST Blocked during Reset Time)


Reclosing – Sequence shown for a permanent fault

39

Typical Coordinating Time Intervals (CTI) that Avista generally uses between protective devices.Other utilities may use different times.

DEVICES: CTI (Sec.)

Relay – Fuse Total Clear 0.2Relay – Series Trip Recloser 0.4Relay – Relayed Line Recloser 0.3Lo Side Xfmr Relay – Feeder Relay 0.4Hi Side Xfmr Relay – Feeder Relay 0.4Xfmr Fuse Min Melt – Feeder Relay 0.4

Coordinating Time Intervals10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

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MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1


2



4


5

5f


6


A


B



40


DEVICES: CTI (Sec.)


Coordinating Time Intervals10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

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1






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1


2



4


5

5f


6


A


B



10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

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TIME-CURRENT CURVES @ Voltage 13.8 kVRAT 231 MID C390R By JDH

For Huetter feeder 142 (2008) Rat feeder 231 (2009+) Midline Recloser C390R No.

Comment Saved in: HUE142-RAT231 MID C390R 2007base.olr Date 2-25-08

HUETTER FDR 142 PROTECTION

RATHDRUM FDR 231 PROTECTION

HUE 142 MID C270R PROTECTION

MIDLINE 390R PROTECTION

CALLED "HUE 142 LINE" IN POWERBASERECLOSE IS 1SEC, 12SEC, LO, 120 SEC RESET

RECLOSE IS 1SEC, 12 SEC, LO, 180 SEC RESET

LARGEST DOWNSTREAM FUSE FROM 390R MIDLINE

1

1. C390R MID 51P CO-9 CO-9 TD=2.500CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s

2

2. C390R MID 50P CO-9 INST TD=1.000CTR=300/5 Pickup=6.A No inst. TP@5=0.048s

3 3. C390R MID 50N CO-9 INST TD=1.000CTR=300/5 Pickup=5.A No inst. TP@5=0.048s

4

4. C390R MID 51N CO-9 CO-9 TD=2.500CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s

5

5. HUE 142 50P 251 INST TD=1.000CTR=160 Pickup=7.A No inst. TP@5=0.048s

6

6. HUE 142 51P 251 SEL-EI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.4072s

7

7. HUE ML C270R 50P INST TD=1.000CTR=400/5 Pickup=8.1A No inst. TP@5=0.048s

8

8. HUE ML C270R 51P CO-11 TD=1.500CTR=400/5 Pickup=7.A No inst. TP@5=0.3766s

9

9. HUE 142 50N 251 INST TD=1.000CTR=160 Pickup=3.A No inst. TP@5=0.048s

10

10. HUE 142 51N 251 SEL-EI TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0858s

11

11. HUE 142 51Q 251 SEL-EI TD=4.000CTR=160 Pickup=5.2A No inst. TP@5=1.0858s

12

12. HUE ML C270R 50N INST TD=1.000CTR=400/5 Pickup=3.A No inst. TP@5=0.048s

13

13. HUE ML C270R 51N CO-11 TD=7.000CTR=400/5 Pickup=3.A No inst. TP@5=1.8052s

14

14. RAT 231 50N CO-11 INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

15

15. RAT 231 51N CO-11 CO-11 TD=5.000CTR=800/5 Pickup=3.A No inst. TP@5=1.3192s

16

16. RAT 231 50P CO-11 INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

17

17. HUE 142 FUSE S&C Link 65TTotal clear.

18

18. RAT 231 FUSE S&C Link100TTotal clear.

19

19. RAT 231 51P CO-11 CO-11 TD=1.800CTR=800/5 Pickup=6.A No inst. TP@5=0.4533s

A

A. Conductor damage curve. k=0.08620 A=133100.0 cmilsConductor ACSR AWG Size 2/0MIDLINE C390 FED FROM RAT 231 - MINIMUM CONDUCTOR TO PROTECT

B

B. Conductor damage curve. k=0.06710 A=556000.0 cmilsConductor AACMIDLINE C390 FED FROM HUE 142 - MINIMUM CONDUCTOR TO PROTECT

41


DEVICES: CTI (Sec.)


Coordinating Time Intervals

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

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1

TIME-CURRENT CURVES @ Voltage 13.8 kVIDR FEEDER 251 By JDH

For Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007base.olr No.Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A Date 11-25-2008



MAXIMUM FEEDER FUSE

HI-SIDE CT'S

LO-SIDE CT'S

MINIMUM FAULT TO DETECT:3LG=2460A, SLG=1833A, L-L=2130A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s H=8.33

2

2. IDR A-777 51G 351 SEL-VI TD=1.000CTR=600/5 Pickup=1.A No inst. TP@5=0.258s H=8.33

3

3. IDR A-777 51N 351 SEL-EI TD=5.700CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s

4

4. IDR A-777 51Q 351 SEL-EI TD=4.200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s H=8.33

5

5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

6

6. IDR 251 50P 351S INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

7

7. IDR 251 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s

8

8. IDR 251 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

9

9. IDR 251 51G 351S SEL-EI TD=3.900CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s

10

10. IDR 251 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11


A

A. Conductor damage curve. k=0.08620 A=133100.0 cmilsConductor ACSR AWG Size 2/0FEEDER 251 SMALLEST CONDUCTOR TO PROTECT

B


42


DEVICES: CTI (Sec.)



10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

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TIME-CURRENT CURVES @ Voltage 13.8 KV By

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1


2

2. LAT421 50P CO-9 INST TD=1.000CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s

3

3. LAT421 51P CO-9 CO-9 TD=1.900CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s

4

4. LAT421 50N CO-9 INST TD=1.000CTR=600/5 Pickup=3.A No inst. TP@5=0.048s

5

5. LAT421 51N CO-9 CO-9 TD=3.000CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s

6

6. LAT421 FUSE S&C Link100TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 6.9 percent.Latah Jct 13.8kV - LATAHJCT115 115.kV T

B

B. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 1/0

43


DEVICES: CTI (Sec.)



