INVESTIGATION OF SHELL-TYPE SMOOTHING REACTORS FOR MINIMUM WEIGHT OF EQUIVALENT MATERIALS REFERRED TO UNIT POWER

I. D. Kutyavin, L. I. Piletskii

The consumption of equivalent active materials per unit power in the design and manufacture of smoothing reactors plays an essential role and depends on the choice of the principal geometrical dimensions. In the present article an attempt is made to derive certain recommendations for the most advantageous consumption of materials per unit power of the reactor.

The reactor under investigation has an air gap equal to the height of the core window. The magnetic system of the reactor is divided into \(n = 6 \div 12\) U-shaped laminated packets enclosing the winding on all sides. It is assumed that the magnetic flux passes only through the steel and closes through the air gap.

The starting points are the expression for the reactor power [1]:

\[ S = K \frac{\kappa_0^2 b^2 h^2 x y P (D-b)^2}{b_1 H (x+i)(y+\delta)^2} \tag{1} \]

and the current density in the winding \(\Delta\), obtained from the heat-balance equation of the coil [1]:

\[ \Delta = \sqrt{\frac{\alpha (x+i)\,\overline{P}}{x y b_1}} . \tag{2} \]

Other initial quantities for determining the equivalent weight of the active materials are the length of the mean turn of the winding \(l_{\mathrm{m}}\):

\[ l_{\mathrm{m}} = \pi (D-b) \tag{3} \]

and the calculated length of the core steel, referred to \(q_c\):

\[ l_c = H + 2r + 2\delta_0 + 0.5D . \tag{4} \]

To determine the consumption of equivalent active materials per unit power in the reactor, let us express certain quantities through its power.

Copper cross-section \(q_{\mathrm{m}}\):

\[ q_{\mathrm{m}} = \frac{b_1 S H (y+\delta)} {K \kappa_0 b h P (D-b)^2}. \tag{5} \]

Weight of copper \(Q_{\mathrm{m}}\):

\[ Q_{\mathrm{m}}=\frac{\gamma_{\mathrm{m}}\cdot \pi \cdot b_1 \cdot S \cdot H \cdot (y+\delta)} {K\cdot k_0 b h P (D-b)^2}, \tag{6} \]

where \(\gamma_{\mathrm{m}}\) is the specific weight of the winding material.

The weight of copper, referred to unit reactor power, is

\[ \frac{Q_{\mathrm{m}}}{S}= \frac{\pi \gamma_{\mathrm{m}} b_1 H (y+\delta)} {K k_0 b h P (D-b)^2}. \tag{7} \]

The weight of steel per unit reactor power is

\[ \frac{Q_{\mathrm{c}}}{S}=\frac{q_{\mathrm{c}} l_{\mathrm{c}} \gamma_{\mathrm{c}}}{S}. \tag{8} \]

Here \(\gamma_{\mathrm{c}}\) is the specific weight of steel.

Substituting the values of \(q_{\mathrm{c}}\) and \(l_{\mathrm{c}}\) and assuming

\[ \frac{x}{x+i}\approx \frac{x+i}{x}\approx 1, \]

we obtain:

\[ \frac{Q_{\mathrm{c}}}{S}= \frac{0.1\pi^2\gamma_{\mathrm{c}}(D-b)(h+l)}{B_{\mathrm{c}}} \sqrt{\frac{\alpha}{KSH}} + \frac{0.1\pi^2\gamma_{\mathrm{c}}D(D-b)}{2B_{\mathrm{c}}} \sqrt{\frac{\alpha}{KSH}} + 2\frac{(0.1\pi^2)^2\gamma_{\mathrm{c}}\alpha(D-b)^2} {B_{\mathrm{c}}^2 K H\cdot k_{\mathrm{c}}\cdot n D \operatorname{tg}\frac{180^\circ}{n}}. \tag{9} \]

In expression (9)

\[ h+l=H+2\delta_0. \tag{10} \]

On the basis of (7) and (9), the reduced weight of active materials per unit power is

\[ \frac{Q_{\mathrm{p}}}{S}=\frac{Q_{\mathrm{c}}}{S}+\beta\frac{Q_{\mathrm{m}}}{S}, \tag{11} \]

where \(\beta\) is the ratio of the specific calculated cost of the winding material to the cost of steel in the product [2].

If the products of the constant quantities entering into (11) are replaced by constant coefficients, we finally obtain:

\[ \varphi= \frac{Q_{\mathrm{p}}}{S} \left[ a_1(D-b)(h+l)+a_2D(D-b) \right]\frac{1}{\sqrt{SH}} +a_3\frac{(D-b)^2}{DH} + a_4\frac{H(y+\delta)}{bhP(D-b)^2}. \tag{12} \]

In expression (12)

\[ \left. \begin{aligned} a_1&=\frac{0.1\pi^2\gamma_{\mathrm{c}}}{B_{\mathrm{c}}}\sqrt{\frac{\alpha}{K}}; &\qquad a_2&=\frac{0.1\pi^2\gamma_{\mathrm{c}}}{2B_{\mathrm{c}}}\sqrt{\frac{\alpha}{K}};\\[0.6em] a_3&=\frac{2(0.1\pi^2)^2\gamma_{\mathrm{c}}\alpha} {K\cdot B_{\mathrm{c}}^2\cdot k_{\mathrm{c}} n \operatorname{tg}\frac{180^\circ}{n}}; &\qquad a_4&=\beta\frac{b_1\pi\gamma_{\mathrm{m}}}{K\cdot k_0}. \end{aligned} \right\} \tag{13} \]

Expression (12), for a given power \(S\), is a function of four variables: \(D\), \(b\), \(h\), and \(y\).

