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Protection Relays Settings
Chapter 19
Protection Relays Settings INTRODUCTION This chapter introduces relay settings for different types of relays used in power system, with real worked examples.
Subchapter 19.1
Overcurrent Protection Settings 19.1.1 INTRODUCTION Overcurrent protection is used in extra-high voltages and high-voltages networks as the main protection, and in medium networks it is also used as backup protection for power transformers. Time grading for overcurrent relays is used in the radial system. The minimum possible time delay is selected for the relay which protects the line farthest from the source. The grading time between the successive relays will be in the order of 0.40.5 seconds, and with modern relays it can be of the order of 0.35 seconds. Current settings for an overcurrent relay should consider the following factors: 1. Maximum possible loading; 2. Transient currents due to switching of transformers, inrush current, heavy motor restarting currents after clearing external faults; 3. Drop off/pickup ratio, which is called the reset ratio. I set 5 ðK 3 RÞ=F 3 I L Max where: K 5 safety factor 5 1.2 R 5 restarting factor, which depend on the load profile Set to 2 for industrial loads Practical Power System and Protective Relays Commissioning. DOI: https://doi.org/10.1016/B978-0-12-816858-5.00019-8 © 2019 Elsevier Inc. All rights reserved.
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Set to 1.5 for residential loads; F 5 reset ratio 5 0.85 for electromechanical relays and set to 0.90.95 for static and digital relays; IL Max 5 maximum expected load; Iset 5 22.5 IL Max. The performance of overcurrent relays is checked by calculating the sensitivity factor, S, which is the ratio of the minimum short circuit current following at the relay in the end protected line in a radial system to the relay setting as follows: S 5 I sc minimum=I set should be . 1:5
19.1.2 TRANSMISSION LINE OVERCURRENT PROTECTION Fig. 19.1.1 illustrates the overcurrent protection of a high-voltage transmission line The circuit data include: Overcurrent relay type MCGG22. Application: Backup protection 500 kV, fault level at end B 5 12,000 MVA, current setting of relay at A 5 800 A. Is 5 800 A current transformer (CT) ratio 5 1600 1 Is 5 800
1 5 0:5A 1600
Plug setting multiplier (PSM) PSM 5 IF 5
If MVAF -I F 5 Is O3 3 V
12; 000 5 13:856 KA O3 3 500
FIGURE 19.1.1 Overcurrent protection of a high-voltage transmission line.
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IF 5 13,856 A IS 5 800 A PSM 5
13; 856 5 17:32 800
Using a standard inverse curve of the relay then we have the relay operating time TR: TR 5
0:14 0:14 5 0:0221 PSM 17:320:0221
TR 5 2.385 Assuming that the relay should operates on 1 second then we have: t.operation 5 TM.TR Where TM 5 time multiplier TM 5
1 5 0:419 2:385
Relay (A) setting: Iset 5 800 A (Primary) Iset secondary 5 0.5 A PSM 5 17 TM 5 0.4 Operating time for a fault at end B will be: t:operation 5 0:4 3
0:14 170:0221
t.operation 5 0.96 seconds
19.1.3 TRANSFORMER OVERCURRENT PROTECTION Transformer overcurrent protection is illustrated in Fig. 19.1.2. Transformer data: 250 MVA 220/66 kV ZTr% 5 10% on MVAb 5 100 Relay R2 setting on low voltage (LV) feeder F1:
FIGURE 19.1.2 Transformer overcurrent protection.
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MVAF 5
MVAb 3 100 100 3 100 5 5 1000MVA Z% 10
If of HV side of Tr 5
MVAF 1000 5 5 2624A O3 3 220 O3 3 220
IF of LV side of Tr 5
1000 5 8747A O3 3 66
Iset 5 600 A, CT 5 500/1 TM 5 0.4 PSM 5 8747 600 5 14:58 Relay operating time 5 TR 5
0:14 14:580:0221
TR 5 2.54 For a fault of 8747 A the operating time will be: t.op 5 TR.TM 5 2.54 3 0.4 5 1.016 second With a grading time step 5 0.4 second. Then the operating time of R1 should be more than 5 1.016 1 0.4 5 1.416 second. Relay R1 setting on the HV side of Tr IF on HV side 5 2624 A We can set the relay as follows: Iset 5 1.45 3 ITr HV side MVA 250 3 103 5 pffiffiffi ITr HV side 5 pffiffiffi 5 656A 3 3 VHV 3 3 220 Iset 5 1.45 3 656 5 951.2 A PSM 5
IFault 2624 5 2:76 5 951:2 Iset
Set top 5 1.5 second, as seen above it should be more than 1.416 second. Top 5 TM.TR TR 5
0:14 SI 2 Curve 2:760:0221
TM 5
top 1:5 5 0:219 5 TR 6:825
TR 5 6.825
R2 Setting Iset 5 600 A PSM 5 14.58 TM 5 0.4 CT 5 500/1 R1 Setting Iset 5 951.2 A PSM 5 2.76 TM 5 0.219 CT 5 1000/1
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Subchapter 19.2
Feeder Backup Overcurrent Protections 19.2.1 INTRODUCTION Overcurrent protection here is used as backup protection, with a considerable time delay, which means it works as a second line of defense to protect the line if the main or primary protection fails to isolate the line for the fault instantaneously.
19.2.2 WORKED EXAMPLE Fig. 19.2.1 gives an example of line backup overcurrent protection. Line AB 500 kV maximum load current 5 IL Max 5 1200 A For OC relay at A (R): Iset 5 1.2 3 IL Max 5 1.2 3 1200 1000 5 1.44 A PSM calculation: IF MVAF andI F 5 IS O3 3 V I F 5 80 KA I S 5 1200 A 80; 000 5 80 PSM 5 1000 PSM 5
Using a standard inverse curve of the OC relay we then have: TR 5
0:14 0:14 5 0:0221 5 1:528 0:0221 PSM 80
where TR 5 relay operation time assuming that the relay will operates on 1 second for this fault then t.operation 5 TM.TR, where TM 5 time multiplier.
FIGURE 19.2.1 Line backup overcurrent protection.
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Then TM 5
t:operation 1 5 0:654 5 TR 1:528
Relay setting: Iset 5 1440 A 5 1.44 A (secondary) PSM 5 80 TM 5 0.65 Operation time at a fault after end B will be: t 5 0:65
0:14 5 0:993s 800:0221
For an earth fault relay at end A connected to the residual connection of CTs at end A: %ZS 5
MVAb 3 100 100 3 100 5 5 0:144 MVAF 80 3 O3 3 500
For a single phase fault at end B we draw the sequence network diagram as shown in Fig. 19.2.2. PSM calculation: IF 5 29.3 KA Iset 5 800 A
FIGURE 19.2.2 Sequence networks.
Z t 5 3Z s 1 2Z 1 1 Z 0 5 3 3 0:144 1 2 3 0:3 1 0:15 5 0:432 1 0:6 1 0:15 5 1:182% 100:100 5 9:769 KA 1:182 3 O3 3 500 I F 5 3I0 5 3 3 9:769 5 29:3 KA I set 5 800 A 800 5 0:8 A ðsecondaryÞ I set 5 1000 I0 5
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PSM 5
287
IF 29:3 3 103 5 36:625 5 IS 800
Assuming that the relay will operates on 1 second for a fault on end B: t:operation TM 5 TR 0:14 0:14 TR 5 5 0:0221 5 1:884 0:0221 PSM 36 TM 5
1 5 0:530 1:884
Relay setting: Iset 5 800 A 5 0.8 A secondary PSM 5 36 TM 5 0.53 0:14 t:operation 5 0:53 0:0221 36 5 0.998 second
Subchapter 19.3
Feeder Unit Differential Protection 19.3.1 INTRODUCTION This protection is used to protect overhead transmission lines and is commonly used as main 2 in utilities. The disadvantage of this protection is the blind zone, which is not protected by this protection—this means the zone between the circuit breaker (CB) and the current transformer in one end—as this zone is protected by backup protections (Fig. 19.3.1).
FIGURE 19.3.1 Line differential protection.
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19.3.2 WORKED EXAMPLE Worked example: Line AB 500 kV B 5 1.877 CT 5 1200/1 at both sides choose the relay characteristic shown in Fig. 19.3.2
FIGURE 19.3.2 Line differential protection characteristic.
jIdiff j 5 jIA 1 IB j jIbias j 5 0:5½jIA1 1 IB1 j Line charging current 5
B 3 1000 V 3 O3 1:877 3 1000 I ch 5 500 3 O3
5 2.167 A I set min 5 2:5 3 2:167 5 5:4175 A Iset 5 0.2 IN I set 5 0:2 3 1200 5 240 A Iset . 2.5 Ich IS1 5 240 A primary 240 5 0:2 A 1200 I S2 5 2I N 5 2 3 1200 5 2400 A
I S1 5
I S2 5 2400 A primary 2400 5 2 A secondary I S2 5 1200 K 1 2 slop1 5 30 % K 2 2 slop2 5 150 %
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Subchapter 19.4
Setting of Distance Protection 19.4.1 INTRODUCTION Distance protection is widely used as the main protection—primary protection—in overhead transmission lines and has the advantage that it gives backup protection for adjacent lines. However, it has the disadvantage that it does not have accurate measurements for close-up faults near the remote end of the line due to the errors in CT and PT measurements and the transient nature of the short circuit current, also due to the tower arc fault resistance—what we call distance protection overreaching and underreaching—that it is not accurate in measurement in short transmission lines due to the effect of the system impedance ratio.
19.4.2 WORKED EXAMPLES Worked example 1 Most distance protection is set to have three zones and one backup start zone. The setting of the first zone is as follows for a relay at end A: Z1 5 0.85 ZL, t1 5 0 instantaneously where ZL 5 line impedance. Setting of the second step zone 2 is as follows for a relay at end A (Fig. 19.4.1): The Z2 setting is the lower value of the following: Z 2 5 0:85 ðZ L11 0:85 Z L2 Þ Z 2 5 0:85 ðZ L1 1 Z T Þ
FIGURE 19.4.1 Distance protection application setting: Example 1.
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Worked example 2
FIGURE 19.4.2 Distance protection application setting: Example 2.
For the following network configuration, the zone 2 setting will be as follows for a relay at end A:where II21 is called the infeed factor “F” I2 3 0:85Z L2 Z 2 5 0:85 Z L1 1 2I 1 IT Z 2 5 0:85 Z L1 1 ZT I1 In general: Z2 $ 1:2-Z 2 5 120% Z L1 Z L1 t2 is set to be 0.5 second. The setting of zone 3 in general is: Z3 $ 1:5 Z 3 5 150% Z L1 Z L1 t3 is set to be 0.9 second. Final (Backup Step) The current starter is used for simple networks where the minimum short circuit current is higher than the maximum load currents. An impedance starter is more commonly used as this start impedance stage should not work in the load area and also should not work for the transformer low-voltage faults. The lower value of Zst should be chosen.
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1. Zst # 0.5 ZL min; this means less than half of the minimum load impedance; 2. Zst $ 1.25 Z3. tst 5 1.3 second. The Compensation Factor (K0)
FIGURE 19.4.3 Distance protection compensation factor (K0) for earth fault measurement.
The voltage measured by the relay at A is: 0 1 Z 2 Z 0 1A V RA 5 Z 1 @I A 1 3I 0 3Z1 I R 5 I A 1 K 0 3I 00 Where K 0 5 @
1
z0 2 z1 A 3z1
The compensation factor is set in the relay by the values of Z1 and Z0 to have the correcting value of distance relay measure the line positive impedance, proportional to its length irrespective of the kind of fault (earth faults or phase faults). Distance Protection Setting Z1F 5 80% of the protected line; Z2F 5 100% of the protected line 1 50% of the shortest adjacent line; Z3F 5 100% of the protected line 1 125% of the longest adjacent line; Z3R 5 25% of the protected line; F 5 forward direction, R 5 reverse direction. Worked example 3
FIGURE 19.4.4 Distance protection application setting: Example 3.
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Data: AB 5 protected line length 5 60 km 220 kV CT 5 800/1 PT 5 220;000 100 Z 1P 5 0:0412 1 J 0:302 Ω = km 3 60 Z 10 5 0:25 1 J 0:95 Ω = km 3 60 BC 5 shortest line adjacent to the protected line AB length 5 50.5 km Z 2P5 0:0412 1 J 0:302 Ω = km 3 50:5 BD 5 longest line adjacent to the protected line AB length 5 68 km Z 3P 5 0:0412 1 J 0:302 Ω = km 3 68 CT ratio 5 800/1 5 800 Rated voltage 5 220 kV Rated current 5 800 A PT 5 220;000 100 Ratio 5 2200 Z p line AB
5 ð0:0412 1 J 0:302Þ50:5 5 2:0806 1 J 15:251 5 15:356 +82:231
Z P line BC
Z P line BD
Z 1 Reach
5 60ð0:0412 1 J 0:302Þ 5 2:472 1 J 18:12 5 18:287+82:231 Ω
5 ð0:0412 1 J 0:302Þ68 5 2:8016 1 J 20:536 5 20:726+82:231
5 0:8 Z P line AB 5 0:8 3 18:287 + 82:231 5 14:629 + 82:231 Ω 5 14:629 Ω
Z 2 Reach
5 Z P line AB 1 0:5 Z P line BC 5 ð18:287 1 0:5 3 15:356Þ+82:231 5 25:965 +82:231 Ω 5 25: 965 Ω
Z 3 Reach
Z P line AB 1 1:25 Z P line BD 5 ð18:287 1 1:25 3 20:726Þ+82:231 5 44:1945 +82:231 5 44:1942 Ω
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Z 3 reverse
5 0:25 3 Z P line AB 5 0:25 3 18:287 +82:231 5 4:571 + 82:231 Ω 5 4:571 Ω
The earth fault residual compensation factor is calculated as: K0 5
Z10 2 Z1P 3Z1P
Z 0 line AB 5 ð0:25 1 J 0:95Þ 3 60 Z 0 line AB 5 15 1 57J 5 58:94 + 75:25 1 K0 5
58:94+75:25 2 18:287+82:231 3 3 18:287+82:231 15 1 J56:99 2 2:472 2 J18:12 5 54:861+82:231 12:528 1 J38:87 5 54:861+82:231 5
40:839+72:135 54:861+82:231
K0 5 0.7444 +10.096 K0 5 0.744 The setting secondary values are: Z 1 Reach
CTratio PTratio 800 5 14:629 3 2200 5 14:629 3 0:363 5 5:319Ω; t1 instantaneous 5 14:629 3
Z 2 Reach
5 25:965 3 0:363 5 9:425 Ω ; t2 5 500 ms
Z 3 Reach
5 44:1945 3 0:363 5 16:042 Ω ; t3 5 900 ms
Z 3 Reverse Power Swing Blocking Settings
5 4:571 3 0:363 5 1:659 Ω
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Z PSB
Z PSB
5 1:3 Z 3 F 5 1:3 3 16:042 5 20:854 Ω ðsecondary valueÞ 5 20:854 Ω
where ZPSB 5 power swing blocking impedance setting, Z3 5 relay third zone. Load encroachment into Z3 reach Z Load min
0:8 rated voltage 1:2 rated current O3 0:8 3 220000 5 1:2 3 800 3 O3
5
5 105:85 Ω Z Load min secondary 5 105:85 3 0:363 5 38:422 Ω Z 3 Reach , 0:9 Z Load min 16:041 , 0:9 3 38:422 16:041 , 34:58; the above criteria are correct:
Subchapter 19.5
Differential Protection Setting 19.5.1 INTRODUCTION Transformer protection can be high-impedance differential protection or lowimpedance differential protection. Both have setting calculations as shown in the following examples.
19.5.2 LOW-IMPEDANCE DIFFERENTIAL PROTECTION 19.5.2.1 Worked Example 1 (Fig. 19.5.1) Tr 66/11 kV 25MVA Dy1 Tap changer 5 6 10% of 11 kV side nominal voltage. Calculation: HV side: 25 3 103 I 1 5 pffiffiffi 5 219 A 3 3 66 Choose CT ratio 5 250/1
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FIGURE 19.5.1 Transformer low-impedance differential protection.
i1 5 219
1 5 0:876 A 250
where I1 5 primary current at the HV side of the transformer; i1 5 secondary current at the HV side of the transformer. 1 HV correction factor 5 i11 5 0:876 5 1:145 LV side: 25 3 103 5 1312 A I 1 5 pffiffiffi 3 3 11 Choose CT ratio 5 1300/1 i2 5 1312
1 5 1:009A 1300
where I2 5 primary current at the LV side of the power transformer; i2 5 secondary current at the LV side of the power transformer. LV correction factor 5
1 5 0:991 1:009
For phase correction and zero-sequence current filtering, we use interposing CT as follows: HV side we use Yy0 transformer LV side we use Yd11 transformer as shown in Fig. 19.5.2. Now the current input to the relay is 1 A +0 degree after the interposing CT ratio and phase correction at the HV side and the LV side of the power transformer. Minimum Bias Current The maximum voltage is calculated at the low-voltage side. V max 5 11 3 1:1 5 12:1 kV
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FIGURE 19.5.2 Interposing transformer in transformer differential protection.
25 3 103 5 1193 A I 2 5 pffiffiffi 3 3 12:1 I R 5 relay current 5 0:95 3 Min Bias
1193 5 0:871 A 1300
5 2 I Relay set Normal I Relay at max voltage 5 2 ð1 0:871Þ 5 0:256
The minimum voltage is calculated at the low-voltage side. V min 5 11 3 0:9 5 9:9 kV 25 3 103 5 1458 A I 2 5 pffiffiffi 3 3 9:9 I R 5 relay current 5 0:95 3 Min Bias
1458 5 1:065 A 1300
5 2 I Relay at min voltage I Relay set at nominal voltage 5 2 ð1:065 2 1Þ 5 0:13
Set the minimum bias at 26%. Minimum bias is the highest calculated value of bias at the maximum voltage and the minimum voltage at the low-voltage side of the power transformer. Pickup or Id min set to 10% In I d min 5 0:1 3 1 5 0:1 A First slope bias 5 26% Second slope bias 5 80%as shown in Fig. 19.5.3.
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FIGURE 19.5.3 Relayoperatebias characteristic of used differential protection.
19.5.2.2 Worked Example 2 A transformer with three-winding Yyd11 500/220/11 kV 500 MVA/400 MVA/100 MVA Tap changer 5 6 10% of 220 kV side and tap charger 5 6 10% of 11 kV side calculation. HV side: 500 kV 500 3 103 I 1 5 pffiffiffi 5 577:35 A 3 3 500 Choose CT 5 600/1 i1 5
577:35 5 0:962 A 600
HV correction factor 5
1 5 1:039 0:962
LV side: 220 kV side 400 3 103 I 2 5 pffiffiffi 5 1049:72 A 3 3 220 Choose CT 5 1000/1 i2 5
1049:72 5 1:0497 A 1000
LV correction factor 5
1 5 0:952 1:0497
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Tertiary side 11 kV side: 100 3 103 I 3 5 pffiffiffi 5 5248:63 A 3 3 11 Choose CT 5 5000/1 i3 5
5248:63 5 1:0497 A 5000
1 Tertiary side correction factor 5 1:0497 5 0:952 For phase correction and zero 5 sequence current filtering we use the following interposing “auxiliary” transformer for current as follows (refer to Fig. 19.5.4): HV side: 500 kV Yd11 ICT; LV side: 220 kV Yd11 ICT; Tertiary side: 11 kV Yy0 ICT. Refer to Fig. 19.5.5 for a detailed connection diagram of three-winding transformer-biased differential protection. Minimum bias current (220 kV side) Maximum voltage at the low-voltage side:
V max 5 220 kV 3 1:1 5 242 kV
FIGURE 19.5.4 Interposing current transformers.
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FIGURE 19.5.5 Detailed connection diagram of three-winding transformer differential protection.
400 3 103 5 954:3 A I 2 5 pffiffiffi 3 3 242 I Relay5 0:95 3 Min Bias
954:3 5 0:906 A 1000
5 2 ðI R at nominal I R at Vmax Þ 5 2 ð1 0:906Þ 5 0:188
Minimum voltage at the low-voltage side: V min 5 220 kV 3 0:9 5 198 kV 400 3 103 5 1166:4 A I 2 5 pffiffiffi 3 3 198 I R at min voltage 5 0:95 3 Min Bias
1166:4 5 1:108 A 1000
5 2 ðI R at min voltage I R at nominal voltage Þ 5 2 ð1:108 1Þ 5 0:218
For the low-voltage side choose bias 5 22%.
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Minimum bias current for the 11 kV side with maximum voltage at the tertiary side is calculated as follows: V max 5 11 3 1:1 5 12:1 kV 100 3 103 5 4771:5 A I 3 5 pffiffiffi 3 3 12:1 I R at Vmax 5 0:95 3 Min Bias
4771:5 5 0:906 5000
5 2 ðI R at Vnominal I R at Vmax Þ 5 2 ð1 0:906Þ 5 0:188
Minimum voltage at the tertiary side (11 kV): V min 5 11 3 0:9 5 9:9 kV 100 3 103 5 5832 A I 3 5 pffiffiffi 3 3 9:9 I R at Vmin5 0:95 3 Min Bias
5832 5 1:108 A 5000
5 2 ðI R at Vmin 2 I R at Vnominal Þ 5 2 ð1:108 1Þ 5 0:216
Choose bias 5 22% for the 11 kV side. We then choose a bias slop of 22% for the differential relay (as shown in Fig. 19.5.6): Idiff. min 5 0.1 A Note: second slope 5 80%, to give more stability for the relay operation on external faults (through faults).
FIGURE 19.5.6 Relayoperatebias characteristic of a three-winding transformer biased differential protection.
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19.5.3 HIGH-IMPEDANCE DIFFERENTIAL PROTECTION SETTING 19.5.3.1 Worked Example 1 (Fig. 19.5.7)
FIGURE 19.5.7 500/220 kV transformer Yd1 500 MVA. Protected by high-impedance differential protection.
The data for this example include: Lead resistance: R 1 5 2 Ω R2 5 2 Ω R3 5 3 Ω R4 5 3 Ω RCT1 5 1.5 Ω RCT2 5 2 Ω The relay stability limit is normally, 16 In then Ithrough fault (IF) 5 16 In: 500 3 103 I n 5 pffiffiffi 5 577:35 A 3 3 500 I F through5 16 3 577:35 5 9237:6 A Assuming one CT is saturated then we calculate the stability voltage of the relay (RV) as follows: VS 5 stability voltage 5 Loop1 Resistance Loop2 Resistance Ithrough
fault
Ithrough fault 3 loop resistance CT 5 R1 1 R2 1 RCT1 5 2 1 2 1 1:5 5 5:5 Ω 5 R3 1 R4 1 RCT2 53131258 Ω
500 kV side 5 9237.6 A.
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Ithrough fault 220kV side 5 9237:6 V S1
V S2 5
500 5 20994:5A 220
I through fault 500 kV side :RLoop1 CT HV side 9237:6 5 5:5 5 84:678 V 5 600 5
I through fault 220 kV side 20; 994:5 8 5 167:956 V 5 1000 CT LV side
VS2 . VS1 Then choose VS 5 VS2. Use the relay with set voltage 25175 V with 25-V taps. Choose VS 5 170 V: RV burden 5 1 VA Also choose CT knee point at least 2VS VK min
5 2 3 170 V 5 340 V
Choose VK 5 400 V for CT, when another relay (current relay) R1 is used with the voltage relay in parallel and given that the burden of current relay is 600 Ω and its current is 40 mA. Then the voltage of the current relay (R1) is 600 3 0.040 5 24 V. Then the series setting resistor of the current relay R1 can be calculated as follows (as shown in Fig. 19.5.8): ðRS ÞSeries resistor ofRI 5
ð170 2 24Þ 5 3650Ω 0:04
The relay circuit consists of the following relays: RV 5 voltage-operated relay; RI 5 current-operated relay; RS 5 series setting resistor of RI; Rshunt 5 shunt resistor.
