BRAKING OF SHORT-CIRCUITED MOTORS OF STANDARD DESIGN

I. G. KULEEV

Docent, Candidate of Technical Sciences

If a motor is disconnected from the network, then, in the presence of a considerable moment of inertia of the rotating masses, the motor coasting may continue for a rather long time.

The time required for it to stop by self-braking can be determined as follows.

The kinetic energy of the rotor connected with the rotating mechanism, at angular velocity \(\omega_2\), is equal to

\[ A_k = J \frac{\omega_2^2}{2}. \]

In the process of coasting (self-braking), the kinetic energy is expended on the useful work performed by the mechanism and on the work of friction forces.

Fig. 1

For \(M_c=\mathrm{const}\) we may write

\[ -\frac{dA_k}{dt} = -J\omega_2\frac{d\omega_2}{dt} = \left(M_c+M'_m\right)\omega_2, \]

where \(M'_m\) is the moment of friction forces.

Assuming, in a first approximation, \(M'_m=\mathrm{const}\), separating the variables and integrating, we obtain the duration of self-braking

\[ t = -\frac{J}{M_c+M'_m} \int_{\omega_2}^{0} d\omega_2 = \frac{J\omega'_2}{M_c+M'_m}. \tag{1} \]

For \(M_c=k\omega_2^2\) we shall have

\[ -\frac{J\omega_2\,d\omega_2}{dt} = M'_m\omega_2+k\omega_2^3; \]

whence

\[ t = -\frac{J}{k} \int_{\omega_2}^{0} \frac{d\omega_2}{\dfrac{M'_m}{k}+\omega_2^2}. \]

Integrating, we obtain

\[ t = \frac{J}{\sqrt{M'_m k}} \operatorname{arctg} \frac{\omega'_2}{\sqrt{\dfrac{M'_m}{k}}}. \tag{2} \]

In a whole series of cases, rapid and smooth stopping of a motor is required; to achieve this effect, the following purely electrical methods of braking are used:

1. Braking by the plugging method, in which two stator phases are switched over and the motor is forced to rotate against the rotation of the field. The motor then operates in the regime of an electromagnetic brake.

2. Dynamic braking, which is carried out under the condition that the stator winding is excited by direct current. In this case, for a star connection of the stator windings, the circuit of Fig. 1 is most often used; for a delta connection—Fig. 2.

Fig. 2

If the active resistance of the stator winding is denoted by \(r_1\), the strength of the direct current by \(I\), and the direct-current voltage by \(u\), then we shall have: for connection of the windings in \(Y\),

\[ I_Y = \frac{2}{3}\,\frac{u}{r_1}, \]

for connection of the windings in \(\Delta\),

\[ I_\Delta = \frac{3}{2}\,\frac{u}{r_1}. \]

The current, flowing around the stator winding, creates a magnetomotive force, under the action of which a stationary magnetic field arises.

When the rotor rotates in this field, an e.m.f. is induced in the rotor windings; the current produced by it will heat both the winding and the section of the rheostat connected into it.

Thus, in the process of braking, the kinetic energy of the moving parts of the drive is converted into thermal energy. The asynchronous motor operates in this case as a synchronous generator of decreasing frequency.

1. Braking with recovery of electrical energy. This method, in normal circuits, is possible only at speeds exceeding the synchronous speed.

Below, the transient regimes of asynchronous motors during braking by the first and second methods will be considered.

Braking of a Motor by the Plugging Method

1. Dependence of Speed on Time During Braking of an Unloaded Motor

The torque developed by an asynchronous machine in the braking regime is determined from the Kloss formula

\[ M = M_{\max}\,\frac{2+\beta S_m}{\dfrac{S}{S_m}+\dfrac{S_m}{S}+\beta S_m}. \]

In addition, for this regime we have

\[ S=\frac{\omega_1+\omega_2}{\omega_1} \quad \text{and} \quad \omega_2=\omega_1(S-1). \]

We shall assume that the mechanism rotated by the motor during coasting does no useful work and that the torque created by it …

resistance from the friction force is close to zero. Then the equation of motion for the run-down process may be written in the following form:

\[ M_{\max}\frac{2+\beta S_m}{\dfrac{S}{S_m}+\dfrac{S_m}{S}+\beta S_m} +J\omega_1\frac{dS}{dt}=0. \]

