Efficiency of Single Phase Transformers | Electrical Machines - Electrical Engineering (EE) PDF Download

Introduction

The performance of a single phase transformer, from a user point of view, is characterised by several parameters. One of the most important of these is the efficiency, denoted here by the Greek letter ν. Efficiency is defined as the ratio of useful output power to input power.

ν = output power / input power

Since input power equals output power plus losses, efficiency may also be written as:

ν = output power / (output power + total losses)

Losses in a Single Phase Transformer

Transformer losses can be divided into two main categories:

• Core (iron) loss - loss in the core due to hysteresis and eddy currents; represented in the equivalent circuit by a resistance R0. Core loss depends on applied voltage and frequency and is effectively constant for a transformer connected to a nominal, fixed-voltage mains supply.
• Copper (winding) loss - loss in the windings due to their resistance; equals I2² Req (or I2'² R'eq when referred to one side). Copper loss varies with load current.

Expression for Output, Core and Copper Losses

The active output (real) power delivered to a load with current I2 and power factor angle φ is

Pout = V2 I2 cos φ

The core (iron) loss is given by

Pcore = V1² / R0

The copper loss (referred to the same side) is

Pcu = I2'² Req = I2² R'eq

For a practical transformer the variation of V2' with I2' is usually very small and the load power factor may be assumed constant for many calculations. Thus the efficiency depends mainly on the load current magnitude.

Condition for Maximum Efficiency

With V2 and cos φ assumed constant, efficiency can be expressed as

ν = V2 I2 cos φ / (V2 I2 cos φ + Pcore + I2² Req)

To find the value of load current I2 at which ν is maximum, differentiate the denominator (or ν) with respect to I2 and set the derivative to zero. This yields the condition that at maximum efficiency the variable copper loss equals the fixed core loss.

Pcore = I2² Req = Pcu (at that load)

Thus the load current at which maximum efficiency occurs is the one for which copper loss equals core loss.

Practical Use of Test Data to Obtain Losses

Open-circuit (no-load) test performed on the low-voltage side yields the core loss (wattmeter reading) and the no-load current. Short-circuit test performed (typically on the high-voltage side) yields the full-load copper loss (wattmeter reading at rated current). Therefore the two tests together provide Pcore and Pcu (full-load copper loss).

Fractional Load Formulation

Let x denote the fractional load (ratio of actual load KVA to rated KVA). The copper loss varies as x² of the full-load copper loss. If Pcu denotes full-load copper loss and P i denotes core loss, then the efficiency at fractional load x and power factor cos φ is

ν(x) = x × (rated KVA) × cos φ / ( x × (rated KVA) × cos φ + P i + x² Pcu )

The fractional load x at which the efficiency is maximum satisfies

x = √(P i / Pcu)

This result is independent of the power factor; it determines the loading (as a fraction of rated load) for maximum efficiency.

Design Implication: Transmission vs Distribution Transformers

When designing transformers one usually tries to make the maximum-efficiency loading coincide with the typical operating loading. For transmission transformers, which are often operated close to full load, the design aim is to have maximum efficiency near full load. Distribution transformers, however, spend much of the day lightly loaded or on no-load. For distribution transformers the maximum-efficiency point is therefore often designed to occur at a reduced load (typically around 60-65% of rated load).

All-Day Efficiency (Energy Efficiency over a Day)

Since transformer loading varies over the day, a single operating-point efficiency does not capture the overall performance. The all-day efficiency is defined as the ratio of total energy delivered by the transformer in a day to the total energy input to the transformer in that day:

All-day efficiency = (Energy output in 24 h) / (Energy input in 24 h)

If the transformer experiences different load levels x1, x2, ... each for specified durations, the energy output is the sum of (KVArated × xi × cos φi × hours_i). Total energy loss is the sum of iron loss for 24 h plus copper losses for each interval (xi² Pcu × hours_i). Then

Eout = Σ (xi × KVArated × cos φi × hours_i)

