Status of Abort System

M. Kikuchi, KEKB Review, 24Feb2015

Abort system, overview

magnete+	  beamT.	  Mimashie-­‐	  beamAbort system, Kickers

T.	  Mimashi

Horizontal
KickerPulsedquad

Pulsedquad

Pulse compression circuit islocated at each H-kickers.

Pulse compression circuit is locatedat each of H-kicker
and the pulse-Quad.
VerticalKicker

LERHER

Abort	  system:	  Waveform	  of	  the	  output	  current	  Current

200nsec

T.	  Mimashi

10μsecVoltage

Current waveform of the horizontal kicker

Current waveform of the vertical kicker

Radia>on	  shield	  of	  pulse	  compression	  circuit

Abort	  system:	  installa>on	  T.	  Mimashi

Layout of abort kicker magnets & pulse compression circuit at HER

Top View

Side View

Horizontal
KickerMagnets

Radiation shield: 7mm Pb + 200mm polyethylene+ 1mm Pb

Pulse
Compression
circuitT.	  Mimashi•Radiation shield is necessary to protect sensitive semiconductors in the pulse
compression circuit which is placed just below the beam line.
• In KEKB operation– Diodes	  for	  shaping	  the	  current	  waveform	  has	  survived,	  which	  have	  been	  placed	  just	  below	  the	  beam	  line.
• The radiation shield is designed to reduce the radiation to 1/10
at the surface of diodes (for both of γ and neutrons).

Abort system: Water-cooled ceramic chamber

HER abort kicker ceramic chamberT.	  Mimashi

100μm Cu-coating applied
inside the Kovar sleeve

Cu sleeve is directly connected tothe ceramic

LER abort kicker ceramic chamber

Abort system: Fabrication status and plan

T.	  Mimashi

Fabrica'on	  status	  and	  plan
R&D,	  	  Design

Fabrica>on

Installa>on

HER

LER

HERLERHERLERH	  kickers○○○2015

20152016V	  kickers○○2015

2016

-­‐○

-­‐

2015

-­‐2016○○△(Ti	  coat)

○

2015

2016

○○

△

△

2015

2016

○○

△

2015

2015

2016

-­‐

-­‐

×2015

-­‐-­‐Pulsed	  quads
CeramicChambers
Power	  supplies
Radia>onshieldSystemTuning△(Mar,	  2015) Mar,	  2015

Abort system: Summary	  Summary	  of	  the	  construc'on	  status	  of	  abort	  system•	  New	  power	  supplies	  for	  kickers	  were	  successfully	  developed.
-­‐ Pulse	  compression	  using	  the	  magne'c	  switch	  -­‐ Sa'sﬁes	  the	  rise	  'me	  of	  200	  ns.
-­‐ Radia'on	  shield	  was	  developed	  to	  protect	  the	  diodes	  in	  the	  pulse	  compressor.
•	  New	  water-­‐cooled	  ceramic	  chambers	  have	  been	  developed.
-­‐ HER:	  cooling	  channels	  in	  the	  ceramic,	  Cu-­‐coated	  Kovar	  sleeve.
-­‐ LER:	  double	  concentric	  ceramic	  pipes	  connected	  both	  ends,	  Cu	  sleeve.
•	  Fabrica'on	  is	  steadily	  going	  on,	  coping	  with	  'ght	  schedule.	  
9Dynamic Stress in the Extraction Windowof SuperKEKB

M. Kikuchi, KEKB Review, 24Feb2015

Introduction

Temperature rise
1
∆T =C�

1 dE
ρ dx

�

N
2πσx σyionization loss for Ti

1 dE
2= 1.98 MeV/(g/cm )ρ dx
σy ∼ 18 µm

σx ∼ 0.3 mm

then∆T = 4 × 105 K

C = 0.54 J/K/geN = 36 µC (LER)!!

•We must reduce the power density• vertical beam-scan: 15 mm /10μs• enlarge σx by a pulsed-quad(LER),
1
∆T =C

�

1 dE
ρ dx

�

or sext w/offset(HER): σx −→ 1.23 mm

N√
2πσx h∆T ∼ 300 K

Dynamic Stress in the Extraction Window

YRelated parameters• Beam parameters
σx = 1.23 mm,•Bunch space :σy = 18 µm∆t = 4 ns•Scan speed :X∆y/∆t = 15 mm/(2500∆t) = 6 µm/4 ns• Window material : Ti
E = 106 GPa,

• Speed of sound3

ν = 0.34, ρ = 4.5 g/cm
α = 8.4 × 10−6�c = E/ρ/(1 − ν 2 ) = 5160 m/s•Propagation length / bunch :c∆t = 21 µm

Dynamic Stress in the Extraction Window•Maximum temperature rise for a single bunch
T0 =1C

�

1 dE
ρ dx

�

N
2πσx σyT0 = 39 K/buncheN = 14.8 nC/bunch•Thermal Stress
αT0 Eσ0 ≡= 52.6 MPa1−ν•Temperature due to multi-bunch
T (y) = T0M−1
�
n=0

�

exp −(y − n∆y)
2σy22

�

� T0

•Temperature saturates at 293 K in several bunches
Tmax = T0√

2πσy= 7.5T0 = 293 K∆y√

2πσyerf(y/σy )∆yDynamic Stress in the Extraction Window• Maximum thermal stress due to multi-bunch
Tmax
σmax =σ0 = 7.5σ0 = 395 MPa ?
T0

Obviously this is incorrect because,• Stress wave moves much faster than the scanned beam
Basic question:• What is the dynamic stress for a Gaussian beam?
• How does it behave in the flat Gaussian beam?

•Summary of related parameters
SLERLER

Beam current

I

3.6

1.64

A

Number of bunches

nb25001584

Charge per bunch

eN

14.8

10.4nC/bunch

Hor. emittance

εx

3.2

18nm

Ver. emittance

εy8.6

150pm

Hor. beam-size at window

σx

1.23

0.70mmVer. beam-size at window

σy1873μm

Bunch spacing

Δt

46ns

propagation/bunch

cΔt

21

31.5μmScan heighth1515mmscan height / bunchΔy69μmTemp. rise per bunch

T03912deg

Maximum temp. riseTmax

293

230

deg

Model I.
Model 1.

•Thin disk
•Cylindrical symmetry•Single Gaussian beam

2a2b- Instantaneous stress induced by abrupt changeof temperature is propagated as a ‘stress-wave’with a speed of soundPosition deviation U and stresses σr , σθ must satisfy the Eq. of motiond2 U∂σr
σr − σθρ 2 =+dt
∂rr

From Hooke’s law
1εr − αT = (σr − νσθ )E1εθ − αT = (σθ − νσr )E(1)U(r,t) : (radial) position deviation(2)

T(r) : temperature(3)and definitions of strains
U∂U, εθ =εr =∂rr

(4)

Ref.
P. Sievers, Elastic stress waves in matter due
to rapid heating by an intense high-energy
particle beam, LAB. II/BT/74-2, 26 June
1974

Model I. Eq. of Motion
The Eq. of motion is rewritten as
∂2U
1 ∂U
UdT
1 ∂2U+− 2 − (1 + ν)α= 2 22∂rr ∂rrdrc ∂t(5)c=d2 ū 1 dūū
dT+− 2 − (1 + ν)α=02drr drrdr(6)

u (time-dependent part of U) satisfies the wave-eq.:

Solve (7) with initial conditions,
U (r, 0) = 0,i.e.,and boundary conditions,u(b, t) = ū(b, t) = 0.

u(r, 0) = −ū(r)E(1 − ν 2 )ρ: Speed of soundPut U = u + ū with a static solution ū of∂ 2 u 1 ∂uu
1 ∂2u+− 2 = 2 22∂rr ∂r
rc ∂t

�

(7)

�∂u ��=0�
∂t t=0(9)(8)

Model I. Solution
Solution:u(r, t) = (1 + ν)αT0∞
�cn J1 (βn r/b) cos(βn ct/b)

βn : n-th zero of J1 (x)

(10)

n=0

Coefficient cn is given by2(1 + ν)αT0 cn = − 2J0 (βn )� 1tū(bt)J1 (βn t)dt

(11)

0Corresponding stresses are��
J1 (z)
�σr =cn J0 (z) − (1 − ν)cos(βn ct/b)
1 − ν n=0zz=βn r/b
�
�∞
�
αT0 E
J1 (z)�σθ =cn νJ0 (z) + (1 − ν)cos(βn ct/b)
1 − ν n=0zz=βn r/b∞
�
αT0 E(12)(13)

c�n = cn βn /b• We denote ‘total’ stresses as:pr ≡ σr + σ̄r ,pθ ≡ σθ + σ̄θ• At t = 0 , εr = εθ = 0,then from (2)(3),αET (r)pr (r, 0) = pθ (r, 0) = −1−ν

Initial stress = thermal stressModel I. Solution / Gaussian
For a Gaussian beam�

2rT (r) = T0 exp − 22a�%'&&'$(b/a = 10)

T&'#

Static solution:ū&'"ū = (1 + ν)αT0 r [f (r/a) − f (b/a)](14)&'!

!

