Power transformer calculations 3

[Luiseng]

Hi all,
I have a 3ph 2700KVA continuous rating or 12677KVA pulsed at 36KV/670V transformer, I received the test results sheet from the manufacturer and I am trying to go trough the calculations to check it out. They also kindly supplied the formulas that they use for their calculations.
The problem I have is the I^2R losses calculations. I got the resistance measured for each winding (primary and secondary) so according to their formulas I should start by getting the resistance for the primary and secondary. The formulas given are:
For a delta connection which the primary is, R should be the sum of the resistances divided by 2.
For a Star connection which the secondary is, R should be the sum of the resistances divided by 6
Then the I^2R losses are calculated as, the sum of the Irms squared multiplied by the resistance. The way I have done it is to use the rated primary current squared multiplied by primary resistance and the secondary rated current squared multiplied by the secondary resistance and then I added the values together.
The problem is that the values do not tie in at all, so or they supplied the wrong values or I am doing something wrong. I favour the later as all the other calculations on the sheet match with my results when I use another method of calculating the values.
Any help is appreciated.
Best regards
Luis

 

[edison123]

Try for delta connection R = 2/3 of resistance between lines

for star connection R = 1/2 of resistance between lines

* Algebra - The weapon of math destruction *

 

[Luiseng]

I Edison123,
It still does not tie in...
rated currents are 43.3A/2327A, primary resistance is 1.164Ohms and secondary is 0.0005497Ohms and the claimed I^2R losses are 9348W.
I'm lost...
thanks for the help.
best regards
Luis

 

[edison123]

I assume both the resistances were measured line to line

For primary,

the phase resistance is 2/3*1.164 = 0.776 ohms

I^2 loss = 3*43.3*43.3*0.776 = 4365 W app

For secondary,

the phase resistance is 1/2*0.0005497 = 0.00027485 ohms

I^2 loss = 3*2327*2327*0.00027485 = 4465 W app

Total loss = 8830 W

9348 W could be with temperature correction




* Algebra - The weapon of math destruction *

 

[chris8410]

Hi Luis,
A few comments regarding load loss calculations for transformers:

1.
When calculating the I2R for delta windings, the current that must be used is the LEG current, not the LINE current. Leg current is used since the I2R is a function of the current that passes through the winding. Line current is not the same current that passes through the winding, it is the current that passes through the bushings, before it enters the windings at the corners of the delta. The leg current of a delta winding is equal to line current divided by 1.732. In your case, leg current will be 43.3A/1.732 = 25A. Use 25A in the I2R calculations. Note that I am assuming 43.3A is the nameplate current for the delta winding set. Nameplate current is always line current.

For wye connected windings, LEG current is the same as LINE current. That is, the current that flows through the winding isthe same as the current that passes through the associated bushing. Therefore, use the stated nameplate current for the LV I2R calculation.

2.
I am not sure what resistance values you are stating in your message. Are these the raw test data, or sum of 3 phases in series, etc, etc. The resistance that must be used is the LEG resistance x 3, for both the delta and wye windings. There is alot more that can be said about resistance measurement data, especially for delta windings. I will stop here for now.

3.
Normally load losses are stated at a referance temperature. Very common in the USA is a referance temperature of 75C for 55C rise machines, and 85C for 65C rise machines. The method of correcting is to apply a correction factor to the measured resistance. For copper, K = (234.5 + Tr)/(234.5 + Tm). Tr is usually either 75C or 85C. Tm is the temperature of the copper during test. K is usually around 1.2, for a Tm of about 20C. Therefore, the correction factor is significant and must be considered.

4.
Load loss of a transformer has 2 components: (1) Copper loss (I2R) and (2) Stray & Eddy (SE). So, when you figure out your I2R value, that will not equal the load loss stated on test report. SE can be as must as 30% of the stated load loss. SE can only be determined by test, and is inversly proportional to temperature (i.e. "K" from above).

Good Luck.

 

[Luiseng]

Hi Edison123,
You are a Star, this seems to be it. I have used the formula in the 60076 standard for temperature compensation for oil-immersed transformers with reference temperature of 75ºC R75 = Rr*(310/(235+tr)), where Rr is the calculated resistance from the measured resistances and tr is the temperature at which the measurements were made.
this gave me a value of 9605W which is close to the value given.
thank you very much for your help
Best regards
Luis

 

[Luiseng]

Hi Chris8410,
I have also tried what you sugested, and it does not tie in with the values given by the manufacturer.
the resistance I gave is the average resistance value on one winding. I probably done something wrong...
thanks
best regards
Luis

 

[fredpar]

Hi Luis:
I did use what Chris suggested and I think I came closer:

Losses=3x[Ip^2Rgivenprimx(3/2) + Is^2Rgivensec(1/2)]*K

Losses=3x[25^21.164*(3/2) + 2327^2*0.0005497(1/2)]*1.2

NOTE: I'M ASSUMING 1.2 FOR TEMPERATURE FACTOR

Losses=9286W

Thanks

 

[jghrist]

You neglected Chris' fourth point. The losses include stray and eddy current losses. Transformer losses are not calculated from the measured resistance, they are measured directly. The measured resistance is used to adjust losses for temperature.

 

[edison123]

Luiseng,

For the primary losses, fredpar & chris8410 got it right.

The phase resistance is 3/2 of measured line resistance. (not 2/3rd as I had stated in a moment of madness)

The phase current is 0.577 of line current. (I completely missed that one).

This would give primary loss as 3274 W. Secondary loss is correct at 4465 W. With 20% increase for temperature (which is normal), the total loss is about 9286 watts

I apologize for the mistake.

fredpar,

I give you a star for spotting my mistake.

* Algebra - The weapon of math destruction *