44


10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

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TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor M15 A-172 CS with fdr 515 with Kearney 140T Fuse No.Comment Three Phase Fault Date 12/12/05

1


2


3



3LG Fault Coordination Example:

Top – Ckt Swr w/ Phase E-M relay

----- 0.4 sec. ------

Middle - E-M Phase relay for a 500 Amp Feeder

----- 0.2 sec. ------

Bottom – 140T Feeder Fuse (Total Clear)

45


10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

2

3457

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TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS with fdr 515 with 140T fuses No.Comment Single Line to Ground Fault Date 12/12/05

1


2


3


4 4. M15 A-172 GND TIME CO-11 TD=4.000CTR=240 Pickup=4.A No inst. TP@5=1.0192sI= 5346.5A (22.3 sec A) T= 0.83s


SLG Fault Coordination Example:

Top – Ckt Swr w/ E-M Phase relay

----- 0.4 sec. ------

Middle - E-M relays (Phase & Ground) for a 500 Amp Feeder

----- 0.2 sec. ------


46


10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

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3457

10

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TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS A-172 with fdr 515 with 140T Fuse No.Comment Line to Line Fault Date

1


2


3


FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L LL Type=B-C

L-L Fault Coordination Example:

Top – Ckt Swr w/ E-M relay

----- 0.4 sec. ------

Middle - E-M (Phase) relay for a 500 Amp Feeder

----- 0.2 sec. ------


Mike’s Turn

47

48

The distribution transformer relay settings have to (should) meet the following criteria (there will be several compromises here): NOTE: These are for an outdoor bus arrangement and not switchgear which also uses a bus breaker between the transformer relaying and the feeder relaying. This also doesn’t include the transformer sudden pressure relay.

• Protect the feeder conductor from thermal damage in case the feeder breaker can’t trip. This normally can’t be done completely since the transformer relaying has to be set higher than the feeder relaying.• Protect the transformer from thermal damage.• Detect as low a fault current as possible. It would be desirable to be just as sensitive as the feeder relaying in case the feeder breaker can’t trip but this can’t always be done.• For all relays other than the phase overcurrent, this means to set the pickup as sensitive as we can and the time lever as low as we can and still coordinate with the highest downstream device by the desired CTI (Table 6). • Coordinate with all downstream devices which is the feeder breaker relaying.• Carry normal maximum load (phase overcurrent only).• After an outage, pickup the station cold load (≅ 2 times maximum normal, phase overcurrent only).

Distribution Transformer Electromechanical Relays - Setting Criteria

49

Transformer - The normal transformer relaying will consist of 3 high side phase and 1 low side ground overcurrent relays plus a sudden pressure. We could also use a high side ground but generally haven’t because it won’t detect ground faults through the transformer.


50

1. Phase overcurrent (on 115 kV transformer BCT’s):

• 51P pickup – This has to be able to pick up the entire transformer load cold load or ≅ 2*normal. Our typical rule of thumb for this is 2.4*highest MVA rating. For a 12/16/20 (5 amps per MVA) this is 2.4*20*5 = 240 amps (1920 amps at 13.8 kV). For an 18/24/30 this is 2.4*30*5 = 360 amps (2880 amps at 13.8 kV). The pickup is so high that we can’t be as sensitive as the feeder phase relay so can’t reach to the same fault points on the feeder. Thus the ability to detect multi phase faults and protect the feeder conductor are compromised.

• 51P time lever – This relay has to coordinate with the feeder phase relays for a maximum feeder fault with 0.4 sec coordination time (Table 6). Avista uses 0.4 sec coordination time because multiple feeders can create a discrepancy in the actual ratio of amps seen on the high side vs. the low side. The worst case coordination occurs for a Ø-Ø fault since the feeder phase relays see 86.6% of what they saw for a 3Ø fault but yet one of the high side relays see the same as for a 3Ø fault because of the delta/wye connection (see figure 10). A typical TL would be ≅ 1.2 – 2.0.

• 50P – This is a direct trip element and is set to not detect a maximum fault on the 13.8 kV bus. Since the inst unit is a hinged armature device, it responds to current ‘DC Offset’ so the setting we have used is 1.7*MAX 13.8 kV fault. A typical setting is 900 to 1500 amps primary.


51

2. Ground overcurrent (on 13.8 kV transformer BCT’s):

• 51N pickup – This relay will be set to coordinate with the feeder phase relays in case the feeder ground relay is out of service. The minimum pickup to coordinate with a 500 amp feeder phase relay is 960 amps. We again compromise our ability to detect feeder faults since the feeder ground is twice as sensitive as the transformer ground.

• 51N time lever – This would normally be ≅ 4.0 – 6.0. This could be set to coordinate with 0.3 seconds margin since we are on the same voltage. However, a little more conservative setting would still use the 0.4 seconds because the transformer unbalance can be greater than any one feeder so it might see slightly more current for a fault. However, the phase OC relay also sees at least some load that the ground relay doesn’t see so the 0.3 seconds may be acceptable. Avista prefers the 0.4 seconds for all elements.

• 50N – We can’t use this element because we can’t coordinate it with the feeder relays.


52

Distribution Transformer & Feeder Protectionwith Microprocessor Relays

The following is at a new station where we can use microprocessor relaying on all of the feeders plus the transformer. We will use SEL 351 and 587 relays for our example since that is Avista’s present standard. Of course there are some general advantages to using MP relaying over E-M. Some of these are:1. The taps are selectable in very small increments.2. There are several independent units contained in the relay (up to 6 instantaneous and 2 time

delay).3. Internal logic programming capability.4. Less CT burden (CT’s can saturate with high burdens and high fault values and won’t produce

a good secondary current to the relay).5. Event reporting.6. Remote Communication capability.

53

Transformer - The normal transformer relaying will consist of a SEL 351S connected to the 115 kV BCT’s and the neutral element (single coil) connected to the 13.8 kV BCT’s (or possibly a neutral CT). In addition to the normal tripping elements, the SEL 351S will also use ‘fast’ (≅ 4 cycles) elements that are blocked by the feeder time pickup elements. That is; if an overcurrent element picks up on a feeder, it will block the 351S ‘fast’ overcurrent elements (known as a Fast Trip Blocking Scheme, FTB). Of course this can’t be used at a station that has any feeders that use E-M relays.

