Single Phase Transformer Confusion

Discussion in 'Homework Help' started by Jess_88, Aug 21, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
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    1
    Hey guys

    I have been asked to determine the equivalent circuit for a single phase transformer connection with the following measurements.

    Given the transformer rating = (125/50 V)(1095 – 115) 50Hz

    Short Circuit Test Measurements - LOW VOLTAGE side
    voltage = 5.6V
    current = 2.1A
    power = 9.9W

    Open Circuit Test Measurements - HIGH VOLTAGE side
    voltage = 238.6V
    current = 37.5mA
    power = 2.625W

    Open Circuit Test Measurements - LOW VOLTAGE side
    voltage = 99.3V
    current = 0.3A
    power = 8.9W

    My attempt
    [​IMG]

    For Short circuit

    θ = cos^(-1)(P/V*I)
    =32.665°

    Z = V/I <32.665°
    = 2.7<32.665° = 2.2729 + j1.4572

    (referred to primary side)
    Re = 2.2729
    Xe = j1.4572

    For Open circuit

    θ = cos^(-1)(P/V*I)
    =72.939°

    Y = I/V <72.939°
    = 157.166<72.939° = 4.612 x10 ^(-5) + j1.5024 x 10^(-4)

    (Referred to the Secondary)
    Rc = 1/4.612 x10 ^(-5)
    = 21.6 kΩ
    Xm = 1/1.5024 x 10^(-4)
    = 6.65 kΩ


    a = Vo/Vin
    = 238.6/99.3
    = 2.4

    Rc (Ref to Primary) = a^(2)* Rc
    =124.416 kΩ
    ... Seems a little high :confused:

    Xm (Ref to Primary) = a^(2)* Xm
    =38.304 kΩ

    These values for Rc' and Xm' seem way out. Where did I go wrong :confused:

    Also If I have a resistive load at the output. How would I go about calculating the Power at the load?

    Would I Refer the load to the Primary??? using R' = a^(2)*R ?

    Thanks guys :) .... I'm pretty confused.
     
  2. Kermit2

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    Feb 5, 2010
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  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Can you confirm the information supplied in your post is exactly the same as it was presented to you.

    There are some curious values ...

    Quote:

    "Given the transformer rating = (125/50 V)(1095 – 115) 50Hz"

    What does this mean?

    It's difficult also to interpret what the various tests conducted were and where the measurements were taken.

    Presumably:

    • The short circuit test values would be observed on the primary side with the primary energized and the secondary shorted.
    • The open circuit test would be conducted with the primary energized at rated voltage with current & power measured on the primary side.
    So what is the third test meant to represent? Which side is energized and where were the current & power measurements observed? Was the open secondary voltage 99.3V with 238.6V applied on the primary side? If the secondary was open where was the 0.3A flowing? Can you clarify what the third test means.
     
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  4. Jess_88

    Thread Starter Member

    Apr 29, 2011
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    "Given the transformer rating = (125/50 V)(1095 – 115) 50Hz"
    I believe the 125/50V is actually to do with the turns ratio... but I don't think that even matters, because wouldn't I just use Vout/Vin?

    The Short Circuit (No Load Test) was performed by energising the secondary and shorting the primary. Measurements were then made across the low voltage side... which seems weird... if thats the side being energised??? I'm thinking the "low voltage" side, in this case, is primary... maybe.


    The Open Circuit Test was performed by energising the primary and measurements made on Secondary (Low Voltage). Measurements were also made on the Primary (High Voltage).

    "Was the open secondary voltage 99.3V with 238.6V applied on the primary side?"
    That is correct.

    " If the secondary was open where was the 0.3A flowing? Can you clarify what the third test means."

    From my understanding, test three is really only to determine the turns ratio of the transformer.

    238.6/99.6 = 2.4
    quite close to the "rated" voltage....thing
    125/50 = 2.5

    is that better?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The question is still unclear if not ambiguous. Your own uncertainty about the nature of the tests is interesting.

    An open circuit test is normally done on the primary side - the primary voltage is usually readily available whereas the secondary may not be.

    In an open circuit test (apart from the open circuit secondary voltage) you measure the important parameters on the side which is excited by the known voltage. Measured values are the excitation current and power loss from which one can deduce the magnetizing inductance and associated core losses. Knowing the applied primary voltage & the open circuit secondary volatge leads you to the turns ratio - as you indicated.

    Short circuit testing is normally done in like manner - albeit with the secondary shorted and the applied voltage greatly reduced below the nominal value (say by using a Variac) . Again the parameters are usually measured on the primary side and from which one derives the series leakage reactance and resistance.

    Do you actually have answers which you can cross check your solution - which by the way is on the right track.
     
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  6. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    Sorry for the confusion. I have left out some parts because I think it would just make things even more confusing.

    These are real measured values. The method in which they were obtained is hard to follow because in the lab method it just refers to points (e.g. a1,a2, LowVoltage side ext) and not weather it is primary or secondary. So I don't have any "solutions" because these are real measurements.

    I'm pretty sure my measurements are correct for the method I was given, as my circuitry was checked by my lab supervisor with each test made.

    Would measured input Voltage/Power/Current with load attached help confirm anything?

    ps. A I did use a Variac was used in the short circuit test.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If the transformer rating is 125/50 (presumably meaning 125 volts on one side and 50 volts on the other), why would you perform an open circuit test with 238.6 volts applied to the high voltage side and with 99.3 volts applied to the low voltage side?

    I find it hard to believe that the measured wattage is as low as you report with applied voltages nearly twice the rated values. I suppose it's possible that 125/50 isn't the manufacturer's rating; maybe it's really a 240VAC transformer, since you are in Australia.

