Nikolai Nekrutenko 9/25/2024


**Slightly compressible fluids** are liquids already in liquid form, such as water, solvents, etc that behave under bulk modulus.

**Fully compressible fluids** are those such as gasses like air, nitrogen, etc that behave under Boyle’s law.

Outlined below is a comparison of slightly compressible liquids for depth of up to 2000m vs fully compressible liquids:

## Slightly compressible fluids:

Slightly compressible liquids exhibit an extremely low compressibility, but it still is there:

He-4 in its liquid form from around 1.2 to 4.2 kelvin exhibits the best compressibility behavior out of all of the liquids at 2 to 5 MPa yet this is not feasible for our sensor realistically speaking.

Bulk modulus is defined as the following expression, where V is the initial volume, and dP/dV represents the derivative of pressure with respect to volume:


Below is a table of bulk moduli for common liquids:

| Liquid | Bulk Modulus (GPa) |
| -------- | -------- |
| Sea Water | 2.34 |
| Water | 2.09 |
| Methanol | 0.823 |
| Ethyl Alcohol | 1.07 |
| Acetone | 0.92 |
| Mercury | 28.5 |

Calculating the change in volume for several candidate liquids:

At a depth of 2000m, this is about 200 bar = 2 * 10^7 Pa relative to the atmospheric pressure of 1 bar that is 10^5 Pa.

**For Distilled Water:**

$\frac{\Delta V}{V_0} = -\frac{\Delta P}{K}$

$\frac{\Delta V}{V_0} = -\frac{1.99 \times 10^7}{2.09 \times 10^9} \approx -0.0085 = 0.952 \%$ change in volume from 0 to 2000m

**For Ethyl Alcohol:**

$\frac{\Delta V}{V_0} = -\frac{\Delta P}{K}$

$\frac{\Delta V}{V_0} = -\frac{1.99 \times 10^7}{1.07 \times 10^9} \approx -0.0085 = 1.86 \%$ change in volume from 0 to 2000m

> This change according to the bulk modulus is linear with increasing pressure, thus we are equivalently senstive throughout the whole range

Could actually be done if the cavity is long enough to measure this change in volume.

For instance, if the

## Fully compressible fluids (gasses):

Boyle's law is as follows:

$$P_1 V_1 = P_2 V_2$$

Thus, if the pressure goes from 1bar (0m) to 200bar (2000m), the volume changes by the following amount:

$$ \frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{1}{200} = 99.5\%$$ decrease in volume relative to the start.

> Additionally, this change is not linear and inversely proportional, thus we have higher sensitivity at shallower depths