I. G. Kuleev.

ENERGY LOSSES IN TRANSFORMERS UNDER TRANSIENT OPERATING CONDITIONS

The Stakhanovite movement is breaking old views of technology, breaking old technical standards, old design capacities, old production plans, and demands the creation of new, higher technical standards, design capacities, and production plans. Welding transformers, operating exclusively under transient conditions, do not have a theoretically substantiated expression for the power used to cover the losses that occur in the transformer itself. Without knowing the losses, it is impossible to calculate the transformer and, consequently, impossible to obtain from it another, higher design capacity.

The present work sets itself the aim of filling this gap.

The power delivered to a transformer operating at no load is expended within it on covering the copper losses of the primary winding and the iron losses due to hysteresis and Foucault currents. The first losses are due to the presence of active resistance in the primary winding. Hysteresis losses are due to the work that must be expended on the cyclic remagnetization of the iron. Losses from Foucault currents are due to those parasitic currents that are induced by the alternating magnetic flux in the sheets of transformer iron.

Let us consider each component of the losses separately. For simplicity, we shall first consider them during no-load operation of the transformer. In this latter case, the magnetization of the transformer iron is produced only by the primary winding.

COPPER LOSSES

The active component of the no-load current is so small in comparison with the magnetizing component that it may be neglected. Consequently, it may be considered that the losses at no load of the transformer are caused only by the magnetizing current.

Therefore, in the following formulas for calculating losses at no load, only the magnetizing current will enter. The dependence between current and flux, or current and induction, is given by the following authors.

1) Bergtold (lit. 1):

\[ i = C_1 \frac{B}{10^4} + C_2 \left(\frac{B}{10^4}\right)^\gamma \tag{1} \]

Here \(C_1\), \(C_2\), and \(\gamma\) are constants depending on the grade of sheet steel, \(B\) is the induction in gausses, and \(i\) is the current in amperes.

2) Ollendorff (lit. 2):

\[ h = \sinh b \tag{2} \]

Here \(h\) is the magnetic-field intensity and \(b\) is the relative induction.

For transformers Ollendorff gives

\[ b = \frac{B}{2500}. \]

Here \(B\) is the maximum value of the magnetic induction in gausses.

3) Summer’s (ref. 3):

\[ h=A_1\frac{b}{B_m}+A_3\,\operatorname{sn}3\,\arcsin\frac{b}{B_m} +A_5\,\operatorname{sn}5\,\arcsin\frac{b}{B_m}+\cdots . \tag{3} \]

Here

\[ A_1=H_{Bm}-(-A_3+A_5-A_7+A_9+\cdots), \]

\[ A_3-A_9=\frac{1}{3}(\gamma_2-\gamma_4+\gamma_6), \]

\[ A_5=\frac{1}{5}(\gamma_2-\gamma_4+\gamma_6-\gamma_8+\gamma_{10}), \]

\(b\)—instantaneous value of the induction,
\(B_m\)—maximum value of the induction,
\(H\)—maximum value of the field strength, and
\(\gamma\)—see Fig. 1.

Ollendorff more simply expresses the dependence of \(b\) on \(h\). Therefore in what follows I shall use this latter formula.

Fig. 1.

When a transformer is switched on at the moment when the voltage does not pass through a maximum, the current reaches a certain maximum value, after which it decays to its normal magnitude. The decay of the current depends on the saturation of the transformer iron. At greater saturations, owing to the small inductive resistance, the decay of the current takes place rather quickly (Fig. 2).

For comparison, Fig. 3 presents an oscillogram of the switching current in a large but moderately saturated transformer. In order to find the dependence of the current on time in explicit form, we shall use the differential equation of the transformer:

\[ W\omega\,10^{-8}\frac{d\Phi}{d\alpha}+iR=V_m\operatorname{sn}\alpha . \]

Here \(\Phi\)—magnetic flux, \(\alpha=\omega t\), \(i\)—instantaneous value of the current, \(R\)—active resistance, \(V_m\)—maximum value of the voltage, \(W\)—number of turns, \(\omega\)—angular velocity, and \(t\)—time.

Since \(h \equiv i\), we may write:

\[ i=i_0\frac{\sinh b}{\sinh b_0} \tag{4} \]

Here \(i_0\)—instantaneous value of the no-load current, \(b_0\)—relative induction at no load

Fig. 2.

and \(b\) is the value of the relative induction at switching-on.

Substituting formula (4) into the differential equation of the transformer, we obtain:

\[ \frac{d\Phi}{d\alpha} = \frac{10^8}{W_\omega} \left( V_m \operatorname{sn}\alpha - i_0 R \frac{\operatorname{snh} b}{\operatorname{snh} b_0} \right) \tag{5} \]

The integration of this differential equation can be carried out only approximately. I consider the simplest method, giving accuracy sufficient for practice, to be the graphical Euler–Cauchy method (Sheet 4), which consists in the following. By straight lines parallel to the axis of ordinates, we divide the axis of abscissas into \(k\) equal parts (Fig. 4), and to the left of the origin of coordinates we lay off the segment \(OP\), equal to the unit value. In my case \(OP\) is equal to one radian. At \(\alpha=\alpha_0=0\), \(\dfrac{d\Phi_0}{d\alpha_0}=0\). Substituting in the right-hand side of equation (4) \(\alpha=\alpha_1\), we determine the value \(\dfrac{d\Phi_1}{d\alpha_1}\) and lay off on the axis of ordinates the segment \(OA_1\), equal to \(\dfrac{d_1\Phi}{d\alpha_1}\). The segment \(PA_1\) will have an angular coefficient equal to \(\dfrac{d\Phi_1}{d\alpha_1}\) and will therefore be parallel to the tangent to the curve at the point \(M_1\). From the point \(M_1\) we draw a ray parallel to \(PA_1\), until it intersects, at the point \(M_2\), the straight line \(\alpha=\alpha_2\), parallel to the axis \(O\Phi\). Proceeding analogously for other values, we construct a broken line which approximately represents the required integral curve. In Fig. 5 are shown the curves \(\Phi=f(\alpha)\). They were constructed by the method indicated above for a transformer of 50 KVA, 50 periods; \(8000/191\) volts, \(\Phi=1.131 \cdot 10^6\) lines of force, \(B=12600\) gauss, \(R=13.5\ \Omega\), and \(I_0=0.2\) A.

Fig. 3.

Fig. 4.

In order to find the dependence of the magnetic flux on time and to observe its changes without resorting to the determination of a large number of points, I have constructed, by the method indicated above, curves of the variation of the magnetic flux with time for the following data (Fig. 5): curves 1, 2, and 3—the switching-on takes place at the moment when the voltage passes through zero, and the resistances of the primary winding are assumed, respectively, equal to:

\[ R_1=13.5;\quad R_2=2500 \quad \text{and} \quad R_3=1000\ \Omega. \]

Curve (4) corresponds to the case when, at the moment of switching-on, the voltage is

\[ V=V_m \operatorname{sn} 67^\circ 30', \]

and the resistance of the primary winding is assumed equal to

\[ R_4=1000\ \Omega. \]

For convenience of observation, for the cases indicated in Fig. 5, I have drawn curves passing through the maximum values of the magnetic flux (Fig. 6). The resultant magnetic flux consists, as is known, of the flux of the steady-state regime and the flux of the transient regime.

Fig. 5.

Fig. 6.