NOTE: One of the major differences between using all microprocessor relays and E-M relays is that with the microprocessor relays, you can coordinate like elements with like elements only. For example, with the E-M we coordinate the transformer ground with the feeder phase because the feeder ground could be out of service. With microprocessors, the entire relay would be out of service and not just one element. Therefore, we can coordinate the transformer ground with just the feeder ground and obtain a more sensitive setting.


54

Transformer - continued

Typical settings for both a 12/16/20 MVA and an 18/24/30 MVA transformer coordinating with the 500 amp feeder above would be:

1. SEL 351 Phase overcurrent (on 115 kV transformer BCT’s):

• 51P pickup – This is identical to the E-M relay setting or 240 amps for a 20 MVA and 360 amps for a 30 MVA.

• 51P TL – This has to coordinate with the feeder phase for a maximum feeder fault with 0.4 seconds coordination. In this case we could consider this to be a 3Ø fault since we will have negative sequence relays on the feeder to take care of the Ø-Ø fault.

• 50P1 pickup – This is the direct trip phase element and is set to not detect a 13.8 kV fault similar to the E-M instantaneous setting. However, this does not respond to DC offset so our setting is ≅ 1.3*MAX 13.8 kV fault.

• 50P2 pickup – This is the torque controlled element. We have set this to account for inrush and still be as sensitive as possible.


55


2. SEL 351 ground (residual) overcurrent (on 115 kV BCT’s) NOTE: This is calculated from the phase quantities:

• 51G pickup – We can set this very low since it doesn’t have to coordinate with anything. We have been setting ≅ 120 amps primary.

• 51G TL – Also can set low. Have been setting at TL 1.0.

• 50G pickup – This can also be set low. We have been choosing a setting equal to the 50P2 TC setting simply because we don’t feel we need more sensitivity but it really can be set as low as the time unit since the inrush to the delta doesn’t contain any 3I0 current. A typical setting is 120 to 360 amps.


56


3. SEL 351 neutral overcurrent (on 13.8 kV BCT’s):

• 51N pickup – We can set as low as possible and still coordinate with the feeder ground settings (not phase). Since the transformer can have more unbalance than the feeder we have been setting ≅ 1.33 times the feeder ground pickup or 640 amps when coordinate with 500 amp feeders.

• 51N TL – We coordinate with the feeder ground with 0.4 seconds margin.

• 50N1 – We can’t set a direct trip element so we use this as a FTB element only. Again we need to set above the feeder ground setting and to be a little conservative we have been setting ≅ 720 amps with 4 cycles time delay.


57

• Inrush – The current seen when energizing a transformer.

For this unit we need to consider transformer inrush current since it can operate ‘fast’ (≅ 4 cycles time delay). A rule of thumb we have used is that a typical transformer inrush RMS current can be as high as ≅ 8 times the transformer base rating (old E-M differential relays had the unrestrained overcurrent trip fixed at ten times the TAP of the restrained trip and the TAP would be set at ≅ transformer full load).

Note that the SEL 351 only responds to the 60 HZ fundamental and that this fundamental portion of inrush current is ≅ 60% of the total. So to calculate a setting for a SEL 351, we could use the 8 times rule of thumb along with the 60% value. For a 12/16/20 MVA transformer, the calculation would be 8*12*5*0.6 = 288 amps. For an 18/24/30 it would be 8*18*5*0.6 = 432 amps. Generally, we would set this unit at about 1.5 times the time unit to be safe. So for a 12/16/20 we would set 1.5*240 = 360 amps and for an 18/24/30 we would set 1.5*360 = 540 amps.


58

Transformer inrush UNFILTEREDcurrent. The maximum peak current is about 1800 amps.

Transformer inrush FILTERED current (filtered by SEL digital filters to show basically only 60 HZ). Maximum peak is about 700 amps.


59

Feeder – The normal relaying will use the SEL 351 phase, negative sequence and ground elements. We use one phase and ground instantaneous element. NOTE: We don’t use the instantaneous negative sequence because of the possibility of false trips due to motor contributions for external faults. The time delay pickup units for phase, negative sequence and ground are used to operate an output contact to block some low set fast elements on the 50/51 transformer relay. The relay is also programmed to provide a breaker failure output in case the breaker fails to open. Again, for a 500 amp feeder we have:

1. Phase overcurrent:• 51P pickup - Set the pickup at 960 amps for the same reasons as the E-M relays.• 51P TL – Set the same as the E-M relays for the same reasons.• 50P pickup – Set similar to the E-M 50PF pickup.

NOTE: This unit has a higher dropout characteristic than the E-M relays so it could probably be set closer to the 51P pickup.

2. Ground overcurrent:• 51G pickup – This is actually the residual element in the 351 which is obtained by adding the

3 phase currents. There is also a neutral element (single coil) that is not used. The pickup is the same as the E-M relay or 480 amps.

• 51G TL – This is the same as the E-M relay.• 50G – This is the same as the E-M relay = 480 amps.


60

Feeder – Continued

3. Negative Sequence:• 51Q pickup – This element can be set for similar criteria to the ground element since it

doesn’t respond to load. Similar to the ground which responds to 3I0, this unit responds to 3I2. That is; it should be set as sensitive as possible with the lowest time lever and still coordinate with the 140T with 0.2 seconds margin. The advantage to this element is that it is more sensitive than the phase element for the Ø-Ø fault.

• 51Q TL – The criteria is the same as for the ground unit.

• 50Q pickup – We don’t use this element because of the possibility of false trips. 4. Reclosing – This is the same as the E-M reclosing. The major difference is that we can

choose the time from lockout to reset to allow more time for the cold load to subside and drop out the inst unit. We have been choosing 30 seconds.


61

• SEL 587 – This relay is connected to both the high and low side transformer BCT’s.

• EM Differential RelaySince the distribution transformer is connected delta – wye the transformer CT’s have to be set wye – delta to compensate for the phase shift.

