TRANSIENT OPERATING MODES OF MOTORS WITH A SPECIAL-DESIGN ROTOR

I. G. Kuleev

The transient operating modes of squirrel-cage motors of normal design have been studied to one degree or another. No one has dealt with the transient operating modes of motors with a Boucherot rotor. The author set himself the task of investigating them.

As a result of the investigations, formulas suitable for practical purposes were obtained, making it possible to calculate the duration of starting and braking of an unloaded motor.

The case of acceleration and braking of a loaded motor was considered; however, it did not lead to a positive result. The equations drawn up for this case proved unsolvable.

A. Transient operating modes of a motor with a deep slot

The rotor of an electric motor with a deep slot has one squirrel-cage winding, consisting of copper bars placed in deep narrow slots. At the beginning of the motor starting process, when the current frequency of the rotor winding is equal to the frequency of the supply network, the current flowing through the bar tends to concentrate in its upper part, since the upper part is linked with a smaller number of magnetic lines of force of the leakage flux than the lower part. An increase in current density in the upper part of the bar is equivalent, on the one hand, to an increase in the active resistance of the bar, and, on the other hand, to a decrease in its reactive resistance. An increase in the active resistance of the rotor bar leads to an increase in the initial torque of the motor, while an increase in the reactive resistance of the rotor winding limits the magnitude of the starting current of the motor.

As the rotational speed of the rotor increases, the frequency of the secondary circuit decreases; the difference in the magnitudes of the reactive resistances of the upper and lower parts of the bar also decreases; the current is distributed over the cross-section of the bar more and more uniformly, and at normal rotational speed the motor becomes an ordinary squirrel-cage motor, with only the difference that the motor has an increased reactive resistance of the rotor winding and, consequently, a somewhat worsened $\cos \varphi$ and a reduced $M_{\max}$ in comparison with a squirrel-cage motor of normal design. The active resistance of the rotor may be represented as the sum

\[ r_2 = r_{\mathrm{e}} + K_r r_n, \tag{1} \]

where $r_{\mathrm{e}}$ is the active resistance of the part of the rotor winding located outside the slot iron, $r_n$ is the resistance of the part of the conductor lying in the slot, and $K_r$ is a coefficient taking into account the increase in active resistance. Similarly, for the reactive resistance of the rotor one may write

\[ x_2 = x_{\mathrm{e}} + K_x x_n, \tag{2} \]

where \(x_{п}\) is the reactance of the part of the winding lying in the slot, \(x_{в}\) is the reactance of the part of the winding located outside the slot, and \(K_x\) is a coefficient allowing for the decrease in reactance.

Fig. 1

The values of \(K_r\) and \(K_x\) may be determined from the following approximate formulas:

\[ K_r = h\sqrt{S}. \tag{3} \]

Here \(h\) is the height of the rotor bar in cm,

\[ K_x = \frac{3}{2h\sqrt{S}} . \tag{4} \]

As follows from Fig. 1, at small values of \(S\), \(K_r\) and \(K_x\) remain practically constant. For \(h\sqrt{S}<1\), \(K_r\) and \(K_x\) should be taken equal to 1. Consequently, the torque of the motor, when \(K_r\) and \(K_x\) are greater than 1, can be expressed as

\[ M= -\frac{ m u_1^2\left(h r'_{п}\sqrt{S}+r'_{в}\right) }{ \omega_1 S \left[ \left( r_1+\frac{h r'_{п}}{\sqrt{S}}+\frac{r'_{в}}{S} \right)^2 + \left( x_1+x'_{в}+\frac{x'_{п}\,3}{2h\sqrt{S}} \right)^2 \right] }. \tag{5} \]

and when \(K_r\) and \(K_x\) are equal to 1,

\[ M= -\frac{ m u_1^2 r'_2 }{ \omega_1 S \left[ \left( r_1+\frac{r'_2}{S} \right)^2+x_k^2 \right] }. \tag{6} \]

