CN105510742A - Experiment method and analysis calculation method for testing transformer volt-ampere characteristic by using low-frequency power supply - Google Patents

Abstract

The invention provides an experiment method and analysis calculation method for testing a transformer volt-ampere characteristic by using a low-frequency power supply. The method is characterized in that: a low-frequency power supply is used to carry out excitation test on a tested sample; a high-speed measuring device is used to measure testing winding voltage current waveforms u1(t) and I (t) and open circuit side voltage waveform u2(t) in testing, a circuit port voltage equation is written (wherein the u2(t) is imputed to a test side), and due to open circuit, the open circuit side voltage waveform u2(t) is approximately equal to the induced electromotive force e(t) of the winding. The method has the advantages that: the test can be completed in a condition far lower than a frequency voltage, the capacity needed by the test is reduced greatly, and the safety risk of the persons and a tested product in the test process is effectively reduced; the transformer test side winding leakage inductance can be calculated by using the voltage and current data in the test process, an accurate circuit equivalent model with leakage inductance is established, and the influence of the leakage inductance in a derivation process is considered; and the mass and volume of test equipment are reduced, the test wire connection can be simplified, and the test efficiency is effectively improved.

Description

A kind of utilize low-frequency power to test transformer volt-ampere characteristic test method and analysis calculation method

Technical field

The present invention relates to electric power transformer, mutual inductor, reactor carry out volt-ampere characteristic (or excitation property) test, be specially adapted to carry out volt-ampere characteristic test to the more insignificant transformer of leakage reactance.

Technical background

The ferromagnetic wires coil elements such as transformer, reactor and mutual inductor are as critical elements in electric system, along with the electric pressure of the increase transformer of power transmission and transformation capacity and capacity increase gradually, the capacity of the shunt reactor compensated along with the growth of transmission distance is also increasing, the electric pressure of current transformer and voltage transformer (VT) is more and more higher, and the transient characterisitics under nonserviceabling to it propose new requirement.Therefore these performance qualities with the ferromagnetic element of iron core and loop construction are related to the safe and stable operation of electric system, and are the important technical checking its performance to the performance test of these ferromagnetic elements, by finding that transformer exists defective insulation between siliconized plate to the test of transformer noload losses and no-load current, interpolar unshakable in one's determination, partial short-circuit scaling loss between sheet, punching bolt or colligation steel band, pressing plate, the insulated part of upper yoke etc. damages, form short circuit, in magnetic circuit, siliconized plate loosens, dislocation, air gap is too large, multipoint earthing unshakable in one's determination, coil has turn-to-turn, layer short circuit or the parallel branch number of turn are not etc., ampere-turn is uneven, misapply high consumption siliconized plate inferior or the defect such as designing and calculating is wrong, by can designing by effecting reaction mutual inductor mutual inductor performance test unshakable in one's determination, manufacturing process defect, the quality of material and performance, such as: Secondary Winding turn-to-turn or layer short circuit, unshakable in one's determination or coil loosens, the situations such as displacement.

Along with the increase of ferromagnetic element capacity, equipment heavy, lifting difficulty when making site test, potential safety hazard is many and work efficiency is low.It is even higher up to tens thousand of volt that novel transient protection mutual inductor can reach flex point electromotive force; considerably beyond mutual inductor turn-to-turn and the insulation of tolerance over the ground; thus conventional power frequency excitation property test method can not ensure the safety of test specimen and secondary wiring terminals; and process of the test equipment needed thereby capacity is larger; apply voltage higher, there is the person and device security hidden danger.

At present, existing researchist proposes the test method and the compensation calculation method that adopt low-frequency variable-frequency power source to measure voltage-current characteristic of ferromagnetic element, the basic theories of the method is still based on the low-frequency power test method(s) of IEC60044-6, in addition analysis and calculation is carried out on the magnetic hysteresis loss of iron core and the impact of eddy current loss, made test findings under low frequency more close to industrial frequency experiment result.But there is the problem of two aspects in this method: the research and development difficulty of the sine-wave power that 1) frequency range is comparatively large, output capacity is large is large, and cost is high; 2) carried out some to the model of mutual inductor to simplify, do not considered the impact of transformer leakage inductance, cause test findings to have difference.

