STARTING A SQUIRREL-CAGE MOTOR OF STANDARD DESIGN

I. G. Kuleev

Docent, Candidate of Technical Sciences

The acceleration process of a motor is determined by the differential equation of the moments acting on the motor shaft during starting,

\[ M-(M_c+M_0)=J\frac{d\omega_2}{dt} \tag{1} \]

Here \(M\) is the torque developed by the squirrel-cage motor,
\(M_c\) is the useful load torque on the shaft,
\(M_0\) is the resisting torque due to the mechanical losses of the motor,
\(J\) is the moment of inertia of the rotating masses,
\(\omega_2\) is the angular velocity of the rotor,
\(t\) is time.

In investigating the acceleration process of a motor, the torque \(M_0\), because of its small magnitude, is usually neglected. Consequently, equation (1) may be rewritten in the form

\[ M-M_c=J\frac{d\omega_2}{dt}, \tag{2} \]

Below we shall consider the starting of an unloaded motor and its starting under load for the most frequently encountered load torques

\[ M_c=\mathrm{const}\quad \text{and}\quad M_c=k'\omega_2^2 \]

Acceleration of an Unloaded Motor

In this case \(M_c=0\). If \(M_0\) is neglected, then, according to (1),

\[ M=J\frac{d\omega_2}{dt}, \tag{2} \]

or, taking into account that \(\omega_2=\omega_1(1-s)\)

\[ M=-J\omega_1\frac{ds}{dt}, \tag{3} \]

where \(s\) is the slip, and \(\omega_1\) is the angular velocity of the field.

Substituting for \(M\) its value from Kloss’s formula, we obtain

\[ M_{\max}\frac{2+\beta S_m}{\dfrac{S}{S_m}+\dfrac{S_m}{S}+\beta S_m} = -J\omega_1\frac{ds}{dt}, \tag{4} \]

where

\[ \beta=\frac{2'r}{c_1 r_2'}, \]

\(r_1\) — active resistance of the stator winding,
\(r_2'\) — reduced active resistance of the rotor winding,

\[ C \simeq \frac{U_1}{E_{s0}}, \]

\(U_1\) — primary phase voltage,
\(E_{s0}\) — e.m.f. of the stator winding at synchronous rotation,
\(M_{\max}\) — maximum torque on the shaft,
\(S_m\) — slip corresponding to \(M_{\max}\).

From equation (4), the duration of acceleration of the motor is

\[ T_p=\frac{J\omega_1}{M_{\max}}\frac{1}{2+\beta S_m} \int_{1}^{S_1}\left[\frac{S}{S_m}+\frac{S_m}{S}+\beta S_m\right]\,dS \tag{5} \]

Here \(S_1\) is the slip up to which the acceleration of the motor continues. Integration gives

\[ T_p=\frac{J\omega_1}{M_{\max}}\frac{1}{2+\beta S_m} \left[ \frac{1}{2}\frac{1-S_1^2}{S_m}+(1-S_1)\beta S_m+S_m\ln\frac{1}{S_1} \right]. \tag{6} \]

Replacing in equation (6) \(T_p\) and \(S_1\) by the current coordinates \(t\) and \(S\), we obtain the acceleration equation for an unloaded motor

\[ t=\frac{J\omega_1}{M_{\max}}\frac{1}{2+\beta S_m} \left[ \frac{1}{2}\frac{1-S^2}{S_m}+(1-S)\beta S_m+S_m\ln\frac{1}{S} \right]. \tag{7} \]

If, in equation (7), we neglect in the numerator the term in which \(\beta S_m\) enters as a factor, and in the denominator the term \(\beta S_m\), because of their relative smallness, then the motor acceleration equation takes the following form

\[ t=\frac{J\omega_1}{2M_{\max}} \left[ \frac{1-S^2}{2S_m}+S_m\ln\frac{1}{S} \right]. \tag{8} \]

Acceleration of a loaded motor \(M_c=\mathrm{const}\)

