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MPS/CO Note 73-76 23 November, 1973 CAN ONE 10 13 PROTON BURST MELT THE PS VACUUM CHAMBER? Ch. Steinbach INTRODUCTION When the PS accelerates 1013 protons/pulse, if the beam gets out of control before utilization, it will hit the vacuum chamber. The energy deposited in this burst is so important that it is feared that the chamber may locally melt. This calculation tries to give as good an estimation as possible of the phenomenon and the temperature rise on the inside surface of the vacuum vessel for one pulse lost. /ed The assumptionsmade are the following a) The beam is lost horizontally, spiralling towards the chamber with a normal B (positive or negative) and no RF acceleration. b) The loss point does not protrude too much in the aperture of the chamber (see § 3.1) c) 1013 protons are contained in a 1.5 TI mm.mrad vertical emittance and a 3 r. horizontal emittance (Ref. 1).

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MPS/CO Note 73-76 23 November, 1973

CAN ONE 1013 PROTON BURST MELT THE PS VACUUM CHAMBER?

Ch. Steinbach

INTRODUCTION

When the PS accelerates 1013 protons/pulse, if the beam gets out of

control before utilization, it will hit the vacuum chamber. The energy

deposited in this burst is so important that it is feared that the chamber

may locally melt.

This calculation tries to give as good an estimation as possible of

the phenomenon and the temperature rise on the inside surface of the vacuum

vessel for one pulse lost.

/ed

The assumptionsmade are the following

a) The beam is lost horizontally, spiralling towards the chamber

with a normal B (positive or negative) and no RF acceleration.

b) The loss point does not protrude too much in the aperture of

the chamber (see § 3.1)

c) 1013 protons are contained in a 1.5 TI mm.mrad vertical emittance

and a 3 r. horizontal emittance (Ref. 1).

- 2 -

1. BEAM SIZE

The loss according to a) is most likely to happen in a focussing section.

The vertical beam size is 2 11.5 x 12 = 8,5 mm.

Assuming a Gaussian distribution, the density at the center is 1.6 bigger

than the average density (2a emittance). With respect to the density of a 1 cm

bright beam, the density at the center is then larger by a factor

10 8,5 x 1.6 = 1.88

The horizontal size of the beam is

2 13 x 21.7 = 16,1 nun

We will make the assumption of a rectangular distribution, which is pes­

simistic since a Gq.ussian distribution would rise the temperature faster in

the middle of the burst and give way to more conduction of heat transversally

through the chamber.

2. SPIRALISATION OF THE BEAl-1

In a focusing straight section

For a

size due to

~ p

6B = ~ B p

1 6r 10-s = f).r

r .027 x 1.16 .027 x 1.16

-3 of .4 x 10 (Ref. 1), the contribution of horizontal beam

momentum spread is therefore :

-3 .027 x 1.16 6r = .4 x 10 x ----- == 1.25 nun

10-5

- 3 -

The maximum betatron amplitude is then

A la.os 2 - .672 41=- 8 mm

The duration of the burst is for B 2.3 Tesla/s and 24 GeV/c (when

B = 1.14 Tesla)

10-s 8 1.14 6t = ------ x .6 x -- =

.027 x 1.16 2.3 1. 36 msec

The penetration of protons for a Q-value of 6.25 is equal to the jump

achieved during the time of 4 revolutions

x 4 8.6 1 -3 -3 0

= x 1 .36

x 2. x 10 = 53 x 10 mm

3. ENERGY DEPOSITION BY THE BEAM

3.1 Possibility of multitraversal

Condition b) is stated because multitraversal can occur when a

short obstacle is protruding inside the chamber.

For instance, if a 5 mm obstacle is hit by protons, the a for the . . 15 r:s- d . . scattering angle is 24 '\J1":77 = .33 mra • The betatron amplitude is

increased from a maximum of 8 mm to

1 8 mm x ----------------

( . 33 x

721.7) cos arc tg ---------

11.4 mm

and multitraversal occurs often if the obstacle is 3 mm inside the average

aperture of the chamber. But the second traversal takes place well inside

the metal, and is spread so that it heats up regions where the temperature

effect of the first traversal is hardly noticible during the time of spill.

,

)c

- 4 -

If the obstacle is less than 1/100 mm, then multitraversal can

become harmful because several traversals can hit the same initial

penetration depth. As an example, this situation corresponds to a

rectangular edge tilted by 45°.

In summary, multitraversal does not come in this c~lculation,

except for sharp edges that may be melted to smoother curves by the

process.

3.2 Energy deposition by the cascade

Monte-Carlo calculations by J. Rauft (Ref. 2) give the density of

energy deposition in the nucleon-mesoncascade initiated by protons and

show that for iron the maximum of the density occurs after 3 or ~ centi--3

meters and is equal to 1.1 10 of the total energy. But it also shows

that in our case, at least a fourth of the energy escapes from the sur­

face in this region, so the maximum heat for 1013 protons is

3 -3 -11 3 4 x 1.1 x 10 x 23.1 GeV/cm x 3.8 x 10 cal/GeV x 101

== 7.2 cal/cm

Taking into account the factor of § 1, the flux of heat penetrating

from the surface into the thickness of the chamber is

7.2 cal/cm2 x 1.88 13.5 cal/cm2

4. TEMPERATURE RISE

4.1 Equation of conduction

The only relevant cause of cooling during a short burst of the order

of milliseconds is conduction. We simplify the problem to a semi-infinite

- 5 -

Ho solid with heat density H cal/cm3s

0

deposition in a thickness x , x being 0

the transverse co-ordinate in cm,

t the time.

The differential equation for the temperature T is

dT H L d2T -+ dx2 dt pC pC

where p is the density of the metal in g/cm3

c its specific heat in cal/ 0 g

y its conductibility in cal/0 cm s

H = H for 0 < x < x 0 0

.... ~ H = 0 for x > x ......

0

4.2 Analytic solution

If one makes the approximation x = O, the analytic solution is

simple (Ref. 3) and the surface temperature rise is given by

= ~ ~ ho

y

For our case, .we find

LIT 2 x 13.5 =

.2x 1. 36 x 10-3

AT = 951°C

} h being the flux in

c:l/cm2 s

J .2 x 1.36 x .12 x 7.86 1T

-3 10

- 6 -

If this approximation is not made,

is also given in Ref. 3 :

i.e. x ~ O, an analytical solution 0

where

and

/J.T = ho t

pC x 0

13.5 = ~~~~~~~~~ .12 x 7.86 x .0053 [

-3 ·2 5.3 x 10

1 - 4 1 erf c c \(2 x 1. 36 x 10 .12 x 7.86

i2 erfc(x) = j 00

ierfc(<) d<

x

erfc(x) = 1 - erf(x) = 2 1--n

integration by parts giving

4 i 2 erfc(x) = (1+2x2) erfc(x) - -1.._ H -x2

x e

In ourcase, we obtain from this formula the value

!J.T = 827°C

4.3 Numerical solution

The equation of§ 4.1 can be solved by numerical methods. The

spatial differential is replaced by the approximation

d2T # T(n+l) + T(n-1) - 2T(n)

dx (!J.x)2

and the integration is carried out by a Runge-Kutta library subroutine

(subroutine INTSTF) modified in its dimension for a better spatial

resolution (101 instead of 20).

- 7 -

Unfortunately, the method diverges unless the step in time is very

small. 6t = 1 psec had to be taken and computer time is long. The

result obtained for the case of § 4.2 is

6t = 826°C

The same result is obtained with two different spatial resolutions

with 3 or 6 bins respectively heated. The repartition of the temperature

after 1.36 msec of irradiation is given by the program and plotted on

Fig. 1.

The radiation cooling at the inner surface of the chamber can be

introduced in the calculation and, as expected, the effect is negligible

and the correction is smaller than 1°c.