Summary of the invention

Under such background condition, the object of the invention is to propose a kind of test method and the analysis calculation method that adopt the volt-ampere characteristic of miniwatt and low-voltage variable-frequency power sources measuring transformer, this method is also applicable to transformer, reactor has iron core and the power equipment of loop construction.The result of this test method and power frequency (50Hz or 60Hz) measured result have consistance.

Technical solution of the present invention is:

The test method utilizing low-frequency power to test transformer volt-ampere characteristic and an analysis calculation method, is characterized in that:

1) carry out test connection by shown in Fig. 1, utilize low-frequency power to carry out exciting test to test specimen.Winding voltage current waveform u is tested when utilizing high-speed measuring device to measure test ₁(t), i (t), and open circuit side voltage waveform u ₂(t), row write circuit port voltage equation (wherein u ₂(t) reduction to testing side), due to open circuit, open circuit side voltage u ₂t () is approximately equal to induced potential e (t) of winding;

2) set up the equivalent-circuit model of Circuit Fault on Secondary Transformer, consider that the leakage inductance of transformer directly hinders, magnetic hysteresis loss and eddy current loss; This circuit by magnetizing inductance Lm, be eddy current loss equivalent resistance R _e, hysteresis loss resistance R _hparallel connection again with test side leakage sense L σ and direct current resistance R _dcbe composed in series; The straight resistance R of ferromagnetic element before test _dc, leakage inductance L _σvalue can be calculated by ferromagnetic element nameplate, u ₁t () is test side voltage, e (t) is transformer experiment side induction electromotive force, i _mt () flows through main excitation L _mthe electric current of branch road, i _e(t) eddy current equivalent current, i _ht () is magnetic hysteresis loss equivalent current, i (t) is the total exciting current of test, and P is test active power, P _tit is winding iron core loss;

3) different frequency f is applied to tested ferromagnetic element ₁, f ₂voltage, make core sataration, measure active-power P ₁and P ₂, total exciting current i ₁(t) and i ₂t (), can calculate test side active power by the electric current and voltage temporal pattern gathered:

$P=\frac{1}{T}{\∫}_{-\frac{T}{2}}^{\frac{T}{2}}{u}_1\left(t\right)\·i\left(t\right)dt---\left(1\right)$

The now induction electromotive force of winding:

$e\left(t\right)=u_2\left(t\right)=u_1\left(t\right)-i\left(t\right)\·R_{dc}+L_{\mathrm{\σ}}\frac{di\left(t\right)}{dt}---\left(2\right);$

4) the core loss P under different frequency is calculated by formula (3) _t1, P _t2, wherein I ₁, I ₂for exciting current effective value under different frequency;

$\left\{\begin{array}{c}{P}_{T1}=P-I_1^2\·R_{dc}\\ P_{T2}=P-I_2^2\·R_{dc}\end{array}\right.---\left(3\right);$

5) core loss and frequency meet relational expression (4), to same transformer, under same magnetic is close, and W _e, W _hconstant, W _hfor each cycle magnetic hysteresis loss (W/Hz); W _ef is each cycle eddy current loss (W/Hz).W can be tried to achieve according to twice test core loss _e, W _hvalue;

$\left\{\begin{array}{c}{P}_{T1}=W_h{f}_1+W_e{f}_1^2\\ P_{T2}=W_h{f}_2+W_e{f}_2^2\end{array}\right.---\left(4\right)$

$\left\{\begin{array}{c}{W}_e=\frac{P_{T1}\×f_2^2-P_{T2}\×f_1^2}{f_1\×f_2(f_2-f_1)}\\ W_h=\frac{P_{T2}\×f_1-P_{T1}\×f_2}{f_1\×f_2(f_2-f_1)}\end{array}\right.$

6) mutual inductor is demagnetized, make iron core residual flux ψ ₀=0; Apply the constant voltage of frequency f by low-frequency power to mutual inductor Secondary Winding, according to the situation of tested mutual inductor, the frequency f that power supply exports should be suitable, and power supply capacity is enough to make transformer core saturated; Slow boosted voltage, until iron core to reach the degree of depth saturated, in the process, use the instrument of high-speed sampling, measure and record instantaneous voltage u (t) and exciting current instantaneous value i (t) of applying;

7) test figure under power supply low frequency f is converted to power frequency f ₀under data; Eddy loss currents i can be calculated according to induced potential and eddy current, magnetic hysteresis equivalent resistance _e(t), computing formula is as follows:

