Vacuum I G. Franchetti CAS - Bilbao 30/5/2011 G. Franchetti 1
Index Introduction to Vacuum Vacuum and the Beam Flow Regimes Creating Vacuum 30/5/2011 G. Franchetti 2
Vacuum in accelerators All beam dynamics has the purpose of controlling a charged beam particle Structure of magnets and RF structure have the purpose of creating a proper guiding (E,B) structure However in an accelerator are present a jungle of unwanted particles which creates a damaging background for beam operation 30/5/2011 G. Franchetti 3
Characteristic of Vacuum Vacuum is considered as a Gas which is characterized by Macroscopic Quantities Pressure Temperature Density Composition Aim: Minimize the interaction of beam with vacuum gas Microscopic Quantities Kinetic Theory 30/5/2011 G. Franchetti 4
Typical numbers Particles m -3 Atmosphere 2.5 x 10 25 Vacuum Cleaner 2 x 10 25 Freeze dryer 10 22 Light bulb 10 20 Thermos flask 10 19 TV Tube 10 14 Low earth orbit (300km) 10 14 H 2 in LHC ~10 14 SRS/Diamond 10 13 Surface of Moon 10 11 Interstellar space 10 5 R.J. Reid 30/5/2011 G. Franchetti 5
Particle Wall collision m The velocity after wall collision depends on the particle-wall interaction v 2 interaction v 1 Momentum is propagated to the wall 30/5/2011 G. Franchetti 6
Pressure SI Units: Low vacuum Atm. pressure to 1 mbar m Medium vacuum High Vacuum (HV) Ultrahigh vacuum (UHV) Extreme high vacuum (XHV) 1 to 10-3 mbar 10-3 to 10-8 mbar 10-8 to 10-12 mbar less than 10-12 mbar other units: 30/5/2011 G. Franchetti 7
Particle Particle interaction v 1 v 1 In a gas large number of collisions Most likely process P-P collision v 2 Interaction m v 2 For elastic collisions: 1) Energy conservation 2) Momentum conservation Temperature 30/5/2011 G. Franchetti 8
Typical numbers Air at T= 20 o C 20% O 2 M O = 8x2 g-mole = 8x2/N A = 2.65 x 10-23 g 80% N 2 M N = 7x2 g-mole = 7x2/N A = 2.32 x 10-23 g Therefore Molecules run fast! But the average velocity is 30/5/2011 G. Franchetti 9
f Velocity Distribution When a gas is at equilibrium the distribution of the velocity follows the Maxwell-Boltzmann distribution 1 0.8 0.6 max velocity ave. velocity 0.4 0.2 30/5/2011 G. Franchetti 10 0 0 1 2 3 4 v/v a
Equation of state An ideal gas in a container of volume V satisfies the equation of state For a gas in equilibrium SI units P Pressure [Pa] V Volume in [m 3 ] n moles [1] T absolute temperature [K] R 0 = 8.314 [Nm/(mole K)] universal constant of gas k B = 1.38x10-23 [JK -1 ] Boltzmann constant N A = 6.022x10 23 [1] Avogadro s Number 30/5/2011 G. Franchetti 11
Mean free path free path = path of a particle between two collisions 30/5/2011 G. Franchetti 12
Mean free path v Effective area covered by the rest atoms Density of atoms Cross-section Transverse area v # particles at l + Δl that did not collide N(l) = particle at l Particles that did not collide 30/5/2011 G. Franchetti 13
Mean free path How many particles collide between and? Therefore dn particles travelled a distance l and then collided Probability that a particle travel l and then collide is Mean free path In a gas 30/5/2011 G. Franchetti 14
Example Air at T = 20 0 C and P = 1 atm From equation of state atom/m 3 Diameter of molecule of air Mean free path 30/5/2011 G. Franchetti 15
N 2 at T = 20 0 C, d = 3.15 10-10 m 30/5/2011 G. Franchetti 16
Vacuum and Beam 30/5/2011 G. Franchetti 17
Beam lifetime After a beam particle collides with a gas atom, it gets ionized and lost because of the wrong charge state with respect to the machine s optics Beam of particles going through a vacuum gas survives according to As beam particle have a velocity v 0, then From the equation of state Beam lifetime Beam lifetime sets the vacuum constrain 30/5/2011 G. Franchetti 18
