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diff --git a/source/know/concept/hydrostatic-pressure/index.md b/source/know/concept/hydrostatic-pressure/index.md new file mode 100644 index 0000000..4add09e --- /dev/null +++ b/source/know/concept/hydrostatic-pressure/index.md @@ -0,0 +1,211 @@ +--- +title: "Hydrostatic pressure" +date: 2021-03-12 +categories: +- Physics +- Fluid mechanics +- Fluid statics +layout: "concept" +--- + +The pressure $p$ inside a fluid at rest, +the so-called **hydrostatic pressure**, +is an important quantity. +Here we will properly define it, +and derive the equilibrium condition for the fluid to be at rest, +both with and without an arbitrary gravity field. + + +## Without gravity + +Inside the fluid, we can imagine small arbitrary partition surfaces, +with normal vector $\vu{n}$ and area $\dd{S}$, +yielding the following vector element $\dd{\va{S}}$: + +$$\begin{aligned} + \dd{\va{S}} + = \vu{n} \dd{S} +\end{aligned}$$ + +The orientation of these surfaces does not matter. +The **pressure** $p(\va{r})$ is defined as the force-per-area +of these tiny surface elements: + +$$\begin{aligned} + \dd{\va{F}} + = - p(\va{r}) \dd{\va{S}} +\end{aligned}$$ + +The negative sign is there because a positive pressure is conventionally defined +to push from the positive (normal) side of $\dd{\va{S}}$ to the negative side. +The total force $\va{F}$ on a larger surface inside the fluid is +then given by the surface integral over many adjacent $\dd{\va{S}}$: + +$$\begin{aligned} + \va{F} + = - \int_S p(\va{r}) \dd{\va{S}} +\end{aligned}$$ + +If we now consider a *closed* surface, +which encloses a "blob" of the fluid, +then we can use the divergence theorem to get a volume integral: + +$$\begin{aligned} + \va{F} + = - \oint_S p \dd{\va{S}} + = - \int_V \nabla p \dd{V} +\end{aligned}$$ + +Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$ +on all its constituent volume elements $\dd{V}$, +we arrive at the following relation: + +$$\begin{aligned} + \boxed{ + \dd{\va{F}} + = - \nabla p \dd{V} + } +\end{aligned}$$ + +If the fluid is at rest, then all forces on the blob cancel out +(otherwise it would move). +Since we are currently neglecting all forces other than pressure, +this is equivalent to demanding that $\dd{\va{F}} = 0$, +which implies that $\nabla p = 0$, i.e. the pressure is constant. + +$$\begin{aligned} + \boxed{ + \nabla p = 0 + } +\end{aligned}$$ + + +## With gravity + +If we include gravity, then, +in addition to the pressure's *contact force* $\va{F}_p$ from earlier, +there is also a *body force* $\va{F}_g$ acting on +the arbitrary blob $V$ of fluid enclosed by $S$: + +$$\begin{aligned} + \va{F}_g + = \int_V \rho \va{g} \dd{V} +\end{aligned}$$ + +Where $\rho$ is the fluid's density (which need not be constant) +and $\va{g}$ is the gravity field given in units of force-per-mass. +For a fluid at rest, these forces must cancel out: + +$$\begin{aligned} + \va{F} + = \va{F}_g + \va{F}_p + = \int_V \rho \va{g} - \nabla p \dd{V} + = 0 +\end{aligned}$$ + +Since this a single integral over an arbitrary volume, +it implies that every point of the fluid must +locally satisfy the following equilibrium condition: + +$$\begin{aligned} + \boxed{ + \nabla p + = \rho \va{g} + } +\end{aligned}$$ + +On Earth (or another body with strong gravity), +it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction, +in which case the above condition turns into: + +$$\begin{aligned} + p + = \rho g_0 z +\end{aligned}$$ + +Where $g_0$ is the magnitude of the $z$-component of $\va{g}$. +We can generalize the equilibrium condition by treating +the gravity field as the gradient of the gravitational potential $\Phi$: + +$$\begin{aligned} + \va{g}(\va{r}) + = - \nabla \Phi(\va{r}) +\end{aligned}$$ + +With this, the equilibrium condition is turned into the following equation: + +$$\begin{aligned} + \boxed{ + \nabla \Phi + \frac{\nabla p}{\rho} + = 0 + } +\end{aligned}$$ + +In practice, the density $\rho$ of the fluid +may be a function of the pressure $p$ (compressibility) +and/or temperature $T$ (thermal expansion). +We will tackle the first complication, but neglect the second, +i.e. we assume that the temperature is equal across the fluid. + +We then define the **pressure potential** $w(p)$ as +the indefinite integral of the density: + +$$\begin{aligned} + w(p) + \equiv \int \frac{1}{\rho(p)} \dd{p} +\end{aligned}$$ + +Using this, we can rewrite the equilibrium condition as a single gradient like so: + +$$\begin{aligned} + 0 + = \nabla \Phi + \frac{\nabla p}{\rho} + = \nabla \Phi + \dv{w}{p} \nabla p + = \nabla \Big( \Phi + w(p) \Big) +\end{aligned}$$ + +From this, let us now define the +**effective gravitational potential** $\Phi^*$ as follows: + +$$\begin{aligned} + \Phi^* \equiv \Phi + w(p) +\end{aligned}$$ + +This results in the cleanest form yet of the equilibrium condition, namely: + +$$\begin{aligned} + \boxed{ + \nabla \Phi^* + = 0 + } +\end{aligned}$$ + +At every point in the fluid, despite $p$ being variable, +the force that is applied by the pressure must have the same magnitude in all directions at that point. +This statement is known as **Pascal's law**, +and is due to the fact that all forces must cancel out +for an arbitrary blob: + +$$\begin{aligned} + \va{F} + = \va{F}_g + \va{F}_p + = 0 +\end{aligned}$$ + +Let the blob be a cube with side $a$. +Now, $\va{F}_p$ is a contact force, +meaning it acts on the surface, and is thus proportional to $a^2$, +however, $\va{F}_g$ is a body force, +meaning it acts on the volume, and is thus proportional to $a^3$. +Since we are considering a *point* in the fluid, +$a$ is infinitesimally small, +so that $\va{F}_p$ dominates $\va{F}_g$. +Consequently, at equilibrium, $\va{F}_p$ must cancel out by itself, +which means that the pressure is the same in all directions. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |