Ch. 3. Pulsed and Water‐Cooled Magnets

T. J. Dolan

Magnetic field calculations Coil forcesCoil forces RLC circuit equations Distribution of  J and BE tEnergy storage Switching and transmissionMagnetic flux compression Component reliability Power and cooling requirements Coil design considerations

1

Coil design considerations  

Dolan, IPR 2010

BackgroundChinese Discovered magnetism and invented compass.

Mapped the world using compass and astronomical navigation.Zheng He brought knowledge to Europe in 1434

Oersted deflection of compass by current in wire

Ampere interaction of current carrying wiresAmpere interaction of current-carrying wires

Faraday magnetic induction

Maxwell equations of electromagnetism

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Required magnetic fieldequ ed ag e c e d

T = 2x108 K, n = 2x1020 m-3, = 0.1 B = 5.9 T

Field at coil is larger: Bcoil = Bo(Ro/Rcoil) coil o( o coil)

Ro = 6 m, Rcoil = 2.5 m B il = 14 T

Rcoil

RBcoil 14 T Ro

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Advantages of Water-Cooled MagnetsAdvantages of Water Cooled Magnets

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(eh and fg are equal and opposite.)

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Toroidal Magnetic Field RippleToroidal Magnetic Field Ripple

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Law of Biot-Savart

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Field from a Circular Current RingField from a Circular Current Ring

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k2 K(k) E(k) k2 K(k) E(k)k2 K(k) E(k) k2 K(k) E(k)

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Example – Field of Circular CoilCircular coil, a = 0.5 m, I = 100 kA. Find B( r = 0.4 m, z = 0.6 m )

k2 = 0.683707K(k) = 2.05215 E(k) = 1.25125

Br = 0.0025 T

Bz = 0.027 T

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Coil Forces

F = dF = JxB dVF = dF = JxB dV F = dF = I dℓxB for thin wireswiresB1 (at I2) = o I1/2rdF/dL = I B = I I /2r

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dF/dL = I2B1 = o I1 I2 /2r

Force between Circular Loops

F 2 aI BFz = 2aI2Br1

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Force between Circular LoopsExample: Two coaxial circular coilswith a = 1 m, separated by z = 1 m. , p yFind F.

k2 = 0 8k = 0.8

K(k) = 2.257, E(k) = 1.178, Br = 0.0697 T

F = 4.38x105 N

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Tensile Stress in Long Solenoid Coil

Example Case:Example Case: r1 = 1 m, ∆r = 0.2 mB = 10 T.

= 212 MPa

Yield stress of copper= 280 MPa

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Force on Torsatron CoilsForce on Torsatron Coils

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Force Reduced Torsatron Coils

Optimum Pitch angle ~ 42ogR/ac ~ 7

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TF Coil Design ConsiderationsgTF coil forces tend to:

i il diincrease coil radius acdecrease major radius Rcbend coils (due to interaction with vertical field)

Considerstress concentrationsfatiguefatiguecreepthermal stress

TF coils shaped like “D” have lower stress than circular coilsTF coils shaped like “D” have lower stress than circular coils.

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Reduction of Field Errors

Coil winding accuracy

Coil alignment

Coil supports to minimize motion

Series connection to equalize currentsSeries connection to equalize currents

Stray B fields from current leads

Stray B fields from ferrous objects

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Componentsp

Energy storageEnergy storage

Switches

Transmission lines

Coils

Diagnostics and controlsDiagnostics and controls

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RLC Circuit EquationsqR = total resistanceL t t l i d tL = total inductance

L(d2q/dt2) +R(dq/dt)+q/C =0L(d2q/dt2) +R(dq/dt)+q/C =0

q(0) = CVo (dq/dt)o = 0

q(t) = CVo e-at [cost + (a/)sint]

I(t) = (Vo/L) e-at sint

a=R/2L = [(1/LC) a2]Dolan, IPR 2010 25

a=R/2L = [(1/LC) – a2]

Current vs TimeCurrent vs. Time

“UndercriticallyUndercritically damped circuit”

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“Crowbar” Switch S2

Close S1 at t=0

Close S2 at t=tmax

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Resistance of Wire, Rod, Plate, TubeResistance of Wire, Rod, Plate, Tubeℓ

