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Thread: Calculating voltage drop across transormer using P.U. method

  1. #1
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    Calculating voltage drop across transormer using P.U. method

    I am interested in how you would calculate the voltage drop across a transformer resulting from a motor inrush current due to the transforers impedance using the P.U. method. What if for example I have the following hypothetical setup.

    Inside my plant on the secondary of the utility transformer I have a 4.16kV-480V 1500kVA transformer with a 5% impedance. On the secondary of this transformer I have a 300hp motor and assume that this 300hp is equivelent to about 300kVA. Lets assume that the starting current for this motor is 2800A.

    I know that I must convert everything to a p.u. impedance. Since I am interested in the VD across the transformer do I simply use .05 as the P.U. impedance or must I covert this to a different kVA base since motor and transformer are two different kVA bases?

    Can someone please help me with how I would go about this?

  2. #2
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    mull982:

    As an approximation.

    As seen on the secondary side full load current is 1,500,000 / 480 = 3125. Thus, 5*2800/3125 = 4.5% is the approximate voltage drop. That is internal to the transformer voltage drop. The, output voltage seen across the output terminals may differ substantially from what you would predict from this internal drop.

    .

  3. #3
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    Quote Originally Posted by gar View Post
    100824-1322 EST

    mull982:

    As an approximation.

    As seen on the secondary side full load current is 1,500,000 / 480 = 3125. Thus, 5*2800/3125 = 4.5% is the approximate voltage drop. That is internal to the transformer voltage drop. The, output voltage seen across the output terminals may differ substantially from what you would predict from this internal drop.

    .
    Why are you using the full load current on the secondary? Are you implying that the 5% rating of the transformer means that there will be a 5% voltage drop across the transformer in the circuit when putting out full load secondary current?

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    mull982:

    Yes. See "Alternating-Current Machinery", Bailey and Gault, 1951, McGraw-Hill, page 22. Or any other comparable book. See a useful vector diagram on p 23.

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  5. #5
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    Quote Originally Posted by mull982 View Post
    Why are you using the full load current on the secondary? Are you implying that the 5% rating of the transformer means that there will be a 5% voltage drop across the transformer in the circuit when putting out full load secondary current?
    That follows from the per unit math. Approximate voltage drop through the transformer is per unit amps times per unit impedance ( V=IZ). At full load per unit amps = 100% = 1.0.
    Voltage drop is then (%Z x 100) times 1.0 = .05 x 1 = 5%.

    To find approximate voltage drop at any other load (neglecting power factor) just ratio the current as gar did.


    Another way to look at it: convert the motor inrush current to per unit.

    First select the per unit base: to make it easy, select the trasnformer base: MVA base = 1500 kVA = 1.5 MVA

    Select the voltage level you are looking at, Vbase= 480V.

    Calculate the base current: Ibase = (MVA base /(kV base x 1.732)) x 1,000 = 1.5 x 1000/(.480 x 1,732) = 1804 Amps.

    Using the base current, determine your per unit motor starting current: Istart = 2800/1804 = 1.55

    Voltage drop = Ipu x Zpu = 1.55 x .05 = 0.0775 = 7.75%. ( Either gar or I are confused by root 3).

    Another way. Motor HP = KVA = 300 KVA. Per unit = 300/1500=0.20. I = Ibase x 0.2 = 360A. You said inrush current is 7.75 times FLA, so inrush MVApu = 7.75 x 0.2 = 1.55.

    Vdrop = 1.55 x 5% = 7.75%. (If pu MVA through the transformer is 1.55 the current is 1.55 pu).
    Bob Wilson

  6. #6
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    I was trying to make it as simple as possible and assumed single phase. Maybe I should have stated that.

    .

  7. #7
    What about the transformer X/R ratio vs the motor starting power factor?

  8. #8
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    IslandSparky:

    Current times impedance gives you the voltage drop across said impedance. This voltage drop is internal to the transformer, and has some phase angle relative to the transformer source voltage.

    If you want the transformer terminal voltage then you need to know the load and draw vector diagrams, or do calculations to determine the terminal voltage.

    Even after you do this there may be substantial errors if the load current is non-sinusoidal.

    If the load is resistive on the transformer and there is a lot of leakage flux in the transformer relative to the transformer equivalent series resistance, then the transformer internal voltage drop subtracted from the transformer equivalent source voltage is not a good predictor of the terminal or load voltage.

    If X and R of the transformer are unknown, you have a phase locked oscillator synced to the secondary voltage with no load, and de-sync at the time of load application, then for a short time after load application it should be possible determine the phase shift resulting from the load current and thus determine X and R of the transformer.

    Another way is to used different load resistances, measure load voltage, and solve some simultaneous equations.

    All these methods require a stationary process for the duration of the experiments. If line source voltage is jumping around, then you might need to wait until it is more stable. This morning over one 1 minute period my voltage on a Fluke 27 was min to max of 124.4 to 124.7, a separate 30 second interval it was 124.6 to 124.6, then it changed to 124.3 .

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  9. #9
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    Quote Originally Posted by rcwilson View Post
    That follows from the per unit math. Approximate voltage drop through the transformer is per unit amps times per unit impedance ( V=IZ). At full load per unit amps = 100% = 1.0.
    Voltage drop is then (%Z x 100) times 1.0 = .05 x 1 = 5%.

    To find approximate voltage drop at any other load (neglecting power factor) just ratio the current as gar did.


    Another way to look at it: convert the motor inrush current to per unit.

    First select the per unit base: to make it easy, select the trasnformer base: MVA base = 1500 kVA = 1.5 MVA

    Select the voltage level you are looking at, Vbase= 480V.

    Calculate the base current: Ibase = (MVA base /(kV base x 1.732)) x 1,000 = 1.5 x 1000/(.480 x 1,732) = 1804 Amps.

    Using the base current, determine your per unit motor starting current: Istart = 2800/1804 = 1.55

    Voltage drop = Ipu x Zpu = 1.55 x .05 = 0.0775 = 7.75%. ( Either gar or I are confused by root 3).

    Another way. Motor HP = KVA = 300 KVA. Per unit = 300/1500=0.20. I = Ibase x 0.2 = 360A. You said inrush current is 7.75 times FLA, so inrush MVApu = 7.75 x 0.2 = 1.55.

    Vdrop = 1.55 x 5% = 7.75%. (If pu MVA through the transformer is 1.55 the current is 1.55 pu).
    O.k. so using this 7.75% as an aproximate how to I translate this into an actual voltage? If the entire circuit was at 480V I know I would take 7.75% of 480V to arrive at the voltage drop but since the primary of the transformer is 4.16kV I'm not sure. Do I still simply take 7.75% of 480V to aproximate the VD across the transformer, and thus the voltage on the seconary terminals of the transformer?

  10. #10
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    Quote Originally Posted by rcwilson View Post
    Select the voltage level you are looking at, Vbase= 480V.
    I may have found the answer to my own question. Since we are interested on the 480V side and the resulting voltage on the 480V side is this why we choose 480V as our base voltage?

    If we were interested in something on the 4.16kV side would have we chosen 4.16kV as our base voltage?

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