Magnetic Field in the Air Gap of a Transverse-Field EMA with a Smooth Armature and a Concentrated Compensating Winding

A. N. Aifert, A. I. Vilner

(Presented by the scientific seminar of the Departments of Electrical Machines and General Electrical Engineering)

At present, extensive work is being carried out in the Soviet Union and abroad on the creation of electrical machines with a smooth armature. Interest in them is due to their high commutational stability and better dynamic properties. However, all published works concern the creation and investigation of DC motors for various purposes, and it does not appear possible to extend the recommendations given there to the design of EMAs with a smooth armature because of the features inherent in them.

Experience in creating such EMAs shows that, in comparison with existing standard EMAs, they possess a number of advantages: with good commutation and a gain factor at the level of standard amplifiers—better speed of response, practically no noise or vibration of magnetic origin [1, 2]; and with a concentrated compensating winding [3], in addition, greater output power in the same volume, linearity of the output characteristics under voltage forcing, and a certain saving of winding copper.

To study the operating properties of an EMA with a smooth armature and questions concerning the design of these machines, it is necessary to know the pattern of distribution of the magnetic field in the air gap. In the literature there are a number of works devoted to calculating magnetic fields in the gap of DC motors with a smooth armature [4, 5], etc. The most complete investigations were carried out in [5]. The author succeeded in obtaining a solution of the problem using the method of G. A. Grinberg [6]. The region of the air gap of an EMA with a smooth armature is more complex (Fig. 1), because of the presence of split poles, a number of additional windings participating in the creation of the machine’s magnetic field (the compensating and magnetizing windings), and also the presence of armature reaction along the longitudinal and transverse axes. Therefore, in the present work a derivation is given of the basic equations for calculating the magnetic field, taking into account the special features of the EMA. As the method of investigation, as in [5], the method of G. A. Grinberg is adopted.

In solving the problem, the following basic assumptions have been made:

1. The field in the air gap is plane-parallel.

2. The magnetic permeability of the steel is infinitely large in comparison with the permeability of air.

3. The curvature of the stator and rotor surfaces is neglected. The shape of the poles is rectangular.

4. The windings lying in the large and medium slots are contracted onto one

Fig. 1

line, so that the total currents of the windings are located in an infinitely thin layer; moreover, it is assumed that the current layer is located over the entire pole height, in view of the insignificance of the slot zone not occupied by the winding.

1. The superposition principle is applied; that is, the magnetic field of each winding is considered separately.

2. The brushes are located on the geometrical neutral, since in a TFEA with a smooth armature a brush shift is inadvisable.

The adopted assumptions do not differ from those usually used in investigations of magnetic fields in electrical machines. Under these assumptions, the magnetic field in the region of the air gap occupied by the current layer will satisfy Poisson’s differential equation

\[ \frac{\partial^{2} A(x,y)}{\partial x^{2}}+ \frac{\partial^{2} A(x,y)}{\partial y^{2}} = -\mu_{0} I(x,y), \tag{1} \]

and, in the regions not occupied by the current layer, Laplace’s equation

\[ \frac{\partial^{2} A(x,y)}{\partial x^{2}}+ \frac{\partial^{2} A(x,y)}{\partial y^{2}} = 0, \tag{2} \]

where

\[ A(x,y) \]
is the vector potential of the magnetic field,

\[ I(x,y) \]
is the law of variation of the current density in the section with the current layer,

\[ \mu_{0} \]
is the magnetic constant of vacuum.

Since the pattern of distribution of the magnetic field in the air gap is repeated, with a change of polarity, under each pole, it is sufficient to consider the magnetic field of the machine over one pole pitch. Figures 3a and 3b show the regions under investigation for the armature reaction of the first and second amplification stages. As can be seen from the figures, these regions differ only in the arrangement of the large and medium slots. Therefore it is sufficient to consider the field of any armature reaction, for example that of the second stage (Fig. 3b). By changing the dimensions of the corresponding regions and the value of the current density, the magnetic field of the first amplification stage can be calculated from the same equations.

The same can be said of the fields produced by the stator windings. In this case it is sufficient to consider, for example, the field of the control winding (Fig. 2b). The fields produced by the other windings can be obtained by changing the dimensions of the slots occupied and not occupied by the winding, and also by changing the magnitudes of the magnetomotive forces (m.m.f.) produced by the winding.

Fig. 2a

Fig. 2b

Fig. 3a

Fig. 3b

Magnetic Field of the Control Winding

The investigated region of the air gap is shown in Fig. 2b. Let us divide the region under consideration into rectangular sections, as shown in the figure. On the basis of the superposition principle, the current density in the armature winding is equal to zero, and the magnetic field throughout the entire region of the air gap satisfies equation (2). The boundary conditions in each of the rectangles have the following form:

Rectangle I

\[ \begin{gathered} H_x\big|_{y=\delta}=0,\\ H_y\big|_{x=0}=H_y\big|_{x=\tau}=0,\\ H_x\big|_{y=0}=f_1(x)= \begin{cases} f_3(x), & 0<x<a;\\ 0, & a<x<c;\\ f_2(x), & c<x<\tau-c;\\ 0, & \tau-c<x<\tau-a;\\ f_4(x), & \tau-a<x<\tau. \end{cases} \end{gathered} \tag{3} \]

Rectangle II

\[ \begin{gathered} H_y\big|_{x=c}=H_y\big|_{x=\tau-c}=H_x\big|_{y=-h}=0,\\ H_x\big|_{y=0}=f_2(x). \end{gathered} \tag{4} \]

Rectangle III

\[ \begin{gathered} H_y\big|_{x=0}=H_x\big|_{y=-l}=0,\\ H_y\big|_{x=a}=\frac{F_{\mathrm{o.y.}}}{l},\\ H_x\big|_{y=0}=f_3(x). \end{gathered} \tag{5} \]

Rectangle IV

\[ \begin{gathered} H_y\big|_{x=\tau}=H_x\big|_{y=-l}=0,\\ H_y\big|_{x=\tau-a}=\frac{F_{\mathrm{o.y.}}}{l},\\ H_x\big|_{y=0}=f_4(x). \end{gathered} \tag{6} \]

Here \(f_1(x), f_2(x), f_3(x)\), and \(f_4(x)\) are unknown functions of the distribution of the tangential components of the magnetic-field intensity on the boundaries of the rectangles at \(y=0\);

\[ H_x=\frac{1}{\mu_0}\frac{\partial A(x,y)}{\partial y} \tag{7} \]

is the tangential component of the magnetic-field intensity;

\[ H_y=-\frac{1}{\mu_0}\frac{\partial A(x,y)}{\partial x} \tag{8} \]

is the normal component of the magnetic-field intensity;
\(F_{\mathrm{o.y.}}\) is the m.m.f. of the control winding.

Let us determine the functions characterizing the distribution of the vector potential in each of the rectangles.

Rectangle I.