Eloss = 24 × P i + Σ (xi² × Pcu × hours_i)

ν_all-day = Eout / (Eout + Eloss)

Numerical Illustration of All-Day Efficiency

Consider a transformer with full-load copper loss Pcu and core loss P i. Suppose the daily loading pattern is:

• 25% loading (x = 0.25) for 6 hours, power factor 0.8
• 85% loading (x = 0.85) for 12 hours, power factor 0.9
• 100% loading (x = 1.0) for 2.4 hours, power factor 0.95
• No-load for 3.6 hours

The energy output over the day (kWh) for a transformer of rated KVA is

Eout = 0.25 × KVArated × 0.8 × 6 + 0.85 × KVArated × 0.9 × 12 + 1.0 × KVArated × 0.95 × 2.4 + 0 × KVArated × 3.6

Total energy loss is

Eloss = 24 × P i + (0.25)² × Pcu × 6 + (0.85)² × Pcu × 12 + (1.0)² × Pcu × 2.4

Then the all-day efficiency is

ν_all-day = Eout / (Eout + Eloss)

The only inputs required are the full-load copper loss Pcu (from short-circuit test), the core loss P i (from open-circuit test) and the loading pattern with power factors.

Temperature Rise and Need for Combined Loss Testing

Transformer temperature rise in service depends on the total loss (core + copper) dissipated simultaneously. Open-circuit test subjects the transformer primarily to core loss; short-circuit test subjects it primarily to copper loss. In neither test are both losses present at rated proportions. Therefore a special test is required to simulate simultaneous iron and copper losses and to measure temperature rise under realistic conditions. This is the back-to-back (Sumpner's) test.

Back-to-Back (Sumpner's) Test - Principle and Procedure

Two (preferably identical) transformers are connected so that their primaries are in parallel to a supply and their secondaries are connected in series with polarity such that voltages oppose. A variable source is applied across the series-connected secondaries to circulate the required load current.

When only the primary supply is applied, the two transformers draw essentially the combined no-load currents and the measured power equals combined core losses.

When the secondary variable source alone is applied (primaries shorted), the circulating currents produce copper heating and the measured power equals combined copper losses.

When both supplies are applied with the secondaries driven to produce the desired load current, the transformers experience both core and copper losses simultaneously and the measured total power equals the sum of core and copper losses. The temperature rise under these conditions is the same as in actual service. Since the power to the variable secondary source is mostly circulating between the two transformers, the test allows simulation of full-load conditions without drawing large real power from the mains.

Worked Example: Finding Equivalent Circuit Parameters and Efficiency

Given: single-phase transformer rated 200 V / 400 V, 4 kVA.

No-load test (performed on LV side): applied V = 200 V, I0 = 0.7 A, wattmeter reading W0 = 35 W.

Short-circuit test (performed on HV side): applied Vsc = 30 V, Isc = 10 A, wattmeter reading Wsc = 90 W.

Find the approximate equivalent circuit parameters referred appropriately and express some in per unit.

Open-circuit (no-load) branch - shunt parameters (referred to LV side)

The no-load power factor cos φ0 is

cos φ0 = W0 / (VLV × I0) = 35 / (200 × 0.7)

The magnetizing and core-loss component currents are

Im = I0 sin φ0

Ic = I0 cos φ0

Therefore

Xm = VLV / Im

R0 = VLV / Ic

Applying the given numbers:

cos φ0 = 35 / (200 × 0.7) = 35 / 140 = 0.25

I0 = 0.7 A

Ic = I0 cos φ0 = 0.7 × 0.25 = 0.175 A

Im = √(I0² - Ic²) = 0.7 × sin φ0 = 0.7 × √(1 - 0.25²) ≈ 0.679 A

Xm = VLV / Im = 200 / 0.679 ≈ 294.5 Ω

R0 = VLV / Ic = 200 / 0.175 ≈ 1142.9 Ω ≈ 1.143 kΩ

Both R0 and Xm are referred to the LV side because the open-circuit test was performed on LV side.

