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Formula for Work to create a vacuum.

  1. Jan 13, 2014 #1

    Can someone give me some guidance on how to calculate the work or energy required to create a vacuum.

    For example, I would like to create a 2.5psi vacuum in a tube of a specific length and diameter using a pneumatic cylinder to remove the air.

    How would I go about calculating the energy required to do that please?

    From that answer I want to calculate the power of an electric motor required to perform that task.


  2. jcsd
  3. Jan 13, 2014 #2
    The work to create difference in pressure is

    W = dP*V*t

    dP is the pressure difference
    V is the volume of the air container
    t is the time for what you have achieved the pressure difference

    Here the heat transfer is not taken in account.
  4. Jan 13, 2014 #3
    Why do you need time here ?
  5. Jan 13, 2014 #4

    Thanks for that formula.9-))

    What units would the work be in please?

    Am I right in saying then that with a cylinder of 200 x 50mm V= 10000 mm3

    Pressure difference = 2.5psi

    So 10000 x 2.5 = 25000 units?

  6. Jan 13, 2014 #5
    I have confused work with power here.

    The power is

    Power = dP*V / t

    so work is only

    W = dP*V

    My mistake. Sorry
  7. Jan 13, 2014 #6
    The formula has to be

    W = dP*V

    t has no place there. My mistake

    in SI system, units are

    W => Joules
    dP => Pascals
    V => m3
  8. Jan 13, 2014 #7


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    Science Advisor

    Not quite.

    Consider the ideal case of a cylinder and a piston. You have a volume V of an ideal gas at ambient pressure Pamb and you want to exhaust enough gas to end up with the same volume V at reduced pressure Pnew.

    That means that you need to push the piston to exhaust a volume V * (1 - Pnew/Pamb) of gas at ambient pressure. Pushing against atmospheric pressure takes work equal to Pamb * V * (1 - Pnew/Pamb) = V * (Pamb - Pnew)

    But this leaves you with a volume V * Pnew/Pamb of gas at ambient pressure with a total volume V available to expand into. That represents work that can be reclaimed. The formula for the work gained with the isothermal expansion of an ideal gas is Vinitial * Pinitial * ln(Pinitial/pfinal). The formula also works if you swap "initial" with "final" and negate the sign.

    Putting that together gives you a required work of:

    W = V * ( Pamb - Pnew ) + V * Pnew * ln(Pnew/Pamb)

    As long as Pnew is lower than Pamb, the second term will be negative, so it subtracts from the work required.

    If you are using a coherent system of units such as SI with volume in cubic meters, pressure in Pascals and work in Joules then the formula will work as is. If you are using pounds per square inch and cubic millimeters then some unit conversion factors will be required.
  9. Jan 13, 2014 #8
    Hi Jbriggs,

    Thanks for the input.:cool:)

    So by my calculations :-

    W = 0.00001*(101325 - 84088) + 0.00001 * 84088 * ln(84088/101325)

    W = 0.87020288682 Joules.

    Please excuse my lack of knowledge of algebra but is the component ln the designation for the result of (Pnew/Pamb)?

    I have converted the units to Pascals and M3.

    V= 0.00001 m3, Pamb = air pressure 101325 Pascals, Reduction of 2.5psi in Pascals =17237

    So Pamb(101325) - Reduction of 17237 = Pnew = 84088

    Have I got that right so far please?

    Can you also advise me where I can find a link to these equations so I can attempt to learn more?

    Last edited: Jan 13, 2014
  10. Jan 14, 2014 #9


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    Science Advisor

    "ln" is natural logarithm. The natural log of 84088/101325 is approximately -0.19

    Looks good.

    The equation for the work done by the (hypothetical) piston pushing out against the ambient atmosphere was trivial. It comes from first principles. You multiply atmospheric pressure by area to get the force on the piston. You multiply force by distance the piston moved to get work done. Work is the product of force (area x pressure) and distance. This can also be expressed as (area x distance) times pressure. Since area x distance = change in volume, it follows that work done is atmospheric pressure times delta V

    The equation for the work done by the expanding gas inside the cylinder came from a Google search and is derived in the same way. But the pressure of the expanding gas in the cylinder declines as it expands. So instead of a straight multiple you need an integral. The pressure of the gas decreases as the inverse of the volume, so one is integrating [some multiple of] 1/V and the resulting integral is some multiple of ln(Vfinal) - ln(Vinitial). Equivalently, this is ln(Vfinal/Vinitial)

    Google "work done by expanding gas at constant temperature". e.g. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/isoth.html You'll find any number of derivations. Most of them express the multiplier in terms of moles and Boltzmann's constant, but one can substitute that out with the equation: pV=nRt.

    I fiddled with the sign conventions to make sure that the contributions from the two terms were correct.
  11. Jan 14, 2014 #10
    Hi Jbriggs,

    Excellent info, thanks very much.

    A bit complex for me but I am getting the hang of it.:cool:)

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