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diff --git a/source/know/concept/meniscus/index.md b/source/know/concept/meniscus/index.md new file mode 100644 index 0000000..1373d63 --- /dev/null +++ b/source/know/concept/meniscus/index.md @@ -0,0 +1,189 @@ +--- +title: "Meniscus" +date: 2021-03-11 +categories: +- Physics +- Fluid mechanics +- Fluid statics +- Surface tension +layout: "concept" +--- + +When a fluid interface, e.g. the surface of a liquid, +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 $y = d$. + +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 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 $d > 0$): + +$$\begin{aligned} + \int_0^d \rho g y \dd{y} + = \frac{1}{2} \rho g d^2 +\end{aligned}$$ + +If you were wondering about the units, +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: + +$$\begin{aligned} + \alpha + = \alpha \sin\phi + \frac{1}{2} \rho g d^2 +\end{aligned}$$ + +We isolate this relation for $d$ +and use some trigonometric magic to rewrite it: + +$$\begin{aligned} + d + = \sqrt{\frac{\alpha}{\rho g}} \sqrt{2 (1 - \sin\phi)} + = \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)} +\end{aligned}$$ + +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 $d < 0$): + +$$\begin{aligned} + \boxed{ + d + = 2 L_c \sin\!\Big(\frac{\pi/2 - \phi}{2}\Big) + } +\end{aligned}$$ + +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 $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$, +and is governed by: + +$$\begin{aligned} + \dv{x}{s} + = \cos\theta + \qquad + \dv{y}{s} + = \sin\theta + \qquad + \dv{\theta}{s} + = \frac{1}{R} +\end{aligned}$$ + +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: + +$$\begin{aligned} + p_0 - \rho g y + = p_0 - \frac{\alpha}{R} +\end{aligned}$$ + +Rearranging this yields that $R = L_c^2 / y$. +Inserting this into the curvature equation gives us: + +$$\begin{aligned} + \dv{\theta}{s} + = \frac{y}{L_c^2} +\end{aligned}$$ + +By differentiating this equation with respect to $s$ +and using $\idv{y}{s} = \sin\theta$, we arrive at: + +$$\begin{aligned} + \boxed{ + L_c^2 \dvn{2}{\theta}{s} = \sin\theta + } +\end{aligned}$$ + +To solve this equation, we multiply it by $\idv{\theta}{s}$, +which is nonzero close to the wall: + +$$\begin{aligned} + L_c^2 \dvn{2}{\theta}{s} \dv{\theta}{s} + = \dv{\theta}{s} \sin\theta +\end{aligned}$$ + +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: + +$$\begin{aligned} + \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2 + = 1 - \cos\theta +\end{aligned}$$ + +Isolating this for $\idv{\theta}{s}$ and using a trigonometric identity then yields: + +$$\begin{aligned} + \dv{\theta}{s} + = \frac{1}{L_c} \sqrt{2 (1 - \cos\theta)} + = \frac{1}{L_c} \sqrt{4 \sin^2\!\Big( \frac{\theta}{2} \Big)} + = - \frac{2}{L_c} \sin\!\Big( \frac{\theta}{2} \Big) +\end{aligned}$$ + +We use trigonometric relations on the equations +for $\idv{x}{s}$ and $\idv{y}{s}$ to get $\theta$-derivatives: + +$$\begin{aligned} + \dv{x}{\theta} + &= \dv{x}{s} \dv{s}{\theta} + = \bigg( 1 - 2 \sin^2\!\Big( \frac{\theta}{2} \Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1} + = L_c \sin\!\Big( \frac{\theta}{2} \Big) - \frac{L_c}{2 \sin(\theta/2)} + \\ + \dv{y}{\theta} + &= \dv{y}{s} \dv{s}{\theta} + = \bigg( 2 \sin\!\Big(\frac{\theta}{2}\Big) \cos\!\Big(\frac{\theta}{2}\Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1} + = - L_c \cos\!\Big( \frac{\theta}{2} \Big) +\end{aligned}$$ + +Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall. +Then, by integrating the above equations, we get the following solutions: + +$$\begin{gathered} + \boxed{ + \frac{x}{L_c} + = 2 \cos\!\Big(\frac{\theta_0}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta_0}{4}\Big) \bigg| + - 2 \cos\!\Big(\frac{\theta}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta}{4}\Big) \bigg| + } + \\ + \boxed{ + \frac{y}{L_c} + = - 2 \sin\!\Big(\frac{\theta}{2}\Big) + } +\end{gathered}$$ + +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 $d$: + +$$\begin{aligned} + d + = y(\theta_0) + = - 2 L_c \sin\!\Big(\frac{\theta_0}{2}\Big) + = 2 L_c \sin\!\Big( \frac{\pi/2 - \phi}{2} \Big) +\end{aligned}$$ + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |