Losses, Efficiency and Performance Prediction

Introduction

Electrical machines do not convert energy perfectly. Part of the input power becomes heat, vibration, acoustic noise and other non-useful effects. These losses determine efficiency, temperature rise, insulation life, operating cost and the machine’s compliance with energy-efficiency regulations.

Performance prediction combines electromagnetic equivalent circuits, empirical loss models, standardized tests and thermal constraints. A credible prediction must state its operating point—voltage, frequency, load, power factor, speed, waveform and temperature—because machine losses are not universally constant.

Learning Objectives

After studying this chapter, students should be able to:

• calculate copper, core, mechanical, stray-load and harmonic losses;
• construct loss budgets and loss-representing equivalent circuits;
• predict efficiency, voltage regulation, torque and power factor;
• interpret transformer and rotating-machine test data;
• determine conditions for maximum efficiency;
• account for nonsinusoidal supply effects;
• apply efficiency standards and lifecycle-costing methods; and
• recognize uncertainty and limitations in performance predictions.

1. Power Flow and Loss Classification

For a motor, electrical input power is converted into mechanical output power. For a generator, the direction is reversed. In either case, conservation of energy requires

where P_in is input power, P_out is useful output power and P_loss is total loss, all in watts.

The principal loss categories are summarized below.

A loss budget lists each loss at a specified operating condition. It should not combine measured, assumed and calculated quantities without identifying their origins.

2. Copper Losses

2.1 Direct-current resistance loss

For a winding carrying direct current, or sinusoidal current when resistance is interpreted as effective AC resistance,

Here m is the number of phases, I_ph is phase RMS current in amperes, and R_ph is phase resistance in ohms at the operating temperature. For a balanced three-phase winding, m=3.

Equivalent line-based formulas depend on the connection. For example, in a balanced star connection I_L=I_ph, whereas in delta I_L=√3I_ph. The phase formula avoids ambiguity.

Winding resistance changes with temperature. Over the normal operating range, an engineering approximation is

where θ₁ and θ₂ are conductor temperatures in degrees Celsius. For copper, k≈234.5°C; for aluminium, a commonly used value is approximately 225°C. Values required by the applicable test standard should be used in compliance work.

Example—temperature correction. A copper phase winding has a resistance of 0.40 Ω at 20°C. At 100°C,

At 30 A per phase, the three-phase copper loss is

Using the cold resistance would underestimate the loss by approximately 24%.

2.2 AC resistance effects

Skin effect and proximity effect make AC resistance exceed DC resistance:

The factor k_ac depends on conductor dimensions, frequency, strand arrangement, slot leakage field and transposition. It becomes especially important in large conductors, high-frequency machines and inverter-fed windings. Finite-element analysis or validated empirical factors may be required where current distribution is strongly nonuniform.

3. Core Loss

Alternating flux produces hysteresis and eddy-current losses. A frequently used empirical separation is

where f is frequency in hertz, B_pk is peak flux density in teslas, V_c is active core volume, and k_h, k_e, k_ex and n are material-dependent coefficients. The final term represents excess or anomalous loss in one commonly used model. Coefficients must come from measurements over the relevant flux-density and frequency range; this equation is not a universal physical law.

For sinusoidal transformer excitation, peak flux is approximately proportional to V/f. Consequently, maintaining constant V/f approximately maintains flux. Overvoltage or underfrequency can cause saturation, sharply increasing magnetizing current, loss and noise.

Manufacturers’ Epstein-frame or single-sheet data may underpredict assembled-core loss because joints, punching, residual stress, burrs and rotational flux introduce additional losses. A design allowance is therefore normally required.

4. Mechanical, Stray-Load and Harmonic Losses

4.1 Mechanical loss

Mechanical loss is usually written as

Bearing and brush losses depend on load, lubrication and speed. Windage rises strongly with speed and fan geometry. A single constant-loss assumption may be reasonable over a narrow speed range but is inappropriate for wide-speed drives.

4.2 Stray-load loss

Stray-load loss includes load-dependent losses not captured by ordinary winding and core-loss calculations. Causes include leakage-field-induced eddy currents in conductors, structural parts and the core; tooth pulsations; and local current redistribution. Standards prescribe test or assigned-loss procedures because direct measurement is difficult.

For preliminary studies, it is sometimes approximated by

where x is per-unit load current and P_stray,r is rated-load stray loss. This is an engineering approximation, not a general law.

4.3 Harmonic losses

For an RMS current spectrum, winding loss can be estimated as

where I_h and R_ac,h are RMS current and effective resistance at harmonic order h. Because AC resistance generally rises with frequency, using total RMS current with fundamental-frequency resistance can underestimate loss.

Voltage harmonics also increase core loss, while harmonic air-gap fields can cause rotor heating, torque ripple, acoustic noise and parasitic torques. Inverter-fed motors may therefore require harmonic loss allowances, filters, derating or designs intended for converter duty.

If supply voltage is nearly sinusoidal, total power factor may be expressed as

where DPF is fundamental displacement power factor, I₁ is fundamental RMS current and I_rms is total RMS current. This special relation should not be applied directly when voltage distortion is substantial.

5. Transformer Loss and Performance Model

A transformer’s core loss is represented by resistance R_c, and magnetizing current by reactance X_m. Referred winding resistance and leakage reactance are represented by R_eq and X_eq. In the common approximate equivalent circuit, the shunt branch is moved to the supply terminals.

Figure: Approximate transformer equivalent circuit with the core-loss and magnetizing branches at the supply terminals and a finite load impedance at the output.

With quantities referred to one side, the input current is approximately

The referred load-path current satisfies

These equations correspond to the illustrated input-shunt approximation. In the exact equivalent circuit, primary impedance remains ahead of the exciting branch, and the ideal transformer is retained explicitly.

5.1 Transformer efficiency

At load fraction x and output power factor cosφ,

where S_r is rated apparent power, P₀ is approximately constant no-load loss, P_cu,r is rated copper loss, and P_other contains any separately included losses.

If constant loss is P_c and variable loss is exactly x²P_v,r, differentiation shows that efficiency is maximum when

Thus,

This condition applies only to the stated constant-plus-quadratic loss model and fixed power factor.

5.2 Voltage regulation

Transformer voltage regulation is

where V_NL and V_FL are secondary no-load and full-load terminal voltages at unchanged primary voltage.

For small series voltage drops, approximate per-unit regulation is

The plus sign applies to lagging current and the minus sign to leading current under the usual load-power-factor convention. The approximation becomes less accurate when series drops are large.

Worked Example 1—Transformer efficiency and regulation

A 50 kVA transformer has a no-load loss of 350 W and rated copper loss of 900 W. It supplies 80% rated kVA at 0.90 lagging power factor.

The output power is

Copper loss is

Therefore,

Maximum efficiency occurs at

If R_eq=0.012 pu and X_eq=0.040 pu, approximate full-load regulation at 0.8 lagging power factor is

Thus regulation is approximately 3.36%. At 0.8 leading power factor, it is approximately 0.0096-0.024=-0.0144, or −1.44%.

6. Induction-Motor Equivalent Circuit and Power Flow

In the approximate input-shunt induction-motor circuit, the magnetizing branch is moved to the supply terminals. Stator and rotor quantities are per phase, with rotor parameters referred to the stator.

Figure: Approximate per-phase induction-motor circuit. The input magnetizing branch is consistent with calculations based directly on phase voltage.

For this approximation,

and

The input current is I₁≈I₀+I₂'. Slip is

where n_s and n_r are synchronous and rotor speed in revolutions per minute, and p is the number of poles.

Air-gap power, rotor copper loss and converted mechanical power are

The electromagnetic torque is

where ω_s and ω_m are synchronous and rotor mechanical angular speeds in rad/s. Shaft output subtracts mechanical and other rotational losses from P_conv.

Worked Example 2—Induction-motor performance

A 400 V, three-phase, star-connected, 50 Hz, four-pole motor has per-phase parameters R₁=0.40Ω, X₁=0.80Ω, R₂'=0.30Ω, X₂'=0.60Ω, X_m=15Ω and R_c=120Ω. At slip 0.04,

The series impedance is

Therefore,

so |I₂'|=28.78A. The exciting current is

Thus I₁=30.27-j20.42A, giving |I₁|≈36.51A and input power factor approximately 0.829 lagging.

The air-gap power is

Rotor copper loss is 0.04(18.65)=0.746kW, and converted mechanical power is approximately 17.90 kW. If mechanical and stray-load losses total 650 W, shaft output is 17.25 kW.

Input power is approximately 20.97 kW, including stator copper and modeled core loss. Hence,

Synchronous speed is 1500 r/min, so

This result is based on the input-shunt approximate circuit; an exact circuit or test-calibrated model is preferred for precision design.

7. Other Machine Performance Relations

For a cylindrical-rotor synchronous machine with negligible armature resistance, balanced sinusoidal operation and per-phase quantities,

where V is terminal phase voltage, E is internal phase EMF, X_s is synchronous reactance and δ is the torque angle. Salient-pole machines require an additional reluctance-power term.

For a DC machine,

where Φ is pole flux, ω is angular speed and I_a is armature current. In SI-consistent formulations, the numerical torque and EMF constants are equal when defined compatibly. Brush drop, armature reaction, saturation and commutation limits must also be considered.

8. Measurement and Performance Prediction

Common standardized tests include:

• No-load test: estimates core, friction and windage losses.
• Blocked-rotor or short-circuit test: estimates series impedance and load loss.
• DC resistance test: establishes winding resistance, followed by temperature correction.
• Direct input–output test: determines efficiency from measured powers.
• Calorimetric test: determines losses from heat flow, useful for high-efficiency machines where input and output powers are close.
• Retardation test: supports separation of rotational losses in some rotating machines.

Efficiency found by subtracting two large measured powers can have substantial relative uncertainty in the small loss difference. If P_loss=P_in-P_out and the measurements are independent, approximate standard uncertainty is

Test reports should state instrument accuracy, temperature, waveform quality, stabilization time and the applicable standard.

Duty-cycle energy efficiency

Peak efficiency alone may not minimize energy consumption. For operating intervals i,

where t_i is interval duration in hours, producing energy in watt-hours or kilowatt-hours when consistent power units are used. Transformers with long lightly loaded periods may be assessed using all-day efficiency, while variable-speed motors require efficiency maps over torque and speed.

9. Detailed Application: Lifecycle Selection of an Industrial Motor

Illustrative case. A water utility must choose between two 75 kW motors. Both satisfy torque, starting and thermal requirements. The pump normally requires 60 kW shaft output for 6000 h per year.

Background and problem

Motor A has 94.5% efficiency at the operating point. Motor B has 96.0% efficiency but costs $2600 more. Electricity costs $0.12/kWh. The project life is 15 years, the nominal discount rate is 7% and electricity-price escalation is 2% per year.

Analysis

Input powers are

Annual energy saving is

First-year monetary saving is approximately

Simple payback is 2600/714≈3.64 years. Simple payback ignores savings after recovery and the time value of money.

The present value of an escalating annual saving is

where r=0.07, g=0.02 and N=15. This formula assumes the first saving occurs after one year and r≠g. It gives a present value of approximately $7320. The incremental net present value is therefore about $4720.

At an illustrative grid-emission factor of 0.40 kg CO₂e/kWh, annual avoided emissions are approximately 2.38 t CO₂e. Actual reporting must use an approved location- or market-based factor.

Approach and outcome

Motor B is economically preferable if reliability, drive compatibility and maintenance performance are comparable. Procurement should specify efficiency at the actual operating point rather than relying only on rated-load class. System-level checks should also confirm that the motor is correctly sized and that pump throttling cannot be replaced by efficient speed control.

