Eng. V. K. Shcherbakov.

Calculation of Voltages and Power Flow Distribution in Complex Electric Power Systems under Normal Operating Conditions

I.

The task of the present work is to examine the question of the present state of the techniques and methods for calculating voltages and power flow distribution in complex electric power systems. At the same time, a method is proposed—one not yet covered in print—for using the nodal-equation method, already known earlier in the practice of calculating low-voltage networks, in calculations of voltages and power flow distribution in complex high-voltage systems.

A complex electric power system is understood to mean an aggregate of power stations and substations connected by high-voltage transmission lines. By connecting individual stations and substations, high-voltage transmission lines, as a rule, form closed circuits which, taken together, may be called the high-voltage network of complex electric power systems.

The calculation of networks of complex systems is, naturally, made more complicated by the need to take into account the self-inductance and capacitance of the lines, since the lines are, as a rule, overhead lines having a predominant inductive resistance, while the high voltages presently used for the transmission of electric energy determine very considerable capacitive currents.

But it is not this latter circumstance that creates difficulties in calculating the networks of complex systems.

As is known, taking account of the capacitive and inductive components of the total line resistances does not introduce fundamental changes into the methods of network calculation, but only complicates the calculation process. What is new in the problem is introduced by the initial conditions of the calculations: the loads at the supply nodes are specified, and it is required to find the voltages at these and other nodes; whereas in calculations of ordinary distribution networks the opposite is true: the voltages at the supply nodes are specified, and it is required to find the loads of these and other nodes.

However, even with this formulation of the question, the problem could easily be solved by the known method of nodal equations—exactly—and by the transformation method—approximately—if the loads at the nodes were specified in currents. It is easy to see that this latter is impossible, if only for the reason that the voltages, which together with the currents determine the powers at the nodes, are the unknown quantities.

Thus, in this case it is impossible to write the current balance for each node of the network. It is possible to write the power balance, but as a result, with respect to the node voltages, a system of second-order equations is obtained. The exact solution of a system of second-order equations is a very complex and so laborious a process that this, in principle such a simple problem, is practically insoluble—at least if one has in mind an absolutely exact solution of it.

It is not difficult to show that any exact solution of the problem of calculating voltages and power flow distribution in complex systems, whatever met—

...data it may be carried out on, will ultimately lead to a system of equations of the second degree. And, consequently, on the path to every exact solution one encounters the difficulties indicated above, associated with solving a system of equations of the second degree.

Avoiding this difficulty, the practice of calculations—first of all in America—has proceeded along the line of creating models of complex systems, on which, by selecting from ohmic, inductive, and capacitive resistances an equivalent circuit of the system being calculated, one finds the distribution of powers in the individual links of the system and the voltages at the nodes, the latter differing from the actual values by a definite scale factor.

Such a method of calculating systems has recently been gaining wide dissemination both abroad and here in the USSR, thereby emphasizing the unquestionable advantages of this method of calculation over others.

However, only individual design and production units can make use of network models, since these models are very costly and significantly cumbersome. This is why the search for ways and methods of calculating complex systems without resorting to the aid of calculating models has not slackened recently; and it must be assumed that, in the end, these searches will result in a number of quite satisfactory solutions to the problem. But, as was said above, these solutions apparently must inevitably be associated with a number of assumptions, and, consequently, the results of these solutions, however closely they may approximate the true value of the quantities sought, will nevertheless always differ from the values of the latter. This assertion is all the more justified because solving a system of equations of the second degree will certainly always be more complicated than solving a system of equations of the first degree; and the latter, especially with a significant number of nodes in the system, already leads to cumbersome computations, which the practice of calculating networks of complex systems will avoid at the first opportunity.

Until very recently one could note only two proposals on calculating voltages and power-flow distribution in complex systems that had been published in print. These are, first, the calculation method that became widespread in the MOGES system and was described in the journal Elektrichestvo for 1932 by engineers Baskakov and Solov’ev, and then supplemented by engineer Baskakov in 1934. This method of calculating systems is an approximate method, making very successful use of the concept of the average line power; in view of the difficulty, noted above, of obtaining an exact solution of the problem, it must be acknowledged that the method of average powers is a comparatively simple and convenient method and at the same time gives calculation accuracy, in a very large number of cases, sufficient for practical purposes. This is, secondly, engineer Pevzner’s proposal to calculate the closed networks of complex systems as radial ones, with subsequent superposition of additional powers obtained at the point of cutting the network from the difference of potentials when calculating the latter as an open network. This latter method has not yet received practical testing, and at present it cannot be said that this method has gained dissemination even within any single organization.

Very recently a book appeared in print, written by Prof. A. A. Glazunov jointly with Prof. A. Ya. Ryabkov and Docent F. P. Lashkov, in which the authors set forth in great detail and thoroughly the possible methods and procedure for calculating the networks of power systems; at the same time, as the principal method for calculating complex systems they recommend the method of network transformation modernized by the authors (if one may so express it).

Looking somewhat ahead, it should be noted that the method of converting complex networks into simple ones, or the so-called method of network transformation, is the principal method for calculating low-voltage networks. If this method directly...

to apply it to the calculation of complex systems, it is necessary to make at least two essential assumptions: first, to adopt a Π-shaped equivalent circuit for the lines of complex systems and, consequently, constancy of the current magnitude along the entire line (which in the vast majority of cases is quite admissible), and, second, to express the loads at the nodes, which are usually given in terms of powers, in terms of currents, having first assumed at all nodes of the network certain presumed voltage values, or the same average voltage at all nodes (for example, the nominal voltage of the network). In the latter operation, i.e., in expressing loads in currents, in many cases an inaccuracy is obtained which leads to results of network calculation that to a considerable degree do not correspond to reality and therefore cannot be used for theoretical and practical purposes.

Prof. A. A. Glazunov and others, in the book mentioned above, proved that the transformation of networks of power systems can also be carried out in the case when the loads at the nodes are expressed in powers. The assumptions that have to be made when using the transformation method for calculating networks of complex systems make this calculation method approximate; however, the errors in the results of the calculations, as is shown in the examples below, are of a magnitude that permits this method of calculating complex systems to be used in the majority of cases of practical calculations. In this new capacity, the transformation method is not inferior in the accuracy of the calculation results to the method of average powers, while at the same time it has a number of advantages over it from the point of view of calculation technique.

In the present work, on the contrary, it is proposed not to abandon the solution of the basic nodal equations, but to obtain this solution by an indirect route, whereby by this method an accuracy of solution of any order can be obtained. This proposal, while not claiming to exclude the above-mentioned methods of solving the problem, may nevertheless in some cases prove to be the only possible way of obtaining results for the calculation of voltages and loads in complex systems that are very close to the true ones.

The present work, like other similar works devoted to the calculation of complex systems, is called forth by the enormous growth of complex electrical systems in the Union, both in terms of the capacities they interconnect and in terms of the extent of the networks of these complex systems. A justification for the appearance of the proposed work may also be the statement by Prof. A. A. Glazunov, made by him in a private letter, that “questions of calculating the electrical networks of systems have only just begun to be developed, and therefore any work that gives something new in this field is of great importance.”

In concluding the introductory part, it is necessary to clarify some of the terminology that must be used here, and also to point out that throughout the work only the normal operating condition of complex systems is considered.

The concept of “complex systems” includes the aggregate of stations, substations, and transmission lines; the concept of “networks of complex systems” includes only the aggregate of transmission lines connecting stations and substations. By analogy with the calculation of low-voltage distribution networks, one may speak of the calculation of “networks of complex systems,” since here too the question is that of finding the power-flow distribution in the network.

But the calculation of “networks of complex systems,” or, in other words, “networks of power systems,” has the feature that this calculation, on the one hand, is specified by the condition of the stations—the power they deliver—and, on the other hand, this same calculation determines the operating condition of the stations—the operating voltage. Moreover, since the voltage at the feeding nodes of the network of a complex system depends on the voltage drop in the mains supplying energy from the stations

to the nodes, these mains are also included in the calculation. Thus, here one may, as it were, speak of calculating the voltages, power-flow distribution, and loads of the system as a whole. It is only necessary to specify that the discussion concerns the calculation of voltages and power-flow distribution in complex systems under normal operating conditions, i.e., the consideration of short-circuit conditions and of the stability of parallel operation is excluded here.

II.

Of all methods for calculating networks, the method of nodal equations is unquestionably the most general method of calculation. The essence of the method consists in setting up, according to Kirchhoff’s law, a system of equations for the equilibrium of currents at the nodes of the network. Expressing the loads along the links of the network through the voltages and constants of the links, one obtains a system of equations with the unknown voltages at the nodes.

Having in mind that the condition of equilibrium of currents at each node of the network may be interpreted as the condition of equilibrium of the powers flowing into the node and taken from it, one may, in the most general case, apply the method of nodal equations also to the calculation of complex systems.

Fig. 1.

Let Fig. 1 represent a part of the network of a complex system; then for node “2” one may write the condition of power equilibrium in the form:

\[ \overline{W}_{\mathrm{I}(k)}-\overline{W}_{\mathrm{II}(n)}=\overline{W}_2 \qquad\ldots\ldots\ldots\ldots (1) \]

Here: \(\overline{W}_{\mathrm{I}(k)}\)—the power at the end of link I,

\[ \overline{W}_{\mathrm{II}(n)}\text{—the power at the beginning of link II.} \]

The voltages and currents at the beginning and end of each link, and, for example, of links 1 and 2, are connected by the equations:

\[ \left. \begin{aligned} \overline{E}_1&=\overline{A}_1\cdot\overline{E}_2+\overline{B}_1\cdot\overline{J}_{\mathrm{I}(k)}\\ \overline{J}_{\mathrm{I}(n)}&=\overline{C}_1\cdot\overline{E}_2+\overline{D}_1\cdot\overline{J}_{\mathrm{I}(k)} \end{aligned} \right\} \qquad\ldots\ldots\ldots\ldots (2) \]

—for link I, and

\[ \left. \begin{aligned} \overline{E}_2&=\overline{A}_{\mathrm{II}}\cdot\overline{E}_3+\overline{B}_{\mathrm{II}}\cdot\overline{J}_{\mathrm{II}(k)}\\ \overline{J}_{\mathrm{II}(n)}&=\overline{C}_{\mathrm{II}}\cdot\overline{E}_3+\overline{D}_{\mathrm{II}}\cdot\overline{J}_{\mathrm{II}(k)} \end{aligned} \right\} \qquad\ldots\ldots\ldots\ldots (3) \]

—for link II.

From the first equation of expression (2), we find

\[ \overline{J}_{\mathrm{I}(k)}= \frac{\overline{E}_1-\overline{A}_1\cdot\overline{E}_2}{\overline{B}_1} \]

and

\[ \overline{W}_{\mathrm{I}(k)} = \overline{J}_{\mathrm{I}(k)}\cdot\hat{E}_2 = \frac{\hat{E}_2\left(\overline{E}_1-\overline{A}_1\cdot\overline{E}_2\right)}{\overline{B}_1} \qquad\ldots\ldots\ldots (4) \]

Then from the equations of expression (3) we obtain:

\[ \overline{J}_{\mathrm{II}(n)} = \frac{\left(\overline{C}_{\mathrm{II}}\cdot\overline{B}_{\mathrm{II}}-\overline{A}_{\mathrm{II}}\cdot\overline{D}_{\mathrm{II}}\right)\cdot\overline{E}_3+\overline{D}_{\mathrm{II}}\overline{E}_2} {\overline{B}_{\mathrm{II}}}; \]

Expression

\[ \overline{C}_{\mathrm{II}}\cdot \overline{B}_{\mathrm{II}} - \overline{A}_{\mathrm{II}}\cdot \overline{D}_{\mathrm{II}} = -1 \]

and therefore

\[ \overline{J}_{\mathrm{II}(n)} = \frac{-\overline{E}_3+\overline{D}_{\mathrm{II}}\cdot \overline{E}_2} {\overline{B}_{\mathrm{II}}} \]

and, consequently,

\[ \overline{W}_{\mathrm{II}(n)} = \frac{ \hat{E}_2\cdot\left(-\overline{E}_3+\overline{D}_{\mathrm{II}}\cdot \overline{E}_2\right) } {\overline{B}_{\mathrm{II}}} \qquad . . . . . . . . (5) \]

Now nodal equation (1) will take the form:

\[ \frac{ \hat{E}_2\left(\overline{E}_1-\overline{A}_1\cdot\overline{E}_2\right) } {\overline{B}_{\mathrm{I}}} - \frac{ \hat{E}_2\left(-\overline{E}_3+\overline{D}_{\mathrm{II}}\cdot\overline{E}_2\right) } {\overline{B}_{\mathrm{II}}} = \overline{W}_2 \qquad . . . . (6) \]

Equations similar to (6) can be written for all nodes of the system; however, bearing in mind the possibility of taking the voltage of one node as known, one may confine oneself to \((n-1)\) equations, if the number of nodes is denoted by “\(n\).” The exclusion of the equation of one of the nodes, and necessarily a supply node, is also necessary because the load of all nodes of the system, with the exception of at least one, may be known, since the losses in the system network are still unknown.

Thus, for any complex system one can set up \((n-1)\) equations, with \((n-1)\) unknown voltages.

Solving this system of equations, we shall find the voltages, and then, very simply, the power flows in the links.

It is necessary to note the generality of equation (6). Like equations (2) and (3), which represent in general form the relation between the voltages and currents of a transmission line, equation (6) relates, in general form, the electrical state of two adjacent links of the network of a complex system. The form of equation (6) does not change in the case when several links are conjugated at a node (more than 2); in this case only new terms are added to its left-hand side, each representing the additional links.

Thus, the method of nodal equations is a general method for calculating the electrical state of closed electric systems. And the determination of voltages and of the distribution of power flows in complex systems is reduced, i.e., to the solution of a system of nodal equations. However, difficulties arise precisely here, for it is easy to see that this will be a system of equations of the 2nd order, and solving a system of equations of the 2nd order by a direct method is an extremely difficult task.

The system of nodal equations can be solved comparatively simply by an indirect method, if the loaded state of the electric system is regarded as the result of the superposition of several partial states, of which the first is represented by arbitrarily chosen load currents at the nodes, greater or less than the actual ones, and the subsequent ones by currents gradually supplementing the current of each node to its actual value, corresponding to the specified power at the node and to the voltage calculated for the preceding partial state.

Each stage of the system load, being represented by definite currents at the nodes, makes it possible, by means of the nodal equations, simply to find the voltages corresponding to the given stage of load; here one has to deal with a system of linear equations, the solution of which presents no particular difficulties.

Let the complex system (Fig. 1) be loaded at the \((n-1)\) nodes by arbitrary currents:

\[ \bar J'_2,\ \bar J'_3,\ \bar J'_4 \ldots \text{ etc.} \]

Then the electrical state of the system can be expressed by the set of equations, according to (6),

\[ \left. \begin{aligned} &\frac{\bar E_1-\bar A_1\cdot \bar E_2}{\bar B_1} -\frac{-\bar E_3+\bar D_{\mathrm{II}}\cdot \bar E_2}{\bar B_{\mathrm{II}}} = \bar J'_2,\\[4pt] &\frac{\bar E_2-\bar A_{\mathrm{II}}\cdot \bar E_3}{\bar B_{\mathrm{II}}} -\frac{-\bar E_4+\bar D_{\mathrm{III}}\cdot \bar E_3}{\bar B_{\mathrm{III}}} -\frac{\bar E_5+\bar D_{\mathrm{V}}\cdot \bar E_3}{\bar B_{\mathrm{V}}} = \bar J'_3,\\[4pt] &\frac{\bar E_3-\bar A_{\mathrm{III}}\cdot \bar E_4}{\bar B_{\mathrm{III}}} -\frac{-\bar E_5+\bar D_{\mathrm{IV}}\cdot \bar E_4}{\bar B_{\mathrm{IV}}} = -\bar J'_4,\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &\text{etc.} \end{aligned} \right\} \qquad \ldots (7) \]

Solving equations (7), whether by substitution and elimination, or by the method of determinants, we obtain expressions for the voltages at the nodes in the form:

\[ \left. \begin{aligned} \bar E'_1 &= \bar E_{10}+\bar K_{21}\cdot \bar J'_2+\bar K_{31}\bar J'_3+\bar K_{41}\cdot(-\bar J'_4)+\ldots\\ \bar E'_2 &= \bar E_2=\mathrm{const}\ldots\ldots \text{(taken as known)}\\ \bar E'_3 &= \bar E_{30}+\bar K_{23}\cdot \bar J'_2+\bar K_{33}\cdot \bar J'_3+\bar K_{43}\cdot(-\bar J'_4)+\ldots\\ \bar E'_4 &= \bar E_{40}+\bar K_{24}\cdot \bar J'_2+\bar K_{34}\cdot \bar J'_3+\bar K_{44}(-\bar J'_4)+\ldots \end{aligned} \right\} \qquad \ldots (8) \]

In expression (8):

\(\bar E_{10}, \bar E_{30}, \bar E_{40}\ldots\) etc. are the true values of the voltages at the nodes under no-load operation of the system, i.e. when \(\bar J_2, \bar J_3, \bar J_4\)—etc.—\(=0\), while \(\bar J_1 \ne 0\). \(\bar K_{21}, \bar K_{23}, \bar K_{24}\) etc. are coefficients depending on the constants of the links of the system network; the indices attached to the coefficients indicate the relation to the corresponding current and voltage.

\(\bar J'_4\) is taken with a minus sign because the 4th node is a supply node. The voltage of the 2nd node (any other node could have been chosen) is taken as constant and coinciding with the axis of real quantities.

The powers at the nodes corresponding to the adopted current loads at the nodes and to the voltages found from expression (8) will be:

\[ \left. \begin{aligned} \bar W'_2 &= \bar J'_2\cdot E_2\\ \bar W'_3 &= \bar J'_3\cdot \hat E'_3\\ \bar W'_4 &= \bar J'_4\cdot \hat E'_4 \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots (9) \]

etc.

Thus, the system is underloaded (or overloaded) at each node by a power equal to the difference between the actual power of the nodes and that calculated from expression (8), i.e.

at the 2nd node by \(\bar W_2-\bar W'_2\),

at the 3rd node by \(\bar W_3-\bar W'_3\),

etc., or, in currents:

by

\[ \frac{\bar W_2-\bar W'_2}{E_2}=\bar i_2(\ne 0), \]

by

\[ \frac{\overline{W}_3-\overline{W}'_3}{\hat{E}'_3}=\bar{i}_3 \]

\[ \cdots\cdots\cdots\cdots \]

and so on.

We shall add these missing currents \(\bar{i}_2, \bar{i}_3, \bar{i}_4\), etc., respectively to the load currents of the first stage and obtain the loads in currents of the second stage of approximation:

\[ \begin{aligned} \bar{J}''_2&=\bar{J}'_2+\bar{i}_2\\ \bar{J}''_3&=\bar{J}'_3+\bar{i}_3\\ \bar{J}''_4&=\bar{J}'_4+\bar{i}_4 \end{aligned} \qquad \cdots\cdots\cdots\cdots \tag{10} \]

\[ \cdots\cdots\cdots\cdots \]

and so on.

But, as is not difficult to verify, these same load currents can be obtained directly by dividing the actual powers at the nodes by the voltages of the first stage of approximation. Indeed:

\[ \left. \begin{aligned} \bar{J}''_2&=\bar{J}'_2+\bar{i}_2 =\frac{\overline{W}'_2}{E_2} +\frac{\overline{W}_2-\overline{W}'_2}{E_2} =\frac{\overline{W}_2}{E_2} \\[6pt] \bar{J}''_3&=\bar{J}'_3+\bar{i}_3 =\frac{\overline{W}'_3}{\hat{E}'_3} +\frac{\overline{W}_3-\overline{W}'_3}{\hat{E}'_3} =\frac{\overline{W}_3}{\hat{E}'_3} \\[6pt] \bar{J}''_4&=\bar{J}'_4+\bar{i}_4 =\frac{\overline{W}'_4}{\hat{E}'_4} +\frac{\overline{W}_4-\overline{W}'_4}{\hat{E}'_4} =\frac{\overline{W}_4}{\hat{E}'_4} \end{aligned} \right\} \qquad \cdots\cdots \tag{11} \]

\[ \cdots\cdots\cdots\cdots \]

Loading the system with the new currents \(\bar{J}''_2, \bar{J}''_3, \bar{J}''_4\), etc., we obtain, respectively, new values of the voltages at the nodes \(\overline{E}''_1, \overline{E}''_3, \overline{E}''_4\), etc.

It is quite obvious that with each new stage of recalculation we shall come closer and closer to the true values of the voltages at the nodes. As will be shown below, the approximation to the true voltage values is obtained very rapidly, and already the third stage of approximation gives voltage values differing from the true ones by fractions or tenths of a percent.

III.

Let us show by an example the application of the nodal-equation method to the calculation

Fig. 2.

of voltages and power-flow distribution in a complex system. Suppose we have a system specified by Fig. 2.