FIGURE 19.5.8 High-impedance differential relay circuit.
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FIGURE 19.5.9 Current transformer magnetizing curve.
Metrosil resistance (nonlinear resistor) We then check whether we need to use a shunt resistor for the relay or if this is not required as follows. From the CT magnitude curve the value of Imagnitising (Im) can be checked at a setting voltage VS for RV at VS 5 175 V. Im will be 40 MA, as shown in Fig. 19.5.9. The total operating current of the relay 5 ICT HV side mag 1 ICT low voltage side mag 1 IRv 1 IRI I RV 5
VA 1 5 5 5:88 mA V 170
IRI 5 40 mA ICT HV 5 40 mA, ICT LV 5 40 mA Total operating current 5 40 1 40 1 5.88 1 40 5 125.88 mA I set
5 0:2 I N HV side 500 3 103 3 1 5 0:192 5 192:45 mA 5 0:2 3 pffiffiffi 600 3 3 500
Then a shunt resistor is required to increase the total operating current of the relay to 192.45 mA: Current of shunt resistor 5 192.45125.88 5 66.57 mA. Shunt resistor 5
170V 5 2552:55Ω 0:0666
Total operating current for setting value
5 I CT HV mag 1 I CT LV mag 1 I RV 1 I RI 1 I R shunt 5 40 1 40 1 5:88 1 40 1 66:57 5 192:45 mA
Based on the available Metrosil resistance in the market, the following equation is used: V metrosil 5 ðI metrosil Þβ :C
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where β 5 0.2 C 5 450 I 5 30 A Vmetrosil 5 (30)002 . 450 5 888 V So, for CT with 1 A secondary use metrosil with C 5 450, β 5 0.2 Imetrosil will be 30 A for 2 seconds Vmetrosil 5 888 V
19.5.3.2 Worked Example 2 (Fig. 19.5.10) Autotransformer 500 MVA 500/220 kV HV side CT1 600/1 RCT1 5 1.5 Ω LV side CT2 1600/1 RCT2 5 2.5 Ω Neutral CT3 1000/1 RCT3 5 2.5 Ω Relay stability limit 5 16 In 5 Ithrough fault V s 5 stability voltage 5 V S1 HV side
I through fault 3 loop resistance CT
500 3 103 5 16 3 pffiffiffi ðR1 1 R2 1 RCT1 Þ 3 3 500 5
16 500 3 103 3 pffiffiffi ð2 1 2 1 1:5Þ 600 3 3 500
5 15:396 ð5:5Þ 5 84:678 V
FIGURE 19.5.10 500/220 kV autotransformer 500 MVA protected by high-impedance differential protection.
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V S2 LV side
5
305
16 3 500 3 103 pffiffiffi ðR3 1 R4 1 RCT2 Þ 1000 3 3 3 220
5 20:994 ð3 1 3 1 2:5Þ 5 20:994 ð8:5Þ 5 178:449 V For an autotransformer: INeutral5ILV sideIHV side V S neutral side
5 ð20:994 15:396ÞðR5 1 R6 1 RCT3 Þ 5 5:598 ð1:5 1 1:5 1 2:5Þ 5 5:598ð5:5Þ 5 30:789 V
Then, VS2 . VS1 . VS3. Choose Vs2 as VS 5 180 V. Use voltage relay 20200 V with 20-V taps. Set VS 5 180 V, burden 5 1 VA. Then VK min should be 2VS. VK min 5 2 3 180 5 360 V. Use CT with VK 5 400 V ,a current relay with burden 5 500 Ω, and current 5 30 mA. The voltage across the current relay RI will be as follows: V 5 500 3 0.03 5 15 V Then a series resistor is required to adjust the setting of the current relay RI calculated as follows (as shown in Fig. 19.5.11): RS ðsetting series resistorÞ 5 RS 5 5500 Ω Check if shunt resistance is needed or not.
FIGURE 19.5.11 Relay circuit.
180 2 15 0:03
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I set
5 0:2 I N HV side 500 3 103 1 3 5 0:2 3 pffiffiffi 600 3 3 500 5 0:1924 A 5 192:24 mA
From the CT magnitude curve check the value of Im at VS 5 180 V-Imag 5 45 mA. The total operating current of the relay 5 IRv 1 IRI 1 ICT HV side mag 1 ICT LV side mag 1 IR CT neutral mag and total operating current 5 Itotal. I RV 5
VA 1 5 5 5:55 mA V 180 I RI 5 30 mA
I total 5 5:55 1 30 1 45 1 45 1 45 5 170:55 mA Then we need a shunt resistor calculated as follows: I of R shunt 5 192.24170.55 5 21.69 mA Rshunt 5 I total
180 5 8219:17 Ω 0:02169
5 I RV 1 I RI 1 I CT HV mag 1 I CT LV mag 1 I CT neutral mag 1 I R shunt 5 5:55 1 30 1 45 1 45 1 45 1 21:69 5 192:24 mA
Subchapter 19.6
Generator Protection Setting 19.6.1 INTRODUCTION As generators are a very costly component of the power system they should have a good protection system. The complexity of this protection system varies depending on the size of the units. The settings of these protections are set at different values depending on the utilities setting polices. Here we provide some examples of generator protection settings.
19.6.2 WORKED EXAMPLES Worked Example 1 (as shown in Fig. 19.6.1) Power station data: Generator unit (G)
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FIGURE 19.6.1 Generator and transformer data.
333 MW xd 5 191% xd0 5 33% Xdv 5 22% F 5 50 Hz I2 5 15% I2 .t 5 10 S CT 5 13,000/122 kV Power Factor (PF) 5 0.8 S 5 417 MVA Unit transformer (T2) 25/15/15 MVA 22/6.4/6.4 kV Z% 5 8% at 15 MVA Generator transformer (T1) 420 MVA 225/22 kV Z% 5 14% Station transformer (T3) 25/15/15 MVA 220/6.4/6.4 kV Z% 5 8.2% at 15 MVA CTs 5 13,000/1 1. Stator earth fault protection 95% stator earth fault relay (R1) Maximum fault current in neutral 5 20 A Iset 5 (10.95) 20.15 5 0.2 A t to be set to 5 0.05 second. Tripping action: 1. 2. 3. 4.
HV CB of transformer T1; T2 6.4 kV incoming feeder tripping; Turbine tripping; Generator excitation tripping.
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2. 100% stator earth fault protection R2 and R3 will be two relays as follows: 1. Undervoltage relay 27 works on third harmonic (R2). 2. Overvoltage relay (R3) works on fundamental frequency. The overlap between 27 and 59 covers 100% of the stator winding. 3. Stator OC protection Iset 5 1.05 IN P 5 333 MW then S 5 417 MVA at a power factor of 0.8. IN 5
417 3 103 5 10; 943 A O3 3 22
I N secondary
5 10; 943 3
I set
1:05 I N 1:05 3 0:842 5 0:884 A SetI set
1 5 0:842 A 13; 000
5 0:9 A t 5 6 seconds
Action: trip of CB of transformer (T1) 220 kV side. 4. Negative-phase sequence current I2 protection (46) Second stage (trip): I set I set Set I set
5 0:15 I N 5 0:15 3 0:842 5 0:1263 A 5 0:127 A 5 12%
I22.t 5 10 where K 5 I22.t, K factor 5 10. First stage (alarm): Set I alarm 5 0:9 I set
5 0:9 3 0:12 5 0:108 5 10:8%
Set talarm 5 3.0 seconds Action: trip 220 kV side CB of T1 second stage. Alarm in first stage. 5. Loss of excitation protection Loss of excitation leads to loss of synchronism and running at higher than synchronous speed. The following impedance relay is used (Fig. 19.6.2). X1 1 x2 5 xd X1 5 xdx2 xd’ 5 33% on 417 MVA base.
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FIGURE 19.6.2 Loss of excitation impedance protection for generators.
X2 5
xd 0 2
Xd 5 191% on 417 MVAb. Z ðΩÞ 5 Xd 0 ðΩÞ 5 X2 5
Z% MVAb:100 ðVLÞ2
33 3 417:100 5 28:43 Ω 100ð22Þ2
xd 0 28:43 5 14:21 Ω 5 2 2
X1 5 xdx2 Xd ðΩÞ 5
191:417 3 100 5 164 Ω 100:ð22Þ2
X1 5 16414.21 5 150.35 Ω X1 5 150.35 Ω primary, secondary x1 5 887.065 Ω x2 5 14.21 Ω primary, secondary x2 5 83.839 Ω t 5 1 second; Z s 5 Z P VT ratio 5
NV 5 5:9 Z P NC
220; 000=O3 5 2200 5 NV 100=O3
CT ratio 5 13,000/1 5 NC 6. Underimpedance protection This relay is used as a backup protection for external faults on the highvoltage side (220 kV) of a transformer (T1), as shown in Fig. 19.6.3.
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FIGURE 19.6.3 Backup impedance protection for generators.
Z set 5 0:7 3 0:36 5 0:252- 5 25:2% Z set ðΩÞ
5
25:2 3 417 3 100 100ð220Þ2
5 0:217 Ω ðprimaryÞ Z set ðΩÞ secondary Z set
5 5:9 Z primary 5 5:9 3 0:217 5 1:2803 Ω
Z set 5 0:7 Z Load Z Load min 5 XG 1 XT 5 Xd} 1 XT 5 0:22 1 o:14 5 0:36 Set time 5 3 seconds. Action: 1. Turbine tripping; 2. Tripping of 6.4 K unit transformer incoming; 3. Generator excitation tripping. 7. Generator frequency protection A. Underfrequency protection This protection is required for load-shedding to allow unit frequency recovery.
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Relay set 5 47 Hz F 5 50 Hz tset 5 4 seconds Action: 1. Trip 220 kV CB. This protection works when load-shedding of the network does not work. 8.Overfrequency protection Under this condition the unit overspeed and trip are set as follows. Relay set 5 54 Hz, F 5 50 Hz Action: 1. Trip of 220 kV CB. 8. Undervoltage stator protection This protection works as a backup for an HV network in two stages. Stage 1 has a definite time uv set to 0.7 Vn of the generator and trips the HV CB. Stage 2 trips the unit after 2.5 seconds. Stage 1: V set 5 0:7V n 5 0:7 3 100 5 70 V tset 5 1.2 second Action trip CB of 220 kV transformer (T1). Stage 2: Vset 5 70 V tset 5 2.5 seconds Action: 1. Turbine tripping; 2. Tripping of 6.4 kV transformer incoming; 3. Generator excitation tripping. 9. Overvoltage protection This protection works when the Automatic Voltage Regulator (AVR) maloperates or for manual operation of the excitation system. First stage V set
5 1:1 V n 5 1:1 3 100 5 110 V
t 5 4.5 seconds Action: alarm Second stage V set 5 1:2 V n 5 120 V tset 5 3 seconds
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Action: 1. 2. 3. 4.
HV CB tripping of T1; Incoming of transformer T2 6.4 kV trip; Turbine tripping; Generator excitation tripping.
10. Generator differential protection A biased differential relay is used in the following circumstances (Fig. 19.6.4)
FIGURE 19.6.4 Biased differential protection for generators.
(IN 5 generator nominal current): P 5 O3 VL. IN. cos [ IN
P 333 3 103 5 pffiffiffi 5 pffiffiffi 3V L :cos φ 3V L :cos φ
333 3 103 5 10; 924 A 5 pffiffiffi 3 3 22 3 0:8 10; 924 5 0:84 A I N secondary 5 13; 000 I set 5 0:1 I N 5 0:1 3 0:84 5 0:084 A Set bias stop 5 20% 11. Reverse power protection PReverse set 5 0.01 3 333 5 3.33 MW
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Pset I set on Relay 5 pffiffiffi 3V L :cos φ 3:33 3 103 5 pffiffiffi 3 3 22 3 0:8 5 109:237 A 109:237 I set secondary 5 13000 5 8:4 mA tset 5 15 seconds Action: 1. Trip HV CB 220 kV of T1; 2. Generator excitation trip; 3. Incoming 6.6 kV of T2 trip.