Transformer Differential Protection

62

87R1

87OP

87R2

PRI I

PRI I

SEC I

SEC I

EXTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN THE SAME DIRECTION AND THENCIRCULATE BACK THROUGH THE CT'S. THEY DO NOTFLOW THROUGH THE OPERATE COIL

CURRENT FLOW THROUGH AN E/M 87 DIFFERENTIAL RELAY FOR AN INTERNAL AND EXTERNAL FAULT

BCT'S

BCT'S

CT POLARITY MARK

Transformer Differential Protection – External Fault

63

87R1

87OP

87R2

PRI I

PRI I

INTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN OPPOSITE DIRECTIONS, ADD AND THEN FLOW THROUGH THE OPERATE COIL ANDBACK TO THE RESPECTIVE CT'S

SEC I

SEC I

BCT'S

BCT'S

CT POLARITY MARK

Transformer Differential Protection – Internal Fault

64


• Where do we start the protection design?– What we know about the feeder:

• The feeder rating is 500 Amps. Relay phase pickup?– From the feeder setting criteria: Pickup the feeder in a cold load

condition (≅ 2 times maximum normal load) or pickup ½ of the next feeder load (≅ 1.0 +2*1/2 = 2 times normal load, phase overcurrent only).

• The fault duty at the bus is 5,188 for a 3Ø fault and 5,346 for a SLG

– Tells the highest feeder fuse size to coordinate with will be a140T so the ground pickup set 480 amps or higher. From Table 1 fuse protection for temporary faults, 140T – 3500 amps, 200T – 5500 amps

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Where do we start (Cont’d)

• What other information do we have/need?– This feeder already has a line recloser and the Field Engineer

wants to keep the same location. Reason for coordination was replacing the older recloser with a newer relayed recloser.

– We then have to run feeder fault studies (normally done by a Distribution Engineer) and look at the feeder configuration so we know what fusing and relay settings to use.

– We then generally start near the end of the feeder and work back towards the substation. A lot of times though we have to go back and forth because our first choice of a fuse size may not work correctly when we figure out the protection towards the substation and we have to make compromises.

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How do we proceed?

• At each coordination point– Temporary Faults - From Table 1

• What can a fuse be protected up to

– Fuse-to-Fuse Coordination - From Table 2• What will coordinate with the downstream device

– Fuse Loading – From Table 3• Are we above the fuse rating

– Conductor Protection - From Table 7• minimum conductor that can be protected by the feeder settings or

fuse

– Fault Detection• Can we detect the fault by our 2:1 margin

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Begin at the END

• NOTE: The actual Moscow feeder 515 does not look exactly like this even though the substation and line recloser settings we will arrive at are the actual settings.

• POINT 8: This is a customers load and we are using 3-250 KVA transformers to serve the load. The full load of this size of bank is 31.4 amps. The Avista transformer fusing standard says to use a 65T on this transformer so that’s what we’ll choose.

• POINT 7: This is the end of a 3Ø lateral and the fault duty is: 3Ø = 322 amps and SLG = 271 amps

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POINT 6B:• What we know:

– 3Ø lateral feeding to a 65T at the end– #4 ACSR exists.

• Coordination:– Temporary Faults - From Table 1 we see that an 80T fuse can be protected up to

2,050 amps for temporary faults and the 3Ø fault is 1,907 amps so we could choose an 80T or higher from that standpoint

– Fuse-to-Fuse Coordination – From Table 2, we see that we need an 80T, or larger, fuse to coordinate with the downstream 65T fuses (at fault duties < 1400A w/preload)

– Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a 100T or smaller fuse

– Fault Detection - Under the Relay Setting Criteria we want to detect the minimum end fault with a 2:1 margin. The SLG is 463 so the maximum fuse we could use would be: 463/2 (2:1 margin) = 231 amps so the largest fuse we could use is 231/2 (blows at twice the rating) = 115 so we could use a 100T or smaller fuse for this criteria

– Loading – Assume the load at the end is all there is so no need to worry

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Point 6B - So what fuse size should we use?

• 80T

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POINT 5 – Midline Recloser:• What we know:

– 3Ø line of #4 ACSR feeding to point 6– 3LG = 3453 A, 1LG = 2762 A

• Coordination:– Temporary Faults - The fault duties at the recloser are: 3Ø = 3,453 amps

and SLG = 2,762. We will assume we have a fused lateral just beyond the line recloser and we want to protect it for temporary faults. From Table 1 we see that we can protect a 100T fuse up to 2,650 and a 140T up to 3,500 amps. Therefore, from this standpoint we could use a 140T fuse beyond the recloser

– Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a 100T or smaller fuse. That means that we won’t be able to set the line recloser to coordinate with a 140T and still protect the conductor. Therefore, we have to compromise and we will sacrifice the fuse protection for temporary faults in order to gain protection for the conductor. NOTE: This normally isn’t too bad because most faults are SLG and are somewhat less than the calculated value because of fault resistance and distance to the fault out on the lateral

– Coordination – We will try and coordinate with a 100T fuse and still protect the #4 ACSR conductor. Once we determine the overcurrent pickup values, we will choose a Time Lever to provide 0.2 seconds (see Table 6) coordination with the fuse and then will recheck to determine if we are satisfied with the conductor protection

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POINT 5 (Cont’d):

1. Fault Detection & Loading - The loading information we obtained from the Distribution Group is 84 amps normal and cold load ≅ 2 times = 168 amps.Therefore, we can set the phase PU at?

• 200 amps which would carry the load and pick it up cold load.• However, #4 ACSR can carry around 140 amps (see Table 4) so we would probably

want to set about 300 amps to allow for load growth. NOTE: We do not set the phase overcurrent to protect the conductor from normal or emergency loading

2. The Ø-Ø fault at point 6 is 0.866*1907 = 1,651 so our margin to detect that fault would be 1651/300 = 5.5:1 so no problem with the 300 amps Ø PU from that standpoint (used Ø-Ø because it’s the minimum multi phase fault)

3. SLG Fault Detection – The SLG at point 6 is 1,492 so we could set the ground up to 1492/2 = 745 amps and still detect the fault. However, our criteria says to set as low as possible and still coordinate with the largest downstream device and from above we’re trying to use a 100T. Again this is 300 amps so in this case the ground PU will be set the same as the phase

4. Instantaneous Units – The phase instantaneous will be set at 350 A and the ground instantaneous unit will be set at 300 A. Note that they can’t reach all the way to point 7 so they can’t protect the point 6C fuse from temporary faults occurring near the end of the lateral

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POINT 3C:• What we know:

– The fault duty at point 3 is: 3Ø = 3,699 amps and the SLG is 3,060– #4 ACSR 3Ø trunk feeding to the line recloser with a SLG of 2,762 amps

• Coordination:1. Temporary Faults - From Table 1 we see that a 140T fuse can be protected up to

3,500 amps for temporary faults and the 3Ø fault is 3,699 amps. This again appears to indicate we would want to use a 200T fuse

2. Coordination – We set the line recloser up to coordinate with a 100T fuse so we wouldn’t be able to coordinate with the recloser by using a 140T so this would indicate we need a 200T

3. Conductor Protection - From Table 7 we see that #4 ACSR requires a 100T fuse or smaller to be protected. OOPS

4. The above coordination and conductor protection dilemma means that we can’t solve this problem by fusing this lateral. We also can’t remove the fuse because the substation setting of 960 A can’t protect the conductor either.

