Conductance of an aperture

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n Vacuum φ A 4 How many molecules travel through A in a time t? P P A onductance of an aperture 3 q φa 3.64 0 ( T / M) na Volume of gas passing through A? Q Q Q / ΔP 3.64 ( T / M ) 3 dv/dt q/n 3.

1 n Vacuum φ A 4 How many molecules travel through A in a time t? P P A onductance of an aperture 3 q φa ( T / M) na Volume of gas passing through A? Q Q Q / ΔP 3.64 ( T / M ) 3 dv/dt q/n ( T / M) Q P ( T / M) A( P ) A l/s A 3 [ cm /sec] 3 Q(,) P (,) dv / dt ( T / M ) AP(,) µ bar nv ave indipendent from P [ mol/sec] µ-scopic 3 cm /sec Μ-scopic dv dv Q / ΔP ( P P ) / ΔP dt dt q / n φa / n vave A va 4 4 A 0 hort tube conductance l 0: / e A 0 is the cross section of the upstream vessel / / + / st lt st lt + lt e e st hort tube serie of long tube and aperture [ + ] ltk" lt lt e lt e π T M T M / / A/ A BL [ A/ A ] 0 Diaphragm effect A A K" BL A 0 K correction factor alled of Knudsen

2 orrection factors Ø hort tube : v Knudsen factor molecule impinging on the surface of the tube come out with a cos θ ( K ) spatial distribution. v lausing factor Higher probability for the molecule to come out in the flow direction ( K ). Described as : P r (Roth), / 0 or (equivalent) a' v A 4 (O Hanlon) where 0 is the conductance of the aperture. lausing factor Accuracy of calculations lausing % Assuming the big dimension of chamber upstream and downstream Knudsen 5 % / / + / st lt A

3 st (3.09) fig. 3.8 Analitical description for tubes / 3 '' 3.8( T / M ) ( D L) K" { +.33 ( D / L) [ ( D / D ) ]} K / lt K" / v If the volume is bigger respect to the tube diameter {.33( D/ )} / 3 M ) D /[ L +. D] K'' / + L 3 ( D L) / lt 3.8( T / M ) / ( ) 3.8( T / 33 K' 5( L / D) + ( L / D) ( L / D) + ( L / D) [ (3/ 4)( L' / )] { + / [ 3 (6/ 7)( L/ D) ]} P r / + D L ' L + anteler 986 Knudsen lausing 93 (3.07) fig. 3.8 omparison Ø ilindrical tube l 0 cm, d cm st π 4 / T A A M BL BL G units. Ø H 300 K ü Long tube 4.67 l/s 0 % ü Knudsen short tube 4. l/s 5 % ( T M ) / a' A 3 cm 4 ü lausing 3.96 l/s % sec 3

4 Monte arlo Methods Molecules randomized P r / 0 match well with the Monte arlo Methodes K Knudsen correction factor K lausing correction factor 4

Example-exercise: He lamp He discharge He I :. ev at Torr HeII: 40.8 ev at 0. Torr I II 0 - III 0-4 chamber 0-8 Torr eff of 500 l/s in the vacuum chamber P base 0-0 Torr without He flow.

5 Example-exercise: He lamp He discharge He I :. ev at Torr HeII: 40.8 ev at 0. Torr I II 0 - III 0-4 chamber 0-8 Torr eff of 500 l/s in the vacuum chamber P base 0-0 Torr without He flow. a) alculate the length of a capillary (d mm) in order to have P work 0-8 Torr in the chamber in working condition of Torr discharge. b) Which are required in II e III stages if the capillary lengths are L II-III L I-II 5 cm? c) Once the length of the whole capillary is designed, calculate the pressures in any stage for the discharge at 0. Torr. heck λ/d for He and flow regime? t c 5 o Viscous (Flow) just mention Ø QpvA where v velocity and Qp Q/ΔP Q through an aperture A, adiabac expansion f(p, P,γ,T, M) For air at 0 0 A/(-P /P ) if > P /P > A l/(s cm ) if P /P < of a tube of diameter D and length L, tube [πd 4 /(8ηL)] P ave ; [D][L ]cm, [P ave ] dyne/cm and [] cm 3 /s For air at 0 tube 8 D 4 /L P ave ; D & L [cm], P ave [Torr], and [l/s] 5

