QUESTIONS OF THE GENERAL THEORY OF COMBINED ELECTRIC MACHINES

A. I. SKOROSPESHKIN

(Recommended by the seminar of the departments of electrical machines and general electrical engineering.)

The necessity of developing a general theory of electric machines is dictated by the need to solve practical problems of electrical machine building, to further develop the theory of electric machines, and to compare various classes of electric machines.

It should be noted that the development of the general theory is being conducted as applied to machines of normal design (asynchronous, synchronous, direct-current machines). Moreover, each of these machines is usually reduced to an idealized two-pole and two-phase machine, and for these machines all the necessary analysis is carried out \[1, 2, 3, 4\].

Combined machines of the type of converters and brushless electromechanical amplifiers, as far as we know, are not considered from the standpoint of the general theory of electric machines. Therefore we have attempted to examine them on the basis of the general theory of electric machines.

The brushless electromechanical amplifiers (BEMA) of direct and alternating current and converters being developed by us are two-stage. Both stages are combined in one magnetic core, while the conditions of combination are satisfied \[5, 6, 7\]. The main ones among them are:

1) absence of forces of unilateral magnetic attraction, which is satisfied when

\[ P_2 \pm P_1 \ne \pm 1, \tag{1} \]

where $P_1$ and $P_2$ are the numbers of pole pairs, respectively, of the first and second stages;

2) absence of transformer couplings between two windings located in one magnetic core.

The fulfillment of such conditions makes it possible first to consider each stage as a separate machine, subsequently establishing the interrelation between the individual stages.

A general approach to the consideration of combined electric machines, in our opinion, can be developed on the basis of the theory of Maxwell’s electromagnetic phenomena and Lagrange’s equations \[8, 9, 10\]. In this case it appears possible to consider an electrical system (machine) consisting of two groups of windings: one of them (2–3 windings) is stationary and is located on the stator, the other (1–2 windings) is rotating and is located on the rotor. In this case each of the windings may be considered separately, and for it the Lagrange equations—

second-kind Lagrange equations are represented in the form of two equations:

for electromechanical forces

\[ M_{\text{em}}=\frac{d}{dt}\frac{\partial\left(\frac{1}{2}m_{ab}i_a i_b\right)}{\partial \dot h} -\frac{\partial\left(\frac{1}{2}m_{ab}i_a i_b\right)}{\partial h} =-\frac{1}{2}i_a i_b\frac{\partial m_{ab}}{\partial h}; \tag{2} \]

for electromotive forces

\[ e_k=\frac{\partial\left(\frac{1}{2}i_k^2 r_k\right)}{\partial i_k} +\frac{d}{dt}\frac{\partial\left(\frac{1}{2}m_{ab}i_a i_b\right)}{\partial i_k} -\frac{\partial\left(\frac{1}{2}m_{ab}i_a i_b\right)}{\partial q_k} = \tag{3} \]

\[ =i_k r_k+\frac{d(l_k i_k)}{dt}+m_{ka}\frac{di_a}{dt} +i_a\frac{dm_{ka}}{dt}. \]

Here \(M_{\text{em}}\) is the generalized external electromechanical force,
\(e_k\) is the generalized e.m.f.,
\(h\) is the generalized geometric coordinate,

\[ \dot h=\frac{dh}{dt} \]

is the generalized geometric velocity,

\(q_k\) is the generalized electrical coordinate,

\[ \dot q_k=\frac{dq_k}{dt}=i_k \]

is the circuit current (generalized velocity),

\(r_k, e_k, m_{ab}, m_{ka}, i_a, i_b, i_k\) are the resistance, inductances, mutual inductances, and currents of the windings under consideration.

In general, for each cascade of a combined machine one may write the equations of electromotive forces and electrical powers for each of the windings, and the equations of electromechanical forces and electromechanical powers, taking into account the interaction of the stator and rotor windings.

The general equations are written in the following form:

1. Equations of e.m.f.

\[ e_k=i_k r_k+\frac{d(l_k i_k)}{dt}+\sum\frac{d(m_{ka}i_a)}{dt}. \tag{4} \]

1. Equations of electrical powers

\[ P_{\text{el}}=e_k i_k=i_k^2 r_k+i_k\frac{d(l_k i_k)}{dt} +i_k\sum\frac{d(m_{ka}i_a)}{dt}. \tag{5} \]

1. Equations of electromechanical forces

\[ M_{\text{em}}=-\frac{1}{2}i_a i_b\frac{\partial m_{ab}}{\partial h}. \tag{6} \]

1. Equations of electromechanical powers

\[ P_{\text{em}}=M_{\text{em}}\frac{dh}{dt} =-\frac{1}{2}i_a i_b\frac{\partial m_{ab}}{\partial h}\frac{dh}{dt}. \tag{7} \]

In electrical machines the angle of rotation \(\alpha\) is chosen as the generalized geometric coordinate \(h\); then, in steady-state operation,

\[ h=\alpha=\omega t. \tag{8} \]

In this case the generalized electromechanical force will be the torque.

Since the inductances depend on the geometric coordinates, and an electric machine is characterized by one geometric coordinate \(x\), the partial derivatives \(\dfrac{\partial m}{\partial x}\) will be equal to the total derivatives \(\dfrac{dm}{dx}\). Therefore (6) and (7) take the form:

\[ M_{\text{em}}=-\frac{1}{2} i_a i_b \frac{dm_{ab}}{dx}; \tag{9} \]

\[ P_{\text{em}}=-\frac{1}{2} i_a i_b \frac{dm_{ab}}{dt}. \tag{10} \]

It should be noted that in a nonsalient-pole machine the inductances and mutual inductances between phases may be regarded as constant, and only the mutual inductances between the stator and rotor windings vary, which is a necessary condition for converting mechanical energy into alternating-current electrical energy, or conversely.

From the system of equations (4)—(7) one can determine the currents, torques, powers, and thus establish the interrelation between the individual cascades.

In the present article, questions are considered that are connected only with the calculation of the magnetic field in the air gap and the determination of the magnetic energy.

The possibility of considering each winding separately makes it possible to calculate the magnetic field created by each winding, and then to determine the resultant magnetic field of the gap and carry out the calculation of the magnetic circuit.

When calculating the magnetic field in the gap, the following assumptions may be adopted:

1. Saturation is absent. The magnetic material of the stator and rotor has a linear characteristic with very high magnetic permeability.

2. The effect of the slots is negligible, and the reserve of magnetic energy exists only in the air gap.

3. To determine the field strength and magnetic induction in the air gap, the surfaces of the stator and rotor may be considered smooth, and the current sheets (linear loads) created by each of the windings may be considered separately, assuming that only one of the windings is supplied and the others are open-circuited.

We shall consider only the fundamental harmonic components in all curves.

Under these conditions, the magnetic field in the air gap may be regarded as potential \((\operatorname{rot}\,\vec H=0)\), described by the Laplace equation

\[ \frac{\partial^2 U_m}{\partial x^2}+\frac{\partial^2 U_m}{\partial l^2}=0, \tag{11} \]

where \(U_m\) is the scalar magnetic potential, with

\[ \vec H=-\operatorname{grad} U_m. \tag{12} \]

The solution of the Laplace equation by the Fourier method can be represented by the general expression

\[ U_m=\sum_n A_n \begin{matrix} \sin\\ \cos \end{matrix} mx \begin{matrix} \operatorname{sh}\\ \operatorname{ch} \end{matrix} my + \sum_n B_n \begin{matrix} \operatorname{sh}\\ \operatorname{ch} \end{matrix} mx \begin{matrix} \sin\\ \cos \end{matrix} my. \tag{13} \]

When the boundary conditions are taken into account, the necessary functions are selected from (13). To calculate the magnetic field in the air gap of an electrical machine, it is sufficient to choose only one term

\[ U_m=\sum_n B_n \begin{matrix} \operatorname{sh}\\ \operatorname{ch} \end{matrix} my\, \begin{matrix} \sin\\ \cos \end{matrix} mx. \tag{14} \]

We shall carry out the calculation of the magnetic field using as an example a brushless electromagnetic amplifier, or converter, with adjustable frequency (Fig. 1). The designations in the figure are: 1 — control winding,

Fig. 1.

3 — rotor winding of cascade II, 2 — rotor winding of cascade I, 4 — output winding of the stator of cascade II.

Winding 1

The current of the control winding

\[ I_1=\frac{U}{R}, \tag{15} \]

where \(U\) is the constant voltage applied to the control winding;
\(R\) is the resistance of the control winding.

Linear loading

\[ A_{1m}=\frac{I_1 W_1}{\tau_1}, \tag{16} \]

where \(W_1\) is the number of turns of the control winding,
\(\tau_1\) is the pole pitch.

The field of the control winding rotates either by means of a semiconductor or a mechanical commutator with angular velocity \(\omega_1\). For the linear loading the following expression is written:

\[ A_{1\mathrm{res}}=A_{1m}\sin(\omega_1 t-nx), \tag{17} \]

here \(n=\dfrac{\pi}{\tau_1}\). To determine the boundary conditions, we choose the coordinate axes on the stator (Fig. 2).

Laplace’s equation

\[ \frac{\partial^2 U_1}{\partial x^2}+\frac{\partial^2 U_1}{\partial l^2}=0. \tag{18} \]

Field strength

\[ \overline{H}_1=-\operatorname{grad} U_1 . \tag{19} \]

The solution of Laplace’s equation for the first harmonic in the general form is determined by expression (14).

Boundary conditions

\[ H_x=-\left.\frac{\partial U_1}{\partial x}\right|_{y=0}=A_{1\text{rez}}; \]

\[ H_x=-\left.\frac{\partial U_1}{\partial x}\right|_{y=\delta}=0;\qquad -\frac{\partial U_1}{\partial x}=S\operatorname{ch}ny\sin nx+C\operatorname{sh}ny\cos nx . \tag{20} \]

Fig. 2.

The number of terms in (20) is chosen taking into account the known boundary conditions. The constants \(S\) and \(C\) are determined from the boundary conditions:

\[ S=\frac{A_{1m}\sin(\omega_1 t-nx)}{\sin nx}, \]

\[ C=-\frac{A_{1m}\sin(\omega_1t-nx)\operatorname{ch}n\delta}{\operatorname{sh}n\delta\cos nx}. \tag{21} \]

Substituting (21) into (20) and integrating with respect to \(x\), we obtain

\[ U_1=-\frac{A_{1m}\cos(\omega_1t-nx)}{n}\operatorname{ch}ny+ \frac{A_{1m}\cos(\omega_1t-nx)\operatorname{ch}n\delta}{n\operatorname{sh}n\delta}\operatorname{sh}ny . \tag{22} \]

The radial component of the magnetic-field strength in the air gap is

\[ H_1=-\frac{\partial U_1}{\partial y} =\left(A_{1m}\operatorname{sh}ny-\frac{A_{1m}\operatorname{ch}n\delta}{\operatorname{sh}n\delta}\operatorname{ch}ny\right)\cos(\omega_1t-nx). \tag{23} \]

Winding 2

Currents in the phases

\[ i_{21}=I_{2m}\sin\omega_2t, \]

\[ i_{22}=I_{2m}\sin\left(\omega_2t-\frac{2\pi}{3}\right), \tag{24} \]

\[ i_{23}=I_{2m}\sin\left(\omega_2t-\frac{4\pi}{3}\right). \]

Linear loads

\[ A_{21}=\frac{I_{2m}W_2}{\tau_2/3}\sin\omega_2t\cos\frac{\pi}{\tau_2}x, \]

\[ A_{22}=\frac{I_{2m}W_2}{\tau_2/3}\sin\left(\omega_2t-\frac{2\pi}{3}\right) \cos\left(\frac{\pi}{\tau_2}x-\frac{2\pi}{3}\right), \tag{25} \]

\[ A_{23}=\frac{I_{2m}W_2}{\tau_2/3}\sin\left(\omega_2t-\frac{4\pi}{3}\right) \cos\left(\frac{\pi}{\tau_2}x-\frac{4\pi}{3}\right), \]

\[ A_{2\mathrm{res}}=\frac{3}{2}\frac{I_{2m}W_2}{\tau_2/3}\sin\left(\omega_2 t-\frac{\pi}{\tau_2}x\right) =A_{2m}\sin(\omega_2t-ax), \]

where \(W_2\) is the number of turns of the phase,

\[ \tau_2 \]
is the pole pitch, \(a=\dfrac{\pi}{\tau_2}\),

\(\omega_2\) is the angular velocity of rotation of the field of winding 2.