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Point 3C - Solutions?

• Move Recloser• Reconductor

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POINT 1:• What we know:

– This is a 500 amp feeder so must set to carry load. NOTE: The load is 325 amps and cold load 650

– The fault duty at point 1 is: 3Ø = 5,158 amps and SLG = 5,346– We want a Fuse Protection Scheme.

• Coordination:1. Temporary Faults – We have decided that a 140T fuse is the maximum fuse we

will use on the feeder even though it can’t be protected at the maximum fault duties.

2. Coordination – We will coordinate with a 140T fuse at the maximum SLG fault duty of 5,346 amps with 0.2 seconds coordination time per Table 6. We will also coordinate with the line recloser settings with 0.3 seconds coordination time again per Table 6

3. Conductor Protection – The main trunk is 556 ACSR and from Table 7 a 500 amp feeder setting of 960 A can protect 1/0 ACSR or higher so no problem since all laterals are fused.

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POINT 1 (Cont’d):4. Fault Detection & Loading – We will set the phase pickup to carry 500 amps

normally plus pick up cold load so will set at 960 amps.• The Ø-Ø fault at point 5 is 0.866*3453 = 2990 so our margin to detect that

fault would be 2990/960 = 3.1:1 so no problem• SLG Fault Detection – Again per our criteria we will set as low as possible

and still coordinate with the maximum downstream devices with the same coordination times as above

• The SLG at point 5 is 2,762 so our margin to detect that fault is: 2762/480 = 5.7:1 so no problem

• Instantaneous Units – The phase instantaneous will be set at 1,120A and the ground instantaneous unit will be set at 480 A

76

Questions?

Routine testing is a necessity for electromechanical relays, because they have no mechanism to inform a user of malfunctions. Other issues that contribute to the need for testing include component drift, component failure, frozen bearings, etc.Digital relays include a self‐test feature that monitors the health of certain aspects of the device. Self‐test routines cannot test some areas, such as analog inputs, digital inputs, and output contacts. However, digital relays also include event reports that can be used to determine the health of the areas not covered by the self test. As a result, an argument can be made that a digital relay does not need to be subjected to routine testing if the alarmbe made that a digital relay does not need to be subjected to routine testing if the alarm contact is monitored and all captured event reports are reviewed.

6

Static tests are simple tests where the input quantities are slowly manipulated to determine that the relay operates within its published tolerances. A simple example is applying current to an instantaneous element to determine at what level it picks up, or operates.Multistate tests are more complicated tests that often require the test set to monitor the output contact of the device under test. A simple example is using an output contact to stop a test in order to determine the operating time of the device. A more complicated example is testing the reclosing sequence of an overcurrent relay and its associatedexample is testing the reclosing sequence of an overcurrent relay and its associated multishot recloser. Multistate tests can also be used to more closely simulate an actual fault on the power system. As an example, normal load currents can be applied, then fault current applied until the relay operates. For these types of tests, the fault currents that are used should be those that are predicted by a short‐circuit study.End‐to End Tests: With the proper equipment, the relays at both ends of the line can be tested in unison. These tests are helpful in properly testing line current differential or directional comparison pilot schemes.System faults are often the best test of a protection system. When available, event reports can help determine if the protection system is working properly.Fault Playback: Digital relays and digital fault recorders generally have the capability of providing a COMTRADE file of a captured system event. The COMTRADE files can be played back by test sets. If a system event resulted in a relay misoperation, playing back the event into the relay, after its settings have been modified, can assist in validating that the newinto the relay, after its settings have been modified, can assist in validating that the new settings are appropriate.

7

When designing a test, consider any security features of the element/scheme. You do not want a test to fail because loss‐of‐potential (LOP) was asserted, or the ratio of positive‐sequence to negative‐sequence current is too low, etc.If any current supervision, e.g. fault detector, is being used, you need to make sure the injected current is greater than the pickup of the supervising element.Likewise, if any element or logic is used in multiple schemes, you need to make sure that you do not mistakenly test the wrong element/scheme.The logic diagrams from manufacturers’ instruction manuals are helpful in determiningThe logic diagrams from manufacturers instruction manuals are helpful in determining security features that may impact a test.

8

The operating principle of the element/scheme must be understood so that any preoperating conditions, such as memory voltage and recloser reset times, are met. You do not want a poorly designed test that causes wasted time trying to track down a nonexistent problem.When testing elements, you want to make sure that all supervisory conditions are met, so that the element—not the supervisory condition— is tested.When testing schemes, you want to create a test that exercises each enabling and blocking element so that the scheme is proven for every operating condition it may encounterelement so that the scheme is proven for every operating condition it may encounter.When you are performing multistate tests, you need to determine the method for controlling the transition between states. Will settings need to be modified? How are connections made to the test set? You want to make sure that you leave the system properly configured when the testing is complete. You do not want to introduce any errors into the system.

9

When applying secondary current to a relay, you must consider the thermal limits of the inputs. If currents can be kept below the steady‐state rating, then no other consideration is required. However, it is often necessary to apply current above the steady‐state limit. In such cases, sufficient time must be allowed between tests to allow the input to cool.When running multiple tests, allow ample time between tests to allow all elements to reset.The relay burden will impact the magnitude of current that can be injected into a relay and may limit the tests that can be performedmay limit the tests that can be performed.If performing single‐state tests, the “turn on” time of the test set can impact the type of tests that can be successfully performed. Performing a single‐state three‐phase test may not be practical because the test set, in all likelihood, will not start the different phase currents at the same time. As a result, the relay operation and targeting may not be as expected.Likewise, for multistate tests, the switching time of the test set may impact the type of tests that can be performed.

