How to Calculate Work to produce a Vacuum?

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SUMMARY

The discussion focuses on calculating the power requirements for an electric motor used to evacuate a vacuum to 2.5 psi (5" Hg) using a pneumatic cylinder with a 50 mm bore and stroke. Key calculations involve determining the force needed based on the differential pressure of 12 psi across the cylinder, which translates to approximately 37.7 pounds of force. The torque required cannot be calculated without details on how the motor is coupled to the pneumatic cylinder. Additionally, the ongoing operational costs are influenced by the leakage rate that must be compensated for during vacuum maintenance.

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Arnak
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Hi Folks,

Could someone please help me?:blushing:

I want to work out the power requirements for the torque or wattage required in an electric motor when evacuating a vacuum to a certain level.

The tube to be evacuated is 200 mm x 50 mm and the evacuation would be done using a pneumatic cylinder of 50 mm bore and 50 mm stroke.

The vacuum to be achieved and maintained is 2.5 psi or 5" Hg.

The pneumatic cylinder will be driven by an electric motor but how to calculate the motor requirements?

To keep it simple I will ignore losses due to friction etc.

Could someone could please help me with the formula to work that out?:confused:

Thanks,

Arnak
 
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It's the same as the work required to reduce pressure.
Opposite of the work required to generate the pressure.

Depends a great deal on how the pressure is being changed.
It will not be easy to model from first principles - you should probably just look it up.

See how other people solve the same problems.
http://www.gastmfg.com/vphb/vphb_s4.pdf
http://lpc1.clpccd.cc.ca.us/lpc/tswain/chapt3.pdf
 
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I believe the cylinder rod area is unimportant if the rod end of the cylinder is open to atmospheric pressure. The 50mm pump cylinder will have 2.5 psi on one face and 14.5 psi atmospheric on the other. That makes a differential pressure of 12 psi. The area of a 50mm piston is about Pi square inches so you need to apply a force of Pi *12 psi = 37.7 pounds force to the rod.

Torque cannot yet be calculated because you have not explained how you will couple the rod to the motor.

The ongoing cost of maintaining the 2.5 psi absolute pressure will be determined by the leakage rate you must make up.

The motor power requirements will be decided by the time you have available to pull the initial partial vacuum.
The rate of air removal must be faster than the leakage rate you need to make up.
 
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Hi,

Thanks for all the advice, complicated subject isn't it.8-((

I will look up the links and see what I can learn.

The connection to the motor will be via a 2.5" throw crank to the motor shaft.

Good point about the leakage rate.8-))

Arnak
 

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