Formula to work out DC Power Requirements for 20,000 PSI Air Pump - Page 1
Forum section: electronics / projects

author: hyteqsolutions
author_rank: Newbie
timestamp: May 30, 2015, 12:25:31 pm
content: Good Evening All,I have a pet project on the way, very long story on this one.Basically-I have an air pump that puts out at 20,000 PSI and a smallish chamber that can hold said 20,000 PSI that is held in place by a metal flap.What I am trying to do now, is build a circuit board that can provide enough power to the PSI air pump and the metal flap (magnetic).I cannot find or remember any formula for working this out. As I have zero info on the air pump, I have spent a couple of weeks trying to find some sort of formula to help me work out what power is required for the air pump, once I have that, I can work the rest out myself.Does anyone know where I can find this formula? or can someone help me work this out?Thank You,- HyTeQ-Solutions

author: Ian.M
author_rank: Super Contributor
timestamp: May 30, 2015, 12:58:42 pm
content: Impossible to tell. You only have the maximum pressure, you'd also need to know the flow rate and how it varies with pressure to know how much work is being done, then you'd need data on its efficiency, again over the pressure/volume range of interest.Air at 20K PSI in a small chamber retained by a flap sounds very much like a rocket. Its certainly high energy, comparable to the peak pressure in a rifle chamber. I do hope you have extensive experience in high pressure testing . . .

author: radioFlash
author_rank: Regular Contributor
timestamp: May 30, 2015, 01:05:25 pm
content: To figure out the amount of work being done, you would start with the thermodynamics of an ideal gas. You should have a look at this article:http://en.wikipedia.org/wiki/Adiabatic_process

author: max_torque
author_rank: Super Contributor
timestamp: May 30, 2015, 01:12:01 pm
content: I assume you mean TWENTY Psi, not TWENTY THOUSAND Psi??

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: May 30, 2015, 01:18:14 pm
content: Err, if you have the pump, just measure it?

author: Ian.M
author_rank: Super Contributor
timestamp: May 30, 2015, 01:22:47 pm
content: Unfortunately at 20K PSI, air is a supercritical fluid, so the theoretical approach is going to get real messy really fast due to the transition between gas and supercritical fluid.Does the pump have manufacturer's data?If you want your control board to survive, its going to need to be able to handle at least twice the motor's max rated running current and have over-current protection in case the motor stalls. Even if the usual motor current is much less, unless you have a soft start circuit, it *WILL* pull a transient peak current comparable to its stall current on startup.@max_torque: I hope you are right. A 20K PSI air pump sounds unlikely and highly specialised (not to mention <EXPLETIVE> dangerous) . . . . .

author: KJDS
author_rank: Super Contributor
timestamp: May 30, 2015, 01:30:26 pm
content: Pressure is when you're bursting to go, flow is when you get to go. If you have long enough to build up the pressure you'll need very little power. It's the flow that takes pumping.

author: rikkitikkitavi
author_rank: Contributor
timestamp: May 30, 2015, 03:14:28 pm
content: 20000psi? Ie 1500 bar. If you compress 20C air that much isentropic it will reach about 2500K, in reality way more, probably 3000K. Or doing it real slow, isothermally. But then you need to cool it, and it will liquefy meanwhile.Anyhow it is impossible to say what power your compressor willuse without more data ie flow, efficiency etc.

author: NiHaoMike
author_rank: Super Contributor
timestamp: May 30, 2015, 10:08:44 pm
content: Quote from: KJDS on May 30, 2015, 01:30:26 pmIn basic electronics, the common analogy is that voltage is pressure and current is flow. Beware, however, that given the way positive displacement pumps/compressors and motors work, pressure is roughly proportional to torque which is roughly proportional to current, while flow is roughly proportional to speed which is roughly proportional to voltage.