Principles of Operation and Calculation of Generator Protection Against External and Internal Phase-to-Phase Short Circuits Using Differential Current Transformers

Part 2. Calculation of a Differential Current Transformer with Two Separate Secondary Windings

N. A. Dulzon

(Presented by the scientific seminar of the Department of Electric Power Stations and Networks)

The equivalent circuit for this variant is shown in Fig. 3.
When the primary currents are equal, \(I_1 = I_2\), which corresponds to the through current in the generator, the currents in the relay are determined from expressions (5) and (6). The corresponding magnetizing currents can be determined from (4).

According to the “Guidelines for Relay Protection,” the operating current of the overcurrent relay, referred to the primary side, should be

\[ I_{\mathrm{cp}1}=\frac{K_{\mathrm{n}}}{K_{\mathrm{v}}} I_{\mathrm{n}}, \tag{18} \]

where \(K_{\mathrm{n}}\) is the reliability factor, taken equal to \(1.1 \div 1.2\);
\(K_{\mathrm{v}}\) is the relay return factor;
\(I_{\mathrm{n}}\) is the rated current of the protected generator.

By the condition of selectivity from external short circuits, the operating current of the relay protecting against internal faults must be taken, with a certain margin, greater than the current in it during an external short circuit.

\[ I_{\mathrm{cp}2}=K_{\mathrm{z}} I_{\mathrm{kz}} \tag{19} \]

where \(K_{\mathrm{z}}\) is the margin factor;
\(I_{\mathrm{kz}}\) is the maximum current in the generator during an external short circuit.

Taking expressions (18) and (19) into account, formulas (5) and (6) take the form

\[ i_{\mathrm{cp}1}= \frac{ K_{\mathrm{n}} I_{\mathrm{n}} \dot{Y}_3\left(\dot{Y}_4+\dot{Y}_{02}\right) }{ K_{\mathrm{v}}\Delta }; \tag{20} \]

\[ i_{\mathrm{cp}2}= \frac{ K_{\mathrm{z}} I_{\mathrm{kz}} \dot{Y}_4\left(\dot{Y}_3+\dot{Y}_{01}\right) }{ \Delta }. \tag{21} \]

The currents in the relay during an internal short circuit are determined by formulas (7) and (8), and the magnetizing currents of the core halves—

according to formulas (9). If the operating current of the protection against internal faults is specified as \(I_{c3}\), and the operating current of the protection against through currents during an internal fault is taken as a fraction \(\alpha\) of the current \(I_{c3}\), then the operating currents of the relays of these protections will be

\[ \dot I_{\mathrm{cp}1}= \frac{ \alpha I_{c3}\dot Y_3 \left[ \left(\frac{\dot Y_{\mathrm{osh}}}{\dot Y_{0s2}}+1\right)(\dot Y_4+\dot Y_{02})+\dot Y_{\mathrm{osh}} \right] }{\Delta}; \tag{22} \]

\[ \dot I_{\mathrm{cp}2}= \frac{ I_{c3}\dot Y_4 \left[ \frac{\dot Y_{\mathrm{osh}}}{\dot Y_{0s1}}(\dot Y_3+\dot Y_{01})+\dot Y_{\mathrm{osh}} \right] }{\Delta}. \tag{23} \]

The corresponding magnetizing currents at operation of the protection against internal faults are

\[ \dot I_{01}= \frac{ I_{c3}Y_{01} \left[ \left(\frac{\dot Y_{\mathrm{osh}}}{\dot Y_{0s2}}+1\right)(\dot Y_4+\dot Y_{02})+\dot Y_{\mathrm{osh}} \right] }{\Delta}; \tag{24} \]

\[ \dot I_{02}= \frac{ I_{c3}Y_{02} \left[ \frac{\dot Y_{\mathrm{osh}}}{\dot Y_{0s1}}(\dot Y_3+\dot Y_{01})+\dot Y_{\mathrm{osh}} \right] }{\Delta}. \tag{25} \]

The simultaneous solution of equations (20--25) is complicated because of the nonlinearity of the admittances \(\dot Y_{01}\) and \(\dot Y_{02}\). To simplify the problem, we shall take \(\dot Y_{01}=\dot Y_{02}=\dot Y_{oc}\), which corresponds to linearization of the magnetization curve of the DTT core material. In addition, taking into account that during an internal short circuit the magnetic flux closes mainly through steel, we shall take, for expressions (22--25), \(Y_{\mathrm{osh}}=\infty\) and \(Y_{0s}=\infty\); during an external short circuit, in formulas (20, 21) we shall take \(\dot Y_{0s1}=\dot Y_{0s2}=\dot Y_{0s}\). Taking these assumptions into account, we obtain the system of equations

\[ \left. \begin{aligned} \dot I_{\mathrm{cp}1}&=\frac{K_{\mathrm{n}}I_{\mathrm{n}}\dot Y_3(\dot Y_4+\dot Y_{oc})}{\Delta K_{\mathrm{v}}};\\[4pt] \dot I_{\mathrm{cp}2}&=\frac{K_3 I_{k3}\dot Y_4(\dot Y_3+\dot Y_{oc})}{\Delta};\\[4pt] \dot I_{\mathrm{cp}1}&=\frac{\alpha I_{c3}\dot Y_3}{2\dot Y_{oc}+\dot Y_3+\dot Y_4};\\[4pt] \dot I_{\mathrm{cp}2}&=\frac{I_{c3}\dot Y_4}{2\dot Y_{oc}+\dot Y_3+\dot Y_4};\\[4pt] I_{01}=I_{02}=I_{oc}&=\frac{I_{c3}\dot Y_{oc}}{2\dot Y_{oc}+\dot Y_3+\dot Y_4}, \end{aligned} \right\} \tag{26} \]

where

\[ \Delta= \left(\frac{2\dot Y_{\mathrm{osh}}}{\dot Y_{0s}}+1\right) (\dot Y_3+\dot Y_{oc})(\dot Y_4+\dot Y_{oc}) +\dot Y_{\mathrm{osh}}(2\dot Y_{oc}+\dot Y_3+\dot Y_4). \]