We seek the solution of the boundary-value problem for equation (2), with boundary conditions (3), in the form

\[ A_n^{(1)}(x,y)=\sum_{n=1}^{\infty} A'_n(y)\cos \frac{n\pi}{\tau}x, \tag{9} \]

where

\[ A_n^{(1)}(y)=\frac{2}{\tau}\int_0^\tau A_n^{(1)}(x,y)\cos \frac{n\pi}{\tau}x\cdot dx. \]

According to the method of G. A. Grinberg, multiply equation (2) by

\[ \frac{2}{\tau}\cos \frac{n\pi}{\tau}x \cdot dx \]

and integrate from \(0\) to \(\tau\). As a result of the transformations [5] we obtain the ordinary second-order differential equation

\[ \frac{d^2 A_n^{(1)}(y)}{dy^2}-\left(\frac{n\pi}{\tau}\right)^2 A_n^{(1)}(y)=0 . \tag{10} \]

Carrying out analogous transformations with expressions (3), we obtain for equation (10) the following boundary conditions

\[ H_x\big|_{y=\delta}=0, \tag{11} \]

\[ H_x\big|_{y=0}=\frac{2}{\tau}\int_0^\tau f_1(x)\cdot \cos \frac{n\pi}{\tau}x \cdot dx=E_n^{(1)} . \]

The solution of equation (10) has the form

\[ A_n^{(1)}(y)=B_n^{(1)}\operatorname{sh}\frac{n\pi}{\tau}y +C_n^{(1)}\operatorname{ch}\frac{n\pi}{\tau}y . \tag{12} \]

The constants \(B_n^{(1)}\) and \(C_n^{(1)}\) are determined from the boundary conditions (11). After finding the constants, for the distribution function of the vector potential in rectangle I we obtain

\[ A_n^{(1)}(x,y)=-\sum_{n=1}^{\infty}\mu_0\frac{\tau}{n\pi}E_n^{(1)} \frac{\operatorname{ch}\dfrac{n\pi}{\tau}(y-\delta)} {\operatorname{sh}\dfrac{n\pi}{\tau}\delta} \cdot \cos \frac{n\pi}{\tau}x . \tag{13} \]

For the further solution it is necessary to know the distribution of the vector potential and its first derivative on the common boundary of the rectangles (at \(y=0\)).

From equation (13) we have

\[ A_n^{(1)}(x,0)=-\sum_{n=1}^{\infty}\mu_0\frac{\tau}{n\pi}E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot \cos \frac{n\pi}{\tau}x , \tag{14} \]

\[ H_x\big|_{y=0}= \frac{1}{\mu_0}\frac{\partial A_n^{(1)}(x,y)}{\partial y}\bigg|_{y=0} =\sum_{n=1}^{\infty}E_n^{(1)}\cos \frac{n\pi}{\tau}x . \tag{15} \]

Rectangle II

Just as in the preceding case, we shall seek the solution of equation (2), under the boundary conditions (4), in the form of a series

\[ A_k^{(2)}(x,y)=\sum_{k=1}^{\infty}A_k^{(2)}(y)\cdot \cos \frac{k\pi}{b}(x-c), \tag{16} \]

where

\[ A_k^{(2)}(y)=\frac{2}{b}\int_c^{\tau-c}A_k^{(2)}(x,y)\cdot \cos \frac{k\pi}{b}(x-c)\cdot dx . \]

Multiplying equation (2) and the boundary conditions (4) by

\[ \frac{2}{b}\cos \frac{k\pi}{b}(x-c)\cdot dx \]

and integrating from \(c\) to \(\tau-c\), after transformations we obtain the differential equation

\[ \frac{d^{2} A_k^{(2)}(y)}{dy^{2}}-\left(\frac{k\pi}{b}\right)^{2} A_{1k}^{(2)}(y)=0 \tag{17} \]

under the boundary conditions

\[ H_x\big|_{y=-h}=0, \]

\[ H_x\big|_{y=0}=\frac{2}{b}\int_{c}^{\,l-c} f_2(x)\cdot \cos \frac{k\pi}{b}(x-c)\cdot dx=E_k^{(2)}. \tag{18} \]

The solution of equation (17) has the form

\[ A_k^{(2)}(y)=B_k^{(2)}\operatorname{sh}\frac{k\pi}{b}y+C_k^{(2)}\operatorname{ch}\frac{k\pi}{b}y . \tag{19} \]

Determining from the boundary conditions (18) the constants \(B_k^{(2)}\) and \(C_k^{(2)}\) for the distribution function of the vector potential in rectangle II, we obtain

\[ A_k^{(2)}(x,y)=\sum_{k=1}^{\infty}\mu_0\frac{b}{k\pi}E_k^{(2)} \frac{\operatorname{ch}\dfrac{k\pi}{b}(y+h)} {\operatorname{sh}\dfrac{k\pi}{b}h} \cos \frac{k\pi}{b}(x-c). \tag{20} \]

From expression (20) we find the value of the function and of its first derivative at the common boundary of the rectangles (for \(y=0\)):

\[ A_k^{(2)}(x,0)=\sum_{k=1}^{\infty}\mu_0\frac{b}{k\pi}E_k^{(2)} \operatorname{cth}\frac{k\pi}{b}h\cdot \cos \frac{k\pi}{b}(x-c), \tag{21} \]

\[ H_x\big|_{y=0} =\left.\frac{1}{\mu_0}\frac{\partial A_k^{(2)}(x,y)}{\partial y}\right|_{y=0} =\sum_{k=1}^{\infty}E_k^{(2)}\cos \frac{k\pi}{b}(x-c). \tag{22} \]

Rectangle III

We seek the solution of equation (2) in rectangle III under boundary conditions (6) in the form of a series

\[ A_p^{(3)}(x,y)=\frac{A_0^{(3)}(y)}{2} +\sum_{p=1}^{\infty} A_p^{(3)}(y)\cdot \cos \frac{p\pi}{a}x, \tag{23} \]

where

\[ A_p^{(3)}(y)=\frac{2}{a}\int_{0}^{a} A_p^{(3)}(x,y)\cdot \cos \frac{p\pi}{a}x\cdot dx, \]

\[ A_0^{(3)}(y)=\frac{2}{a}\int_{0}^{a} A_0(x,y)\cdot dx. \]

Multiplying equation (2) and the boundary conditions (5) by

\[ \frac{2}{a}\cos \frac{p\pi}{a}x\cdot dx \]

and integrating from 0 to \(a\), we obtain the following differential equation

\[ \frac{d^{2}A_p^{(3)}(y)}{dy^{2}} -\left(\frac{p\pi}{a}\right)^{2}A_p^{(3)}(y) =-\frac{2}{a}\frac{F_0y}{l}\mu_0\cos p\pi, \tag{24} \]

under the boundary conditions

\[ \left. H_x \right|_{y=-l}=0, \]

\[ \left. H_x \right|_{y=0}\frac{2}{a}\int_0^a f_3(x)\cdot \cos\frac{p\pi}{a}x\cdot dx=E_p^{(3)}. \tag{25} \]