Assuming that at the beginning of the braking process \(S=2\), we obtain

\[ t=-\frac{J\omega_1}{M_{\max}(2+\beta S_m)} \int\left(\frac{S}{S_m}+\frac{S_m}{S}+\beta S_m\right)dS \]

or, after integration, we obtain

\[ t=\frac{J\omega_1}{M_{\max}(2+\beta S_m)} \left(\frac{3}{2S_m}+\beta S_m+0.695\,S_m\right). \]

Expressing time and slip through the current coordinates \(t\) and \(S\), we shall have

\[ t=\frac{J\omega_1}{M_{\max}(2+\beta S_m)} \left[ \frac{4-S^2}{2S_m}+\beta S_m(2-S)+S_m\ln\frac{2}{S} \right]. \tag{3} \]

2. Dependence of speed on time during braking of a loaded motor \(M_c=\mathrm{const}\)

In this case the equation of motion for the run-down process has the form

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

whence the duration of braking is

\[ t=-J\omega_1\int_{2}^{1} \frac{(S_m^2+S^2+\beta S_m^2S)\,dS} {M_{\max}(2+\beta S_m)S_mS+M_c(S_m^2+S^2+\beta S_m^2S)}. \]

After reducing the integrand to a proper fraction, we shall have

\[ t=-\frac{J\omega_1}{M_c}\int_{2}^{1}dS +\frac{J\omega_1M_{\max}S_m}{M_c^2} \int_{2}^{1} \frac{S\,dS} {S^2+S_m^2+S\left[\beta S_m^2\left(\frac{M_{\max}}{M_c}+1\right)+\frac{2M_{\max}S_m}{M_c}\right]}. \]

Denoting

\[ \beta S_m^2\left(\frac{M_{\max}}{M_c}+1\right)+\frac{2M_{\max}S_m}{M_c} \]

by \(b'\), we obtain

\[ t=-\frac{J\omega_1}{M_c}\int_{2}^{1}dS +\frac{J\omega_1}{M_c^2}M_{\max}S_m(\beta S_m+2) \int_{2}^{1}\frac{S\,dS}{S^2+S_m^2+b'S}. \]

Integration gives

\[ t=\frac{J\omega_1}{M_c}(2-S) +\frac{a}{2}\ln\frac{S+b'S+S_m^2}{4+2b'+S_m^2} +\frac{ab'}{\sqrt{4S_m^2-b'^2}} \operatorname{arctg} \frac{(4-S)\sqrt{4S_m^2-b'^2}} {4(S_m^2+S)+b'(4+S)}. \tag{4} \]

3. Dependence of speed on time when braking a loaded motor

\(M_c = k'\omega_2^2\)

In this case the equation of motion for the run-down process has the form

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

whence the braking duration is

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

Denoting \(\beta S_m^2-2\) by \(a_1\), \(1+S_m^2-2\beta S_m^2\) by \(b_1\), and

\[ \beta S_m^2-2S_m^2+\frac{(2+\beta S_m^2)M_mS_m}{k} \]

by \(c'\), we obtain

\[ t=-\frac{J\omega_1}{k} \int_{2}^{1} \frac{(S^2+S_m^2+\beta S_m^2S)\,dS} {S^4+S^3a_1+S^2b_1+c'S+S_m^2}. \tag{6} \]

Integration gives

\[ t=-\frac{J\omega_1}{k} \left[ A\ln\frac{S-f/2-d}{2-f/2-d} + B\ln\frac{S-f/2+d}{2-f/2+d} +\right. \]

\[ \left. +\frac{C}{2}\ln\frac{S^2+fS+f^2/4+e^2} {2+2f+f^2/4+e^2} - \frac{2D-Cf}{2e} \operatorname{arctg} \frac{4e}{4e^2+2S(2+f)+2f+f^2} \right]. \tag{7} \]

For the coefficients \(A, B, C, D, d, f,\) and \(e\), see the author’s work: “Starting of Short-Circuited Motors of Standard Design,” Izvestiya TPI (this same volume).

Figure 3 shows an oscillogram of speed during braking by the counter-switching method, \(M_c=0\).

