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I am looking at generating a vacuum from two interconnected pistons, a driver piston from high pressure and the drivee creating a negative pressure in a container.

I am looking at improving the amount of vacuum I can get with the use of Pascal's principle but am unsure if I am trying to get something for nothing (I think I might be)

Taking the picture below, lets say I have two pistons, with D2 being double D1. If I put high pressure air into the inlet at p1, due to the surface areas being double (and via Pascal's principle) will I create double the force at the D2 and therefore allow the piston to move the same distance, but as the force has doubled create double the amount of vacuum'd pressure at D2.

enter image description here

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Consider you've achieved some pressure in the left cylinder, the system has moved and reached a state of rest. Lets find the pressure in the right cylinder. I'm not sure what exactly you mean by

I have two pistons, with D2 being double D1

I assume you mean the areas of the pistons, so lets say that the area of the left piston is $A_1$ and the area of the right piston is $A_2 = 2A_1$. The pressure acting on them I will call $P_1$ and $P_2$ respectively, and the resulting forces: $F_1$ and $F_2$.

Consider the picture:State at rest

We said that the system is in rest, so all forces must cancel out, so $F_1 = -F_2$. We know that the pressure force is equal to the product of the pressure and the area it acts on, so: $F=PA$ When we substitute this in $F_1 = -F_2$ we get $$P_1A_1 = -P_2A_2$$ Then we subsitute $A_2 = 2A_1$ and we get $$P_1A_1 = -2P_2A_1$$ With a bit of algebra $$P_2 = -\frac{P_1}{2}$$

So we reach the conclusion that the pressure in the right tank would be negative half of the pressure in the left tank. So if you have $1kPa$ presure in the left tank, you would have $-0.5kPa$ in the right.

I think the key mistake you make is applying the Pascal's principle wrong

in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container.

For the Pascal's principle to work the medium must be a fluid. In your example the medium is a solid - the pistons and rod, connecting them.

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  • $\begingroup$ FYI, it seems most likely that $D_1$ and $D_2$ were for diameters. Diameter is generally important in piston calculations, so it is often specified along with area. Doesn't really change how the analysis is done obviously. $\endgroup$ – JMac Feb 22 '19 at 12:59

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