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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?
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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
EIEN20 Design of Electrical Machines, IEA, 2016 2
Avo R Design of Electrical Machines 5
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Rotary machinerotating magnetic field
Origin of forces?
Magnetic coupling?
Avo R Design of Electrical Machines 6
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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
EIEN20 Design of Electrical Machines, IEA, 2016 3
Avo R Design of Electrical Machines 9
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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
Avo R Design of Electrical Machines 12
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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
EIEN20 Design of Electrical Machines, IEA, 2016 4
Avo R Design of Electrical Machines 13
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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
Avo R Design of Electrical Machines 16
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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
EIEN20 Design of Electrical Machines, IEA, 2016 5
Avo R Design of Electrical Machines 17
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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
EIEN20 Design of Electrical Machines, IEA, 2016 6
Avo R Design of Electrical Machines 21
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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
EIEN20 Design of Electrical Machines, IEA, 2016 7
Avo R Design of Electrical Machines 25
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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
EIEN20 Design of Electrical Machines, IEA, 2016 8
Avo R Design of Electrical Machines 29
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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
Avo R Design of Electrical Machines 32
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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
EIEN20 Design of Electrical Machines, IEA, 2016 9
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
Avo R Design of Electrical Machines 36
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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
EIEN20 Design of Electrical Machines, IEA, 2016 10
Avo R Design of Electrical Machines 37
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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
EIEN20 Design of Electrical Machines, IEA, 2016 11
Avo R Design of Electrical Machines 41
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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