1

Vacuum technology

n Vacuum

A

[ ]/seccm )/(1064.3 3213 AMTq/ndV/dt ⋅==

)()/(1064.3 21213

21 PPAMTQQQ −⋅=−=

C =Q /ΔP = 3.64 (T /M )1

2A l/s

P1 P2

A

Conductance of an aperture

/seccm )/(1064.3/Q 3)2,1(

213

)2,1((1,2) barAPMTdtdVP µ⋅==

avenv41 =φ

How many molecules travel through A in a time t? [ ]mol/sec )/(1064.3 2

13 nAMTAq ⋅==φVolume of gas passing through A?

µ-sco

pic

Μ-sc

opic

←→ ←→ ←→ ←→

C indipendent from P

dtdVP

dtdVPPPQC =Δ−=Δ= /)(/ 21 AvAvnAnq ave 4

141// ≡==φ

Vacuum technology

l/s )/(64.3/ 21AMTPQC ⋅=Δ=P1 P2

A

Conductance of a diaphragm C indipendent from P

•  Si ottiene

1→2→3 1C1−>3

= 1CA0

+ 1C2+ 1Ce

1C3−>1

= 1CA+ 1C23→2→1

⎟⎠⎞

⎜⎝⎛ −

=

01 AA

CC Ae

1 2 3

A0 A

22

1111110

CCCCCC AeA+=++=

Q3 = P3C3−>1Q1 = P1C1−>3

Q =Q1 −Q3 =P1=P3

P1C1−>3 −P3C3−>1 ≠ 0Ci sarebbe flusso anche se P1=P3 uguali,

quindi C1->3 = C3->1

•  A<<Ao : Ce ~ CA (Apertura) •  A=Ao : Ce = ∞ ( no resistenza al flusso), •  A=0.5 Ao: Ce =2 CA (effetto diaframma)

2

Conductance of a long tube Ø Knudsen derived assuming molecules moving

randomly, and number collisions on unit surface and unit time = ¼ nvave (*).

Clt =8

3 π2kTm

⎛

⎝⎜

⎞

⎠⎟

12 A2

BL⎛

⎝⎜

⎞

⎠⎟=

3.44 ⋅104

πTM⎛

⎝⎜

⎞

⎠⎟

1/2 A2

BL⎛

⎝⎜

⎞

⎠⎟

in CGS units, A is the area and B the boundary.

Clt = 3.81 T M( )1

2 ⋅ D3

L( )D cm[ ], L cm[ ] e C l/s[ ].

LDClt3

1.12 ⋅=

For a cilindrical Tube: A=πD2/4 and B=πD

Air at 20° C:

(*) A. Roth Vacuum Technology (Elsevier Science , Amsterdam, 1990 Third edition)

Vacuum technology

eltst CCC /1/1/1 +=

e/Cl 1:0→

Cst =Clt 1+ CltCe( )⎡

⎣⎢⎤⎦⎥;

Short tube = serie of long tube and aperture

[ ]02/1

3

22/14

/1/1064.3

1044.3

AAAMT

BLA

MT

CC

e

lt

−⎟⎠

⎞⎜⎝

⎛⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⋅

=π

[ ] 11" −+== eltltltst CCCKCC

1

0

133.51"−

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=AA

BLAK

K’’ correction factor Called of Knudsen

Diaphragm effect

A0

A0 is the cross section of the upstream vessel

Short tube conductance

3

Vacuum technology

Correction factors Ø Short tube :

v Knudsen factor molecule impinging on the surface of the tube come out with a cos θ ( K”) spatial distribution.

v Clausing factor Higher probability for the molecule to come out in the flow direction ( K’).

Described as : (Roth), or (equivalent) (O’Hanlon)

where C0 is the conductance of the aperture.

