T. J. Dolan Magnetic field calculations Coil forces Coil forces RLC - PowerPoint PPT Presentation

Ch. 3. Pulsed and Water ‐ Cooled Magnets T. J. Dolan Magnetic field calculations Coil forces Coil forces RLC circuit equations Distribution of J and B Energy storage E t Switching and transmission Magnetic flux compression Component reliability Power and cooling requirements Coil design considerations Coil design considerations Dolan, IPR 2010 1

Background Chinese 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 wires Ampere interaction of current-carrying wires Faraday magnetic induction Maxwell equations of electromagnetism Dolan, IPR 2010 2

Required magnetic field equ ed ag e c e d T = 2x10 8 K, n = 2x10 20 m -3 ,  = 0.1  B = 5.9 T Field at coil is larger: B coil = B o (R o /R coil ) o ( coil ) coil o R coil R o = 6 m, R coil = 2.5 m  B B coil il = 14 T 14 T R R o Dolan, IPR 2010 3

Advantages of Water-Cooled Magnets Advantages of Water Cooled Magnets Dolan, IPR 2010 4

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(eh and fg are equal and opposite.) Dolan, IPR 2010 8

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Toroidal Magnetic Field Ripple Toroidal Magnetic Field Ripple Dolan, IPR 2010 10

Law of Biot-Savart Dolan, IPR 2010 11

Field from a Circular Current Ring Field from a Circular Current Ring Dolan, IPR 2010 12

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k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k) Dolan, IPR 2010 14

Example – Field of Circular Coil Circular coil, a = 0.5 m, I = 100 kA. Find B( r = 0.4 m, z = 0.6 m ) k 2 = 0.683707 K(k) = 2.05215 E(k) = 1.25125 Br = 0.0025 T Bz = 0.027 T Dolan, IPR 2010 15

Coil Forces F =  d F =  J x B dV F =  d F =  J x B dV F =  d F = I  d ℓ x B for thin wires wires B 1 (at I 2 ) =  o I 1 /2  r dF/dL = I 2 B 1 =  o I 1 I 2 /2  r =  I I /2  r dF/dL = I B Dolan, IPR 2010 16

Force between Circular Loops F z = 2  a I 2 B r1 F 2 a I B Dolan, IPR 2010 17

Force between Circular Loops Example: Two coaxial circular coils with a = 1 m, separated by z = 1 m. , p y Find F. k 2 = 0 8 k = 0.8 K(k) = 2.257, E(k) = 1.178, B r = 0.0697 T F = 4.38x10 5 N Dolan, IPR 2010 18

Tensile Stress in Long Solenoid Coil Example Case: Example Case: r 1 = 1 m, ∆ r = 0.2 m B = 10 T.  = 212 MPa Yield stress of copper = 280 MPa Dolan, IPR 2010 19

Force on Torsatron Coils Force on Torsatron Coils Dolan, IPR 2010 20

Force Reduced Torsatron Coils Optimum Pitch angle ~ 42 o g R/a c ~ 7 Dolan, IPR 2010 21

TF Coil Design Considerations g TF coil forces tend to: i increase coil radius a c il di decrease major radius R c bend coils (due to interaction with vertical field) Consider stress concentrations fatigue fatigue creep thermal stress TF coils shaped like “D” have lower stress than circular coils TF coils shaped like “D” have lower stress than circular coils. Dolan, IPR 2010 22

Reduction of Field Errors Coil winding accuracy Coil alignment Coil supports to minimize motion Series connection to equalize currents Series connection to equalize currents Stray B fields from current leads Stray B fields from ferrous objects Dolan, IPR 2010 23

Components p Energy storage Energy storage Switches Transmission lines Coils Diagnostics and controls Diagnostics and controls Dolan, IPR 2010 24

RLC Circuit Equations q R = total resistance L L = total inductance t t l i d t L(d 2 q/dt 2 ) +R(dq/dt)+q/C =0 L(d 2 q/dt 2 ) +R(dq/dt)+q/C =0 q(0) = CV o (dq/dt) o = 0 q(t) = CV o e -at [cos  t + (a/  )sin  t] I (t) = (V o /  L) e -at sin  t a=R/2L  = [(1/LC) – a 2 ]  = [(1/LC) a=R/2L a 2 ] Dolan, IPR 2010 25

Current vs Time Current vs. Time “Undercritically Undercritically damped circuit” Dolan, IPR 2010 26

“Crowbar” Switch S 2 Close S1 at t=0 Close S2 at t=t max Dolan, IPR 2010 27

Resistance of Wire, Rod, Plate, Tube Resistance of Wire, Rod, Plate, Tube ℓ R =  dx  / S  d R / S 0 Example: Copper tube r 1 = 0.02 m, r 2 = 0.025 m, ℓ = 3 m  = 2x10 -6 Ohm-m  2x10 Ohm m S =  (r 2 2 -r 1 2 ) = 0.000707 m 2 R =  ℓ / S = 0.00849 Ohm Dolan, IPR 2010 28

