Accounting for Saturation in Calculating the Intermittent Excitation of an Impact Generator

G. A. SIPAILOV, A. I. VERKHOTUROV

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

In circuits with steel, the inductance retains a constant value only within the limits of the straight-line portion of the magnetization curve. Therefore, the relations given in [1] for calculating the permissible forcing coefficients under intermittent excitation, which are valid for constant inductance, cannot be applied in calculating transients with a saturated magnetic circuit; and since impact generators may operate in regimes with strong saturation, the assumption of constant inductance may lead to significant errors in the calculation.

In the present article the problem is posed of taking into account saturation of the magnetic circuit in calculating the permissible forcing coefficient and of estimating the error that results when the calculation is made with constant inductance.

For calculating transients in excitation circuits with allowance for saturation, the method of piecewise-linear approximation is used most widely [2].

Replacing the characteristic of a nonlinear resistance by segments of straight lines makes it possible to pass from a nonlinear differential equation to several linear differential equations differing only in the values of their coefficients.

In practice, the magnetization curve is replaced by two straight lines \(aa\) and \(ab\) (Fig. 1). In this case there are constant inductances in each segment.

Neglecting the leakage flux between the damper and excitation windings, for connecting the excitation winding to a constant voltage \(ku_2\) we obtain the following system of differential equations for the saturated part of the no-load characteristic:

\[ \left. \begin{aligned} r_2 i_2 + L_{2\alpha}\frac{di_2}{dt} + \sqrt{L_{2\alpha}\cdot L_{3\alpha}}\,\frac{di_3}{dt} &= ku_2,\\ r_3 i_3 + L_{3\alpha}\frac{di_3}{dt} + \sqrt{L_{2\alpha}\cdot L_{3\alpha}}\,\frac{di_2}{dt} &= 0. \end{aligned} \right\} \qquad (1) \]

where

\[ r_2,\ r_3 \]
— active resistances of the excitation and damper windings,

\[ L_{2\alpha},\ L_{3\alpha} \]
— inductances of the same windings in the unsaturated region,

Fig. 1. Accounting for saturation by the method of piecewise-linear approximation

\(k\) — the multiple of the voltage applied to the excitation winding as compared with the rated value.

The solution of system (1):

\[ i_2=\frac{u_2}{r_2}\left[k+(n-k)B e^{-\frac{t}{T_\alpha}}\right], \tag{2} \]

\[ i_3=-\frac{u_2}{r_2}(1-B)(n-k)e^{-\frac{t}{T_\alpha}}, \tag{3} \]

where

\[ T_{2\alpha}=\frac{L_{2\alpha}}{r_2}, \qquad T_{3\alpha}=\frac{L_{3\alpha}}{r_3} \]

are the time constants of the excitation winding and of the damping winding in the unsaturated region,

\[ T_\alpha=T_{2\alpha}+T_{3\alpha} \]

is the total rotor time constant in the unsaturated region,

\[ B=\frac{T_{2\alpha}}{T_\alpha} \]

is the ratio of the excitation-winding time constant to the total time constant in the unsaturated region.

\(n\) is a coefficient indicating from what value of the voltage at the generator terminals forcing begins.

The value of \(n\) varies within the limits \(0\le n\le m\), where

\[ m=\frac{I_k}{I_0}. \]

The flux that is transferred electromagnetically from the rotor to the stator is proportional to the total current of the rotor windings and to the coefficient of mutual induction between the rotor and the stator \(M_\alpha\).

where

\[ \Phi_\alpha=\Phi_{0\alpha}\left[k-(k-n)e^{-\frac{t_1}{T_\alpha}}\right], \tag{4} \]

\[ \Phi_{0\alpha}=\frac{u_2}{r_2}\,\frac{M_\alpha}{w_1}, \]

\(w_1\) is the number of turns of the stator winding.