$I_e=\frac{W_e\·f^2}{E}---\left(6\right);$

8) by frequency relation, the eddy loss currents under low frequency is converted under power frequency, I _enfor conversion is to industrial frequency experiment amount:

I _en＝I _e·f _n/f(7)

Overall test electric current after compensation under power frequency is I _exn:

I _exn＝I _ex-I _e+I _en(8)；

9) owing to needing to compensate according to compensating rear exciting current instantaneous value the voltage drop directly hindered and in leakage inductance, directly exciting current instantaneous value can be converted in proportion:

i _exn(t)＝k·i _ex(t)(9)

In formula, k is scale-up factor k=Iexn/Iex, and finally just can calculate the field voltage under power frequency, induction electromotive force directly can be converted by frequency, and the electric current after the pressure drop needs compensation of straight resistance and leakage inductance calculates:

$u_n\left(t\right)=e\left(t\right)\·f_n/f+i_{exn}\left(t\right)\·R_{dc}+L_{\mathrm{\σ}}\frac{{\mathrm{di}}_{exn}\left(t\right)}{dt}---\left(10\right);$

10) current and voltage data be scaled under power frequency being calculated its effective value, take voltage as ordinate, and electric current is horizontal ordinate, just can make the volt-ampere characteristic under industrial frequency experiment, complete the test of volt-ampere characteristic.

Beneficial effect of the present invention is:

1. can complete test under far below the condition of power-frequency voltage, significantly reduce the capacity required for test, effectively reduce the security risk for personnel and test specimen in process of the test;

2. utilize electric current and voltage data in process of the test can calculate transformer test side winding leakage reactance, establish the exact circuitry equivalent model containing leakage inductance, in derivation, consider the impact of leakage reactance;

3. reduce quality and the volume of testing equipment, can short form test wiring, effectively improve test efficiency.

Accompanying drawing explanation

Fig. 1 volt-ampere characteristic test equivalent electrical circuit;

Fig. 2 parameter recognition principle figure;

Fig. 3 volt-ampere characteristic.

Embodiment

Below in conjunction with accompanying drawing, the invention will be further described.

The test method utilizing low-frequency power to test transformer volt-ampere characteristic and an analysis calculation method, feature of the present invention is:

1) carry out test connection by shown in Fig. 1, utilize low-frequency power to carry out exciting test to test specimen; Winding voltage current waveform u is tested when utilizing high-speed measuring device to measure test ₁(t), i (t), and open circuit side voltage waveform u ₂(t), row write circuit port voltage equation (wherein u ₂(t) reduction to testing side), due to open circuit, open circuit side voltage u ₂t () is approximately equal to induced potential e (t) of winding;

2) equivalent-circuit model of Circuit Fault on Secondary Transformer is set up, as shown in Figure 2, wherein R _dcfor winding directly hinders, L σ is test side leakage sense (the straight resistance R of the front ferromagnetic element of test _dc, leakage inductance L _σvalue can be calculated by ferromagnetic element nameplate), Lm is magnetizing inductance, R _efor eddy current loss equivalent resistance, R _hfor hysteresis loss resistance.U ₁t () is test side voltage, e (t) is transformer induction electromotive force, i _mt () flows through main excitation L _mthe electric current of branch road, i _e(t) eddy current equivalent current, i _ht () is magnetic hysteresis loss equivalent current, i (t) is the total exciting current of test, and P is test active power, P _tit is winding iron core loss;

3) different frequency f is applied to tested ferromagnetic element ₁, f ₂voltage, make core sataration, measure active-power P ₁and P ₂, total exciting current i ₁(t) and i ₂t (), can calculate test side active power by the electric current and voltage temporal pattern gathered:

$P=\frac{1}{T}{\∫}_{-\frac{T}{2}}^{\frac{T}{2}}{u}_1\left(t\right)\·i\left(t\right)dt---\left(1\right)$

The now induction electromotive force of winding:

$e\left(t\right)=u_2\left(t\right)=u_1\left(t\right)-i\left(t\right)\·R_{dc}+L_{\mathrm{\σ}}\frac{di\left(t\right)}{dt}---\left(2\right);$

4) the core loss P under different frequency is calculated by formula (3) _t1, P _t2, wherein I ₁, I ₂for exciting current effective value under different frequency;