R d / SR = dx / S0

Example:Copper tube r1 = 0.02 m, r2 = 0.025 m, ℓ = 3 m= 2x10-6 Ohm-m 2x10 Ohm m

S = (r22-r1

2) = 0.000707 m2

R = ℓ / S = 0.00849 Ohm

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Inductance of N-turn SolenoidInductance of N turn SolenoidLength ℓ , Radii r1 and r2

= r2/r11 2

Example: N=20,ℓ = 1 m r = 5 mℓ = 1 m, r1 = .5 m,r2 = .8 m

= 1.6, = 2, L/N2r1 = 1.2x10-6

= ℓ /r1

L = 2.4x10-4 Henry

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Parallel Plate Transmission LineParallel Plate Transmission Line

L=oS ℓ Ksh / h

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Graph of K h vs (s/h)Graph of Ksh vs. (s/h)

If /h 1If s/h << 1,

then Ksh = 1

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Coaxial Cable or TubesCoaxial Cable or Tubes

L = ℓ ln(b/a)/2L = o ℓ ln(b/a)/2

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Distribution of J and BDistribution of J and B

J/t = 2J/J/t J/B/t = 2B/

Assume 2B ≈ B/2

B/t ≈ B

Then B ≈ B/2 ≈

Ski d th (2/ )1/2Skin depth = (2/)1/2

In copper at 1 MHz, = 0.07 mm

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pp

Structural Support of CoilStructural Support of Coil

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Distribution of J and B in coilDistribution of J and B in coil

Actual

Approximate

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Axial Distribution of B in Solenoids

Single-turn

Uniform J

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Coil Melting and YieldingYielding at B ~ By[(r2-r1)/2r1]1/2

B (Cu SS) = 25 T B (Ta) = 32 TBy (Cu, SS) = 25 T By(Ta) = 32 T

Melting at B ~ B /1/2 ~ 3Melting at B ~ Be/1/2 ~ 3 Be(Cu, SS) = 90 T Be(Ta) = 137 T

Fatigue failures after many shots

Sudden B > 70 T Coil explodes

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Energy Storage Systems

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Scyllac CapacitorScyllac Capacitor

60 kV, 1.85 F

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Energy Storage System CostsEnergy Storage System Costs

Fusion experimentsPower gridsS lSolar powerWind power

flywheel

~1980 values

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Inductive Energy Storage

Opening switch S1 forces current to flow throughplasma confinement coil.

S1 : difficult to prevent arcing

Transfer efficiency = LsL/(Ls+L)2 25%

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50 MJ Homopolar GeneratorEr = vxBz

U of Texas

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U. of Texas

Flywheel Energy Storage

Can store about 500 MJ/m3Can store about 500 MJ/m

Princeton Plasma Physics LaboratoryPrinceton Plasma Physics Laboratory

200 MW 3motor-generator 200 MW, 3 s

More expensive than homopolar system

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Spark Gap Switchp p

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Los Alamos Dual Spark Gap Switch

Low “jitter”:3240 switchesfired within 10 ns.

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Exploding Foil Switch

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Marx Bank

+400 kV

Capacitors are charged in parallel

100 kV 100 kV 100 kV 100 kV

Capacitors are charged in parallelThen discharged in series to give high voltage

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High Voltage Coaxial Cable

Scyllac experiment had 250 km of these cablesScyllac experiment had 250 km of these cables. 105 shot reliability.

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Magnetic Flux Compression

“Bellows” type

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Imploding Metallic LinerImploding Metallic Linerexplosive

linerDebris

lowflux

Debris

Liner

Compressed

flux

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Failure Ratesfjdt = failure probability of item j during dt

tFailure probability between 0 and t: p = dt f (t)Failure probability between 0 and t: pf = dt fj(t)

0

Probability of not failing before time t = 1 pProbability of not failing before time t = 1-pf

“Failure Rate” at time t: rj(t) = pf/(1-pf)

Total failure rate r(t) = rj(t) (failures per second)jj

Estimated time between failures ETBF = 1/r(t)

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Component Reliability

Capacitors, cables,automobilesautomobiles, …

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Estimated Time to Next Failure

Example:

100 capacitors with j = 10-4 per shot and

600 cables with j = 2x10-4 per shot

Find ETNFFind ETNF

ETNF = 1 / [100x10-4 + 600x2x10-4] = 7.7 shots

Similar analysis for laser systems, automobiles, etc.

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Coil Power Requirements

= (copper volume)/(coil volume) dP = Jc2dV

Circular coils: P = Jc2 dz 2rdr over coil volume

fField at center of solenoid with length L, radii r1 and r2:

Bz = 3/2 o g()(P/r1)1/2z o g( )( 1)

where = r2/r1 = L/2r1

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Relation of B to Input PowerRelation of Bz to Input PowerBz = 3/2 o g()(P/r1)1/2Bz o g()(P/r1)

= 2x10-8 Ohm-m

Example: = 0.9r1 = 0.1 m, P = 100 kW Find optimum coil Bz

Optimum g() =0.142 atOptimum g() 0.142 atr2 = 3r1, L = 4r1.

B = 1 1 TDolan, IPR 2010 55

Bz = 1.1 T

Coil Power RequirementsCoil Power Requirements

Given r1 = 3 m, B = 10 T, find P

Result: P = 240 MWResult: P 240 MW

This is why big experiments use superconducting coils.

Liquid N2 coolant (77 K) can lower and required power.

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Heat Removal RateP =CmT(dV/dt)

C = specific heat of coolant = mass density of coolantm mass density of coolantT = temperature rise of coolantdV/dt = volumetric flow rate of coolant = AwvA l t h lAw = coolant channel areav = coolant flow speed

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Pumping Powerp gReynold’s Number

Re = dVm/

p = f Lcmv2/2D (Pa)

Pumping power: Pc = p (dV/dt)/p p = pump efficiency

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Friction F tFactor

Re = dVm/

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Example – Pumping Power100 kW coil, 16 coolant channels in parallel, each 30 m long, 4.6 mm diameter, t = 60 K. Find dV/dt v p PFind dV/dt, v, p, Pc

Total dV/dt = P/CmT = 3.99x10-4 m3/s O h l (dV/d )/16 2 49 10 5 3/One channel = (dV/dt)/16 = 2.49x10-5 m3/sAc = 1.66x10-5 m2

v = (dV/dt)/A = 1.50 m/s( )Re = 6872 f = 0.035 from graphp = f L v2/2D = 2 56x105 Pap = f Lcmv /2D = 2.56x10 PaPc = p (dV/dt)/p = 128 W.

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Coil Winding MethodsgHollow Conductor

Tape-wound coil vzConductor

“Pancake coil”, v

coil vz

Bitter magnetvr

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Safe Value of Current DensitySafe Value of Current Density

Heat dissipated in volume Vch cooled by one channelPch = J2Vch

Equate to heat removed by coolant: P =Cmt(dV/dt)

Solve for safe value of J

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Maximum Safe CurrentAssuming

T = 60 K

p 0 41 MPap = 0.41 MPa

Match coilresistance topower supply

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Coil Electrical ResistanceRc = Lc/AcExample: 16 coils in series,Example: 16 coils in series, each Lc = 30 m,Square copper with a = 8 64 mm D = 4 66 mm

Da = 8.64 mm, D = 4.66 mm.

Ac = a2 – D2/4 = 5.76x10-5 m2

a

One winding: Rc = 0.0104 Ohm16 in Series R = 1.67 Ohm.

Joints add resistance.

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Coil WindingCoil WindingWind around a formHi h t i t b d t hHigh tension to bend copper to shapeAccurate position of copper important

Fiberglass insulation between layersEpoxy to hold conductor rigidlyBrazed jointsBrazed joints

Aluminum joints unreliable

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Summary – Pulsed MagnetsSummary Pulsed MagnetsComponents and circuit calculations are simple, but

diff i f J d B ldiffusion of J and B are more complex.

Capacitor banks are widely used for pulsed magnetsy g

Inductive storage and flywheels less expensive atvery high energyvery high energy

Magnetic flux compression very high B

Component reliability for millions of shots is difficult

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Summary – Water-Cooled Magnets

Cryogenic insulation refrigeration not neededCryogenic insulation, refrigeration not neededBrazed joints reliableBolted joints easy to assemble and disassembleC fCan tolerate high neutron fluencesDoes not require stabilizationTechnology well developed, reliablegy p ,

Calculations of current, B field, coil power,and pumping power simple and reliableand pumping power simple and reliable.

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