Ch. 3. Pulsed and Water Cooled Magnets. T. J. Dolan. Magnetic field calculations

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

2 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 carrying wires Faraday magnetic induction Maxwell equations of electromagnetism Dolan, IPR

3 Required ed magnetic field 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 ) coil o( o coil) R coil R o = 6 m, R coil = 2.5 m B coil = 14 T R o Dolan, IPR

4 Advantages of Water-Cooled Magnets Dolan, IPR

5 Dolan, IPR

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7 Dolan, IPR

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

9 Dolan, IPR

10 Toroidal Magnetic Field Ripple Dolan, IPR

11 Law of Biot-Savart Dolan, IPR

12 Field from a Circular Current Ring Dolan, IPR

13 Dolan, IPR

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

15 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 = K(k) = E(k) = Br = T Bz = T Dolan, IPR

16 Coil Forces F= df = JxB dv F = df = I dlxb for thin wires B 1 (at I 2 ) = o I 1 /2 r df/dl = I 2 B 1 = o I 1 I 2 /2 r Dolan, IPR

17 Force between Circular Loops F z = 2 ai 2 B r1 Dolan, IPR

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

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

20 Force on Torsatron Coils Dolan, IPR

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

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

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

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

25 RLC Circuit Equations R = total resistance L = total t inductance 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 ] Dolan, IPR

26 Current vs. Time Undercritically damped circuit Dolan, IPR

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

28 Resistance of Wire, Rod, Plate, Tube l R = dx /S 0 Example: Copper tube r 1 = 0.02 m, r 2 = m, l = 3 m = 2x10-6 Ohm-m m S = (r 22 -r 12 ) = m 2 R = l / S = Ohm Dolan, IPR

29 Inductance of N-turn Solenoid Length l, Radii r 1 and r 2 = r 2 /r Example: N=20, l =1m 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

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

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

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

33 Distribution of J and B J/ t = 2 J/ B/ t = 2 B/ Assume 2 B B/ 2 B/ t B Then B B/ 2 Skin depth = (2/ ) ) 1/2 In copper at 1 MHz, = 0.07 mm Dolan, IPR

34 Structural Support of Coil Dolan, IPR

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

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

37 Coil Melting and Yielding Yielding at B ~ B y [(r 2 -r 1 )/2r 1 ] 1/2 B y (Cu, SS) = 25 T B y (Ta) = 32 T Melting at B ~ B e / 1/2 ~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

38 Energy Storage Systems Dolan, IPR

39 Scyllac Capacitor 60 kv, 1.85 F Dolan, IPR

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

41 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

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

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

44 Spark Gap Switch Dolan, IPR

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

46 Exploding Foil Switch Dolan, IPR

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

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

49 Magnetic Flux Compression Bellows type Dolan, IPR

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

51 Failure Rates f j dt = failure probability of item j during dt t Failure probability between 0 and t: p f = dt f j (t) 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) 0 Dolan, IPR

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

53 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 ETNF = 1 / [100x x2x10-4 ] = 7.7 shots Similar analysis for laser systems, automobiles, etc. Dolan, IPR

54 Coil Power Requirements = (copper volume)/(coil volume) dp = J c2 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 : B z = 3/2 g( )(P/ r 1/2 o 1 ) where = r 2 /r 1 = L/2r 1 Dolan, IPR

55 Relation of B z to Input Power B = 3/2 1/2 z o g( )(P/ r 1 ) = 2x10-8 Ohm-m Example: = 0.9 r 1 = 0.1 m, P = 100 kw Find optimum coil B z Optimum g( ) =0.142 at r 2 = 3r 1, L = 4r 1. B z =11T 1.1 Dolan, IPR

56 Coil Power Requirements Given r 1 = 3 m, B = 10 T, find P Result: P = 240 MW This is why big experiments use superconducting coils. Liquid N 2 coolant (77 K) can lower and required power. Dolan, IPR

57 Heat Removal Rate P =C m T(dV/dt) C = specific heat of coolant m = mass density of coolant T = temperature rise of coolant dv/dt = volumetric flow rate of coolant = A w v A w = coolant channel area v = coolant flow speed Dolan, IPR

58 Reynold s Number Re = dv m / p = f L c m v 2 /2D (Pa) Pumping Power Pumping power: P c = p (dv/dt)/ p p = pump efficiency Dolan, IPR

59 Friction Factor Re = dv m / Dolan, IPR

60 Example Pumping Power 100 kw coil, 16 coolant channels in parallel, each 30 m long, 4.6 mm diameter, t = 60 K. Find dv/dt, v, p, P c Total dv/dt = P/C m T = 3.99x10-4 m 3 /s One channel = (dv/dt)/16 = 2.49x10-5 m 3 /s A c = 1.66x10-5 m 2 v = (dv/dt)/a = 1.50 m/s Re = 6872 f = from graph p =fl 2 5 c m v /2D = 2.56x10 Pa P c = p (dv/dt)/ p = 128 W. Dolan, IPR

61 Coil Winding Methods Hollow Conductor Pancake coil, v Tape-wound coil v z Bitter magnet v r Dolan, IPR

62 Safe Value of Current Density Heat dissipated in volume V ch cooled by one channel P ch = J 2 V ch Equate to heat removed by coolant: P =C m t(dv/dt) Solve for safe value of J Dolan, IPR

63 Maximum Safe Current Assuming T = 60 K p = 0.41 MPa Match coil resistance to power supply Dolan, IPR

64 Coil Electrical Resistance R c = L c /A c Example: 16 coils in series, each L c = 30 m, Square copper with a=864mm 8.64 mm, D=466mm 4.66 mm. A c = a 2 D 2 /4 = 5.76x10-5 m 2 One winding: R c = Ohm 16 in Series R = 1.67 Ohm. Joints add resistance. a D Dolan, IPR

65 Coil Winding Wind around a form High tension to bend copper to shape Accurate position of copper important Fiberglass insulation between layers Epoxy to hold conductor rigidly Brazed joints Aluminum joints unreliable Dolan, IPR

66 Summary Pulsed Magnets Components and circuit calculations are simple, but diffusion i of J and B are more complex. Capacitor banks are widely used for pulsed magnets Inductive storage and flywheels less expensive at very high energy Magnetic flux compression very high B Component reliability for millions of shots is difficult Dolan, IPR

67 Summary Water-Cooled Magnets Cryogenic insulation, refrigeration not needed Brazed joints reliable Bolted joints easy to assemble and disassemble Can tolerate high neutron fluences Does not require stabilization Technology well developed, reliable Calculations of current, B field, coil power, and pumping power simple and reliable. Dolan, IPR

68 Dolan, IPR

Chapter 30 Inductance and Electromagnetic Oscillations

Chapter 30 Inductance and Electromagnetic Oscillations Units of Chapter 30 30.1 Mutual Inductance: 1 30.2 Self-Inductance: 2, 3, & 4 30.3 Energy Stored in a Magnetic Field: 5, 6, & 7 30.4 LR Circuit: 8,