L10 – Thermal Design - IEA - Lund University · L10 – Thermal Design EIEN20 Design of Electrical Machines, IEA, 2016 3 Avo R Design of Electrical Machines 9 Industrial Electrical

L10 – Thermal Design

EIEN20 Design of Electrical Machines, IEA, 2016 1

Heat dissipation

Life span of the electrical machines

Avo R Design of Electrical Machines 2

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Forced cooling• Cooling channel

– 236 mm long 1 mm wide– 1.5 mm parallel plates

• Convection – 140-160 W/m2K– Empiric vs FEM

• Flow rate– Previous

1-8 m3/min• Temperature

– 2D FEM conjugate heat transfer

– P/Q=constant for out=100oC


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PreviouslyEnergy loss in electriccircuitsqe=ρJ2

Energy loss in magnetic circuitsqΦ=ChB2f+Ce(Bf)2

Energy loss in mechanic circuitsqω=


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NextConductioncooling

Convectioncooling

Radiativecooling

Thermal

circu

it an

d th

ermal

desi

gn

Hea

t sou

rces,

hea

t sin

ks an

dhe

at fl

ow.




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nContent

• Heat transfer (transport) vs heat and mass transfer– Conduction– Convection and advection – Radiation

• Temperature distribution and limitations– Insulation systems and realisations

• Thermal design– Coolant and cooling ducts– Conduction vs convection


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Equivalent circuit relations

P=Q·R

G=·A/l

Q=v·A

P=·l

Cooling circuit

=Q·RN·I=Φ·RU=I·ROhm’s Law

G=λ·A/lG=μ·A/lG=γ·A/lConductive element

Q=q·AΦ=B·AI=J·AFlow

=G·lN·I=H·lU=E·lPotential

Thermal circuit

Magnetic circuit

Electrical circuitRelation


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Thermal conductivity

• Conduction is heat transfer by diffusion in a stationary medium due to a temperature gradient. The medium can be a solid, a liquid or gas

• Diffusion through the substance

x

1

2

λ Q

A l

21

21

lAQ

lAQ


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Thermal conductivity

10SmCo

9NdFeB0.2Avg.ins.system

200-220Aluminum0.64Bonding epoxy

360Copper0.4-0.6Mica

20-40Laminated iron0.12Kapton

25-30Stainless-steel0.11Nomex

40-46Cast iron0.025-0.035Air

λ [W/mK]Materialλ [W/mK]Material




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nConvection

ambn hknq

• Convection is heat transfer between either a hot surface and a cold moving fluid or a cold surface and a hot moving fluid. Convection occurs in liquids and gases

• Movement of the substance

x

1

2

α1

Q

A

amb

hot

α2

l

ambhot

amb

lAQ

AQ

21

22

11


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Transport of heatQ - the required flow rate, m3/s, Ph - required cooling power, W, ρ - the density of the heat carrier, kg/m3, c - the specific heat capacity, J/kg°C, Δ - the temperature difference between incoming and outgoing temperature °C

Natural convection

Forced cooled plane surface by air speed v

Empirical cooling capability

cP

Q h

2255mK

W

78.06.0208.7 v

25.21mkW

AP

cool

loss


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High Performance Cooling

• Electronics-cooling.com• Spray and jet cooling, continuous and fluctuating• Single-phase and two-phase flows, phase changing materials• Micro and minicahnnels, higher intensity cooling


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Conjugate heat transfer

dcool

dcond

L Lh

out in

win

cQPcool

cool

heat

hAP

• Heat transfer and pressure dropin the cooling channel is determined by flow

• Flow characterisation– Development: laminar, unstable

or transitional or turbulent– entrance length, – boundary layer

• Dimensionless quantities• Reynolds number characterizes

the flow and Mach number illustrates the compressibility of the flow.