The time of flux rise to point “a” is determined from the condition that the total current of the windings is equal to

\[ I_k=m\frac{u_2}{r_2}. \]

\[ t_1=T_\alpha\ln\frac{k-n}{k-m}. \tag{5} \]

The energy losses in the steel of the machine over the time interval from \(0\) to \(t_1\) are

\[ Q_{c\alpha}=\frac{P_0'}{\Phi_{\mathrm n}^{\,2}}\int_0^{t_1}\Phi_\alpha^2\cdot dt, \tag{6} \]

where

\[ \Phi_{\mathrm n}=\frac{u_2}{r_2w_1}\,[M_\beta+m(M_\alpha-M_\beta)] =\frac{u_2M_\alpha}{r_2w_1}\,[c+m(1-c)]. \tag{7} \]

\[ c=\frac{M_\beta}{M_\alpha}=\frac{\tg\beta}{\tg\alpha} \]

is the ratio of the mutual inductances in the saturated and unsaturated regions,

\(P_0'\) is the no-load loss in the steel of the machine.

After substituting into (6) the value of the flux \(\Phi_\alpha\) from (4) and integrating, we obtain

\[ Q_{c\alpha}= \frac{P_0'T_\alpha}{[c+m(1-c)]^2} \left[ k^2\ln\frac{k-n}{k-m} -(m-n)(k+0.5m+0.5n) \right]. \tag{8} \]

For the energy of losses in the copper of the excitation winding and the damper winding,

\[ Q_{v\alpha}=\int_0^{t_1} i_2^2\cdot r_2\cdot dt = \frac{u_2^2}{r_2}T_\alpha \left\{ k^2\ln\frac{k-n}{k-m} -(m-n)B[k(2-B)+0.5B(m+n)] \right\}, \tag{9} \]

\[ Q_{d\alpha}=\int_0^{t_1} i_3^2\cdot r_3\cdot dt = -\frac{u_2^2}{r_2}(1-B)BT_\alpha(m-n)(k-0.5m-0.5n). \tag{10} \]

After the magnetic system passes into the saturated state, the system of differential equations has the previous form (1), except that, instead of the unsaturated quantities with subscript \(\alpha\), saturated quantities with subscript \(\beta\) enter. The solution of this system, taking into account the new initial conditions at \(t_2=0\),

\[ i_2+i_3=m\frac{u_2}{r_2} \tag{11} \]

has the form:

\[ i_2=\frac{k u_2}{r_2}\left(1-B\,\frac{k-m}{k}\,e^{-\frac{t_2}{T_\beta}}\right), \tag{12} \]

\[ i_3=-\frac{u_2(k-m)}{\sqrt{L_{2\beta}\cdot L_{3\beta}(\delta_{2\beta}+\delta_{3\beta})}}\, e^{-\frac{t_2}{T_\beta}}, \tag{13} \]

\[ \left. \begin{aligned} \delta_{2\beta}&=\frac{r_2}{L_{2\beta}}\\ \delta_{3\beta}&=\frac{r_3}{L_{3\beta}} \end{aligned} \right\} \quad \text{— damping coefficients of the currents in the excitation and damper windings in the saturated region,} \]

\[ L_{2\beta},\ L_{3\beta} \quad \text{— inductances of the excitation and damper windings in the saturated region,} \]

\[ T_\beta=T_{2\beta}+T_{3\beta} \quad \text{— total rotor time constant in the saturated region,} \]

\[ B=\frac{T_{2\beta}}{T_\beta}=\frac{T_{2\alpha}}{T_\alpha} \quad \text{— ratio of the time constant of the excitation winding to the total time constant in the saturated and unsaturated regions.} \]

The total current of the excitation and damper windings is

\[ i_2+i_3=\frac{k u_2}{r_2}\left(1-\frac{k-m}{k}\,e^{-\frac{t_2}{T_\beta}}\right). \tag{14} \]

The rise time of the total current from \(m\,\dfrac{u_2}{r_2}\) to the steady-state value can be determined from the condition that the total current is equal to \(\dfrac{u_2}{r_2}\).

\[ t_2=T_\beta \ln \frac{k-m}{k-1}. \tag{15} \]

The rise time of the total current from the value corresponding to point \(a\) (Fig. 1) to \(0.95\,\dfrac{u_2}{r_2}\)

\[ t'_2=T_\beta \ln \frac{k-m}{k-0.95}. \tag{16} \]