The solution of equation (24) has the form

\[ A_p^{(3)}(y)=B_p^{(3)}\operatorname{sh}\frac{p\pi}{a}y+ C_p^{(3)}\operatorname{ch}\frac{p\pi}{a}y+ \frac{2a\mu_0E_0\cdot y}{l(p\pi)^2}\cos p\pi . \tag{26} \]

For \(p=0\), equation (24) is rewritten as

\[ \frac{d^2A_0^{(3)}(y)}{dy^2} =-\frac{2}{a}\frac{F_{0.y}}{l}\mu_0 . \tag{27} \]

The solution of equation (27) has the form

\[ A_0^{(3)}(y)=B_0+C_0y-\frac{\mu_0F_{0.y}}{a\cdot l}y^2 . \tag{28} \]

The constants \(B_p^{(3)}\) and \(C_p^{(3)}\) are determined from the boundary conditions (25); \(C_0\) is determined by substituting \(p=0\) into (25), while \(B_0\) takes no direct part in solving the problem, and therefore there is no need to find it. Having determined the constants entering equations (26) and (28) for the distribution of the vector potential in rectangle III, we obtain

\[ \begin{aligned} A_p^{(3)}(x,y) &=-\frac{\mu_0E_0}{2}y+\frac{\mu_0E_0}{4l}y^2+\frac{B_0}{2}+ \\ &\quad+\sum_{p=1}^{\infty} \left[ \mu_0\frac{a}{p\pi}E_p^{(3)} \frac{\operatorname{ch}\frac{p\pi}{a}(y+l)} {\operatorname{sh}\frac{p\pi}{a}l} -\mu_0\frac{E_0a^2}{l(p\pi)^2}\cos p\pi \right]\cos\frac{p\pi}{a}x . \end{aligned} \tag{29} \]

Here

\[ E_0=\frac{C_0}{\mu_0}=-\frac{2}{a}F_{0.y}. \]

The vector-potential function and its first derivative at the common boundary (for \(y=0\)) have the following form:

\[ A_p^{(3)}(x,0)=\frac{B_0}{2} +\sum_{p=1}^{\infty} \left[ \mu_0\frac{a}{p\pi}E_p^{(3)}\operatorname{cth}\frac{p\pi}{a}l -\mu_0\frac{E_0a^2}{l(p\pi)^2}\cos p\pi \right]\times \]

\[ \times \cos\frac{p\pi}{a}x, \tag{30} \]

\[ \left. H_x \right|_{y=0} = \left.\frac{1}{\mu_0}\frac{\partial A_p^{(3)}(x,y)}{\partial y}\right|_{y=0} = \frac{E_0}{2}+ \sum_{p=1}^{\infty}E_p^{(3)}\cdot \cos\frac{p\pi}{a}x . \tag{31} \]

Rectangle IV

The derivation of the function satisfying Laplace’s equation (2) for the vector potential in rectangle IV under the boundary conditions (6) is carried out analogously to how this was done for rectangle III, and has the following form:

\[ \begin{aligned} A_{\mathrm p}^{(4)}(x,y) &= \frac{\mu_0 E_0}{2}\,y -\frac{\mu_0 E_0}{4l}\,y^2 +\frac{B_0}{2} \\ &\quad +\sum_{p=1}^{\infty} \left[ \mu_0 \frac{a}{p\pi} E_p^{(4)} \frac{\operatorname{ch}\dfrac{p\pi}{a}(y+l)} {\operatorname{sh}\dfrac{p\pi}{a}l} +\mu_0 \frac{E_0 a^2}{l(p\pi)^2} \right] \cos \frac{p\pi}{a}(x-\tau+a), \end{aligned} \tag{32} \]

where \(E_0,\ E_p^{(4)}\) are functions representing the unknown boundary conditions for equations of type (24) and (27) in rectangle IV at \(y=0\).

The vector-potential function and its first derivative on the common boundary (at \(y=0\)) have the form

\[ A_{\mathrm p}^{(4)}(x,0) = \frac{B_0}{2} +\sum_{p=1}^{\infty} \left[ \mu_0 \frac{a}{p\pi} E_p^{(4)} \operatorname{cth}\frac{p\pi}{a}l +\frac{\mu_0 E_0 a^2}{l(p\pi)^2} \right] \cos \frac{p\pi}{a}(x-\tau+a), \tag{33} \]

\[ \left. H_x \right|_{y=0} = \left. \frac{1}{\mu_0}\, \frac{\partial A_{\mathrm p}^{(4)}(x,y)}{\partial y} \right|_{y=0} = \frac{E_0}{2} + \sum_{p=1}^{\infty} E_p^{(4)} \cos \frac{p\pi}{a}(x-\tau+a). \tag{34} \]

To solve the problem it is necessary to determine the components of the magnetic-field intensity on the surface of the stator and armature (at \(y=0\) and \(y=\delta\) in rectangle I). The unknown coefficients \(E_n^{(1)}\), entering into the vector-potential function of rectangle I (13), can be determined from the matching conditions—the equality of the functions and of their first derivatives on the common boundary of rectangles II, III, and IV with rectangle I.

From the condition of equality of the functions on the common boundary, using expressions (14), (21), (30), and (33), one may write

\[ \frac{B_0}{2} + \sum_{p=1}^{\infty} \left[ \mu_0 \frac{a}{p\pi} E_p^{(3)} \operatorname{cth}\frac{p\pi}{a}l - \mu_0 \frac{E_0 a^2}{l(p\pi)^2}\cos p\pi \right] \cos \frac{p\pi}{a}x = \]

\[ = -\sum_{n=1}^{\infty} \mu_0 \frac{\tau}{n\pi} E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot \cos \frac{n\pi}{\tau}x, \qquad 0<x<a, \tag{35} \]

\[ \sum_{k=1}^{\infty} \mu_0 \frac{b}{k\pi} E_k^{(2)} \operatorname{cth}\frac{k\pi}{b}h \cdot \cos \frac{k\pi}{b}(x-c) = \]

\[ = -\sum_{n=1}^{\infty} \mu_0 \frac{\tau}{n\pi} E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot \cos \frac{n\pi}{\tau}x, \qquad c<x<\tau-c, \tag{36} \]

\[ \frac{B_0}{2} + \sum_{p=1}^{\infty} \left[ \mu_0 \frac{a}{p\pi} E_p^{(4)} \operatorname{cth}\frac{p\pi}{a}l + \mu_0 \frac{E_0 a^2}{l(p\pi)^2} \right] \cos \frac{p\pi}{a}(x-\tau+a) = \]

\[ = - \sum_{n=1}^{\infty} \mu_0 \frac{\tau}{n\pi} E_n^{(1)} \operatorname{cth} \frac{n\pi}{\tau}\delta \cdot \cos \frac{n\pi}{\tau}x, \qquad \tau-a < x < \tau . \tag{37} \]

Multiplying both sides of expression (35) by \(\dfrac{2}{a}\cos \dfrac{p\pi}{a}x \cdot dx\) and integrating from \(0\) to \(a\), the unknown coefficients \(E_p^{(3)}\) can be expressed in terms of \(E_n^{(1)}\):

\[ E_p^{(3)} = \frac{E_0}{l p\pi}\cos p\pi \cdot \operatorname{th}\frac{p\pi}{a}l - \frac{2\tau}{a^2}\operatorname{th}\frac{p\pi}{a}l \sum_{n=1}^{\infty}\frac{p}{n}E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot d_{pn}^{(3)}, \tag{38} \]

where

\[ d_{pn}^{(3)} = \int_{0}^{a} \cos \frac{p\pi}{a}x \cdot \cos \frac{n\pi}{\tau}x \cdot dx = \]

\[ = \frac{ \sin \pi\left(p-n\frac{a}{\tau}\right) }{ 2\left(\dfrac{p\pi}{a}-\dfrac{n\pi}{\tau}\right) } + \frac{ \sin \pi\left(p+n\frac{a}{\tau}\right) }{ 2\left(\dfrac{p\pi}{a}+\dfrac{n\pi}{\tau}\right) } \quad \text{for } \frac{n\pi}{\tau} \ne \frac{p\pi}{a}; \]

\[ d_{pn}^{(3)}=\frac{a}{2} \quad \text{for } \frac{n\pi}{\tau}=\frac{p\pi}{a}. \tag{39} \]

In an analogous way one can express the coefficients \(E_k^{(2)}\) and \(E_p^{(4)}\) in terms of \(E_n^{(1)}\):

\[ E_k^{(2)} = - \frac{2\tau}{B_2} \operatorname{th}\frac{k\pi}{b}h \cdot \sum_{n=1}^{\infty}\frac{k}{n}E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot d_{kn}^{(2)}, \tag{40} \]

where

\[ d_{kn}^{(2)} = \int_{c}^{\tau-c} \cos \frac{k\pi}{b}(x-c)\cdot \cos \frac{n\pi}{\tau}x \cdot dx = \]

\[ = \frac{ \sin \frac{\pi}{2}\left(k+n\frac{b}{\tau}\right)\cdot \cos \frac{\pi}{2}(k+n) }{ \dfrac{k\pi}{b}+\dfrac{n\pi}{\tau} } + \frac{ \sin \frac{\pi}{2}\left(k-n\frac{b}{\tau}\right)\cdot \cos \frac{\pi}{2}(k-n) }{ \dfrac{k\pi}{b}-\dfrac{n\pi}{\tau} } \]

\[ \text{for } \frac{n\pi}{\tau}\ne \frac{k\pi}{b}, \]

\[ d_{kn}^{(2)} = \frac{b}{2}\cos \frac{\pi}{2}(k-n) \quad \text{for } \frac{n\pi}{\tau}= \frac{k\pi}{b}, \tag{41} \]

\[ E_p^{(4)} = - \frac{E_0}{l p\pi}\operatorname{th}\frac{p\pi}{a}l - \frac{2\tau}{a^2}\operatorname{th}\frac{p\pi}{a}l \cdot \sum_{n=1}^{\infty}\frac{p}{n}E_n^{(1)} \operatorname{cth}\frac{n\pi}{\tau}\delta \cdot d_{pn}^{(4)}, \tag{42} \]

where

\[ d_{pn}^{(4)}= \int_{\tau-a}^{\tau} \cos \frac{p\pi}{a}(x-\tau+a)\cos \frac{n\pi}{\tau}x\cdot dx = \]

\[ = \frac{\sin \pi\left(n-n\frac{a}{\tau}\right)} {2\left(\frac{p\pi}{a}-\frac{n\pi}{\tau}\right)} - \frac{\sin \pi\left(n-n\frac{a}{\tau}\right)} {2\left(\frac{p\pi}{a}+\frac{n\pi}{\tau}\right)} \quad \text{when} \quad \frac{n\pi}{\tau}\ne \frac{p\pi}{a}, \tag{43} \]

\[ d_{pn}^{(4)}=\frac{a}{2}\cos \pi(p-n) \quad \text{when} \quad \frac{n\pi}{\tau}= \frac{p\pi}{a}. \]

On the other hand, for the tangential component of the magnetic-field intensity on the common boundary, on the basis of the boundary conditions (3), one may write

\[ \frac{2}{\tau}\int_{0}^{\tau} f_1(x)\cdot \cos \frac{n\pi}{\tau}x\cdot dx = \frac{2}{\tau}\int_{0}^{a} f_3(x)\cdot \cos \frac{n\pi}{\tau}x\cdot dx+ \]

\[ +\frac{2}{\tau}\int_{c}^{\tau-c} f_2(x)\cdot \cos \frac{n\pi}{\tau}x\cdot dx+ \frac{2}{\tau}\int_{\tau-a}^{\tau} f_4(x)\cdot \cos \frac{n\pi}{\tau}x\cdot dx. \tag{44} \]

Here it has been taken into account that on the sections \(a<x<c\) and \(\tau-c<x<\tau-a\) the tangential component of the magnetic-field intensity becomes zero. Substituting into equality (44), instead of the functions \(f_1(x)\), \(f_2(x)\), \(f_3(x)\), and \(f_4(x)\), their values from equations (15), (22), (31), and (34), we obtain

\[ \frac{2}{\tau}\int_{0}^{\tau} \left(\sum_{n=1}^{\infty} E_n^{(1)}\cos \frac{n\pi}{\tau}x\right) \cos \frac{n\pi}{\tau}x\cdot dx = \]

\[ = \frac{2}{\tau}\int_{0}^{a} \left(\frac{E_0}{2}+\sum_{p=1}^{\infty} E_p^{(3)}\cos \frac{p\pi}{a}x\right) \cos \frac{n\pi}{\tau}x\cdot dx+ \]

\[ +\frac{2}{\tau}\int_{c}^{\tau-c} \left[\sum_{k=1}^{\infty} E_k^{(2)}\cos \frac{k\pi}{b}(x-c)\right] \cos \frac{n\pi}{\tau}x\cdot dx+ \tag{45} \]

\[ +\frac{2}{\tau}\int_{\tau-a}^{\tau} \left[\frac{E_0}{2}+\sum_{p=1}^{\infty} E_p^{(4)}\cos \frac{p\pi}{a}(x-\tau+a)\right] \cos \frac{n\pi}{\tau}x\cdot dx. \]

After transformations, expression (45) can be reduced to the form

\[ E_n^{(1)} = \frac{2}{\tau}\sum_{p=1}^{\infty} E_p^{(3)}d_{pn}^{(3)} + \frac{2}{\tau}\sum_{k=1}^{\infty} E_k^{(2)}d_{kn}^{(2)} + \frac{2}{\tau}\sum_{p=1}^{\infty} E_p^{(4)}d_{pn}^{(4)} + \]

\[ +\frac{2E_0}{n\pi}\sin \frac{n\pi}{\tau}a. \tag{46} \]

