Calculation of the Inductance of the End Windings of a Shock Generator in the Presence of Conducting Shields
Abstract Generated abstract
This paper develops an analytical method for estimating the leakage inductance of stator end windings in shock generators when the end conductors are enclosed by conducting shields. Treating the shield as an ideal conductor and using a symmetric rectangular geometry, the author solves Maxwell’s equations through the magnetic vector potential, with Fourier expansion of the current density and boundary conditions at the conductor, shield, and intervening regions. The resulting expressions give the magnetic induction, field energy, and inductance per unit length as functions of conductor dimensions and shield spacing. The analysis indicates that shielding effectiveness depends chiefly on the gap between the conductor and shield and on the conductor aspect ratio, with experimental measurements showing satisfactory agreement with the calculated dependence.
Full Text
CALCULATION OF THE INDUCTANCE OF THE END WINDINGS OF A SHOCK GENERATOR IN THE PRESENCE OF CONDUCTING SHIELDS
L. O. OGANYAN
(Submitted by the scientific seminar of the Departments of Electrical Machines
and General Electrical Engineering)
The leakage inductive reactance in the zone of the end windings of the stator winding of a synchronous generator constitutes a significant part of its subtransient reactance. Reduction of the end-winding leakage fluxes of the stator winding in shock generators is achieved either by selecting a type of winding in which the leakage flux is minimal, or by enclosing the end parts of the machine in shields made of materials with high electrical conductivity, when, owing to the induction of eddy currents in the shield walls, the leakage field in the end-winding zone is weakened, which leads to a reduction in the leakage inductance.
An estimate of the influence of conducting shields on the leakage inductance of the stator winding of a shock generator is of practical interest at the stage of its design.
In the present article this problem is solved by integrating Maxwell’s equations for the electromagnetic field in the shielded space [1]:
\[ \operatorname{rot}\overline{B} = \mu \overline{\delta}, \tag{1} \]
\[ \operatorname{div}\overline{B} = 0. \tag{2} \]
We neglect displacement currents in the end-winding zone. The scheme of the system under consideration is shown in Fig. 1,a. A bar of height \(b\) and width \(a\) is installed in a rectangular shield with internal dimensions \(H\) and \(T\). The distance between the bar and the inner surface of the shield is determined by the insulation thickness \(\Delta\). In solving the problem, the following assumptions are adopted:
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The shield material is an ideal conductor. By an ideal conductor is meant a conductor whose specific conductivity is equal to infinity \((\gamma=\infty,\ \mu=1)\). In this case the current on the shield surface confines the magnetic field within the shielded region.
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In order to balance the electromagnetic forces between the bar and the shield, a symmetrical placement of the bar in the shield is adopted [2].
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The specific conductivity of the bar material is taken equal to 1 \((\mu=1)\).
With the adopted configuration of the system under consideration, it is convenient to consider the electromagnetic field in a Cartesian coordinate system. And since the system is arranged symmetrically with respect to the \(x\)-axis,
Fig. 1. Computational scheme for solving the problem: a) relative arrangement of the elements: 1—bar, 2—shield; b) distribution of induction over the surface of the shield
then it is sufficient to calculate and investigate the field in only one of the parts of the space under consideration.
The investigation is carried out with the aid of the vector-potential function of the magnetic field. Equations (1) and (2) are written in the form
\[ \operatorname{rot}\overline{A}=\overline{B}, \tag{1a} \]
\[ \operatorname{div}\overline{A}=0. \tag{2a} \]
The distribution of the induction in the shielded space can be found by first determining the vector potential of the magnetic field as a function of the coordinates.
In the chosen system, equations (1a) and (2a) are written as follows:
\[ \frac{\partial^2 A}{\partial x^2}+\frac{\partial^2 A}{\partial y^2}=-\mu_0\delta, \tag{3} \]
\[ \frac{\partial^2 A}{\partial x^2}+\frac{\partial^2 A}{\partial y^2}=0, \tag{4} \]
where \(\delta\) is the current density.
Assumption 1 does not make it possible to investigate the field inside the body of the shield. The vector potential is determined in two regions into which the space under study is divided [3].
In region I, Poisson’s equation (3) is valid; in region II, Laplace’s equation (4). For the indicated position of the system under consideration relative to the coordinate axes, the current density in the bar is represented in the form of a trigonometric series
\[ \delta=\sum_{n=1}^{\infty}\delta_n(y)\sin kx, \tag{5} \]
where the coefficient \(\delta_n(y)\) in the general case depends on \(y\) and is determined by the well-known Fourier formulas:
\[ \delta_n(y)=\frac{1}{T}\int_{-T}^{T}\delta_0\sin kx\,dx =\frac{4}{n\pi}\frac{I}{ab}\cos k\bar{x}\quad(n=1,3,5..), \tag{6} \]
\[ k=\frac{n\pi}{T}, \]
\(T\) — the half-period of the expansion;
\(ab\) — the cross section of the rod;
\(I\) — the current in the rod.
Consequently, expression (5) is written in the form
\[ \delta=\frac{4I}{\pi ab}\sum_{n=1,3}\frac{1}{n}\cos k\Delta \sin kx . \tag{7} \]
The vector potential of the magnetic field is a periodic function of \(x\) with the same period as \(\delta\); the general expression for \(A\) will be
\[ A=\sum_{n=1}^{\infty} A_n(y)\sin kx . \tag{8} \]
Substituting (7) and (8) into (3), and (8) into (4), we obtain expressions for determining \(A_n(y)\), respectively, in regions I and II,
\[ A_n''(y)-k^2 A_n(y)=-\mu_0\frac{4I}{n\pi ab}\cos k\Delta , \tag{9} \]
\[ A_n''(y)-k^2 A_n(y)=0 . \tag{10} \]
We obtain the solution of equations (9) and (10) in the form
\[ A_n(y)=C_1 e^{ky}+C_2 e^{-ky}+\frac{4\mu_0}{(\pi n)^3}\frac{I}{ab}\,T^2\cos k\Delta , \tag{11} \]
\[ A_n(y)=F_1 e^{ky}+F_2 e^{-ky}. \tag{12} \]
In determining the constants of integration we proceed from the following boundary conditions:
- In the plane \(y=0\) there is no tangential component of the induction,
\[ \frac{\partial A_{\mathrm{I}}}{\partial y}=0. \]
- At the boundary of the regions, the tangential and normal components of the magnetic-field induction must be continuous
\[ \text{for } y=-\frac{b}{2}. \]
\[ \frac{\partial A_{\mathrm{I}}}{\partial y} = \frac{\partial A_{\mathrm{II}}}{\partial y}; \quad \frac{\partial A_{\mathrm{I}}}{\partial x} = \frac{\partial A_{\mathrm{II}}}{\partial x}. \]
- For \(y=-\frac{H}{2}\), the value of the vector potential is equal to 0.
The boundary conditions considered above give a system of four algebraic equations for determining the coefficients of integration \(C_1, C_2, F_1\), and \(F_2\).
Solving the system and substituting the values of the coefficients into equations (11) and (12), we obtain the following expressions for \(A_n(y)\):
in region I
\[ A_n(y)=\frac{4\mu_0}{(\pi n)^3}\frac{I}{ab}\cos k\Delta \left( 1-\frac{\operatorname{ch} k\Delta}{\operatorname{ch} k\frac{H}{2}}\,\operatorname{ch} ky \right), \tag{13} \]
in region II
\[ A_n(y)=\frac{4\mu_0}{(\pi n)^3}\frac{I}{ab}\cos k\Delta\, \frac{\operatorname{sh} k\frac{b}{2}}{\operatorname{ch} k\frac{H}{2}}\, \operatorname{sh} k\left(\frac{H}{2}-y\right). \tag{14} \]
Accordingly, substituting the values \(A_n(y)\) from (13) and (14) into formula (8), we obtain the general expression for the vector potential function:
for region I
\[ A=-\frac{4\mu_0}{\pi^3}\frac{I}{ab}T^2 \sum_{n=1,3,5}\frac{1}{n^3}\cos k\Delta \left(1-\frac{\operatorname{ch} k\Delta}{\operatorname{ch} k\frac{H}{2}}\operatorname{ch}ky\right)\sin kx, \tag{15} \]
for region II
\[ A=\frac{4\mu_0}{\pi^3}\frac{I}{ab}T^2 \sum_{n=1,3,5}\frac{1}{n^3}\cos k\Delta\, \frac{\operatorname{sh} k\frac{b}{2}}{\operatorname{ch} k\frac{H}{2}}\, \operatorname{sh} k\left(\frac{H}{2}-y\right)\sin kx. \tag{16} \]
From expressions (15) and (16) one can obtain formulas for calculating all the technical characteristics of the system under consideration: the distribution of induction, the field energy, the inductance, the interaction forces between the rod and the shield, etc.