The flux in section \(ab\) is equal to the flux at the bend point \(m\,\dfrac{u_2}{r_2}\,\dfrac{M_\alpha}{w_1}\), plus the increment due to the difference of the currents \(\left(i_2-\dfrac{u_2}{r_2}m\right)\dfrac{M_\beta}{w_1}\) (Fig. 1); then

\[ \begin{aligned} \Phi_\beta &=\frac{m u_2 M_\alpha}{r_2 w_1} +\frac{k u_2}{r_2 w_1}M_\beta \left(1-\frac{k-m}{k}\,e^{-\frac{t_2}{T_\beta}}\right) -\frac{m u_2 M_\beta}{r_2 w_1} \\ &=\frac{M_\alpha u_2}{r_2 w_1} \left\{(1-c)m+c\left[k-(k-m)e^{-\frac{t_2}{T_\beta}}\right]\right\}. \end{aligned} \tag{17} \]

The dependence of the energy of losses in the machine steel on the forcing coefficient over the section \(ab\):

\[ Q_{c\beta}=\frac{P_0'}{\Phi_{\mathrm{н}}^2}\int_0^{t_2}\Phi_\beta^2\cdot dt = \frac{P_0'T_\beta}{[c+m(1-c)]^2} \left\{[(1-c)m+ck]^2 \times \right. \]

\[ \left. {}\times \ln\frac{k-m}{k-1} +(m-1)c(2m-1.5mc+ck+0.5c) \right\}. \tag{18} \]

The energy released in the form of losses in the excitation winding over the section \(ab\) is equal to:

\[ Q_{v\beta}=\int_0^{t_2} i_2^2\cdot r_2\cdot dt = \frac{u_2^2}{r_2}T_\beta \left[ k^2\ln\frac{k-m}{k-1} +(m-1)B(2k+ \right. \]

\[ \left. {}+0.5Bm+0.5B-Bk) \right]. \tag{19} \]

In the damper winding:

\[ Q_{d\beta}=\int_0^{t_2} i_3^2\cdot r_3\cdot dt = \frac{u_2^2}{r_2}T_\beta(m-1)B(k+0.5m+0.5). \tag{20} \]

Adding the right-hand sides of equations (19) and (20), we obtain the energy of losses released in the excitation winding and the damper winding during the time \(t_2\).

\[ Q_{\beta m}= \frac{u_2^2}{r_2}T_\beta \left[ k^2\ln\frac{k-m}{k-1} +(m-1)B(k+0.5m+0.5) \right]. \tag{21} \]

Knowing the dependence of the energy of losses in the copper of the excitation winding on the forcing coefficient over two sections, one can find the permissible forcing coefficient from the condition of equality of the energies of losses released in the excitation winding under intermittent forced excitation and under constant excitation over one and the same time. However, it will be more correct to determine the permissible forcing coefficient for the entire operating cycle, taking into account the second stage of intermittent excitation—forced extinction.

Impact generators have a powerful damping winding, the damping circuits of which are short-circuited. Therefore, even after complete disconnection of the current in the excitation winding, there remains a magnetic flux that is considerable in magnitude and decays slowly. The only possibility of carrying out field extinction is to act upon the magnetic field of the damping circuits by some other magnetic field of opposite direction [2, 3].

Let us consider the qualitative aspect of the extinction process. Neglecting the leakage flux between the damping and excitation windings, we obtain the following system of differential equations for the saturated part of the no-load characteristic:

\[ r_2 i_2+L_{2\beta}\frac{di_2}{dt} +\sqrt{L_{2\beta}\cdot L_{3\beta}}\frac{di_3}{dt} = -ku_2, \]

\[ r_3 i_3+L_{3\beta}\frac{di_3}{dt} +\sqrt{L_{3\beta}\cdot L_{2\beta}}\frac{di_2}{dt} = 0. \tag{22} \]

The solution of this system of equations has the form:

\[ i_2=\frac{u_2}{r_2}\left[(1+k)B e^{-\frac{t_3}{T_\beta}}-k\right], \tag{23} \]

\[ i_3=-\frac{u_2}{r_2}(1+k)(1-B)e^{-\frac{t_3}{T_\beta}} . \tag{24} \]