Expressing the complex quantities entering this system of equations in terms of their moduli gives

\[ \left. \begin{aligned} I_{\mathrm{cp}1} &= \frac{K_i I_n Y_3 (Y_4+Y_{\mathrm{oc}})} {K_{\mathrm{B}}|\Delta|};\\[4pt] I_{\mathrm{cp}2} &= \frac{K_3' I_{\mathrm{kz}}Y_4 (Y_3+Y_{\mathrm{oc}})} {|\Delta|}; \end{aligned} \right\} \]

\[ \left. \begin{aligned} I_{\mathrm{cp}1} &= \frac{a I_{\mathrm{c}3}Y_3} {2Y_{\mathrm{oc}}+Y_3+Y_4};\\[4pt] I_{\mathrm{cp}2} &= \frac{I_{\mathrm{c}3}Y_4} {2Y_{\mathrm{oc}}+Y_3+Y_4};\\[4pt] I_{\mathrm{oc}} &= \frac{I_{\mathrm{c}3}Y_{\mathrm{oc}}} {2Y_{\mathrm{oc}}+Y_3+Y_4}, \end{aligned} \right\} \tag{27} \]

where

\[ |\Delta|= \sqrt{ \left(\frac{2Y_{\mathrm{on}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^2 (Y_3+Y_{\mathrm{oc}})^2(Y_4+Y_{\mathrm{oc}})^2 + Y_{\mathrm{osh}}^2(2Y_{\mathrm{oc}}+Y_3+Y_4)^2 + } \]

\[ \overline{ \frac{2(2Y_{\mathrm{osh}}+Y_{\mathrm{os}})}{Y_{\mathrm{os}}} (Y_3+Y_{\mathrm{oc}})(Y_4+Y_{\mathrm{oc}}) (2Y_{\mathrm{oc}}+Y_3+Y_4)Y_{\mathrm{osh}}\sin\varphi }. \]

In deriving the last formulas it was assumed that the phase angles of the secondary circuits are approximately equal to one another,

\[ \varphi_3 \simeq \varphi_4=\varphi, \]

and that the sum of the phase angle of the secondary circuit and the angle of the losses in the steel is close to \(90^\circ\), that is,

\[ \sin(\gamma_c+\varphi)=1. \]

Thus, we have obtained a system of equations into which the sought quantities \(w_3\), \(w_4\), \(a\), and \(\delta\) enter in the following form:

\[ Y_{\mathrm{oc}}=\frac{aw_c l_c\sqrt{2}\cdot10^8}{\omega B_cab}; \tag{28} \]

\[ Y_3=\frac{w_3^2}{Z_3}; \tag{29} \]

\[ Y_4=\frac{w_4^2}{Z_4}; \tag{30} \]

\[ Y_{\mathrm{osh}}=f(\delta), \tag{31} \]

where \(Z_3\) and \(Z_4\) are the active resistances of the secondary circuits.

In addition, for the relay operating currents reduced to the primary side, we have

\[ I_{\mathrm{cp}1}=I_{\mathrm{cp}1}'w_3; \tag{32} \]

\[ I_{\mathrm{cp}2}=I_{\mathrm{cp}2}'w_4, \tag{33} \]

where \(I_{\mathrm{cp}1}'\), \(I_{\mathrm{cp}2}'\) are, respectively, the actual operating currents of the protection relays against external and internal faults.

Let us note that, for a specified actual operating current of the protection relay against internal faults \(I_{\mathrm{cp}2}'\), the operating current of the relay

protection against through currents \(I'_{\mathrm{cp}1}\) cannot be specified arbitrarily and must be determined.

We solve the system of equations (27). Using the first equation in this system, we eliminate the quantity \(|\Delta|\) (and consequently also the unknown \(\delta\)) from the remaining equations:

\[ \left. \begin{aligned} \frac{I_{\mathrm{cp}2}}{I_{\mathrm{cp}1}} &= \frac{A Y_4 (Y_3+Y_{\mathrm{oc}})}{Y_3 (Y_4+Y_{\mathrm{oc}})};\\[6pt] I_{\mathrm{cp}1} &= \frac{a I_{\mathrm{c3}} Y_3}{2Y_{\mathrm{oc}}+Y_3+Y_4};\\[6pt] I_{\mathrm{cp}2} &= \frac{I_{\mathrm{c3}} Y_4}{2Y_{\mathrm{oc}}+Y_3+Y_4};\\[6pt] I_{\mathrm{oc}} &= \frac{I_{\mathrm{c3}} Y_{\mathrm{oc}}}{2Y_{\mathrm{oc}}+Y_3+Y_4}, \end{aligned} \right\} \tag{34} \]

where

\[ A=\frac{K_3 K_{\mathrm{B}} I_{\mathrm{kz}}}{K_{\mathrm{n}} I_{\mathrm{n}}}. \]

Solving the system of equations (34) gives

\[ w_4= \frac{\dfrac{\alpha A}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}}{I'_{\mathrm{cp}2}}; \tag{35} \]

\[ a= \frac{P_3 \sqrt{2}\cdot 10^8} {B_{\mathrm{c}}\omega b\left(\dfrac{\alpha A}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right)}; \tag{36} \]

\[ w_3= \left(\frac{\alpha A}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right) \sqrt{ \frac{ Z_3\left(\dfrac{1}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right) }{ P_2\left(\dfrac{\alpha A}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right) } }; \tag{37} \]

\[ I'_{\mathrm{cp}1} = a\sqrt{ \frac{ P_2\left(\dfrac{1}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right) }{ Z_3\left(\dfrac{\alpha A}{\alpha A+1} I_{\mathrm{c3}}-I_{\mathrm{oc}}\right) } }. \tag{38} \]