Subchapter 19.7
Switchgear (Busbar) Protection Settings 19.7.1 INTRODUCTION The busbars in substations are a very important part of the system and should have adequate protection, however this protection is very expensive in 11 kV and 22 kV, and the faults in busbars are rare, therefore it is used in 66 kV systems and above as essential protection. The most commonly used scheme for busbar protection is highimpedance differential protection.
19.7.2 WORKED EXAMPLES FOR BUSBAR PROTECTION One hundred and thirty-two kiloVolts busbars have two sections and a bus coupler CB, each busbar has one feeder and an incomer transformer (as shown in Fig. 19.7.1). CT ratios: CT1, CT2, CT3, CT4, CT7, CT8, CT9, CT10 -1200/1 RCT 5 2.9 Ω Imag 5 40 mA at V 5 130 V Imag 5 35 mA at V 5 125 V CT ratios: CT5, CT6-1500/1 RCT 5 3.1 Ω Imag 5 45 mA at V 5 130 V Lead resistances: R1 5 0.3 Ω R4 5 0.32 Ω R7 5 0.33 Ω
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FIGURE 19.7.1 Two-section busbar with two feeders and two incomers.
R2 5 0.35 Ω R5 5 0.36 Ω R8 5 0.31 Ω R3 5 0.4 Ω R6 5 0.48 Ω R9 5 0.33 Ω R10 5 0.39 Ω We use three relays as follows: (M1) Discriminating main 1 relay to protect busbar section A1; (M2) Discriminating main 2 relay to protect busbar section A2. Check zone relay (CH) to protect busbar A1 and A2 to confirm that the fault exists on the busbars. IF Relay setting voltage 5 CT Rloop max Rloop max 5 maximum lead resistance Assuming saturation in one CT: M1 (main relay setting) Rlead 1
5 2 3 R1 1 RCT1 5 2 3 0:3 1 2:9 5 3:5 Ω
Rlead 2
5 2 3 R2 1 RCT3 5 2 3 0:35 1 2:9 5 3:6 Ω
Rlead 3
5 2 3 R3 1 RCT6 5 2 3 0:4 1 3:1 5 3:9 Ω
Rlead max 5 Rlead 3 5 3.9 Ω Maximum through fault current 5 40 kA
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VS
315
IF : Rloop max CT 40; 000 3:9 5 130 V 5 1200 5
Choose voltage relay RV as follows: Range 25175 V IRV 5 35 mA Set VS 5 130 V Then the knee point voltage of CT should be as follows: V K min
5 2 3 VS 5 2 3 130 5 260 V
Choose VK 5 300 V Use a current relay with burden 5 3500 Ω I 5 30 mA Then the voltage across the current relay RI is as follows: 3500 3 0.03 5 105 V. Then, the series resistor is required to adjust the setting of the current relay RI and is calculated as follows (as shown in Fig. 19.7.2): RS ðSetting ResistorÞ 5
130 2 105 5 833 Ω 0:03
M1 relay circuit—busbar protection We then need to check whether we need shunt resistance or not: Iset 5 0.15 Ifault min Assume the single-phase minimum fault 5 12 kA, and the three-phase minimum fault 5 14kA: I set 5 0:15 3 12 5 1:8 kA I set secondary 5
1800 5 1:5 A 1200
Itotal operating current 5 IRV 1 IRI 1 ICT1 IRV 5 35 mA IRI 5 30 mA
FIGURE 19.7.2 Relay circuit.
mag 1 ICT3 mag 1 ICT6 mag
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FIGURE 19.7.3 Current transformer magnetization curve.
By checking from the magnetization curve of CT, the magnetization current at VS 5 130 V (Fig. 19.7.3). Itotal 5 35 1 30 1 40 1 40 1 45 1 IR shunt 5 190 mA 1 IR shunt Itotal 5 Iset 5 1.5 A 5 1500 mA IR shunt 5 1500190 5 1310 mA Rshunt 5
VS 130 5 99:23 Ω 5 I R shunt 1310 =1000
VS 5 130 V Choose Metrosil as follows: Vmetrosil 5 (Imetrosil)β C Choose C 5 450 β 5 0.25 I 5 30 A V metrosil
25
5 ð30Þ00 3 450 5 1053 V
Imetrosil will be 30 A for 2 seconds. Main 2 relay setting M2 Rlead1
Rlead
5 2 3 R6 1 RCT9 5 2 3 0:48 1 2:9 5 3:86 Ω
Rlead 2
5 2 3 R5 1 RCT7 5 2 3 0:36 1 2:9 5 3:62 Ω
Rlead3
5 2 3 R4 1 RCT5 5 2 3 0:32 1 3:1 5 3:74 Ω
max 5 Rlead 1 5 3.86
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IF Rloop max CT 40; 000 3:86 5 128:66 V 5 1200
VS 5
5
VS 5 130 V Choose relay RV 5 130 V, IRV 5 35 mA Range: 25175 V Knee point of CT is as follows: V K min 5 2V S 5 2 3 130 5 260 V Choose CT with Vk 5 300 V Use a current relay with burden 3500 Ω, I 5 30 mA. As before in the M1 calculation: Rsh 5 99.23 Ω Vmetrosil 5 1053 V Rseries Resistor (RS) 5 833 Ω C 5 450, β 5 0.25 Imetrosil 5 30A for 2 seconds Then check the zone relay CH calculation: Rlead 1
5 2 3 R7 1 RCT2 0:33 3 2 1 2:9 5 3:56 Ω
Rlead 2
5 2 3 R8 1 RCT4 2 3 0:31 1 2:9 5 3:52 Ω
Rlead 3
5 2 3 R9 1 RCT8 5 2 3 0:33 1 2:9 5 3:56 Ω 5 2 3 R10 1 RCT10 5 2 3 0:39 1 2:9 5 3:68
Rlead 4 Rlead
max 5 Rlead 4 5 3.68
VS
IF RL max CT 40; 000 3:68 5 1200 5 122:66 5
Set VS 5 125 V Choose the relay range: 25175 V Vset 5 125 V IRV 5 35 mA The knee point voltage of CT should be as follows: V K min 5 2 3 V S 5 2 3 125 5 250 V Choose VK 5 300 V of CT using a current relay with burden 5 3500 Ω
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FIGURE 19.7.4 Check the zone (CH) relay circuit.
I 5 30 mA Then, the voltage across the current relay RI will be as follows: 5 3500 3 0:03 5 105V The series setting resistor RS is calculated as follows (as shown in Fig. 19.7.4): RS 5
125 2 105 5 666 Ω 0:03
Then check whether we need the shunt resistance or not: I set
5 0:15 I fault min 5 0:15 3 12 5 1:8 KA
I set secondary 5
1:8 3 103 5 1:5 A 1200
Itotal operation 5 IRV 1 IRI 1 ICT2 mag 1 ICT4 mag 1 ICT8 mag 1 ICT10 mag 1 IR shunt Itotal 5 35 1 30 1 35 1 35 1 35 1 35 5 205 mA Imag 5 40 mA for CT2, CT4, CT8, CT10 from VImag. The curve of CT at VS 5 125 V Imag 5 35 mA Itotal 5 205 1 IR shunt I total 5 I set 5 1:5 3 103 5 1500 mA IR
shunt
5 1500 205 5 1295 mA Rshunt 5
VS 125 5 IRshunt 1295=1000
Rshunt 5 96.525 Ω Choose metrosil resistance with voltage: Vmetrosil 5 (Imetrosil)β C C 5 450, β 5 0.25, I 5 30 A Vmetrosil 5 (30)0.25 450 5 1053 V Relay setting
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Main 1 zone (discriminating zone)—M1 RV Vrange 25175 V, IRV 5 35 mA, Vset 5 130 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 833 Ω Rshunt 5 99.23 Ω, Metrosil C 5 450 β 5 0.25 I 5 30 A at 2 seconds. Main 2 zone (discriminating zone)—M2 RV Vrange 5 25175 V, IRV 5 35 mA, Vset 5 130 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 833 Ω Rshunt 5 99.23 Ω, Metrosil C 5 450, β 5 0.25, I 5 30 A at 2 seconds. Check zone (CH) RV Vrange 5 25175 V, IRV 5 35 mA, Vset 5 125 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 666 Ω Rshunt 5 96.525 Ω, Metrosil C 5 450, β 5 0.25, I 5 30 A at 2 seconds.
19.7.3 INTRODUCTION TO BREAKER FAILURE PROTECTION This protection is used as backup protection to isolate a fault if the circuit breaker of the feeder has a mechanical or other problem preventing it from opening the circuit. This protection then isolates all sources which feed the fault and also the remote end of the line.
19.7.4 WORKED EXAMPLE FOR BREAKER FAILURE PROTECTION CB failure time tCB 5 110180 ms Fault clearance time should not exceed 300 ms Worked example (Figs. 19.7.5 and 19.7.6): Breaker Failure (BF) relay trip logic BF relay operating times Refer to Fig. 19.7.7 for the breaker failure time and the total fault clearance time. Iset BF 5 0.5 IN
FIGURE 19.7.5 Breaker failure relay connection on an overhead transmission line.
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FIGURE 19.7.6 Breaker failure relay operation logic.
FIGURE 19.7.7 Breaker failure time and total fault clearance time.
5 0.5 3 1200 5 600 A tCB 5 120 ms Action: The back trip for all feeders connected on busbar A through busbar protection trip relays for the section which this feeder connected to it should not exceed 300 ms. The remote trip for the breaker is at end B. The local retrip for the breaker is at end A.
Subchapter 19.8
Motor Protection Setting 19.8.1 INTRODUCTION Here we provide some examples for motor protection settings.
19.8.2 WORKED EXAMPLES 19.8.2.1 Motor Faults 1. 2. 3. 4.
Winding faults Overloading; Reduced or loss of supply voltage; Phase reversal;
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5. Phase unbalance; 6. Out-of-step operation for synchronous motors; 7. Loss of excitation for synchronous motors.
19.8.2.2 Phase Fault Protection A high set Over Current (OC) unit is used. Differential protection is also used as it is not affected by the motor starting current. For the H set OC relay the setting should be more than the motor starting current: Iset . Istart.
19.8.2.3 Earth Fault Protection An inverse or very inverse OC relay is connected to the residual CT circuit (as shown in Fig. 19.8.1).
FIGURE 19.8.1 Earth fault protection for motors.
Iset 5 0.25 IF
min,
where IF
min 5 minimum
fault current.
19.8.2.4 Locked Rotor Protection We use the OC relay with characteristics higher than the starting current of the motor and less than the locked rotor current (as shown in Fig. 19.8.2), where ts 5 motor starting time; tR 5 locked rotor time. Also, the OC (51) relay characteristic is used to detect the locked rotor condition of the motor or impedance relay is used to detect this condition, as shown below in Fig. 19.8.3.
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FIGURE 19.8.2 Locked rotor protection for motors.
FIGURE 19.8.3 Impedance relay for locked rotor condition.
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19.8.2.5 Overload Protection Thermal replica relays are used to protect the motor from overloads using a separate relay in each phase of the motor. Thermal relays are also used with thermal replica relays (resistance temperature detectors embedded in machine winding) to give complete protection for the motor.
19.8.2.6 Low-Voltage Protection Undervoltage time delay relay is used as follows: Vset 5 0.5 V nominal Action: Trip after a few seconds.
19.8.2.7 Phase Rotation Protection A negative-phase sequence voltage relay is used. v2 set 5 10% V2
19.8.2.8 Phase Unbalance Protection Detect I2 and set as follows: Iset 5 0.15 IN To protect the motor against single-phasing condition: I22 t 5 40 in general
19.8.2.9 Out-of-Step Protection for Synchronous Motors A relay is used to detect the condition in which the motor draws a heavy current in a low-power factor. Therefore, this relay will be a power factor relay and will issue a signal to disconnect the field winding and let the motor run as an induction motor.
19.8.2.10 Loss of Excitation Protection for Synchronous Motors Undercurrent relay is inserted in a field circuit.
Protection Relays Settings INTRODUCTION This chapter introduces relay settings for different types of relays used in power system, with real worked examples.