Substituting into equation (1), instead of \(M\), its value from Kloss’s formula, we obtain

\[ M_{\max}\frac{2+\beta S_m}{\dfrac{S_m}{S}+\dfrac{S}{S_m}+\beta S_m} -M_c = -J\omega_1\frac{dS}{dt}, \tag{9} \]

whence the acceleration duration is

\[ T_p = -\int_{1}^{S_1} \frac{ J\omega_1\left(S_m^2+S^2+\beta S_m^2S\right)\,dS }{ M_{\max}(2+\beta S_m)S_mS - M_c\left(S_m^2+S^2+\beta S_m^2S\right) }. \tag{10} \]

Integration gives, for the current values \(t\) and \(S\),

\[ t=\frac{J\omega_1}{M_c}(S-1) +\frac{a}{2}\ln \frac{S^2+bS+S_m^2}{1-b+S_m^2} + \frac{ab}{\sqrt{4S_m^2-b^2}} \arctg \frac{(S-1)\sqrt{4S_m^2-b^2}} {2(S_m^2+S)-b(1+S)} \tag{11} \]

where

\[ \left. \begin{aligned} a&=J\omega_1\,\frac{M_{\max}S_m}{M_c^2}\left(\beta S_m+2\right),\\ b&=\beta S_m^2\left(\frac{M_{\max}}{M_c}-1\right)+\frac{2M_{\max}S_m}{M_c}. \end{aligned} \right\} \tag{12} \]

Assuming the active resistance of the stator winding \(r_1=0\), which is very often done when using Kloss’s formula, for \(a\) and \(b\), entering into equation (11), we find other values, namely, at \(\beta=\dfrac{2r_1}{c_1 r_2'}=0\),

\[ a=\frac{J\omega_1 M_{\max}S_m(\beta S_m+2)}{M_c^2} =\frac{2M_{\max}S_m J\omega_1}{M_c^2}, \tag{13} \]

\[ b=\beta S_m^2\left(\frac{M_{\max}}{M_c}-1\right) +\frac{2M_{\max}}{M_c}S_m =\frac{2M_{\max}S_m}{M_c^2}. \tag{14} \]

Operation of a Loaded Motor \(M_c=k'\omega_2^2\)

Expressing the angular velocity of the working machine \(\omega_2\) through the slip, we obtain the following expression for the torque of the working machine:

\[ M_c=k'\omega_2^2=k'\omega_1^2(1-S)^2. \tag{15} \]

Here: \(k'\) is the constant of the working machine,
\(\omega_1\) is the rotational speed of the field.

Combining the constant quantities, we obtain

\[ M_c=k(1-2S+S^2), \tag{16} \]

where

\[ k=k'\omega_1^2. \tag{17} \]

Substituting equation (16) into equation (9), we obtain the equation of the torques acting on the motor shaft,

\[ M_{\max}\, \frac{2+\beta S_m} {\dfrac{S}{S_m}+\dfrac{S_m}{S}+\beta S_m} = k\left(1-2S+S^2\right)-J\omega_1\frac{dS}{dt}. \tag{18} \]

Clearing the fractions, we shall have

\[ M_{\max}(2+\beta S_m)S_mS -k(1-2S+S^2)\left(S^2+\beta S_m^2S+S_m^2\right) = -J\omega_1\left(S^2+\beta S_m^2S+S_m^2\right)\frac{dS}{dt}, \]

whence the duration of motor acceleration is

\[ T_n=J\omega_1 \int_{1}^{S_1} \frac{\left(S^2+\beta S_m^2S+S_m^2\right)dS} {k(1-2S+S^2)(S^2+\beta S_m^2S+S_m)-(2+\beta S_m)M_{\max}S_mS}. \]

After multiplying out, collecting like terms, and taking \(k\) outside the integral sign, we shall have

\[ T_n=\frac{J\omega_1}{k} \int_{1}^{S_1} \frac{\left(S^2+\beta S_m^2S+S_m^2\right)dS} {S^4+S^3\left(\beta S_m^2-2\right) +S^2\left(1+S_m^2-2\beta S_m^2\right) +S\left[3S_m^2-2S_m^2-\frac{(2+\beta S_m)M_{\max}S_m}{k}\right]} \tag{19} \]