Substituting into (46), instead of the unknown coefficients \(E_p^{(3)}\), \(E_k^{(2)}\), and \(E_p^{(4)}\), their values from (38), (40), and (42), to determine the coefficients \(E_n^{(1)}\), we obtain the following infinite system of equations:

\[ \begin{aligned} E_n^{(1)}={}&-\frac{4}{b^2}\sum_{m=1}^{\infty}\frac{E_m^{(1)}}{m}\operatorname{cth}\frac{m\pi}{\tau}\,\delta \sum_{k=1}^{\infty} k\cdot \operatorname{th}\frac{k\pi}{b}\,h\cdot d_{kn}^{(2)}\cdot d_{km}^{(2)} \;-\; \\[3pt] &-\frac{4}{a^2}\sum_{m=1}^{\infty}\frac{E_m^{(1)}}{m}\operatorname{cth}\frac{m\pi}{\tau}\,\delta \sum_{p=1}^{\infty} p\cdot \operatorname{th}\frac{p\pi}{a}\,l\cdot d_{pn}^{(3)}\cdot d_{pm}^{(3)} \;-\; \\[3pt] &-\frac{4}{a^2}\sum_{m=1}^{\infty}\frac{E_m^{(1)}}{m}\operatorname{cth}\frac{m\pi}{\tau}\,\delta \sum_{p=1}^{\infty} p\cdot \operatorname{th}\frac{p\pi}{a}\,l\cdot d_{pn}^{(4)}\cdot d_{pm}^{(4)} \;+\; \\[3pt] &+\sum_{p=1}^{\infty}\frac{2aE_0}{lp\pi}\operatorname{th}\frac{p\pi}{a}\,l\cdot \cos p\pi\cdot d_{pn}^{(3)} \;- \\[3pt] &-\sum_{p=1}^{\infty}\frac{2aE_0}{lp\pi}\operatorname{th}\frac{p\pi}{a}\,l\cdot d_{pn}^{(4)} +\frac{2E_0}{n\pi}\sin\frac{n\pi}{\tau}\,a . \end{aligned} \tag{47} \]

As investigations show, the solution of the problem takes place for
\(n=1,3,5,\ldots,\; k=1,3,5,\ldots,\; p=1,2,3,4,\ldots\). In this case the terms on the right-hand side of the system that pertain to rectangles III and IV are equal. Therefore the system of equations (47) can be represented as

\[ \begin{aligned} E_n^{(1)}={}&-\frac{4}{b^2}\sum_{m=1}^{\infty}\frac{E_m^{(1)}}{m}\operatorname{cth}\frac{m\pi}{\tau}\,\delta \sum_{k=1}^{\infty} k\cdot \operatorname{th}\frac{k\pi}{b}\,h\cdot d_{kn}^{(2)}\cdot d_{km}^{(2)} \;- \\[3pt] &-\frac{8}{a^2}\sum_{m=1}^{\infty}\frac{E_m^{(1)}}{m}\operatorname{cth}\frac{m\pi}{\tau}\,\delta \sum_{p=1}^{\infty} p\cdot \operatorname{th}\frac{p\pi}{a}\,l\cdot d_{pn}^{(3)}\cdot d_{pm}^{(3)} \;+ \\[3pt] &+\frac{4aE_0}{\pi l}\sum_{p=1}^{\infty}\frac{\operatorname{th}\dfrac{p\pi}{a}\,l}{p}\cdot d_{pn}^{(3)} +\frac{2E_0}{n\pi}\sin\frac{n\pi}{\tau}\,a . \end{aligned} \tag{48} \]

The components of the magnetic-field strength in the air gap, which give the final solution of the problem, are determined by formulas (7) and (8), using the expression for the vector potential in rectangle I (13):

\[ H_x\big|_{y=\delta}=0, \]

\[ H_x\big|_{y=0}=\sum_{n=1}^{\infty} E_n^{(1)}\cos\frac{n\pi}{\tau}\,x, \]

\[ H_y\big|_{y=\delta} =\sum_{n=1}^{\infty} E_n^{(1)} \frac{1}{\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} \cdot \sin\frac{n\pi}{\tau}\,x, \tag{49} \]

\[ H_y\big|_{y=0} =\sum_{n=1}^{\infty} E_n^{(1)}\operatorname{cth}\frac{n\pi}{\tau}\,\delta \cdot \sin\frac{n\pi}{\tau}\,x . \]

Thus, after determining the unknown coefficients \(E_n^{(1)}\) from the infinite system of equations (48), one can use equations (49) to calculate the field strength of the control winding.

Magnetic field of the armature reaction

The region of the air gap under investigation is shown in Fig. 3b. As in the case of the field of the control winding, we divide the region under consideration into rectangular sections. In contrast to the preceding problem, the main air gap (rectangle I) must be subdivided into three rectangles, separately isolating the current layer (Fig. 3b).

rectangle I–2), since in it the magnetic field must satisfy not Laplace’s equation (2), but Poisson’s equation (1).

The boundary conditions in each of the rectangles have the form:

Rectangle I-1.

\[ \begin{gathered} H_y\big|_{x=0}=H_y\big|_{x=\tau}=0,\\ H_x\big|_{y=\delta}=0,\\ H_x\big|_{y=\alpha+\beta}=f_{21}(x). \end{gathered} \tag{50} \]

Rectangle I-2

\[ \begin{gathered} H_y\big|_{x=0}=H_y\big|_{x=\tau}=0,\\ H_x\big|_{y=\alpha+\beta}=f_{21}(x),\\ H_x\big|_{y=\alpha}=f_{23}(x). \end{gathered} \tag{51} \]

Rectangle I-3.

\[ \begin{gathered} H_y\big|_{x=0}=H_y\big|_{x=\tau}=0,\\ H_x\big|_{y=\alpha}=f_{23}(x),\\ H_x\big|_{y=0}=f_1(x)= \begin{cases} f_3(x), & 0<x<a;\\ 0, & a<x<c;\\ f_2(x), & c<x<\tau-c;\\ 0, & \tau-c<x<\tau-a;\\ f_4(x), & \tau-a<x<\tau. \end{cases} \end{gathered} \tag{52} \]

Rectangle II

\[ \begin{gathered} H_y\big|_{x=c}=H_y\big|_{x=\tau-c}=H_x\big|_{y=-h}=0,\\ H_x\big|_{y=0}=f_3(x). \end{gathered} \tag{53} \]

Rectangle III

\[ \begin{gathered} H_y\big|_{x=0}=H_y\big|_{x=a}=H_x\big|_{y=-l}=0,\\ H_x\big|_{y=0}=f(x). \end{gathered} \tag{54} \]

Rectangle IV

\[ \begin{gathered} H_y\big|_{x=\tau-a}=H_y\big|_{x=\tau}=H_x\big|_{y=-l}=0,\\ H_x\big|_{y=0}=f_4(x). \end{gathered} \tag{55} \]

Here, as in the preceding problem, the functions \(f_1(x)\), \(f_2(x)\), \(f_3(x)\), \(f_4(x)\), \(f_{21}(x)\), \(f_{23}(x)\) characterize the unknown distribution of the tangential components of the magnetic-field intensity on the boundaries separating the rectangles.

