
#1
Jan1314, 11:30 AM

P: 6

Hi,
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. Thanks, Arnak 



#2
Jan1314, 11:55 AM

P: 117

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. 



#3
Jan1314, 12:10 PM

P: 101





#4
Jan1314, 01:26 PM

P: 6

Formula for Work to create a vacuum.
Hi,
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? Arnak 



#5
Jan1314, 02:07 PM

P: 117

The power is Power = dP*V / t so work is only W = dP*V My mistake. Sorry 



#6
Jan1314, 02:13 PM

P: 117

W = dP*V t has no place there. My mistake in SI system, units are W => Joules dP => Pascals V => m^{3} 



#7
Jan1314, 02:43 PM

P: 751

Consider the ideal case of a cylinder and a piston. You have a volume V of an ideal gas at ambient pressure P_{amb} and you want to exhaust enough gas to end up with the same volume V at reduced pressure P_{new}. That means that you need to push the piston to exhaust a volume V * (1  P_{new}/P_{amb}) of gas at ambient pressure. Pushing against atmospheric pressure takes work equal to P_{amb} * V * (1  P_{new}/P_{amb}) = V * (P_{amb}  P_{new}) But this leaves you with a volume V * P_{new}/P_{amb} 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 V_{initial} * P_{initial} * ln(P_{initial}/p_{final}). 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 * ( P_{amb}  P_{new} ) + V * P_{new} * ln(P_{new}/P_{amb}) As long as P_{new} is lower than P_{amb}, 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. 



#8
Jan1314, 05:58 PM

P: 6

Hi Jbriggs,
Thanks for the input.8)) 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? Martin 



#9
Jan1414, 07:48 AM

P: 751

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(V_{final})  ln(V_{initial}). Equivalently, this is ln(V_{final}/V_{initial}) Google "work done by expanding gas at constant temperature". e.g. http://hyperphysics.phyastr.gsu.edu...rmo/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. 



#10
Jan1414, 01:45 PM

P: 6

Hi Jbriggs,
Excellent info, thanks very much. A bit complex for me but I am getting the hang of it.8)) Arnak 


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