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% Impedance of Power Transformer

Elaborate on % impedance of power transformer.

By ramesh on 5 August, 2008 - 1:12 am

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By Ananymous on 6 August, 2008 - 11:28 pm

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"The percentage impedance of a transformer is the volt drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage."

"It is also the percentage of the normal terminal voltage at on side required to circulate full-load current under short circuit conditions on other side."

The impedance of a transformer has a major effect on system fault levels. It determines the maximum value of current that will flow under fault conditions.

It is easy to calculate the maximum current that a transformer can deliver under symmetrical fault conditions. By way of example, consider a 2 MVA transformer with an impedance of 5%. The maximum fault level available on the secondary side is:

2 MVA x 100/5 = 40 MVA

and from this figure the equivalent primary and secondary fault currents can be calculated.

A transformer with a lower impedance will lead to a higher fault level (and vice versa).

The figure calculated above is a maximum. In practice, the actual fault level will be reduced by the source impedance, the impedance of cables and overhead lines between the transformer and the fault, and the fault impedance itself.

By N.S. Narayanan on 3 November, 2008 - 1:52 am

My client has stipulated that the PERCENTAGE IMPEDANCE VOLTAGE for the 60MVA - 132/66KV Power Transformer shall be MORE THAN 20%.

My questions are as follows:

[1] If the Percentage Impedance is more, the FAULT LEVEL CURRENT will less. Thus it will be useful.

[2] But are there any ADVERSE EFFECTS if the PERCENTAGE IMPEDANCE is more?

[3] Is there any ceiling for the PERCENTAGE IMPEDANCE VALUE? If yes, how much it should be?

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Responding to Points [2] & [3] in Narayanan's 03-Nov-08 (01:52) query:

Typically a transformer's impedance is sufficient to withstand maximum fault exposure. Also, for most manufacturers, 20% is considered an upper limit. However, voltage-regulation will suffer and non-normal or additional taps may be be required as compensation. Beyond 20%, physical constraints can result in a model far removed from good engineering practice.

In my opinion, it is far more prudent to install an impedance in the transformer's primary supply circuit than choose a one-of-a-kind transformer design.

Regards, Phil Corso (cepsicon@aol.com)

By Dilip on 28 June, 2011 - 3:27 pm

@Phil,

i am purchasing a transformer for Submerged Arc Furnace. Rating 45 MVA,33KV Primary, 170-300 v secondary ,, cooling OFWF. On Load tao changer on 33 KV side with 32 tap positions. Bidders are quoting impedance in range 6%-8%.

What should be my requirement to ask for more impedance on less? pl elaborate with calculations. You may send mail on mel_em-at-rediffmail.com

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Dilip... for the assistance you seek, please contact me off-list at:

cepsicon[at]AOL[dot]com

Regards, Phil Corso

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Dilip... further to my earlier reply, I like to know why you never responded to Control.Com Thread #:

http://www.control.com/thread/1257876382 "Parallel Operation of Transformers"

Phil Corso

By Sdhir B. Pednekar on 1 April, 2012 - 1:09 pm

> Responding to Points [2] & [3] in Narayanan's 03-Nov-08 (01:52) query:

> Typically a transformer's impedance is sufficient to withstand maximum fault exposure. Also, for most manufacturers,

> 20% is considered an upper limit. However, voltage-regulation will suffer and non-normal or additional taps may be

> be required as compensation. Beyond 20%, physical constraints can result in a model far removed from good engineering practice.

> In my opinion, it is far more prudent to install an impedance in the transformer's primary supply circuit

> than choose a one-of-a-kind transformer design.

> Regards, Phil Corso (cepsicon@aol.com)

Phil's answer is totally right answer.

Hello dear,

1. If percentage impedance of transformer increase, then losses will increase and efficiency will decrease.

2. Your 20% figure is abnormal to me. Z% is a trade off between economics and performance.

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Randhir,

1) While a higher impedance will result in an increase in percent regulation, it doesn't necessarily increase losses.

2) 10% is usually the preferred maximum limit, but manufacturers can go higher.

3) I agree % Z is a trade off between economics and performance, but aren't most designs?

