The T.M.Haja Sahib Transformer efficiency thread

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T.M.Haja Sahib

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The simple point I was making is that the notion of over rating a transformer to obtain better efficiency, as you have suggested simply, doesn't hold water.
It will do in relation to load growth.
 

iwire

Moderator
Staff member
Location
Massachusetts
Yes.Do not load it fully,if you do not have protective relays that would trip the transformer to prevent its loss of service life should its hot spot temperature rises above a certain limit.In the absence of such protective relays,you may safely load the transformer to 80% of its capacity.

That is just a design choice.

It is not a design choice.:dunce:


Spare capacity is a concept of modern electrical design.


Is your position that spare capacity is a design issue or not?




(Please note that instead of saying 'see post 7, 8 & 9 I took the time to post them here and each has a link back to the original context, that makes it easier for all. Try it out, it is not hard to do. )
 
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Besoeker

Senior Member
Location
UK
0 load: 3960.7 = 1.00%
1/2 load: 3960.7+1160.2= 1.28%
3/4 load: 3960.7+2610.4= 1.64%
full load: 8597.6/400000= 2.15%

By gosh the efficiency IS a lot less at lower loads. But someone PLEASE tell me what significance this has to the use of a transformer? By this reckoning, we should use ALL transformers at their most "efficient" point: 0% load!
I understand the point you're making. Operation on no load has the least losses. But not the best efficiency. In fact, at that point it's zero. Power in and no power out.
I hope you don't mind that I've calculated efficiencies based on your figures:

400kVALosses.jpg


What it shows is that the efficiency doesn't vary very much with load over the likely operating range and is best at rated load thus negating the argument that operating below full load rating delivers better operating efficiency.
 

mike_kilroy

Senior Member
Location
United States
I understand the point you're making. Operation on no load has the least losses. But not the best efficiency. In fact, at that point it's zero. Power in and no power out.
I hope you don't mind that I've calculated efficiencies based on your figures:

400kVALosses.jpg


What it shows is that the efficiency doesn't vary very much with load over the likely operating range and is best at rated load thus negating the argument that operating below full load rating delivers better operating efficiency.

Excellent analysis showing the OVERALL power out/ power in efficiency IS better loaded higher.

This was my in post 50:

to follow up on this, what DOES happen if you use a double size xfmr to make it "more efficient?" If one adds the actual WATTS losses, they will see this more efficient xfmr actually is LESS efficient at 1/2 load would he not? 1.28% of 800kva= 10,240watts loss........ 2.15% above is 8597.6watts loss on the 400kva transformer for the same load? Which is "MORE" efficient?

efficiency is power out/power in (tnx for pointing out typo) so just to state a transformers different load points by itself is silly.

Sorry TM, but this seems to contradict your post pointing to a post that pointed to another post that pointed to another:

May be misleading.See post#55.
where u said 2pc 1000kva xfmrs loaded to 50% each is more efficient than 1pc loaded to 100%, Unless u can provide core and cu losses to back up that particular comment. I know we can design transformers to trade losses between core and cu as needed so do not come back with a different design 100kva for the single use model - all must be the same for valid comparison. If u give design proof showing better power out/power in here I will ask our transformer design wizards to check your math.

That was what I was trying to state: effy by itself of the transformer is a micro picture; the macro picture is the overall power out/ power in - in which case probably most times a transformer loaded less than 100% will be less efficient than a larger transformer at a given load.

I feel arguing with you about this is beating a dead horse TM, but the OP was asking about % transformer load.
 
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T.M.Haja Sahib

Guest
Excellent analysis showing the OVERALL power out/ power in efficiency IS better loaded higher.

This was my in post 50:



Sorry TM, but this seems to contradict your post pointing to a post that pointed to another post that pointed to another:


where u said 2pc 1000kva xfmrs loaded to 50% each is more efficient than 1pc loaded to 100%, Unless u can provide core and cu losses to back up that particular comment. I know we can design transformers to trade losses between core and cu as needed so do not come back with a different design 100kva for the single use model - all must be the same for valid comparison. If u give design proof showing better power out/power in here I will ask our transformer design wizards to check your math.

That was what I was trying to state: effy by itself of the transformer is a micro picture; the macro picture is the overall power out/ power in - in which case probably most times a transformer loaded less than 100% will be less efficient than a larger transformer at a given load.

I feel arguing with you about this is beating a dead horse TM, but the OP was asking about % transformer load.

My suggestion is you may leave all math to your wizards to check but have a fair idea of principles behind it.Here the principle involved is the transformer efficiency is maximum when its copper loss equals its iron loss.Based on it,the following equation derives:

KVA at maximum efficiency=KVA at full load* SQRT[ Iron loss/Full load copper loss].

Substituting your data in above equation,

KVA at maximum efficiency=400*SQRT[ 3960.7/4636.9]=400*0.924=369.6 or 370KVA.

So the maximum efficiency occurs at 93% and not at full load in this case.

So it may not be a general purpose transformer for which maximum efficiency is at 35%.
 
T

T.M.Haja Sahib

Guest
No it isn't.

My post #77


Your post #78


My post #79
OK.


Here are the figures again:

400kVALosses.jpg


You claimed that my figures were wrong.
I made a reasonable request that you post corrected figures.
Care to do so now?
First recheck your calculations.The efficiency at 100% loading is 400/(400+0.85976) =400/400.85976=99.7855% and not 97.85%.
 

topgone

Senior Member
First recheck your calculations.The efficiency at 100% loading is 400/(400+0.85976) =400/400.85976=99.7855% and not 97.85%.

Excuse me. Your math is off.

Eff = 400/(400+8.5976) = 0.978958 or 97.90%

If you insist that your efficiency figure is correct, please tell me where people can buy lots of these 99.79% efficient transformers?
 
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