Drive beam linac surface lines building db mb 500 gev bypass for the db power for the db
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Drive Beam Linac , Surface lines & building DB+MB, 500 GeV Bypass for the DB, Power for the DB. B. Jeanneret, CTC, 24th August 2010 With material thanks to G. McMonagle , E.Jensen , R. Wegner , A. Aksoy , D. Siemaszko & D. Nisbet , J. Osborne, N. Baddams

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Drive beam linac surface lines building db mb 500 gev bypass for the db power for the db

Drive BeamLinac, Surface lines & building DB+MB,500 GeVBypass for the DB,Power for the DB

B. Jeanneret,

CTC, 24th August 2010

Withmaterialthanks to

G. McMonagle ,E.Jensen, R. Wegner, A. Aksoy, D. Siemaszko & D. Nisbet,

J. Osborne, N. Baddams

L. Rinolfi, D.Schulte


Db linacs
DB Linacs


Db linac parameters
DB Linacparameters


Cell structure r wegner e jensen a aksoy d schulte
Cell structureR. Wegner, E. Jensen, A, Aksoy, D. Schulte

  • RF active section : 20 cells × 0.1m = 2m

  • Coupler, interconnect : 0.5 m

  • Quad, MBCO, BPM : 0.6 m

  • TOTAL : 3.1 m

  • May needadjustement ( Vacuum equip. for pressure p ~ 5 10-11 Torr, … )

15 MW klystrons

Acc. V/module 3.2 MV

819 cells / linac

Total length 2540 m

Plug-power for RF (2 linacs):

290 MW

( 0.115 MW/m)

3100

Linac 1


Magnets quadrupoles
Magnets – quadrupoles

  • N=819 Q, spaced by 3.1m for eachLinac (total Linaclength 2540m)

  • R = 3.7 10-2Ω, Imax = I0 = 414 A, pmax = 6.3 kW

  • Gradient must follow the beamenergyincrease

    • I(n) = nI0/N

  • Power schemesimilar to DECEL_Q advisable

  • Total power : P = Npmax/3 = 4 MW

I(n)

I0 = 414 A

1 (s=0)

N=819 (s=2540m)


Power cooling
Power & Cooling

  • Total plug power for RF : Ptot = 290 MW

    • 140 MW to beam150 MW to heat

  • Total power for magnets (Quad + Sol): 9 MW to heat

  • Total installed power :

    • Pinstalled = (1+ ε) × ( 290 + 9 )

    • withε the 18kV conversion yield (seewith Daniel/Cesary)

    • Add for light, cranes, safetysystems, …

  • Total cooling power (tunnel & surf. Blg) :

    • Pcool = (1+ ε) × ( 150 + 9 )


Db linac tunnel
DB Linac, tunnel

  • Total lengthnow : 2540 m

  • conflictwith EHN1

  •  move towards top of the drawing, seebelow

Water cooling pipes,

Cabletrays

Loads

1200

3000

1000

1000

800

6000


B jeanneret ctc 24th august 2010 with material thanks to

2m

DB Linac building

Top view

(half)

3.1m

Width: 23.5m

Height: 9m ?

Length : 2540 m

0.75m

3m

5m

3m

Klystrons are transported

along the central path.

A technical area without RF

must beforeseenat

crane extremities

End view

(half)

11.75m

3.5m ?

3m

1.5m

2.5m

4m

2m


Top view of db linac
Top View of DB Linac

xx

  • A small building isneeded for 18kV station ~ everyyy = 40 m ?

  • Size : xx by xx , xx = ?

  • Installed power

    4.6 MW/40m

xx

Awaiting data

From RF and EL

  • Still to beworked-out

  • withCesary

~yy

23500


Top view of db linac buildings
Top view of DB Linac buildings

2140 m

23 m

30 m

Fixed point

(beforedelayloops)

420 m

Underground & surface (slide 6 & 7)

Underground,

Same section as in slide 6

(CDR, to berevised for TDR)


Surface layout
Surface layout


Site implementation
Site implementation

DB Linac (undergroud and surface)

- Long segment : 2140 m

- Short segment : 420 m

Widthnow 23m

But annexblgs for

18kV missing

 Use 30m as average


500 gev delayed transfer for the drive beam

500 GeVdelayedtransfer for the drive beam

B.Jeanneret , July 2010


Input data
Input data

  • Time delaybetween e+ & e- DB train :

    • The two sets of 5 trains each 240 ns are produced one after the otherwith a single Linac

    • Δt = 5 × 24 × 240 ns = 28800 ns Δ = 8640 m

  • The trains must besynchronouswhen passing along the end of the Main Linac


B jeanneret ctc 24th august 2010 with material thanks to

Option I : Coming back to surface

Case 1

G

D

C

I

H

F

TA

  • Case 1 :

    • e+ trains come first and go through : AFGHI-TA-IHGFBCD

    • e- trains go through : AFGHI

  • Case 2 :

    • e- trains come first and go through : ABCD-TA-DCBFGHI

    • e+ trains go through : ABCD

r

x

From

CR2

B

A

Blue : additionalbeam line

Red : additional tunnel

D,I : end of ML

Case 2

G

D

C

I

H

F

x

TA

r

From

CR2

B

A

  • AG = 770

  • GH = 1449

  • HI = CD = 25 (500GeV)

  • AC = 2117

  • AB = 2r = 524

  • TA = 146

  • With y = AG+GH+AC+2*HI

  • x = (Δ + 2r – y –TA)/2

  • x(case1) = x(case2) = 2316 m

See ~/clic/CDR/500gev/bypath.sce


Option ii use a by path around ip preferred case
Option II : Use a by-patharound IP-- Preferred case --

  • D,I : end of ML (500 GeV)

  • e+ in D & e-in I : synchronous

  • Δdb = 8640

  • ID = 2×1870 , TA = 146

  • δbp = 14

  • x = (Δdb - δbp - ID –TA)/2 = 2361 m

  • x / Lsector = 2361/877 = 2.70 sectors

IP

I

D

C

x

TA

From

CR2

A

see :/Users/Bernard/clic/CDR/500gev/bypath_ugnd.sce


Impact for hardware underground by path
Impact for hardware / underground by-path

  • x / Lsector = 2361/877 = 2.7 sectors

    • one extra DB line to behoused in the main tunnel

  • Civ. Eng. :

    • 374 m of tunnel to beadded for by-path

    • 1 additionnalturn-around , of type ‘DB’

  • Additionalbeamlines:

    • 3740 m ( DI ) of type ‘DB-LTL’

    • 1 dedicatedturn-around , of type ‘DB’

  • Beamlines not needed :

    • Surface loop & descent down (2460 m)

  • Advantage : disconnectedfrom surface changes for Main Beam

Alexey,

Michele


Experimental area and by path
Experimental area and by-path




Summary for drive beam power end linac enter decelerator
Summary for drive beam powerEnd-Linac Enter Decelerator