$\left\{\begin{array}{c}{P}_{T1}=P-I_1^2\·R_{dc}\\ P_{T2}=P-I_2^2\·R_{dc}\end{array}\right.---\left(3\right);$

5) core loss and frequency meet relational expression (4), to same transformer, under same magnetic is close, and W _e, W _hconstant, W _hfor each cycle magnetic hysteresis loss (W/Hz); W _ef is each cycle eddy current loss (W/Hz).W can be tried to achieve according to twice test core loss _e, W _hvalue;

$\left\{\begin{array}{c}{P}_{T1}=W_h{f}_1+W_e{f}_1^2\\ P_{T2}=W_h{f}_2+W_e{f}_2^2\end{array}\right.---\left(4\right)$

$\left\{\begin{array}{c}{W}_e=\frac{P_{T1}\×f_2^2-P_{T2}\×f_1^2}{f_1\×f_2(f_2-f_1)}\\ W_h=\frac{P_{T2}\×f_1-P_{T1}\×f_2}{f_1\×f_2(f_2-f_1)}\end{array}\right.---\left(5\right);$

6) mutual inductor is demagnetized, make iron core residual flux ψ ₀=0; Apply the constant voltage of frequency f by low-frequency power to mutual inductor Secondary Winding, according to the situation of tested mutual inductor, the frequency f that power supply exports should be suitable, and power supply capacity is enough to make transformer core saturated; Slow boosted voltage, until iron core to reach the degree of depth saturated, in the process, use the instrument of high-speed sampling, measure and record instantaneous voltage u (t) and exciting current instantaneous value i (t) of applying;

7) test figure under power supply low frequency f is converted to power frequency f ₀under data; Eddy loss currents i can be calculated according to induced potential and eddy current, magnetic hysteresis equivalent resistance _e(t), computing formula is as follows:

$I_e=\frac{W_e\·f^2}{E}---\left(6\right);$

8) by frequency relation, the eddy loss currents under low frequency is converted under power frequency, I _enfor conversion is to industrial frequency experiment amount:

I _en＝I _e·f _n/f(7)

Overall test electric current after compensation under power frequency is I _exn:

I _exn＝I _ex-I _e+I _en(8)；

9) owing to needing to compensate according to compensating rear exciting current instantaneous value the voltage drop directly hindered and in leakage inductance, directly exciting current instantaneous value can be converted in proportion:

i _exn(t)＝k·i _ex(t)(9)

In formula, k is scale-up factor k=Iexn/Iex, and finally just can calculate the field voltage under power frequency, induction electromotive force directly can be converted by frequency, and the electric current after the pressure drop needs compensation of straight resistance and leakage inductance calculates:

$u_n\left(t\right)=e\left(t\right)\·f_n/f+i_{exn}\left(t\right)\·R_{dc}+L_{\mathrm{\σ}}\frac{{\mathrm{di}}_{exn}\left(t\right)}{dt}---\left(10\right);$

10) current and voltage data be scaled under power frequency is calculated its effective value U ₀, I ₀, take voltage as ordinate, electric current is horizontal ordinate, just can make the volt-ampere characteristic under industrial frequency experiment, complete the test of volt-ampere characteristic.

Claims (1)

1. the test method utilizing low-frequency power to test transformer volt-ampere characteristic and an analysis calculation method, is characterized in that:

1) low-frequency power is utilized to carry out exciting test to test specimen; Winding voltage current waveform u is tested when utilizing high-speed measuring device to measure test ₁(t), i (t), and open circuit side voltage waveform u ₂(t), row write circuit port voltage equation (wherein u ₂(t) reduction to testing side), due to open circuit, open circuit side voltage u ₂t () is approximately equal to induced potential e (t) of winding;

2) set up the equivalent-circuit model of Circuit Fault on Secondary Transformer, consider that the leakage inductance of transformer directly hinders, magnetic hysteresis loss and eddy current loss; This circuit by magnetizing inductance Lm, be eddy current loss equivalent resistance R _e, hysteresis loss resistance R _hparallel connection again with test side leakage sense L σ and direct current resistance R _dcbe composed in series; The straight resistance R of ferromagnetic element before test _dc, leakage inductance L _σvalue can be calculated by ferromagnetic element nameplate, u ₁t () is test side voltage, e (t) is transformer experiment side induction electromotive force, i _mt () flows through main excitation L _mthe electric current of branch road, i _e(t) eddy current equivalent current, i _ht () is magnetic hysteresis loss equivalent current, i (t) is the total exciting current of test, and P is test active power, P _tit is winding iron core loss;