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The rate of change in the test parameters must take into account the characteristics of the relay under test.The delta value should be related to the sensitivity/setting of the relay under test. If the delta value is too small, you will run many tests that are extremely similar and provide no significant value. As a general rule, 3 to 10 steps within the tolerance range is sufficient. You do not want a step size too large, because the transient overreach response of the relay may provide misleading information when trying to determine a set point.As an example you would have the following for a 10 Amp relay with a +/– 5% tolerance:As an example, you would have the following for a 10 Amp relay with a +/ 5% tolerance:

• 3 steps: (10 * .05) / 3 = 167 mA• 10 steps: (10 * .05) / 10 = 50 mA

Therefore, a delta value between 50 and 150 mA is sufficient.The delta time between tests must be long enough for the relay to respond. If the step time is too short, the test set will change to a new value before the relay has a chance to respond to the present test.Because the relay response time is related to the magnitude of the operating quantity, the delta time needs to be selected based on the expected operating time, with adequate margin, at the given test value.

11

How do you test an electromechanical overcurrent relay?

12

How do you test a multifunction overcurrent relay?

15

The basic function of a synchronous machine rotor is to produce the rotating magnetic field necessary to induce voltage in the stator windings. The source for this magnetic field is the dc current that circulates through the field winding. Synchronous machine rotors can be divided in two groups: round rotors and salient pole rotors. Round rotors are found in machines with high nominal speeds (1800–3600 rpm @ 60 Hz), while salient pole rotors are common to machines with lower speeds (1200 rpm and below @ 60 Hz). There are some specific applications where salient pole rotors are required to operate at 1800 rpm (4‐pole machine @ 60 Hz)required to operate at 1800 rpm (4 pole machine @ 60 Hz).

22

ANSI Device Number Functions:

21 Backup Mho Phase Distance Element

24 Volts/Hertz Element

25 Synchronism Check

27 Undervoltage

32 Reverse/Low‐Forward Power

40 Loss‐of‐Field

46 N i S O46 Negative‐Sequence Overcurrent

49 Thermal Element

50N/51N Neutral Overcurrent

50P Phase Overcurrent

51C/51V Voltage‐Controlled/Voltage‐Restrained Time‐Overcurrent

59 Overvoltage

64G 100% Stator Ground Elements64G 100% Stator Ground Elements

64F Field Ground

78 Out‐of‐Step

81 Over‐/Underfrequency Elements

81 AC Abnormal Frequency Scheme

87 Current Differential Elements

87N Ground Differential Element

30

A stator or transformer core is composed of laminated sheets of stacked iron, separated with a coating of thin insulation, and bound together. At the end of the core are nonlaminated components that help keep the core together. The magnetic flux density of the core is proportional to the voltage applied and inversely proportional to the frequency applied.

32

A stator or transformer core is composed of laminated sheets of stacked iron separated with a coating of thin insulation and bound together. At the end of the core are non‐laminated components that help keep the core together.

The magnetic flux density of the core is proportional to the voltage applied and inversely proportional to the frequency applied. When the core flux density becomes excessive, the core becomes magnetically saturated. When the core saturates, the magnetic flux begins to flow in paths not designed to carry it, such as the core‐end non‐laminated components. flow in paths not designed to carry it, such as the core end non laminated components. Damage to the core can occur within seconds of the overexcitation event. Saturation is most damaging at the ends of the core where the magnetic field can induce high eddy currents in the solid core assembly components and at the end‐or‐core laminations. Eddy currents can circulate in the iron due to the induced voltage. These currents result in higher I2R losses and heating in those components. In addition to the higher temperatures, eddy currents also cause interlaminar voltages hi h ld f th d d th thi i l ti f th l i ti O th i l ti iwhich could further degrade the thin insulation of the laminations. Once the insulation is

broken down by the high temperatures and voltages, severe core iron damage can result. These high temperatures and voltages can result in damage to the core within seconds. After the damage occurs the core is useless. The damage is more severe than most winding failures, and the repair may require removal of the entire winding and restacking a portion of the core. Downtime will be significant and costly.

33

ANSI/IEEE C50.13 sets the standards for continuous overexcitation limits for generators. The limits for continuous generator overexcitation are 1.05 p.u. on the generator base at no load and at full load. ANSI/IEEE C57.12.00 sets the standards for continuous transformers overexcitation limits. The limits set for continuous overexcitation of the transformer are 1.10 p.u. on the transformer base at no load and 1.05 p.u. on the transformer secondary base at rated load and 0.8 pf or greater. The primary side of the transformer is connected to the generator, so the secondary side would be the system or load side. C b d l d hi h d fi th li it th it d d d ti f it tiCurves can be developed which define the limits on the magnitude and duration of overexcitationevents (V/Hz). Manufacturer curves show the limits of permissible operation in terms of percent of normal V/Hz versus time. Often these curves are hard to find for existing equipment, but for new equipment curves must be part of the equipment specifications. Manufacturers will not generally provide this information unless it is specified at the time of the purchase. Typical curves for generators and transformers are shown above. In setting the V/Hz protection for a generator, it is important that the permissible operating curves for generators and transformers use a common voltage base This is necessary because in somefor generators and transformers use a common voltage base. This is necessary because in some cases the voltage rating of the transformers low‐voltage winding is slightly less than the generator’s. The resulting turns ratio partially compensates for the voltage drop across the bank due to load flow. Normally the generator’s rated voltage is used as the voltage base since the PTs typically used for the relay voltage signal are connected to the unit between the generator and the transformer. The transformer can be put on the generator base using the following formula. (V/Hz)tg = (Vg/Vt) * (V/Hz)t where (V/Hz)t is the permissible V/Hz rating of the transformer on the transformer base, Vg is the rated generator voltage Vt is the rated voltage of the primary side of the transformer andis the rated generator voltage, Vt is the rated voltage of the primary side of the transformer and (V/Hz)tg is the transformer permissible V/Hz rating converted to the generator voltage base.

35

Generator Data:

Rated MVA: 252.4 MVARated Voltage: 13.8 kVRated Current: 10560 A

38

Improper synchronizing of a generator to a system can result in damage to the generator step‐up transformer and any type of generating unit.

