p34x ct requirements

Current Transformer Requirements (AP) 6 Application Notes

P34x/EN AP/K96 Page (AP) 6-149

4 CURRENT TRANSFORMER REQUIREMENTS

The current transformer requirements for each current input will depend on the protection function with which they are related and whether the line current transformers are being shared with other current inputs. Where current transformers are being shared by multiple current inputs, the kneepoint voltage requirements should be calculated for each input and the highest calculated value used.

The P342/P343/P344/P345 is able to maintain all protection functions in service over a wide range of operating frequency due to its frequency tracking system (5 - 70 Hz).

When the P342/P343/P344/P345 protection functions are required to operate accurately at low frequency, it will be necessary to use CT’s with larger cores. In effect, the CT requirements need to be multiplied by fn/f, where f is the minimum required operating frequency and fn is the nominal operating frequency.

4.1 Generator Differential Function

4.1.1 Biased Differential Protection

The kneepoint voltage requirements for the current transformers used for the current

inputs of the generator differential function, with settings of s1 = 0.05 n, k1 = 0%,

s2 = 1.2 n, k2 = 150%, and with a boundary condition of through fault current 10 n, is:

For phase-earth faults

Vk 50 n (Rct + 2RL + Rr) with a minimum of 60

n for X/R <120 If <10 In


n for X/R < 40 If < 10 In

Where the generator is impedance earthed and the maximum secondary earth fault

current is less than n then the CT knee point voltage requirements are:

For phase-phase, 3 phase faults

Vk 25 n (Rct + RL + Rr) with a minimum of 60



n for X/R <100 If <10In, X/R <120 If

<5In



Where:

Vk = Minimum current transformer kneepoint voltage for through fault stability

n = Relay rated current

Rct = Resistance of current transformer secondary winding ( )

RL = Resistance of a single lead from relay to current transformer ( )

Rr = Resistance of any other protective relays sharing the current transformer

( )

If = maximum through fault current

(AP) 6 Application Notes Current Transformer Requirements

Page (AP) 6-150 P34x/EN AP/K96

For Class-X current transformers, the excitation current at the calculated kneepoint

voltage requirement should be less than 2.5 n (<5% of the maximum perspective fault

current 50 n, on which these CT requirements are based). For IEC standard protection class current transformers, it should be ensured that class 5P are used.

4.1.2 High Impedance Differential Protection

If the generator differential protection function is to be used to implement high impedance differential protection, then the current transformer requirements for phase faults are:

Rs = [1.5 * ( f) * (RCT + 2RL)] / S1

Vs = 1.5 If (RCT + 2RL)

VK 2 * s1 * Rs = 2 Vs

Where:

Rs = Value of stabilizing resistor (ohms)

f = Maximum secondary through fault current level (amps)

VK = CT knee point voltage (volts)

S1 = Current setting of differential element (amps)

RCT = Resistance of current transformer secondary winding (ohms)

RL = Resistance of a single lead from relay to current transformer (ohms)

Vs = Stability voltage

4.2 Generator-Transformer Differential Function

4.2.1 Biased Differential Protection

The kneepoint voltage requirements for the current transformers used for the current

inputs of the generator-transformer differential function, with settings of s1 = 0.2 n, k1 =

30%, s2 = 1.0 n, k2 = 80%, and with a boundary condition of through fault current 16

n, is:

For phase-earth faults


n for X/R <120 If <16 In, X/R <600 If

<10 In


n for X/R < 120 If < 10 In



Where the generator is impedance earthed and the maximum secondary earth fault

current is less than n then the CT knee point voltage requirements are:

For phase-phase, 3 phase faults






n for X/R <100 If <10 In, X/R <120 If <

5 In



Where:






( )

If = maximum through fault current

For Class-X current transformers, the excitation current at the calculated kneepoint


current 50 n, on which these CT requirements are based). For IEC standard protection class current transformers, make sure that class 5P are used.

4.3 Voltage Dependent Overcurrent, Field Failure, Thermal Overload, Pole Slipping, Underimpedance and Negative Phase Sequence Protection Functions

When determining the current transformer requirements for an input that supplies several protection functions, it must be ensured that the most onerous condition is met. This has been taken into account in the formula given below. The formula is equally applicable for current transformers mounted at either the neutral-tail end or terminal end of the generator.

Vk 20 n (Rct + 2RL + Rr)

Where:






( )

For class-X current transformers, the excitation current at the calculated kneepoint

voltage requirement should be less than 1.0 n. For IEC standard protection class current transformers, make sure that class 5P are used.



4.4 Sensitive Directional Earth Fault Protection Function Residual Current Input

4.4.1 Line Current Transformers

With reference to section 2.17, the sensitive directional earth fault input current transformer could be driven by three residually connected line current transformers.