Calculations show that the flux of the transient regime can be expressed by the following formula proposed by me:

\[ \Phi_n=\Phi_m e^{-1.48 b_0 \sqrt{R}\, t}. \tag{6} \]

Here \(\Phi_n\) is the transient component of the magnetic flux, \(\Phi_m\) is the greatest value of the transient component of the magnetic flux, \(t\) is time, and \(e\) is the base of natural logarithms. In this case, the resulting magnetic flux will be equal to (p. 5):

\[ \Phi=\Phi_m\left[-\cos(\alpha+\psi)+\cos\psi\,e^{-1.48 b_0\sqrt{R}\,t}\right] \tag{7} \]

Here \(\Phi_m\) is the maximum value of the steady-state magnetic flux; \(\psi\) determines the instantaneous value of the voltage at the moment of switching on.

Substituting into this formula the values \(R\), and also \(b_0,\Phi_m,\psi\), and \(t\), corresponding to the switching cases occurring in Fig. 5, we obtain the values of the magnetic flux marked in Fig. 6 by points. We see that the flux found by the graphical method differs very little from the magnetic flux calculated by formula (7), i.e. the formula (6) proposed by me represents with sufficient accuracy the transient component of the magnetic flux.

Fig. 7.

For the purpose of comparing the curve of the magnetic flux found from formula (7) with the actual curve, I recorded an oscillogram of the switching current (Fig. 7) of a coil having active resistance \(R=0.7\) ohm. From the oscillogram (Fig. 7) and from the magnetization curve (curve 2, Fig. 9), I constructed the curve of the resulting magnetic flux (Fig. 8). The points above the curve show the value of the magnetic flux calculated from formula (7). On the basis of the comparisons made, I come to the conclusion that formula (7) may quite well be used for quantitative calculations connected with the transient operating condition of a transformer.

Fig. 8.

Since \(\Phi \equiv b\), formula (7) may be transformed as follows:

\[ b=b_0\left[-\cos(\alpha+\psi)+\cos\psi\,e^{-1.48 b_0\sqrt{R}\,t}\right] \]

Consequently, the mean effective value of the current during the switching interval is

\[ I=\frac{I_0}{\sinh b_0} \sqrt{ \frac{1}{t_1} \int_0^{t_1} \sinh^2 b_0\left[-\cos(\alpha+\psi)+\cos\psi\,e^{-1.48 b_0\sqrt{R}\,t}\right]\,dt } \]

(Here \(I_0\) is the effective value of the no-load current),

or, replacing \(\Sinh[\ldots]\),

\[ I=\frac{I_0}{\sinh b_0} \sqrt{ \frac{1}{2t_1} \int_0^{t_1} \cosh 2b_0\left[-\cos(\alpha+\psi)+\cos\psi\,e^{-1.48 b_0\sqrt{R}\,t}\right]\,dt -\frac{1}{2t_1}\int_0^{t_1}dt } \tag{8} \]

Here \(t_i\) is the switching-on time, i.e., the time from the moment of switching on to the moment of switching off.

This integral can be calculated only approximately, for example, by Simpson’s formula (sheet 6)

\[ \int_a^b f(x)\,dt=\frac{b-a}{3k}\left[f(x_0)+\varphi f(x_1)+2f(x_2)+\varphi f(x_3)+\ldots f(x_k)\right] \tag{9} \]

(Here \(k\) is the number of terms, equal to an even number);

Fig. 9.

or by the following transformations of the integrand expression

\[ \frac{1}{2t_i}\int_0^{t_i}\cosh 2b_0\left[-\cos(\alpha+\psi)+\cos\psi\,e^{-1.48b_0\sqrt{R}\,t}\right]\,dt = \]

\[ =\frac{1}{2.2t_i}\int_0^{t_i} e^{2b_0\left[-\cos(\alpha+\psi)-\cos\psi\,e^{-1.48b_0\sqrt{R}\,t}\right]}\,dt+ \]

\[ +\frac{1}{2.2t_i}\int_0^{t_i} e^{-2b_0\left[-\cos(\alpha+\psi)-\cos\psi\,e^{-1.48b_0\sqrt{R}\,t}\right]}\,dt = \]

\[ =\frac{1}{4t_i}\int_0^{t_i} e^{2b_0\cos(\alpha+\psi)}\cdot e^{2b_0\cos\psi\,e^{-1.48b_0\sqrt{R}\,t}}\,dt+ \]

\[ +\frac{1}{4t_i}\int_0^{t_i} e^{2b_0\cos(\alpha+\psi)}\cdot e^{-2b_0\cos\psi\,e^{-1.48b_0\sqrt{R}\,t}}\,dt. \]

If the following abbreviated notations are introduced

\[ 2b_0\cos\psi=a;\qquad 1.48b_0\sqrt{R}=c, \]

then the equation may also be written as follows:

\[ \frac{1}{2t_i}\int_{0}^{t_i}\cosh 2b_0\left[-\cos(\alpha+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt}}\right]\,dt = \]

\[ = \frac{1}{4t_i}\int_{0}^{t_i} e^{-2b_0\cos(\alpha+\psi)}\cdot e^{ae^{-ct}}\,dt + \frac{1}{4t_i}\int_{0}^{t_i} e^{2b_0\cos(\alpha+\psi)}\cdot e^{-ae^{-ct}}\,dt . \]

From the theory of Bessel functions (Ref. 7) it is known that

\[ e^{2b_0\cos(\alpha+\psi)} = J_0(2b_0)+2J_1(2b_0)\cos(\alpha+\psi) +2J_2(2b_0)\cos(2\alpha+\psi)+\cdots \]

\[ e^{-2b_0\cos(\alpha+\psi)} = J_0(2b_0)-2J_1(2b_0)\cos(\alpha+\psi) +2J_2(2b_0)\cos(2\alpha+\psi)-\cdots . \]

and from the theory of series it is known (Ref. 8) that

\[ e^{ae^{-ct}}=1+\frac{ae^{-ct}}{1!}+\frac{a^2e^{-2ct}}{2!}+\frac{a^3e^{-3ct}}{3!}+\cdots \]

\[ e^{-ae^{-ct}}=1-\frac{ae^{-ct}}{1!}+\frac{a^2e^{-2ct}}{2!}-\frac{a^3e^{-3ct}}{3!}+\cdots . \]

therefore,

\[ \frac{1}{2t_i}\int_{0}^{t_i}\cosh 2b_0\left[-\cos(\alpha+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt}}\right]\,dt = \]

\[ = \frac{1}{4t_i}\int_{0}^{t_i} \left[J_0(2b_0)-2J_1(2b_0)\cos(\alpha+\psi)+ \right. \]

\[ \left. +2J_2(2b_0)\cos(2\alpha+\psi)+\cdots \right] \left[ 1+\frac{ae^{-ct}}{1!}+\frac{a^2e^{-2ct}}{2!}+\cdots \right]\,dt+ \]

\[ +\frac{1}{4t_i}\int_{0}^{t_i} \left[J_0(2b_0)+2J_1(2b_0)\cos(\alpha+\psi)+\cdots\right]\times \]

\[ \times\left[ 1-\frac{ae^{-ct}}{1!}+\frac{a^2e^{-2ct}}{2!}-\cdots \right]\,dt . \]

