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Problem for Course H9

Problem for Course H9. W. T. Weng August 3, 2012. Problem - 1. S how that for periodic system of wavelength β c and beam pipe radius, a, acceptance is A=4a 2 / β c , Where β c is Courant and Snyder invariant and satisfy ellipse. Δ y. 2 α. 2 α. 2 α.

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Problem for Course H9

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  1. Problem for Course H9 W. T. Weng August 3, 2012

  2. Problem - 1 • Show that for periodic system of wavelength βc and beam pipe radius, a, acceptance is A=4a2/βc , Where βc is Courant and Snyder invariant and satisfy ellipse Δy 2α 2α 2α

  3. Problem - 2 • Show that the mismatch factor (as defined above) between the two ellipses is given by

  4. Problem-- 3 Consider space between RFQ and beam line is about 100 mm without any optical elements. RFQ and beam line design for zero current, with the 100 mm space. Beam is match into the beam line for zero current . For 100 mA beam will be mismatched into the beam line. Show that due to this mismatch, emittance growth will be more than 100% in all three planes,given following initial conditions Ion: proton T=2 MeV Freq=400 MHz Transverse emittance =25 pi mm-mrad Longitudinal emittance=700 pi kev-deg Beam radius : x=y=z = 2 mm Beam divergence : xp=yp=12.5 mrad, dw=47.44 keV

  5. Problem-- 4 • Changing one parameter at a time, how to raise SNS proton beam power from 1.0 MW to 2.0 MW by varying a. Beam energy b. Beam intensity c. Beam rep-rate

  6. Solution to P-3 Ion proton Rest mass 938.37 e6 eV KE 2.0e6 eV γ 1.0021 β 0.0652 Frequency 400e6 λ 0.75 m ε0 8.8542e-12 F/m Π 3.14159 Beam Beam current 0.1 A Beam radius xy 2,2, mm z 2 mm Beam divergence x,y, 12.5,12.5, mrad z 11.873 mrad Emittamce X,y 25 π mm mrad Z 23.745 mm mrd IntialTwiss Parameter βT(x,y) 22 /25=0.16 mm/mrad γT(x,y) 12.52 /25=6.25 mrad/mm αT(x,y) (0.16*6.15-1)0.5 = 0 Bunch length Z ( deg) =z(mm)*360/(βλ(mm)) 2*360/(0.0652*750=)14.72 deg βT(z) 22/23.745 =0.1685 mm/mrad Momentum spread (Δp/P)= 11.873e-3*2=0.023746 MeV/c Energy spread ΔE/E =(γ+1)/γ * Δp/p=47.492 γT(z) 11.8732/23.745 =5.937 mrad//mm αT(z) 0 The Twiss parameter at start αT(x,y,z) 0 βT(x,y) 0.16 mm/mrad βT(z) 0.1685 mm/mrad γT(x,y) 6.25 mrad/mm γT(z) 5.937 mrad/mm

  7. Ellipse transformation for zero current at distance ΔS Δs=0.1 m βT(x,y)= 0.16 -2*0.1*0.0+0.12 *6.25=0.2225 αT(x,y) =0-0.1*6.25 =-0.625 γT(x,y) =6.25 βT(z)= 0.1685 -2*0.1*0.0+0.12 *5.937=0.2279 αT(z) =0-0.1*5.937 =-0.5937 γT(z) =5.937 This Twiss parameters are for matched condition βC(z)=0.2279 αC(z) =-0.5937 γC(z) =5.937 βC(x,y)=0.2225 αC(x,y) =-0.625 γC(x,y) =6.25

  8. Ellipse transformation with space charge • Generally space charge kick should be apply every βλ (=0.0652*750=48.9 mm), but here for simplicity we will apply one kick only at middle of drift space • Transport ellipse to 50 mm • Calculate new beam size • Calculate electric field and space charge kick • Calculate new Twiss parameters • Transport new ellipse to end • Calculate MMF with this ellipse and ellipse transported with zero current • Calculate emittance growth