It has been assumed that the sensitive directional earth fault protection function will only be applied when the stator earth fault current is limited to the stator winding rated current or less. Also assumed is that the maximum X/R ratio for the impedance to a bus earth fault will be no greater than 10. The required minimum kneepoint voltage will therefore be:

Vk 6 n (Rct + 2RL + Rr)

Where:




RL = Resistance of a single lead from relay to current transformer ( ).


( ).



current 20 n, on which these CT requirements are based). For IEC standard protection class current transformers, make sure that class 5P are used.

4.4.2 Core Balanced Current Transformers

Unlike a line current transformer, the rated primary current for a core balanced current transformer may not be equal to the stator winding rated current. This has been taken into account in the formula:

Vk > 6N n (Rct + 2RL + Rr)

Where:


N = Stator earth fault current

Core balanced current transformer rated primary current





( )

Note N should not be greater than 2. The core balance current transformer ratio should be selected accordingly.



4.5 Stator Earth Fault Protection Function

The earth fault n current input is used by the stator earth fault protection function.

4.5.1 Non-Directional Definite Time/IDMT Earth Fault Protection

CT requirements for time-delayed earth fault overcurrent elements

VK cn/2 * (RCT + 2RL + Rrn)

4.5.2 Non-Directional Instantaneous Earth Fault Protection

CT requirements for instantaneous earth fault overcurrent elements

VK sn (RCT + 2RL + Rrn)

Where:

VK = Required CT knee-point voltage (volts)

cn = Maximum prospective secondary earth fault current or 31 times > setting (whichever is lower) (amps)

sn = Earth fault setting (amps)



Rrn = Impedance of the relay neutral current input at n (ohms)

4.6 Restricted Earth Fault Protection

4.6.1 Low Impedance

VK 24 * n * (RCT + 2RL) for X/R < 40 and f < 15 n

VK 48 * n * (RCT + 2RL) for X/R < 40, 15 n < f < 40 n

and 40 <X/R < 120, f < 15 n

Where:

Vk = VA x ALF

n + ALF x n x Rct

VK = Required CT knee point voltage (volts)

n = rated secondary current (amps)

RCT = Resistance of current transformer secondary winding ( )


f = Maximum through fault current level (amps)

4.6.2 High Impedance

The High Impedance Restricted Earth Fault element shall maintain stability for through faults and operate in less than 40 ms for internal faults provided the following equations are met in determining CT requirements and the value of the associated stabilizing resistor:

Rs = ( f) * (RCT + 2RL) / S1



Vs = 1.5 If (RCT + 2RL)

VK 4 * s1 * Rs = 4 Vs

Where:

Rs = Value of Stabilizing resistor (ohms)

f = Maximum secondary through fault current level (amps)

VK = CT knee point voltage (volts)

S1 = Current setting of REF element (amps)



Vs = Stability voltage

4.7 Reverse and Low Forward Power Protection Functions

For both reverse and low forward power protection function settings greater than 3% Pn, the phase angle errors of suitable protection class current transformers will not result in any risk of mal-operation or failure to operate. However, for the sensitive power protection if settings less than 3% are used, it is recommended that the current input is driven by a correctly loaded metering class current transformer.

4.7.1 Protection Class Current Transformers

For less sensitive power function settings (>3%Pn), the phase current input of the P34x should be driven by a correctly loaded class 5P protection current transformer.

To correctly load the current transformer, its VA rating should match the VA burden (at rated current) of the external secondary circuit through which it is required to drive current.

4.7.2 Metering Class Current Transformers

For low Power settings (<3%Pn), the n Sensitive current input of the P34x should be driven by a correctly loaded metering class current transformer. The current transformer accuracy class will be dependent on the reverse power and low forward power sensitivity required. The table below indicates the metering class current transformer required for various power settings below 3%Pn.

To correctly load the current transformer, its VA rating should match the VA burden (at rated current) of the external secondary circuit through which it is required to drive current. Use of the P34x sensitive power phase shift compensation feature will help in this situation.

Reverse and low forward power settings %Pn Metering CT class

0.5 0.1

0.6

0.8 0.2

1.0

1.2

1.4

1.6 0.5

1.8

2.0



Reverse and low forward power settings %Pn Metering CT class

2.2

2.4

2.6

2.8

3.0 1.0

Table 19 - Sensitive power current transformer requirements

4.8 100% Stator Earth Fault Protection Function 20 Hz Inputs

4.8.1 Line Current Transformers

4.8.1.1 Generator Earthed via a Primary Resistor in Generator Starpoint

It has been assumed that the 100% stator earth fault protection function will only be applied when the stator earth fault current is limited to <2x rated current or less as the linear range of the sensitive current input is 2 In. The required minimum kneepoint voltage is:

Vk fn/20 x 2 n (Rct + 2RL + Rr)

Where:




RL = Resistance of a single lead from relay to current transformer ( ).