In formulas (35–36), \(P_2\) denotes the power in the secondary circuit of the protection against internal faults when it operates, that is,

\[ P_2=(I'_{\mathrm{cp}2})^2 Z_4. \tag{39} \]

From the first equation of the system (27) we find

\[ Y_{\mathrm{osh}} = \frac{Y}{ \dfrac{4Y^2}{Y_{0s}^{2}}+\dfrac{4Y\sin\varphi}{Y_{0s}}+1 } \times \left\{ \sqrt{ \left[ \frac{K_3 I_{\mathrm{kz}}(\alpha A+1)}{I_{\mathrm{c3}}\alpha A} \right]^2 \left[ \frac{4Y^2}{Y_{0s}^{2}}+\frac{4Y\sin\varphi}{Y_{0s}}+1 \right] -\cos^2\varphi } - \left( \frac{2Y}{Y_{0s}}+\sin\varphi \right) \right\}; \tag{40} \]

where

\[ Y=\frac{I_{\mathrm{sc}}\,aA\left(\dfrac{\alpha A}{\alpha A+1}I_{\mathrm{sc}}-I_{\mathrm{oc}}\right)} {P_{2}(\alpha A+1)^{2}}. \]

To calculate by this formula it is necessary to know the leakage admittance \(Y_{os}\), which depends in a complicated way on various factors. Analytical determination of this admittance is difficult, and therefore it is most expedient to determine it experimentally. The latter is possible, since the basic parameters of the DCT can be determined independently of the value of \(Y_{os}\) by formulas (35--38), and, consequently, a model of the DCT can be made. However, in most cases there is no need for this, since the magnetic resistance of the leakage paths is usually large and, as practical calculations show, the calculation error is not great if \(Y_{os}=\infty\) is assumed. Under this assumption formula (40) will be

\[ Y_{\mathrm{err}} = Y\left\{ \sqrt{ \left[ \frac{K_{3}I_{\mathrm{sc}}(\alpha A+1)} {I_{\mathrm{sc}}^{\circ}A} \right]^{2} -\cos^{2}\varphi } -\sin\varphi \right\}. \tag{41} \]

From the known admittance \(Y_{\mathrm{err}}\) (proportional to the magnetic resistance of the gap between the primary current conductors of the DCT), it is possible to find the required air gap at the middle leg of the DCT.

Calculations show that, for powerful generators, the required gap value proves to be greater than the width of the DCT window. In this case the middle leg is not made, and two identical additional windings are wound on the outer legs of the magnetic core; these are connected in series so that, under normal generator operation, the emfs in them add, while for internal short circuits they subtract.

The schematic diagram of a DCT with a short-circuited winding is shown in Fig. 1, a. In Fig. 1, b the equivalent circuit of this DCT is shown, obtained by the method set forth in [4]. In this circuit \(z_{k}\) is the complex impedance of the short-circuited-winding circuit. If the same assumptions are adopted as in the preceding case, and, in addition, the reactive impedances of the secondary circuits and of the short-circuited winding are neglected (which is permissible, since for the protection of powerful generators it is necessary to use a crude relay with low inductive resistance), then the magnitudes of the currents in the secondary circuits of the corresponding modes

a) \(M\), \(I_{1}\), \(I_{2}\), \(D\).
b) \(I_{1}\), \(z_{os1}\), \(z_{\mathrm{err}}\), \(z_{k}\), \(z_{01}\), \(z_{3}\), \(z_{4}\), \(z_{02}\), \(z_{os2}\), \(I_{2}\).

Fig. 1.

will be described by the same equations as in system (27), with the difference that the denominator of the first two equations of the system will be

\[ |\Delta'| = \sqrt{ \left(\frac{2Y_{\mathrm{osh}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^2 \left[(Y_3+Y_{\mathrm{oc}})(Y_4+Y_{\mathrm{oc}}) +Y_{\mathrm{k}}(2Y_{\mathrm{oc}}+Y_3+Y_4)\right]^2 +Y_{\mathrm{osh}}^2(2Y_{\mathrm{oc}}+Y_3+Y_4)^2 }. \tag{42} \]

The magnitude of the current in the short-circuited winding when there is through current in the generator is \(I_1\),

\[ I_{\mathrm{k}}= \frac{I_1Y_{\mathrm{k}}(2Y_{\mathrm{oc}}+Y_3+Y_4)}{|\Delta'|} = \frac{I_1Y_{\mathrm{k}}}{ \sqrt{ \left(\frac{2Y_{\mathrm{osh}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^2 (Y+Y_{\mathrm{k}})^2+Y_{\mathrm{osh}}^2 } }. \tag{43} \]

In equations (42) and (43), \(Y_{\mathrm{k}}\) is the admittance of the short-circuited winding.

From the first equation of system (27), taking (42) into account, we find the required value of the admittance of the short-circuited winding for a specified admittance \(Y_{\mathrm{osh}}\) (corresponding to the absence of the middle core of the DTT magnetic circuit):

\[ Y_{\mathrm{k}}= Y\left\{ \frac{Y_{\mathrm{os}}}{2Y_{\mathrm{osh}}+Y_{\mathrm{os}}} \sqrt{ \left[ \frac{K_z I_{\mathrm{kz}}(\alpha A+1)}{I_{\mathrm{sz}}\alpha A} \right]^2 - \left(\frac{Y_{\mathrm{osh}}}{Y}\right)^2 } -1 \right\}, \tag{44} \]

where the value of \(Y\) is the same as in formula (40).