Example It is important to know the cross-section of the interaction beam-vacuum Example LHC Of what is formed the vacuum? Does it depends on the energy? H 2 at 7 TeV at T = 5 0 K for Nuclear scattering cross section at 7 TeV for different gases and the corresponding densities and equivalent pressures for a 100 hours beam lifetime LHC Design Report 30/5/2011 G. Franchetti 19
O. Gröbner, CAS 2007 30/5/2011 G. Franchetti 20
Impingement rate For a Maxwell-Boltzmann distribution J units: [#/( s m 2 )] and 30/5/2011 G. Franchetti 21
More on collision mean free path mean free path 30/5/2011 G. Franchetti 22
Knudsen Number Flow Regimes Continuous Regime Particles collide mainly among them the fluid is continuous SI units: [1] Transitional Regime This is a regime in between continuous and molecular Molecular Regime particles collide mainly with walls 30/5/2011 G. Franchetti 23
The Throughput Vessel with a gas at Number of particles The quantity is proportional to the number of particles is called throughput SI units: [Pa m 3 /s] 30/5/2011 G. Franchetti 24
Conductance In absence of adsorption and desorption processes the throughput does not change 1 T = constant 2 equal Conductance SI units: [m 3 /s] 30/5/2011 G. Franchetti 25
Composition (Series) C 1 C 2 P u1 P d1 =P u2 P d2 equal throughput T = constant Conductance of 1+2 30/5/2011 G. Franchetti 26
Composition (Parallel) C 1 Vessel 1 Vessel 2 Conductance of 1+2 P u P d C 2 T = constant 30/5/2011 G. Franchetti 27
Molecular Flow Regime dominated by Particle Wall interaction K n > 0.5 P < 1.3 x 10-3 mbar D = 0.1 m Classical Molecular Mechanics regime m v 2 v 2??? v 1 v 1 30/5/2011 G. Franchetti 28
Cosine-Law 1. A particle after a collision with wall will loose the memory of the initial direction 2. After a collision a particle has the same velocity v 3. The probability that particles emerge in a certain direction follow the cosine law = solid angle 30/5/2011 G. Franchetti 29
Consequences Concept of Transmission probability 1 A gas particle can return back! 2 In the molecular regime two fluxes in opposite direction coexist particle flux = N 1 = particle per second through 1 N 2 = particle per second through 2 30/5/2011 G. Franchetti 30
Conductance of an Aperture Gas current Volumetric flow Downstream Upstream Throughput Conductance 30/5/2011 G. Franchetti 31
Conductance of a long tube long tube = no end effect d stationary state L 0 make α which is wrong: This result is valid for L long The dependence of 1/L is consistent with the rule of composition of conductance 30/5/2011 G. Franchetti 32
Continuum flow regime K n < 0.01 P > 6.5 x 10-2 mbar V = 0.1 m Roughly more than 100 collision among particles before wall collision Local perturbation of propagate through a continuum medium Collision among particles create the Viscosity Collision with walls cancel the velocity zero velocity at the walls Max. velocity 30/5/2011 G. Franchetti 33
Viscous/Laminar regime Reynold Number Laminar Regime ρ = density of gas [Kg/m 3 ] v = average velocity [m/s] D h = hydraulic diameter [m] Turbulent Regime A = cross sectional area B = perimeter η = fluid viscosity [Pa-s] M. Lesiuer, Turbulence in Fluids, 4 th Edition, Ed. Springer 30/5/2011 G. Franchetti 34
Reynold Number as function of Throughput For air (N 2 ) at T=20C η = 1.75 x 10-5 Pa-s with Therefore the transition to a turbulent flow takes place at the throughput of For a pipe of d = 25 mm [mbar l /s] [mm] At P = 1 atm this threshold corresponds to v = 0.389 m/s 30/5/2011 G. Franchetti 35
Laminar Regime 1. Fluid is incompressible: true also for gas when Ma < 0.2 2. Fluid is fully developed 3. Motion is laminar Re < 2000 4. Velocity at Walls is zero Fully developed flow entry length velocity distribution at injection in the pipe velocity distribution of a fully developed flow 30/5/2011 G. Franchetti 36
