Losses and Efficiency in Electric Machines GATE EE Exam Quick Notes

Introduction

Energy Conversion in Electric Machines

  • Electric machines convert energy from one form to another

  • During conversion, some energy is always lost as heat

  • Efficiency indicates how well a machine converts energy

  • Understanding losses is crucial for:

    • Machine design optimization

    • Performance analysis

    • Economic operation

Types of Losses

Classification of Losses

  • Total Losses = Fixed Losses + Variable Losses

  • Fixed Losses = Core Losses + Mechanical Losses

  • Variable Losses = Copper Losses + Stray Losses

Key Point for GATE:

  • Fixed losses remain constant with load

  • Variable losses change with load (usually proportional to \(I^2\))

Fixed Losses

Core Losses (Iron Losses)

Components:

  • Hysteresis Loss: Due to magnetic domain reversal

  • Eddy Current Loss: Due to induced currents in core

Formulas:

\[\begin{aligned} P_h &= K_h \cdot f \cdot B_m^{1.6} \cdot V \text{ (Hysteresis)} \\ P_e &= K_e \cdot f^2 \cdot B_m^2 \cdot t^2 \cdot V \text{ (Eddy Current)} \\ P_{core} &= P_h + P_e \end{aligned}\]

Simplified for GATE:

\[P_{core} = K_1 f B_m^2 + K_2 f^2 B_m^2\]

Where: \(f\) = frequency, \(B_m\) = max flux density, \(t\) = lamination thickness

Mechanical Losses

Components:

  • Friction Loss: Bearing friction, brush friction

  • Windage Loss: Air resistance to rotation

Characteristics:

  • Depends on speed and machine construction

  • Generally proportional to speed squared for windage

  • Relatively constant for friction at constant speed

GATE Note: Usually given as constant value in problems

Variable Losses

Copper Losses (I²R Losses)

Definition: Resistive losses in windings due to current flow

General Formula:

\[P_{cu} = I^2 R\]

For Different Machines:

  • Transformer: \(P_{cu} = I_1^2 R_1 + I_2^2 R_2\)

  • DC Machine: \(P_{cu} = I_a^2 R_a + I_f^2 R_f\)

  • Induction Motor: \(P_{cu} = P_{s1} + P_{r2}\)

Important for GATE:

  • Varies with square of current

  • Dominant loss at high loads

Stray Losses

Definition: All other losses not accounted for in main categories

Components:

  • Stray load losses due to non-uniform current distribution

  • Losses due to harmonic fields

  • Surface losses

GATE Approximation:

  • Usually taken as 0.5% to 1% of output power

  • Can be determined by residual method

  • Often neglected in simple problems

Efficiency Calculations

Efficiency Definition

General Definition:

\[\eta = \dfrac{\text{Output Power}}{\text{Input Power}} = \dfrac{P_{out}}{P_{in}}\]

Alternative Forms:

\[\begin{aligned} \eta &= \dfrac{P_{out}}{P_{out} + P_{losses}} \\ \eta &= \dfrac{P_{in} - P_{losses}}{P_{in}} = 1 - \dfrac{P_{losses}}{P_{in}} \end{aligned}\]

Percentage Efficiency:

\[\eta\% = \eta \times 100\%\]

Transformer Efficiency

Transformer Efficiency

Total Losses:

\[P_{losses} = P_{core} + P_{cu}\]

At Any Load (x = fraction of full load):

\[P_{cu} = x^2 \times P_{cu(fl)}\]

Efficiency Formula:

\[\eta = \dfrac{x \cdot S \cdot \cos\phi}{x \cdot S \cdot \cos\phi + P_{core} + x^2 \cdot P_{cu(fl)}}\]

Where:

  • \(x\) = fraction of full load (0 to 1)

  • \(S\) = apparent power rating (kVA)

  • \(\cos\phi\) = power factor

  • \(P_{cu(fl)}\) = full load copper loss

Maximum Efficiency in Transformers

Condition for Maximum Efficiency:

\[P_{core} = P_{cu}\]

Loading for Maximum Efficiency:

\[x_{max} = \sqrt{\dfrac{P_{core}}{P_{cu(fl)}}}\]

Maximum Efficiency Value:

\[\eta_{max} = \dfrac{S \cdot \cos\phi}{S \cdot \cos\phi + 2P_{core}}\]

GATE Tip: Maximum efficiency is independent of load power factor but its value depends on power factor

All-Day Efficiency

Definition: Efficiency over 24-hour period

Formula:

\[\eta_{all-day} = \dfrac{\text{kWh delivered in 24 hours}}{\text{kWh input in 24 hours}}\]
\[\eta_{all-day} = \dfrac{\sum P_{out} \times t}{\sum P_{out} \times t + P_{core} \times 24 + \sum P_{cu} \times t}\]

