EFFECT OF EXCITATION FORCING ON THE MAGNITUDE OF ENERGY LOSSES IN AN IMPACT GENERATOR

G. A. SIPAILOV, A. I. VERKHOTUROV

(Presented by the scientific seminar of the Departments of Electrical Machines and General Electrical Engineering)

Impact-power generators, whose normal operating mode is connection for a time approximately equal to one period of the emf to a load with a small resistance, comparable with the magnitude of the generator’s subtransient inductive reactance, under continuous excitation have higher electromagnetic no-load losses in comparison with turbogenerators of the same rated power. Therefore, during prolonged operation of an impact generator even at no load it may overheat; and for generators of interrupting and impact power the maximum permissible time of no-load operation at rated voltage is specially specified. For example, for an interrupting-power generator made by ASEA (Sweden), with a symmetrical short-circuit power of 2500 MVA, the maximum permissible time is 10 min. [1].

To reduce no-load energy losses in such generators, when the intervals between current pulses are 15–30 min, the voltage is reduced to zero for the period of the dead pause [2].

The expediency of using intermittent excitation when the pauses between stator-current pulses are longer than the duration of the cycle under intermittent excitation without forcing is obvious, since the no-load energy losses released in the generator during such a time interval, under constant excitation, will be much greater than under intermittent excitation without forcing.

For impact generators there may occur such modes in which the duration of the pauses between pulses is less than or equal to the duration of the cycle under intermittent excitation without forcing. It is therefore necessary to assess the possibility of using forced intermittent excitation in such modes.

In the present article, part of intermittent excitation is considered—the forced rise of the magnetic flux from zero or another final value to the prescribed value.

With a machine excitation system, it is considerably more convenient to carry out the forcing on the excitation side of the exciter.

In the calculations, however, we shall assume that the forcing is performed directly in the rotor circuits. Such an assumption is justified by the fact that the time constant of the exciter \(T_{\mathrm{в}}\) is considerably smaller than the total time constant of the rotor circuits.

To establish the indicated dependences, let us consider the transient process of switching the field winding onto the constant voltage \(k u_2\):

\[ \left. \begin{aligned} i_2 r_2 + L_2 \frac{di_2}{dt} + M_{23}\frac{di_3}{dt} &= k u_2,\\ i_3 r_3 + L_3 \frac{di_3}{dt} + M_{23}\frac{di_2}{dt} &= 0 \end{aligned} \right\}, \tag{1} \]

where

\[ \begin{aligned} r_2,\ L_2 &\text{ are the parameters of the excitation winding,}\\ r_3,\ L_3 &\text{ are the parameters of the damper winding,}\\ M_{23} &\text{ is the mutual inductance of the rotor windings.} \end{aligned} \]

Let us assume that the coupling coefficient between the windings is equal to, or close to, unity. For turbogenerators the value of this coefficient is 0.96 [3].

In this case \(M_{23}=\sqrt{L_2L_3}\), and the expressions for the currents have the form

\[ i_2=\frac{k u}{r_2}+A e^{-\frac{t}{T}}, \tag{2} \]

\[ i_3=-\frac{r_2 L_3 A}{r_3\sqrt{L_2L_3}}\,e^{-\frac{t}{T}}, \tag{3} \]

where \(T=T_2+T_3\) is the total time constant of the rotor circuits. The constant of integration \(A\) is determined from the condition that the sum of the currents in both windings is equal to \(n\frac{u_2}{r_2}\) at \(t=0\),

\[ i_2+i_3=\frac{k u_2}{r_2}+A+\frac{r_2 L_3 A}{r_3\sqrt{L_2L_3}} = n\frac{u_2}{r_2}, \tag{4} \]

where \(n\) is a coefficient indicating from what value of the excitation current the forcing begins; the value of this coefficient varies within the limits \(0 \leq n \leq 1\), whence

\[ A=\frac{k u_2 r_3\sqrt{L_2L_3}(n-k)} {r_2\left(r_3\sqrt{L_2L_3}+r_2L_3\right)}. \tag{5} \]

When the analysis is carried out in relative units, \(L_2 \approx L_3\), and the expression for \(A\) is substantially simplified,

\[ A=\frac{k u_2 B(n-k)}{r_2}, \tag{6} \]

where

\[ B=\frac{\delta_3}{\delta_2+\delta_3}=\frac{T_2}{T}. \]

After substituting the constant of integration into the expressions for the currents, we obtain:

\[ i_2=\frac{u_2}{r_2}\left[k+(n-k)B e^{-\frac{t}{T}}\right], \tag{7} \]

\[ i_3=\frac{u_2}{r_2}(n-k)(1-B)e^{-\frac{t}{T}}. \tag{8} \]

