CN105510742A

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

Info

Publication number: CN105510742A
Application number: CN201510900250.5A
Authority: CN
Inventors: 刘涛; 梁仕斌; 王俊凯; 刘鑫; 彭庆军; 王磊; 田庆生; 廖圣
Original assignee: YUNNAN ELECTRIC TECHNOLOGIES Co Ltd; Chongqing University; Electric Power Research Institute of Yunnan Power System Ltd; Yunnan Electric Power Test and Research Institute Group Co Ltd
Current assignee: YUNNAN ELECTRIC TECHNOLOGIES Co Ltd; Chongqing University; Electric Power Research Institute of Yunnan Power System Ltd; Yunnan Electric Power Test and Research Institute Group Co Ltd
Priority date: 2015-12-08
Filing date: 2015-12-08
Publication date: 2016-04-20
Anticipated expiration: 2035-12-08
Also published as: CN105510742B

Abstract

A test method and analysis and calculation method for testing the volt-ampere characteristics of a transformer by using a low-frequency power supply, which is characterized in that: using a low-frequency power supply to conduct an excitation test on a tested product; using a high-speed measuring device to measure the test winding voltage and current waveform u ₁ (t ), i(t), and the open circuit side voltage waveform u ₂ (t), write the circuit port voltage equation (where u ₂ (t) has been attributed to the test side), due to the open circuit, the open circuit side voltage u ₂ (t) Approximately equal to the induced potential e(t) of the winding, etc., its beneficial effect is: the test can be completed under the condition of far lower than the power frequency voltage, the capacity required for the test is greatly reduced, and the test process is effectively reduced. The safety risk of the product; the leakage reactance of the winding on the test side of the transformer can be calculated by using the voltage and current data during the test, and an accurate circuit equivalent model containing the leakage inductance is established, and the influence of the leakage reactance is considered in the derivation process; the cost of the test equipment is reduced. The quality and volume can simplify the test wiring and effectively improve the test efficiency.

Description

A test method and analysis and calculation method for testing the volt-ampere characteristics of a transformer using a low-frequency power supply

技术领域technical field

本发明涉及电力用变压器、互感器、电抗器进行伏安特性(或励磁特性)试验，特别适用于对漏抗较大不可忽略的变压器进行伏安特性试验。The invention relates to a volt-ampere characteristic (or excitation characteristic) test of a power transformer, a transformer, and a reactor, and is particularly suitable for carrying out a volt-ampere characteristic test on a transformer whose leakage reactance is large and cannot be ignored.

技术背景technical background

变压器、电抗器和互感器等铁磁线圈元件作为电力系统中重要元件，随着输变电容量的增大变压器的电压等级和容量逐渐增大，随着输电距离的增长补偿用的并联电抗器的容量也越来越大，电流互感器和电压互感器的电压等级越来越高，对其在故障状态下的暂态特性提出新的要求。因此这些具有铁心和线圈结构的铁磁元件的性能优劣关系到电力系统的安全稳定运行，而对这些铁磁元件的性能试验是检查其性能的重要技术手段，通过对变压器空载损耗和空载电流的测试可以发现变压器存在硅钢片间绝缘不良，铁心极间、片间局部短路烧损，穿芯螺栓或绑扎钢带、压板、上轭铁等的绝缘部分损坏、形成短路，磁路中硅钢片松动、错位、气隙太大，铁心多点接地，线圈有匝间、层间短路或并联支路匝数不等、安匝不平衡等，误用了高耗劣质硅钢片或设计计算有误等缺陷，通过对互感器铁心性能试验能够有效反应互感器设计、制造工艺缺陷、材料和性能的优劣，比如：二次绕组匝间或层间短路，铁心或线圈松动、移位等情况。Ferromagnetic coil components such as transformers, reactors and transformers are important components in power systems. With the increase of transmission and transformation capacity, the voltage level and capacity of transformers gradually increase. With the increase of transmission distance, shunt reactors for compensation The capacity of the transformer is getting larger and larger, and the voltage level of the current transformer and voltage transformer is getting higher and higher, which puts forward new requirements for its transient characteristics under fault conditions. Therefore, the performance of these ferromagnetic components with core and coil structure is related to the safe and stable operation of the power system, and the performance test of these ferromagnetic components is an important technical means to check their performance. The current-carrying test can find that the transformer has poor insulation between silicon steel sheets, local short-circuit burning between the poles and sheets of the core, damage to the insulation parts of the core bolts or binding steel strips, pressure plates, upper yokes, etc. The silicon steel sheet is loose, misplaced, the air gap is too large, the core is grounded at multiple points, the coil has inter-turn, interlayer short circuit, or the number of parallel branch turns is unequal, and the ampere-turn is unbalanced. If there are defects such as errors, the performance test of the transformer core can effectively reflect the design, manufacturing process defects, material and performance of the transformer, such as: secondary winding inter-turn or inter-layer short circuit, core or coil looseness, displacement, etc. .

