l2 – introsurvey · rotary-to-linear • electromechanical energy conversion is ‘efficient’...

L2 – IntroSurvey

EIEN20 Design of Electrical Machines, IEA, 2016 1

Industrial Electrical Engineering and AutomationLund University, Sweden

L2: Variety of electrical machines

Machine construction: an overviewTransformer is it machine too?

Avo R Design of Electrical Machines 2

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Today’s goal

• Introduction to machine construction– Constructional layouts and mechanical arrangements– Supply & control and Application & integration– A few examples

• Introduction to machine analysis– Methods and tools– Analysis: parameterisation, magnetic, thermal– Implementation: example and home assignment


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Previous lecture

• Magnetic field aroundcurrent carrying coil

• The principle of operation of any rotating electric motor is derived from Lorenz force

A- A+

I

A+

I

A+

I

M

F

M M

A+ A- I

A+ A- I

A+ A- I

A+ A- I

M M

F

F

F

F

Magnetic

coupling

Origin of fo

rces


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L2 – IntroSurvey



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Rotary machinerotating magnetic field

Origin of forces?

Magnetic coupling?


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Classification of electrical machines

• Machine – device that uses energy to perform useful work = energy conversion, also transformation

EM

Gap-coil orientation (s) Type of excitation Power character

Axial flux

Transversal flux

Induced, Reluctance

Permanent/Electromagnet

Hybrid

AC, DC, modulated

Rotary, Linear

Radial flux

Circumferential flux


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• DC

• AC

• Switched power

Supply & Control

u

t

u

t

Constant or slowly

varying supply

Constant or slowly

varying supply in

frequency or/and

amplitude


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Origin of forces

DC machine

synchronous mach

ine

RM machine

asynchron

ous machine

L2 – IntroSurvey



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nMachine (excitation) types

• Separate excitation– PM excitation – magnet

reluctance dominates, PMSM, BLDCM, DCM

– PM hybrids + reluctance modulation

– Field winding – small gaps and good permeability is useful, single or doubly fed

• Excitation via armature– Minimize magnetization

current – small air-gap and good permeability essential

– Induction, reluctance and stepping machines

• Combined excitation– Hybrid excitation


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Constructional layoutCirular, Rotary

Angular,

Spherical Planar,

Disk

Cannular,

Sandwiched

Tubular


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Air-gap versus flux path• Machine construction with respect to of the air-gap

surface and the direction of the armature field.

2-10 Nm/kg1-3 Nm/kg1-3 Nm/kg1-2 Nm/kg

TransversalCircumferentialAxialRadial


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Application orientation

• Mechanical arrangement– Inner/outer rotor – Short/long mover

• Type of mechanical motion– Rotary– Linear– Reciprocative or oscillatory– Combined

L2 – IntroSurvey



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nFew examples

• Linear drive: actuator vs ”machine”• Transfersal flux machine: claw-pole• Electrically magnetised synchronous machinewithout sliprings


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Rotary-to-Linear

• Electromechanical energy conversion is ‘efficient’ at a higher rotation speed than 0

• Usually a linear movement is needed in industries


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Linear actuator

1

10

100

1000

10000

100000

1 10 100 1000 10000 100000 Distance, mm

Force, N

PneumaticCylinders Pneumatic

RodlessCylinders

HydraulicCylinders

Electrical Linear Actuators

Electrical Linear Motors

ScrewDrives

Belt Drives

Rack-and-Pinion


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Electrical linear actuator

1

10

100

1000

10000

100000

1 10 100 1000 10000 100000 Distance, mm

Force, N

Linear synchronousPM Motors

Magnetostrictive& thermal actuators

Solenoids

Linear Induction Motors

Hybrid steppersVoice coil Actuators

Electrostatic Actuators

ReluctanceMotors

L2 – IntroSurvey



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nLinear hybrid machine

• Digital-to-mechanical stepper machine

• Tubular configuration Ø30/114 L…/260 mm

• The primary part has 2winding and a PM

• Relatively high force and precision


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Claw-pole motor


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Electrically magnetised synchronousmachine without sliprings

