TDI impedance and power loss O. Aberle, F. Caspers, A. Grudiev, E. Metral, N. Mounet, B. Salvant

Context TDI power loss Follow up of E. Metral’s talk at LCE meeting 11/06/2004 for Flat chamber with large aspect ratio  Form factor for longitudinal = 1 Formula for multilayer round pipe without approximation inner radius=4.6 mm for  y =43 m Inner radius=7.7 mm for  y =118 m 1 st Block (2.8 m) Vacuum hBN (54 mm) Ti 3  m 2 nd Block (0.6 m) Vacuum Al (54 mm) Cu 10  m 3 rd Block (0.7 m) Vacuum Cu (54 mm)

Material properties Copper  DC = 17 10^-9 Ω.m  rel = 1  rel = 1  H 1 = 0 relaxationTime= 27 10^-15 s Titanium  DC = 58 10^-8 Ω.m  rel = 1  rel = 1  H 1 = 0 relaxationTime= 0 Hexagonal Boron Nitride (hBN)  DC = 4 10^12 Ω.m  rel = 5  rel = 1  H 1 = 0 relaxationTime= 0 Aluminum  DC = 28 10^-9 Ω.m  rel = 1  rel = 1  H 1 = 0 relaxationTime= 0

Total longitudinal impedance (  y =43 m) Linear scale Log scale Losses occur mainly in the first block Ztotal 1 st block (Ti+hBN+vacuum) 2 nd block (Cu+Al+vacuum) 3 rd block (Cu+vacuum) Z(2 nd block)~Z(3 rd block) Ztotal~Z(1 st block) P loss (1 st block) ~ 162 W P loss (2 nd block) ~ 0.6 W P loss (3 rd block) ~ 0.7 W P loss (total) ~ 163 W

Total longitudinal impedance (  y =118 m) Linear scale Log scale Losses occur mainly in the first layer Ztotal 1 st block (Ti+hBN+vacuum) 2 nd block (Cu+Al+vacuum) 3 rd block (Cu+vacuum) Z(2 nd block)~Z(3 rd block) Ztotal~Z(1 st block) P loss (1 st block) ~97 W P loss (2 nd block) <1 W P loss (total) ~ 98 W

1st block (Ti-hBN-Vacuum)  Z 1 layer (Ti) 2 layers (Ti+hBN) 3 layers (Ti+hBN+vacuum) Infinite thick wall (Ti)

Power loss in the first block From F. Ruggiero, Single-beam collective effects in the LHC CERN-SL AP (1995) Formula assumes a gaussian bunch for P loss = 163 W P loss ~ 98 Wfor 1 layer (Ti) 2 layers (Ti+hBN) 3 layers (Ti+hBN+vacuum) Infinite thick wall (Ti) Significant losses in the hBN?

Losses in the hBN 1 layer (Ti) 2 layers (Ti+hBN) 3 layers (Ti+hBN+vacuum) Infinite thick wall (Ti) 2 layers (Ti+vacuum) At f=10 10 Hz, the skin depth in titanium is ~ 3  m… A single layer of titanium surrounded with vacuum leads to P loss ~ 0.04 W This 3  m layer surrounded with 54 mm of hBN leads to P loss ~ 162 W This means that all the power (162 W) is lost in the hBN.

Effect of hBN conductivity on impedance of the 1 st block  (hBN)= Ω.m  (hBN)= Ω.m  (hBN)= Ω.m  No effect of hBN conductivity on the power loss (in this Ω.m range…)

Effect of hBN permittivity on impedance of the 1 st block  r (hBN)=1  r (hBN)=1.1  r (hBN)=2  r (hBN)=5  strong effect of hBN permittivity on the power loss, but only if  r ~ 1. If  r >2, the effect is small, as Alexej already observed

Conclusion Significant power loss dissipated in the hBN (162 W)  r = 1 leads to suppressing almost all the losses in the hBN (and therefore everywhere). However  r >2 for instance f leads to very small changes (P=160W instead of 163W) for P loss ~ 163 W P loss ~ 98 Wfor In surprising agreement with previous estimates (Ploss ~ 165 W and Ploss ~ 100 W )