ANALYTICAL METHOD FOR CALCULATING THE STARTING PARAMETERS OF A WOUND-ROTOR INDUCTION MOTOR AS A FUNCTION OF CURRENT

V. M. OSIPOV

For driving mechanisms with a large initial static torque, wound-rotor induction motors are usually used. The principal advantage of these motors, as is known, is a high starting torque and a small starting current. The starting torque, as well as the starting current, depends on the magnitude of the additional resistance inserted into the rotor circuit.

Wound-rotor motors often have automatic starting control, implemented according to various schemes. Schemes based on limiting the current in the rotor circuit have become widespread (starting as a function of current). The calculation of the starting parameters is carried out by a graphical-analytical method and consists in determining the following quantities: the additional resistance \(R_g\) inserted into the rotor circuit, the current in the stator circuit at the initial instant of starting \((I_{\text{start}})\), the slip \(s_{\text{per}}\) at which the additional resistance is to be short-circuited, and the value of the current in the stator circuit before the instant of short-circuiting the additional resistance, to which the switching relay must be adjusted.

The starting condition is the equality of the currents at the initial instant of starting and after short-circuiting the additional resistance (or the equality of the corresponding torques).

An approximate value of the slip \(s_{\text{per}}\) is found by graphical constructions, after which the further calculation is carried out analytically with the aid of simple formulas [1]. The proposed method makes it possible to determine the slip \(s_{\text{per}}\) analytically with accuracy sufficient for practice.

Starting with single-stage acceleration

The dependence \(M_{\text{em}} = f(s)\), called the mechanical characteristic of the motor, is expressed by the equation:

\[ M_{\text{em}}=\frac{2B}{\dfrac{s}{s_{kp}}+\dfrac{s_{kp}}{s}}, \tag{1.1} \]

where \(B\) is the coefficient of overload capacity of the motor,
\(s_{kp}\) is the critical slip.

In the absence of additional resistance \(R_g\) in the rotor circuit, the mechanical characteristic of the motor (1.1) is called the natural characteristic; for \(R_g > 0\), it is called an artificial characteristic.

When starting the motor, when \(s=1\), from (1.1) we have:

\[ M'_{\text{em,start}}=\frac{2Bs_{kp}}{1+s_{kp}^{2}}. \tag{1.2} \]

According to the starting condition, the torque after switching must be equal to the starting torque. Suppose the starting torque is chosen in accordance with (1.2); then

\[ \frac{2B}{\dfrac{s_{\text{sw}}}{s_{kp}}+\dfrac{s_{kp}}{s_{\text{sw}}}} = \frac{2Bs_{kp}}{1+s_{kp}^{2}} . \tag{1.3} \]

After transformations we obtain the quadratic equation with respect to \(s_{\text{sw}}\):

\[ s_{\text{sw}}^{2}-s_{\text{sw}}(1+s_{kp}^{2})+s_{kp}^{2}=0 . \tag{1.4} \]

The solution of equation (1--4) gives the roots:

\[ s_{\text{sw}1}=1;\qquad s_{\text{sw}2}=s_{kp}^{2}. \]

Thus, for \(M_{\text{t.start}}=M'_{\text{t.start}}\), the switching slip can be determined very simply:

\[ s_{\text{sw}}=s_{kp}^{2}. \tag{1.5} \]

Usually the motor is started with \(M_{\text{t.start}}\) different from \(M'_{\text{t.start}}\).

Let

\[ \frac{M_{\text{t.start}}}{M'_{\text{t.start}}}=m . \tag{1.6} \]

Then

\[ s_{\text{sw}}= \frac{ \dfrac{1+s_{kp}^{2}}{m} - \sqrt{\left(\dfrac{1+s_{kp}^{2}}{m}\right)^{2}-4s_{kp}^{2}} }{2}. \tag{1.7} \]

Obviously, for \(m=1\), \(s_{\text{sw}}=s_{kp}^{2}\);

for \(m=m_{\text{lim}}=\dfrac{1+s_{kp}^{2}}{2s_{kp}}\), \(s_{\text{sw}}=s_{kp}\);

for \(m>m_{\text{lim}}\), \(s_{\text{sw}}\) is expressed by a complex number, which has no physical meaning.