The time of field extinction in the first interval can be determined from the condition that the total current at the inflection point is equal to \(m \dfrac{u_2}{r_2}\):

\[ t_3=T_\beta \ln \frac{k+1}{k+m}. \tag{25} \]

In the section \(ab\) the flux varies according to the relation:

\[ \Phi_\beta=\frac{u_2 M_\alpha}{r_2 w_1} \left\{ m(1-c)+c\left[(1+k)e^{-\frac{t_3}{T_\beta}}-k\right] \right\} \tag{26} \]

The energy released in the steel of the machine during the time \(t_3\) is equal to:

\[ Q_{c\beta г}= \frac{P_0 T_\beta}{[c+m(1-c)]^2} \left\{ \ln \frac{k+1}{k+m}\,[m(1-c)-ck]^2+ \right. \]

\[ \left. +(1-m)c(2m-1.5mc-kc+0.5c) \right\}. \tag{27} \]

In the excitation winding,

\[ Q_{в\beta г}= \int_0^{t_3} i_2^2 \cdot r_2 \cdot dt = \frac{u_2^2 T_\beta}{r_2} \left\{ k^2 \ln \frac{k+1}{k+m}+ \right. \]

\[ \left. +(m-1)B[2k-B(k+0.5m+0.5)] \right\}. \tag{28} \]

In the damper winding,

\[ Q_{д\beta г}= \int_0^{t_3} i_3^2 \cdot r_3 \cdot dt = -\frac{u_2^2}{r_2}(1-B)T_\beta(m-1)\times(k+0.5m+0.5). \tag{29} \]

Adding the right-hand sides of equations (28) and (29), we obtain the energy of the losses released in the damper winding and the excitation winding:

\[ Q_{в\beta г}+Q_{д\beta г} = \frac{u_2^2}{r_2}T_\beta \left[ k^2 \ln \frac{k+1}{k+m} + B(m-1)(k-0.5m-0.5) \right]. \tag{30} \]

After the transition of the magnetic system into the unsaturated state, the system of differential equations will have the same form (22), but instead of quantities with the index \(\beta\), quantities with the index \(\alpha\) will enter into them. Taking into account the new initial conditions:

\[ i_2(0)+i_3(0)=m\frac{u_2}{r_2} \tag{31} \]

the expressions for the currents are equal to

\[ i_2=\frac{u_2}{r_2}\left[(k+m)Be^{-\frac{t_4}{T_\alpha}}-k\right], \tag{32} \]

\[ i_3=\frac{u_2}{r_2}(k+m)(1-B)e^{-\frac{t_4}{T_\alpha}} \tag{33} \]

The extinction time is determined from the condition

\[ i_2+i_3=n\frac{u_2}{r_2}, \tag{34} \]

whence

\[ t_4=T_\alpha \ln \frac{k+m}{k+n}. \tag{35} \]

The flux in section \(oa\):

\[ \Phi_\alpha=\frac{u_2M_\alpha}{r_2u_1}\left[(k+m)e^{-\frac{t_4}{T_\alpha}}-k\right]. \tag{36} \]

The dependence of the iron losses on \(k\) for section \(oa\):

\[ Q_{c\alpha}= \frac{P'_0T_\alpha}{[c+m(1-c)]^2} \left[ k^2\ln\frac{k+m}{k+n}+(m-n)(0{,}5m+0{,}5n-k) \right], \tag{37} \]

The energy of the losses released in the excitation winding in section \(oa\) is equal to

\[ Q_{\text{в} \alpha \text{г}}=\int_0^{t_4} i_2^2\cdot r_2\cdot dt =\frac{u_2^2T_\alpha}{r_2} \left[ k^2\ln\frac{k+m}{k+n} + \right. \]

\[ \left. +(m-n)B(0{,}5mB+0{,}5nB+Bk-2k) \right]. \tag{38} \]

In the damper winding

\[ Q_{\text{д} \alpha \text{г}}=\int_0^{t_4} i_3^2\cdot r_3\cdot dt = \frac{u_2^2}{r_2}T_\alpha(1-B)B(m-n)\cdot(0{,}5m+0{,}5n+k). \tag{39} \]

The total losses in the rotor windings for section \(oa\):

\[ Q_{\text{в} \alpha \text{г}}+Q_{\text{д} \alpha \text{г}} = \frac{u_2^2T_\alpha}{r_2} \left[ k^2\ln\frac{k+m}{k+n} + B(m-n)(0{,}5m+0{,}5n-k) \right]. \tag{40} \]

Knowing the distribution of the losses in the copper and iron of the machine as a function of the forcing coefficient under forced excitation and de-excitation, it is possible to find the total losses over the operating cycle.

