Massachusetts Institute of Technology 22.68J/2.64J Superconducting Magnets ⇓ February 27, 2003 • Lecture #4 – Magnetic Forces and Stresses 1 Forces • For a solenoid, energy stored in the magnetic field acts equivalent to an internal pressure: E B2 = Vol 2 µ o B – Long, thin solenoid – For B = 6 T ⇒ Internal pressure equivalent ≡ 140 atm – (typical working pressure of a gas cylinder) – For B = 10 T ⇒ P = 400 atm, the yield stress of copper Impact of Forces • Can break the structure and destroy the magnet. • Can damage insulation. • Can damage superconductor including overstraining brittle Nb3Sn wire. • Can degrade magnet performance – motion → release energy → instability → quench → training (?) 2 F = J×B Compute the field throughout the winding volume • Combine computation of forces and stress analysis with calculation of B ¾ Usually done by computer. ¾ Simple analytical formulations are useful for early stages of design. ¾ Must have knowledge of materials properties: – Mechanical – Thermal (COE for thermal stresses) – From room temperature (RT) to 4K 3 Forces and Stresses in Solenoids • B is highest in axial field direction, Bz, and current is azimuthal, Jθ, so largest forces are radial → circumferential hoop stresses • Br is high at coil ends → axial compression 4 Analysis • Simplest assumption: Each turn acts independently • The a tension develops in each turn: T = B(r )Ir • Overall hoop stress σθ σ θ = B( r )Jr [1] Where both J and σθ are averaged over the winding pack • Thus stresses increase with B, J and r (size) 5 • Field distribution across a solenoid is often approximately linearly decreasing from inner radius to outer radius: B= (a2 − r )B1 + (r − a1 )B2 (a2 − a1 ) B( ρ ) = (α − ρ )B1 + ( ρ − 1)B2 (α − 1) [2] Where ρ = r/a1 , α = a2/a1 • For the special case of an infinitely long solenoid, we have B2 = 0 and B1 = µo J (a2 – a1) so that ⎛ B12 ⎞ 2(α − ρ )ρ ⎟⎟ σ θ = BJr = ⎜⎜ 2 ⎝ 2 µ o ⎠ (α − 1) [3] Note that B2/2µo is a scaling factor! 6 • • • Often the approximation that turns act independently is a poor approximation. Adjacent turns pressing on each other develop a radial stress which modifies the hoop stress. Assume elastic and isotropic properties – Then there is an analytic solution based on distributed forces in a cylinder (Timoshenko, S. (1956), Strength of Materials, Part 2) – Let u be the local displacement in the radial direction – Reduce the equilibrium condition between σr and σθ and BJr to a single equation (Lontai & Marston (1965); and Montgomery (1969)) ( 1 d ⎧ du ⎫ u 1 −υ 2 ) BJ ⎨r ⎬ − 2 = − r dr ⎩ dr ⎭ r E [4a] With σθ = E ⎧u du ⎫ + υ ⎨ ⎬ dr ⎭ 1 −υ 2 ⎩ r σr = E ⎧ du u⎫ υ + ⎨ ⎬ r⎭ 1 − υ 2 ⎩ dr [4b] [4c] Where E = Young’s Modulus ν = Poisson’s Ratio and Tensile Stress is defined to be positive (+) 7 • For convenience, let r = ρa, a2 = αa1, w = uE/[a1(1-ν2)] and BJa1 = K - Mρ where, (αB1 − B2 )Ja1 K= (α − 1) M= and (B2 − B1 )Ja1 (α − 1) So that 1 d ⎧ dw ⎫ w ⎨ρ ⎬ − 2 = − K + Mρ ρ dρ ⎩ dρ ⎭ ρ σθ = σr = w [5a] dw dρ [5b] dw w +υ dρ ρ [5c] ρ +υ The solution of Eqn. [5a] is: K ρ 2 Mρ 3 w = Cρ + − + 3 8 ρ D [6] where the coefficients C and D are obtained by the boundary conditions at r = a1 and a2, i.e., ρ = 1 and α. 8 • Depending on coil design there may be radial stress at the inner and outer boundaries, e.g.: – Inner winding mandrel – Outer pre-stress winding or structural support cylinder or ring. • Most usual condition, though, is that the boundaries are free, i.e., σr = 0 at ρ =1 and