Using the condition

\[ \left. \begin{aligned} \frac{\partial\varphi}{\partial D}&=0; &\qquad \frac{\partial\varphi}{\partial h}&=0;\\[0.6em] \frac{\partial\varphi}{\partial b}&=0; &\qquad \frac{\partial\varphi}{\partial y}&=0 \end{aligned} \right\} \tag{14} \]

for determining the optimal values of these variables is not possible, since the function \(\varphi\) has no extremum with respect to some variables. Thus, function (12) has no extremum with respect to the variable \(b\). Let us reduce the number of variables by imposing certain restrictions. From (1) we express \(D\) through \(h\), taking

\[ \frac{b_1 (y+\delta)^2}{yP \cdot K \cdot k_0^2} = Z \quad \text{and} \quad P = (k_{\text{п}} b_1 + y). \tag{15} \]

Then

\[ \begin{aligned} (D-b)^2 &= \frac{SHZ}{b^2 h^2},\\ D-b &= \frac{\sqrt{SHZ}}{bh},\\ D &= b + \frac{\sqrt{SHZ}}{bh}. \end{aligned} \tag{16} \]

Expression (12), taking into account (15) and (16), will take the form

\[ \varphi = a_1 \frac{\sqrt{Z}}{bh}(h+l) + a_2 \frac{\sqrt{Z}}{bh} \left(\frac{\sqrt{SHZ}}{bh}+b\right) + a_3 \frac{SZ}{bh\left(\sqrt{SHZ}+b^2h\right)} + a_4 \frac{\sqrt{H}(y+\delta)}{P\sqrt{SZ}} . \tag{17} \]

With respect to the variable \(b\), expression (17) has no extremum. As \(b\) increases, the specific consumption of equivalent materials decreases.

The optimal value \(y_0\) obtained as a result of the investigation by the graphical-analytical method tends toward nonconstructive dimensions (for \(S = 10^3\) kVA, \(y_0 < 0.1\) cm, and for \(S = 10^6\) kVA, \(y_0 \simeq 0.3\) cm). With increasing power, \(y_0\) increases only slightly. To determine the optimal value \(h_0\), we shall use the condition

\[ \frac{\partial \varphi}{\partial h}=0, \]

\[ \frac{\partial \varphi}{\partial h} = A_4 h^2 - A_1 \sqrt{H} - A_2 \left(3+\frac{2l}{h}\right) - A_3 \frac{4b^2h\sqrt{SHZ}+SZ(3h+l)} {b\left(\sqrt{SHZ}+b^2h\right)^2} =0, \tag{18} \]

where

\[ \begin{aligned} A_1 &= \frac{2}{\sqrt{S}}\left(a_1\frac{l}{b}+a_2\right); & A_2 &= a_2 \frac{\sqrt{Z}}{b^2}; \\ A_3 &= a_3; & A_4 &= a_4 \frac{(y+\delta)}{PSZ}. \end{aligned} \tag{19} \]

Solving (18) with respect to \(h\) is a difficult task; therefore we shall use the graphical-analytical method for determining the roots of equation (18).

After \(h_0\) is determined, the remaining quantities affecting the minimum consumption of equivalent materials are found relatively simply.

Table 1 shows the influence of power on the minimum consumption of equivalent materials at the ratio \(b/D \simeq 0.45\).

Table 1

The investigations were carried out for \(\alpha = 1.35 \cdot 10^4\); \(K_p = 0.78\); \(b_1 = 9\ \text{cm}\); \(K_0 = 0.9\); \(B_c = 16500\ \text{gf}\); \(\delta = 1\ \text{cm}\); \(y = 1.4\ \text{cm}\); \(l = 40\ \text{cm}\); \(\beta = 3.5\); \(\varepsilon = 0.16\ \text{W}/\text{cm}^2\).

The investigations showed that, as the ratio \(\dfrac{b}{D}\) increases up to 0.5, the specific consumption of equivalent materials decreases.

[Graph: vertical axis \(\varphi\), \(\text{g}/\text{kVA}\), with marks 1000 and 2000; horizontal axis \(\lg S\), \(\text{kVA}\), with marks \(10^3\), \(10^4\), \(10^5\), \(10^6\).]

Fig. 1

Figure 1 shows the dependence of the specific consumption of equivalent materials \(\varphi\) on the power \(S\) at \(\dfrac{b}{D} = 45\).

REFERENCES

1. A. I. Piletskii, I. D. Kutyavin. On the limiting power of smoothing reactors with a linear characteristic (see above).

2. I. D. Kutyavin. On determining the optimal dimensions of three-phase two-winding transformers. Proceedings of the Tomsk Polytechnic Institute, vol. 130, 1964.