Let us set up, for the given example, the system of nodal equations according to (6):

\[ \left\{ \begin{aligned} &\frac{\dot E_2\left(\overline E_1-\overline A_{\mathrm I}\cdot \overline E_2\right)}{\overline B_{\mathrm I}} - \frac{\dot E_2\left(-\overline E_3+\overline D_{\mathrm{II}}\cdot \overline E_2\right)}{\overline B_{\mathrm{II}}} = \overline W_2, \\[6pt] &\frac{\dot E_3\left(\overline E_2-\overline A_{\mathrm{II}}\cdot \overline E_3\right)}{\overline B_{\mathrm{II}}} - \frac{\dot E_3\left(-\overline E_4+\overline D_{\mathrm{III}}\cdot \overline E_3\right)}{\overline B_{\mathrm{III}}} = -\overline W_3, \\[6pt] &\frac{\dot E_4\left(\overline E_3-\overline A_{\mathrm{III}}\cdot \overline E_4\right)}{\overline B_{\mathrm{III}}} - \frac{\dot E_4\left(-\overline E_1+\overline D_{\mathrm{IV}}\cdot \overline E_4\right)}{\overline B_{\mathrm{IV}}} = \overline W_4. \end{aligned} \right. \]

or:

\[ \left\{ \begin{aligned} &\overline B_{\mathrm{II}}\cdot \overline E_1 - \left(\overline A_{\mathrm I}\cdot \overline B_{\mathrm{II}} + \overline B_{\mathrm I}\cdot \overline D_{\mathrm{II}}\right)\cdot \overline E_2 + \overline B_{\mathrm I}\cdot \overline E_3 = \frac{\overline W_2}{\dot E_2}\cdot \overline B_{\mathrm I}\cdot \overline B_{\mathrm{II}}, \\[6pt] &\overline B_{\mathrm{III}}\cdot \overline E_2 - \left(\overline A_{\mathrm{II}}\cdot \overline B_{\mathrm{III}} + \overline B_{\mathrm{II}}\cdot \overline D_{\mathrm{III}}\right)\cdot \overline E_3 + \overline B_{\mathrm{II}}\cdot \overline E_4 = -\frac{\overline W_3}{\dot E_3}\cdot \overline B_{\mathrm{II}}\cdot \overline B_{\mathrm{III}}, \\[6pt] &\overline B_{\mathrm{IV}}\cdot \overline E_3 - \left(\overline A_{\mathrm{III}}\cdot \overline B_{\mathrm{IV}} + \overline B_{\mathrm{III}}\cdot \overline D_{\mathrm{IV}}\right)\cdot \overline E_4 + \overline B_{\mathrm{III}}\cdot \overline E_1 = \frac{\overline W_4}{\dot E_4}\cdot \overline B_{\mathrm{III}}\cdot \overline B_{\mathrm{IV}}. \end{aligned} \right. \]

Taking, furthermore, as the basis a \(\Pi\)-type equivalent circuit, we calculate the constants \(\overline A\), \(\overline B\), and \(\overline D\) by the formulas:

\[ \overline A = 1 + \frac{\overline y \cdot \overline Z}{2}, \]

\[ \overline B = \overline Z \]

\[ \overline D = 1 + \frac{\overline y \cdot \overline Z}{2} = A^{-*} \]

The results of the calculations are summarized in Table 1.

Table No. 1.

Link	\(\overline A\)	\(\overline B\)
I	\(0{,}994 \angle 14'\)	\(26{,}2 \angle 54^\circ\)
II	\(0{,}977 \angle 48'\)	\(48{,}8 \angle 58^\circ 30'\)
III	\(0{,}987 \angle 32'\)	\(39{,}6 \angle 54^\circ\)
IV	\(0{,}949 \angle 1^\circ 38'\)	\(67{,}8 \angle 61^\circ 54'\)

We substitute the values of the constants into the nodal equations and finally obtain the system of calculation equations in the form:

\[ \left\{ \begin{aligned} &\overline E_1\cdot 48{,}8 \angle 58^\circ 30' - \overline E_2\cdot 74{,}07 \angle 57^\circ 23' + \overline E_3\cdot 26{,}2 \angle 54^\circ = \frac{\overline W_2}{\dot E_2}\cdot 1278 \angle 112^\circ 30', \\[6pt] &\overline E_2\cdot 39{,}6 \angle 54^\circ - \overline E_3\cdot 86{,}7 \angle 57^\circ 7' + \overline E_4\cdot 48{,}8 \angle 58^\circ 30' = -\frac{\overline W_3}{\dot E_3}\cdot 1932 \angle 112^\circ 30', \\[6pt] &\overline E_3\cdot 67{,}8 \angle 61^\circ 54' - \overline E_4\cdot 104{,}1 \angle 60^\circ 0' + \overline E_1\cdot 39{,}6 \angle 54^\circ = \frac{\overline W_4}{\dot E_4}\cdot 2685 \angle 115^\circ 54'. \end{aligned} \right. \]

The voltage at node “2” is taken as known and equal to 105 kV.

* \(A\) and \(D\) are equal in our particular case, where the \(\Pi\)-type equivalent circuit is symmetrical.

Assuming the current loads at the nodes to be known, let us solve the system of equations with respect to the voltages \(\overline{E}_1\), \(\overline{E}_3\), and \(\overline{E}_4\). We shall carry out the solution by means of determinants, for which we write the system of three complete linear equations in the general form:

\[ \begin{aligned} \overline{a}_{11}\cdot \overline{E}_1+\overline{a}_{13}\cdot \overline{E}_3+\overline{a}_{14}\cdot \overline{E}_4&=\overline{b}_1,\\ \overline{a}_{21}\cdot \overline{E}_1+\overline{a}_{23}\cdot \overline{E}_3+\overline{a}_{24}\cdot \overline{E}_4&=\overline{b}_2,\\ \overline{a}_{31}\cdot \overline{E}_1+\overline{a}_{33}\cdot \overline{E}_3+\overline{a}_{34}\cdot \overline{E}_4&=\overline{b}_3 \end{aligned} \]

and the solutions of this system of equations through determinants:

\[ \overline{E}_1= \frac{\overline{\Delta}_{11}}{\overline{\Delta}}\cdot \overline{b}_1+ \frac{\overline{\Delta}_{21}}{\overline{\Delta}}\cdot \overline{b}_2+ \frac{\overline{\Delta}_{31}}{\overline{\Delta}}\cdot \overline{b}_3; \]

\[ \overline{E}_3= \frac{\overline{\Delta}_{12}}{\overline{\Delta}}\cdot \overline{b}_1+ \frac{\overline{\Delta}_{22}}{\overline{\Delta}}\cdot \overline{b}_2+ \frac{\overline{\Delta}_{32}}{\overline{\Delta}}\cdot \overline{b}_3; \]

\[ \overline{E}_4= \frac{\overline{\Delta}_{13}}{\overline{\Delta}}\cdot \overline{b}_1+ \frac{\overline{\Delta}_{23}}{\overline{\Delta}}\cdot \overline{b}_2+ \frac{\overline{\Delta}_{33}}{\overline{\Delta}}\cdot \overline{b}_3; \]

Here: \(\overline{\Delta}\) is the determinant of the system of equations, in the present case equal to

\[ \overline{\Delta}= \begin{vmatrix} \overline{a}_{11} & \overline{a}_{13} & \overline{a}_{14}\\ \overline{a}_{21} & \overline{a}_{23} & \overline{a}_{24}\\ \overline{a}_{31} & \overline{a}_{33} & \overline{a}_{34} \end{vmatrix}; \]

\(\overline{\Delta}_{11}\), \(\overline{\Delta}_{21}\), \(\overline{\Delta}_{31}\), etc., are the adjugates of the principal determinant \(\overline{\Delta}\), in the present case equal to:

\[ \overline{\Delta}_{11}=(-1)^{1+1}\cdot \begin{vmatrix} \overline{a}_{23} & \overline{a}_{24}\\ \overline{a}_{33} & \overline{a}_{34} \end{vmatrix}, \]

\[ \overline{\Delta}_{21}=(-1)^{2+1}\cdot \begin{vmatrix} \overline{a}_{13} & \overline{a}_{14}\\ \overline{a}_{33} & \overline{a}_{34} \end{vmatrix}, \]

\[ \overline{\Delta}_{31}=(-1)^{3+1}\cdot \begin{vmatrix} \overline{a}_{13} & \overline{a}_{14}\\ \overline{a}_{23} & \overline{a}_{24} \end{vmatrix}, \]

and so on.

From comparison of the system of nodal equations and the system of three linear equations in general form, we have:

\[ \begin{aligned} \overline{a}_{11}&=48.8\angle 58^\circ 30',& \overline{a}_{21}&=0,\\ \overline{a}_{13}&=26.2\angle 54^\circ,& \overline{a}_{23}&=-86.7\angle 57^\circ 7',\\ \overline{a}_{14}&=0,& \overline{a}_{24}&=48.8\angle 58^\circ 7',\\ \overline{a}_{31}&=39.6\angle 54^\circ,& \overline{b}_1&=\overline{J}_2\cdot 1278\angle 112^\circ 30' + 105\cdot 74.07\angle 57^\circ 23',\\ \overline{a}_{33}&=67.8\angle 61^\circ 54',& \overline{b}_2&=-\overline{J}_3\cdot 1932\angle 112^\circ 30' - 105\cdot 39.6\angle 54^\circ,\\ \overline{a}_{34}&=-104.2\angle 60^\circ 00',& \overline{b}_3&=\overline{J}_4\cdot 2685\angle 115^\circ 54' + 0. \end{aligned} \]

Where:

\[ \overline{J}_2=\frac{\overline{W}_2}{\hat{E}_2}, \]

\[ \overline{J}_3=\frac{\overline{W}_3}{\hat{E}_3}, \]

\[ \overline{J}_4=\frac{\overline{W}_4}{\hat{E}_4}. \]

After substituting the values of the coefficients of the unknowns into the expressions for the determinant and the adjoints, we obtain:

\[ \overline{\Delta}=-329.6\cdot 10^3 \angle -7^\circ 18'; \]

\[ \overline{\Delta}_{11}=9260\angle 117^\circ 58' - 3310\angle 120^\circ 24; \]

\[ \overline{\Delta}_{21}=2730\angle 114^\circ 00' \]

\[ \overline{\Delta}_{31}=1278\angle 112^\circ 30'; \]

\[ \overline{\Delta}_{12}=1932\angle 112^\circ 30'; \]

\[ \overline{\Delta}_{22}=-5085\angle 118^\circ 30'; \]

\[ \overline{\Delta}_{32}=-2381\angle 117^\circ; \]

\[ \overline{\Delta}_{13}=3433\angle 111^\circ 7'; \]

\[ \overline{\Delta}_{23}=-3308\angle 120^\circ 24' + 1037\angle 108^\circ \]

\[ \overline{\Delta}_{33}=-4231\angle 115^\circ 37' \]

Now the expressions for the unknown voltages take the form:

\[ \overline{E}_1 =100.09\angle 1^\circ 30' +23.05\,\overline{J}_2\angle 57^\circ 16' -16.0\,\overline{J}_3\angle 53^\circ 48' +10.42\,\overline{J}_4\angle 55^\circ 42' \]

\[ \overline{E}_3 =109.7\angle -1^\circ 18' +7.5\,\overline{J}_2\angle 52^\circ 18' -29.79\,\overline{J}_3\angle 238^\circ 18' +19.41\,\overline{J}_4\angle 240^\circ 12', \]

\[ \overline{E}_4 =109.5\angle -1^\circ 12' +13.10\,\overline{J}_2\angle 50^\circ 55' +13.50\,\overline{J}_3\angle 65^\circ 48' +34.36\,\overline{J}_4\angle 238^\circ 49' \]

Load at the nodes:

\[ \overline{W}_2=89.45\angle -26^\circ 36'; \]

\[ \overline{W}_3=111.9\angle -26^\circ 34'; \]

\[ \overline{W}_4=94.4\angle -31^\circ 48'; \]

The load currents of the first approximation stage, under the assumption that the average voltage of 105 kV is present at all nodes:

\[ \overline{J}_2= \frac{89.45\angle -26^\circ 36'}{105} =0.852\angle -26^\circ 36'\ \mathrm{kA}, \]

\[ \overline{J}_3= \frac{111.9\angle -26^\circ 34'}{105} =1.065\angle -26^\circ 34'\ \mathrm{kA}, \]

and

\[ \overline{J}_4= \frac{94.4\angle -31^\circ 48'}{105} =0.899\angle -31^\circ 48'\ \mathrm{kA}. \]

The corresponding voltages are obtained as:

\[ \overline{E}'_3 =109.7\angle -1^\circ 18' +7.50\angle 52^\circ 18'\cdot 0.852\angle -26^\circ 36' \]
\[ -29.79\angle 238^\circ 18'\cdot 1.065\angle -26^\circ 34' +19.41\angle 240^\circ 12'\cdot 0.899\angle -31^\circ 48' =127.3\angle 3^\circ 54'\ \mathrm{kV}; \]

\[ \overline{E}'_4 = 109{,}5 \angle -1^\circ 12' + 13{,}10 \angle 50^\circ 55' \cdot 0{,}852 \angle -26^\circ 36' + \]
\[ {}+ 13{,}50 \angle 65^\circ 48' \cdot 1{,}065 \angle -26^\circ 34' + 34{,}36 \angle 238^\circ 49' \cdot 0{,}899 \angle -31^\circ 48' = 103{,}3 \angle -1^\circ 24' \text{ kV}. \]

Note: The voltage at the first node \(\overline{E}_1\) does not participate in the preliminary calculation, since the load at the first node is not given.

The load currents of the second approximation stage are:

\[ \overline{J}''_2 = \overline{J}_2 = 0{,}852 \angle -26^\circ 36' \text{ kA}. \]

\[ \overline{J}''_3 = \frac{\overline{W}_3}{\overline{E}'_3} = \frac{111{,}9 \angle -26^\circ 34'}{127{,}3 \angle -3^\circ 54'} = 0{,}879 \angle -22^\circ 40' \text{ kA}; \]

\[ \overline{J}''_4 = \frac{\overline{W}_4}{\overline{E}'_4} = \frac{94{,}4 \angle -31^\circ 48'}{103{,}3 \angle 1^\circ 24'} = 0{,}914 \angle -33^\circ 12' \text{ kA}; \]

And the corresponding voltages:

\[ \overline{E}''_3 = 120{,}75 \angle 3^\circ 36' \text{ kV}, \]

\[ \overline{E}''_4 = 100{,}03 \angle -1^\circ 48' \text{ kV}. \]

In the same way one can obtain the voltage values for the 3rd, 4th, etc. approximation stages. The results of the calculations are given in Table 2.

Table No. 2.

	1st	2nd	3rd	4th	5th
Voltage at node \(\overline{E}^{(k)}_3\) kV	\(127{,}3 \angle 3^\circ 54'\)	\(120{,}75 \angle 3^\circ 36'\)	\(121{,}76 \angle 3^\circ 30'\)	\(121{,}5 \angle 3^\circ 42'\)	\(121{,}5 \angle 3^\circ 42'\)
\(\dfrac{E_3 - E^{(k)}_3}{E_3}\,100\)	\(4{,}7\%\)	\(1{,}03\%\)	\(0{,}2\%\)	\(0\%\)	\(0\%\)
\(\angle \sigma_3 - \angle \sigma^{(k)}_3\)	\(\angle 12'\)	\(\angle 6'\)	\(\angle 12'\)	\(\angle 0'\)	\(\angle 0'\)
Voltage at node \(\overline{E}^{(k)}_4\) kV	\(103{,}3 \angle -1^\circ 24'\)	\(100{,}08 \angle -1^\circ 48'\)	\(99{,}550 \angle -1^\circ 42'\)	\(99{,}4 \angle -1^\circ 48'\)	\(99{,}4 \angle -1^\circ 48'\)
\(\dfrac{E_4 - E^{(k)}_4}{E_4}\,100\)	\(3{,}93\%\)	\(0{,}68\%\)	\(0{,}15\%\)	\(0\%\)	\(0\%\)
\(\angle \sigma_4 - \angle \sigma^{(k)}_4\)	\(\angle 24'\)	\(\angle 0'\)	\(\angle 6'\)	\(\angle 0'\)	\(\angle 0'\)

Here:

a) \(\overline{E}^{(k)}_3\) and \(\overline{E}^{(k)}_4\) are the voltages of the given, \(k\)-th, approximation stage;

b) \(\overline{E}_3\) and \(\overline{E}_4\) are the true values of the voltages at the nodes.

Since the sixth approximation stage gives voltage values that do not differ from those in the case of the fifth approximation stage, it is necessary to regard

\[ \overline{E}^{V}_3 = 121{,}5 \angle 3^\circ 42' \text{ kV} \]

\[ \overline{E}^{V}_4 = 99{,}40 \angle -1^\circ 48' \text{ kV} \]

as the true values of the voltages at nodes 3 and 4. \({}^{2)}\)

\({}^{2)}\) Taking into account the accepted degree of calculation accuracy.

Now we immediately find the voltage at the first node

\[ \overline{E}_1 = 100.09 \angle 1^\circ 30' + 23.05 \angle 57^\circ 16' \cdot 0.852 \angle -26^\circ 36' - \]

\[ -16.00 \angle 53^\circ 48' \cdot 0.922 \angle -22^\circ 52' + 10.42 \angle 55^\circ 42' \]

\[ \cdot 0.950 \angle -33^\circ 36' = 113.63 \angle 4^\circ 0' \ \text{kv} \]

We shall find the distribution of powers over the links of the network of the complex system, according to expressions (4) and (5), as

\[ \overline{W}_{\mathrm{I}(н)} = \frac{-\dot{E}_1 \cdot \overline{E}_2}{\overline{B}_{\mathrm{I}}} + \frac{\overline{D}_{\mathrm{I}}}{\overline{B}_{\mathrm{I}}} \cdot E_1^2 = \frac{-113.63 \angle -4^\circ 0' \cdot 105}{26.2 \angle 54^\circ} + \]

\[ + \frac{0.994 \angle 14'}{26.2 \angle 54^\circ} \cdot 113.63^2 = 49.5 - j 8.0 \ \text{mva} \]

\[ \overline{W}_{\mathrm{I}(к)} = \frac{\dot{E}_2 \cdot \overline{E}_1}{\overline{B}_{\mathrm{I}}} - \frac{\overline{A}_{\mathrm{I}}}{\overline{B}_{\mathrm{I}}} \cdot E_2^2 = \frac{105 \cdot 113.63 \angle 4^\circ 0'}{26.2 \angle 54^\circ} - \]

\[ - \frac{0.994 \angle 14'}{26.2 \angle 54^\circ} \cdot 105^2 = 46.4 - j 12.2 \ \text{mva} \]

\[ \overline{W}_{\mathrm{II}(н)} = \frac{-\dot{E}_2 \cdot \overline{E}_3}{\overline{B}_{\mathrm{II}}} + \frac{\overline{D}_{\mathrm{II}}}{\overline{B}_{\mathrm{II}}} \cdot E_2^2 = \frac{-105 \cdot 121.5 \angle 3^\circ 42'}{48.8 \angle 58^\circ 30'} + \]

\[ + \frac{0.977 \angle 48'}{48.8 \angle 58^\circ 30'} \cdot 105^2 = -(33.0 - j 27.4)\ \text{mva} \]

\[ \overline{W}_{\mathrm{II}(к)} = \ldots \ldots \ldots = -(36.0 - j 18.0)\ \text{mva} \]

\[ \overline{W}_{\mathrm{III}(н)} = \ldots \ldots \ldots = 64.4 - j 32.5\ \text{mva} \]

\[ \overline{W}_{\mathrm{III}(к)} = \ldots \ldots \ldots = 56.0 - j 30.5\ \text{mva} \]

\[ \overline{W}_{\mathrm{IV}(н)} = \ldots \ldots \ldots = -(24.5 - j 18.3)\ \text{mva} \]

\[ \overline{W}_{\mathrm{VI}(к)} = \ldots \ldots \ldots = -(26.2 - j 20.7)\ \text{mva} \]

The voltages found at the nodes and the loads in the links are entered on the diagram of Fig. 3.

Under no-load conditions, when

\[ \overline{J}_2 = 0, \]

\[ \overline{J}_3 = 0, \]

\[ \overline{J}_4 = 0, \]

and

\[ \overline{J}_1 \ne 0, \]

the voltages at the nodes will be:

\[ \overline{E}_{10} = 100.09 \angle 1^\circ 30' \ \text{kv} \]

\[ \overline{E}_{20} = 105 \ \text{kv} \]

\[ \overline{E}_{30} = 109.7 \angle -1^\circ 18' \ \text{kv} \]

\[ \overline{E}_{40} = 109.5 \angle -1.12' \ \text{kv} \]

Fig. 3.

The power drawn from the first node, and consequently required to supply the entire system at no load, will be:

\[ \overline{W}_{I(\mathrm{xx})} = -\left(\overline{W}_{IV(k)\mathrm{XX}}-\overline{W}_{I(k)\mathrm{XX}}\right) = \]

\[ = -\left[ \left( \frac{\dot{E}_{10}\cdot \overline{E}_{40}}{\overline{B}_{IV}} - \frac{\overline{A}_{IV}}{\overline{B}_{IV}}\cdot E_{10}^{2} \right) - \left( \frac{-\overline{E}_{20}\cdot \dot{E}_{10}}{\overline{B}_{I}} + \frac{\overline{D}_{I}}{\overline{B}_{I}}\cdot E_{10}^{2} \right) \right] \]

\[ = -\left[(0.00+j\,24.3)-(19.5+j\,12.00)\right] = 19.50+j\,36.30\ \text{mva}. \]

It should be noted that the voltages and the load at \(X X\) could also have been calculated under the condition of supplying the system simultaneously at nodes 1 and 3.

IV.

In many cases calculations may be carried out with less accuracy than is obtained when the line constants are taken as distributed, or, in any case, when calculations are performed on the basis of equivalent constants. If the calculation is based on a \(\Pi\)-shaped equivalent circuit for the sections of the system, and the capacitive currents are taken into account through a possible average voltage at the nodes of the system, then the entire process of determining voltages and loads in the system by the method of nodal equations is considerably simplified, while at the same time the accuracy of the calculation will not suffer too much.

Fig. 4.

The equivalent circuit of the complex system shown in Fig. 1 will now have the form shown in Fig. 4.

Note: The currents of active conductance, if they cannot be neglected, may also be taken into account by distributing the conductance of each section to its ends.

