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1m cubic container with sucking piston to vacuum

Consider I have a cubic container with its volume is 1m cubic as in the simple picture above. It is sealed nicely, it is air proof, no air leaking. I would like to pump out (to vacuum) the container to make the pressure inside 0 ATM, or very close to it, like 0.01ATM. The pump I will use will be a kind of piston which displacement volume is 1L or 1,000cc. It is directly coupled to the cube container. By a simple calculation, I need to pump out 1,000 times to make the cube empty. But I don't think that it is true. I am pretty sure it is not that way to calculate as the air will expand every time the cube is pumped out.

Then my question are:

  1. How many times I need to pump out the cube to make the cube 0 ATM, or very close to it.
  2. As every time the cube is pumped out the cube is less pressure, the required power to pump out the cube will be higher, need stronger. So, how to calculate the required power to vacuum it to 0 ATM, or very close to it.

Here we assume that the cube container's structure is strong enough to retain the low pressure inside. If required, also assume that the outside pressure is 1 ATM.

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2 Answers 2

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Let's say your pump is pumping 1 liter per second so the pressure left in the tank is $ P_t*0.999 $ we call the remaining tank pressure $P_{t}$

$$dP/dt=0.999*P_{t} $$

$$P_{t} =\text{ momentary pressure of the gas in the tank}$$

Solving this differential equation we get a decreasing exponential function.

$$P_{t}=e^{-0.999*t}$$

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  • $\begingroup$ What about the required power, or force, to pump the air out? As in my opinion, the less the pressure, the more power/force we need to pump the air out. $\endgroup$ Feb 18, 2023 at 9:52
  • $\begingroup$ @AirCraftLover It gets harder and harder actually as the vacuum gets higher. it's not easy to dump the discharge of a very low-pressure pump into the ambient atmosphere! $\endgroup$
    – kamran
    Feb 18, 2023 at 17:27
  • $\begingroup$ Of course, it will be harder and harder every time it is pumped out. If not, not so many question here. $\endgroup$ Feb 18, 2023 at 18:35
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Following on from @kamran's answer and ignoring the slight change in pressure as the 1 L piston moves, you can calculate the force required at each step as $ F = \Delta P \times A $ (pressure differential by piston area). The work done on each stroke will be $W = F \times d$.

Can you work it out from there? I've got a bad headcold and wouldn't trust myself.

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  • $\begingroup$ What is the d in your formula? The piston's stroke? $\endgroup$ Nov 16, 2023 at 3:23
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Transistor
    Nov 16, 2023 at 9:26

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