随着铁磁元件容量的增大，使得现场试验时设备笨重、吊装困难，安全隐患多和工作效率低下。新型暂态保护用互感器可达拐点电势高达数万伏甚至更高，远远超过互感器匝间和对地的耐受绝缘，因而常规工频励磁特性试验方法已经不能保证被试品及二次接线端子的安全，并且试验过程所需设备容量较大，施加电压较高，存在人身和设备安全隐患。With the increase of the capacity of ferromagnetic components, the field test equipment is bulky, difficult to hoist, many safety hazards and low work efficiency. The inflection point potential of the new transformer for transient protection can be as high as tens of thousands of volts or even higher, which far exceeds the withstand insulation between the turns of the transformer and the ground. The safety of the secondary terminals, and the equipment required for the test process has a large capacity and a high applied voltage, which poses potential safety hazards for people and equipment.

目前，已有研究人员提出了采用低频变频电源测量铁磁元件伏安特性的试验方法和补偿计算方法，该方法的基本理论仍然基于IEC60044-6的低频电源试验法，此外还对铁芯的磁滞损耗和涡流损耗的影响进行了分析和计算，使得低频下的试验结果更加接近于工频试验结果。但是这种方法存在两个方面的问题：1)频率变化范围较大、输出容量大的正弦波电源的研发难度大，成本高；2)对互感器的模型进行了一些简化，没有考虑变压器漏感的影响，导致试验结果具有差异。At present, researchers have proposed a test method and a compensation calculation method for measuring the volt-ampere characteristics of ferromagnetic components using a low-frequency variable-frequency power supply. The basic theory of this method is still based on the low-frequency power supply test method of IEC60044-6. The influence of hysteresis loss and eddy current loss is analyzed and calculated, which makes the test results at low frequency closer to the power frequency test results. However, this method has two problems: 1) It is difficult to develop a sine wave power supply with a large frequency range and large output capacity, and the cost is high; 2) The model of the transformer is simplified, and the transformer leakage Sensitive influences lead to differences in test results.

发明内容Contents of the invention

在这样的背景条件下，本发明的目的是提出了一种采用小功率和低电压变频电源测量变压器的伏安特性的试验方法和分析计算方法，这种方法也适用于变压器、电抗器具有铁芯和线圈结构的电力设备。这种试验方法的结果与工频(50Hz或60Hz)实测结果具有一致性。Under such background conditions, the purpose of the present invention is to propose a test method and analysis and calculation method for measuring the volt-ampere characteristics of a transformer using a low-power and low-voltage variable-frequency power supply. This method is also applicable to transformers and reactors with iron Electrical equipment with core and coil structures. The results of this test method are consistent with the measured results of power frequency (50Hz or 60Hz).

本发明技术方案为：Technical scheme of the present invention is:

一种利用低频电源对变压器伏安特性测试的试验方法和分析计算方法，其特征是：A test method and analysis and calculation method for testing the volt-ampere characteristics of a transformer using a low-frequency power supply, which is characterized by:

1)按图1所示进行试验接线，利用低频电源对被试品进行励磁试验。利用高速测量装置测量试验时试验绕组电压电流波形u₁(t)、i(t)，以及开路侧电压波形u₂(t)，列写电路端口电压方程(其中u₂(t)已归算至试验侧)，由于开路，开路侧电压u₂(t)近似等于绕组的感应电势e(t)；1) Conduct the test wiring as shown in Figure 1, and use a low-frequency power supply to conduct an excitation test on the tested product. Use the high-speed measuring device to measure the test winding voltage and current waveforms u ₁ (t), i(t), and the voltage waveform u ₂ (t) on the open circuit side during the test, and write the circuit port voltage equation (wherein u ₂ (t) has been reduced to the test side), due to the open circuit, the voltage u ₂ (t) on the open circuit side is approximately equal to the induced potential e(t) of the winding;

