HARMONICS--TRANSFORMER DERATING

Daniel W. Egolf Student Member, IEEE Washington State University Pullman, WA

Abstract - Transformers supplying nonsinusoidal load current often need to be derated. The ANSMEEE C57.110-1986 standard provides a method by which derating may be performed. However, derating only provides an estimate of remaining transformer capacity, since subsequent load may significantly change the relative harmonic current values. The contribution of this paper is to provide a clear guide that can be used in applying the standard. ms paper is not meant to replace the standard but to supplement it by providing correct examples and simplified equations.

INTRODUCTION

Non-linear loads produce harmonic current. In many cases the transformer was not designed to handle this harmonic current and as a result it must be derated. The American National Standard (ANSUIEEE) C57.110-1986 Recommended Practice for Establishing Transformer CaDabilitv When SuDplying Nonsinusoidal Load Currents [ 11 provides a derating procedure that applies to most power transformers. This paper presents a clear guide that can be used in applying the standard.

ANWIEEE DERATING

For transformers, the winding insulation is usually the most sensitive part to heat. In addition, there are hot spots in the windings--parts of the windings heat up faster than others. Because the hottest spot is very critical to winding life [2], detailed information is needed about the transformer characteristics. A complete harmonic spectrum analysis of the load current also must be available. A power analyzer or spectrum analyzer can be used to obtain the load current information. The per unit power loss density of the winding hot spot must either be supplied by

0-7803-1877-3/94/$3.00 @ 1994 IEEE 79

Alfred J. Flechsig Senior Member, IEEE Washington State University Pullman, WA

the manufacturer (often unavailable) or calculated from available information (certified test report data or nameplate data). The following equations were developed from the IEEE standard [ 13 equations--see Appendix A for a compar- ison.

PLL(pu) is the load loss density at the winding hot spot under actual conditions. It is with reference to the winding PR loss density (which is not dependent on harmonics--the standard [l] ignores slun effect).

P,,@u) is the winding eddy current loss density at the winding hot spot under rated (no harmonic) conditions. It is with reference to the winding 12R loss density. This is the number that is often unavailable.

PLL.R@~) is the load loss density at the winding hot spot under rated conditions (no harmonics). It is with reference to the winding 12R loss density.

I,-@) = THDF = Transformer Harmonic Derating Factor = per-unit of full load available.

Note that the per unit quantities are defined as the loss density at the winding hot spot, and not the overall loss density.

The following table is an example of a harmonic magnitude spectrum analysis based on I,, (total line current). This is not the only way data may be found and care must be taken to make sure data are in the correct form. For this example, the measured line current is 1389.8 amps:

h I,(rms) I, (DUI

1 1,345 A .9678 3 350 A .2519

Per unit is referenced to I,-. For example:

- - l M 5 = 0.9678 pu 1345 - 1345 - Z- Jm 1.389.8 - -

Examde One. ComDutation of Useable kVA:

PEC-,@U) = 0.15 pu at the point of maximum loss density, as supplied by the manufacturer. PLL-R@u) = 1 + 0.15 = 1.15 PU To calculate the summation in PLL@u) make the following table:

h I,(pu) I,(Du)~ h2 I,(Du)~ * h2

1 .9678 .9366 1 3 .2518 .0634 9

1

PLL@u) = 1 + (0.15 * 1 Therefore,

and,

.9366 .5708 1,5074

5074) = 1.226 pu

If the nameplate rated value is 500 kVA then:

Useable kVA = 500 * 0.968 = 484 kVA

Any number of harmonics may be included--in the example, only the fundamental and third harmonic are of importance. Useable kVA is only an estimate of total transformer capacity. If future load significantly changes the harmonic per-unit values, useable kVA should be recalculated.

ExamDle Two, Computation of Useable kVA:

PEC, = 0.15 pu at the point of maximum loss density, as supplied by the manufacturer.