The damage incurred can be slipped couplings, increased shaft vibration, a change in bearing alignment, loosened stator windings, loosened stator laminations and fatigue damage to shafts and other mechanical parts.

39

Secondary voltage magnitude differences between the two sides of the generator breaker can be associated with differences in transmission voltage levels and differences in the PT connections and ratios.

Secondary voltage angle differences between the two sides of the generator breaker are caused by power transformer and PT connections.

Synchronizing secondary voltage limits must be set to ensure that proper voltage exists onSynchronizing secondary voltage limits must be set to ensure that proper voltage exists on each side of the generator breaker before the generator breaker is closed.

Synchronizing secondary angle limits must be set to ensure that proper phase angles exist on each side of the generator breaker before the generator breaker is closed.

Synchronizing slip limits are usually placed on the control device to ensure that the generator and system are very close to the same frequency before synchronizing.

41

Unbalanced current in the stator windings generally implies the existence of negative‐sequence currents. The negative‐sequence current produces a rotating field in the machine gap, which rotates at about synchronous speed, but in the reverse direction compared to the normal positive‐sequence field. This new field induces currents in the rotor, whose frequency is twice the synchronous frequency. These currents produce rotor overheating and can eventually cause rotor damage.

44

Rotor and positive sequence flux rotate counter clockwise at synchronous speed. Negative‐sequence flux rotates clockwise at synchronous speed. Frequency of the induced currents on the rotor is twice the synchronous frequency.

45

The rotor wedging, the rotor rings, and the iron core can heat up to intolerable values. The resulting damage can cause a prolonged generator outage and significant repair time.

47

Unbalanced short circuits are not the only cause of imbalances in the generator windings. Series faults, such as those one or two open conductors, can also produce a relatively large amount of negative sequence.Different conditions, such as unbalanced loads and transmission line asymmetry, can also cause imbalances in the stator windings.

48

The diagram shows the simplest differential scheme using an instantaneous overcurrent relay. This is sometimes called a differential overcurrent scheme.Differential protection is one of the most effective means of protecting power equipment. Put simply, differential protection operates on the difference between the measured current entering and exiting the protected zone. Because it operates on the difference only and not the total current flowing in the circuit, a relay featuring this protection can have greater sensitivity to faults than other types of relays. And because the zone of protection is precisely defined by the location of the CTs surrounding the protected zone the relay isis precisely defined by the location of the CTs surrounding the protected zone, the relay is highly selective. With high selectivity, a differential relay can trip quickly with no coordinating time interval.

52

The diagram shows the behavior of the simplest differential scheme using an instantaneous overcurrent relay during an external fault. If the current transformers are considered ideal and identical, the primary and secondary currents at both sides of the protected equipment are equal. There will not be an operation for differential current.

53

For an internal fault, the primary currents allow the secondary currents to produce a differential current through the overcurrent relay. If this differential (operation) current is larger than the relay’s pickup, then the relay will trip both circuit breakers instantaneously. The characteristics of differential protection can be summarized as follows:Principle:

• Measure current entering and exiting the zone of protection• If currents are not equal, a fault is present

Provides:Provides:• High sensitivity• High selectivity

Result:• Relatively high speed

54

All differential protection must deal with the challenge of remaining secure for large through‐faults. During a severe external fault, a CT may saturate and supply less than its ratio current. In this case, the currents do not completely cancel, and a false differential current results.

55

The figure shows the results of a simulation produced with a program property of Schweitzer Engineering Laboratories, Inc.

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Transmission Line Protection Objective

General knowledge and familiarity with transmission protection schemes


Transmission Line Protection Topics

Primary/backup protection

Coordination

Communication-based schemes

Breaker failure protection

Out-of-step relaying

Synchronism checking


Primary Protection Function

Trip for abnormal system conditions that mayEndanger human life

Damage system equipment

Cause system instability


Protection Zones

Primary protectionfirst line of defense

Backup protectionoperates when primary fails


How Can A Protection System Fail?

Current or voltage signal supply

Tripping voltage supply

Power supply to the relay

Protective relay

Tripping circuit

Circuit breaker


Two Types of Backup

Remote backupLocated at different station

No common elements

Local backupLocated at same station

Few common elements– Separate relays

– Independent tripping supply and circuit

– Different current and voltage inputs


Remote vs Local Backup

SpeedRemote is slower

SelectivityRemote disconnects larger part of the system

PriceLocal requires additional equipment


Primary and Backup Coordination

Best selectivity with minimum operating time

Achieved through settingsPick up values

Time delays


Coordination Types Considered

Time-Overcurrent

Time-Stepped Distance

Communication-Aided Schemes


Time-Overcurrent Relays

Definite-Time OvercurrentOperate in a settable time delay when the current exceeds the pickup value

Instantaneous operation – no intentional time delay

Inverse-Time OvercurrentOperating curve defined by limiting the total fault energy (Kd = I2·t)


Time-Overcurrent CoordinationInverse-Time Overcurrent

Pickup of relay A set low enough to see the fault shown and backup relay C

Pickup should be above emergency load conditions (phase relays)

Time delay of relay A should allow relay C to clear the fault first


Time-Overcurrent Coordination

S - Selectivity time delay (aka CTI):Breaker operating time

Overtravel (impulse) time (E/M relays)

Safety margin

0.2s ≤ CTI ≤ 0.4s


Time-Overcurrent CoordinationLooped Systems

Directionality required for most relays

1-2-3-4-5

a-b-c-d-e

Relays at ‘5’ and ‘e’ can be nondirectional


Instantaneous Overcurrent Protection

Inverse-Time O/C coordination may result in long time delays

Instantaneous O/C relays set to trip for faults for ~80% of the line section

Significantly reduced tripping times for many faults


Time-Stepped Distance Protection

Coordination similar to that of inverse-time O/C

Relay at A set to trip instantaneous for faults in its Zone 1 (reaching ~80% of the line section)

Relay at A backs up relay at C after Zone 2 timer times out

Faults at the end of the line also cleared in Zone 2 time


Communication-Based ProtectionRationale

Distance protection can clear faults instantaneously for 60% to 80% of the line length