6 Another example: towards the experience Ø Base Pressure v P base mbar Ø Max Pressure allowed v P max mbar ü Find the max troughput which can be supported by the system. of turbo for N 600 l/s, and is connected by a tube of D50 mm and L 8-73 mm. Dimension of the chamber for next steps 6

7 Dimensioning vacuum system: in stationary condition Ø Maximum allowed (required) pressure in the chamber? ᴥ P ch Q/ ch ᴥ / ch / hv +/ hv ch hvp hv hv-out bp ᴥ P hv-outlet maximum allowed pressure at the out-let of HV pump. ᴥ bp conductance between HV pump and back pump. ᴥ P bhv-out Q/ hv-out ᴥ but ᴥ / hv-out / bp +/ bp P bhv-out P bp bp bp 5 m 3 h - Ø We have two possible model of HV pumps one requires at the outlet a pressure below 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 m. heck which turbo pump is required? 7

8 In which regime are we? d dimension of ducts, chamber tate of the gas k Flow regime R e Viscous d/λ >0 Turbulent 00 Air admi7ed in vacuum Laminar 00; Transi8on < d/λ < 0 Intermediate Rarefied d/λ < Molecular d P > cm Torr < d P < cm Torr d P < cm Torr omparison between viscous and molecular regimes Molecular regime A ( T ) A; [ A] cm, [ ] l/s M For air at 0 A.6 A l/s, A 9.6 D l/s, Pumping speed through an aperture: Q/P Q/P (P -P )/P for P <0.. P (usual case):.6 A l/s The maximum pumping speed in molecular regime is.6 A l/(s cm ), if compared to the viscous regime: 0 A l/(s cm ) 8

9 Intermediate regime Ø (c D 4 P ave +c D 3 )/L c π/8 η c (π/6)[(π/)(r 0 T/M) / [(-f)/f] f fraction of molecules absorbed an reemitted from the surface If the pressure P ave is sufficiently high the term D 4 dominates (viscous flow). When the two terms are equal: P ave P t c /(c D) transition pressure. When the pressure becomes low according : D. P ave < Torr cm molecular regime J v / m 9

10 Evacuation from the atmosphere, Only yseful in case of big chamber, or frequently opened Ø Evacuation from a volume V P Q Q dp dt V P d( PV ) dt Q P P e 0 P t V Evacuation in viscous regime P p P p // p +/ (π/8)[d 4 /(ηl)] P ave P E (P+P p )/ ave (P+P p )/ QP-V(dP/dt) Equation can be solve an plotted by using as a parameter D 4 /L(8/π )η E 0

11 V 00 l D cm L 00 cm D 4 /L If p l/s, then we have t/v6 then t 600 sec Assuming L 0, we have t/v4.5 t 450 sec Through-put(Q) sources in vacuum systems Ø The following sources of throughput appear in a vacuum system

12 Q sources: vaporization In a closed system aturated vapour pressure Liquid Open system / Vacuum system Vapour Pressure Liquid Φ ( molecules s ) nv n P/ kt ave / A / 4 v 8kT ave π m P v A M T Φ molecular flow coming out from a liquid of surface A I units: [P v ] Pa, [A] m [T] K 0 order K Q sources: desorption Outgassing of gases adsorbed in atm and or finale stage of diffusion and permeation Tipical law of rate of lost proportional to the initial /( N0kT ) d( t) e K ( t) ( t) dt E d τ 0 Ed / N0kT τ r τ 0e τ r residence time decreases ( t) e 0 Kt e 0 t /τ r increasing T. (t) exponentially decreased increasing T. 0 order H - H+H - H d dt d dt ( t) ( t) k ( ) t k ( + K t) 0 0

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

13 Q sources: diffusion Process slower than desorption Γ dn dx D / + D n n d q 0 ( ) exp t 0 Dt / D t 0 : q 0 t t Initial concentration t : q D e d πdt d Bulk thickness t -/ -/ e at The diffusion constant is function of the activation energy and T D 0 D e E kt E Activation Energy Then increasing the T speed up the process of diffusion in the bulk. To speed up the time of reduction of outgassin : Bake-out Increasing T diffusion coefficient D increases exponetially! D D 0 e E D / kt Baking-out a vacuum chamber we can get less outgassing in a very short time. 3

