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1 files changed, 34 insertions, 34 deletions
diff --git a/source/know/concept/meniscus/index.md b/source/know/concept/meniscus/index.md index e032a4e..a53b5e9 100644 --- a/source/know/concept/meniscus/index.md +++ b/source/know/concept/meniscus/index.md @@ -15,19 +15,19 @@ touches a flat solid wall, it will curve to meet it. This small rise or fall is called a **meniscus**, and is caused by surface tension and gravity. -In 2D, let the vertical $y$-axis be a flat wall, -and the fluid tend to $y = 0$ when $x \to \infty$. -Close to the wall, i.e. for small $x$, the liquid curves up or down -to touch the wall at a height $y = d$. +In 2D, let the vertical $$y$$-axis be a flat wall, +and the fluid tend to $$y = 0$$ when $$x \to \infty$$. +Close to the wall, i.e. for small $$x$$, the liquid curves up or down +to touch the wall at a height $$y = d$$. Three forces are at work here: -the first two are the surface tension $\alpha$ of the fluid surface, -and the counter-pull $\alpha \sin\phi$ of the wall against the tension, -where $\phi$ is the contact angle. +the first two are the surface tension $$\alpha$$ of the fluid surface, +and the counter-pull $$\alpha \sin\phi$$ of the wall against the tension, +where $$\phi$$ is the contact angle. The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient inside the small portion of the fluid above/below the ambient level, which exerts a total force on the wall given by -(for $\phi < \pi/2$ so that $d > 0$): +(for $$\phi < \pi/2$$ so that $$d > 0$$): $$\begin{aligned} \int_0^d \rho g y \dd{y} @@ -35,7 +35,7 @@ $$\begin{aligned} \end{aligned}$$ If you were wondering about the units, -keep in mind that there is an implicit $z$-direction here too. +keep in mind that there is an implicit $$z$$-direction here too. This results in the following balance equation for the forces at the wall: $$\begin{aligned} @@ -43,7 +43,7 @@ $$\begin{aligned} = \alpha \sin\phi + \frac{1}{2} \rho g d^2 \end{aligned}$$ -We isolate this relation for $d$ +We isolate this relation for $$d$$ and use some trigonometric magic to rewrite it: $$\begin{aligned} @@ -52,10 +52,10 @@ $$\begin{aligned} = \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)} \end{aligned}$$ -Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$, -yielding an expression for $d$ -that is valid both for $\phi < \pi/2$ (where $d > 0$) -and $\phi > \pi/2$ (where $d < 0$): +Here, we recognize the definition of the capillary length $$L_c = \sqrt{\alpha / (\rho g)}$$, +yielding an expression for $$d$$ +that is valid both for $$\phi < \pi/2$$ (where $$d > 0$$) +and $$\phi > \pi/2$$ (where $$d < 0$$): $$\begin{aligned} \boxed{ @@ -66,9 +66,9 @@ $$\begin{aligned} Next, we would like to know the exact shape of the meniscus. To do this, we need to describe the liquid surface differently, -using the elevation angle $\theta$ relative to the $y = 0$ plane. -The curve $\theta(s)$ is a function of the arc length $s$, -where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$, +using the elevation angle $$\theta$$ relative to the $$y = 0$$ plane. +The curve $$\theta(s)$$ is a function of the arc length $$s$$, +where $$\dd{s}^2 = \dd{x}^2 + \dd{y}^2$$, and is governed by: $$\begin{aligned} @@ -82,23 +82,23 @@ $$\begin{aligned} = \frac{1}{R} \end{aligned}$$ -The last equation describes the curvature radius $R$ -of the surface along the $x$-axis. +The last equation describes the curvature radius $$R$$ +of the surface along the $$x$$-axis. Since we are considering a flat wall, there is no curvature in the orthogonal principal direction. Just below the liquid surface in the meniscus, we expect the hydrostatic pressure and the [Young-Laplace law](/know/concept/young-laplace-law/) -to agree about the pressure $p$, -where $p_0$ is the external air pressure: +to agree about the pressure $$p$$, +where $$p_0$$ is the external air pressure: $$\begin{aligned} p_0 - \rho g y = p_0 - \frac{\alpha}{R} \end{aligned}$$ -Rearranging this yields that $R = L_c^2 / y$. +Rearranging this yields that $$R = L_c^2 / y$$. Inserting this into the curvature equation gives us: $$\begin{aligned} @@ -106,8 +106,8 @@ $$\begin{aligned} = \frac{y}{L_c^2} \end{aligned}$$ -By differentiating this equation with respect to $s$ -and using $\idv{y}{s} = \sin\theta$, we arrive at: +By differentiating this equation with respect to $$s$$ +and using $$\idv{y}{s} = \sin\theta$$, we arrive at: $$\begin{aligned} \boxed{ @@ -115,7 +115,7 @@ $$\begin{aligned} } \end{aligned}$$ -To solve this equation, we multiply it by $\idv{\theta}{s}$, +To solve this equation, we multiply it by $$\idv{\theta}{s}$$, which is nonzero close to the wall: $$\begin{aligned} @@ -123,16 +123,16 @@ $$\begin{aligned} = \dv{\theta}{s} \sin\theta \end{aligned}$$ -We integrate both sides with respect to $s$ -and set the integration constant to $1$, -such that we get zero when $\theta \to 0$ away from the wall: +We integrate both sides with respect to $$s$$ +and set the integration constant to $$1$$, +such that we get zero when $$\theta \to 0$$ away from the wall: $$\begin{aligned} \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2 = 1 - \cos\theta \end{aligned}$$ -Isolating this for $\idv{\theta}{s}$ and using a trigonometric identity then yields: +Isolating this for $$\idv{\theta}{s}$$ and using a trigonometric identity then yields: $$\begin{aligned} \dv{\theta}{s} @@ -142,7 +142,7 @@ $$\begin{aligned} \end{aligned}$$ We use trigonometric relations on the equations -for $\idv{x}{s}$ and $\idv{y}{s}$ to get $\theta$-derivatives: +for $$\idv{x}{s}$$ and $$\idv{y}{s}$$ to get $$\theta$$-derivatives: $$\begin{aligned} \dv{x}{\theta} @@ -156,7 +156,7 @@ $$\begin{aligned} = - L_c \cos\!\Big( \frac{\theta}{2} \Big) \end{aligned}$$ -Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall. +Let $$\theta_0 = \phi - \pi/2$$ be the initial elevation angle $$\theta(0)$$ at the wall. Then, by integrating the above equations, we get the following solutions: $$\begin{gathered} @@ -172,9 +172,9 @@ $$\begin{gathered} } \end{gathered}$$ -Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall, -and $x = 0$ for $\theta = \theta_0$. -This result is consistent with our earlier expression for $d$: +Where the integration constant has been chosen such that $$y \to 0$$ for $$\theta \to 0$$ away from the wall, +and $$x = 0$$ for $$\theta = \theta_0$$. +This result is consistent with our earlier expression for $$d$$: $$\begin{aligned} d |