I consider Simpson’s method to be less cumbersome and, consequently, more suitable for solving equation (8). As applied to my case, Simpson’s formula will take the following form:

\[ i=\frac{I_0}{\operatorname{snh} b_0} \sqrt{ \frac{1}{2t_i}\int_{0}^{t_i} \cosh 2b_0\left[-\cos(\alpha+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt}}\right]\,dt -\frac{1}{2t_i}\int_{0}^{t_i}dt } = \]

\[ =\frac{J_0}{\operatorname{snh} b_0} \sqrt{ \frac{1}{2.3\,k} \left\{ \cosh 2b_0\left[-\cos(\alpha_0+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt_0}}\right]+ \right. } \]

\[ \left. +4\cosh 2b_0\left[-\cos(\alpha_1+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt_1}}\right] +2\cosh 2b_0\left[-\cos(\alpha_2+\psi)+ \right. \]

\[ \left. +\cos\psi\, e^{-1.48b_0\sqrt{Rt_2}}\right] +\cdots+ \cosh 2b_0\left[-\cos(\alpha_k+\psi)+\cos\psi\, e^{-1.48b_0\sqrt{Rt_k}}\right] \right\} -\frac{1}{2} \tag{11} \]

The relative induction \(b_0=\dfrac{B_m}{x}\), corresponding to the normal voltage, must be chosen so that the current

\[ i=i_0\frac{\operatorname{snh} b}{\operatorname{snh} b_0} \tag{4} \]

corresponds to the value obtained from the magnetization curve.

Before proceeding to determine, by formula (11), the making current for the above-mentioned coil, I carried out a selection of \(b_0\), which consists in the following: assigning an arbitrary value of \(b_0\), I found from formula (4) such a current which, for the assigned value of \(b_0\), corresponded to the current found from the magnetization curve. In order to compare the currents obtained from formula (4) with the currents found from the magnetization curve, the currents I found from formula (4) are indicated in Fig. 9 by points with the corresponding indices.

The points with index I near the magnetization curve correspond to the case when a voltage of 135 volts is applied to the coil, \(B_m=10500\) gauss and \(b_0=3.18\). The points with index II refer to the case when a voltage of 168 volts is applied to the coil; \(B_m=13000\) gauss and \(b_0=5.05\). The points with index III refer to the case when a voltage of 188 volts is applied to the coil; \(B_m=14600\) gauss and \(b_0=5.37\).

When the coil is switched on at the moment when the voltage passes through zero, the current surge reaches its greatest value. When the coil is switched on at the moment when the voltage passes through a maximum, there will be no current surge at all (Fig. 10). When the coil is switched on at a moment different from those indicated above, the current surge will have some intermediate value. Since the occurrence of one or another switching case is equally probable, the equivalent value of the making current for the 3 switching cases noted above will be determined by formula (9)

Fig. 10.

\[ I_{\mathrm{э}}= \sqrt{ \frac{I_1 I_2 t_1+I_2 I_3 t_2+\ldots}{t_1+t_2+\ldots} + \frac{I_{12}^{2}t_1+I_{23}^{2}t_2}{3(t_1+t_2+\ldots)} } \tag{12} \]

Here \(t_1\) and \(t_2\) are the time intervals between two adjacent ordinates; in my case they are equal, i.e. \(t_1=t_2\);

\[ I_{12}=I_1-I_2;\qquad I_{23}=I_2-I_3; \]

\(I_1\)—the effective value of the making current if it occurred at the moment when the voltage passes through zero;

\(I_2\)—the effective value of the making current if switching occurred at the moment when the voltage \(v=v_m\sin45^\circ\);

\(I_3\)—the effective value of the making current if switching occurred at the moment when the voltage passes through its maximum.

The equivalent making current will be the greater, the greater the induction. Table 1 gives the values of the equivalent making currents, calculated by formulas (11) and (12), for different values of the voltage applied to the coil with active resistance \(R=0.7\) ohm.

Table 1.

	Voltage	Voltage	Voltage
	135 V	168 V	188 V
\(B_m\)	10500	13000	14000
\(b_0\)	3.18	−5.05	5.35
\(I_1\)	1.4	9.75	23.7
\(I_2\)	1.24	4.95	9.3
\(I_3\)	0.6	0.9	1.55
\(I_{\mathrm{e}}\)	1.2	5.25	13.3
\(k\)	160	124	100
\(t_1\)	0.0025	0.0025	0.0025
\(t\)	0.4	0.31	0.25
\(P_{\mathrm{med}}\)	1 watt	24.2 watts	123 watts

Here \(t_1=t_2=t_3=\ldots\) are the time intervals between two adjacent ordinates; \(t_1\) is the switching-on time; \(I_{\mathrm{e}}\) is the equivalent switching-on current; \(P_{\mathrm{med}}\) is the copper loss, calculated by the formula \(I_{\mathrm{e}}^2 R\). The number of ordinates in each case was chosen so that the current at the end of switching on amounted to no more than \((1–5\%)\) of the current surge at switching on.

It is evident from Table 1 that the copper losses are the greater, the greater the induction. With sufficient accuracy for practice they may be calculated by the formula I propose:

\[ P_{\mathrm{med}} = R\left(\frac{B_m}{9000}\right)^{\gamma} \tag{13} \]

Indeed, substituting the values of \(\gamma\) given below, we obtain, for

\(B = 10500\) gauss,

\[ P_{m_1}=0.7\left(\frac{10500}{9000}\right)^{10.0}=3.2\ \text{watts}, \]

for \(B = 13000\) gauss

\[ P_{m_2}=0.7\left(\frac{13000}{9000}\right)^{10.0}=28\ \text{watts}; \]

and for \(B = 14600\) gauss

\[ P_{m_3}=0.7\left(\frac{14600}{9000}\right)^{10.4}=115\ \text{watts}. \]

The considerable discrepancy in the first case is not of great significance, since at low saturation the copper losses amount to only 5%.

Fig. 11.

from the iron losses. In the second case, as we shall see below, there is no discrepancy with the experimental data. With another number of switchings per second the exponent has a different value. In order to be able to use this formula for the most varied number of switchings per second, on the basis of formulas 11 and 12 I have calculated and constructed the curve shown in Fig. 11, representing the dependence of the exponent \(\gamma\) on the number of switchings per second \(n'\).

LOSSES DUE TO FOUCAULT CURRENTS

In the steady operating regime, Foucault currents in the transformer are produced by the electromotive force induced by changes in the magnetic flux. When the transformer is switched on, in its iron, in addition to the steady current \(I_{\phi y}\), there appears also a transient component \(I_{\phi n}\). The latter has the following expression (l. 10)

\[ I_{\phi n}=-A_3 e^{-\frac{t}{T_h}}+A_4 e^{-\frac{t}{T_s}} . \tag{13} \]

\(T_s\)—time constant of the leakage fields of the Foucault currents, \(T_s=L_s:R_\phi\),

\(L_s\)—coefficient of self-inductance of leakage due to Foucault currents;

\(R_\phi\)—resistance to Foucault currents,

\(T_h\)—time constant of the main field of the Foucault currents,

\(A_3\) and \(A_4\)—constants, the values of which see below.