Sum of the currents in the windings

\[ i_2+i_3=\frac{u_2}{r_2}\left[k+(n-k)e^{-\frac{t}{T}}\right] \tag{9} \]

The flux that is transmitted electromagnetically from the rotor to the stator is proportional to the total current of the windings and to the coefficient of mutual induction between the rotor and stator \(M_{12}\):

\[ \Phi=\Phi_0\left[k+(n-k)e^{-\frac{t}{T}}\right], \tag{10} \]

\[ \Phi_0=\frac{u_2M_{12}}{r_2w_1} \]

\(W_1\) is the number of turns of the stator winding.

The time for the flux to rise to the steady-state value can be determined from the condition that the total current of the windings is equal to the quantity \(\dfrac{u_2}{r_2}\):

\[ \frac{u_2}{r_2}\left[k+(n-k)e^{-\frac{t}{T}}\right]=\frac{u_2}{r_2}, \tag{11} \]

whence

\[ t=T\ln\frac{k-n}{k-1}. \tag{12} \]

Expression (12), for \(k=1\), becomes infinite, i.e., without forcing of the excitation the flux reaches its steady-state value at \(t=\infty\). In practical calculations it is assumed that the flux rises to the steady-state value in the time \(t=3T\); then expression (12) can be rewritten in the following form:

\[ t'=T\ln\frac{k-n}{k-0.95}. \tag{13} \]

The energy losses in the steel of the machine are proportional to the square of the flux. Knowing the steady-state no-load flux and the losses in the steel corresponding to this flux, one can calculate the losses corresponding to any other value of the flux. Therefore, the losses released in the steel of the machine over the time interval \(t\) under forced excitation can be calculated as:

\[ Q_{\mathrm{cf}}=\frac{P'_0}{\Phi_0^2}\int_0^t \Phi^2dt = P'_0T\left[(n-1)(k+0.5n+0.5)-k^2\ln\frac{k-1}{k-n}\right]. \tag{14} \]

Knowing the law of variation of the current in the excitation and damper windings with time, one can determine the losses released in these windings for any forcing coefficient \(k\) over the time \(t\).

For the excitation winding

\[ Q_{2k}=\int_0^t i_2^2r_2dt = \frac{u_2^2T}{r_2} \left\{ B(n-1)\left[2k+B(0.5n-k+0.5)\right] -k^2\ln\frac{k-1}{k-n} \right\} \tag{15} \]

For the damping winding,

\[ Q_{3k}=\int_{0}^{t} i_3^2 r_3\,dt =\frac{u_2^2}{r_2}(1-B)BT(n-1)(0.5n+0.5-k). \tag{16} \]

The total losses in the rotor windings can be obtained by adding the right-hand sides of equations (15) and (16):

\[ Q_{\phi k}=\frac{u_2^2}{r_2}T\left[ B(n-1)(k+0.5n+0.5)-k^2\ln\frac{k-1}{k-n} \right]. \tag{17} \]

If, in equations (14), (15), (16), and (17), the value \(t^1\) from (13) is substituted for \(t\), then the equations for the loss energies in the steel, the excitation winding, and the damping winding can be written, respectively, as

\[ Q'_{c\phi}=P'_0T\left[(n-0.95)(k+0.5n+0.475)-k^2\ln\frac{k-0.95}{k-n}\right], \tag{18} \]

\[ Q'_{2k}=\frac{u_2^2}{r_2}T\left\{ B(n-0.95)[2k+B(0.5n+0.5-k)]-k^2\ln\frac{k-0.95}{k-n} \right\}, \tag{19} \]

\[ Q'_{3k}=\frac{u_2^2}{r_2}T(0.5n+0.5-k)B(1-B)(n-0.95). \tag{20} \]

The total energy losses released in the rotor windings during the time \(t^1\) can be obtained by adding the right-hand sides of equations (19) and (20):

\[ Q'_{\phi k}=\frac{u_2^2}{r_2}T\left[ B(n-0.95)(k+0.5n+0.5)-k^2\ln\frac{k-0.95}{k-n} \right], \tag{21} \]

In order to prove the advisability of using excitation forcing to reduce no-load energy losses calculated per one stator-current pulse, it is necessary to compare the energy losses in the machine steel and in the copper of the rotor windings under forced excitation and under constant excitation over the same time, but for different forcing coefficients.