For the purpose of shortening the notation, let us introduce the following designations:

\[ \beta S_m^2 - 2 = a_1;\qquad 1 + S_m^2 - 2\beta S_m^2 = b_1; \]

\[ \beta S_m^2 - 2S_m^2 - \frac{(2+\beta S_m)M_{\max}S_m}{k}=c_1 . \]

Then equation (19) will take the following form:

\[ T_{\mathrm{p}}=\frac{J\omega_1}{k}\int \frac{(S^2+\beta S_m^2S+S_m^2)\,dS} {S^4+a_1S^3+b_1S^2+c_1S+S_m^2}. \tag{20} \]

The expression under the integral sign is a rational fraction, whose integration is possible when the denominator is decomposed into simple factors. We carry out the decomposition by Ferrari’s method.

Putting

\[ S=x-\frac{a_1}{4} \]

we obtain

\[ S^4+a_1S^3+b_1S^2+c_1S+S_m^2 = x^4+px^2+qx+r, \tag{21} \]

where

\[ p=-\frac{3}{8}a_1^2+b_1;\qquad q=\frac{a_1^3}{8}-\frac{b_1a_1}{2}+c_1. \]

\[ r=-\frac{3}{256}a_1^4+\frac{b_1a_1^2}{16}-\frac{c_1a_1}{4}+S_m^2. \]

To find the roots of the right-hand side of equation (21), it is first necessary to solve the following cubic equation:

\[ y^3+2py+(p^2-4r)y-q^2=0 \tag{22} \]

Putting in it \(y=z-\frac{2p}{3}\), we obtain the reduced cubic equation of the form

\[ z^3+p'z+q'=0, \tag{23} \]

where

\[ p'=-\frac{p^2}{3}-4r, \]

\[ q'=\frac{8}{3}pr-q^2-\frac{2}{27}p^3, \]

The roots of equation (23) are determined by Cardano’s formulas

\[ Z_1=U+V;\qquad Z_2=W_1U+W_2V;\qquad Z_3=W_2U+W_1V \tag{25} \]

where

\[ W_1=\frac{-1+j\sqrt{3}}{2};\qquad W_2=\frac{-1-j\sqrt{3}}{2} \]

\[ u^3=-\frac{q'}{2}+\sqrt{\frac{q'^2}{4}+\frac{p'^3}{27}}; \qquad v^3=-\frac{q'}{2}-\sqrt{\frac{q^2}{4}+\frac{p'}{27}}. \]

The roots of the reduced fourth-degree equation (21) are found from the formulas

\[ \begin{aligned} x_1&=\frac{1}{2}\left(\sqrt{Z'_1}+\sqrt{Z'_2}+\sqrt{Z'_3}\right),\\ x_2&=\frac{1}{2}\left(\sqrt{Z'_1}-\sqrt{Z'_2}-\sqrt{Z'_3}\right),\\ x_3&=\frac{1}{2}\left(-\sqrt{Z'_1}+\sqrt{Z'_2}-\sqrt{Z'_3}\right),\\ x_4&=\frac{1}{2}\left(-\sqrt{Z'_1}-\sqrt{Z'_2}+\sqrt{Z'_3}\right). \end{aligned} \tag{25 a--d} \]

Substituting the values \(Z'_1\), \(Z'_2\), and \(Z'_3\), we find

\[ x_1=\frac{1}{2}\left( \sqrt{u'+v} + \sqrt{\frac{-u'+j\sqrt{3}u}{2}+\frac{-v-j\sqrt{3}v}{2}} + \sqrt{\frac{-u'-j\sqrt{3}u}{2}+\frac{-v+j\sqrt{3}v}{2}} \right). \tag{26} \]

