Violin modes in tungsten wire

1. Goran Skoro Violin modes in tungsten wire

2. See: vm_real_target.mpg Peak stress (beam on axis): 150 MPa Surface displacement oscillation frequencies: ~120 kHz (radial stress); ~10 kHz (longitudinal stress); ~2.5 kHz (bending). Tungsten target: 3x20 cm Beam: 4 MW, 50 Hz, 6 GeV, 4x2ns, 10 s Reminder: Violin modes in tungsten target Beam radius = rod radius; Beam offset = 1/2 radius. (3) (2) (1) (3) (1) (2) Energy deposition:MARS beam beam beam Peak Stress = 170 MPa Peak Stress = 175 MPa Peak Stress = 182 MPa LS-DYNA (3D)

3. TUNGSTEN Reminder: Single wire case Initial temperature = 2000 K Geometry • 0.5 mm diameter; 5 cm long wire Loads • Current pulse: 5 -7.5 kA, 800 ns long strain • Energy density; temperature rise across the wire; • Lorenz force induced pressure wave

4. How to induce transversal (violin) modes of wire oscillation (and put additional stress into the wire) Ideas: 1) By pulsing the current through 2 parallel wires* 2) By pulsing the current through single bent wire** strain *) Additional force between the wires should induce their bending **) Bending the wire (before pulsing) should broke the (cylindrical) symmetry and force the wire to bend additionally

5. Case 1. Two parallel wires y +a dl’ d 0 x c dl -a Additional force = ? • “Very long” wires (1) d – distance between wires • If finite lengths -> force depends on position (c) and wires lengths (l=2a) …after a few simple transformations and integration… X-component: Y-component: if a >> d then X-comp -> (1), Y-comp -> 0

6. Case 1. Parallel wires Von Mises Stress 7500 A 1 cm distance 3 cm distance 5 cm distance LS-DYNA l = 5 cm

7. See: vm_2wires.mpg Case 1. Parallel wires Radial displacement 7500 A 5 cm distance LS-DYNA l = 5 cm

8. Case 1. Parallel wires NAS For calculation of the SWS for 0.5mm diameter, 5cm long tungsten wire at 2000 K a quick, approximate formula can be used: SWS [MPa] = 6xI[kA]xI[kA]. Additional stress as a function of the peak current and distance between the wires For example, doubling the stress using the 2nd wire (NAS = 1) means that F = 3.6. For I = 6 kA (let’s say), distance between wires should be d = 1 cm. LS-DYNA Additional stress (AS) normalized to the single wire stress (SWS) for corresponding peak current I = peak current [kA], d = distance between wires [mm].

9. Case 2. Bent wire 2 characteristic points 2 1 wire length = 5 cm bent at the middle (b) wire radius (r) = 0.25 mm b Peak current = 5 kA b/r = 1/8, 1/4, …, 4, 8 LS-DYNA

10. See: vm_bent_wire.mpg Bending frequency: ~ 1 kHz Case 2. Bent wire 2 1 wire length = 5 cm wire radius (r) = 0.25 mm b Peak current = 5 kA b/r = 2 b/r = 2 C LS-DYNA

11. Case 2. Bent wire b/r 1 2 NAS 1 2 Comparing this with Slide (8) one can see that both approach can be used for inducing violin modes of oscillation and putting additional stress into the wire, but ‘bent wire’ approach looks less complicated. ~ constant Relatively small bending gives similar results as ‘the 2 wire approach’. Additional stress as a function of the wire bending LS-DYNA Additional stress (AS) normalized to the straight wire stress (SWS) for 5 kA peak current