Parameters of the NF Target Proton Beam pulsed10-50 Hz pulse length1-2  s energy 2-30 GeV average power ~4 MW Target (not a stopping target) mean power dissipation1 MW energy dissipated/pulse20 kJ (50 Hz) energy density0.3 kJ/cm 3 (50 Hz) 2 cm 20 cm beam

Schematic diagram of the RAL radiation cooled rotating toroidal tantalum target rotating toroid proton beam solenoid magnet toroid at 2300 K radiates heat to water-cooled surroundings toroid magnetically levitated and driven by linear motors

The New Programme Since NuFact04 the R&D programme has altered significantly: 1. Simulate shock by passing a pulsed current through a wire. 2. Measure the radial motion of the wire to evaluate the constitutive equations (with 3.). 3. Use a commercial package, LS-DYNA to model the behaviour. 4. Life time/fatigue test. 50 Hz for 12 months. 5. Investigate the possibility of widely spaced micro- bunches of proton beam to reduce the shock impact.

Proposed R&D (2004) 1.Calculate the energy deposition, radio-activity for the target, solenoid magnet and beam dump. Calculate the pion production (using results from HARP experiment) and calculate trajectories through the solenoid magnet. 2. Model the shock a) Measure properties of tantalum at 2300 K. b) Model using hydrocodes developed for explosive applications at LANL, LLNL, AWE, etc. c) Model using dynamic codes developed by ANSYS. 3. Radiation cooled rotating toroid a) Calculate levitation drive and stabilisation system. b) Build a model of the levitation system. 4. Individual bars a) Calculate mechanics of the system. b) Model system. 5. Continue electron beam tests on thin foils, improving the vacuum. 6. In-beam test at ISOLDE pulses. 7. In-beam tests at ISIS – 10 7 pulses. 8. Design target station.

Shock, Pulse Length and Target Size When a solid experiences a temperature rise the material expands. Because of mass inertia there will always be a slight lag in the expansion. This causes pressure waves to ripple through the material. When the temperature rise is relatively large and fast, the material can become so highly stressed that there is permanent distortion or failure - shock. Short high intensity beam pulses will give rise to shock in a target. The shock wave travels through matter at the speed of sound, where E is Young's modulus of elasticity and ρ is the density.

The time taken for the wave to travel from the outer surface to the centre is given by If the beam pulse ( τ p ) is long compared to the characteristic time τ s, then little energy goes into the target in this time and the shock wave in the target is reduced. If the target is small compared to the beam pulse length the shock is reduced. If No problem! Must have sufficient pulsed energy input!

The Proton Pulse macro-pulse micro-pulse Proton beam “macro-pulses” and “micro-pulses”. Traditionally we have considered the micro-pulses as ~1 ns wide and the macro-pulses as ~1  s wide. The temperature rise per macro-pulse is ΔT ~ 100 K. For the tantalum bar target, radius 1 cm and length 20 cm, then:  The time for the shock wave to travel a radius is 3 ms  The time for the shock wave to travel a half the length is 30 ms However, in the RAL proton driver scheme with ~10 micro- pulses, it is likely that they could be spaced apart by ~40  s, thus reducing the effective thermal shock to only ΔT ~ 10 K.

Effect of 10 micro-pulses in 1 and 5  s long macro-pulses. NF Target: 1 cm radius 20 cm long  = 3  s Total temperature rise:  T = 100 K Goran Skoro, Sheffield University

Pulse Heating Pulsed ohmic-heating of wires can replicate pulsed proton beam induced shock. current pulse tantalum wire Energy density in the wire needs to be ε 0 = 300 J cm -3 to correspond to 1 MW dissipated in a target of 1 cm radius and 20 cm in length at 50 Hz.

Transient Conditions Assume an electric field E is instantaneously applied across a conducting wire. Apply Maxwell’s equations. This produces a diffusion equation: In cylindrical coordinates, where j is the current density. The solution is:  = 1/  0 

The Velocity of the Shock Wave Shock travels at the Speed of Sound. Can only use small diameter wires of up to ~0.4 mm to get current penetration into the wire. “Choose a better material”

Characteristic Times for the shock to travel across the radius and for the current to penetrate the wire (square pulse). τ s τ I ns a, mm

Doing the Test The ISIS Extraction Kicker Pulsed Power Supply Time, 100 ns intervals Voltage waveform Rise time: ~100 nsVoltage peak: ~40 kVRepetition rate up to 50 Hz. Flat Top: ~300 nsCurrent Peak: ~8 kAThere is a spare available for use. Exponential with 20 ns rise time fitted to the waveform

Solution for the case of an exponential rise in current

r/a j/j 0 30 ns 10 ns 1 ns 100 ns 1000 ns Current density versus radius at different times. Wire radius, a = 0.3 mm.

Current density at r = 0 versus time (t, s), for different wire radii (a, mm). 0.1 mm0.2 mm0.3 mm0.4 mm 0.6 mm j/j0j/j0, s

The average current density over the wire cross section versus time (t, s), for various wire radii, (a, mm). 1 mm

Effect of different pulse lengths (shown by arrows), for exponential current rise of 30 ns. Goran Skoro, Sheffield University.

Effect of different pulse lengths (shown by arrows), for exponential current rise of 30 ns. Goran Skoro, Sheffield University.

The energy density dissipated at the centre of the wire, within a timeτ s for different wire radii, withI 0 = A. Energy Density in the Wire

Pulse Current Requirements The current, I 0, required to dissipate 300 MJ m -3 at the centre of the wire within a time τ s, for different wire radii.

to vacuum pump or helium gas cooling heater test wire insulators to pulsed power supply water cooled vacuum chamber Schematic diagram of the test chamber and heater oven.