2)建立变压器二次侧的等效电路模型，考虑变压器的漏感直阻、磁滞损耗和涡流损耗；该电路由激磁电感Lm、为涡流损耗等效电阻R_e、磁滞损耗电阻R_h并联再与试验侧漏感Lσ和直流电阻R_dc串联组成；试验前铁磁元件的直阻R_dc，漏感L_σ的值可由铁磁元件铭牌计算得出，u₁(t)为试验侧端电压，e(t)是变压器实验侧感应电动势，i_m(t)是流过主励磁L_m支路的电流，i_e(t)涡流等效电流，i_h(t)为磁滞损耗等效电流，i(t)是试验总励磁电流，P是试验有功功率，P_T是绕组铁芯损耗；2) Establish the equivalent circuit model of the secondary side of the transformer, considering the leakage inductance direct resistance, _hysteresis loss and _eddy current loss of the transformer; It is connected in parallel with the test side leakage inductance Lσ and DC resistance R _dc in series; the value of the direct resistance R _dc and leakage inductance L _σ of the ferromagnetic component before the test can be calculated from the nameplate of the ferromagnetic component, and u ₁ (t) is the test side Terminal voltage, e(t) is the induced electromotive force on the experimental side of the transformer, i _m (t) is the current flowing through the main excitation L _m branch, i _e (t) is the eddy current equivalent current, i _h (t) is the hysteresis loss Equivalent current, i(t) is the total excitation current of the test, P is the active power of the test, and P _T is the loss of the winding core;

3)对被试铁磁元件施加不同的频率f₁、f₂的电压，使铁芯饱和，测量有功功率P₁和P₂、总励磁电流i₁(t)和i₂(t)，通过采集的电压电流瞬时波形能够计算出试验侧有功功率：3) Apply voltages of different frequencies f ₁ and f ₂ to the tested ferromagnetic element to saturate the iron core, measure active power P ₁ and P ₂ , total excitation current i ₁ (t) and i ₂ (t), and pass The collected instantaneous waveform of voltage and current can calculate the active power of the test side:

$P P = = \frac{11}{T T} {&Integral; &Integral;}_{- - \frac{T T}{22}}^{\frac{T T}{22}} {u u}_{11} ((t t)) \cdot &Center Dot; i i ((t t)) d d t t - - - - - - ((11))$

此时绕组的感应电动势：At this time, the induced electromotive force of the winding is:

$e e ((t t)) = = {u u}_{22} ((t t)) = = {u u}_{11} ((t t)) - - i i ((t t)) \cdot \cdot {R R}_{d d c c} + + {L L}_{σ σ} \frac{d d i i ((t t))}{d d t t} - - - - - - ((22));;$

4)按式(3)计算不同频率下的铁芯损耗P_T1、P_T2，其中I₁、I₂为不同频率下励磁电流有效值；4) Calculate the core losses _PT1 and _PT2 at different frequencies according to formula (3), where I ₁ and I ₂ are the effective values of the excitation current at different frequencies;

$\{\begin{matrix} {P P}_{T T 11} = = P P - - {I I}_{11}^{22} \cdot \cdot {R R}_{d d c c} \\ {P P}_{T T 22} = = P P - - {I I}_{22}^{22} \cdot \cdot {R R}_{d d c c} \end{matrix} - - - - - - ((33));;$

5)铁心损耗和频率满足关系式(4)，对同一台变压器，在相同磁密下，W_e、W_h是常数，W_h为每个周期磁滞损耗(W/Hz)；W_ef为每个周期涡流损耗(W/Hz)。根据两次试验铁心损耗可求得W_e、W_h的值；5) The core loss and frequency satisfy the relationship (4). For the same transformer, under the same magnetic density, W _e and W _h are constants, and W _h is the hysteresis loss per cycle (W/Hz); W _e f Eddy current loss (W/Hz) for each cycle. The values of W _e and W _h can be obtained according to the core loss of the two tests;