Applying the formula for extracting the square root of a complex number, we obtain for the second term of equation (26)

\[ \sqrt{-\frac{u'+v}{2}+\frac{\sqrt{-3(u-v)^2}}{2}} = \sqrt{ -\frac{u'+v}{2\cdot 2} +\frac{1}{2} \sqrt{ \left(\frac{u+v}{2}\right)^2+\frac{3}{4}(u-v)^2 } } + \sqrt{ -\frac{u'+v}{4} -\frac{1}{2} \sqrt{ \left(\frac{u+v}{2}\right)^2+\frac{3}{4}(u-v)^2 } }. \]

Expanding the parentheses, we shall have

\[ \sqrt{-\frac{u'+v}{2}+\frac{\sqrt{-3(u-v)^2}}{2}} = \sqrt{ -\frac{u'+v}{4} +\frac{1}{2}\sqrt{u_1^2-u_1v+v^2} } + \sqrt{ -\frac{u'+v}{4} -\frac{1}{2}\sqrt{u_1^2-u_1v+v^2} }. \tag{27} \]

Introducing the notation

\[ \left. \begin{aligned} \sqrt{ -\frac{u'+v}{4} +\frac{1}{2}\sqrt{u_1^2-u_1v+v^2} } &=d,\\ \sqrt{ -\frac{u'+v}{4} -\frac{1}{2}\sqrt{u_1^2-u_1v+v^2} } &=je, \end{aligned} \right\} \tag{28} \]

we obtain

\[ \sqrt{Z'_2} = \sqrt{ -\frac{u'+v}{2} +\frac{\sqrt{-3(u-v)^2}}{2} } =d+je. \tag{29} \]

Applying the formula for extracting the square root of a complex number to the third term of equation (25a), we find

\[ \sqrt{Z'_3} = \sqrt{ -\frac{u'+v}{2} -\frac{j\sqrt{3(u-v)^2}}{2} } =d-je. \tag{30} \]

Substituting equations (29 and 30) into equation (25a), we obtain:

\[ x_1=\frac{1}{2}\left(\sqrt{u'+v}+2d\right)=\frac{f}{2}+d. \tag{31} \]

Substituting the values \(Z'_1\), \(Z'_2\), \(Z'_3\) into equations (25b—d), we obtain:

\[ \left. \begin{aligned} x_2&=\frac{1}{2}\left(\sqrt{u'+v}-d-je-d+je\right)=\frac{f}{2}-d,\\ x_3&=\frac{1}{2}\left(-\sqrt{u'+v}+d+je-d+je\right)=-\frac{f}{2}+je,\\ x_4&=\frac{1}{2}\left(-\sqrt{u'+v}-d-je+d-je\right)=-\frac{f}{2}-je. \end{aligned} \right\} \tag{32} \]

Here

\[ f=\sqrt{u'+v} \tag{33} \]

Substituting the values \(x_1, x_2, x_3\) and \(x_4\) into equation (20), we obtain

\[ T_{\mathrm{п}}=\frac{J\omega_1}{k}\int_{1}^{s_1} \frac{(s^2+\beta s_m^2s+s_m^2)\,ds} {\left(s-\frac{f_1}{2}-d\right)\left(s-\frac{f_1}{2}+d\right)\left(s+\frac{f_1}{2}-je\right)\left(s+\frac{f_1}{2}+je\right)} . \]

After multiplying the last two factors of the denominator, we obtain

\[ T_{\mathrm{п}}=\frac{J\omega_1}{k}\int_{1}^{s_1} \frac{(s^2+\beta s_m^2s+s_m^2)\,ds} {\left(s-\frac{f_1}{2}-d\right)\left(s-\frac{f_1}{2}+d\right)\left(s^2+f_1s+\frac{f_1^2}{4}+e^2\right)} . \tag{34} \]

The integrand is a rational fraction, the denominator of which contains factors of the second degree, but not repeated. Let

\[ \frac{s^2+\beta s_m^2s+s_m^2} {\left(s-\frac{f_1}{2}-d\right)\left(s-\frac{f_1}{2}+d\right)\left(s^2+sf_1+\frac{f_1^2}{4}+e^2\right)} = \frac{A}{s-\frac{f_1}{2}-d} + \frac{B}{s-\frac{f_1}{2}+d} + \frac{Cs+D}{s^2+f_1s+\frac{f_1^2}{4}+e^2}. \tag{35} \]

