Transformer Load Regulation Old Exam Question Help

AI Thread Summary
The discussion revolves around solving a transformer load regulation problem involving a 10,000V to 415V transformer with specific resistance and reactance values. The participant initially calculated the turns ratio and referred the secondary impedance to the primary side, but faced uncertainty regarding the power factor. Clarifications were made that the resistance and reactance values are implied to be secondary, and the importance of calculating the power factor was emphasized. Ultimately, the participant confirmed they found the correct answer after following the advice provided. The conversation highlights the significance of understanding impedance transformation and power factor in load regulation calculations.
Omar7177
Messages
4
Reaction score
2
•• moved from technical forum to homework, so template is missing ••
Hi, I was wondering if anybody would be willing to point me in the right direction for solving the following question.

The questions is as follows:
A 10000V RMS (primary) to 415V RMS (secondary) transformer rated 400kVA has the following resistance and reactance:
Rs = 3 ohms
Xs = 10 ohms

A load of (0.4 + 0.4j) ohms is applied across the 415V secondary, what is the load regulation?

Edit: parameters changed, since I don't want to lift the same exact question.

What I did first was determine the turns ratio N1/N2 by doing V1/V2.

Then I referred impedance at the secondary to the primary.

After that, I determined I2 by doing S/V2 although I'm not sure if I should work out power factor, but I assumed unity since I wasn't sure how to work it out.

The proceeded to determine I2(N2/N1) and used that value to calculate a very high voltage drop which seems wrong.

Any help is extremely appreciated!
 
Last edited:
Physics news on Phys.org
Here's a little snippet from https://www.quora.com/How-power-factor-of-load-effects-voltage-regulation
Looks like you need to find what the Power Factor is.
loadreg.jpg
 

Attachments

  • loadreg.jpg
    loadreg.jpg
    28.6 KB · Views: 1,043
Omar7177 said:
Hi, I was wondering if anybody would be willing to point me in the right direction for solving the following question.

The questions is as follows:
A 10000V RMS (primary) to 415V RMS (secondary) transformer rated 400kVA has the following resistance and reactance:
Rs = 3 ohms
Xs = 10 ohms

A load of (0.4 + 0.4j) ohms is applied across the 415V secondary, what is the load regulation?

Edit: parameters changed, since I don't want to lift the same exact question.

What I did first was determine the turns ratio N1/N2 by doing V1/V2.

Then I referred impedance at the secondary to the primary.

After that, I determined I2 by doing S/V2 although I'm not sure if I should work out power factor, but I assumed unity since I wasn't sure how to work it out.

The proceeded to determine I2(N2/N1) and used that value to calculate a very high voltage drop which seems wrong.

Any help is extremely appreciated!
Hi Omar7177. :welcome:

First things first...is it made clear in the question that Rs and Xs are the values referred to the secondary?
 
NascentOxygen said:
Hi Omar7177. :welcome:

First things first...is it made clear in the question that Rs and Xs are the values referred to the secondary?
Hi! The question is not terribly clear, but I believe that's implied.
Anyway I think I have solved it, I'm getting the right answer, so that was rather short lived haha.
 
  • Like
Likes dlgoff
In my opinion, a power factor is always positive. But X could be negative if it is capacitive. Fortunately, in your case the reactance is inductive that means positive.
Let's say the high voltage -10 kV- does not change with the load so what you have to do is to calculate the total impedance-viewed from primary terminals-and to divide the primary voltage by total impedance in order to get the current. Power factor will be the resistive part divided by total.
Do not forget to transfer the load impedance from secondary to primary side.
 
  • Like
Likes Omar7177
Babadag said:
In my opinion, a power factor is always positive. But X could be negative if it is capacitive. Fortunately, in your case the reactance is inductive that means positive.
Let's say the high voltage -10 kV- does not change with the load so what you have to do is to calculate the total impedance-viewed from primary terminals-and to divide the primary voltage by total impedance in order to get the current. Power factor will be the resistive part divided by total.
Do not forget to transfer the load impedance from secondary to primary side.
Yep, that's pretty much the way I did it, and I've got the answer. Thanks!:biggrin:
 

Similar threads

Replies
1
Views
3K
Replies
1
Views
2K
Replies
2
Views
4K
Replies
16
Views
8K
Replies
33
Views
9K
Replies
48
Views
11K
Replies
11
Views
7K
Replies
6
Views
3K
Back
Top