Let us denote

\[ k_{4\phi}=\frac{Q_{c\phi}}{P'_0t}, \]

where \(P'_0t\) is the no-load energy loss in the steel under constant excitation over the time \(t\); then

\[ k_{4\phi}=k^2+\frac{(n-1)(k+0.5n+0.5)} {\ln\frac{k-n}{k-1}}. \tag{22} \]

The results of the calculation by (22) are presented in Fig. 1 for \(n=0\) and \(n=0.5\). It is evident from the figure that, as the forcing coefficient increases, the energy losses in the steel decrease, and for \(k=2,\ n=0\), the losses are 2.8 times smaller than the energy losses in the steel at no load and constant excitation. With further increase in the forcing coefficient, the reduction in energy losses occurs to a lesser degree.

The ratio of the energy losses under intermittent and constant excitation in the excitation winding and damping winding over the same time is denoted by \(k_{5\phi}\):

\[ k_{5\phi}=\frac{Q_{\phi k}}{\dfrac{u_2^2}{r_2}t} =k^2+\frac{B(n-1)(k+0.5n+0.5)} {\ln\frac{k-n}{k-1}}. \tag{23} \]

Fig. 1. Dependence of \(k_{\mathrm{ef}}\) on the forcing coefficient

The results of calculation by (23) are presented in Fig. 2, a and b, for \(n=0\) and \(n=0.5\), for various ratios of the time constants of the excitation winding and the damping winding. The calculations show that, as the forcing coefficient increases, the total energy losses in the rotor windings increase; however, at low values of the forcing coefficients the energy losses are smaller than with constant excitation. With further increase of the forcing coefficient, there comes a moment when the losses under forced excitation are equal over one and the same time interval; after this, as the forcing coefficient increases, the energy losses under forced excitation exceed the energy losses under constant excitation over one and the same time interval. From Figs. 2, a and b, it is seen that, as the forcing coefficient increases, the increase in the energy losses proceeds the faster, the larger the ratio

\[ \frac{T_{\mathrm{d}}}{T_2}. \]

By analogy, for the ratio of the energy losses in the excitation winding under intermittent and constant excitation one may write

\[ k_{\mathrm{ef}} = \frac{Q_{2k}}{\dfrac{u_2^2}{r_2}t} = \kappa^2 + \frac{ B(n-1)\,[2\kappa + B(0.5n-\kappa+0.5)] }{ \ln \dfrac{\kappa-n}{\kappa-1} }. \tag{24} \]

The results of calculation by (24) are presented in Fig. 3, a and b, for \(n=0\) and \(n=0.5\), for various ratios between the excitation winding and the damping winding. From the figures it is seen that the character of the dependences is the same as in Fig. 2, a and b.

From an analysis of the figures it is seen that the permissible forcing coefficients may be chosen from two conditions:

a) from the condition of permissible total losses in the rotor windings, for which equation (23) must be set equal to unity, that is, so that the losses in the rotor windings under constant and intermittent excitation over one and the same time interval are identical:

\[ \kappa^2 + \frac{ B(n-1)(\kappa+0.5n+0.5) }{ \ln \dfrac{\kappa-n}{\kappa-1} } = 1. \tag{25} \]

Fig. 2. Dependence of the ratio of losses in the rotor windings under forced excitation and under constant excitation over the same time on the forcing coefficient, for \(n=0\) and \(n=0.5\)

Fig. 3. Dependence of the ratios of losses in the excitation winding under forced and constant excitation over the same time on the forcing coefficient, for \(n=0\) and \(n=0.5\)

Or, after simplification:

\[ \frac{k^2-1}{\ln \frac{k-n}{k-1}}+B(n-1)(k+0.5n+0.5)=0. \tag{26} \]

For given \(B\) and \(n\), this transcendental equation can be solved approximately with any desired degree of accuracy with respect to \(k\). The solution of the equation is shown in Fig. 4, \(a\), for \(n=0\) and \(n=0.5\);

b) from the condition of permissible losses in the excitation winding, for which equation (24) must be set equal to unity, that is, so that the losses in the excitation winding under constant and intermittent excitation over the same time are identical:

\[ k^2+\frac{B(n-1)[2k+B(0.5n+0.5-k)]}{\ln \frac{k-n}{k-1}}=1. \tag{27} \]

Or, after simplification:

\[ (k^2-1)/n \, \frac{k-n}{k-1} + B(n-1)[2k+B(0.5n-k+0.5)] = 0 . \tag{28} \]

The solution of the last equation is presented in Fig. 4, b.

