On the Protection of Generators Against Ground Faults Using Cable Current Transformers

I. D. Kutyavin

In connection with the decision of the Technical Department of the Ministry of Electric Power Stations on the grounding mode of neutral points and protection against ground faults in 2–35 kV networks (see the chronicle in No. 10 of Electric Stations for 1949), protection of generators against ground faults using cable current transformers should become very widespread.

As is known, the implementation of this type of protection for generators connected to switchgear by several cables or buses has encountered difficulties. These difficulties, apparently, were overcome by the staff of the Institute of Electrical Engineering of the Academy of Sciences of the Ukrainian SSR (see, for example, the article by I. M. Sirota, “Testing a New Generator Protection Against Ground Faults,” in No. 6 of Electric Stations for 1949) by applying

Fig. 1

a common cable transformer enclosing the entire bundle of cables. Such a transformer, as I. M. Sirota indicates, can also be used successfully for generators connected to switchgear by buses.

Along with this, protection using individual cable transformers is also finding application. An analogous protection was described by V. E. Kazansky (see No. 3 of Electric Stations for 1949), who recommends a very original circuit with biasing from a zero-sequence transformer. However, the simplest circuit with individual transformers in the form of differential current protection of the zero sequence, shown in Fig. 1, can also be used successfully. For the implementation of this ...

of protection for a generator with a grounded neutral, identical cable transformers with equal transformation ratios are placed on the grounding conductor and on each cable connecting the generator with the distribution device. The secondary windings of these transformers are connected for current summation and are connected to the protection relay.

The equivalent circuit of the protection is shown in Fig. 2, where \(z_n\) is the impedance of the magnetizing branch of the transformers; \(z_1, z_2 \ldots\) are the impedances of the connecting leads and secondary windings of the transformers; \(z_p\) is the impedance of the relay winding; \(i_1\) is the current in the grounding conductor; \(i_2, i_3\) are the zero-sequence currents in the cables; and \(i_{p1}\) is the current in the relay. All quantities are referred to the secondary number of turns of the transformers.

Fig. 2

Unbalance Current of the Protection

The unbalance current of the circuit under consideration consists of three components: the unbalance current \(i_{nb1}\), caused by the generator load current; the current \(i_{nb2}\), caused by errors of the cable transformers in transforming the through active zero-sequence current (this occurs only for generators with a grounded neutral); and the current \(i_{nb3}\), equal to the transformed capacitive zero-sequence current determined by the capacitance of the generator circuit located within the protection zone.

The current \(i_{nb1}\) during normal operation of the generator is very small; therefore it may be neglected if the protection is provided with blocking against overcurrents. In the absence of such blocking, the current \(i_{nb1}\) during through overcurrents in the primary circuit may reach large values and substantially affect the value of the pickup current. However, as the number of generator cables increases, and in the event of a through overcurrent, the current \(i_{nb1}\) will decrease because the unbalance currents of the individual transformers, having arbitrary phase, will compensate one another.

The currents \(i_{nb2}\) and \(i_{nb3}\) are produced by zero-sequence currents and do not depend on the generator load. The total value of these two currents can be determined quite simply from experience. For this purpose it is necessary to ground each phase of the generator in turn outside the protection zone, raise its voltage to the rated value, and measure the unbalance current. The largest of the three currents obtained should be taken into account when adjusting the protection.

The unbalance currents \(i_{nb2}\) and \(i_{nb3}\) can also be found experimentally by passing through the primary circuit the corresponding currents equal to the currents that produce them.

Using the equivalent circuit of Fig. 2, one can also obtain analytical expressions for the currents \(i_{nb2}\) and \(i_{nb3}\). Let us first consider the particular case in which, in the primary circuit, current flows only through the grounding conductor. Then the following equation may be written:

\[ (i_1-i'_1)z_{n1}=i'_1\left[z_1+\frac{z_x z_p}{z_x+(m-1)z_p}\right], \tag{1} \]

where \(z_x=z_{n1}+z_1 \simeq z_{n2}+z_2 \simeq z_{n3}+z_3=\ldots\) is the transformer no-load impedance as seen from the secondary winding, and \(m\) is the number of current transformers in the circuit.

On the basis of (1), the secondary current of the first transformer can be determined:

\[ i'_1=\frac{\hat{i}_1 z_{n1}\left[z_x+(m-1)z_p\right]}{z_x(z_x+mz_p)}. \tag{2} \]

The current in the relay from the zero-sequence transformer is

\[ i_p=\frac{i'_1 z_x}{z_x+(m-1)z_p} =\frac{i_1 z_{n1}}{z_x+mz_p}. \tag{3} \]

Analogous expressions for the relay currents can also be obtained for the other transformers. Then the total current in the relay on the basis of the equivalent circuit of Fig. 2 is

\[ i_{nb2}=i_{p1}-(i_{p2}+i_{p3}+\ldots) =\frac{i_1 z_{n1}-(i'_2 z_{n2}+i'_3 z_{n3}+\ldots)}{z_x+mz_p}. \tag{4} \]

In expression (4), the sum of the currents \(i'_2, i'_3, i'_4,\ldots\) is equal to the current \(i_1\), since the current \(i_{nb2}\) is produced by the through active zero-sequence current.

