Transformer voltage regulation and calculations

Click For Summary
SUMMARY

The forum discussion centers on calculating the secondary winding resistance of a 415:11000 V step-up transformer rated at 200 kVA with a 2% voltage regulation. The output voltage across the transformer terminals is 10780 V under load conditions. Participants debate whether to use the nominal voltage (11000 V) or the loaded voltage (10780 V) for resistance calculations, leading to different resistance values of 605 Ohms and 581.04 Ohms, respectively. The discussion also touches on the importance of reflected resistance and leakage reactance in determining the total impedance of the transformer.

PREREQUISITES
  • Understanding of transformer voltage regulation concepts
  • Knowledge of electrical resistance and reactance calculations
  • Familiarity with the kVA rating and its implications for power factor
  • Ability to apply Ohm's Law in AC circuits
NEXT STEPS
  • Study transformer voltage regulation calculations in detail
  • Learn about reflected impedance in transformers
  • Explore the impact of leakage reactance on transformer performance
  • Investigate the differences between single-phase and three-phase transformer calculations
USEFUL FOR

Electrical engineers, students studying power systems, and professionals involved in transformer design and analysis will benefit from this discussion.

  • #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.
 
Physics news on Phys.org
  • #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)?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
10
Views
5K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
9K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K