How Long Does It Take to Change Air in a Box Using a 700 mbar Vacuum Pump?

[hick]

dear reader,

given:

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



question:

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



solution:

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,


sincerly,


Hick

 

[Chi Meson]

Something appears to be missing. Specifically, the power of the pump. Is there any other given or implied information?

 

[quark]

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.

 

[hick]

Hello,

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,

greets

Hick

 

[quark]

700 mbar is the pressure(vacuum) that is possible to achieve 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.