If it is assumed that the change in the transformer voltage occurs according to a sine law, then the mean effective value of the transient current during the switching time will be determined by the formula

\[ I_{\phi n} = \sqrt{ \frac{1}{t_i\pi} \int_0^{t_i}\int_0^\pi \left[ -A_3 e^{-\frac{t}{T_h}}\cos\psi + A_4 e^{-\frac{t}{T_s}}\cos\psi \right]^2 \,d\psi\,dt } = \]

\[ = \sqrt{ \frac{1}{t_i\pi} \int_0^{t_i}\int_0^\pi \left( A_3^2 e^{-\frac{2t}{T_h}} - 2A_3A_4 e^{-\frac{t}{T_h}}e^{-\frac{t}{T_s}} + A_4^2 e^{-\frac{2t}{T_s}} \right) \cos^2\psi\,d\psi\,dt } = \]

\[ = \sqrt{ \frac{1}{2t_i\pi} \int_0^{t_i}\int_0^\pi \left( A_3^2 e^{-\frac{2t}{T_h}} - 2A_3A_4 e^{-\frac{t}{T_h}}e^{-\frac{t}{T_s}} + A_4^2 e^{-\frac{2t}{T_s}} \right) (\cos 2\psi+1)\,d\psi\,dt } = \]

\[ = \sqrt{ \frac{T_h A_3^2}{4t_i} \left(1-e^{-\frac{2t_i}{T_h}}\right) + \frac{T_s}{4t_i}A_4^2 \left(1-e^{-\frac{2t_i}{T_s}}\right) - A_3A_4 \frac{T_hT_s}{T_h+T_s} \left(1-e^{-t_i\frac{T_h+T_s}{T_hT_s}}\right) } \tag{15} \]

In order to determine \(R_\phi\), let us use the following considerations.

Fig. 12 represents the cross-section of the transformer core. By Stokes’ theorem (l. 11), \(\int A\,ds\), taken along the curve \(S\), will be equal to the sum of the same integrals taken along the periphery of the individual areas, for along the line of contact of two adjacent areas \(A\,ds\) will have opposite values, since \(A\) preserves its direction, while \(ds\) has opposite values for adjacent areas. Consequently, the magnetic acti-

the action of the Foucault currents \(\Phi_f kO\), passing through the core, may be replaced by their action when passing through an iron turn located on the periphery of the core, whose active resistance we shall find as follows:

Losses from Foucault currents

\[ P_f = I_f^2 R_f, \]

whence

\[ R_f=\frac{P_f}{I_f^2} \tag{16} \]

Fig. 12.

G. N. Petrov (ref. 12) gives the losses from Foucault currents, for an iron sheet thickness of \(0.35\ \text{mm}\), as eight times smaller than the hysteresis losses, and for an iron sheet thickness of \(0.5\ \text{mm}\) the losses from Foucault currents as 5 times smaller than the hysteresis losses, i.e.

\[ P_f=\left(\frac{1}{6}\div\frac{1}{9}\right)P_{\text{ж}} \tag{17} \]

Here \(P_{\text{ж}}\) denotes losses in the iron.

Consequently

\[ R_f=\frac{P_f}{I_f^2} =\left(\frac{1}{6}\div\frac{1}{9}\right)\frac{P_{\text{ж}}}{I_{\text{ж}}^2} = \]

\[ =\left(\frac{1}{6}\div\frac{1}{9}\right) \frac{P_{\text{ж}}}{\left(\dfrac{P_f}{V}\right)^2} = \left(\frac{1}{6}\div\frac{1}{9}\right) \frac{P_{\text{ж}}} {\left[\left(\dfrac{1}{6}\div\frac{1}{9}\right)\dfrac{P_{\text{ж}}}{V}\right]^2} = \frac{6\div 9}{P_{\text{ж}}}V^2\ \text{ohm}. \]

Height of the turn

\[ H_v=2h_k+2l_j+2h_j \tag{18} \]

Here \(h_k\) is the height of the core leg, \(l_j\) is the length of the yoke between the axes of the legs, and \(h_j\) is the height of the yoke. The thickness of the turn is equal to the thickness of the iron sheets of the transformer; the length of the air gap in which the leakage flux passes is taken equal to \(\delta=1\ \text{cm}\).

The coefficient of self-induction corresponding to the leakage flux is

\[ L_s=\frac{0.4\pi w^2}{H_v}\left(\frac{C}{3}+\frac{\delta}{2}\right)U \tag{19} \]

Here \(U\) is the length of the iron turn.

For transformers, \(U\) is equal to twice the width of the iron sheet of the leg or yoke, if the leg and yoke have identical dimensions. In our case \(U=2.3.6=7.2\ \text{cm}\).

The time constant of the leakage flux from the Foucault currents is

\[ T_s=\frac{L_s}{R_f} = \frac{0.4\pi w^2 P_{\text{ж}}U} {2(h_k+l_j+h_j)(6\div 9)V^2} \left(\frac{C}{3}+\frac{\delta}{2}\right) \]

Substituting \(C=0.85\ \text{mm}\), we obtain

\[ T_s= \frac{0.4\pi\,35.7.2}{2.41.6.9.168^2} \left(\frac{0.035}{3}+0.5\right) = 7.92\cdot 10^{-6}\ \text{sec}. \]

Here \(W=1\), for \(C=0.5\ \text{cm}\); \(T_s=10.3\cdot 10^{-6}\ \text{sec}\).

In formula (6), by analogy with unsaturated circuits, one may take

\[ 1.48\, b_0 \sqrt{R}=\frac{1}{T}. \tag{20} \]

Here \(T\) is the time constant.

In this case the time constant of the main field of the primary winding is

\[ T_1=\frac{1}{1.48\, b_0 \sqrt{R}} =\frac{1}{1.48 \cdot 5.05 \sqrt{0.7}} =0.16 \text{ sec.} \]

The time constant of the main field of the Foucault currents at \(V=168\)

\[ T_\phi= \frac{1}{1.48 \cdot 5.05 \sqrt{\dfrac{9 \cdot 168^2}{35}}} =1.57 \cdot 10^{-3} \text{ sec.} \]

In order to determine the constants \(A_3\) and \(A_4\) entering equation (13), let us use the differential equations of two windings inductively linked by a common magnetic flux. For a short circuit on the side of the secondary winding of the transformer, they take the following form:

\[ L_1 \frac{di_1}{dt}+M\frac{di_2}{dt}+R_1 i_1=E \]

\[ L_2 \frac{di_2}{dt}+M\frac{di_1}{dt}+R_2 i_2=0 \tag{21} \]

If equation (21) is differentiated with respect to \(t\) and \(i_2\) is eliminated from the equations thus obtained, then the following linear second-order differential equation is obtained for \(i_1\), the right-hand side of which is set equal to zero, since we are interested only in the decaying currents of the transient regime (l. 13):

\[ \frac{d^2 i_{1n}}{dt^2}+(a_1+a_2)\frac{di_{1n}}{dt}+a_1 a_2 \tau i_{1n}=0. \]

Here

\[ a_1=\frac{R_1}{L_1 \tau}; \qquad a_2=\frac{R_2}{L_2 \tau}; \qquad \tau=1-\frac{M^2}{L_1 L_2}, \]

\(R_1, R_2, L_1, L_2\) are, respectively, the active resistances and coefficients of self-induction of the primary and secondary windings of the transformer; \(M\) is the coefficient of mutual induction; \(i_{1n}\) is the transient component of the current of the primary winding.

The above differential equation has the following solution:

\[ i_{1n}=A_1 e^{-a_1 t}+A_2 e^{-a_2 t}. \tag{22a} \]

Exactly the same differential equation as for \(i_{1n}\) exists also for \(i_{2n}\). Thus, for the current flowing in the secondary winding, one obtains

\[ i_{2n}=A_3 e^{-a_1 t}+A_4 e^{-a_2 t}. \tag{22b} \]

Here \(t\) is time in seconds;

\[ \left. \begin{aligned} \alpha_1&=\frac{a_1+a_2}{2}-\sqrt{\frac{(a_1+a_2)^2}{4}-a_1a_2\tau},\\ \alpha_2&=\frac{a_1+a_2}{2}+\sqrt{\frac{(a_1+a_2)^2}{4}-a_1a_2\tau}. \end{aligned} \right\} \tag{23} \]

\(A_1, A_2, A_3, A_4\) are constants of integration.

