Conductance of an aperture

Transcription

1 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 ( T / M) ( P ) [ cm /sec] Q(,) P (,) dv / dt.64 0 ( T / M ) P(,) µ bar nv ave indipendent from P [ mol/sec] µ-scopic cm /sec Μ-scopic dv dv Q / ΔP ( P P ) / ΔP dt dt q / n φ / n vave v 4 4 P P onductance of a diaphragm Q / ΔP.64 ( T / M ) > e > + l/s indipendent from P 0 Q P > + Si ottiene Q P > Q Q Q P P P > P > e e 0 << o : e ~ (pertura) o : e ( no resistenza al flusso), 0 i sarebbe flusso anche se P P uguali, quindi -> -> 0.5 o : e (effetto diaframma)

2 onductance of a long tube Ø Knudsen derived assuming molecules moving randomly, and number collisions on unit surface and unit time ¼ nv ave (*). lt 8 kt π m T BL π M / BL in GS units, is the area and B the boundary. For a cilindrical Tube: πd /4 and BπD ir at 0 : lt.8 T M ( ) ( ) D L D[ cm], L [ cm] e [ l/s].. D lt (*). Roth Vacuum Technology (Elsevier Science, msterdam, 990 Third edition) L 0 Short tube conductance l 0: / e 0 is the cross section of the upstream vessel / / + / st st lt lt ( ) + lt e ; e st Short tube serie of long tube and aperture [ + ] ltk" lt lt e lt e.44 0 π T M T M / / / BL [ / ] 0 Diaphragm effect K" + 5. BL 0 K correction factor alled of Knudsen

3 orrection 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 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 4 (O Hanlon) where 0 is the conductance of the aperture. lausing factor ccuracy of calculations lausing % ssuming the big dimension of chamber upstream and downstream Knudsen 5 % / / + / st lt

4 st nalitical description for tubes / ''.8( T / M ) ( D L) K" { +. ( D / L) [ ( D / D ) ]} K / lt K" / v If the volume is bigger with respect to the tube diameter (.09) fig..8 {.( D/ )} / M ) D /[ L +. D] K'' / + L ( D L) / lt.8( T / M ) / ( ).8( T / K' 5( L / D) + ( L / D) 0 + 8( L / D) + ( L / D) [ (/ 4)( L' / )] { + / [ (6/ 7)( L/ D) ]} P r / + D L ' L + Santeler 986 Knudsen lausing 9 (.07) fig..8 omparison Ø ilindrical tube l 0 cm, d cm st.44 0 π 4 / T + 5. M BL BL GS units. Ø H 00 K ü Long tube 4.67 l/s 0 % ü Knudsen short tube 4. l/s 5 % ( T M ) / a' cm 4 ü lausing.96 l/s % sec 4

5 Monte arlo Methods Molecules randomized P r / 0 match well with the Monte arlo Methodes K Knudsen correction factor K lausing correction factor 5

6 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 S 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 S 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 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? 6

7 In which regime are we? d dimension of ducts, chamber State of the gas k Flow regime R e Viscous d/λ >0 Turbulent 00 ir admi,ed 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 Viscous (Flow) just mention Ø Qpv where v velocity and QpS Q/ΔP Q through an aperture, adiabac expansion f(p, P,γ,T, M) For air at 0 0 /(-P /P ) if > P /P > 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 /s For air at 0 tube 8 D 4 /L P ave ; D & L [cm], P ave [Torr], and [l/s] 7

8 omparison between viscous and molecular regimes Molecular regime ( T ) ; [ ] cm, [ ] l/s..64 M For air at 0.6 l/s, 9.6 D l/s, Pumping speed through an aperture: S Q/P S Q/P (P -P )/P for P <0.. P (usual case): S.6 l/s The maximum pumping speed in molecular regime is.6 l/(s cm ), if compared to the viscous regime: S 0 l/(s cm ) Intermediate regime Ø (c D 4 P ave +c D )/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 8

9 J v / m Evacuation from the atmosphere, Only yseful in case of big chamber, or frequently opened Ø Evacuation from a volume V P S Q Q dp dt V P d( PV ) dt Q P S P e 0 P S S t V 9

10 Evacuation in viscous regime S P p P S p /S/S p +/ (π/8)[d 4 /(ηl)] P ave P E (P+P p )/ ave (P+P p )/ QPS-V(dP/dt) Equation can be solve an plotted by using as a parameter D 4 /L(8/π )η E V 00 l D cm L 00 cm D 4 /L If S p l/s, then we have t/v6 then t 600 sec ssuming L 0, we have t/v4.5 t 450 sec 0

11 nother example 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. S of turbo for N 600 l/s, and is connected by a tube of D50 mm and L 8-7 mm. Dimensioning vacuum system: in stationary condition Ø Maximum allowed (required) pressure in the chamber? ᴥ P ch Q/S ch ᴥ /S ch /S hv +/ hv S ch S hvp hv S 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/S hv-out ᴥ but ᴥ /S hv-out /S bp +/ bp P bhv-out P bp S bp S bp 5 m h -

12 Dimension of the chamber for next steps Through-put(Q) sources in vacuum systems Ø The following sources of throughput appear in a vacuum system

13 Q sources: vaporization In a closed system Saturated vapour pressure Liquid Open system / Vacuum system Vapour Pressure Liquid Φ ( molecules s ) nv n P/ kt ave / / 4 v 8kT ave π m P v M T Φ molecular flow coming out from a liquid of surface SI units: [P v ] Pa, [] 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

14 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 ctivation 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. 4

15 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 bsorption 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 S t V + i S Q i Q D gasdesorbed & /or diffused. Q p Q Q V L leaks vapor permeation 0 gg. 0 5 aa 5

16 Evacuation in molecular regime. Q S P S p P ( P P ) e i S u {( S V ) t [ ( S ) ]} + Transition P eff S p p p P e i + P u { Seff V } t, ( S + ) time evolution P ( t ) Q ( G t ) ultimate pressure P( t) p Q G S S q q n t α Real Surfaces 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. 6

17 Outgassing of materials Table for other materials 7

18 ontinuing practice system. Steinless 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 S eff (effective pumping speed) and then P (t) base pressure after 0 h or P u ultimate pressure (after bake-out treatments). 8

19 inlet nd part outlet Max P inlet (in order to have Q max ) Max P outlet 9

20 Pumping speed of back pump rd part Ø Dimensioning the back pump depends on the needs. inlet Max P inlet (in order to have Q max ) outlet v S 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. 0

21 Specification of T 600 and TW 600 Table of formula for rough design of vacuum system usually tube or circle aperture perture Long tube Short tube viscous 0 /(-P /P) molecular.64 (T\M)/ 9.6 D [πd4/(8ηl]pave.8 (T\M)/ (D\L) 8 D4\L Pave. D\L \st\lt +\ Pave average pressure In red air at T 0 o L, D in cm, in cm, T in K and in l/s