Question

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.

Answer 1

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.