0 / CCPr =

AvaC4

' ⋅=

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Clausing factor Accuracy of calculations

Clausing 1 % Assuming the big

dimension of chamber upstream and downstream

Knudsen 5 %

Altst CCC /1/1/1 +=

4

Vacuum technology

Analitical description for tubes

( ) K"LDMTKCC ltst ⋅== /)/(81.3'' 32/1

[ ]{ }2)/(1)/(33.11/1 vDDLDK" −⋅⋅+=

If the volume is bigger with respect to the tube diameter { })/(33.11/1 LDK'' +=

( )LDMTClt /)/(81.3 32/1=

[ ]( )DLDMTC 33.1/)/(81.3 32/1 +=

2

2

)/(12)/(3820)/(12)/(15'DLDL

DLDLK++

+=

[ ])/')(4/3(1/1 DLPr +=

[ ]{ })/)(7/6(3/11' DLLL ++=

Knudsen

Clausing 1932

Santeler 1986

3.18 fig. )109.3(

3.18 fig. )107.3(

Vacuum technology

Comparison

Ø Cilindrical tube l = 10 cm, d =1 cm

Ø H2 300 K

ü Long tube 4.67 l/s 10 % ü Knudsen short tube 4.11 l/s 5 %

ü Clausing 3.96 l/s 1 %

units. CGS

3.511044.3 122/14 −

⎥⎦

⎤⎢⎣

⎡ +⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⋅=

BLA

BLA

MTCst π

( ) seccm 4

'1045.1 3

2/14 AaMTC ⋅=

5

Vacuum technology

Monte Carlo Methods

0/CCPr =

Molecules randomized

match well with the Monte Carlo Methodes

Vacuum technology

K’’ Knudsen correction factor

K’ Clausing correction

factor

6

Vacuum technology

I 1

II 10-2

III 10-4

chamber 10-8 Torr

He I : 22.21 eV at 1 Torr HeII: 40.8 eV at 0.1 Torr

Seff of 500 l/s in the vacuum chamber Pbase = 10-10 Torr without He flow. a) Calculate the length of a capillary (d = 2 mm) in order to have

Pwork = 10-8 Torr in the chamber in working condition of 1 Torr discharge. b) Which S are required in II e III stages if the capillary lengths are LII-III=LI-II= 5 cm?

c) Once the length of the whole capillary is designed, calculate the pressures in any stage for the discharge at 0.1 Torr.

He discharge Example-exercise: He lamp

Check λ/d for He and flow regime? tc = 25 oC

Ø We have two possible model of HV pumps one requires at the outlet a pressure below 5 10-1 mbar, the other a pressure below 8-9 mbar.

Ø For the connection between the HV pump

and the back pump assume a 40 mm in diameter tube having a length of 1 m.

Check which turbo pump is required?

Vacuum technology

7

In which regime are we?

Vacuum technology Vacuum technology

d•P > 5 10 -1 cm Torr5 10 -3 < d•P < 5 10 -1 cm Torr

d•P < 5 10 -3 cm Torr

Stateofthegas k Flowregime Re

Viscous d/λ>110 Turbulent 2100 Airadmi,edinvacuumLaminar 1100;

Transi8on 1<d/λ<110 Intermediate

Rarefied d/λ <1 Molecular

d = dimension of ducts, chamber

Vacuum technology

Viscous (Flow) just mention Ø Q=pvA where v velocity and Q=pS

•  QthroughanapertureA,adiaba2cexpansion•  C=Q/ΔP

•  C=f(P1,P2,γ,T,M)• Forairat20°CC=20 A/(1-P2/P1) if 1 > P2/P1 > 0.525

•  Ctube=[πD4/(128ηL)]Pave; [D]=[L ]=cm,[Pave]=dyne/cm2and[C]=cm3/s

• C=20Al/(scm2)if P2/P1 < 1•  C of a tube of diameter D and length L,

•  Forairat20°C Ctube = 182 D4/L Pave; D & L [cm], Pave [Torr], and C [l/s]

8

Comparison between viscous and

molecular regimes

Vacuum technology

( ) [ ] [ ] l/s. ,cm ;64.3 221

==⋅⋅= CAAMTCA

Mol

ecul

ar re

gim

e

•  For air at 20 °C CA =11.6 A l/s, CA =9.16 D2 l/s,

Pumping speed through an aperture: S= Q/P1

S= Q/P1 =C(P1 -P2 )/P1 for P2 <0.1 . P1 (usual case):

S =C=11.6 A l/s

The maximum pumping speed in molecular regime is 11.6 A l/(s cm2), if compared to the viscous regime: S=C= 20 A l/(s cm2)

If the pressure Pave is sufficiently high the term D4 dominates (viscous flow).

