# Transformer efficiency



## Omer (Aug 3, 2017)

I am posting this problem since I usually get it wrong first time.

this post will help me remember not to make the same mistake always.

Try it,

A single phase 100 kVA, 4.6/0.480 kV transformer no-load loss is measured to be 300 W. The total

loss at 75% loading is measured at 1200 W. Transformer efficiency at 25% load is approximately:


----------



## rg1 (Aug 3, 2017)

Omer, it would be a good idea to know where exactly you get it wrong. But talking for the whole of the question, the no load loss remains constant of the load. (It is dependant at the Voltage at which it is carried out, which is rated V in most experiments). The game is only in the Load losses, which are I**2 X R. If you assume Ifl as full load current , then FL losses will be Ifl**2 X R and at any other say y percent of full load it will be (y X Ifl/100)**2 X R. Rest is maths. If you mention where exactly you loose track, we can discuss that part.


----------



## Omer (Aug 3, 2017)

@rg1

yes, calculation is straight, however sometimes I miss it in the math part.

can you try it, I just want to see your solution before I tell my confusion.


----------



## rg1 (Aug 3, 2017)

Here it is. 

View attachment 9927


----------



## Omer (Aug 3, 2017)

rg1 said:


> Here it is.
> 
> View attachment 9927


interestingly, you did the same two mistakes I did.

can you figure them out?

I think you will..


----------



## rg1 (Aug 3, 2017)

Oh yes; 25% load. Lol. Good, Very good @Omer for raising this question. 25000/25400=98.4% What else? Thanks Omer. I was getting complacent last few weeks. This has shattered my overconfidence. Thanks for bringing me to ground again. I will start preparing again. Thanks a lot.


----------



## rg1 (Aug 3, 2017)

rg1 said:


> Oh yes; 25% load. Lol. Good, Very good @Omer for raising this question. 25000/25400=98.4% What else? Thanks Omer. I was getting complacent last few weeks. This has shattered my overconfidence. Thanks for bringing me to ground again. I will start preparing again. Thanks a lot.


I knew that I had reached early peak of the preparation. This is too bad. I was talking to my wife; this is not a good sign but then it happened.  IMHO you should reach your  0.99 pu of peak not even 1 pu; on the day of the exam to perform best.


----------



## Omer (Aug 4, 2017)

rg1 said:


> Oh yes; 25% load. Lol. Good, Very good @Omer for raising this question. 25000/25400=98.4% What else? Thanks Omer. I was getting complacent last few weeks. This has shattered my overconfidence. Thanks for bringing me to ground again. I will start preparing again. Thanks a lot.


yes, I think very difficult to notice the 25% after using it to calculate the losses and you are ready to calculate for the efficiency. sometimes it is better to slow down and not rushing fast into solving problems familiar to you. unfortunately, I do.

Ok, for the next part, I need you to confirm it for me.

Calculating for efficiency, it is: output/input . for this problem and generally the transformer rating, is it input or output.

in your calculation and also mine, you take it as output. However, I think the right one should be input.

calculation should be : (25000-400)/25000

instead of 25000/25400.

for this problem results are close but generally speaking it makes difference.

I like the formula : 1 - losses/input


----------



## TNPE (Aug 4, 2017)

Omer said:


> yes, I think very difficult to notice the 25% after using it to calculate the losses and you are ready to calculate for the efficiency. sometimes it is better to slow down and not rushing fast into solving problems familiar to you. unfortunately, I do.
> 
> Ok, for the next part, I need you to confirm it for me.
> 
> ...


You'd be right.  I noticed that error as well, that he used a higher "input" than what was actually there.


----------



## rg1 (Aug 4, 2017)

For all machines the power rating given is Output power. I remember  having  come across many times. I will share with you the legislation if I find one, which I will.


----------



## rg1 (Aug 4, 2017)

http://www.trafomodern.com/php/terms_and_definitions,21338.html.

Then, this Wildi question attached. It must be available in all book dealing with Machines. 

View attachment 9930


----------



## TNPE (Aug 4, 2017)

We're working with a XFMR here, not a motor.


----------



## knight1fox3 (Aug 4, 2017)

TNPE said:


> We're working with a XFMR here, not a motor.


The induction motor is fundamentally a transformer in which the stator is the primary and the rotor is short circuited secondary. :thumbs:


----------



## TNPE (Aug 4, 2017)

I understand that, but he's referring to shaft power.  A XFMR doesn't have a shaft  :thumbs: ... So no need to take that into account.  Only electrical power at this point.


----------



## knight1fox3 (Aug 4, 2017)

TNPE said:


> I understand that, but he's referring to shaft power.  A XFMR doesn't have a shaft  :thumbs: ... So no need to take that into account.  Only electrical power at this point.


Just giving you a hard time. Been lightly monitoring the good discussions here lately. Which have been absent the past few exam cycles. It's somewhat refreshing to see some activity again.


----------



## TNPE (Aug 4, 2017)

I figured you were rattling my cage, just keeping the discussion going is all.


----------



## rg1 (Aug 4, 2017)

knight1fox3 said:


> Just giving you a hard time. Been lightly monitoring the good discussions here lately. Which have been absent the past few exam cycles. It's somewhat refreshing to see some activity again.


