112.5 KVA Transformer heat rejection
- Thread starter dav5y
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Phil Corso
Senior Member
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- Boca Raton, Fl, USA
Dav5y...
How accurate an answer do you want or need?
Regards, Phil Corso
How accurate an answer do you want or need?
Regards, Phil Corso
In the absence of any better information I usually work it at around 2% for transformers.
You need to know its efficiency and power factor. Say its efficiency 95% and power factor is 0.9. Then its heat rejection would be 112.5x0.9x0.05=5.06 Kilojoules/second under steady state condition.
You might want to rethink that.
Instead of beating around the bush simply ask the manufacturer for the total losses which is the n0 load loss plus the full load loss unless this is a test question.
You have overlooked stating what the temperature rise is also which makes a very big difference.
For a 112.5kva, 600v class 150degC rise transformer expect of about 3800 watt, 115degC rise about 3200 watts, and an 80degC rise about 2900 watts. K-factor transformers would be different yet for example you may expect a K-13 80degC rise to be around 2000 watts and a K-4 150deg C rise to be around 3500 watts.
I did??
Lemme add:
@ 50 watts per kVA, it would be 50 x 112kVA = 5.6kW.
5.6/112 x 100 = 5%!
Not that I believe this to be exact!
Just want to give the OP a ballpark figure.
Interesting that the table gives heat loss as BTu. You need to assume a rate to get power from that. So, now we have 2%, 3%, and 5%. And I didn't see where you took that all important temperature rise into account.
Anyway, this is where my 2% comes from:
(p) At normal voltage and ratio :
(1) losses at no load :0.3 %
(2) losses at 50% full load :0.75 %
(3) losses at 75% full load :1.0 %
(4) losses at full load normal tapping:1.75 %
It was a real transformer for a real application and the system it went into (a VSD) was tested to ensure that our declared efficiency met what we declared it would. It did.
mivey
Senior Member
2% sounds more like a valid number. Actually I would expect it to be less than that so I would think 2% is on the conservative side. At least that is the case for utility distribution transformers.
beanland
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R + jX Impedance
R + jX Impedance
The best way is to get the transformer impedance. Typical small dry type may be 2% R + 3% X. Thus, the heat losses at full load are 2% (the R part) of the rating. Bigger transformers typically have lower R%, 1% or less. Small ones can be 5%.
R + jX Impedance
The best way is to get the transformer impedance. Typical small dry type may be 2% R + 3% X. Thus, the heat losses at full load are 2% (the R part) of the rating. Bigger transformers typically have lower R%, 1% or less. Small ones can be 5%.
If you didn't what is the temperature rise of the transformer you are refering to? Unless it is your belief that doesn't affect the full load losses which it appears to be as such.
And am I to assume that "heat rejection" is with reference to heating wats of a fully loaded 112.5kva transformer which is commonly started as total lossed in watts which is the no load loss plus the full load loss. If so this value will be different for 150, 115, and 80 degc rise transformers.
That statement has the implicit assumption that transformers are designed with the same parameters for an 80C rise as for a 150C rise.
They aren't.
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
Dav5y...
Essentially, ignoring the small losses caused by hysteresis, eddy-current, and skin-effect, leaves two principal losses: the magnetic-core loss, whch is taken as constant from no-load to full-load; and the winding-resistance loss, which of course varies as the square of load-current!
The former can be found from the open-circuit test (xfmr's secondary disconnected from load!) The latter is determined from the short-circuit test (varying primary-voltage untll rated-current flows in the secondary!)
Regards, Phil Corso
Essentially, ignoring the small losses caused by hysteresis, eddy-current, and skin-effect, leaves two principal losses: the magnetic-core loss, whch is taken as constant from no-load to full-load; and the winding-resistance loss, which of course varies as the square of load-current!
The former can be found from the open-circuit test (xfmr's secondary disconnected from load!) The latter is determined from the short-circuit test (varying primary-voltage untll rated-current flows in the secondary!)
Regards, Phil Corso
If I implied what you said regarding "That Statement" which post of mine are you referring to? If any of my replies were misleading I will correct it as in no way were any of my replies based on "assumptions." Without specifying a temperature rise losses don't mean a thing because they are infact different with each 112.5kva transformer design.
I provided some "real" manufacturer's data bases upon typical 150, 115, and 80degC rise 112.5kva transformers and well as K-factor transformers.
The one I quoted.
Are you saying that they are not diffent but are the same??!!! How would that relate to my post #6?
is what you said.
Here's a thought experiment for you.
Operate a 500W heater in a closed metal box. Check the internal temperature of the box.
Repeat this test with a box half the size. I think you would agree that it runs at a higher temperature. But there's no difference in the losses.
I'm not suggesting that transformers of equal ratings but different temperature classifications will have the same losses. I'm just pointing out that you can necessarily infer the differences, if any, from just that.
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