CT+Selection+requirements

CT REQUIREMENTS FOR GE MULTILIN RELAYS. (R2) Introduction: The operation of protections is influenced by distortion in the measuring quantities that occurs when CTs saturate. Since it is not practically possible to avoid CTs saturation for all fault conditions, measurements must be taken to obtain a certain degree of saturation, which still allows proper protection operation. Different considerations should be taken depending on the application of the protection and also on the setting range available for each specific protection. For high voltage applications and protections operating instantaneously, as distance, line differential, transformer differential or instantaneous over current, the transient response of the CT must be considered. An over dimensioning factor, taking into account the primary system time constant, secondary circuit time constant and minimum time of un-saturated signal required by the relay will be calculated. For medium and low voltage applications and protections operating as definite time over current or inverse time over current, the transient response is usually neglected, due to the lower requirements in terms of operating times. Definitions: The following terms are used in this document: In: Nominal secondary current. Pn: Nominal Accuracy Power of the CT n: Nominal Accuracy limit factor (ALF) of the CT Pr: Power associated to the relay burden, at In. Internal Power consumption of the CT at In. Pct: Rct : Internal secondary CT resistance. n’: Real Accuracy limit factor of the CT, associated to its real burden. nt: Transient Accuracy limit factor of the CT. Ktf : CT Over-dimensioning factor. It is used instead n’ in t: Response time of the protection relay. Tp: Primary time constant. Secondary circuit time constant. Ts: If: Secondary Fault current. Is: Pick up setting in the relay (secondary). Leads resistance. RL: Rp: Protection relay resistance. Rb: (RL + Rp): Total burden Vkp: Voltage at knee-point. Maximum voltage in the linear region of operation of the CT. For a given CT, 20VA 5P20: Accuracy Power: Pn = 20 VA Accuracy Limit Factor: n = 20 Accuracy Class: 5P, meaning that it is a CT for protection applications, with a transformation error lower than 5% for currents lower than 20 times the nominal current. The real accuracy limit factor (n’) describes which is the real limit of the linear region of the CT for a given burden in the CT, different from the nominal burden.

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The formula is as follows: n' = n . (Pn + Pct) / (Pr + Pct) Transient accuracy limit factor takes into account the transient response of the CT and the decaying DC component of the fault current. The Annex 2 shows all the theory that support the use of the different Over-dimensioning factors. In General we can say that the Ktf factors recommended are as follows: -

Transmission lines(*): Ktf = 10 (It can be reduced to 5 in case of L90 relay)

-

Step-up Power Transformers: Ktf = 11

-

Other Power transformers: Ktf = 6

-

Generators: Ktf = 15 (It can be reduced to 10 in case of G60 and SR489-Firmware revision 150.000 and up)

-

Busbars: Ktf = 8. Ktf = 5 for voltages lower than 132 kV.

-

Internal faults with instantaneous trip: Ktf = 3

Note 1: The above criteria is only valid for low impedance relays, because in low impedance, operation in the linear portion of the saturation curve is critical during external faults. During internal faults, the CT´s could saturate and only we need to assure that the relay will receive enough signal level to operate. In general, we need to take care only of the time inverse over current relays that could be affected in their coordination with other over current relays, if the saturation is too high. In this last case, additional setting calculation could be needed or the adoption of other measures as voltage restrain. Additional comments on this subject are given in the Annex 1. Note 2: When the CT saturation is analyzed, it is necessary to establish some limits as the minimum saturation time. In general, there are no fixed rules, but as general approach a minimum saturation time of 0.1 of cycle is the one established and guarantee by the most complex applications as the Low impedance Bus protection. (*) Note 3: In transmission lines, it is considered normal practice to apply a reduction factor (RF) from 0.5 to 0.7 of CT requirements for voltages under 145 kV, as only 60% or less DC is encountered in most of the faults because it is more frequently to have a fault when the voltage is maximum (and the DC current component is minimum), than when the voltage is zero (maximum offset or DC component) and also because the X/R ratio is lower as the nominal voltage drops. This reduction factor must not be generalized for all voltage levels. The reduction factors are recommended to be applied as follows: 66 to 145 kV: 0.7 Under 66 kV: 0.5 Note 4: In the Annex 4 we describe the criteria used to calculate the CT´s for GE high impedance relays.

