ON THE LIMITING POWER OF SHUNT REACTORS*

I. D. Kutyavin,
N. P. Kostritskaya

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

The design of powerful high-voltage reactors must take into account the conditions of their transportation from the manufacturing plant to the installation site.

The problem of determining the limiting power of shunt reactors on the basis of transport restrictions is currently of importance because of the growth of reactor voltages and powers.

This article presents a method for determining the limiting power of shunt reactors by maximizing the expression for reactor power while introducing the necessary dimensional restrictions. A reactor with a laminated core, made of eight C-shaped elements, with a coil winding and an air gap equal to the height of the core window, is considered.

The notation for dimensions and the basic relations are given in [1].

For the study we shall use the expression for the power of a single-phase reactor in kva:

\[ S = 4.44 f \cdot 10^{-11}\psi I, \tag{1} \]

where \(f\) is the network frequency,
\(\psi\) is the total flux linkage,
\(I\) is the reactor current.

For the reactor design under consideration, the total flux linkage may be written as:

\[ \psi = \pi B_{з} w \left[ \frac{(D - 2b)^2}{4} + \frac{b}{6}(2D - 3b) \right], \tag{2} \]

where the induction in the air gap is

\[ B_{з} = \frac{0.4\pi \sqrt{2}\, Iw}{H}. \tag{3} \]

Substituting into (2) the value of \(B_{з}\) from (3) and, instead of \((Iw)^2\), the value \((\Delta q_{м})^2\)—see [1], we write the power formula in the following form:

\[ S = K_s \frac{P h x y h^2}{k_r H (x+i)(y+\delta)^2} \left[ \frac{(D - 2b)^2}{4} + \frac{b}{6}(2D - 3b) \right], \tag{4} \]

where

\[ K_s = 1.776\sqrt{2}\,\pi^2 f a \cdot 10^{-11}. \tag{5} \]

* In contrast to [1], the present article uses the refined expressions (2), (4), (21).

The values of \(D\) and \(h\) are determined by the transport dimensions of the reactor. Therefore the reactor rating is a function of the variables: \(b, x, y\).

To determine the critical values of the variables, we shall use the condition:

\[ \frac{\partial S}{\partial b}=0;\qquad \frac{\partial S}{\partial x}=0;\qquad \frac{\partial S}{\partial y}=0. \tag{6} \]

As a result we obtain the following equations:

\[ 4b^{2}-4Db+D^{2}=0, \tag{7} \]

\[ k_{\mathrm{p}}i-x(x+i)\cdot k'_{\mathrm{r}(x)}=0, \tag{8} \]

\[ k_{\mathrm{r}}\bigl[\delta(y+P)-k_{\mathrm{p}}by\bigr]-k'_{\mathrm{r}(y)}P(y+\delta)=0, \tag{9} \]

where \(k'_{\mathrm{r}(x)}\) and \(k'_{\mathrm{r}(y)}\) are the derivatives of \(k_{\mathrm{r}}\) with respect to the corresponding variables.

From (7) we find the optimum value of the radial dimension of the winding:

\[ b_{0}=0.5D. \tag{10} \]

In practice \(b_{0}\) cannot be realized, since it is necessary to provide for the high-voltage lead from the middle of the winding to the outside. Therefore \(b\) is determined as

\[ b=0.5(D-d), \tag{11} \]

where \(d\) is the inner diameter of the winding, determined by the dimensions of the line lead and by the required insulation distance from the lead shield to the winding.

Solving (8) and (9) simultaneously, we find the optimum values of the dimensions of the winding conductor. Equation (8) is of the fourth degree with respect to \(x\) and cannot be solved analytically.

Therefore \(x_{0}\) is determined graphically-analytically from the relation

\[ \sqrt{\frac{3(K_{\mathrm{p}}b-\delta)}{K_{\mathrm{p}}b^{2}ci}} = U_{1} = \frac{v}{\sqrt[4]{2v+3}}, \tag{12} \]

where

\[ v=\frac{x}{i}. \tag{13} \]

In Fig. 1 a curve is shown constructed from the right-hand side of expression (12).

Having determined \(U_{1}\) (for known \(b\)) from the left-hand side of (12), from the curve in Fig. 1 we find \(v\) and, consequently, \(x_{0}\). From equation (9) we find the optimum value of the axial dimension of the winding conductor:

\[ y_{0} = \frac{k_{\mathrm{p}}b\delta(x+i)} {k_{\mathrm{p}}b(x+2i)-\delta(2x+3i)}. \tag{14} \]

If the adopted value of \(y\) differs substantially from the value found, then the optimum value \(x_{0}\) is determined from equation (8) or from curve 2 in Fig. 1, constructed from the right-hand side of the expression:

\[ \frac{3(y+\delta)}{biy} = U_{2} = \frac{v^{2}\sqrt{2v+3}}{v+1}, \tag{15} \]

From consideration of the geometry of the yoke, the volume of steel of the upper and lower horizontal elements of the yoke can be written as:

\[ V_{\mathrm{c}1} = 2k_{\mathrm{c}}nr_{1} \left\{ \tau\left[ r_{2}+\delta_{0} +\frac{D}{2}\left(1-\cos\frac{180}{n}\right) \right] -\frac{\pi}{n} \left(\frac{d-2\delta_{0}}{2}\right)^{2} \right\}, \tag{16} \]

Fig. 1.

where

\[ \tau = D \cdot \sin \frac{180}{n}. \tag{17} \]

The same, but for the vertical elements of the yoke:

\[ V_{c_2} = k_c n r_2 D \sin \frac{180}{n}\, H. \tag{18} \]

The thickness of the horizontal and vertical elements of the yoke is taken to be equal:

\[ r_1 = r_2 = r. \tag{19} \]