Thus, there exists a quite definite value \(m=m_{\text{lim}}\) at which

\[ s_{\text{sw}}=s_{\text{sw.max}}=s_{kp} \]

\[ m_{\text{lim}}=\frac{1+s_{kp}^{2}}{2s_{kp}} . \tag{1.8} \]

To determine \(s_{\text{sw}}\) for \(m\ne1\), formula (1.7) is inconvenient; it can be considerably simplified, taking into account that, at small slips, the natural mechanical characteristic of the motor is almost rectilinear. Then

\[ s_{\text{sw}}\simeq ms_{kp}^{2}. \tag{1.9} \]

Indeed, from the drawing (Fig. 1) we have:

\[ \frac{b'a'}{a'd} = \frac{b''a''}{a''d+c''b''} \simeq \frac{b''a''}{a''d}; \tag{1.10} \]

\[ c''b''\simeq0;\qquad b'a'=M'_{\text{t.start}};\qquad a'd=s'_{\text{sw}}=s_{kp}^{2}, \]

\[ b''a''=mM'_{\text{t.start}};\qquad a''d=s_{\text{sw}}. \]

Substituting the values of the segments into proportion (1.10), we obtain:

\[ \frac{M'_{v.\,start}}{s_{kr}^{2}} \simeq \frac{m M'_{v.\,start}}{s_{per}}, \]

whence

\[ s_{per} \simeq m s_{kr}^{2}. \]

The error thereby allowed does not exceed the error in the graphical determination of \(s_{per}\), and it is smaller the closer \(m\) is to unity.

Fig. 1. Starting diagram of a wound-rotor induction motor as a function of current with single-stage acceleration

The referred additional resistance inserted in the rotor circuit [1]:

\[ R'_g = s'_{kr}\sqrt{r_{ec}^{\prime 2}+x_k^2} - r'_p = s'_{kr}\frac{r'_p}{s_{kr}} - r'_p = r'_p\left(\frac{s'_{kr}}{s_{kr}}-1\right), \tag{1.11} \]

where \(r'_p\) is the referred active resistance of the rotor, and \(s'_{kr}\) is the value of the critical slip when an additional resistance \(R_g\) is present in the rotor circuit:

\[ s'_{kr} = \frac{M_{v.\,max}}{M_{v.\,start}} = \frac{B}{m M'_{v.\,start}}. \tag{1.12} \]

Substituting (1.2) into (1.12), we obtain:

\[ s'_{kr} = \frac{B(1+s_{kr}^{2})}{m \cdot 2B s_{kr}} = \frac{1+s_{kr}^{2}}{m \cdot 2s_{kr}}. \]

or, taking equality (1.8) into account, we shall have:

\[ s'_{kp}=\frac{m_{pr}}{m}. \tag{1.13} \]

Substituting the obtained value \(s'_{kp}\) into (1.11), we find

\[ R'_g=r'_p\left(\frac{m_{pr}}{m\,s_{kp}}-1\right). \tag{1.14} \]

The actual resistance is

\[ R_g=\frac{R'_g}{K^2}, \tag{1.15} \]

where \(K\) is the winding transformation ratio.

The other quantities are determined in the usual way.

Starting with Two-Stage Acceleration

The results obtained can easily be applied to the analytical calculation of starting with two-stage acceleration. Indeed, from the drawing (see Fig. 2) we have:

Fig. 2. Starting with two-stage acceleration

\[ \frac{M_{v.per}}{s_{per12}}=M_{v.pusk} \tag{2.1} \]

and

\[ \frac{M_{v.pusk}}{s_{per12}}=\frac{M_{v.per}}{s_{per23}}. \tag{2.2} \]

The error introduced in this case is quite small, since in their initial portion the mechanical characteristics of the motor are almost rectilinear.

Solving (2.1) and (2.2) simultaneously, we obtain

\[ s_{per12}=\sqrt{s_{per23}}. \tag{2.3} \]

The slip \(s_{per23}\) will be equal to the slip for single-stage starting, that is,

\[ s_{per23}=m s_{kr.3}^{2}, \tag{2.4} \]

where \(s_{kr.3}\) is the critical slip at \(R_g=0\) (curve 3).

On the basis of (1.14), the reduced additional resistance included in the rotor circuit is

\[ R'_g=r'_p\left(\frac{m_{pr}}{m s_{kr3}}-1\right). \tag{2.5} \]

The first stage of the additional resistance \((R'_{g1})\) can be determined from the following considerations. The critical slip \(s_{kr1}\) in the presence of the additional resistance \(R'_g\) in the rotor circuit may be found from the known formula [1]

\[ s_{kr1}=\frac{r'_p+R'_g}{\sqrt{r_{2c}^{2}+x_k^{2}}}. \tag{2.6} \]

The critical slip in the presence of the additional resistance \(R_{g2}\) in the rotor circuit (after the first switching)

\[ s_{kr2}=\frac{r'_p+R'_{g2}}{\sqrt{r_c^{2}+x_k^{2}}} =\frac{r'_p+R'_g-R'_{g1}}{\sqrt{r_c^{2}+x_k^{2}}}. \tag{2.7} \]

From (2–7) we have:

\[ R'_{g1}=r'_p+R'_g-s_{kr2}\sqrt{r_c^{2}+x_k^{2}}. \]

Using equality (2–6), we obtain

\[ R'_{g1}=(r'_p+R''_g)\cdot\left(1-\frac{s_{kr2}}{s_{kr1}}\right). \tag{2.9} \]

From the drawing (Fig. 2) we have

\[ \frac{M_{v.\max}}{s_{kr2}}\simeq \frac{M_{v.\,pusk}}{s_{per12}}, \quad \text{whence} \]

\[ \frac{s_{kr2}}{\dfrac{M_{v.\max}}{M_{v.\,pusk}}} =s_{per12}=\frac{s_{kr2}}{s_{kr1}}. \tag{2.10} \]

Then

\[ R'_{g1}=(r'_p+R'_g)(1-s_{per12}). \tag{2.11} \]

Reasoning analogously, we can obtain

\[ R'_{g2}=r'_p\left(\frac{m_{pr}}{m s_{kr3}}\cdot s_{per12}-1\right) =r'_p\left(\frac{m_{pr}}{\sqrt{m}}-1\right); \]

obviously,

\[ R'_g=R'_{g1}+R'_{g2}. \tag{2.12} \]

For the rest, the calculation does not differ from the usual one.

Starting with three-stage acceleration

In this case the procedure for deriving the formulas is analogous to the preceding one. The calculation formulas take the following form:

1. Switching slips

\[ s_{per12}=\sqrt{s_{per23}}; \tag{3.1} \]

\[ s_{per23}=\sqrt{s_{per34}^{3}}; \tag{3.2} \]

\[ s_{per34}=m s_{kr4}^{2}. \tag{3.3} \]

2. Critical slips

\[ s_{kp1}=\frac{m_{pr}}{m}; \tag{3.4} \]

\[ s_{kp2}=\frac{m_{pr}}{m}-s_{per12}, \tag{3.5} \]

\[ s_{kp3}=\frac{m_{pr}}{m}-s_{per23}, \tag{3.6} \]

\[ s_{kp4}=\frac{r'_p}{\sqrt{r_c^2+x_k^2}}. \tag{3.7} \]

3. Resistances of the stages

\[ R'_g=r'_p\left(\frac{m_{pr}}{m\,s_{kp4}}-1\right), \tag{3.8} \]

\[ R'_{g1}=(r'_p+R'_g)(1-s_{per12}), \tag{3.9} \]

\[ R'_{g2}=(r'_p+R'_g)(1-S_{per23})-R'_{g1}, \tag{3.10} \]

\[ R'_{g3}=r'_p\left(\frac{m_{pr}}{m\,s_{kp4}}\cdot s_{per23}-1\right). \tag{3.11} \]

The subsequent calculation is carried out with the aid of the usual formulas.

It should be noted that the multiplicity used above,

\[ m=\frac{M_{v.\,pusk}}{M'_{v.\,pusk}}, \]

introduces a certain convention, since usually the starting torque is specified by the multiplicity \(a\) with respect to the rated torque, i.e.

\[ M_{v.\,pusk}=a\cdot M_{v.\,nom} \quad \text{or} \quad M_{v.\,pusk*}=a. \tag{3.12} \]

This convention disappears if \(m\) is expressed through \(a\).

