Compressor Question

Hire an instrumentation consultant.

Your "requirements" are unintelligible.

"partially filled water tank first we have to increase 10kg/cm2 pressure inside the tank"???

"suddenly it should reduce to 5kg/cm"???

Much more explanation of what you really want to do is in order.

Define "very very fast". Is that days. or picoseconds?

"I have to design a control system for response analysis."

Sounds like your over your head. Out source it! As the information you just provided is little to none.

That isn't a tank, Mildred. It's a pressure vessel.

What is wrong with a simple pressure relief valve?

What is wrong with regulating the incoming supply so the vessel doesn't fart so much?

1- you should define the time interval to fill and the one to empty.

2- depending on those values you could define the size of compressor and of valves to obtain them.

How you put the question it is impossible to give you any serious answer and that you can feel by the answers you already got.

As scheme it is simple: you need a compressor and a series of valves and that is all with respect to hard ware but every component needs to be dimensioned thus the need for a better definition.

Have a pneumatic engineer or technician design and construct a pneumatic computer for you. Yes there are such things. They were used before PLC's and PC's for the control of processes. They are not difficult to design and make. They are also very simple to operate.

The function is the computer senses the pressure in the tank. At high pressure it pressurizes the actuator of the dump valve (a pneumatic valve that opens and closes quickly and is controlled by a pilot, sort of like a pneumatic operated solenoid valve) which opens and dumps the pressure in the tank. When the pressure lowers down to the lower pressure the computer shuts off the pilot pressure which closes the dump valve.

These pneumatic computers are not difficult to design of build. Parker Pneumatics (used to be Parker-Hanaffin) has several old books that are training manuals and guide books. These have been put on their web site

http://www.parker.com/portal/site/PARKER/menuitem.223a4a3cce02eb6315731910237ad1ca/?vgnextoid=c00aeea74775e210VgnVCM10000048021dacRCRD&vgnextfmt=default&vgnextfmt=EN

Now you know what you want, have someone else who will design and make it correctly. Don't do it on your own. You will waste a lot of time, money and components and it usually won't work the first several times you bread board it.

Don't forget-A pneumatics engineer of technician, not you!

Good Luck, Old Salt

Why are you trying to manipulate the water pressure by means of air pressure above it? Why not directly by a positive displacement pump?

Mount two cylinders mechanically linked so that as one cylinder extends the other retracts. Size the cylinders so that the volume of each full bore end when fully retracted is equal to the ullage in the tank when water is compressed to 10kg/cm². Compensate for the connection pipe work. (Volumes can easily be matched by raising or lowering the water level in the tank) Port both full bore ends to the top of the water tank via a two position valve triggered by the cylinder linkage. It is critical that the assembly is fully gas tight. Have one cylinder fully retracted with the valve open to that cylinder. Pre-charge the tank (and the one open port cylinder) to 5kg/cm². Extend the retracted cylinder compressing all the air into the tank. The closed cylinder will simultaneously retract creating a vacuum in the full bore end. Because the volume is halved the pressure will double to 10kg/cm². Trigger the valve. The air will flood the vacuum and equalise the pressure at 5kg/cm². Repeat the process by extending the second cylinder. Expansion into the vacuum will absorb much of the compression heat but not all. Fit a cooling coil into the tank and maintain the water/air temperature using a PID controller. This will prevent the fixed volume of air from expanding with temperature and increasing the mean pressure.

If the size of your tank allows it, connect a single pneumatic cylinder with piston of suitable size such that at full stroke the pressure is 10. Then withdraw piston until pressure drops to five.

Connect a motor with adjustable crank of length to match piston stroke.

Run motor at what speed you like.

Moving a single cylinder back and forth would work, but you need to compensate for the increase in heat. If driven from a crank, the cylinder motion would be sinusoidal, which does not match the Ops criteria of "suddenly" dropping to 5kg/cm². Using two linked cylinders and triggering the valve opening and cylinder reversal from the linkage rather than sensing the pressure allows you to set the trip point slightly forward of the end of stroke to nest the valve reaction times into the last part of the advance stroke so that there is no plateau at 10kg/cm². This would shave milliseconds from each cycle. The pressure profile would be nearer to a saw tooth which is what the OP implies is his requirement. The up slope of the pressure ramp is approximately linear and adjustable within the limits of the drive compressor output. Driving the annular ends of the cylinders reduces the air volume requirement and allows increased speed for the same input. The down stroke is a decay curve based on the port sizes and the pressure differential, so the slope would be made much steeper by large ports dumping into a vacuum.

Attach a large manometer to the tank. By large I mean that the tubing has a diameter of 2" to 6" ( intuitive guess only ) and fit an even larger "bucket" to the top of the tubing. Adjust the length of the water column to give you your max pressure just as the water column reaches the "bucket". The bolus of air will escape thereby allowing the column to collapse back down the tubing, You can derive mathematically the required volume of water and then work backwards to give you the diameter of the tubing required. Hint; the tubing on the tank side can be a different diameter to the tubing on the "bucket" side.

Jim

So you are going from 9.8 Bar to 4.9 Bar. Sounds like you are building a water jet gun of some type? Planning on a water pistol war? You did not state how fast. Here is what I think about this: (1) tank is 3/4 full with water at 9.8 Bar, (2) tank (pressure vessel) is 3/8 full with water at 4.9 Bar. Use level switches that have a fast response for the high and low level points, a constant pressure water supply of well in excess of 9.8 Bar, say 20 Bar. The level switches switch relays that in turn active dump or fill solenoid valves, but you must not dump too fast, or the entrapped air will also escape from the bottom. Good luck getting this contraption to do what you want it to, as long it has some productive purpose. I hope you are not re-inventing someone else's wheel, but I have to ask really what this is?

Is it a pneumatic powered water cooler? I would hate to see the expression the face of the first gent to step up for a drink!!! Is it a control system for a water fountain display? I think you are engineering a fountain to show off.... what do I win, if correct?