• Flow rate Q [L/min]• Flow speed v=Q/A [m/s]• Re=inertia force / viscous

force




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nEstimation of heat transfer

• The character of flow is described by Reinoldsnumber,

• the heat transfer is expressed by Nusseltnumber

• and the coolant is described by Prandtl number

• The hydraulic diameter is related to the geometric layout of the cooling channel

hin

h DAQvD

1Re

bulkwall

hh k

qDDkhNu

kcpPr

perimeterareaDh

4


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Coolant

• Ideal coolant = high thermal capacity & low viscosity– Hydrogen is used in large turbo generators

• Ability to store and carry heat = mass density times specific heat capacity reduces with temperature

– Coolant steam

Tr OilH20C02H2

200100019141182318, uPas10211168659424162271783326λ,mW/mK8168799469991.361.830.060.080.891.20, kg/m3

2.111.714.254.190.940.8514.514.21.011.00c, kJ/kgK1202012020120201202012020, degC

Air


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Radiation

• Radiation is heat transfer between cooling surface A at temperature 2 and ambience at temperature ambvia electromagnetic waves

amb

ambrad

rad

ambrad

c

AcQ

2

442

2

442

100100

100100

x

1

2

α1

Q

A

amb

hot

α2

l


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Transient heat flow

• Steady state temperature• Heating time constant• Temperature rise during

the transient heating

x

1

α1

A

amb

hot

α2

l

QP QS QD

2

2AdtdcVP

RdtdCP

QQQ

thth

DSP

t

ambmamb

ththth

thm

th

e

AcVRC

APRP

1

2

2




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nTransient heat flow

• Thermal model representing a physical model

• Mathematical formulation• Many simplifications and

approximations• Heat is not internally

generated in the body• Losses are applied to

specific node-point

1 Rth 2

Cth P

2

1

12

121

av

ththth CRCP

dtd


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Heat transfer problem formulation for electrical devices - machines

• Heat sources and sinks • Temperature distribution and limits


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Design target - Thermal limits

• The most critical component in the electrical machine is insulation and temperature dependent is magnet.

• Insulation lifetime is shortened radically if temperature exceeds the limit and that is due to accelerated oxidation process in the insulation material.

• Δ=100K -> ½ lifetimeAvo R Design of Electrical Machines 20

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Temperature dependence

• Materials’ temperature dependence is taken account with material thermal coefficients

coilcoilcoilcoilcoil 00 1

magnmagnBrmagnmagnRRmagn BB 00 1

magnmagnHcmagnmagnCCmagn HH 00 1




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nEvaluation of thermal loading

• Heat transfer– Input: heat sources and

cooling conditions– Outcome: temperature

distribution• Computational tools

– Analytic, empiric, numeric– FEA, CFD, lumped circuits

for heat transfer and fluid flow

• Material characterization• Sub-model validation


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Machine slots

• Total conductor area 225.7 mm2 insulated slot area 508.4 mm2

• Specific conductor losses 4 W/mm3 reduced for winding 1.77 W/mm3

• Slot impregnation 0.21 W/mK selected equivalent thermal conductivity 0.4 W/mK


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Complexity

• Electrical machine is– A complex 3D electromagnetic structure– A complex spatial fluid dynamic structure with cooling

medium

• In order to determine the temperature distribution– A good estimate of losses has to be known– Properties of the cooling process has to be known– The thermal characteristics and properties has to be known

• An optimized thermal design can help increase machine rated power substantially


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Thermal design

• Good estimate of losses –the spatial and temporal distribution of heat sources– Waveform of a loss origin– Distribution of heat sources– Duty cycle – operational

cycle time often muchshorter than thermal time constant

– Short time operation– Intermittent

• Thermal characteristics of materials– Temperature dependence– Temperature limits

• Heat dissipation – thermal circuit and cooling system– Thermal efficiency– Cooling conditions (normal,

forced)• Maximum allowed loading

according to the thermal limits at cooling capability




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nHeat transfer

• Steady state and transient• Heat transfer problem

according to temperature (potential) and heat balance between source, sink and storage

• heat transfer convection-diffusion equation

• incompressible Navier-Stokes equations for fluid dynamics

tcQ

zyx

Qzyx

pzyx

zyx

2

2

2

2

2

2

2

2

2

2

2

2

0

Qckt

c pp u

0

2

u

Fuuuu pt


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Thermal circuit at steady state

• Node points i, Qi [W], i [K]5. Coil loss and temperature4. Tooth loss and temperature6. Yoke loss and temperature7. 8. Ambience temperature

• Thermal conductivity elements Gij [W/K]– From coil to tooth G54

– From coil to yoke G56

– From tooth to yoke G46

– From yoke to ambience G67

cooling

heating


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Equivalent circuit


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Thermal modelling example I

• Determine heat sources – in regions• Specify cooling conditions – over cooling surfaces• Find heat balance i.e. temperature distribution




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nThermal circuit – thermal contacts

• A bad electric conductor is usually also a bad thermal conductor

• No air-gaps in electrical circuit, many air-gaps in thermal circuit

• Thermal contact between stator core and housing– 0.1 mm +5K– 0.2 mm +10K


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Thermal circuit – heat carrier

