SIMPLIFIED DETERMINATION OF SHORT-CIRCUIT CURRENTS IN NETWORKS WITH VOLTAGES UP TO 1000 VOLTS BASED ON THE KNOWN LENGTH AND CROSS-SECTION OF THE SUPPLY LINE

I. F. ANTONOV, B. I. KUDRIN

(Presented by the scientific seminar of the departments of electric power stations and electrical systems and networks)

To verify the stability of busbars, switching and protective apparatus, and the effectiveness of grounding systems, the Rules for Electrical Installations [1, Chs. 1—43] provide for determining the magnitude of three-phase and single-phase short-circuit currents (s.c.c.). When determining the magnitude of the s.c.c., it is necessary to take into account the active and inductive resistances of all elements of the short-circuited circuit, which considerably complicates the calculations, making them very labor-intensive.

G. V. Mirer and M. K. Tulchin [2] developed a method for simplified calculation of the magnitude of s.c.c. for supply and distribution networks (of residential and public buildings) made with cables or conductors laid in pipes and ducts. In developing the method it was assumed that the active components of voltage losses and the design load current are known for each section of the line.

The distribution electrical networks of large shops are formed, as a rule, by trunk lines with a large number of branches to bus assemblies, distribution points, etc. In established design practice, because of the great labor intensity of the calculations, the voltage loss at each electrical receiver is not determined. To ensure the required voltage at all consumers, the voltage loss is determined only up to the most remote electrical receivers. Thus, when using the method [2], it is necessary first to determine the magnitude of the active component of the voltage loss in each section and only then to determine the magnitude of the s.c.c., which practically does not reduce the duration of the calculations.

The proposed method for determining the magnitudes of s.c.c. for shop distribution networks can be applied in the design of networks of industrial enterprises and is based on the use of known lengths and cross-sections of supply lines.

In calculating the s.c.c., the following assumptions are adopted:

a) the resistance of the high-voltage network is not taken into account; the power of the electric system is considered infinitely large, and the periodic component of the s.c.c. nondecaying;

b) algebraic addition of total impedances is used;

c) the nominal voltage of the network is taken to be the nominal voltage of the electrical receivers, and not the no-load voltage of the transformers;

d) contact transition resistances are not taken into account in the calculation.

The assumptions adopted act in the direction of both a certain increase and a decrease of the short-circuit current. The error of the proposed calculation method, in comparison with generally accepted methods, does not exceed 3–5%.

Simplified determination of a three-phase short-circuit current

The periodic component of the short-circuit current may be determined by the formula [2]:

\[ I_v^{(3)}=\frac{100}{\displaystyle \sum_{i=1}^{n}\frac{k_1\Delta U_i}{I_{pi}}+\frac{U_k\%}{I_{\mathrm{nt}}}}, \tag{1} \]

where

\(k_1\) is a coefficient characterizing the ratio of the voltage drop on the given section of the line to the active component of the voltage loss on this same section;

\[ k_1=\frac{\Delta U}{\Delta U_a}=\frac{z}{r\cos\varphi}=\frac{z_0}{r_0\cos\varphi}; \]

\(I_p\) is the calculated current of the given section of the network, A;
\(U_k\%\) is the short-circuit voltage of the transformer, %;
\(I_{\mathrm{nt}}\) is the rated current of the transformer, A;
\(z_0, r_0\) are the total and active resistance of one kilometer of line, ohm/km;
\(\cos\varphi\) is the power factor of the electrical receiver.

The coefficient \(k_1\) is practically independent of the cross-section of the feeder line because of the insignificant value of the inductive resistance of cables and wires. In accordance with [2], it is recommended that the value of \(k_2\) be taken from the data of Table 1. From this table it is evident that the product of the power factor by the coefficient \(k_1\) is close to unity.

Table 1

Let us take the value of the active component of the voltage loss on each section to be equal to

\[ \Delta U_a=\frac{I_p r\cos\varphi\sqrt{3}}{100U_n},\%, \tag{2} \]

where

\(U_n\) is the rated voltage of the network, kV;
\(r\) is the active resistance of the network section, ohm.

Express the value of the active resistance through the known length and cross-section of the network section:

\[ r=\frac{l}{\gamma S},\ \text{ohm}, \tag{3} \]

\(l\) is the length of the network section, m;
\(S\) is the cross-section of the network section, sq. mm;

\(\gamma\) — the specific conductivity of the network material, \(m/ohm \cdot sq.\ mm\). For copper and aluminum, the specific conductivity is taken as equal to 53 and 32, respectively.

Substituting the values (2) and (3) into (1), taking the product of the coefficient \(k_1\) and the power factor as equal to unity, and carrying out transformations, we obtain

\[ I_V^{(3)}= \frac{10^5} {\displaystyle \sum_{i=1}^{n}\frac{100l\sqrt{3}}{U_n\gamma S}+\frac{U_k}{I_{nt}}\cdot 10^3}, \ A. \tag{4} \]

To simplify the analysis of (4), introduce the coefficients:

\[ k_2=\frac{100\sqrt{3}}{U_n\gamma}, \]

\[ k_3=\frac{U_k}{I_{nt}}\cdot 10^3. \]

For networks with a voltage of \(380\ V\), made with wires and cables with aluminum conductors, \(k_2=14.2\); the same with copper conductors gives \(k_2=8.6\).

The coefficient \(k_3\) depends on the nameplate characteristics of the transformer and, for standard oil transformers with ratings of 630, 750, and \(1000\ kVA\) at a lower voltage of \(0.4\ kV\), is respectively equal to 5.8; 5.1; 3.8. Large shops of modern industrial enterprises are characterized by branched distribution networks, in which the sum of the ratios of the lengths of the sections to the cross sections of the network sections is sufficiently large in comparison with the coefficient \(k_3\), and therefore its value may be neglected. In this case formula (4) takes the form

\[ I_V^{(3)}= \frac{10^5} {\displaystyle k_2\sum_{i=1}^{n}\frac{l_i}{S_i}}, \ A. \tag{5} \]

The amplitude value of the peak current of a three-phase short circuit is determined by the formula

\[ i_y=k_y\sqrt{2}\,I_V^{(3)},\ A, \tag{6} \]

\(k_y\) — peak coefficient.

It may be taken as equal to 1.3 for a short circuit on distribution boards up to \(525\ V\), supplied from transformers of \(560\text{--}1000\ kVA\) with a short-circuit voltage of \(5.5\%\), and as equal to 1.0 at remote points of the distribution network.

The largest rms value of the total short-circuit current is determined by the formula

\[ I_y^{(3)}=I_V^{(3)}\sqrt{1+2(k_y-1)^2},\ A. \tag{7} \]

We determine the magnitude of the three-phase short-circuit current at the network point in Fig. 1 for a typical diagram of the supply and distribution network of a shop, taking into account the resistances of all elements of the short-circuit circuit and according to the proposed simplified method. The active and inductive resistances of the power transformer, of the coils (releases) of the maximum-current circuit breakers, of the cables and wires, and the active resistance of the transient contacts of the circuit breakers are found in the usual way [3] and are summed as series-connected. The resulting total resistance

for the example in Fig. 1 is 390 milliohms. The three-phase short-circuit current by the usual method is

\[ I_V^{(3)}=\frac{U_{\text{av}}\cdot 1000}{z_{\text{res}}\sqrt{3}} =\frac{400\cdot 1000}{390\sqrt{3}}=595\ \text{A}. \]

Fig. 1. Calculation diagram for determining the short-circuit current.

We determine the three-phase short-circuit current by the simplified method (5)

\[ I_V^{(3)}=\frac{10^5}{k_2\displaystyle\sum_{i=1}^{n}\frac{l_i}{S_i}} =\frac{10^5}{14.2\left(\frac{80}{120}+\frac{120}{50}+\frac{50}{6}\right)} =612\ \text{A}. \]

The error in determining the short-circuit current is 2.9%, which provides the necessary engineering accuracy.

Simplified determination of a single-phase short-circuit current

The periodic component may be determined by the formula [4]

\[ I_V^{(1)}= \frac{1}{ \displaystyle\sum_{i=1}^{n}\frac{k_4\Delta U_a\%}{100 I_p} +\frac{z_t}{3U_\phi} },\ \text{A}, \tag{8} \]

where

\(k_4\) — a coefficient depending on the conductor cross section, the method of laying the network, and the load power factor;

\(z_t\) — the total impedance of the transformer to a fault current to the enclosure, ohms;

\(U_\phi\) — the nominal phase voltage of the network, V;

The active component of the voltage loss is determined by the formula

\[ \Delta U_a\%=\frac{I_p r_{\phi 0} l\cos\varphi\ 100}{U_\phi}, \tag{9} \]

\(r_{\phi 0}\) — active resistance of 1 km of a phase conductor, ohm/km;

\(l\) — length of the network section, km.

For the purpose of simplification, introduce the coefficient \(k_5\), which depends on the design of the network, and \(k_6\), which depends on the transformer rating and the connection method of its windings:

\[ k_5 = k_4 \cos \varphi, \]

\[ k_6 = \frac{z_{\mathrm{t}}}{3U_{\phi}} . \]

Table 2

Table 3

The coefficients \(k_5\) and \(k_6\) are taken respectively from Table 2 and Table 3. Substituting into (8) the value (9) for a nominal phase voltage of the network of 220 V, after transformation we obtain the simplified formula for determining the periodic component of the short-circuit current:

\[ I_{V}^{(1)} = \frac{b}{k_5 \sum_{i=1}^{n} \frac{l_i}{S_i} + k_6}, \ \mathrm{A}, \tag{10} \]

where

\(b = 7000\) when aluminum conductors are used and \(11200\) for copper conductors.

Conclusions

1. For branched shop distribution networks of alternating current with voltages up to 1000 V, calculation of short-circuit current is possible from the known length and cross-section of the supply line with accuracy sufficient for practice.

2. When the power of the supply transformer exceeds 560 kVA, the short-circuit current may be calculated without taking the transformer resistance into account.

3. The accuracy of the calculation increases as the short-circuit point is moved farther away and as the cross-section of the line feeding the electrical receiver is reduced.

References

1. Rules for Electrical Installations. Gosenergoizdat. Moscow, 1966.

2. G. V. Mirer, I. K. Tulchin. Simplified determination of short-circuit current in AC networks with voltages up to 1000 V. Promyshlennaya energetika, 1966, No. 8.

3. Ya. M. Bolsham et al. Handbook on Electrical Installations of Industrial Enterprises, vol. 1, part I, Gosenergoizdat, 1963.

4. G. V. Mirer, I. K. Tulchin. Simplified determination of single-phase short-circuit currents in lighting networks. Svetotekhnika, 1969, No. 10.