THERMAL VERIFICATION OF A MOTOR IN A MEDIUM-POWER ION DRIVE WITH PULSE CONTROL OF ROTATIONAL SPEED

Yu. M. ACHKASOV, V. A. BEINAROVICH, A. I. ZAITSEV

(Presented by the scientific seminar of the Electromechanical Faculty)

Pulse control of motor speed is based on the fact that energy is supplied to the motor armature in the form of individual pulses; the rotational speed of the motor depends on the amount of energy in the pulse and on the pulse repetition frequency.

The advantages of this method of control are described in a preceding paper of the present collection and in [1, 2].

An interesting but little-studied question remains the heating of a motor operating in a pulse-control system when supplied from a controlled multiphase ionic rectifier.

In most cases, medium-power motors are supplied from rectifiers with a number of phases equal to two or more. The motor can operate both in the region of discontinuous current and in the region of continuous current. Therefore, when selecting the motor power and checking it for heating, it is necessary to take into account the number of phases and the operating mode of the rectifier.

Let us consider the operation of a motor in pulse mode when it is supplied from a three-phase rectifier circuit with a neutral conductor (Fig. 1).

We shall adopt the following assumptions:

1) the voltage drop in the valve arc is constant;
2) the ionization and deionization times are not taken into account;
3) the load closing the circuit consists of a counter e.m.f. and ohmic resistance and inductance that are constant in magnitude.

Having adopted these assumptions, the operating rectifier–motor circuit may be replaced by an equivalent circuit (Fig. 2), where \(U_{\mathrm{m}}\) is the amplitude of the alternating voltage; \(E_{\mathrm{d}}\) is the voltage drop in the arc; \(K\) is a switch that closes and opens the circuit depending on the control signal in the automatic speed-control circuit; \(i\) is the instantaneous value of the current in the motor circuit; \(E\) is the motor counter e.m.f.

To simplify the derivations and for convenience in applying them, let us express the principal quantities in relative units.

The relative values of the voltage are

\[ \varepsilon=\frac{E+E_{\mathrm{d}}}{U_{\mathrm{m}}}. \tag{1} \]

The relative value of the current through the valve is

\[ i=\frac{iR}{U_m}=\frac{i}{I_m}, \tag{2} \]

where \(I_m\) is the base current,

\[ I_m=\frac{U_m}{R}. \tag{3} \]

The electromagnetic time constant of the circuit is \(T=\dfrac{L}{R}\). The load parameter is

\[ \tan \Theta=\frac{\omega L}{R}=\omega T, \tag{4} \]

Fig. 1.

Fig. 2.

where \(\omega=2\pi f\) is the angular frequency of the alternating-current network.

\[ \Theta=\arc \tg \omega T. \tag{5} \]

Current in the valve circuit can arise if the opening pulse is applied at a positive anode voltage, that is, under the condition

\[ U_{\mathrm{m}}\sin \vartheta_{\mathrm{з}}-E-E_{\mathrm{д}}\geq 0, \]

or

\[ \sin \vartheta_{\mathrm{з}}\geq \frac{E+E_{\mathrm{д}}}{U_{\mathrm{m}}}=\varepsilon . \]

The ignition angle \(\vartheta_{\mathrm{з}}\) is measured from the beginning of the sinusoid of the alternating voltage and is determined for the first conducting anode by

\[ \vartheta_{\mathrm{з}}=\arcsin \varepsilon . \tag{6} \]

The conduction angle of the valve is equal to the difference between the extinction and ignition angles,

\[ \lambda=\vartheta_{\mathrm{п}}-\vartheta_{\mathrm{з}} . \]

For the circuit of one phase one may write the equation of electrical equilibrium

\[ U_{\mathrm{m}}\sin(\vartheta_{\mathrm{з}}+\omega t)=E+E_{\mathrm{д}}+iR+L\frac{di}{dt} \tag{7} \]

or, in relative units,

\[ \sin(\vartheta_{\mathrm{з}}+\omega t)=\varepsilon+i+T\frac{di}{dt}, \tag{8} \]

the solution of this differential equation for the current pulse through one valve has the form [3]

\[ i=\left[\cos\Theta\sin(\vartheta_{\mathrm{з}}-\Theta+\omega t)-\varepsilon\right] +\left[\varepsilon-\cos\Theta\sin(\vartheta_{\mathrm{з}}-\Theta)\right]e^{-\omega t\operatorname{ctg}\Theta}. \tag{9} \]

This equation is valid for the region of discontinuous currents at \(i>0\).