The capacitive currents at the nodes are taken into account in the form of additional leading reactive powers added to the specified loads at the nodes. Thus new loads are obtained at the nodes:

\[ \overline{W}'_{2},\quad \overline{W}'_{3},\quad \overline{W}'_{4}\ldots\ldots\ \text{etc.} \]

The currents in the sections now do not vary along the length of the sections and are determined as:

\[ \overline{J}_{I} = \frac{\overline{E}_{1}-\overline{E}_{2}}{\overline{Z}_{I}}, \]

\[ \overline{J}_{II} = \frac{\overline{E}_{2}-\overline{E}_{3}}{\overline{Z}_{II}} \]

\[ \ldots\ldots \]

\[ \text{etc.} \]

The system of nodal equations will have the form (as applied to Figs. 1 and 4):

\[ \left. \begin{aligned} &\frac{\hat E_2\cdot(\overline E_1-\overline E_3)}{\overline Z_{\mathrm I}} - \frac{\hat E_2(\overline E_2-\overline E_3)}{\overline Z_{\mathrm{II}}} = \overline W'_2 \\[6pt] &\frac{\hat E_3\cdot(\overline E_2-\overline E_3)}{\overline Z_{\mathrm{II}}} - \frac{\hat E_2(\overline E_3-\overline E_4)}{\overline Z_{\mathrm{III}}} - \frac{\hat E_3(\overline E_3-\overline E_5)}{\overline Z_{\mathrm V}} = \overline W'_3 \\ &\cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \\ &\cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \\ &\text{etc.} \end{aligned} \right\} \qquad \cdot\ \cdot\ \cdot\ \cdot\ (12) \]

A further simplification is achieved by assuming equality of the ratios \(\dfrac{R}{X}\) in the branches connected at the given node. This identical ratio \(\dfrac{R}{X}\) for all branches connected at the given node is taken as the mean of the actual ratios. That is, in other words, if

\[ \overline Z_{\mathrm I}=Z_{\mathrm I}\angle\varphi_{\mathrm I} \]

and

\[ \overline Z_{\mathrm{II}}=Z_{\mathrm{II}}\angle\varphi_{\mathrm{II}} \]

are the total impedances of two branches connected at node “2,” then it is assumed that:

\[ \overline Z'_{\mathrm I}=Z_{\mathrm I}\angle\varphi_{\mathrm{av}(2)} \]

and

\[ \overline Z'_{\mathrm{II}}=Z_{\mathrm{II}}\angle\varphi_{\mathrm{av}(2)}, \]

where:

\[ \varphi_{\mathrm{av}(2)}=\frac{\varphi_{\mathrm I}+\varphi_{\mathrm{II}}}{2}. \]

In the same way we obtain, for the following nodes:

\[ \frac{\varphi_{\mathrm{II}}+\varphi_{\mathrm{III}}+\varphi_{\mathrm V}}{3} = \varphi_{\mathrm{av}(3)} \]

\[ \frac{\varphi_{\mathrm{III}}+\varphi_{\mathrm{IV}}}{2} = \varphi_{\mathrm{av}(4)} \qquad \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ (13) \]

and so on.

Taking the last simplification into account, equations (12) are rewritten in the form:

\[ \left. \begin{aligned} &\frac{\overline E_1-\overline E_2}{Z_{\mathrm I}} - \frac{\overline E_2-\overline E_3}{Z_{\mathrm{II}}} = \frac{\overline W'_2}{\hat E}\angle\varphi_{\mathrm{av}(2)} \\[6pt] &\frac{\overline E_2-\overline E_3}{Z_{\mathrm{II}}} - \frac{\overline E_3-\overline E_4}{Z_{\mathrm{II}}} - \frac{\overline E_3-\overline E_5}{Z_{\mathrm V}} = \frac{\overline W'_3}{\hat E_3}\angle\varphi_{\mathrm{av}(3)} \\[6pt] &\frac{\overline E_3-\overline E_4}{Z_{\mathrm{III}}} - \frac{\overline E_4-\overline E_5}{Z_{\mathrm{IV}}} = - \frac{\overline W'_4}{\hat E_4}\angle\varphi_{\mathrm{av}(4)} \\ &\cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \\ &\cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \end{aligned} \right\} \qquad \cdot\ \cdot\ (14) \]

and so on.

In this form the equations are solved comparatively easily, since the coefficients at the unknowns are represented by simple numbers.

As the subsequent calculations will show, the error in the values of the nodal voltages calculated by this simplified method of nodal equations is comparatively small.

Let us show the application of the simplified method of nodal equations to the calculation of a specific complex system, Fig. 5¹. All lines are double-circuit. The lengths of the branches are:

\[ \begin{aligned} L_{\mathrm{I}} &= 150\ \text{km} & L_{\mathrm{V}} &= 30\ \text{km} \\ L_{\mathrm{II}} &= 150\ \text{''} & L_{\mathrm{VI}} &= 25\ \text{''} \\ L_{\mathrm{III}} &= 35\ \text{''} & L_{\mathrm{VII}} &= 125\ \text{''} \\ L_{\mathrm{IV}} &= 100\ \text{''} \end{aligned} \]

Fig. 5.

The average value of the capacitive conductance for all branches is taken as equal to

\[ 2.85 \cdot 10^{-6}\ \mho/\text{km}/\text{circuit}. \]

The additional reactive powers at the nodes due to the capacitive currents of the branches, at an average voltage of 105 kV, are obtained as:

\[ \begin{array}{ll} \text{For the 2nd node} & j9.45\ \text{MVA},\\ \text{'' 3rd ''} & j5.84\ \text{''},\\ \text{'' 4th ''} & j8.2\ \text{''},\\ \text{'' 5th ''} & j4.1\ \text{''},\\ \text{'' 6th ''} & j1.73\ \text{''}. \end{array} \]

Calculated powers at the nodes:

\[ \begin{array}{rcl} \text{at the 2nd node} & 75 - j60 + j9.45 &= 75 - j50.6\ \text{MVA},\\ \text{'' 3rd ''} & -100 + j40 + j5.85 &= -100 + j45.84\ \text{''},\\ \text{'' 4th ''} & 125 - j35 + j8.2 &= 125 - j26.8\ \text{''},\\ \text{'' 5th ''} & 20 - j15 + j4.1 &= 20 - j10.9\ \text{''},\\ \text{'' 6th ''} & 30 - j30 + j1.73 &= 30 - j28.27\ \text{''}. \end{array} \]

Average angles of the vectors of total impedances by nodes:

\[ \text{For the 2nd node}\quad \frac{ \arctg \frac{30}{15}+\arctg \frac{30}{15} }{2} =64^\circ \]

\[ \text{For the 3rd node}\quad \frac{ \arctg \frac{30}{15}+\arctg \frac{7.5}{4.3} }{2} =62^\circ \]

\[ \text{For the 4th node}\quad \frac{ \arctg \frac{7.5}{4.3}+\arctg \frac{20}{10}+\arctg \frac{25}{15} }{3} =61^\circ \]

\[ \text{For the 5th node}\quad \frac{ \arctg \frac{20}{10}+\arctg \frac{6}{3} }{2} =64^\circ \]

For the 6th node

$$ \frac{\operatorname{arctg}\frac{6}{3}+\operatorname{arctg}\frac{5}{4}}{2}=58^\circ $$

The nodal equations, according to (14), will be written as:

$$ \left. \begin{aligned} \frac{\overline{E}_1-\overline{E}_2}{33{,}6} -\frac{\overline{E}_2-\overline{E}_3}{33{,}6} &= \frac{\overline{W}'_2}{\hat{E}_2}\angle 64^\circ = \overline{J}_2\angle 64^\circ \\[4pt] \frac{\overline{E}_2-\overline{E}_3}{33{,}6} -\frac{\overline{E}_3-\overline{E}_4}{8{,}6} &= -\frac{\overline{W}'_3}{\hat{E}_3}\angle 62^\circ = -\overline{J}_3\angle 62^\circ \\[4pt] \frac{\overline{E}_3-\overline{E}_4}{8{,}6} -\frac{\overline{E}_4-\overline{E}_5}{22{,}3} -\frac{\overline{E}_4-\overline{E}_1}{29{,}2} &= \frac{\overline{W}'_4}{\hat{E}_4}\angle 61^\circ = \overline{J}_4\angle 61^\circ \\[4pt] \frac{\overline{E}_4-\overline{E}_5}{22{,}3} -\frac{\overline{E}_5-\overline{E}_6}{6{,}6} &= \frac{\overline{W}'_5}{\hat{E}_6}\angle 64^\circ = \overline{J}_5\angle 64^\circ \\[4pt] \frac{\overline{E}_5-\overline{E}_6}{6{,}6} -\frac{\overline{E}_6-\overline{E}_1}{6{,}4} &= \frac{\overline{W}'_6}{\hat{E}_6}\angle 58^\circ = \overline{J}_6\angle 58^\circ \end{aligned} \right\} $$

Or, after some transformations,

$$ \left. \begin{aligned} \overline{E}_1-2\overline{E}_2+\overline{E}_3 &=33{,}6\,\overline{J}_2\angle 64^\circ \\ \overline{E}_2-4{,}91\overline{E}_3+3{,}9\overline{E}_4 &=-33{,}6\,\overline{J}_3\angle 62^\circ \\ \overline{E}_1+3{,}38\overline{E}_3-5{,}7\overline{E}_4+1{,}3\overline{E}_5 &=29{,}2\,\overline{J}_4\angle 61^\circ \\ \overline{E}_4-4{,}38\overline{E}_5+3{,}38\overline{E}_6 &=22{,}3\,\overline{J}_5\angle 64^\circ \\ \overline{E}_5-2{,}03\overline{E}_6+1{,}03\overline{E}_1 &=6{,}6\,\overline{J}_6\angle 58^\circ \end{aligned} \right\} $$

Bearing in mind that the voltage of the 4th node is given and equal to 105 kv, we thus obtain a system of five equations with five unknowns.

By the method of successive elimination of unknowns ⁴) we find the solution of the nodal equations in the form:

$$ \overline{E}_1 = 106{,}6 +5{,}95\overline{J}_2\angle 64^\circ -11{,}8\overline{J}_3\angle 62^\circ +13{,}2\overline{J}_4\angle 61^\circ +4{,}84\overline{J}_5\angle 64^\circ + $$

$$ +2{,}38\overline{J}_6\angle 58^\circ\ \text{kv}; $$

$$ \overline{E}_2 = 46{,}6 +0{,}556\overline{E}_1 -18{,}7\overline{J}_2\angle 64^\circ +3{,}82\overline{J}_3\angle 62^\circ\ \text{kv}; $$

$$ \overline{E}_3 = 93{,}2 +0{,}112\overline{E}_1 -3{,}8\overline{J}_2\angle 64^\circ +7{,}64\overline{J}_3\angle 62^\circ\ \text{kv}; $$

$$ \overline{E}_4=105\ \text{kv}; $$

$$ \overline{E}_5 = 218{,}3 -1{,}06\overline{E}_1 +9{,}88\overline{J}_2\angle 64^\circ -19{,}86\overline{J}_3\angle 62^\circ +22{,}4\overline{J}_4\angle 61^\circ\ \text{kv}; $$

$$ \overline{E}_5 = 0{,}296\left(4{,}38\overline{E}_5-105+22{,}3\overline{J}_5\angle 64^\circ\right)\ \text{kv}; $$

⁴) This method often has an advantage over the method using determinants, because in the given case one usually has to deal with a system of incomplete equations.

Next, by successive approximation we obtain the values of the node voltages.

The results of the calculations are given in Table 3.

Table No. 3

Voltages	First approximation	Second approximation	Third approximation
$\bar E_1 \ldots$	$113{,}8 \angle 4^\circ 06'$	$114{,}8 \angle 4^\circ 06'$	$114{,}8 \angle 4^\circ 06'$
$\bar E_2 \ldots$	$99{,}55 \angle -48'$	$98{,}4 \angle -48'$	$98{,}15 \angle -12'$
$\bar E_3 \ldots$	$109{,}45 \angle 2^\circ 06'$	$109{,}95 \angle 2^\circ 06'$	$109{,}9 \angle 2^\circ 06'$
$\bar E_4 \ldots$	$105 \angle 0^\circ$	$105 \angle 0^\circ$	$105 \angle 0^\circ$
$\bar E_5 \ldots$	$105{,}6 \angle 2^\circ 24'$	$107{,}75 \angle 2^\circ 00'$	$107{,}82 \angle 2^\circ 00'$
$\bar E_6 \ldots$	$107{,}8 \angle 2^\circ 54'$	$110{,}2 \angle 3^\circ 12'$	$110{,}4 \angle 3^\circ 00'$

Attention is drawn to the quite insignificant difference in the voltage values of the 2nd and 3rd stages of approximation.

We shall also find the distribution of loads over the links²:

\[ \text{Link 1: a) }\ \bar W_{\mathrm{I}(н)} = \frac{\bar E_1 - \bar E_2}{\bar Z_{\mathrm{I}}}\cdot \hat E_1 = \]

\[ = \frac{114{,}8 \angle 4^\circ 06' - 98{,}15 \angle -12'}{33{,}6 \angle 64^\circ} \cdot 114{,}8 \angle -4^\circ 06' = 48 - j40{,}0\ \text{mva} \]

\[ \text{c) }\ \bar W_{\mathrm{I}(к)} = \frac{\bar E_1 - \bar E_2}{\bar Z_{\mathrm{I}}}\cdot \hat E_2 = \]

\[ = \frac{114{,}8 \angle 4^\circ 06' - 98{,}15 \angle -12'}{33{,}6 \angle 64^\circ} \cdot 98{,}15 \angle 12' = 43{,}5 - 32{,}0\ \text{mva} \]

\[ \text{Link 2: a) }\ \bar W_{\mathrm{II}(н)} = \ldots = -(28{,}0 - j23{,}0)\ \text{mva} \]

\[ \text{c) }\ \bar W_{\mathrm{II}(к)} = \ldots = -(30 - j25{,}0)\ \text{''} \]

\[ \text{Link 3: a) }\ \bar W_{\mathrm{III}(н)} = \ldots = 74 - j30\ \text{mva} \]

\[ \text{c) }\ \bar W_{\mathrm{III}(к)} = \ldots = 71 - j25\ \text{''} \]

\[ \text{Link 4: a) }\ \bar W_{\mathrm{IV}(н)} = \ldots = -(19{,}0 + j1)\ \text{mva} \]

\[ \text{c) }\ \bar W_{\mathrm{IV}(к)} = \ldots = -(22{,}0 - j2)\ \text{''} \]

\[ \text{Link 5: a) }\ \bar W_{\mathrm{V}(н)} = \ldots = -(45 - j11)\ \text{''} \]

\[ \text{c) }\ \bar W_{\mathrm{V}(к)} = \ldots = -(48 - j14)\ \text{''} \]

\[ \text{Link 6: a) }\ \bar W_{\mathrm{VI}(н)} = \ldots = -(75 - j38)\ \text{''} \]

\[ \text{c) }\ \bar W_{\mathrm{VI}(к)} = \ldots = -(78 - j40)\ \text{''} \]

\[ \text{Link 7: a) }\ \bar W_{\mathrm{VII}(н)} = \ldots = -(45{,}0 - j18{,}0)\ \text{''} \]

\[ \text{c) }\ \bar W_{\mathrm{VII}(к)} = \ldots = -(42{,}0 - j16{,}0)\ \text{''} \]

The electrical condition of the calculated system is presented in Fig. 6.

The results of the calculation (Fig. 6) show that the power imbalance at the nodes in this case may be considerable. True, here not everything can be attributed to the simplification of the method of nodal equations; here part of the error is due to the slide rule. Nevertheless, in order to be certain that the nodal voltages obtained by the simplified method of nodal equations do not contain a significant error, let us calculate, by this same method, the example of Figure 2, Chapter III. This example was previously calculated by the exact method of nodal equations, and therefore it is possible to compare the results of the calculations in both cases.

Fig. 6.

These results are given in Table 4.

Table No. 4.

Calculation methods	$E_1$	$E_2$	$E_3$	$E_4$
Exact method of nodal equations	113.63	105	121.5	99.4
Simplified method of nodal voltages	114.2	105	121.85	100

As we see, the deviation from the true values of the nodal voltages, even in this example, where the links are 200 km long and more, has proved to be insignificant.

V.

The method of conversions, or transformations, in the form in which it is applied to the calculation of low-voltage networks, is based on several propositions, proposed by Frick and Kennelly, which make it possible to transform a network from a complex one into a simple one, to find for this latter the distribution—still fictitious—and then, going backward, i.e. retrans-forming the network, to find the actual current distribution in the links of the network and, consequently, the voltages at the nodes.

Fig. 7.

Proposition 1⁶⁾. Between points $A$ and $B$ in Fig. 7, two wires with resistances $R_{\mathrm{I}}$ and $R_{\mathrm{II}}$ may be replaced by one with resistance $R$ such that, when one and the same current flows from point $A$ to point $B$, the voltage drop between these points will be the same in both cases.

This proposition is satisfied by the relation—

\[ R=\frac{R_{\mathrm{I}}\cdot R_{\mathrm{II}}}{R_{\mathrm{I}}+R_{\mathrm{II}}}\ \Omega \]

⁶⁾ Prof. A. A. Glazunov, Theory and Calculation of Distribution Networks.

Proposition 2. A current load (Fig. 8) at point “C” can be transferred to two other points, “a” and “b,” so that the currents \(a\), and, consequently, the voltage drops in sections \(Ab\) and \(Bb\), remain unchanged.

Fig. 8.

The parts of the current \(J_c\) at points “a” and “b” are calculated from the relations:

\[ J_{a(c)} = J_c \frac{R_{cb}}{R_{ac}+R_{cb}}, \]

\[ J_{b(c)} = J_c \frac{R_{ca}}{R_{ac}+R_{cb}}, \]

where

\[ J_{a(c)} + J_{b(c)} = J_c . \]

Proposition 3. The current \(J\) flowing through a conductor with resistance \(R\) (Fig. 9) can be represented as the sum of two currents \(J_{\mathrm{I}}\) and \(J_{\mathrm{II}}\), flowing through two conductors with resistances \(R_{\mathrm{I}}\) and \(R_{\mathrm{II}}\), so that the voltage drop between points \(A\) and \(B\) remains unchanged.

Fig. 9.

The currents \(J_{\mathrm{I}}\) and \(J_{\mathrm{II}}\) are calculated by the formulas:

\[ J_{\mathrm{I}} = J \frac{R}{R_{\mathrm{I}}}; \]

\[ J_{\mathrm{II}} = J \frac{R}{R_{\mathrm{II}}}. \]

Proposition 4. A part of a network having the form of a triangle (Fig. 10) can be replaced by an equivalent star, the resistances of the star arms being calculated by the formulas:

\[ r_{\mathrm{I}} = \frac{R_{\mathrm{I}}\cdot R_{\mathrm{III}}} {R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}}}; \]

\[ r_{\mathrm{II}} = \frac{R_{\mathrm{I}}\cdot R_{\mathrm{II}}} {R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}}}; \]

\[ r_{\mathrm{III}} = \frac{R_{\mathrm{II}}\cdot R_{\mathrm{III}}} {R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}}}. \]

Fig. 10.

In the inverse transformation of a star into a triangle, the conductance of the sides of the equivalent triangle is determined as follows:

\[ y_{\mathrm{I}} = \frac{\lambda_{\mathrm{I}}\cdot \lambda_{\mathrm{II}}} {\lambda_{\mathrm{I}}+\lambda_{\mathrm{II}}+\lambda_{\mathrm{III}}}; \]

\[ y_{\mathrm{II}} = \frac{\lambda_{\mathrm{II}}\cdot \lambda_{\mathrm{III}}} {\lambda_{\mathrm{I}}+\lambda_{\mathrm{II}}+\lambda_{\mathrm{III}}}; \]

\[ \lambda_{\mathrm{III}} = \frac{\lambda_{\mathrm{I}}\cdot \lambda_{\mathrm{III}}} {\lambda_{\mathrm{I}}+\lambda_{\mathrm{II}}+\lambda_{\mathrm{III}}}. \]

By means of the indicated four propositions, a complex network can be reduced to a line fed from two sides (Fig. 11), or to the case of a ring network (Fig. 12). If the voltages of the supply points A and B are identical, as is assumed in most calculations of low-voltage networks, then both of these cases are identical, and the currents flowing out of the supply points—for example, Fig. 12 (the ring)—can be determined as

Fig. 11.

\[ J_{\mathrm{I}} = J_c \frac{R_{\mathrm{II}}}{R_{\mathrm{I}} + R_{\mathrm{II}}}, \]

\[ J_{\mathrm{II}} = J_c \frac{R_{\mathrm{I}}}{R_{\mathrm{I}} + R_{\mathrm{II}}}. \]

Fig. 12.

All the propositions given, on the basis of which a complex network can be transformed and then calculated, apply to the case of direct current or to the case of alternating current when the load is non-inductive and when the network possesses neither capacitance nor self-inductance. In calculating an alternating-current network in the general case, and especially with a high-voltage alternating-current network in mind, one has to deal with loads that include reactive components, and with conductors possessing capacitance and self-inductance.

Adopting a Π-shaped equivalent circuit and including the capacitive currents in the loads of the nodes, in calculating networks one may have in view only the case of the flow of currents having reactive components through conductors possessing inductance. Can a complex network in this case be transformed into a simple one on the basis of the propositions given above? Yes, it can. Propositions 1, 3, and 4, as applied to alternating current and to conductors possessing ohmic and inductive resistances, are proved in theoretical electrical engineering¹). In the same way, in manuals on network calculation²), the possibility is proved of determining the current distribution in a ring network and in a line fed from two sides by formulas entirely analogous to the case of direct current; only in these formulas, as in the formulas of propositions 1, 2, and 4, the resistances and currents are expressed by complex numbers. Proposition 2 is likewise fully applicable to the calculation of alternating-current networks on the basis of a Π-shaped equivalent circuit; but since proofs of this proposition, as applied to the case under consideration, are usually not given, we shall prove this proposition as applied to an alternating-current network with all the peculiarities following from this.

In Fig. 13 a line fed from two sides is shown, having a load at point C. It is required to carry the current \(\bar{J}_c\) into points “a” and “b” in such a way that the current and voltage drop in the sections Aa and Bb do not change. In this case, the current and the resistances of the line sections are complex numbers.

Fig. 13.

¹) Krug, K. A., Fundamentals of Electrical Engineering.

²) Prof. A. A. Glazunov, Theory and Calculation of Distribution Networks.