On the basis of (1–12) and (1–13) we have (using (3.12))

\[ s_{kp1}=\frac{M_{v.\,max*}}{M_{v.\,pusk*}}=\frac{m_{pr}}{m}=\frac{B}{a}, \]

whence

\[ m=\frac{a}{B}\,m_{pr}. \tag{3.13} \]

Let us show the application of the formulas obtained in specific examples.

Example 1. A wound-rotor induction motor has automatic starting as a function of current with one-stage acceleration.

Motor data¹:

Overload capacity

\[ B=\frac{M_{v.\,max}}{M_{v.\,nom}}=3. \]

Active resistance of the stator \(r_c=0.56\ \Omega\).

Active resistance of the rotor \(r'_p=0.52\ \Omega\).

Reactance \(x_k=x_c+x'_p=3.5\ \Omega\).

Transformation ratio

\[ K=\frac{U_{c.\phi}}{U_{p.\phi}}=2. \]

Starting torque \(M_{\mathrm{v.start}}=2M_{\mathrm{v.nom}}\), or \(M_{\mathrm{v.start}*}=2\). Determine the slip \(s_{\mathrm{tr}}\) and the magnitude of the additional resistance \(R_g\).

Calculation.

Critical slip

\[ s_{\mathrm{cr}}=\frac{r'_p}{\sqrt{r_c^2+x_k^2}} =\frac{0.52}{\sqrt{0.56^2+3.5^2}}=0.147. \]

Limiting ratio

\[ m_{\mathrm{lim}}=\frac{1+s_{\mathrm{cr}}^2}{2s_{\mathrm{cr}}} =\frac{1+0.147^2}{2\cdot 0.147}=3.47. \]

Starting-torque ratio

\[ m=\frac{a}{B}\cdot m_{\mathrm{lim}}=\frac{2}{3}m_{\mathrm{lim}} =0.667m_{\mathrm{lim}}=2.35. \]

Slip \(s_{\mathrm{tr}}\)

\[ s_{\mathrm{tr}}=m\cdot s_{\mathrm{cr}}^2 =2.35\cdot 0.147^2=0.05. \]

Additional resistance

\[ R'_g=r'_p\left(\frac{m_{\mathrm{lim}}}{m\cdot s_{\mathrm{cr}}}-1\right) =0.52\left(\frac{1}{0.667\cdot 0.147}-1\right)=4.78\ \Omega. \]

Actual resistance

\[ R_g=\frac{R'_g}{K^2}=\frac{4.78}{2^2}=1.19\ \Omega. \]

The results obtained agree well with the graphical calculation.

Example 2. A mill motor has automatic current-dependent starting with three-stage acceleration. Calculate the starting parameters.

Motor data\(^{1)}\):

Rated power \(P_n=368\ \mathrm{kW}\).

Rated voltage \(U_n=3000\ \mathrm{V}\).

Active stator resistance \(r_c=0.461\ \Omega\).

Active rotor resistance \(r'_p=0.364\ \Omega\).

Reactive resistance \(x_k=x_c+x'_p=5.6\ \Omega\).

Transformation ratio \(K=5.55\).

Magnetizing current \(I_n\simeq I_{xx}=26\ \mathrm{A}\).

Connection of the stator and rotor windings—star.

Starting torque \(M_{\mathrm{v.start}*}=M_{\mathrm{v.start.max}}=B\).

Calculation.

According to the starting mode, \(m=m_{\mathrm{lim}}\).

1. Critical slip

\[ s_{\mathrm{cr}4}=\frac{r'_p}{\sqrt{r_c^2+x_k^2}} =\frac{0.364}{\sqrt{0.461^2+5.6^2}}=0.065. \]

\(^{1)}\) The data are taken from the book by I. A. Syromyatnikov, Operating Conditions of Induction Motors, GEI, 1950.

