METHOD FOR DETERMINING LOSSES IN THE AT-3 TRANSFORMER

I. G. Kuleev

COPPER LOSSES

The active component of the no-load current is a very small quantity in comparison with the magnetizing component. Consequently, it may be assumed that the copper losses under no-load operation of the transformer are caused only by the magnetizing current. Therefore, in the formulas below for calculating copper losses under no-load operation, only the magnetizing current will enter.

Many authors have studied the transient regime of a transformer and similar electromagnetic apparatus. The first reference to a current inrush occurs in 1878: L. Elphinston and M. Vinston (1), while studying the lifting force of an electromagnet, first observed this phenomenon.

In 1888 Sumpner (2), in an article on the variable coefficient of self-induction, anticipates the possibility of an inrush and even constructs a graph of the variation of current with time.

In 1892 Fleming (3), in studying the properties of transformers, drew attention to the current inrush and presented experiments to clarify its nature.

In 1894 and 1898 Hay (4) published the results of his investigations of the current inrush, in which the element of chance had already been eliminated. The author gives not only an exhaustive theory of the inrush for a circuit with constant permeability, but also clarifies the physical picture of the transient regime for a circuit with iron.

In 1905 Johnn (5) gives a method for constructing the current-inrush curve, the magnitude of which is found by successive approximation.

In 1909 Schweiger (6) proposed another graphical method, giving the result directly, but not possessing sufficient accuracy.

Subsequent authors, essentially speaking, repeat the above-mentioned method of successive approximation. These include: Steinmetz (7), Bragstad and la Cowr (8), and Jensen (9).

Below are given the principal forms of the analytical expression of the magnetization curve.

1. Bergtold’s equation (10) has the following form

\[ i=c_1\frac{B}{10^4}+c_2\left(\frac{B}{10^4}\right)^\gamma, \tag{1} \]

where \(c_1\), \(c_2\), and \(\gamma\) are constants depending on the grade of sheet steel, \(B\) is the induction in gausses, and \(i\) is the current in amperes.

1. Lamont (11) proposed an equation with exponential dependence

\[ B=M(i-e^{-kH}). \tag{2} \]

1. F. Mullendorf (12) derived the following two formulas for the dependence of \(B\) on \(H\):

\[ B = a e^{-\frac{\beta}{H^{\delta}+H}}, \tag{3} \]

\[ B = a \left[1-\frac{1}{(1+bH^{m})^{n}}\right] + H, \tag{4} \]

where \(a\) is the saturation induction, and \(\beta, \delta, b, m\), and \(n\) are constants characterizing the material.

The formulas of Bergtold, Lamont, and E. Mullendorf can be used only when the constants characterizing the properties of the material are known.

1. Summers’s equation (13)

\[ h = A_1 \frac{b}{B_m} + A_3 \sin 3 \arc \sin \frac{b}{B_m} + \]
\[ + A_5 \sin 5 \arc \sin \frac{b}{B_m} + \cdots ; \tag{5} \]

here

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

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

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

\(B\) is the instantaneous value of induction, \(B_m\) is the maximum value of induction, \(H\) is the field strength, and \(\gamma\) is indicated in Fig. 1.

Fig. 1.

It is quite obvious that the use of Summers’s formula in solving problems associated with the transient regime leads to a very cumbersome expression.

1. The Frolich–Kenneli formula (14)

\[ B - H = \frac{H}{a+bH}. \tag{6} \]

For small values of \(H\) in comparison with \(B\), this formula is written as

\[ B = \frac{H}{a+bH} \tag{7} \]

and is called Frolich’s formula.

It is not difficult to see that expression 7 is the equation of a rectangular hyperbola with asymptotes parallel to the coordinate axes. One of the asymptotes is at a distance \(\frac{1}{b}\) from the \(H\)-axis, and the other at a distance \(\frac{a}{b}\) from the \(B\)-axis (Fig. 2).

Frolich’s formula is very simple and convenient for practical application; however, in terms of accuracy it leaves much to be desired. For example, at \(H = 20\), according to Frolich’s formula (15):

for cast iron \(B = 8042\) gauss; difference against measurements . . 34%

for wrought iron \(B = 49100\) gauss; difference against measurements . . 260%

Quite successful results have very recently been obtained by A. Koepseł (16) by a simple transformation of Frolich’s formula, namely, taking \(B - \lg B\), and consequently Koepseł’s formula will be written

\[ \lg B = \frac{H}{a + bH} \tag{8} \]

or

\[ B = e^{\frac{H}{a+bH}}, \tag{9} \]

where \(e\) is the base of the Naperian logarithms.

From equation 9 one can find all quantities relating to the main curve. To determine the point of inflection of the curve, we take the second derivative of \(B\) with respect to \(H\) and set it equal to zero

Fig. 2.

\[ \frac{d^{2}B}{dH^{2}} = e^{\frac{H}{a+bH}} \left[ \frac{a^{2} - 2ab(a+bH)}{(a+bH)^{4}} \right] = 0, \]

whence

\[ a^{2} = 2ab(a+bH) = 0 \]

and the magnetic-field intensity corresponding to the point of inflection of the curve is

\[ H = \frac{a}{2b^{2}}(1 - 2b). \]

To compute the constants \(a\) and \(b\), it is necessary to obtain by measurement the values of \(B\) corresponding to two, even if closely neighboring, values of \(H\), and, substituting these values of \(B\) and \(H\) into the formula

\[ \lg B = \frac{H}{a + bH}, \]

to form a system of two equations with two unknowns \(a\) and \(b\). For example, Tables 1 and 2 give values of \(B\) calculated by Koepseł’s formula and obtained by direct measurement (for soft iron, Table 1, and for hardened steel, Table 2).

