Solution Manual to Chapter 13 of Power System Protection and Switchgear PDF

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this is a solution manual of power system protection and switchgear by oza, nair, mehta and makwana which is helpful for the students in solving the numericals and this is for electrical engineering...

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13

PROTECTIVE CURRENT AND POTENTIAL TRANSFORMERS

1. A 6.6 kV generating station contains 4 generators each rated at 10 MVA and having a reactance of 30%. Each of these is connected to a 6.6/66 kV, 10 MVA transformer with 10% reactance. The units so formed are connected to a busbar, which feed a number of feeders. If the protective current transformers on the feeders have each a transformation ratio of 150/5 amp and if saturation occurs at secondary voltage of 100 volts, determine the maximum permissible impedance burden on each C.T. secondary. 10 MVA 6.6 kV Generators 30%

6.6/66 kV 10 MVA Transformers 10%

150/5

F

Fault current for fault at F is to be calculated. Base MVA is taken as 10 MVA.

0.3

0.3

0.3

0.3

0.1

0.1

0.1

0.1

F (a)

0.1

F (b)

Fig. 13.15

Power System Protection and Switchgear

2

Fault MVA = 10/0.1 = 100 MVA 100 × 103 ________ = 874.77 amp Fault current If = __ ÷3 × 66 Secondary reproduction of this current, 874.77 if = ______ = 29.16 amp 150/5 if Zmax = KPV 29.16 × Zmax = 100 Zmax = 3.43 W 12. A C.T. of ratio 300/5 is installed at a point in a power system at which the fault level is 2400 amp. The knee point voltage of the C.T. is 140 volts. Determine the maximum permissible impedance burden on the secondary and the corresponding VA. If =2400 amp Secondary equivalent of this fault current, 2400 if = _____ = 40 amp 300/5 if Zmax = KPV 40 × Zmax = 140

13.

(a) (b) (c) (i) (ii) (iii)

Zmax = 3.5 W \ Rated burden of C.T., is2 × Zmax = 52 × 3.5 = 87.5 VA A small power plant consists of three 5 MVA, 6.6 kV, 3-phase, 50 Hz alternators with 20% reactance each connected to a common busbar. The busbar supplies a number of feeders, on each of which are installed C.T.s with details as follows: C.T. ratio: 600/5 amp C.T. secondary resistance: 0.8 ohm Burden impedance: 1.2 ohm Neglecting the lead burden and assuming that no generating source exists at the far ends of the feeders, specify: The secondary voltage up to which the C.T.s should not saturate (knee point voltage) Accuracy limit factor Rated burden of the C.T. 5 MVA, 6.6 kV Generators 20%

600/5

F

Fig. 13.16

Electromagnetic Relays

Fault current for fault at F is to be calculated. Base MVA is taken as 5 MVA.

0.2

0.2

(a)

0.2

F

0.067

F (b)

Fig. 13.17 Fault MVA = 5/0.067 = 74.63 MVA 74.63 103 __ = 6528.43 amp Fault current If = ________ ÷3 × 6.6 Secondary reproduction of this current, 6528.43 _______ = 54.4 amp if = 600/5 KPV = if Z = 54.4 × (1.2 + 0.8) = 108.8 volts Fault current ALF = _______________ C.T. primary rating 6528.43 _______ 600 = 10.88 i.e. 15 Rated burden of C.T., = is 2 Z = 25 × 2 = 50 VA Maximum load current and fault current on a feeder are 87 and 1290 amp, respectively. Specify: C.T. ratio Accuracy limit factor Accuracy class, and Rated short time current of a protective C.T. to be installed on the feeder. As rated current is 87 amp, C.T. ratio shall be 100/1 amp. Fault current ALF = _______________ C.T. primary rating =

14. (i) (ii) (iii) (iv)

3

Power System Protection and Switchgear

4

=

15. (i) (ii) (iii)

1290 ____ 100

= 12.9 i.e. 15 Accuracy class selected is 5P or C.T. of 5P15 can be selected. Rated short time current = 1290 amp for 3 seconds. Figure 13.14 shows a single line diagram for protection of a 220/66 kV, 100 MVA transformer. Find out: C.T. ratio of all the CTs Accuracy limit factor of CT 3 Knee point voltage of CT 2 and CT 3 Assume Rct < 5 ohms, Rl < 1.5 ohms 220/66 kV 100 MVA, 3-Phase %Z = 10% CT1

CT2

CT3

51

Fig. 13.14 3 100 ________ __ × 10 = 262.43 amp ÷3 × 220 200 ___ = 874.77 amp Rated secondary current = 262.43 × 66 \ C.T. ratio of CT 1 and CT 2 = 300/1 amp C.T. ratio of CT 3 = 1000/1 amp Considering primary of transformer to be connected to infinite bus, fault current on secondary side, 100 ___ Ifs = 874.77 × = 8747.7 amp 10 \ Fault current on primary side, 66 ___ = 2624.3 amp If p = 8747.7 × 220

Rated primary current =

Fault current ALF of CT 3 = _______________ C.T. primary rating 8747.7 = ______ = 8.747 i.e. 10 1000 KPV of CT for differential protection = 2if (RCT + RL) 2624.3 ______ (5 + 1.5) = 113.7 volts KPV of CT 2 = 2 × 300 8747.7 KPV of CT 3 = 2 × ______ (5 + 1.5) = 113.7 volts 1000...