By analogy with the solution of a similar problem for the case of direct current, let us assume that:

\[ \bar J_{a(c)}=\frac{\bar Z_{Cb}}{\bar Z_{aC}+\bar Z_{Cb}}\cdot \bar J_C, \]

\[ \bar J_{b(c)}=\frac{\bar Z_{aC}}{\bar Z_{aC}+\bar Z_{bC}}\cdot \bar J_C . \]

Then in section \(Aa\), after transformation of the network, the current will be

\[ \bar J_{A'}= \frac{\bar J_{(C)a}(\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB})+\bar J_{b(C)}\cdot \bar Z_{bB}} {\bar Z_{Aa}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB}} = \]

\[ = \frac{ \bar J_C\frac{\bar Z_{Cb}}{\bar Z_{aC}+\bar Z_{Cb}}\cdot(\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB}) +\bar J_C\frac{\bar Z_{aC}}{\bar Z_{aC}+\bar Z_{Cb}}\cdot \bar Z_{bB} } {\bar Z_{aA}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB}} = \]

\[ = \frac{ \bar J_C\cdot\bar Z_{Cb}\cdot\bar Z_{aC} +\bar J_C\cdot\bar Z_{Cb}\cdot\bar Z_{Cb} +\bar J_C\cdot\bar Z_{Cb}\cdot\bar Z_{bB} +\bar J_C\cdot\bar Z_{aC}\cdot\bar Z_{bB} } {(\bar Z_{aC}+\bar Z_{Cb})(\bar Z_{Aa}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB})}. \]

Before the transformation, in section \(Aa\) the current flowing was

\[ \bar J_A= \frac{\bar J_C\cdot(\bar Z_{Cb}+\bar Z_{CB})} {\bar Z_{Aa}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB}}, \]

or, after multiplying the numerator and denominator by \((\bar Z_{aC}+\bar Z_{CB})\),

\[ \bar J_A= \frac{\bar J_C\cdot\bar Z_{Cb}+\bar J_C\cdot\bar Z_{bB}} {\bar Z_{Aa}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB}} \cdot \frac{\bar Z_{aC}+\bar Z_{Cb}}{\bar Z_{aC}+\bar Z_{Cb}} = \]

\[ = \frac{ \bar J_C\cdot\bar Z_{Cb}\cdot\bar Z_{aC} +\bar J_C\cdot\bar Z_{Cb}\cdot\bar Z_{Cb} +\bar J_C\bar Z_{bB}\cdot\bar Z_{aC} +\bar J_C\bar Z_{bB}\cdot\bar Z_{Cb} } {(\bar Z_{aC}+\bar Z_{Cb})(\bar Z_{Aa}+\bar Z_{aC}+\bar Z_{Cb}+\bar Z_{bB})}, \]

i.e.

\[ \bar J_{A'}=\bar J_A. \]

Let us apply the transformation method to determining the voltages and the distribution of power flows in the system (Fig. 5), in which the voltages at the nodes and the powers in the links have already been found above by the method of nodal equations. To express the loads at the nodes in currents, we shall take as the average voltage \(110\ kv\).

Then the nodal loads in currents, taking into account the capacitive currents of the links, will be:

\[ \text{At node 2}\quad \frac{75-j50.6}{110}=0.682-j0.460\ KA \]

\[ \text{At node 3}\quad \frac{100-j45.84}{110}=0.910-j0.417\ KA \]

\[ \text{At node 4}\quad \frac{125-j26.8}{110}=1.135-j0.244\ KA \]

\[ \text{At node 5}\quad \frac{20-j10.9}{110}=0.182-j0.0992\ KA \]

\[ \text{At node 6}\quad \frac{30-j28.27}{110}=0.273-j0.257\ KA \]

The obtained current loads, with the error due to assuming the average voltage at all nodes, and also the resistances of the network links, are shown in Fig. 14.

Fig. 14.

First of all, according to the statement of the 2nd transformation method, we transfer the loads of the 2nd and 3rd nodes to nodes “1” and “4”. The magnitude of the additional fictitious loads at nodes “1” and “4” is determined as:

\[ \overline{J}_{1(2,3)} = \frac{ \overline{J}_{2}(\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) - \overline{J}_{3}\cdot \overline{Z}_{\mathrm{III}} }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} } \]

and

\[ \overline{J}_{4(2,3)} = \frac{ \overline{J}_{2}(\overline{Z}_{\mathrm{I}}) - \overline{J}_{3}\cdot(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}) }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} } = \overline{J}_{2}+(-\overline{J}_{3})-\overline{J}_{1(2,3)}. \]

If the expression for determining \(\overline{J}_{1(2,3)}\) is transformed*), then the active and reactive components of the current \(\overline{J}_{1(2,3)}\) can conveniently be found separately from the expressions:

\[ J_{1(2,3)a} = G_{0}\sum\left(J_{a(n)}\cdot R'_{n}+J_{r(n)}\cdot X'_{n}\right) + B_{0}\sum\left(J_{a(n)}\cdot X'_{n}+J_{r(n)}\cdot R'_{n}\right); \]

\[ J_{1(2,3)r} = -G_{0}\sum\left(J_{a(n)}\cdot X'_{n}-J_{r(n)}\cdot R'_{n}\right) + B_{0}\sum\left(J_{a(n)}\cdot R'_{n}+J_{r(n)}\cdot X'_{n}\right). \]

Here: \(G_{0}\) and \(B_{0}\) are the active and reactive conductances of the part of the contour between the points to which the loads are transferred; the products \(J_{a(n)}\cdot R'_{n}\), \(J_{r(n)}\cdot X'_{n}\), etc., are the products of the active and reactive components of the loads to be transferred by the corresponding ohmic or inductive resistance up to the opposite point (i.e., in the present case, up to point “4”).

\[ G_{0} = \frac{ R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}} }{ (R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}})^{2} + (X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}})^{2} } = \]

\[ = \frac{ 15+15+4.3 }{ (15+15+4.3)^{2}+(30+30+7.5)^{2} } = 0.00598\ \mho; \]

\[ B_{0} = \frac{ X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}} }{ (R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}})^{2} + (X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}})^{2} } = \]

\[ = \frac{ 30+30+7.5 }{ (15+15+4.3)^{2}+(30+30+7.5)^{2} } = 0.01175\ \mho; \]

\[ \sum\left(J_{a(n)}\cdot R'_{n}+J_{r(n)}\cdot X'_{n}\right) = (0.682\cdot19.3-0.910\cdot4.3+0.460\cdot37.5-0.417\cdot7.5) = \]

\[ =23.37; \]

\[ \sum\left(J_{a(n)}\cdot X'_{n}-J_{r(n)}\cdot R'_{n}\right) = (0.682\cdot37.5-0.910\cdot7.5+0.460\cdot19.3-0.417\cdot4.3) = \]

\[ =11.99. \]

Consequently:

\[ J_{1(2,3)a} = 0.00598\cdot23.37+0.01175\cdot11.69 = 0.277\ \mathrm{kA} \]

\[ J_{1(2,3)r} = -0.00598\cdot11.69+0.01175\cdot23.37 = 0.204\ \mathrm{kA} \]

\[ \text{*) See, for example, Prof. A. A. Glazunov, “Theory and Calculation of Electric Distribution Networks.”} \]

The part of the current assigned to the 4th node will now be—

\[ J_{4(2,3)}=0.862-j0.460-0.910+j0.417-0.277+j0.204= \]

\[ =-(0.505-j0.161)\ \mathrm{kA}. \]

The minus sign before the current \(J_{4(2,3)}\) indicates that this current flows into the 4th node.

In transformed form the network is shown in Fig. 15.

The sum of the loads at the 4th node is—

\[ J'_4=1.135-j0.244+[-(0.505-j0.161)]= \]

\[ =0.630-j0.083\ \mathrm{kA} \]

Fig. 15.

Equivalent resistance of branches VII and I, II, III

\[ \overline{Z}_{\mathrm{e}}= \frac{\overline{Z}_{\mathrm{VII}}\cdot(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}})} {\overline{Z}_{\mathrm{VII}}+(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}})} = \]

\[ =\frac{(15+j25)\cdot(15+j30+15+j30+4.3+j7.5)} {15+j25+15+j30+15+j30+4.3+j7.5} =10.45+j18.35\ \Omega \]

Now the network is a ring with supply at point “1.” The current distribution in this ring will be found by the formula just used for distributing the load of the 2nd and 3rd nodes.

\[ G'_0= \frac{R_{\mathrm{e}}+R_{\mathrm{IV}}+R_{\mathrm{V}}+R_{\mathrm{VI}}} {(R_{\mathrm{e}}+R_{\mathrm{IV}}+R_{\mathrm{V}}+R_{\mathrm{VI}})^2+(X_{\mathrm{e}}+X_{\mathrm{IV}}+X_{\mathrm{V}}+X_{\mathrm{VI}})^2} = \]

\[ =\frac{10.45+10+3+4} {(10.45+10+3+4)^2+(18.35+20+6+5)^2} =0.00862\ \mho \]

\[ B'_0= \frac{X_{\mathrm{e}}+X_{\mathrm{IV}}+X_{\mathrm{V}}+X_{\mathrm{VI}}} {(R_{\mathrm{e}}+R_{\mathrm{IV}}+R_{\mathrm{V}}+R_{\mathrm{VI}})^2+(X_{\mathrm{e}}+X_{\mathrm{IV}}+X_{\mathrm{V}}+X_{\mathrm{VI}})^2} = \]

\[ =\frac{18.35+20+6+5} {(10.45+10+3+4)^2+(18.35+20+6+5)^2} =0.01545\ \mho \]

The calculation of the sum and difference of the products of the active and reactive components of the currents of the nodal loads (4th, 5th, and 6th) by the corresponding ohmic and inductive resistances is placed in Table 5.

Table No. 5.

Node	\(J_{a(n)}\)	\(J_{r(n)}\)	\(R'_n\)	\(X'_n\)	\(J_{a(n)}\!\cdot R'_n\)	\(J_{a(n)}\!\cdot X'_n\)	\(J_{r(n)}\!\cdot R'_n\)	\(J_{r(n)}\!\cdot X'_n\)
4 . . . . .	0.630	0.083	17	31	11.05	20.15	1.410	2.570
5 . . . . .	0.182	0.0992	7	11	1.27	2.00	0.694	1.090
6 . . . . .	0.273	0.257	4	5	1.09	1.363	1.027	1.283
\(\Sigma\) . . . . .	1.085	0.4392	27.45	49.35	13.41	23.513	3.131	4.943

Therefore

\[ \sum (J_{a(n)}\cdot R'_n+J_{r(n)}\cdot X'_n)=13.41+4.943=18.353; \]

\[ \sum (J_{a(n)}\cdot X'_n-J_{r(n)}\cdot R'_n)=23.513-3.131=20.382 \]

and

\[ J_{1(4,5,6)a}=0.00862\cdot18.353+0.01545\cdot20.382=0.474\ KA; \]

\[ J_{1(4,5,6)r}=0.00862\cdot20.382+0.01545\cdot18.353=0.109\ KA; \]

i.e.

\[ \overline{J}_{1(4,5,6)}=0.474-j0.109\ KA, \]

The part of the current \(\overline{J}_{1(4,5,6)}\) flowing through branch 1 may be found according to the provision of the 3rd transformation method

\[ \overline{J}'_1=\overline{J}_{1(4,5,6)}\cdot \frac{\overline{Z}_3}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}} =(0.474-j0.109)\cdot \]

\[ \cdot \frac{10.45+j18.35}{(15+j3)+(15+j30)+(4.3+j7.5)} =0.130-j0.0367\ KA \]

The actual current flowing through branch 1 will be obtained if, to the current just calculated, the load of the first node is added, which resulted from the allocation of the currents of nodes “2” and “3”.

\[ \overline{J}_1=\overline{J}'_1-\overline{J}_{1(2,3)} =0.130-j0.0367+0.277-j0.204 =0.407-j0.241\ KA \]

The loads of all the other branches will be found by Kirchhoff’s law, subtracting (algebraically) from the currents of the preceding branches the loads of the corresponding nodes.

\[ \overline{J}_{\mathrm{II}}=\overline{J}_1-\overline{J}_2 =0.407-j0.241-(0.682-j0.460) =-(0.275-j0.219)\ KA \]

\[ \overline{J}_{\mathrm{III}}=\overline{J}_{\mathrm{II}}-\overline{J}_3 =-(0.275-j0.219)-[-(0.910-j0.417)] = \]

\[ =0.635-j0.198\ KA \]

\[ \overline{J}_{\mathrm{VII}}=\overline{J}_{1(4,5,6)}-\overline{J}'_1 =0.474-j0.109-(0.13-j0.367) =0.344-j0.072)\ KA \]

\[ \overline{J}_{\mathrm{IV}}=\overline{J}_{\mathrm{VII}}+\overline{J}_{\mathrm{III}}-\overline{J}_4 =0.344-j0.072+0.635-j0.198-(1.135-j0.244)= \]

\[ =-(0.156+j0.026)\ KA \]

\[ \overline{J}_{\mathrm{V}}=\overline{J}_{\mathrm{IV}}-\overline{J}_5 =-(0.156+j0.026)-(0.182-j0.0992)= \]

\[ =-(0.338-j0.073)\ KA \]

\[ \overline{J}_{\mathrm{VI}}=\overline{J}_{\mathrm{V}}-\overline{J}_6 =-(0.338-j0.073)-(0.273-j0.267)= \]

\[ =-(0.611-j0.330)\ KA. \]

Let us check the current in branch VI from the relation:

\[ \overline{J}_{\mathrm{VI}}=(\overline{J}'_4+\overline{J}_5+\overline{J}_6)-\overline{J}_{1(4,5,6)} =(0.630-j0.083+0.182-j0.0992+ \]

\[ +0.273-j0.257)-(0.474-j0.109) =0.611-j0.330\ KA. \]

The current distribution in the network, found by the transformation method, is plotted in Fig. 16; moreover, from the station of node “1” the current taken is

\[ \overline{J}_1=\overline{J}_1+\overline{J}_{\mathrm{VII}}+\overline{J}_{\mathrm{VI}} +\frac{1}{2}(\overline{J}_{c\mathrm{I}}+\overline{J}_{c\mathrm{VII}}+\overline{J}_{c\mathrm{VI}}) =0.407-j0.241+ \]

\[ +0.344-j0.072+0.611-j0.330+jJ_{1(c)} =1.362-j0.643+jJ_{1(c)}\ ^*) \]

\[ \overline{\phantom{+0.344-j0.072}} \]

\(^*)\ \(jJ_{1(c)}\) is the current at the first node due to the capacitance of branches I, VII, and VI.

It is now not difficult to find the voltages at the nodes, under the condition that

\[ \overline{E}_4 = E_4 = 105\ kv \]

Fig. 16.

\[ \overline{E}_3 = \overline{E}_4+\overline{J}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{III}} = 105+(0.635-j0.198)\cdot(4.3+j7.5) = 109.20+j3.93; \]

and

\[ E_3=\sqrt{109.20^2+3.93^2}=109.25\ kv. \]

\[ \overline{E}_2 = \overline{E}_3-\overline{J}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{II}} = 109.20+j3.93-(0.276-j0.219)\cdot(15+j30) = \]

\[ =98.49-j1.06; \]

and

\[ E_2=\sqrt{98.49^2+1.06^2}=98.50\ kv. \]

\[ \overline{E}_1 = \overline{E}_2+\overline{J}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{I}} = 98.49+j1.06+(0.407-j0.241)\cdot(15+j30) = \]

\[ =111.81+j7.37; \]

and

\[ E_1=\sqrt{111.81^2+7.37^2}=112.06\ kv. \]

\[ \overline{E}_5 = \overline{E}_4+\overline{J}_{\mathrm{IV}}\cdot\overline{Z}_{\mathrm{IV}} = 105+(0.156+j0.026)\cdot(10+j20) = 106.05+j3.40; \]

and

\[ E_5=\sqrt{106.05^2+3.40^2}=106.1\ kv. \]

\[ \overline{E}_6 = \overline{E}_5+\overline{J}_{\mathrm{V}}\cdot\overline{Z}_{\mathrm{V}} = 109.05+j3.40+(0.338-0.073)\cdot(3+j6) = \]

\[ =107.51+j5.21; \]

and

\[ E_6=\sqrt{107.51^2+5.21^2}=107.63\ kv. \]

\[ \overline{E}_1 = \overline{E}_6+\overline{J}_{\mathrm{VI}}\cdot\overline{Z}_{\mathrm{VI}} = 107.51+j5.21+(0.611-j0.330)\cdot(4+j5) = 111.6+j7.0; \]

and

\[ E_1=\sqrt{111.6^2+7.0^2}=111.85\ kv. \]

Some discrepancies in the values of \(E_1\) calculated from branches I and VI are explained by the inaccuracy of the calculations on the slide rule.

The found voltage values are plotted on the network diagram, Fig. 17.

Fig. 17.

The voltage values found by the transformation method, when the loads at the nodes are taken into account in currents through the average nodal voltage—

... at 110 \(kv\), for comparison with the same values determined by the method of nodal equations, are entered in Table 6.

Table 6.

Nodal voltages	\(E_1\)	\(E_2\)	\(E_3\)	\(E_4\)	\(E_5\)	\(E_6\)
By the transformation method	112.0	98.5	109.25	105	106.1	107.63
By the method of nodal equations	114.8	98.15	109.9	105	107.82	110.4
Difference in %%	2.67%	0.35%	0.6%	0.0%	1.6%	2.5%

As can be seen from the table, the error by the transformation method (if the true values of the voltages at the nodes are taken to be those calculated by the method of nodal equations) comes out to be more than two percent. Such an error in approximate calculations may be permissible with respect to voltages; however, it must be taken into account that an error of 2% in the values of the nodal voltages is associated with a much greater error in the matter of current distribution in the branches. Thus, for example, for branch IV (at its end, if going counterclockwise) we have the following values of power flows:

a) by the transformation method . . . \(16.7 - j2.23\) \(mva\)
b) by the method of nodal equations . . \(22.6 + j8\) ”

Difference in active powers . . . \(26\%\), and
in reactive powers . . . \(70\%\).

In addition to all this, it must also be taken into account that all these inaccuracies still depend on the successful choice of the average voltage by which the calculated current loads are determined.

Obviously, it is necessary to conclude that the transformation method in this form cannot be recommended for broad application or for final calculations.

Strictly speaking, the process of transforming the network itself does not introduce an error into the results of the network calculation. The error was introduced at the very outset into the basic calculation data—in the loads. The current loads at the nodes were taken as not corresponding to reality, since the powers specified at the nodes were referred to one and the same “average” voltage of 110 \(kv\). As a result of the calculation, in some nodes voltages were obtained that were entirely different from this “average” voltage. Thus, for example, at node 2 a voltage of 98.5 \(kv\) was obtained, etc.; the calculation should have been based on the current

\[ \bar{J}_2^{\,1}=\frac{75-j50.6}{98.5}=0.762-j0.514, \]

instead of

\[ \bar{J}_2=0.682-j0.460\ KA. \]

In addition, it would also have been necessary to take into account the phase difference of the voltage vectors at all nodes.

But it turns out, as has recently been shown by Prof. A. A. Glazunov, Prof. A. Ya. Ryabkov, and Docent F. P. Lashkov,¹ that networks can be transformed with loads at the nodes expressed in powers, etc.; in the process of calculating the network there is no need to know the nodal voltages beforehand. This circumstance is extremely important, since it makes it possible to expand to a great degree the possibilities in the field of calculations of complex systems.

¹ Prof. A. A. Glazunov, Prof. A. Ya. Ryabkov and Docent F. P. Lashkov, “Calculation of Electric Networks of Power Systems.”

We shall give here proofs of the fundamental propositions that make it possible to use the transformation method in this new capacity.

Proposition 1. Between points A and B (Fig. 18) of the network, powers \(\overline{W}_\mathrm{I}\) and \(\overline{W}_\mathrm{II}\) flow along two parallel lines with resistances \(\overline{Z}_\mathrm{I}\) and \(\overline{Z}_\mathrm{II}\). It is required to find the value of the resistance of a single line along which, between points A and B, there would flow a power equal to the sum of \(\overline{W}_\mathrm{I}\) and \(\overline{W}_\mathrm{II}\), while the voltage drop between points A and B would remain unchanged.

Fig. 18.

Since the equality

\[ \frac{\overline{W}_\mathrm{I}}{\overline{E}_B} + \frac{\overline{W}_\mathrm{II}}{\overline{E}_B} = \frac{\overline{W}}{\overline{E}_B} \]

is equivalent to the equality

\[ \overline{J}_\mathrm{I}+\overline{J}_\mathrm{II}=\overline{J}, \]

it is obvious that

\[ \overline{Z}_e= \frac{\overline{Z}_\mathrm{I}\cdot \overline{Z}_\mathrm{II}} {\overline{Z}_\mathrm{I}+\overline{Z}_\mathrm{II}}. \]

On the same basis, for the process of inverse transformation one may write:

\[ \overline{W}_\mathrm{I} = \overline{W}\frac{\overline{Z}_e}{\overline{Z}_\mathrm{I}}, \]

\[ \overline{W}_\mathrm{II} = \overline{W}\frac{\overline{Z}_e}{\overline{Z}_\mathrm{II}}. \]

Proposition 2. There is a ring network (Fig. 19) with a single load at point “b” and a supply point at point “A.” Let us determine the power-flow distribution in this network.

It may be written:

\[ \frac{\overline{W}_{\mathrm{I}(b)}}{\overline{E}_b}\cdot \overline{Z} = \frac{\overline{W}_{\mathrm{II}(b)}}{\overline{E}_b}\cdot \overline{Z}_\mathrm{II}, \]

where \(\overline{W}_{\mathrm{I}(b)}\) and \(\overline{W}_{\mathrm{II}(b)}\) are the assumed power flows along branches I and II near point “b,” and consequently

\[ \frac{\overline{W}_{\mathrm{I}(b)}}{\overline{W}_{\mathrm{II}(b)}} = \frac{\overline{Z}_\mathrm{II}}{\overline{Z}_\mathrm{I}}. \]

Fig. 19.

