My 1Ph dry type transformer weighs..

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ptonsparky

ptonsparky

Tom
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NE (9.06 miles @5.9 Degrees from Winged Horses)
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EC - retired
43 lbs. What KVA is it? JK.

I do have one minus a label. Any chance a guy could even come close by physical size etc?

I back fed it in my shop to get single phase 480. I could program a drive if I needed. It moved with me.

I have 30 amp 120v generators and want 120/240. Or I could buy the proper generator. That's too simple though.
 
So you know it can handle 480V, thus likely a 120/240 to 240/480 transformer.

I'd measure the wire size on the 480V side, and try to guess from that.

Jon
 
220612-1736 EDT

I have a rather standard small transformer that I make various measurements on. The primary is two 120 V windings in parallel. This primary is what I will look at in the following data with no load on the secondary. So I am looking at transformer winding and core loss at no load. Pretty much little of the loss is winding resistance until we get to higher voltages.

Data:

Volt ..... AMP ....... Sat V

060 ..... 0.018

070 ..... 0.022

080 ..... 0.030

090 ..... 0,047 ....... Start see some

100 ..... 0.079

110 ..... 0.133
115 ..... 0.166

120 ..... 0.195
125 ..... 0.240 ....... 1 A peak at 90 deg phase angle

130 ..... 0,291
135 ..... 0.343

140 ..... 0.405 ....... 10 A peak at 90 deg phase angle

Transformer peak magnetizing current occurs at excitation voltage zero crossing as is expected. There is virtually no saturation current until excitation voltage reaches about 90 V. Whether most transformers are designed with this approximate point I don't know.

You can study any winding of your transformer with the above approach. This will allow you to estimate the voltage rating of any other winding. Once you can find a single winding that is rated 120 or some other voltage, then you can measure voltage on other windings to determine their approximate ratings.

Once you determine what is the total primary winding, then you can approximately assume the transformer is designed for 1/2 half its losses are in the primary. Next you assume the transformer is approximately 95 % efficient.

Next you make some assumption of insulation temperature rating, and then measure winding temperature rise under estimate fully load, and run a test.

.
 
ptonsparky said:
I have 30 amp 120v generators and want 120/240.
Click to expand...
ptonsparky said:
Let's say the xfmr is indeed 2kva.
Back feeding it with 120v on X side I get 480 on H. Still 2kva.
Click to expand...
Yes, the transformer will still handle 2kVA when backfed from 120V to 480V.
However, to get your desired 120/240 from 120V only the X1,X2 and X3,X4 windings will be usable, with one as the primary and one as the secondary to provide the opposite phase of 120V. You will be limited by the current handling capability of the windings used and not the size of the transformer core. And so you'll only be able to get 1kVA from the transformer in this case.
 
synchro said:
Yes, the transformer will still handle 2kVA when backfed from 120V to 480V.
However, to get your desired 120/240 from 120V only the X1,X2 and X3,X4 windings will be usable, with one as the primary and one as the secondary to provide the opposite phase of 120V. You will be limited by the current handling capability of the windings used and not the size of the transformer core. And so you'll only be able to get 1kVA from the transformer in this case.
Click to expand...
Thank you. I managed to get the desired output voltage and was wondering how that affected the KVA.

I applied 120 to the 240 configured secondary (x) It gave me the 120/240 (h) I was looking for.
The current draw is considerably less.
 
220618-2238 EDT

ptonsparky:

To address your question I am making the following assumptions as they simplify the analysis:

Suppose the 2000 VA transformer is wound to provide 120 V at full load, and that there is about a 5 % internal voltage drop at full load. This means turns ratio is about 480 / (120*1.05) = 3.8095 . No load secondary voltage is 480 / 3.8095 = 126.0 V. At full load secondary current is 2000 / 120 = 16.667 A based on VA rating. As viewed on the secondary side the transformer internal impedance is R = ( 126.0 - 120 ) / 16.667 = 0.36 ohms. Total power loss in transformer is about 16.7^2 * 0.36 = 100 watts = Ploss. This value makes sense since current remains close to constant for the change in voltage. I will assume I can put that power loss in any distributed way between primary and secondary for calculation purposes.

Impedance or R reflects between between primary and secondary on a turns ratio squared basis for a tightly coupled transformer. This is true for either forward or reverse operation of the transformer. Our turns ratio is 3.8095 and not 4 to achieve 120 V from 480 V with an internal voltage drop of 5% at full load. Winding resistance does not change with change in direction of operation..

When the transformer is run in reverse from 120 V, then unloaded secondary output voltage will be about 120 * 3.8095 = 457 V. Loaded this will drop to about 434 V. Same power dissipation is allowed in the transformer in reverse as in forward. This dissipation is 16.7^2 * 0.36 = 100 W from above. This means total internal resistance as seen at the 480 winding is R480 = 0.36 * 3.8095^2 = 5.224 . Power loss from above is 100 W. Thus, secondary current for full load is 100 = I^2 * R = I^2 * 5.224 or I = 4.375 A, and maximum power output is 434 * 4.375 = 1899 W. Thus, about 90% of transformer rating as could be guessed from 0.95^2 = 0.9025 .

Assuming I made no mistakes.

.
 
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