Let us determine the functions characterizing the distribution of the vector potential in each of the rectangles. The problem of finding a function satisfying Laplace’s equation in rectangles II, III, IV under the boundary conditions (53)—(55) is analogous to that already considered in the case of the control field (derivation of the function for rectangle II); therefore, for the required functions one may write

\[ A_k^{(2)}(x,y)= \sum_{k=1}^{\infty} \mu_0\,\frac{b}{k\pi}\,F_k^{(2)} \frac{ \operatorname{ch}\dfrac{k\pi}{b}(y+h) }{ \operatorname{sh}\dfrac{k\pi}{b}h } \cos\frac{k\pi}{b}(x-c), \tag{56} \]

\[ A_p^{(3)}(x,y)=\sum_{p=1}^{\infty}\mu_0\frac{a}{p\pi}F_p^{(3)} \frac{\operatorname{ch}\frac{p\pi}{a}(y+l)} {\operatorname{sh}\frac{p\pi}{a}l} \cos\frac{p\pi}{a}x, \tag{57} \]

\[ A_p^{(4)}(x,y)=\sum_{p=1}^{\infty}\mu_0\frac{a}{p\pi}F_p^{(4)} \frac{\operatorname{ch}\frac{p\pi}{a}(y+l)} {\operatorname{sh}\frac{p\pi}{a}l} \cos\frac{p\pi}{a}(x-\tau+a). \tag{58} \]

where the coefficients \(F_k^{(2)}, F_p^{(3)}\) and \(F_p^{(4)}\) have the same meaning as \(E_k^{(2)}, E_p^{(3)}\) and \(E_p^{(4)}\) in the case of the control field.

The derivation of functions satisfying equation (2) in rectangles I-1 and I-3 is analogous to the derivation of function (13) in the case of the control field. The distribution of the vector potential in these rectangles is determined by the following expressions:

\[ A_n^{(\mathrm{I}-1)}(x,y)= -\sum_{n=1}^{\infty}B_n^{(1)} \frac{\operatorname{ch}\frac{n\pi}{\tau}(y-\delta)} {\operatorname{sh}\frac{n\pi}{\tau}\gamma} \cos\frac{n\pi}{\tau}x, \tag{59} \]

\[ A_n^{(\mathrm{I}-3)}(x,y)= \sum_{n=1}^{\infty} \left( \mu_0\frac{\tau}{n\pi}F_n^{(1)}\operatorname{sh}\frac{n\pi}{\tau}y + C_n^{(3)}\operatorname{ch}\frac{n\pi}{\tau}y \right) \cos\frac{n\pi}{\tau}x, \tag{60} \]

where the coefficients \(B_n^{(1)}, C_n^{(3)}\), and \(F_n^{(1)}\) are determined by the unknown boundary conditions at \(y=\alpha+\beta\), \(y=\alpha\), and \(y=0\), respectively.

Thus, the special feature of the problem under consideration is the determination of the vector-potential distribution function in rectangle I-2, which must satisfy Poisson’s equation (1). Multiplying equation (1) by \(\frac{2}{\tau}\cos\frac{n\pi}{\tau}x\cdot dx\) and integrating over the limits from 0 to \(\tau\), one can obtain the following ordinary second-order differential equation [5]:

\[ \frac{d^2A_n(y)}{dy^2} - \left(\frac{n\pi}{\tau}\right)^2 A_n(y) = Q_n, \tag{61} \]

where

\[ Q_n= -\mu_0\frac{2}{\tau}\int_0^\tau I(x,y)\cos\frac{n\pi}{\tau}x\cdot dx = \frac{8j\mu_0\tau}{(n\pi)^2b_{\mathrm{sh}}} (-1)^{\frac{n+1}{2}} \cdot \sin\frac{n\pi}{\tau}\cdot\frac{b_{\mathrm{sh}}}{2}. \tag{62} \]

Here, as in [5], it is assumed that the current density \(J\) over the height of the conductors does not change and that, within the limits from \(0\) to \(\frac{\tau}{2}-\frac{b_{\mathrm{sh}}}{2}\), it is equal to \(+j\); from \(\frac{\tau}{2}-\frac{b_{\mathrm{sh}}}{2}\) to \(\frac{\tau}{2}+\frac{b_{\mathrm{sh}}}{2}\), it varies according to a rectilinear law from \(+j\) to \(-j\); and from \(\frac{\tau}{2}+\frac{b_{\mathrm{sh}}}{2}\) to \(\tau\), it is equal to \(-j\).

The solution of equation (61) has the form

\[ A_n^{(I-2)}(x,y)=\sum_{n=1}^{\infty}\left[ B_n^{(2)}\operatorname{sh}\frac{n\pi}{\tau}(y-\alpha) +C_n^{(2)}\operatorname{ch}\frac{n\pi}{\tau}(y-\alpha) -\left(\frac{\tau}{n\pi}\right)^2 Q_n \right]\cos\frac{n\pi}{\tau}x, \tag{63} \]

where the coefficients \(B_n^{(2)}\) and \(C_n^{(2)}\) are determined from the boundary conditions at \(y=\alpha\) and \(y=\alpha+\beta\).

The functions (59), (60), and (63) can be expressed solely in terms of the unknown coefficients \(F_n^{(1)}\), using the matching conditions for the functions (equality of the functions and of their first derivatives) at the boundaries of the division of rectangles I-1, I-2 and I-2, I-3. After transformations these functions are written in the form

\[ A_n^{(I-1)}(x,y) = -\sum_{n=1}^{\infty} \left[ F_n'\, \frac{\operatorname{ch}\dfrac{n\pi}{\tau}(y-\delta)} {\operatorname{sh}\dfrac{n\pi}{\tau}\delta} + Q_n' \cdot N_n \operatorname{ch}\frac{n\pi}{\tau}(y-\delta) \right]\cos\frac{n\pi}{\tau}x, \tag{64} \]

\[ A_n^{(I-2)}(x,y) = -\sum_{n=1}^{\infty} \left[ F_n'\, \frac{\operatorname{ch}\dfrac{n\pi}{\tau}(y-\delta)} {\operatorname{sh}\dfrac{n\pi}{\tau}\delta} + Q_n' M_n\operatorname{ch}\frac{n\pi}{\tau}y - Q_n'\operatorname{ch}\frac{n\pi}{\tau}(y-\alpha) + Q_n' \right]\cos\frac{n\pi}{\tau}x, \tag{65} \]