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Instantaneous or Definite Time Over current Protection: Definite time over current protections operate on a fixed set time for currents greater than the value of the pick up setting, independently on the value of the current. (Instantaneous is considered to be a particular case of Definite Time, with intentional delay set to zero). Definite Time Over current elements are encountered in main and backup protection relays, as well as in breaker failure relays, etc. To evaluate if a given CT is adequate for a definite time over current protection application, n’ must be greater than the maximum current that should be measured by the relay to warranty its operation, divided by the nominal current. If the maximum fault current at the location of the relay is If, and the setting of the relay is Is, then n’ must be greater than Is / In or If / In, whichever is smaller. It is recommended to use a security margin of 1.5 or 2, that is in line with the factor Ktf = 3 using a relay setting of at least 0.5 the minimum short circuit current, which means that: n' > 1.5 . min ( Is / In ; If / In ) If the setting of the relay is unknown at the moment of selecting the CTs, the maximum value settable in the relay must be selected, or a reasonable overestimated setting. NOTE: MODERN DIGITAL RELAYS PROVIDE VERY WIDE SETTING RANGES. Due to this flexibility, taking the maximum value settable in a modern relay may lead to a greater CT than which would correspond to a simpler relay, with lower values settable. This, of course, does not mean that modern digital relays require greater CTs. To directly compute Vkp for this application, the formula is as follows: Ix = min ( Is ; If ) Vkp = 1.5 . Ix . (Rct + RL + Rp) and then, the power of the CT will be as follows: P = [Vkp / (n . In ) – Rct ]. In2 Application to specific GE relays, with a 5P20 CT, In = 1 A: in Internal faults RELAY TYPE MIF F650 DBF

50 SETTING RANGE 0,1 to 30 x In 0,05 to 160 A 0,2 to 2,4 x In

Pn 3.5 VA 3.5 VA 2 VA

Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A). For F650, the value of 30A has also been considered, as a reasonable overestimation of the setting for an over current application (30 x In) In case 160A would be used, the required power would be greater than 20VA. Please note that it is not the F650 relay who requires that high power, but the high current that the user may want to measure. To obtain lower CT requirements use a lower reasonable pickup value.

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Inverse Time Overcurrent Protection: The operating time of an inverse time over current protection is a function of the current value, and then, it depends very much on the correct measuring of the fault current. The associated CT must ensure a linear response over the complete range of the curve, or up to the maximum fault current, whichever is smaller. To evaluate if a given CT is adequate for an inverse time over current protection application, n’ must be greater than: n' > 1.5 . Imaxcurve / In

or

n’ > 1.5 . If / In

where Imaxcurve is the limit value for which the relay starts responding as definite time. (A security margin of 1.5 has been applied) NOTE: MODERN DIGITAL RELAYS PROVIDE VERY WIDE SETTING RANGES. Due to this flexibility, taking the maximum value settable in a modern relay may lead to a greater CT than which would correspond to a simpler relay, with lower values settable. This, of course, does not mean that modern digital relays require greater CTs. To directly compute Vkp for this application, the formula is as follows: Vkp = 1.5 . Imaxcurve . (Rct + RL + Rp) or Vkp = 1.5 . If . (Rct + RL + Rp) and then, taking the minimum of these two voltages, the power of the CT will be as follows: P = [Vkp / (n . In ) – Rct ]. In2 Application to specific GE relays, with a 5P20 CT, In = 1 A: In internal faults RELAY TYPE MIF F650

51 SETTING RANGE (Is) 0,1 to 2,4 x In 0,05 to 160 A

Curve limit (Imaxcurve) 20 x Is 48 x Is

Pn 3,5 VA 3,5 VA

Note: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A). For MIF and F650, a maximum current of 30A has been considered.

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Distance protection: Distance relays are used in applications more critical than over current relays, and must operate very fast, during the transient period (first cycles) of the fault. Due to the nature of this application, the use of the transient over dimensioning factor is required. Two points of operation will be checked: Beginning of the line. Reach limit of Zone 1. For each point, the over dimensioning factor will be computed, and the maximum value will be used to compute the required Vkp and Power of the CT. From Annex 1, we can say that Ktf = 10 for faults in the reach point of the relay. The knee voltages would be: At the reach limit of Zone 1: Vkp = ktf . If1 . (Rct + RL + Rp) and the power of the CT would be: P = [Vkp / (n . In ) – Rct ]. In2 For modern distance protection relays as ALPS and D60 use 10 ms as response time (t), then in this case, Ktf will be = 3 for internal faults before the reach point. Equations shown above correspond to the worst condition possible for the fault current containing maximum DC offset. It is normal practice to apply a reduction factor (RF) from 0.5 to 0.7 to these requirements, as only 60% or less DC is encountered in most of the faults. The reduction factors are recommended to be applied as follows: 66 to 145 kV: 0.7 Under 66 kV: 0.5 Vkp = ktf . If . (Rct + RL + Rp) . RF Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A).

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Transformer Differential protection: Differential relays must remain stable for through fault current until the corresponding protective relay clears the fault (i.e. 100 ms), and must operate rapidly for internal faults. Transient response is critical, and the transient over dimensioning factor must be taking into account. To check if a selected CT is acceptable for this application, the general condition that should be fulfilled, for internal and through faults As this condition would lead to very over dimensioned CTs, current transformers are calculated to ensure stability for through faults. Protections do not require receiving unsaturated currents from both sides of the power transformer to ensure operation for internal faults. For through faults, the mentioned condition must be fulfilled at both sides of the power transformer. For relays using a percentage restraint principle, experience shows that the requirement may be normally reduced by a factor of 2, as the percentage restraint algorithm helps to avoid the operation, and maximum DC offset is not normally encountered. Because in general the differential algorithm takes around 15 ms to detect the fault, Ktf = 4 is recommended for internal faults. For external faults, we need to differentiate between Step-up Power transformers (Power transformers located close the generators) and other transformers. In the first case, a value of Ktf = 22 is recommended (As we commented before, because the restrain and the decrement curve in Generators, this value can become Ktf = 11). In case of other transformers the Ktf = 10 is recommended (the value because the percentage restrain, can become to Ktf = 5 or 6 (*) to Big Transformers). To estimate the power of the CTs, the following calculations are done on each side of the power transformer, to ensure stability for external faults. Knowing the Ktf value, The knee voltage needed to ensure stability for through faults will be: Vkp = ktf . If. (Rct + RL + Rp) (*) and the power of the CT would be: P = [Vkp / (n . In ) – Rct ]. In2

Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A). (*) In the formulae use the value already corrected. In this case 5 or 6 for external faults.

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Line Differential protection: Line differential protection is a critical application that requires a relay operating in a very fast and secure manner. To compute the power requirements for the associated CTs, the transient performance must be taken into account. This is done by using the transient over dimensioning factor, previously discussed, in the CT size calculation process In our case for L90 we can use Ktf = 5 or a maximum of 6 for EHV transmission lines. The knee voltage would be: Vkp = ktf . If . (Rct + RL + Rp) and the power of the CT would be: P = [Vkp / (n . In ) – Rct ]. In2 Equations shown above correspond to the worst condition possible for the fault current containing maximum DC offset. It is usually a normal practice to apply a reduction factor (RF) from 0.5 to 0.7 to these requirements, as only 60% or less DC is encountered in most of the faults. The reduction factors are recommended to be applied as follows: 66 to 145 kV: 0.7 Under 66 kV: 0.5 Vkp = ktf . If . (Rct + RL + Rp) . RF Please check the recommendations given in the instruction manual of the relay, if applicable, for more detailed information. Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A).

