BEE108 Rotating Electric Machines: Electro-mechanical Energy Conversion

Unit Overview: BEE108 Rotating Electric Machines

  • Course Title: Bachelor of Science (Electrical Engineering) BEE108: Rotating Electric Machines.

  • Specific Focus: Tutorial on Electro-mechanical Energy Conversion Principles.

  • Software Requirements:     - MATLAB/SIMULINK     - PSCAD

  • Hardware Requirements: N/A.

Learning Outcomes

  • Fundamental Principles: Explain the fundamental principles behind electro-mechanical energy conversion and calculate the relation between motor rated power and torque (Bloom's Level 3).

  • DC Machines: Describe the principle of operation of DC machines, the function of the commutator, and the different types of DC machines based on field winding connections (Bloom's Level 2).

  • 3-Phase Synchronous Machines: Explain the theory of 3-phase synchronous machines and the relation between excitation and power factor (Bloom's Level 2).

  • Induction Machines: Explain the fundamental theory of 3-phase and single-phase induction machines, the various types of machines, and their control aspects (Bloom's Level 2).

  • Efficiency and Construction: Discuss the construction of AC machines and the ways of enhancing motor efficiency (Bloom's Level 4).

  • Testing Procedures: Describe the procedure for testing AC machines and perform these tests in a simulated environment to calculate motor parameters (Bloom's Level 3).

Assessments Summary

  • Assessment 1: Weekly Quizzes     - Type: Multiple-choice quiz questions.     - Weighting: 10%10\% of total unit marks.     - Timing: Weekly.     - Learning Outcomes Assessed: All (Topics 2 to 11).

  • Assessment 2: Invigilated Test     - Type: Short/long answer questions and numerical problems.     - Weighting: 20%20\% of total unit marks.     - Timing: During Topic/Week 7.     - Learning Outcomes Assessed: 1, 2, 4 (Topics 1 to 6).

  • Assessment 3: Practical (Report) & Pre-recorded Presentation     - Type: Practical project using software and research on state-of-the-art solutions.     - Weighting: 30%30\% of total unit marks.     - Timing: End of Topic/Week 11.     - Learning Outcomes Assessed: All (All topics).

  • Assessment 4: Invigilated Exam     - Type: Mixed theoretical answers and engineering problems.     - Weighting: 40%40\% of total unit marks.     - Timing: Exam Week.     - Learning Outcomes Assessed: All (All topics).

Unit Content Topics

  • Topic 1: Electro-mechanical energy conversion principles     - Comparison between electrical and magnetic circuits.     - Energy and Co-energy.     - Induction.     - Forces and torques in magnetic fields and balance equations.

  • Topic 2: DC Machines Part 1 (Generators)     - Separately excited DC generators.     - Self-excited DC generators.     - Shunt generators.     - Load characteristics of DC generators.

  • Topic 3: DC Machines Part 2 (Motors)     - Working principles and types of DC motors.     - Speed control.     - Faults, inspection, and maintenance.

  • Topic 4: Synchronous Machines Part 1 (Generators)     - General aspects and operation.     - Synchronizing methods (single phase & three phase).     - Classification.

  • Topic 5: Synchronous Machines Part 2 (Motors)     - Operation, construction, and excitation.     - V-curves and inverted V-curves.     - Characteristics.

  • Topic 6: Induction Machines - Part 1     - Principles of single-phase AC machines.     - Equivalent circuits.     - Zero starting torque issues and split winding construction.     - Universal motors (based on DC series motors).

  • Topic 7: Induction Machines - Part 2     - Principles of 3-phase induction motors/generators.     - Equivalent circuits, leakage resistance, and reactance.     - Starting behavior and the role of back EMF.

  • Topic 8: Control of 3-phase motors and Industrial Applications - Part 1     - Torque-speed characteristics of cage and slip ring motors.     - Crawling and cogging problems.     - Impact of voltage reduction on motor torque.

  • Topic 9: Control of 3-phase motors and Industrial Applications - Part 2     - Assisted starting in weak systems.     - Motor starting methods and braking.     - Selection based on characteristics.     - Direct-on-line (DOL) and reversible motor circuits.

  • Topic 10: Testing of Motors and Generators     - DC tests, No-load tests, Blocked rotor tests.     - Heat Run tests.     - Component tests.

