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    Transformer Question

    I am studdying for the PE exam next week and have a question on one of the practice problems. I have the answer but it is twice as big as i thought it was... Is that because I^2 * R = P so (2)^2 * R = P so its 4 times not twice?

    A power transformer has the following test data at rated voltage:
    % of Name Plate kVA || Total Losses (W)
    0 || 460
    50 || 2370

    The total transformer losses (W) at 100% of nameplate kVA are most nearly:

    A 5200
    B 8100
    C 9480
    D 11320


    Solution

    Poc = 460 W

    P50% = 2370 - 460 = 1910 W
    P100% = 4 * 1910 = 7640 W

    Pt = 7640 + 460 = 8100W

    #2
    Originally posted by spikes2020 View Post
    I am studdying for the PE exam next week and have a question on one of the practice problems. I have the answer but it is twice as big as i thought it was... Is that because I^2 * R = P so (2)^2 * R = P so its 4 times not twice?

    A power transformer has the following test data at rated voltage:
    % of Name Plate kVA || Total Losses (W)
    0 || 460
    50 || 2370

    The total transformer losses (W) at 100% of nameplate kVA are most nearly:

    A 5200
    B 8100
    Ci320


    Solution

    Poc = 460 W

    P50% = 2370 - 460 = 1910 W
    P100% = 4 * 1910 = 7640 W


    Pt = 7640 + 460 = 8100W
    Yes. At half power the transformer will be carrying half of the full load current, since the output voltage can be considered constant.
    But the internal resistance is constant (ignoring temperate coefficient) so the internal losses will increase as the square of the load current.
    There will be a very small correction because the magnetizing current in the primary will cause the full load current to be less than twice the 50% load current, but I am sure the question ignores this. Which is fine if the load has a high PF or that the resistive losses are all in the secondary only.
    Last edited by GoldDigger; 10-16-13, 03:06 PM.

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