Transformer open circuit test

Discussion in 'General Electronics Chat' started by Nikoz121, Mar 18, 2015.

  1. Nikoz121

    Thread Starter New Member

    Mar 17, 2015
    4
    0
    Hello,

    I had to do an experiment about a single phase transformer 230V/230V, 50 Hz, 100 VA , where the transformer secondary was open-circuit and there was an universal power supply which goes from 0% to 100%. I had to record the primary and secondary voltage and the primary current and these were my results.

    % ____Primary Voltage(V) __Secondary Voltage(V) _Primary Current(mA)
    0_______17.915______________19.403 _____________ 6.556
    10_______42.100_____________45.590______________10.423
    20_______66.200_____________71.940______________12.870
    30_______88.69 _____________96.32_______________14.901
    40_______112.29_____________122.16_______________16.792
    50_______137.70_____________150.16_______________18.870
    60_______162.62_____________176.96_______________21.36
    70_______187.24_____________203.8________________24.60
    80_______210.4______________229_________________29.12
    90_______235.5______________256.2________________36.91
    100_______240.8_____________260.21________________38.28

    From those results I got that the turns-ratio is about 1.08 but I have to compare and comment on the measured transformer secondary voltage considering the rating of the transformer but I'm not too sure what do I have to say.

    Also, I have plotted the primary voltage against the primary (magnetising) current as it can seen below but I'm not sure if the shape of the curve is right. Is it normal to get a shape like this one, if it is, why?

    Untitled.png

    Thank you for your help!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    @Nikoz121
    It's a bit unusual that you plot the dependent variable (Current) on the horizontal axis rather than the vertical axis - the usual convention being that one plots the independent variable (in this case Voltage) on the horizontal axis.
    Notwithstanding that small matter, one can readily observe that the magnetizing current is not linearly related to the excitation voltage over the entire voltage range. You should be able to explain what's happening based on your understanding of the magnetic properties of transformer grade steels. One element of your explanation would presumably include some discussion of magnetic saturation, reluctance etc.
    It's also curious that your experiment seems to overlook several other important parameters encountered in understanding transformer behavior and properties. It's difficult to envisage how a comprehensive discussion about the transformer in question might be developed in the absence of the "missing parts of the jigsaw".
     
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  3. DickCappels

    Moderator

    Aug 21, 2008
    2,647
    632
    It is more or less what one would expect. If you flip and rotate it as t_n_k indicated, it looks like a typical magnetizing current cuve.On a different day the curve might shift to the left or the right, depending upon the flux in the core when power was last removed.

    [​IMG]
     
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  4. uwed

    Member

    Mar 16, 2015
    64
    17
    Some interpretation:
    For open secondary we can simplify:
    U = 2*pi*f*Lm* I
    Lm is proportional to ur (rel.permeability of core)
    Because of saturation of the core (see B-H-curve of core material): ur is approximately constant and high below a certain current (saturation), and then drops to low values above a certain current (saturation)
    --> U proportional ur(I) * I
    ... which is a U(I)-curve with a linear rise with a steep slope below saturation, and then with a flat slope above saturation.
    As you can see from your curve, this change of slope happens above 200V which makes sense considering it is a 230V-transformer (otherwise the transformer design would be quite bad).
    This is exactly the curve you have measured.
     
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  5. Nikoz121

    Thread Starter New Member

    Mar 17, 2015
    4
    0
    Thank you for all your responses

    Yes, you're right.

    This is only one part of the experiment and again thank you very much but I still don't know what to say about the measured transformer secondary voltage considering the rating of the transformer
     
  6. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    some transformer manufacturers put a little more in the secndary ( 1.108 to 1) to compensate for the resistive losses when you draw current from the secondary (makes the ratio more like 1 to 1 underf load)
     
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  7. Nikoz121

    Thread Starter New Member

    Mar 17, 2015
    4
    0
    @alfacliff @uwed @t_n_k @DickCappels
    Hello all, I've got another question regarding this experiment:

    This part of the experiment will examine the relationship between the transformer primary and secondary currents. The Switched Three-Phase Resistive 67-142 is to be used to load the transformer secondary. The three switches in this unit can be used to connect the three resistors 950 ohms, 1950 Ohms and 3770 Ohms in parallel.

    Calculate the load resistance for the transformer that would be required to draw rated current from the secondary. Hence calculate the settings of the three-phase load resistor switches that would give approximately (within a few percent) rated current.

    And this is what I did:
    1/Rt = 1/950 + 1/1950 + 1/3770
    Rt=546 Ohms

    I= V/R = 230/546= 421.2 mA

    Then I had to set the transformer primary voltage to 230V and with an open-circuit secondary,re-record the primary current using the power analyser. Switch in the rated load resistance as calculated in the previous exercise, and record the primary and secondary transformer parameters.

    Off-load Primary current (mA) 34.90

    Full-load Primary Current (mA) 482

    Full Load Secondary Current (mA) 421


    The ideal transformer equation relates the primary and secondary currents using its turns-ratio. Using this equation, compute the predicted value of the transformer, on-load primary current based on the measured on-load secondary current.

    So Ip/ Is = turns-ratio where the turns-ratio was calculated earlier as 1.08
    therefore,
    Is= Ip*1.08 = 482*1.08= 446.29 mA

    Compare and comment on the predicted and measured values of the on-load secondary current, and if required, use further calculations to justify any discrepancies.

    Am I doing it right? what further calculations can I use and what comments can I talk about?
     
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