Calculating Transformer Losses

[GJRS]

I have a question regarding calculating transformer losses. I report each month on revenue metering variances and one comparison is on our station load which we use the revenue meter from the utility company versus our in house meter - simple enough. The problem I am having is the revenue meter and our in house meter are on the line and load side respectively. This being the case I need to account for the losses across the transformer and add them to our in house meter.

I have the following information from the transformer OEM:

High Voltage - 18000 V Delta
Low Voltage - 4160 V Wye
KVA - 15000
Phase 3
Hertz 60

MEASURED TRANSFORMER DATA
Winding Resistance @ 85 degrees C reference temperature:
-high voltage OHMS = 0.2685
-low voltage OHMS = 0.9946
No load loss @ 100% of rated voltage WATTS = 12760
Exciting current @ 100% of rated voltage % = 0.18
Load Loss @ 85C ref temp WATTS = 58055
Impedance @ 85 degrees C ref temp % = 8.56

CALCULATED DATA
Regulation at reference temp 85 C
and 1.0 PF % = 0.75
and 0.8 PF % = 5.65
Efficiencies @ 100% full load % = 99.53
75% full load % = 99.60
50% full load % = 99.64
25% full load % = 99.56

The load side varies based on plant conditions so what I want to do is calculate the real time losses - ideally I would implement the algorithm in the DCS to provide a corrected measurement.

Do I have enough information to perform such a calculation? I think from what I have I can correct for when the unit is @ 100% full load - can I use the efficiencies to calculate the losses throughout the load profile?

I would appreciate any help with this - my background is process controls so this is not my area of expertise.

 

[jghrist]

The instantaneous losses are proportional to the square of the instantaneous current, so you would have to have the complete load profile, calculate the loss at each instant, and then integrate over the billing period.

 

[DTR2011]

You may also want to check if the meter can perform these calculations for you. Many modern meters do.

 

[GuntherA]

As jghrist said the Load Loss is proportional to the square of the current. No-Load losses are there all the time.
The figure quoted for load loss at 85 degrees is at rated current, so if you know what current you are drawing, it's a mater of (Measured_I / Rated_I)^2 * 58055 + 12760

It won't be 100% but it will be a pretty fair approximation.

 

[LSpark]

If you're lucky you'll be able to programme the meter with the transformer losses - it you can then remember to adjust the load losses to the full load current of the meter (Isq ratio). This current is normally higher than the rated current of the transformer.

 

[prc]

It may be almost impossible to estimate load losses correctly on real time.In such a power transformer,load loss will consists of I2R plus stray losses that may be 15-20 % I2R loss.Copper loss will increase linearly with temperature.Stray loss will vary inversely to temperature.Copper temperature will vary based on ambient temperature and oil temperature rise,that will vary based on square of load current.It is because of these variants that standards call for specifying load losses corresponding to a reference temperature-IEC on 75C and IEEE on 85C