Fig. 4. For the calculation of permissible forcing coefficients

It is necessary to compare the total energy losses, calculated per one stator-current pulse, when operating with permissible forcing coefficients selected from the first and second conditions. In the comparison we take the ratio of the time constants to be

\[ \frac{T_3}{T_2}=0.8, \]

then, when operating with the permissible coefficient selected from the condition of equality of the total losses in the rotor windings, \(k_{\mathrm{per}}=1.28\) (Fig. 4, a), under constant and intermittent excitation, the energy losses in the steel of the machine are reduced by a factor of 2.17 compared with constant excitation over the same time. The electromagnetic losses in the UGM-100-2 impact generator are distributed at no load as follows:

Stator steel losses — 346.2 kW.
Rotor steel losses — 281 kW,
Total steel losses — 627.2 kW.
Excitation losses — 120 kW.

The ratio of the total no-load energy losses under constant and forced excitation over one and the same time is equal to:

\[ \frac{289t+120t}{627.2t+120t}=0.55, \tag{29} \]

that is, the energy losses when the flux rises to one and the same value are reduced by a factor of 1.8 in comparison with the case in which the excitation is left constant; in this case the forcing duration will be \(t=1.52T\) sec. At \(n=0.5\) the loss ratio is 0.72, that is, the energy losses when the flux rises to one and the same value under forced excitation are reduced by a factor of 1.38 in comparison with constant excitation over one and the same time; in this case the cycle duration will be \(1.25T\) sec. If one compares the energy of the losses released at \(n=0.5\) and \(n=0\), this ratio is 0.926, that is, approximately equal to unity; however, at \(n=0.5\) the forcing time is reduced by approximately a factor of 1.2.

When selecting the permissible forcing coefficient \(k_{d2\phi}\) from the condition of equality of the energy losses in the excitation winding, for

\[ \frac{T_3}{T_2}=0.8 \]

and \(n=0\), from Fig. 4, b, we find \(k_{d2\phi}=1.49\).

The ratio of the total energy losses under forced and constant excitation for \(k_{d2\phi}=1.49\) is 0.55 when the forcing duration is \(t=1.1T\).

If one finds the ratio of the energy losses under forced excitation with \(k_{d2\phi}=1.28\) and under forced excitation with \(k_{d2\phi}=1.49\), it is equal to 1.55; that is, the energy losses under forced excitation with \(k_{d2\phi}=1.49\) are 1.55 times smaller than under forced excitation with \(k_{d2\phi}=1.28\).

At \(n=0\) the permissible forcing coefficient is \(k_{d2\phi}=1.24\). Let us compare the ratio of the total energy losses at \(k_{d2\phi}=1.49\) and \(n=0\) with the energy losses at \(k_{d2\phi}=1.24\) and \(n=0.5\). This ratio is 0.97, i.e., approximately equal to unity.

Consequently, when selecting the permissible coefficient from the second condition, the operating variants with \(n=0\) and \(n=0.5\) are equivalent.

The calculated results were checked on an experimental setup consisting of a specially designed generator, exciter, and load. A description of the setup is given in [4].

The rise of current in the excitation winding was oscillographed when it was switched on to voltages of 6.7, 13.4, and 20.1 V, and the rise of voltages was also oscillographed. When the excitation winding was switched on to a voltage of 6.7 V, the voltages at the generator terminals up to 84 V were recorded when each of the voltages was applied to the excitation winding. When the excitation winding was switched on to a voltage of 6.7 V, the voltage at the terminals of the stator winding reached 84 V in the steady-state regime. From the oscillograms of the voltage at the terminals of the stator winding as a function of time, the areas of the obtained curves were compared with the areas that would be obtained at a constant excitation current of 55 A and a constant voltage of 84 V, if they acted for the same time as under forced excitation when the voltage rose to 84 V.

The results of processing the oscillograms are presented in Table 1. At the final voltage of 84 V the magnetic system is not saturated, since the rated voltage of the generator is 287 V.

Thus, the permissible forcing coefficient must be selected from the condition of equality of the no-load energy losses under forced excitation and under constant excitation over one and the same

time. Experimental verification showed satisfactory agreement between the calculated and experimental data.

Table 1

LITERATURE

1. Enseie Arnold, Slamecka Ernst. A new experimental laboratory for breaking capacities of the Siemens—Schuckertwerke firm. Express Information. Electrical Machines and Apparatus, 17, 1962.

2. A. I. Vorontsov, G. M. Khutoretskii. A breaking-capacity generator rated at 2500 MVA. Coll. “Electrosila,” 20, 1960.

3. O. B. Bron. Automatic devices for extinguishing the magnetic field, Gosenergoizdat, 1961.

4. G. A. Sigailov, A. I. Verkhoturov, K. A. Khorkov, A. V. Loos. Model of an impact-power generator, Proceedings of TPI, vol. 160, 1966.