Substituting equations (22) into equation (21), and equating the right-hand side to zero, we obtain

\[ \left. \begin{aligned} L_1A_1\alpha_1+MA_3\alpha_1-R_1A_1&=0,\\ L_1A_2\alpha_2+MA_4\alpha_2-R_1A_2&=0,\\ L_2A_3\alpha_1+MA_1\alpha_1-RA_3&=0,\\ L_2A_4\alpha_2+MA_2\alpha_2-RA_4&=0. \end{aligned} \right\} \tag{24} \]

Let us assume that switching on occurs at the moment when the voltage passes through zero, i.e. (sheet 14)

\[ i_{1n}=-I_1,\quad i_{2n}=-I'_2, \tag{25} \]

\[ A_1+A_2=-i_1,\quad A_3+A_4=-i'_2. \tag{26} \]

Then equation (24), together with equation (26), gives the necessary and sufficient number of equations for determining the constants of integration from \(A_1\) to \(A_4\):

\[ A_1=\frac{I_1\alpha_1(1-\alpha_2T_1)}{\alpha_2-\alpha_1} -\frac{M\alpha_1\alpha_2 I'_2}{R(\alpha_2-\alpha_1)}, \]

\[ A_2=-\frac{I_1\alpha_2(1-\alpha_1T_1)}{\alpha_2-\alpha_1} +\frac{M\alpha_1\alpha_2 I'_2}{R(\alpha_2-\alpha_1)}, \]

\[ A_3=\frac{(1-\alpha_2T_1)L_1I_1(1-T_1\alpha_1)}{(\alpha_2-\alpha_1)MT_1} +\frac{I'_2\alpha_1(1-T_1\alpha_2)}{\alpha_2-\alpha_1}. \]

\[ A_4=-\frac{(1-\alpha_1T_1)L_1I_1(1-T_1\alpha_2)}{(\alpha_2-\alpha_1)MT_1} +\frac{I'_2\alpha_1(1-T_1\alpha_2)}{\alpha_2-\alpha_1}. \]

The leakage coefficient \(J\) is a very small quantity. Taking \(\tau=0\), we obtain for the damping constant \(\alpha_1\) an indeterminate form of the type \(\frac{0}{0}\), which, however, after evaluation by l’Hôpital’s rule (sheet 15), becomes

\[ \alpha_1=\frac{1}{T_1+T_2} \]

and, in addition,

\[ \alpha_2=\infty. \]

Substituting the value of \(\alpha_1\) into \(\alpha_2\) and the equation for \(A_3\), I obtain an indeterminacy of the type \(\frac{\infty}{\infty}\), which, after evaluation by l’Hôpital’s rule, becomes

\[ A_3=(-I_1+I'_2)\frac{T_2}{T_1+T_2}. \]

Since \(-I_1+I'_2=-I_\mu\), then

\[ A_3=-I_\mu\frac{T_2}{T_1+T_2} \tag{26a} \]

Other constants of integration are not sought, since they are not required. Here \(T_2\) is the time constant of the secondary circuit. In the case under consideration \(T_2 = T_{\phi}\), since inductively linked by the common magnetic flux will be the primary winding of the transformer and the iron core, through which the eddy currents pass, while the secondary winding is open. Consequently, in final form the formula will be written as:

\[ A_3 = - I_{\mu}\frac{T_{\phi}}{T_1 + T_{\phi}} . \]

Neglecting the component due to the leakage flux, the rms value of the transient component of the eddy currents during switching, according to formula (15), is

\[ I_{\phi n} = \frac{A_3}{2}\sqrt{\frac{T_h}{t_1}\left(1 - e^{-\frac{2t_1}{T_h}}\right)} . \]

Substituting the values of \(A_3\) and \(T_h = T_1 + T_{\phi}\) (p. 16), we obtain

\[ I_{\phi n} = - \frac{I_{\mu}}{2}\frac{T_{\phi}}{T_1 + T_{\phi}} \sqrt{\frac{T_1 + T_{\phi}}{t_1} \left(1 - e^{-\frac{2t_1}{T_1 + T_{\phi}}}\right)} . \]

The component of the eddy currents in the steady-state regime is

\[ I_{\phi y} = \frac{P_{\mathrm{ж}}}{9V}. \]

The resultant value of the eddy currents during switching is

\[ I_{\phi v} = \sqrt{I_{\phi y}^{2} + I_{\phi n}^{2}} . \tag{27} \]

The losses from eddy currents in the non-steady-state operating regime are

\[ P_{\phi v} = I_{\phi v}^{2} R_{\phi} \]

The losses from eddy currents in the non-steady-state operating regime depend on the number of switchings per second and are given in Table 2.

Table 2.

Switching time in seconds	\(I_{\phi n}\)	\(I_{\phi y}\)	\(I_{\phi v}\)	\(P_{\phi v}\)	\(P_{\phi y}\)	\(V\)	\(\dfrac{P_{\phi v}}{P_{\phi y}}\)
0.31	\(3.16 \cdot 10^{3}\)	\(8.32 \cdot 10^{2}\)	\(2.34 \cdot 10^{2}\)	4.0	3.9	168	1.03
0.02	\(5.9 \cdot 10^{3}\)	\(2.32 \cdot 10^{2}\)	\(2.38 \cdot 10^{2}\)	4.12	3.9	168	1.08

In Table 2, \(I_{\mu}\) is taken from the magnetization curve (Fig. 9), \(T_{\phi}\) is calculated by formula (20), and the losses from eddy currents in the steady-state operating regime are calculated by the formula

\[ P_{\phi y} = \frac{P_{\mathrm{ж}y}}{9}. \]

Here \(P_{\mathrm{ж}y}\) is the loss in the iron in the steady-state operating regime. From Table 2 it follows that, if the number of switchings per second is less than one hundred, then the transient component of the losses from eddy currents may be neglected.

In this case, the losses from Foucault currents under transient operating conditions will be expressed in the same way as under steady-state operating conditions, i.e.

\[ P_{\phi в}=\frac{P_{жv}}{6 \div 9} \]

Hysteresis losses

Until now no one has dealt with the question of hysteresis losses in large displaced cycles. The work of J. D. Ball (ref. 17) concerns small displaced cycles. In the book by V. K. Arkad’ev (ref. 18), likewise, hysteresis losses in large displaced cycles are not considered.

Hysteresis losses are proportional to the loop of the hysteresis cycle. In most cases of transformer switching, the hysteresis cycle is displaced relative to the axis of field intensity (Figs. 13 and 15). When a transformer is switched on at the instant when the voltage passes through zero, the magnetic flux, and consequently the induction, reach twice their value (Fig. 13). When a transformer is switched on at the instant when the voltage passes through a maximum, the magnetic flux in the iron of the transformer during switching will not exceed the normal value (Fig. 14), i.e. it will be the same as that which occurs under steady-state conditions. In all other cases of transformer switching the hysteresis cycles will

Fig. 13.

Fig. 14.

occupy intermediate positions between those indicated in Figs. 13 and 14, for example, Fig. 15. Thus, the maximum induction when the transformer is switched on will reach values within the range from \(B_n\) to \(2B_n\). Here \(B_n\) is the normal value of the induction.

