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.

 

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

 

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

 

dear Mr Corso

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

regards ...

 

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

 

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

 

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)

 

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

Regards, Phil Corso

 

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?

 

> 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 ([email protected])

Phil's answer is totally right answer.

 

>[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 %

 

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

 

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

 

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.

 

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)²] !

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

 

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?

 

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?

 

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 ([email protected])

 

AT... Additional to my previous post:

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

Phil

 

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

Phil