In computing the constants \(a\) and \(b\) for the case of Table 1, three field values were taken: \(H = 99\) and \(H = 258\), and also \(H = 99\) and \(H = 720\), whereby on average we have

\[ a = \frac{H_{1}H_{2}(\lg B_{2} - \lg B_{1})} {(H_{2} - H_{1}) \lg B_{1}\lg B_{2}} = 0.42, \]

\[ b = \frac{H_{2}\lg B_{1} - H_{1}\lg B_{2}} {(H_{2} - H_{1}) \lg B_{1}\lg B_{2}} = 0.2307. \]

For hardened steel (Table 2), \(a\) and \(b\) are computed from the value of \(B\) for two fields:

\[ H = 232 \text{ and } H = 860 \]

\[ a = 1.81 \text{ and } b = 0.235 \]

Table 1

H	B, obtained by measurement	B, obtained by formula	Difference in percent	H	B, obtained by measurement	B, obtained by formula	Difference in percent
99	18020	18030	+ 0.06	503	21260	20940	− 1.50
128	18440	18800	+ 2.00	630	21370	21000	− 1.50
157	18810	19270	+ 2.60	720	21440	21100	− 1.50
214	19480	19320	− 0.80	809	21460	21130	− 1.60
258	19860	20180	+ 1.6	1411	21570	21400	− 0.8
307	20280	20370	+ 0.45	3014	21570	21630	+ 0.3
412	20890	20700	− 0.95	—	—	—	—

Table 2

H	B, obtained by measurement	B, obtained by formula	Difference in percent	H	B, obtained by measurement	B, obtained by formula	Difference in percent
232	13020	13200	+ 1.40	1090	16810	16800	− 0.06
318	14080	—	—	1710	17390	17250	− 0.08
430	14890	15140	+ 1.70	2420	17730	17460	− 1.50
860	16440	16520	+ 0.50	2630	17810	17490	− 1.30

As for Koepsel’s equation, in terms of accuracy it is, to be sure, superior to all the others, but the possibility of its practical application is almost excluded, since all the operations that have to be carried out in order to obtain \(a\) and \(b\) can hardly, in respect of speed, replace the necessary measurements—especially since at least two points of the principal curve must in any case be obtained by measurement.

Finally, in computing the constants \(a\) and \(b\), the greatest accuracy must be observed, for the slightest deviations from it may, when the equation is used, lead to errors not only no smaller than, but even exceeding, the errors incurred when using Froelich’s formula (15).

b) For a qualitative analysis of phenomena in circuits with saturated iron, Ollendorff (18) used the formula

\[ h = \sinh b ; \tag{10} \]

here \(h\) is the field strength 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.

Ollendorff’s formula, obviously, expresses the dependence of \(h\) on \(b\) more simply, giving, as we shall see below, with a suitable choice of the values of \(b\), accuracy sufficient for practical purposes. Therefore, in what follows we shall use this latter formula.

Since \(h=i\), the effective value of the current during switching-on is

\[ I=\frac{I_0}{\sin h b_0} \sqrt{ \frac{1}{t_1}\int_0^{t_1} \sin h^2 b_0 \left[ -\cos(\alpha+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt}} \right]\,dt } \tag{11} \]

This integral can be solved only approximately. Of the large number of methods for approximate evaluation of integrals, Simpson’s method is less cumbersome and gives accuracy sufficient for practical purposes.

Applied to the present case, Simpson’s formula takes the following form:

\[ \begin{aligned} I&=\frac{I_0}{\sin h b_0} \sqrt{ \frac{1}{t_1}\int_0^{t_1} \sin h^2 b_0 \left[ -\cos(\alpha+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt}} \right]\,dt } =\\[4pt] &=\frac{I_0}{\sin h b_0} \sqrt{ \frac{1}{3_k} \Bigg\{ \sin h^2 b_0 \left[ -\cos(\alpha_0+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_0}} \right] +\\[4pt] &\qquad\qquad +4\sin h^2 b_0 \left[ -\cos(\alpha_1+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_1}} \right] +\\[4pt] &\qquad\qquad +2\sin h^2 b_0 \left[ -\cos(\alpha_2+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_2}} \right] +\cdots\\[4pt] &\qquad\qquad +\sin h^2 b_0 \left[ -\cos(\alpha_k+\psi)+\cos\psi e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_k}} \right] \Bigg\} }. \end{aligned} \tag{12} \]

The relative induction

\[ b_0=\frac{B_m}{x}, \]

corresponding to the normal voltage, must be so chosen that the current

\[ i=i_0\frac{\sin h b}{\sin h b_0} \tag{13} \]

corresponds to the value obtained from the magnetization curve. Here \(i_0\) is the no-load current at normal voltage.

Before proceeding to determine the switching current by formula 12, a selection of \(b_0\) was carried out for various voltages at the terminals of transformer AT-3.

The magnetization curve (1) and the curve of consumed power (2), which are shown in Fig. 3. The selection consists in the following: by assigning an arbitrary \(b_0\), we find from formula 13 such a current which, for the assigned value of \(b_0\), is equal to the current found from the magnetization curve.

In Fig. 3, the currents found from formula 13 are marked with the corresponding indices.

The points with index I refer to the case when a voltage of \(220\text{ V}\) is applied to the transformer, \(B_m=6750\) gauss and \(b_0=2.18\).

*) Concerning the decaying component entering into this equation, see the author’s work “Energy Losses in Transformers under Non-Steady Operating Conditions,” Izvestiya TII, No. 56, 1938.

Points with index II refer to the case when a voltage of 360 V is applied to the transformer, \(B_m = 10300\) gauss and \(b_0 = 3.65\).

Points with index III refer to the case when a voltage of 380 V is applied to the transformer, \(B_m = 12000\) gauss and \(b_0 = 3.9\).

To avoid recalculations in all computations and experiments, the transformer plug will be in the fourth position.

If, when calculating the switching-on current by formula 12, each period is divided into 4 parts, then the terms containing \((\alpha+\psi)=0^\circ\), \((\alpha+\psi)=90^\circ\), \((\alpha+\psi)=270^\circ\), will be so small in comparison with the terms containing \((\alpha+\psi)=180^\circ\) that they may be neglected (Table 3). In this case the formula becomes less cumbersome, containing only the terms \((\alpha+\psi)=180\).