For the iron:

\[ Q_{cц}= \frac{P'_0T_\alpha}{[c+m(1-c)]^2} \left\{ k^2\ln \frac{(k-n)(k+m)}{(k-m)(k+n)} + c[(1-c)m+ \right. \]

\[ \left. +ck]^2\ln\frac{k-m}{k-1} + c[m(1-c)-ck]^2\ln\frac{k+1}{k+m} + 2k[c^3(m-1)-m+n] \right\} \tag{41} \]

For the copper of the excitation winding:

\[ Q_{M\mathrm{exc}} = \frac{u_2^2 T_\alpha}{r_2} \left\{ k^2 \ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} + 2Bk(B-2)[(m-n)-c(m-1)] \right\}. \tag{42} \]

The energy of the losses released during the operating cycle in the excitation winding and in the damping winding is

\[ Q_{M\mathrm{c}} = \frac{u_2^2 T_\alpha}{r_2} \left\{ k^2 \ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} + 2Bk[c(m-1)-m+n] \right\}. \tag{43} \]

The rise and decay times of the magnetic field, which determine the duration of the operating cycle of an impact generator under intermittent excitation, can be obtained by adding expressions (5), (15), (25), and (35):

\[ t_{\mathrm{c}} = T_\alpha \ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} . \tag{44} \]

It is of interest to examine the dependence of the individual components of the energy losses under intermittent excitation and under constant excitation for the same time, for various forcing factors.

Let us denote

\[ k_{4\mathrm{n}} = \frac{Q_{\mathrm{st}}}{P'_0 t_{\mathrm{c}}}, \]

where \(P'_0 t_{\mathrm{c}}\) is the energy loss in the steel of the machine over a time equal to the operating cycle under constant excitation,

\[ k_{4\mathrm{n}} = \frac{ k^2 \ln \frac{(k-n)(k+m)}{(k-m)(k+n)} + N \ln \frac{k-m}{k-1} + F \ln \frac{k+1}{k+m} }{ [c+m(1-c)]^2 \ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} + \frac{2k[c^3(m-1)-m+n]}{\longrightarrow} }, \tag{45} \]

where

\[ N = c[(1-c)m+ck]^2 \]

\[ F = c[m(1-c)-ck]^2 . \]

The results of the calculation by (45) are presented in Fig. 2 for the following parameters, taking account of saturation of the machine [4]:

\[ m=0.5;\quad c=0.317;\quad \text{for } n=0 \text{ and } n=0.5. \]

Under intermittent excitation, for \(n=0\), as the forcing factor is increased to two, the steel losses decrease by almost a factor of 3 in comparison with constant excitation for the same time. With a further increase in the forcing factor, the energy losses in the steel remain almost constant, equal to the losses at \(k=2\). For \(n=0.5\) and \(k=2\), the energy of the losses in the steel of the machine, when calculated for one stator-current pulse, decreases by approximately a factor of 1.3 in comparison with constant excitation; however, the cycle duration for the same forcing factor at \(n=0.5\) decreases by almost a factor of 2 in comparison with the cycle duration at \(n=0\).

Fig. 2. Dependence of the ratio of energy losses in the steel under intermittent and constant excitation on the forcing coefficient.