Here \(r'_2\) is the referred active resistance of the rotor winding, and \(x_k\) is the reactance of the motor at short circuit. When the motor is accelerated at no load, when \(K_r\) and \(K_x\) are greater than 1, the equation of the moments acting on the motor shaft is

\[ -\frac{ m u_1^2\left(h r'_{п}\sqrt{S}+r'_{в}\right) }{ \omega_1 S \left[ \left( r_1+\frac{h r'_{п}}{\sqrt{S}}+\frac{r'_{в}}{S} \right)^2 + \left( x_1+x'_{в}+\frac{3x'_{п}}{2h\sqrt{S}} \right)^2 \right] } = -J\omega_1\frac{dS}{dt} \]

and when \(K_r\) and \(K_x\) are equal to 1,

\[ \frac{ m u_1^2 r'_2 }{ \omega_1 S \left[ \left( r_1+\frac{r'_2}{S} \right)^2+x_k^2 \right] } = -J\omega_1\frac{dS}{dt}. \]

Consequently, the starting duration is

\[ t=-\frac{J\omega_1^2}{m u_1^2} \left\{ \int_{0}^{S_1} \frac{ \left[ \left( r_1+\frac{h r'_n}{\sqrt{S}}+\frac{r'_{\mathrm{в}}}{S} \right)^2+ \left( x_1+x'_{\mathrm{в}}+\frac{3x'_n}{2h\sqrt{S}} \right)^2 \right]S\,dS }{ h r'_n\sqrt{S}+r'_{\mathrm{в}} } + \int_{S_1}^{S_2} \frac{ \left[ \left(r_1+\frac{r'_2}{S}\right)^2+x_k^2 \right]S\,dS }{ r'_2 } \right\}. \tag{7} \]

Here \(S_1\) is the slip at which \(h\sqrt{S}=1\), and \(S_2\) is the slip up to which the motor is accelerated.

The solution of integral (7) has the following form:

\[ \begin{aligned} t=\frac{2J\omega_1^2}{m u_1^2}\Bigg[ &A\left(S_1^{3/2}-1\right) +B\left(S_1-1\right) +C\left(\sqrt{S}-1\right) \\ &+D\ln \frac{ \sqrt{S_1}+\dfrac{r'_{\mathrm{в}}}{h r'_n} }{ 1+\dfrac{r'_{\mathrm{в}}}{h r'_n} } +r'_{\mathrm{в}}\ln\left(\sqrt{S_1}-1\right) \\ &+\frac{r_1^2+x_k^2}{4r'_2}\left(S_2^2-S_1^2\right) +r_1\left(S_2-S_1\right) +\frac{r'_2}{2}\ln\frac{S_1}{S_2} \Bigg]. \tag{8} \end{aligned} \]

Here

\[ A=\frac{r_1^2+x_1^2}{3h r'_n}; \qquad B=r_1+\frac{x_1+x'_n}{h^2 r'_n} -r'_{\mathrm{в}}\frac{r_1^2+x_1^2}{2h^2 r_n'^2}; \]

\[ C=\frac{r_{\mathrm{в}}'^2\left(r_1^2+x_1^2\right)}{h^3 r_n'^3} -r'_n\left(\frac{2r_1}{h r'_n}+\frac{3x_1x'_n}{h^3 r_n'^2}\right) +\frac{2r'_{\mathrm{в}}r_1}{h r'_n} +h r'_n +\frac{9x_n'^2}{4h^3 r'_n}; \]

\[ D=r_{\mathrm{в}}'^2\left(\frac{2r_1}{h^2 r_n'^2} +\frac{3x_n'^2}{h^4 r_n'^3}\right) -\frac{r_{\mathrm{в}}'^3\left(r_1^2+x_1^2\right)}{h^4 r_n'^4} -r'_{\mathrm{в}}\left(\frac{2r'_{\mathrm{в}}r_1}{h^2 r_{\mathrm{в}}'^2} +1+\frac{9x_n'^2}{4h^4 r_n'^2}\right). \]

Braking of the motor by the plugging method

The analytical expression for the dependence of speed on time during braking by the plugging method differs from equality (7) only in the limits of integration. Consequently, the duration of reversal can be expressed as

\[ t_{\mathrm{п}}=-\frac{J\omega_1^2}{m u_1^2} \left\{ \int_{2}^{S_1} \frac{ \left[ \left( r_1+\frac{h r'_n}{\sqrt{S}}+\frac{r'_{\mathrm{в}}}{S} \right)^2+ \left( x_1+x'_{\mathrm{в}}+\frac{3x'_n}{2h\sqrt{S}} \right)^2 \right]S\,dS }{ h r'_n\sqrt{S}+r'_{\mathrm{в}} } + \int_{S_1}^{S_2} \frac{ \left[ \left(r_1+\frac{r'_2}{S}\right)^2+x_k^2 \right]S\,dS }{ r'_2 } \right\}. \tag{9} \]