Vacuum technology

Ø C=(c1D4Pave+c2D3)/L Intermediate regime

c1=π/128 η c2=(π/16)[(π/2)(R0 T/M)1/2 [(2-f)/f]

f= fraction of molecules absorbed an reemitted from the surface

When the two terms are equal: Pave = Pt = c2 /(c1 D) transition pressure.

When the pressure becomes low according : D . Pave < 5 .10-3 Torr cm molecular regime

9

Vacuum technology

J=C

v /C

m

Vacuum technology

Ø Evacuation from a volume V

Evacuation from the atmosphere, Only yseful in case of big chamber, or frequently opened

tVS

SPQ

ePP

SPVdtdP

dtPVdQ

−

=

=⇔

⇔=−=

=−=

0

)(

P

Q S

10

Vacuum technology

Evacuation in viscous regime 1/S=1/Sp+1/C C=(π/128)[D4/(ηL)] Pave

C= E (P+Pp)/2

P S

Sp C

Pp Pave=(P+Pp)/2

Q=PS=-V(dP/dt)

Equation can be solve an plotted by using as a parameter

D4/L=(128/π )η E

Vacuum technology

V = 100 l D = 2 cm

L = 200 cm D4/L = 8.10-2

If Sp = 2 l/s, then we have t/V=6 then

t = 600 sec

Assuming L = 0, we have t/V=4.5

t = 450 sec

11

Vacuum technology

Another example

Ø Base Pressure v  Pbase =5 .10-8 mbar

Ø Max Pressure allowed v Pmax =5 .10-4 mbar

ü  Find the max troughput which can be supported by the system.

S of turbo for N2 1600 l/s, and is connected by a tube of D=250 mm and L= 128 - 173 mm.

Ø  Maximum allowed (required) pressure in the chamber? ᴥ  Pch=Q/Sch

ᴥ  1/Sch=1/Shv +1/Chv

ᴥ  Phv-outlet = maximum allowed pressure at the out-let of HV pump.

ᴥ  Cbp = conductance between HV pump and back pump.

ᴥ  Pbhv-out = Q/Shv-out

ᴥ  but ᴥ  1/Shv-out=1/Sbp +1/Cbp

Dimensioning vacuum system: in stationary condition

Vacuum technology

Shvp

Sch Chv

Pbhv-out

Cbp

Sbp Pbp Sbp= 15 m3 h-1

Shv-out

12

Vacuum technology

Dimension of the chamber for next steps

Vacuum technology

Ø The following sources of throughput appear in a vacuum system

Through-put(Q) sources in vacuum systems

13

Vacuum technology

Q sources: vaporization

( )TMAPAnvsmoleculesΦ v

kTPn

v

ave

mkTave

24/

1063.24

/2/1

8

×===

⎟⎠

⎞⎜⎝

⎛=π

Saturated vapour pressure Liquid

Vapour Pressure

Liquid

SI units: [Pv] = Pa, [A] = m2 [T] = K

In a closed system Open system / Vacuum system

Φ molecular flow coming out from a liquid of surface A

Vacuum technology

Q sources: desorption

10 order ( ) ( ) )(

0

)/(

1

0

tCetCKdttdC kTNEd

τ

−

−=−=

20 order( ) ( )22 tCkdttdC

−=

kTNEr

deK

0/0

1

1 ττ == τr residence time decreases increasing T.