By analogy, denote

\[ k_{5\mathrm{n}}=\frac{Q_{\mathrm{mc}}}{\frac{u_2^2}{r_2}t_{\mathrm{c}}}, \]

where

\[ \frac{u_2^2 t_{\mathrm{c}}}{r_2} \]

is the energy loss in the excitation winding and the damper under constant excitation over the operating cycle.

\[ k_{5\mathrm{n}}=k^2+ \frac{2Bk\left|c(m-1)-m+n\right|} {\ln \dfrac{(k+1)^c(k-m)^{c-1}(k-n)} {(k-1)^c(k+m)^{c-1}(k+n)}}. \tag{46} \]

The results of calculation by equation (46) are presented in Fig. 3, a and b, for \(n=0\) and \(n=0.5\). The value \(B\) depends on the ratio \(\dfrac{T_3}{T_2}\); in the calculation this ratio was varied within the limits \(0.2 \div 1\).

From Fig. 3, a and b, it follows that, for \(n=0\), \(\dfrac{T_3}{T_2}=0.8\), equality of the energy losses in the rotor under constant and intermittent excitation is reached at \(k=1.38\), and for \(n=0.5\) at \(k=1.2\).

However, for \(k=1.2\) and \(n=0.5\), the cycle duration will be 2.3 times less than for \(n=0\) and \(K=1.38\); in this case the losses in the steel for \(n=0\) and \(K=1.38\) are 2.7 times less than with constant excitation, while for \(n=0.5\) and \(K=1.2\) they are 1.23 times less than with constant excitation over the same time.

Let the ratio of the energy losses in the excitation winding under intermittent excitation to the energy losses in the excitation winding under constant excitation over the operating cycle be denoted by \(k_{6\mathrm{n}}\); then

\[ k_{6\mathrm{n}}=k^2+ \frac{2Bk(B-2)\left[(m-n)-c(m-1)\right]} {\ln \dfrac{(k+1)^c(k-m)^{c-1}(k-n)} {(k-1)^c(k+m)^{c-1}(k+n)}}. \tag{47} \]

In Fig. 3: vertical axis \(k_{BN}\); horizontal axis \(K\). Panels: \(b)\) with \(m=0.5,\ C=0.317,\ n=0.5\); \(a)\) with \(m=0.5,\ C=0.317,\ n=0\). Curves are marked \(T_3/T_2=0.2,\ 0.4,\ 0.6,\ 0.8,\ 1\).

Fig. 3. Dependence of \(k_{BN}\) on the forcing coefficient

The results of calculations from equation (47) are presented in Fig. 4, \(a\) and \(b\), for \(n=0\) and \(n=0.5\).

From an analysis of the dependences presented, it is evident that, as the forcing ratio increases, the energy losses in the steel, calculated for one impulse, decrease in comparison with the losses under constant excitation over the same time. On the other hand, the losses in the rotor windings increase. If it is assumed that the losses in the field winding under intermittent excitation are equal to the losses released in the winding over the same time under continuous excitation, then the admissible forcing coefficient can be determined from the expression

\[ \frac{ \dfrac{u_2^2 T_{\alpha}}{r_2} \left\{ k^2 \ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} + 2Bk(B-2)\big[(m-n)-c(m-1)\big] \right\} }{ \dfrac{u_2^2}{r_2} T_{\alpha}\ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} } =1 \tag{48} \]

After simplifications,

\[ (k^2-1)\ln \frac{(k+1)^c (k-m)^{c-1}(k-n)} {(k-1)^c (k+m)^{c-1}(k+n)} + 2Bk(B-2)\big[(m-n)-c(m-1)\big]=0. \tag{49} \]

For known machine parameters this transcendental equation can be solved approximately with respect to \(K\). The results of calculations from equation (49) are presented in Fig. 5 for \(n=0\) and \(n=0.5\).

Fig. 4. Dependence of the ratio of loss energy
in the field winding under intermittent and
constant excitation on \(K\)

Fig. 5. For calculating the allowable
forcing coefficient with allowance
for saturation

