CALCULATION OF HIGH-VOLTAGE LINES WITH UNIFORMLY DISTRIBUTED LOAD

A. I. ZAITSEV and I. D. KUTYAVIN

The conversion of extended sections of railroads to electric traction, as well as the electrification of agricultural regions of the country, require the application of a special simplified method for calculating a high-voltage network with uniformly distributed load. Indeed, the supply of traction substations on extended sections with a uniform profile will have almost ideal conditions of uniformly distributed load.

The loads of some substations—on mountain and suburban sections—will differ from the load of main-line substations, but they will not have a decisive influence [3] on the operating regime of the line.

The points of power take-off may be spaced 15–30 km apart, depending on the method of servicing and the type of rectifier installations at the traction substations [2]. Smaller distances will probably be used with complete automation of traction substations equipped with a single rectifier unit with single-anode rectifiers.

For calculating high-voltage lines with uniformly distributed load, one may use the known equations of energy transmission with uniformly distributed constants [1]:

\[ u_{\phi 1}=u_{\phi 2}Ch\sqrt{z_0y_0}\,l+I_2\sqrt{\frac{z_0}{y_0}}\,Sh\sqrt{z_0y_0}\,l; \tag{1} \]

\[ I_1=u_{\phi 2}\sqrt{\frac{y_0}{z_0}}\,Sh\sqrt{z_0y_0}\,l+I_2Ch\sqrt{z_0y_0}\,l. \tag{2} \]

Or one may also use approximate formulas obtained from (1) and (2) by expanding the hyperbolic functions in a series:

\[ u_{\phi 1}=u_{\phi 2}\left(1+\frac{z_0y_0}{2}l^2\right)+I_2z_0l\left(1+\frac{z_0y_0}{6}l^2\right); \tag{3} \]

\[ I_1=u_{\phi 2}y_0l\left(1+\frac{z_0y_0}{6}l^2\right)+I_2\left(1+\frac{z_0y_0}{2}l^2\right). \tag{4} \]

When applying these equations for calculating lines with uniformly distributed load, it is necessary to include in their conductance \(y_0\), in addition to the natural conductance of the line, also the conductance of the load referred to one kilometer of length, which may be determined from the specific active load:

\[ g_k=\frac{P}{u_p^2}\cdot 10^{-3}\ \text{siemens/km}, \tag{5} \]

Reactive conductance of the load, referred to one kilometer of line length:

\[ b_{n}=\frac{Q}{u_{p}^{2}}\cdot 10^{-3}\ \text{sim/km}, \tag{6} \]

where \(u_{p}\) is the design voltage in kV, for which one may take the voltage at the end or the average operating voltage of the line,

\(P\) — active load in kW/km, and
\(Q\) — reactive load in kvar/km.

The total impedance of a phase of the line, referred to one kilometer of length:

\[ z_{0}=r_{0}+jx_{0}=z_{0}e^{j\varphi_{z}}. \tag{7} \]

The total admittance of a phase of the line, referred to one kilometer of length:

\[ y_{0}=(g_{l}+g_{n})+j(b_{c}-b_{n})=y_{0}e^{j\varphi_{y}}, \tag{8} \]

where \(r_{0}\) and \(x_{0}\) are the impedances of one kilometer of line in ohms,
\(g_{l}\) and \(b_{c}\) are the admittances of one kilometer of line in siemens,
\(g_{n}\) and \(b_{n}\) are the admittances of the load per one kilometer of line.

In the absence of power transit along the line, one may take \(I_{2}=0\). Then the calculation formulas (1–4) are simplified and take the form:

\[ u_{\phi 1}=u_{\phi 2}\operatorname{Ch}\sqrt{z_{0}y_{0}}\,l; \tag{9} \]

\[ I_{1}=u_{\phi 2}\sqrt{\frac{y_{0}}{z_{0}}}\operatorname{Sh}\sqrt{z_{0}y_{0}}\,l; \tag{10} \]

\[ u_{\phi 1}=u_{\phi 2}\left(1+\frac{z_{0}y_{0}}{2}l^{2}\right); \tag{11} \]

\[ I_{1}=u_{\phi 2}y_{0}l\left(1+\frac{z_{0}y_{0}}{6}l^{2}\right). \tag{12} \]

Formulas (1–4) and (9–12), as applied to lines with uniformly distributed load, give relationships of the type:

\[ u_{\phi 1}=f(l)\quad \text{and}\quad I_{1}=\varphi(l). \]

Using these formulas, we can determine the voltage and current at any point of a high-voltage transmission line with uniformly distributed load without any intermediate calculations.

Let us consider, as an example, the case in which the loads are specified as concentrated loads (Fig. 1).

As is known, a concentrated load will cause the same voltage loss in the line if it is applied at the center of the uniformly distributed load of the given section. We shall replace the concentrated loads by uniformly distributed ones (Fig. 1).

Fig. 1

When determining voltage losses, a fictitious span equal to \(\dfrac{l}{2}\) must be added to the length of the line, since when replacing a concentrated load—

of a uniformly distributed load, point 1 shifts to the right by \(\frac{l}{2}\) (point \(1'\)), which is equivalent to lengthening the line.

When determining the magnitude of the increment current, the fictitious span need not be added, since the current at point \(2'\), according to the preceding conclusions, will correspond to the current at the beginning of section \(1—2\).

Table 1 gives a comparison of calculation methods for a radial line 300 km long, AS—150 conductor, voltage at the terminal substation \(u_{\phi 2}=104.5\) kV, power of each substation \(P_{nc}=6000\) kW, \(\cos\varphi=0.95\), distance between substations \(l_{nc}=30\) km. The line impedances per 1 km are \(r_0=0.21\ \Omega/\text{km}\); \(x_0=0.42\ \Omega/\text{km}\). The line conductances are \(g_l=0\) and \(b_c=2.7\cdot10^{-6}\ \text{S}/\text{km}\).