• Windings– Distributed– Concentrated

• Permanent magnets– Discrete– Multi-pole

• Magnetic core– Stack of laminations (2D)– Iron powder (3D)

• Principle of operation– Base on excitation


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Electromagnetic device calculations I

• Methods– Simple approach– Equivalent circuit method (Magnetic EC, Thermal EC

1D elements describing 3D object)– Finite element method (multiphysics 2D, 3D)

• Tools– Matlab (Analytic model, ECM, Design environment)– FEMM (Magnetism + Heat transfer + …)

L2 – IntroSurvey



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nElectromagnetic device calculations II

• Procedure– Geometry and dimensions– Materials and their properties– Physical processes and sources– Boundaries and symmetry – Mathematical description of the physical process

related to the geometry and medium


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Geometry simplification

Can “flow” be seen only on a plane? Simplify 3D problem to 2DSymmetric repetition or reflection? Solve a section from the whole


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Geometry parameterization

• It is advisable to describe a device geometry by a number of parameters:

– proportions, – (Variable) dimensions or

numbers

• Electromagnetic, thermal, etc formulation and calculations base on a parametric geometry input

length

height lc

lc

ins

g


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Electromagnetic circuit

Fn

+N·I -N·I

phi m-core

iron bar

air-gap

• Start from static and continue with dynamics

• Ampere’s circuital law applied to magnetic circuit – ΣHL = NI

• Maxwell stress concept – forces in magnetic field – F=1/(2μ0)B2A

L2 – IntroSurvey



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nThermal circuit

0 QDt

• Heat flows from hotter to cooler regions

• Heat sources and sinks• Fourier’s heat

conduction in the materials

• Newton’s convection boundary condition

J2ρKf

qn=h(-amb)

J2ρKf


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Example• Electric conductors with fill

factor of Kf =0.6 [-] in a cross-section of 0.02x0.05 [m]

• Current density J=2e+6 [A/m2]• Resistivity ρ 2.4e-8 [Ωm]• Specific loss q=J2ρKf [W/m3]• Ambient temperature =20 [C]• Thermal conductivity λ =0.2

[W/mK]• Heat transfer coefficient α=20

[W/m2K]Length

Width height


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Analytic model

• Symmetric part of geometry

• 1D heat equation

• Solution

02

2

qdxd

x

222/42

2/ rdqdqrr

222/2

2/ xdqdqxx

02

2

qdrd

r


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Equivalent circuit method amb

ambcoil

Gth1

Gth2

1

2 3

• Guess heat flow paths for the symmetric part of the geometry

• Estimate heat conductivity elements

• Formulate relations between temperatures at the node points and heat flow into the node point

• Solve the equation system

L2 – IntroSurvey



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nFinite Element Analysis

• Pre-processing– Geometry– Material properties and heat sources– Boundary conditions– Discretization – FE-mesh

• Processing• Post-processing

– Field distribution, flow density, etc


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Pre-Processor• Drawing the endpoints of

the lines and arc segments for a region,

• Connecting the endpoints with either line segments or arc segments to complete the region,

• Defining material properties and mesh sizing for each region,

• Specifying boundary conditions on the outer edges of the geometry.

coil


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Post-Processor

• Estimate hot-spot and average coil temperature for a given loss density and cooling conditions

• Temperature distribution indicates the thermal loading

Density Plot: Temperature (K)

1.207e+002 : >1.249e+0021.165e+002 : 1.207e+0021.123e+002 : 1.165e+0021.080e+002 : 1.123e+0021.038e+002 : 1.080e+0029.958e+001 : 1.038e+0029.535e+001 : 9.958e+0019.113e+001 : 9.535e+0018.690e+001 : 9.113e+0018.268e+001 : 8.690e+0017.845e+001 : 8.268e+0017.423e+001 : 7.845e+0017.000e+001 : 7.423e+0016.578e+001 : 7.000e+0016.155e+001 : 6.578e+0015.733e+001 : 6.155e+0015.310e+001 : 5.733e+0014.887e+001 : 5.310e+0014.465e+001 : 4.887e+001<4.042e+001 : 4.465e+001


