Table 4α   Error limits for measurement current transformers

 Class % current error at the given proportion of rated current shown below % phase error at the given proportion of the  rated current shown below 2.0* 1.2 1.00 0.50 0.20 0.10       0.05 2.0* 1.2 1.0 0.5 0.2 0.1 0.05 0.1 0.1 0.1 0.2 0.25 5 5 8 10 0.2 0.2 0.2 0.35 0.50 10 10 15 20 0.5 0.5 0.5 0.75 1.00 30 30 45 60 1.0 1.0 1.0 1.5 2.00 60 60 - 90 120 - 3.0 3.0 3.0 - -       - - _ 120 - 120 - - - 0.1 0.1 0.1 0.2 0.25 0.4 5 - 5 8 10 15 0.2 ext 0.2 0.2 0.35 0.50 0.75 10 - 10 15 20 30 0.5 ext 0.5 0.5 0.75 1.00 1.5 30 - 30 45 60 90 1.0 ext 1.0 1.0 1.5 2.00 60 - 60 - 90 120 - 3.0 ext 3.0 - - 3.0 - - - 120 - - 120 - - -

*ext = 200 %

Table 4b  Error limits for protection current transformers

 Accuracy Class +/- percentage Current ratio error +/- Phase error (minutes) % Current 5 20 100 120 5 20 100 120 0.1 0.4 0.2 0.1 0.1 15 8 5 5 0.2 0.75 0.35 0.2 0.2 30 15 10 10 0.5 1.5 0.75 0.5 0.5 90 45 30 30 1.0 3 1.5 1.0 1.0 180 90 60 60

Total error for nominal error limit current and nominal load is five per cent for 5P and 5Ρ ext CTs and ten per cent for 10P and 10P ext CTs.

The cross-sectional area of metal and the saturation flux density are sometimes difficult to obtain.

The latter can be taken as equal to 100 000 lines/Cm2, which is a typical value for modern transformers. To use the formula, V is determined from eqn. 4.1 and Bmax. is then calculated using eqn. 2. If Bmax.

Exceeds the saturation density, there could be appreciable errors in the secondary current and the CT selected would not be appropriate.

Example 1.

Assume that a CT with a ratio of 2000/5 is available, having a steel core of high permeability, a cross-sectional area of 3.25 In cm2 and a secondary winding with a resistance of 0.31 Ω. The impedance of the relays, including connections, is 2 Ω. Determine whether the CT would be saturated by a fault of 35 000 A at 50 Hz.

Solution

If the CT is not saturated, then the secondary current, IL, is

35 000x 5/2000=87.5 A. N= 2000/5 = 400 turns

And Vs=87.5x (0.31+2) =202.1 V. Using eqn. 4.2, Bmax, can now be calculated:

Bmax = 202.1X108/4.44X50X3.25X400=70 030 lines/ cm2

Since the transformer in this example has a steel core of high permeability, this relatively low value of flux density should not result in saturation.

Using the magnetization curve

Typical CT excitation curves which are supplied by manufacturers state the r.m.s. current obtained on applying an r.m.s. voltage to the secondary winding, with the primary winding open-circuited.

The curves give the magnitude of the excitation current required order to obtain a specific secondary voltage.

The method consists of producing a curve which shows the relationship between the primary and secondary currents for one tap and specified load conditions, such as shown in Figure 4.9.

Starting with any value of secondary current, and with the help of the magnetisation curves, the value of the corresponding primary current can be determined. The process is summarized in the following steps:

(a) Assume a value for IL.

(b) Calculate Vs in accordance with eqn. 4.1.

(c)  Locate the value of Vs on the curve for the tap selected, and find the associated value of the magnetization current, Ie.

(d) Calculate IH/n (=IL + Ie) and multiply this value by n to refer it to the primary side of the CT.

(e)  This provides one point on the curve of IL against IH, and the process is then repeated to obtain other values of IL and the resultant values of IH. By joining the points together the curve of IL against IH  is obtained.  Figure 4.9 using the magnetization curve

a- assume a value for IL.

b-Vs = IL (ZL +ZC+ZB)

c - find Ie from the curve

d - IH=n(I1,+Ie )

e - draw the point on the curve

This method incurs an error in calculating IH /n by adding Ie and IL together arithmetically and not vectorially, which implies not taking account of the load angle and the magnetizations branch of the equivalent circuit. However, this error is not great and the simplifica­tion snakes it easier to carry out the calculations.

After construction, the curve should be checked to confirm that the maximum primary fault current is within the transformer saturation zone. If not, then it will be necessary to repeat the process, changing the tap until the fault current is within the linear part of the characteristic.

In practice it is not necessary to draw the complete curve because it is sufficient to take the known fault current and refer to the secondary winding, assuming that there is no saturation for the tap selected.

This converted value can be taken as IL initially for the process described earlier. If the tap is found to be suitable after finishing the calculations, then a value of IH can be obtained which is closer to the fault current.

Accuracy classes established by the ANSI standards

The ANSI accuracy class of a CT (Standard C57.13) is described by two symbols — a letter and a nominal voltage; these define the capability of the CT.

C indicates that the transformation ratio can be calculated, and T indicates that the transformation ratio can be determined by means of tests. The classification C includes those CTs with uniformly distributed windings and other CTs with a dispersion flux which has a negligible effect on the ratio, within defined limits.

The classification T includes those CTs with a dispersion flux which considerably affects the transformation ratio.

For example, with a CT of class C—100 the ratio can be calculated, and the error should not exceed ten per cent if the secondary current does not go outside the range of 1 to 20 times the nominal current and if the load does not exceed 1Ω (1Ω x 5 Ax 20=100 V) at a minimum power factor of 0.5.

These accuracy classes are only applicable for complete windings. When considering a winding provided with taps, each tap will have a voltage capacity proportionally smaller, and in consequence it can only feed a portion of the load without exceeding the ten per cent error limit. The permissible load is defined as ZB= (NP  Vc) / 100, where ZB, is the permissible load for a given tap of the CT, NP, is the fraction of the total number of turns being used and Vc is the ANSI voltage capacity for the complete CT.

2.6 DC saturation

Up to now, the behavior of a CT has been discussed in terms of a steady state, without considering the DC transient component of the

DC saturation is particularly significant in complex protection schemes since, in the case of external faults, high fault currents circulate through the CTs.

If saturation occurs in different CTs associated with a particular relay arrangement, this could result in the circulation of unbalanced secondary currents which would cause the system to malfunction.

2.7 Precautions when working with CTs

Working with CTs associated with energized network circuits can be extremely hazardous. In particular, opening the secondary circuit of a CT could result in dangerous over voltages which might harm operational staff or lead to equipment being damaged, because the current transformers are designed to be used in power circuits which have impedance much greater than their own.

As a consequence, when secondary circuits are left open, the equivalent primary-circuit impedance is almost unaffected but a high voltage will be developed by the primary current passing through the magnetizing impedance Thus, secondary circuits associated with CTs must always he kept in a closed condition or short-circuited in order to prevent these adverse situations occurring. To illustrate this, an example is given next using typical data for a CT and a 13.2 kV feeder.

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