$\{\begin{matrix} {P P}_{T T 11} = = {W W}_{h h} {f f}_{11} + + {W W}_{e e} {f f}_{11}^{22} \\ {P P}_{T T 22} = = {W W}_{h h} {f f}_{22} + + {W W}_{e e} {f f}_{22}^{22} \end{matrix} - - - - - - ((44))$

$\{\begin{matrix} {W W}_{e e} = = \frac{{P P}_{T T 11} \times \times {f f}_{22}^{22} - - {P P}_{T T 22} \times \times {f f}_{11}^{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \\ {W W}_{h h} = = \frac{{P P}_{T T 22} \times \times {f f}_{11} - - {P P}_{T T 11} \times \times {f f}_{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \end{matrix}$

6)对互感器进行退磁，使得铁芯剩磁通ψ₀＝0；通过低频电源对互感器二次绕组施加频率f恒定的电压，根据被试互感器的状况，电源输出的频率f应合适，电源容量足以使变压器铁芯饱和；缓慢升高电压，直至铁芯达到深度饱和，在此过程中，使用高速采样的仪器，测量并记录施加的电压瞬时值u(t)和励磁电流瞬时值i(t)；6) Demagnetize the transformer so that the core residual flux ψ ₀ = 0; apply a voltage with a constant frequency f to the secondary winding of the transformer through a low-frequency power supply. According to the condition of the transformer under test, the frequency f of the power output should be appropriate , the power supply capacity is enough to saturate the transformer iron core; slowly increase the voltage until the iron core reaches deep saturation, during this process, use a high-speed sampling instrument to measure and record the instantaneous value u(t) of the applied voltage and the instantaneous value of the excitation current i(t);

7)将电源低频f下的试验数据转换至工频f₀下的数据；根据感应电势和涡流、磁滞等效电阻可以计算出涡流损耗电流i_e(t)，计算公式如下：7) Convert the test data at the low frequency f of the power supply to the data at the power frequency _f0 ; the eddy current loss current i _e (t) can be calculated according to the induced potential, eddy current, and hysteresis equivalent resistance, and the calculation formula is as follows:

${I I}_{e e} = = \frac{{W W}_{e e} \cdot &Center Dot; {f f}^{22}}{E E.} - - - - - - ((66));;$

8)按频率关系对低频下的涡流损耗电流折算至工频下，I_en为折算至工频试验量：8) Convert the eddy current loss current at low frequency to power frequency according to the frequency relationship, and I _en is the test value converted to power frequency:

I_en＝I_e·f_n/f(7)I _en =I _e ·f _n /f(7)

补偿之后工频下的总试验电流为I_exn：The total test current at power frequency after compensation is I _exn :

I_exn＝I_ex-I_e+I_en(8)；I _exn =I _ex -I _e +I _en (8);

9)由于需要根据补偿后励磁电流瞬时值来补偿直阻和漏感上的电压降，可直接按比例折算励磁电流瞬时值：9) Since the voltage drop on the direct resistance and leakage inductance needs to be compensated according to the instantaneous value of the excitation current after compensation, the instantaneous value of the excitation current can be directly converted proportionally:

i_exn(t)＝k·i_ex(t)(9)i _exn (t) = k · i _ex (t) (9)

式中，k为比例系数k＝Iexn/Iex，最后就可以计算出工频下的励磁电压，感应电动势可按频率直接折算，直阻和漏感的压降需要用补偿后的电流进行计算：In the formula, k is the proportional coefficient k=Iexn/Iex, and finally the excitation voltage under the power frequency can be calculated, the induced electromotive force can be directly converted according to the frequency, and the voltage drop of the direct resistance and leakage inductance needs to be calculated with the compensated current:

${u u}_{n no} ((t t)) = = e e ((t t)) \cdot \cdot {f f}_{n no} / / f f + + {i i}_{e e x x n no} ((t t)) \cdot \cdot {R R}_{d d c c} + + {L L}_{σ σ} \frac{{di di}_{e e x x n no} ((t t))}{d d t t} - - - - - - ((1010));;$

10)将换算至工频下的电流电压数据计算其有效值，以电压为纵坐标，电流为横坐标，就可以作出工频试验下的伏安特性曲线，完成伏安特性的测试。10) Calculate the effective value of the current and voltage data converted to the power frequency, take the voltage as the ordinate and the current as the abscissa, then the volt-ampere characteristic curve under the power frequency test can be drawn to complete the test of the volt-ampere characteristic.