α. • Substituting these conditions into [5c] and [6] allows definition of C and D and thus the stress functions: (1 + 2υ )(α + 1)⎫ K (2 + υ ) ⎧ 2 α2 + + + − σθ = α α 1 ρ ⎨ (2 + υ ) ⎬⎭ 3(α + 1) ⎩ ρ2 α 2 (1 + 3υ ) 2 ⎫ M (3 + υ ) ⎧ 2 ρ ⎬ − ⎨α + 1 + 2 − (3 + υ ) ⎭ 8 ρ ⎩ [7a] ⎫ α2 K (2 + υ ) ⎧ 2 ( ) + + − − + α α 1 α ρ 1 σr = ⎬ ⎨ 3(α + 1) ⎩ ρ2 ⎭ α2 M (3 + υ ) ⎧ 2 2⎫ − ⎨α + 1 − 2 − ρ ⎬ 8 ρ ⎩ ⎭ [7b] 9 • For the particular case of an Infinite Solenoid: B2 = 0 and µoJa1(α-1) = B1 and assuming ν ≈ 1/3 (usual for most materials) [7a] and [7b] reduce to: ⎡ B1 2 ⎤ 2 ⎧ 7α ⎫ ⎡ 2 ⎤ α2 5 ( ) + + + − + 1 σθ = ⎢ α α α 1 ρ ⎬⎢ ⎥ ⎥ 2 ⎨ ( ) + 2 9 1 µ α ρ2 7 ( ) − 1 α ⎩ ⎭⎣ ⎦ ⎣ o⎦ 5⎡ 2 α 2 3 2⎤ − ⎢α + 1 + 2 − ρ ⎥ 12 ⎣ ρ 5 ⎦ [8a] ⎡ B1 2 ⎤ 2 ⎧ 7α ⎫ ⎡ 2 ⎤ α2 ( ) + + − − + σr = ⎢ α α 1 1 ρ α ⎬⎢ ⎥ ⎥ 2 ⎨ ( ) + 2 µ 9 α 1 ρ2 ( ) − α 1 ⎩ ⎭⎣ ⎦ ⎣ o⎦ ⎤ α2 5⎡ 2 − ⎢α + 1 − 2 − ρ 2 ⎥ 12 ⎣ ρ ⎦ • [8b] See Figure for σ vs ρ at different α. 10 • Comments Thin Coil α = 1.3 – Large hoop stress for unsupported turn case – σr is effective in spreading out hoop stress and lowering peak value. Medium Coil α = 1.8 – Special case where σ θ = σ’θ’ at ρ = 1, i.e., the peak hoop stress is not affected by σr Fat Coil α = 4 – Radial stress becomes tensile at inner layers. This results in doubling the hoop stress at ρ = 1. 11 Conclusions – Radial stress is beneficial in reducing hoop stress in solenoids if it is compressive. – Radial stress makes matters worse if it is tensile. – Also, tensile stress is bad for insulation – film and epoxy resins cannot take much tension before cracking or separating. • Could cause winding delamination • Could lead to energy release and quenching. – Often prestress is applied at RT during coil fabrication to maintain only radial compression under all conditions of cool-down and operating energization. – Fig. 4.3 shows criteria derived by Middleton and Trowbridge to provide σr compression. 12 Other Considerations • For thick windings, divide the coil into several thinner, mechanically separate, concentric sections to prevent radial tension. • The assumption of isotropic properties (elastic) is often not good: – E for metals is usually much higher than E for insulating materials. – This makes windings ‘spongy’ in the radial direction. • Axial forces always cause compressive stresses and do not interact with σr and σθ . – Compute independently and sum them. – Insulating materials are often strong (or at least adequate) in compression. • A special case is a split pair solenoid which requires extra structure to bridge the gap. 13 Strains • Mechanical and Thermal: εθ = εr = σθ Eθ σr Er − υθr − υ rθ σr Er σθ Eθ − δθ 4K θ δ θ = ∫300 α K T dT −δr 4K r δ r = ∫300 α K T dT • Condition of orthotropy: du εr = dr u εθ = r υ rθ Eθ = υθr Er • Consider forces and stress distribution under FAULT conditions, e.g. internal short in the coil – This can lead to substantial, and often unsupported stresses and coil damage or destruction. 14 Lorentz Forces in Solenoids 15 Hoop Stress in an Infinite Solenoid 16 Radial Stress in an Infinite Solenoid 17 Hoop Stress in a Finite Solenoid 18 Radial Stress in a Finite Solenoid 19 Axial Stress Comparison Infinite vs Finite 20 ITER Central Solenoid Model Coil B = 13 T Internal pressure ≈ 660 atm ≈10,000 psi 21 Dipoles SSC Prototype Dipole 22 Ideal Tori with Circular Cross-Section G G G G ∫ H ⋅ dl = ∫∫ J ⋅ nda Bφ µo ⋅ 2πr = NI µ o NI Bφ = 2πr Inside Bφ = 0 Outside 23 Ideal Tori with Non-Circular Cross-Section G G G G ∫ H ⋅ dl = ∫∫ J ⋅ nda Bφ ⋅ 2πr = NI µo Bin = Bmax Bmax r1 µ o NI = 2πr1 Bout = 0 r Inside Inside Outside 24 PF Coils and Functions • OH Coils – Ohmic Heating – – – – • Plasma initiation Plasma heating by transformer action High flux linkage with plasma current No or low vertical field in plasma EF Coils – Equilibrium Field – Create vertical field at plasma for radial equilibrium • Elongation Coils – Elongation of Plasma – Double or single null • Shaping Coils – Shape plasma to D, bean or triangular shape in conjunction with elongation coils • Control Coils – Control plasma position both vertically and radially – Either inside Vacuum Vessel (VV) for fast control or outside VV for slow control – Used in conjunction with other PF coil functions • • • All coils are coaxial solenoids or Ring Coils All coils excited independently with time-varying currents (often bipolar) during plasma cycle – Fields from coils are superposed so current supply may follow complex waveform for combining functions Seldom use magnetic materials – Exception: OH transformer, e.g. JET, Tore Supra (but not with SC magnets) – Sometimes used for TF ripple correction 25 ITER Coils 26 ITER Coils 27 CS Currents Operation Scenario 28 PF Currents Operation Scenario 29 CS and PF Currents Operation Scenario ITER CS 15MA Senario 30 20 Plasma 10 CSU3 CSU2 CSU1 0 0 500 1000 1500 2000 CSL1 CSL2 -10 CSL3 -20 -30 ITER PF 15MA Senario 20 15 Series1 10 PF1 PF2 5 PF3 PF4 PF5 0 0 500 1000 1500 2000 PF6 -5 -10 30 Out-Of-Plane (OOP) Forces in TF Coils • Under usual operating conditions there are no out-of-plane loads on a TF coil due to other TF coils • Watch out for fault conditions (e.g. unequal currents) 31 TF Forces Figure 3 32 OOP in TF Coils 33 Force Calculation dF = 1 ( I × B )ds 2 NI ⎛ 1 + γ ⎞ Fz = µ o ⎟ ⎜ 4π ⎝ 1 − γ ⎠ ρ minor radius = where γ = rav major radius 2 Note : γ < 1 Fr = µ o NI 2 ⎡ 2 ⎤ ⎢1 − 2 ⎥ (1 − γ )⎥⎦ ⎢⎣ 1 Note : Since γ < 1 ⎯ ⎯→ Fr < 0 (Centering Force) 34 ρ = kr 2T 4πT k= = IBm r1 µ o NI 2 ⎡ ⎛ dr ⎞ ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎝ dz ⎠ ⎦ ⎣ ρ =± d 2r dz 2 2 32 Combine the 3 equations: d r 1 ⎡ ⎛ dr ⎞ ⎤ r 2 = ± ⎢1 + ⎜ ⎟ ⎥ k ⎣ ⎝ dz ⎠ ⎦ dz r1 = ro e − k , r2 = roe +k 2 2 32 ro = r1r2 1 k = ln(r2 r1 ) 2 35 Comments About Constant Tension Tori NFz = Bm2 πr12 µo ln(r2 r1 ) Fz = Z - directed force/coil half N = number of coils in torus • This shows why structural design becomes more difficult as size and magnetic field increase Fz ∝ (Bm r1 ) 2 • Note also that Fz depends only on r1 and r2, not on shape of torus. µo I 2 • For equilibrium Tc − Td = 4π (kC − k D ) 36 Comments About Constant Tension Tori • Methods of connection and supports affect constant T characteristics and generate bending moments. • Each coil segment is constant T, but not momentless unless coil reactions respond to loads in a particular way: – Must account for shape change in response to EM loads + reactions. – Must take account of actual coil and structure stiffness and cross-sections. – Must account for discreteness of coils in toroidal direction. – Actual location and size of coil supports is affected by overall Tokamak design: • Size, location of plasma, VV, shields, ports, maintenance access, etc. – In general, a 3D analysis of loads and stresses must be made. 37 Alcator C-Mod Alcator C-Mod Cross Section/Elevation View - Structural Elements Sliding finger joints in Alcator CMod 38 3D FEA FIRE Model 39

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