P,,@u) = 1 + 0.15 = 1.15 pu To calculate the summation in PLL(pu) make the following table:

1 1,672 A 0.9778 292 A 0.1708 5

7 185 A 0.1082 11 75 A 0.0439 13 48 A 0.0281 17 26 A 0.0152 19 17 A 0.0099

0.9561 1 0.9561 25 0.7290 0.0292

0.0117 49 0.5735 0.00192 121 0.23277 0.000788 169 0.13317 0.000231 289 0.06681 0.000099 361 0.03568

1 2.72708

Therefore,

PLL@u) = 1 + (0.15 * 2.72708) = 1.4091 PU

If the nameplate rated value is 500 kVA then:

Useable kVA = 500 * 0.9034 = 451 kVA

Method for Calculatine P , . m

The critical number is PEc&u)--transformers designed for harmonic loads will have a low number. If P,,@u) at the point of maximum loss density is unavailable, it must be calculated from certified test report data. Several assump- tions are made and the following method only applies to common power transformers. In specialty transformers, the construction is often different and the standard [l] does not apply.

The following data are needed:

1) Rated current in H.V. side = I,,

2) Rated current in L.V. side = 12,

3) Equivalent wye resistance measured between tw0H.V. terminals = kbHv = RI (AppendixB)

4) Equivalent wye resistance measured between two L. V . terminals = kbLv = R, (Appendix B)

5) Total load losses for the transformer = P,,

If the three phase transformer data shows the resistance of the three phases in series calculate RI and R, as follows (Appendix B):

Delta Winding: R, or R, = (2/9) * R,, Wye Winding: RI or R, = (2/3) * R-

If all that is available is transformer per unit resistance (careful, nameplates give per unit impedance but generally not resistance), calculate RI and R, in ohms as follows [3]:

This assumes equal per unit resistances in the high side and low side windings--this is not necessarily the case, but without measured information it is the best that can be done. There is no factor of two in the equation since the division by two (equal resistance per high side and low side windings) is canceled by looking through two phase windings (Appendix B).

Constants X and K must now be selected (See Appendix C for a flowchart). If the transformer has a self-cooled current rating less than 1,OOO Amps or the tums ration is 4:l or less, then X = 2.4, otherwise X = 2.8. For a three phase transformer, K = 1.5. For a single phase transform- er, K = 1.0.

The needed equations, as given in the standard [l], are now: PEC-R = PLL - K * ((ZZlR * R,) + (Z2, * 4)) watts and

(4)

The constant X converts an average loss density to a maximum loss density. The low voltage (inner winding) is assumed to have 60% or 70% of the total PEC-,, and the point of maximum eddy-current loss density is assumed to be 400 % of the average value [ 13. Therefore, the constant X = 2.4 or 2.8.

Example Three, Calculation of PEC.&ul. if not Available from Manufacturer:

Given the following transformer data:

I,, = 600 amps delta winding I,, RI = 0.0368340hm, R,,, R, = 0.001926 o h m , R3, P,, = 9,096 watts

= 1,390 amps wye winding

Since the given resistances are for three phases in series, the values of resistance to use are:

R, = (2/9) * 0.036834 = 0.0081853 o h m

R, = (2/3) * 0.001926 = 0.001284 ohms

The tums ratio is less than 4:l so X = 2.4. It is a three phase transformer so K = 1.5.

Using equations (4) and (5):

PE,-, = 9,096 - 1.5 * (600 * .0081853 + 1,390, * .001284) = 954.7 watts

P,.,@u) = (2.4 * 954.7) / (1.5 * .001284 * 1,3902) = 0.616 pu

Once PEC-,@u) is calculated, use equations (l), (2), and (3) just as if the manufacturer had supplied PEC.&u). The answer is not as accurate, but it is the best that can be done without the manufacturers figure.