Conductance in Laminar Regime long tube = no end effect D P u P d L Conductance: where The conductance now depends on the pressure! 30/5/2011 G. Franchetti 37
Conductance in Turbulent Regime long tube = no end effect D P u P d L The throughput becomes a complicated relation (derived from the Darcy-Weisbach formula) Darcy friction factor, dependent from the Reynold number 30/5/2011 G. Franchetti 38
Sources of vacuum degradation 30/5/2011 G. Franchetti 39
P [atm] Evaporation/Condensation At equilibrium P E = saturate vapor pressure condensation Example: water vapor pressure 1 evaporation 0.8 0.6 0.4 0.2 0 0 20 40 60 80 100 T [C] 30/5/2011 G. Franchetti 40
Outgassing The outgassing is the passage of gas from the wall of the Vessel or Pipe to the vacuum Mean stay time Θ = fraction of sites occupied E d [Kcal/mole] Cases τ d [s] 0.1 Helium 1.18 x 10-13 1.5 H 2 physisorption 1.3 x 10-12 3-4 Ar,CO,N 2,CO 2 physisorption 1.6 x 10-11 10-15 Weak chemisorption 2.6 x 10-6 Throughput due to outgassing 20 H 2 chemisorption 66 25 3.3 x 10 5 (~half week) 30 CO/Ni chemisorption 1.6 x 10 9 (~50 years) 40 4.3 x 10 16 (~half age of earth) N s = A x 3x10 15 A = surface in cm 2 150 O.W chemisorption 1.35 x 10 98 (larger than the age of universe) P. Chiggiato, CAS 2007 30/5/2011 G. Franchetti 41
Leaks High vacuum system Q L < 10-6 mbar l /s Q L < 10-5 mbar l /s Q L < 10-4 mbar l /s very tight tight leaks Small Channels Diameter hole 0.01 mm 10-10 m Leaks rate 10-2 mbar-l/s 10-12 mbar-l/s K. Zapfe, CAS 2007 317 years for leakage of 1 cm 3 30/5/2011 G. Franchetti 42
Pumps and Gas flow Ideal Pump particles that enter in this chamber never returns back!! Pumping speed S 0 [m 3 /s] but S 0 is Here Example N 2 at T=20C Take D=0.1m S 0 = 0.92 m 3 /s 30/5/2011 G. Franchetti 43
Pumps and Conductance Pumping speed S [m 3 /s] Pumping speed of ideal pump S 0 d conductance C As the throughput is preserved Example: if the pipe is long l=100d m 3 /s 30/5/2011 G. Franchetti 44
Pumping Process -- Pump-down Time - - Ultimate Pressure hole Pump throughput due to evaporation throughput due to outgassing throughput due to leaks effective throughput 30/5/2011 G. Franchetti 45
Pumping Process -- Pump-down Time - - Ultimate Pressure Therefore P/P 0 1 Ultimate pressure equilibrium between sources of throughput and pumps throughput t [s] Pump down time 30/5/2011 G. Franchetti 46
Multistage pumps First stage pump down Second stage P/P 0 1 t [s] 30/5/2011 G. Franchetti 47
Creating the Vacuum: Pumps Examples of vacuum in some accelerators LHC 10-10 10-11 mbar (LHC Design Report) FAIR HEBT 10-9 mbar J.Y. Tang ICA01, TUAP062 SIS100 10-12 mbar A. Kraemer EPAC2006, TUPCH175 30/5/2011 G. Franchetti 48
Positive Displacement Pumps Principle: A volume of gas is displaced out of the Vessel Vessel Out Piston Pumps volume V P out Rotary Pumps Liquid ring pumps P in Dry vacuum pump A compression of the volume V is always necessary to bring the pressure from P in to a value larger then P out 30/5/2011 G. Franchetti 49
Piston Pumps Vessel P i P o P o Minimum volume Vmin 30/5/2011 G. Franchetti 50
Piston Pumps Vessel P i P o P Here the pressure P is smaller than P o but larger than P i 30/5/2011 G. Franchetti 51
Piston Pumps Vessel P i P o P i At this point the internal pressure becomes equal to P i P i V i = P 0 V min 30/5/2011 G. Franchetti 52
Piston Pumps Vessel P i P o The internal pressure remains P i but the volume increase to V max Effective gas volume entered into the pump equal to V max V i : this is the gas which will be expelled 30/5/2011 G. Franchetti 53
Piston Pumps Vessel P i P o P o At the Volume V c the internal pressure becomes equal to P o and the valve opens. P o V c = P i V max and from now on the gas entered into the pump goes out 30/5/2011 G. Franchetti 54