Learning points

• Small efficiency differences can produce material lifecycle savings at high utilization.
• Rated efficiency does not replace duty-cycle analysis.
• The motor, converter and driven load should be optimized as a system.
• Discounted cash flow is more informative than purchase price or simple payback alone.

10. Regulation, Governance and Contemporary Practice

IEC 60034-30-1 defines efficiency classes for covered line-operated AC motors, while IEC 60034-2-1 provides relevant loss and efficiency test methods. Transformer and motor minimum-efficiency requirements also exist under national and regional legislation, including US Department of Energy and European ecodesign frameworks. Scope, dates, power ranges and exemptions change; engineers must consult the current legally applicable edition rather than relying on a class label alone.

Good professional practice includes:

• specifying the test standard and efficiency determination method;
• preventing selective use of favorable test tolerances;
• retaining traceable calibration and test records;
• considering repair effects on efficiency;
• evaluating embodied materials, operational energy and end-of-life recovery; and
• avoiding misleading environmental claims based only on nameplate efficiency.

Digital twins and data-driven models can update loss estimates from voltage, current, temperature, vibration and speed measurements. Artificial intelligence may help detect bearing degradation or abnormal harmonic loss, but predictions require validated sensors, representative training data and engineering oversight. A statistical warning is not a substitute for insulation, thermal or mechanical protection.

11. Design Challenges and Limitations

Equivalent circuits use lumped parameters even though saturation, temperature and flux distribution are spatially nonuniform. Core-loss coefficients derived from simple alternating-flux tests may not represent rotational flux. Mechanical losses change with lubrication and ventilation, while converter switching introduces frequency-dependent effects.

Best practice is therefore to combine analytical models, finite-element analysis, thermal-network calculations, prototype tests and uncertainty analysis. Model complexity should match the decision: preliminary sizing may tolerate empirical factors, whereas regulatory certification and thermal-limit verification require standardized, traceable methods.

Chapter Summary

Machine performance prediction begins with a complete power balance. Copper loss depends on RMS current, effective resistance and operating temperature. Core loss arises principally from hysteresis, eddy-current and excess-loss mechanisms, while mechanical, stray-load and harmonic losses require speed-, load- and waveform-sensitive treatment.

Transformer and induction-motor equivalent circuits represent loss mechanisms through resistive elements. When the exciting branch is moved to the supply terminals, circuit diagrams and equations must consistently use that approximation. Efficiency, regulation, torque and power factor can then be predicted from circuit quantities, subject to model limitations.

Maximum efficiency occurs when variable and constant losses are equal only under a constant-plus-quadratic loss model. Practical selection should use duty-cycle energy, standards-compliant test data, uncertainty evaluation and discounted lifecycle cost rather than purchase price alone.

Key Terms

AC resistance: Effective resistance under alternating current, including skin and proximity effects.

Air-gap power: Electromagnetic power transferred from a stator field to an induction-motor rotor.

All-day efficiency: Energy-output-to-energy-input ratio over an extended duty cycle.

Core loss: Power dissipated through hysteresis, eddy-current and related magnetic effects.

Copper loss: Resistive winding loss, commonly calculated as phase count times current squared times resistance.

Displacement power factor: Cosine of the phase angle between fundamental voltage and current.

Efficiency: Ratio of useful output power to input power.

Harmonic loss: Additional loss caused by nonsinusoidal voltage, current or air-gap fields.

Lifecycle cost: Total discounted ownership cost, including acquisition, energy, maintenance and disposal.

Loss budget: Structured accounting of machine losses at a stated operating condition.

Mechanical loss: Bearing, brush, seal and windage losses in a rotating machine.

Power factor: Ratio of real power to apparent power.

Rotor copper loss: Resistive loss in an induction-motor rotor, equal to slip times air-gap power.

Slip: Fractional difference between synchronous and rotor speeds.

Stray-load loss: Load-dependent loss not captured by conventional copper and core-loss calculations.

Temperature correction: Adjustment of winding resistance or loss to a reference temperature.