Regards, Phil Corso (cepsicon@aol.com)

Hello,

I would like to include the transformer tap (real value) in the load flow and optimisation. For this, i need the realistic data of resistance and reactance of a transformer (5 MVA, 33/0.4 kV). The rating of a transformer could be higher as well. Next, how can i incorporate the transformer data(resistance and reactance) with the transmission line between two buses?

bishnu (getbishnu100 [at] yahoo.com)

By pinaki chatterjee on 28 September, 2012 - 3:59 am

>[3] Is there any ceiling for the

>PERCENTAGE IMPEDANCE VALUE? If yes, how

>much it should be?

the value should be as per ISS 20 %

By aj on 26 December, 2011 - 1:57 am

this % impedance is calculated at 50 hz , what will be impedance at 5hz. 10 hz will it be linear i.r for 5hz divide by 10 and 10hz divide by 5 and so on. As R is not affected due to frequency change only L will be changed.

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AJ... the impedance formula is:

%Z = SQRT[ (%R)^2 + (%X)^2 ] at 50Hz.

And, as you deduced, only the X term is affected by frequency. Thus, the %Z formula, for any frequency, f, is:

%Z = SQRT [ (%R)^2 + (%X x f/50)^2 ]

Please note the formula is for typical power transformers! It ignores frequency-sensitive elements found in low-impedance transformer designs, high-efficiency transformers, non-linear core-effects, harmonics due to saturation, interwinding capacitancs, and winding-to-core capacitances.

Regards, Phil Corso

By Jeffrey Alden Olson on 10 January, 2012 - 7:31 am

when recording audio, is the voltage AC or DC? and does that matter when i am Trying to figure Impedance? To tell you the truth, I am still VERY lost, and confused.. If i have a voltometer and I would like to measure whether the salvaged transformers I have are Step-Up, Step-Down, or even for isolation (As you can see, i know very little of what the internet has provided me with) ** I am also quite young. Please any help will do.

j.alden.olson [at] gmail [dot] com

By shiv on 30 March, 2012 - 3:34 am

Hello,

I am a trainee. I need to simulate the power system in a software called power world simulator.

For that, I need R and X specifications of transformer and transmission line.

In transformer being used here at my workplace, %impedence volt is provided. Its given like 6.25% +/- IS TOL. What does it mean? How can I derive R & X values from this. Load Loss is also provided.

For the following ratings, I need R & X values -

1) 1000kva

2) 6300KVA

3) 1600KVA

4)630KVA

5) 400KVA

For Transmission line, XLPE cable is used. Where to find its R & X values? In general, what are the values for it?

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Responding to Ramesh's 5-Aug-08 (00:12) request... following is a simplified discussion:

A transformer's impedance (also referred to as impedance-voltage) is equal to the voltage, in % of its rated-voltage, that when applied to the primary-winding of a transformer will cause rated-current to flow in its shorted secondary-winding. For example, consider a transformer rated 500kVA, 6kV/400V, and 4.0% impedance. It means that when 240V is applied to the primary-winding, then the rated current, 720A, will flow in the shorted secondary.

To elaborate, consider a two-winding transformer having an equal number of turns in its primary and secondary windings. The effective transformer resistance is the sum of the primary and secondary resistances. Similarly, the effective transformer reactance is the sum of the primary and secondary leakage reactances. Finally, the effective transformer impedance is the vector addition of its effective resistance and reactance.

If additional information is required, let me know!

Regards, Phil Corso (cepsicon@aol.com)

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Bishnu... your question is a good one. Although, the effect taps have on load-flow and fault-duty studies is seldom considered it is clear tap-changers can influence accuracy of those studies.

To give you an idea of the variation that influences the theoretical effects, following are several design factors that must be taken into account: a) core or shell construction; b) 3 or 5-limb magnetic structure; c) location of tap, i.e., terminal-end or neutral-end of a wye-winding ; d) tap-changer steps that are symmetrical about the nominal rating or skewed to either the low higher voltage; and e) others!

All is not lost however. Most system planners or designers use either of two approaches to "adjusting" the reactance component of the transformer's nominal impedance: A) proportional to the step-interval; or B) proportional to the square of the step-interval. If you would like additional detail contact me off-forum.

