Massachusetts Institute of Technology 22.68J/2.64J Superconducting Magnets. February 27, Lecture #4 Magnetic Forces and Stresses
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1 Massachusetts Institute of Technology.68J/.64J Superconducting Magnets February 7, 003 Lecture #4 Magnetic Forces and Stresses 1
2 Forces For a solenoid, energy stored in the magnetic field acts equivalent to an internal pressure: E B = Vol 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 (?)
3 F = J B Compute the field throughout the winding volume Combine computation of forces and stress analysis with calculation of B Usually done by computer. Simple analytical formulations are useful for early stages of design. Must have knowledge of materials properties: Mechanical Thermal (COE for thermal stresses) From room temperature (RT) to 4K 3
4 Forces and Stresses in Solenoids B is highest in axial field direction, B z, and current is azimuthal, J θ, so largest forces are radial circumferential hoop stresses B r is high at coil ends axial compression 4
5 Analysis Simplest assumption: Each turn acts independently The a tension develops in each turn: T = B (r )Ir Overall hoop stress σ θ σ θ = B (r )Jr [1] Where both J and σ θ are averaged over the winding pack Thus stresses increase with B, J and r (size) 5
6 Field distribution across a solenoid is often approximately linearly decreasing from inner radius to outer radius: B = ( a r ) B 1 + ( r a 1) B ( a a 1 ) B(ρ ) = (α ρ ) B + (ρ 1) B 1 (α 1) [] Where ρ = r/a 1, α = a /a 1 For the special case of an infinitely long solenoid, we have B = 0 and B 1 = µ o J (a a 1 ) so that σ θ = BJr = B 1 (α ρ µ o (α 1) ρ ) [3] Note that B /µ o is a scaling factor! 6
7 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 ) Let u be the local displacement in the radial direction Reduce the equilibrium condition between σ r and σ θ and BJr to a single equation (Lontai & Marston (1965); and Montgomery (1969)) 1 d (1 υ ) r du u = BJ [4a] r dr dr r E With σ = θ E u du + υ 1 υ r dr [4b] σ r E du u = + υ 1 υ dr r [4c] Where E = Young s Modulus ν = Poisson s Ratio and Tensile Stress is defined to be positive (+) 7
8 For convenience, let r = ρ a, a = α a 1, w = ue/[a1(1-ν )] and BJa 1 = K - Mρ where, K = (α B B ) Ja and M = ( B B 1 ) Ja (α 1) (α 1) So that 1 d dw ρ w = K + Mρ [5a] ρ dρ dρ ρ σ θ = w dw ρ +υ dρ σ = dw +υ w r dρ ρ [5b] [5c] The solution of Eqn. [5a] is: 3 D Kρ Mρ w = Cρ + + [6] ρ 3 8 where the coefficients C and D are obtained by the boundary conditions at r = a 1 and a, i.e., ρ = 1 and α. 8
9 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., σ = 0 at ρ =1 and α. r Substituting these conditions into [5c] and [6] allows definition of C and D and thus the stress functions: K ( +υ) α α + α ρ ( 1 + υ α + 1) )( σ θ = 3(α + 1) ρ ( +υ) M ( 3 +υ) α 1 ( 1 + 3υ) [7a] α ρ ( 3 +υ) ρ σ = K ( +υ) r α 3(α + 1) ρ α + α + 1 (α + 1)ρ M ( 3 +υ) α 1 [7b] α + ρ 8 ρ 9
10 For the particular case of an Infinite Solenoid: α B = 0 and µ oja 1 ( -1) = B 1 and assuming ν 1/3 (usual for most materials) [7a] and [7b] reduce to: σ θ = B 1 7 α α 5 α + α ( α o 9( α + 1 ) µ ( α ) ρ 7 ) ρ 5 α 1 α ρ 3 ρ 5 [8a] σ r = B 1 7α α α α ( α+ )ρ ( α + ) ρ µ o ( α 1) α 1 α [8b] 1 ρ ρ + See Figure for σ vs ρ at different α. 10
11 Comments Thin Coil α = 1.3 Large hoop stress for unsupported turn case σ r 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 σ r Fat Coil α = 4 Radial stress becomes tensile at inner layers. This results in doubling the hoop stress at ρ = 1. 11
12 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 σ r compression. 1
13 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 σ r and σ θ. 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
14 Strains Mechanical and Thermal: σ r 4 K θ δ θ E r ε θ = σ θ υ θ r δ θ = 300 K α T dt E θ σ 4 K r r r r T E E θ ε = r υ rθ σ θ δ δ = 300 K α dt r υ rθ υ θ r Condition of orthotropy: Eθ du ε r = dr ε θ = u r = E r 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
15 Lorentz Forces in Solenoids 15
16 Hoop Stress in an Infinite Solenoid 16
17 Radial Stress in an Infinite Solenoid 17
18 Hoop Stress in a Finite Solenoid 18
19 Radial Stress in a Finite Solenoid 19
20 Axial Stress Comparison Infinite vs Finite 0
21 ITER Central Solenoid Model Coil B = 13 T Internal pressure 660 atm 10,000 psi 1
22 Dipoles SSC Prototype Dipole
23 Ideal Tori with Circular Cross-Section G H d l G G G = J n da B φ πr = NI µ o B φ = µ NI o πr B φ = 0 Inside Outside 3
24 Ideal Tori with Non-Circular Cross-Section G G G G H d l = J n da B φ πr = NI µ o r B in = B 1 max Inside r µ o NI B max = πr 1 Inside B out Outside 4
25 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 5
26 ITER Coils 6
27 ITER Coils 7
28 CS Currents Operation Scenario 8
29 PF Currents Operation Scenario 9
30 CS and PF Currents Operation Scenario ITER CS 15MA Senario Plasma CSU3 CSU CSU1 CSL1-10 CSL CSL ITER PF 15MA Senario Series1 PF1 PF 5 PF3 PF PF5 PF
31 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
32 TF Forces Figure 3 3
33 OOP in TF Coils 33
34 where Force Calculation df = 1 (I B) ds NI 1+γ =µ o 4π 1 γ ρ minor radius γ = = r major radius F z av Note : γ < 1 F r = µ NI o γ ( ) Note : Since γ <1 F r <0 (Centering Force) 34
35 ρ = kr T 4 π T k = = IB m r 1 µ NI o dr ρ = ± dz d r dz Combine the 3 equations: d r 1 dr 3 r = ± 1 + dz k dz k + k r, = re r 1 = o e r o k r o = = r r 1 ln 1 ( r r ) 1 35
36 Comments About Constant Tension Tori NF z = B mπr 1 ln(r r ) 1 µ o F z = 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 F (B m r ) z 1 Note also that F z depends only on r 1 and r, not on shape of torus. o For equilibrium T T d = µ I c (k C kd ) 4π 36
37 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
38 Alcator C-Mod Alcator C-Mod Cross Section/Elevation View - Structural Elements Sliding finger joints in Alcator C- Mod 38
39 3D FEA FIRE Model 39
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