Integration gives

\[ t_n=\frac{2J\omega_1^2}{m u_1^2} \left[ B(S_1-2)+C(\sqrt{S}-\sqrt{2})+ D\ln \frac{\sqrt{S_1}+\dfrac{r'_e}{h r'_n}} {\sqrt{2}+\dfrac{r'_e}{h r'_n}} \right. \]

\[ \left. {}+r'_e\ln(\sqrt{S}-\sqrt{2}) +\frac{r_1^2+x_k^2}{4r'_e}\left(S_2^2-S_1^2\right) +r_1(S_2-S_1) +\frac{r'_2}{2}\ln\frac{S_1}{S_2} \right]. \tag{10} \]

Dynamic braking

The rotor losses during dynamic braking may be expressed as

\[ m I_2^2\left(r_e+r_n h\sqrt{S}\right) = \frac{ m\varepsilon^2 I_1^2\left(r_e+r_n h\sqrt{S}\right) }{ S\left[ \left(\dfrac{r_e+r_n h\sqrt{S}}{S}\right)^2 + \left(x_e+\varepsilon+\dfrac{3x_n}{2h\sqrt{S}}\right)^2 \right] }. \tag{11} \]

Here \(\varepsilon=\dfrac{U}{I_0}\), \(I_0\) is the no-load current, and \(I_1\) is the stator current. The torque developed by the motor is

\[ M= -\frac{ m\varepsilon^2 I_1^2\left(r_e+r_n h\sqrt{S}\right) }{ \omega_1 S\left[ \left(\dfrac{r_e+r_n h\sqrt{S}}{S}\right)^2 + \left(x_e+\varepsilon+\dfrac{3x_0}{2h\sqrt{S}}\right)^2 \right] }. \tag{12} \]

The equation of the torques acting on the motor shaft during no-load braking is

\[ -\frac{ m\varepsilon^2 I_1^2\left(r_e+r_n h\sqrt{S}\right) }{ \omega_1 S\left[ \left(\dfrac{r_e+r_n h\sqrt{S}}{S}\right)^2 + \left(x_e+\varepsilon+\dfrac{3x_0}{2h\sqrt{S}}\right)^2 \right] } = J\omega_1\frac{dS}{dt}. \]

Whence the braking time is

\[ t_b= \frac{2J\omega_1^2}{m\varepsilon^2 I_1^2} \left[ A(2S_1-1) + B\left(\sqrt{S_1}-\sqrt{1-S_1}\right) + C\ln \frac{\sqrt{S_1}-\dfrac{r_e}{h r_n}} {\sqrt{1-S_1}+\dfrac{r_e}{h r_n}} \right. \]

\[ \left. {}+ r_e\ln\left(\sqrt{S_1}-\sqrt{1-S_1}\right) + \frac{x_2}{4r_2}\left(S_2^2-S_1^2\right) + \frac{r_2}{2}\ln\frac{S_1}{S_2} \right]. \tag{13} \]

Here

\[ A=\frac{x_n}{h^2 r_n}-r_e; \qquad B=r_n\left(h r_n+\frac{9x_n^2}{4h^3 r_n^2}\right); \]

\[ D=r_e^2\left(\frac{3x_n^2}{h^4 r_n^3}\right) + \left(1-\frac{9x_n^2}{4h^4 r_n^2}\right)r_e. \]

Figures 2 and 3 show oscillograms of the stator current for run-up and braking by the counter-switching method. According to the oscillogram (Fig. 2), the starting time is 0.9 sec; by formula (8), 0.96 sec. According to the oscillogram (Fig. 3), the duration of reversing is 1.65 sec; by formula (10), 1.72 sec. The current oscillograms were taken for a motor of 5 kW, 1440 rpm,

Fig. 2

Fig. 3

380 V, 10.5 A, winding connection Δ/Δ, \(Z_k = 3.5\,\Omega\), \(X_k = 2.8\,\Omega\), \(r_k = 2.11\,\Omega\); \(h = 3.5\,\text{cm}\Omega\); \(r'_в = 0.16\,\Omega\); \(X_в = 0.72\,\Omega\).

B. Transient operating modes of a motor with a Boucherot rotor

For various special cases in which, when starting from rest, the resisting torque is sufficiently high and remains almost constant throughout the starting period, an electric motor with a Boucherot rotor is used. These include mechanisms with large static and dynamic resisting torques (crushers, conveyors, reciprocating compressors, ball mills, centrifuges, and others). In considering transient regimes it was assumed that the number of bars in the starting \(A\) and working \(B\) cages is the same, that the bars are located in the slots one above the other (Fig. 4), and that they are connected to their own rings.

By analogy with the expression for the torque of a machine of normal design, for a motor with a Boucherot rotor we may write

\[ M = \frac{m u_1^2 r_{2f}} {\omega_1 S \left[\left(r_1 + \frac{r_{2f}}{S}\right)^2 + \left(x_1 + x_{2f}\right)^2\right]} . \tag{14} \]

Here \(U_1\) is the primary voltage, \(\omega_1\) is the angular frequency of the stator, \(r_1\) is the active resistance of the primary winding, \(x_1\) is the reactance of the per-

winding, \(S\) is the slip, \(r_2\) is the active resistance of the rotor winding, \(x_{2f}\) is its reactance.

\[ r_{2f}=\frac{r_B(r_A+r_B)+S^2x_B^2}{(r_A+r_B)^2+S^2x_B^2}\cdot r_A \tag{15} \]

\[ x_{2f}=\frac{r_A^2x_B}{(r_A+r_B)^2+S^2x_B^2} \tag{16} \]

Substituting in equality (14) the values of \(r_{2f}\) and \(x_{2f}\) from equalities (15 and 16), we shall have

\[ M=\frac{mu_1^2}{\omega_1 S} \left[ \frac{x_1^2+r_1^2}{r_A}S+2r_1+ \frac{(Ar_B+S^2x_B^2)r_A(A^2+S^2x_B^2)S}{B_1S^4+C_1S^2+D_1} \right]. \]

Here \(A=r_A+r_B\); \(B_1, C_1, D_1\) are coefficients.

It is seen from Fig. 4 that the current flowing in the internal bar \(A\) creates a leakage flux \(\Phi_B\) linked with bar \(B\). This drawing also shows the leakage flux \(\Phi_{AB}\), embracing both bars \(A\) and \(B\). The flux \(\Phi_{AB}\) is created by the geometric sum of the currents flowing in the bars of the upper and lower cages.

Fig. 4

Fig. 5

If the resistance \(X_{AB}\), which corresponds to the flux \(\Phi_{AB}\), and also the reactance of cage \(A\), as very small quantities, are not taken into account, then the equivalent circuit for the Boucherot rotor may be represented in the following form (Fig. 5), where \(r_A\) is the resistance of cage \(A\) to direct current; \(r_B\) is the resistance of cage \(B\) to direct current; \(X_B\) is the reactance of cage \(B\) at a frequency of \(50\,Hz\). Substituting the expression for \(M\) into the equation of the moments acting on the motor shaft during no-load acceleration, we shall have the following expression for the starting duration

\[ t=\frac{J\omega_1^2}{mu_1^2} \left[ \int_{1}^{S_1}\frac{r_A}{(x_1^2+r_1^2)S}\,dS +\int_{2}^{S_1}\frac{dS}{2r_1} + \int_{1}^{S_1} \frac{(B_1S^4+C_1S^2+D_1)dS}{(Ar_B+S^2x_B^2)r_A(A^2+S^2x_B^2)S} \right]. \tag{17} \]

Integration gives:

\[ t=\frac{J\omega_1^2}{m u_1^2}\left\{ \frac{1-S}{2r_1} +\frac{r_A}{x_1^2+r_1^2}\ln\frac{1}{S_1} +B\ln\frac{1}{S_1} +\frac{C}{2x_B^2 r_A}\ln \frac{1+\dfrac{A r_B}{x_B^2}} {S_1^2+\dfrac{A r_B}{x_B^2}} +\frac{E}{2x_B^2}\ln \frac{1+\dfrac{A r_B}{x_B^2}} {S_1+\dfrac{A r_B}{x_B^2}} \right\}. \tag{18} \]

