Transformer impedance test calculations

[Mondy]

Hi All

Can anyone help me with the calculations involved in a transformer short circuit impedance test. I have the following data:

Transformer rated voltage =230v
Transformer rated power = 960VA
Secondary short circuited during test
ambient temperature during test = 22 deg.C

Measured impedance voltage Ez = 3.55v at 50% of rated current and at the transformers rated frequency (50Hz)

Supplied current = 2.188 average amps

Measured copper loss CuW = 7.51W at 50% of rated current and at the transformers rated frequency (50Hz)

Impedance voltage (Ez) and Load Loss (CuW) corrected to 100% of the rated current and a reference temperature of 80deg.C resulting in:

Short circuit resistance Er = 3.84%
Short circuit impedance Ez = 3.87%
Short circuit reactance Ex = 0.51%
Copper Loss CuW @ 100% load = 36.84W

I assume that the Ez of 3.84% is calculated from the 3.55v @ 50% of rated current (although I cannot get it to add up) and then the reactance Ex = 0.51 derived from

Sqr(Ez 3.87^2 - Er 3.84^2)

and Er = 36.84W * 10 /960VA

as usual, but I cant get it to add up. Or is the Ex = 0.51 calculated and then added in quadrature to come up with the Ez = 3.87%

Can anyone get these figures to make sense as I cannot get them to correlate and I can only assume I am not taking into account something in the correction of the parameters measured at 50% of rated load to 100% rated load and 80 deg.C reference temperature?

Cheers
Ray

 

[davidbeach]

I don't know about the 50% rated current business, but (E_r²+E_x²)^1/2=E_z as it should.

 

[ScottyUK]

I think your problem is likely to be in the correction from 22C to 80C ambient. How are you doing this?

Is this a college assignment? It certainly smells like one.


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If we learn from our mistakes,
I'm getting a great education!

 

[Mondy]

Hi Dave & Scotty

Thanks for your replies. No this is not a college assignment (although it feels a bit like one!) It is part of the analysis of a Norwegian transformer that is not performing at all well in the UK (at 240v + 10%) and is part of a large underfloor heating plan. These are the test figures for the transformer by the manufacturer and I need to have a good grounding on their tests as there is about 10 grand resting on the issue!

I always thought that the measured impedance voltage Ez = 3.55v was used to find the percentage impedance by

Impedance volts / rated input volts * 100 as a %

I am not sure why they didn't just measure the voltage at 100% of the current as the transformer is only small. I assumed then that Ez = 2* 3.55v / 230v = 3.087% which does not match up with the 3.87% stated. I dont think this is temperature dependant as the percentage resistance would be and here I just used the following to come up with the corrected copper loss

1+((80-22)*0.00393) * (7.51 * 4) = 36.89W

which I thought was pretty close, but maybe you guys know how to approach this correctly regarding both correction of Ez voltage as well as CuW!

I thought the process was to find the impedance voltage by measurement and then use this to determine the overall impedance %. Then use the watts measured to determine the percentage resistance Er using

CuW/rated power * 100 as a %

and then calculate the reactive percentage by

Sqrt(Ez^2 - Er^2)

But if the above calculations are correct this would work out as a lower overall percentage impedance Ez = 3.087% than Er = 3.84% and so you couldn't extract the Ex % from the above equation.

Can any of you good people advise me?

Cheers
Ray

 

[ScottyUK]

Hi Ray,

Are you sure about 240V + 10%? That is way outside normal UK tolerance at 264V. The old UK spec was 240 +/- 6% which gave a top limit of about 254V. 230V + 10% is the new top limit which is 253V. Either way, something is wrong.

If you are really pushing the transformer that hard, the iron core will be well in to saturation, where it will draw a high and non-sinusoidal magnetising current and get very hot. The secondary voltage will be distorted. This is quite possible even if the transformer was designed for 230V service if the designer was ungenerous with the core sizing.
This tends to happen quite a lot, especially with small transformers. So much for European harmonisation!

What are the symptoms that the transformer is exhibiting when it is "not performing at all well"? There might be a simpler answer rather than diving deep into the open- and short-circuit characteristics.




----------------------------------

If we learn from our mistakes,
I'm getting a great education!

 

[Mondy]

Hi Scotty

Yes I agree with what your saying and this is almost definately the case the location is rural and +10% over volts is a regular occurrence for them apparantly, the iron saturation and hence the peaky input current is causing the issue, but there are other factors and that is why I need to be sure I can calculate the varuious parameters form the test results and this is why I posted. Do you have any experience in transformer impedance/short circuit tests or can you point me in the right direction?

 

[garyvin]

I might be able to help you out here but I need to know some more information.

To properly calculate everyhing I need to know the HV and LV resistance of the windings and the primary and secondary voltage. Also I would assume the short circuit test were done by shorting the LV and applying voltage to the HV until they got 50% or rated current. I don't know either why they stopped at 50%. 4 amps is nothing.

 

[Mondy]

Thanks SparksAlot

If you could help me on this I would be eternally gratefull!

OK here goes... all are measured values

Hv (pri) resistance @ 22deg.C = 0.4978R
Lv (sec) resistance as above = 0.0116R

Voltage ratio Pri: 230v Sec: 25.5V = ratio 9.109 (24v rated on load at 40A or 960VA

No-Load current @ 230v input 50Hz = 0.409A
No-Load Loss (FeW) = 24.7W
Total Losses @ 100% load and reference temperature (80deg.C) = 61.54W

The idea I had was to put together a spreadsheet so that I can enter all test data and then calculate the Efficiencies and regulation at various loads etc and so that I can work out short circuit currents etc. If you can show me the light as to how it's all done (and particularly how values corrected to the reference temperature and 100% load) then the beers will be on me for a considerable time!!

Cheers
Ray

 

[garyvin]

You know that is very interesting... I do not know how they cam up with what they got either. Here is why I get using the values you gave me.

Short Test Info
Rated current @ 230v and 960VA = 4.173 A
Ratio up test voltage at rated current = 6.772 V
Ratio up tested wattage = 27.329 W

@ 80 deg C
%IZ 3.5665
%IR 3.4864
%IX 0.751709297
%EX.I 1.0225
Total Losses at 80 deg. C= 57.23 Watts

What I don't get it something is wrong in either the resistance or the tested copper loss or rated voltage. With my calculations it makes the stray losses at 80 C .0786 and that just can't be. I usually deal with 60Hz not 50Hz but if your data was tested at 50Hz if shouldn't make a difference.

What is the rated primary voltage for that unit? if I make it 220V the numbers make more sense.

%IZ 3.8260
%IR 3.7368
%IX 0.821599625
%EX. I 1.0225

Total Loss = 59.642 W

What are your thoughts?

 

[Mondy]

Hi SparksAlot

Many thanks for your time it is very much appreciated. Can you show me the equations you used to get your results. As I see it you came to the following using:

Rated current @ 230v and 960VA = 960/230= 4.173 A

Ratio up test voltage at rated current =
(4.173A/2.188A)* 3.55V = 6.772V

Ratio up tested wattage =
7.51 * (4.173A/2.188A)^2 = 27.329W

But how did you come up with the percentages corrected to 80deg.C?

This transformer has a primary with three taps with the following turns:
0 - 144 - 163 - 182ts
0 - 184 - 207 - 230v

The seconadry is 20ts

This is used by applying 230v to the 182ts normally, and the seconadry then should give 24v @ 40A. If 28v or 30v is required then you apply 230v to the other taps accordingly. The data is for the 182ts @ 230v input as the voltage ratio would seem to agree with this

What are your thoughts and if you could show me your workings it would be much appreciated?

Once again thanks my friend!

Ray

 

[garyvin]

960 VA
H.V. Volts 230
L.V. Volts 25.5
H.V. Amps 4.174
L.V. Amps 37.647
Resistance HV 0.4978
LV 0.0116
Tested Watts 27.3295
Test IZ volts 6.772117
core loss watts 24.7
current 0.409

Test Data Calculations
Test Temp 22 oC
Perf Temp 80 oC

CU Loss at 22 oC
I2R at 22 oC Cu= 234.5 if AL =225
L.V. 16.4406 K=1.226120858
H.V. 8.67247 k= (80+234.5)/(22+234.5)
SUM= 25.1131 WATTS
SUM Cor=30.7917 AT 80 oC =sum * k

Stray at 22 oC
W= 27.329
I2R= 25.113-
STRAY= 2.2163 WATTS
TOTAL Loss at 80 oC
I2R at 80 oC 30.7917
STRAY 1.8076+ =2.21638*K
WM CU. 32.5994+
CORE 23.7688+
TOTAL 56.3682 WATTS
IMPEDANCE
%IZ2 AT 22 oC 8.6695 =(6.772/230*100)^2
%IR2 AT 22 oC 8.1044- =(27.3295/960*100)^2
%IX2 AT 0.5651
%IR2 AT 80 oC 11.5312 =(32.5994/960*100)^2
%IZ2 AT 80 oC 12.0963 = 11.5312+.5651
TEST RESULTS AT 80 oC
%IZ 3.4780 These three are sqrt of the last three
%IR 3.3958
%IX 0.7517
%EX. I 1.0864 = .409/37.647*100

I hope you can make sense of my progression. and sorry but this messes up my spacing when I submit it. I also have an excel spread sheet that does this, its crude but it works if you want it.

 

[Mondy]

Hi SparkAlot

Thanks for that my friend your effort is very much appreciated. I will have a good go with this. Your spreadsheet will be much appreciated if you would be so kind!

Cheers
Ray

 

[garyvin]

The trick here is how am I to send you the file without posting an email address since we are not supposed to do that?

 

[Mondy]

Thats a good point, Is there no send facility via the forum as they have our addresses?

Cheers buddy
Ray

 

[Mondy]

Hi Sparks Alot

I think I have got most of it down, just a couple of questions.

1. How did you get to CORE = 23.7688+ from the measured was 24.7W?

2. The stray loss is 2.2163 WATTS @ 22deg.C and reduces due to the rise in resistance with temperature to 2.2163/k =1.8076W at 80deg.C is this correct?

Cheers
Ray

P.S : Does anyone know how to send someone an attachment file on this forum?

 

[ScottyUK]

Hi Mondy,

Sorry for the slow reply - busy week at work. I'm doing this at home!


3.55V circulates 2.188A through the transformer impedance. The impedance is therefore 3.55 [÷] 2.188 = 1.622[Ω]

Copper loss is 7.51W at 2.188A. Resistance is therefore 7.51 [÷] 2.188² = 1.569[Ω]

Leakage reactance X_L = SqRt (Z² - R²) = 0.413[Ω]

Resistance temperature correction factor is (234.5 + 80) [÷] (234.5 + 22) = 1.226.
Leakage reactance is not temperature dependent, so no correction is required.

Resistance at 80C R₈₀ = R₂₂ x 1.226 = 1.923[Ω]

Impedance at 80C Z₈₀ = SqRt (R₈₀² + X_L²) = 1.967[Ω]

Rated current = VA [÷] V = 4.174A

Copper loss at 80C P₈₀ = 1.923 x 4.174² = 33.50W

The ohmic values of R, X, and Z can be converted to % values by establishing a base impedance. Normally this is calculated from:

Z_Base = V² [÷] VA = 55.10[Ω]

R₈₀ = 1.923[Ω] = 3.49%
Z₈₀ = 1.967[Ω] = 3.57%
X_L = 0.413[Ω] = 0.75%
P₈₀ = 33.50W







----------------------------------

If we learn from our mistakes,
I'm getting a great education!

 

[garyvin]

For the Core Loss I was under the assumption that is was also tested at 22 deg and I corrected it to 80. If it was alreayd corrected then just add it in as it is. The ANSI formula for correction is 1+ PU * (Pt-Tt) = corr. factor

Where PU = 0.00065
Tt = perf temp
Pt = test temp

Then simply multiply your tested wattage by the correction factor.


Yes the stay goes down as the temp goes up the opposite of the resistance.

 

[Mondy]

Hi SparksAlot

Thanks my friend. From this am I right in saying that as the temperature goes up the core restivity increases and so the eddy currents reduce?

Also what does the 0.00065 refer to?

Cheers
Ray

 

[garyvin]

Yes I believe that's how that works as the temp rises the stay loss goes down.

I understand that PU is the constant for the Per unit change of core loss per one deg C. I think that is from ANSI std C57.12.90 if you want to check it out.

 

[Mondy]

Just wanted to say thanks to both SparksAlot and ScottyUK for your invaluable help with this information and for the time you both spent helping me out.

It is much appreciated and I owe you both a drink!!

Cheers
Ray