1. Limiting ratio

\[ m_{pr}=\frac{1+s_{kr}^{2}}{2s_{kr}}=\frac{1+0.0042}{2\cdot 0.065}=7.7. \]

1. Switching slips

\[ s_{per\,34}=s_{kr\,4}=0.065; \]

\[ s_{per\,23}=\sqrt{s_{per\,34}^{3}}=\sqrt{0.065^{3}}=0.165; \]

\[ s_{per\,12}=\sqrt{s_{per\,23}}=\sqrt{0.165}=0.405. \]

1. Total additional resistance

\[ R'_g=r'_p\left(\frac{m_{pr}}{ms_{kr4}}-1\right)=0.364\left(\frac{1}{0.065}-1\right)=5.25\ \Omega. \]

Actual resistance

\[ R_g=\frac{R'_g}{K^2}=\frac{5.25}{5.55^2}=0.174\ \Omega. \]

1. Resistance of the first stage

\[ R'_{g1}=(r'_p+R'_g)(1-s_{per\,12})=5.61(1-0.405)=3.34\ \Omega. \]

Actual resistance

\[ R_{g1}=\frac{R'_{g1}}{K^2}=\frac{3.34}{5.55^2}=0.111\ \Omega. \]

1. Resistance of the second stage

\[ R'_{g2}=(r'_p+R'_g)(1-s_{per\,23})-R_{g1}=5.61(1-0.165)-3.34=1.36\ \Omega. \]

Actual resistance

\[ R_{g2}=\frac{R'_{g2}}{K^2}=\frac{1.36}{5.55^2}=0.045\ \Omega. \]

1. Actual resistance of the third stage

\[ R_{g3}=R_g-(R_{g1}+R_{g2})=0.174-(0.111+0.045)=0.018\ \Omega. \]

1. Current in the rotor circuit before the moment of short-circuiting the first resistance stage

\[ I'_{p\,per1}= \frac{U_{\phi}} {\sqrt{\left(r_c+\frac{r'_p+R'_g}{s_{per\,12}}\right)^2+x_k^2}} = \]

\[ =\frac{\dfrac{3000}{\sqrt{3}}} {\sqrt{\left(0.461+\dfrac{5.61}{0.405}\right)^2+5.6^2}} =117\ \mathrm{A}. \]

According to the starting condition

\[ I'_{p\,per1}=I'_{p\,per2}=I'_{p\,per3}=I'_{p\,per}=117a. \]

1. Current in the stator circuit (operating current of the current relay)

\[ I_c=\sqrt{I_{p\,per}^{\prime 2}+I_{2n}^{2}}=\sqrt{117^2+26^2}=120a. \]

1. Current in the rotor circuit after the first switching

\[ I'_{p1}= \frac{U_{\phi}} {\sqrt{\left(r_c+\frac{r'_p+R'_g-R'_{g1}}{s_{\text{per}12}}\right)^2+x_k^2}} = \]

\[ = \frac{\dfrac{3000}{\sqrt{3}}} {\sqrt{\left(0.461+\frac{5.61-3.34}{0.405}\right)^2+5.6^2}} =210\ \text{A}. \]

1. Current in the rotor circuit at the initial instant of starting

\[ I'_{p.\text{start}}= \frac{U_{\phi}} {\sqrt{(r_c+r'_p+R'_g)^2+x_k^2}} = \]

\[ = \frac{\dfrac{3000}{\sqrt{3}}} {\sqrt{(0.461+5.61)^2+5.6^2}} =210\ \text{A}. \]

According to the starting condition,

\[ I'_{p1}=I'_{p.\text{start}}=210\ \text{A}. \]

The graphical calculation gives results that agree well with those obtained.

Conclusion

In the author’s opinion, the proposed calculation method makes it possible:

1. To speed up the calculation in comparison with the graphical-analytical method, especially in the case of multistage starting.

2. To obtain results no less accurate; and in two particular cases, when \(m=1\) and \(m=m_{pr}\), the error is due only to the fact that the natural mechanical characteristic of the motor (1–1) is derived under the assumption that \(r_c=0\).

References

1. Solov’ev I. I. Automation of Power Systems. Gosenergoizdat, 1950.
2. Syromyatnikov I. A. Operating Conditions of Induction Motors. Gosenergoizdat, 1950.

1. The motor data are taken from the book by I. I. Solov’ev, Automation of Power Systems. GEI, 1950. ↩