Torque angle: Angular displacement between relevant synchronous-machine voltage or field phasors.

Voltage regulation: Change in terminal voltage between specified load conditions at unchanged excitation or supply.

Review Questions

Short Answer Questions

1. Why must winding resistance be corrected to operating temperature?
2. Distinguish core loss from stray-load loss.
3. Define induction-motor slip and air-gap power.
4. Why can total power factor be lower than displacement power factor?
5. What information should accompany a reported efficiency value?

Long Answer Questions

1. Explain the physical origins and operating dependencies of the principal electrical-machine losses.
2. Derive the maximum-efficiency condition for a machine having constant and current-squared losses.
3. Explain the approximate transformer equivalent circuit and its use in predicting efficiency and regulation.
4. Develop the power-flow relationships of an induction motor from stator input to shaft output.
5. Discuss how inverter supplies affect losses, torque, power factor and machine thermal design.

Analytical / Application Questions

1. A 100 kVA transformer has a 500 W no-load loss and 1.6 kW rated copper loss. Calculate its efficiency at 60% load and 0.85 power factor, and determine the load fraction for maximum efficiency.
2. Using a stated set of induction-motor equivalent-circuit parameters, calculate current, power factor, air-gap power, rotor copper loss, torque and efficiency at two slips. Explain the differences.
3. Compare two motors using an annual load profile, electricity-price escalation, discount rate, maintenance cost and carbon factor. Recommend an option and identify the most influential assumptions.

Multiple Choice Questions with Answers

Question 1

A balanced three-phase winding carries 20 A per phase and has an operating-temperature phase resistance of 0.50 Ω. What is its copper loss?

A. 200 W

B. 400 W

C. 600 W

D. 1200 W

Correct Answer: C. 600 W

Question 2

Under a constant-loss plus current-squared variable-loss model, maximum efficiency occurs when:

A. output power equals input power

B. variable loss equals constant loss

C. core loss is zero

D. power factor is unity

Correct Answer: B. variable loss equals constant loss

Question 3

A transformer has 300 W constant loss and 1200 W rated variable loss. At what per-unit load is efficiency maximum?

A. 0.25 pu

B. 0.50 pu

C. 0.75 pu

D. 1.00 pu

Correct Answer: B. 0.50 pu

Question 4

In the transformer equivalent circuit illustrated in the chapter, where is the approximate exciting branch connected?

A. Across the supply terminals

B. In series with the load

C. Across the series reactance only

D. After an output short circuit

Correct Answer: A. Across the supply terminals

Question 5

A four-pole, 50 Hz induction motor runs at 1440 r/min. What is its slip?

A. 0.02

B. 0.04

C. 0.06

D. 0.10

Correct Answer: B. 0.04

Question 6

An induction motor has an air-gap power of 20 kW at a slip of 0.03. What is its rotor copper loss?

A. 0.30 kW

B. 0.60 kW

C. 6.00 kW

D. 19.40 kW

Correct Answer: B. 0.60 kW

Question 7

Which effect most directly causes AC winding resistance to exceed DC resistance?

A. Power-factor correction

B. Voltage regulation

C. Skin and proximity effects

D. Reduced synchronous speed

Correct Answer: C. Skin and proximity effects

Question 8

For a transformer with R_eq=0.01 pu and X_eq=0.05 pu supplying a 0.8 lagging-power-factor load, what is the approximate full-load voltage regulation?

A. −2.2%

B. 0.8%

C. 3.8%

D. 6.0%

Correct Answer: C. 3.8%

Question 9

A motor delivers 45 kW while drawing 50 kW. What is its efficiency?

A. 80%

B. 85%

C. 90%

D. 95%

Correct Answer: C. 90%

Question 10

Why can direct input–output efficiency measurement have significant loss uncertainty in a high-efficiency machine?

A. Input and output powers are both zero

B. Loss is obtained as the small difference between two large measured powers

C. Resistance is independent of temperature

D. Mechanical output cannot be measured

Correct Answer: B. Loss is obtained as the small difference between two large measured powers