5. CASE OF THE ELECTROSTATIC SEPTUM

This piece of equipment is protruding into the vacuum chamber and

its thickness (.1 llllll) is small compared to the penetration in a block as shown

on Fig. 1

5.1 Stainless steel septum

If the same assumptions are made as in the case of the vacuum chamber,

one can conduct the numerical calculation for a strip of iron . . l mm thick,

and this leads to the result at the inner surface of the septum

6t 1500°c

Introducing radiation on both sides of the septum decreases this 0 result by 1 C

6t = 1499°c

- 8 -

These calculations do not take into account multitraversals, so

the temperature could increase beyond the values calculated if the losses

occur on the septum without resonance but with the local bumps powered.

One can expect, however, that the presence of the thin septum magnet two

sections downstream of ES 83 will prevent the multitraversal process, if

it is placed in its normal operating position. In most cases, it should

even prevent the electrostatic septum to be the first obstacle to be

encountered with, because the phase relation with and without resonance

are different in 83 and 85.

One may add that anyway in the future one could decrease the local

bump slightly before the end of the flat-top to avoid losses in 83 in

the most dangerous case, namely when most of the beam is still present

a~ the end of the flat-top.

The repartition of temperature through the thickness of the septum

as given by the program is shown on Fig. 2.

5.2 Molybdenum s eptum

The figure of heat dissipation given by the Monte-Carlo program

(Ref. 2) is 1.58 10-3

of the total energy, which following the calcu­

lation of § 3 coresponds to 19.5 cal/cm2. The calculation is done like

in 5.1 and leads to the temperature rises

~t = 2608°C without radiation

~t 2594°C with radiation on both sides.

The temperature repartition is given in Fig. 2 .

-9-

CONCLUSION

The results are obviously given with the accuracy of the degree C only

for the sake of comparison between methods of calculation. They should be

considered as approximate results under the assumptions stated in the intro­

duction. Because the Monte-Carlo cascade programs (Ref. 2) do not apply in

their present available form to the fine analysis of a cascade initiated by

a very thin beam, the figure of heat dissipation used in the present report

is on the pessimistic side. Therefore, all results are upper limits of expec­

tations for real cases.

There seems then to be no danger of melting the PS chamber in one pulse 0 (stainless steel melts around 1500 C). If the emittances become smaller than

those assumed, one would however get nearer to a critical situation. Moreover

stresses in metal due to temperature rises of the order of aoo0 c during 1 or

2 msec should be looked into. Outgazing of the chamber may become very serious too.

Small obstacles or sharp angles should be avoided all along the aperture

of the chamber.

As for the electrostatic septa, one is likely to reach just about their

melting point, for stainless steel and for molybdenum. To verify the argu­

ments of § 5.1, it is suggested to conduct tests at the present normal inten­

sity (2.10 12 ) with various abnormal operational conditions, observing the beam

loss detectors. They should aim to assess what safety measure has to be taken

as far as the power supplies of the slow extraction are concerned, to protect

these delicate pieces of apparatus.

ACKNOWLEDGEMENTS

The author wishes to thank J. Boillot, R. Gouiran, L. Henny, W. Kubischta,

D. Lewis, J.P. Potier and R. Tinguely for their corrections and comments on

this study.

- 10 -

R E F E R E N C E S

1) 0. Barbalat - C.P.S. Beam Emittances - MPS/DL/Note 71-16

2) J. Ranft, J. Routti - Fluka and Magka, Monte-Carlo Programs for

calculating nucleon-meson cascades in cylindrical geometries -

Lab.II-RA/ 71-4

3) H.S. Carslaw: and J.C. Jaeger - Conduction of heat in solids,

Oxford University Press.

Distribution

MAC Members MST PSS H. Bargmann, 300 J. Boillot D. Dekkers J.Y. Freeman c. Germain R. Gouiran L. Henny u. Jacob w. Kubis ch ta D. Lewis w. Middelkoop, 300 R. Tinguely p, Sievers, 300

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