本发明的有益效果为：The beneficial effects of the present invention are:

1.可以在远低于工频电压的条件下完成试验，大幅降低试验所需要的容量，有效降低了试验过程中对于人员和被试品的安全风险；1. The test can be completed under the condition of far lower than the power frequency voltage, which greatly reduces the capacity required for the test, and effectively reduces the safety risks to personnel and the tested product during the test;

2.利用试验过程中电压电流数据可计算出变压器试验侧绕组漏抗，建立了含有漏感的精确电路等效模型，在推导过程中考虑漏抗的影响；2. Using the voltage and current data during the test, the leakage reactance of the winding on the test side of the transformer can be calculated, and an accurate circuit equivalent model containing leakage inductance is established, and the influence of the leakage reactance is considered in the derivation process;

3.减小了试验设备的质量和体积，可以简化试验接线，有效提高试验效率。3. The mass and volume of the test equipment are reduced, the test wiring can be simplified, and the test efficiency can be effectively improved.

附图说明Description of drawings

图1伏安特性试验等效电路；Figure 1 Equivalent circuit of volt-ampere characteristic test;

图2参数识别原理图；Fig. 2 Schematic diagram of parameter identification;

图3伏安特性曲线。Figure 3 Volt-ampere characteristic curve.

具体实施方式detailed description

下面结合附图对本发明作进一步说明。The present invention will be further described below in conjunction with accompanying drawing.

一种利用低频电源对变压器伏安特性测试的试验方法和分析计算方法，本发明特征是：A test method and analysis and calculation method for testing the volt-ampere characteristics of a transformer using a low-frequency power supply, the features of the invention are:

1)按图1所示进行试验接线，利用低频电源对被试品进行励磁试验；利用高速测量装置测量试验时试验绕组电压电流波形u₁(t)、i(t)，以及开路侧电压波形u₂(t)，列写电路端口电压方程(其中u₂(t)已归算至试验侧)，由于开路，开路侧电压u₂(t)近似等于绕组的感应电势e(t)；1) Carry out the test wiring as shown in Figure 1, and use a low-frequency power supply to conduct an excitation test on the tested product; use a high-speed measuring device to measure the test winding voltage and current waveforms u ₁ (t), i(t), and the open-circuit side voltage waveform during the test u ₂ (t), write the circuit port voltage equation (where u ₂ (t) has been attributed to the test side), due to the open circuit, the voltage u ₂ (t) on the open circuit side is approximately equal to the induced potential e(t) of the winding;

2)建立变压器二次侧的等效电路模型，如图2所示，其中R_dc为绕组直阻，Lσ为试验侧漏感(试验前铁磁元件的直阻R_dc，漏感L_σ的值可由铁磁元件铭牌计算得出)，Lm为激磁电感，R_e为涡流损耗等效电阻，R_h为磁滞损耗电阻。u₁(t)为试验侧端电压，e(t)是变压器感应电动势，i_m(t)是流过主励磁L_m支路的电流，i_e(t)涡流等效电流，i_h(t)为磁滞损耗等效电流，i(t)是试验总励磁电流，P是试验有功功率，P_T是绕组铁芯损耗；2) Establish the equivalent circuit model of the secondary side of the transformer, as shown in Figure 2, where R _dc is the direct resistance of the winding, Lσ is the leakage inductance of the test side (the direct resistance R _dc of the ferromagnetic element before the test, the leakage inductance L _σ The value can be calculated from the ferromagnetic component nameplate), Lm is the excitation inductance, _Re is the equivalent resistance of eddy current loss, and R _h is the hysteresis loss resistance. u ₁ (t) is the test side terminal voltage, e(t) is the transformer induced electromotive force, i _m (t) is the current flowing through the main excitation L _m branch, i _e (t) eddy current equivalent current, i _h ( t) is the hysteresis loss equivalent current, i(t) is the total excitation current of the test, P is the active power of the test, and P _T is the winding core loss;

