Physics #20: Fluid Dynamics

by Qubit Factory

Mass Density

The mass density of a fluid (liquid or gas) is defined as

$\rho \equiv \frac{m}{V}$

where m is the mass of the liquid and V is the volume of the liquid.

Pressure

Pressure is defined as the net force per area:

$P \equiv \frac{F}{A}$

Note that this can reduce to

$P \equiv \frac{ma}{A} \equiv \frac{m}{A}\,\frac{\mathrm{d}^2 s}{\mathrm{d}t^2}.$

Now, given $\rho$ it follows that $m \equiv \rho V$ , therefore:

$P \equiv \frac{\rho Va}{A}.$

Note that $P \in \mathbb{R}.$ (thus not a vector).

Also note that the units for pressure are $1 \frac{\text{ N}}{\text{ m}^2} \equiv 1 \text{ Pa}$ where the latter stands for Pascal.

Lastly,

$F \equiv PA$

$\therefore\, ma \equiv PA$

$\therefore \frac{\mathrm{d}^2 s}{\mathrm{d}t^2} \equiv \frac{PA}{m}.$

Water Pressure

Given density $\rho,$ depth $h,$ and gravitational acceleration $\mathbf{g},$ the pressure differential between two points underwater is given by

$\Delta P \equiv \rho gh.$

I have realized something:

$\rho g$

is defined in some textbooks as the weight density, but I redefine it as:

$\frac{mg}{V},$

which can be seen as the net gravitational pull (acceleration) per volumetric area. Now, multiplying by h causes the following:

$\frac{mg}{lwh}\, h \implies \frac{mg}{lw}$

which can be seen as the net gravitational pull per rectilinear area. Since force is mass times acceleration, the pressure equation literally reduces into “mass times acceleration per area” in the form of gravitational pull over area.

Liquid Pressure

$P_t \equiv P_l + P_{\text{atm}}$

Pascal’s Principle

A perturbation of pressure at one point in a fluid will diffuse to all other points within the fluid, as well as the enclosure. This is mathematically stated as:

$\frac{F_1}{A_1}\equiv\frac{F_2}{A_2}.$