Parameters of the NF TargetProton Beam pulsed 10-50 Hz pulse length 1-2 s energy 2-30 GeV average power ~4 MW

Target (not a stopping target)

mean power dissipation 1 MW energy dissipated/pulse 20 kJ (50 Hz) energy density 0.3 kJ/cm3 (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 ProgrammeSince 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 - 105 pulses.7. In-beam tests at ISIS – 107 pulses.8. Design target station.

Shock, Pulse Length and Target SizeWhen 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.

Es

The time taken for the wave to travel from the outer surface to the centre is given by

If the beam pulse (τp) is long compared to the characteristic time τs, 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!

psd

sd

s

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 HeatingPulsed ohmic-heating of wires can replicate pulsed proton

beam induced shock.

current pulse

tantalum wire

Energy density in the wire needs to be ε0 = 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:

rj

rrj

tj

zz 112

2

0

1 1

00

221n nn

ntzz aJ

rJe

ajj n

= 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”

151- s cm 3.3x10 ns μm 3.3 -Es

2aI v

as

Characteristic Times for the shock to travel across the radius and for the current to penetrate the wire (square pulse).

Ips

τsτIns

a, mm 10.1

10

100

10000

1000

Doing the TestThe ISIS Extraction Kicker Pulsed Power Supply

Time, 100 ns intervals

Voltage waveform

Rise time: ~100 ns Voltage peak: ~40 kV Repetition rate up to 50 Hz.

Flat Top: ~300 ns Current Peak: ~8 kA There 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

tejj 10

12

1

0

2

0

2

0

0

2

21n n

nn

nt

e

aJ

a

earJ

ae

aJ

rJ

jj

tan

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

r/a

j/j0

30 ns

10 ns 1 ns

100 ns1000 ns

Current density versus radius at different times. Wire radius, a = 0.3 mm.

0 2 10 8 4 10 8 6 10 8 8 10 8 1 10 70

0.2

0.4

0.6

0.8

1

t0 2 10 8 4 10 8 6 10 8 8 10 8 1 10 7

0

0.2

0.4

0.6

0.8

1

0 2 10 8 4 10 8 6 10 8 8 10 8 1 10 70

0.2

0.4

0.6

0.8

1

t

Current density at r = 0 versus time (t, s), for different wire radii (a, mm).

0.1 mm 0.2 mm 0.3 mm 0.4 mm

0.6 mm

te 1j/j0

, s

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

a

e drtarrjM0

),,(2

1 10 9 1 10 8 1 10 7 1 10 60

0.2

0.4

0.6

0.8

1

t1 10 9 1 10 8 1 10 7 1 10 60

0.2

0.4

0.6

0.8

1

t

te 1

1 mm

0.5

0.4

0.3

0.2

0.1

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.

Radiusmm

Characteristic timeτs, ns

Energy densityMJ m-3

0.1 30 3230.2 60 690.3 90 210.4 120 80.5 150 3.40.6 180 1.60.7 210 0.80.8 240 0.40.9 270 0.21.0 300 0.1

The energy density dissipated at the centre of the wire, within a time τs fordifferent wire radii, with I0 = 10000 A.

t e dtjE0

2

Energy Density in the Wire

Radiusmm

Characteristic timeτs, ns

CurrentI0, kA

0.1 30 100.2 60 210.3 90 370.4 120 610.5 150 940.6 180 1380.7 210 1970.8 240 2730.9 270 3731.0 300 499

02

0 jaI

Pulse Current Requirements

The current, I0, required to dissipate 300 MJ m-3 at the centre of the wire within a time τs, for different wire radii.

to vacuum pumpor helium gas cooling

heater

test wire

insulators

to pulsed power supply

to pulsed power supply

water cooled vacuum chamber

Schematic diagram of the test chamber and heater oven.