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Analytic vs. numeric model

• Differential equations that describe physics can be solved for a simple geometries

• Equivalent circuit method has a low number of elements, which makes it relatively fast and inaccurate

• Finite element method has a high number of elements, which makes it relatively slow and accurate

• The sources in EC are concentrated and FE model distributed

L2 – IntroSurvey


Industrial Electrical Engineering and AutomationLund University, Sweden

Home assignment

Analysis of heat transfer in a single-phase transformer


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1 transformer

• Primary winding is magnetically loaded by secondary winding

• Ideally no power losses, no voltage or magnetomotiveforce drop across the corresponding circuits


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Assignment goals

• Estimate transferred power capability by studying the limits for cooling power at maximum temperature

• Learn to use finite element (FE) and equivalent circuit (EC) model for heat transfer

• Select one of the TRAMO-ETV transformers to dimension and validate your model


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Performance equations

• Apparent power

• Voltage

• Magnetic flux

• Current

• Ability to transfer power

• Losses

• Core losses are estimated from the specific loss curves

• Flux density is constant

mmmeeeloss qlAqlAP

memmmm AAJBIUS 21

21

mm IUS21

dt

tdNdt

tdtetUtu m cos

tABtN

Utt mmm

m

sinsinsin

tNAJ

tN

NItIti

em

mm

cos

coscos

L2 – IntroSurvey



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nParameterization

• A single phase shell type of transformer

• The influence of end turns are excluded

• The same electric loading is assumed in primary and secondary coil


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Program structure

Parameterization ― Lengt ― Geometric proportions ― Magnetic loading ― Material properties

Geometric modelling ― Derive geometry in

respect with parameters

2D Finite Element Method ― Heat transfer (Mirage))

2D Equivalent circuit method― Thermal circuits

Parametric change ― Sensitivity study ― Proportion between

magnetic and electric circuits

Transformer specification

Objective ― Estimate electric loading

• The goal of the calculation is to find optimal relation between magnetic and thermal circuit

• The FE model of heat transfer is established in lua script

• The EC model of heat transfer is established in m script


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Iterative calculations

Initialization― Cooling conditions ― Magnetic: B -> pfe ― Electric: Jk,ρ0 -> pcu

Find temperature― Coil hot-spot max ― Coil average ave

Obtain current density ― if target - max < - 40

then Jk+1=Jk x 0.5 ― if target - max > 40

then Jk+1=Jk x 2 ― else

Jk+1=Jk W(target - max)

Target― |target - max|≤0.05 ― iter ≥ max_iter

Obtain new values ― ρ= ρ0(1+α(ave-0)) ― pcu=0.5 ρ (Jk+1)2 Kf ― iter=iter+1

Result visualization― Temperature plot (bmp) ― Electric loading (txt)

• The goal of the computation routine is to estimate current loadingwithin the thermal limit

• Thermal dependence of copper is taken into account


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Thermal equivalent circuit

1

2 3

5

4

6

9

7

1011

12

8

12

3 4

56

7

89

10

11

12 13

1415

16

• Thermal conductivity network of 11 elements

• Symmetric part of core• Copper losses are

applied to node 2 and 3, core losses to node 1

• Convection elements are 4 and 11

L2 – IntroSurvey



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nAssignment

• Compare the estimated current density from Femm and Matlab (e.g. Excel table)

• Interpret/validate the methods and results– What geometric proportions between the electric

circuit and magnetic gives the highest transferred power and which the highest efficiency

– What is the difference between ECM and FEM estimations and what might be the reason

– How realistic you think the models and estimated results are

l2 – introsurvey · rotary-to-linear • electromechanical energy conversion is ‘efficient’...

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