Piston Pumps Therefore if the pumps make N c cycles per second Conclusion: the pumping speed depends on the ratio of outlet/inlet pressure When the inlet pressure is too low the pump stops pumping S S 0 P limit P o 30/5/2011 G. Franchetti 55
If the gas compression/expansion is isentropic then all transformation follow the law hence Clearly the dependence affects the ultimate pressure even in absence of sources of throughput Ultimate pressure 30/5/2011 G. Franchetti 56
General Pump Pumping speed of ideal pump S 0 But in real pumps there is a backflow Pump Inlet Outlet P i P 0 Pump throughput Backflow 30/5/2011 G. Franchetti 57
Zero Load Compression Rate Inlet P i0 P 0 Outlet Zero Load compression rate Closed Backflow The backflow is found 30/5/2011 G. Franchetti 58
Rotary Pumps S = 1-1500 m 3 /h Pl = 5 x 10-2 mbar (1stage) Pl = 10-3 mbar (2stage) Gas displaced in the moving vane Gas Ballast During the compression there can be gas component (G), which partial pressure P G can be too high condensation But the maximum pressure during compression do not exceed P 0 therefore by injectionng non condensable gas during the compression rate P G is lowered blow the condensation point 30/5/2011 G. Franchetti 59
Liquid Ring Pumps S = 1 27000 m 3 /h P = 1000 mbar 33 mbar The gas enter in the cavity which expand. When the pressure in the cavity reaches the sature vapor pressure P s the water boils. During the compression the vapor bubbles implode creating the CAVITATION At T = 15 0 C P s = 33 mbar which sets the P limit =Ps 30/5/2011 G. Franchetti 60
Dry Vacuum Pumps: Roots pumping speed: 75 30000 m 3 /h operating pressure: 10 10-3 mbar 30/5/2011 G. Franchetti 61
Kinetic Vacuum Pumps Principle The pump gives a momentum to the gas particles so that they flows out from the outlet Molecular drag pump Turbo molecular Pump Diffusion Ejector pump 30/5/2011 G. Franchetti 62
Molecular Drag Pumps Based on the molecular drag effect m U 30/5/2011 G. Franchetti 63
w h U Volumetric flow: Taking into account of the backflow K 0 = zero load compression rate, 30/5/2011 G. Franchetti 64
P i P o U The zero load compression rate is But for a long tube if U ~ v a Example L = 250 mm, h = 3 mm S 0 /C > 10 and K 0 >> 1 A. Chambers, Modern Vacuum Physics, CRC, 2005 30/5/2011 G. Franchetti 65
Example of Molecular drag pump pumping speed: 7-300 l/s operating pressure: 10-3 10 3 Pa ultimate pressure 10-5 to 10-3 Pa 30/5/2011 G. Franchetti 66
Turbo-molecular pump Principle v a φ U General particle velocity distribution U tilted blades 30/5/2011 G. Franchetti 67
Turbo-molecular pump In the reference frame of the rotating blades U v a - U φ General particle velocity distribution tilted blades 30/5/2011 G. Franchetti 68
Turbo-molecular pump In the reference frame of the rotating blades U v a - U General particle velocity distribution 30/5/2011 G. Franchetti 69
Turbo-molecular pump Returning in the laboratory frame U U The vacuum particle has received a momentum that pushes it down but the component of velocity tangential becomes too large 30/5/2011 G. Franchetti 70
Multistage U U rotor stator U U rotor stator The rotational component is lost and another stage can be placed 30/5/2011 G. Franchetti 71
Turbo-molecular pump Probability of pumping The maximum probability W max is found when P i = P o And we also find that K 0 = compression rate at zero load 30/5/2011 G. Franchetti 72
The compression rate at zero load Therefore therefore different species have different pumping probability In addition the maximum pumping probability W max is Therefore we find that the maximum pumping speed independent on the gas mass is 30/5/2011 G. Franchetti 73
Maximum compression as function of foreline pressure Pumping speed as function of the inlet pressure (Blazers Pfeiffer) pumping speed: 35-25000 l/s ultimate pressure 10-8 to 10-7 Pa (Blazers Pfeiffer) 30/5/2011 G. Franchetti 74
End of Vacuum I 30/5/2011 G. Franchetti 75