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SIS 100 Magnet cooling and cryogenic distribution

SIS 100 Magnet cooling and cryogenic distribution. SIS100. Each sextant consists: 18 Dipoles 28 Quadrupoles 24 Correctors ? Scraper 4 warm sections Length: 180 m. No collimators at cryogenic temperature. SIS100 cooling scheme. Maximal D p = ? => dipole

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SIS 100 Magnet cooling and cryogenic distribution

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  1. SIS 100 Magnet cooling and cryogenic distribution

  2. SIS100 Each sextant consists: 18 Dipoles 28 Quadrupoles 24 Correctors ? Scraper 4 warm sections Length: 180 m No collimators at cryogenic temperature

  3. SIS100 cooling scheme • Maximal Dp = ? => dipole • All other users have to be adopted to this pressure drop Dp by: • introducing an orifice at the inlet (small rage due to manufacturing accuracy • and/ or • Combination of different users

  4. SIS100 cooling scheme • System requires nearly no pressure drop along the supply lines to achieve the same pressure head along the sector for every component. • Dpsupply < 50 mbar • Dpsuction< 20 mbar • dsupply = 0,038 m (0,045m) • dsuction= 0,070 m (0,060 m)

  5. SIS100 cooling dipole Maximal flow given by the geometry <=> flow impedance Minimal flow given by flow regime: slug or plug flow have to be avoided to reach equal temperature distribution within the coil pressure control Vapour fraction at yoke outlet should be kept between .9 and 1., to achieve an energy efficient operation

  6. Internal heat exchangerRecooler between supply line and magnet cooling Approx. 50mdipole/200msextant= 0,25 => Qmax,supply=0.1 W/m

  7. SIS100 inlet conditions = 0.1W/m

  8. Tests Two traditional Nuclotron magnets:

  9. SIS100 cooling dipole Curved double layer dipole Cycle 2c B= 0 -> 1.9 T B=4T/s tcycle=1.8 s .

  10. SIS100 cooling Pressure drop calculations Hermann, 2.104 <Re < 2.106

  11. SIS100 cooling Pressure drop measurement

  12. SIS100 cooling Tin=4.5K; Pin=1.56 bar; Psuction=1.1 bar; m=2.45g/s; tcoil= 63s, tYoke= 12 s

  13. SIS100 cooling Tin=4.5K; Pin=1.56 bar; Psuction=1.1 bar; m=2.45g/s; tcoil= 66s, tYoke= 12 s Tin=4.5K; Pin=1.51 bar; Psuction=1.12 bar; m=2.18g/s; tcoil= 68s, tYoke= 13 s

  14. SIS100 cooling Tin=4.5K; Pin=1.56 bar; Psuction=1.1 bar; m=2.45g/s; tcoil= 66s, tYoke= 12 s Tin=4.4K; Pin=1.56 bar; Psuction=1.1 bar; m=2.56g/s; tcoil= 66s, tYoke= 12 s

  15. SIS100 cooling Tin = 4.51K

  16. SIS100 cooling Tin = 4.51K

  17. SIS100 cooling Tin = 4.51K

  18. SIS100 cooling Tin = 4.5K

  19. SIS100 cooling

  20. SIS100 cooling • Curved double layer option • There is a common range of pressures of operation, where both magnets, the first and the last will work, but • The heat load onto the supply line should not exceed .08 W/m in the magnet cryostats. This should be possible be special shielding from the return line. • To achieve a similar operational field, following groups for the correction elements are proposed by H. Khodzhibagiyan: • Module M1: a quadrupole, pickup and steerer • Module M2: a quadrupole, pickup and multipole • Module M3: a quadrupole, multipole and collimator • Module M4: a quadrupole, steerer and collimator • The beam pipe cooling should be handled as a separate consumer and be put into the modules M1..M4

  21. SIS100 cooling Dipole: static load 7W, load in cycle 2c: 35.7 W Quadrupole: static load 4W, load in cycle 2c: 18.7 W Correctors: ? Beam pipe cooling:? Scraper: ?

  22. 180 m

  23. Feed box on tunnel level in niche • => short reaction time for the control valves • Transfer lines through the tunnel • as few valves as possible in the non accessible area

  24. Distribution box on tunnel level in niche • => Smaller space requirements in the transfer section • Transfer lines through the tunnel • valves in the non accessible area

  25. SIS100 cooling -> To Refrigerator

  26. Pressure drop versus heat load on the coil for a constant overall heat load of 30 W.

  27. T-s-Diagram for three different load distribution onto the coil. The pressure drop variation caused by this load variation is below 2 mbar (or .5%).

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