On the other hand, the admittance of the short-circuited winding is equal to

\[ Y_{\mathrm{k}}=\frac{w_{\mathrm{k}}^2}{\sqrt{r_{\mathrm{k}}^2+x_{\mathrm{k}}^2}}, \tag{45} \]

where \(r_{\mathrm{k}}\), \(x_{\mathrm{k}}\), and \(w_{\mathrm{k}}\) are, respectively, the active resistance, leakage reactance, and number of turns of the short-circuited winding.

If the leakage inductance of one turn is denoted by \(L_{\mathrm{k}}\), the mean length of a turn by \(l_{\mathrm{k}}\), and the cross section of the wire of the short-circuited winding by \(q_{\mathrm{k}}\), then

\[ r_{\mathrm{k}}=2\rho\frac{l_{\mathrm{k}}w_{\mathrm{k}}}{q_{\mathrm{k}}}; \qquad x_{\mathrm{k}}=2\omega L_{\mathrm{k}}w_{\mathrm{k}}, \]

where \(\rho\) is the resistivity;

\(\omega\) is the angular frequency.

Taking these quantities into account, from expression (41) one can find the number of turns

\[ w_{\mathrm{k}}= \frac{2Y_{\mathrm{k}}\sqrt{\rho^2l_{\mathrm{k}}^2+\omega^2L_{\mathrm{k}}^2q_{\mathrm{k}}^2}}{q_{\mathrm{k}}}. \tag{46} \]

If the inductance \(L_{\mathrm{k}}\) is neglected, then

\[ w_{\mathrm{k}}=\frac{2Y_{\mathrm{k}}\rho l_{\mathrm{k}}}{q_{\mathrm{k}}}. \tag{47} \]

This expression can be used to determine the required number of turns of the short-circuited winding after specifying its cross section. Let us determine the current density in the short-circuited winding. The current in this winding, referred to the primary side, is determined by formula (43). Accordingly, the current density is

\[ j_{\mathrm{k}}= \frac{I_{\mathrm{k}}}{q_{\mathrm{k}}w_{\mathrm{k}}} = \frac{I_1Y_{\mathrm{k}}}{ q_{\mathrm{k}}w_{\mathrm{k}} \sqrt{ \left(\frac{2Y_{\mathrm{osh}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^2 (Y+Y_{\mathrm{k}})^2+Y_{\mathrm{osh}}^2 } }. \tag{48} \]

Replacing the quantity \(Y_{\mathrm{k}}\) in the numerator of this expression by means of (47) gives

\[ j_{\mathrm{k}} = \frac{I_1}{ 2\rho l_{\mathrm{k}} \sqrt{ \left(\frac{2Y_{\mathrm{osh}}+Y_{os}}{Y_{os}}\right)^2 (Y+Y_{\mathrm{k}})^2+Y_{\mathrm{osh}}^2 } }. \tag{48a} \]

As we see, the current density in the short-circuited winding, for a specified value of \(Y_{\mathrm{k}}\), does not depend on the adopted conductor cross section.

If the current density obtained from formula (48) proves to be within the permissible limits, the calculation may be considered complete. Otherwise it becomes necessary to reduce the current density to the permissible value \(j_{\mathrm{dop}}\). As is seen from (48), the latter can be achieved only by increasing \(Y_{\mathrm{k}}\). An attempt to reduce the current density by increasing the specific resistance of the short-circuited winding does not lead to the desired result, since in this case the permissible current density simultaneously decreases proportionally.

Using expression (48), we find the required value of the conductance of the short-circuited winding, proceeding from the permissible current density,

\[ Y_{\mathrm{k\,dop}} = \frac{Y_{os}}{2Y_{\mathrm{osh}}+Y_{os}} \sqrt{ \left(\frac{I_1}{2\rho l_{\mathrm{k}}j_{\mathrm{dop}}}\right)^2 -Y_{\mathrm{osh}}^2 } -Y . \tag{49} \]

Increasing the conductance \(Y_{\mathrm{k}}\) to the value \(Y_{\mathrm{k.dop}}\) is associated with an increase in the safety factor \(K_z\), the new value of which can be found from (44). It is understood that the new value of \(K_z\) requires refinement of all the data of the DTT according to formulas (35--41). We note that this method cannot always yield an acceptable solution, since an increase in the safety factor is associated with the need to increase the sensitivity of the overcurrent relay, while the sensitivity of commercially manufactured relays is limited.

Another way of reducing the current density in the short-circuited winding consists in introducing an additional resistance \(R_{\mathrm{d}}\) into this circuit.

In considering this question we shall neglect the inductive resistance of the short-circuited winding and of the additional resistance. The conductance of the short-circuited circuit is equal to

\[ Y_{\mathrm{k}}=\frac{w_{\mathrm{k}}^2}{R_{\mathrm{k}}}, \tag{50} \]

where \(R_{\mathrm{k}}\) is the resistance of the short-circuited winding circuit;

\[ R_{\mathrm{k}}=2\rho\,\frac{l_{\mathrm{k}}w_{\mathrm{k}}}{q_{\mathrm{k}}}+R_{\mathrm{d}} . \]

From (50) we find that

\[ w_{\mathrm{k}}q_{\mathrm{k}} = Y_{\mathrm{k}} \left\{ \sqrt{\frac{R_{\mathrm{d}}q_{\mathrm{k}}^2}{Y_{\mathrm{k}}}+\rho^2l_{\mathrm{k}}^2} +\rho l_{\mathrm{k}} \right\}. \tag{51} \]