Subchapter 19.1
Overcurrent Protection Settings 19.1.1 INTRODUCTION Overcurrent protection is used in extra-high voltages and high-voltages networks as the main protection, and in medium networks it is also used as backup protection for power transformers. Time grading for overcurrent relays is used in the radial system. The minimum possible time delay is selected for the relay which protects the line farthest from the source. The grading time between the successive relays will be in the order of 0.40.5 seconds, and with modern relays it can be of the order of 0.35 seconds. Current settings for an overcurrent relay should consider the following factors: 1. Maximum possible loading; 2. Transient currents due to switching of transformers, inrush current, heavy motor restarting currents after clearing external faults; 3. Drop off/pickup ratio, which is called the reset ratio. I set 5 ðK 3 RÞ=F 3 I L Max where: K 5 safety factor 5 1.2 R 5 restarting factor, which depend on the load profile Set to 2 for industrial loads Practical Power System and Protective Relays Commissioning. DOI: https://doi.org/10.1016/B978-0-12-816858-5.00019-8 © 2019 Elsevier Inc. All rights reserved.
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Set to 1.5 for residential loads; F 5 reset ratio 5 0.85 for electromechanical relays and set to 0.90.95 for static and digital relays; IL Max 5 maximum expected load; Iset 5 22.5 IL Max. The performance of overcurrent relays is checked by calculating the sensitivity factor, S, which is the ratio of the minimum short circuit current following at the relay in the end protected line in a radial system to the relay setting as follows: S 5 I sc minimum=I set should be . 1:5
19.1.2 TRANSMISSION LINE OVERCURRENT PROTECTION Fig. 19.1.1 illustrates the overcurrent protection of a high-voltage transmission line The circuit data include: Overcurrent relay type MCGG22. Application: Backup protection 500 kV, fault level at end B 5 12,000 MVA, current setting of relay at A 5 800 A. Is 5 800 A current transformer (CT) ratio 5 1600 1 Is 5 800
1 5 0:5A 1600
Plug setting multiplier (PSM) PSM 5 IF 5
If MVAF -I F 5 Is O3 3 V
12; 000 5 13:856 KA O3 3 500
FIGURE 19.1.1 Overcurrent protection of a high-voltage transmission line.
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IF 5 13,856 A IS 5 800 A PSM 5
13; 856 5 17:32 800
Using a standard inverse curve of the relay then we have the relay operating time TR: TR 5
0:14 0:14 5 0:0221 PSM 17:320:0221
TR 5 2.385 Assuming that the relay should operates on 1 second then we have: t.operation 5 TM.TR Where TM 5 time multiplier TM 5
1 5 0:419 2:385
Relay (A) setting: Iset 5 800 A (Primary) Iset secondary 5 0.5 A PSM 5 17 TM 5 0.4 Operating time for a fault at end B will be: t:operation 5 0:4 3
0:14 170:0221
t.operation 5 0.96 seconds
19.1.3 TRANSFORMER OVERCURRENT PROTECTION Transformer overcurrent protection is illustrated in Fig. 19.1.2. Transformer data: 250 MVA 220/66 kV ZTr% 5 10% on MVAb 5 100 Relay R2 setting on low voltage (LV) feeder F1:
FIGURE 19.1.2 Transformer overcurrent protection.
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MVAF 5
MVAb 3 100 100 3 100 5 5 1000MVA Z% 10
If of HV side of Tr 5
MVAF 1000 5 5 2624A O3 3 220 O3 3 220
IF of LV side of Tr 5
1000 5 8747A O3 3 66
Iset 5 600 A, CT 5 500/1 TM 5 0.4 PSM 5 8747 600 5 14:58 Relay operating time 5 TR 5
0:14 14:580:0221
TR 5 2.54 For a fault of 8747 A the operating time will be: t.op 5 TR.TM 5 2.54 3 0.4 5 1.016 second With a grading time step 5 0.4 second. Then the operating time of R1 should be more than 5 1.016 1 0.4 5 1.416 second. Relay R1 setting on the HV side of Tr IF on HV side 5 2624 A We can set the relay as follows: Iset 5 1.45 3 ITr HV side MVA 250 3 103 5 pffiffiffi ITr HV side 5 pffiffiffi 5 656A 3 3 VHV 3 3 220 Iset 5 1.45 3 656 5 951.2 A PSM 5
IFault 2624 5 2:76 5 951:2 Iset
Set top 5 1.5 second, as seen above it should be more than 1.416 second. Top 5 TM.TR TR 5
0:14 SI 2 Curve 2:760:0221
TM 5
top 1:5 5 0:219 5 TR 6:825
TR 5 6.825
R2 Setting Iset 5 600 A PSM 5 14.58 TM 5 0.4 CT 5 500/1 R1 Setting Iset 5 951.2 A PSM 5 2.76 TM 5 0.219 CT 5 1000/1
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Subchapter 19.2
Feeder Backup Overcurrent Protections 19.2.1 INTRODUCTION Overcurrent protection here is used as backup protection, with a considerable time delay, which means it works as a second line of defense to protect the line if the main or primary protection fails to isolate the line for the fault instantaneously.
19.2.2 WORKED EXAMPLE Fig. 19.2.1 gives an example of line backup overcurrent protection. Line AB 500 kV maximum load current 5 IL Max 5 1200 A For OC relay at A (R): Iset 5 1.2 3 IL Max 5 1.2 3 1200 1000 5 1.44 A PSM calculation: IF MVAF andI F 5 IS O3 3 V I F 5 80 KA I S 5 1200 A 80; 000 5 80 PSM 5 1000 PSM 5
Using a standard inverse curve of the OC relay we then have: TR 5
0:14 0:14 5 0:0221 5 1:528 0:0221 PSM 80
where TR 5 relay operation time assuming that the relay will operates on 1 second for this fault then t.operation 5 TM.TR, where TM 5 time multiplier.
FIGURE 19.2.1 Line backup overcurrent protection.
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Then TM 5
t:operation 1 5 0:654 5 TR 1:528
Relay setting: Iset 5 1440 A 5 1.44 A (secondary) PSM 5 80 TM 5 0.65 Operation time at a fault after end B will be: t 5 0:65
0:14 5 0:993s 800:0221
For an earth fault relay at end A connected to the residual connection of CTs at end A: %ZS 5
MVAb 3 100 100 3 100 5 5 0:144 MVAF 80 3 O3 3 500
For a single phase fault at end B we draw the sequence network diagram as shown in Fig. 19.2.2. PSM calculation: IF 5 29.3 KA Iset 5 800 A
FIGURE 19.2.2 Sequence networks.
Z t 5 3Z s 1 2Z 1 1 Z 0 5 3 3 0:144 1 2 3 0:3 1 0:15 5 0:432 1 0:6 1 0:15 5 1:182% 100:100 5 9:769 KA 1:182 3 O3 3 500 I F 5 3I0 5 3 3 9:769 5 29:3 KA I set 5 800 A 800 5 0:8 A ðsecondaryÞ I set 5 1000 I0 5
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PSM 5
287
IF 29:3 3 103 5 36:625 5 IS 800
Assuming that the relay will operates on 1 second for a fault on end B: t:operation TM 5 TR 0:14 0:14 TR 5 5 0:0221 5 1:884 0:0221 PSM 36 TM 5
1 5 0:530 1:884
Relay setting: Iset 5 800 A 5 0.8 A secondary PSM 5 36 TM 5 0.53 0:14 t:operation 5 0:53 0:0221 36 5 0.998 second
Subchapter 19.3
Feeder Unit Differential Protection 19.3.1 INTRODUCTION This protection is used to protect overhead transmission lines and is commonly used as main 2 in utilities. The disadvantage of this protection is the blind zone, which is not protected by this protection—this means the zone between the circuit breaker (CB) and the current transformer in one end—as this zone is protected by backup protections (Fig. 19.3.1).
FIGURE 19.3.1 Line differential protection.
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19.3.2 WORKED EXAMPLE Worked example: Line AB 500 kV B 5 1.877 CT 5 1200/1 at both sides choose the relay characteristic shown in Fig. 19.3.2
FIGURE 19.3.2 Line differential protection characteristic.
jIdiff j 5 jIA 1 IB j jIbias j 5 0:5½jIA1 1 IB1 j Line charging current 5
B 3 1000 V 3 O3 1:877 3 1000 I ch 5 500 3 O3
5 2.167 A I set min 5 2:5 3 2:167 5 5:4175 A Iset 5 0.2 IN I set 5 0:2 3 1200 5 240 A Iset . 2.5 Ich IS1 5 240 A primary 240 5 0:2 A 1200 I S2 5 2I N 5 2 3 1200 5 2400 A
I S1 5
I S2 5 2400 A primary 2400 5 2 A secondary I S2 5 1200 K 1 2 slop1 5 30 % K 2 2 slop2 5 150 %
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Subchapter 19.4
Setting of Distance Protection 19.4.1 INTRODUCTION Distance protection is widely used as the main protection—primary protection—in overhead transmission lines and has the advantage that it gives backup protection for adjacent lines. However, it has the disadvantage that it does not have accurate measurements for close-up faults near the remote end of the line due to the errors in CT and PT measurements and the transient nature of the short circuit current, also due to the tower arc fault resistance—what we call distance protection overreaching and underreaching—that it is not accurate in measurement in short transmission lines due to the effect of the system impedance ratio.
19.4.2 WORKED EXAMPLES Worked example 1 Most distance protection is set to have three zones and one backup start zone. The setting of the first zone is as follows for a relay at end A: Z1 5 0.85 ZL, t1 5 0 instantaneously where ZL 5 line impedance. Setting of the second step zone 2 is as follows for a relay at end A (Fig. 19.4.1): The Z2 setting is the lower value of the following: Z 2 5 0:85 ðZ L11 0:85 Z L2 Þ Z 2 5 0:85 ðZ L1 1 Z T Þ
FIGURE 19.4.1 Distance protection application setting: Example 1.
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Worked example 2
FIGURE 19.4.2 Distance protection application setting: Example 2.
For the following network configuration, the zone 2 setting will be as follows for a relay at end A:where II21 is called the infeed factor “F” I2 3 0:85Z L2 Z 2 5 0:85 Z L1 1 2I 1 IT Z 2 5 0:85 Z L1 1 ZT I1 In general: Z2 $ 1:2-Z 2 5 120% Z L1 Z L1 t2 is set to be 0.5 second. The setting of zone 3 in general is: Z3 $ 1:5 Z 3 5 150% Z L1 Z L1 t3 is set to be 0.9 second. Final (Backup Step) The current starter is used for simple networks where the minimum short circuit current is higher than the maximum load currents. An impedance starter is more commonly used as this start impedance stage should not work in the load area and also should not work for the transformer low-voltage faults. The lower value of Zst should be chosen.
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1. Zst # 0.5 ZL min; this means less than half of the minimum load impedance; 2. Zst $ 1.25 Z3. tst 5 1.3 second. The Compensation Factor (K0)
FIGURE 19.4.3 Distance protection compensation factor (K0) for earth fault measurement.
The voltage measured by the relay at A is: 0 1 Z 2 Z 0 1A V RA 5 Z 1 @I A 1 3I 0 3Z1 I R 5 I A 1 K 0 3I 00 Where K 0 5 @
1
z0 2 z1 A 3z1
The compensation factor is set in the relay by the values of Z1 and Z0 to have the correcting value of distance relay measure the line positive impedance, proportional to its length irrespective of the kind of fault (earth faults or phase faults). Distance Protection Setting Z1F 5 80% of the protected line; Z2F 5 100% of the protected line 1 50% of the shortest adjacent line; Z3F 5 100% of the protected line 1 125% of the longest adjacent line; Z3R 5 25% of the protected line; F 5 forward direction, R 5 reverse direction. Worked example 3
FIGURE 19.4.4 Distance protection application setting: Example 3.