Inductance of N-turn Solenoid Inductance of N turn Solenoid Length ℓ ,  = r 2 /r 1 Radii r 1 and r 2 1 2 Example: N=20, ℓ = 1 m r = 5 m ℓ = 1 m, r 1 = .5 m, r 2 = .8 m  = 1.6,  = 2, L/N 2 r 1 = 1.2x10 -6 L = 2.4x10 -4 Henry  = ℓ /r 1 Dolan, IPR 2010 29

Parallel Plate Transmission Line Parallel Plate Transmission Line L=  o S ℓ K sh / h Dolan, IPR 2010 30

Graph of K h vs (s/h) Graph of K sh vs. (s/h) If /h If s/h << 1, 1 then K sh = 1 Dolan, IPR 2010 31

Coaxial Cable or Tubes Coaxial Cable or Tubes L =  ℓ ln(b/a)/2  L =  o ℓ ln(b/a)/2  Dolan, IPR 2010 32

Distribution of J and B Distribution of J and B  J/  t =  J/  t  J/   2 J/   B/  t =  2 B/  Assume  2 B ≈ B/  2  B/  t ≈  B Then  B ≈ B/  2     ≈  Skin depth  = (2/    ) 1/2 th  Ski d (2/ ) 1/2 In copper at 1 MHz,  = 0.07 mm pp Dolan, IPR 2010 33

Structural Support of Coil Structural Support of Coil Dolan, IPR 2010 34

Distribution of J and B in coil Distribution of J and B in coil Actual Approximate Dolan, IPR 2010 35

Axial Distribution of B in Solenoids Single-turn Uniform J Dolan, IPR 2010 36

Coil Melting and Yielding Yielding at B ~ B y [(r 2 -r 1 )/2r 1 ] 1/2 B (Cu SS) = 25 T B y (Cu, SS) = 25 T B y (Ta) = 32 T B (Ta) = 32 T Melting at B ~ B /  1/2 Melting at B ~ B e /  1/2  ~ 3  ~ 3 B e (Cu, SS) = 90 T B e (Ta) = 137 T Fatigue failures after many shots Sudden B > 70 T  Coil explodes Dolan, IPR 2010 37

Energy Storage Systems Dolan, IPR 2010 38

Scyllac Capacitor Scyllac Capacitor 60 kV, 1.85  F Dolan, IPR 2010 39

Energy Storage System Costs Energy Storage System Costs Fusion experiments Power grids S l Solar power Wind power flywheel ~1980 values Dolan, IPR 2010 40

Inductive Energy Storage Opening switch S1 forces current to flow through plasma confinement coil. S 1 : difficult to prevent arcing Transfer efficiency = L s L/(L s +L) 2  25% Dolan, IPR 2010 41

50 MJ Homopolar Generator E r = v  xB z U of Texas U. of Texas Dolan, IPR 2010 42

Flywheel Energy Storage Can store about 500 MJ/m 3 Can store about 500 MJ/m Princeton Plasma Physics Laboratory Princeton Plasma Physics Laboratory motor-generator 200 MW, 3 s 200 MW 3 More expensive than homopolar system Dolan, IPR 2010 43

Spark Gap Switch p p Dolan, IPR 2010 44

Los Alamos Dual Spark Gap Switch Low “jitter”: 3240 switches fired within 10 ns. Dolan, IPR 2010 45

Exploding Foil Switch Dolan, IPR 2010 46

Marx Bank +400 kV 100 kV 100 kV 100 kV 100 kV Capacitors are charged in parallel Capacitors are charged in parallel Then discharged in series to give high voltage Dolan, IPR 2010 47

High Voltage Coaxial Cable Scyllac experiment had 250 km of these cables Scyllac experiment had 250 km of these cables. 10 5 shot reliability. Dolan, IPR 2010 48

Magnetic Flux Compression “Bellows” type Dolan, IPR 2010 49

Imploding Metallic Liner Imploding Metallic Liner explosive liner Debris Debris low Liner flux Compressed flux Dolan, IPR 2010 50

Failure Rates f j dt = failure probability of item j during dt t Failure probability between 0 and t: p =  dt f(t) Failure probability between 0 and t: p f =  dt f j (t) 0 Probability of not failing before time t = 1 p Probability of not failing before time t = 1-p f “Failure Rate” at time t: r j (t) = p f /(1-p f ) Total failure rate r(t) =  r j (t) (failures per second) j Estimated time between failures ETBF = 1/r(t) Dolan, IPR 2010 51

Component Reliability Capacitors, cables, automobiles automobiles, … Dolan, IPR 2010 52

Estimated Time to Next Failure Example: 100 capacitors with  j = 10 -4 per shot and 600 cables with  j = 2x10 -4 per shot Find ETNF Find ETNF ETNF = 1 / [100x10 -4 + 600x2x10 -4 ] = 7.7 shots Similar analysis for laser systems, automobiles, etc. Dolan, IPR 2010 53

Coil Power Requirements  = (copper volume)/(coil volume) dP =  J c 2 dV Circular coils: P =   J c 2  dz  2  rdr over coil volume Field at center of solenoid with length L, radii r 1 and r 2 : f B z =  3/2  o g(  )(P/  r 1 ) 1/2  o g(  )(  1 ) z where  = r 2 /r 1  = L/2r 1 Dolan, IPR 2010 54