Enginerds, I have a fluids question

[Al Kaholick]

I was wondering if I could make a poor mans hydraulic pump to power a ram by making a tank and filling it partially with hydraulic fluid with an air nipple on top to pressurize the tank and the hydraulic fitting on the bottom. It seems to me that if I pressurized a tank with a cross-sectional area of say 10 square inches and used a hydraulic line with say a 1 inch cross-sectional area (no i dont plan on using that big of a line, but no point in making the math cumbersome to check an idea) then 100 PSI of air would yield me 1000 PSI of line pressure. Is my reasoning correct? If not is there a way to refine the idea so that it would work? Im hoping to do something like this to power my ram for the bender since plan A didnt work out.

[87GMCJimmy]

Nope doesn't work like that. In a tank the air/fluid boundry is the area over which the force is applied and they will always be equal.

Mike

[CTENG]

Theoretically it would work, although you would want to add a piston or diaphram between the air and hydraulic fluid to prevent cavitation.

[HalfFastFord]

won't the air pressure drop as fluid leaves the resevior? drops the line pressure?

[bigdreamin]

That will work fine until the system encounters a counter force. The reason hyd. systems work is due to the fact that hyd. fluid doesn't compress or expand anywhere else in the system (for the most part) air how ever is easily comprssed. Think of it like this, in your orgianal system you put 100 psi of force to the fluid to move the fluid through the line, when you fluid encounters a force of a 100 psi the system will equalize. This is b/c the 100psi of counterforce will ultimatly be exerted on the compressed air used to orginaly move the fluid, thus equalizing the pressure in the resivor.

My $.02 try to find a used log splitter or otehr hyd. eqp. and pirate the motor, resivor, control valve and whatever else you may need.

[fivetenben]

I was wondering if I could make a poor mans hydraulic pump to power a ram by making a tank and filling it partially with hydraulic fluid with an air nipple on top to pressurize the tank and the hydraulic fitting on the bottom. It seems to me that if I pressurized a tank with a cross-sectional area of say 10 square inches and used a hydraulic line with say a 1 inch cross-sectional area (no i dont plan on using that big of a line, but no point in making the math cumbersome to check an idea) then 100 PSI of air would yield me 1000 PSI of line pressure. Is my reasoning correct? If not is there a way to refine the idea so that it would work? Im hoping to do something like this to power my ram for the bender since plan A didnt work out.


The air has 10 sq in, but its acting on 10 sq in of hydro fluid, so it cancels. 100psi in the tank is 100psi in the line. For it to work like you want, youd need a 10 sq in air cyl mechanically linked to a 1 sq in slave cyl. 10 sq in @ 100 psi = 1000 pounds of push. apply that force with to a 1 sq in hydro cyl and youd get 1000psi line pressure.

[CrustyJeep]

The air has 10 sq in, but its acting on 10 sq in of hydro fluid, so it cancels. 100psi in the tank is 100psi in the line. For it to work like you want, youd need a 10 sq in air cyl mechanically linked to a 1 sq in slave cyl. 10 sq in @ 100 psi = 1000 pounds of push. apply that force with to a 1 sq in hydro cyl and youd get 1000psi line pressure.Exactly.

[bigdreamin]

The air has 10 sq in, but its acting on 10 sq in of hydro fluid, so it cancels. 100psi in the tank is 100psi in the line. For it to work like you want, youd need a 10 sq in air cyl mechanically linked to a 1 sq in slave cyl. 10 sq in @ 100 psi = 1000 pounds of push. apply that force with to a 1 sq in hydro cyl and youd get 1000psi line pressure.

said that a hell of alot better than me