Then

\[ \begin{aligned} T_{\mathrm{п}} &=\frac{J\omega_1}{k}\int_{1}^{s_1} \frac{(s^2+\beta s_m^2s+s_m^2)\,ds} {\left(s-\frac{f_1}{2}-d\right)\left(s-\frac{f_1}{2}+d\right)\left(s^2+sf_1+\frac{f_1^2}{4}+e^2\right)} \\ &=\frac{J\omega_1}{k}\left[ \int_{1}^{s_1}\frac{A\,ds}{s-\frac{f_1}{2}-d} + \int_{1}^{s_1}\frac{B\,ds}{s-\frac{f_1}{2}+d} + \int_{1}^{s_1}\frac{(Cs+D)\,ds}{s^2+f_1s+\frac{f_1^2}{4}+e^2} \right]. \end{aligned} \tag{36} \]

Here \(A, B, C\), and \(D\) are constants to be determined.
Integrating the right-hand side of equation (36) gives

\[ T_{\mathrm{п}}=\frac{J\omega_1}{k}\left[ A\ln \frac{s_1-\frac{f_1}{2}-d}{1-\frac{f_1}{2}-d} + B\ln \frac{s_1-\frac{f_1}{2}+d}{1-\frac{f_1}{2}+d} + \frac{c}{2}\ln \frac{s^2+f_1s+\frac{f_1^2}{4}+e^2} {1+f_1+\frac{f_1^2}{4}+e^2} + \frac{2D-cf_1}{2e}\operatorname{arctg} \frac{(s_1-1)4e}{4e^2+2s(2+f_1)+2f_1+f_1^2} \right]. \tag{37} \]

Replacing \(T_{\mathrm{п}}\) and \(s_1\) by the current coordinates \(t\) and \(s\), we obtain the equation for the motor run-up

\[ t=\frac{J\omega_1}{k}\left[ A\ln \frac{s-\frac{f_1}{2}-d}{1-\frac{f_1}{2}-d} + B\ln \frac{s-\frac{f_1}{2}+d}{1-\frac{f_1}{2}+d} + \frac{c}{2}\ln \frac{s^2+f_1s+\frac{f_1^2}{4}+e^2} {1+f_1+\frac{f_1^2}{4}+e^2} + \frac{2D-cf_1}{2e}\operatorname{arctg} \frac{(s-1)4e}{4e^2+2s(2+f_1)+2f+f_1^2} \right. \]

Assuming the active resistance of the stator winding \(r_1=0\) for the coefficients \(b_1\) and \(c_1\) entering equation (20), we shall find other values, namely, with

\[ \beta=\frac{2r_1}{c_1r_2}=0 \]

\[ \left. \begin{aligned} a_1&=\frac{\beta s_m^2-2}{4}=-\frac{1}{2};\\ b_1&=1+s_m^2-2\beta s_m^2=1+s_m^2;\\ c_1&=\beta s_m^2-2S_m^2-\frac{(2+\beta s_m)M_mS_m}{k} =-2s_m^2-\frac{2M_mS_m}{k}. \end{aligned} \right\} \tag{38} \]

The coefficients \(A, B, C, D\) entering equations (36 and 37) may be determined as follows.

Clearing the fractions in equation (35), we obtain

\[ s^2+\beta S_m^2S+s_m^2 =A\left(s-\frac{f_1}{2}+d\right) \left(s^2+f_1s+\frac{f_1^2}{4}+e^2\right)+ \]

\[ +B\left(s-\frac{f_1}{2}-d\right) \left(s^2+f_1s+\frac{f_1^2}{4}+e^2\right)+ \]

\[ +(cs+D)\left(s-\frac{f_1}{2}-d\right) \left(s-\frac{f_1}{2}+d\right). \tag{39} \]