2f (x) = (1 − e−x /2 )/x2

thenc�n = −2J02 (βn )λ"r/a��� 1−λ2 /2−λ2 t2 /21−eJ0 (βn ) − βneJ1 (βn )dt2(15)0For λ � 1��2222−βn /2/λ−λ /2c�n = − 2e−eJ0 (βn ) ,2J0 (βn )λ−λ2 /2) .
• (16) is accurate in order of o(e• Maximum number of terms N in (12)(13)

must be such that βN � λ#λ ≡ b/a(λ � 1)(16)$%&

Model I. Solution / Gaussian / Examples
λ = b/a = 10r = 4a

λ = b/a = 10

pr /σ0pr /σ0

r=0ct/act/a = (0, 2, 4, 6, 8),ct/a
λ = 10• Multiple echoes.
• ‘Wave packet‘ with a width ofpr /σ0beam-size.
• Never exceeds the initial stress.
This is a remarkable feature of theGaussian beam. Rectangular (stepfunction) beam, for example, shows
catastrophe (infinite stress) at theorigin (P. Sievers, 1974).
r/aExample: Step-function caseExample : Step function

T0

T (r) = T0 (0 < r < a)
= 0 (a < r < b)r=0

λ = b/a = 5
inwardreflected

pr /σ0inwardoutwardreflected

ct/a0abr.
• Multiple echoes• Inward wave generated at r=aconcentrates at the origin.
Catastrophe (infinite stress) at theorigin (P. Sievers, 1974).
• The first inward-wave is
independent on the boundaryconditions.
• Step-wise distribution isdangerous: collimation should be
avoidedModel II.
Model II.

•Thin disk
•2-D Gaussian beam with an aspect ratio m2a2ma•Wider Gaussian is represented as a superpositionof narrower Gaussians
� ∞x2(x−ξ)2− 2−e 2a1 =g(ξ)e 2a2 dξ
−∞

�2ξa1
�
exp −g(ξ) = √222(a21 − a2 )2πa a1 − a�

•A 2-D Gaussian is a superposition of round 1-DGaussians, for which we know the response.
y2x2− 2 − 2a22a1e� ∞�

(x − ξ)2 + y=g(ξ) exp −2a2−∞

�2yσr

σθdξr
θ
ξ

xModel II.
Stresses in Cartesian coordinates� ∞��g(ξ)dξ
22px =(x − ξ) pr + y pθ ,22−∞ (x − ξ) + y� ∞� 2�g(ξ)dξ2py =y pr + (x − ξ) pθ ,22−∞ (x − ξ) + y� ∞g(ξ)dξ[(x − ξ)y(pr − pθ )] .
τxy =22−∞ (x − ξ) + y(pr , pθ ): Stresses in 1-DGaussian beam

• From symmetry τxy = 0• In calculation we used a finite summation with ∆ξ = a , for m � 1•This model is correct until the waves

reach the boundary, because thegeometry of the boundary for each ‘microbunch’ is different each other.

Model II. Example IExample 1: m = 5,

λ = 50�2xyT (x, y) = T0 exp − 2 2 − 22m a2aτ ≡ ct/a�py
τ =5

τ = 10τ = 15

τ = 10τ = 15y/aτ =02x/apx

y/aτ =0x/aτ =5Model II. Example 1
Py on x-axisτ = (0, 5, 10, 15, 20, 25)

py /σ0py /σ0

Py on y-axis τ = (0, 5, 10, 15, 20, 25)y/ax/aPx on x-axisτ = (0, 5, 10, 15, 20, 25)

px /σ0px /σ0Px on y-axis τ = (0, 5, 10, 15, 20, 25)y/a• Py propagates in almost one-dimensional manner• Px stays longer time around the origin• Waves along x-axis is very smallx/aModel II. SLER single bunch
Example 2: m =√2 × 68.3 (Actual beams of SuperKEKB LER, w/ tilted window)

Py on y-axis

Py on x-axisτ = (0 ∼ 300)

py /σ0py /σ0τ = (0 ∼ 40)

y/a

Px on y-axis

x/aPx on x-axisτ = (0 ∼ 300)

px /σ0px /σ0τ = (0 ∼ 40)y/a• Py propagates in almost one-dimensional manner• Px stays longer time around the origin• Waves along x-axis is very small

x/apy ∼ 0.5 σ0px ∼ 0.65 σ0Model III. SLER multi-bunch
Model III.