Regards, Phil Corso (cepsicon [at] aol [dot] com)

By George Mattam on 12 January, 2011 - 2:29 pm

If a lower than standard impedance voltage is required for a power transformer while designing, how is it achieved?

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George... a large number of factors influence a transformer's impedance.

Key factors are core material selection such as permeability, flux density, laminate thickness, and coating. Others are conductor characteristics (resistance, shape, Ampacity, and insulation.) Still others are core geometry such construction-type (core or shell,) number of limbs or legs (3 or 5) limb shape (circular, square, hexagonal, or octagonal) coil arrangement on the core-limb, and others.

Fortunately, there are a number of manufactures that will gladly provide cost increments based on the desired deviation from industrial standards.

In closing, if you are contemplating purchase of a transformer having an off-norm impedance to compensate for power supply inadequacies, or to mitigate harmonic effects, there are alternatives, as well as restraints, that should be considered. One such concern is the impact on short-circuit duty. On the other hand, if basic knowledge is your goal, then I was glad to help.

Regards, Phil Corso

By Ronnie on 8 February, 2011 - 4:29 am

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I'm using 45/55 MVA ONAN/ONAF transformer. Suppose I'm using an impedance of 12.5% for the purpose of calculation...then this impedance which corresponds to 55MVA would be different for transformer operation at 45 MVA? Please advice. Also advice if there's any IEC code which states sth related to this problem.

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Ronnie... Unless shown differently on the transformer's nameplate, the impedance is based on its OA (45 MVA) rating. Furthermore, impedance is not altered for operation at other capacities. However, tap setting does affect reactance.

Regards, Phil Corso (cepsicon [at] aol [dot] com)

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Phil Corso wrote:

> ...if you are contemplating purchase of a transformer having an

> off-norm impedance to compensate for power supply inadequacies, or to

> mitigate harmonic effects, there are alternatives, as well as restraints,

> that should be considered. One such concern is the impact on short-circuit duty...

I am considering three 20/110kV step-up transformers rated at 30, 45 and 55 MVA. The 30MVA transformer substation 110kV busbar is connected to a distribution network via 19km of 630mm˛ trefoil XLPE. The 45MVA is connected to the same 110kV busbar for a 5km remote generator. The 55MVA is connected to the same 110kV busbar via 13km of 150mm˛ trefoil cable.

Impedances are expected to be less than 10%, I need to do some calculations for this but I'm actually interested in mitigating harmonic effects and also in dealing with charging currents for such long underground HV lines.

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Phil Level...; to avoid mis-interpretation of your data, please forward a simple SLD (Single-Line-Diagram) or hand-drawn sketch to me at:

cepsicon[at]AOL[dot]com

Regards, Phil Corso

By mustafa on 15 February, 2012 - 2:49 pm

dear Mr Corso

i am asking why it was called percentage impedance (%Z) while it is % of voltages?

regards ...

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Mustafa... your question is really an excellent one. In fact, %Z (impedance) and %V (% of rated primary-voltage to produce rated secondary-current with secondary shorted) are not exactly equal! But, approximations during testing are negligible in the interest of practicality!

Let's start with the definition of "Exact"" impedance. Impedance is the total opposition to AC current flow in an electrical apparatus, thus 'involved' with losses. But, in order to simplify calculations, testing parameters are carried out to just a few significant figures.

For example consider the "Open-Circuit" test. Impedance voltage is usually lower than full-load impedance voltage. Thus, excitation losses for the part-voltage test are just a few percent less than when the test is carried out at full-voltage

Now consider the "Short-Circuit" test. If the primary-current is at rated voltage, but exciting-current ignored, then secondary-current is also at its rated value.

I hope the above is adequate for your purposes. If not, let me know. Regards, Phil Corso

By Sunil on 16 February, 2012 - 9:31 am

I need help from you guys,

We have two transformers (20MVA, 10% IMPEDANCE, 16KA fault amaps) feeding to 6.6KV switchgear (single bus). Now there is an issue with CT knee point voltage, before the consultant has designed the C.T as per transformer fault calculation. but now new setting has been issued by considering B/C close condition (2 transformer in parallel) and the incomer CT's for REF protection are to be replaced with New setting with Knee point voltage double then before.

Now my question is how does the CT sizing for the incomers matter with the B/C Open or close condition.

