Calculate Transformer Losses

[bonding jumper]

Stats:
2500kVA
13.2kv
480Y/277V
%Z 5.62
Operating at 1700kVA

Does anybody know a formula I can use to calculate transformer losses based on this data? I would like to be able to calculate for multiple transformers with the same basic ratings but variable load.
Thanks in advance.

 

[kc8dxx]

I recommend contacting the transformer manufacturer. Our transformers come with test data indicating losses at 0%, 25%, 50%, 75%, 100%, and 120% of rated load. Your particular transformer may or may not have had test data taken, however the OEM should still be able to advise typical data.

 

[mpross]

xfmr loss

xfmr loss

I will try to take a stab at this...

Z = 5.62% is the same as Z = 0.0562 (per unit)

calculating Z(actual) by the following:

Z(per unit) = Z(actual) / Z(base)

where Z(base) = V^2 / S(base) = 13200^2 / 2500*10^3 = 69.70 Ohms

...so Z(actual) = 3.92 Ohms

so for operating at 1700kVA...
I = 1700 / 13.2 = 128.79 A

and the loss would be...
I^2*Z(actual) = (128.79)^2*(3.92) = 65 kVA

65/1700 = 3.8% loss... ...fully loaded at 2500 kVA would be a 5.6% loss.

maybe try checking this against some data if you have it. I am curious.

 

[kc8dxx]

FWIW I checked over some data we had taken recently for a 10KV primary 480/277 secondary 600KVA transformer, with 5.53%Z. The losses were 2.2KW, 3.2KW, 5.8KW, 10.2KW, 16.4KW, and 22.6KW at 0%, 25%, 50%, 75%, 100%, and 120% of rated load.

 

[bonding jumper]

I recommend contacting the transformer manufacturer. Our transformers come with test data indicating losses at 0%, 25%, 50%, 75%, 100%, and 120% of rated load. Your particular transformer may or may not have had test data taken, however the OEM should still be able to advise typical data.

Thanks for all your input. I appreciate it very much. The transformers in question are all General Electric. I put in a call to them to find a curve but we'll see, I know they are difficult to get answers from.

 

[bob]

...so Z(actual) = 3.92 Ohms

so for operating at 1700kVA...
I = 1700 / 13.2 = 128.79 A

and the loss would be...
I^2*Z(actual) = (128.79)^2*(3.92) = 65 kVA

65/1700 = 3.8% loss... ...fully loaded at 2500 kVA would be a 5.6% loss.

mpross
Great try. However that Z you have caculated is mostly inductive. The reactance portion of Z is about 10 times the resistance portion. The resistance value is what you are looking for. This may be available from the MFG. Transformers used to have a value of "watts copper loss" that came with the transformer, but I don't think its available now.

 

[bonding jumper]

Not to take anything away from mpross, but I put in KC8DXX's numbers into mpross's functions and came out a little off

At 25% = 2.1kW
At 50% = 8.3kW
At 75% = 18.7kW
At 100% = 33.2kW

 

[mpross]

calc

calc

yeah, my calc is for a VA loss. and I do agree that the reactive loss is about 10x higher! It helps if you have a given value for R and also for X

thanks

-Matt

 

[kingpb]

matt,

Might want to check your voltage that you used to calc the current at 1700KVA. I think you are missing the sqrt 3 in the denominator.

Good effort though, at least you gave it a shot.

As Bob stated the X/R ratio is going to be around 10. Therefore, R=X/10, or as calculated by Matt ~= 0.392 ohm.

Irated= 2500KVA/(sqrt3*13.2KV) = 110 Amps

I^2*R=4.69kW at full load

 

[mpross]

Thanks

Thanks

kingpb,

Thanks for the heads up. I did forget the sqrt of 3!

I used this method over the summer for approximating reactive power losses through a bunch of transformers, but I used some X values that came from a PSS/E model.

-Matt

 

[jtester]

Many transformer makers will give losses in two values, no load and load losses at some load value. The no load represent losses incurred in energizing the transformer, and they are constant 24 hours a day, 365 days a year. The load losses are based on some value of loading and obviously can be different. This method allows you to evaluate a transformer's efficiency much more closely than most methods.

This is a method that utilities often use.

Jim T

 

[bob]

I used this method over the summer for approximating reactive power losses through a bunch of transformers, but I used some X values that came from a PSS/E model.
-Matt

What do you mean by reactive power loss? Are you caculating voltage drop
within the transformer?

 

[mpross]

Reactive Power Loss

Reactive Power Loss

I have calculated the instantaneous reactive power loss of several transformers on our system. I am developing a display for the system operators that monitors real-time reactive power reserves within our control area. Many of the measurements that we have are after the step-up transformer, so I need to approximate the reactive power loss for another part of my calcs.

Q(loss) = I^2*X : Reactive Power Loss of Xfmr

Another way to put it is VAR consumption of the xfmr.

-Matt