Fig. 3

Figure 4 shows an oscillogram of speed during braking by the counter-switching method, \(M_c=\mathrm{const}\).

Figure 5 shows an oscillogram of speed during braking by the counter-switching method, \(M_c=k'\omega_2^2\).

For ease of comparison, the following table gives the values of \(t\) obtained from the formulas and taken from the oscillograms (Figs. 3, 4, 5).

Table of values of \(t\).

| \multicolumn{2}{c}{\(M_c = 0\)} | \multicolumn{2}{c}{\(M_c = \mathrm{const}\)} | \multicolumn{2}{c}{\(M_c = k'\omega_2^2\)} |
|---|---|---|---|---|---|
| By formula | From oscillogram | By formula | From oscillogram | By formula | From oscillogram |
| 0.65 | 0.66 | 0.86 | 0.88 | 1.11 | 1.12 |

Calculations and oscillographic records were made for a motor having the following data: type T-750, \(U_1 = 220\text{ V}\), \(E_2 = 145\text{ V}\), \(I_1 = 38\text{ A}\),

Fig. 4

\(I_2 = 45.5\text{ A},\quad n = 715\ \text{rpm},\quad \cos\varphi = 0.8,\quad r_1 = 0.4\ \Omega,\quad r_2 = 0.08\ \Omega\)
\(r_k = 1.1\ \Omega,\ x_1 = 1.35\ \Omega,\ x_2 = x_1,\ Z_k = 2.92\ \Omega,\ S_m = 0.213,\ J = 0.05\ \text{kgm sec}^2\)
\(M_n = 11\ \text{kgm},\ M_{\max} = 30\ \text{kgm},\ M_c = 6.6\ \text{kgm}.\)

Fig. 5

Dynamic braking

4. Dependence of speed on time when braking an unloaded motor

Braking by the plugging method, as is known, has major shortcomings, the chief of which are: considerable energy losses in the motor windings during braking, the need to insulate the rotor windings for double voltage (in the case of a slip-ring rotor), and the need for precise disconnection of the motor at the end of braking, since any delay causes the motor to accelerate in the reverse direction. Dynamic braking is free of all these drawbacks. In this method of braking, after the motor stator has been disconnected from the supply network, it is connected to a special direct-current network, and as a result a magnetic field stationary in space is produced in the motor.

This field induces an alternating current in the conductors of the rotor rotating by inertia. Thus, the motor is transformed into a synchron-

…generator, producing a braking torque. The braking work, except for friction losses, is converted into heat in the rotor circuit.

Fig. 6 gives the principal connection diagrams of the stator windings to a direct-current network and indicates the currents flowing through the individual phases.

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 m.m.f. can be determined by the geometric addition of the m.m.f.’s of the individual phases. Provided that identical ampere-turns of the stator and an identical rotating field are produced, the direct current can be replaced by an equivalent alternating current. The magnitude of this equivalent current \(I_1\), flowing in a stator phase, can be found from the equality

Fig. 6

turns of the stator and an identical rotating field are produced, the direct current can be replaced by an equivalent alternating current. The magnitude of this equivalent current \(I_1\), flowing in a stator phase, can be found from the equality

\[ AW_n = \frac{3}{2}\sqrt{2}\, I_1 W_1, \tag{9} \]

where \(AW_n\) is the total stator ampere-turns produced by the direct current, and \(\frac{3}{2}\sqrt{2}\, I_1 W_1\) is the amplitude of the total stator ampere-turns when it is supplied with alternating current.

Knowing \(AW_n\), from formula (9) one can find the equivalent current \(I_1\); for example, for circuit a:

\[ I_1=\frac{\sqrt{2}}{3W_1}AW_n=I_n\frac{\sqrt{2}}{\sqrt{3}}, \]

whence

\[ I_n=I_1\frac{\sqrt{3}}{\sqrt{2}}. \]

Similarly, for circuit b:

\[ I_1=I_n\frac{2\sqrt{2}}{3}. \]

For circuit c:

\[ I_1=\frac{I_n\sqrt{2}}{2}. \]

For circuit d:

\[ I_1=I_n\frac{\sqrt{2}}{3}. \]

For circuit e:

\[ I_1=\frac{I_n}{\sqrt{2}\sqrt{3}}. \tag{10 a–e} \]

The total ohmic resistances for the various circuits are respectively

\[ R_a=2R_1,\qquad R_b=3R_1,\qquad R_c=\frac{3R_1}{2},\qquad R_d=\frac{2}{3}R_1,\qquad R_e=\frac{1}{2}R_1, \]

where \(R_1\) is the active resistance of one phase of the stator winding.