I feel the sizing for the CT's should be changed only for out going feeders not for incoming side. but I want to know whether I am right or wrong, please give your thoughts.

Regards

Sunil Kumar

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Sunil... location and ratios of the phase and neutral CTs used for the REF scheme is very important. Can you provide a simple Single-Line-Diagram (SLD) or a hand-drawn sketch of system?

Regards, Phil Corso (cepsicon[at]AOL[dot]com)

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Sunil... based on the SLD you forwarded to me, your instinct that a CT change out is unwarranted, is correct!

Regards, Phil Corso

By Jacky on 20 February, 2011 - 4:36 am

i'm doing my uni's project. About open-circuit test for a transformer, why does increasing the input voltage, Xm (reactance i suppose) will decrease? is there an explanation behind it?

and also, why is Rc constant throughout the experiment?

You can email me at jackyk1988 [at] hotmail.com

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Jacky...

(a) Reur Xm comment:

I don't quite understand your comment that, as test voltage increases, Xm decreases. Can you provide a calculation illustrating your observation?

(b) Reur Rc comment:

Because Rc represents the winding resistance, as well as the core loss.

Regards, Phil Corso

By AT on 28 March, 2014 - 6:27 pm

Hello Phil:

I have a question. I have a 1000kVA Delta-Wye transformer with 13.8kV/0.48kV, Z% = 5.75%, loading = 80%, and pf = 0.8 lagging.

I have to find an appropriate tap setting in 2.5% ) steps (All tap settings = 5%, 2.5%, 0, -2.5%, 5%) on the HV winding so that the voltage at the secondary terminal is 480V +/- 5%.

For this, I am told I have to consider the effect of the internal voltage drop within the transformer (Z% effect) as a function of % loading.

What would be the best way to calculate internal voltage drop of transformer with above information? and how would I calculate the appropriate tap setting on the primary side to make sure that the voltage on secondary is +/- 480V?

V_primary = 13800

Am i correct to this?

V-percent_impedance = 0.0575 * 13800 = 793.5

I_rated = (1000kVA*0.8 + j*(1000kVA*sin(cos-1(0.8)))) / (sqrt(3)*13800) = 33.469 + j*25.102

Z = V-percent_impedance / I_rated = 15.17 - j11.37

sin(cos-1(0.8))=0.6

V_secondary = (V_primary - (((1000kVA*0.8 + j*1000kVA*0.6)*0.8)/(sqrt(3)*V_primary))*Z)*480/13800 = 465.87 - j10.2

V_primary_required = 480*(13800/480) + ((1000kVA*0.8 + j*1000kVA*0.6)/(sqrt(3)*V_primary))*Z = 14593.145 - j0.25102

CHG = (abs(V_primary_required) - abs(V_primary)) / abs(V_primary) = 0.05746

and select the appropriate tap setting based on CHG.

Therefore selecting a tap of +5% change..

Am i correct.. I am very confused and need to know that I am on the right track.

Thanks

If I am incorrect, then kindly let me know

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AT... I hope you don't mind my asking the following just to be sure about your query:

1) A Xfmr's Nameplate never lists Load- Characteristics!

2) Thus, are you looking for the mathematical procedure to determine what the Voltage-Regulation (VR) is, for the Load you defined as 800kVA at 0.8pf, lagging?

Regards,

Phil Corso

By AT on 30 March, 2014 - 3:06 am

Hello Phil:

Yes, the loading information is not listed on the Xfmr's Nameplate. Yes, I am trying to look for a mathematical procedure that would allow me to calculate the voltage drop across the transformer with the information and allow me to calculate appropriate tap setting to achieve desired voltage on the secondary side.

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AT... OK! Let's start with basics:

A) Given: kVA(L) = 800; PF(L) = 0.8; kV(L) =0.48; Xfmr's % Z = 5.75 !

B) Find: Load Current Magnitude: I(L) = kVA(L) / [ SQRT(3) * kV(L) ] = 1,155A !

C) Find: Load Current in Complex Units; Ir(L): I(L) * PF(L) + j * I(L)*SQRT[1-PF(L)˛] !

D) Find: Voltage-Regulation! But, what is Voltage Regulation? The answer::

It is the change in Secondary voltage (Vsec) from its No-Load value, Vs(o) to its Under-Load value, Vs(u), with the Primary voltage (Vpri) held constant! Expressed as a percent, it is:

Vr = 100 * [ Vs(o) – Vs(u) ] / Vs(u), in %!

Although it appears simple, its accuracy depends very much on the Xfmr's efficiency, thus more detail is required! Certainly, one can use the Rule-of Thumb (RoT)... Vr is proportional to its % impedance! The basic problem is that the No-Load secondary voltage, Vs(o), when 13.8kV is applied to the Xfmr's primary, is not known, and not given on the Xfmr's nameplate! Using RoT, the Vr would be 80% of 5.75% or about 4.6% , yielding 458V!

In reality, the actual Voltage-Drop is not proportional to %Z, but instead varies dis-proportionately with exciting-current, which is ignored in the RoT approach!

AT, at this point I must ask the question, "How much accuracy is required?"

Regards,

Phil

By AT on 30 March, 2014 - 6:08 pm

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Hello Phil:

Thank you for the speedy reply.

I would like as much accuracy as possible from the information that I am given.

But to recap...

So, because the 1000 kVA transformer is 80% loaded, we have determined the load kVA = 800 kVA with PF (L) = 0.8, kV (L) = 0.48, %Z = 5.75%.

Now you said to find the load current and gave the answer to be 1155A.

When I did the calculation =>

I(L) = 800 kVA / [ SQRT(3) * 480] = 962.250 A

am i wrong? please tell me.... I must have made a mistake somewhere..

Load current in complex units => I(L)*0.8 + j*I(L)*0.6 = 769.800 + j 577.350

Vr = 100*[Vs(o) - Vs(u)] / Vs(u)

80 percent of 5.75 is 4.6, therefore,

Vr = 4.6%

Therefore,

4.6 = 100* [Vs(o) - Vs(u)] / Vs(u)

4.6/100 = [Vs(o) - Vs(u)] / Vs(u)

0.046 = Vs(o)/Vs(u) - 1

0.046 + 1 = Vs(o)/Vs(u)

1.046 = Vs(o)/Vs(u)

Assuming Vs(o) = 480V

1.046 * Vs(u) = 480

Vs(u) = 480/1.046 = 458.89

As far as accuracy is concerned I tried a different approach and I think that I got similar results. Now I am hoping that it is not a fluke.

As far as I have read, I think that the percent impedance is calculated by shorting out the secondary of a transformer and injecting voltage into the primary until the rated current is measured in the primary or secondary. And the voltage at which the primary current is measured is then divided by the rated voltage to generate the percent impedance value.

So with that in mind:

%Z = Vz / Vrated

5.75% = Vz / 13800

0.0575 * 13800 = Vz

Vz = 793.5 V

Irated_primary = 1000kVA / [ Sqrt(3)*13800]

Irated_primary = 41.837 A

therefore,

%Z = Vz/Vrated and

Z = Vz/Irated_primary

Z = 793.5/41.837 = 18.966 ohms

Assuming V_primary = 13800 V

cos(theta) = 0.8

sin (cos-1(theta)) = 0.6

Therefore, after the voltage drop due to the transformer impedance =>

I_excit = [((1000k*0.8) + j*(1000k*0.6))*0.8] / [Sqrt(3)*13800] = 26.776 + j20.082 Amps

V_primary_induced = (13800*0.8 + j*13800*0.6) - (26.776 + j20.082)*18.966 = 10532.17274 + j 7899.129558 = 13165.21593 <36.86989

V_secondary = 13165.216 <36.87 * 480/13800 = 457.9205 <36.869

the V secondary that I calculated is pretty close to:

80 percent of 0.0575 = 0.046

1-0.046 = 0.954

Vsec = 480 * 0.954 = 457.92 V

The above is I think what you calculated in your original reply.

Do you think that my second approach (subtracting voltage drop from primary and converting it to secondary) is appropriate? Or is my approach totally flawed and i got that answer by mere fluke?

