10.Thermal design - IEA · 2020-02-27 · Thermal design purpose is to bear out power rating and safety margin in respect to thermal limits of construction and used materials •

27-FEB-2020EIEN20

Design of Electrical Machines

10.Thermal design Thermal loads, heat paths/barriers

Cooling integration

Lund University / LTH / IEA / Avo Reinap / EIEN20 / 2020-02-27 2

L10: Thermal design•

Heat and

mass transfer

–

Conduction–

Convection and advection

–

Radiation –

Phase change

•

Temperature

distribution and limitations–

Insulation systems and realisations

•

Thermal design–

Coolant and cooling ducts

–

Conduction vs

convection

c


Transfer, capability, integration


Heat sourcesEnergy loss in electric

circuits qe

=ρJ2

Energy loss in magnetic

circuits

qΦ

=Ch

B2f+Ce

(Bf)2

Energy loss in mechanic

circuits

qω

=


Heat dissipationConduction

coolingConvection

cooling

Radiative cooling


Thermal design•

Thermal design purpose is to bear out power rating and safety margin in respect to thermal limits of construction and used materials

•

Topologies and concepts for thermal circuits

•

Physical processes related to heat transfer

•

Materials: thermal properties and limits

•

Calculation methods and models

•

Cooling integration


Cooling

Concepts•

Structure / construction

–

Where the energy conversion, heat generation, transfer and temperature (Δ)

takes place

•

Heat sources–

Energy converted to heat

•

Heat sinks–

Heat dissipation

•

Cooling concepts –

arrangement of heat sources, paths and sinks

–

Indirect Cooling (high Δ)–

Direct cooling (low Δ)


Equivalent circuits & relations

Relation Electrical circuit

Magnetic circuit

Thermal circuit

Cooling circuit

Potential U=E·l N·I=H·l =G·l P=·l

Flow I=J·A Φ=B·A Q=q·A Q=v·A

Conductive element G=γ·A/l G=μ·A/l G=λ·A/l G=·A/l

Ohm’s Law U=I·R N·I=Φ·R =Q·R P=Q·R


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

Al

21

21

lAQ

lAQ


Thermal conductivityMaterial λ

[W/mK] Material λ

[W/mK]

Air 0.025-0.035 Cast iron 40-46

Nomex 0.11 Stainless- steel 25-30

Kapton 0.12 Laminated iron 20-40

Mica 0.4-0.6 Copper 360Bonding epoxy 0.64 Aluminum 200-220

Avg.ins.syst em 0.2 NdFeB 9

SmCo 10


Convection

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


Transport of heat

Q -

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


High Performance Cooling

•

Electronics-cooling.com•

Spray and jet cooling, continuous and fluctuating

•

Single-phase and two-phase flows, phase changing materials

•

Micro and mini-cahnels, higher intensity cooling


Conjugate heat transfer

dcool

dcond

L Lh

out in

win

cQPcool

cool

heat

hAP

•

Heat transfer

and pressure drop

in 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


Estimation of heat transfer•

The character of flow

is

described by Reinolds number,

•

the heat transfer

is expressed by Nusselt

number

•

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


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

Air H2 C02 H2

0 Tr

Oil, degC 20 120 20 120 20 120 20 120 20 120c, kJ/kgK 1.00 1.01 14.2 14.5 0.85 0.94 4.19 4.25 1.71 2.11, kg/m3 1.20 0.89 0.08 0.06 1.83 1.36 999 946 879 816λ,mW/mK 26 33 178 227 16 24 594 686 111 102, uPas 18 23 8 11 14 19 1000 200


Radiation

•

Radiation is heat transfer between cooling surface A at temperature 2 and ambience at temperature amb via electromagnetic waves

amb

ambrad

rad

ambrad

c

AcQ

2

442

2

442

100100

100100

x

1

2

α1

Q

A

amb

hot

α2

l


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


Heat transfer problem objective

•

Heat sources and sinks

•

Temperature distribution and limits


Thermal limits of materials

•

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

lifetime


Temperature dependence•

Materials’

temperature dependence

is taken

account with material thermal coefficients

coilcoilcoilcoilcoil 00 1

magnmagnBrmagnmagnRRmagn BB 00 1

magnmagnHcmagnmagnCCmagn HH 00 1


Evaluation 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


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


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


Heat 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


Thermal design A•

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

much shorter

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


Thermal design B•

Cooling surfaces and cooling conditions

•

Speed and magnetization -> specific core losses pfe

=Ch

Bhf+Ce

B2f2+Ca

B1.5f1.5

•

Load requirement -> specific conductor losses pcu

=ρJ2()•

Geometric layout and material choice for more internal thermal transport


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


Equivalent circuit


Thermal 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


Thermal

modelling

example

I

•

Determine

heat sources

–

in regions•

Specify

cooling

conditions

–

over cooling

surfaces

•

Find

heat balance

i.e. temperature

distribution


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


Reduced

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


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


Thermal

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


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

,,,,,,


Thermal

modelling

example

III

•

Directly cooled laminated windings


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


Cooling integration

0 100 200 300 400 500 600 700 800 900 10000

50

100

150

200

250

100100

100 100

300

300

300300

500

500

500

700

700

700

900900

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


•

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


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

700

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

80

100

120

140

160

180

200

220

200

200

200

400

400

400

600

600

600

800

800

800

1000

1000

1000

1200

1200

1200

1400

1400

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

8.2

8.4

8.6


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

20002000

40004000

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

100100

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

]



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

]

10001000

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

8000

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

1000

1000

1000

1500

15001500

2000

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 in0 1000 2000 3000 4000 5000 6000 7000 8000

0.2

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]


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/s

L=25 mmc=0.2 mm

L=50 mmc=0.4 mm

L=100 mmc=0.6 mm

L=200 mmc=0.8 mm


Thermal 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


Summary•

Thermal constrains and dependences

•

Thermal circuits, heat sources and cooling options

•

Heat transfer model and modelling

•

Learning skills from the assignments


Outlook

Documents

10.Thermal design - IEA · 2020-02-27 · Thermal design purpose is to bear out power rating and safety margin in respect to thermal limits of construction and used materials •