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Data: AB 5 protected line length 5 60 km 220 kV CT 5 800/1 PT 5 220;000 100 Z 1P 5 0:0412 1 J 0:302 Ω = km 3 60 Z 10 5 0:25 1 J 0:95 Ω = km 3 60 BC 5 shortest line adjacent to the protected line AB length 5 50.5 km Z 2P5 0:0412 1 J 0:302 Ω = km 3 50:5 BD 5 longest line adjacent to the protected line AB length 5 68 km Z 3P 5 0:0412 1 J 0:302 Ω = km 3 68 CT ratio 5 800/1 5 800 Rated voltage 5 220 kV Rated current 5 800 A PT 5 220;000 100 Ratio 5 2200 Z p line AB
5 ð0:0412 1 J 0:302Þ50:5 5 2:0806 1 J 15:251 5 15:356 +82:231
Z P line BC
Z P line BD
Z 1 Reach
5 60ð0:0412 1 J 0:302Þ 5 2:472 1 J 18:12 5 18:287+82:231 Ω
5 ð0:0412 1 J 0:302Þ68 5 2:8016 1 J 20:536 5 20:726+82:231
5 0:8 Z P line AB 5 0:8 3 18:287 + 82:231 5 14:629 + 82:231 Ω 5 14:629 Ω
Z 2 Reach
5 Z P line AB 1 0:5 Z P line BC 5 ð18:287 1 0:5 3 15:356Þ+82:231 5 25:965 +82:231 Ω 5 25: 965 Ω
Z 3 Reach
Z P line AB 1 1:25 Z P line BD 5 ð18:287 1 1:25 3 20:726Þ+82:231 5 44:1945 +82:231 5 44:1942 Ω
Protection Relays Settings Chapter | 19
Z 3 reverse
5 0:25 3 Z P line AB 5 0:25 3 18:287 +82:231 5 4:571 + 82:231 Ω 5 4:571 Ω
The earth fault residual compensation factor is calculated as: K0 5
Z10 2 Z1P 3Z1P
Z 0 line AB 5 ð0:25 1 J 0:95Þ 3 60 Z 0 line AB 5 15 1 57J 5 58:94 + 75:25 1 K0 5
58:94+75:25 2 18:287+82:231 3 3 18:287+82:231 15 1 J56:99 2 2:472 2 J18:12 5 54:861+82:231 12:528 1 J38:87 5 54:861+82:231 5
40:839+72:135 54:861+82:231
K0 5 0.7444 +10.096 K0 5 0.744 The setting secondary values are: Z 1 Reach
CTratio PTratio 800 5 14:629 3 2200 5 14:629 3 0:363 5 5:319Ω; t1 instantaneous 5 14:629 3
Z 2 Reach
5 25:965 3 0:363 5 9:425 Ω ; t2 5 500 ms
Z 3 Reach
5 44:1945 3 0:363 5 16:042 Ω ; t3 5 900 ms
Z 3 Reverse Power Swing Blocking Settings
5 4:571 3 0:363 5 1:659 Ω
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Z PSB
Z PSB
5 1:3 Z 3 F 5 1:3 3 16:042 5 20:854 Ω ðsecondary valueÞ 5 20:854 Ω
where ZPSB 5 power swing blocking impedance setting, Z3 5 relay third zone. Load encroachment into Z3 reach Z Load min
0:8 rated voltage 1:2 rated current O3 0:8 3 220000 5 1:2 3 800 3 O3
5
5 105:85 Ω Z Load min secondary 5 105:85 3 0:363 5 38:422 Ω Z 3 Reach , 0:9 Z Load min 16:041 , 0:9 3 38:422 16:041 , 34:58; the above criteria are correct:
Subchapter 19.5
Differential Protection Setting 19.5.1 INTRODUCTION Transformer protection can be high-impedance differential protection or lowimpedance differential protection. Both have setting calculations as shown in the following examples.
19.5.2 LOW-IMPEDANCE DIFFERENTIAL PROTECTION 19.5.2.1 Worked Example 1 (Fig. 19.5.1) Tr 66/11 kV 25MVA Dy1 Tap changer 5 6 10% of 11 kV side nominal voltage. Calculation: HV side: 25 3 103 I 1 5 pffiffiffi 5 219 A 3 3 66 Choose CT ratio 5 250/1
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FIGURE 19.5.1 Transformer low-impedance differential protection.
i1 5 219
1 5 0:876 A 250
where I1 5 primary current at the HV side of the transformer; i1 5 secondary current at the HV side of the transformer. 1 HV correction factor 5 i11 5 0:876 5 1:145 LV side: 25 3 103 5 1312 A I 1 5 pffiffiffi 3 3 11 Choose CT ratio 5 1300/1 i2 5 1312
1 5 1:009A 1300
where I2 5 primary current at the LV side of the power transformer; i2 5 secondary current at the LV side of the power transformer. LV correction factor 5
1 5 0:991 1:009
For phase correction and zero-sequence current filtering, we use interposing CT as follows: HV side we use Yy0 transformer LV side we use Yd11 transformer as shown in Fig. 19.5.2. Now the current input to the relay is 1 A +0 degree after the interposing CT ratio and phase correction at the HV side and the LV side of the power transformer. Minimum Bias Current The maximum voltage is calculated at the low-voltage side. V max 5 11 3 1:1 5 12:1 kV
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FIGURE 19.5.2 Interposing transformer in transformer differential protection.
25 3 103 5 1193 A I 2 5 pffiffiffi 3 3 12:1 I R 5 relay current 5 0:95 3 Min Bias
1193 5 0:871 A 1300
5 2 I Relay set Normal I Relay at max voltage 5 2 ð1 0:871Þ 5 0:256
The minimum voltage is calculated at the low-voltage side. V min 5 11 3 0:9 5 9:9 kV 25 3 103 5 1458 A I 2 5 pffiffiffi 3 3 9:9 I R 5 relay current 5 0:95 3 Min Bias
1458 5 1:065 A 1300
5 2 I Relay at min voltage I Relay set at nominal voltage 5 2 ð1:065 2 1Þ 5 0:13
Set the minimum bias at 26%. Minimum bias is the highest calculated value of bias at the maximum voltage and the minimum voltage at the low-voltage side of the power transformer. Pickup or Id min set to 10% In I d min 5 0:1 3 1 5 0:1 A First slope bias 5 26% Second slope bias 5 80%as shown in Fig. 19.5.3.
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FIGURE 19.5.3 Relayoperatebias characteristic of used differential protection.
19.5.2.2 Worked Example 2 A transformer with three-winding Yyd11 500/220/11 kV 500 MVA/400 MVA/100 MVA Tap changer 5 6 10% of 220 kV side and tap charger 5 6 10% of 11 kV side calculation. HV side: 500 kV 500 3 103 I 1 5 pffiffiffi 5 577:35 A 3 3 500 Choose CT 5 600/1 i1 5
577:35 5 0:962 A 600
HV correction factor 5
1 5 1:039 0:962
LV side: 220 kV side 400 3 103 I 2 5 pffiffiffi 5 1049:72 A 3 3 220 Choose CT 5 1000/1 i2 5
1049:72 5 1:0497 A 1000
LV correction factor 5
1 5 0:952 1:0497
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Tertiary side 11 kV side: 100 3 103 I 3 5 pffiffiffi 5 5248:63 A 3 3 11 Choose CT 5 5000/1 i3 5
5248:63 5 1:0497 A 5000
1 Tertiary side correction factor 5 1:0497 5 0:952 For phase correction and zero 5 sequence current filtering we use the following interposing “auxiliary” transformer for current as follows (refer to Fig. 19.5.4): HV side: 500 kV Yd11 ICT; LV side: 220 kV Yd11 ICT; Tertiary side: 11 kV Yy0 ICT. Refer to Fig. 19.5.5 for a detailed connection diagram of three-winding transformer-biased differential protection. Minimum bias current (220 kV side) Maximum voltage at the low-voltage side:
V max 5 220 kV 3 1:1 5 242 kV
FIGURE 19.5.4 Interposing current transformers.
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FIGURE 19.5.5 Detailed connection diagram of three-winding transformer differential protection.
400 3 103 5 954:3 A I 2 5 pffiffiffi 3 3 242 I Relay5 0:95 3 Min Bias
954:3 5 0:906 A 1000
5 2 ðI R at nominal I R at Vmax Þ 5 2 ð1 0:906Þ 5 0:188
Minimum voltage at the low-voltage side: V min 5 220 kV 3 0:9 5 198 kV 400 3 103 5 1166:4 A I 2 5 pffiffiffi 3 3 198 I R at min voltage 5 0:95 3 Min Bias
1166:4 5 1:108 A 1000
5 2 ðI R at min voltage I R at nominal voltage Þ 5 2 ð1:108 1Þ 5 0:218
For the low-voltage side choose bias 5 22%.
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Minimum bias current for the 11 kV side with maximum voltage at the tertiary side is calculated as follows: V max 5 11 3 1:1 5 12:1 kV 100 3 103 5 4771:5 A I 3 5 pffiffiffi 3 3 12:1 I R at Vmax 5 0:95 3 Min Bias
4771:5 5 0:906 5000
5 2 ðI R at Vnominal I R at Vmax Þ 5 2 ð1 0:906Þ 5 0:188
Minimum voltage at the tertiary side (11 kV): V min 5 11 3 0:9 5 9:9 kV 100 3 103 5 5832 A I 3 5 pffiffiffi 3 3 9:9 I R at Vmin5 0:95 3 Min Bias
5832 5 1:108 A 5000
5 2 ðI R at Vmin 2 I R at Vnominal Þ 5 2 ð1:108 1Þ 5 0:216
Choose bias 5 22% for the 11 kV side. We then choose a bias slop of 22% for the differential relay (as shown in Fig. 19.5.6): Idiff. min 5 0.1 A Note: second slope 5 80%, to give more stability for the relay operation on external faults (through faults).
FIGURE 19.5.6 Relayoperatebias characteristic of a three-winding transformer biased differential protection.
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19.5.3 HIGH-IMPEDANCE DIFFERENTIAL PROTECTION SETTING 19.5.3.1 Worked Example 1 (Fig. 19.5.7)
FIGURE 19.5.7 500/220 kV transformer Yd1 500 MVA. Protected by high-impedance differential protection.
The data for this example include: Lead resistance: R 1 5 2 Ω R2 5 2 Ω R3 5 3 Ω R4 5 3 Ω RCT1 5 1.5 Ω RCT2 5 2 Ω The relay stability limit is normally, 16 In then Ithrough fault (IF) 5 16 In: 500 3 103 I n 5 pffiffiffi 5 577:35 A 3 3 500 I F through5 16 3 577:35 5 9237:6 A Assuming one CT is saturated then we calculate the stability voltage of the relay (RV) as follows: VS 5 stability voltage 5 Loop1 Resistance Loop2 Resistance Ithrough
fault
Ithrough fault 3 loop resistance CT 5 R1 1 R2 1 RCT1 5 2 1 2 1 1:5 5 5:5 Ω 5 R3 1 R4 1 RCT2 53131258 Ω
500 kV side 5 9237.6 A.
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Ithrough fault 220kV side 5 9237:6 V S1
V S2 5
500 5 20994:5A 220
I through fault 500 kV side :RLoop1 CT HV side 9237:6 5 5:5 5 84:678 V 5 600 5
I through fault 220 kV side 20; 994:5 8 5 167:956 V 5 1000 CT LV side
VS2 . VS1 Then choose VS 5 VS2. Use the relay with set voltage 25175 V with 25-V taps. Choose VS 5 170 V: RV burden 5 1 VA Also choose CT knee point at least 2VS VK min
5 2 3 170 V 5 340 V
Choose VK 5 400 V for CT, when another relay (current relay) R1 is used with the voltage relay in parallel and given that the burden of current relay is 600 Ω and its current is 40 mA. Then the voltage of the current relay (R1) is 600 3 0.040 5 24 V. Then the series setting resistor of the current relay R1 can be calculated as follows (as shown in Fig. 19.5.8): ðRS ÞSeries resistor ofRI 5
ð170 2 24Þ 5 3650Ω 0:04
The relay circuit consists of the following relays: RV 5 voltage-operated relay; RI 5 current-operated relay; RS 5 series setting resistor of RI; Rshunt 5 shunt resistor.
FIGURE 19.5.8 High-impedance differential relay circuit.
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FIGURE 19.5.9 Current transformer magnetizing curve.
Metrosil resistance (nonlinear resistor) We then check whether we need to use a shunt resistor for the relay or if this is not required as follows. From the CT magnitude curve the value of Imagnitising (Im) can be checked at a setting voltage VS for RV at VS 5 175 V. Im will be 40 MA, as shown in Fig. 19.5.9. The total operating current of the relay 5 ICT HV side mag 1 ICT low voltage side mag 1 IRv 1 IRI I RV 5
VA 1 5 5 5:88 mA V 170
IRI 5 40 mA ICT HV 5 40 mA, ICT LV 5 40 mA Total operating current 5 40 1 40 1 5.88 1 40 5 125.88 mA I set
5 0:2 I N HV side 500 3 103 3 1 5 0:192 5 192:45 mA 5 0:2 3 pffiffiffi 600 3 3 500
Then a shunt resistor is required to increase the total operating current of the relay to 192.45 mA: Current of shunt resistor 5 192.45125.88 5 66.57 mA. Shunt resistor 5
170V 5 2552:55Ω 0:0666
Total operating current for setting value
5 I CT HV mag 1 I CT LV mag 1 I RV 1 I RI 1 I R shunt 5 40 1 40 1 5:88 1 40 1 66:57 5 192:45 mA
Based on the available Metrosil resistance in the market, the following equation is used: V metrosil 5 ðI metrosil Þβ :C
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where β 5 0.2 C 5 450 I 5 30 A Vmetrosil 5 (30)002 . 450 5 888 V So, for CT with 1 A secondary use metrosil with C 5 450, β 5 0.2 Imetrosil will be 30 A for 2 seconds Vmetrosil 5 888 V
19.5.3.2 Worked Example 2 (Fig. 19.5.10) Autotransformer 500 MVA 500/220 kV HV side CT1 600/1 RCT1 5 1.5 Ω LV side CT2 1600/1 RCT2 5 2.5 Ω Neutral CT3 1000/1 RCT3 5 2.5 Ω Relay stability limit 5 16 In 5 Ithrough fault V s 5 stability voltage 5 V S1 HV side
I through fault 3 loop resistance CT
500 3 103 5 16 3 pffiffiffi ðR1 1 R2 1 RCT1 Þ 3 3 500 5
16 500 3 103 3 pffiffiffi ð2 1 2 1 1:5Þ 600 3 3 500
5 15:396 ð5:5Þ 5 84:678 V
FIGURE 19.5.10 500/220 kV autotransformer 500 MVA protected by high-impedance differential protection.
Protection Relays Settings Chapter | 19
V S2 LV side
5
305
16 3 500 3 103 pffiffiffi ðR3 1 R4 1 RCT2 Þ 1000 3 3 3 220
5 20:994 ð3 1 3 1 2:5Þ 5 20:994 ð8:5Þ 5 178:449 V For an autotransformer: INeutral5ILV sideIHV side V S neutral side
5 ð20:994 15:396ÞðR5 1 R6 1 RCT3 Þ 5 5:598 ð1:5 1 1:5 1 2:5Þ 5 5:598ð5:5Þ 5 30:789 V
Then, VS2 . VS1 . VS3. Choose Vs2 as VS 5 180 V. Use voltage relay 20200 V with 20-V taps. Set VS 5 180 V, burden 5 1 VA. Then VK min should be 2VS. VK min 5 2 3 180 5 360 V. Use CT with VK 5 400 V ,a current relay with burden 5 500 Ω, and current 5 30 mA. The voltage across the current relay RI will be as follows: V 5 500 3 0.03 5 15 V Then a series resistor is required to adjust the setting of the current relay RI calculated as follows (as shown in Fig. 19.5.11): RS ðsetting series resistorÞ 5 RS 5 5500 Ω Check if shunt resistance is needed or not.
FIGURE 19.5.11 Relay circuit.
180 2 15 0:03
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I set
5 0:2 I N HV side 500 3 103 1 3 5 0:2 3 pffiffiffi 600 3 3 500 5 0:1924 A 5 192:24 mA
From the CT magnitude curve check the value of Im at VS 5 180 V-Imag 5 45 mA. The total operating current of the relay 5 IRv 1 IRI 1 ICT HV side mag 1 ICT LV side mag 1 IR CT neutral mag and total operating current 5 Itotal. I RV 5
VA 1 5 5 5:55 mA V 180 I RI 5 30 mA
I total 5 5:55 1 30 1 45 1 45 1 45 5 170:55 mA Then we need a shunt resistor calculated as follows: I of R shunt 5 192.24170.55 5 21.69 mA Rshunt 5 I total
180 5 8219:17 Ω 0:02169
5 I RV 1 I RI 1 I CT HV mag 1 I CT LV mag 1 I CT neutral mag 1 I R shunt 5 5:55 1 30 1 45 1 45 1 45 1 21:69 5 192:24 mA
Subchapter 19.6
Generator Protection Setting 19.6.1 INTRODUCTION As generators are a very costly component of the power system they should have a good protection system. The complexity of this protection system varies depending on the size of the units. The settings of these protections are set at different values depending on the utilities setting polices. Here we provide some examples of generator protection settings.
19.6.2 WORKED EXAMPLES Worked Example 1 (as shown in Fig. 19.6.1) Power station data: Generator unit (G)
Protection Relays Settings Chapter | 19
FIGURE 19.6.1 Generator and transformer data.
333 MW xd 5 191% xd0 5 33% Xdv 5 22% F 5 50 Hz I2 5 15% I2 .t 5 10 S CT 5 13,000/122 kV Power Factor (PF) 5 0.8 S 5 417 MVA Unit transformer (T2) 25/15/15 MVA 22/6.4/6.4 kV Z% 5 8% at 15 MVA Generator transformer (T1) 420 MVA 225/22 kV Z% 5 14% Station transformer (T3) 25/15/15 MVA 220/6.4/6.4 kV Z% 5 8.2% at 15 MVA CTs 5 13,000/1 1. Stator earth fault protection 95% stator earth fault relay (R1) Maximum fault current in neutral 5 20 A Iset 5 (10.95) 20.15 5 0.2 A t to be set to 5 0.05 second. Tripping action: 1. 2. 3. 4.
HV CB of transformer T1; T2 6.4 kV incoming feeder tripping; Turbine tripping; Generator excitation tripping.
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2. 100% stator earth fault protection R2 and R3 will be two relays as follows: 1. Undervoltage relay 27 works on third harmonic (R2). 2. Overvoltage relay (R3) works on fundamental frequency. The overlap between 27 and 59 covers 100% of the stator winding. 3. Stator OC protection Iset 5 1.05 IN P 5 333 MW then S 5 417 MVA at a power factor of 0.8. IN 5
417 3 103 5 10; 943 A O3 3 22
I N secondary
5 10; 943 3
I set
1:05 I N 1:05 3 0:842 5 0:884 A SetI set
1 5 0:842 A 13; 000
5 0:9 A t 5 6 seconds
Action: trip of CB of transformer (T1) 220 kV side. 4. Negative-phase sequence current I2 protection (46) Second stage (trip): I set I set Set I set
5 0:15 I N 5 0:15 3 0:842 5 0:1263 A 5 0:127 A 5 12%
I22.t 5 10 where K 5 I22.t, K factor 5 10. First stage (alarm): Set I alarm 5 0:9 I set
5 0:9 3 0:12 5 0:108 5 10:8%
Set talarm 5 3.0 seconds Action: trip 220 kV side CB of T1 second stage. Alarm in first stage. 5. Loss of excitation protection Loss of excitation leads to loss of synchronism and running at higher than synchronous speed. The following impedance relay is used (Fig. 19.6.2). X1 1 x2 5 xd X1 5 xdx2 xd’ 5 33% on 417 MVA base.
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FIGURE 19.6.2 Loss of excitation impedance protection for generators.
X2 5
xd 0 2
Xd 5 191% on 417 MVAb. Z ðΩÞ 5 Xd 0 ðΩÞ 5 X2 5
Z% MVAb:100 ðVLÞ2
33 3 417:100 5 28:43 Ω 100ð22Þ2
xd 0 28:43 5 14:21 Ω 5 2 2
X1 5 xdx2 Xd ðΩÞ 5
191:417 3 100 5 164 Ω 100:ð22Þ2
X1 5 16414.21 5 150.35 Ω X1 5 150.35 Ω primary, secondary x1 5 887.065 Ω x2 5 14.21 Ω primary, secondary x2 5 83.839 Ω t 5 1 second; Z s 5 Z P VT ratio 5
NV 5 5:9 Z P NC
220; 000=O3 5 2200 5 NV 100=O3
CT ratio 5 13,000/1 5 NC 6. Underimpedance protection This relay is used as a backup protection for external faults on the highvoltage side (220 kV) of a transformer (T1), as shown in Fig. 19.6.3.
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FIGURE 19.6.3 Backup impedance protection for generators.
Z set 5 0:7 3 0:36 5 0:252- 5 25:2% Z set ðΩÞ
5
25:2 3 417 3 100 100ð220Þ2
5 0:217 Ω ðprimaryÞ Z set ðΩÞ secondary Z set
5 5:9 Z primary 5 5:9 3 0:217 5 1:2803 Ω
Z set 5 0:7 Z Load Z Load min 5 XG 1 XT 5 Xd} 1 XT 5 0:22 1 o:14 5 0:36 Set time 5 3 seconds. Action: 1. Turbine tripping; 2. Tripping of 6.4 K unit transformer incoming; 3. Generator excitation tripping. 7. Generator frequency protection A. Underfrequency protection This protection is required for load-shedding to allow unit frequency recovery.
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Relay set 5 47 Hz F 5 50 Hz tset 5 4 seconds Action: 1. Trip 220 kV CB. This protection works when load-shedding of the network does not work. 8.Overfrequency protection Under this condition the unit overspeed and trip are set as follows. Relay set 5 54 Hz, F 5 50 Hz Action: 1. Trip of 220 kV CB. 8. Undervoltage stator protection This protection works as a backup for an HV network in two stages. Stage 1 has a definite time uv set to 0.7 Vn of the generator and trips the HV CB. Stage 2 trips the unit after 2.5 seconds. Stage 1: V set 5 0:7V n 5 0:7 3 100 5 70 V tset 5 1.2 second Action trip CB of 220 kV transformer (T1). Stage 2: Vset 5 70 V tset 5 2.5 seconds Action: 1. Turbine tripping; 2. Tripping of 6.4 kV transformer incoming; 3. Generator excitation tripping. 9. Overvoltage protection This protection works when the Automatic Voltage Regulator (AVR) maloperates or for manual operation of the excitation system. First stage V set
5 1:1 V n 5 1:1 3 100 5 110 V
t 5 4.5 seconds Action: alarm Second stage V set 5 1:2 V n 5 120 V tset 5 3 seconds
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Action: 1. 2. 3. 4.
HV CB tripping of T1; Incoming of transformer T2 6.4 kV trip; Turbine tripping; Generator excitation tripping.
10. Generator differential protection A biased differential relay is used in the following circumstances (Fig. 19.6.4)
FIGURE 19.6.4 Biased differential protection for generators.
(IN 5 generator nominal current): P 5 O3 VL. IN. cos [ IN
P 333 3 103 5 pffiffiffi 5 pffiffiffi 3V L :cos φ 3V L :cos φ
333 3 103 5 10; 924 A 5 pffiffiffi 3 3 22 3 0:8 10; 924 5 0:84 A I N secondary 5 13; 000 I set 5 0:1 I N 5 0:1 3 0:84 5 0:084 A Set bias stop 5 20% 11. Reverse power protection PReverse set 5 0.01 3 333 5 3.33 MW
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Pset I set on Relay 5 pffiffiffi 3V L :cos φ 3:33 3 103 5 pffiffiffi 3 3 22 3 0:8 5 109:237 A 109:237 I set secondary 5 13000 5 8:4 mA tset 5 15 seconds Action: 1. Trip HV CB 220 kV of T1; 2. Generator excitation trip; 3. Incoming 6.6 kV of T2 trip.
Subchapter 19.7
Switchgear (Busbar) Protection Settings 19.7.1 INTRODUCTION The busbars in substations are a very important part of the system and should have adequate protection, however this protection is very expensive in 11 kV and 22 kV, and the faults in busbars are rare, therefore it is used in 66 kV systems and above as essential protection. The most commonly used scheme for busbar protection is highimpedance differential protection.
19.7.2 WORKED EXAMPLES FOR BUSBAR PROTECTION One hundred and thirty-two kiloVolts busbars have two sections and a bus coupler CB, each busbar has one feeder and an incomer transformer (as shown in Fig. 19.7.1). CT ratios: CT1, CT2, CT3, CT4, CT7, CT8, CT9, CT10 -1200/1 RCT 5 2.9 Ω Imag 5 40 mA at V 5 130 V Imag 5 35 mA at V 5 125 V CT ratios: CT5, CT6-1500/1 RCT 5 3.1 Ω Imag 5 45 mA at V 5 130 V Lead resistances: R1 5 0.3 Ω R4 5 0.32 Ω R7 5 0.33 Ω
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FIGURE 19.7.1 Two-section busbar with two feeders and two incomers.
R2 5 0.35 Ω R5 5 0.36 Ω R8 5 0.31 Ω R3 5 0.4 Ω R6 5 0.48 Ω R9 5 0.33 Ω R10 5 0.39 Ω We use three relays as follows: (M1) Discriminating main 1 relay to protect busbar section A1; (M2) Discriminating main 2 relay to protect busbar section A2. Check zone relay (CH) to protect busbar A1 and A2 to confirm that the fault exists on the busbars. IF Relay setting voltage 5 CT Rloop max Rloop max 5 maximum lead resistance Assuming saturation in one CT: M1 (main relay setting) Rlead 1
5 2 3 R1 1 RCT1 5 2 3 0:3 1 2:9 5 3:5 Ω
Rlead 2
5 2 3 R2 1 RCT3 5 2 3 0:35 1 2:9 5 3:6 Ω
Rlead 3
5 2 3 R3 1 RCT6 5 2 3 0:4 1 3:1 5 3:9 Ω
Rlead max 5 Rlead 3 5 3.9 Ω Maximum through fault current 5 40 kA
Protection Relays Settings Chapter | 19
VS
315
IF : Rloop max CT 40; 000 3:9 5 130 V 5 1200 5
Choose voltage relay RV as follows: Range 25175 V IRV 5 35 mA Set VS 5 130 V Then the knee point voltage of CT should be as follows: V K min
5 2 3 VS 5 2 3 130 5 260 V
Choose VK 5 300 V Use a current relay with burden 5 3500 Ω I 5 30 mA Then the voltage across the current relay RI is as follows: 3500 3 0.03 5 105 V. Then, the series resistor is required to adjust the setting of the current relay RI and is calculated as follows (as shown in Fig. 19.7.2): RS ðSetting ResistorÞ 5
130 2 105 5 833 Ω 0:03
M1 relay circuit—busbar protection We then need to check whether we need shunt resistance or not: Iset 5 0.15 Ifault min Assume the single-phase minimum fault 5 12 kA, and the three-phase minimum fault 5 14kA: I set 5 0:15 3 12 5 1:8 kA I set secondary 5
1800 5 1:5 A 1200
Itotal operating current 5 IRV 1 IRI 1 ICT1 IRV 5 35 mA IRI 5 30 mA
FIGURE 19.7.2 Relay circuit.
mag 1 ICT3 mag 1 ICT6 mag
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FIGURE 19.7.3 Current transformer magnetization curve.
By checking from the magnetization curve of CT, the magnetization current at VS 5 130 V (Fig. 19.7.3). Itotal 5 35 1 30 1 40 1 40 1 45 1 IR shunt 5 190 mA 1 IR shunt Itotal 5 Iset 5 1.5 A 5 1500 mA IR shunt 5 1500190 5 1310 mA Rshunt 5
VS 130 5 99:23 Ω 5 I R shunt 1310 =1000
VS 5 130 V Choose Metrosil as follows: Vmetrosil 5 (Imetrosil)β C Choose C 5 450 β 5 0.25 I 5 30 A V metrosil
25
5 ð30Þ00 3 450 5 1053 V
Imetrosil will be 30 A for 2 seconds. Main 2 relay setting M2 Rlead1
Rlead
5 2 3 R6 1 RCT9 5 2 3 0:48 1 2:9 5 3:86 Ω
Rlead 2
5 2 3 R5 1 RCT7 5 2 3 0:36 1 2:9 5 3:62 Ω
Rlead3
5 2 3 R4 1 RCT5 5 2 3 0:32 1 3:1 5 3:74 Ω
max 5 Rlead 1 5 3.86
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IF Rloop max CT 40; 000 3:86 5 128:66 V 5 1200
VS 5
5
VS 5 130 V Choose relay RV 5 130 V, IRV 5 35 mA Range: 25175 V Knee point of CT is as follows: V K min 5 2V S 5 2 3 130 5 260 V Choose CT with Vk 5 300 V Use a current relay with burden 3500 Ω, I 5 30 mA. As before in the M1 calculation: Rsh 5 99.23 Ω Vmetrosil 5 1053 V Rseries Resistor (RS) 5 833 Ω C 5 450, β 5 0.25 Imetrosil 5 30A for 2 seconds Then check the zone relay CH calculation: Rlead 1
5 2 3 R7 1 RCT2 0:33 3 2 1 2:9 5 3:56 Ω
Rlead 2
5 2 3 R8 1 RCT4 2 3 0:31 1 2:9 5 3:52 Ω
Rlead 3
5 2 3 R9 1 RCT8 5 2 3 0:33 1 2:9 5 3:56 Ω 5 2 3 R10 1 RCT10 5 2 3 0:39 1 2:9 5 3:68
Rlead 4 Rlead
max 5 Rlead 4 5 3.68
VS
IF RL max CT 40; 000 3:68 5 1200 5 122:66 5
Set VS 5 125 V Choose the relay range: 25175 V Vset 5 125 V IRV 5 35 mA The knee point voltage of CT should be as follows: V K min 5 2 3 V S 5 2 3 125 5 250 V Choose VK 5 300 V of CT using a current relay with burden 5 3500 Ω
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FIGURE 19.7.4 Check the zone (CH) relay circuit.
I 5 30 mA Then, the voltage across the current relay RI will be as follows: 5 3500 3 0:03 5 105V The series setting resistor RS is calculated as follows (as shown in Fig. 19.7.4): RS 5
125 2 105 5 666 Ω 0:03
Then check whether we need the shunt resistance or not: I set
5 0:15 I fault min 5 0:15 3 12 5 1:8 KA
I set secondary 5
1:8 3 103 5 1:5 A 1200
Itotal operation 5 IRV 1 IRI 1 ICT2 mag 1 ICT4 mag 1 ICT8 mag 1 ICT10 mag 1 IR shunt Itotal 5 35 1 30 1 35 1 35 1 35 1 35 5 205 mA Imag 5 40 mA for CT2, CT4, CT8, CT10 from VImag. The curve of CT at VS 5 125 V Imag 5 35 mA Itotal 5 205 1 IR shunt I total 5 I set 5 1:5 3 103 5 1500 mA IR
shunt
5 1500 205 5 1295 mA Rshunt 5
VS 125 5 IRshunt 1295=1000
Rshunt 5 96.525 Ω Choose metrosil resistance with voltage: Vmetrosil 5 (Imetrosil)β C C 5 450, β 5 0.25, I 5 30 A Vmetrosil 5 (30)0.25 450 5 1053 V Relay setting
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Main 1 zone (discriminating zone)—M1 RV Vrange 25175 V, IRV 5 35 mA, Vset 5 130 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 833 Ω Rshunt 5 99.23 Ω, Metrosil C 5 450 β 5 0.25 I 5 30 A at 2 seconds. Main 2 zone (discriminating zone)—M2 RV Vrange 5 25175 V, IRV 5 35 mA, Vset 5 130 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 833 Ω Rshunt 5 99.23 Ω, Metrosil C 5 450, β 5 0.25, I 5 30 A at 2 seconds. Check zone (CH) RV Vrange 5 25175 V, IRV 5 35 mA, Vset 5 125 V RI burden 5 3500 Ω, IRV 5 30 mA, RS 5 666 Ω Rshunt 5 96.525 Ω, Metrosil C 5 450, β 5 0.25, I 5 30 A at 2 seconds.
19.7.3 INTRODUCTION TO BREAKER FAILURE PROTECTION This protection is used as backup protection to isolate a fault if the circuit breaker of the feeder has a mechanical or other problem preventing it from opening the circuit. This protection then isolates all sources which feed the fault and also the remote end of the line.
19.7.4 WORKED EXAMPLE FOR BREAKER FAILURE PROTECTION CB failure time tCB 5 110180 ms Fault clearance time should not exceed 300 ms Worked example (Figs. 19.7.5 and 19.7.6): Breaker Failure (BF) relay trip logic BF relay operating times Refer to Fig. 19.7.7 for the breaker failure time and the total fault clearance time. Iset BF 5 0.5 IN
FIGURE 19.7.5 Breaker failure relay connection on an overhead transmission line.
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Practical Power System and Protective Relays Commissioning
FIGURE 19.7.6 Breaker failure relay operation logic.
FIGURE 19.7.7 Breaker failure time and total fault clearance time.
5 0.5 3 1200 5 600 A tCB 5 120 ms Action: The back trip for all feeders connected on busbar A through busbar protection trip relays for the section which this feeder connected to it should not exceed 300 ms. The remote trip for the breaker is at end B. The local retrip for the breaker is at end A.
Subchapter 19.8
Motor Protection Setting 19.8.1 INTRODUCTION Here we provide some examples for motor protection settings.
19.8.2 WORKED EXAMPLES 19.8.2.1 Motor Faults 1. 2. 3. 4.
Winding faults Overloading; Reduced or loss of supply voltage; Phase reversal;
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5. Phase unbalance; 6. Out-of-step operation for synchronous motors; 7. Loss of excitation for synchronous motors.
19.8.2.2 Phase Fault Protection A high set Over Current (OC) unit is used. Differential protection is also used as it is not affected by the motor starting current. For the H set OC relay the setting should be more than the motor starting current: Iset . Istart.
19.8.2.3 Earth Fault Protection An inverse or very inverse OC relay is connected to the residual CT circuit (as shown in Fig. 19.8.1).
FIGURE 19.8.1 Earth fault protection for motors.
Iset 5 0.25 IF
min,
where IF
min 5 minimum
fault current.
19.8.2.4 Locked Rotor Protection We use the OC relay with characteristics higher than the starting current of the motor and less than the locked rotor current (as shown in Fig. 19.8.2), where ts 5 motor starting time; tR 5 locked rotor time. Also, the OC (51) relay characteristic is used to detect the locked rotor condition of the motor or impedance relay is used to detect this condition, as shown below in Fig. 19.8.3.
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Practical Power System and Protective Relays Commissioning
FIGURE 19.8.2 Locked rotor protection for motors.
FIGURE 19.8.3 Impedance relay for locked rotor condition.
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19.8.2.5 Overload Protection Thermal replica relays are used to protect the motor from overloads using a separate relay in each phase of the motor. Thermal relays are also used with thermal replica relays (resistance temperature detectors embedded in machine winding) to give complete protection for the motor.
19.8.2.6 Low-Voltage Protection Undervoltage time delay relay is used as follows: Vset 5 0.5 V nominal Action: Trip after a few seconds.
19.8.2.7 Phase Rotation Protection A negative-phase sequence voltage relay is used. v2 set 5 10% V2
19.8.2.8 Phase Unbalance Protection Detect I2 and set as follows: Iset 5 0.15 IN To protect the motor against single-phasing condition: I22 t 5 40 in general
19.8.2.9 Out-of-Step Protection for Synchronous Motors A relay is used to detect the condition in which the motor draws a heavy current in a low-power factor. Therefore, this relay will be a power factor relay and will issue a signal to disconnect the field winding and let the motor run as an induction motor.
19.8.2.10 Loss of Excitation Protection for Synchronous Motors Undercurrent relay is inserted in a field circuit.