Fig. 3.

For example, in determining the effective value of the current during a switching-on time equal to 0.25 sec., the following will be required:

\[ \frac{0.25-0.01}{0.02}+1=13 \text{ terms}. \]

The minus 0.01 is introduced to account for the first term, which is separated from the instant of switching on by 0.01 seconds.

The 0.25 seconds has been taken because, after the indicated time has elapsed, the switching-on current of transformers having normal induction reaches approximately 5% of the inrush current (Table 3).

Calculation of the current by formula 12 with such a number of terms will not require much time, and therefore in this form it can quite well be used for calculating the current under non-steady-state operating conditions.

The welding time depends on the thickness of the sheets being welded, the current intensity, and the physical constants of the metal being welded. F. I. Kislyuk (19) gives the following formula for calculating the welding time:

\[ t=\frac{T_k q A}{I_2^2 R_{1k}}; \tag{14} \]

where

\[ \begin{aligned} T_k &\text{ — temperature at the contact, equal to the temperature of the plastic state } +1480^\circ\text{C},\\ q &\text{ — area of the electrode end face,}\\ A &\text{ — coefficient taking into account the physical data of the metal,}\\ A &= \sqrt{\pi c\gamma\lambda},\\ C &\text{ — heat capacity of the welded metal — watt sec}/^\circ\text{C},\\ \gamma &\text{ — specific weight of the material in g/cm}^3,\\ \lambda &\text{ — thermal conductivity of the metal, watt/cm}^\circ\text{C},\\ I_2 &\text{ — welding current,}\\ R_{1k} &\text{ — internal resistance of the contact } R_{1k}=\frac{r_{1k}}{p^\alpha},\\ r_{1k} &\text{ — unit resistance of the welding contact for the conditions}\\ &\text{\quad of a cleaned contact and parallelism of its faces.} \end{aligned} \]

On the basis of the experiments of F. I. Kislyuk, for low-carbon steel

\[ r_{1k}=0.002 \quad \text{and} \quad \alpha=0.7. \]

It follows from formula (14) that the smaller the welding current, the longer the welding time. In order for the transformer to heat up in a short time,

we shall assume that iron sheets of such thickness are being welded that the welding time is equal to the time during which the current reaches a value equal to approximately 5% of the current surge.

As an example, Tables 3, 4, 5, 6, and 7 are given; at the end of these tables there are values of the switching currents calculated by formula 18.

In Table 3 the first six readings are taken at intervals of 90°, in order to show how large the terms containing \(-\cos(\alpha+\psi)=1\) are relative to the others.

Table 3.

For calculating the switching current of transformer AT-3 at \(b_0=3.9,\ U=380,\ \psi=0,\)
\(B=12000\) gauss, \(I_0=2.4\ \text{A}\). The plug is in the fourth position.

\(k\)	\(t\)	\(4\sinh^2(\ldots)\)	\(2\sinh^2(\ldots)\)	\(-\cos(\alpha+\psi)\)	\(\cos\psi\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt}}\)
0	0	0	0	\(-1\)	1
1	0,005	2400		0,98	0,98
2	0,01		\(2,2\ 10^6\)	\(+1\)	0,96
3	0,015	1600		0	0,95
4	0,02		0,5	\(-1\)	0,93
5	0,025	1220		0	0,91
6	0,03		\(1,28\ 10^6\)	\(-1\)	0,89
10	0,05		\(0,78\ 10^6\)	\(+1\)	0,83
14	0,07		\(0,5\ 10^6\)	\(+1\)	0,77
18	0,09		\(0,32\ 10^6\)	\(+1\)	0,71
22	0,011		\(0,212\ 10^6\)	1	0,66
26	0,013		\(0,145\ 10^6\)	1	0,61
30	0,015		\(0,09\ 10^6\)	1	0,57
34	0,017		0,070	1	0,525
38	0,019		\(0,55\ 10^6\)	1	0,49
42	0,021		\(0,042\ 10^6\)	1	0,45
46	0,023		\(0,029\ 10^6\)	1	0,42
50	0,025		\(0,025\ 10^9\)	1	0,39

Total . . . \(5,678\ 10^6\)

If we neglect the terms for \(k\) equal to an odd number and the terms for \(k\) equal to an even number but not included in this table, then

\[ I_1=\frac{I_0}{\sinh b_0} \sqrt{ \frac{2}{3k} \left\{ \sinh^2 b_0 \left[ -\cos(\alpha_0+\psi) +\cos\psi\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_0}} \right] \right. } + \]

\[ \left. +4\sinh^2 b_0 \left[ -\cos(\alpha_1+\psi) +\cos\psi\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_1}} \right] \right. + \]

\[ \left. +2\sinh^2 b_0 \left[ -\cos(\alpha_2+\psi) +\cos\psi\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{Rt_2}} \right] +\ldots \right\} \]

\[ =\frac{2,4}{25}\sqrt{\frac{5,67810^6}{3\cdot 50}}=18,7\ \text{A}; \]

here \(I_0\) is taken from the magnetization curve (Fig. 3).

Table 4

For calculating the inrush current of transformer AT-3 at \(b_0=3.9\), \(U=390\), \(\psi=45^\circ\), \(I_0=2.4\text{ A}\), \(B=12000\) gauss. The plug is in the fourth position.