The active conductance of the load per 1 km of line, on the basis of (5), is:

\[ g_n=\frac{P_{nc}}{l_{nc}u_p^2}\,10^{-3} =\frac{6000}{30\cdot104.5^2}\,10^{-3} =18.3\cdot10^{-6}\ \text{S}/\text{km}. \]

The reactive conductance of the load per 1 km of line, on the basis of (6), is:

\[ b_n=\frac{Q_{nc}}{l_{nc}u_p^2}\cdot10^{-3} =\frac{P_{nc}\operatorname{tg}\varphi}{l_{nc}u_p^2}\cdot10^{-3} =\frac{6000\cdot0.328}{30\cdot104.5^2}\cdot10^{-3} =6.01\cdot10^{-6}\ \text{S}/\text{km}. \]

Columns 2, 3, and 4 of Table 1 give the results of calculating the line shown in Fig. 1 by formulas (1) and (2), taking into account the stepped nature of the loads. The data in columns 5, 6, and 7 were obtained by the most widely used formulas, composed on the basis of equivalent circuits. Finally, the data in columns 8, 9, and 10 were obtained by formulas (9) and (10), and those in columns 11, 12, and 13 by formulas (11) and (12), taking into account the uniformly distributed load. The data in columns 2 and 3, as control values, were computed with a high degree of accuracy.

Comparing the data in columns 2, 5, 8, and 11 with one another, one may conclude that the latter three methods (columns 5, 8, and 11) give results very close to the first. We observe only small discrepancies in the voltages in the third significant figure; the difference in the values of the currents turned out to be more substantial. Thus, for example, the simplified calculation method based on the equivalent circuit gives a current underestimated by 10% (column 6), while calculation with allowance for the uniformly distributed load (columns 9 and 12) gives currents overestimated by 9%. The latter is explained by the fact that, in calculating the load conductances, the voltage at the end of the line, 104.5 kV, was taken in the calculation instead of the average operating voltage, 110 kV; therefore these conductances turned out to be somewhat overestimated.

However, the proposed method for calculating high-voltage lines with uniformly distributed load is very simple. This method made it possible to compute the data of any row of Table 1 without computing the data of the preceding rows, whereas the stepped method requires the successive calculation of the line in each span between loads. Thus, for example, to obtain the data of the last row of Table 1 in the stepped calculation (columns 2 and 3), it was necessary to apply formulas (1) and (2) ten times; while to obtain the data of the last row with allowance for uniformly distributed load, it was necessary to apply, for columns 8 and 9, formulas (9) and (10) once, and for columns 11 and 12, formulas (11) and (12) also once.

The calculation of more complex networks will be reduced, as usual, to determining the current-division point under equal voltages at the supply nodes, cutting the network at this point into elementary sections, and calculating these elementary sections by equations (1), (2), (9—12).

Table 1

After, as is known, the equalizing currents caused by the inequality of the actual voltages at the nodes are determined, these currents are superimposed on the line and new current-division points are determined, etc.

Let us consider, as an example, the calculation procedure for a line with two-sided supply, shown in Fig. 2.

This line has a uniformly distributed load on section \(AB\), while on section \(BC\) there is no power take-off. It is required to find the distribution of voltage and currents in this line if the voltages at nodes \(A\) and \(C\) must be equal, respectively, to \(u_a\) and \(u_c\).

Fig. 2

Let us first assume that the voltages at points \(A\) and \(B\) are identical and equal to \(u_a\); then the current-division point will be at point \(m\), in the middle of section \(AB\).

Using equations (1), (2), (9—12) and proceeding from the specified voltage, under these conditions we determine the voltages at points \(m\) and \(C\), and the currents flowing out of points \(A\), \(B\), and \(C\). If in this case the voltage at point \(C\) proves to differ from the specified voltage \(u_c\), the equalizing current is determined which must flow out of point \(C\) in order to change the voltage at this point. After superimposing the equalizing current on line \(AC\), we find the new current-division point \(n\) and again determine the voltages at points \(n\), \(B\), and \(C\) for the specified voltage \(u_a\), as well as the currents flowing out of \(A\), \(B\), and \(C\). If this time, too, the voltage at \(C\) differs from the actual one, it is necessary once again to determine an additional equalizing current flowing out of point \(C\), and again find the new current-division point \(p\) and determine the voltages \(u_p\), \(u_b\), and \(u_c\), and the currents flowing out of \(A\), \(B\), and \(C\). Thus, using the method of successive approximation, the indicated line can be calculated with a degree of accuracy sufficient for practice.

CONCLUSIONS

1. The method presented above for calculating high-voltage lines with uniformly distributed load is not inferior in accuracy to calculations by equivalent circuits.

2. This method can significantly accelerate and simplify the design of high-voltage networks with uniformly distributed load.

3. This method is applicable in all cases where power is taken from the line in approximately equal portions and at equal intervals. The minimum number of take-offs at which sufficient accuracy is still obtained is 3—4.

4. Formulas (11 and 12), which do not require the use of hyperbolic functions and provide sufficient accuracy, are more convenient for calculation.

REFERENCES

1. Krug K. A. — Fundamentals of Electrical Engineering, vol. II, pp. 503—513, 1946.
2. Rosenfeld V. E. — Electric Traction of Trains, pp. 581—586.
3. Serdinov O. M. — Electricity, No. 9, p. 3, 1947.