But since

\[ \overline{W}_{\mathrm{I}(b)}+\overline{W}_{\mathrm{II}(b)} = \overline{W}_b, \]

then

\[ \overline{W}_\mathrm{I} = \overline{W}_b \frac{\overline{Z}_\mathrm{II}} {\overline{Z}_\mathrm{I}+\overline{Z}_\mathrm{II}} \]

and

\[ \overline{W}_\mathrm{II} = \overline{W}_b \frac{\overline{Z}} {\overline{Z}_\mathrm{I}+\overline{Z}_\mathrm{II}}. \]

That is, the formulas have a form similar to the case when the load of the ring is specified in amperes.

\[ \overline{W}'_\mathrm{III}\cdot \overline{Z}_\mathrm{III} - \left(\frac{W'_\mathrm{III}}{E_A}\right)^2 \cdot Z^2_\mathrm{III} + \overline{W}'_\mathrm{III}\cdot \overline{Z}_\mathrm{II} - \overline{W}_c\cdot \overline{Z}_\mathrm{II} - \Delta \overline{W}_\mathrm{III}\cdot \overline{Z}_\mathrm{II} - \overline{W}_b\cdot \overline{Z} - \]

\[ - \overline{W}_c\cdot \overline{Z}_\mathrm{I} - \Delta \overline{W}_\mathrm{I}\cdot \overline{Z}_\mathrm{I} - \Delta \overline{W}_\mathrm{II}\cdot \overline{Z}_\mathrm{I} - \Delta \overline{W}_\mathrm{III}\cdot \overline{Z}_\mathrm{I} + \overline{W}'_\mathrm{III}\cdot \overline{Z}_\mathrm{I} + \]

However, the derived formulas do not give an idea of what power flows depart from point “A.” These powers must cover the load at point “b” and, in addition, the losses in the ring, i.e.

\[ \overline{W}'_{\mathrm{I}}=\overline{W}_{\mathrm{I}(b)}+\Delta\overline{W}_{\mathrm{I}(b)} \]

and

\[ \overline{W}'_{\mathrm{II}}=\overline{W}_{\mathrm{II}(b)}+\Delta\overline{W}_{\mathrm{II}(b)} . \]

To determine \(\overline{W}'_{\mathrm{I}}\) and \(\overline{W}'_{\mathrm{II}}\), we write the relation:

\[ \frac{\overline{W}'_{\mathrm{I}}}{\hat{E}_{A}}\cdot \overline{Z}_{\mathrm{I}} = \frac{\overline{W}'_{\mathrm{II}}}{\hat{E}_{A}}\cdot \overline{Z}_{\mathrm{II}}, \]

or:

\[ \frac{\overline{W}'_{\mathrm{I}}}{\overline{W}'_{\mathrm{II}}} = \frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}}; \]

But since:

\[ \overline{W}_{A} = \overline{W}'_{\mathrm{I}}+\overline{W}'_{\mathrm{II}} = \overline{W}_{\mathrm{I}(b)}+\overline{W}_{\mathrm{II}(b)} +\Delta\overline{W}_{\mathrm{I}(b)} +\Delta\overline{W}_{\mathrm{II}(b)} = \overline{W}_{(b)} + \left(\Delta\overline{W}_{\mathrm{I}(b)} +\Delta\overline{W}_{\mathrm{II}(b)}\right), \]

then:

\[ \overline{W}'_{\mathrm{I}} = \overline{W}_{A}\frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} = \overline{W}_{b}\frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} + \left(\Delta\overline{W}_{\mathrm{I}(b)} +\Delta\overline{W}_{\mathrm{II}(b)}\right) \frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} \]

and

\[ \overline{W}'_{\mathrm{II}} = \overline{W}_{A}\frac{\overline{Z}_{\mathrm{I}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} = \overline{W}_{b}\frac{\overline{Z}_{\mathrm{I}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} + \left(\Delta\overline{W}_{\mathrm{I}(b)} +\Delta\overline{W}_{\mathrm{II}(b)}\right) \frac{\overline{Z}_{\mathrm{I}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}}. \]

The derived formulas point to the very important circumstance that the losses in the ring network are distributed according to the very same law as the main power flow.

And since, from comparing the relations:

\[ \frac{\overline{W}_{\mathrm{I}(b)}}{\overline{W}_{\mathrm{II}(b)}} = \frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}} \]

and

\[ \frac{\overline{W}_{\mathrm{I}}}{\overline{W}_{\mathrm{II}}} = \frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}} = \frac{\overline{W}_{\mathrm{I}(b)}+\Delta\overline{W}_{\mathrm{I}(b)}}{\overline{W}_{\mathrm{II}(b)}+\Delta\overline{W}_{\mathrm{II}(b)}} \]

it follows that also

\[ \frac{\Delta\overline{W}_{\mathrm{I}(b)}}{\Delta\overline{W}_{\mathrm{II}(b)}} = \frac{\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{I}}}, \]

or

\[ \frac{\Delta\overline{W}_{\mathrm{I}(b)}}{\Delta\overline{W}_{\mathrm{II}(b)}} = \frac{\overline{W}_{\mathrm{I}(b)}}{\overline{W}_{\mathrm{II}(b)}}, \]

the calculation and allocation of losses around the ring are therefore considerably simplified, since—as it turns out—the losses are proportional to the power flows in the ring found without taking losses into account.

Let us examine to what extent the rule derived here for determining the distribution of power flows in a ring network, using the example of a single load, is applicable to the case of several loads.

A ring network is given (Fig. 20) with two points of power take-off. Assuming that at point “b” there is a power-flow split, and also taking into account the notation of Fig. 20, one may write:

$$ \frac{\overline{W}'_{\mathrm{III}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{III}} + \frac{\overline{W}'_{\mathrm{II}}}{\widehat{E}_C}\cdot \overline{Z}_{\mathrm{II}} - \frac{\overline{W}'_{\mathrm{I}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{I}} =0 $$

But since

$$ \overline{W}'_{\mathrm{II}}=\overline{W}'_{\mathrm{III}}-\overline{W}_c-\Delta\overline{W}_{\mathrm{III}} $$

and

$$ \overline{W}'_{\mathrm{I}} = \overline{W}_b+\overline{W}_c+\Delta\overline{W}_{\mathrm{I}} +\Delta\overline{W}_{\mathrm{II}} +\Delta\overline{W}_{\mathrm{III}} -\overline{W}'_{\mathrm{III}}, $$

where $\Delta\overline{W}_{\mathrm{I}}$, $\Delta\overline{W}_{\mathrm{II}}$ and $\Delta\overline{W}_{\mathrm{III}}$ are the losses in the corresponding sections of the network, the voltage-balance equation in the ring will be rewritten as follows:

Fig. 20.

$$ \frac{\overline{W}'_{\mathrm{III}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{III}} + \frac{\overline{W}'_{\mathrm{III}}-\overline{W}_c-\Delta\overline{W}_{\mathrm{III}}}{\widehat{E}_C}\cdot \overline{Z}_{\mathrm{II}} - \frac{ \overline{W}_b+\overline{W}_c+\Delta\overline{W}_{\mathrm{I}} +\Delta\overline{W}_{\mathrm{II}}+\Delta\overline{W}_{\mathrm{III}} -\overline{W}'_{\mathrm{III}} }{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{I}} =0. $$

Further, bearing in mind that

$$ \overline{E}_C=\overline{E}_A-\frac{\overline{W}'_{\mathrm{III}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{III}}, $$

we rewrite the basic relation in the form:

$$ \frac{\overline{W}'_{\mathrm{III}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{III}} + \frac{\overline{W}'_{\mathrm{III}}-\overline{W}_c-\Delta\overline{W}_{\mathrm{III}}} {\widehat{E}_A-\dfrac{\overline{W}'_{\mathrm{III}}}{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{III}}} \cdot \overline{Z}_{\mathrm{II}} - \frac{ \overline{W}_b+\overline{W}_c+\Delta\overline{W}_{\mathrm{I}} +\Delta\overline{W}_{\mathrm{II}}+\Delta\overline{W}_{\mathrm{III}} -\overline{W}'_{\mathrm{III}} }{\widehat{E}_A}\cdot \overline{Z}_{\mathrm{I}} =0 $$

or

$$ \overline{W}'_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{III}} - \left(\frac{\overline{W}'_{\mathrm{III}}}{E_A}\right)^2\cdot \overline{Z}^{\,2}_{\mathrm{III}} + \overline{W}'_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{II}} - \overline{W}_c\cdot \overline{Z}_{\mathrm{II}} - \Delta\overline{W}_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{II}} - \overline{W}_b\cdot \overline{Z}_{\mathrm{I}} - $$

$$ - \overline{W}_c\cdot \overline{Z}_{\mathrm{I}} - \Delta\overline{W}_{\mathrm{I}}\cdot \overline{Z}_{\mathrm{I}} - \Delta\overline{W}_{\mathrm{II}}\cdot \overline{Z}_{\mathrm{I}} - \Delta\overline{W}_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{I}} + \overline{W}'_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{I}} + $$

$$ + \frac{\overline{W}'_{\mathrm{III}}}{E_A^{2}} \left( \overline{W}_b+\overline{W}_c+\Delta\overline{W}_{\mathrm{I}} +\Delta\overline{W}_{\mathrm{II}}+\Delta\overline{W}_{\mathrm{III}} -\overline{W}'_{\mathrm{III}} \right)\cdot \overline{Z}_{\mathrm{I}}\cdot \overline{Z}_{\mathrm{III}} =0. $$

In this equation the term

$$ \frac{\overline{W}'_{\mathrm{III}}}{E_A^{2}} \left( \overline{W}_b+\overline{W}_c+\Delta\overline{W}'_{\mathrm{I}} +\Delta\overline{W}_{\mathrm{II}}+\Delta\overline{W}_{\mathrm{III}} -\overline{W}'_{\mathrm{III}} \right)\cdot \overline{Z}_{\mathrm{I}}\cdot \widehat{E}_{\mathrm{III}} $$

is equal to the expression—

$$ \frac{\widehat{W}'_{\mathrm{III}}}{E_A^{2}}\cdot \overline{W}'_{\mathrm{I}}\cdot \overline{Z}_{\mathrm{I}}\cdot \widehat{Z}_{\mathrm{III}}, $$

which, in magnitude, may without great detriment to the accuracy of the entire conclusion be replaced by the expression

\[ \frac{\widehat{W}_{\mathrm{III}}}{E_A^2}\,\overline{W}_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{III}}\cdot \widehat{Z}_{\mathrm{III}} = \left(\frac{\overline{W}_{\mathrm{III}}}{E_A}\right)^2\cdot Z'_{\mathrm{III}} . \]

If such a replacement is made, then the principal expression will have the form:

\[ \overline{W}'_{\mathrm{III}}\cdot(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) - \overline{W}_c(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}) - \overline{W}_b\cdot \overline{Z}_{\mathrm{I}} - \]

\[ - \Delta \overline{W}_{\mathrm{III}}(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}) - (\Delta \overline{W}_{\mathrm{I}}+\Delta \overline{W}_{\mathrm{II}})\cdot \overline{Z}_{\mathrm{I}} =0. \]

Whence we find—

\[ \overline{W}'_{\mathrm{III}} = \frac{ \overline{W}_c(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}) + \overline{W}_b\cdot \overline{Z}_{\mathrm{I}} }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} } + \frac{ \Delta \overline{W}_{\mathrm{III}}(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}) + (\Delta \overline{W}_{\mathrm{I}}+\Delta \overline{W}_{\mathrm{II}})\cdot \overline{Z}_{\mathrm{I}} }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} }. \]

By similar reasoning one may obtain that

\[ \overline{W}'_{\mathrm{I}} = \frac{ \overline{W}_c\cdot \overline{Z}_{\mathrm{III}} + \overline{W}_b(\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} } + \frac{ \Delta \overline{W}_{\mathrm{III}}\cdot \overline{Z}_{\mathrm{III}} + (\Delta \overline{W}_{\mathrm{I}}+\Delta \overline{W}_{\mathrm{II}})(\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}} }. \]

The power-flow distribution in the ring, in this form, obeys the same law as in the case of a single load. This law, evidently, can now be extended to the case of any number of loads.

Proposition 3-e. In a ring network (Fig. 21), the load applied at point “b” is to be distributed between points “a” and “c” in such a way that the power flows, neglecting losses in the network, on sections I and IV remain unchanged.

It can be shown that such a distribution of the load \(\overline{W}_b\) is equivalent to determining the power-flow distribution in the ring network if this network is supplied at equal voltage from points “a” and “c”; that is, that the component of the load \(\overline{W}_b\) applied at point “a” will be equal to

\[ \overline{W}_{a(b)} = \frac{\overline{W}_b\cdot \overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}} \]

and that the component of the load \(\overline{W}_b\) applied at point “c” will be equal to

\[ \overline{W}_{c(b)} = \frac{\overline{W}_b\cdot \overline{Z}_{\mathrm{III}}}{\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}}. \]

Fig. 21.

To prove the proposition that on sections I and IV the power-flow distribution has not changed as a result of such a distribution of the load \(\overline{W}_b\), let us compare the power flows, neglecting losses, for example on section I, before and after the transformation of the network.

Before the transformation

\[ \overline{W}_{\mathrm{IV}} = \frac{ \overline{W}_a(\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{I}}) + \overline{W}_b(\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{I}}) + \overline{W}_c\cdot \overline{Z}_{\mathrm{I}} }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}} }; \]

after transformation

\[ \overline{W}'_{\mathrm{IV}}= \frac{ \left(\overline{W}_a+\frac{\overline{W}_b\cdot\overline{Z}_{\mathrm{II}}}{\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}}\right) \cdot(\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{I}}) + \left(\overline{W}_c+\frac{\overline{W}_b\cdot\overline{Z}_{\mathrm{III}}}{\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}}\right) \overline{Z}_{\mathrm{I}} }{ \overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}} } = \]

\[ = \frac{ \overline{W}_a\cdot\overline{Z}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{III}} +\overline{W}_a\cdot\overline{Z}^{\,2}_{\mathrm{II}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{III}} +\overline{W}_b\cdot\overline{Z}^{\,2}_{\mathrm{II}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{II}} }{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) } + \]

\[ + \frac{ \overline{W}_a\cdot\overline{Z}^{\,2}_{\mathrm{III}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{I}} +\overline{W}_c\cdot\overline{Z}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{I}} +\overline{W}_c\cdot\overline{Z}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{I}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{I}} }{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) } \]

If now the expression for the power flow in the transformed network is multiplied and divided by the sum

\[ (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) \]

then we obtain:

\[ \overline{W}_{\mathrm{IV}}= \frac{ \overline{W}_a\cdot\overline{Z}_{\mathrm{III}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_a\cdot\overline{Z}^{\,2}_{\mathrm{III}} +\overline{W}_a\cdot\overline{Z}^{\,2}_{\mathrm{II}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{III}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_a\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{III}} }{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) } + \]

\[ + \frac{ \overline{W}_b\cdot\overline{Z}^{\,2}_{\mathrm{II}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{II}}\cdot\overline{Z}_{\mathrm{III}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_b\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{III}} +\overline{W}_c\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{II}} +\overline{W}_c\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{III}} }{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) }. \]

From the resulting comparison of the expressions for the flows in the transformed and untransformed network it is seen that the magnitude of the flows is the same.

Proposition 4 concerns the transformation of a triangle of impedances into a star of impedances, and conversely. The course of reasoning in deriving the transformation formulas, and the transformation formulas themselves, are entirely identical with the case of a current load, since both in the first and in the second case the condition of equality of voltage drops between the corresponding corner points of the triangle in the transformed and untransformed network is satisfied.

The four propositions given here make it possible to reduce a complex network to the form of a simple one, to find the distribution of power flows in this simple network, and then, transforming the network back, to determine also the actual flow distribution in all branches of the network.

The derivation of all four propositions was quite rigorous, with the sole exception of the second proposition, where the following expressions were allowed to be equated:

\[ \left(\frac{W'_{\mathrm{III}}}{E_A}\right)^2 Z^2_{\mathrm{III}} \quad \text{and} \quad \frac{\dot{W}'_{\mathrm{III}}}{E_A^2}\cdot \overline{W}'_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{I}}\cdot\overline{Z}_{\mathrm{III}}. \]

These terms are completely identical if the condition is satisfied:

\[ \frac{\overline{W}'_{\mathrm{I}}}{\overline{W}'_{\mathrm{III}}} = \frac{\overline{Z}_{\mathrm{III}}}{\overline{Z}_{\mathrm{I}}}. \]

This condition was indeed satisfied in the case of a single load. In the case of two loads, however, this condition will be satisfied only in a special case. Nevertheless, approximate calculations show that even in the case where this proportion is far from being observed, the error obtained is insignificant. At the same time, equating the above-mentioned expressions makes it poss—

made it possible to give a very simple formulation to the rule for the distribution of losses in the network.

As Prof. A. A. Glazunov has shown, the losses in a network can usually be distributed over the branches, proceeding from the point of power-flow division against the direction of the active-power flows.

However, if there is one or several additional points of flow division in the network, then, as a result of such a distribution of losses, we arrive, as it were, at a dead end (Fig. 22, point “3”). If at the station of node “3” we had a reserve of some power, then the losses in branches “II” and “III” would be covered from this reserve. But the station of node 3 is taken as a station operating according to a strictly prescribed schedule and, consequently, the losses in branches II and III can be covered only by the station of node 1. But in this case we must know the law of distribution of losses in the network, if the distribution of these losses cannot be carried out by the simple method indicated above. Such a law is given by Proposition 2.

Fig. 22.

The complete application of the transformation method for the case when loads are accounted for in powers is shown below in two different examples.

VII.

It is required to determine the power-flow distribution in the branches and the voltages at the nodes of the network (Fig. 23). The reactive powers due to the capacitances of the branches have been included in the corresponding nodal loads; in this case the reactive components of the nodal loads decreased somewhat, while the reactive component of the power of the station (node “3”) increased somewhat.

Fig. 23.

In order to reduce the given network to the form of a ring network, we distribute the loads of the 2nd and 3rd nodes to nodes “1” and “4”.

According to Proposition 3 of the transformation method, the component of the powers \(\overline{W}_3\) and \(\overline{W}_2\) referred to node “1” will be equal to—

\[ \overline{W}_{1(2,3)} = \frac{\overline{W}_2(\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}})-\overline{W}_3 \cdot \overline{Z}_{\mathrm{III}}} {\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}}. \]

The minus sign before the term \(\overline{W}_3 \cdot \overline{Z}_{\mathrm{III}}\) is explained by the fact that the power \(\overline{W}_3\) enters the network and therefore has a direction opposite to the case when power is taken from the network.

The expression for \(\overline{W}_{1(2,3)} = P_{1(2,3)} - jQ_{1(2,3)}\) can be transformed so that the values of the active and reactive components may be obtained separately. In transformed form the expressions for \(P_{1(2,3)}\) and \(Q_{1(2,3)}\) will have the form:

\[ P_{1(2,3)} = G_0\{[P_2(R_{\mathrm{II}}+R_{\mathrm{III}})-P_3 \cdot R_{\mathrm{III}}]+[Q_2(X_{\mathrm{II}}+X_{\mathrm{III}})-Q_3 \cdot X_{\mathrm{III}}]\} + \]

\[ +B_0\{[P_2(X_{\mathrm{II}}+X_{\mathrm{III}})-P_3 \cdot X_{\mathrm{III}}]-[Q_2(R_{\mathrm{II}}+R_{\mathrm{III}})-Q_3 \cdot R_{\mathrm{III}}]\}. \]

and

\[ Q_{1(2,3)}=-G_0\{[P_2(X_{\mathrm{II}}+X_{\mathrm{III}})-P_3X_{\mathrm{III}}]-[Q_2(R_{\mathrm{II}}+R_{\mathrm{III}})-Q_3R_{\mathrm{III}}]\}+ \]
\[ +B_0\{[P_2(R_{\mathrm{II}}+R_{\mathrm{III}})-P_3R_{\mathrm{III}}]+[Q_2(X_{\mathrm{II}}+X_{\mathrm{III}})-Q_3X_{\mathrm{III}}]\}. \]

Or, in general form:

\[ P=G_0\cdot \Sigma(P_nR'_n+Q_nX'_n)+B_0\Sigma(P_nX'_n-Q_nR'_n) \]

and

\[ Q=-G_0\Sigma(P_nX'_n-Q_nR'_n)+B_0\Sigma(P_nR'_n+Q_nX'_n). \]

Here: \(G_0\) and \(B_0\) are the active and reactive conductances of the part of the network between the points into which the loads are distributed.

In the present case

\[ G_0=\frac{R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}}}{(R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}})^2+(X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}})^2}=0{,}00598\ \mho \]

and

\[ B_0=\frac{X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}}}{(R_{\mathrm{I}}+R_{\mathrm{II}}+R_{\mathrm{III}})^2+(X_{\mathrm{I}}+X_{\mathrm{II}}+X_{\mathrm{III}})^2}=0{,}01175\ \mho \]

The results of the calculations, according to the operations in parentheses, lowercase and uppercase, are concentrated in Table 7.

Table 7.

Nodes	\(P_n\)	\(Q_n\)	\(R'_n\)	\(X'_n\)	\(P_nR'_n\)	\(P_nX'_n\)	\(Q_nR'_n\)	\(Q_nX'_n\)
2	75	50,6	19,3	37,5	1446	2810	976	1895
3	—100	—45,84	4,3	7,5	—430	—750	—197,5	—344
\(\Sigma\)	—25	4,76			1016	2060	779	1551

\[ \Sigma(P_nR'_n+Q_nX'_n)=1016+1551=2567; \]

\[ \Sigma(P_nX'_n-Q_nR'_n)=2060-779=1281. \]

Consequently:

\[ P_{1(2,3)}=0{,}00598\cdot2567+0{,}01175\cdot1281=30{,}38\ \mathrm{MW} \]

and

\[ Q_{1(2,3)}=-0{,}00598\cdot1281+0{,}01175\cdot2567=22{,}62\ \mathrm{MVAr} \]

i.e.

\[ \overline{W}_{1(2,3)}=P_{1(2,3)}-jQ_{1(2,3)}=30{,}38-j22{,}62\ \mathrm{MVA}. \]

Now we easily obtain:

\[ \overline{W}_{4(2,3)}=\overline{W}_2+\overline{W}_3+\overline{W}_{1(2,3)}=-25-j4{,}76-(30{,}38-j22{,}62)= \]

\[ =-(55{,}38-j17{,}86)\ \mathrm{MVA}. \]

Let us check the calculation of \(\overline{W}_{1(2,3)}\) and \(\overline{W}_{4(2,3)}\)

\[ \overline{W}_{4(2,3)}=P_{4(2,3)}-jQ_{4(2,3)}, \]

where

\[ P_{4(2,3)}=G_0\{[P_2R_{\mathrm{I}}-P_3(R_{\mathrm{I}}+R_{\mathrm{II}})]+[Q_2X_{\mathrm{I}}-Q_3(X_{\mathrm{I}}+X_{\mathrm{II}})]\}+ \]

\[ +B_0\{[P_2X_{\mathrm{I}}-P_3(X_{\mathrm{I}}+X_{\mathrm{II}})]-[Q_2R_{\mathrm{I}}-Q_3(R_{\mathrm{I}}+R_{\mathrm{II}})]\} \]

and

\[ Q_{4(2,3)}=-G_0\{[P_2X_{\mathrm{I}}-P_3(X_{\mathrm{I}}+X_{\mathrm{II}})]-[Q_2R_{\mathrm{I}}-Q_3(R_{\mathrm{I}}+R_{\mathrm{II}})]\}+ \]

\[ +B_0\{[P_2R_{\mathrm{I}}-P_3(R_{\mathrm{I}}+R_{\mathrm{II}})]+[Q_2X_{\mathrm{I}}-Q_3(X_{\mathrm{I}}+X_{\mathrm{II}})]\} \]

The results of the computations for the quantities enclosed in square brackets are presented in Table 7a.

