Current transformer selection for Allen-Bradley series overcurrent and differential relays

Transcription

1 pplicatti ion otte Current transformer selection for llen-bradley series overcurrent and differential relays ron core current transformers (CT) are accurate in amplitude and phase when used near their nominal values. t very low and at very high currents they are far from ideal. For overcurrent and differential protection, the actual performance of CTs at high currents must be checed to ensure correct function of the protection relay. 1. CT classification according EC , 1996 CT model Figure 1 CT equivalent circuit. Lm is the saturable magnetisation inductance, L is secondary of an ideal current transformer, RCT is resistance of the CT secondary winding, RW is resistance of wiring and RL is the burden i.e. the protection relay. Composite error C ron core current transformers (CT) are accurate in amplitude and phase when used near their nominal values. t very low and at very high currents they are far from ideal. For overcurrent and differential protection, the actual performance of CTs at high currents must be checed to ensure correct function of the protection relay. 1 T ( K is ip ) dt T 0 C 100% (eq. 1) P T = Cycle time K = Rated transformation ratio Primary/secondary is = instantaneous secondary current ip = instantaneous primary current P = Rms value of primary current

2 pplication ote ote: ll current based protection functions of llen-bradley relays, except arc protection, thermal protection and nd harmonic blocing functions, are using the base frequency component of the measured current. The EC formulae include an RMS value of the current. That is why the composite error defined by EC is not ideal for llen-bradley relays. However the difference is not big enough to prevent rough estimation. Standard accuracy classes t rated frequency and with rated burden connected, the amplitude error and phase error and composite error of a CT shall not exceed the values given in the following table. ccuracy class mplitude error at rated primary current (%) Phase displacement at rated primary current ( ) Composite error C at rated accuracy limit primary current (%) 5P P 3-10 Maring: The accuracy class of a CT is written after the rated power. E.g. 10 V 5P10, 15 V 10P10, 30 V 5P0 ccuracy limit current L Current transformers for protection must retain a reasonable accuracy up to the largest relevant fault current. Rated accuracy limit current is the value of primary current up to which the CT will comply with the requirements for composite error C. ccuracy limit factor The ratio of the accuracy limit current to the rated primary current. = L/. (eq. ) The standard accuracy limit factors are 5, 10, 15, 0 and 30. Maring: ccuracy limit factor is written after the accuracy class. E.g. 10 V 5P10, 15 V 10P10, 30 V 5P0. Si S (eq. 3) S S i Figure This figure of equation 3 shows that it is essential to now the winding resistance RCT of the CT if the load is much less than the nominal. 10 V 5P10 CT with 5% load gives actual values from when the winding resistance varies from 0.5 to = ccuracy limit factor at rated current and rated burden Si = nternal secondary burden. (Winding resistance RCT in Figure 1 S = Rated burden of the CT S = ctual burden including wiring and the load. The actual accuracy limit factor depends on the actual burden. (Figure )

3 pplication ote 3 f the current is an asymmetric short circuit current lie in Figure 3 the needed accuracy limit factor should be multiplied by coefficient DC: 1 (eq. 4) DC = ngle frequency f = Time constant of the short circuit current 15 Current () Time (s) Figure 3 symmetric short circuit current with time constant = 50ms. CT requirements for differential protection When the through current equals and exceeds x there may be enough secondary differential current to trip a relay although there is no inzone fault. This is because the CTs are unique and they do not behave equally when saturating. To avoid false tripping caused by heavy through faults the actual accuracy limit factor of the CTs should obey equation: Tra c DC (eq. 5) CT c = Safety factor. See Table 1. = Maximum through fault short circuit current Tra = Rated current of the transformer CT = Rated primary current of the CT DC = Extra coefficient for decaying dc component according equation 4.

4 pplication ote 4 Table 1 Safety factor c for accuracy limit factor. Using slightly smaller safety factor than indicated in the table will increase the setting inaccuracy. Protection application Safety factor c Overcurrent 1.4 Transformer differential, -winding or unearthed Y-winding 3 Transformer differential, earthed Y-winding 4 Generator differential 3 Formula to solve needed CT power rating S By replacing the complex power terms with corresponding resistances in equation 3 we get R R CT (eq. 6) RCT RW RL where the nominal burden resistance is S R (eq. 7) CT sec RCT = Winding resistance (See Figure 1) RW = Wiring resistance (from CT to the relay and bac) RL = Resistance of the protection relay input S = ominal burden of the CT CTsec = ominal secondary current of the CT By solving S and substituting according equation 5, we get S c R R R R DC Tra CT W L CT CT sec (eq. 8) CT Example 1 Transformer: 16 MV Yd11 Z = 10% 110 V / 1 V (84 / 440 ) CTs on HV side: 100/1 5P0 15 V Winding resistance RCT = 0.3 (RCT depends on the CT type, CT and power rating. Let's say that the selected CT type, 100 and an initial guess of 15 V, yields to 0.3.) Safety factor c = 4. (Table1. Transformer differential, earthed Y.) CTs on LV side: 500/1 5P0 15 V Winding resistance RCT = 0.4 (RCT depends on the CT type, CT and power rating. Let's say that the selected CT type, 500 and an initial guess of 15 V, yields to 0.4.) Safety factor c = 3. (Table 1. Transformer differential,.) Maximum through fault short circuit current =10 x