Here

\[ B=\frac{r_B-r_A}{A}; \qquad E=-\frac{x_1^2+r_1^2}{r_A}\,2A^2 -\frac{x_B^2 r_A}{A+r_B} \left(3r_B+\frac{A^3}{r_A^2}+A\right) + \]

\[ +x_1^2 r_A\left(\frac{r_B}{A}+\frac{2A}{A+r_B}\right) +\frac{2A}{A+r_B} \left(\frac{r_1 A^3}{r_A}-x_1 r_A x_B+x r_A^2 r_B\right). \]

\[ C=(x_1^2+r_1^2)(3A r_B+A^2) +\frac{x^2 b r_A r_B}{A} \left( 1-\frac{2r_B}{A} +\frac{3r_B r_A^2+A^3-A r_A}{r_A^2(A+r_B)} \right) - \]

\[ -\frac{r_A r_B\,2A}{A(A+r_B)} \left( \frac{r_1^2 A^2}{r_B} -x_1 r_A x_B +x r_A^2 r_B \right) +A x_B^2+2x r_A^2 x_B . \]

If the integration in equality (17) is carried out from \(S_1\) to 1, then we obtain the braking duration by the plugging method.

Dynamic braking

In dynamic braking, after being disconnected from the supply network, the motor stator is connected to a direct-current network; as a result, a magnetic field stationary in space is formed in the motor. This field induces alternating current in the conductors of the rotor rotating by inertia. Thus the motor is converted into a synchronous alternating-current generator, producing a braking torque. In Fig. 6

Fig. 6

the principal circuits for connecting the stator windings to a direct-current network are given, and the currents flowing through the individual phases are indicated. As with alternating current, the magnetomotive forces of the individual phases are displaced from one another in space by 120 electrical degrees; therefore the resultant MMF can be determined by geometrical addition of the MMFs of the individual phases.

Analogously to normal operation, for the rotor current we may write—

\[ I'_2=\frac{E'_2 S}{\sqrt{(r'_{2f})^2+(x'_{2f}S)^2}} . \tag{19} \]

If saturation is neglected, one may assume

\[ E'_2=\varepsilon I_0 \]

Here \(I_0\) is the magnetizing current, \(\varepsilon\) is the proportionality coefficient

\[ \varepsilon=\frac{E'_2}{I_0}=\frac{E_1}{I_0}. \]

Here \(E_1\) is the stator EMF. Substituting \(E'_2\) in formula (19), we obtain

\[ I'_2=\frac{\varepsilon I_0 S}{\sqrt{(r'_{2f})^2+(x'_{2f}S)^2}}. \tag{20} \]

Substituting the value of \(I'_2\) into the equality of the torques acting on the motor shaft under dynamic braking, we obtain

\[ M_m=\frac{m\varepsilon^2 I_0^2 r'_{2f}} {\omega_1 S\left[\left(\frac{r'_{2f}}{S}\right)^2+x_{2f}^2\right]}. \tag{21} \]

Replacing \(I_0\) by its value from the current vector diagram, we obtain

\[ M_m=\frac{m\varepsilon^2 I_1^2 r'_{2f}} {\omega_1 S\left(\frac{r'_{2f}}{S}\right)^2+\left(x'_{2f}+\varepsilon\right)^2}. \tag{22} \]

Replacing \(r'_{2f}\) and \(x'_{2f}\) by their values from equations 14 and 15, we obtain—

\[ M_m=\frac{m\varepsilon^2 I_1^2\left[(r_BA+Sx_B^2)r_A(A^2+S^2x_B^2)\right]S} {\omega_1\left\{[r_BA+S^2x_B^2]r_A+\left[r_A^2x_B+\varepsilon(A^2+S^2x_B^2)\right]^2S^2\right\}}. \tag{23} \]

From the equation of the torques acting on the motor shaft during braking of an unloaded motor, we find the braking duration—

\[ t_m=-\frac{J\omega_1^2}{m\varepsilon I_1^2} \left[ \int_1^{S_1} S\frac{\varepsilon^2}{r_A}\,dS+ \int \frac{(B_1S^4+C_1S^2+D_1)\,ds} {(r_BA+S^2x_B^2)r_A(A^2+S^2x_B^2)S} \right]. \]

Here \(B_1\), \(C_1\), and \(D_1\) are coefficients. Integration gives

\[ t_m=-\frac{J\omega_1^2}{m\varepsilon I^2} \left[ \frac{\varepsilon^2}{2r_A}\left(1-S_1^2\right) +B\ln\frac{1}{S_1} +\frac{C}{2x_B^2}\ln\frac{1}{S_1} \right]. \tag{24} \]

Here

\[ B=\frac{r_Ar_B}{A}; \qquad C=x_B^2r_A^2\left(2\varepsilon+\frac{x_B}{A}\right)+\varepsilon^2\left(r_A^2+r_Ar_B\right). \]