Post on 18-Apr-2020
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
' ⋅=
Vacuum technology
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
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