GATE Note:

  • Core losses occur for full 24 hours

  • Copper losses occur only when loaded

  • Important for distribution transformers

DC Machine Efficiency

DC Machine Efficiency

Generator:

\[\eta = \dfrac{V \cdot I_L}{V \cdot I_L + I_a^2 R_a + P_{field} + P_{core} + P_{mech}}\]

Motor:

\[\eta = \dfrac{V \cdot I_L - I_a^2 R_a - P_{field} - P_{core} - P_{mech}}{V \cdot I_L}\]

Where:

  • \(V\) = terminal voltage

  • \(I_L\) = line current

  • \(I_a\) = armature current

  • \(R_a\) = armature resistance

  • \(P_{field}\) = field losses

Induction Motor Efficiency

Induction Motor Power Flow

Power Flow:

\[P_{in} \rightarrow P_{gap} \rightarrow P_{mech} \rightarrow P_{out}\]

Relationships:

\[\begin{aligned} P_{gap} &= P_{in} - P_{s1} - P_{core} \\ P_{mech} &= P_{gap} - P_{r2} = P_{gap}(1-s) \\ P_{out} &= P_{mech} - P_{mechanical} - P_{stray} \end{aligned}\]

Efficiency:

\[\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{P_{mech} - P_{mechanical} - P_{stray}}{P_{in}}\]

Where: \(s\) = slip, \(P_{s1}\) = stator copper loss, \(P_{r2}\) = rotor copper loss

Important Formulas Summary

Key Formulas for GATE

Parameter Formula
Efficiency \(\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{P_{out}}{P_{out} + P_{losses}}\)
Copper Loss \(P_{cu} = I^2 R\) (varies as \(x^2\))
Core Loss Constant (independent of load)
Max Efficiency When \(P_{core} = P_{cu}\)
Transformer Efficiency \(\eta = \dfrac{x S \cos\phi}{x S \cos\phi + P_{core} + x^2 P_{cu(fl)}}\)
Loading for Max Eff \(x = \sqrt{\dfrac{P_{core}}{P_{cu(fl)}}}\)
All-day Efficiency \(\eta_{ad} = \dfrac{\sum P_{out} \times t}{\sum P_{out} \times t + P_{core} \times 24 + \sum P_{cu} \times t}\)

GATE Problem Types

Typical GATE Problems

1. Efficiency Calculation:

  • Given losses, find efficiency at different loads

  • Given efficiency, calculate unknown losses

2. Maximum Efficiency:

  • Find loading for maximum efficiency

  • Calculate maximum efficiency value

3. Loss Separation:

  • Separate core and copper losses from test data

  • No-load test and short-circuit test analysis

4. All-day Efficiency:

  • Calculate efficiency over 24-hour period

  • Compare with conventional efficiency

Sample GATE Problem

Problem: A 100 kVA, 2000/400 V transformer has core loss of 1000 W and full load copper loss of 1600 W. Find:

  1. Efficiency at half load and 0.8 power factor

  2. Loading for maximum efficiency

  3. Maximum efficiency at 0.8 power factor

Solution:

\[\begin{aligned} \text{1. } P_{cu(half)} &= (0.5)^2 \times 1600 = 400 \text{ W} \\ P_{out} &= 0.5 \times 100 \times 0.8 = 40 \text{ kW} \\ \eta &= \dfrac{40000}{40000 + 1000 + 400} = 96.6\% \\ \text{2. } x_{max} &= \sqrt{\dfrac{1000}{1600}} = 0.79 \\ \text{3. } \eta_{max} &= \dfrac{100 \times 0.8}{100 \times 0.8 + 2 \times 1} = 97.6\% \end{aligned}\]

Quick Tips for GATE

  • Remember: Core losses constant, copper losses vary with \(I^2\)

  • Maximum efficiency: Variable losses = Fixed losses

  • Transformers: Use \(x = \sqrt{\dfrac{P_{core}}{P_{cu(fl)}}}\) for max efficiency

  • Units: Always check units in calculations (W, kW, MW)

  • Power factor: Affects efficiency value but not max efficiency condition

  • All-day efficiency: Core losses for 24 hours, copper losses only when loaded

  • Induction motors: Remember slip relation: \(P_{mech} = P_{gap}(1-s)\)

  • Test data: No-load test \(\to\) core losses, SC test \(\to\) copper losses

Common GATE Mistakes to Avoid

  • Don’t confuse: kVA rating with kW output

  • Don’t forget: To square the load fraction for copper losses

  • Don’t mix: Core losses (constant) with copper losses (variable)

  • Don’t ignore: Power factor in efficiency calculations

  • Don’t calculate: All-day efficiency same as conventional efficiency

  • Don’t assume: Maximum efficiency occurs at full load

  • Remember: Efficiency is always less than 1 (or 100%)