Dependence of Current on Time During Starting of Standard-Design Squirrel-Cage Motors

I. G. Kuleev
Associate Professor, Candidate of Technical Sciences

Dependence of Current on Time During No-Load Starting

In determining the dependence of current on time, we shall start from the equation of acceleration of the motor, which, for no-load starting, has the form

\[ M = -J\omega_1 \frac{ds}{dt}. \tag{1} \]

Substituting

\[ M = \frac{m_2 r_2 I_2^2}{\omega_1 s}, \]

we obtain

\[ \frac{m_2 r_2 I_2^2}{9{,}81\,\omega_1 s} = -J\omega_1 \frac{ds}{dt}. \tag{2} \]

The dependence of \(s\) on \(I_2\) may be found as follows. From the relation for the starting torque

\[ M_{\text{st}} = \frac{m_2 r_2 I_{2\text{st}}^2}{9{,}81\,\omega_1} \tag{3} \]

to the torque of rotation on the shaft, we find

\[ \frac{M_{\text{st}}}{M} = \frac{I_{2\text{st}}^2 S}{I_2^2}. \tag{4} \]

Here \(I_{2\text{st}}\) is the rotor current at starting.

Dividing the numerator and denominator of the left-hand side of equation (4) by \(M_{\max}\), we obtain

\[ \frac{M_{\text{st}}}{M} = \frac{\dfrac{M_{\text{st}}}{M_{\max}}} {\dfrac{M}{M_{\max}}}. \tag{5} \]

Replacing \(\dfrac{M_{\text{st}}}{M_{\max}}\) and \(\dfrac{M}{M_{\max}}\) by their values from Kloss’s equation, we obtain

\[ \frac{M_{\text{st}}}{M} = \frac{S_m^2 + S^2}{(S_m^2 + 1)S}. \tag{6} \]

Substituting the value \(-\dfrac{M_n}{M}\) into equation (4), we obtain

\[ \frac{I_{2n}^{2}S}{I_2^{2}}= -\frac{S_m^{2}+S^{2}}{(S_m^{2}+1)S}. \tag{7} \]

Equation (7) is a quadratic equation of the form

\[ S^{2}\left[I_{2n}^{2}(1+S_m^{2})-I_2^{2}-I_2^{2}S_m^{2}\right]=0, \tag{8} \]

whence

\[ S=\pm \frac{S_m I_2}{\sqrt{I_{2n}^{2}(1+S_m^{2})-I_2^{2}}}; \tag{9} \]

\[ ds=\pm \frac{S_m dI_2\sqrt{I_{2n}^{2}(1+S_m^{2})-I_2^{2}}} {I_{2n}^{2}(1+S_m^{2})-I_2^{2}} + \frac{S_m I_2^{2}dI_2} {\left[I_{2}^{2}(1+S_m^{2})-I_2^{2}\right]^{3/2}} \tag{10} \]

(\(+\) refers to motor operation, \(-\) to generator operation).

Reducing to a common denominator, we find

\[ ds= \frac{I_{2n}^{2}(S_m+S_m^{3})\,dI_2} {\left[I_{2n}^{2}(1+S_m^{2})-I_2^{2}\right]^{3/2}}. \tag{11} \]

Substituting the value of \(S\) from equation (9) and \(ds\) from equation (10) into equation (2), we find the expression for the dependence of current on time in the form

\[ t_1= \frac{9.81J\omega_1S_m^{2}}{m_2r_2} \int_{I_{2n}}^{I_2'} \frac{-I_{2n}^{2}(1+S_m^{2})\,dI_2} {I_2\left[I_{2n}^{2}(1+S_m)^{2}-I_2^{2}\right]^{2}}. \tag{12} \]

Integration gives

\[ t= -\frac{9.81J\omega_1^{2}S_m}{2m_2r_2} \left[ \frac{1}{\varepsilon}\ln\frac{S_m^{2}I_2^{2}}{\varepsilon-I_2^{2}} + \frac{I_2^{2}-I_{2n}^{2}}{I_{2n}^{2}S_m(\varepsilon-I_2^{2})} \right], \tag{13} \]

where

\[ \varepsilon=I_{2n}^{2}(1+S_m^{2}). \]

Dependence of current on time during starting under load \(M_c=\mathrm{const}\).

In this case the equation of the torques acting on the motor shaft will have the form

\[ M=\frac{m_2r_2I_2^{2}}{9.81\,\omega_1S} = M_c-J\omega_1\frac{ds}{dt}. \tag{14} \]

Replacing in equation (14) \(S\) and \(ds\) by their values from equations (9) and (10), we obtain

\[ -J\omega_1 \frac{S_m\varepsilon\,dI_2}{dt(\varepsilon-I_2^{2})^{3/2}} = \frac{m_2r_2I_2\sqrt{\varepsilon-I_2^{2}}}{9.81\,\omega_1S_m} - M_c, \tag{15} \]

whence the dependence of current on time is

\[ t= -J\omega_1^{2}S_m^{2}\varepsilon \int_{I_{2n}}^{I_2'} \frac{dI_2} {(\varepsilon-I_2^{2})^{3/2} \left( \dfrac{r_2I_2m_2\sqrt{\varepsilon-I_2^{2}}}{9.81} - M_c\omega_1S_m \right)} \tag{16} \]

