Transformer Heat Release given kVA, Z% and X/R 1

[bhellmer]

Transformer kVA, Z% and X/R are commonly available.

Proposing the following equation for transformer heat release:

kW = kVA * Z% / SQRT((X/R)² + 1)

Is this equation legitimate?

Equation givens:
Q = X/R and thus X = Q * R
Z² = X² + R²

Combining given equations to eliminate X:
Z² = (Q * R)² + R²

Solving for R:
Z² = (Q² * R²) + R²
Z² = (Q² + 1) * R²
R² = Z² / (Q² + 1)
R = Z / SQRT(Q² + 1)

Reintroducing X/R:
R = Z / SQRT((X/R)² + 1)

Revise equation for percents:
R% = Z% / SQRT((X/R)² + 1)

Transformer heat release:
kW = kVA * Z% / SQRT((X/R)² + 1)

 

[prc]

Yes it will be correct for I2 R losses. But remember there are some additional heat loss inside transformer -from hysteresis and eddy loss inside core + stray losses in windings and clamping structures from leakage flux from windings.

 

[7anoter4]

I agree with you prc.

 

[bhellmer]

My understanding is that Z% and X/R are based upon the transformer's total impedance at rated conditions, and thus the Thevenin equivalent of all copper and iron impedances.

prc: Are you suggesting that Z% and X/R do not represent the transformer's iron impedances?

I would of course prefer to use manufacturer's no-load and load-dependent heat-release properties. But they can be difficult to get, especially for old out-of-production transformers.

 

[waross]

Temperature matters.
Z% is tested with the transformer at operating temperature.
Test results on small dry type transformers at room temperature by students yielded results that were noticeably lower than the nameplate ratings.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bhellmer]

waross: We use transformer heat release to design its cooling system and want transformer properties at maximum allowable operating temperature.

 

[bhellmer]

Forgot to convert Z% to a simple ratio by dividing by 100.

kW = kVA * (Z% / 100) / SQRT((X/R)² + 1)

 

[dpc]

Are you suggesting that Z% and X/R do not represent the transformer's iron impedances?

Correct. The measured R or %Z will not include any core losses - eddy current or hysteresis.

 

[bacon4life]

The IEEE nameplate impedance is typically corrected to 85C, which could be considerably cooler than the maximum operating temperature of the transformer winding.

Note that if you have a transformer test report with winding resistance measurements, the winding resistance measurement in ohms is a DC resistance rather than an AC resistance. Your equation is correct for the series portion of the losses.

If the transformer has taps, losses can vary dramatically between taps.

If the transformer has forced cooling fans, such auxiliary losses are typically listed separately on the test report.

 

[waross]

Impedance is measured by exciting the transformer with a sufficient AC potential to cause rated current to flow in the windings with the secondary short circuited. This current will be supplying all the losses.
The question is:
How is the R of the X/R ratio determined. The DC resistance may not be used to calculate the total losses.
However the X/R ratio is published to facilitate fault current calculations and DC offset. As such I would expect it to be the AC equivalent resistance and so would include the losses.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[prc]

waross- Thanks for correcting me. My statement "But remember there are some additional heat loss inside transformer -from hysteresis and eddy loss inside core + stray losses in windings and clamping structures from leakage flux from winding" is to be modified. The second term is to be deleted. R will include stray losses in winding and clamping structures as during Z measurement, rated currents are flowing through windings. But it will not include iron losses.

In fact it may include a small portion of iron loss also but standards are not asking for any correction for this. Let us take a transformer of Z=10 %. Then Z measurement is made by applying approx 10 % of rated voltage on primary with secondary shorted to get rated current flow on both windings. Power input is taken as load loss. At 10 % voltage, flux density will be 10 % of rated conditions. But iron losses at 10 % of rated flux will not be 10 % of rated voltage condition, but only 1 % which is neglected.

Temperature correction is also not done for Z measurement esp for power transformers as R in modern transformers is 1.5 % -0.1 % ( 100 kVA- 500,000 kVA).But measured load loss is corrected for temperature as stray losses will be 10-30 % in power transformers. I2R component will increase with temperature and stray losses will decrease with increase in temperature.
Please do not estimate losses of old transformers as bheller proposed. During the past 30 years, both load and no-load losses of transformers have come down by more than 50 %.

