OPEN CIRCUIT TEST ON TRANSFORMER

In the previous article, we have seen that there are four tests of the transformer.

1) Polarity test

2) Open circuit test

3) Short circuit test

4) Sumpner’s test or back to back test

In this article, we will explain the open circuit test of the transformer.

Open circuit test should be conducted at rated flux in the transformer core. The rated flux can be achieved by keeping other winding terminals open and keeping the existing transformer at rated voltage and frequency.

The objectives of open circuit test are:

a) To find the constant losses of the transformer.

b) To find shunt branch parameters of equivalent circuit R₀ and X₀.

c) To separate iron losses and find hysteresis loss and eddy current loss.

This test is conducted on the LV side of the transformer because at LV side low range volt meter and watt meter are sufficient to conduct the test. Another reason is that the magnitude of the no-load current is more on the LV side. So, we can measure no-load current more accurately compared to the HV side. Therefore, in most of the cases, this test is conducted on the LV side of the transformer.

Generally, for this test LPF wattmeter is used because the power factor is at nearly 0.2 lag. The measurements are primary voltage (V₁), no-load current (I₀) and no-load power (W₀). This entire test can be divided into three parts.

1) Find out R₀ and X₀:

Above circuit represent approximate equivalent circuit of the transformer. From this circuit,

R₀ = V₁/I_w

X₀ = V₁/I_u

Where, I_w = core loss current = I₀ cos ф_o

and I_u = magnetizing current = I_u sin ф_o

W₀ = No load power = V₁I₀cosф_o

Cos ф_o = W₀/V₁I₀

From the measurement, we have no-load current (I₀), no-load power (W₀) and voltage (V₁). First, we have to find Iw and Iu. From the above equation, we can find out R₀ and X₀.

2) To find out constant losses:

W₀ = losses in transformer under no load condition

W₀ = Iron loss + dielectric loss + No load primary copper loss

The iron loss and dielectric loss is considered as a constant loss. The dielectric loss is negligible in this case.

Iron loss = W₀ – I₀²R₁

Now, by test measurement, we have a value of W₀ and I₀. But we don’t have the value of R₁. Therefore, the Kelvin double bridge method is used to measure the value of resistance R₁. In this condition, we neglect the primary copper loss and dielectric loss. so, watt meter reading during open circuit test is approximately considered as an iron loss.

3) Separation of iron losses: The iron loss can be divided into two parts; the hysteresis loss and eddy current loss. This test should be conducted with variable frequency and voltage, in such a way that the ratio of voltage and frequency (V₁/f) remains constant.

The first thing to do here is to apply rated voltage and rated frequency and note down the reading of watt meter and frequency meter. After that, apply to reduce voltage and set frequency in such a way that V₁/f ratio remains same and note down the reading of wattmeter and frequency meter.

By repeating this procedure with changes in voltage and frequency, we can draw a graph for (W_i/f) vs frequency. The find constant A and B, where A is the y-intercept and B is the slope of the curve.

Now from this you can calculate the hysteresis loss and eddy current loss at rated frequency and voltage by below equations.

Hysteresis loss at rated frequency = A x rated frequency

Eddy current loss at rated frequency = B x square of the rated frequency.

W_i = Af + Bf²

W_i/f = A + Bf

m = slop of straight line = tan θ = B