After carrying out a solution analogous to the preceding one, we find the radial component of the magnetic-field strength in the air gap from winding 2:

\[ H_2=\frac{A_{2m}\operatorname{ch} ay}{\operatorname{sh} a\delta}\cos(\omega_2t-ax). \tag{26} \]

By analogy with winding 2, for winding 3 we obtain

\[ H_3=\frac{A_{3m}\operatorname{ch} by}{\operatorname{sh} b\delta}\cos(\omega_3t-bx). \tag{27} \]

Here \(\omega_3\) is the angular velocity of rotation of the field of winding 3,

\[ b=\frac{\pi}{\tau_3}. \]

For winding 4, located on the stator, we obtain

\[ H_4=\left(A_{4m}\operatorname{sh} cy-\frac{A_{4m}\operatorname{ch} c\delta}{\operatorname{sh} c\delta}\operatorname{ch}cy\right)\cos(\omega_4t-cx), \tag{28} \]

where

\[ c=\frac{\pi}{\tau_4}, \]

\(\omega_4\) is the angular velocity of rotation of the field of winding 4.

The instantaneous value of the resultant magnetic-field strength in the air gap is equal to

\[ H_\delta=H_1+H_2+H_3+H_4. \tag{29} \]

The instantaneous value of the resultant induction in the air gap will be equal to

\[ B_\delta=\mu_0H_\delta. \tag{30} \]

With a stationary field of the control winding, only the expression for the field strength of the control winding changes somewhat. It takes the form

\[ H_1=\left(A_{1m}\operatorname{sh} ny-\frac{A_{1m}\operatorname{ch} n\delta}{\operatorname{sh} n\delta}\operatorname{ch} ny\right)\sin nx. \tag{31} \]

The remaining expressions for the field strengths remain unchanged, bearing in mind that the numerical values of the angular velocities of rotation and of other quantities will be different.

After determining the field strength and induction in the air gap, one can proceed to the calculation of the magnetic circuit, taking into account the superposition of two magnetic fields in one magnetic core.

The above determination of the field strength and induction in the air gap, created by each winding, makes it possible to determine the stored magnetic energy by means of a spatial integral over the volume of the air gap.

\[ W_m=\int_0^l\int_0^\delta\int_0^{2\pi R}\frac{1}{2}BH\,dz\,dy\,dx . \tag{32} \]

Then the instantaneous value of the energy created by winding 1 is obtained as

\[ \begin{aligned} W_{m1}={}&\frac{1}{2}\mu_0 l \left[ \pi R+\frac{1}{2n}\cos(\omega_1 t-\pi R n)\sin 2\pi R n \right]\times\\ &\times \left[ A_{1m}^{2}\frac{\operatorname{th}^{2} n\delta}{2} \left(\frac{1}{2n}\operatorname{sh}2n\delta+\delta\right) -\frac{A_{1m}^{2}}{4}\left(\operatorname{sh}2n\delta+2\delta\right) \right]; \end{aligned} \]

for winding 2

\[ \begin{aligned} W_{m2}={}&\frac{1}{4}\mu_0 l\frac{A_{2m}^{2}}{\operatorname{sh}^{2} a\delta} \left[ \pi R+\frac{1}{2a}\cos(\omega_2 t-\pi R a)\sin 2\pi R a \right]\times\\ &\times \left[ \frac{1}{2a}\operatorname{sh}2a\delta+\delta \right]; \end{aligned} \tag{33} \]

for windings 3 and 4:

\[ \begin{aligned} W_{m3}={}&\frac{1}{4}\mu_0 l\frac{A_{3m}^{2}}{\operatorname{sh}^{2} b\delta} \left[ \pi R+\frac{1}{2b}\cos(\omega_3 t-\pi R b)\sin 2\pi R b \right]\times\\ &\times \left[ \frac{1}{2b}\operatorname{sh}2b\delta+\delta \right]; \end{aligned} \]

\[ \begin{aligned} W_{m4}={}&\frac{1}{2}\mu_0 l \left[ \pi R+\frac{1}{2c}\cos(\omega_4 t-\pi R c)\sin 2\pi R c \right]\times\\ &\times \left[ A_{4m}^{2}\frac{\operatorname{th}^{2} c\delta}{2} \left(\frac{1}{2c}\operatorname{sh}2c\delta+\delta\right) -\frac{A_{4m}^{2}}{4}\left(\operatorname{sh}2c\delta+2\delta\right) \right]. \end{aligned} \]

The total energy in the air gap is

\[ W_m=W_{m1}+W_{m2}+W_{m3}+W_{m4}. \tag{34} \]

Thus, the general approach to combined electrical machines on the basis of the Lagrange–Maxwell equations makes it possible to solve a whole series of theoretical and practical questions connected with the investigation of operating modes and with the design of combined electrical machines.

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