Integration gives

\[ \begin{aligned} t = -\,\frac{J\omega_1 S_m}{M_c}\Bigg[ &-\,\frac{I'_2}{a_1a_2\sqrt{\varepsilon-I_2'^2}} +\frac{I_{2n}}{a_1a_2\sqrt{\varepsilon-I_{2n}^{\,2}}} +\frac{a_1+a_2}{a_1^2a_2^2}\ln \frac{I_{2n}\sqrt{\varepsilon-I_2'^2}}{I'_2\sqrt{\varepsilon-I_{2n}^{\,2}}} \\ &-\frac{a_1^2+1}{a_1^2(a_2-a_1)} \ln \frac{I_{2n}\left(\sqrt{\varepsilon-I_2'^2}-I'_2a_1\right)} {I'_2\left(\sqrt{\varepsilon-I_{2n}^{\,2}}-I_{2n}a_1\right)} \\ &+\frac{a_2^2+1}{a_2^2(a_2-a_1)} \ln \frac{I_{2n}\left(\sqrt{\varepsilon-I_2'^2}-I'_2a_2\right)} {I'_2\left(\sqrt{\varepsilon-I_{2n}^{\,2}}-I_{2n}a_2\right)} \Bigg], \end{aligned} \tag{17} \]

where

\[ a_{1,2}=\frac{H\varepsilon}{2M_c}\pm \sqrt{\left(\frac{H\varepsilon}{2M_c}\right)^2-1}; \qquad H=\frac{m_2r_2}{9.81\,\omega_1S_m}. \]

Dependence of current on time during starting under load

\[ M_c=k'\omega_2^2 \]

In this case the equation of the moments acting on the motor shaft will have the form

\[ \frac{m_2r_2I_2^2}{9.81\,\omega_1S} -k(1-S)^2 = -J\omega_1\frac{dS}{dt}, \tag{18} \]

where

\[ k=k'\omega_1^2. \]

Substituting the values of \(S\) and \(dS\) from equations (9, 10) into equation (18), we obtain

\[ \frac{m_2r_2I_2\sqrt{\varepsilon-I_2^2}}{9.81\,\omega_1S_m} -k\left(1-\frac{S_mI_2}{\sqrt{\varepsilon-I_2^2}}\right)^2 = -J\omega_1 \frac{S_m\varepsilon\,dI_2}{dt\,(\varepsilon-I_2^2)^{3/2}}, \]

whence the dependence of current on time is

\[ t=-J\omega_1S_m\varepsilon \int_{I_{2n}}^{I'_2} \frac{dI_2} {(\varepsilon-I_2^2)^{1/2} \left[ I_2H\sqrt{\varepsilon-I_2^2} -k\left(1-\frac{S_mI_2}{\sqrt{\varepsilon-I_2^2}}\right)^2 \right]}, \]

where

\[ H=\frac{m_2r_2}{9.81\,\omega_1S_m}. \]

Reducing to a common denominator, we obtain

\[ t=-J\omega_1S_m\varepsilon \int_{I_{2n}}^{I'_2} \frac{dI_2} {(\varepsilon-I_2^2)^{1/2} \left\{ I_2H(\varepsilon-I_2^2)^{3/2} -k\left[(\varepsilon-I_2^2) -2S_mI_2\sqrt{\varepsilon-I_2^2} +S_m^2I_2^2\right] \right\}} \tag{19} \]

The integral entering the right-hand side of equation (19) is solved by substituting a new variable.

Let

\[ \varepsilon-I_2^2=Z^2I_2^2, \tag{20} \]

then

\[ -2I_2\,dI_2=I_2^2\,2Z\,dZ+Z^2\,2I_2\,dI_2, \]

whence

\[ -d I_2=\frac{I_2 Z\,dZ}{1+Z^2} \]

From equation (20) we find

\[ I_2^2=\frac{E}{1+Z^2}. \]

Substituting the values of \(I_2\) and \(dI_2\) into equation (19), we obtain

\[ t=J\omega_1 S_m \varepsilon \int_{I_{2n}}^{I_2'} \frac{(1+Z^2)\,dZ}{H Z^3 \varepsilon^2-k(1+Z^2)(Z^2\varepsilon-2S_m Z+S_m^2\varepsilon)} . \]

After collecting like terms and taking \(\frac{1}{\varepsilon k}\) outside the integral sign, we shall have

\[ t=-\frac{J\omega_1 S_m}{k}\int_{I_{2n}}^{I_2'} \frac{(1+Z^2)\,dZ}{Z^4-Z^3\left(\frac{H\varepsilon}{k}+2S_m\right)+Z^2(1+S_m^2)-2ZS_m+S_m^2}. \tag{21} \]

For the purpose of shortening the notation, denote \(\frac{H\varepsilon}{k}+2S_m\) by \(h\), and \(1+S_m^2\) by \(g\); then equation (21) will take the form

\[ t=-\frac{J\omega_1 S_m}{k}\int_{I_{2n}}^{I_2} \frac{(1+Z^2)\,dZ}{Z^4-Z^3h+Z^2g-2ZS_m+S_m^2}. \tag{22} \]

The expression under the integral sign is a rational fraction, whose integration is possible by decomposing the denominator into simple factors.

Putting \(Z=x+\frac{h}{4}\), we obtain

\[ Z^4-Z^3h+Z^2g-2ZS_m+S_m^2=x^4+px^2+qx+r. \tag{23} \]

Here

\[ p=-\frac{3}{8}h^2+g,\qquad q=-\frac{1}{8}h^3+\frac{gh}{2}-2S_m, \]

\[ r=-\frac{3}{256}h^4+\frac{gh^2}{16}-\frac{S_m h}{2}+S_m^2. \]

To find the roots of equation (23), it is first necessary to solve the following equation of the third degree

\[ y^3+2py^2+(p^2-4r)g-q^2=0. \]

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

\[ S^3+p'S+q'=0, \tag{24} \]

where

\[ p'=-\frac{1}{3}p^2-4r, \]

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

Having determined the roots of equation (24), \(S_1, S_2\), and \(S_3\), by Cardano’s formulas, we find the roots of equation (23):

\[ x_1=\frac{f}{2}+d;\qquad x_2=\frac{f}{2}-d;\qquad x_3=-\frac{f}{2}+je;\qquad x_4=-\frac{f}{2}-je \]

where

\[ d=\sqrt{-\frac{u+v}{4}+\frac{1}{2}\sqrt{u^2-uv+v^2}}, \]

\[ je=\sqrt{-\frac{u+v}{4}-\frac{1}{2}\sqrt{u^2-uv+v^2}},\qquad f=\sqrt{u+v}. \]

Substituting the values \(x_1, x_2, x_3, x_4\) into equation (22), we obtain

\[ t=-\frac{J\omega_1 S_m}{k} \int_{I_{2n}}^{I_2'} \frac{(1+Z^2)\,dZ} {\left(Z-\frac{f}{2}-d\right)\left(Z-\frac{f}{2}+d\right) \left(Z^2+Zf+\frac{f^2}{4}+e^2\right)} \tag{25} \]

Integration gives

\[ \begin{aligned} t=-\frac{J\omega_1 S_m}{k}\Bigg[& A\ln \frac{\sqrt{\frac{\varepsilon}{I_2'^2}-1}-\frac{f}{2}-d} {\sqrt{\frac{\varepsilon}{I_{2n}^2}-1}-\frac{f}{2}-d} \\ &+B\ln \frac{\sqrt{\frac{\varepsilon}{I_2'^2}-1}-\frac{f}{2}+d} {\sqrt{\frac{\varepsilon}{I_{2n}^2}-1}-\frac{f}{2}+d} \\ &+\frac{C}{2}\ln \frac{ \frac{\varepsilon}{I_2'^2}-1 +f\sqrt{\frac{\varepsilon}{I_2'^2}-1} +\frac{f^2}{4}+e^2 }{ \frac{\varepsilon}{I_{2n}^2}-1 +f\sqrt{\frac{\varepsilon}{I_{2n}^2}-1} +\frac{f^2}{4}+e^2 } \\ &+\frac{2D-Cf}{2e} \left( \arctg \frac{2\sqrt{\frac{\varepsilon}{I_2'^2}-1}+f}{2e} - \arctg \frac{2\sqrt{\frac{\varepsilon}{I_{2n}^2}-1}+f}{2e} \right) \Bigg] \end{aligned} \]

Fig. 1

where

\[ A= \frac{ 1+\frac{f^2}{4}+f\,2d+d^2 }{ 2d\left(f^2+2fd+d^2+e^2\right) }; \qquad B= \frac{ 1+\frac{f^2}{4}-fd+d^2 }{ 2d\left(f^2-2fd+d^2+e^2\right) }; \]

\[ D=1-A\left(1+\frac{f}{2}\right) +B\left(d-\frac{f}{2}+cf\right); \qquad C=-A-B . \]

The value of $I$ during motor starting, obtained by formulas (13, 17, 26) and taken from oscillograms 1, 2, 3 for the same instants of time, is given in the following table:

Table

Fig. 2

Fig. 3

Calculations and oscillographic records were made for a motor having the following data: type T-750, $U_1 - 220\text{ V}$, $E_2 = 145\text{ V}$, $I_1 = 38\text{ A}$, $I_2 = 45{,}5\text{ A}$; $n = 715$ rpm, $\cos \varphi = 0{,}8$, $r_1 = 0{,}4$ ohm, $r_2 = 0{,}08$ ohm, $r_k = 1{,}1$ ohm, $x_1 = 1{,}35$ ohm, $x_2^1 = x_1$, $Z_k = 2{,}92$ ohm, $S_m = 0{,}213$, $J = 0{,}05\ \mathrm{kgm\,sec^2}$. $M_k = 11$ kgm, $M_{\max} = 30$ kgm, $M_c = 6{,}6$ kgm.

ERRATA NOTED