After multiplication and collection of like terms we shall have

\[ s^2+\beta S_m^2S+S_m^2 =S^3(A+B+C)+S^2\left(Ad+\frac{Af_1}{2}+\frac{Bf_1}{2}-Bd-cf_1+D\right)+ \]

\[ +S\left(Adf_1-\frac{Bf_1^2}{4}-Ae^2-Bdf_1+Be^2+\frac{cf_1^2}{4}-cd^2-Df_1\right) -\frac{f_1^2}{8}(A+B)+ \]

\[ +\frac{df_1^2}{4}(A-B)-\frac{f_1e^2}{2}(A+B)+de^2(A-B) +D\left(\frac{f_1^2}{4}-d^2\right) \]

Equating coefficients of like powers, we obtain

\[ \left. \begin{aligned} A+B+C&=0\\ Ad+\frac{Af_1}{2}+\frac{Bf_1}{2}-Bd-cf_1+D&=1\\ Adf_1-\frac{Af_1^2}{4}-\frac{Bf_1^2}{4}+Ae^2-Bdf_1+Be^2+cd^2-Df_1&=\beta s_m^2\\ -\frac{f_1^3}{8}(A+B)+\frac{df_1^2}{4}(A-B)-\frac{fe^2}{2}(A+B)+de^2(A-B) +D\left(\frac{f_1^2}{4}-d^2\right)&=S_m^2 \end{aligned} \right\} \tag{40a--d} \]

From equations (40a and 40b) we find

\[ \left. \begin{aligned} C&=-B-A\\ D&=1-A\left(d+\frac{f_1}{2}\right)+B\left(d-\frac{f_1}{2}\right)+cf_1 \end{aligned} \right\} \tag{41} \]

Substituting into equation (39)

\[ S=\frac{f_1}{2}+d, \]

we find

\[ B=-\frac{f_1^2-4f_1d+4d^2+2\beta S_m^2f_1-4\beta S_m^2d+4S_m^2} {8d\left(f_1^2+d^2-2f_1d+e^2\right)}. \tag{42} \]

Substituting \(S=-\dfrac{f_1}{2}+d\) into equation (39), we obtain

\[ A= \frac{ f_1^2+4f_1d+4d^2+4\beta S_m^2 d+2\beta S_m^2+4S_m^2 }{ 8d\left(f_1^2+2f_1d+d^2+e^2\right) } \tag{43} \]

Assuming the active resistance of the stator winding \(r_1=0\), we obtain: for

\[ \beta=\frac{2r_1}{c_1 r'_2}=0 \]

Fig. 1

Fig. 2

Fig. 3

\[ \left. \begin{aligned} A&=-\frac{f_1^2+4f_1d+4d^2+4S_m^2} {8d\left(f_1^2+2f_1d+d^2+e^2\right)},\\[6pt] B&=-\frac{f_1^2-4f_1d+4d^2+4S_m^2} {8d\left(f_1^2+d^2-2f_1d+e^2\right)},\\[6pt] D&=B\left(d-\frac{f_1}{2}\right)-A\left(d+\frac{f_1}{2}\right)+cf_1+1 \end{aligned} \right\} \tag{44} \]

For convenience of comparison, the following table gives values obtained from the formulas and taken from the oscillograms (Figs. 1, 2, 3).

Table of values \(T_{\text{p}}\)

Calculations and oscillographic records were made for a motor having the following data:

\[ \begin{gathered} \text{Type T-750},\quad U = 220\,\mathrm{V},\quad E_2 = 145\,\mathrm{V},\quad I_1 = 38\,\mathrm{A},\quad n = 715\ \mathrm{rpm},\\ \cos\varphi = 0.8,\quad r_1 = 0.40\,\Omega,\quad r_2 = 0.08\,\Omega;\quad r_k = 1.1\,\Omega,\quad X_1 = 1.35\,\Omega,\\ X_2 = X_1,\quad Z_k = 2.92\,\Omega,\quad S_m = 0.213,\quad J = 0.05\ \mathrm{kg\cdot m\cdot s^2},\\ M = 11\ \mathrm{kg\cdot m},\quad M_{\max} = 30\ \mathrm{kg\cdot m},\quad M_c = 6.6\ \mathrm{kg\cdot m}. \end{gathered} \]