14 Permeation Ideal behavior of permea8on rate as func8on of 8me, assuming we expose a chamber (absolute vacuum) a t0 to an external pressure. d : wall thickness D : diffusion coefficient Diffusion through the wall of the chamber Absorption on The extenal surface of the vacuum chamber Desorption from the internal wall. Permeation is a three processess phenomenon ontribution to the throughput P P e 0 t V + i Q i Q D gasdesorbed & /or diffused. Q p Q Q V L leaks vapor permeation 0 gg aa 4

15 Evacuation in molecular regime. Q P p P ( P P ) e i u {( V ) t [ ( ) ]} + Transition P eff p p p P e i + P u { eff V } t, ( + ) time evolution P ( t ) Q ( G t ) ultimate pressure P( t) p Q G q q n t α Real urfaces 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 q thorughput per area and its time relation after h Or time relation after 0 h q n q t 0 α0 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. 5

16 Outgassing of materials Table for other materials ont. 6

ontinuing practice system. teinless steel chamber Turbo Pumps TW 600, Back Pump Ecodry 5 Which is the base(ultimate) pressure we can reach after 0 h of pumping (or in-situ bake-out)?

17 ontinuing practice system. teinless steel chamber Turbo Pumps TW 600, Back Pump Ecodry 5 Which is the base(ultimate) pressure we can reach after 0 h of pumping (or in-situ bake-out)? Use the yellow labelled data in previous tables. st part Ø Estimate Q G due to the material of the chamber. Ø eff (effective pumping speed) and then P (t) base pressure after 0 h or P u ultimate pressure (after bake-out treatments). 7

18 inlet nd part outlet Max P inlet (in order to have Q max ) Max P outlet 8

19 Pumping speed of back pump 3 rd part Ø Dimensioning the back pump depends on the needs. inlet Max P inlet (in order to have Q max ) outlet v p 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 K 0? v K 0 P outlet / P inlet This is ideal parameter for Q0, use this for cross-check P u. 9

20 pecification of T 600 and TW 600 tationary systems: distributed throughput Ø Accelerators, beam pipes, ecc. dq qd Bdx Q through ( dx : Q L ) dp ( ) dx dq L d P dx d ( ) dx : d P qd B L dx dp q B x + k D L dx For x L : (dp dx )L 0 k qd B dp q B x + q D B D L dx P [qd B L ] x + (qd B ) x + k For x 0 : P0 q D BL p k [ P qd B L p + (x ) x (L )]; [(x ΔP Px P0 qd B PL P0 qd BL [ ] PL qd B L p + ( ) ) x (L )] ( ) 0

21 Distributed pumping Ø Pressure drop Δp is indipendent from p. x x+dx L Q( x) throughput Pa m outgassing / A 3 - q(x) Pa m s superficieper unità di l A m conduttanza specificac m s 4 s c L dp dx x± L 0 Boundary condition P(0) Q(0) Q(0) BqL - [ ] dq dx Bq Q ( x) cdp Bq c d P dx dx... Distributed pumping P( x) Lx x Bq c + L P ave L L L P( x) dx Bq + L 3c 0 P max L Bq P min L Bq + 3c L L L P min Bq Bq 3c 3 It is possible to have a more homogeneous pressure, where P max can be controlled by the pumping speed the distance in between the pumps and the conductance of the tube. Bq 3 L + L

22 Table of formula for rough design of vacuum system usually tube or circle aperture viscous molecular Aperture A 0 A/(-P /P ) 3.64 (T\M) / A 9.6 D Long tube [πd 4 /(8ηL]P ave 3.8 (T\M) / (D 3 \L) 8 D 4 \L P ave. D 3 \L hort tube \ st \ lt +\ A P ave average pressure In red air at T 0 o L, D in cm, Ain cm, T in K and in l/s

Conductance of an aperture

Conductance of an aperture n Vacuum φ 4 How many molecules travel through in a time t? P P onductance of an aperture q φ.64 0 ( T / M) n Volume of gas passing through? Q Q dv/dt q/n.64 0 ( T / M) Q P Q / ΔP.64 (T / M ) l/s.64 0