$P P = = \frac{11}{T T} {&Integral; &Integral;}_{- - \frac{T T}{22}}^{\frac{T T}{22}} {u u}_{11} ((t t)) \cdot \cdot i i ((t t)) d d t t - - - - - - ((11))$

$\{\begin{matrix} {P P}_{T T 11} = = P P - - {I I}_{11}^{22} \cdot &Center Dot; {R R}_{d d c c} \\ {P P}_{T T 22} = = P P - - {I I}_{22}^{22} \cdot &Center Dot; {R R}_{d d c c} \end{matrix} - - - - - - ((33));;$

$\{\begin{matrix} {W W}_{e e} = = \frac{{P P}_{T T 11} \times \times {f f}_{22}^{22} - - {P P}_{T T 22} \times \times {f f}_{11}^{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \\ {W W}_{h h} = = \frac{{P P}_{T T 22} \times \times {f f}_{11} - - {P P}_{T T 11} \times \times {f f}_{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \end{matrix} - - - - - - ((55));;$

I_en＝I_e·f_n/f(7)I _en =I _e ·f _n /f(7)

I_exn＝I_ex-I_e+I_en(8)；I _exn =I _ex -I _e +I _en (8);

i_exn(t)＝k·i_ex(t)(9)i _exn (t) = k · i _ex (t) (9)

${u u}_{n no} ((t t)) = = e e ((t t)) \cdot &Center Dot; {f f}_{n no} / / f f + + {i i}_{e e x x n no} ((t t)) \cdot &Center Dot; {R R}_{d d c c} + + {L L}_{σ σ} \frac{{di di}_{e e x x n no} ((t t))}{d d t t} - - - - - - ((1010));;$

10)将换算至工频下的电流电压数据计算其有效值U₀、I₀，以电压为纵坐标，电流为横坐标，就可以作出工频试验下的伏安特性曲线，完成伏安特性的测试。10) Calculate the effective values U ₀ and I ₀ from the current and voltage data converted to the power frequency, and take the voltage as the ordinate and the current as the abscissa, then the volt-ampere characteristic curve under the power frequency test can be drawn to complete the volt-ampere characteristic test.

Claims

1. A kind of test method and analytical calculation method utilizing low-frequency power supply to transformer volt-ampere characteristic test, it is characterized in that:

1) Use a low-frequency power supply to perform an excitation test on the tested product; use a high-speed measuring device to measure the test winding voltage and current waveforms u ₁ (t), i(t), and the open-circuit side voltage waveform u ₂ (t), and write the circuit The port voltage equation (where u ₂ (t) has been attributed to the test side), due to the open circuit, the voltage u ₂ (t) on the open circuit side is approximately equal to the induced potential e(t) of the winding;

2) Establish the equivalent circuit model of the secondary side of the transformer, considering the leakage inductance direct resistance, _hysteresis loss and _eddy current loss of the transformer; It is connected in parallel with the test side leakage inductance Lσ and DC resistance R _dc in series; the value of the direct resistance R _dc and leakage inductance L _σ of the ferromagnetic component before the test can be calculated from the nameplate of the ferromagnetic component, and u ₁ (t) is the test side Terminal voltage, e(t) is the induced electromotive force on the experimental side of the transformer, i _m (t) is the current flowing through the main excitation L _m branch, i _e (t) is the eddy current equivalent current, i _h (t) is the hysteresis loss Equivalent current, i(t) is the total excitation current of the test, P is the active power of the test, and P _T is the loss of the winding core;

3) Apply voltages of different frequencies f ₁ and f ₂ to the tested ferromagnetic element to saturate the iron core, measure active power P ₁ and P ₂ , total excitation current i ₁ (t) and i ₂ (t), and pass The collected instantaneous waveform of voltage and current can calculate the active power of the test side:

P P = = \frac{11}{T T} {&Integral; &Integral;}_{- - \frac{T T}{22}}^{\frac{T T}{22}} {u u}_{11} ((t t)) \cdot &Center Dot; i i ((t t)) d d t t - - - - - - ((11))