\(k\)	\(t\)	\(2\sinh^2(\ldots)\)	\(-\cos(\alpha+\psi)\)	\(\cos\psi\, e^{-\left(1+\cos \dfrac{\psi}{2}\right)}\, b\sqrt{Rt}\)
	0	0	\(-0,707\)	0,707
1	0,0025		0
2	0,0075	\(2,44\cdot10^5\)	1	0,68
6	0,0275	\(1,62\cdot10^5\)	1	0,63
10	0,0475	\(1,2\cdot10^5\)	1	0,587
14	0,0675	\(0,8\cdot10^5\)	1	0,54
18	0,0875	\(0,63\cdot10^5\)	1	0,505
22	0,1075	\(0,48\cdot10^5\)	1	0,47
26	0,1275	\(0,36\cdot10^5\)	1	0,432
30	0,1475	\(0,27\cdot10^5\)	1	0,40
34	0,1675	\(0,22\cdot10^5\)	1	0,37
38	0,1875	\(0,18\cdot10^5\)	1	0,342
42	0,2075	\(0,15\cdot10^5\)	1	0,322
46	0,2275	\(0,12\cdot10^5\)	1	0,294
50	0,2475	\(0,1\cdot10^5\)	1	0,272

Total . . . \(8,57\cdot10^5\)

\[ I_2=\frac{2,4}{25}\sqrt{\frac{8,57\cdot10^5}{3\cdot50}}=7,2\ \text{A}. \]

Equivalent inrush current

\[ I_{\mathrm{e}}= \sqrt{ \frac{18,7\cdot7,2+7,2\cdot2,4}{2} + \frac{11,5^2+4,8^2}{6} } =10,8; \]

here \(2.4\text{ A}\) is the inrush current if it occurred at the moment when the voltage passes through its maximum.

Table 5

For calculating the inrush current of transformer AT-3 at \(b_0=3.65\), \(\psi=0^\circ\), \(I_0=1.65\text{ A}\), \(U=360\), \(B=10300\) gauss. The plug is in the fourth position.

\(k\)	\(t\)	\(2\sinh^2(\ldots)\)	\(-\cos(\alpha+\psi)\)	\(\cos\psi\, e^{-\left(1+\cos \dfrac{\psi}{2}\right)}\, b_0\sqrt{Rt}\)
0	0	0	1	1
2	0,01	\(8,5\cdot10^5\)	1	0,97
6	0,03	\(5,4\cdot10^5\)	1	0,9
10	0,05	\(3,52\cdot10^5\)	1	0,84
14	0,07	\(2,43\cdot10^5\)	1	0,78
16	0,09	\(1,54\cdot10^5\)	1	0,73
22	0,11	\(0,9\cdot10^5\)	1	0,67
26	0,13	\(0,76\cdot10^5\)	1	0,63
30	0,15	\(0,5\cdot10^5\)	1	0,58
34	0,17	\(0,42\cdot10^5\)	1	0,55
38	0,19	\(0,31\cdot10^5\)	1	0,53
42	0,21	\(0,23\cdot10^5\)	1	0,49
46	0,23	\(0,18\cdot10^5\)	1	0,44
50	0,25	\(0,15\cdot10^5\)	1	0,41
54	0,27	\(0,11\cdot10^5\)	1	0,38

Total . . . \(24,95\cdot10^5\)

\[ I_1=\frac{I_0}{\sin h\, b_0}\sqrt{\frac{24{,}95\cdot 10^5}{3\cdot k}} =\frac{1{,}65}{19{,}2}\sqrt{\frac{24{,}95\cdot 10^5}{3\cdot 54}} =10{,}7\ \mathrm{A} \]

Table 6

for calculating the switching-on current of transformer AT—3 at \(b_0=3{,}65\), \(\psi=45^\circ\), \(I_0=1{,}65\), \(U=360\), \(B=10300\); the fork is in the fourth position.

\(k\)	\(t\)	\(\dfrac{2\sin^2 b_0}{(\ldots)}\)	\(-\cos(\alpha+\psi)\)	\(\cos\psi e-\left(1+\cos\dfrac{\psi}{2}\right)b_0\sqrt{R}\,t\)
	0	0	1	1
2	0,0075	\(11{,}50\cdot 10^4\)	1	0,692
6	0,0275	\(8{,}00\cdot 10^4\)	1	0,642
10	0,0475	\(5{,}72\cdot 10^4\)	1	0,6
14	0,0675	\(5{,}00\cdot 10^4\)	1	0,568
18	0,0875	\(3{,}36\cdot 10^4\)	1	0,523
22	0,1075	\(3{,}50\cdot 10^4\)	1	0,487
26	0,1275	\(2{,}00\cdot 10^4\)	1	0,452
30	0,1475	\(1{,}60\cdot 10^4\)	1	0,424
34	0,1675	\(1{,}28\cdot 10^4\)	1	0,394
38	0,1875	\(1{,}06\cdot 10^4\)	1	0,367
42	0,2075	\(0{,}88\cdot 10^4\)	1	0,341
46	0,2275	\(0{,}77\cdot 10^4\)	1	0,318
52	0,2475	\(0{,}62\cdot 10^4\)	1	0,296
54	0,2675	\(0{,}54\cdot 10^4\)	1	0,275

Total . . . \(44{,}33\cdot 10^4\)

\[ I_2=\frac{I_0}{\sin h b_0}\sqrt{\frac{44{,}83\cdot 10^4}{3\cdot k}} =\frac{1{,}65}{19{,}2}\sqrt{\frac{44{,}83\cdot 10^4}{3\cdot 54}} =4{,}54\ \mathrm{A}. \]

Equivalent switching-on current

\[ I_{\mathrm{e}}=\sqrt{\frac{10{,}7\cdot 4{,}54+4{,}54\cdot 1{,}65}{2} +\frac{6{,}12^2+2{,}89^2}{6}} =6\ \mathrm{A}. \]

Table 7

for calculating the switching-on of transformer AT—3 at \(b_0=2{,}18\), \(\psi=0\), \(I_0=0{,}45\mathrm{A}\), \(U=220\), \(B=6750\); the fork is in the fourth position.