( ).

fn = fundamental frequency 50 or 60 Hz (fn/20 is to account for operation at 20 Hz)



current 2 n, on which these CT requirements are based). For IEC standard protection class current transformers, it should be ensured that class 5P is used; a 15 VA 5P10 CT will be adequate for most applications.

4.8.1.2 Generator Earthed via Earthing Transformer and Secondary Resistor at the Terminals or Star Point of the Generator

A 400/5A CT can be ordered for this application, Vk = 720 V (50/60 Hz)

4.8.2 Earthing Transformers

To prevent the secondary load resistance from becoming too small (it should be > 0.5

!! where possible) a high secondary voltage, such as 500 V, should be chosen for the neutral or earthing transformer.

It is important that the earthing transformer never becomes saturated otherwise ferroresonance may occur. It is sufficient that the transformer knee point voltage be equal to the generator rated line voltage, Vn.



4.8.2.1 Generator Earthed via a Primary Resistor in Generator Starpoint

Voltage transformer rating: Vn/ 3 / 500 V, 3000 VA (for 20 s) class 0.5 (non-saturated up to Vn, Generator)

Vn = rated generator line voltage (phase-phase)

4.8.2.2 Generator Earthed via Earthing Transformer and Secondary Resistor at the Terminals of the Generator

Voltage transformer rating: Vn/ 3 / 500/3 V (non-saturated up to Vn, Generator)

The transformer VA rating for 20 s per phase = 1.3 x 1/3 x If x Vn x 3 x 10/ 20

for 3 single phase transformers.

If = primary fault current

The 1.3 accounts for an overvoltage factor from field forcing.

The 10/ 20 increases the rating from 10 to 20 s.

For a 3 phase transformer the VA rating is 3 times higher.

4.8.2.3 Generator Earthed via Earthing Transformer and Secondary Resistor at the Starpoint of the Generator

Voltage transformer rating: Vn/ 3 / 500 V (non-saturated up to Vn, Generator)

The transformer VA rating for 20 s per phase = 1.3 x If x Vn x 3 x 10/ 20

The 1.3 accounts for an overvoltage factor from field forcing.

The 10/ 20 increases the rating from 10 to 20 s.

4.9 Converting an IEC185 Current Transformer Standard Protection Classification to a Kneepoint Voltage

The suitability of an IEC standard protection class current transformer can be checked against the kneepoint voltage requirements specified previously.

If, for example, the available current transformers have a 15 VA 5P 10 designation, then an estimated kneepoint voltage can be obtained as follows:

Vk = VA x ALF

n + ALF x n x Rct

Where:

Vk = Required kneepoint voltage

VA = Current transformer rated burden (VA)

ALF = Accuracy limit factor

n = Current transformer secondary rated current (A)


If Rct is not available, then the second term in the above equation can be ignored.

Example: 400/5 A, 15 VA 5P 10, Rct = 0.2

Vk =15 x 10

5 + 10 x 5 x 0.2

= 40 V



4.10 Converting IEC185 Current Transformer Standard Protection classification to an ANSI/IEEE Standard Voltage Rating

The Px40 series protection is compatible with ANSI/IEEE current transformers as specified in the IEEE C57.13 standard. The applicable class for protection is class "C", which specifies a non air-gapped core. The CT design is identical to IEC class P, or British Standard class X, but the rating is specified differently.

The ANSI/IEEE “C” Class standard voltage rating required will be lower than an IEC knee point voltage. This is because the ANSI/IEEE voltage rating is defined in terms of useful output voltage at the terminals of the CT, whereas the IEC knee point voltage includes the voltage drop across the internal resistance of the CT secondary winding added to the useful output. The IEC/BS knee point is also typically 5% higher than the ANSI/IEEE knee point.

Therefore:

Vc = [ Vk - Internal voltage drop ] / 1.05

= [ Vk - (In . RCT . ALF) ] / 1.05

Where:

Vc = “C” Class standard voltage rating

Vk = IEC Knee point voltage required

n = CT rated current = 5 A in USA

RCT = CT secondary winding resistance (for 5A CTs, the typical resistance is 0.002 ohms/secondary turn)

ALF = The CT accuracy limit factor, the rated dynamic current output of a "C" class CT (Kssc) is always 20 x In

The IEC accuracy limit factor is identical to the 20 times secondary current ANSI/IEEE rating.

Therefore:

Vc = [ Vk - (100 . RCT ) ] / 1.05

p34x ct requirements

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