I went through earlier posts, when I joined the forum for knowing so many things. I also felt the same. At  one point in time I wanted to point out that the quality of discussions were very good in something around 2011.


----------



## rg1 (Aug 4, 2017)

TNPE said:


> I understand that, but he's referring to shaft power.  A XFMR doesn't have a shaft  :thumbs: ... So no need to take that into account.  Only electrical power at this point.


What I mean is, all machines ( including Transformer) are rated at their Output, is the convention. The motor was one example. I will get specific legislation  for Xmer too from these books. If someone come accross something else should let know. The link attached with that post is for transformer.


----------



## TNPE (Aug 4, 2017)

And the output is the rated input minus losses, not added.  So, it's out/in with in at 25... you can see the rest.


----------



## TNPE (Aug 4, 2017)

I see what you're saying, and I could be wrong (been a while since I've looked at this), but, to me, it is not intuitive if solved in that way.  It makes sense, and I could see why, but it still ain't intuitive.  

The answers are very close to the same, but it's not the same.


----------



## TNPE (Aug 4, 2017)

Just looked it up.... your way seems to be the consensus.


----------



## rg1 (Aug 4, 2017)

TNPE said:


> Just looked it up.... your way seems to be the consensus.


I knew. except that I jumped to full load while calculating the eff. at 25% there was nothing wrong. I did it by actually finding the resistance  (the value is somewhere 3.58 Ohms) too when Omer said two mistakes. I was sure there was no other mistake. That was mistake of, being out of touch, loss of concentration. Lol.


----------



## TNPE (Aug 4, 2017)

Dont know why this hasn't come to me till now, maybe a cold beer on a Friday evening does the trick, but looking from the output back makes sense.  To evaluate any equipment, you need to know how it responds to a certain condition.  The only way to do this is to expose it to said condition and record its behavior.  

I feel like a dipshidiot!!!


----------



## Omer (Aug 5, 2017)

I did some research and generally in transformer efficiency problems they give you the LOAD at which you are required to find the efficiency at. Thus, it is the output not the input as said by @rg1.

One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this case VA rating is used to calculate currents and losses , however, efficiency calculations is done on the Watts ratings only in this case. I thought it is worth  mentioning.


----------



## rg1 (Aug 5, 2017)

Omer said:


> I did some research and generally in transformer efficiency problems they give you the LOAD at which you are required to find the efficiency at. Thus, it is the output not the input as said by @rg1.
> 
> One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this case VA rating is used to calculate currents and losses , however, efficiency calculations is done on the Watts ratings only in this case. I thought it is worth  mentioning.


@Omer Some misunderstanding there. I never meant the  power ratings of a xmer are input ratings. In my solution I took it as output and in my posts too I mention that for all machines, the Power ratings are output of that machine. The second part of your post is really worth mentioning, I agree. Good point. For losses we  have too keep in mind what is the current through the Xmer because they purely depend on Input Voltage and current , I will just extend it a little further. If the pf improvement is done on feeding side (Primary side) of Xmer we take the current from VA of load and Power in W but if pf improvement is done on Load side of the Xmer (Secondary side) then current with improved pf needs to taken flowing through the Xmer. Just think on it. It is interesting.


----------



## Omer (Aug 5, 2017)

rg1 said:


> @Omer Some misunderstanding there. I never meant the  power ratings of a xmer are input ratings. In my solution I took it as output and in my posts too I mention that for all machines, the Power ratings are output of that machine. The second part of your post is really worth mentioning, I agree. Good point. For losses we  have too keep in mind what is the current through the Xmer because they purely depend on Input Voltage and current , I will just extend it a little further. If the pf improvement is done on feeding side (Primary side) of Xmer we take the current from VA of load and Power in W but if pf improvement is done on Load side of the Xmer (Secondary side) then current with improved pf needs to taken flowing through the Xmer. Just think on it. It is interesting.


Yes, I meant you got it right, it is the output, not the input.

Nice scenario and indeed interesting..


----------



## rg1 (Aug 5, 2017)

Omer said:


> Yes, I meant you got it right, it is the output, not the input.
> 
> Nice scenario and indeed interesting..


LOl. My bad. Sorry.I got  your English wrong. it is the output not the input as said by.........


----------



## TNPE (Aug 5, 2017)

The "anomaly" with the math is simply that the problem statement explicitly says the XFMR is loaded at 25%.  It doesn't say the output is loaded at any percent.  Loaded, to me, means input.  I get it.  I know the math, and I understand it clearly and can see why it's done that way, but the language and math do not jibe.  Loads are depleted by series impedance, looking toward the load, not the other way around.  Hence, why I adamantly maintain it's not necessarily intuitive.  But yes, if you run across this type of problem, solve it by the approach used in academia.  I'm not saying academia is wrong. Hell, I've been there and done it and have gone through this whole process.  I'm merely saying the verbiage and approach do not mesh.  That's all.


----------



## P-E (Aug 5, 2017)

My son just broke an Optimus Prime he grabbed from my parents house - probably a collecor's item.   It didn't take him long.


----------



## knight1fox3 (Aug 7, 2017)

P-E said:


> My son just broke an Optimus Prime he grabbed from my parents house - probably a collecor's item.   It didn't take him long.


LOL


----------

