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Q: How do I calculate the approximate output pressure of a water piston ( Answered ,   1 Comment )
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 Subject: How do I calculate the approximate output pressure of a water piston Category: Science > Physics Asked by: keenan-ga List Price: \$10.00 Posted: 08 Jul 2005 09:09 PDT Expires: 07 Aug 2005 09:09 PDT Question ID: 541262
 ```I would like to build a water piston that will significantly increase the output PSI in relation to the pistons input pressure. Is there a basic formula to figure this out? EXAMPLE: Piston cylinder and piston are 12 inches in diameter. The liquid being compressed is water and has a stroke of 3 feet. The output nozzle is 1/4 inch in diameter. Disregarding frictional losses, etc...if I apply a pressure of 100 lbs to the piston head, what will be the resultant PSI at I can expect at the output of the piston? I'd like to experiment with differing cylinder sizes and pressures, so I need a formula to roughly calculate the output with varying input & output parameters. Also, what would be some good websites to take a look at? (remember, I'm no brainchild!) Basis sites that explain fluid dynamics and how to calculate stuff relating to hydrolics and the building of a practical water piston.```
 ```Hello keenan, I think what you are trying to build is what is known as a pressure intensifier. However, what you describe will not increase your output pressure. I will try and explain. The area of your 12 inch diameter piston is: Area = pi x r^2 = 3.14 x 36 = 113 sq. in. You say that you are applying 100 lbs of pressure to the piston. I assume you mean 100 lbs of force and not 100 psi (if I am wrong, please let me know). So, to calculate the pressure of the water in the cylinder we have: Pressure = force/area = 100 lbs/113 sq. in. = .88 psi The water will dribble out of your 1/4 inch nozzle. If you mean that you are applying a pressure of 100 psi to the piston, then you have exactly 100 psi in your water. You have done nothing to increase the pressure. To build a pressure intensifier you need cylinders of different diameters whose pistons are connected by a rod. You apply a small pressure to the larger cylinder and it will give you a higher pressure in the smaller cylinder. This may be hard for you to understand. I found a couple of drawings that might help you: http://www.allstar.fiu.edu/aero/Hydr22.htm Example: Suppose that the actuating cylinder (used to move the control surface) normally gets hydraulic fluid at a pressure of p = 1000 psi and that this pressure would act on a piston surface area A = 10 square inches. The total force developed F would be p times A or 10,000 lb. But, suppose we need a force of 60,000 lb to move the control surface... So we add a booster. If the inlet side area A1 is 3 square inches, then the force, F, developed by the hydraulic fluid is Since the pistons of the booster are connected, and we assume that they have negligible inertia, then the force developed on the smaller piston is the same as the force on the larger piston. If the outlet side area A2 is 1/2 square inch, then the pressure developed on the outlet side would be At the actuating cylinder the force developed would be the pressure times the area of the actuating cylinder or (6000 lb/sq. in.) x (10 sq. in.) = 60,000 lb http://www.newport-scientific.com/HydPressIntens.htm Hydraulic Pressure Intensifiers are used for creating pressures greater than those ordinarily obtained directly from either Hand Operated Pumps or Motor Driven Pumps. While not intended for high volumetric capacity, their simplicity of design makes then an economical means of obtaining high pressure. The intensifier consists of a double piston-cylinder arrangement as shown in the sectional view. Oil is pumped into the large cylinder by a hand hydraulic pump or other means, and oil is expelled from the small cylinder at a higher pressure, dependent upon the area ration and friction. The large piston is fitted with an O-ring for maximum efficiency at the pressures encountered in the cylinder. The small piston is lapped to an extremely close fit and is packed with a combination packing on the principle of the unsupported area. Due to the necessary close fit of the pistons in their respective cylinders, the friction created does not permit 100% efficiency as calculated from the ratio. Hydraulic intensifiers are manufactured in two basic models for producing pressures of 50,000 and 100,000 psi. Their construction is similar except that the 100,000 psi model is built with heavier cylinders and pistons. You may have more questions after reading this and I will be glad to help you out with more answers. Please ask for a clarification and I will try and explain until you fully understand. Redhoss``` Request for Answer Clarification by keenan-ga on 08 Jul 2005 14:49 PDT ```Hey Redhoss... Thanks for the explaination! After I submitted the request, I quickly found a site that helped me to understand that the psi remained the same. DUH!!! I'm trying to figure out how to artifically create ~200 to 400 ft. of H20 head pressure using a simple water piston. Sorry for sounding so stupid, but wouldn't a greater pressure on the piston (say 100-300 pounds) create this head. Intuitively, I think you've already showed me that it doesn't do this in your previous explanation. Any ideas would be appreciated...and Thanks so much for your prompt reply to my goofy question. Keenan``` Clarification of Answer by redhoss-ga on 08 Jul 2005 16:59 PDT ```Now that you understand that what you were talking about won't work, we can look at a way that would work. If you could make a 3 inch diameter cylinder and piston just like your 12 inch cylinder, you could connect the cylinders and connect the two pistons with a rod. Then you could use an air compressor to develop 100 psi (make sure that the 12 in cylinder is rated for 100 psi) behind the piston in the 12 inch cylinder. This will develop a force on the 3 inch piston (transmitted thru the rod) of 113 x 100 = 11,300 lbs. Then the pressure in the 3 inch cylinder would be 11,300 / 7 = 1,614 psi. Now that is some serious pressure. Of course you don't have near the volume of water, but you can get the pressure you are looking for. Or, if you went with a 4 inch cylinder instead of a 3 inch you would get: Pressure = 11,300 / (pi x 2^2) = 11,300 / 12.5 = 904 psi Hope this gives you some ideas. Thanks for the 5 stars, Redhoss```
 keenan-ga rated this answer: ```Thanks for helping me to understand the error of my ways! You were very articulate in your answer & the supporting URL info help me to clearly see the reality of the situation.```
 ```Ok I am working on a similar problem.... I wan tto magnify the force of a lever to create more pressure to force a liquid through a hole. Based on this discussion I drew the following highly simplified diagram.. www.andrewjacobs.itgo.com/piston can you let me know if I understood ok. Cheers```