    Furthermore, even though the voltages you report as having been applied for the open circuit test are nearly double the apparent rated values, their ratio is approximately 2.5:1, which is the rated voltage ratio for this transformer. If you perform an open circuit test on each winding, with the applied voltages being in the same ratio as the rated voltage ratio, the measured wattage for these two situations should be nearly identical. The open circuit test wattage is essentially just the core loss, which should be the same if the same volts per turn is applied to the windings in the tests.

    However, if you were to apply substantially more volts per turn on one winding than on the other, the core would be driven further into saturation and the core loss would be greater for the case with the higher applied volts per turn.

    You report a wattage of 2.625W for one case and 8.9W for the other case even though the applied voltages are in the ratio 2.4, which is close to the rated 2.5 ratio (125/50). This should not be, and leads me to think that the turns ratio is not really 2.5:1.

    As t_n_k suggests, these numbers you report are rather peculiar.
     
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  8. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    ok, so I have been reading carefully the procedure in which my measurements were made and this is what I have. Maybe it will be easier to see what I have done wrong.

    Open Circuit Test
    The secondary was open circuit and power applied the the primary.
    Volage current and power were then measured across the primary.
    Voc = 238.6V
    Ioc = 37.5mA
    Poc = 2.625W

    and voltage at the output was noted for the turns ratio
    Vs oc = 99.3

    giving the ratio 2.5:1

    From these values I can determine
    θ = cos^(-1)(Psc/Isc*Vsc)
    Rc = Voc/ Ioc cos θ
    =21.686 kΩ
    Xm = Voc/ Ioc sin θ
    =6.655 kΩ

    Short Circuit Test
    The Primary was short circuited and rated current (2.1A) applied to the Secondary.

    Voltage Power and Power Factor were then measured across the Secondary.
    Vs sc = 5.6V
    Ps sc = 9.9W
    PF = 0.839

    From these values
    θ = cos^(-1)(Psc/Isc*Vsc)
    Req = (Vsc/Isc)cos θ
    =2.2729Ω
    Xeq = (Vsc/Isc)sin θ
    =j1.45726Ω

    Dose this seem at all more clear?

    I'm pretty sure The method in which I obtained the results is ok. I'm having the most trouble determining which side my measurements are with reference to.

    Which values (Xeq, Req, Rc or Xm) would I need to refer to the primary of to calculate core and copper losses???

    To calculate efficiency what values of 'V' and 'I' would I be using for
    η = (VI cosθ)/(Pc + Pcu + VI cosθ)

    thanks so much for your help guys :)
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One can't help being confused when the facts aren't clear. Thanks for clearing up the matter - mostly.

    There's obviously still some confusion about the turns ratio.

    You presumably "calculated" the turns ratio from V_prim/V_sec with an open circuit secondary condition. I believe 238.6/99.3=2.4 - which is what you deduced in an earlier post. Why have you returned to the nominal 2.5 value?

    The open circuit test was done on the primary side so the core loss and magnetizing inductance (reactance) will already be known with respect to the primary side.

    Since the short circuit test was done on the secondary side you would need to refer both the copper loss and leakage inductance (reactance) to the primary if you want everything referred to the primary side. While this referencing is highly convenient, it isn't mandatory - the equivalent circuit two port model would give the same results provided you keep in mind which "side" you are considering during any analysis.

    Also keep in mind these lumped values don't exactly represent the true state of affairs. For instance no attempt is made to split the primary and secondary copper and leakage losses. The model takes no account of any non-linear magnetic core properties. But the result is good enough to provide a useful model of the transformer.
     
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  10. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    ah yeah my mistake. Turns ratio should be 2.4
    So I get and efficiency of 89.24%. Dose that sound about right?

    Also, with a load resistance how do I determine the power deliver to the load?
     
  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You've reported these results:

    "Short Circuit Test Measurements - LOW VOLTAGE side
    voltage = 5.6V
    current = 2.1A
    power = 9.9W

    Open Circuit Test Measurements - HIGH VOLTAGE side
    voltage = 238.6V
    current = 37.5mA
    power = 2.625W

    Open Circuit Test Measurements - LOW VOLTAGE side
    voltage = 99.3V
    current = 0.3A
    power = 8.9W"

    I explained to you that the power measured in the two open circuit tests should be the same.

    Just to make sure that's the case, I found a transformer that is the same size as the one you're using. It's a 200 VA transformer, with 120/240 VAC primary and 24 VAC, 8 amp secondary.

    I measured the open circuit power consumption (this is the core loss) from both the high voltage side and the low voltage side. In both cases, the power was 9.5 watts.

    Furthermore, in a typical commercial transformer of this size, the core and copper losses should be about the same. I measured the short circuit power and got 8.5 watts (this is the copper loss), which is fairly close to the 9.5 watt core loss.

    Your measurement for the open circuit power from the high voltage side is 2.625 watts and 8.9 watts from the low voltage side. This can't be; you have made a measurement error. Note that your copper loss is 9.9 watts, and this is almost the same as the 8.9 watts you measured open circuit from the low voltage side; these two numbers are believable.

    It's just not possible for the two open circuit power measurements to be as different as what you have measured, and I believe that the 2.625 watt figure is the one which is in error. If you use that number for your calculations, you will get an efficiency that is too high.
     
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  12. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ok, I think I know where I went wrong.
    In the measurements for some of my tests, I increased the windings around my watt meter to increase the current reading (it was to low to measure using my equipment). I though I had the correct number of turns used in my measurements... but this must be the cause of error.

    thanks for all your help.
     
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