  • Topic 11: Special Machines     - Stepper motors, Permanent-Magnet DC motors.     - Servo motors, Brushless DC motors.     - Brushless Synchronous generators.

  • Topic 12: Unit Review     - Final review of content, student work review, and clarification of outstanding issues.

Electromechanical Energy Conversion Foundations

  • Definition: The process of changing the form of energy from electrical to mechanical or vice versa is called electromechanical energy conversion.

  • Generators: Devices that convert mechanical energy into electrical energy to provide electricity.

  • Motors: Devices that convert electrical energy into mechanical motion and work.

  • Electromechanical System Components:     - Electrical systems and mechanical systems are joined by a coupling field.     - Coupling Field: Can be electrical or magnetic.     - Losses: Losses occur at every stage of the system; therefore, it is never an ideal system.     - Efficiency: A critical consideration during machine selection due to existing losses.

Questions & Discussion: Machine Selection and Operation

  • Question: What basic factors should be considered while selecting a machine?     - Nature of the electric supply.     - Types of drives.     - Types of loads.     - Electrical characteristics.     - Service capacity and rating (specifically Size and Cost).

  • Question: Compare motor and generator in terms of torque.     - In a Motor: Electromagnetic torque is produced by the interaction of electric current with a magnetic field. This torque is used to drive a load. The load provides a mechanical torque that opposes the electromagnetic torque.     - In a Generator: Electric current is generated when mechanical torque is applied to turn conductors through a magnetic field. This mechanical torque is opposed by an electromagnetic torque resulting from the current's interaction with the magnetic field.

  • Question: What are the basic methods to produce an EMF?     - 1. An electromagnetic force (EMF) is created by the interaction of two magnets.     - 2. A coil possessing magnetic flux can create an EMF similar to a permanent magnet.

  • Question: Where does maximum energy storage take place in a DC machine?     - Options: a) Stator; b) Rotor; c) Air gap.     - Answer: Maximum energy storage occurs in the Air gap. The predominant energy storage occurs here because the properties of the magnetic circuit are determined by the air gap dimensions.

  • Question: Explain methods for innovating conversion systems.     - Modern techniques utilize new motor types and modern power controllers.     - Controllers using power electronics offer energy-efficient, user-friendly, and high-performance drives.     - Advantages: Considerable energy savings; ability to limit currents to pre-recorded values during starting, overload, or unbalanced supply; prolonged equipment life.

Magnetic Circuit and Electromagnetic Principles

  • Comparison of Circuits:     - Electric Circuit: Applied voltage causes a current II to flow.     - Magnetic Circuit: Applied magnetomotive force (F\mathcal{F}) causes flux (ϕ\phi) to be produced.

  • Magnetic Circuit Variables:     - F\mathcal{F} = Magnetomotive Force (measured in Amp.Turn or A.TA.T).     - ϕ\phi = Flux of the circuit (measured in Weber or WbWb).     - R\mathcal{R} = Reluctance of the circuit (measured in A.T/WbA.T/Wb).     - Fundamental Equation: F=ϕR\mathcal{F} = \phi \mathcal{R}.

  • Key Laws:     - Ohm’s Law: Electromotive force (EMF) produces a current proportional to the path's conductivity.     - Ohm’s Law for Magnetic Circuits: Electric current constitutes a magnetomotive force (MMF) producing magnetic flux proportional to the path's permeability.     - Lenz’s Law: The polarity of induced EMF in a coil causes a current that opposes any change in the magnetic flux.     - Motor Action: A magnetic field exerts a force on a current-carrying conductor.     - Generator Action (Faraday’s Law): A conductor moving through a magnetic field (or a magnetic flux changing with time through a coil) will have an EMF induced in it.

Mathematical Principles of Induction and Force

  • Induced EMF Calculation:     - For a conductor of length ll moving at speed vv in flux density BB: e=lv×Be = lv \times B.     - The direction is determined by the "right-hand rule" for cross products.     - For a coil of NN turns: e=(dϕdt)e = -(\frac{d\phi}{dt}). Note: The minus sign indicates Lenz's law opposition and is often neglected in magnitude calculations.     - General expression related to inductance (LL) and current (II): e=dϕdt=L(didt)+i(dLdx)(dxdt)e = \frac{d\phi}{dt} = L(\frac{di}{dt}) + i(\frac{dL}{dx})(\frac{dx}{dt}).     - In a magnetically linear system, self-inductance is independent of current but is a function of displacement xx in systems with moving parts.