Per-unit values on LV base

Base impedance on LV side:

Zbase_LV = VLV_rated² / Sbase = 200² / 4000 = 40,000 / 4000 = 10 Ω

Therefore

R0_p.u. = R0 / Zbase_LV = 1142.9 / 10 = 114.29 p.u.

Xm_p.u. = Xm / Zbase_LV = 294.5 / 10 = 29.45 p.u.

Short-circuit branch - series parameters (referred to HV side)

On short-circuit the shunt branch may be neglected and the equivalent series impedance magnitude is

Zsc = Vsc / Isc = 30 / 10 = 3.0 Ω

Short-circuit power gives the copper loss at the test current:

Psc = Isc² Req = 90 W

Hence

Req = Psc / Isc² = 90 / 10² = 0.9 Ω

Then reactance

Xeq = √(Zsc² - Req²) = √(3.0² - 0.9²) ≈ √(9 - 0.81) = √8.19 ≈ 2.862 Ω

Both Req and Xeq are referred to the HV side because the short-circuit test was performed on the HV side.

Per-unit values on HV base

Base impedance on HV side:

Zbase_HV = VHV_rated² / Sbase = 400² / 4000 = 160,000 / 4000 = 40 Ω

Therefore

Req_p.u. = Req / Zbase_HV = 0.9 / 40 = 0.0225 p.u.

Xeq_p.u. = Xeq / Zbase_HV = 2.862 / 40 = 0.0716 p.u.

Equivalent Circuit Summary

The approximate per-unit equivalent circuit therefore has large shunt branch values (R0_p.u., Xm_p.u.) on the LV side and series branch values (Req_p.u., jXeq_p.u.) on the HV side. To use a single-sided per-unit model the referred values should be converted to the same side if required for further calculations.

Example: Efficiency Calculations for the Given Transformer

Given: rated KVA = 4 kVA. From tests we have core loss P i = 35 W and full-load copper loss Pcu = 90 W.

Full-load efficiency at 0.8 power factor

Output power at full load: Pout = 4,000 × 0.8 = 3,200 W

Total losses at full load: Ploss = P i + Pcu = 35 + 90 = 125 W

Therefore

ν_full = 3200 / (3200 + 125) = 3200 / 3325 ≈ 0.9623 = 96.23%

Half-load efficiency at the same power factor (0.8)

Output at half load: Pout = 0.5 × 4,000 × 0.8 = 1,600 W

Copper loss at half load: (0.5)² × 90 = 0.25 × 90 = 22.5 W

Total losses at half load: Ploss = P i + 22.5 = 35 + 22.5 = 57.5 W

Therefore

ν_half = 1600 / (1600 + 57.5) = 1600 / 1657.5 ≈ 0.9653 = 96.53%

Load for Maximum Efficiency

From the condition for maximum efficiency x = √(P i / Pcu)

Substitute P i = 35 W and Pcu = 90 W:

x = √(35 / 90) ≈ √0.3889 ≈ 0.624

So maximum efficiency occurs at about 62.4% of rated load. This is consistent with the typical design objective for distribution transformers where maximum efficiency is desired at reduced load.

Graphical Note on Efficiency vs Load

Efficiency curves plotted against percentage load for different power factors have peaks at the same fractional load x determined by √(P i / Pcu). The absolute value of maximum efficiency depends on the power factor, but the loading at which the peak occurs is independent of power factor.

Concluding Remarks

Key practical points:

• Open-circuit and short-circuit tests provide the core and full-load copper losses respectively, which are sufficient to compute efficiency at any load and the all-day efficiency when a load profile is known.
• The load at which maximum efficiency occurs is given by the square root relation x = √(P i / Pcu) and does not depend on power factor.
• Back-to-back (Sumpner's) test is required to measure temperature rise under combined core and copper losses in realistic proportions.
• Transformer design (transmission vs distribution) targets different points of maximum efficiency according to typical loading patterns.