Transformer voltage regulation and calculations

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The discussion revolves around calculating the secondary winding resistance of a 415:11000 V transformer rated at 200 kVA with 2% voltage regulation. The user is uncertain whether to use the nominal voltage of 11,000 V or the loaded voltage of 10,780 V for their calculations, as both yield different resistance values. Participants clarify that the calculations should reflect the secondary voltage under load conditions, emphasizing the importance of accounting for reflected resistances and reactances from the primary side. The conversation also touches on the complexities introduced by leakage reactances and the definitions of kVA ratings in relation to real loads. Ultimately, the focus is on determining the maximum allowable secondary resistance for maintaining the specified voltage regulation.
  • #31
What you have done is essentially correct. You have approx. 40Ω secondary reactance which, even though essentially inductive rather than resistive, will still drop about 18A x 40Ω = 720V which well exceeds the alloted 220V.

So, there are but two possibilities:

1. you're confusing transformers. This transformer is to have only the 0.01Ω primary resistance, with no leakage impedance. In that case the secondary internal resistance is about 7Ω, putting the internal secondary drop to only 7Ω x 18A = 126V, well within the allotted 220V drop.

2. You have been given an impossible assignment.
 
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  • #32
Thank you rude man.

Following your reasoning, I have a total secondary impedance of (##605\Omega##), minus the impedance of the load (##592.9\Omega##), minus the reflected primary impedance (##7.0257\Omega##).

As the question asked for the maximum value of the secondary winding resistance, I am left with ##5.0743\Omega##

That look good?
 
  • #33
Numbskull said:
Thank you rude man.

Following your reasoning, I have a total secondary impedance of (##605\Omega##), minus the impedance of the load (##592.9\Omega##), minus the reflected primary impedance (##7.0257\Omega##).

As the question asked for the maximum value of the secondary winding resistance, I am left with ##5.0743\Omega##

That look good?
That looks fine.
 
  • #34
Numbskull said:

Homework Statement


A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance and leakage reactance when referred to the primary are 0.014 Ω and 0.057 Ω respectively.

(a) Determine the % regulation of the transformer at 0.8 power factor lagging.

(b) In designing a particular 415V to 11 kV, 200 kVA transformer, the primary winding resistance is to be 10 mΩ. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations


##R's = Rs + \frac{Rp}{n^2} \leftarrow## this is to calculate the resistance reflected from the primary to the secondary winding.

##L's = Ls + \frac{Lp}{n^2} \leftarrow## this is to calculate the reactance reflected from the primary to the secondary winding.

##n^{2} = \frac{415}{11000} = 0.0014233471 \leftarrow## this is the value of ##n^{2}##

The Attempt at a Solution


Firstly, with unity power factor, I believe that I can effectively treat this as 200kW because with a part reactive load we would have a non-unity power factor. Therefore calculations are based upon this assumption.

To calculate the primary (reflected to the secondary) impedance we use ##R's = Rs + \frac{Rp}{n^2}## which using ##\frac{Rp}{n^2}## and thus ##\frac{10 mΩ}{0.00142...}## gives us only the reflected portion of the impedance which is 40.06 (X) - we have not added the Rs

Similarly, the primary (reflected to the secondary) reactance is given by ##L's = Ls + \frac{Lp}{n^2}## which using ##\frac{Lp}{n^2}## and thus ##\frac{0.057Ω}{0.00142...}## gives us only the reflected portion of the reactance which is 7.0256 (R) - we have not added the Ls

As my previous method of calculation appears to be incorrect, I'm not sure what to do now. All I can see is that to achieve a power of 200kW at 11kV, would require a current of 18.181818 Amps. This gives a total impedance across the winding (i.e. the load) of 605 Ohms.

I have the feeling that I'm missing some key relationship between the reflected impedance and the only possible (maximum) value of the secondary winding. I don't know what the load value is, or the secondary winding resistance and reactance, so obviously the necessary data is in the given information.

There have been several topics on this question in this forum, all of which I have tried to learn from, but inevitably they become protracted or focus on a different part of the question. And like those other posters have said, the material supplied is often confusing, incomplete or simply inadequate.

Any help greatly appreciated!
Numbskull said:

Homework Statement


A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance and leakage reactance when referred to the primary are 0.014 Ω and 0.057 Ω respectively.

(a) Determine the % regulation of the transformer at 0.8 power factor lagging.

(b) In designing a particular 415V to 11 kV, 200 kVA transformer, the primary winding resistance is to be 10 mΩ. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations


##R's = Rs + \frac{Rp}{n^2} \leftarrow## this is to calculate the resistance reflected from the primary to the secondary winding.

##L's = Ls + \frac{Lp}{n^2} \leftarrow## this is to calculate the reactance reflected from the primary to the secondary winding.

##n^{2} = \frac{415}{11000} = 0.0014233471 \leftarrow## this is the value of ##n^{2}##

The Attempt at a Solution


Firstly, with unity power factor, I believe that I can effectively treat this as 200kW because with a part reactive load we would have a non-unity power factor. Therefore calculations are based upon this assumption.

To calculate the primary (reflected to the secondary) impedance we use ##R's = Rs + \frac{Rp}{n^2}## which using ##\frac{Rp}{n^2}## and thus ##\frac{10 mΩ}{0.00142...}## gives us only the reflected portion of the impedance which is 40.06 (X) - we have not added the Rs

Similarly, the primary (reflected to the secondary) reactance is given by ##L's = Ls + \frac{Lp}{n^2}## which using ##\frac{Lp}{n^2}## and thus ##\frac{0.057Ω}{0.00142...}## gives us only the reflected portion of the reactance which is 7.0256 (R) - we have not added the Ls

As my previous method of calculation appears to be incorrect, I'm not sure what to do now. All I can see is that to achieve a power of 200kW at 11kV, would require a current of 18.181818 Amps. This gives a total impedance across the winding (i.e. the load) of 605 Ohms.

I have the feeling that I'm missing some key relationship between the reflected impedance and the only possible (maximum) value of the secondary winding. I don't know what the load value is, or the secondary winding resistance and reactance, so obviously the necessary data is in the given information.

There have been several topics on this question in this forum, all of which I have tried to learn from, but inevitably they become protracted or focus on a different part of the question. And like those other posters have said, the material supplied is often confusing, incomplete or simply inadequate.

Any help greatly appreciated!
I'm actually on the same question now. What was your working out for question (a)?
 

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