\(k\)	\(t\)	\(\dfrac{2\sin^2 b_0}{(\ldots)}\)	\(-\cos(\alpha+\psi)\)	\(\cos\psi e-\left(1+\cos\dfrac{\psi}{2}\right)b_0\sqrt{R}\,t\)
	0	0	1	1
2	0,01	2800	1	0,98
6	0,03	2340	1	0,94
10	0,05	2040	1	0,9
14	0,07	1460	1	0,86
18	0,09	1370	1	0,82
22	0,11	1240	1	0,79
26	0,13	1120	1	0,76
30	0,15	940	1	0,73
34	0,17	820	1	0,70
38	0,19	720	1	0,67

Continuation of Table 7

$k$	$t$	$\dfrac{2\sin^{2} b_{0}}{(\ldots)}$	$-\cos(a+\psi)$	$\cos\psi e-\left(1+\cos\dfrac{\psi}{2}\right)b_{0}\sqrt{R}\,t$
42	0.21	630	1	0.64
46	0.23	560	1	0.61
50	0.25	570	1	0.59
54	0.27	450	1	0.56
58	0.29	408	1	0.54
62	0.31	375	1	0.52
66	0.33	332	1	0.49
70	0.35	292	1	0.47
74	0.37	278	1	0.45
78	0.39	262	1	0.44
82	0.41	232	1	0.41
86	0.43	220	1	0.40
90	0.45	208	1	0.38

Total . . . 19607

\[ I=\frac{0.45}{4.42}\sqrt{\frac{19607}{3\cdot 90}}=0.87\,A \]

When the transformer is switched on at the moment when the voltage passes through zero, the current surge reaches its greatest value. When the transformer is switched on at the moment when the voltage passes through a maximum, there will be no current surge at all. When the transformer 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 value of the equivalent switching current for the noted 3 switching cases can be determined by the formula

(20)

\[ I_{\mathrm{e}}=\sqrt{\frac{I_{1}I_{2}t_{1}+I_{2}I_{3}t_{2}+\ldots}{t_{1}+t_{2}} +\frac{I_{12}^{2}t_{1}+I_{23}^{2}t_{2}+\ldots}{3(t_{1}+t_{2}+\ldots)}} , \tag{15} \]

where $t_{1}$ and $t_{2}$ are the time intervals between two adjacent ordinates; in the present case they are equal, i.e. $t_{1}=t_{2}$,

\[ I_{12}=I_{1}-I_{2}, \qquad I_{23}=I_{2}-I_{3}, \]

$I_{1}$ is the rms value of the switching current if switching occurred at the moment when the voltage passes through zero,

$I_{2}$ is the rms value of the switching current if switching occurred at the moment when the voltage $U=U_{\max}\sin45^\circ$,

$I_{3}$ is the rms value of the switching current if it occurred at the moment when the voltage passes through a maximum. The values of the equivalent switching current for the AT-3 transformer, calculated by formula 15 for various voltages, are given at the end of Tables 4, 6, and 7.

Having calculated the value of the equivalent current by formula 15, we find the copper losses by the formula

\[ P_{\mathrm{m}}=I_{\mathrm{e}}R. \tag{16} \]

HYSTERESIS LOSSES

In most cases of transformer switching, the hysteresis cycle is shifted relative to the axis of field intensity (Figs. 4 and 6). Upon switching—

of the transformer at the moment when the voltage passes through zero, the magnetic flux, and therefore the induction, reach a doubled value (Fig. 4); when the transformer is switched on at the moment when the voltage passes through a maximum, the magnetic flux in the transformer iron during switching-on will not exceed the normal value (Fig. 5), i.e., it will be the same as that which occurs in the steady-state regime. In all other cases of switching on the transformer, the hysteresis cycles will occupy intermediate positions between those indicated in Figs. 4 and 5, for example, Fig. 6; i.e., the maxima of induction when the transformer is switched on will lie within the limits from \(B_n\) to \(2B_n\). Here \(B_n\) is the maximum value of induction in the symmetrical cycle.

In order to establish the quantitative changes in hysteresis losses for large displaced cycles, in the electrical-measurement laboratory

Fig. 4.

Fig. 5.

of the Tomsk Industrial Institute, using a Koppels apparatus, for the above-noted moments of switching on, the hysteresis loop was recorded (Figs. 4, 5, and 6). The span of the loop in all cases remained one and the same. The investigated core was made of ordinary iron.

With the aid of a planimeter the following areas were found: (Fig. 4) \(s_1 = 19.5\ \text{cm}^2\); (Fig. 6) \(s_2 = 17\ \text{cm}^2\), and (Fig. 5) \(s_n = 15.1\ \text{cm}^2\). 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, the area increased by

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

times. When the maximum induction of the displaced cycle was increased relative to the maximum of the symmetrical cycle by

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

times, the area increased by

\[ \frac{17.0}{15.1} = 1.125 \]

times.

In Fig. 7 a graph is presented of the variation

\[ \frac{s}{s_n} = f\!\left(\frac{B}{B_n}\right). \]

Here \(B\) is the maximum value of induction in the asymmetrical cycle. As follows ...

gives from Fig. 7, \(\dfrac{s}{s_n}=f\left(\dfrac{B}{B_n}\right)\)—a straight line, whose analytical expression is

\[ \frac{s}{s_n}=1+a\frac{B}{(1+\cos\psi)B_n}; \tag{17} \]

here \(\alpha=\dfrac{s-s_n}{s_n}\), \(\psi\) is the angle determining the instant of switching on.

The area of the loop is proportional to the hysteresis losses; consequently, the graph \(\dfrac{s}{s_n}=f\left(\dfrac{B}{B_n}\right)\) represents the variation of hysteresis losses

Fig. 6.

Fig. 7.

as a function of the change in the maximum induction under asymmetric magnetization; therefore we can write

\[ \frac{P_g}{P_{gy}}=1+\frac{aB}{(1+\cos\psi)B_n}; \tag{18} \]

here \(P_{gy}\)—hysteresis losses for a symmetric cycle,
\(P_g\)—hysteresis losses for an asymmetric cycle.

