Transformer Load Regulation Old Exam Question Help

In summary, the transformer has a resistance of 3 ohms, and an impedance of 10 ohms. The transformer is rated at 400kVA, and a load of 0.4 + 0.4j ohms is placed across the 415V secondary. The load regulation is 1.4%.
  • #1
Omar7177
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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:
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  • #2
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
 

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  • #3
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?
 
  • #4
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.
 
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  • #6
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.
 
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Likes Omar7177
  • #7
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:
 

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