In order to establish the quantitative changes in hysteresis losses for large displaced cycles, I used, in the electrical-measurements laboratory of the Tomsk Industrial Institute, a Koepsel apparatus for deter-

for the above-mentioned switching instants, the hysteresis loop was recorded (Figs. 13, 14 and 15). The sweep of the loop in all cases remained the same. The rod under investigation was made of ordinary iron.

Using a planimeter, the area was found to be: Fig. 13—\(S_1=19.5\ \text{cm}^2\); Fig. 15—\(S_2=17\ \text{cm}^2\), and Fig. 14—\(S_3=15.1\ \text{cm}^2\). That is, when the maximum induction of the displaced cycle was increased relative to the maximum of the symmetrical cycle by

\[ \frac{14500}{7500}=1.94 \]

times (Fig. 13), the area increased by

\[ \frac{19.5}{15.1}=1.29 \]

times; when the maximum induction of the displaced cycle was increased by

\[ \frac{10500}{7500}=1.4 \]

times (Fig. 15), the area increased by

\[ \frac{17}{15.1}=1.125 \]

times. The area of the hysteresis loop represents the hysteresis losses. Consequently, we may conclude that the hysteresis losses under a one-sided increase of induction grow proportionally to

\[ \left(\frac{B}{B_y}\right)^{0.375}. \]

Here \(B_y\) is the maximum value of induction in the steady-state regime, and \(B\) is the maximum value of induction in the transient regime.

Fig. 15.

Indeed, for

\[ \left(\frac{B}{B_y}\right)=1.94 \]

\[ \left(\frac{B}{B_y}\right)^{0.375}=(1.94)^{0.375}=1.282 \]

\[ \frac{S_1}{S_3}=\frac{19.5}{15.1}=1.29; \]

for

\[ \frac{B}{B_y}=1.4 \]

\[ \left(\frac{B}{B_y}\right)^{0.375}=(1.4)^{0.375}=1.134 \]

\[ \frac{S_2}{S_3}=1.125. \]

Consequently, the first hysteresis cycle after switching on will cause losses

\[ P_{1z}=P_y'\left(\frac{B_1}{B_y}\right)^{0.375}. \]

Here \(P_y\) is the hysteresis loss for one cycle under steady-state operation; \(B_1\) is the maximum value of the induction in the first cycle after switching on. The subsequent cycles, owing to damping, will have a smaller value of the maximum induction. Finally, there will come a moment when the hysteresis cycle is symmetric with respect to the time axis. The hysteresis losses are composed of the losses caused by the individual hysteresis cycles; consequently, their rms value will be equal to

\[ P_2=\sqrt{\frac{1}{f}\sum_{0}^{f} P_{\mathrm{ц}2}^{\,2}} . \]

Here \(f\) is the number of cycles; \(P_{\mathrm{ц}2}\) is the hysteresis loss for one cycle, equal to

\[ P_{\mathrm{ц}2}=P_y\left(\frac{B}{B_y}\right)^{0.375}. \tag{28} \]

The maximum value of the induction at each given instant of the non-steady regime consists of the steady-state component and the transient component determined by the instant of switching on, i.e.,

\[ B_f=B_y\left(1+\cos \psi \, e^{-\alpha T f}\right). \]

Here \(T\) is the time of one cycle; \(f\) is the number of cycles; \(\psi\) determines the instantaneous value of the voltage at switching on; and \(B_f\) is the maximum value of the induction in the \(f\)-th cycle. If it is assumed that switching on occurs at \(\cos \psi=1\) and formula (29) is substituted into formula (28), then the hysteresis losses for \(f_1\) cycles will be equal to

\[ P_2=P_y\sqrt{\frac{1}{f_1}\int_{0}^{f_1}\left(1+e^{-\alpha T f}\right)^{0.75}\,df}. \]

Here \(P_{2y}\) is the hysteresis loss under steady-state operation, and

\[ \alpha=1.48\,b_0\sqrt[3]{R}. \]

Let

\[ 1+e^{-\alpha T f}=u. \]

Then

\[ e^{-\alpha T f}=u-1;\qquad -f=\frac{\ln (u-1)}{\alpha T} \]

\[ -df=\frac{du}{(u-1)\alpha T}, \]

and

\[ \sqrt{\frac{1}{f_1}\int_{0}^{f_1}\left(1+e^{-\alpha T f}\right)^{0.75}\,df} = \sqrt{-\frac{1}{\alpha T f_1}\int_{0}^{f_1}\frac{u^{3/4}}{u-1}\,du}. \]

Suppose

\[ u=z^4. \]

then

\[ du=4z^3\,dz. \]

\[ \sqrt{-\frac{1}{\alpha T f_1}\int_{0}^{f_1}\frac{u^{3/4}}{u-1}\,du} = \sqrt{-\frac{4}{\alpha T f_1}\int_{0}^{f_1}\frac{z^6\,dz}{z^4-1}}; \qquad z^6 : z^4-1 \Rightarrow z^2+\frac{z^2}{z^4-1}. \]

Consequently,

\[ \sqrt{-\frac{4}{\alpha T f_1}\int_0^{f_1}\frac{z^6\,dz}{z^4-1}} = \sqrt{-\frac{4}{\alpha T f_1}\int_0^{f_1} z^2\,dz-\frac{4}{\alpha T f_1}\int_0^{f_1}\frac{z^2}{(z^2)^2-1}\,dz}. \]

Decomposing \(\dfrac{z^2}{(z^2)^2-1}\) into partial fractions, we obtain

\[ \frac{z^2}{(z-1)(z+1)(z^2+1)} = \frac{A}{z-1}+\frac{B}{z+1}+\frac{Cz+D}{z^2+1}. \]

Consequently,

\[ \sqrt{-\frac{4}{\alpha T f_1}\int_0^{f_1} z^2\,dz-\frac{4}{\alpha T f_1}\int_0^{f_1}\frac{z^2}{(z^2)^2-1}\,dz} = \]

\[ = \sqrt{ -\frac{4}{\alpha T f_1} \left[ \frac{z^3}{3} -\frac{1}{4}\ln(z-1) +\frac{1}{4}\ln(z+1) -\frac{1}{2}\operatorname{arctg} z \right]_0^{f_1} }. \]

Substituting

\[ z=u^{1/4}=\left(1+e^{-\alpha T f}\right)^{1/4}, \]

we find

\[ P_z=P_y \sqrt{ \frac{1}{\alpha T f_1} \left\{ -\frac{4}{3}\left(1+e^{-\alpha T f_1}\right)^{3/4} +\ln\left[\left(1+e^{-\alpha T f_1}\right)^{2/4}+1\right] \right. } \]

\[ \left. -\ln\left[\left(1+e^{-\alpha T f_1}\right)^{1/4}-1\right] -2\operatorname{arctg}\left(1+e^{-\alpha T f_1}\right)^{1/4} +1.53 \right\}. \]

The expression obtained for the hysteresis losses under non-steady operating conditions is inconvenient for use. If, using this formula, one plots the curve, it turns out to differ very little from a straight line. I shall not make a large error if I assume that the change in hysteresis losses from the instant of switching on and up to the steady state occurs along a straight line. Consequently, for the instant of switching on, when the voltage passes through zero, the equivalent hysteresis losses over the non-steady period, according to formula (12), will be equal to

\[ P_{z0}=P_{zy}\sqrt{1.29\cdot 1+0.29^2}=1.14. \]

Here \(1\) determines the losses under the steady operating condition, and \(1.29\) determines the maximum losses at switching on, when the voltage passes through zero.