If the impedances \(z_{n2}, z_{n3}, z_{n4},\ldots\) are taken equal to some average value \(z_{nav}\), then for \(i_{nb2}\) we obtain the following expression:

\[ i_{nb2}=\frac{i_1(z_{n1}-z_{nav})}{z_x+mz_p} =\frac{I_{gm}v(z_{n1}-z_{nav})}{n_T(z_x+mz_p)}. \tag{5} \]

By the same method one can obtain an expression for \(i_{nb3}\), having the following form:

\[ i_{nb3}=\frac{i''_2 z_{n2}+i''_3 z_{n3}+\ldots+z_{nm}i_m}{z_x+mz_p} = \]

\[ =\frac{i_{cz} z_{nav}}{z_x+mz_p} =\frac{I_{cz}v z_{nav}}{n_T(z_x+mz_p)}, \tag{6} \]

where

\[ \begin{aligned} I_{gm} &\text{ is the maximum grounding current due to the resistance,}\\ I_{cz} &\text{ is the capacitive current flowing into the protection zone for a fault outside the zone,}\\ v &\text{ is the relative number of turns shorted to ground,}\\ n_T &\text{ is the transformation ratio of the transformers,}\\ i''_2,i''_3,\ldots &\text{ are the capacitive components of the currents } i_2,i_3,i_4,\ldots,\\ i_{cz}=i''_2+i''_3+i''_4+\ldots &\text{ is the total capacitive current due to the capacitance of the generator circuit.} \end{aligned} \]

In view of the fact that the currents \(i_{nb1}\) and \(i_{nb2}\) may have an arbitrary phase, the total unbalance current may be equal to the sum of all three components:

\[ i_{nb}=i_{nb1}+i_{nb2}+i_{nb3}. \tag{7} \]

If the protection is provided with blocking against through currents, the current \(i_{nb1}\) in expression (7) may be disregarded. In addition, for generators with an isolated neutral the current \(i_{nb2}\) is equal to zero.

The current \(i_{nb3}\) can also be reduced to a very small value by excluding the generator cables from the protection zone. Technically this is accomplished by installing the transformers at the generator ends of the cables, while the grounding of the cable armor is passed through the transformer window. Furthermore, such an installation of the transformers reduces the length of the connecting wires to a minimum and thereby reduces the current \(i_{nb2}\).

Calculation of the Protection

The pickup current of the relay of the protection under consideration must be greater than the maximum unbalance current appearing in the relay when there is no fault in the protection zone:

\[ i_{pm}=K_n i_{nb}, \tag{8} \]

where \(K_n>1\) is the reliability factor.

In the case of a ground fault in the protected zone, the current in the relay will have the following expression (it can be obtained in the same way as (4)):

\[ i_p=\frac{v z_n I_d}{n_T (z_x+m z_p)}-i_{nb1}, \tag{9} \]

where \(I_d\) is the current in the arc at the point of insulation breakdown,
\(v\) is the number of turns shorted to ground, and
\(n_T\) is the transformation ratio of the cable transformers.

In expression (9) one should take into account the unbalance current \(i_{nb1}\) corresponding to the rated load of the generator, whereas in expressions (7) and (8) this current \(i_{nb1}\) must be substituted, when blocking against through currents is present, by the value corresponding to the blocking current, and, when blocking is absent, by the value corresponding to the through current in the primary circuit. It should be emphasized here that none of the authors considering the calculation of ground-fault protection, nor the Guidelines for Relay Protection of 1948, make this distinction in accounting for the unbalance current for a ground fault inside and outside the protection zone, which is an error.

In view of the fact that at rated generator load the current \(i_{nb1}\) is very small, it may be disregarded in expression (9).

The maximum ground-fault current required to ensure a dead zone \(v_o\) at \(i_p=i_{pm}\) is

\[ I_{zm}=\frac{i_{pm} n_T (z_x+m z_p)}{v_o z_n}. \tag{10} \]

The minimum dead zone of the protection at a ground-fault current equal to \(I_{dm}\) is

\[ v_{o.m}=\frac{i_{pm} n_T (z_x+m z_p)}{z_n I_{dm}}, \tag{11} \]

where \(I_{dm}\) is the maximum current in the arc.

The optimum resistance of the relay winding is

\[ z_p=\frac{z_x}{m}. \tag{12} \]

Example. Calculate the protection of a generator having 10 cables and a grounded neutral. The cable current transformers have the constants \(A = 40\) and \(n_T = 25\). The maximum unbalance current during a through short circuit with simultaneous ground fault is \(20\ \mathrm{mA}\). For the protection it is proposed to use a relay with \(z_p = 10\ \Omega\) and with limits \(i_{pm} = 5 \div 100\ \mathrm{mA}\). The relay pickup current at \(K_n = 2\) is:

\[ i_{pm} = 2 \cdot 0.02 = 0.04\ \mathrm{A}. \]

The voltage at the relay terminals at pickup current is

\[ U_p = i_{pm} z_p = 10 \cdot 0.04 = 0.4\ \mathrm{V}. \]

The resistance of the transformer magnetizing branch is

\[ z_n = A U_p^{1/3} = 40 \cdot 0.4^{1/3} = 30\ \Omega. \]

The ground-fault current required to ensure \(v_0 = 0.3\) is

\[ I_{z m} = \frac{0.04 \cdot 25(30 + 11 \cdot 10)} {0.3 \cdot 30} = 15.6 . \]