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Diffstat (limited to 'source/know/concept/hydrostatic-pressure')
| -rw-r--r-- | source/know/concept/hydrostatic-pressure/index.md | 66 |
1 files changed, 33 insertions, 33 deletions
diff --git a/source/know/concept/hydrostatic-pressure/index.md b/source/know/concept/hydrostatic-pressure/index.md index 2e55246..020fc75 100644 --- a/source/know/concept/hydrostatic-pressure/index.md +++ b/source/know/concept/hydrostatic-pressure/index.md @@ -9,7 +9,7 @@ categories: layout: "concept" --- -The pressure $p$ inside a fluid at rest, +The pressure $$p$$ inside a fluid at rest, the so-called **hydrostatic pressure**, is an important quantity. Here we will properly define it, @@ -20,8 +20,8 @@ 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 $\dd{\va{S}}$: +with normal vector $$\vu{n}$$ and area $$\dd{S}$$, +yielding the following vector element $$\dd{\va{S}}$$: $$\begin{aligned} \dd{\va{S}} @@ -29,7 +29,7 @@ $$\begin{aligned} \end{aligned}$$ The orientation of these surfaces does not matter. -The **pressure** $p(\va{r})$ is defined as the force-per-area +The **pressure** $$p(\va{r})$$ is defined as the force-per-area of these tiny surface elements: $$\begin{aligned} @@ -38,9 +38,9 @@ $$\begin{aligned} \end{aligned}$$ 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 $\dd{\va{S}}$: +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 $$\dd{\va{S}}$$: $$\begin{aligned} \va{F} @@ -57,8 +57,8 @@ $$\begin{aligned} = - \int_V \nabla p \dd{V} \end{aligned}$$ -Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$ -on all its constituent volume elements $\dd{V}$, +Since the total force on the blob is simply the sum of the forces $$\dd{\va{F}}$$ +on all its constituent volume elements $$\dd{V}$$, we arrive at the following relation: $$\begin{aligned} @@ -71,8 +71,8 @@ $$\begin{aligned} If the fluid is at rest, then all forces on the blob cancel out (otherwise it would move). Since we are currently neglecting all forces other than pressure, -this is equivalent to demanding that $\dd{\va{F}} = 0$, -which implies that $\nabla p = 0$, i.e. the pressure is constant. +this is equivalent to demanding that $$\dd{\va{F}} = 0$$, +which implies that $$\nabla p = 0$$, i.e. the pressure is constant. $$\begin{aligned} \boxed{ @@ -84,17 +84,17 @@ $$\begin{aligned} ## With gravity If we include gravity, then, -in addition to the pressure's *contact force* $\va{F}_p$ from earlier, -there is also a *body force* $\va{F}_g$ acting on -the arbitrary blob $V$ of fluid enclosed by $S$: +in addition to the pressure's *contact force* $$\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}$$ -Where $\rho$ is the fluid's density (which need not be constant) -and $\va{g}$ is the gravity field given in units of force-per-mass. +Where $$\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} @@ -116,7 +116,7 @@ $$\begin{aligned} \end{aligned}$$ On Earth (or another body with strong gravity), -it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction, +it is reasonable to treat $$\va{g}$$ as only pointing in the downward $$z$$-direction, in which case the above condition turns into: $$\begin{aligned} @@ -124,9 +124,9 @@ $$\begin{aligned} = \rho g_0 z \end{aligned}$$ -Where $g_0$ is the magnitude of the $z$-component of $\va{g}$. +Where $$g_0$$ is the magnitude of the $$z$$-component of $$\va{g}$$. We can generalize the equilibrium condition by treating -the gravity field as the gradient of the gravitational potential $\Phi$: +the gravity field as the gradient of the gravitational potential $$\Phi$$: $$\begin{aligned} \va{g}(\va{r}) @@ -142,13 +142,13 @@ $$\begin{aligned} } \end{aligned}$$ -In practice, the density $\rho$ of the fluid -may be a function of the pressure $p$ (compressibility) -and/or temperature $T$ (thermal expansion). +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 +We then define the **pressure potential** $$w(p)$$ as the indefinite integral of the density: $$\begin{aligned} @@ -166,7 +166,7 @@ $$\begin{aligned} \end{aligned}$$ From this, let us now define the -**effective gravitational potential** $\Phi^*$ as follows: +**effective gravitational potential** $$\Phi^*$$ as follows: $$\begin{aligned} \Phi^* \equiv \Phi + w(p) @@ -181,7 +181,7 @@ $$\begin{aligned} } \end{aligned}$$ -At every point in the fluid, despite $p$ being variable, +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 @@ -193,15 +193,15 @@ $$\begin{aligned} = 0 \end{aligned}$$ -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 $a^3$. +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, +$$a$$ 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. |