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Thread: No load losses - normal vs reverse fed

  1. #91
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    GE white paper on backfeeding dry type
    Attached Images Attached Images

  2. #92
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    Quote Originally Posted by jminer99er View Post
    GE white paper on backfeeding dry type
    Good find!
    Si hoc legere scis nimium eruditionis habes.

  3. #93
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    Experimental results:

    1. Transformer is a P-3197, probably Stancor, from later 1940s, but before 1950. Rubber insulated wire. Predates PVC. Input 117 V 60 CPS, and rated 80 W.

    2. Primary resistance 4.7 ohms.

    3. Secondary resistance 5.9 ohms.

    4. Using no load to determine turns ratio.
    Primary input 110.1 V secondary 116.2 V, ratio secondary to primary 1.055 .
    Secondary input 116.1 V primary 110.1, ratio secondary to primary 1.054 .
    Good correlation.

    5. Using a Kill-A-Watt for source measurement, and a Fluke 27 for output voltage. An 0.2 V difference in calibration between the two instruments was corrected for.

    6. First powering the primary with no secondary load. Then power input to secondary and no load on primary.

    Vin = 110.1 V, Iin = 0.09 A, Pin = 4.5 W, VAin = 10.1 VA, PF = 0.41 --- Vout = 116.2 V.
    Vin = 116.1 V, Iin = 0.09 A, Pin = 4.6 W, VAin = 11.0 VA, PF = 0.41 --- Vout = 110.1 V.

    Using these values both tests had close to the same core flux density. To obtain a more accurate comparison more accurate instruments would be required and a more stable voltage source than my home line voltage.

    The results, as should be expected, are essentially the same. If you do not work at the same flux density, then you should not expect core losses to be the same.

    If the coils were wound side by side, then you would expect to see the resistance ratio about the square of the turns ratio. But that would make the secondary 5.2 ohms and not 5.9 ohms. Thus, the conclusion is that the secondary is wound on top of the primary.

    With a 75 W incandescent as a secondary load the results are:
    Vin = 110.1 V, Iin = 0.66 A, Pin = 72.7 W, VAin = 72.9 VA, PF = 0.99 --- Vout = 108.9 V.

    .

  4. #94
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    Thanks, good info. Strange transformer. Now repeat with 75 kva 480-120/208
    Ethan Brush - East West Electric & Specialty Construction. NY, WA, RI, MA

  5. #95
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    Gar...

    Can you perform a Short-Circuit Test? Also what are xfmr's physical dimensions?

    Phil

  6. #96
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    electrofelon:

    It is not a strange transformer. It is what you should expect. This is a single phase two winding transformer.

    This is a small transformer 80 W on nameplate, thus 80 VA is possibly a better descriptor.

    I don't have a 75 kVA to test or an appropriate load. But, transformers whether large or small follow pretty much the same theory.

    .

  7. #97
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    Phil Corso:

    I have not run a short circuit test. I shall try to do that.

    However, after my post last night I did some calculations from the available data and did not get a good correlation on power. But there is a reason that does not change the core loss measurement which is basically correct.

    I made a determination of the transformer turns ratio from the open circuit voltage measurements that resulted in a ratio of 1.055 . This is incorrect relative to the actual turns ratio because of the percentage of flux lines that do not couple the two coils, leakage flux.

    If I use as the turns ratio the value calculated from the open circuit voltage ratio, which is low because of the leakage flux, to predict the secondary current under load from the 75 W bulb, then I predict a higher secondary current than actually occurs, calculated load power from V*I is 108.9*0.66/1.055 = 68.1 W. The measured input power is 72.7 W. The difference 72.7 - 68.1 = 4.6 W, just the ballpark for core loss, nothing for I*R loss. This results from using a turns ratio that is low from ignoring the leakage flux.

    Today I did an additional experiment to determine the actual secondary currrent. With the same 75 W bulb load and 110.1 V into the transformer the primary was 0.66 A, as before. and the measured secondary current was 0.58 A. Turns ratio calculated based on the current ratio is 0.66/0.58 = 1.14 . To be expected based on there being leakage flux.

    Now the load power calculates to 0.58*108.9 = 63.1 W. The sum of the losses is now 0.66*0.66*4.7 = 2.04 W (primary I*I*R), core 4.5 W, 0.58*0.58*5.9 = 1.98 W, and summed = 8.5 W. The difference between input and load is 72.7 - 63.1 = 9.6 W. Now we have a better correlation, and the results make sense. More accurate measurements should provide a better correlation.

    If I used a high peramability material tape wound, close to no air gap, then the input to output voltage ratio would be much closer to the actual turns ratio.

    .

  8. #98
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    ////////.m,m,m,.
    Last edited by Electric-Light; Yesterday at 05:19 PM. Reason: doubled up
    Light Emitting Decorations

  9. #99
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    Quote Originally Posted by gar View Post
    This is a small transformer 80 W on nameplate, thus 80 VA is possibly a better descriptor.

    I don't have a 75 kVA to test or an appropriate load. But, transformers whether large or small follow pretty much the same theory.
    You're reporting the samething I reported on a 600VA transformer, except even smaller. The GE whitepaper notes transformers over 3kVA are generally uncompensated. Sometimes the most difficult journey is implementing theory into reality.

    Quote Originally Posted by gar
    If I used a high peramability material tape wound, close to no air gap, then the input to output voltage ratio would be much closer to the actual turns ratio.
    Toroidal one?
    Light Emitting Decorations


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