When the ratio of the maximum of the displaced cycle to the maximum of the symmetric cycle changes, the hysteresis losses change linearly (Fig. 7); consequently, in order to determine the hysteresis losses during the transformer switching-on time, it is necessary to find the average value of the losses over this time interval. Since the change of the hysteresis loop when the transformer is switched on at the instant when the voltage does not pass through a maximum occurs along an exponential curve, then, substituting into formula 18

\[ B=B_n\left(1+\cos\psi e^{-\left(1+\frac{\cos\psi}{2}\right)b_0\sqrt{R}\,t}\right), \]

find the average value of the hysteresis losses during switching-on

\[ \begin{aligned} P_{2c} &= \frac{P_{2y}}{t_1} \int_0^{t_1} \left[ 1+\frac{\alpha}{1+\cos\psi} \left( 1+\cos\psi\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{\bar R t}} \right) \right]\,dt =\\ &=P_{2y}+\frac{P_{2y}\alpha}{1+\cos\psi} -\frac{\alpha\cos\psi\,P_{2y}} {(1+\cos\psi)\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{\bar R t_1}}\, e^{-\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{\bar R t}} +\\ &\quad +\frac{\alpha\cos\psi\,P_{2y}} {(1+\cos\psi)\left(1+\cos\frac{\psi}{2}\right)b_0\sqrt{\bar R t_1}} . \end{aligned} \tag{19} \]

Since the occurrence of one or another switching-on case is equally probable, the hysteresis losses under the non-steady operating condition will be equal to

\[ P_2=\frac{P_{2c\,90^\circ}+P_{2c\,45^\circ}+P_{2c\,0^\circ}}{3}; \tag{20} \]

here \(P_{2c\,0^\circ}\) are the hysteresis losses at switching-on when \(\psi=0\),
\(P_{2c\,45^\circ}\) are the hysteresis losses at switching-on when \(\psi=45^\circ\),
\(P_{2c\,90^\circ}\) are the hysteresis losses at switching-on when \(\psi=90^\circ\).
\(P_{2c\,90^\circ}=P_{2y}\) are the hysteresis losses under the established operating condition.

Using formulas 19 and 20, hysteresis losses have been calculated for various values of \(t\) (Table 9), on the basis of which the curve

\[ \frac{P_2}{P_{2y}}=f(t) \]

(Fig. 8) has been constructed for transformer AT-3 at \(U=380\) volts and \(b_0=3.9\). Here \(t\) is the switching-on time.

Fig. 8.

It follows from Fig. 8 that the hysteresis losses, when the switching-on time changes, vary only slightly and can be expressed as

\[ P_2=kP_{2y}, \tag{21} \]

where \(k=1.1 \div 1.175\) (Fig. 9).

For a small number of switchings-on, the losses due to Foucault currents \((P_\phi)\) under the established condition are equal to \(^{1}\).

Table 9

\(t\)	\(P_{2c}\ 0^\circ\)	\(P_{2c}\ 45^\circ\)	\(P_{2c}\ 90^\circ\)	\(P_2\)	\(\dfrac{P_2}{P_{2y}}\)
0.01	156	146	120	141	1.175
0.1	152	144	120	139	1.16
0.5	145	140	120	135	1.12
1.00	141	138	120	133	1.11
2.00	140	137	120	132	1.10

\(^{1}\) See the author’s work “Energy Losses in Transformers under a Non-Steady Operating Condition,” Izvestiya TTI No. 56, 1938.

Consequently, the total losses in the transformer under a non-steady operating condition will be equal to

\[ P_n = I_3^2 R + P_{\phi} + kP_{gy}. \tag{22} \]

EXPERIMENTAL PART

For the purpose of experimentally confirming formula (22), the following experiments were carried out in the electrical-machine laboratory of the Tomsk Industrial Institute.

The AT-3 transformer, for which it was desired to determine the losses by formula (22), operated for a period of time under a non-steady condition. At the end of the experiment the power consumed by the transformer was found; this was then compared with the power calculated by formula (22).

The data for the AT-3 transformer are as follows (21).

The voltage of the primary winding is \(U_1 = 380\) volts. The voltage of the secondary winding is \(U_2 = 1.16;\ 130;\ 1.49\) and \(1.74\) volts. The cross-section of the core is \(62\ \mathrm{cm}^2\); the active resistance of the primary winding, when the plug is in the fourth position, is \(R = 0.435\) ohm; the magnetization curve (1) and the curve of the power consumed at no load (2) are given in Fig. 3. The induction, when the plug is in the 4th position, is \(B = 12000\) gauss. During the experiment the transformer was immersed in transformer oil. In order to reduce the amount of heat given off to the surrounding medium, the tank with the oil was insulated from the latter by a layer of cotton wool.

During the experiment, the energy absorbed by the transformer,

\[ Pt = t(I^2R + P_{\phi} + P_g) \tag{23} \]

is expended: on heating the oil, the winding, the iron, and the vessel in which the transformer was placed, and a small part of it goes into the surrounding medium.

In formula (23), \(P\) is power, \(t\) is time in seconds, \(I\) is current, \(R\) is the active resistance of the winding, \(P_{\phi}\) is the loss due to Foucault currents, and \(P_g\) is the hysteresis loss; \(P\) is power.

The heat-balance equation has the following form:

\[ Pt = c\tau + A\tau_{cp}t, \tag{24} \]

where \(\tau\) is the increase in oil temperature during the experiment, \(\tau_{cp}\) is the mean excess of the oil temperature over the ambient temperature, \(c\) is the total heat capacity, and \(A\) is the heat transfer. Determining the total heat capacity \(c\) and the heat transfer \(A\) by calculation is difficult. To determine these quantities, the following method was used.

For 50 minutes the above-mentioned transformer was connected to the network, consuming 117 watts. During the time the transformer was connected to the network, the oil temperature rose from \(20.5^\circ\mathrm{C}\) to \(22.5^\circ\mathrm{C}\).