Table 7a

Nodes	$P_n$	$Q_n$	$R''_n$	$X''_n$	$P_n R''_n$	$P_n X''_n$	$Q_n R''_n$	$Q_n X''_n$
2	75	50,6	15	30	1125	2250	758	1516
3	—100	—45,84	30	60	—3000	—6000	—1377	—2754
$\Sigma$	—25	4,76			—1875	—3750	—619	—1238

\[ \Sigma \left(P_n R''_n + Q_n X''_n\right) = -1875 - 1238 = -3113; \]

\[ \Sigma \left(P_n X''_n - Q_n R''_n\right) = -3750 - (-619) = -3131. \]

Consequently:

\[ \underline{P}_{4(2,3)} = 0{,}00598 \cdot (-3113) + 0{,}01175 \cdot (-3131) = -55{,}38 \ \text{MW} \]

and

\[ \underline{Q}_{4(2,3)} = -0{,}00598 \cdot (-3131) + 0{,}01175 \cdot (-3113) = -17{,}85 \ \text{MVA}. \]

As we see, the computations were made correctly.

The load of the 4th node, taking into account the component of the load of the 2nd and 3rd nodes, is

\[ \underline{\overline{W}}_4' = \underline{\overline{W}}_4 + \underline{\overline{W}}_{4(2,3)}' = 125 - j26{,}8 + [-(55{,}38 - j17{,}86)] = 69{,}62 - j8{,}94 \ \text{MVA}. \]

Fig. 24.

The network in its transformed form is shown in Fig. 24.

The equivalent impedance of the two parallel branches between nodes “1” and “4” is

\[ \underline{\overline{Z}}_3 = \frac{ \overline{Z}_{\mathrm{VII}} \cdot \left(\overline{Z}_{\mathrm{I}} + \overline{Z}_{\mathrm{II}} + \overline{Z}_{\mathrm{III}}\right) }{ \overline{Z}_{\mathrm{VII}} + \overline{Z}_{\mathrm{I}} + \overline{Z}_{\mathrm{II}} + \overline{Z}_{\mathrm{III}} } = 10{,}45 + j18{,}35 \ \Omega \]

In the ring network thus obtained, we shall find the power-flow distribution.

Table 8.

Nodes	$P_n$	$Q_n$	$R'_n$	$X'_n$	$P_n R'_n$	$P_n X'_n$	$Q_n R'_n$	$Q_n X'_n$
4	69,62	8,94	17	31	1180	2156	151,7	276,5
5	20,00	10,90	7	11	140	220	76,3	120,0
6	30,00	28,27	4	5	120	150	113,0	141,2
$\Sigma$	119,62	48,11			1448	2526	341,0	537,7

Table 8 presents the results of the basic computations for determining the power flow leaving node “1” along the equivalent branch (here $\overline{W}_{1(2,3)}$ is not yet taken into account).

The watt conductance of the ring is

\[ G'_0 = \frac{ R_3 + R_{\mathrm{IV}} + R_{\mathrm{V}} + R_{\mathrm{VI}} }{ \left(R_3 + R_{\mathrm{IV}} + R_{\mathrm{V}} + R_{\mathrm{VI}}\right)^2 + \left(X_3 + X_{\mathrm{IV}} + X_{\mathrm{V}} + X_{\mathrm{VI}}\right)^2 } = 0{,}00862 \ \text{S} \]

The wattless conductance of the ring

\[ B'_0=\frac{X_{\mathrm{e}}+X_{\mathrm{VI}}+X_{\mathrm{V}}+X_{\mathrm{VI}}}{(\Sigma R)^2+(\Sigma X)^2}=0.01545\ \mho \]

\[ \Sigma \left(P_n R'_n+Q_n X'_n\right)=1440+537.7=1977.7 \]

\[ \Sigma \left(P_n X'_n-Q_n R'_n\right)=2526-341.0=2185.0. \]

Consequently:

\[ P_{\mathrm{e}}=0.00862\cdot 1977.7+0.01545\cdot 2185=50.80\ \mathrm{MW} \]

and

\[ Q_{\mathrm{e}}=-0.00862\cdot 2185+0.01545\cdot 1977.7=11.73\ \mathrm{MVAr}, \]

i.e.:

\[ \overline{W}_{\mathrm{e}}=P_{\mathrm{e}}-jQ_{\mathrm{e}}=50.80-j\,11.73\ \mathrm{MVA}. \]

Now it is possible to find the actual power flow in branch “VI,” without taking losses into account, as—

\[ \overline{W}_{\mathrm{VI}} = \overline{W}'_4+\overline{W}_5+\overline{W}_6-\overline{W}_{\mathrm{e}} = 69.62-j\,8.94+20-j\,10.9+30 \]

\[ -j\,28.27-(50.80-j\,11.73) =-(68.82-j\,36.38)\ \mathrm{MVA}\ ^*). \]

Let us check the computation of \(\overline{W}_{\mathrm{VI}}\) directly by the formula for flow distribution of power in the ring (Table 9).

Table 9.

Node	\(P_n\)	\(Q_n''\)	\(R_n''\)	\(X_n''\)	\(P_n R_n''\)	\(P_n X_n''\)	\(Q_n R_n''\)	\(Q_n X_n''\)
4	69.62	8.94	10.45	18.35	728.0	1276	93.5	164.0
5	20.00	10.90	20.45	38.35	409.0	767.0	222.6	418.0
6	30.00	28.27	23.45	44.35	704.0	1330	663.0	1253
\(\Sigma\)	119.62	48.11			1841.0	3373	979.1	1835

\[ \Sigma \left(P_n R''_n+Q_n X''_n\right)=1841+1835=3676 \]

\[ \Sigma \left(P_n X''_n-Q_n R''_n\right)=3373-979.1=2393.9. \]

Hence:

\[ P_{\mathrm{VI}}=0.00862\cdot 3676+0.01545\cdot 2394=68.72\ \mathrm{MW} \]

and

\[ Q_{\mathrm{VI}}=-0.00862\cdot 2394+0.01545\cdot 3676=36.25\ \mathrm{MVAr}. \]

Thus, the discrepancy between the results of the calculations in the first and in the second case is quite insignificant.

Next, let us find the power flow in the 1st branch of the transformed network.

\[ \overline{W}_{\mathrm{I}} = \overline{W}_{\mathrm{e}} \frac{\overline{Z}_{\mathrm{e}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}} = (50.80-j\,11.73)\frac{10.45+j\,18.35}{34.3+j\,67.5} \]

\[ =13.92-j\,3.93\ \mathrm{MVA}. \]

The power flow in the 1st branch of the untransformed network is

\[ \overline{W}_{\mathrm{I}} = \overline{W}'_{\mathrm{I}}+\overline{W}_{\mathrm{I}(2,3)} = 13.92-j\,3.93+30.38-j\,22.52 \]

\[ =44.30-j\,26.45\ \mathrm{MVA}. \]

* The direction of powers from left to right is taken as positive.

South. coll. TII. 18.

Power flow in link VII

\[ \overline{\overline{W}}_{\mathrm{VII}} = \overline{\overline{W}}_{e} - \overline{\overline{W}}'_{1} = 50.80 - j\,11.73 - (13.92 - j\,3.93) = 36.88 - j\,7.80\ \text{mva}. \]

The same power flow is obtained directly from the formula for distribution of the equivalent power:

\[ \overline{\overline{W}}_{\mathrm{VII}} = \overline{\overline{W}}_{e}\, \frac{\overline{Z}_{e}}{\overline{Z}_{\mathrm{VII}}} = (50.80 - j\,11.73)\, \frac{10.45 + j\,18.35}{15 + j\,25} = 37.05 - j\,7.7\ \text{mva} \]

The discrepancy of the results is permissible.

It is now possible simply (by Kirchhoff’s law) to find the power flows in all links of the untransformed network.

\[ \overline{\overline{W}}_{\mathrm{II}} = \overline{\overline{W}}_{1} - \overline{\overline{W}}_{2} = 44.30 - j\,26.45 - (75 - j\,59.5) = \]

\[ = -(30.70 - j\,24.15)\ \text{mva}. \]

\[ \overline{\overline{W}}_{\mathrm{III}} = \overline{\overline{W}}_{\mathrm{II}} - \overline{\overline{W}}_{3} = -(30.70 - j\,24.15) - [-(100 - j\,45.84)] = \]

\[ = 69.30 - j\,21.69\ \text{mva}; \]

\[ \overline{\overline{W}}_{\mathrm{IV}} = \overline{\overline{W}}_{\mathrm{VII}} + \overline{\overline{W}}_{\mathrm{III}} - \overline{\overline{W}}_{4} = 36.88 - j\,7.80 + 69.30 - j\,21.69 - (125 - j\,26.8) = \]

\[ = -(18.82 + j\,2.72); \]

\[ \overline{\overline{W}}_{\mathrm{V}} = \overline{\overline{W}}_{\mathrm{IV}} - \overline{\overline{W}}_{5} = -(18.82 + j\,2.72) - (20 - j\,10.9) = -(38.82 - j\,8.18); \]

\[ \overline{\overline{W}}_{\mathrm{VI}} = \overline{\overline{W}}_{\mathrm{V}} - \overline{\overline{W}}_{6} = -(38.82 - j\,8.18) - (30 - j\,28.27) = -(68.82 - j\,36.45). \]

The power-flow distribution thus found in the network without allowance for losses is plotted on the network diagram, Fig. 25.

Fig. 25.

As already noted, the power-flow distribution obtained does not take losses into account, and therefore the actual voltages at the nodes cannot be determined from these flows. Nevertheless, in order to clarify the magnitude of the errors obtained when calculating voltages from flows without allowance for losses, let us find the values of the nodal voltages according to the power-flow distribution of Fig. 25.

It is most convenient to calculate voltages when the load of a link is given in mw or mva, by the formula: *)

\[ \overline{E}_{n+1} = \left( E_n + \frac{P \cdot R + QX}{E_n} \right) + j\,\frac{PX - QR}{E_n}, \]

whence the modulus of the voltage vector is

\[ E_{n+1} = \sqrt{ \left( E_n + \frac{P \cdot R + QX}{E_n} \right)^2 + \left( \frac{PX - QR}{E_n} \right)^2 } \]

) Prof. A. A. Glazunov et al., Calculation of networks of electric power systems, 1935. See also Prof. A. Ya. Ryabkov, “Methods for determining voltages in power transmissions and networks of 35, 110, and 220 kv,” journal Electric Stations*, No. 1, 1935.

The voltage of the 4th node is specified and equal to

\[ \bar{E}_4 = E_4 = 105\ kv, \]

The voltages (moduli) of the remaining nodes will be:

\[ E_3 = \sqrt{ \left(105+\frac{69{,}30\cdot 4{,}3+21{,}69\cdot 7{,}5}{105}\right)^2 + \left(\frac{69{,}30\cdot 7{,}5-21{,}69\cdot 4{,}3}{105}\right)^2 } = \]

\[ =\sqrt{109{,}38^2+4{,}06^2}=109{,}45\ kv. \]

\[ E_2 = \sqrt{ \left(109{,}45-\frac{30{,}70\cdot 15+24{,}15\cdot 30}{109{,}45}\right)^2 + \left(\frac{30{,}70\cdot 30-24{,}15\cdot 15}{109{,}45}\right)^2 } = \]

\[ =\sqrt{98{,}63^2+5{,}13^2}=98{,}80\ kv, \]

\[ E_1 = \sqrt{ \left(98{,}80+\frac{44{,}30\cdot 15+26{,}45\cdot 30}{98{,}80}\right)^2 + \left(\frac{44{,}30\cdot 30-26{,}45\cdot 15}{98{,}80}\right)^2 } = \]

\[ =\sqrt{113{,}55^2+9{,}45^2}=113{,}80\ kv. \]

\[ E_5 = \sqrt{ \left(105+\frac{18{,}82\cdot 10-2{,}72\cdot 20}{105}\right)^2 + \left(\frac{18{,}82\cdot 20+2{,}72\cdot 10}{105}\right)^2 } = \]

\[ =\sqrt{106{,}27^2+3{,}84^2}=106{,}4\ kv\ ^*) \]

\[ E_6 = \sqrt{ \left(106{,}4+\frac{38{,}82\cdot 3+8{,}18\cdot 6}{106{,}4}\right)^2 + \left(\frac{38{,}82\cdot 6-8{,}18\cdot 3}{106{,}4}\right)^2 } = \]

\[ =\sqrt{107{,}98^2+1{,}5^2}=107{,}99\ kv. \]

\[ E_1 = \sqrt{ \left(107{,}99-\frac{63{,}82\cdot 4+36{,}38\cdot 5}{107{,}99}\right)^2 + \left(\frac{68{,}82\cdot 5-36{,}38\cdot 4}{107{,}99}\right)^2 } = \]

\[ =\sqrt{112{,}25^2+1{,}84^2}=112{,}29\ kv. \]

In traversing the network we did not obtain a zero sum of voltage drops for the closed network. The values of the same voltage of the 1st node proved to differ comparatively, depending on the direction from which the network was traversed in computing the nodal voltages.
When moving from the 4th node through nodes 3 and 2, the result was: \(E_1 = 113{,}80\ kv\).
When moving from the 4th node through nodes 5 and 6, the result was: \(E_1 = 112{,}29\ kv\).
Such an imbalance is entirely legitimate, as has been shown in the book, mentioned repeatedly by us, by Prof. A. A. Glazunov, Prof. A. Ya. Ryabkov, and Docent F. P. Lashkov.

Indeed. The power flow in section III (Fig. 21), without taking losses in the network into account, will be equal to

\[ \bar{W}_{\mathrm{III}} = \frac{ \bar{W}_c\left(\bar{Z}_{\mathrm{II}}+\bar{Z}_{\mathrm{I}}\right) + \bar{W}_b \cdot \bar{Z}_{\mathrm{I}} }{ \bar{Z}+\bar{Z}_{\mathrm{II}}+\bar{Z}_{\mathrm{III}} }. \]

\(^*\) In the calculation the signs in the formula changed, because the reactive power of branch IV has the signs \(+\).

Keeping in mind that

\[ \overline{W}_{c}=\overline{W}_{\mathrm{III}}-\overline{W}_{\mathrm{II}}, \]

and

\[ \overline{W}_{b}=\overline{W}_{\mathrm{II}}+\overline{W}_{\mathrm{I}}, \]

we obtain:

\[ \overline{W}_{\mathrm{III}}(\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{III}}) = (\overline{W}_{\mathrm{III}}-\overline{W}_{\mathrm{II}}) (\overline{Z}_{\mathrm{II}}+\overline{Z}_{\mathrm{I}}) + (\overline{W}_{\mathrm{II}}+\overline{W}_{\mathrm{I}})\cdot \overline{Z}_{\mathrm{I}} \]

or

\[ \overline{W}_{\mathrm{III}}\overline{Z}_{\mathrm{III}} + \overline{W}_{\mathrm{II}}\overline{Z}_{\mathrm{II}} = \overline{W}_{\mathrm{I}}\overline{Z}_{\mathrm{I}}. \]

But the last relation can hold only in the case when:

\[ \frac{\overline{W}_{\mathrm{III}}}{\overline{E}_{0}}\overline{Z}_{\mathrm{III}} + \frac{\overline{W}_{\mathrm{II}}}{\overline{E}_{0}}\overline{Z}_{\mathrm{II}} = \frac{\overline{W}_{\mathrm{I}}}{\overline{E}_{0}}\overline{Z}_{\mathrm{I}}, \]

i.e., when the currents in all branches are obtained by dividing the power flows in these branches by one and the same voltage. Consequently, only in this case can a voltage balance be obtained in a closed network if, in determining the power-flow distribution, the losses in the network have not been taken into account.

Let us next find the power losses in the branches of the network. An exact determination of the power loss in each branch would be possible in the case of an equally exact knowledge of the current magnitude in each branch. However, using the transformation method, we have no possibility of knowing the currents in the branches exactly in advance, for two reasons.

First, until the network calculation has been carried through to the end, the nodal voltages corresponding to the full power flows in the branches cannot be known; consequently, the currents in the branches also cannot be determined.

Second, as long as the losses in the branches are unknown, it is impossible to determine the full power flows in the branches, since, in an exact determination of the losses, it would be necessary to take into account not only the power flows in the branches due to the nodal loads, but also the additional powers due to losses in the branches located farther from the supply point.

However, since the power losses in the branches usually amount to only a few percent, and the voltage drop in the branches enters only as fractions of the total value of the nodal voltage, it is recommended, in order to shorten the calculation process, to compute the losses by determining the currents in the branches from the nominal or average voltage and from the power flows in the branches without taking into account the losses in the preceding branches. Otherwise, in order to achieve high accuracy in the calculations, repeated recalculation would have to be performed.

The losses in any branch are

\[ \Delta\overline{W}_{n} = \Delta P_{n} - j\Delta Q_{n} = J_{n}^{2}\cdot R_{n} - jJ_{n}^{2}\cdot X_{n}, \]

and since

\[ J_{n}^{2} = J_{a}^{2} + J_{r}^{2} = \frac{P_{n}^{2}+Q_{n}^{2}}{E_{n}^{2}}, \]

then

\[ \Delta\overline{W}_{n} = \frac{P_{n}^{2}+Q_{n}^{2}}{E_{n}^{2}}\cdot R_{n} - j\frac{P_{n}^{2}+Q_{n}^{2}}{E_{n}^{2}}\cdot X_{n}. \]

The results of calculating the power losses in the branches of the network are presented in Table 10.

Table 10.

Link	Power of the link without allowance for losses	Resistance of the link	Active power losses	Reactive power losses
I	\(44.30 - j\,26.45\)	\(15 + j\,30\)	3.31	6.63
II	\(30.70 - j\,24.15\)	\(15 + j\,30\)	1.89	3.78
III	\(69.30 - j\,21.69\)	\(4.3 + j\,7.5\)	1.865	3.26
IV	\(18.82 + j\,2.72\)	\(10 + j\,20\)	0.300	0.600
V	\(38.82 - j\,8.18\)	\(3 + j\,6\)	0.390	0.780
VI	\(68.82 - j\,36.38\)	\(4 + j\,5\)	2.02	2.50
VII	\(36.88 - j\,7.80\)	\(15 + j\,25\)	1.68	2.80

In distributing the additional power flows in the network on account of the calculated losses in the links, one can proceed in two ways. One may add the power loss in that link alone to the power flow of each link and, disregarding a certain imbalance of powers at the nodes due to this operation, calculate the node voltages on the basis of the newly obtained link power flows, while calculating the load of the station with a free schedule with allowance for the power losses in all links. Such a calculation would considerably simplify the accounting of losses in the network, but it would undoubtedly introduce a certain error into the final results of the calculation. This error would arise, first, because of a certain underloading of each subsequent link by the losses of all preceding links, if one moves from the point of flow division toward the feeding node, and, second, because, as was noted above, the additional power flows through the links due to losses are not always distributed proportionally to the power flows in the links obtained without allowance for losses. In the case where there is not one point of flow division (not counting the node of the station with a free schedule), but several such points, this disproportion is precisely what occurs.

In our case there are two points of power-flow division (Fig. 25): points 2 and 4. If the usual method of distributing losses in the network were used here, then, proceeding from points 2 and 4 toward the flows of active powers in links II and III, we would obtain at node 3 an additional demand for active and reactive power, which we cannot cover at node 3, since the power of the 3rd node has already been distributed. Consequently, this demand can be satisfied by the station of the 1st node, but then the useful delivered power of the station of the 3rd node through links II and III will correspondingly decrease, while the power flows of links I, VII, and IV will increase in total by the same amount. In particular, the loading of link I will be increased by as much as the loading of link II is decreased.

Such a mechanical transfer of load from one link to another should, logically, have led equally, for example, to the following absurdity: the power of the 3rd node has fallen to zero; consequently the power flow of link II will become equal to zero, and that of link I equal to the load of the 2nd node, i.e. \(74 - j50.6\) MVA.

It is far more correct in such cases to use the general law of distributing losses in a network, substantiated in statement 2 of the transformation method. According to this law, the power losses of each link must be regarded as an additional load, and the point of its application is the end of the given link, if the direction of the active-power flow in this link is taken.

In our case, therefore, it is necessary to apply the losses of links I and II at node 2, the losses of links III, IV, and VII at node 4, those of link V at node 5, and those of link VI at node 6.

[[unclear: link]]	[[unclear: value]]	[[unclear: value]]	[[unclear: value]]	[[unclear: value]]
V	\(41.28 - j\,12.61\)	\(0.39 - j\,0.78\)	\(40.89 - j\,11.83\)	\(30 - j\,20.21\)
IV	\(20.89 - j\,0.93\)	\(0.30 - j\,0.60\)	\(20.59 - j\,0.33\)	\(20 - j\,10.9\)

Power of the station of the 1st node

Next, according to the same law, it is necessary to distribute this additional load among the branches of the network according to the general rule of the transformation method.

Fig. 26.

In Fig. 26 a network is presented with loads at the nodes due to losses, where:

\[ \Delta \overline{W}_4 = \Delta \overline{W}_{\mathrm{III}} + \Delta \overline{W}_{\mathrm{IV}} + \Delta \overline{W}_{\mathrm{VII}} \]

\[ = +1,865-j3,26+0,30-j0,60+ \]

\[ 1,68-j2,80 = 3,85-j6,66 \ \text{MVA} \]

and

\[ \Delta \overline{W}_2 = \Delta \overline{W}_{\mathrm{I}} + \Delta \overline{W}_{\mathrm{II}} = 1,89-j3,78+ \]

\[ 3,31-j6,63 = 5,20-j10,41 \ \text{MVA} \]

The procedure for the direct and inverse transformation of the network is the same as in the preceding case.