Multiplying this resistance by \(I_n\), found from equations (10 a–e), we find the direct-current voltages \(U_n\) applied to the stator terminals. For example, for circuit a:

\[ U_n=\sqrt{3}\sqrt{2}\, I_1 R_1 = 2.44 I_1 R_1. \]

It is also easy to find the direct-current power

\[ P_n = I_n U_n. \]

The rotor current creates the corresponding ampere-turns, which will be stationary in space, although the currents in the individual conductors change their direction and magnitude; however, different conductors approach a given point in space, always carrying at that instant one and the same current. Therefore the rotor ampere-turns add to the stator ampere-turns and give the resultant magnetizing ampere-turns.

Denoting by \(I_2'\) the rotor current referred to the stator, and by \(I_0\) the magnetizing current, we shall have \(I_0 = I_1 + I_2'\).

In contrast to the normal operation of an induction motor, when the magnetizing current and flux remain approximately unchanged, while the current \(I_1\) changes with changes in slip, in dynamic braking the current \(I_1\) remains unchanged, while \(I_0\) and the flux change with changes in \(s\).

If the rotor were rotating in a stationary field at the synchronous speed \(n_0\), then the number of periods of its current would be

\[ f = p \frac{n_0}{60}. \]

When, however, during braking it rotates at a speed \(n\), the frequency of the current in the rotor is

\[ f_2 = \frac{pn}{60} = f_1 \frac{n}{n_0}. \]

Therefore, by analogy with normal operation, let us denote by

\[ s=\frac{n}{n} \]

the slip during dynamic braking.

Analogously to normal operation, one may write:

\[ I_2' = \frac{E_{20}'s}{\sqrt{(R_2')^2 + (x_2's)^2}}, \]

where \(E_{20}'\) is the rotor e.m.f., referred to the stator, at the synchronous speed \(n_0\), equal to the e.m.f. of a stationary rotor when the stator is supplied with three-phase current; \(R_2'\) and \(x_2'\) are the referred active and reactive resistances of the rotor circuit, and \(x_2'\) corresponds to the frequency \(f_1\).

If saturation is neglected, then one may take

\[ E_{20}' = b I_0, \]

where the coefficient of proportionality \(b\) may be found from the expression

\[ b = \frac{E_{20}'}{I_0} = \frac{E_1}{I_0}, \tag{11} \]

where \(E_1\) is the e.m.f. of the stator winding at the magnetizing current.

Approximately,

\[ b = \frac{U_n}{I_0}. \]

Therefore

\[ I_2' = \frac{b I_0 s}{\sqrt{(R_2')^2 + (x_2's)^2}}. \tag{12} \]

From the vector diagram of the currents, taking equality (12) into account, we find

\[ I_0 = I_1 \sqrt{\frac{(R_2')^2 + (x_2' S)^2}{(R_2')^2 + (x_2' + b)^2 S^2}} . \tag{13} \]

Substituting the value of \(I_0\) into equality (12), we obtain

\[ I_2' = \frac{b I_1}{\sqrt{\left(\dfrac{R_2'}{S}\right)^2 + (x_2' + b)^2}} . \tag{14} \]

The power converted into heat in the rotor circuit during braking is

\[ \frac{M_m n \cdot 1000}{975} = 3 I_2^2 R_2, \]

where \(M_m\) is the braking torque developed by the motor, and \(I_2\) and \(R_2\) are the current and active resistance of the rotor circuit. Introducing here the value of \(S\), we obtain

\[ \frac{M_m S n_0}{0.975} = 3 I_0 R_2 . \]

Since

\[ I_2' R_2' = I_2 R_2, \]

then

\[ M_m = \frac{3 I_2'^2 R_2' \cdot 0.975}{S n_0}. \tag{15} \]