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Generator Differential protection: Differential relays must remain stable for through fault current until the corresponding protective relay clears the fault (i.e. 100 ms), and must operate rapidly for internal faults. Transient response is critical, and the transient over dimensioning factor must be taking into account. To check if a selected CT is acceptable for this application, the general condition that should be fulfilled, for internal and through faults As this condition would lead to very over dimensioned CTs, current transformers are calculated to ensure stability for through faults. Protections do not require receiving unsaturated currents from both sides of the generator to ensure operation for internal faults. For through faults, the mentioned condition must be fulfilled at both sides of the generator. For relays using a percentage restraint principle, experience shows that the requirement may be normally reduced by a factor of 2, as the percentage restraint algorithm helps to avoid the operation, and maximum DC offset is not normally encountered. Because in general the differential algorithm takes around 15 ms to detect the fault, Ktf = 4 is recommended for internal faults. For external faults, the value recommended for relays as G30 is Ktf = 30. Because the percentage restrain principle and the current decrement during the short circuit, this value finally can become to Ktf = 15 (*). In case of G60 and SR489 (firmware revision 150.000 and up -releases after 2002), because the directional algorithm, a Ktf = 10 (*) is the most suitable. To estimate the power of the CTs, the following calculations are done on each side of the power transformer, to ensure stability for external faults. Knowing the Ktf value, The knee voltage needed to ensure stability for through faults will be: Vkp = ktf . If. (Rct + RL + Rp) (*) and the power of the CT would be: P = [Vkp / (n . In ) – Rct ]. In2

Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A). (*) In the formulae use the value already corrected. In this case 15 or 10 for external faults.

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Low Impedance Busbar protection: Differential relays must remain stable for through fault current until the corresponding protective relay clears the fault (i.e. 100 ms), and must operate rapidly for internal faults. Transient response is critical, and the transient over dimensioning factor must be taking into account. To check if a selected CT is acceptable for this application, the general condition that should be fulfilled, for internal and through faults As this condition would lead to very over dimensioned CTs, current transformers are calculated to ensure stability for through faults. Protections do not require receiving unsaturated currents to ensure operation for internal faults. For through faults, the mentioned condition must be fulfilled for external faults. For relays using a percentage restraint principle, experience shows that the requirement may be normally reduced by a factor of 2, as the percentage restraint algorithm helps to avoid the operation, and maximum DC offset is not normally encountered. Because in general the differential algorithm takes less than 10 ms to detect the fault, Ktf = 3 is recommended for internal faults. For external faults, the value recommended for relays B90 and B30 is Ktf = 8. because the directional algorithm and the stability provided by the restrain on the differential algorithm (max of the currents, that in practice is equivalent to the fault current) that usually is set higher than 60%. The over dimensioning factor can be as low as Ktf = 5 (*), particularly for voltages lower than 132 kV. To estimate the power of the CTs, the following calculations are done on each side of the power transformer, to ensure stability for external faults. Knowing the Ktf value, The knee voltage needed to ensure stability for through faults will be: Vkp = ktf . If. (Rct + RL + Rp) (*) and the power of the CT would be: P = [Vkp / (n . In ) – Rct ]. In2

Notes: RCT burden considered = 2 Ω (RL=1Ω (200m of 4mm2 copper cable); Rp=0,2Ω). Rct will depend of the CT ratio and the secondary rated value (1 A or 5 A). (*) In the formulae use the value already corrected. In this case 5 for external faults.

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ANNEX 1 Criteria for CT´s verification during internal faults The following simulation shows the response of the measuring algorithms of relays sampling at 64 samples per cycle and 16 samples per cycle to a saturated fault current, with 90% of DC offset. CT has been dimensioned based on the real setting of the definite time pickup of the relay, Is, equal to 5A. The simulation shows that the CT will provide signal just to ensure the pickup of the relay. A horizontal line is shown in the graphics to indicate the pickup level. Inverse time elements would operate incorrectly, as the relay cannot measure the real magnitude of the fault current, but just a value slightly higher than the pickup setting of the definite time o/c element. As mentioned in this document, a security margin of 2 is suggested, to ensure relay operation, so the CT would not be dimensioned based on Is, but on 2 x Is.

Ktd = Is = Is peak=

40 30 20 10 0 -10 -20 -30 0,000

1 5 7,071068

INPUT PARAMETERS: ENTER: Inverse of sat. curve slope = S= 20 --Vs volts RMS voltage at 10A exc. 20 Vs = rms current = Turns ratio = n2/1= N = 240 --Ve Winding resistance = Rw = 2,000 ohms volts Burden resistance = Rb = 2,000 ohms rms Burden reactance = Xb = 0,000 ohms 9,0 --System X/R ratio = XoverR = Per unit offset in primary 0,80 1