Fifteen hours 55 minutes after the transformer was disconnected, the oil temperature had fallen by \(2.8^\circ\mathrm{C}\) with a mean excess of the oil temperature over the ambient temperature of

\[ \tau_{cp} = \frac{22.5 + 19.7}{2} - 19.9 = 1.2^\circ\mathrm{C}; \]

here \(19.9\) is the mean ambient temperature.

The excess of the oil temperature over the surrounding temperature

of the medium is insignificant; therefore the heat losses to the surrounding medium are negligibly small. Neglecting the latter, we shall have

\[ c=\frac{Pt}{\tau}=\frac{117\cdot 3000}{2{,}5}=140000\ \frac{\text{watt sec}}{^\circ\mathrm{C}}. \]

After the transformer is disconnected from the network, the heat-balance equation will take the form

\[ c\tau=A\tau_{cp}t'; \tag{25} \]

here \(t'\) is the time during which the transformer cooled.

Substituting into formula 25 the value \(c=140000\ \text{watt sec}/^\circ\mathrm{C}\), we find

\[ A=\frac{c\tau}{\tau_{cp}t'}=\frac{140000\cdot 2{,}8}{1{,}2\cdot 57300}=5{,}7\ \text{watt}/^\circ\mathrm{C}. \]

By the method of successive approximations we finally find

\[ c=138500\ \text{watt sec}/^\circ\mathrm{C}\quad\text{and} \]

\[ A=5{,}65\ \text{watt}/^\circ\mathrm{C}. \]

In determining \(A\) and \(C\), the power supplied to the transformer was measured by a System Weston wattmeter No. 1359 for 20 A; the ambient temperature was measured with a BKU thermometer with a scale from 0 to 100°C. The active resistance of the primary winding of the transformer was determined with the aid of a control ammeter MEI No. 815 and a System Weston voltmeter No. 26969 for 6 volts.

When taking the no-load curve (1) and the curve of power consumed at no load (2) (Fig. 3), a MEI voltmeter No. 819 with an additional resistance No. 712, a MEI ammeter No. 896, and a MEI wattmeter No. 717 with an additional resistance No. 902 were used.

The comparison of the losses calculated by formula (22) with those obtained from the experiment was made on the basis of three experiments.

In all three experiments the time during which the transformer was connected to the network amounted to 75% of the cycle time, i.e., 75% of the time from one switching-on to the next switching-on. The switching-on time in each experiment corresponds to that value at which the current of the non-steady-state regime was calculated by formula 17. In this case the heat-balance equation for the duration of the experiment is expressed as

\[ 0{,}75P_0t=c\tau+A\tau_{cp}t'; \tag{26} \]

here \(t\) is the time during which the experiment lasted, \(P_0\) is the equivalent power of the non-steady-state regime, and \(t'\) 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 transformer was carried out with the aid of a direct-current machine with a wooden pulley, 0.75 of whose circumference was covered with iron. In the first experiment the number of revolutions of the machine was

\[ n_1=100\ \text{rev/min} \]

or, per second,

\[ \frac{n_1}{60}=\frac{100}{60}=1{,}67\ \text{rev/sec}. \]

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

\[ t_1=\frac{0{,}75}{1{,}67}=0{,}45\ \text{sec}, \]

in the second experiment

\[ t_l=\frac{0{,}75\cdot 60}{167}=0{,}27\ \text{sec}, \]

in the third experiment

\[ t_l=\frac{0{,}75\cdot 60}{180}=0{,}25\ \text{sec}. \]

The data of the experiments are summarized in Table 10.

Table 10

	1st experiment	2nd experiment	3rd experiment
$\theta_{mn}$	21,9°C	19,9°C	17,5°C
$\theta_{oc}$	22°C	18,2°C	17,3°C
$t_n$	11 h 57 min	4 h 45 min	12 h 55 min
$n$	100 rev/min	167 rev/min	180 rev/min
$t_k$	12 h 57 min	5 h 45 min	13 h 55 min
$\theta_{mk}$	22,5°C	21,7°C	20°C
$t''$	13 h 55 min	6 h 00 min	16 h 35 min
$U$	220 volts	360 volts	390 volts
$\tau'$	0,50°C	3,5°C	2,7°C
$\tau_0$	0,10°C	1,7°C	0,2°C
$\tau_{cp}$	0,20°C	2,6°C	1,45°C
$\tau$	0,60°C	1,8°C	2,5°C
$P_0$	34,0 watts	115,0 watts	167,0 watts
$t'$	7080 sec	4500 sec	13200 sec
$t$	3600 sec	3600 sec	3600 sec

Here $\theta_{mn}$ is the oil temperature at the beginning of the experiment;

$\theta_{oc}$ — ambient temperature;

$t_n$ — time at the beginning of the experiment;

$n$ — number of revolutions per minute of the direct-current machine;

$t_k$ — time of switching off the transformer;

$\theta_{mk}$ — oil temperature at the end of the experiment;

$t''$ — the time when the oil temperature reached the steady state;

$U$ — voltage;

$\tau'$ — excess of the oil temperature over the ambient temperature at the steady state of the oil temperature,

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

$\tau_0$ — excess of the oil temperature over the ambient temperature at the beginning of the experiment,

\[ \tau_0=\theta_{mn}-\theta_{oc}; \]

$\tau_{cp}$ — average excess of the oil temperature over the ambient temperature,

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

$\tau$ — increase of the oil temperature during the experiment;

$t'$ — time from the beginning of the experiment to the moment when the oil temperature reached the steady state;

$P_0$ — losses found from the experiment,

\[ P_0=\frac{c\tau+A\tau_{cp}t'}{0{,}75\,t}. \]

In experiments 2 and 3 (Table 10), the occurrence of one or another switching case is assumed to be equally probable, which occurs only at negligible voltage. Otherwise, owing to breakdown of the air gap of the switching device, there may be certain deviations. To avoid the noted phenomenon, experiment No. 1 was carried out with the transformer switched on at the instant when the voltage passes through zero.