  • Lorentz Force Law:     - Determines force (FF) on a particle of charge qq in electric (EE) and magnetic (BB) fields: F=q(E+v×B)\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}).     - Pure Electric System: F=qEF = qE.     - Pure Magnetic System: F=q(v×B)F = q(v \times B).     - Force Density (ff): For continuous charge distribution, where ρ\rho is charge density (charge/volumecharge/volume) and JJ is current density (J=ρvJ = \rho v): f=ρE+J×Bf = \rho E + J \times B.

  • Torque Factors:     - Torque requires two forces/coupling forces acting at different angles (not 180180^{\circ}).     - Depends on: Current through the coil, length of the current-carrying coil, and the magnetic field passing through the coils.

The Energy Method and Energy Balance

  • Conservation of Energy Principle: Energy is neither created nor destroyed, only changed in form.

  • Lossless Magnetic Energy Storage System:     - Electric terminal: Variables are voltage (ee) and current (II).     - Mechanical terminal: Variables are force (ff) and position (xx).     - Energy Balance Equation: dWelec=dWmech+dWflddW_{elec} = dW_{mech} + dW_{fld}.     - dWelec=eidt=idλdW_{elec} = ei dt = i d\lambda (Differential electric energy input).     - dWmech=fflddxdW_{mech} = f_{fld} dx (Differential mechanical energy output).     - dWflddW_{fld} = Differential change in magnetic stored energy.

  • State Variables: Magnetic stored energy (WfldW_{fld}) is a state function uniquely specified by values of flux linkages (λ\lambda) and position (xx).     - dWfld=idλfflddxdW_{fld} = i d\lambda - f_{fld} dx.     - Determining Force: ffld=Wfld(λ,x)xf_{fld} = -\frac{\partial W_{fld}(\lambda, x)}{\partial x}.     - For linear systems where λ=L(x)i\lambda = L(x)i: Wfld=12L(x)i2W_{fld} = \frac{1}{2} L(x) i^2.

  • Multiple Electrical Terminals:     - Independent variables: θ,λ1,λ2\theta, \lambda_1, \lambda_2.     - dWfld(λ1,λ2,θ)=i1dλ1+i2dλ2TflddθdW_{fld}(\lambda_1, \lambda_2, \theta) = i_1 d\lambda_1 + i_2 d\lambda_2 - T_{fld} d\theta.     - Linear system Co-energy (WfldW'_{fld}): Simplified torque/force determination using current as a terminal variable.

Numerical Worked Examples

  • Example 1: Magnetic Circuit and Reluctance     - Problem: Core path length =40cm= 40\,cm, gap =0.05cm= 0.05\,cm, area =12cm2= 12\,cm^2, μr=4000\mu_r = 4000, N=400N = 400. Fringing increases air gap area by 5%5\%. Find total reluctance and current required for 0.5T0.5\,T in gap.     - Solution Details:         - Effective gap area: 1.05×12=12.6cm21.05 \times 12 = 12.6\,cm^2.         - Total Reluctance: R<em>total=R</em>core+Rgap\mathcal{R}<em>{total} = \mathcal{R}</em>{core} + \mathcal{R}_{gap}.         - Equation used: NI=BARNI = B A \mathcal{R}.

  • Example 2: Induced EMF     - Given: N=220N = 220, A=0.09m2A = 0.09\,m^2, B1=0.2Wb/m2B_1 = 0.2\,Wb/m^2, B2=0.05Wb/m2B_2 = 0.05\,Wb/m^2, t=0.02st = 0.02\,s.     - Calculation: e=NAB2B1dt=220×0.09×0.050.20.02=148.5Ve = -N A \frac{B_2-B_1}{dt} = -220 \times 0.09 \times \frac{0.05-0.2}{0.02} = 148.5\,V.

  • Example 3: Force on a Conductor     - Given: B=4×103TB = 4 \times 10^{-3}\,T, l=55cm=0.55ml = 55\,cm = 0.55\,m, I=6AI = 6\,A, θ=90\theta = 90^{\circ}.     - Calculation: F=BIlsin(θ)=4×103×6×0.55×sin(90)=0.0132NF = BIl \sin(\theta) = 4 \times 10^{-3} \times 6 \times 0.55 \times \sin(90^{\circ}) = 0.0132\,N.