At this time, the induced electromotive force of the winding is:

e e ((t t)) = = {u u}_{22} ((t t)) = = {u u}_{11} ((t t)) - - i i ((t t)) \cdot &Center Dot; {R R}_{c c t t} + + {L L}_{σ σ} \frac{d d i i ((t t))}{d d t t} - - - - - - ((22));;

4) Calculate the core losses _PT1 and _PT2 at different frequencies according to formula (3), where I ₁ and I ₂ are the effective values of the excitation current at different frequencies;

\{\begin{matrix} {P P}_{T T 11} = = P P - - {I I}_{11}^{22} \cdot &Center Dot; {R R}_{d d c c} \\ {P P}_{T T 22} = = P P - - {I I}_{22}^{22} \cdot &Center Dot; {R R}_{d d c c} \end{matrix} - - - - - - ((33));;

5) The core loss and frequency satisfy the relationship (4). For the same transformer, under the same magnetic density, W _e and W _h are constants, and W _h is the hysteresis loss per cycle (W/Hz); W _e f is the eddy current loss of each cycle (W/Hz); the value of W _e and W _h can be obtained according to the core loss of two tests;

\{\begin{matrix} {P P}_{T T 11} = = {W W}_{h h} {f f}_{11} + + {W W}_{e e} {f f}_{11}^{22} \\ {P P}_{T T 22} = = {W W}_{h h} {f f}_{22} + + {W W}_{e e} {f f}_{22}^{22} \end{matrix} - - - - - - ((44))

\{\begin{matrix} {W W}_{h h} = = \frac{{P P}_{T T 11} \times \times {f f}_{22}^{22} - - {P P}_{T T 22} \times \times {f f}_{11}^{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \\ {W W}_{e e} = = \frac{{P P}_{T T 22} \times \times {f f}_{11} - - {P P}_{T T 11} \times \times {f f}_{22}}{{f f}_{11} \times \times {f f}_{22} (({f f}_{22} - - {f f}_{11}))} \end{matrix} - - - - - - ((55));;

6) Demagnetize the transformer so that the core residual flux ψ ₀ = 0; apply a voltage with a constant frequency f to the secondary winding of the transformer through a low-frequency power supply. According to the condition of the transformer under test, the frequency f of the power output should be appropriate , the power supply capacity is enough to saturate the transformer iron core; slowly increase the voltage until the iron core reaches deep saturation, during this process, use a high-speed sampling instrument to measure and record the instantaneous value u(t) of the applied voltage and the instantaneous value of the excitation current i(t);

7) Convert the test data at the low frequency f of the power supply to the data at the power frequency _f0 ; the eddy current loss current i _e (t) can be calculated according to the induced potential, eddy current, and hysteresis equivalent resistance, and the calculation formula is as follows:

{I I}_{e e} = = \frac{{W W}_{e e} \cdot \cdot {f f}^{22}}{E E.} - - - - - - ((66));;

8) Convert the eddy current loss current at low frequency to power frequency according to the frequency relationship, and I _en is the test value converted to power frequency:

I _en =I _e ·f _n /f(7)

The total test current at power frequency after compensation is I _exn :

I _exn =I _ex -I _e +I _en (8);

9) Since the voltage drop on the direct resistance and leakage inductance needs to be compensated according to the instantaneous value of the excitation current after compensation, the instantaneous value of the excitation current can be directly converted proportionally:

i _exn (t) = k · i _ex (t) (9);

In the formula, k is the proportional coefficient k=Iexn/Iex, and finally the excitation voltage under the power frequency can be calculated, the induced electromotive force can be directly converted according to the frequency, and the voltage drop of the direct resistance and leakage inductance needs to be calculated with the compensated current:

{u u}_{n no} ((t t)) = = e e ((t t)) \cdot \cdot {f f}_{n no} / / f f + + {i i}_{e e x x n no} ((t t)) \cdot &Center Dot; {R R}_{d d c c} + + {L L}_{σ σ} \frac{{di di}_{e e x x n no} ((t t))}{d d t t} - - - - - - ((1010));;

10) Calculate the effective value of the current and voltage data converted to the power frequency, take the voltage as the ordinate and the current as the abscissa, then the volt-ampere characteristic curve under the power frequency test can be drawn to complete the test of the volt-ampere characteristic.