CN110118904A - A kind of k-factor transformer equivalent load conversion method - Google Patents
A kind of k-factor transformer equivalent load conversion method Download PDFInfo
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- CN110118904A CN110118904A CN201910349516.XA CN201910349516A CN110118904A CN 110118904 A CN110118904 A CN 110118904A CN 201910349516 A CN201910349516 A CN 201910349516A CN 110118904 A CN110118904 A CN 110118904A
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- G01R—MEASURING ELECTRIC VARIABLES; MEASURING MAGNETIC VARIABLES
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Abstract
The present invention relates to transformer technology fields, and in particular to a kind of k-factor transformer equivalent load conversion method, comprising the following steps: step A: measures the D.C. resistance R of power transformer product winding at normal temperatureDC, then calculate D.C. resistance and P be lostDC；Step B: it measures in frequency f1With the respective load loss P1 and Px of the product under frequency fx, and winding eddy current loss Pwe1 and structural member stray loss Pse1 are decomposited；Step C: the total eddy-current loss Pwe of winding is calculated according to K value and eddy-current loss Pwe1；Step D: overload current multiple I is calculatedN；Step E: simulation measures winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.The phenomenon that for the present invention is with respect to UL1561 standard, more reasonably defining k-factor transformer equivalent load test method, avoid for the stray loss Pse in product structure part being mixed into winding, causing the equivalent loss virtual height of product and tested by transition.
Description
Technical field
The present invention relates to transformer technology fields, and in particular to a kind of k-factor transformer equivalent load conversion method.
Background technique
Currently, with the rapid development of power electronic technique, more and more nonlinear loads are applied to every field.
Due to its nonlinear characteristic, the Electro Magnetic Compatibility of equipment is reduced, Severe distortion occurs for net side input current, and generates a large amount of
Harmonic wave be discharged into power grid or headend equipment, bring harm to power grid and other electrical equipments.It is present in power grid
Various harmonic voltages or nonlinear load can generate additional harmonic loss in the transformer coil of its front end, when these are humorous
The temperature that will lead to transformer when wave is lost sufficiently large increases, and even more than it allows running temperature, not only reduces transformer
Service life, it is also possible to cause transformer to burn in a short time.
In recent years, external some mechanisms (such as UL) propose K-Rated transformation for the transformer with nonlinear load
The concept of device, i.e. k-factor transformer.According to the size of harmonic content in load current, k-factor is introduced, defines K=(∑ Ih2×
h2)/(∑Ih2), wherein h and Ih is each harmonic number and the subharmonic virtual value (not including fundamental wave).K value is bigger, shows to bear
Load current harmonic content is higher, and representative value has K=1, K=4, K=9, K=13, K=20, K=30 etc..
It is varied since the harmonic content of k-factor transformer is from Hz to KHz, it is to be difficult to find identical harmonic wave
Load carrys out test product.It is directed to the test of k-factor transformer in the industry at present or certification authority is all made of UL1561 standard, now will
The test method abstract of UL1561 standard is as follows:
Step 1: product winding D.C. resistance R is measuredDC, according to I2RDCCalculate straight resistance loss PDC；
Step 2: load loss P of the product winding at fundamental wave (such as 50Hz) is measuredAC(short-circuit method)；
Step 3: the winding temperature coefficient T estimated under equilibrium temperature is calculatedC；
Tc=(Ts+Tk)/(Tm+Tk)；
Ts: transformer winding equilibrium temperature (DEG C) is estimated；
Tm: the environment temperature (DEG C) when test；
Tk: 234.5 DEG C of copper, 225 DEG C of aluminium；
Step 4: the rated condition winding per unit eddy-current loss P of calculation assumptionEC(the I of per unit nominal load2R damage
Consumption)；
As capacity≤300KVA: PEC=(PAC-PDC)/(PDC×Tc2)；
As capacity > 300KVA: PEC=C × (PAC-PDC)/(PDC-I×Tc2)；
PDC-IIt is lost for the D.C. resistance of winding in transformer, when turn ratio is greater than 4:1 or has more than one winding
C=0.7, C=0.6 in the case of other when rated current is greater than 1000A.
Step 5: it calculates the equivalent total load of winding under k-factor harmonic wave and P is lostLL；
PLL=PDC×(1+K×PEC)×Tc；
In the experiment for measuring equivalent temperature rise, total losses P is maintainedLLIt is constant, it is generally tested using short-circuit method, finally obtains production
The equivalent temperature rise of product.