Requiring that, at a primary current of the DTT equal to the rated current of the generator \((I_1=I_{\mathrm{n}})\), the permissible current density \(j_{\mathrm{dop}}\) be ensured in the short-circuited winding, we find from (48)

\[ w_{\mathrm{k}}q_{\mathrm{k}} = \frac{I_{\mathrm{n}}Y_{\mathrm{k}}}{ j_{\mathrm{dop}} \sqrt{ \left(\frac{2Y_{\mathrm{osh}}+Y_{os}}{Y_{os}}\right)^2 (Y+Y_{\mathrm{k}})^2+Y_{\mathrm{osh}}^2 } }. \tag{52} \]

Equating the right-hand sides of equations (51) and (52), we find the relation between the added resistance and the cross section of the short-circuited winding in the form

\[ R_{\mathrm{d}}=\frac{P_{\mathrm{dc}}}{j_{\mathrm{allow}}^{2}q_{\mathrm{k}}^{2}} . \tag{53} \]

Thus, if the resistivity of the material of the short-circuited winding \(\rho\) is specified, the average length of a turn of this winding \(l_{\mathrm{k}}\) is determined structurally, and its cross section \(q_{\mathrm{k}}\) is adopted, then, using formulas (51) and (53), one can find the number of turns \(w_{\mathrm{k}}\) and the value of the added resistance \(R_{\mathrm{d}}\). The power dissipated in the added resistance at rated generator load will be

\[ P_{\mathrm{dc}}=I_{\mathrm{k}}^{2}R_{\mathrm{d}}=j_{\mathrm{allow}}^{2}R_{\mathrm{d}}q_{\mathrm{k}}^{2}; \]

\[ P_{\mathrm{dc}} = \frac{I_{\mathrm{n}}^{2}Y_{\mathrm{k}}} {\left(\dfrac{2Y_{\mathrm{err}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^{2}(Y+Y_{\mathrm{k}})^{2}+Y_{\mathrm{err}}^{2}} - \]

\[ -\frac{2I_{\mathrm{n}}Y_{\mathrm{k}}l_{\mathrm{k}}\rho j_{\mathrm{allow}}} {\sqrt{\left(\dfrac{2Y_{\mathrm{err}}+Y_{\mathrm{os}}}{Y_{\mathrm{os}}}\right)^{2}(Y+Y_{\mathrm{k}})^{2}+Y_{\mathrm{err}}^{2}}}. \tag{54} \]

Thus, by formulas (35--38), (40), (46), or (51) and (53), it is possible to find the necessary stack thickness of the DTT core, the number of turns of its secondary windings, and, if necessary, the number of turns of the additional short-circuited winding and the added resistance in the circuit of this winding. In doing so, it is necessary to specify the induction in the core \(B_{\mathrm{c}}\) at the tripping current of the internal-fault protection, which, apparently, for the sake of better utilization of the steel, should be taken as corresponding to the maximum magnetic permeability of the core material. In addition, it is also necessary to specify the values of the reserve factor \(K_{\mathrm{z}}\), the power in the circuit of the internal-fault protection at the tripping current of this protection \(P_{2}\), and the ratio of the primary tripping current of the protection against through currents during an internal fault to the tripping current of the internal-fault protection \(\alpha\). Since the protection against through currents in the generator serves as backup for the internal-fault protection, the value of \(\alpha\) should expediently be taken less than or equal to unity. However, for turbine generators this value may also be taken greater than unity, but so as to ensure a sufficient sensitivity factor for the backup protection.

The necessary selectivity of the protection against internal short circuits during through overcurrents in the generator is ensured by the choice of the design reserve factor \(K_{\mathrm{z}}\). Through overcurrents in the generator may arise during external short circuits, when the generator is connected for parallel operation by the self-synchronization method, and when it is incorrectly switched on under precise synchronization. In the first two cases, the maximum possible currents are identical and equal to \(I_{\mathrm{sc}}\). If selectivity is to be ensured only in these cases, the reserve factor should be taken on the order of 1.2--1.5. In this case the internal-fault protection will disconnect the generator without a time delay if it is incorrectly switched on in phase opposition by the precise-synchronization method. If the latter is undesirable, then the reserve factor should be taken greater than 2, since the maximum possible current during such switching-on of the generator may reach twice the short-circuit current. It should be noted that the coeffi-

the safety factor can be increased only up to a certain limit. Indeed, as is seen from formulas (37) and (38), maximum protection is feasible only when

\[ \frac{I_{\mathrm{c}3}}{\alpha A+1}-I_{\mathrm{oc}}>0, \]

that is, when

\[ \alpha K_3<\frac{K_{\mathrm{n}} I_{\mathrm{n}}}{K_{\mathrm{B}} I_{\mathrm{k}3}} \left(\frac{I_{\mathrm{c}3}}{I_{\mathrm{oc}}}-1\right). \tag{55} \]

During operation, the resistance of the secondary circuits may change somewhat as a result of contact oxidation. This increase in resistance is associated with a certain decrease in the conductance of the given circuit. Let us consider the effect of such a slight decrease in conductance on the protection parameters.

The safety factor (see the system of equations (27)) is equal to

\[ K_3=\frac{I_{\mathrm{cp}2}\,|\Delta|} {I_{\mathrm{k}3}Y_4\left(Y_3+Y_{\mathrm{oc}}\right)}. \]

For simplicity, let us take the phase angle of the secondary circuits to be zero and neglect the leakage of the primary winding \((Y_{0s}=\infty)\); then:

\[ K_3=\frac{I_{\mathrm{cp}2}}{I_{\mathrm{k}3}} \sqrt{ \left(1+\frac{Y_{\mathrm{oc}}}{Y_4}\right)^2 + Y_{\mathrm{on}}^2 \left[ \frac{2Y_{\mathrm{oc}}+Y_3+Y_4} {Y_4\left(Y_3+Y_{\mathrm{oc}}\right)} \right]^2 }. \]

From this expression it is evident that a decrease in the conductance \(Y_3\) (or \(Y_4\)) leads to an increase in the safety factor, though to a somewhat lesser degree. Thus, the increase in the resistance of the secondary circuit that is possible during operation is associated with an increase in the safety factor and, consequently, from this point of view is not dangerous. Moreover, in a three-phase implementation of the protection, a simultaneous increase in the transition resistances in all three phases is unlikely, and for the “healthy” phases the ratios will remain as calculated.