• Experience from A3 A good electric conductor is usually also a good thermal conductor

• Interested in hotspots: 100% conductor in the middle of winding

• Heat is taken from end-windings: conduction, convection or both


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Thermal model

• Geometry of a PMSM• Material & thermal loading

– Winding– Permanent magnets

• Surface & cooling– Natural convection

• Temperature nodes– Nodes of interest

• Thermal circuits– Heat transfer rather than flow

network• Thermal resistances

– Focus on thermal ”air-gaps”

pm

win

surf

amb

pm

win

surf

amb

mwmwms

mwmwsws

mswswsasa

sasa

kkkkkkkkkk

kk

00

00

pm

win

QQ

00


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Model development

• Sorces and loads– Conductor losses– Convection cooling

• 2D heat transfer– Approximate rating– Extraction of elements

• 3D heat transfer– Extrucion from 2D– Focus on end turns

• Heat exchange through end-turns– Thermal conduction




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nThermal modelling example II

• Calculating flux (and current) density waveform • Estimating losses densities in the symmetric part of machine• Calculating temperature distribution according to heat sources and sinks


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Multi-physics → FEM

• Different problems in physics ‘share’ the same geometry

• Calculate for a single element– The variation of loss origin– RMS power loss– MEAN temperature

• A field equation is solved for the finite size of volume

• boundaries suppose to specify a potential (essential), flow naturally given.

N 1 (x 1 ,y 1 )

N 2(x 2 ,y 2 )

N 3 (x 3 ,y 3 )

u 1

u 3

u 5

u 2

u 4

u 6

1

3

2

x

y

fe

cu

ptzyxBptzyxJ

,,,,,,


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Thermal modelling example III

• Directly cooled laminated windings


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Thermal design

0 100 200 300 400 500 600 700 800 900 10000

50

100

150

200

250

100

100100 100

300

300

300300

500

500

500

700

700

700

900

900

900

1100

1100

1300

1300

1500

flow rate, Q [L/min]

wal

l tem

pera

ture

, ou

t [ C

]

Cooling power, p=cpQ(out-in) [W]

• Peak heat sources– Jm=22.3…28.8 A/mm2

– p=10.0…16.6 W/cm3

– P=2.9 kW• Thermal management

– Limit winding, wall and outlet temperature

– 100 L/min = 1.25 m/s per div

• FEM heat transfer– Contribution from

conduction and natural convection




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• Ideal coil geometry and cooling conditions• non cooled spots overheated – terminal leads & small cross-section

layers close to the air-gap• cooling intensity -- flow rate -- control over hot-spot temperatures

Heat transfer analysis


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Mapping operation points

• Driving parameters for cooling P=f(out,Q) at in

• Flow (Re) and coolant (Pr) characterization

• Heat transfer – correlations (Nu) and – coefficient h

• Wall and winding temperature• Pressure across cooling channel

– Power for supply• Expected cooling power

P=f(w,Q) at in

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

100100

100100 100

300

300

300300

500

500

500

500

700

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700

700

900

900

900

1100

1100

1100

1300

1300

1300

1500

1500


outle

t tem

pera

ture

, ou

t [ C

]


0 100 200 300 400 500 600 700 800 900 100020

40

60

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200

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1000

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1200

1200

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1600


outle

t tem

pera

ture

, ou

t [ C

]

Reynolds number, Re=2dhQ/(A) [-]

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

6.6

6 .6

6.6

6.8

6.8

6.8

7

7

77.

2

7.2

7.2

7.4

7.4

7.4

7.6

7.6

7.6

7.8

7.8

7.8

8

8

8

8.2

8.28.

48.6


outle

t tem

pera

ture

, ou

t [ C

]

Nusselts number, Nu=f(Re,Pr) [-]

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

340

360

360

360

380

380

380

380

400

400

400

400

420

420

420

440


outle

t tem

pera

ture

, ou

t [ C

]

Heat transfer coefficient, h=Nu k/Dh [W/(m2K)]

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

20

20

2020

40

40

40

40

60

60

60

80

80

80

100

100

120

120

140

160


outle

t tem

pera

ture

, ou

t [ C

]

Temperature across boundary, Pcool/(hAcool) [C]

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

20002 0 00

40 00400 0

60006000

8000

8000

800010000

10000

12000

1200014000


outle

t tem

pera

ture

, ou

t [ C

]

Pressure drop, dP [Pa]

0 100 200 300 400 500 600 700 800 900 100020

40

60

80

100

120

140

160

180

200

220

5050

50

10 0100

100

150150

150

200200


outle

t tem

pera

ture

, ou

t [ C

]

Ideal cooling supply power, dPQ [-]

0 100 200 300 400 500 600 700 800 900 10000

50

100

150

200

250

100

100100 100

300

300

300300

500

500

500

700

700

700

900

900

900

1100

1100

1300

1300

1500


wal

l tem

pera

ture

, ou

t [ C

]



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7kW@120oC&4m3/min

0 1000 2000 3000 4000 5000 6000 7000 800020

40

60

80

100

120

140

160

outle

t tem

pera

ture

, ou

t [ C

]

1 0001000

1000

1000 1000

2000

2000

2000

2000

3000

3000

3000

3000

4000

4000

4000

4000

5000

5000

5000

5000

6000

6000

6000

7000

7000

7000

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8000

8000


cooling power, p=cpQ(out-in) [W]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

500500

500

100 0

1000

100 0

1 50 0

15001500

2 000

20002000

25002500

2500

30003000

3000


Reynolds number, Re=2dhQ/(A) [-]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

77

78

8

8

88

9

9

9

9

10

10

10

11

11

11

12

12

13


Nusselts number, Nu=f(Re,Pr) [-]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

80

80

100

100

100

100

120

120

120

120

140

140

140

140

160

160160

160

200200

200200

250 250250 250

300 300 300 300

500 500 500


heat transfer coefficient, h=Nu k/Dh [W/(m2K)]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

10

10

10

10

20

20

20

20

30

30

30

30

40

40

40

50

50

60

60

70


temperature drop across boundary layer, Pcool/(hAcool) [K)]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

4040

40

100

100

100

200

200

200

200

400

400

400

400

1000

1000

1000

1000

2000

2000

20002000

4000

40004000

4000

1000010000

10000 10000


pressure drop, dP [Pa]

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

40

40

40

40

100

100

100

100

200

200

200

200

400

400

400

400

10001000

1000

2000

20002000

40004000 4000

1000010000


ideal cooling supply power, dPQ [-]

Defining designing cooling channels

• Driving parameters for cooling P=f(out,Q) at in

• Flow (Re) and coolant (Pr) characterization

• Heat transfer – correlations (Nu) and – coefficient h

• Wall and winding temperature• Pressure across cooling channel

– Power for supply• Expected cooling power

P=f(w,Q) at in

0 1000 2000 3000 4000 5000 6000 7000 80000.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

chan

nel h

eigh

t, d

[mm

]

130

130

130

130

140

140

140

140

150

150

150

150

160

160

160

170

170

180

190


winding temperature, Tw =Tout+Pcool/(hAcool) [C]


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Parallel plates, laminar flow, …• Narrow cooling channels allow higher surface speed, thus higher

cooling capability for the same flow rate • Narrow channels results higher pressure drop and is difficult to secure

in production • Lack of cooling (flow leakage) results high risk for overheating

– Slide sow: L from 25 mm to 200 mm @ 12 m/sL=25 mmc=0.2 mm

L=50 mmc=0.4 mm

L=100 mmc=0.6 mm

L=200 mmc=0.8 mm




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nThermal circuit – cooling circuits

• Natural and Forced• Integrated cooling as a

result of machine integrated construction

• Slotted stator operates as a cooling circuit

• Directly cooled heat sources– Cooling ducts, cooling

jackets, cooling channels

• Cooling capability– Maximize the cooling

surface area– Improve cooling medium

parameters and velocity

• Smallest temperature rise is the goal when designing a thermal circuit


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Cooling Concepts• Structure

– Where the energy conversion, heat transfer and temperature drop (Δ)takes place

• Heat sources– Energy converted to heat

• Cooling sources– Heat dissipation

• Cooling concepts – arrangement of heating and

cooling sources– Indirect Cooling (high Δ)– Direct cooling (low Δ)


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Summary

• Thermal constrains and dependences

• Thermal circuits, heat sources and cooling options

• Heat transfer model and modelling

• Learning skills from the assignments

L10 – Thermal Design - IEA - Lund University · L10 – Thermal Design EIEN20 Design of Electrical Machines, IEA, 2016 3 Avo R Design of Electrical Machines 9 Industrial Electrical

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