Comparison of low-frequency non-sinusoidal measurement for transformer no-load characteristics based on different interpolation methods

Abstract

As the calculation of transformer no-load characteristics measured by low-frequency test method is not perfect, different interpolation methods are explored to fit and convert the transformer's volt-ampere characteristics under power frequency (50 Hz) excitation to find the best interpolation fitting conversion method. The no-load test of transformer is carried out by using 10 Hz, 15 Hz, 20 Hz low-frequency square wave and triangular wave excitation. Then the measured excitation voltage, excitation current and no-load loss are converted to those of 50 Hz sine wave excitation. The converted volt-ampere characteristic curve is calculated by the proposed algorithm, and compared with the measured results under the 50 Hz sine wave excitation. Experiments show that under the condition that the magnetic flux amplitudes are equal at each frequency of non-sinusoidal and sine waves, whether it is square wave or triangular wave excitation, the volt-ampere characteristic curve converted to the 50 Hz sine wave excitation by the proposed method can better approximate the curve of the measured 50 Hz sine wave excitation. The average values of the relative errors of the converted excitation current of the three frequencies are within 4%, and those of the converted excitation voltage of the square wave are within 1.1%, and those of the converted excitation voltage of the triangular wave are within 0.47%. Compared with the generalized Steinmetz equation method and the multi-frequency method, the proposed method has a better conversion effect and can almost replace the no-load test results of the 50 Hz sine wave, which has practical value in engineering.

Appendices

Appendix A

For sine wave excitation, it is assumed that the induced electromotive force at both ends of the core is:

$$ e(t) = E_{m} \sin (\omega t) $$

(A1)

where E_m is the peak value of electromotive force.

From formula (1), the magnetic flux density B(t) is:

$$ B(t) = B_{m} \cos (\omega t) $$

(A2)

$$ B_{m} = \frac{{E_{m} }}{NA \cdot 2\pi f} = \frac{E}{NA \cdot \sqrt 2 \pi f} $$

(A3)

In formulas (A2) and (A3), B_m is the magnitude of the magnetic flux density, and E is the effective value of the electromotive force.

Similarly, for square wave excitation, it is assumed that the induced electromotive force and magnetic flux density at both ends of the core are:

$$ e(t) = \left\{ {\begin{array}{*{20}c} {E_{m} } &\quad {0 < t < T/2} \\ { - E_{m} } &\quad {T/2 < t < T} \\ \end{array} } \right. $$

(A4)

$$ B(t) = \left\{ {\begin{array}{*{20}l} {B_{m} (4t/T - 1)} \hfill &\quad {0 < t < T/2} \hfill \\ {B_{m} (3 - 4t/T)} \hfill & \quad{T/2 < t < T} \hfill \\ \end{array} } \right. $$

(A5)

$$ B_{m} = \frac{{E_{m} }}{4NAf} = \frac{E}{4NAf} $$

(A6)

From equations (A3) and (A6), it can be seen that to keep the amplitude of the magnetic flux density B_m generated by the excitation voltages of square wave and sine wave at the same frequency equal, and it is necessary to make:

$$ \frac{{E_{\sin } }}{{E_{{{\text{squ}}}} }} = \frac{\pi }{2\sqrt 2 } \approx 1.11 $$

(A7)

where E_sin is the effective value of sine wave, and E_squ is the effective value of square wave.

Similarly, for triangular wave excitation, it is assumed that the induced electromotive force and magnetic flux density at both ends of the core are:

$$ e(t) = \left\{ {\begin{array}{*{20}c} {E_{m} (4t/T - 1)} & \quad{0 < t < T/2} \\ {E_{m} (3 - 4t/T)} &\quad {T/2 < t < T} \\ \end{array} } \right. $$

(A8)

$$ B(t) = \left\{ {\begin{array}{*{20}l} {\frac{{E_{m} }}{NA}(2t^{2} /T - t)} \hfill & \quad{0 < t < T/2} \hfill \\ {\frac{{E_{m} }}{NA}( - 2t^{2} /T + 3t - T)} \hfill & \quad{T/2 < t < T} \hfill \\ \end{array} } \right. $$

(A9)

$$ B_{m} = \frac{{E_{m} }}{8NAf} = \frac{\sqrt 3 E}{{8NAf}} $$

(A10)

It can be seen from formulas (A3) and (A10) that to keep the amplitude of the magnetic flux density B_m generated by the excitation voltages of triangular wave and sine wave at the same frequency equal, and it is necessary to make:

$$ \frac{{E_{\sin } }}{{E_{{{\text{tri}}}} }} = \frac{\sqrt 6 \pi }{8} \approx 0.962 $$

(A11)

where E_sin is the effective value of sine wave, and E_tri is the effective value of triangular wave.

For the transformer, ignoring the voltages of the leakage inductance and DC resistance, so that the effective value of the sine wave excitation voltage is 1.11 times that of the square wave, then the amplitude of magnetic flux density B_m generated under the two waveform excitations can be equal. This is also the basis of the volt-ampere characteristic curve and core loss under square wave excitation that can be converted to a sine wave. At the same time, it can be known from Eqs. (5) and (8) that, for voltage excitations at different frequencies of the same waveform, as long as the ratio of the effective value of the applied voltage is equal to the ratio of the frequencies, the generated B_m is equal.

Appendix B

In reference [33], the core loss was calculated and derived, and the equivalent formula for eddy current loss was obtained:

$$ P_{e} = \frac{1}{{8\pi \rho N^{2} A}}\frac{1}{T}\int\limits_{0}^{T} {e(t)^{2} {\text{d}}t} $$

(B1)

where N is the number of turns, A is the area of the cross section and ρ is the resistivity of the core material.

Considering the effective value of electromotive force:

$$ E = \sqrt {\frac{1}{T}\int\limits_{0}^{T} {e(t)^{2} {\text{d}}t} } $$

(B2)

Combining equations (B1) and (B2), it can be concluded that:

$$ P_{e} = \frac{{E^{2} }}{{8\pi \rho N^{2} A}} $$

(B3)

Combining equations (A3) and (B3), the eddy current loss P_e-sin under sine wave excitation can be obtained as follows:

$$ P_{e - \sin } = \frac{\pi A}{{4\rho }}B_{m}^{2} f^{2} $$

(B4)

Combining equations (A6) and (B3), it can be concluded that the eddy current loss under square wave excitation P_e-squ is:

$$ P_{{e - {\text{squ}}}} = \frac{2A}{{\pi \rho }}B_{m}^{2} f^{2} $$

(B5)

According to formulas (A10) and (B3), the eddy current loss P_e-tri under triangular wave excitation is as follows:

$$ P_{{e - {\text{tri}}}} = \frac{8A}{{3\pi \rho }}B_{m}^{2} f^{2} $$

(B6)

From equations (B4), (B5) and (B6), it can be seen that the eddy current loss is still proportional to the square of B_m and f, no matter for sine wave, square wave or triangular wave excitation, only the difference of proportion coefficient.

Appendix C

See Tables 2, 3 and 4.

Table 2 The effective values of voltage and current of the test sine wave with different frequencies (10 Hz, 15 Hz, 20 Hz, 50 Hz)

Table 3 The effective values of voltage and current of the test triangular wave with different frequencies (10 Hz, 15 Hz, 20 Hz)

Table 4 The effective values of voltage and current of the test square wave with different frequencies (10 Hz, 15 Hz, 20 Hz)