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TDI longitudinal impedance simulation with CST PS A.Grudiev 20/03/2012.

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Presentation on theme: "TDI longitudinal impedance simulation with CST PS A.Grudiev 20/03/2012."— Presentation transcript:

1 TDI longitudinal impedance simulation with CST PS A.Grudiev 20/03/2012

2 Geometry All metal and dielectric parts are from PEC. No losses. No ferrites are included. Magnetic wall BC is applied at the horizontal plane PML BCs are applied at the up/downstream ends

3 Mesh, sigma_z=500mm

4 Longitudinal Wake, sigma_z=500mm

5 Longitudinal impedance, sigma_z=500mm

6

7 Mesh, sigma_z=200mm

8 Longitudinal Wake, sigma_z=200mm

9 Longitudinal impedance, sigma_z=200mm

10

11 Mesh, sigma_z=100mm

12 Longitudinal Wake, sigma_z=100mm

13 Longitudinal impedance, sigma_z=100mm

14

15 Mesh, sigma_z=50mm

16 Longitudinal Wake, sigma_z=50mm

17 Longitudinal impedance, sigma_z=50mm

18 Mesh, sigma_z=20mm

19 Longitudinal Wake, sigma_z=20mm

20 Longitudinal impedance, sigma_z=20mm

21 Longitudinal Wake, Summary plots

22

23 Longitudinal Impedance, Summary plots

24

25

26

27 Different beam locations: b0, b1, b2 b0; X=0 b1; X=-8mm b2; X=-68mm

28 Longitudinal Wake, σ z =100mm: b0, b1, b2

29 Longitudinal Impedance, σ z =100mm: b0, b1, b2

30 Longitudinal Impedance, real part, σ z =100mm: b0, b1, b2

31 Longitudinal Impedance, imaginary part, σ z =100mm: b0, b1, b2 Half gap = 8mm b0: Z/n = 155 Ohm/250MHz * 400.8MHz/35640 = 7.0 mOhm b1: Z/n = 150 Ohm/250MHz * 400.8MHz/35640 = 6.7 mOhm b2: Z/n = 70 Ohm/200MHz * 400.8MHz/35640 = 3.9 mOhm

32 Longitudinal Wake, σ z =100mm: b0 PML8 -> PML16

33 Longitudinal Impedance, real part, σ z =100mm: b0, PML8 -> PML16 Almost no difference

34 Longitudinal Wake, σ z =100mm: b0 beam pipe length: 200mm -> 100mm and 300mm

35 Longitudinal Impedance, σ z =100mm: b0, beam pipe length 200mm -> 100mm and 300mm

36 Beam pipe length of 300 mm is better, but the difference is only at f ~ 0 And the negative offset of the ReZl is always there at the same level.

37 Ti coating of hBN blocks Dear all, Here is a coating report from Wil (please follow the link), for a batch of BN coated in 2010. The specifications we had been asked to meet were Rsquare<0.5 Ohm. For a thickness of about 5 µm that means a resistivity of about 250 e-8 Ohm.m, larger than the nominal Ti value. This is likely due to the large amount of outgassing from the porous BN material. Cheers, Sergio & Wil See EDMS link https://edms.cern.ch/document/1085514/1https://edms.cern.ch/document/1085514/1 For this coating skin depth in the range from 10 MHz to 1 GHz is 250 um to 25 um which is bigger than the coating thickness of 5 um.