General Considerations:

Derating is not very helpful if the cooling vents are blocked. Without proper cooling, any transformer can overheat. Be sure to mark clearly the modified kVA rating of the trans- former, or else it may be overloaded at a future date. Remember that modified kVA is just an estimate. If subsequent load produces heavy current harmonics, the transformer may need additional derating. If the formulas require true rms, use a true rms instrument or else the data may be invalid. Reference [2] is useful in attempting to estimate the remaining insulation life of a transformer. Reference [4] provides additional information on transformer derating as well as the effects of harmonics on other power system components. Reference [5] presents general harmon- ic considerations. Reference [6] takes a more detailed look at transformer winding temperature.

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APPENDIX A

There are three major differences between the equations in this paper and the ANSMEEE [ 11 equations. The following explains the differences and gives the derivation.

EOUATION 1:

h = l

corresponds to ANSIIIEEE (Eq 7).

However, provided enough terms are calculated (as has to be done anyway for accuracy):

h-h ,

Zh(pU)2 = 1 h = l

EOUATION 3:

corresponds to ANSUIEEE (Eq 8).

h=h-

h-1 1 + h - L c f:

h=l

The numerator is the same, the problem is the denominator. It must be shown that:

h=&

* Zh2 = PLL@U) (Q 1) h-I 1 + h-h,

EX L=l

This can be done in the following way:

h-1 = (Zh(pu)2*h2) h - L A = l .. - . c f:

h = l

but,

h - L h - l g 1

h-I - h - L

'hbU)' 1 h - h m

h = l h - 1 z l@U)2 z l@U)2 h = l

h - h- zh@u)Zh2

h = l

since, h-h- -

Zh@U)2 = 1 h = l

EOUATIONS (4) and (5):

K is the same, but X does not appear in the ANWIEEE standard [ 11; however, it is used and calculated as a number in ANSIIIEEE (Eq 12) and (Eq 13).

APPENDIX B

R, and R2 are measured between two terminals. Their values will depend on the resistance of an individual winding (R) and the way the transformer is connected (delta or wye). Note that the equivalent Wye diagram is just a mathematical tool to represent a transformer that is physically hooked up in Delta. If the transformer is physically hooked up in wye, replace R/3 with R and omit the 1/3 in equation B1.

If the resistance for the three phases in series, R- is given:

If the transformer is physically connected in delta and the

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i

Figure 1: Winding Resistance for Delta and the Wye Equivalent

resistance for the three phases in series is given, then the resistance between terminal a and b is:

However, if the per-unit value of R is' given (the real part of the per-unit impedance), it is almost certainly the single phase resistance of the transformer. A summary of three phase per-unit equations is given in [3].

APPENDIX C

Self -Cooled Current -Ra t in9 less Than 1,000 A ?

Turns 4 1 or less? h

Figure 2: Flowchart to Calculate X and K

REFERENCES

1. IEEE-PES Transformer Committee, "Recommend- ed Practice for Establishing Transformer Capability When Supplying Nonsinusoidal Load Currents," C57.110-1986.

2. W. J. McNutt, "Insulation Thermal Life Consider- ations for Transformer Loading Guides," IEEE Transac- tions on Power DeZivety, Vol. 7, No. 1, pp. 392-398, Jan. 1992.

3. and Design, PWS Publishers, Boston, pg. 99, 1987.

J. D. Glover, M. S a m , Power System Analysis

4. D. E. Rice, "Adjustable Speed Drive and Power Rectifier Harmonics--Their Effect on Power Systems Components, " IEEE Transactions on Industry Applications, Vol. 1A-22, No. 1, pp. 161-177, Jan./Feb. 1986.

5. T. M. Gruzs, "Uncertainties in Compliance with Harmonic Current Distortion Limits in Electric Power Systems, " IEEE Transactions on Industry Applications, Vol. 27, NO. 4, pp. 680-685, July/August 1991.

6. M. D. Hwang, W. M. Grady, H. W. Sanders, Jr., "Calculation of Winding temperatures in Distribution Transformers Subjected to Harmonic Currents, " ZEEE Transactions on Power Delivery, Vol. 3 , No. 3, pp. 1074- 1079, July 1988.

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