I was thinking about trying to find the exact R + jX value from percent impedance to calculate a more accurate voltage drop by calculating percent reactance by using formula:

%X = [((kV)*(kV))/kVA ] * %Z/100

%X = [(13800*13800) / 1000kVA ] * (5.75/100)

%X = 10.9503

%R = Sqrt[(%Z*%Z) - (%X*%X)]

%R = Sqrt[(18.966*18.966) - (10.9503*10.9503)]

%R = 15.485

I_excit = [((1000k*0.8) + j*(1000k*0.6))*0.8] / [Sqrt(3)*13800] = 26.776 + j20.082 Amps

V_primary_induced = (13800*0.8 + j*13800*0.6) - (26.776 + j20.082)*(15.485 + j10.9503) = 10845.27756 + j7675.824997 = 13286.7792 < 35.2892

Vsecondary = 13286.779 <35.2892 * 480/13800 = 462.149

Am i at all on the right track in the above two methods of calculations or is my brain out for lunch. Which of the two methods is more accurate in your opinion?

By AT on 30 March, 2014 - 6:36 pm

Hello Phil:

Thank you for the speedy reply.

I would like as much accuracy as possible from the information that I am given.

But to recap...

So, because the 1000 kVA transformer is 80% loaded, we have determined the load kVA = 800 kVA with PF (L) = 0.8, kV (L) = 0.48, %Z = 5.75%.

Approach in your answer:

Now you said to find the load current and gave the answer to be 1155A.

When I did the calculation =>

I(L) = 800 kVA / [ SQRT(3) * 480] = 962.250 A

am i wrong? please tell me.... I must have made a mistake somewhere..

Load current in complex units => I(L)*0.8 + j*I(L)*0.6 = 769.800 + j 577.350

Vr = 100*[Vs(o) - Vs(u)] / Vs(u)

80 percent of 5.75 is 4.6, therefore,

Vr = 4.6%

Therefore,

4.6 = 100* [Vs(o) - Vs(u)] / Vs(u)

4.6/100 = [Vs(o) - Vs(u)] / Vs(u)

0.046 = Vs(o)/Vs(u) - 1

0.046 + 1 = Vs(o)/Vs(u)

1.046 = Vs(o)/Vs(u)

Assuming Vs(o) = 480V

1.046 * Vs(u) = 480

Vs(u) = 480/1.046 = 458.89

As far as accuracy is concerned I tried a different approach and I think that I got similar results. Now I am hoping that it is not a fluke.

My approach # 1:

As far as I have read, I think that the percent impedance is calculated by shorting out the secondary of a transformer and injecting voltage into the primary until the rated current is measured in the primary or secondary. And the voltage at which the primary current is measured is then divided by the rated voltage to generate the percent impedance value.

So with that in mind:

%Z = Vz / Vrated

5.75% = Vz / 13800

0.0575 * 13800 = Vz

Vz = 793.5 V

Irated_primary = 1000kVA / [ Sqrt(3)*13800]

Irated_primary = 41.837 A

therefore,

%Z = Vz/Vrated and

Z = Vz/Irated_primary

Z = 793.5/41.837 = 18.966 ohms

Assuming V_primary = 13800 V

cos(theta) = 0.8

sin (cos-1(theta)) = 0.6

Therefore, after the voltage drop due to the transformer impedance =>

I_excit = [((1000k*0.8) + j*(1000k*0.6))*0.8] / [Sqrt(3)*13800] = 26.776 + j20.082 Amps

V_primary_induced = (13800*0.8 + j*13800*0.6) - (26.776 + j20.082)*18.966 = 10532.17274 + j 7899.129558 = 13165.21593 <36.86989

V_secondary = 13165.216 <36.87 * 480/13800 = 457.9205 <36.869

the V secondary that I calculated is pretty close to:

80 percent of 0.0575 = 0.046

1-0.046 = 0.954

Vsec = 480 * 0.954 = 457.92 V

The above is I think what you calculated in your original reply.

Do you think that my second approach (subtracting voltage drop from primary and converting it to secondary) is appropriate? Or is my approach totally flawed and i got that answer by mere fluke?

