Compressibility/Expansion in a closed system

• acm11216
In summary, the pressure at the bottom of the cylinder would be: P_2 = 500\ psi + \rho gh' + \frac{mg-F}{A} + \rho_i gh'' Where ##mg## is the piston weight, ##A## is the cylinder area and ##\rho_i## is the density of the fluid inside the control line and cylinder. After closing the valve and introducing ##F##: P_2 = 500\ psi + \rho gh' + \frac{mg-F}{A} + \rho_i gh'' and ##P_1## would remain the same: P_1 = P_2 - \rho_i gh As the
acm11216
Howdy. Long time reader first time posting. Usually try to solve these problems on my own but this one has caused a few sleepness nights already so looking for some insight.

The attached shows a simplified schematic of what I am dealing with. It is basically a hydraulic cylinder with the piston side control line running up to a valve on the "surface" and filled with hydraulic fluid. If I were to apply 500 psi at the surface the pressure below the piston will be 500 psi + head pressure (pgh). For simplicity assume no movement in the piston (resting on shoulder and weight greater than pressure end load from below). Will there be any effect to P2 when closing the surface valve? I don't believe so but unsure the role of hydraulic head in a closed system.

With the surface valve closed I now have a hydraulic lock scenario. However, because the fluid is not perfectly incompressible, I will achieve some movement when pulling up on the hydraulic cylinder. Assuming no seal leakage, this movement will be maxed out when the pressure drops to 0 psi. I can use bulk modulus of elasticity for the control fluid to relate volume and pressure changes but how do I apply in this case? Is the maximum pressure drop 500 psi, 500 psi + head pressure, or somewhere in between? In the long run I am working to determine piston stroke vs F and P1/P2 relation.

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More accurately, the pressure at the bottom of the cylinder would be:
$$P_2 = 500\ psi + \rho gh' + \frac{mg}{A} + \rho_i gh''$$
$$P_1 = P_2 - \rho_i gh$$
Where ##mg## is the piston weight, ##A## is the cylinder area and ##\rho_i## is the density of the fluid inside the control line and cylinder. ##h'## and ##h''## are respectively the heights over the piston and below the piston.
(##h = h' + h'' + piston\ thickness##)

After closing the valve and introducing ##F##:
$$P_2 = 500\ psi + \rho gh' + \frac{mg-F}{A} + \rho_i gh''$$
and ##P_1## would remain the same:
$$P_1 = P_2 - \rho_i gh$$
As the volume of the cylinder increases, ##\rho_i## will decrease.

For the stroke, relating to ##P_2## and using compressibility:
$$\beta = -\frac{dV}{VdP} = -\frac{AdS}{(V_0 + AS)dP} = -\frac{dS}{\left(\frac{V_0}{A}+S\right)dP}$$
Where ##S## is the stroke and ##V_0## is the volume of the control line. Then:
$$-\frac{dS}{\frac{V_0}{A}+S} = \beta dP$$
Integrating from ##a## to ##b##:
$$\ln\left(S_a + \frac{V_0}{A}\right) - \ln\left(S_b + \frac{V_0}{A}\right) = \beta\left(P_b - P_a\right)$$
Where the initial ##S_a## will be equal to the initial cylinder volume ##V_{cyl0}## divided by the cylinder area ##A##. The stroke increase ##S## due to pressure going from ##P_a## to ##P_b## will be ##S_b - S_a##, giving:
$$S = \frac{Ae^{\beta\left(P_b-P_a\right)}}{V_{cyl0} + V_0} - \frac{V_{cyl0} + V_0}{A}$$
If the fluid inside the cylinder begins to boil due to the pressure being too low, then the pressure-volume relationship becomes the ideal gas law, i.e. ##PV = mRT##.

I think I did not make mistakes.

1. What is compressibility and expansion in a closed system?

Compressibility and expansion in a closed system refer to the ability of a substance to be compressed or expand when subjected to pressure or changes in temperature within a closed container.

2. How does compressibility and expansion affect the behavior of gases?

Compressibility and expansion play a crucial role in the behavior of gases, as they determine how the gas molecules will behave in response to changes in pressure and temperature. As a gas is compressed, its molecules become more closely packed together, resulting in an increase in pressure and a decrease in volume. Expansion, on the other hand, occurs when the gas molecules are allowed to spread out, leading to a decrease in pressure and an increase in volume.

3. How is compressibility and expansion related to the ideal gas law?

The ideal gas law, which states that the product of pressure and volume is directly proportional to the number of moles of a gas and its absolute temperature, is directly related to compressibility and expansion. As a gas is compressed, its volume decreases, and according to the ideal gas law, this should result in an increase in pressure. Similarly, as a gas expands, its volume increases, leading to a decrease in pressure.

4. What factors affect the compressibility and expansion of a substance?

The compressibility and expansion of a substance are influenced by several factors, including the type of substance, its temperature, and the amount of pressure applied. Generally, gases are more compressible and expandable than liquids and solids, and the higher the temperature, the greater the degree of expansion and compressibility. Additionally, the amount of pressure applied to a substance can also impact its compressibility and expansion.

5. Why is understanding compressibility and expansion important in science?

Compressibility and expansion are essential concepts in science, particularly in chemistry and physics, as they help explain the behavior of matter in different conditions. By understanding how substances respond to changes in temperature and pressure, scientists can accurately predict and control the behavior of gases in various applications, such as in industrial processes or in the design of new materials.

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