We remove the load of the 2nd node (in the 3rd node the load is zero):

\[ \Delta P_{1(2)} = 0,00598(5,20\cdot 19,3+10,41\cdot 37,5)+ \]

\[ +0,01175(5,20\cdot 37,5-10,41\cdot 19,3) = 2,86 \ \text{MW} \]

\[ \Delta Q_{1(2)} = -0,00598(5,20\cdot 37,5-10,41\cdot 19,3)+ \]

\[ +0,01175(5,20\cdot 19,3+10,41\cdot 37,5) = 5,80 \ \text{MVAr}, \]

Consequently:

\[ \Delta \overline{W}_{1(2)} = 2,86-j5,80 \ \text{MVA} \]

and

\[ \Delta \overline{W}_{4(2)} = \Delta \overline{W}_2 - \Delta \overline{W}_{1(2)} = \]

\[ = 5,20-j10,41-(2,86-j5,80) = 2,34-j4,61 \ \text{MVA}. \]

The load of the 4th node of the transformed network is

\[ \Delta \overline{W}'_4 = \Delta \overline{W}_4 + \Delta \overline{W}_{4(2)} = \]

\[ = 3,85-j6,66+2,34-j4,61 = 6,19-j11,27 \ \text{MVA}. \]

Power-flow distribution in the ring. The results of the main computations are in Table 11.

Table 11.

Nodes	\(\Delta P_n\)	\(\Delta Q_n\)	\(R'_n\)	\(X'_n\)	\(\Delta P_n R'_n\)	\(\Delta P_n X'_n\)	\(\Delta Q_n R'_n\)	\(\Delta Q_n X'_n\)
4	6,19	11,27	17	31	105,20	192,00	191,80	350,0
5	0,39	0,78	7	11	2,73	0,43	5,46	8,57
6	2,02	2,50	4	5	8,08	10,10	10,00	12,50
\(\Sigma\)	8,60	14,55			116,01	202,53	207,26	371,07

\[ \Sigma(\Delta P_n R'_n+\Delta Q_n X'_n) = 116,01+371,07 = 487,08 \]

\[ \Sigma(\Delta P_n X'_n-Q\Delta_n R'_n) = 202,53-207,26 = -4,73 \]

\[ \Delta P_3 = 0,00862\cdot 487,08 + 0,01545\cdot(-4,73) = 4,12 \ \text{MW} \]

\[ \Delta Q_3 = 0,00862\cdot(-4,73) + 0,01545\cdot 487,08 = 7,56 \ \text{MVAr}. \]

i.e.

\[ \Delta \overline{W}_3 = 4,12-j7,56 \ \text{MVA} \]

Consequently, the power flow issuing from node I along link VI, to cover the corresponding part of the losses in the network, is

\[ \Delta \overline{W}_{VI} = \Delta \overline{W}_{1} + \Delta \overline{W}_{5} + \Delta \overline{W}_{6} - \Delta \overline{W}_{э} = 4.48 - j7.00 \ \text{MVA}. \]

The power flow issuing from node I along link I, to cover the corresponding part of the losses in the network, is

\[ \Delta \overline{W}_{I} = \Delta \overline{W}_{э} \frac{\overline{Z}_{э}} {\overline{Z}_{I}+\overline{Z}_{II}+\overline{Z}_{III}} + \Delta \overline{W}_{1(2)} = \]

\[ = 3.08 - j7.94 \ \text{MVA}. \]

The power flow issuing from node I along link VII, to cover the corresponding part of the losses, is

\[ \Delta \overline{W}_{VII} = \Delta \overline{W}_{э} - \Delta \overline{W}_{э} \frac{\overline{Z}_{э}} {\overline{Z}_{I}+\overline{Z}_{II}+\overline{Z}_{III}} = \Delta \overline{W}_{э} \frac{\overline{Z}_{э}} {\overline{Z}_{VII}} = \]

\[ = 3.08 - j5.42 \ \text{MVA}. \]

The total power flows in links I, VI, VII, allowing for losses in the network, are

\[ \overline{W}'_{I} = \overline{W}_{I} + \Delta \overline{W}_{I} \; *) = \]

\[ = 44.30 - j26.45 + 3.90 - j7.94 = 48.20 - j34.39 \ \text{MVA}. \]

\[ \overline{W}'_{VI} = \overline{W}_{VI} + \Delta \overline{W}_{VI} = \]

\[ = 68.82 - j36.38 + 4.48 - j7.00 = 73.30 - j43.38 \ \text{MVA}. \]

\[ \overline{W}'_{VII} = \overline{W}_{VII} + \Delta \overline{W}_{VII} = \]

\[ = 36.88 - j7.80 + 3.08 - j5.52 = 39.96 - j13.22 \ \text{MVA}. \]

The power flows, allowing for losses, at the beginning and end of each link of the entire network (taking as the beginning the node from which the active power issues) are presented in Table 12.

Table 12.

Link	Power at the beginning of the link	Power losses in the link	Power at the end of the link	Load of the node in which the links converge
I	\(48.20 - j34.39\)	\(3.31 - j6.63\)	\(44.99 - j27.76\)	\(\left\{\begin{array}{l}75 - j50.6\\ -(100 - j45.84)\end{array}\right.\)
II	\(32 - j26.62\)	\(1.89 - j3.78\)	\(30.11 - j22.84\)	\(\left\{\begin{array}{l}75 - j50.6\\ -(100 - j45.84)\end{array}\right.\)
III	\(68 - j19.22\)	\(1.87 - j3.26\)	\(66.13 - j15.96\)	\(\left\{\begin{array}{l}75 - j50.6\\ -(100 - j45.84)\end{array}\right.\)
VII	\(39.96 - j13.22\)	\(1.68 - j2.80\)	\(38.28 - j10.42\)
VI	\(73.30 - j43.38\)	\(2.02 - j2.50\)	\(71.28 - j40.88\)	\(\left\{\begin{array}{l}30 - j28.27\end{array}\right.\)
V	\(41.28 - j12.61\)	\(0.39 - j0.78\)	\(40.89 - j11.83\)	\(\left\{\begin{array}{l}30 - j28.27\end{array}\right.\)
IV	\(20.89 - j0.93\)	\(0.30 - j0.60\)	\(20.59 - j0.33\)	\(\left\{\begin{array}{l}20 - j10.9\end{array}\right.\)

Power of the station of the 1st node

\[ \overline{W}'_{1} = \overline{W}'_{I} + \overline{W}'_{VII} + \overline{W}'_{VI} + \overline{W}_{Ic} = 482.0 - \]

\[ -j34.39 + 39.96 - j13.22 + 73.90 - j43.38 + W_{Ic} = 161.46 - j90.99 + jW_{Ic} \; **). \]

*) It must be remembered that \(\Delta \overline{W}_{I}, \Delta \overline{W}_{II}, \ldots\), etc., denote not the losses in links I, II, etc., but the additional powers flowing in these links and covering the losses in the network.

**) \(W_{Ic}\) is the reactive power in the first node due to the capacitances of links I, VII, and VI.

On the other hand, the same power must be obtained as

\[ \overline{W}_1' = \overline{W}_2+\overline{W}_3+\overline{W}_4+\overline{W}_5+\overline{W}_6+\Sigma\Delta\overline{W}+\overline{W}_{1c} = 75-j50{,}6+ \]

\[ +[-(100-j45{,}84)]+125-j26{,}8+20-j10{,}9+30-j28{,}27+ \]

\[ +(3{,}31-j6{,}63+1{,}89-j3{,}78+1{,}865-j3{,}26+0{,}30-j0{,}60+0{,}39-j0{,}78+ \]

\[ +2{,}02-j2{,}50+1{,}68-j2{,}80)+jW_{1c} = 161{,}46-j91{,}08+jW_{1c}. \]

The voltages at the nodes, on the basis of the power-flow distribution found in the network, are obtained as:

\[ E_1 = \sqrt{ \left( E_4+ \frac{P_{\mathrm{VII}}\cdot R_{\mathrm{VII}}+Q_{\mathrm{VII}}\cdot X_{\mathrm{VII}}}{E_4} \right)^2 + \left( \frac{P_{\mathrm{VII}}\cdot X_{\mathrm{VII}}-Q_{\mathrm{VII}}\cdot R_{\mathrm{VII}}}{E_4} \right)^2 } = \]

\[ = \sqrt{ \left( 105+ \frac{38{,}28\cdot 15+10{,}42\cdot 25}{105} \right)^2 + \left( \frac{38{,}28\cdot 25-10{,}42\cdot 15}{105} \right)^2 } = \]

\[ = \sqrt{112{,}98^2+7{,}62^2} = 113{,}23\ \mathrm{kv} \]

\[ E_3 = \sqrt{ \left( E_4+ \frac{P_{\mathrm{III}}\cdot R_{\mathrm{III}}+Q_{\mathrm{III}}\cdot X_{\mathrm{III}}}{E_4} \right)^2 + \left( \frac{P_{\mathrm{III}}\cdot X_{\mathrm{III}}-Q_{\mathrm{III}}\cdot R_{\mathrm{III}}}{E_4} \right)^2 } = \]

\[ = \sqrt{ \left( 105+ \frac{66{,}13\cdot 4{,}3+15{,}96\cdot 7{,}5}{105} \right)^2 + \left( \frac{66{,}13\cdot 7{,}5-15{,}96\cdot 4{,}3}{105} \right)^2 } = \]

\[ = \sqrt{108{,}83^2+4{,}04^2} = 108{,}91\ \mathrm{kv} \]

\[ E_2 = \sqrt{ \left( 108{,}91- \frac{32{,}00\cdot 15+26{,}62\cdot 30}{108{,}91} \right)^2 + \left( \frac{32{,}00\cdot 30-26{,}62\cdot 15}{108{,}91} \right)^2 } = \]

\[ = \sqrt{97{,}14^2+5{,}15^2} = 97{,}3\ \mathrm{kv} \]

\[ E_1 = \sqrt{ \left( 97{,}30+ \frac{44{,}89\cdot 15+27{,}76\cdot 30}{97{,}30} \right)^2 + \left( \frac{44{,}89\cdot 30-27{,}76\cdot 15}{97{,}30} \right)^2 } = \]

\[ = \sqrt{112{,}8^2+9{,}55^2} = 113{,}21\ \mathrm{kv} \]

\[ E_5 = \sqrt{ \left( E_4+ \frac{P_{\mathrm{IV}}\cdot R_{\mathrm{IV}}+Q_{\mathrm{IV}}\cdot X_{\mathrm{IV}}}{E_4} \right)^2 = \left( \frac{P_{\mathrm{IV}}\cdot X_{\mathrm{IV}}-Q_{\mathrm{IV}}\cdot R_{\mathrm{IV}}}{E_4} \right)^2 } = \]

\[ = \sqrt{ \left( 105+ \frac{20{,}59\cdot 10+0{,}33\cdot 20}{105} \right)^2 + \left( \frac{20{,}59\cdot 20-0{,}33\cdot 10}{105} \right)^2 } = \]

\[ = \sqrt{107{,}06^2+3{,}95^2} = 107{,}13\ \mathrm{kv} \]

\[ E_6 = \sqrt{ \left( 107{,}13+ \frac{40{,}89\cdot 3+11{,}83\cdot 6}{107{,}13} \right)^2 + \left( \frac{40{,}89\cdot 6-11{,}83\cdot 3}{107{,}13} \right)^2 } = \]

\[ = \sqrt{108{,}93^2+1{,}95^2} = 108{,}96\ \mathrm{kv} \]

\[ E_1 = \sqrt{ \left( 108{,}96+ \frac{71{,}28\cdot 4+40{,}88\cdot 5}{108{,}96} \right)^2 = \left( \frac{71{,}28\cdot 5-40{,}88\cdot 4}{108{,}96} \right)^2 } = \]

\[ = \sqrt{113{,}44^2+1{,}75^2} = 113{,}42\ \mathrm{kv}. \]

Some discrepancy in the values of the voltage \(E_1\) can be explained, on the one hand, by the insufficient accuracy of the slide-rule calculations and, on the other, by a certain inaccuracy in accounting for losses in the network.

The results of the calculation of the entire system are plotted in Fig. 27.

If we now compare the values of the nodal voltages obtained as a result of calculations by the method of nodal equations, by the transformation method without accounting for losses, and by the transformation method with accounting for losses (Table 13), then one may conclude that the latter calculation results, although they differ from those obtained by the method of nodal equations, have an almost constant character; i.e., at all nodes the voltage values according to the transformation method are lower by approximately \(1\ \mathrm{kv}\) than the voltage values according to the method of nodal equations.

Fig. 27.

Table 13.

Calculation methods	\(E_1\)	\(E_2\)	\(E_3\)	\(E_4\)	\(E_5\)	\(E_6\)
Method of nodal equations	114,8	98,15	109,9	105	107,82	110,4
Transformation method without accounting for losses	113,8	98,80	109,45	105	106,4	107,99
Transformation method with accounting for losses	112,29	97,3	108,91	105	107,13	108,96

This circumstance evidently indicates that, in determining voltage fluctuations at the nodes, the results of calculations by the first and second methods may differ only insignificantly.

However, no final conclusions can be drawn about the degree of accuracy of calculations by the transformation method on the basis only of the calculation results for the example analyzed above, since in this example links shorter than \(100\ \mathrm{km}\) predominate. Therefore we shall carry out one more calculation by the transformation method, in which the length of all links is greater than \(100\ \mathrm{km}\).

Fig. 28.

In Fig. 28 the initial data of the second example of a complex system are presented.

IV . . . . .	\(26,76 - j\,13,06\)	\(1,05 - j\,1,965\)	\(25,71 - j\,11,09\)	\(80 - j\,34,88\)

At a nominal voltage of 110 kV, we find the reactive powers due to the capacitances of the links and assign the powers found to the nodes.

\[ \begin{aligned} \text{For link } I\ldots\quad &110^2 \cdot j\,5.4\cdot 10^{-4}=j\,6.53\ \text{MVAr},\\ \text{For link } II\ldots\quad &110^2 \cdot j\,10.8\cdot 10^{-4}=j\,13.06\ \text{MVAr},\\ \text{For link } III\ldots\quad &110^2 \cdot j\,8.0\cdot 10^{-4}=j\,9.68\ \text{MVAr},\\ \text{For link } IV\ldots\quad &110^2 \cdot j\,17.0\cdot 10^{-4}=j\,20.56\ \text{MVAr}. \end{aligned} \]

Calculated powers at the nodes:

\[ \overline{W}_2=80-j40+j\,\frac{6.53+13.06}{2}=80.0-j30.21\ \text{MVA} \]

\[ \overline{W}_3=-(100-j50)+j\,\frac{13.06+9.68}{2}=-(100-j61.37)\ \text{MVA} \]

\[ \overline{W}_4=80-j50+j\,\frac{9.68+20.56}{2}=80-j34.88\ \text{MVA} \]

Table 14

Nodes	\(P_n\)	\(Q_n\)	\(R_n'\)	\(X_n'\)	\(P_n R_n'\)	\(P_n X_n'\)	\(Q_n R_n'\)	\(Q_n X_n'\)
2	80	30.21	80.7	133.7	6456	10696	2438	4039
3	−100	−61.37	55.2	92.0	−5520	−9200	−3388	−5646
4	80	34.88	32	60.0	2560	4800	1116	2093
\(\Sigma\)	60	3.72	96.2	154.9	3496	6296	166	486

According to the data of Table 14, we have:

\[ \Sigma(P_n R_n' + Q_n X_n')=3496+486=3982; \]

\[ \Sigma(P_n X_n' - Q_n R_n')=6296-166=6130; \]

\[ G_0=\frac{96.2}{96.2^2+154.9^2}=0.00289\ \mho, \]

\[ B_0=\frac{154.9}{96.2^2+154.9^2}=0.00467\ \mho. \]

Hence,

\[ P_1=0.00289\cdot 3982+0.00467\cdot 6130=40.11\ \text{MW} \]

and

\[ Q_1=0.00289\cdot 6130+0.00467\cdot 3981=0.88\ \text{MVAr}. \]

That is,

\[ \overline{W}_1=P_1-jQ_1=40.11-j0.88\ \text{MVA}. \]

and

\[ \overline{W}_{IV}=\overline{W}_2+\overline{W}_3+\overline{W}_4-\overline{W}_1=19.89-j2.84\ \text{MVA}. \]

Let us check the computation of \(\overline{W}_{IV}\) (Table 15).

Table No. 15

Nodes	\(P_n\)	\(Q_n\)	\(R_n''\)	\(X_n''\)	\(P_n R_n''\)	\(P_n X_n''\)	\(Q_n R_n''\)	\(Q_n X_n''\)
2	80	30.21	15.5	21.2	1239	1696	468	641
3	−100	−61.37	41	62.9	−4100	−6290	−2515	−3860
4	80	34.88	64.2	94.9	5130	7590	2232	3300
\(\Sigma\)	60	3.72	96.2	154.9	2268	2996	185	81

\[ \Sigma(P_n \cdot R_n'' + Q_n \cdot X_n'')=2349 \]

\[ \Sigma(P_n \cdot X_n'' - Q_n \cdot R_n'')=2811. \]

Hence:

\[ P_{IV}=0{,}00289\cdot 2349+0{,}00467\cdot 2811=19{,}89\ \text{mw} \]

and

\[ Q_{IV}=-0{,}00289\cdot 2811+0{,}00467\cdot 2349=2{,}83\ \text{mvA}. \]

Now we find:

\[ \overline{W}_{II}=\overline{W}_{1}-\overline{W}_{2}=40{,}11-j0{,}88-(80-j30{,}21)=-(39{,}89-j29{,}33) \]

\[ \overline{W}_{III}=\overline{W}_{II}-\overline{W}_{3}=-(39{,}89-j29{,}33)-[-(100-j61{,}37)]= \]

\[ =60{,}11-j32{,}04\ \text{mvA}. \]

Losses in the branches—see Table 16.

Table No. 16.

Branch	Branch power without allowance for losses	Branch resistance	Power losses, active	Power losses, reactive
I	\(40{,}11-j\ 0{,}88\)	\(15{,}5+j21{,}2\)	2,027	2,86
II	\(39{,}89-j29{,}33\)	\(25{,}5+j41{,}7\)	5,17	8,41
III	\(60{,}11-j32{,}04\)	\(23{,}2+j32\)	8,92	12,27
IV	\(19{,}89-j\ 2{,}84\)	\(32+j60\)	1,05	1,965

Distribution of losses in the network—see Table 17.

Table No. 17.

Nodes	\(\Delta P_n\)	\(\Delta Q_n\)	\(R_n'\)	\(X_n'\)	\(\Delta P_n\cdot R_n'\)	\(\Delta P_n\cdot X_n'\)	\(\Delta Q_n\cdot R_n'\)	\(\Delta Q_n\cdot X_n'\)
2	7,197	11,270	80,7	133,7	580,0	962	910	1507
4	9,970	14,235	32	60	318,2	597	456	853
\(\Sigma\)	18,167	25,505	—	—	898,2	1559	1366	2360

\[ \Sigma(\Delta P_n\cdot R_n' + \Delta Q_n\cdot X_n')=3258{,}2, \]

\[ \Sigma(\Delta P_n\cdot X_n' - \Delta Q_n\cdot R_n')=193. \]

Hence:

\[ \Delta P_I=0{,}00289\cdot 3258{,}2+0{,}00467\cdot 193=10{,}3\ \text{mw} \]

and

\[ \Delta Q_I=0{,}00289\cdot 193+0{,}00467\cdot 3258{,}2=14{,}59\ \text{mvA}, \]

i.e.,

\[ \Delta \overline{W}_I=10{,}3-j14{,}59\ \text{mvA}. \]

The complete power flows (allowing for losses) at the beginning and end of each branch (the beginning of a branch is the node from which active power flows out) are presented in Table 18.

Table No. 18.

Branch	Power at the beginning of the branch	Power losses in the branch	Powers at the end of the branch	Power withdrawal at the node
I . . . . .	\(50{,}41-j15{,}47\)	\(2{,}027-j\ 2{,}86\)	\(48{,}18-j12{,}61\)	\(\} \ 80-j30{,}21\)
II . . . . .	\(36{,}79-j26{,}31\)	\(5{,}17-j\ 8{,}41\)	\(31{,}62-j17{,}90\)	\(\} \ -(100-j61{,}37)\)
III . . . .	\(63{,}21-j35{,}06\)	\(8{,}92-j12{,}27\)	\(54{,}29-j23{,}79\)	\(\} \ 80-j34{,}88\)
IV . . . . .	\(26{,}76-j13{,}06\)	\(1{,}05-j\ 1{,}965\)	\(25{,}71-j11{,}09\)

Node voltages:

\[ E_1=\sqrt{\left(105+\frac{48.38\cdot 15.5+12.61\cdot 21.2}{105}\right)^2+ \left(\frac{48.38\cdot 21.2+12.61\cdot 15.5}{105}\right)^2} \]

\[ =\sqrt{114.69^2+7.89^2}=114.96\ \mathrm{kv} \]

\[ E_2=105\ \mathrm{kv}\quad \text{(specified)} \]

\[ E_3=\sqrt{\left(105+\frac{31.62\cdot 25.5+17.9\cdot 41.7}{105}\right)^2+ \left(\frac{31.62\cdot 41.7+16.9\cdot 25.5}{105}\right)^2} \]

\[ =\sqrt{119.8^2+8.17^2}=120.08\ \mathrm{kv} \]

\[ E_4=\sqrt{\left(120.08-\frac{63.21\cdot 23.2+35.06\cdot 32}{120.08}\right)^2+ \left(\frac{63.21\cdot 32-35.06\cdot 23.2}{120.08}\right)^2} \]

\[ =\sqrt{98.5^2+10^2}=99\ \mathrm{kv} \]

\[ E_1=\sqrt{\left(99+\frac{25.71\cdot 32+11.09\cdot 60}{99}\right)^2- \left(\frac{25.71\cdot 60-11.09\cdot 32}{99}\right)^2} \]

\[ =\sqrt{114.1^2+11.3^2}=114.7\ \mathrm{kv}. \]

The results of the calculation of the second example of a complex system by the transformation method are presented in Fig. 29.

Comparison of the results in Tables 19 and 20 for the calculations of the second example by the transformation method and by the nodal-equation method shows that the transformation method is by no means a calculation method giving a complete representation of the electrical state of the system, since both in the values of the node voltages and in the values of power flows in the branches obtained by this method there are noticeable deviations from the true values. However, precisely from the results of this example, where the branches have lengths exceeding 100 km (some—200 km), it may be concluded that, for the vast majority of practical calculations, the accuracy of the results obtained by calculation by the transformation method will be quite sufficient. Significant deviations of the values of reactive powers in the branches from the true—

Fig. 29.

