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diff --git a/content/know/concept/stokes-law/index.pdc b/content/know/concept/stokes-law/index.pdc new file mode 100644 index 0000000..acc98af --- /dev/null +++ b/content/know/concept/stokes-law/index.pdc @@ -0,0 +1,371 @@ +--- +title: "Stokes' law" +firstLetter: "S" +publishDate: 2021-05-04 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-05-04T19:28:30+02:00 +draft: false +markup: pandoc +--- + +# Stokes' law + +**Stokes' law** describes the size of the drag force $D$ +at low [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \ll 1$ +experienced by a spherical object in a steady, uniform flow at velocity $U$. + + +## Flow field + +Imagine a sphere with radius $a$ sinking in a viscous liquid. +To model this situation, let us pretend that the sphere is fixed instead, +and the fluid comes from infinity at velocity $U$ along the $z$-axis, +flows past the sphere, and continues to infinity at the same $U$. +The Reynolds number is: + +$$\begin{aligned} + \mathrm{Re} + = \frac{2 a U}{\nu} +\end{aligned}$$ + +We assume that $\mathrm{Re} \ll 1$, in which case +the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/) +are reduced to the **steady Stokes equations**: + +$$\begin{aligned} + \nabla p + = \eta \nabla^2 \va{v} + \qquad \quad + \nabla \cdot \va{v} + = 0 +\end{aligned}$$ + +The goal is to solve for $p$ and $\va{v}$. +We make the following ansatz in +[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$, +where $q(r)$, $f(r)$ and $g(r)$ are unknown functions: + +$$\begin{gathered} + p + = \eta U q(r) \cos\theta + \\ + v_r + = U f(r) \cos\theta + \qquad + v_\theta + = - U g(r) \sin\theta + \qquad + v_\phi + = 0 +\end{gathered}$$ + +The fluid hits the sphere head on, +so the solution is taken to be $\phi$-independent due to symmetry. +Note that $\theta$ is the angle to the positive $z$-axis, +which is the direction of $\va{U} = U \vu{e}_z$. +Moreover, note that $\va{U} \cdot \vu{e}_r = U \cos\theta$ +and $\va{U} \cdot \vu{e}_\theta = - U \sin\theta$, +where $\vu{e}_r$ and $\vu{e}_\theta$ are basis vectors. + +To begin with, we insert this ansatz into the incompressibility condition, +yielding: + +$$\begin{aligned} + 0 + = \nabla \cdot \va{v} + &= \pdv{v_r}{r} + \frac{1}{r} \pdv{v_\theta}{\theta} + \frac{2 v_r}{r} + \frac{v_\theta}{r \tan \theta} + \\ + &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \\ + &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) +\end{aligned}$$ + +The parenthesized expression must be zero for all $r$, +leading us to the following relation: + +$$\begin{aligned} + g(r) + = f + \frac{r}{2} \dv{f}{r} +\end{aligned}$$ + +Next, we take the divergence of the first Stokes equation, +and insert incompressibility: + +$$\begin{aligned} + \nabla^2 p + = \eta \nabla \cdot (\nabla^2 \va{v}) + = \eta \nabla^2 (\nabla \cdot \va{v}) + = 0 +\end{aligned}$$ + +This is simply the Laplace equation, +which is as follows for our ansatz $p(r, \theta)$: + +$$\begin{aligned} + 0 + = \nabla^2 p + &= \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{\theta} \Big( \sin\theta \pdv{p}{\theta} \Big) + \\ + 0 + &= \frac{\eta U \cos\theta}{r^2} \dv{r} \Big( r^2 \dv{q}{r} \Big) + - \frac{\eta U q}{r^2 \sin\theta} \pdv{\theta} \Big( \sin^2\theta \Big) + \\ + &= \frac{\eta U \cos\theta}{r^2} \dv{r} \Big( r^2 \dv{q}{r} \Big) + - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta + \\ + &= \eta U \cos\theta \Big( \dv[2]{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) +\end{aligned}$$ + +Again, the parenthesized expression must be zero for all $r$, +meaning it is an ODE for $q(r)$, +whose solution is straightforwardly found to be: + +$$\begin{aligned} + q(r) + = \frac{C_3}{r^2} + C_4 r +\end{aligned}$$ + +Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later). +The pressure is therefore: + +$$\begin{aligned} + p + = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big) +\end{aligned}$$ + +Consequently, its gradient $\nabla p$ in spherical coordinates is as follows: + +$$\begin{aligned} + \nabla p + = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta} + = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big) +\end{aligned}$$ + +According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$. +Let us look at the $r$-component of $\nabla^2 \va{v}$: + +$$\begin{aligned} + (\nabla^2 \va{v})_r + &= \pdv[2]{v_r}{r} + \frac{1}{r^2} \pdv[2]{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r} + + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta + \\ + &= U \cos\theta \Big( \dv[2]{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f + + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) + \\ + &= U \cos\theta \Big( \dv[2]{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) +\end{aligned}$$ + +Substituting $g$ for the expression we found from incompressibility lets us simplify this: + +$$\begin{aligned} + \eta (\nabla^2 \va{v})_r + &= \eta U \cos\theta \Big( \dv[2]{f}{r} + \frac{4}{r} \dv{f}{r} \Big) +\end{aligned}$$ + +The Stokes equation says that this must be equal to the $r$-component of $\nabla p$: + +$$\begin{aligned} + \eta U \cos\theta \Big( \dv[2]{f}{r} + \frac{4}{r} \dv{f}{r} \Big) + = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) +\end{aligned}$$ + +Where we have inserted $\dv*{q}{r}$. +Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$, +satisfied by: + +$$\begin{aligned} + f(r) + = C_1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \frac{C_4 r^2}{10} +\end{aligned}$$ + +Then, thanks to our earlier relation again, +we know that $g(r)$ is as follows: + +$$\begin{aligned} + g(r) + = C_1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} + \frac{C_4 r^2}{5} +\end{aligned}$$ + +So what about $C_1$, $C_2$, $C_3$ and $C_4$? +For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$, +meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$. +This implies that $C_4 = 0$ and $C_1 = 1$, leaving: + +$$\begin{aligned} + f(r) + = 1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \qquad \quad + g(r) + = 1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} +\end{aligned}$$ + +Furthermore, the viscous *no-slip* condition demands +that $\va{v} = 0$ at the sphere's surface $r = a$, so $f(a) = g(a) = 0$ there. +Inserting $a$ into $f$ and $g$, setting them to zero, +and solving the resulting system of equations +yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$. +Therefore the full solution is: + +$$\begin{gathered} + \boxed{ + p + = - \frac{3 \eta U a}{2 r^2} \cos\theta + } + \\ + \boxed{ + v_r + = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) + \qquad + v_\theta + = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) + } +\end{gathered}$$ + + +## Drag force + +From the definition of [viscosity](/know/concept/viscosity/), +we know that there must be shear stresses at the sphere surface, +described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$. +The drag force $\va{D}$ on the surface is: + +$$\begin{aligned} + \va{D} + = \oint \hat{\sigma} \cdot \dd{\va{S}} + = \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi} +\end{aligned}$$ + +Where $\vu{e}_r$ is the sphere's surface normal vector. +The integrand can be expanded as follows: + +$$\begin{aligned} + \hat{\sigma} \cdot \vu{e}_r + = \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r} +\end{aligned}$$ + +To calculate this, we start by taking the gradient of the velocity field $\va{v}$: + +$$\begin{aligned} + \nabla\va{v} + &= \vu{e}_r \vu{e}_r \pdv{v_r}{r} + \vu{e}_r \vu{e}_\theta \pdv{v_\theta}{r} + + \vu{e}_\theta \vu{e}_r \Big( \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big) + \\ + &\qquad + \vu{e}_\theta \vu{e}_\theta \Big( \frac{1}{r} \pdv{v_\theta}{\theta} - \frac{v_r}{r} \Big) + + \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big) +\end{aligned}$$ + +Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$: + +$$\begin{aligned} + \sigma_{rr} + &= - p + 2 \eta (\nabla\va{v})_{rr} + = - p + 2 \eta \pdv{v_r}{r} + \\ + &= \frac{3 \eta U a}{2 r^2} \cos\theta + 2 \eta U \cos\theta \: \Big( \!