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
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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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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 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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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]
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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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
Distribution of J and B in coilDistribution of J and B in coil
Actual
Approximate
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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 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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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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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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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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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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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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