The current in the maximum-protection relay, referred to the primary side, under through currents (27) is equal to

\[ I_{\mathrm{p}1}=\frac{I_1Y_3\left(Y_4+Y_{\mathrm{oc}}\right)}{|\Delta|}. \]

As we see, when the conductance \(Y_3\) (or \(Y_4\)) decreases, the current in the maximum-protection relay decreases. Consequently, the maximum protection of that phase in which the resistance of the secondary circuit has increased will be somewhat desensitized, which is also not dangerous.

The currents in the relay of the protection against internal phase-to-phase short circuits and in the maximum-protection relay under internal faults are respectively equal to (27):

\[ I_{\mathrm{p}1}=\frac{I_1Y_3}{2Y_{\mathrm{oc}}+Y_3+Y_4}; \]

\[ I_{\mathrm{p}2}=\frac{I_1Y_4}{2Y_{\mathrm{oc}}+Y_3+Y_4}. \]

From these expressions it is evident that a decrease in conductance in the maximum-protection circuit \((Y_3)\) leads to a decrease in the current in the maximum-protection relay and to an increase in the current in the relay of the protection against internal faults, that is, to a certain increase in the sensitivity of the protection

from internal faults and to the blocking of the backup overcurrent protection. The increase in the sensitivity of the protection against internal faults is small, since \(Y_4 \gg Y_3\). When the conductance \(Y_4\) is reduced, the protection against internal faults is similarly blocked and the sensitivity of the overcurrent protection is increased. As already noted, all the indicated phenomena occur only in the phase in which the transient resistance has increased. For the remaining phases, the protection parameters remain unchanged. For three-phase protection, a change of resistance in one of the phases does not lead to a change in the parameters of the protection as a whole.

The choice of the power in the circuit of the protection against internal short circuits when it operates, \(P_2\), is determined by the operating current \(I'_{\mathrm{cp}2}\) adopted for the protection relay and by the total resistance of the secondary circuit \(Z_4\) (39). When determining the stack thickness of the DTT core by formula (36), one should take the greatest realistically possible resistance \(Z_{1\mathrm{M}}\) and adopt such an operating current \(I'_{\mathrm{cp}2}\) for which the required stack thickness \(a\) would not exceed structurally acceptable dimensions. The value obtained in this way,

\[ P_2 = (I'_{\mathrm{cp}2})^2 Z_{4\mathrm{M}} \]

is substituted into the remaining calculation formulas.

Under actual conditions, when a manufactured DTT is used, the secondary-circuit resistance will generally be less than \(Z_{4\mathrm{M}}\). In order not to have to change the stack thickness of the DTT core, under these conditions it is necessary to adopt an increased operating current \(I'_{\mathrm{cp}2}\) such that the power \(P_2\) remains equal to the calculated value. This is connected with the need to change the number of turns \(W_4\) (35).

Similarly, when the actual resistance \(Z_3\) differs from the calculated one (37 and 38), the number of turns \(W_3\) and the operating current of the overcurrent protection \(I'_{\mathrm{cp}1}\) should be changed accordingly. Consequently, when manufacturing the DTT, taps must be provided on its secondary windings for adjusting the protection at secondary-circuit resistances that differ from the calculated ones. The required number of taps should be determined in each particular case.

In conclusion, we note that, when determining the conductance of the short-circuited winding by formula (44), the quantity \(Y_{\mathrm{os}}\) should be taken into account; as already indicated, it is most expedient to determine it experimentally.

Example of calculating a DTT for a TVS-30 turbogenerator

Initial data

Turbogenerator type TVS-30.
Rated current \(I_n = 2065\) A.
Subtransient reactance along the direct axis \(X'_d = 0.152\).
DTT core material: E330 steel without additional heat treatment. The maximum magnetic permeability of this steel corresponds to an induction \(B_c = 5700\) G at \(a w_c = 0.445\ \text{A/cm}\) and \(T_c = 55^\circ\). The greatest calculated resistance of the secondary circuits \(Z_{3\mathrm{M}} = Z_{4\mathrm{M}} = 2\ \Omega\). The operating current of the protection against internal faults is taken equal to half the rated current, \(I_{c3} = 1032\) A.

As the relay for protection against internal faults, we take a relay of type ET521/10 with setting \(I'_{\mathrm{cp}2} = 8.4\) A. The phase angle of the secondary circuit is approximately \(\varphi = 0\).

We take the ratio of the pickup current of the overcurrent protection to the pickup current of the protection against internal faults in the case of a phase-to-phase short circuit inside the generator to be equal to unity: \(\alpha = 1\).

The following has been adopted constructively:

width of the DTT window—\(\delta = 13.5\) cm;
width of the steel stack—\(a = 3.5\) cm;
length of the mean magnetic line \(2l_c = 121\) cm.

We take the detuning (margin) coefficient of the protection against internal faults from currents in the relay under through currents in the generator as \(K_3 = 2.2\).

According to condition (55), we check the feasibility of the overcurrent protection

\[ \alpha K_3 = 1 \cdot 2.2 < \frac{K_{\mathrm{n}} I_{\mathrm{n}}}{K_{\mathrm{v}} I_{\mathrm{kz}}} \left( \frac{I_{\mathrm{cz}}}{I_{\mathrm{oc}}} - 1 \right) = \frac{K_{\mathrm{n}} X_d''}{K_{\mathrm{v}}} \left( \frac{I_{\mathrm{cz}}}{\alpha w_c l_c} - 1 \right) = \]

\[ = 1.4 \cdot 0.152 \left( \frac{1032}{27} - 1 \right) = 7.85 . \]

The condition is satisfied. Consequently, the overcurrent protection is feasible.

The required number of turns of the secondary winding for protection against internal phase-to-phase short circuits at \(I'_{\mathrm{cp}2} = 8.4a\), according to formula (35),

\[ w_4 = \frac{ \frac{10.33}{10.33+1}\,4032 - 0.445 \cdot 60.5 }{8.4} = 109 \text{ turns}, \]

where

\[ A = \frac{K_3 K_{\mathrm{v}}}{K_{\mathrm{n}} X_d''} = \frac{2.2}{1.4 \cdot 0.152} = 10.33 . \]

The required thickness of the core stack according to formula (36):

\[ a = \frac{8.4^2 \cdot 2 \cdot \sqrt{2} \cdot 10^8} {5700 \cdot 314 \cdot 3.5 \cdot 918} = 3.5 \text{ cm} \]

The required number of turns of the secondary winding for the overcurrent protection according to formula (37)

\[ w_3 = \left( \frac{10.33}{11.33}\,1032 - 27 \right) \sqrt{ \frac{ Z_3 \left( \frac{1}{11.33}\,1032 - 27 \right) }{ 141 \left( \frac{10.33}{11.33}\,1032 - 27 \right) } } = 20.45 \sqrt{Z_3}. \]

The pickup current of the overcurrent-protection relay according to formula (38)

\[ I'_{\mathrm{cp}1} = \alpha \sqrt{ \frac{ p_3 \left( \frac{1}{\alpha A + 1} I_{\mathrm{cz}} - I_{\mathrm{oc}} \right) }{ Z_3 \left( \frac{\alpha A}{\alpha A + 1} I_{\mathrm{cz}} - I_{\mathrm{oc}} \right) } } = \frac{3.14}{\sqrt{Z_3}} . \]

We shall reduce the dependences of \(w_3\) and \(I'_{\mathrm{cp}1}\) on \(Z_3\) to Table 1.

It follows from Table 1 that, in the considered range of resistances \(Z_3\), an ET-521/6 type relay is applicable for overcurrent protection; its resistance with parallel connection of the coils (for pickup currents from 3 to 6 A) is \(0.01\ \Omega\), and with series connection of the coils (for pickup currents from 1.5 to 3 A) is \(0.04\ \Omega\).

Table 1

For values of the resistance of the secondary circuit of the internal-fault protection differing from the calculated value ($Z_{4\,\mathrm{m}} = 2$ ohms), the power in this circuit when the protection operates must remain constant,

\[ P_2 = (I'_{\mathrm{sr}2})^2 Z_4 = 8.4^2 \cdot 2 = 141 \text{ VA}. \]

From this we find the dependence of the operating current of the relay of this protection on the resistance $Z_4$:

\[ I'_{\mathrm{sr}2}=\sqrt{\frac{141}{Z_4}}. \]

The number of turns $w_4$ in this case (35) varies inversely with the value $I'_{\mathrm{sr}2}$:

\[ w_4=\frac{109\cdot 8.4}{I'_{\mathrm{sr}2}}. \]

The results of the calculation are summarized in Table 2.

Table 2

From the data of Table 2 it is evident that, over the entire range of resistances considered, for the internal-fault protection a relay of type ET521/20 is applicable, with adjustment of the operating current within the range from 5 to 20 A.

We note that the range of resistances considered corresponds to lengths of copper connecting leads from 50 to 270 m for a conductor cross section of 2.5 mm², and from 32 to 185 m for 1.5 mm². The actual lengths of connecting leads at power stations do not exceed these limits.

From Tables 1 and 2 it is evident that the secondary windings of the DTT must have taps that can be selected in accordance with the tables. When setting up the protections, taps should be selected in accordance with Tables 1 and 2 with the nearest smaller number of turns, since this is associated only with some increase in the safety factor, which is quite permissible.

Let us find the required shunt conductance $Y_{\mathrm{sh}}$ from formula (41).

\[ Y_{\mathrm{sh}}= 542\left\{ \sqrt{\left[ \frac{2.2\cdot 2065\cdot 11.33}{1032\cdot 0.152\cdot 10.33} \right]^2-1-0} \right\} = 17200 \ \frac{1}{\mathrm{ohm}}, \]

where

\[ Y= \frac{1032\cdot 1\cdot 1033\cdot 918}{141(11.33)^2} = 542 \ \frac{1}{\mathrm{ohm}}. \]

Calculations using the corresponding formulas [1, 2, 3] show that, in order to obtain the conductance \(Y_{\mathrm{osh}} = 17200\ 1/\mathrm{ohm}\), an air gap in the middle core exceeding the length of this core is required.