According to the equivalent load test method of UL1561 standard presented hereinbefore, there is two o'clock to be worth discussion in step 4:
First, it is separation why capacity, which takes 300KVA,；And in the denominator of capacity > 300KVA Shi Qi formula,
PDC-ITake the D.C. resistance of winding in transformer be lost be again why, the D.C. resistance of interior winding and the purposes of product and its design
Shi Silu is closely related, can not have definite greatly, when denominator comes and go, its result is also Protean；When
C=0.7, C=0.6 in the case of other when turn ratio is greater than 4:1 or has more than one winding rated current to be greater than 1000A,
These coefficients all lack scientific basis, and without convincingness, and the result error tested is very big.
Second, calculating per unit eddy-current loss PECUsing PAC-PDC, i.e., the load loss under rated condition subtracts directly
Resistance loss.And transformer load loss PACCalculation formula it is as follows:
PAC=PDC+Pwe1+Pse1；
PDC: (I is lost in the straight resistance of product winding2RDC)；
Pwe1: product winding eddy current loss；
Pse1: product structure part stray loss；
It can be appreciated that in UL1561 standard, winding per unit eddy-current loss PECActually winding eddy current loss Pwe1
The sum of with structural member stray loss Pse1, it is more proper that transformer winding per unit added losses should be referred to as.See below formula:
PEC=(Pwe1+Pse1)/(PDC×Tc2)；
Next in UL1561 standard step 5, directly by per unit eddy-current loss PECMultiplied by K value, by the formula of step 5
The equivalent total load loss P of winding under k-factor harmonic wave is calculatedLL.It is not difficult to find out that UL1561 standard is by the spuious damage in structural member
Consumption Pse1, which has been substituted into winding, to be gone, and final result is exactly the equivalent loss virtual height of product winding and is excessively tested；In other words,
Meet UL1561 standard, it is necessary to do the product than larger capacity in situation needed for user, need more to invest, occupy more
More spaces, and actually this is not required for user.
Therefore, existing test method has to be modified.
Summary of the invention
The purpose of the present invention is being directed to above-mentioned deficiency in the prior art, a kind of k-factor transformer equivalent load is provided
Conversion method.
The purpose of the present invention is achieved through the following technical solutions: a kind of k-factor transformer equivalent load conversion method, including
Following steps:
Step A: the D.C. resistance R of power transformer product winding is measured at normal temperatureDC, then calculate D.C. resistance and P be lostDC；
Step B: it measures in frequency f1With the respective load loss P1 and Px of the product under frequency fx, and winding vortex is decomposited
Pwe1 and structural member stray loss Pse1 is lost；
Step C: combine the total eddy-current loss Pwe of winding calculated under harmonic wave with eddy-current loss Pwe1 according to K value；
Step D: P is lost according to D.C. resistanceDC, the total eddy-current loss Pwe of winding be calculated it is equivalent under specified harmonic load
Overload current multiple I when for power frequencyN；
Step E: simulation measures winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.
The present invention is further arranged to, and in stepb, calculates P1 and Px by following formula:
P1=PDC+Pwe1+Pse1；
PX=PDC+Pwe1×(fX/f1)2+Pse1×(fX/f1)0.8；Pwe1, Pse1 are solved by upper two formulas simultaneous.
The present invention is further arranged to, and in step C, calculates the total eddy-current loss Pwe of winding by following formula:
PWe=PWe1×K。
The present invention is further arranged to, in step D, comprising the following steps:
D1: the winding temperature coefficient T estimated under equilibrium temperature is calculated according to formula Tc=(Ts+Tk)/(Tm+Tk)C；Wherein
Ts is to estimate transformer winding equilibrium temperature；Virtual environment temperature when Tm is test；Tk is the thermal constant of medium；
D2: overload current multiple I when power frequency equivalent according to winding under following equation calculating k-factor harmonic waveN:
The present invention is further arranged to, and in step d2, the thermal constant of copper is 234.5 DEG C, and the thermal constant of aluminium is 225
℃。
The present invention is further arranged to, will be by test transformer winding in step E using short-circuit method under the conditions of power frequency
Apply INTimes rated current, simulation measure winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.
The present invention is further arranged to, and in stepb, takes fx >=3f1.So that institute's calculated result has quite high essence
Degree.
Beneficial effects of the present invention: for the present invention is with respect to UL1561 standard, k-factor transformer is more reasonably defined
Equivalent load test method avoids for the stray loss Pse in product structure part being mixed into winding, causes the equivalent damage of product
The phenomenon that consuming virtual height and being tested by transition.