For the case when, at switching on, the instantaneous value of the induction reaches \(B=1.5B_y\), the equivalent hysteresis losses over the non-steady period will be equal to

\[ P_{r1.5}=\sqrt{1.16\cdot 1+\frac{0.16^2}{3}}\,P_{zy}=1.08P_{zy}. \]

Here

\[ 1.16=\left(\frac{B}{B_y}\right)^{0.375}=(1.5)^{0.375}. \]

Since the occurrence of either switching-on case is equally probable, the equivalent hysteresis losses from the instant of switching on to the instant of the steady operating condition will be equal to

\[ P_{zv}=P_{zy} \sqrt{ \frac{1.14\cdot 1.08+1.08\cdot 1}{2} + \frac{0.32^2+0.08^2}{6} } =1.075P_{zy}. \]

Here \(1\) determines the losses at switching on, when the voltage passes through its maximum.

If switching on occurs every \(0.02\) sec, i.e., during the switching-on time only one hysteresis cycle is completed, then the hysteresis losses during switching on will be equal to

\[ P_{hv}=P_{hy}\sqrt{\frac{1.29\cdot 1.16+1.16\cdot 1}{2}+\frac{0.13^2+0.16^2}{6}}=1.15P_{hy}. \]

Consequently, it may be assumed that the hysteresis losses in the established operating regime of the transformer are

\[ P_{hv}=(1.075\ldots 1.15)P_{hy} \tag{31} \]

Here \(1.075\) refers to the case when each subsequent switching-on occurs after complete disappearance of the transient component, and \(1.15\) refers to the case when switching-on occurs every \(0.02\) sec.

The total losses in the transformer under an unsteady operating regime, according to formulas 13, 27, and 31, will be equal to

\[ P_n=R\left(\frac{B_m}{9000}\right)^\gamma+I_{\phi v}^{2}R_{\phi}+(1.075\div 1.15)P \tag{32} \]

EXPERIMENTAL PART

For the purpose of experimental confirmation of formula (32), I carried out the following experiments in the electrical-machinery laboratory of the Tomsk Industrial Institute.

A coil (a transformer model) operated for a certain time in an unsteady regime, for which it was desirable to determine the losses by formula (32). At the end of the experiment the power consumed by the coil was found, which was then compared with that calculated by formula (32).

The coil had the following data: core cross-section \(13\ \mathrm{cm^2}\), thickness of the transformer-iron sheets \(0.35\ \mathrm{mm}\), active resistance \(R=0.7\ \Omega\), number of turns \(w=444\); the magnetization curve of the coil (1) and the curve of power consumed at no load (2) are given in Fig. 9. During the experiment the coil was immersed in transformer oil. In order to reduce the amount of heat given off to the surroundings, the tank with oil was insulated from the surrounding medium.

During the experiment the energy absorbed by the coil,

\[ Pt=t(I^2R+P_{\phi}+P_{h}), \tag{33} \]

is expended on heating the oil, the winding, the iron, and the vessel in which the coil with oil was placed, and part of it goes into the surrounding medium.

In formula (33) \(P\) is power, \(t\) is time in seconds, \(I\) is current, \(R\) is the active resistance of the winding, \(P_{\phi}\) is the loss from Foucault currents, and \(P_h\) is the hysteresis loss.

The equation of heat balance has the following form (ref. 9)

\[ Pt=C\tau+A\tau_{\mathrm{avg}}t \tag{34} \]

Here \(\tau\) is the increase in temperature during the experiment, \(\tau_{\mathrm{avg}}\) is the average excess of the oil temperature over the temperature of the surrounding medium, \(C\) is the total heat capacity, and \(A\) is the heat transfer. It is difficult to determine the total heat capacity \(C\) and the heat transfer \(A\) by calculation. To determine these quantities I proceeded as follows.

The coil indicated above was connected for a certain time to a network with a voltage of 230 volts. During the time that the coil was connected to the network, the oil temperature rose to \(36.5^\circ\mathrm{C}\).

Four hours 30 min. after disconnecting the coil, the oil temperature decreased by \(\tau = 6^\circ\mathrm{C}\), with

\[ \tau_{cp}=\frac{36.5+30.5}{2}-18=15.5^\circ\mathrm{C}. \]

Here \(18=\vartheta\) is the ambient temperature. After disconnecting the coil from the network, the heat-balance equation takes the following form

\[ C\tau=A\tau_{cp}t \tag{35} \]

Substituting into this formula the approximate value found by the calculation method,

\[ C \simeq 30000\ \frac{\text{watt}}{^\circ\mathrm{C}}, \]

we find

\[ A=\frac{C\tau}{\tau_{cp}t} =\frac{30000\cdot 6}{15.5\cdot 16200} =0.74\ \frac{\text{watt}}{^\circ\mathrm{C}}. \]

The heat capacity \(C\) has been found approximately. In order to find its more accurate value, the coil indicated above was connected to a network with a voltage of 230 volts for 10080 seconds. Operating under steady-state conditions, the coil consumed a power of 50 watts. During the experiment the oil temperature rose by \(14.5^\circ\mathrm{C}\). The heating of the oil is insignificant. Therefore it may be assumed that the increase in the oil temperature during heating occurs along a straight line and with very small heat losses to the surrounding medium.

In this case

\[ \tau_{cp}=\frac{Rt}{2C}=\frac{50\cdot 10080}{2\cdot 30000}\simeq 8.4^\circ\mathrm{C}, \]

and the more accurate value of the heat capacity is

\[ C=\frac{Pt-A\tau_{cp}t}{\tau} =\frac{50\cdot 10080-0.74\cdot 8.4\cdot 10080}{14.6} =30400\ \frac{\text{watt}}{^\circ\mathrm{C}}. \]

I finally take

\[ C=30400\ \frac{\text{watt}}{^\circ\mathrm{C}} \quad\text{and}\quad A=0.74\ \frac{\text{watt}}{^\circ\mathrm{C}}. \]

The comparison of the losses calculated by formula (32) with the losses obtained from experiment was made using the data of three experiments.

In all three experiments the time during which the coil was connected to the network amounts to 75% of the cycle time, i.e., 75% of the time from one switching-on to the next. The switching-on time in each experiment corresponds to the value at which the current in the transient state was calculated by formula (11). In this case the heat-balance equation of the coil during the experiment is expressed as

\[ 0.75\,P_o t=C\tau-A\tau_{cp}t' \tag{36} \]

Here \(t\) is the time during which the experiment lasted; \(P_o\) is the equivalent power of the transient regime; \(t'\) is the time from the beginning of the experiment to the end of the measurement of the oil temperature. This corresponds to the moment when the oil temperature reaches the steady state.

The periodic switching-on of the coil was carried out by means of a direct-current machine, on whose pulley a wooden ring was mounted, 0.75 of whose circumference was covered with iron. In the first experiment the number of revolutions of the machine was \(n_1 = 112.5\) rev/min, or per second

\[ \frac{n_1}{60}=\frac{112.5}{60}=1.875\ \text{rev/sec}. \]

Consequently, the switching-on time in the first experiment was

\[ t_1=\frac{0.75}{1.87}=0.4\ \text{sec}, \]

in the second experiment

\[ t_1=\frac{0.75\cdot 60}{145}=0.31\ \text{sec}, \]

in the third experiment

\[ t_1=\frac{0.75\cdot 60}{180}=0.25\ \text{sec}, \]

The experimental data are brought together in Table 3.

Table 3.