Fig. 9.

For this purpose, a Huebner washer was connected to the shaft of a synchronous machine manufactured by Siemens Schuckert, \(27.5 \times 3\ A,\ 250\ V\) (Fig. 9); on its axis was placed a wooden drum, \(0.75\) of whose circumference was covered with iron. By changing the position of the brushes on the drum, it was possible to achieve switching-on of the transformer at the instant when the voltage passes through zero.

Fig. 10.

Before the start of the first experiment, the above-described method was used to adjust the installation for switching on when the voltage passes through zero. Fig. 10 shows the oscillogram of the switching current for this case.

In addition, in order to verify the theoretical conclusion regarding losses from Foucault currents, experiment No. 1 was carried out at a reduced voltage (220 V), when no large error in determining copper losses can be allowed, since they have a negligibly small value ($P_{\text{copper}} = 0.32$ watt, which amounts to 1% of the losses at no load).

The energy losses calculated by formula (22) and obtained from the experiment are brought together in Table 11. The last entry gives the number of switchings per second

Table 11

	First experiment	Second experiment	Third experiment
$U$	220 volts	360 volts	380 volts
$B_m$	6750	10300	12000
$h'$	2.22	3.7	4.0
$P_{\text{nsu}}$	30.04 watts	95 watts	105.0 watts
$P_{\text{sr}}$	3.03 watts	10.5 watts	11.7 watts
$P_2$	31.5 watts	97.0 watts	108.0 watts
$P_m$	0.32 watts	15.6 watts	54.7 watts
$P_n$	34.9 watts	124.1 watts	174.4 watts
$P_0$	35.00 watts	118.0 watts	168.0 watts

\[ n'=\frac{1}{k t_1}. \]

Here $k$ is the number of ordinates in Simpson’s formula and $t_1$ is the distance between two adjacent ordinates. $k$ in each experiment is taken to be that which was used when applying Simpson’s formula; $P_{\text{nsu}}$ is taken from curve 2 (Fig. 3); $P_n$ denotes the losses calculated by formula (22), and $P_0$ the losses obtained from the experiment. The losses from Foucault currents entering into Table 11 are taken equal to the losses from Foucault currents in the steady-state regime.

In the first experiment of Table 11 the losses calculated by formula (22) ($P_n$) and those found from the experiment ($P_0$) differ very little from each other ($P_0 - P_n = 0.1$ watt, which amounts to $\sim 0.3\%$ of $P_0$). Consequently, the conclusion that the losses from Foucault currents at a small number of switchings per second and in the steady-state operating regime are approximately equal is quite correct.

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:

\[ i_1=-\frac{I_\mu}{2}e^{-\frac{t}{Th}}-I e^{-\frac{t}{Ts}}, \tag{27} \]

\[ i_2=-\frac{I_\mu}{2}e^{-\frac{t}{Th}}+I e^{-\frac{t}{Ts}}; \tag{28} \]

here $I$ is the load current, $I_\mu$ the magnetizing current, $e$ the base of natural logarithms, $t$ time, $Th$ the time constant of the main flux, and $Ts$ the time constant of the leakage fluxes.

The magnetizing current is supplied, as is evident from equations 27 and 28, by both windings in equal measure, whereas the magnetizing current of the steady state is supplied, as is known, only from the primary side.

The transient component of the current in the secondary winding ultimately reaches only half the height of the magnetizing current corresponding to the superposed transient flux. Therefore, in the secondary winding one should expect only a doubled value of the short-circuit current.

short circuit, then in the primary winding, if the leakage is very small, higher current surges may occur.

In this case, taking the current of the secondary winding to coincide with the voltage, the resultant current of the primary winding will be

\[ I_1 = \sqrt{ I_e^2+\frac{I_{m_2}^{\prime\,2}}{\alpha_1} \int_0^{x_1} \left[ -\cos(\alpha+\psi)+\cos\psi\, e^{-\frac{t}{T_s}} \right]^2 d\alpha }; \]

here \(I'_{m2}\) is the maximum value of the referred secondary current; \(\psi\) 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

\[ I_1=\sqrt{I_e^2+I_2^{\prime\,2}}; \tag{29} \]

here \(I'_2\) is the referred value of the secondary current. Consequently, the losses in the transformer under transient conditions and with the secondary winding short-circuited will be equal to

\[ P_n=I_1^2R_1+I_2^2R_2+P_{cp}+kP_{vy}; \tag{30} \]

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

Conclusions

1. If the number of transformer switchings is small (up to 4–5 switchings per second), then the losses due to eddy currents under steady-state and transient operating conditions are equal to one another.

2. The hysteresis losses under the transient operating condition of the transformer are \(1\text{--}1.17\) of the hysteresis losses under the steady-state operating condition.

3. At no load of the transformer, the copper losses are \(P_m=I_e^2 R_1\), where \(I_e\) is the equivalent current determined by the formula

\[ I_e= \sqrt{ \frac{I_1 I_2 t_1+I_2 I_3 t_2+\ldots}{t_1+t_2} + \frac{I_{23}^2 t_2+I_{12}^2 t'_2+\ldots}{3(t_2+t'_2+\ldots)} }; \]

the currents \(I_1, I_2\), and \(I_3\ldots\) are determined by the formula

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

1. At present, the calculation of the power of welding transformers for a specified \(PKR_x\%\) is performed according to the formula

\[ P_x=P_{100}\sqrt{\frac{100}{PKR_x\%}}. \]

As we can see, this formula does not take into account the frequency and magnitude of the transformer switching current surge; therefore, it is not possible to compare this method of calculating transformer power with the one proposed by the author.

1. The advantage of the proposed method is that it makes it possible, with accuracy sufficient for practice, to determine losses under the most diverse operating conditions of the transformer.

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