It is seen from Fig. 5 that for the ratio \(\dfrac{T_\mathrm{e}}{T_\mathrm{c}}=0.8\) the permissible forcing coefficient is equal to two. For this forcing coefficient, with \(n=0\), the values of the other coefficients are:

\[ k_{4\mathrm{n}}=0.34;\quad k_{5\mathrm{n}}=1.9;\quad k_{6\mathrm{n}}=1, \]

that is, the energy losses in the steel under intermittent forced excitation will be reduced by approximately a factor of 3, and the energy losses in the excitation winding will remain unchanged in comparison with the losses under constant excitation over the same time; however, in this case the total energy losses in the rotor copper will increase by a factor of 1.9 owing to the energy losses in the damper winding. Since the damper winding has no insulation, the losses released in it will pass into the body of the rotor. The question arises: will the rotor heat more strongly under intermittent excitation than under constant excitation over the same time? To answer this question, it is necessary to know the ratio between the losses in the rotor steel and the losses in the excitation winding at no load. For example, for the TI-100-2 generator this ratio is as follows: the losses in the rotor steel are 281 kW, and the losses in the copper of the excitation winding are 120 kW. The energy losses in the rotor steel under constant excitation over a time corresponding to the duration of the operating cycle at \(K=2\) are equal to \(281t_\mathrm{c}\) kJ.

The energy losses in the rotor steel over the operating cycle for \(K=2\) under intermittent excitation are \(0.34\cdot281t_\mathrm{c}\approx95t_\mathrm{c}\) kJ. The energy losses in the excitation winding over the operating cycle at \(K=2\) are \(120t_\mathrm{c}\) kJ. The energy losses in the damper winding over the operating cycle under intermittent excitation are \(0.9\cdot120t_\mathrm{c}=108t_\mathrm{c}\) kJ.

The ratio of the energy of losses in the rotor steel and copper under intermittent excitation to the energy of losses under constant excitation over the operating cycle is equal to:

\[ \frac{95t_\mathrm{c}+108t_\mathrm{c}+120t_\mathrm{c}} {281t_\mathrm{c}+120t_\mathrm{c}}=0.8. \tag{49a} \]

Thus, under intermittent excitation the energy losses in the rotor will be 20 percent less than under constant excitation over the same time.

For \(n>m\), the machine operates on the second segment of the no-load characteristic. In this case the total losses released in the machine steel over the cycle are:

\[ Q_{\mathrm{s}\mathrm{p}}= \frac{P'_0T_\mathrm{e}}{[c+m(1-c)]^2} \left\{ [m(1-c)+ck]^2\ln\frac{k-n}{k-1} + [m(1-c)-ck]^2\ln\frac{k+1}{k+n} +2kc^2(n-1) \right\}. \tag{50} \]

The energy losses over the operating cycle in the excitation winding are:

\[ Q_{\mathrm{m}\mathrm{v}\mathrm{c}}= \frac{u_2^2T_\mathrm{e}}{r_2} \left[ k^2\ln\frac{(k+1)(k-1)}{(k+n)(k-1)} +2Bk(n-1)(2-B) \right]. \tag{51} \]

The total losses in the excitation winding and in the damper winding over the operating cycle are

\[ Q_{\mathrm{m}\mathrm{c}}= \frac{u_2^2}{r_2}T_\mathrm{e} \left[ k^2\ln\frac{(k+1)(k-n)}{(k-1)(k+n)} +2Bk(n-1) \right]. \tag{52} \]

The duration of the operating cycle of the impact generator is

\[ t_\mathrm{c}=T_\mathrm{e}\ln\frac{(k+1)(k-n)}{(k+n)(k-1)}. \tag{53} \]

The dependence of the individual components of energy losses under intermittent and constant excitation over one and the same time interval, for different forcing coefficients, can be represented analogously to equations (45--49).

For steel,

\[ k_{\mathrm{sh}} = \frac{ \left[m(1-c)+ck\right]^2 \ln \frac{k-n}{k-1} + \left[m(1-c)-ck\right]^2 \ln \frac{k+1}{k+n} }{ \left[c+m(1-c)\right]^2 \ln \frac{(k+1)(k-n)}{(k+n)(k-1)} } + \frac{2c^2(n-1)k}{\longrightarrow}. \tag{54} \]

For the energy of losses in the excitation and damper windings,

\[ k_{\mathrm{ed}} = k^2 + \frac{2Bk(n-1)} { \ln \frac{(k+1)(k-n)}{(k-1)(k+n)} }. \tag{55} \]

For the energy of losses in the excitation winding,

\[ k_{\mathrm{e}} = k^2 + \frac{2Bk(n-1)(2-B)} { \ln \frac{(k+1)(k-n)}{(k-1)(k+n)} }. \tag{56} \]

The permissible forcing coefficient can be determined from the condition of equality of the energies of copper losses in the excitation winding under constant and intermittent excitation over one and the same time interval:

\[ (k^2-1)\ln \frac{(k+1)(k-n)}{(k-1)(k+n)} + 2Bk(n-1)(2-B)=0. \tag{57} \]

The results of calculation by (57) are presented in Fig. 5 for \(n=0.8\).

