Transformer voltage regulation and calculations

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Discussion Overview

The discussion revolves around the calculations related to the resistance of the secondary winding of a transformer, specifically a 415:11000 V step-up transformer rated at 200 kVA. Participants explore concepts of voltage regulation, load current, and the effects of reflected resistance and reactance on the calculations. The scope includes theoretical calculations and practical implications for transformer design.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the secondary winding resistance using both the nominal voltage (11000 V) and the loaded voltage (10780 V) due to 2% voltage regulation, leading to different resistance values.
  • Another participant suggests calculating the current on the secondary side and mentions neglecting series inductance to find the secondary resistance based on voltage drop and current.
  • A different viewpoint indicates that the initial calculations may be focused on load resistance rather than transformer resistance, suggesting a need for a clearer distinction between these concepts.
  • Concerns are raised about whether the transformer is single-phase or three-phase, which could affect the calculations.
  • One participant provides equations for calculating reflected resistance and reactance from the primary to the secondary winding, emphasizing the need to account for these in the overall impedance calculations.
  • Another participant discusses the maximum allowable secondary winding resistance under specific conditions, including the impact of power factor and voltage drop.
  • Clarifications are made regarding the voltage drop due to load and its relation to the nominal voltage, with calculations presented for full-load current based on the given transformer specifications.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the secondary winding resistance, with no consensus reached on the methodology or the values to be used. Some participants focus on different aspects of the problem, leading to multiple competing interpretations of the requirements.

Contextual Notes

Participants note that the calculations depend on assumptions regarding the transformer type (single-phase vs. three-phase) and the treatment of reflected resistances and reactances. There is also mention of the potential complexity introduced by including primary resistance and leakage reactances in the calculations.

Who May Find This Useful

This discussion may be useful for students and professionals involved in electrical engineering, particularly those studying transformer design and analysis, as well as those working on related homework or assignments.

  • #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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