My approach # 2:

I was thinking about trying to find the exact R + jX value from percent impedance to calculate a more accurate voltage drop by calculating percent reactance by using formula:

%X = [((kV)*(kV))/kVA ] * %Z/100

%X = [(13800*13800) / 1000kVA ] * (5.75/100)

%X = 10.9503

%R = Sqrt[(%Z*%Z) - (%X*%X)]

%R = Sqrt[(18.966*18.966) - (10.9503*10.9503)]

%R = 15.485

I_excit = [((1000k*0.8) + j*(1000k*0.6))*0.8] / [Sqrt(3)*13800] = 26.776 + j20.082 Amps

V_primary_induced = (13800*0.8 + j*13800*0.6) - (26.776 + j20.082)*(15.485 + j10.9503) = 10845.27756 + j7675.824997 = 13286.7792 < 35.2892

Vsecondary = 13286.779 <35.2892 * 480/13800 = 462.149

Summation:

Am i at all on the right track in the above two approaches (1 and 2) or is my brain out for lunch. Which of the two approaches is more accurate in your opinion? Is any of my approach valid?

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AT... Several comments about your thesis:

1) "When I did the calculation => I(L)=800kVA / [ SQRT(3)- 480]= 962.250A!" You are correct, I erred!

2) "% Z ... shorting sec'y, decreasing pri'y V until rated-current flows in pri'y or sec'y" Pri'y (NO! Sec'y (YES!)

3) "Do you think my 2nd approach (subtracting voltage-drop from priy and converting it to sec'y is appropriate?" (NO!)

4) "Or is my approach totally flawed and I got that (AN) answer by mere fluke?" (YES!)

Now, 'The rest of the Story!' I said that using RoT was an approximation! Furthermore, if you wanted accuracy additional parameters are essential, including the selected 'Equivalent Circuit Model, i.e., 'PI' or 'T'! In fact, Iexciting-current is constant for the 'PI' model, but it is a function of pri'y voltage-drop with the 'T' model!

AT... bringing your query to a conclusion: the exact answer is quite formidable but can be found in a 'Power Transformer Design' Text! The required data can be determined from 'Open-Ckt' and 'Short-Ckt' tests (usually available from the Xfmr Mfgr)!

If you want the answer from me then contact me off-list with your full name, your title, or job function, and your location.

Phil (cepsicon@aol.com)

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AT... Additional to my previous post:

5) " %X = . . . . . = 10.9503 " (NO!) Your calculated %X value is greater than the given %Z value!

Phil

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AT... would you prefer the RoT formula, ignoring impact of X/R ratio and exciting-current, on List?

Phil

By KS on 22 April, 2014 - 7:48 pm

Hi Phil

Can you kindly advise what is the effect of system fault level on transformer design.

I have a 1250kVA, 6.6/0.42kV, 5% transformer for a client who mentioned system fault level at 40kA.

Now the client is asking if the transformer is still OK if the system fault level is 63kA??

You valuable feedback is much appreciated.

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KS...

do you mean the short-circuit level on the 6.6kV level increased from 40 to 63kA?

Regards,

Phil Corso

By Majid Rasheed on 13 July, 2014 - 7:15 pm

Dear Sir,

I am using 63kVA 20/.4kv, 4% impedance transformer. Will it have fault level current = 45.5 amps on primary side?

Am I right?

Or please tell me the actual way to calculate this. also what are the things which may affect the fault level.

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Pritful... are you referring to a distortion of the voltage or current wave-shape?

Regards,

Phil Corso

By Miguel Sanchez on 5 January, 2016 - 6:48 pm

I think it is important the definition of impedance as percentage of nominal voltage necessary to apply to the primary(or secondary) winding so to get nominal current flow thru the transformer with the secondary (or primary) winding terminals connected together in short circuit.

It is also very important to understand what makes a transformer have higher or lower impedance.

Impedance on a power transformer is mainly a function of the inductance. The resistive portion is not totally insignificant but it is irrelevant in this analysis.

In a perfect "Ideal" transformer the magnetic field created by one winding is 100% seen and wrapped by the other winding. Every single turn on both windings is affected by 100% of the magnetic field in the core.

The reason why a real transformer shows a value of impedance, more specifically, "inductance" is due to the fact that not all the field is shared by both windings. When each winding keep a portion of its field for itself they inherently build in the transformer two separate series reactances, one on the primary and one on the secondary. These are the reactors that stay inserted in the current path of the short circuit test or in a true short circuit.