Static hydraulic pressure in a pipe

[russ_watters]

If the pressure is 100,000 psi and if the water is compresses by 15% then the volume change is (v-1) * .15 = (V-2)
Is this correct?

(1-.15)
[Edit]
Your second attempt works.

 

[Lnewqban]

Ok take Pascal's law out of the question.
We start with the compressibility of water.
Our piston will move a distance that is related to the compressibility of the water.
So depending on what the applied pressure is, the total volume of the water in the pipe will be reduced by a percent of its original volume. If the pressure is 100,000 psi and if the water is compresses by 15% then the volume change is (v-1) * .15 = (V-2)
Is this correct?

Please, see:
https://en.m.wikipedia.org/wiki/Properties_of_water#Compressibility

You could calculate how thick the walls and cap of your pipes need to be to hold those 100,000 psi.
Knowing or assuming the diameter of your piston, you could also calculate how much force it needs to be pushed with to generate that pressure.

Then, even after water compresses some under that tremendous pressure, which your pipes resist, and your piston is able to create, the measurable static pressure at the end of both pipes should be exactly the same that at the piston's face.

 

[Pascal pressure]

The parts where you said it acts as a solid and also that it's basically incompressible under 3000 psi (though that second one can be an assumption in some circumstances..

Ok, I agree it only acts like a solid in regards to compression not tension or sheer.
And yes, at some pressure under 3,000 psi the compressibility of water is essentially negligible, as mentioned, in this case it acts similar to a steel rod under compression.

In regard to the main question; the pressure drop in the water when it is under enough pressure to be compressed; So we have our starting volume v-1 which is the volumes of water in the pipe before compressing the water, then we compress the water by 15% (100,000 psi) so (v-2) = (V-1) - [(v-1) * .15]

Oops, wait a minute Huston, we have no unknown variable with this approach, because I'm saying P-1 = 14.7 psi and P-2 =100,000 psi.

Ok got to think about this some more.

 

[jack action]

As I think about it some more, I guess the answer to the question should be fairly straight forward:
(p-1) * (v-1) = (p-2) * (v-2)

Even though I think this equation is used for pneumatics, I think it could be applied to hydraulics

Where:
(v-1) = volume of the water in the pipe at 14.7 psi [(p-1) = 14.7 psi]
(v-2) = volume of the water in the pipe at pressures above 3,000 psi minus % compression
Then pressure drop of the water in the pipe = (p-2)

First, this equation can only be used in pneumatics IF the temperature is kept constant.

Second, it cannot be applied to hydraulics. Here is the correct equation:
$$\frac{p_1 V_1 - p_2 V_1}{V_2-V_1} = E$$
Where $E$ is the Bulk Modulus Elasticity of the liquid.

Reference: https://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

What that equation state is, if a liquid is at pressure $p1$ and gets increased to $p_2$, its volume will go from $V_1$ to $V_2$.

In your comparison, both cases have the same $p_1$ and $p_2$ (which is constant throughout the volume of liquid as the pressure increases from $p_1$ to $p_2$), only the initial $V_1$ is different. Obviously, $V_2$ will be different as well but the ratio $\frac{V_2}{V_1}$ will be constant:
$$\frac{V_2}{V_1} = 1-\frac{p_2-p_1}{E}$$
In your cases (with water):
$$\frac{V_2}{V_1} = 1-\frac{3000-14.7}{312000}=0.99$$
Because one of your pipes is 100 times longer than the other ($=\frac{1000\ ft}{10\ ft}$) and the pipe area is constant throughout the process, then:
$$\frac{L_2}{L_1} = \frac{\frac{V_2}{A_2}}{\frac{V_1}{A_1}} = \frac{V_2}{V_1}\frac{A_1}{A_2} =\frac{V_2}{V_1}(1) = \frac{V_2}{V_1}$$
$$L_{2\ 1000\ ft} = 0.99 \times 1000\ ft = 990\ ft$$
$$L_{2\ 10\ ft} = 0.99 \times 10\ ft = 9.90\ ft$$
So, increasing the pressure from 14.7 psi to 3000 psi in both pipes - everywhere in the pipe - your piston will sink in 0.1 ft with the 10-foot pipe, but it will sink 10 ft in the 1000-foot pipe.

Note also that - as long as it doesn't change during the pressure increase - the area of the pipe doesn't even have to be the same in both cases for this to be true.