Homework help about transformers

Discussion in 'Homework Help' started by shadow3, Sep 23, 2013.

  1. shadow3

    Thread Starter New Member

    Sep 15, 2013
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    **A single phase 50 kVA, 2400/240 volt, 60 Hz distribution transformer is used as a stepdown transformer at the load end of a 2400-volt feeder whose series impedance is 1+j2 ohms. The equivalent series impedance of the transformer is (1+j2.5) ohms referred to the high voltage (primary) side. The transformer is delivering rated load at 0.8 pf lag and at rated secondary voltage. Neglecting the transformer exciting current, deternmine
    a.) the voltage at the transformer primary terminals
    b.) voltage at the sending end of the feeder
    c.) P and Q delivered to the sending end of the feeder..

    I need help on this one :)

    1.) What does "The transformer is delivering rated load at 0.8 pf lag and at rated secondary voltage." exactly mean? I understand the pf lag but by "at rated secondary voltage", does it mean that E2 (voltage across the secondary coil is 240? (which in turn tells me that E1 is 2400)

    Here are my drawings:
    transformer showing both primary and secondary: http://postimg.org/image/5pl47mhq9/
    transformer with low referred to high: http://postimg.org/image/i6j4rm4v5/

    Are my diagrams drawn as they were supposed to be? Also, are my planned steps correct? First, I will use the pf to solve for current phasor that flows through the load. I can use the fact that the current lags the voltage load by 36.87 degrees, and I'll just set one voltage (maybe VL) as the phasor reference voltage... After this, I can easily solve for the primary voltages and others..Also, about the power supplied at the sending end of the line, does it mean the power delivered to 1+j2? thanks.... Please clear the "sending end of the feeder" confusion...
     
  2. shadow3

    Thread Starter New Member

    Sep 15, 2013
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    0
    Actually, when I referred the low voltage to high voltage, then that must mean that the impedance of the load is actually already in the 1+j2.5? If that's the case, then if I let ZL be the impedance of load (without any side being referred yet), then I can solve for ZL such that [((2400/240)^2)*ZL] + 1 + j2 = 1 + j2.5, giving me that ZL is just an inductor...
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,140
    3,054
    That would be my assumption. There's nothing else given in the question that would support any other choice. Just my opinion.
     
  4. misterrobo

    New Member

    Sep 24, 2013
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    That's how I see it. And no, the secondary voltage is not 240 V. The feeder voltage is 2400 V, right?

    You can get it by finding the rated load. Rated load is the maximum power output the transformer was designed to handle.

    Read more: http://www.physicsforums.com

    Oh hell week.
     
    Last edited: Sep 24, 2013
  5. misterrobo

    New Member

    Sep 24, 2013
    2
    0
    Sorry, the secondary voltage is indeed 240.
     
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