75 m 175 so 68 10 8 m 68 nm at atmospheric pressure 68

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Unformatted text preview: havior 10-10 10 • For air molecules, r 1.75 m 1.75 • So 6.8 10 8 m = 68 nm at atmospheric pressure 6.8 10 68 • Note that mean free path is inversely proportional to the number density, which is itself proportional to pressure • So we can make a rule for = (5 cm)/(P in mtorr) (5 mtorr) Winter 2008 Lecture 5 – above 100 mTorr (about 0.00013 atm), air is still collisionally dominated (viscous) • is about 0.5 mm at this point – below 100 mTorr, gas is molecular, and flow is statistical rather than viscous (bulk air no longer pushes on bulk air) 7 Winter 2008 8 2 Vacuum Systems 01/24/2008 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Gas Flow Rates Evacuation Rate • At some aperture (say pump port on vessel), the flow rate is S = dV/dt (liters per second) • A pump is rated at a flow rate: S p = dV/dt at pump inlet • • • • • What you care about is evacuation rate of vessel S = Q/P1 but pump has Sp = Q/P2 Q is constant (conservation of mass) is Q = (P1 P2)C, from which you can get: 1/S = 1/S p + 1/C • The mass rate through the aperture is just: Q = PS (Torr liter per second) • And finally, the ability of a tube or network to conduct gas is C (in liters per second) QP 1 C P2 Q Q pump: Sp • So the net flow looks like the “parallel” combination of parallel” the pump and the tube: – the more restrictive will dominate • Usually, the tub...
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This note was uploaded on 01/30/2014 for the course PHYS 121 taught by Professor Staff during the Winter '08 term at UCSD.

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