Substituting, in place of \(I_2'\), its value according to formula (14), we obtain

\[ \tag{16} \]

\[ M_m = \frac{2.92\, b^2 I_1^2 R_2' S} {n_0 \left[(R_2')^2 + (x_2' + b)^2 S^2\right]} . \tag{17} \]

Differentiating \(M_m\) with respect to \(S\) and equating the derivative to zero, we obtain the value of slip at which the torque reaches a maximum,

\[ S_{m1} = \frac{R_2'}{x_2' + b}. \tag{18} \]

Substituting the value \(S_{m1}\) into equation (17), we obtain the expression for the maximum torque

\[ M_{m\max} = \frac{1.46\, b^2 I_1^2}{n_0 (x_2' + b)} . \tag{29} \]

In accordance with this, equation (19) may be represented in the form

\[ M_m = \frac{2 M_{m\max}} {\dfrac{S_{m1}}{S} + \dfrac{S}{S_{m1}}} . \tag{20} \]

Whence it follows that the torque equation under dynamic braking is analogous to Kloss’s formula; consequently, the equation of motion under dynamic braking of an unloaded motor may be represented as

\[ - \frac{2 M_{m\max}} {\dfrac{S_{m1}}{S} + \dfrac{S}{S_{m1}}} = J \frac{d\omega_2}{dt}. \tag{21} \]

Since \(\omega_2=\omega_1 S\), consequently, \(\dfrac{d\omega_2}{dt}=\omega_1\dfrac{dS}{dt}\), equation (21) can be transformed into the following form:

\[ -\frac{2M_{m\max}}{\dfrac{S_{m1}}{S}+\dfrac{S}{S_{m1}}} = J\omega_1\frac{dS}{dt}, \]

whence the duration of braking is

\[ t=-\frac{J\omega_1}{2M_{m\max}} \int_{1-S_1}^{0.05} \left( \frac{S_{m1}}{S}+\frac{S}{S_{m1}} \right)dS \tag{22} \]

Here \(S_1\) is the slip at which the stator was disconnected from the alternating-current network. Integration gives

\[ t=\frac{J\omega_1}{2M_{m\max}} \left[ \frac{(1-S_1)^2-0.0025}{2S_{m1}} + S_{m1}\ln\frac{1-S_1}{0.05} \right], \tag{23} \]

5. Dependence of speed on time during braking of a loaded motor \(M_c=\mathrm{const}\)

In this case the equation of motion can be written in the following form:

\[ \frac{2M_{m\max}}{\dfrac{S_{m1}}{S}+\dfrac{S}{S_{m1}}} - M_c = J\omega_1\frac{dS}{dt}, \]

whence the braking time is

\[ t=-\frac{J\omega_1}{M_c} \int_{1-S_1}^{0} \frac{(S^2+S_{m1}^2)\,dS} {S^2+S_{m1}^2+\dfrac{2M_{m\max}}{M_c}S_{m1}S}. \]

After reducing the integrand to a proper fraction, we obtain

\[ t=-\frac{J\omega_1}{M_c} \int_{1-S_1}^{0} dS - \frac{2M_{m\max}S_{m1}}{M_c} \int_{1-S_1}^{0} \frac{S\,dS} {S^2+S_{m1}^2+\dfrac{2M_{m\max}}{M_c}S_{m1}S}. \tag{24} \]

Integration gives

\[ t=\frac{J\omega_1}{M_c}(1-S_1-S) + \frac{a}{2} \ln \frac{S^2+bS+S_{m1}^2} {1+S_1(b-2)+S_1^2+b+S_{m1}^2} - \]

\[ -\frac{ab}{\sqrt{4S_{m1}^2-b^2}} \operatorname{arctg} \frac{2(S-1-S_1)\sqrt{4S_{m1}^2-b^2}} {S(4+2b-4S_1)+2b(1-S_1)+4S_{m1}^2}, \tag{25} \]

The duration of dynamic braking obtained by formula (23) and taken from the oscillogram of Fig. 7, for \(M_c=0\) and \(I=50\) amp. for the above-mentioned motor, is as follows.

Fig. 7

By the formula \(t=1.6\) sec. From the oscillogram \(t=1.7\) sec.

Noticed Misprints