The components of the magnetic induction are equal to:
in region I
\[ B_{x\mathrm{I}}=\frac{\partial A_{\mathrm{I}}}{\partial y} =-\frac{4\mu_0}{\pi^2}\frac{I}{ab}T \sum_{n=1,3,5}\frac{1}{n^2}\cos k\Delta\, \frac{\operatorname{ch} k\Delta}{\operatorname{ch} k\frac{H}{2}}\, \operatorname{sh}ky\,\sin kx, \tag{17} \]
\[ B_{y\mathrm{I}}=-\frac{\partial A_{\mathrm{I}}}{\partial x} =-\frac{4\mu_0}{\pi^2}\frac{I}{ab}T \sum_{n=1,3,5}\frac{1}{n^2}\cos k\Delta \left(1-\frac{\operatorname{ch} k\Delta}{\operatorname{ch} k\frac{H}{2}}\operatorname{ch}ky\right)\cos kx; \]
in region II
\[ B_{x\mathrm{II}}=\frac{\partial A_{\mathrm{II}}}{\partial y} =-\frac{4\mu_0}{\pi^2}\frac{I}{ab}T \sum_{n=1,3,5}\frac{1}{n^2}\cos k\Delta\, \frac{\operatorname{sh} k\frac{b}{2}}{\operatorname{ch} k\frac{H}{2}}\, \operatorname{ch} k\left(\frac{H}{2}-y\right)\sin kx, \tag{18} \]
\[ B_{y\mathrm{II}}=-\frac{\partial A_{\mathrm{II}}}{\partial x} =-\frac{4\mu_0}{\pi^2}\frac{I}{ab}T \sum_{n=1,3,5}\frac{1}{n^2}\cos k\Delta\, \frac{\operatorname{sh} k\frac{b}{2}}{\operatorname{sh} k\frac{H}{2}}\, \operatorname{sh} k\left(\frac{H}{2}-y\right)\cos kx. \]
The pattern of the induction distribution over the surface of the shield \(x=0\) for the case \(a=1,\ b=3a,\ \Delta=0.5a\) is shown in Fig. 1b. A similar pattern of induction distribution occurs for the remaining surfaces.
The inductance of the system under consideration is determined through the magnetic-field energy
\[ L=\frac{2}{\mu_0 I^2}\int_V E^2\,dV. \tag{19} \]
After integration we obtain the following expression for calculating the inductance of the shielded end part per unit of its length:
\[ L=\frac{8\mu_0}{\pi^5}\frac{T^4}{(ab)^2} \sum_{n=1,3,5}\frac{1}{n^5}\cos k\Delta\,(2N_I+N_{II}), \tag{20} \]
where
\[ N_I=\frac{kb}{2}+ \frac{\operatorname{ch} k\Delta\, \operatorname{sh} k\frac{b}{2}} {\operatorname{ch} k\frac{H}{2}} \left( \frac{\operatorname{ch}^2 k\Delta\, \operatorname{ch} k\frac{b}{2}} {\operatorname{ch} k\frac{H}{2}} -2 \right), \]
\[ N_{II}= \frac{\operatorname{sh}^2 k\frac{b}{2}} {\operatorname{ch}^2 k\frac{H}{2}} \,\operatorname{sh} 2k\Delta . \]
The obtained formula (20) makes it possible to calculate the inductance of the shielded bars of the end connections of the stator winding as a function of the geometric dimensions of the elements of the system under consideration.
Analysis of the expression obtained for calculating the inductance shows that the shielding effect depends mainly on the distance between the shield and the copper surface of the end part and on the aspect ratio
\[ \frac{b}{a}. \]
Fig. 2. Change of \(L\) as a function of the ratio \(\dfrac{b}{a}\) at a constant bar cross-section and \(\Delta=0.5a\).
In Fig. 2 the dependence of the inductance of the end part on the size ratio
\[ \frac{b}{a} \]
is given for a constant bar cross-section and the ratio
\[ \frac{\Delta}{a}=0.5. \]
The greatest inductance is possessed by the system in which the ratio
\[ \frac{b}{a}=1. \]
The more this ratio differs from 1, the smaller the inductance. At
\[ \frac{\Delta}{a}=6 \]
the inductance decreases by approximately a factor of three.
The shielding effect depends to a large extent on the ratio
Fig. 3. Variation as a function of the distance between the rod and the shield surface at various ratios \(b/a\).
\(\Delta/a\). Figure 3 shows the character of the change in the inductance of shielded rods of the end part as a function of the distance to the shield surface at various ratios \(b/a\). It is obvious that the smaller the value of \(\Delta/a\), the stronger the shielding effect. This is confirmed experimentally. In Fig. 3, the dashed line gives the experimental curve of the dependence of the inductance of the end part on the ratio \(\Delta/a\). It was obtained for the rod-side ratio \(b/a = 1\) and a supply-network frequency of 1000 Hz. The shield thickness is 6 mm. The experimental results gave satisfactory agreement with the theoretical calculations.
Calculations show the exceptional influence of the value \(\Delta\) on the leakage inductance of the end part. Thus, increasing \(\Delta\) from \(0.1a\) (\(\Delta = 0.1a\) corresponds to the minimum permissible insulation thickness of the end part of the winding) to \(0.5a\) leads to an increase in the inductance of the end part by a factor of 2–3. At \(\Delta = 5 \div 6a\), the inductance of the end part reaches its greatest value. This value should be regarded as the self-inductance of the end part in air in the absence of shields. Thus, using the relationships presented, one can estimate, to a first approximation, the damping effect of conductive shields on the end leakage inductance by comparing the inductance at the adopted values of \(\Delta\) with the inductance at \(\Delta_{\max}\).
REFERENCES
- K. Shimon, Theoretical Electrical Engineering. “Mir,” 1964.
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- K. Binns, P. Lawrenson. Analysis and Calculation of Electric and Magnetic Fields. “Energiya,” 1970.