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3 February, 19:51

A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nominal frequency. The short-circuit voltage is found as 100 V and the short-circuit current and power are 110 A and 3200 W, respectively. Find the series impedance of the transformer referred to its low voltage side.

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  1. 3 February, 20:46
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    Given Information:

    Primary secondary voltage ratio = 2500/250 V

    Short circuit voltage = Vsc = 100 V

    Short circuit current = Isc = 110 A

    Short circuit power = Psc = 3200 W

    Required Information:

    Series impedance = Zeq = ?

    Answer:

    Series impedance = 0.00264 + j0.00869 Ω

    Step-by-step explanation:

    Short Circuit Test:

    A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.

    The series impedance in polar form is given by

    Zeq = Vsc/Isc < θ

    Where θ is given by

    θ = cos⁻¹ (Psc/Vsc*Isc)

    θ = cos⁻¹ (3200/100*110)

    θ = 73.08°

    Therefore, series impedance in polar form is

    Zeq = 100/110 < 73.08°

    Zeq = 0.909 < 73.08° Ω

    or in rectangular form

    Zeq = 0.264 + j0.869 Ω

    Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq

    Req = 0.264 Ω

    Xeq = j0.869 Ω

    To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.

    Turn ratio = a = Vp/Vs = 2500/250 = 10

    Zeq2 = Zeq/a²

    Zeq2 = (0.264 + j0.869) / 10²

    Zeq2 = (0.264 + j0.869) / 100

    Zeq2 = 0.00264 + j0.00869 Ω

    Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.
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