 

[bhellmer]

prc: As I previously stated "I would of course prefer to use manufacturer's no-load and load-dependent heat-release properties. But they can be difficult to get, especially for old out-of-production transformers."

Somehow I must come up with heat release values for these old transformers to pass on to the HVAC engineers. I can apply a safety factor to address uncertainties. But I must have some rational basis for my proposed cooling requirements.

 

[edison123]

"Magnetic Flux is current dependent, not voltage dependent"

Wrong. It is voltage and frequency dependent.

If you do not have the original test certificates for these transformers, wouldn't it be easier to use to the oil temperature rise (measurable) and oil volume (available in nameplate) to estimate the total heat loss?

Muthu

http://www.edison.co.in

 

[prc]

bhellmer & Edison, both of you are correct. The magnetic flux in the core varies as voltage and inversely as frequency. The leakage flux coming out from the HV-LV winding gap depends on the current flow in winding. Zero at no-load and maximum at full load. To find the load on HV AC system, best way is as Edison mentioned- find out max oil temperature. Find temperature rise over the then ambient temperature. Let it be T . Measure the tank and radiator surface approximately in square meters. Then W/m2 = 5.15x T raised to 1.25 . At 45 C temperature rise, heat dissipation will be 600 W/m2.

 

[bhellmer]

prc & edison123: I stand corrected with respect to current versus voltage dependency for flux density and core losses.

We do not have original test certificates for these old transformers. Typically we can only get what is on their nameplates. Most manufacturers don't keep this data for their old transformer's, or at least that what they tell us. While some of these transformers are fluid filled, most are dry. These transformer are actively in use and not available for testing; nor does the client want to pay for heat-release testing.

We are simply attempting to get a reasonable approximations for heat release based on limited information.

We are not measuring Z% and X/R value, we are getting them or at least the Z% from nameplates. I do not know the basis for the nameplate values and whether they have accounted for core losses. I will be asking some manufacturers about their nameplate values and whether they currently and historically include adjustments for core losses.

I have actually been using this equation for about 30 years, but only when I could not get manufacturer's heat release or efficiency data. Prior to DOE transformer efficiency mandates, heat release data was quite difficult to obtain.

You may think that this model is crude, but the following from

http://www.engineeringtoolbox.com

is a typical suggestion:

Heat loss for

150 kVA and smaller : 50 Watts/kVA (aprox. 5%)
150 - 500 kVA : 30 Watts/kVA (aprox. 3%)
500 - 1000 kVA : 25 Watts/kVA (aprox. 2.5%)
1000 - 2500 kVA : 20 Watts/kVA (aprox. 2%)
larger than 2500 kVA : 15 Watts/kVA (aprox. 1.5%)

 

[prc]

I can suggest a much better way for evaluating losses - you can always measure the resistance of windings. Correct it for the normal operating temperature. Add to the calculated I2R, 10-30 % for stray losses, depending on rating of transformer. This gives load losses. No load losses are 1/4 to 1/5 (except for auto transformers where it can be double these figures). During the initial days of transformers, this loss ratio was 1-1.2, it increased to 2 by 1920, reached 3-4 in 1970 and currently at 4-6,thanks to the massive reduction in no-load losses compared to load loss reduction.
I have been suggesting to Indian utilities to insist for a disposal plate from transformer manufacturers where they should show, measured no-load and load losses, net copper and electrical steel weight, weight of mild steel and oil, type of painting provided. Some have implemented. Just as losses, copper and steel weights are another factor which users require at the end of life, to estimate correct salvage value.

 

[bhellmer]

prc: We are not going to have these transformers taken out of services and tested for heat release data.

I have attached a typical transformer nameplate.

Notice that the manufacturer has provided "Impedance Volts 5.90 percent at rated volts at 2000 KVA". The manufacturer either loadbank tested their transformers or otherwise made adjustments to account for the transformer's fully-rated state.

Most nameplate do not include X/R and obtaining an authoritative value can be as difficult as obtaining heat release values.

One of the reasons I have been focusing on Z% and X/R is that we must obtain them anyway for short-circuit and arc-flash calculations.

 

[stevenal]

I was able to get some original test reports for Rome built units from GE for a fee. Talk to your rep.

 

[bhellmer]

Stevenal: Do these GE original test reports include heat release data? Would you upload a typical GE original test report?

 

[stevenal]

Yes. See the No Load and Load Loss wattages.