-\! \frac{3 a^3}{2 r^4} + \frac{3 a}{2 r^2} \Big) + \\ + &= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big) +\end{aligned}$$ +$$\begin{aligned} + \sigma_{\theta r} + &= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big) + = \eta \: \Big( \pdv{v_\theta}{r} + \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big) + \\ + &= \eta U \sin\theta \: \Big( \!-\! \frac{3 a^3}{4 r^4} - \frac{3 a}{4 r^2} + - \frac{1}{r} - \frac{a^3}{2 r^4} + \frac{3 a}{2 r^2} + + \frac{1}{r} - \frac{a^3}{4 r^4} - \frac{3 a}{4 r^2} \Big) + \\ + &= - \frac{3 \eta U a^3}{2 r^4} \sin\theta +\end{aligned}$$ + +At the sphere's surface we set $r = a$, so these expressions reduce to the following: + +$$\begin{aligned} + \sigma_{rr} + = \frac{3 \eta U}{2 a} \cos\theta + \qquad \quad + \sigma_{\theta r} + = - \frac{3 \eta U}{2 a} \sin\theta +\end{aligned}$$ + +Now we can finally calculate the effective stress on the surface, +by converting the basis vectors $\vu{e}_r$ and $\vu{e}_\theta$ to Cartesian coordinates: + +$$\begin{aligned} + \hat{\sigma} \cdot \vu{e}_r + &= \vu{e}_r \frac{3 \eta U}{2 a} \cos\theta - \vu{e}_\theta \frac{3 \eta U}{2 a} \sin\theta + \\ + &= \Big( \vu{e}_x \sin\theta \cos\phi + \vu{e}_y \sin\theta \sin\phi + \vu{e}_z \cos\theta \Big) \frac{3 \eta U}{2 a} \cos\theta + \\ + &\qquad - \Big( \vu{e}_x \cos\theta \cos\phi + \vu{e}_y \cos\theta \sin\phi - \vu{e}_z \sin\theta \Big) \frac{3 \eta U}{2 a} \sin\theta + \\ + &= \Big( \vu{e}_z \cos^2\theta + \vu{e}_z \sin^2\theta \Big) \frac{3 \eta U}{2 a} + = \vu{e}_z \frac{3 \eta U}{2 a} +\end{aligned}$$ + +Remarkably, the stress at every point on the sphere is purely in the $z$-direction! +This is not entirely unexpected though: symmetry cancels out all other components. + +With this, we can do the integrals for $\va{D}$, +which reduce to a surface area factor $4 \pi a^2$: + +$$\begin{aligned} + \va{D} + %= \int_0^{2\pi} \!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:r^2 \sin\theta \dd{\theta} \dd{\phi} + = \vu{e}_z \frac{3 \eta U}{2 a} \int_0^{2\pi} \!\!\!\! \int_0^\pi a^2 \sin\theta \dd{\theta} \dd{\phi} + = \vu{e}_z \frac{3 \eta U}{2 a} 2 \pi a^2 \int_0^\pi \sin\theta \dd{\theta} + = \vu{e}_z \: 6 \pi \eta U a +\end{aligned}$$ + +At last, we arrive at Stokes' law, +which simply expresses the magnitude of $\va{D}$: + +$$\begin{aligned} + \boxed{ + D + = 6 \pi \eta U a + } +\end{aligned}$$ + +To arrive at this result, +we assumed that the sphere was fixed, and the fluid was flowing past it. +We can equally well let the fluid be at rest, +with the sphere falling through it at $U$. +The force of gravity then exerts the following force $G$ on it, +subtracting [buoyancy](/know/concept/archimedes-principle/): + +$$\begin{aligned} + G + = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 +\end{aligned}$$ + +Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities, +and $g_0$ is the gravitational acceleration. +Since $D$ acts in the opposite sense of $G$, +after some time, they cancel out: + +$$\begin{aligned} + 6 \pi \eta U a + = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 +\end{aligned}$$ + +This is an equation for the **terminal velocity** $U_t$, +which we find to be as follows: + +$$\begin{aligned} + U_t + = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta} +\end{aligned}$$ + +The falling sphere will accelerate until $U_t$, +and then continue falling at constant speed. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |