Chapter 3: Protection of Transmission Lines by Distance Relays

Example 8.1 Obtain 3 zone settings for (i) a reactance relay, and (ii) a mho relay of 60° MTA from the following data: CT 400/1 A PT 132 kV/110 V Impedance for the first section is 2.5 + j 5.0 ohms (primary) and that for the second section is 3.5 + j 7.0 ohms (primary). The first zone covers 80% of the first section, the second zone covers the first section plus 30% of the second section and the third zone covers the first section plus 120% of the second section. Solution Impedance of first line section

= 2.5 + j 5 (primary)

Impedance of second section

= 5.59 –63.43 ohms = 3.5 + j 7 (primary)

Impedance of first zone Z1

= 0.8 × 5.59 –63.43 ohms

Impedance of second zone Z2

= 5.59 –63.43 + 0.3 × 7.82 –63.43 ohms

= 7.82 –63.43 ohms = 4.472 –63.43 ohms (primary) = 7.936 –63.43 ohms (primary)

Z3

3

Z2

2

1

K3

K2

Z1

K1

63.43 R

Fig. 8.29

Example 8.1

Impedance of third zone Z3

= 5.59 –63.43 + 1.2 × 7.82 –63.43 ohms

= 14.97 –63.43 ohms (primary) These impedances are to be transferred to the relay side (i.e., secondary side of CT and PT) because the relay receives transformed juices of current and voltage. Z1 (secondary) = (CTR/PTR) × Z1 (primary) 400/1 = __________ × 4.472 –63.43 132000/110 = 1.49 –63.43 ohms Similarly, Z2 (secondary) = 2.645 –63.43 ohms Z3 (secondary) = 5 –63.43 ohms As the reactance relay measures the reactance up to the fault (refer Fig. 8.29), 1 = Z1 sin (63.43 ) = 1.49 sin (63.43 ) = 1.33 ohms 2 = 2.36 ohms 3 = 4.47 ohms Now, for the mho relay (refer Fig. 8.29), Z1 K1 = _________ cos (f – q) 1.49 = _______________ = 1.49 ohms cos (63.43 – 60 ) Similarly, K2 = 2.649 ohms K3 = 5 ohms

Example 8.2 A 220 kV long transmission line has an impedance of 2 + j 8 ohms. Suggest suitable distance relays for its protection and determine the settings of the relays for all the three zones given that (i) Zone 1 covers 80% of the line length (ii) Zone 2 covers 150% of the line length (iii) Zone 3 covers 225% of the line length Assume (a) a fault resistance of 2 ohms while deciding settings, and (b) a suitable characteristic angle of the distance relay suggested by you. Relevant data: CT ratio = 1000/1 A PT ratio = 220 kV/110 V Solution As the line to be protected is a long one, mho relays are most suitable. The characteristic angle is taken as 70 , as normally the faulted power factor angle is between 70 , and 90 . Refer Fig. 8.30. Impedance of line = 2 + j 8 ohms = 8.246 – 75.96 ohms Z1 = 0.8 × 8.246 –75.96 = 6.597 –75.96 = 1.6 + j6.4 ohms Considering a fault resistance of 2 ohms, Z1 will be modified. Z ¢1 = 2 + j (1.6 + 6.4) = 3.6 + j 6.4 = 7.34 –60.64

K3

Z3

Z ¢3

Z2

K2 Z ¢2

K1 Z1

Z ¢1 70 67.38

60.64 R

Fig. 8.30

Example 8.2 Z ¢1 (secondary) = (CTR/PTR) × Z ¢1 1000/1 = __________ × 7.34 –60.64 = 3.67 –60.64 ohms 220000/110 3.67 K1 = _______________ = 3.72 ohms cos (60.64 – 70 ) Z2 = 1.5 × 8.246 –75.96

= 12.369 –75.96 = 3 + j 12 ohms Considering a fault resistance of 2 ohms, Z ¢2 = 2 + (3 + j 12) = 13 –67.38 ohms Z ¢2 (secondary) = (CTR/PTR) × Z ¢2 = 6.5 –67.38 ohms 6.5 K2 = _______________ = 6.51 ohms cos (67.38 – 70 ) Now,

Z3 = 2.25 × 8.246 –75.96 = 18.5535 –75.96 = 4.5 + j 18 ohms

Considering a fault resistance of 2 ohms, Z 3¢ = 19.14 –70.145 Z 3¢ (secondary) = 9.57 –70.145 9.57 K3 = ________________ = 9.57 ohms cos (70.145 – 70 )

Example 8.3 A three-section radial feeder, shown in Fig. 8.31, is to be protected by distance relays. The relevant data is as follows: Radial Feeder (i) Impedance of Section I 4 + j 16 ohms (primary) (ii) Impedance of Section II 3 + j 12 ohms (primary) (iii) Impedance of Section III 2 + j 8 ohms (primary) (iv) Rated load current of feeder 1000 A at 0.8 power factor lag (v) Probable overloading 200% of the rated current (vi) Probable voltage dip 10% CT ratio 1000/1 A PT ratio 132 kV/110 V Relay R (i) Transient over-reach 10% (ii) Characteristic angle 60° of mho relay Determine the settings of zones 1, 2 and 3 of the distance relay R. Also determine the reach of all the three zones, for the line to be protected in terms of percentage of impedance of the first section. Solution The reach of the zone 3 is restricted by overloading and voltage dip. The load impedance under overload condition and voltage dip of 10% is __

[(132000/÷3 ) × 0.9 ZL = ________________ ohms 2000 –– 36.87 = 34.29 –36.87 ohms

4+j16

3+j12

Load current 7000 A

R

Gen.

