Vacuum Class -Gas flow - gigin 1
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Transcript of Vacuum Class -Gas flow - gigin 1
Content
2
• 3.Gas Flow
– 3.1 Flow Regimes
– 3.2 Definition of Throughput and Pumping Speed
– 3.3 Conductance• 3.3.1 Definition
• 3.3.2 Combination
• 3.3.3 Long Tubes
• 3.3.4 Orifice
– 3.4 Calculating Pumping Speed at Different Places
3.1 Flow Regime (1)
3
• Flow Regime: Flow that through all different type of flow (because vacuum system)
• Vacuum system start form turbulent flow ,when pressure fall flow change to laminar flow. Both flow are viscous ; molecules striking each other and pushing each other
• Transition from turbulent to viscous flow depend on Reynold number; which function of flow velocity, mass density, tube diameter
3.1 Flow Regime (2)
4
• Knudsen number
• In viscous flow mfp< character dimension ; gas to gas collision ; Kn <0.01
• If pressure reduce then mpf = characteristic dimension; gas to wall interaction
• In region 1>Kn>0.01 transition region;• Kn>1 Flow considered molecular flow; In
molecular flow gas –wall collision predominate, The wall interaction is diffuse reflection
𝐾𝑛 =𝐿
𝑑
L=mean free path ;mfp [L1]
d=characteristic dimension of the system [L1]
Molecular interaction with surface
5
Diffuse reflectance . The length of the arrow for desorption is proportional to the probability of desorption in that direction
Specular reflectance: The angle of reflection equals the angle of incidence
Example Flashlight to mirror
Example Flashlight to wall with mate finish
Wrong: ping pong ball bouncing in the table
Right : Glue many ping pong ball on a table And attempt to bounce to another surface
Wrong: Incoming ball is very large compare to roughness of surface
Right : Consider that molecule do not bounce. Consider absorption, there resident time and desorbs
Cosine distribution
Molecular interaction with surface
Molecular flow
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Molecule arriving from region 1 move region 2
To gases not collide effected one and anotherHigher gas density region 1
More molecule move right than to left
Three different regions with P1>P2>P3
Section A
Molecular Flow; random traversing back & forth of the molecules from wall to wall with progress of molecules through the vacuum lines a matter of statistics
3.2 Definition of Throughput and Pumping Speed
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The rate flow trough the passage is depends on -Capacity of the pump-Geometrical shape-Type of flow-Gas Characteristic
Gas Flow rate – Mass Flow Rate Q- Volumetric Flow Rate S
Mass Flow Rate Q or ThroughputThe net number of molecules passing a given plane per unit timeUnit : pressure volume per unit time, example torr-liter per second
8
Steady state condition : the pressure at given location not changing as function of time
Pseudo-Steady state condition: no leak valve but gas evolving from the walls act similar to leak valve
Volumetric flow Rate
9
Volumetric flow Rate S or pumping Speed: the actual amount of substance which moved a distance d is not specify size that depends on Pressure
Unit volumetric flow is volume per unit time example liter/sec
Volumetric flow rate and mass flow rate are related to pressure by equation
𝑄 = 𝑆 𝑥𝑃
3.3 Conductance
10
• 3.3.1Definition
Conductance property of component which usually fills a certain amount of three dimensional space whereas volumetric flow rate is property of a position in space ( plane)
𝐶 =𝑄
𝑃1 − 𝑃2
Unit for Conductance same as volumetric flow is volume per unit time example liter/sec
Combination
11
• 3.3.2 Combination Q = constant
𝑃1 − 𝑃2 =𝑄
𝐶1𝑃3 − 𝑃4 =
𝑄
𝐶3𝑃2 − 𝑃3 =
𝑄
𝐶2
𝑃1 − 𝑃4 = 𝑄1
𝐶1+
1
𝐶2+
1
𝐶3
𝑃1 − 𝑃4 = 𝑄1
𝐶𝑡
1
𝐶𝑡=
1
𝐶1+
1
𝐶2+
1
𝐶3
𝐶𝑡 = 𝐶1 + 𝐶2 + 𝐶3
Conductance in series
Conductance in parallel
Remember ,Q=Mass Flow Rate/throughput
Long Tubes
12
• 3.3.3 Long Tubes
Long tubes has length significantly greater than its diameter
𝐶𝑣 =3000 < 𝑃 > 𝐷4
𝐿
𝐶𝑚 =80𝐷3
𝐿
Molecular flow od dry at 20oC
Viscous flow od dry at 20oC
<P>= Average Pressure , TorrD= tube diameter, inchesL=Tube Length ,inches
Orifice
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• 3.3.4 Orifice
Conductance of orifice is not infinity
𝐶0 = 11.6 𝐴
1
𝐶𝑡=
1
𝐶𝑜+1
𝐶𝑙
A =Area in cm2
Condutance ?
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• example
A Tube 10 inches long and 0.5 inches diameter
𝐶𝑙 =80(0.5)3
10=1.0 liter/sec
𝐶0 = 11.6 𝜋 0.25 𝑥 2.54 2 =14.7 liter/ sec
1
𝐶𝑡=
1
1.0+
1
14.7=1.07
𝐶𝑡 =0.94 liter/sec
A Tube 4 inches long and 6 inches diameter
𝐶𝑙 =80(4)3
6=853.33 liter/sec
𝐶0 = 11.6 𝜋 2 𝑥 2.54 2 = 940.45 liter/ sec
1
𝐶𝑡=
1
853+
1
940=0.00223
𝐶𝑡 = 447.39 liter/sec
3.4 Calculating Pumping Speed at Different Places
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• f 𝑃1 =𝑄
𝑆𝑙𝑃𝑜 =
𝑄
𝑆𝑝
𝐶 =𝑄
𝑃1 − 𝑃𝑜
𝐶 =1
1𝑆1−
1𝑆𝑝
1
𝑆1=
1
𝑆𝑝+1
𝐶𝑡1
𝑆𝑎=
1
𝑆𝑏+
1
𝐶𝑎𝑏
From equation previously
Conductance equation
Pumping speed
Generalized equation
Example
16
• Example
Pump with pumping speed 200 liters/sec
Attached to chamber via 6 inch tube with 10 inches diameter
Conductance = 951 liters /sec
Pumping speed enter the chamber?
1
𝑆𝑐=
1
200+
1
951
𝑆𝑐=0.936 liters/ sec
6 inches long and 10 inches diameter
𝐶𝑙 =80(6)3
10=1728 liter/sec
𝐶0 = 11.6 𝜋 2 𝑥 2.54 2 = 2116.01 liter/ sec
1
𝐶𝑡=
1
1728+
1
2116.01=
𝐶𝑡 = 951.21 liter/sec