Table No. 19.

Calculation method	Node voltages	\(E_1\)	\(E_2\)	\(E_3\)	\(E_4\)
By the nodal-equation method . . . . .		111.63	105	121.5	99.4
By the transformation method . . . . .		114.96 / 114.7	105	120.08	99.0

Table No. 20.

Link	Position	Transformation method	Method of nodal equations
I	At the beginning	\(50{,}41 - j\,15{,}47\)	\(49{,}5 - j\,8\)
I	At the end	\(48{,}38 - j\,12{,}61\)	\(46{,}3 - j\,12{,}0\)
II	At the beginning	\(36{,}79 - j\,26{,}31\)	\(36{,}0 - j\,18{,}0\)
II	At the end	\(31{,}62 - j\,17{,}90\)	\(33{,}0 - j\,27{,}2\)
III	At the beginning	\(63{,}21 - j\,35{,}06\)	\(64{,}4 - j\,32{,}5\)
III	At the end	\(54{,}29 - j\,23{,}79\)	\(56{,}0 - j\,30{,}5\)
IV	At the beginning	\(26{,}76 - j\,13{,}06\)	\(26{,}2 - j\,2{,}7\)
IV	At the end	\(25{,}71 - j\,11{,}09\)	\(24{,}5 - j\,18{,}3\)

relative values cannot have great practical significance, since the magnitudes of the total power flows are determined mainly by the active components.

VIII.

A successful solution of the question was proposed in the work of engineers Baskakov and Solov’ev.* According to this method, the distribution of powers along the links of the system is first found, and only then are the voltages at the nodes determined. To find the distribution of powers along the links it is necessary to solve a system of equations of the form:

\[ \Sigma(aR + bX) = 0 \]

\[ \Sigma(aX - bR) = 0, \]

where the number of equations is equal to twice the number of closed loops in the system.

The expressions

\[ aR + bX = 0 \]

and

\[ aX - bR = 0 \]

are formed for each link, traversing each closed loop, let us say, from left to right. The symbols “\(a\)” and “\(b\)” represent the so-called average watt and wattless load of the given link, expressed through the known loads at the nodes and the load subject to determination of the line taken as the initial one.

The voltage magnitudes at the nodes are then determined from the expressions:

\[ E_1^2 - E_2^2 = 2(aR + bx)_{\mathrm{I}} \]

\[ E_2^2 - E_3^2 = 2(aR + bx)_{\mathrm{II}} \]

\[ \cdots \]

\[ \cdots \]

and so on.

And, finally, the angles between the vectors are found from the equations:

\[ E_1 \cdot E_2 \cdot \mathrm{Sn}\,\sigma_{12} = (ax - bR)_{\mathrm{I}} \]

\[ E_2 \cdot E_3 \cdot \mathrm{Sn}\,\sigma_{23} = (ax - aR)_{\mathrm{II}} \]

\[ \cdots \]

and so on.

* The derivation of the calculation equations by this method is omitted, as it is already sufficiently well known. Those interested may be referred, for example, to the book by engineer Stepanov, Calculation of Electric Networks, or to the book by Prof. A. A. Glazunov et al., Calculation of Networks of Energy Systems

Since the concept of the average power of a branch is made the basis for deriving the calculation equations by this method, the entire method may also be briefly called the “method of average powers” (although it could also have been called the method of “loop equations”).

The method of average powers is an approximate method, although it must be noted that, if desired and with some additional recalculations, the results of the calculation can be obtained in a quite satisfactory form.

This method is approximate because, first, the derivation of the calculation formulas is based on a Π-shaped equivalent circuit of the branches, which made it possible to take the current of each branch as constant over its entire length. Second, the method of average powers requires a preliminary allocation of the losses in the branches among the nodes, and the entire subsequent calculation process is carried out under the assumption that there are no losses in the branches. Undoubtedly, any assumed determination of losses will introduce into the calculation an error, greater or smaller depending on the success of the assumption. The error can be made arbitrarily small only as the result of repeated recalculations, when each subsequent calculation is performed with allowance for the losses on the basis of the preceding one. Engineers Baskakov and Solov’ev recommend carrying out the calculation by considering the losses in the branches and the reactive powers due to capacitive currents as if already included in the substation loads declared by the consumers. This assumption is justified by the authors on the grounds that consumers’ applications for the power they require are not absolutely exact and have deviations of ±5% from the actual powers consumed at substations.

One should apparently agree with this proposal, but again on the assumption of preliminary calculations. For final computations of the voltages at the nodes, the inaccuracies obtained by neglecting losses and capacitive powers in the branches may sometimes, apparently, be unacceptable. The latter is confirmed by comparing the results of the calculation of an example by the method of average powers, with the indicated assumption, given in the article by Engineers Baskakov and Solov’ev, with the results of calculation by the simplified method of nodal equations (Table 21); moreover, as was shown, the latter results are very close to the true ones on the basis of the Π-shaped equivalent circuit.

Table No. 21

Voltages at nodes	By the method of nodal equations	By the method of average powers
$E_1$	114,8	113,93
$E_2$	98,15	97,63
$E_3$	109,9	109,24
$E_5$	107,82	107,28
$E_6$	110,4	109,33

Comparison of the voltage values at the nodes shows that even in this example, where the branches are mainly shorter than 100 km, the error in calculations by the method of average powers without allowance for losses and capacitive powers in the branches is appreciable. It is apparently necessary to conclude that, in order to obtain quite satisfactory results of calculations by the method of average powers, at least one additional repeated recalculation must be performed.

The next assumption, to which the method of average powers chiefly owes its relative simplicity, consists in setting equal to zero the expression $\sum \dot E_m \cdot E_n \operatorname{Sn}\sigma_{mn}$, owing to which it became possible to obtain the relation

$$ \sum (ax - bR) = 0 $$

It must be acknowledged that the error obtained by virtue of this latter assumption, especially in the light of the assumptions already made earlier, is insignificant. One cannot, however, agree with the proof adduced by the authors in favor of the latter assumption. Indeed, it would seem that the direct path of proving that the expression

\[ \sum E_m \cdot E_n \operatorname{Sn}\sigma_{mn} \]

may be equated to zero should consist in comparing its magnitude with the quantities together with which it enters into the general expression

\[ \sum (ax - bR) = \sum E_m \cdot E_n \operatorname{Sn}\sigma_{mn}, \]

i.e., one should compare, say,

\[ \sum bR + \sum E_m E_n \cdot \operatorname{Sn}\sigma_{mn} \]

and

\[ \sum bR \]

and from this already draw a conclusion as to the possibility of assuming that

\[ \sum bR + \sum E_m \cdot E_n \cdot \operatorname{Sn}\sigma_{mn} \cong \sum bR, \]

i.e.

\[ \sum E_m E_n \operatorname{Sn}\sigma_{mn} \cong 0. \]

However, the authors of the method under consideration see the proof of the possibility that

\[ \sum E_m \cdot E_n \cdot \operatorname{Sn}\sigma_{mn} \cong 0 \]

in the fact that the area of the polygon \(E_1 E_2 E_3 E_4 E_5\) (Fig. 30) is small in comparison with the areas of the triangles: \(OE_1E_2\), \(OE_2E_3\ldots\), etc. It is quite possible that this is so, but what basis follows from this for considering that

\[ \sum (ax - bR) = 0? \]

Fig. 30.

It is also necessary to note that a repeated recalculation, for the purpose of taking into account the power losses in the links, requires the compilation of a new system of equations and, consequently, a new solution of a system of equations with several unknowns.

The number of unknowns depends on the number of closed contours of the system, with each contour represented by two unknowns.

IX.

Eng. B. I. Pevzner in 1933*) proposed applying the superposition method to the calculation of complex systems. It is necessary to dwell in more detail on the analysis of this method of calculating complex systems, for, at first glance, it may seem that by this method of calculation the assertions made at the beginning of the present work are refuted, namely that: “any exact solution of the problem of calculating voltages and power-flow distribution in complex systems, by whatever methods it is carried out, will ultimately lead to a system of equations of the second order,” and that: “these solutions (i.e., those not involving a system of equations of the second order), apparently, must inevitably be associated with a number of assumptions.”

) Eng. Pevzner B. I., “Electrical Calculation of Complex Systems,” Electricity*, 1933, Nos. 6–7.

Eng. Pevzner, however, proposed a calculation method recommended by him as an exact method, not involving the necessity of solving a system of second-order equations. But, as will be shown below, Eng. Pevzner, in the course of his reasoning, made one inaccuracy of a fundamental nature, which places the calculation method he proposes in the category of, at best, approximate ones.

The essence of the proposal is as follows. A closed system (Fig. 31) is cut at a point, let us say, “a,” and is considered at first as a radial network (Fig. 32).

Fig. 31.

Fig. 32.

In this case, if the voltage at point “3” is specified, then by means of the well-known methods for calculating open transmission lines it is possible to find the voltages at points “2” and “1.” Then link III is again connected at point “a,” and the equalizing current in the loop is determined, flowing under the difference of the voltages \(\overline{E}_3\) and \(\overline{E}_1\).

The additional voltage drops in the links, caused by the equalizing current, are added to the voltages found earlier.

The voltages obtained in this way are regarded as true.

It is easy to see that an error has been made here. The first state—the radial state—of the system was determined by the loads at the nodes or, ultimately, by the currents at the nodes. Namely, only completely definite currents at the nodes and, consequently, also in the links caused equally definite voltage drops in the links and, ultimately, voltages at the nodes. The problem would have been solved correctly in that case if the superposition of the equalizing currents did not change the currents of the first state at the nodes of the system. However, this is precisely impossible to imagine.

A change in the voltages at the nodes due to the equalizing currents will necessarily cause a corresponding change in the load currents, provided the powers at the nodes remain unchanged. Thus, the basic condition of superposition—the invariability of the conditions at the nodal points of the circuit—is not preserved. All that has been said may be expressed by the formula: superposition of powers is impossible*). The latter proposition is sufficiently well known and proven.

Furthermore, it can be shown that the calculation method under discussion for voltages and loads in complex systems cannot be recommended even as an approximate calculation method, since it is cumbersome. According to the author, the analytical solution of the problem by his method is excessively cumbersome. Therefore the author proposes that the solution be carried out graphically. The entire calculation process then consists of the following stages:

1) cut the ring, converting the network into a radial one;
2) find the distribution of powers in the radial network;
3) find the voltages at the cut points of the network;
4) find the equalizing current and its distribution over the sections of the network;
5) find the voltage drops due to the equalizing current;
6) construct the voltage diagram for the closed network and determine the voltages at the points of the network.

It must also be taken into account that each closed contour of the system requires carrying out all the operations listed above, and, moreover, the graphical calculation method, while generally having a number of advantages, in this case—

*) If exact calculations are meant.

is often very inconvenient, because of the need to lay off and measure very small angles and vectors.

Nevertheless, the principle of equalizing currents and equalizing powers can be used in many cases of calculations of networks of power systems, as Prof. A. A. Glazunov has shown. Thus, for example, in the case of calculating the network of a complex system having several operating voltages, where the connection of individual parts of the network with different operating voltages is made through transformers (Fig. 33), and the transformation ratios of the latter do not, when multiplied, give unity, i.e., there is no possibility of reducing all the links in the network to one operating voltage—in this case, equalizing currents will occur in the closed network, superimposed on the main currents of the network links.

Fig. 33.

In the network shown in Fig. 33, let us assume that the product of the transformation coefficients “\(K_1\)” and “\(K_2\)” of transformers \(T_{p1}\) and \(T_{p2}\) is not equal to unity; then additional powers, flowing in the network due to an electromotive force equal to

\[ \Delta E = E_n \cdot K_1 \cdot K_2 - E_n, \]

must be superimposed on the powers of the links calculated under the condition that \(K_1 \cdot K_2 = 1\),

where \(E_n\) is the voltage to which the network was reduced in calculating it, without taking into account the equalizing powers.

The principle of equalizing currents can also be used to calculate the value of the reactive power required to raise or lower the voltage at a given node of the network, if operating conditions so require. For example, as a result of the network calculation in Chapter VII (Fig. 29), at the 3rd node a voltage of \(120.08\ kv\) was obtained, which, let us suppose, does not satisfy the operating requirements of the station of the 3rd node. Suppose it is required to bring the voltage at the station of the 3rd node to \(112\ kv\), while preserving the voltage of the station of the 1st node at the level of \(114.96\ kv\), and under this condition to find the voltages of the 2nd and 4th nodes. We reason in this case as follows. The voltage difference between the stations of nodes 1 and 3 can be maintained only by the flow from one of the above-mentioned stations to the other of equalizing power; it is necessary to change the magnitude of this equalizing power so that the difference

\[ E_3 - E_1 = 120.08 - 114.96 = 5.12\ kv \]

would change and become equal to

\[ E'_3 - E_1 = 112.0 - 114.96 = -2.96\ kv, \]

i.e., in other words, it is necessary, in addition to the previously flowing equalizing power, to add still another power flow, flowing from one station to the other, so that, while keeping unchanged the potential level at node 1, at the same time the potential level of node 3 is lowered by

\[ 5.12 + 2.96 = 120.08 - 112 \simeq 8\ kv. \]

And further—since in the present case it is completely unimportant to us what mutual voltage the voltage vectors of nodes 1 and 3 will have, then

we can determine this additional equalizing power as entirely reactive and find its magnitude from the relation

\[ E_1^2=\left[E'_3+\frac{P_{yp}\cdot R+(Q_{yp}+Q'_{yp})\cdot X}{E'_3}\right]^2+ \]

\[ +\left[\frac{P_{yp}\cdot X-(Q_{yp}+Q'_{yp})\cdot R}{E'_3}\right]^2, \]

where:

\(E'_3\)—the desired voltage at point “3”;

\((P_{yp}-jQ_{yp})\)—the flow of equalizing power between nodes “1” and “3” in the initial state of the network;

\(Q'_{yp}\)—the flow of reactive equalizing power additionally transported from node “1” to node “3” in order to change the voltage of node “3”.

This expression may be simplified by making the assumption that the modulus of the voltage vector \(E_1\) is equal to the longitudinal component of this vector, i.e., that

\[ E_1=E_3+\frac{P_{yp}\cdot R+Q_{yp}\cdot X}{E_3}- \]

before the superposition of the equalizing power, and

\[ E_1=E'_3+\frac{P_{yp}\cdot R+(Q_{yp}+Q'_{yp})\cdot X}{E'_3}- \]

after the superposition of the additional reactive power.

From the last two relations we obtain

\[ -(E_3-E'_3)=-\Delta E= \]

\[ =\frac{P_{yp}\cdot R+Q_{yp}\cdot X}{E_3} -\frac{P_{yp}\cdot R+Q_{yp}\cdot X}{E'_3} -\frac{Q'_{yp}\cdot X}{E'_3}. \]

To determine the magnitude of the power flow passing between nodes “1” and “3” and maintaining the potential difference between these nodes, we transfer the loads of nodes “2” and “4” to nodes “1” and “3”. In this last operation it is also necessary to take into account the loads due to losses in the network.

\[ \overline{W}_{3(2)}=\overline{W}_2\frac{\overline{Z}_{\mathrm{I}}}{\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}}} =\frac{26.2\angle54^\circ}{75.2\angle57^\circ}\cdot 85.6\angle-20^\circ42' \]

\[ =29.9\angle-23^\circ42'=27.3-j12.0\ \text{MVA}. \]

\[ \overline{W}_{3(4)}=\overline{W}_4\frac{\overline{Z}_{\mathrm{IV}}}{\overline{Z}_{\mathrm{IV}}+\overline{Z}_{\mathrm{III}}} =87.2\angle-23^\circ36'\frac{67.8\angle61^\circ54'}{107.3\angle59^\circ}= \]

\[ =55.1\angle-20^\circ42'=51.5-j19.4\ \text{MVA}. \]

To the powers found one would still have to add the losses due to transporting these powers over links III and II, but we know the losses in these links corresponding to somewhat different flows, namely, to the flows actually taking place in these links, i.e., taking into account also the equalizing power. However, since in the formula for determining the additional equalizing power the terms containing the power now being determined are of secondary importance and have little effect on the final result of the calculations of the additional equalizing...

power, then one may make use of the losses in links III and II known to us (Table 16).

Thus, the part of the power of node “3” going directly to cover the corresponding parts of the loads at nodes “4” and “2”

\[ \overline{W}'_{3(4,2)} = \overline{W}_{3(4)}+\overline{W}_{3(2)}+\Delta \overline{W}_{\mathrm{II}}+\Delta \overline{W}_{\mathrm{III}} = \]

\[ =51.5-j19.4+27.3-j12+5.17-j8.41+8.92-j12.27= \]

\[ =92.89-j52.08\ \text{mva}. \]

The remainder of the power of node “3” will flow from node “3” to node “1.” This remainder is equal to

\[ \overline{W}_{\mathrm{ur}}=\overline{W}_{3}-\overline{W}'_{3(4,3)}= \]

\[ =-(100-j61.37)-(92.89-j52.08)=-(7.11-j9.29). \]

To lower the voltage at node “3” by \(8\ \text{kv}\), it is necessary to pass through the network between nodes “3” and “1” an additional equalizing power, whose magnitude is determined (with some error) from the expression:

\[ -\Delta E= \frac{P_{\mathrm{ur}}\cdot R+Q_{\mathrm{ur}}\cdot X}{E_{3}} = \frac{P_{\mathrm{ur}}\cdot R+Q_{\mathrm{ur}}\cdot X}{E'_{3}} - \frac{Q'_{\mathrm{ur}}\cdot X}{E'_{3}}. \]

The total resistance of the network between nodes “3” and “1”

\[ \overline{Z} = \frac{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}})\cdot (\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) }{ (\overline{Z}_{\mathrm{I}}+\overline{Z}_{\mathrm{II}})+ (\overline{Z}_{\mathrm{III}}+\overline{Z}_{\mathrm{IV}}) } = \]

\[ = \frac{75.2\angle 57^\circ \cdot 107.3\angle 59^\circ}{182.3\angle 58^\circ 6'} = 44.3\angle 57^\circ 54' = 23.5+j37.6\ \Omega. \]

Consequently:

\[ -\Delta E=-8= \frac{(-7.11)\cdot 23.5+(-9.29)\cdot 37.6}{120.08} - \]

\[ - \frac{(-7.11)\cdot 23.5+(-9.29)\cdot 37.6}{112.0} - \frac{Q'_{\mathrm{ur}}\cdot 37.6}{112} = \]

\[ =-4.23-(-4.53)-0.335\,Q'_{\mathrm{ur}}, \]

whence

\[ Q'_{\mathrm{ur}}= +\frac{8-0.3}{0.335} = +23\ \text{mva}. \]

The sign \(+\) indicates that the reactive power must be transported from node “1” to node “3,” i.e., in other words, the power of the station of the 3rd node, in terms of its reactive component, will not be used by \(23\ \text{mva}\); these \(23\ \text{mva}\) will be supplied by the station of the 1st node.

Next, we divide the additional reactive power, respectively, between the two parallel branches II-I and III-IV

\[ \overline{Q}'_{\mathrm{ur}(\mathrm{II-I})} = \overline{Q}'_{\mathrm{ur}} \frac{\overline{Z}}{\overline{Z}_{\mathrm{II}}+\overline{Z}} = \]

\[ = +23\angle -90^\circ \frac{44.3\angle 57^\circ 54'}{75.2\angle 57^\circ} = \]

\[ = 13.55\angle -89^\circ 6' = 0.20-j13.55\ \text{mva}. \]

\[ \overline{Q}'_{vp(\mathrm{III-IV})} = \overline{Q}'_{ur}\, \frac{\overline{Z}}{\overline{Z}_{\mathrm{IV}}+\overline{Z}_{\mathrm{III}}} = \]

\[ = 23\angle -90^\circ\, \frac{44.3\angle 57^\circ 54'}{107.3\angle 59^\circ} = 9.5\angle -91^\circ 6' = \]

\[ = -0.19-j9.5\ \text{mva}. \]

The new power flows in the network links, taking into account the additional equalizing power, are presented in Table 22:

Table No. 22

Link	Initial power	Additional equalizing power	Link power including additional equalizing power
I (begin.) . . .	\(50.41-j15.47\)	\(0.20-j13.55\)	\(50.61-j29.02\)
II (end) . . .	\(31.62-j17.90\)	\(0.20-j13.55\)	\(31.42-j4.35\)
III (end) . . .	\(54.29-j23.79\)	\(+0.19-j9.5\)	\(54.48-j14.29\)
IV (begin.) . . .	\(26.76-j13.06\)	\(-0.19-j9.5\)	\(26.57-j22.56\)

We shall find the voltages at the nodes, bearing in mind the voltage specified at the 1st node.

\[ E_1=114.96\ \text{kv}, \]

\[ E_2= \sqrt{ \left( 114.96-\frac{50.61\cdot15.5+29.02\cdot21.2}{114.96} \right)^2 + \left( \frac{50.61\cdot21.2-29.02\cdot15.5}{114.96} \right)^2 } \]

\[ =\sqrt{102.70^2+5.4^2}=102.8\ \text{kv}, \]

\[ E_3= \sqrt{ \left( 102.8+\frac{31.42\cdot25.5+4.35\cdot41.7}{102.8} \right)^2 + \left( \frac{31.42\cdot41.7-4.35\cdot25.5}{102.8} \right)^2 } \]

\[ =\sqrt{112.35^2+12^2}=112.95\ \text{kv}. \]

According to the assignment it would have been necessary to obtain 112.08. The inaccuracy of the calculation result is explained by the many assumptions made in the process of determining both the additional reactive power and the total power flows in the links. For example, the change in losses in the links due to changes in the powers in the links at the expense of the additional equalizing power has not been taken into account.

X.

Thus, we have examined the following methods for calculating voltages and power-flow distribution in networks of complex systems, as possible or as known in the literature:

a) the node-equation method,
b) the transformation method,
c) the mean-power method (loop equations),
d) the equalizing-power method.

Each of these calculation methods has both its advantages and its shortcomings. With time, practical workers will give their greatest preference to one of these methods and thereby prove its greatest viability. On the other hand, the methods themselves may, with time, undergo changes in the direction of their improve-

1. The problem statement is borrowed from the article by engineers Baskakov and Solov’ev, Electric Stations, No. 3, 1932, p. 164. ↩

2. These, like many other, calculations here were made on a 25-cm logarithmic slide rule. ↩