Single-Phase Transformer No-Load Test & Iron Loss Separation

Introduction to Transformer No-Load Testing

The no-load test of a transformer is performed by feeding one of its windings with rated voltage and frequency, while the other winding is open-circuited. This test provides the value of iron losses and the no-load current (I₀), allowing for the display of its waveform and observation of its characteristic bell shape.

Test Objectives and Fundamentals

Understanding Iron Losses

The power absorbed by a transformer operating under no-load (or vacuum) conditions primarily represents the iron losses, as copper losses are practically negligible due to the small no-load current. Iron losses in a transformer are composed of two main components:

• P0: Power absorbed under no-load conditions (total iron losses)
• PFe: Total iron losses
• PH: Hysteresis losses
• PF: Eddy current losses

The formulas for these losses are:

PH = kH · Bmx · f

PF = kF · Bm2 · f2

Where:

• kH and kF are constants.
• Bm = Maximum magnetic induction.
• f = Frequency.
• Note: The exponent 'x' for B_m in hysteresis loss typically ranges from 1.6 to 2. Here, we assume it's implicitly handled by k_H for simplicity in the separation method.

Therefore, total iron losses can be expressed as:

PFe = PH + PF = kH · f + kF · f2 (assuming B_m is constant)

Separating Hysteresis and Eddy Current Losses

To separate these losses, the transformer, operating under no-load conditions, will be supplied with two different frequencies (e.g., 40 Hz and 60 Hz) to create a system of equations with two unknowns (k_H and k_F):

PFe(40) = kH · 40 + kF · 402 (Power absorbed at 40 Hz)

PFe(60) = kH · 60 + kF · 602 (Power absorbed at 60 Hz)

Once kH and kF are determined from these equations, the hysteresis losses (P_H) and eddy current losses (P_F) at the nominal frequency (f_n) can be calculated:

PH(fn) = kH · fn

PF(fn) = kF · fn2

The sum of these two components will yield the total iron losses (P_Fe(fn)) at the nominal frequency, which corresponds to the power absorbed during the no-load test at that frequency.

Maintaining Constant Magnetic Induction

To ensure the maximum magnetic induction (B_m) remains constant at different frequencies, the supply voltage must be adjusted proportionally. This is based on the transformer EMF equation:

E1 = 4.44 · f1 · N1 · S · Bm

Where:

• E1: Induced EMF in the primary winding
• N1: Number of turns in the primary winding
• f1: Supply frequency
• S: Cross-sectional area of the magnetic core
• Bm: Maximum magnetic induction

If Bm is constant, then the ratio E1 / f1 must be constant. Assuming the supply voltage U is approximately equal to the induced EMF E1, for constant induction, the following relationship must be maintained:

U / f = constant

Therefore, if Un is the nominal voltage at nominal frequency fn (e.g., 50 Hz), then:

Un / fn = U60 / 60 = U40 / 40

Where:

• Un: Nominal voltage
• fn: Rated frequency (e.g., 50 Hz)
• U60: Supply voltage at 60 Hz
• U40: Supply voltage at 40 Hz

An oscilloscope is used to display the waveform of the no-load current (I₀), which is typically non-sinusoidal due to magnetic saturation.

Required Equipment for No-Load Test

• A single-phase transformer
• A true RMS wattmeter (essential for accurate power measurement with highly reactive no-load current)
• An ammeter
• A voltmeter
• An oscilloscope
• A frequency counter
• A variable voltage and frequency power source