  9. Ellipse transformation for zero current at distance 0.5*ΔS Δs=0.05 m βT(x,y)= 0.16 -2*0.05*0.0+0.052 *6.25=0.1756 αT(x,y) =0-0.05*6.25 =-0.3125 γT(x,y) =6.25 βT(z)= 0.1685 -2*0.05*0.0+0.052 *5.937=0.1833 αT(z) =0-0.05*5.937=-0.2969 γT(z) =5.937 New beam radius X,y (0.1756*25)0.5 = 2.095 mm Z (0.1833*23.745)0.5 = 2.086 mm

  10. Calculate electric field Parameter p P=1.0021*2.086/(2.095*2.095)0.5 = 0.9978 ~ 1. form factor f=1/3p=0.3333 Electric filed =(1/(4*3.1415*8.8542e-12)*(3*.1*.75/3e8)*(1/1.00212)*(1-0.333)/((2.095+2.095)*2.086*1e-6)) =5.1222E5 V/m =(1/(4*3.1415*8.8542e-12)*(3*.1*0.75/3e8)*(0.333)/((2.095*2.095)*1e-6)) =5.1166 V/m

  11. Calculate Space Charge Kick Space charge kick in u (xy,z) Δp(x,y) =5.1222e5*0.1*(1.0021/(1+1.0021)/(938.27e6*1.0021*0.06522) = 6.41 e-3 rad =6.41 mrad Δp(z) =5.1166e5*0.1*(1.0021/(1+1.0021)/(938.27e6*1.0021*0.06522) = 6.4 e-3 rad =6.4 mrad

  12. Calculate Twiss parameter after space charge kick x’=y’=12.5+6.41=18.91 mrad γT(x,y) = 18.912 /25 =14.3035 mrad/mm βT(x,y)=0.1756 mm/rad αT(x,y) = (0.1756*14.3035 -1)0.5 =-1.2295 Z’=11.873+6.4=18.273 mrad γT(z) = 18.2732 /23.745 =14.062 mrad/mm βT(z)=0.1833 mm/rad αT(z) = (0.1833*14.062 -1)0.5 =-1.256

  13. Ellipse transformation for zero current at distance 0.5*ΔS Δs=0.05 m βT(x,y)= 0.1756-2*0.05*1.2295+0.052 *14.3035=0.3343 αT(x,y) =-1.2295-0.05*14.3035 =-1.9446 γT(x,y) =14.3035 βT(z)= 0.1833 +2*0.05*1.256+0.052 *14.062=0.3441 αT(z) =-1.256-0.05*14.062=-1.9591 γT(z) =14.062 Beam radius x,y = (0.3343*25)0.5=2.890 mm Z=(0.3441*23.745)0.5 =2.8584 mm =2.8584*360/(0.0652*750)=21.04 deg x’,y’ =(25*14.3035)0.5=18.91 mrad z’=(23.745*14.062)0.5 =18.273 mrad Δp/p=18.273e-3*2=0.03654 MeV/c ΔE/E =0.03654*(1+1.0021)/1.11021=0.730 MeV These Twiss parameter to be compare with matched parameter

  14. Calculate MMF and emittance growth Rx,y =.2225*14.3035+0.3343*6.25-2*(-0.625)*(-1.9446)=2.8411 MMF(x,y) = (0.5*(2.8411+(2.84112-4)0.5 )0.5 -1=0.5587 Δε(x,y) = (1+0.5587)2 -1 =1.43 =143% Rz = 0.2279*14.062+.3441*5.937-2*(-0.5937)*(-1.9591)=2.9214 MMF(z) =(0.5*(2.9214+(2.92142-4)0.5)0.5-1=0.5892 Δε(z) = (1+0.5892)2 -1 =1.52 =152%

  15. Problem-- 2 • 1. Changing one parameter at a time, how to raise SNS proton beam power from 1.0 MW to 2.0 MW by varying • a. Beam energy • b. Beam intensity • c. Beam rep-rate • 2. What is the best combination( realistic and cost-effective), • by varying all possible parameters to reach 2 MW? • 3. Compare the relative advantages of LAR and RCS design • of CSNS design, in terms of • a. Accelerator performance limitation • b. Required changes of power supply, RF, and Vacuum • c. Cost and reliability

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