It is necessary to estimate the error that is obtained when calculating the cycle duration without allowing for saturation \((L_a)\), and when allowing for saturation by selecting saturated parameter values for the specified mode \((L_{a\mathrm{sr}})\), with the cycle duration calculated with allowance for saturation by replacing the no-load characteristic of the machine by two straight lines.

The cycle duration without allowance for saturation is

\[ t_{\mathrm{c}1}=T_a \ln \frac{(k+1)(k-n)}{(k-1)(k+n)}. \tag{58} \]

The cycle duration when allowing for saturation by selecting saturated parameter values for the specified mode is

\[ t_{\mathrm{c}2}=T_{a\mathrm{sr}} \ln \frac{(k+1)(k-n)}{(k-1)(k+n)}. \tag{59} \]

The cycle duration \(t_{\mathrm{c}}\), when the no-load characteristic is approximated by two straight lines for \(0<n<m\), is determined by (44). For \(K=2\), \(n=0\), \(m=0.5\), \(c=0.317\):

\[ \frac{t_{\mathrm{c}1}}{t_{\mathrm{c}}}=1.8; \qquad \frac{t_{\mathrm{c}2}}{t_{\mathrm{c}}}=1.14. \tag{59a} \]

The error in calculating the cycle duration increases further if forcing begins not from zero voltage, but from some specified value, for example, when \(n>m\).

In this case, when calculating the cycle duration by the method of approximating the no-load characteristic by two straight lines, the time constant is determined by the angle of inclination \(\beta\) of the no-load characteristic (Fig. 1).

\[ t_{\mathrm{ц}3}=T_{\beta}\ln\frac{(k+1)(k-n)}{(k-1)(k+n)}. \tag{60} \]

Then, for the case under consideration, instead of \(\dfrac{t_{\mathrm{ц}2}}{t_{\mathrm{ц}}}=1.14\), we obtain

\[ \frac{t_{\mathrm{ц}2}}{t_{\mathrm{ц}3}}=2. \tag{61} \]

In the calculations, the values of the angles corresponding to the no-load characteristics of existing short-circuit generators [4] were adopted:

\[ \beta=54^\circ;\qquad \alpha_{\mathrm{av}}=70^\circ;\qquad \alpha=77^\circ. \]

Thus, in studying the intermittent excitation of an impact generator it is necessary to take into account saturation of the magnetic circuit; moreover, when calculating the duration of the intermittent-excitation cycle, saturation must be allowed for by approximating the no-load characteristic by at least two straight-line segments, while in calculating the permissible forcing coefficient it is sufficient to take account of saturation of the magnetic circuit by selecting saturated parameter values for the specified operating mode.

The use of intermittent excitation at permissible forcing coefficients makes it possible to reduce significantly the no-load losses in the calculation per one pulse of stator current, as compared with the case in which excitation is carried out continuously. The effectiveness of using forced intermittent excitation is the greater, the smaller the ratio of the time constant of the damper winding to the time constant of the field winding.

REFERENCES

1. G. A. Sipaylov, A. I. Verkhoturov. The influence of excitation forcing on the magnitude of energy losses of an impact generator. See the article in this same collection.

2. O. B. Bron. Magnetic-field quenching automata. Gosenergoizdat, 1961.

3. N. N. Nikiforskii. On the question of forced de-excitation of synchronous machines, Elektrichestvo, 1963, No. 9.

4. L. A. Kilgore, E. J. Hill, C. Flick. A new three-million Kva short circuit generator. I.E.E.E. Transactions on Power Apparatus and Systems, vol. 67, 1963.