Detailed description of the invention
Invention is described further using attached drawing, but the embodiments in the accompanying drawings do not constitute any limitation to the present invention,
For those of ordinary skill in the art, without creative efforts, it can also be obtained according to the following drawings
Its attached drawing.
Fig. 1 is the principle of the present invention figure.
Specific embodiment
The invention will be further described with the following Examples.
Known by Fig. 1；A kind of k-factor transformer equivalent load conversion method described in the present embodiment, comprising the following steps:
Step A: the D.C. resistance R of power transformer product winding is measured at normal temperatureDC, then calculate D.C. resistance and P be lostDC；
Step B: it measures in frequency f1With the respective load loss P1 and Px of the product under frequency fx, and winding vortex is decomposited
Pwe1 and structural member stray loss Pse1 is lost；
Step C: combine the total eddy-current loss Pwe of winding calculated under harmonic wave with eddy-current loss Pwe1 according to K value；
Step D: P is lost according to D.C. resistanceDC, the total eddy-current loss Pwe of winding be calculated it is equivalent under specified harmonic load
Overload current multiple I when for power frequencyN；
Step E: simulation measures winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.
For the present embodiment is with respect to UL1561 standard, k-factor transformer equivalent load test side is more reasonably defined
Method avoids for the stray loss Pse in product structure part being mixed into winding, causes the equivalent loss virtual height of product and by transition
The phenomenon that test.
The present invention is suitable for that the Equivalent Calculation side that the general table of detailed harmonic wave frequency has only estimated its K value size cannot be provided in user
Method.
For the ease of explaining the Computing Principle of the present embodiment, following transformer is now provided, calculates the product in this k-factor item
Overload current multiple I when power frequency is equivalent under partNAnd the equivalent temperature rise of winding；Wherein, the rated capacity of the transformer:
250KVA, rated voltage ratio: 380V/380V, connection group: YNyn0, aluminium foil coiling, short-circuit impedance 4%, H grades of heatproof, environment
30 DEG C of temperature, allow temperature rise 100K Max, K-Rated=30.
Specific calculating is described below, and steps are as follows:
Step A: the D.C. resistance R of power transformer product winding is measured at 30 DEG C of room temperatureDC, then according to I2RDCCalculate direct current
Resistance loss PDC；
Step B: two kinds of frequency f are measured using dual-frequency method1(50Hz) and f4Respective load loss P under (200Hz)1
(50Hz) and P4(200Hz), and decomposite out winding eddy current loss Pwe1 and structural member stray loss Pse1.Measure P1=
2497.4W P4=3270.9W；Further according to formula P1=PDC+Pwe1+Pse1；PX=PDC+Pwe1×(fX/f1)2+Pse1×(fX/
f1)0.8；The Pwe1=41.42W of calculation, Pse1=74.95W.
Step C: formula P is substituted into according to K value and eddy-current loss Pwe1We=PWe1× K calculates total eddy-current loss Pwe；PWe=
PWe1× K=1242.6W.
Step d1: the winding temperature coefficient T estimated under equilibrium temperature is calculated according to formula Tc=(Ts+Tk)/(Tm+Tk)C；
Estimate transformer winding equilibrium temperature Ts=100+30=130 (DEG C)；Tc=(Ts+Tk)/(Tm+Tk)=1.39；
Step d2: overload current multiple I when calculating the equivalent power frequency of winding under k-factor harmonic waveN:
The above calculated result shows:
1, product power frequency rated current when without harmonic wave are as follows: 250/1.732/0.38=379.85 (A)；
2, by under K-Rated=30 harmonic wave operating condition, 9% need to be overloaded by being equivalent to winding current；；
3, by equivalent power current=379.85 of winding × 1.09=414 (A) under K-Rated=30 harmonic wave.
Final step E: using short-circuit method under the conditions of power frequency, will apply 414A rated current by test transformer winding,
I.e. analog measures winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.
And the method computation overload electric current multiple I introduced by the UL1561 in background techniqueNAnd the equivalent temperature rise of winding,
It is as shown below:
Step 1: PDC=2381W；
Step 2: PAC=2497.4W；
Step 3: Tc=1.39；
Step 4: PEC=(PAC-PDC)/(PDC×Tc2)=0.0253；
Step 5: it calculates the equivalent total load of winding under k-factor harmonic wave and P is lostLL；
K-Rated is calculated by the numerical value of the general table of harmonic wave frequency first:
K=(∑ Ih2×h2)/(∑Ih2)
=(∑ (Ix/I1)2×(fx/f1)2)/(∑(Ix/I1)2)
=30
PLL=PDC×(1+K×PEC) × Tc=5821.6W；
In temperature rise experiment, total losses P is maintainedLL=5821.6W is constant, is generally tested using short-circuit method.