The cross-sectional area of the steel of the vertical element of the yoke is

\[ q_{c_2} = k_c r_2 D \sin \frac{180}{n}. \tag{20} \]

The thickness of the yoke, \(r\), can be determined from the condition of equality of the flux in the steel and in the air gap, for a specified induction in the steel, \(B_c\):

\[ B_c q_{c_2} n = \frac{\pi B_\delta}{2} \left[ \frac{(D - 2b)^2}{2} + \frac{b}{3}(3D - 4b) \right]. \tag{21} \]

Taking (20) into account, we write the yoke thickness \(r_2\) from (21):

\[ r_2 = k_{r_2} \frac{h}{D H (y + \delta)} \sqrt{ \frac{P b x y}{k_c (x + i)} \left[ \frac{(D - 2b)^2}{2} + \frac{b}{3}(3D - 4b) \right] }, \tag{22} \]

where

\[ k_{r_2} = \frac{0.4\pi^2 \sqrt{2a}} {k_c n B_c \sin \frac{180}{n}}. \tag{23} \]

The transport dimensions of the reactor are taken from the outlines of the fourth-degree oversize clearance. In height, the railway clearance is used with allowance for transporting the reactor suspended on articulated transport. In this case the clearance between the bottom of the tank and the level of the rail heads is \(250\) mm [2]. From the outlines of the railway clearance \(H_r = 360.8\) cm and \(D_r = 400\) cm, one may pass to the dimensions of the removable part of a \(500\) kV reactor.

Height of the removable part

\[ H_{\mathrm{v}}=360.8-25-35.8=300\ \text{cm}, \]

where \(25\ \text{cm}\) is the height from the rail heads to the bottom of the tank; \(35.8\ \text{cm}\) includes the thickness of the tank bottom and cover, stiffening ribs—\(13\ \text{cm}\), and the distance from the tank to the magnetic core—\(28\ \text{cm}\) (on two sides).

Diameter of the removable part:

\[ D_{\mathrm{v}}=400-50=350\ \text{cm}, \]

where \(50\ \text{cm}\) includes: the thickness of the tank walls—\(2\ \text{cm}\) (on two sides), stiffening ribs \(13\ \text{cm}\), the clearance between the magnetic core and the tank \(30\ \text{cm}\) (on two sides), and \(5\ \text{cm}\) allowance between the outlines of the clearance gauge and the tank.

The dimensions of the removable part, in turn, are related to the main dimensions of the reactor by simple relations:

\[ H_{\mathrm{v}}=h+2l_0+2r_1, \tag{24} \]

\[ D_{\mathrm{v}}=d+2b+2\delta_0+2r_2. \tag{25} \]

Substitute into (25) the value of \(r_2\) from (22):

\[ D_{\mathrm{v}}-d-2\delta_0 = 2b+ \frac{k_{\mathrm{r}}h(3d^2+6bd+4b^2)} {3H(d+2b)(y+\delta)} \times \sqrt{ \frac{(k_{\mathrm{p}}b+y)bxy}{k_{\mathrm{r}}(x+i)} }. \tag{26} \]

The radial dimension of the winding \((b)\) can be found from (26) by a search method. Assigning various values \(b=20;\ 40;\ 60;\ \ldots\), we find \(x_0;\ y_0\).

Table 1 gives the results of calculating a reactor of limiting power by the above method, with a winding made of copper conductor and insulation thickness on both sides: \(i=0.01;\ 0.1;\ 0.2;\ 0.4\). The oil-duct size was taken equal to \(1.2\ \text{cm}\). The insulation distances for a reactor with voltage \(500\ \text{kV}\) were taken as \(\delta_{01}=l=35\ \text{cm}\), \(l_0=10\ \text{cm}\). The remaining general data are: \(\alpha=13.5\cdot10^4;\ c=0.92;\ \gamma_{\mathrm{c}}=7.65;\ \gamma_{\mathrm{m}}=8.9;\ k_{\mathrm{c}}=0.8;\ k_{\mathrm{p}}=0.78;\ B_{\mathrm{c}}=16.5\ \text{kGs};\ f=50\ \text{Hz}\).

Items 10–17 give the geometrical dimensions of the reactor; item 18 gives the power in kVA; items 19–21 give the weights of the active materials.

The weight of steel was determined by the formula:

\[ Q_{\mathrm{c}}=\gamma_{\mathrm{c}}(V_{\mathrm{c}1}+V_{\mathrm{c}2})\cdot10^{-3},\ \text{kg}. \tag{27} \]

Weight of copper

\[ Q_{\mathrm{m}}=\gamma_{\mathrm{m}}q_{\mathrm{m}}l_{\mathrm{m}}\cdot10^{-3},\ \text{kg}. \tag{28} \]

It is seen from Table 1 that reducing the insulation of the winding conductor from \(i=0.1\) to \(i=0.01\) (the use of enamel insulation) leads to a twofold increase in power.

Increasing the insulation from \(i=0.1\) to \(i=0.2\) reduces the power by a factor of 3.

Figure 2 shows the dependence of the maximum power on the geometrical dimensions of the winding conductor, calculated by this method. With a change in the radial dimension of the winding conductor, the power changes sharply. Therefore it is advisable to use a value of \(x\) close to the optimum. A change in the axial dimension of the conductor \(y\) has little effect on the maximum power.

The method presented can also be used to determine the optimum geometry of reactors for a specified power and voltage.

Table 1

Fig. 1.

REFERENCES

1. I. D. Kutyavin, N. P. Kostritskaya. On the limiting power of compensating reactors with an air gap. Izv. TPI, vol. 172, 1967.
2. V. Sh. Anshin and A. G. Kraiz. Assembly of high-power transformers, GEI, 1961.