C(t) exponentially decreased increasing T.

rttK eCeCtC τ/00

1)( −− ==

( )( )220

202

1 tKCCk

dttdC

+−=

Outgassing of gases adsorbed in atm and or finale stage of diffusion and permeation

H2 - H+H - H2

Tipical law of rate of lost proportional to the initial

14

Q sources: diffusion

Vacuum technology

dxdnD 1

1 −=Γ

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+⎟

⎠

⎞⎜⎝

⎛= ∑∞

0

222/1

0 exp)1(1Dtdn

tDCq n

atdDt

eedDCqt

tDCqt

−⎟⎠

⎞⎜⎝

⎛ −

⋅≈=∞→

≈⎟⎠

⎞⎜⎝

⎛=→

1/2-2

1/2-2/1

0

t 2 :

t :0

2π

Process slower than desorption

Initial concentration

Bulk thickness

kTEeDD −= 0

Energy Activation=E

The diffusion constant is function of the activation energy and T

Then increasing the T speed up the process of diffusion in the bulk.

Vacuum technology

To speed up the time of reduction of outgassin : Bake-out

Increasing T diffusion coefficient D increases exponetially! kTEDeDD /

0−=

Baking-out a vacuum chamber we can get less outgassing in a

very short time.

15

Vacuum technology

Permeation

d : wall thickness D : diffusion coefficient

Absorption on The extenal surface

of the vacuum chamber

Diffusion through the wall of the chamber

Desorption from the internal wall.

Idealbehaviorofpermea8onrateasfunc8onof8me,assumingweexposeachamber(absolutevacuum)at=0toanexternalpressure.

Permeation is a three processess phenomenon

Contribution to the throughput

Vacuum technology

•  2’ •  110 gg •  3.2 105 aa

S

QePP i

itVS ∑+=

−

0

permeationvapor

.diffused/or & desorbedgas

=

=

=

=

p

V

D

L

QQ

QleaksQ

16

Evacuation in molecular regime.

( ) ( )[ ]{ }u

CStVSui PePPP pp +⋅−= +− 1)(

{ }

( )CSCSSePP

ppeff

tVSi

eff

+⋅=

⋅= − , Transition

StQtP G )()( evolution time =

P Q S

Sp C

SQtP G=)( pressure ultimate

1

1αtq

qn =

10

10αtq

qn =

Theoretical description quite complex different monolayer coverage. different interaction energy and chemical-physical bonding, dependence on the

concentration. For Vacuum design we “use” tables where are reported

From the tables we estimate (measuring the inner surface of our system) we estimate the throughput from the chamber, usually we call this base

pressure, its depends on the materials used and the processes of bake out Ex-sity or in-situ.

q =thorughput per area and its time relation after 1 h

Or time relation after 10 h

Real Surfaces

17

Outgassing of materials

Table for other materials

18

Continuing practice system.2

Steinless steel chamber Turbo Pumps TW 1600,

Back Pump Ecodry 15

Which is the base(ultimate) pressure we can reach after 10 h of pumping (or in-situ bake-out)?

Use the yellow labelled data in previous tables.

1st part

Ø Estimate QG due to the material of the chamber.

Ø Seff (effective pumping speed) and then

P (t) base pressure after 10 h or Pu ultimate pressure (after bake-out treatments).

19

Vacuum technology

2nd part

Max P inlet (in order to have Qmax)

inlet

outlet

Max P outlet

Vacuum technology

20

Pumping speed of back pump

Vacuum technology

3rd part Ø Dimensioning the back pump depends on the needs.

v Sp required in order to pump down the maximum

troughput and have right pressure required on the out-let of the HV pump.

v Or for UHV another parameter compression factor K0? v K0 = P outlet / P inlet

This is ideal parameter for Q=0, use this for cross-check Pu. Vacuum technology

Max P inlet (in order to have Qmax)

inlet

outlet

21

Specification of T 1600 and TW 1600

Vacuum technology

Table of formula for rough design of vacuum system

usually tube or circle aperture

Vacuum technology

viscous molecular Aperture A 20 A/(1-P2 /P1) 3.64 (T\M)1/2 A

9.16 D2

Long tube [πD4/(128ηL]Pave 3.81 (T\M)1/2 (D3\L)

182 D4\L Pave 12.1 D3\L Short tube 1\Cst=1\Clt +1\CA

Pave= average pressure In red air at T = 20 oC

L, D in cm, Ain cm2, T in K and C in l/s