3) different frequency f is applied to tested ferromagnetic element ₁, f ₂voltage, make core sataration, measure active-power P ₁and P ₂, total exciting current i ₁(t) and i ₂t (), can calculate test side active power by the electric current and voltage temporal pattern gathered:

$P=\frac{1}{T}{\∫}_{-\frac{T}{2}}^{\frac{T}{2}}{u}_1\left(t\right)\·i\left(t\right)dt---\left(1\right)$

The now induction electromotive force of winding:

$e\left(t\right)=u_2\left(t\right)=u_1\left(t\right)-i\left(t\right)\·R_{ct}+L_{\mathrm{\σ}}\frac{di\left(t\right)}{dt}---\left(2\right);$

4) the core loss P under different frequency is calculated by formula (3) _t1, P _t2, wherein I ₁, I ₂for exciting current effective value under different frequency;

$\left\{\begin{array}{c}{P}_{T1}=P-I_1^2\·R_{dc}\\ P_{T2}=P-I_2^2\·R_{dc}\end{array}\right.---\left(3\right);$

5) core loss and frequency meet relational expression (4), to same transformer, under same magnetic is close, and W _e, W _hconstant, W _hfor each cycle magnetic hysteresis loss (W/Hz); W _ef is each cycle eddy current loss (W/Hz); W can be tried to achieve according to twice test core loss _e, W _hvalue;

$\left\{\begin{array}{c}{P}_{T1}=W_h{f}_1+W_e{f}_1^2\\ P_{T2}=W_h{f}_2+W_e{f}_2^2\end{array}\right.---\left(4\right)$

$\left\{\begin{array}{c}{W}_h=\frac{P_{T1}\×f_2^2-P_{T2}\×f_1^2}{f_1\×f_2(f_2-f_1)}\\ W_e=\frac{P_{T2}\×f_1-P_{T1}\×f_2}{f_1\×f_2(f_2-f_1)}\end{array}\right.---\left(5\right);$

6) mutual inductor is demagnetized, make iron core residual flux ψ ₀=0; Apply the constant voltage of frequency f by low-frequency power to mutual inductor Secondary Winding, according to the situation of tested mutual inductor, the frequency f that power supply exports should be suitable, and power supply capacity is enough to make transformer core saturated; Slow boosted voltage, until iron core to reach the degree of depth saturated, in the process, use the instrument of high-speed sampling, measure and record instantaneous voltage u (t) and exciting current instantaneous value i (t) of applying;

7) test figure under power supply low frequency f is converted to power frequency f ₀under data; Eddy loss currents i can be calculated according to induced potential and eddy current, magnetic hysteresis equivalent resistance _e(t), computing formula is as follows:

$I_e=\frac{W_e\·f^2}{E}---\left(6\right);$

8) by frequency relation, the eddy loss currents under low frequency is converted under power frequency, I _enfor conversion is to industrial frequency experiment amount:

I _en＝I _e·f _n/f(7)

Overall test electric current after compensation under power frequency is I _exn:

I _exn＝I _ex-I _e+I _en(8)；

9) owing to needing to compensate according to compensating rear exciting current instantaneous value the voltage drop directly hindered and in leakage inductance, directly exciting current instantaneous value can be converted in proportion:

i _exn(t)＝k·i _ex(t)(9)；

In formula, k is scale-up factor k=Iexn/Iex, and finally just can calculate the field voltage under power frequency, induction electromotive force directly can be converted by frequency, and the electric current after the pressure drop needs compensation of straight resistance and leakage inductance calculates:

$u_n\left(t\right)=e\left(t\right)\·f_n/f+i_{exn}\left(t\right)\·R_{dc}+L_{\mathrm{\σ}}\frac{{\mathrm{di}}_{exn}\left(t\right)}{dt}---\left(10\right);$

10) current and voltage data be scaled under power frequency being calculated its effective value, take voltage as ordinate, and electric current is horizontal ordinate, just can make the volt-ampere characteristic under industrial frequency experiment, complete the test of volt-ampere characteristic.