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Vacuum-pump and air-change

  1. May 16, 2006 #1
    dear reader,


    - a box with a volume of 3/100 l (liter)
    - a vacuum pump of 700 mbar.


    how long takes it, to change the air inside the box with the described vacuum pump.


    i) 1 bar = 10^5 Pascal = 10^5 J/m^3
    ==> 700 mbar = 0.7 bar = 7*10^4 J/m^3
    ii) 1 m^3 = 1000 l (liter)
    ==> 7*10^4 J/m^3 = 7*10^4*10^-3 J/l = 70 J/l
    iii) problem: how can I calculate the needed time, when I have the energy: 70 J/l and the volume: 3/100 l?

    Thank you very much,


  2. jcsd
  3. May 16, 2006 #2

    Chi Meson

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    Something appears to be missing. Specifically, the power of the pump. Is there any other given or implied information?
  4. May 17, 2006 #3
    What you lack there is capacity of the pump. If you know it then use Gaede's equation to get the time duration.

    PS: Not that it is incorrect, the pressure unit of J/cu.mtr is the funniest I have ever seen. It is totally redundant in this context.
  5. May 17, 2006 #4

    the pump sucks with 700 mbar. Is it possible, that this value is the capacity of the pump? The pump sucks 700 mbar/sec and an infinite volume can bee sucked in?

    Thank`s for helping me,


  6. May 18, 2006 #5
    700 mbar is the pressure(vacuum) that is possible to acheive by the pump. What you should know is the flow capacity in liters/sec or any other suitable units.

    If you know the pump capacity, the time of evacuation is calculated, for a leak free system, by

    t = (V/Q) x ln(P1/P2)

    t is time in seconds
    V is volume of the vessel in liters
    Q is actual flowrate of pump in liters/minute
    P1 and P2 are initial and final absolute pressures (in your case P2 is 313.25 mBar and P1 is atmospheric)

    You can use any units consistent to the equation.
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