Protection speed may be critical to maintain system stability

High-speed autoreclosing application


Communication-Based ProtectionCommunication Mediums

Power Line Carrier

Microwave

Fiber-Optics

Private and Leased Pilot Channels


Communication-Based ProtectionScheme Types

Permissive Overreaching Transfer Trip (POTT)

Permissive Underreaching Transfer Trip (PUTT)

Directional Comparison Blocking (DCB)

Directional Comparison Unblocking (DCUB)

Direct Underreaching Transfer Trip (DUTT)

Direct Transfer Trip (DTT)


Permissive Overreaching TTProtective Zones


Permissive Overreaching TT

Permissive signal must be detected from the remote end for the communication-aided trip

Absence of communication channel disables the accelerated tripping


Permissive Overreaching TTComplications and Concerns

Desensitization due to infeed Dependability issue – failure to trip high speed

Current reversalOccurs in parallel lines with sequential tripping

Security issue – coupled with long channel reset times may cause trip of the healthy parallel line


Current ReversalAll Sources In

Z2 at Breaker 1 picks up and sends permissive signal to Breaker 2

Z2 at Breakers 3 and 4 send permissive signals to each other

Z1 at Breaker 4 trips instantaneously


Current ReversalSystem After Breaker 4 Opens

Current reverses through the healthy line

Z2 at Breaker 2 picks up

If the permissive signal has not reset, Breaker 2 trips on POTT


Current ReversalPossible Solution

Timer with instantaneous pickup and time delayed dropout, initiated on reverse Z3

Delay trip with POTT until the timer drops out


Permissive Underreaching TT

Similar to POTT but permissive signal sent by underreaching Z1 elements

At the receiving end, Z2 elements qualify the permissive signal

No problems with current reversal since Z1 doesn’t overreach


Directional Comparison BlockingProtective Zones

Zone 2 elements cover the entire line

Reverse Zone 3 elements must reach further than the opposite Zone 2 overreach


Directional Comparison BlockingBasic Logic

In-section faults will not key transmitter and both ends trip high-speed

Out-of-section fault will key the transmitter at the nearest end to block the trip at the opposite end


Directional Comparison BlockingComplications and Concerns

Coordinating time at fault inceptionZ3 faster than Z2, but channel delay time reduces the margin

Z2 must be slowed down

External fault clearingZ3 and Z2 race to drop out, if Z3 drops out first Z2 overtrips

Z3 operates faster and drops slower

Channel reset time helps

Slower transmitter key dropout time helps



External fault clearing failureLocal backup provided by time-delayed Z3 or external BF relay clears the near bus

Remote backup provided by Z2 clears the line

Stop preference over start



Current reversal

Reach MarginZ3 reaches farther back than remote Z2 by at least 50% of Z2 overreach


Directional Comparison Unblocking

Essentially the same as POTT

Requires FSK

In-section fault may impede communication

In case of channel loss, a 150 ms window is open when permissive signal is bypassed and Z2 allowed to trip high speed


Direct Underreaching TT

Underreaching Z1 elements send direct transfer trip

Noisy channel can cause false trip

Very secure channel required


Pilotless Accelerated Trip Schemes

Communication equipment not justifiable in lower voltage transmission applications

In-section faults may be uniquely determined by system conditions

Detecting these conditions is all that is needed for high speed tripping


Pilotless Accelerated Trip SchemesFaulted System with Breakers Closed

After Breaker 2 opens the only current that can flow is the fault current


Pilotless Accelerated Trip SchemesFaulted System with Breaker 2 Open

Tripping conditions:

Three-phase load was present before the fault

Three-phase current was lost

Current above the threshold detected in at least one phase


Breaker Failure Relaying

Minimize the damage when a breaker fails to clear a fault

Trips all sources locally within the critical clearing time to maintain system stability


Breaker Failure RelayingCommon Causes of Breaker Failure

Main breaker poles failed to clear the fault due to inadequate insulating medium

Open trip coil or trip coil circuit

Loss of tripping dc

Mechanical failure of the breaker trip mechanism


Breaker Failure RelayingOperating Philosophy

Activated only when a trip signal is issued from protective relay

If current is above threshold after a pre-set time period, breaker failure condition is declared


Breaker Failure RelayingConsiderations

Timer settings must take into account the clearing time of the slowest breaker and the reset time of the fault detector

The effect of any opening resistors in the circuit breakers on the reset time of fault detector

Substation bus configuration must be taken into account to trip minimum number of breakers

In multi-breaker schemes, possible transfer trip to the remote end


Out-of-Step Detection and BlockingCauses of Out-of-Step

Power swings result from faults, switching, or big changes in load or generation

Magnitude of the swing depends on the system impedance change during such conditions

Swings can be stable or unstable


Power Transfer Equation

,

sin

where:power transferred from the sending to the receiving endsending end voltagereceiving end voltage

angle by which leads total reactance between the sending and rece

S RS R

S

R

S R

V VP P PX

PVV

V VX

δ

δ

= = =

−−

−−

− iving end


Power Transfer Curves

,


Electrical Quantities During SwingApparent Impedance Trajectories


Electrical Quantities During SwingApparent Impedance Trajectories

Apparent impedance during power swings can enter into the reach of distance relays

If the apparent impedance stays longer than the time delay in a given zone, that distance element will trip as for a fault

To prevent such tripping, out-of-step blocking schemes are employed


Out-of-Step Blocking Distance Elements

If the timer expires between the two zones,out-of-step condition is declared and selected distance elements are blocked


Synchronism Checking

After clearing a fault, one end of the line will reclose to “test” the line

If the test is good, the other end can be closed but only if voltages are close enough and there is a small phase angle difference

If the conditions are not right, the system will undergo a mechanical and electrical shock with a possible unstable swing


Synchronism CheckingMonitored Quantities

Voltage magnitudes

Phase angle difference

Slip frequency


Synchronism CheckingSynchronizer Window


Synchronism CheckingConditions

Both voltage phasors are above 59V setting

Phase angle difference is small

The above conditions are maintained for a short time, ensuring that the slip frequency is small enough and measurements are valid


Synchronism Checking Relay

Phasor difference setting

Timer setting

The measured phasor difference should be below the setting for the given time

Different than a synchronizing relay

25DV 59V sinθ= ⋅

225T360 f

θ⋅=

⋅Δ


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