132/110 V PT

2+j8

Fig. 8.31(a) Example 8.3

Considering a 5% margin for relay error, ZL¢ = 0.95 × 34.29 –36.87 ohms = 32.57 –36.87 ohms

Z3 K3

1000/1 ZL¢ (secondary) = __________ × 32.57 –36.87 ohms 132000/110 = 27.14 –36.87 ohms Therefore,

Z2

K2

Z ¢L K3 = _______________ = 29.51 ohms cos(60 – 36.87 ) Z3 = K3 cos(f – q)

Z¢L Z1

K1

= 29.51 cos (75.96 – 60 ) = 28.37 ohms Impedance of Section I = 4 + j16 = 16.49 –75.96 Considering 10% over-reach of relay, Z1 = (1/1.1) × 16.49 –75.96 ohms = 15 –75.96 ohms Z1 (secondary) = 12.5 –75.96 ohms

75.96

60 36.87

Z1 K1 = _________ = 13.0 ohms cos(f – q) Fig. 8.31(b) Example 8.3 = 4 + j16 + 3 + j12 = 28.86 –75.96 ohms (primary) = 24.05 –75.96 ohms (secondary) As Z3 is more than this figure the relay will cover sections I and section II as required in the zone 3. Taking impedance of the first section plus 50% of the impedance of the second section for the second zone setting Z2 = 16.49 –75.96 + 0.5 (3 + j12) = 22.67 –75.96 ohms (primary) Z2 (secondary) = 18.89 –75.96 ohms Impedance of sections I and II

Z2 K2 = _________ = 19.647 ohms cos (f – q) Determination of the reach of all the three zones, for the line to be protected in terms of percentage impedance of the first section is left as an exercise for the reader.

R

Example 8.4 Figure 8.32 shows the single-line diagram of a portion of a power system. The relevant data is as follows: Line

Impedance ohms km

Distance km

L1 L2 L3 L4 L5

0.03 + j 0.12 0.04 + j 0.16 0.05 + j 0.15 0.08 + j 0.24 0.10 + j 0.30

150 100 50 32 25

Mho relay R is with characteristic angle = 60°, Zone 1 setting K1 = 14.5 ohms, Zone 2 setting K2 = 160% of K1, Zone 3 setting K3 = 360% of K1, CT ratio of 1000/1 and PT ratio of 132 kV/110 V is given. Find out in terms of distance in km: • Zone 1 reach of R from the switchyard A for the line L1. • Zone 2 reach of R from the switchyard B for the line L2. Also find in which zone of relay R, will the faults of lines L3, L4 and L5 will be cleared? Will the lines L3, L4 and L5 be fully protected by the relay R? Solution ZL1/km = 0.03 + j 0.12 ohm/km = 0.1237 –75.96 /km From the data K1 = 14.5 ohms with a characteristic angle of 60 Z1 = K1 × cos (f – q) in secondary terms Z1 (secondary) = 14.5 × cos (76 – 60 ) = 13.938 ohms Z1 (primary) = Z1(secondary) × PTR/CTR 132/66 kV, 100 MVA, Transformers

B

B

L3

A

L1

% = 14%

L4

R L2

L5

Substation

Fig. 8.32

Example 8.4 132000/110 =13.938 × __________ = 16.726 ohms 1000/1

Thus

ZL1 reach for L1 = Z1(primary)/(Z1/km) = 16.726/0.1237 = 135.2 km from the switchyard A K2 = 160% of K1 = 1.6 × 14.5 = 23.2 ohms Z2 (secondary) = 23.2 × cos 16 = 22.3 ohms Z2 (primary) = 26.76 ohms

Z2 – impedance of the line L1 = Z2 – ZL1 = 26.76 – 150 (0.03 + j 0.12) = 8.205 ohms ZL2 = Impedance of the line L2 = 0.04 + j 0.16 ohms/km = 0.1649 –76 /km Zone 2 reach of the relay R from the substation B for the line L2 = (Z2 – ZL1)/ZL2 /km

= 8.205/0.1649 = 49.75 km

Ohmic value of the reactance of the transformer at the 132 kV base = (1322 / 100) × 0.14 = 24.39 ohms For two transformers in parallel, reactance of transformers, X T = j 12.19 ohms ZL1 + XT = 150 (0.03 + j 0.12) + j 12.19 = 4.5 + j 30.19 ohms = 30.52 –81.52 ohms Hence, the transformer fault will be seen in the second zone.

K3 = 360% of K1 = 3.6 × 14.5 = 52.2 ohms ZL1 +

T

+ ZL3

= 4.5 + j 18 + j 12.19 + (132/66)2 (0.05 + j 0.15) × 50 = 61.91 –76.45 (as the impedance of the line ZL3 seen by the relay R is (132/66)2 times the impedance of the line) Z3 (secondary) = 52.2 cos (76.45 – 60 ) = 50.06 ohms Z3 (primary) = 60.07 ohms As Z3 (ZL1 + XT + ZL3), the line L3 will not be fully covered by the zone 3. However, the portion of line to be protected will be covered in the zone 3 of the relay. Similarly,

(ZL1 + X T + ZL4) = 62.66 –76.39 ohms

Line L4 also will not be covered fully by the relay R. In the same manner, (ZL1 + XT + ZL5) = 61.91 –76.45 Line L5 will also not be covered fully by the relay R.

8.5

LIMITATIONS OF DISTANCE PROTECTION FOR TRANSMISSION LINES