38 Longitudinal Wake, σ z =100mm: b0 PEC -> hBN

39 Longitudinal Impedance real part, σ z =100mm: b0, PEC -> hBN

40 Longitudinal Impedance, imaginary part, σ z =100mm: b0, PEC  hBN Half gap = 8mm b0, PEC: Z/n = 155 Ohm/250MHz * 400.8MHz/35640 = 7.0 mOhm b0, hBN: Z/n = 2620 Ohm/400MHz * 400.8MHz/35640 = 73.7 mOhm

41 Longitudinal Impedance, real part, : b0, hBN, σ z =100 - > 50 mm

42 Influence of the ferrite 4S60

43

44 Longitudinal impedance gap 16mm hBN, with and w/o 4S60 NO DIFFERENCE

45 Influence of Mask for RF fingers region

46 Longitudinal impedance gap 16mm hBN, σ z = 100 mm, with and w/o Mask No big difference in CST wakefield solver BUT Saves a lot of mesh in HFSS eigenmode solver

47 Longitudinal impedance gap 16mm hBN, σ z = 50 mm, with and w/o Mask No big difference in CST wakefield solver BUT saves a lot of mesh in HFSS eigenmode solver

48 R/Q estimate from PEC impedance Reminder from classical P. Wilson, SLAC-PUB-4547 For impedance of N modes with Q >> f/df, where df=c/s_max, for PEC Q~∞

49 R/Q estimated from longitudinal impedance, hBN, b0, σ z = 50 mm 4(Zl-Zl0)*df/πf is plotted where Zl0 = 71 Ohm to make the real part positive

50 Go to HFSS results

51 Power estimated from ReZl, hBN, hgap=8mm, σ z = 85 mm, same HWHH: b0,b1,b2

52 Power estimated from ReZl, hBN, hgap=20mm, σ z = 85 mm, same HWHH : b0,b1,b2

53 Power estimated from ReZl, hBN, hgap=55mm, σ z = 85 mm, same HWHH : b0,b1,b2

54 Power estimated from ReZl, hBN, hgap=8->20->55mm, cos^2 bunch, HL-LHC 25 ns beam : b0,b1,b2 hgap=8mm hgap=20mm hgap=55mm The impedance of the low frequency modes (<200MHz) weakly (far from linear) depends on the gap! At fully open jaws position a few 100s of Watts can be dissipated mainly on the block keepers and beam screen. The impedance of the higher frequency modes (> 1 GHz) depends on the gap, roughly linear with the gap. Power dissipation is reduced from a few kilowatts down to the level of 100 Watts.

55 Transverse impedance

56 Transverse impedance dy=2mm, hgap=8mm, b0 different materials for the hBN blocks: PEC and hBN PEC+PEC(pure geometrical): Im{Zy}(f->0) = 600Ω/2mm = 300 kΩ/m PEC+hBN(geometrical+dielactric): Im{Zy}(f->0) = 5400Ω/2mm = 2.7 MΩ/m Non coataed hBN blocks result in 9 times higher transverse BB impedance

57 Transverse impedance dy=2mm, hgap=8mm, b0 hBN blocks with and without ferrite 4S60 No significant difference. Ferrite does not damp transverse modes significantly. Its location is not optimal.

58 Some conclusions The ferrite 4S60 are not very effective in its present location Imaginary part of the Broad band impedance both longitudinal and transverse is increased by ~ factor 9 if no coating is assumed on the hBN blocks Parameters (f0, Q, R/Q) of all significant trapped modes has been calculated in FD using HFSS. R/Q and f0 agree rather well with CST estimate. RF heating estimate based on the CST results for half gap of 8, 20 and 55 mm are made for HL-LHC beam parameters. For cos^2 bunch shape it can reach few kW level if no coating is assumed on the hBN blocks

59 Recommendations Maximum LHC beam parameters are assumed to be 2808 b x 1.15e+11 p/b for the operation between LS1 and LS2 1.Cu coating of hBN blocks of at least 10 um or more if possible 2.Improve cooling of the jaws to be adequate to the RF heating of the absorber block keepers which can reach 1 kW level at injection (half gap 8 mm) or 100 W level at collisions (half gap 55 mm) 3.The stainless steel beam screen must be coated with at 100-200 um of Cu. Adequate cooling if necessary has to be implemented in order to evacuate the RF heating power load of 100 W all along the screen.


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