5 pplication ote 5 RL = 0.05 Typical burden of an llen-bradley relay 1 current input. RWHV = Wiring impedance of high voltage side. (x16 m Cu, 4 mm ) RWLV = Wiring impedance of low voltage side. (x10 m Cu, 4 mm ) f = 50 Hz Frequency = 50 ms DC time constant Equation 4 gives: DC = x 0.05 = 16.7 The needed CT power on HV side will be (eq. 8) S V 15 V is a good choice for HV side! nd on the LV side S V 15 V is a good choice for HV side! 3. CT requirements for overcurrent protection Undirectional overcurrent protection does not set as high requirements for a CT as the differential protection. The nominal primary current should be enough for the maximum short circuit current according equation: CTpri (eq. 9) 100 CTpri = ominal primary current of the CT = Maximum short circuit current The needed minimum value for the actual accuracy limit factor (equation 3.) depends on the highest setting value, the applied delay type definite/inverse and the needed fault current grading for selectivity. reasonable actual accuracy limit factor for most cases should be according equation 9. c (eq. 10) SET DC c = Safety factor. See Table 1. SET = Relative setting of the most coarse overcurrent stage DC = Extra coefficient for decaying dc component according equation 4.

6 pplication ote 6 The needed power rating for the CT is c DC SET S RCT RW RL RCT CT sec (eq. 11) Example etwor: = 30 Maximum short circuit current RL = Typical burden of an llen-bradley relay 5 current input. RW = 0.09 Secondary wiring impedance CT: 1000/5 10P10 30 V (10P10 10% 10x1000 ) RCT = 0.4 Secondary winding resistance Settings of the most coarse overcurrent stage: set = 10 x = Delay type = definite time f = 50 Hz Frequency = 50 ms DC time constant ccording equation 9 the CT primary value is o (30/1000 = 30 and 30 is well under 100). Equation 4 gives: DC = x 0.05 = 16.7 ext we chec if the power rating is adequate (equation 11) S V This is an impractical big CT. We must either use a CT with 1 nominal secondary or ignore the DC-component. The latter choice may increase trip time at heavy faults. Let's ignore the DC effect S V 10 V is enough, but only if we ignore the decaying DC component.

7 pplication ote 7 4. Maximum allowed wiring distance between CT and a relay From equation 11 we can solve the maximum possible wiring resistance: S RW RCT RCT RL max (eq. 1) CT sec c SET This resistance corresponds to a wire length of L R max (eq. 13), where Lmax = Maximum wire length R = Wiring resistance = Cross-sectional area of the wire = unit resistance of the wire The corresponding distance will be half of the wire length, because there are two wires from the CT to the relay. Max. distance = Lmax/ (eq. 14) Example 3 Let us calculate the maximum possible distance between CT and protection relay with in the following case. CT = 500/5 10P10 FL = 10 ccuracy limit factor at rated current and rated burden according CT specification. CTsec = 5 ominal secondary current of the CT S = 15 V Rated burden of the CT RL = Burden of an llen-bradley relay 5 current input RCT = 0.6 Secondary winding resistance c = 1.4 Safety factor. See Table 1. SET = 8x Overcurrent setting Wire =.5 mm Cu Cross-sectional area and material Cu = 17. nm Unit resistance of copper DC = 1 This ignores any decaying DC component From equation 11 we get the maximum allowed wiring resistance R W max and from equation 13 we get the corresponding wire length Lmax 669m Thus the maximum possible distance will be according equation 14 Distancemax = 669/ = 330 m

8 pplication ote 8 Keywords : Current transformer, Current selection, Current requirements, Current accuracy class, Current accuracy limit, Current power rating, Current wiring Medium Voltage Products, 135 Dundas Street, Cambridge, O, 1R 5X1 Canada, Tel: (1) , Fax: (1) , Publication 857-P019-E-P June 011 Copyright 011 Rocwell utomation, nc. ll rights reserved. Printed in Canada.