• Thin disk
• 2-D Gaussian beam
• Multi-bunch, scanned vertically

2a2am- Results of Model II are overlaid with shiftedvertical positions and shifted time.
1 ~ 20 bunches

1 ~ 20 bunches

Px on y-axis

px /σ0py /σ0

Py on y-axis

y/ay/a• Maximum stress is 5.2 σ0 = 274 MPa√•
2πσpy,max �y(c∆t − ∆y)0.5 σ0 = 1.5 σ0px,max �√

2πσy0.65 σ0 = 4.9 σ0∆yModel III. Multi-bunch, von Meses stress• The “von Mises stress” is responsible to yield of materials
p2M =��222222(px − py ) + (py − pz ) + (pz − px ) + 6(τxy + τyz + τzx ) /2

pM /σ0Von Mises stress on y-axis 1 ~ 20 bunches

y/a• Maximum Von Mises stress is 3.2 σ0 = 168 MPa• Taking into account possible overlap with reflected
waves, maximum Von Mises stress is expected to�be(|p| + |p|)2 /2 = 4.9 σ = 258 MPax,maxy,max0• This is larger than the yield strength of pure Ti.

Model III. Multi-bunch, von Meses stress• Ti alloy (Ti-6Al-4V) may be a candidate.

KEKB, LER single bunch
Example 3: m =√2 × 9.6 (Actual beams of KEKB, w/ tilted window)

Py on y-axis

Py on x-axisτ = (0 ∼ 50)

py /σ0py /σ0τ = (0 ∼ 40)

y/a

Px on y-axis

x/aPx on x-axisτ = (0 ∼ 50)

px /σ0px /σ0τ = (0 ∼ 40)y/a• Py propagates in almost one-dimensional manner• Px stays for longer time around the origin• Waves along x-axis is very small

x/apy ∼ 0.45 σ0px ∼ 0.6 σ0

KEKB, LER multi-bunch
σy = 73 µm,σ0 = 16.2 MPac∆t = 31.5 µm,∆y = 9 µm√

2πσypy,max �
0.45 σ0 = 3.72 σ0(c∆t − ∆y)Py on y-axis

px,max �

2πσy0.57 σ0 = 11.6 σ0∆yPx on y-axis

40 bunches

py /σ0px /σ040 bunches√pM /σ0

Py on y-axis

y/ay/a40 bunches

y/a• Maximum stress :• Max. von Meses stress :• Max. von Meses with

11.6 σ0 = 188 MPa
6.9 σ0 = 112 MPa

180 MPa
reflections:• Maximum stress wasvery close to the yieldstrength.
(Ignorance is bliss! 知らぬが仏)SummarySummary• SuperKEKB abort-system uses maneuvers of the verticalscan and horizontal enlargement to effectively reduce thepower dissipation in the extraction window.

• We have developed an analytical method applicable to thedynamic stresses generated by the two-dimensional
Gaussian beam.

• For the Gaussian distribution, the dynamic stress neverexceeds the initial thermal stress.

• Maximum von Mises stress is 260 MPa.
• For KEKB case, von Mises stress is estimated as 180 MPa.
• Ti alloy may be a candidate as the window material.

SparesBeam trajectory at the beam
dump of KEKB HER

Spares
Annealing
1.最大温度 300+25=325℃• 再結晶温度~450 ℃

これを越えて加熱すると結晶粒径が大きくなり、強度が落ちる
1/3∝
結晶粒径（加熱時間）

Tmax = T0

√

2πσy
1=∆yC

�

1 dE
ρ dx

�N
√
2πσx ∆y• σx ∆y は再結晶温度に対して1.4 倍の余裕がある
西村孝、日本鐡鋼協會々誌 70(15),1898-1905,1984-11-01

Sparesc’(n) lambda=100c’(n) lambda=50

λ vs n that maximizes |c’(n)|nmax � 5/16λSpares��2222−βn /2/λ−λ /2c�n = − 2e−eJ0 (βn ) ,2J0 (βn )λは、さらに(λ � 1)

よって、波の主成分周波数はc�n = −22−βn/2/λ2e2J0 (βn )λωmain = βnmax c/b = λ(c/b) = c/aと近似できる。 βn � π(n + 1/4) および漸近式
�2cos(z − π/4)J0 (z) ∼πz

波長 Λ はから、して描くと次図の様になる。

が成立する。したがって
βn −βn2 /2/λ2
�cn � −π 2 eλ�c
これから、 n を最大にする βn は

βnmax = λ�となる。 cn を βn /λ の関数と

βn /λ1c�n λ/π1π∼ βn2J0 (βn )2Λ = 2πa�0.1
�0.2

Out[74]= �0.3

�0.4
�0.5
�0.62345

Spares
この c�n の分布に関して�� 2�
�βn � = π/2 λβn = 2 λ2��
2∆βn ≡ �βn2 � − �βn � = 2 − π/2 λSpares
境界での折り返しの様子
λ = 20,原点での折り返しの様子

τ = {16, 24, .2}

λ = 20,τ = {36, 44, .2}

σr /σ0τ = 19τ = 39λ = 20,τ = {56, 64, .2}r/aτ = {76, 80, .2}

τ = 56

τ = {80, 84, .2}