The limiting mode of operation of an \(m\)-phase rectifier will be the mode in which

\[ i=0 \quad \text{at} \quad \omega t=\frac{2\pi}{m}. \]

If \(i>0\) at \(\omega t=\dfrac{2\pi}{m}\), then we have a continuous-current mode.

In this case equation (9) is supplemented by a term with the initial current value \(i_0\) for the next anode entering operation. The expression for the current will have the form

\[ i=\left[\cos\Theta\sin(\vartheta_{\mathrm{з}}-\Theta+\omega t)-\varepsilon\right] +\left[\varepsilon-\cos\Theta\sin(\vartheta_{\mathrm{з}}-\Theta)+i_0\right]e^{-\omega t\operatorname{ctg}\Theta}. \tag{10} \]

Let us consider the operating mode of the rectifier under pulse regulation of speed in the case where the number of consecutively conducting anodes is much greater than the number of skipped ignitions. This mode will be observed when regulating the motor speed below nominal over a small range (less than 1.6) and may proceed with either discontinuous or continuous current when several anodes conduct one after another.

When the rectifier operates in the region of continuous currents, anode overlap occurs, that is, a section \(\gamma\) of simultaneous operation of two valves (Fig. 3).

In this section the current of one valve decreases to zero, while that of the other increases to a certain value. For practical calculation we assume that the current transfers from one anode to another instantaneously at the current value \(i_0\).

Thus, for the instant at which the current transfers from one anode to another, on the basis of equation (10) one may write

\[ i_0=\left[\cos\Theta\sin\left(\vartheta_3-\Theta+\frac{2\pi}{m}\right)-\varepsilon\right]+ \]

\[ +\left[\varepsilon-\cos\Theta\sin(\vartheta_3-\Theta)\right]e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} +i_0 e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta}, \tag{11} \]

whence

\[ i_0= \frac{ \left[\cos\Theta\sin\left(\vartheta_3-\Theta+\frac{2\pi}{m}\right)-\varepsilon\right] +\left[\varepsilon-\cos\Theta\sin(\vartheta_3-\Theta)\right]e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} }{ 1-e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} }. \tag{12} \]

Substituting the value of \(i_0\) in (10) and carrying out the transformation, we obtain

\[ i=\left[\cos\Theta\sin(\vartheta_3-\Theta+\omega t)-\varepsilon\right]+ \]

\[ +\frac{ 2\cos\Theta\sin\frac{\pi}{m}\cos\left(\vartheta_3-\Theta+\frac{\pi}{m}\right) }{ 1-e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} } e^{-\omega t\operatorname{ctg}\Theta}. \tag{13} \]

Fig. 3.

Let us determine the area of one pulse when the rectifier operates in the region of continuous current:

\[ S_1=\int_0^\lambda i\,d\vartheta = \int_0^\lambda \left\{ \left[\cos\Theta\sin(\vartheta_3-\Theta+\omega t)-\varepsilon\right]+ \right. \]

\[ \left. +\frac{ 2\cos\Theta\sin\frac{\pi}{m}\cos\left(\vartheta_3-\Theta+\frac{\pi}{m}\right) }{ 1-e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} } e^{-\omega t\operatorname{ctg}\Theta} \right\}d\vartheta . \tag{14} \]

We replace \(\omega t=\vartheta\) and, integrating over the limits from 0 to \(\dfrac{2\pi}{m}\), obtain

\[ S_1=\left[\cos \vartheta_3-\cos\left(\vartheta_3+\frac{2\pi}{m}\right)-\varepsilon\frac{2\pi}{m}\right]. \tag{15} \]

If it is assumed that the rectifier passes all current pulses, then it may be written that the average current through the motor over one period when an \(m\)-phase rectifier is operating is

\[ i_{\mathrm{avg}}=\frac{m}{2\pi}S_1 =\frac{m}{2\pi}\left[\cos\vartheta_3-\cos\left(\vartheta_3+\frac{2\pi}{m}\right)-\varepsilon\frac{2\pi}{m}\right]. \tag{16} \]

But when the rectifier operates in a pulse speed-control system, the rectifier may fail to pass some of the current pulses, being blocked. As the motor speed decreases and the supply voltage remains unchanged, the amplitude of the current pulses increases, since the motor emf decreases. Consequently, to overcome the same load torque as the speed is reduced, the number of pulses per period decreases. The average current in the motor then remains unchanged, since it is determined by the magnitude of the load,

\[ i_{\mathrm{avg}}=\frac{i_{\mathrm{avg\,m}}}{d}, \tag{17} \]

where \(d=\dfrac{S_1}{S_{1\mathrm{n}}}\) is the ratio of the area of one current pulse at \(\varepsilon\) corresponding to the given motor speed to the area of one current pulse at \(\varepsilon=\dfrac{E_{\mathrm{n}}+E_{\mathrm{d}}}{U_{\mathrm{m}}}\), where \(E_{\mathrm{n}}\) is the rated emf of the motor. The number of current pulses falling within one period at the given motor speed is

\[ N=\frac{2\pi i_{\mathrm{avg\,m}}}{S_1 d}. \tag{18} \]

To determine the heating-equivalent motor current, we find the quadratic area of one current pulse

\[ S_{1\mathrm{sq}}=\int_0^{\frac{2\pi}{m}} i^2\,d\vartheta . \tag{19} \]

Substituting into equation (19) the value of \(i\) from (13) and integrating the given expression, we obtain

\[ \begin{aligned} S_{1\mathrm{sq}}={}& \frac{\pi}{m}\cos^2\Theta -\frac{1}{2}\cos^2\Theta\sin\frac{2\pi}{m} \cos 2\left(\vartheta_3-\Theta+\frac{\pi}{m}\right) \\ &+2\varepsilon\cos\Theta \left[ \cos\left(\vartheta_3-\Theta+\frac{2\pi}{m}\right) -\cos(\vartheta_3-\Theta) \right] \\ &- \frac{ \sin^2 2\Theta\, \sin\frac{\pi}{m}\, \cos\left(\vartheta_3-\Theta+\frac{\pi}{m}\right) }{ 1-e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} } \Bigg\{ \left[ \operatorname{ctg}\Theta\, \sin\left(\vartheta_3-\Theta+\frac{2\pi}{m}\right) +\cos\left(\vartheta_3-\Theta+\frac{2\pi}{m}\right) \right] e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta} \\ &\qquad -\left[ \operatorname{ctg}\Theta\sin(\vartheta_3-\Theta) +\cos(\vartheta_3-\Theta) \right] \Bigg\} - \end{aligned} \]

\[ -4\varepsilon \sin \Theta \sin \frac{\pi}{m}\cos\left(\vartheta_3-\Theta+\frac{\pi}{m}\right)+\frac{2\pi}{m}\varepsilon^2+ \]

\[ +\sin 2\Theta \sin^2 \frac{\pi}{m}\cos^2\left(\vartheta_3-\Theta+\frac{\pi}{m}\right) \frac{1+e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta}} {1-e^{-\frac{2\pi}{m}\operatorname{ctg}\Theta}}. \tag{20} \]

Let us determine the motor current equivalent with respect to heating:

\[ i_{\mathrm{eq}}=\sqrt{\frac{S_{1\mathrm{sq}}N}{2\pi}}. \tag{21} \]

The motor will operate without overheating if the condition

\[ i_{\mathrm{n}}>(1.1 \div 1.2)\,i_{\mathrm{eq}}, \tag{22} \]

is satisfied, where

\[ i_{\mathrm{n}}=\frac{I_{\mathrm{n}}}{I_{\mathrm{m}}}. \]

The coefficient \((1.1 \div 1.2)\) takes into account the additional losses in the motor associated with operation with a pulsating current.

When the motor operates in the discontinuous-current region, in computing the square area of one pulse it is necessary to omit the value \(t_0\) and to take the integral over the limits from 0 to \(\lambda\), where \(\lambda\)—the conduction angle of one valve—is determined from the diagrams [3].

Conclusions

The method considered for checking the selected motor for heating when operating in a pulse-control system for rotational speed makes it possible to calculate the motor current equivalent with respect to heating under constant load as a function of the number of phases and the operating region of the rectifier.

The same method can be extended to operation of the motor with a phase-controlled rectifier. In this case the rectifier firing angle will be equal to the control angle, and the number of current pulses occurring in one period will be equal to the number of rectifier phases.

LITERATURE

1. D. P. Morozov, M. G. Chilikin, N. G. Lysenkov, and L. M. Tverdin. A new circuit for fast-acting pulse control in systems with ion converters. Elektrichestvo, No. 2, 1958.

2. M. G. Chilikin, D. P. Morozov, L. M. Tverdin. Pulse control of the rotational speed of DC motors. Electric Drive and Automation of Industrial Installations. GEI, 1960.

3. A. A. Bulgakov. Electronic Devices of Automatic Control. GEI, 1958.

4. I. L. Kaganov. Electronic and Ion Converters. GEI, Part I, 1951; Part II, 1955; Part III, 1956.

Received by the editors
in May 1962.