\[ A_n^{(I-3)}(x,y) = -\sum_{n=1}^{\infty} \left[ F_n'\, \frac{\operatorname{ch}\dfrac{n\pi}{\tau}(y-\delta)} {\operatorname{sh}\dfrac{n\pi}{\tau}\delta} + Q_n' M_n\operatorname{ch}\frac{n\pi}{\tau}y \right]\cos\frac{n\pi}{\tau}x, \tag{66} \]

where

\[ M_n= \frac{ \operatorname{sh}\dfrac{n\pi}{\tau}(\beta+\gamma) - \operatorname{sh}\dfrac{n\pi}{\tau}\gamma }{ \operatorname{sh}\dfrac{n\pi}{\tau}\delta }; \]

\[ N_n= \frac{ \operatorname{sh}\dfrac{n\pi}{\tau}(\alpha+\beta) - \operatorname{sh}\dfrac{n\pi}{\tau}\alpha }{ \operatorname{sh}\dfrac{n\pi}{\tau}\delta }; \]

\[ Q_n'=\left(\frac{\tau}{n\pi}\right)^2 Q_n; \qquad F_n'=\mu_0\frac{\tau}{n\pi}F_n^{(1)}. \]

Thus, for the final solution of the problem it is necessary to find the unknown coefficients \(F_n^{(1)}\). For this purpose, as in the case of the field of the control winding, we shall use the matching method.

Let us find expressions for the functions and tangential components of the magnetic-field intensity on the common boundary of rectangles II, III, and IV with rectangle I-3 (at \(y=0\)).

\[ A_n^{(1-3)}(x,0) = -\sum_{n=1}^{\infty} \left( F_n' \operatorname{cth}\frac{n\pi}{z}\,\delta + Q_n' M_n \right) \cos\frac{n\pi}{\tau}\,x, \tag{67} \]

\[ H_x^{(1-3)}\bigg|_{y=0} = \frac{1}{\mu_0} \frac{\partial A_n^{(1-3)}(x,y)}{\partial y} \bigg|_{y=0} = \sum_{n=1}^{\infty} F_n^{(1)}\cos\frac{n\pi}{\tau}\,x, \tag{68} \]

\[ A_k^{(2)}(x,0) = \sum_{k=1}^{\infty} \mu_0 \frac{b}{k\pi} F_k^{(2)} \operatorname{cth}\frac{k\pi}{b}\,h \cdot \cos\frac{k\pi}{b}(x-c), \tag{69} \]

\[ H_x^{(2)}\bigg|_{y=0} = \frac{1}{\mu_0} \frac{\partial A_k^{(2)}(x,y)}{\partial y} \bigg|_{y=0} = \sum_{k=1}^{\infty} F_k^{(2)} \cos\frac{k\pi}{b}(x-c), \tag{70} \]

\[ A_p^{(3)}(x,0) = \sum_{p=1}^{\infty} \mu_0 \frac{a}{p\pi} F_p^{(3)} \operatorname{cth}\frac{p\pi}{a}\,l \cdot \cos\frac{p\pi}{a}\,x, \tag{71} \]

\[ H_x^{(3)}\bigg|_{y=0} = \frac{1}{\mu_0} \frac{\partial A_p^{(3)}(x,y)}{\partial y} \bigg|_{y=0} = \sum_{p=1}^{\infty} F_p^{(3)} \cos\frac{p\pi}{a}\,x, \tag{72} \]

\[ A_p^{(4)}(x,0) = \sum_{p=1}^{\infty} \mu_0 \frac{a}{p\pi} F_p^{(4)} \operatorname{cth}\frac{p\pi}{a} \cdot l\cos\frac{p\pi}{a}(x-\tau+a), \tag{73} \]

\[ H_x^{(4)}\bigg|_{y=0} = \frac{1}{\mu_0} \frac{\partial A_p^{(4)}(x,y)}{\partial y} \bigg|_{y=0} = \sum_{p=1}^{\infty} F_p^{(4)} \cos\frac{p\pi}{a}(x-\tau+a). \tag{74} \]

Equating the functions on the common boundary of rectangles II, III and IV with rectangle 1-3 [expressions (67), (69), (71) and (73)] and carrying out transformations analogous to those performed in deriving expressions (38), (40) and (42), for the coefficients \(F_k^{(2)}\), \(F_p^{(3)}\) and \(F_p^{(4)}\) we obtain

\[ F_k^{(2)} = -\frac{2}{b^2}\frac{k\pi}{\mu_0} \operatorname{th}\frac{k\pi}{b}\,h \sum_{n=1}^{\infty} \left( F_n' \operatorname{cth}\frac{n\pi}{z}\,\delta + Q_n' M_n \right) \cdot d_{kn}^{(2)}, \tag{75} \]

\[ F_p^{(3)} = -\frac{2}{a^2}\frac{p\pi}{\mu_0} \operatorname{th}\frac{p\pi}{a}\,l \sum_{n=1}^{\infty} \left( F_n' \operatorname{cth}\frac{n\pi}{z}\,\delta + Q_n' M_n \right) \cdot d_{pn}^{(3)}, \tag{76} \]

\[ F_p^{(4)} = -\frac{2}{a^2}\cdot\frac{p\pi}{\mu_0} \operatorname{th}\frac{p\pi}{a}\,l \sum_{n=1}^{\infty} \left( F_n' \operatorname{cth}\frac{n\pi}{z}\,\delta + Q_n' M_n \right) \cdot d_{pn}^{(4)}, \tag{77} \]

where the coefficients \(d_{kn}^{(2)}\), \(d_{pn}^{(3)}\) and \(d_{pn}^{(4)}\) are determined by expressions (39), (41) and (43), respectively.

On the other hand, for the tangential component of the magnetic-field intensity in rectangle 1-3 at \(y=0\), equality (44) is valid, since the boundary conditions on this surface have remained unchanged. Therefore, substituting into this equality, instead of the functions \(f_1(x)\), \(f_2(x)\), \(f_3(x)\) and \(f_4(x)\), their values from (68), (70), (72) and (74), after transformations we obtain the following infinite system of equations for determining the unknown coefficients \(F_n^{(1)}\)

\[ F_n^{(1)} = -\frac{4}{b^2} \sum_{m=1}^{\infty} \left( \frac{F_m^{(1)}}{m} \operatorname{cth}\frac{m\pi}{z}\,\delta - \frac{zQ_m}{\mu_0 m\pi}M_m \right) \sum_{k=1}^{\infty} k\cdot \operatorname{th}\frac{k\pi}{b}\,h \cdot d_{kn}^{(2)} \times \]

\[ \times d_{km}^{(2)} - \frac{8}{a^2} \sum_{m=1}^{\infty} \left( \frac{F_m^{(1)}}{m} \operatorname{cth}\frac{m\pi}{z}\,\delta - \frac{zQ_m}{\mu_0 m\pi}M_m \right) \times \]

\[ \times \sum_{p=1}^{\infty} p\cdot \operatorname{th}\frac{p\pi}{a}\,l \cdot d_{pn}^{(3)} \cdot d_{pm}^{(3)}. \tag{78} \]

Here use has been made of the circumstance that the coefficients \(d_{pn}^{(3)}\) and \(d_{pn}^{(4)}\) are equal for identical values of \(p\) and \(n\). The solution, as in the case of the field of the control winding, holds for \(n=1,3,5,\ldots,\ k=1,3,5,\ldots,\ p=1,2,3,4\).