Consequently, the DCT being calculated must be made without a middle core and with an additional short-circuited winding. To calculate the additional short-circuited winding it is necessary to know the leakage conductance \(Y_{cs}\). To determine the latter, a mock-up of the DCT was made, on which the conductances were determined experimentally:

\[ Y_{os}=9100\ 1/\mathrm{ohm} \quad \text{and} \quad Y_{\mathrm{osh}}=8500\ 1/\mathrm{ohm}. \]

Using formula (44), we find the required conductance of the short-circuited winding:

\[ Y_k = 542 \left\{ \frac{9100}{26100} \sqrt{ \left[ \frac{2.2 \cdot 2065 \cdot 11.33} {1032 \cdot 0.152 \cdot 10.33} \right]^2 - \left(\frac{8500}{542}\right)^2 - 1 } \right\} = 4700\ 1/\mathrm{ohm}. \]

If the short-circuited winding is made without an added resistance, then the current density in it at the rated generator current is determined by formula (48a):

\[ j_k = \frac{2065} { 2 \cdot 1.85 \cdot 10^{-2} \cdot 0.19 \sqrt{ \left(\frac{26100}{9100}\right)^2 (542+4700)^2 - (8500)^2 } } = \]

\[ =16.8\ \mathrm{A}/\mathrm{mm}^2, \]

where

\[ \rho = 0.0185\ \frac{\mathrm{ohm}\,\mathrm{mm}^2}{\mathrm{m}} \]

is the resistivity of the copper winding conductor;

\[ l_k = 0.19\ \mathrm{m} \]

is the structurally determined mean length of a turn of the short-circuited winding.

The current density has turned out to be excessively large; therefore the short-circuited winding must be made with an added resistance connected in series.

Let us take the permissible current density in the short-circuited winding as

\[ j_{\mathrm{perm}} = 2\ \mathrm{A}/\mathrm{mm}^2. \]

Then the power of the added resistance, according to formula (54), will be

\[ P_d = \frac{(2055)^2\,4700} { \left(\frac{2 \cdot 8500 + 9100}{9100}\right)^2 (542+4700)^2 + (8500)^2 } - \]

\[ - \frac{ 2 \cdot 2055 \cdot 4700 \cdot 0.0185 \cdot 0.19 \cdot 2 } { \sqrt{ \left(\frac{2 \cdot 8500 + 9100}{9100}\right)^2 (542+4700)^2 + (8500)^2 } } = 120\ \mathrm{W}. \]

Let us take the conductor cross section of the short-circuited winding as

\[ q_k = 1.65\ \mathrm{mm}^2 \]

(the standard conductor diameter is \(1.45\ \mathrm{mm}\)); then the value of the added resistance according to formula (53) is

\[ R_d = \frac{120}{(2 \cdot 1.65)^2} = 11\ \mathrm{ohm}. \]

Thus, a type PE-150 resistor of \(11\ \mathrm{ohm}\) may be used as the added resistance.

The number of turns of the short-circuited winding according to formula (51)

\[ w_{\mathrm{k}}=\frac{4700}{1.65}\left\{\sqrt{\frac{11\cdot(1.65)^2}{4700}+(1.85\cdot10^{-2}\cdot0.19)^2}+(1.85\cdot10^{-2}\cdot0.19)\right\}= \]

\[ =236\ \text{turns}. \]

The calculated DCT can also be made without a short-circuited winding by increasing the value of \(\alpha\) (41). Indeed, if \(\alpha=2.15\) is adopted, then the condition for feasibility of the maximum protection (55) is satisfied:

\[ \alpha K_{\mathrm{z}}=2.15\cdot2.2=4.74<7.85. \]

If \(P_{\mathrm{z}}=149\ \mathrm{VA}\) is adopted, then the required stack thickness of the DCT core (36)

\[ a=\frac{149\sqrt{2}\cdot10^8}{5700\cdot314\cdot3.5\cdot963}=3.5\ \text{cm}. \]

The number of turns and the operating current of the maximum-protection relay according to (37) and (38)

\[ w_{\mathrm{z}}=963\sqrt{\frac{Z_{\mathrm{z}}\cdot17}{149\cdot963}}=10.5\sqrt{Z_{\mathrm{z}}}, \]

\[ I'_{\mathrm{cp1}}=\frac{3.54}{\sqrt{Z_{\mathrm{z}}}}. \]

We summarize the dependence of \(w_{\mathrm{z}}\) and \(I'_{\mathrm{cp1}}\) in Table 3.

Table 3

Here, as in the preceding case, a relay of type ET521/6 is applicable.

For \(P_{\mathrm{z}}=149\ \mathrm{VA}\), the dependence of \(w_4\) and \(I'_{\mathrm{cp2}}\) on \(Z_4\) will be:

Table 4

The required conductance according to formula (41)

\[ Y_{\mathrm{sh}}=276\sqrt{\left[\frac{2.2\cdot2065\cdot23.2}{1032\cdot0.152\cdot22.2}\right]^2-1}-0=8350\ \frac{1}{\Omega}, \]

where

\[ Y=\frac{1032\cdot 2.15\cdot 10.33\cdot 963}{149\cdot(23.2)^2}=276\ \mathrm{ohm}^{-1}. \]

The actual value in the absence of the middle core, \(Y_{\mathrm{err}}=8500\ \mathrm{ohm}^{-1}\), differs little from the calculated value; consequently, the differential current transformer can be made without the middle core and without a short-circuited winding at \(\alpha=2.15\), which corresponds to the operating current of the maximum protection under internal faults,

\[ I_{\mathrm{op\,max}}=\alpha I_{\mathrm{op}}=2.15\cdot 1032=2216\ \mathrm{A}. \]

(at the generator rated current of 2065 A), which is entirely permissible.

References

1. B. K. Bul’. Determination of errors and limits of applicability of formulas for specific magnetic permeances, Elektrichestvo, No. 4, 1960.

2. B. K. Bul’. Investigation of the field near an air gap and calculation of magnetic permeance. Vestnik elektropromyshlennosti, No. 9, 1959.

3. A. V. Gordon, A. G. Slivinskaya. Direct-Current Electromagnets, GEI, 1960.

4. N. A. Dul’zon. Equivalent circuit of differential current transformers, Izvestiya vuzov, Elektromekhanika, No. 1, 1963.