Convert below computation overload electric current multiple under room temperature (due to UL1561 standard do not decomposite eddy-current loss with it is miscellaneous
Loss is dissipated, is not easy conversion to overload current multiple under equilibrium temperature).
Step 6: the equivalent overload current multiple I of winding under k-factor harmonic wave when calculating room temperatureN；
PEC-2=(PAC-PDC)/PDC=0.0489；
PLL-2=PDC×(1+K×PEC)=5874W；
IN=(PLL/PAC)0.5=1.53；
In order to there is comparativity, now by a kind of isolating transformer equivalent load applied under harmonic condition of the invention mentioned
Also computation overload electric current multiple I under room temperature is converted in test method specific embodiment step 5N
The equivalent current that can be seen that two formulas difference 1.53-1.23=30% from upper calculated result, i.e., will meet UL1561
Standard just needs additional increase-volume 30% (75KVA), needs more to invest, occupies more spaces, and actually this is not to use
Required for family.
Finally it should be noted that the above embodiments are merely illustrative of the technical solutions of the present invention, rather than the present invention is protected
The limitation of range is protected, although explaining in detail referring to preferred embodiment to the present invention, those skilled in the art are answered
Work as understanding, it can be with modification or equivalent replacement of the technical solution of the present invention are made, without departing from the reality of technical solution of the present invention
Matter and range.
Claims (7)
1. a kind of k-factor transformer equivalent load conversion method, it is characterised in that: the following steps are included:
Step A: the D.C. resistance R of power transformer product winding is measured at normal temperatureDC, then calculate D.C. resistance and P be lostDC；
Step B: it measures in frequency f1With the respective load loss P1 and Px of the product under frequency fx, and winding eddy current loss is decomposited
Pwe1 and structural member stray loss Pse1；
Step C: combine the total eddy-current loss Pwe of winding calculated under harmonic wave with eddy-current loss Pwe1 according to K value；
Step D: P is lost according to D.C. resistanceDC, the total eddy-current loss Pwe of winding is calculated under specified harmonic load and is equivalent to work
Overload current multiple I when frequencyN；
Step E: simulation measures winding equivalent temperature rise of the product under corresponding K coefficient harmonic wave.
2. a kind of k-factor transformer equivalent load conversion method according to claim 1, it is characterised in that: in step B
In, P1 and Px are calculated by following formula:
P1=PDC+Pwe1+Pse1；
PX=PDC+Pwe1×(fX/f1)2+Pse1×(fX/f1)0.8；Pwe1, Pse1 are solved by upper two formulas simultaneous.
3. a kind of k-factor transformer equivalent load conversion method according to claim 1, it is characterised in that: in step C
In, the total eddy-current loss Pwe of winding is calculated by following formula:
PWe=PWe1×K。
4. a kind of k-factor transformer equivalent load conversion method according to claim 1, it is characterised in that: in step D
In, comprising the following steps:
D1: the winding temperature coefficient T estimated under equilibrium temperature is calculated according to formula Tc=(Ts+Tk)/(Tm+Tk)C；Wherein Ts is
Estimate transformer winding equilibrium temperature；Virtual environment temperature when Tm is test；Tk is the thermal constant of medium；
D2: overload current multiple I when power frequency equivalent according to winding under following equation calculating k-factor harmonic waveN:
5. a kind of k-factor transformer equivalent load conversion method according to claim 4, it is characterised in that: in step d2
In, the thermal constant of copper is 234.5 DEG C, and the thermal constant of aluminium is 225 DEG C.
6. a kind of k-factor transformer equivalent load conversion method according to claim 1, it is characterised in that: in step E
Using short-circuit method under the conditions of power frequency, I will be applied by test transformer windingNTimes rated current, simulation measure product in corresponding K
The equivalent temperature rise of winding under coefficient harmonic wave.
7. a kind of k-factor transformer equivalent load conversion method according to claim 1, it is characterised in that: in step B
In, take fx >=3f1。
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Cited By (1)
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| CN113835051A (en) * | 2020-12-25 | 2021-12-24 | 华北电力大学（保定） | Method for determining stray loss of magnetic conduction component under alternating current-direct current composite excitation |
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