  • Example 4: Aeroplane Wing EMF     - Given: l=22ml = 22\,m, v=45m/sv = 45\,m/s, Bvertical=3×105TB_{vertical} = 3 \times 10^{-5}\,T.     - Calculation: ϵ=Blv=3×105×45×22=0.0297V\epsilon = Blv = 3 \times 10^{-5} \times 45 \times 22 = 0.0297\,V.

  • Example 5: Force and Torque Inclination     - Given: B=0.9TB = 0.9\,T, I=20AI = 20\,A, l=0.3ml = 0.3\,m.     - Scenario 1 (θ=90\theta = 90^{\circ}): F=0.9×20×0.3=5.4NF = 0.9 \times 20 \times 0.3 = 5.4\,N.     - Scenario 2 (θ=30\theta = 30^{\circ}): F=5.4×sin(30)=2.7NF = 5.4 \times \sin(30^{\circ}) = 2.7\,N.

  • Example 6: Relay Stored Energy     - Parameters: N=1000N = 1000, g=2mmg = 2\,mm, d=0.15md = 0.15\,m, l=0.1ml = 0.1\,m, I=10AI = 10\,A.     - Function of position xx: Wfld=236(1xd)J×100W_{fld} = 236(1 - \frac{x}{d})\,J \times 10^0.

  • Example 7: Complex Torque in Multiply-Excited System     - Inductances: L11=(3+cos(2θ))×103L_{11} = (3 + \cos(2\theta)) \times 10^{-3}, L12=0.3cos(θ)L_{12} = 0.3\cos(\theta), L22=30+10cos(2θ)L_{22} = 30 + 10\cos(2\theta).     - Result for i1=0.8A,i2=0.01Ai_1 = 0.8\,A, i_2 = 0.01\,A: T=1.64×103sin(2θ)2.4×103sin(θ)T = -1.64 \times 10^{-3} \sin(2\theta) - 2.4 \times 10^{-3} \sin(\theta).

  • Example 8: Mechanical Power     - Given: τ=50N.m\tau = 50\,N.m, speed =1500r/min= 1500\,r/min.     - Calculation (Watts): P=τω=50×(1500×2π60)=7854WP = \tau \omega = 50 \times (1500 \times \frac{2\pi}{60}) = 7854\,W.     - Calculation (Horsepower): P=7854746=10.5hpP = \frac{7854}{746} = 10.5\,hp.

  • Example 9: Three-Legged Core Flux     - Core Regions: Core divided into four regions (reluctances R<em>1,R2,R3,R4\mathcal{R}<em>1, \mathcal{R}_2, \mathcal{R}_3, \mathcal{R}_4).     - Total Reluctance: Calculated as R</em>total=R<em>1+((R2+R3)R4)\mathcal{R}</em>{total} = \mathcal{R}<em>1 + ((\mathcal{R}_2 + \mathcal{R}_3) \parallel \mathcal{R}_4).     - Flux Divider Rule: Used to find individual leg fluxes: ϕ</em>centre=ϕtotal×R4R2+R3+R4\phi</em>{centre} = \phi_{total} \times \frac{\mathcal{R}_4}{\mathcal{R}_2 + \mathcal{R}_3 + \mathcal{R}_4}.

System Efficiency and Applications

  • Types of Losses During Conversion:     - Core losses (Iron losses).     - Electrical losses (Copper losses).     - Mechanical losses.

  • Examples of Electromechanical Devices:     - Speakers, Alternators, Typewriters.     - Amusement rides, Hydraulic systems.     - High-speed automation production lines, Robotics.     - Systems using Programmable Logic Controllers (PLC).

References

  • Bakshi, U.A. & Bakshi, M.V., Electrical Machines – II.

  • Umans, S., Fitzgerald and Kingsley's Electric Machinery, 7th Ed., 2013.

  • Chapman, S. J., Electric Machinery Fundamentals, 5th Ed., 2012.

  • Gieras, Jacek F., Electrical Machines: Fundamentals of Electromechanical Energy Conversion, 2017.

  • Mukerji, S. K., Khan, A. S., & Singh, Y. P., Electromagnetics for Electrical Machines, 2015.