Gas Flow

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Content

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• 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)

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• 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)

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• 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

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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

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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

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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

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• 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

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• 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

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• 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

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• 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