	1st experiment	2nd experiment	3rd experiment
\(\vartheta_{mn}\)	28.2	19.4	11.8°C
\(\vartheta_{oc}\)	15.7	20.2	14.2°C
\(t_n\)	17 h 27 min	5 h 7 min	12 h 30 min
\(n\)	112.5 rev/min	145 rev/min	180 rev/min
\(t_k\)	18 h 27 min	5 h 45 min	13 h 30 min
\(\vartheta_{mk}\)	28.5°C	23°C	26°C
\(t'\)	18 h 43 min	6 h 5 min	13 h 58 min
\(V\)	135	168	188
\(\tau'\)	12.8°C	2.8°C	11.8°C
\(\tau_o\)	12.5°C	−0.8	−2.4°C
\(\tau_{sr}\)	12.65°C	1°C	9.4°C
\(\tau\)	0.3°C	3.6°C	14.2°C
\(P_o\)	19.6 watts	65.3 watts	173.5 watts

Here

\(\vartheta_{mn}\) — oil temperature at the beginning of the experiment;
\(\vartheta_{oc}\) — ambient temperature;
\(t_n\) — time of the beginning of the experiment;
\(n\) — number of revolutions per minute of the direct-current machine;
\(t_k\) — time of switching off the transformer;
\(\vartheta_{mk}\) — oil temperature at the end of the experiment;
\(t'\) — time when the oil reached the steady state;
\(V\) — voltage;
\(\tau'\) — excess of the oil temperature over the ambient temperature in the steady state

\[ \tau'=\vartheta_{mk}-\vartheta_{oc}; \]

\(\tau_o\) — excess of the oil temperature over the ambient temperature at the beginning of the experiment

\[ \tau_o=\vartheta_{mn}-\vartheta_{oc}; \]

\(\tau_{\mathrm{avg}}\)—the average excess of the oil temperature over the ambient temperature,

\[ \tau_{\mathrm{avg}}=\frac{\tau' + \tau_0}{2}; \]

\(\tau\)—the increase in the oil temperature during the test;
\(P_o\)—the losses found from the test

\[ P_o=\frac{C\tau-A\tau_{\mathrm{avg}}t'}{0.75\,t}. \]

The energy losses calculated by formula (22) and obtained from the test are collected in Table 4. In the table the number of switchings per second is \(n'=\frac{1}{kt_1}\).

Here \(k\) is the number of ordinates in Simpson’s formula, and \(t_1\) is the distance between two adjacent ordinates; in each test \(k\) is taken to be the same as was used when applying Simpson’s formula. \(P_{\mathrm{zhy}}\) is taken from curve 2 (Fig. 9), \(P_n\) is the loss calculated by formula (32), and \(P_o\) is the loss obtained from the test.

Table 4.

	First test \(V=135\)	Second test \(V=168\)	Third test \(V=188\)
\(B_m\)	10500	13000	14600
\(n'\)	2.5	3.22	4.0
\(\gamma\)	10.0	10.0	10.4
\(P_{\mathrm{zhy}}\)	19 watts	35 watts	50 watts
\(P_{\phi}\)	2.1 watts	3.9 watts	5.6 watts
\(P_g\)	18.0 watts	33.5 watts	47.8 watts
\(P_m\)	3.2	28 watts	115 watts
\(P_n\)	23.3 watts	65.4	168.4
\(P_o\)	19.6	65.3 watts	173.5 watts
\(R\)	0.7	0.7	0.7

From Table 4 it is clear that the increase in losses under non-steady-state conditions occurs mainly at the expense of copper losses. They are the greater, the greater the saturation of the transformer iron and the number of switchings per second. For comparison of the losses at different numbers of switchings per second and at voltage \(V=168\) volts, Table 5 was compiled using formulas (13), (27), and (31). In it, when calculating the copper losses, the load current is not taken into account.

Table 5.

\(n'\)	\(P_m\)	\(P_g\)	\(P_{\phi}\)	\(P_{\mathrm{zhy}}\)	\(\dfrac{P_m}{P_{\mathrm{zhy}}}\)
0	3.2	31.1	3.9	35	0.09
4	28	33.5	3.9	35	0.8
50	204	25.8	4.1	35	5.85

If the transformer is switched on under short-circuit conditions, then the transient components of the current in the windings can be expressed by the following equations

\[ i_1=-\frac{I_{\mu}}{2}e^{-\frac{t}{T_h}}-Ie^{-\frac{t}{T_s}} \tag{37a} \]

\[ i_2=-\frac{I_{\mu}}{2}e^{-\frac{t}{T_h}}+Ie^{-\frac{t}{T_s}} \tag{37b} \]

Here \(I\) is the load current, \(I_\mu\) the magnetizing current, \(e\) the base of natural logarithms, \(t\) time, \(T_h\) the time constant of the main flux, and \(T_s\) the time constant of the leakage flux.

The magnetizing current is supplied, as is evident from equation (37), by both windings to an equal extent, whereas the magnetizing current in the steady state is supplied, as is known, only from the primary side.

The transient component of the current in the secondary winding reaches, finally, only one-half the height of the magnetizing current corresponding to the superimposed transient flux. If, therefore, in the secondary winding one should expect only twice the value of the short-circuit current, then in the primary winding, if the leakage is not very small, higher current surges may occur.

If we turn to formula (26a), we shall see that the magnetizing component will be the smaller, the greater the active resistance of the circuit. Since, when a welding transformer is switched on under load, the active resistance of the secondary circuit is large, it may be assumed with sufficient accuracy for practice that in this case also the magnetization of the transformer takes place only from the primary side. In such a case, taking the current of the secondary winding to coincide with the voltage, the resultant current of the primary winding will be

\[ I_1 = \sqrt{ \left(\frac{B_m}{9000}\right)^\gamma + \frac{I_{m2}^{\prime\,2}}{\alpha_1} \int_0^{\alpha_1} \left[ -\cos(\alpha+\psi)+\cos\psi\,e^{-\frac{t}{T_s}} \right]^2 \,d\alpha }. \]

Here \(I'_{m2}\) is the maximum value of the referred secondary current, and \(\psi\) determines the instantaneous value of the current at switching-on. Since the leakage time constant \(T_s\) is a small quantity, it may be assumed that

\[ I_1 = \sqrt{ \left(\frac{B_m}{9000}\right)^\gamma + I_2^{\prime\,2} }. \tag{38} \]

Here \(I'_2\) is the referred value of the secondary current. In this case the losses in the transformer in the transient regime will be equal to

\[ P_p = R_1\left[ \left(\frac{B_m}{9000}\right)^\gamma + I_2^{\prime\,2} \right] + R_2 I_2^2 + R_\phi I_\phi^2 b + (1.075 \div 1.15) P_{m y}. \tag{39} \]

Here \(R_2\) is the resistance of the secondary winding and \(I_2\) the current of the secondary winding.

Formula (39) makes it possible to calculate the losses in the transformer for any switching-on case and for any value of the induction. Consequently, with the aid of this formula a more perfect calculation of the transformer can be carried out and a greater power obtained for the same expenditure.

At the same time it is necessary to note the circumstance that the use of the welding transformer will be better in the case where the induction is small. I shall not dwell on this question in detail, since I propose to resolve this question, and a number of others connected with it, in the next work.

References

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2. F. Ollendorff. Afe, vol. XXI, 1928, p. 6; vol. XXII, 1929, p. 349; and vol. XXIII, 1929, p. 162.
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18. V. K. Arkad’ev. Electromagnetic Processes in Metals, 1935.
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