Having determined the unknown values of the coefficients \(F_n^{(1)}\) from system (78), one can also calculate the values of the normal and tangential components of the magnetic-field intensity in the air gap, which give the final solution of the problem, from the formulas

\[ H_y\bigg|_{y=\delta} = \sum_{n=1}^{\infty} \left( F_n^{(1)} \frac{1}{\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} + \frac{\tau N_n}{\mu_0 n\pi} \right) \sin \frac{n\pi}{\tau}\,x, \]

\[ H_x\bigg|_{y=\delta+\beta} = \sum_{n=1}^{\infty} \left( F_n^{(1)} \frac{\operatorname{sh}\dfrac{n\pi}{\tau}\,\gamma} {\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} + \frac{\tau N_n}{\mu_0 n\pi}\, Q_n\,\operatorname{sh}\frac{n\pi}{\tau}\,\gamma \right) \cos \frac{n\pi}{\tau}\,x, \]

\[ H_y\bigg|_{y=\delta+\beta} = \sum_{n=1}^{\infty} \left( F_n^{(1)} \frac{\operatorname{ch}\dfrac{n\pi}{\tau}\,\gamma} {\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} + \frac{\tau N_n}{\mu_0 n\pi}\, Q_n\,\operatorname{ch}\frac{n\pi}{\tau}\,\gamma \right) \sin \frac{n\pi}{\tau}\,x, \]

\[ H_x\bigg|_{y=\alpha} = \sum_{n=1}^{\infty} \left[ F_n^{(1)} \frac{\operatorname{sh}\dfrac{n\pi}{\tau}(\beta+\gamma)} {\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} + \frac{\tau M_n}{\mu_0 n\pi}\, Q_n\,\operatorname{sh}\frac{n\pi}{\tau}\,\alpha \right] \cos \frac{n\pi}{\tau}\,x, \]

\[ \tag{79} \]

\[ H_y\bigg|_{y=\alpha} = \sum_{n=1}^{\infty} \left[ F_n^{(1)} \frac{\operatorname{ch}\dfrac{n\pi}{\tau}(\beta+\gamma)} {\operatorname{sh}\dfrac{n\pi}{\tau}\,\delta} + \frac{\tau M_n}{\mu_0 n\pi}\, Q_n\,\operatorname{ch}\frac{n\pi}{\tau}\,\alpha \right] \sin \frac{n\pi}{\tau}\,x, \]

\[ H_x\big|_{y=0} = \sum_{n=1}^{\infty} F_n^{(1)}\cos \frac{n\pi}{\tau}\,x, \]

\[ H_y\bigg|_{y=0} = \sum_{n=1}^{\infty} \left( F_n^{(1)}\operatorname{cth}\frac{n\pi}{\tau}\,\delta + \frac{\tau M_n}{\mu_0 n\pi}\,Q_n \right) \sin \frac{n\pi}{\tau}\,x . \]

The method presented makes it possible to calculate the magnetic fields produced by the control windings, the transverse magnetizing winding, the concentrated compensating winding, and also the magnetic field of the armature reaction of the first and second amplification stages. By superposing the fields of the individual windings and the armature reactions of both stages, one can obtain the resultant picture of the magnetic field in the air gap of the machine for any loads and degree of compensation.

Investigations show that, in solving the infinite systems of equations (48) and (78), it is sufficient to restrict oneself to twenty terms. In view of the large volume of computational work, all calculations were performed on the M-20 digital computer.

To investigate the influence of the parameters of the air-gap geometry and of the current sheet on the distribution of the magnetic field, the following were varied: the air gap \((\delta)\), the thickness of the current sheet \((\beta)\), the gap between the current sheet and the armature stack \((\gamma)\), the opening \((b)\) and depth of the stator slots \((l, h)\), and the brush width \((b_{\text{sh}})\). In addition, for a specific machine a series of calculations was carried out with different numbers of turns of the transverse biasing winding, voltage forcing, and a wide variation of load, for subsequent comparison with experimental data.

Figure 4 gives, as an example, the results of calculating the magnetic field of an experimental sample of a transverse-field EMU with a smooth armature and a concentrated compensating winding, having the following parameters:

\[ \begin{aligned} P_{\text{n}} &= 1.2\ \text{kW}, & U_{\text{n}} &= 60\ \text{V}, & n &= 5000\ \text{rpm},\\ \delta &= 1.7\ \text{mm}, & \gamma &= 0.2\ \text{mm}, & \beta &= 0.86\ \text{mm},\\ &&& \tau &= 144\ \text{mm},\\ b &= 20\ \text{mm}, & l &= 32\ \text{mm}, & h &= 13\ \text{mm}. \end{aligned} \]

In the figure, curve 1 is the normal component of the magnetic-field intensity on the stator surface, curve 2 is the normal component of the intensity on the armature surface, and curve 3 is the tangential component of the magnetic-field intensity on the stator surface.

Fig. 4.
a) Field-intensity curve of the control winding.
b) Field-intensity curve of the armature-reaction field of the first cascade.
c) Field-intensity curve of the transverse biasing winding.

c) Curve of the field strength of the armature reaction of the second stage.
d) Curve of the field strength of the compensating winding.

e) Resultant field curve of the machine at no load.
zh) Resultant field curve of the machine at rated load.

REFERENCES

1. A. I. Skorospeshkin, G. G. Konstantinov. Basic characteristics of a transverse-field EMA with a smooth armature. Izv. TPI, vol. 172, 1967.

2. A. I. Skorospeshkin, G. G. Konstantinov, Yu. V. Gorst. Investigation of pulsations and vibroacoustic characteristics of production transverse-field EMAs and EMAs with a smooth armature. Izv. TPI, vol. 190, 1968.

3. A. I. Skorospeshkin, A. I. Vilner, B. I. Kostylev. Selection of a compensating winding in an EMA with a smooth armature. Izv. TPI, vol. 212.

4. L. D. Osnovich, E. A. Osnovich. Calculation of the magnetic field of direct-current electrical machines with a cylindrical laminated or smooth armature. Elektrichestvo, No. 7, 1968.

5. I. I. Kurkalov. Magnetic field in the gap of a direct-current machine with a toothless armature. In: Contactless Electrical Machines, V, Riga, 1967.

6. G. A. Grinberg. Selected problems in the mathematical theory of electrical and magnetic phenomena. Publishing House of the USSR Academy of Sciences, 1948.