Technology & Engineering Division Lead Stabilization Alexey Radovinsky (PSFC, MIT) MICE CC Cryostat Design Review Lawrence Berkeley National Laboratory 29 February 2012

Technology & Engineering Division Current Design Cu CC 1 st Stage Cryocoolers (CC) Cu Current Lead HTS Current Lead Cu Terminals Bracket G10 Insulating Plate G10 Insulating Inserts Radiation Shield (transparent) Stabilized SC Leads (redesigned-?) Cu Current Lead HTS Current Lead Cu CC 1 st Stage G10 Spacer RT Feedthrough

Technology & Engineering Division McFee Optimized Cu Current Leads Cold End: T=T0, Q=Q0 Warm (RT) End: T=T1=300 K, Q=Q1=0 0Lx K(RRR,T)*d 2 T/dx 2 + dK(RRR,T)/dT*(dT/dx) 2 + rho(RRR,T)*(Ic/A) 2 = 0 RRROptimum (L*Ic/A)T0Q0A (Ic=210 A, L=0.3 m) A/mKWmm^ E E E Here K(RRR,T), rho(RRR,T) – Cu thermal conductivity and resistivity by NIST Iterate with boundary conditions, T| x=L =T1, dT/dx| x=L =0, and adjusting the cross section, A, till the cold end acquires the required temperature, T| x=0 = T0, Calculate Q0= - K(RRR,T0)*A*dT/dx| x=0 Cold end temperature, T0, is set to 46 K  Low sensitivity of optimum (L*Ic/A) to RRR, within tolerances on A  Cold end heat load about 10 W per lead (agrees with M.Green)

Technology & Engineering Division McFee Optimized Cu Current Leads (2) RRR=10 RRR=50 RRR=100

Technology & Engineering Division Challengers of CU Current Leads analyses Question is what is the appropriate temperature of the cold end of the Cu Current Lead, T0 Account for the balance of heat loads at the intermediate cooling station integrating 3 cryocoolers on 3 copper plates connected by 2 pairs of copper joint straps, and 2 copper blocks with prongs for terminals for the RT and HTS current leads Cryocooler cooling power is a function of the temperature of its first and second stages Temperatures at the first stage of the Central and Side cryocoolers are different Temperature drop between the cryocooler first stage and Cu Lead cold end is calculated by FEA Heat loads are defined with a wide margin of probability – up to 100% contingency is applied Usually analyses iterate between the cryocooler performance characteristics and the intermediate cooling station temperature distribution FEA A method is developed to converge in one step

Technology & Engineering Division Prior Art (0): References WANG, Li, “Calculations and FEA simulations for MICE/ Muool Coupling Magnet Cryostat Design,” Review at SINAP/Shanghai, Michael A. Green, “Results from the Cooler and Lead Tests,” MICE Note 291, Lawrence Berkeley National Laboratory, 20 June 2010 “Coupling Solenoid Magnet Engineering Design Report,” INSTITUTE OF CRYOGENICS AND SUPERCONDUCTIVITY TECHNOLOGY. HARBIN INSTITUTE OF TECHNOLOGY. CHINA, Solid Works Coupling Coil for 20MIT/ Next 4 slides summarize conclusions and the data from these studies used for the present analyses Highlighted in RED

Technology & Engineering Division Prior Art (1): Heat Loads at the First Stage of the Cryocooler Heat loads from 300K to 60K (W) Previous designCurrent LBL design Copper leads from 300 K19.30 Cold mass supports (intercept T=72-77 K) Radiation Heat to the Shields * Instrumentation wires0.092 He Cooling Tubes Level sensor tube Cooler SS sleeves9.73 Neck shield supports0.88 Heat shield supports (longer support) Sub-total (calculated) Total Heat Load with 50% Contingency Total Heat Load with 100% Contingency Total heat load at the first station from all sources, except for the Current Leads is set to 74 W. It conservatively accounts for an ~ 100% contingency.

Technology & Engineering Division Prior Art (2): Heat Loads at the Second Stage of the Cryocooler Heat load from 60K to 4.2K (W) Previous designCurrent LBL design HTS current leads ( Warm HTS end T = 64 K) K Cold mass supports (intercept T=73-78K) Radiation heat to 4.2K cold mass (Shield Tave =70K) (W) (MLI layer is 20)* Instrumentation Wires He cooling tubes Level sensor tube Cooler SS sleeves** Superconducting Splices at 10 nW per splice and 210 A 0.01 Sub-total (calculated) Total with 50% Contingency Total with 100% Contingency Total heat load at the second station with a 100% contingency is close to the full capacity of 3 PT415 cryocoolers. Certified Performance: with We can assume 4.2 K at the second stage.

Technology & Engineering Division Prior Art (3): Cryocooler Performance Diagram Specification by Cryomech Data taken by Florida State University, confirmed by M. Green P415 = *T *T^ *T^ *T^4 PT415 first stage capacity, P415 [W], vs. temperature, T [K], for the second stage at 4.2 K P415 (W) First stage T (K)

Technology & Engineering Division Prior Art (4): LBL Design and Temperature Distribution at Intermediate Station 60.0 K at the Cryocooler Flange 69.6 K at Cu Lead side 63.5 K at HTS Lead side  G10 spacer with dT≈150 K unaccounted in design  6 bolts connecting Cu Bracket and Cu Plate electrically insulated by G10 inserts.  1-mm thick G10 electrical insulator between the Cu Bracket and Cu Plate.  Too complex and unreliable.  Temperature drop between the first stage of the cryocooler and the cold end of the Copper Lead dT~9 K.  Too much. Alex Zhukovsky proposed redesign of the Cu Brackets mounting on the Cu Plate

Technology & Engineering Division Proposed redesign of the Cu Brackets The Copper Block has no holes in it. It is wrapped into Kapton and clamped to the Copper Plate by 2 SS plates, one below the Copper Block and one above the Copper Plate, tied together by 6 bolts, 3 in a row on each side of the Copper Block. SS plates are insulated from the Copper Block by Kapton and their connection bolts need no insulation from the Copper Plates. A 2-mil ( mm) layer of Kapton with thermal conductivity of 0.15 W/m-K is easily sufficient for a 1-kV max voltage defined by Brad Smith quench analyses The distance between the holes and the edge of the Copper Plate is sufficient for the present bolted connection to the radiation shield. To make this possible the width of the Copper Blocks is reduced from 82 mm to 60 mm. The width of the terminal prongs is scaled in the same proportion, 60/82. This redesign is implemented in the FE model of the intermediate cooling station

Technology & Engineering Division Vector Fields FEM Parts Diagram Bottom ViewTop View Central (c) Copper Plate shown in blue and two Side (s) Copper Plates shown in magenta are treated as separate parts. Below parameters related to these plates are marked by respective extensions, “_c” and “_s”.

Technology & Engineering Division Power Balance Power balances for the Central and the Side Copper Plate subassemblies Central:- P415_c(T415_c) + P_shield_c + 2*P_straps = 0 Side:- P415_s(T415_s) + P_shield_s + P_300k + P_4k - P_straps = 0 Here Cryocooler Heat Capacities P415_c(T415_c) = * T415_c * T415_c ^ * T415_c ^ * T415_c ^4 P415_s(T415_s) = * T415_s * T415_s ^ * T415_s ^ * T415_s ^4 P_300k = 10 W and P_4k = W - power loads from the RT and HTS Current Leads P_straps = dPdT*( T415_s - T415_c), where dPdT= 11.2 W/K from subsidary model Solving nonlinear system of equations calculate a combination of values of the temperature and cooling power of cryocoolers, (T415_c, P415_c) and (T415_s, P415_s), which is expected to provide a power balance between the plates subassemblies with the account of heat transfer through the straps.

Technology & Engineering Division Vector Fields FEM Boundary Conditions BCs Flux_cryo_s and Temp_cryo_s BCs Flux_cryo_c and Temp_cryo_c BC Flux_shield_sBC Flux_shield_c BC Flux_4k BC Flux_300k

Technology & Engineering Division Vector Fields FEM Boundary Conditions (2) Calculated (T415_c, P415_c) and (T415_s, P415_s) are shown in BLUE Heat flux, Flux_shield_c = Flux_shield_s = Flux_shield, is defined by smearing 74 W, the total heat load at the first station from all sources, except for the Current Leads, over the outer perimeter of the Cu Plates contacting the radiation shield. Verification: FEM calculated power transferred through a pair of Straps is W, which is close to the expected |P_straps| = W in the table

Technology & Engineering Division FEM Results Due to the abundance of the cooling power at the first stations of the cryocoolers their temperatures are close to 37 K and the cooling power is evenly shared between 29 W in the Central cryocooler and 32 W in the Side cryocoolers 1.5-cm thick Copper Plates provide sufficient thermal conductivity to limit temperature drops between the cryocooler and any point of the Copper Plate to less than 1 K With a 2-mil Kapton insulator between the Copper Plate and the Copper Block the warmest point of the model, the Copper Current Lead terminal of the Copper Block, is just 2 K warmer than the first station of the cryocooler attached to the Side Copper Plate. Temperature drop across the insulator is a fraction of 1 K.

Technology & Engineering Division FEM Results (2) The purpose of Copper Straps is to allow independent flotation of the Copper Plates to avoid overstressing the structure of the cryocolers between the RT and the first station flanges due to its different thermal contraction. Structural analyses can show if the flexibility of the Straps made of solid copper as in the current design is sufficient for this purpose. One of the alternatives, free of these potential drawbacks can be using flexible thermal links made of laminated Cu sheets like MIT used in LDX L-coil. The spec for choosing these links is that thermal resistance of a connection between any of the Side Plates to the Central Plate at the 40-K temperature has to be no less than 11.2 W/K as defined in this study. A model was analyzed with no Straps connecting the Copper Plates to assess their value for the design. In this case the Central cryocooler delivers 3 times less power than each of the Side cryocoolers. It’s temperature is almost 6 K lower. Central: (T415_c, P415_c) = (33.7 K, W) Side: (T415_s, P415_s) = (39.4 K, W)

Technology & Engineering Division Optimized Cu Leads Sensitivity Conservative Case: 100% contingency on heat load – 74 W, T0-Tcc=9 K temperature drop between cryocooler and Cu lead cold end Optimistic Case: 0% contingency on heat load – 37 W, T0-Tcc=0 K temperature drop between cryocooler and Cu lead cold end Potential power disbalance at the RT end of Cu Lead is of the order of 1 W

Technology & Engineering Division Steady State Heat Leak thru Cu Ic=0 The maximum heat leak occurs when the temperature drop between the RT and the cold ends of the lead is the biggest. The lowest possible temperature at the first stage, 32 K, is defined by the Cryocooler Capacity diagram. Assuming T1=300 K and the cold end at T0=32 K we can calculate corresponding values of the heat leak, Q1= Q0. Q = -A/L*Integral[K(T), { T0, T1}] The maximum heat leak at the Lead RT end is small, about 7 W per lead. It can be removed by thermal conduction to the cryostat or by using a fan.

Technology & Engineering Division Cold End Temperature Rise at Cryocooler Failure If all cryocoolers fail at the same time at full coil current the emergency current dump scenario will be initiated. At this time we are not aware of the specifics of the current vs. time scenario of for this event. Let us assume that it is safe if the temperature of the conductor carrying full current, Ic, grows by deltaT=200 K and evaluate how long it will take for the cold and warm ends of the lead to increase its temperature from the normal operational value, T0 and T1 to T0+deltaT and T1+deltaT respectively. Under the most conservative adiabatic assumption this temperature rise is time0 = dens*(A/Ic)^2*Integral[Cp(T)/rho(T), { T0, T0+deltaT}] time1 = dens*(A/Ic)^2*Integral[Cp(T)/rho(T), { T1, T1+deltaT}] The minimum time for an adiabatic temperature rise of 200 K is about 150 seconds. This is quite sufficient for detecting the failure and quick discharge (124 s from kV at terminals)

Technology & Engineering Division HTS Current Lead Stabilization Design Cu stabilizer for SC Current Lead Insulation Pad Twisted Cu Bridge for Cu Leads (assumed properly sized)

Technology & Engineering Division HTS Current Lead Failure Modes 500 A HTS-110 current leads can go to 80 K before sudden thermal runaway occurs Mode 1: At the HTS Lead warm end all 3 cryocoolers fail simultaneously. How long does it take for the intermediare cooling station to go from 40 K to 80 K? Heat load, P=100 W, copper mass m=54 kg t = m/P*Integral[Cp,{T,40,80}] = 2900 s Mode 2: Quench of LTS conductor between HTS Lead cold end and the coil. Conductor: length, L=30 cm, Cu cross section area, A=1.2 mm 2. Shunted by Cu block: mass, m=0.15 kg How long does it take for the Cu block to go from 4.5 K to 80 K? Heat load, P=rho(T)*(Ic/A)2*(L*A) W t = m*A/(L*Ic 2 )*Integral[rho(T)*Cp(T),{T,4.5,80}] = 73 s This time is insufficient for detecting the failure and quick discharge. The mass of the stabilizer has to be at least doubled