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diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md index e3b4526..3a02a83 100644 --- a/source/know/concept/stokes-law/index.md +++ b/source/know/concept/stokes-law/index.md @@ -9,17 +9,17 @@ categories: layout: "concept" --- -**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 $U$. +**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. +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 $U$. +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: $$\begin{aligned} @@ -27,7 +27,7 @@ $$\begin{aligned} = \frac{2 a U}{\nu} \end{aligned}$$ -We assume that $\mathrm{Re} \ll 1$, in which case +We assume that $$\mathrm{Re} \ll 1$$, in which case the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/) are reduced to the **steady Stokes equations**: @@ -39,10 +39,10 @@ $$\begin{aligned} = 0 \end{aligned}$$ -The goal is to solve for $p$ and $\va{v}$. +The goal is to solve for $$p$$ and $$\va{v}$$. We make the following ansatz in -[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$, -where $q(r)$, $f(r)$ and $g(r)$ are unknown functions: +[spherical coordinates](/know/concept/spherical-coordinates/) $$(r, \theta, \phi)$$, +where $$q(r)$$, $$f(r)$$ and $$g(r)$$ are unknown functions: $$\begin{gathered} p @@ -59,12 +59,12 @@ $$\begin{gathered} \end{gathered}$$ 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. +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: @@ -79,7 +79,7 @@ $$\begin{aligned} &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) \end{aligned}$$ -The parenthesized expression must be zero for all $r$, +The parenthesized expression must be zero for all $$r$$, leading us to the following relation: $$\begin{aligned} @@ -98,7 +98,7 @@ $$\begin{aligned} \end{aligned}$$ This is simply the Laplace equation, -which is as follows for our ansatz $p(r, \theta)$: +which is as follows for our ansatz $$p(r, \theta)$$: $$\begin{aligned} 0 @@ -115,8 +115,8 @@ $$\begin{aligned} &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) \end{aligned}$$ -Again, the parenthesized expression must be zero for all $r$, -meaning it is an ODE for $q(r)$, +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: $$\begin{aligned} @@ -124,7 +124,7 @@ $$\begin{aligned} = \frac{C_3}{r^2} + C_4 r \end{aligned}$$ -Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later). +Where $$C_3$$ and $$C_4$$ are linearity constants ($$C_1$$ and $$C_2$$ appear later). The pressure is therefore: $$\begin{aligned} @@ -132,7 +132,7 @@ $$\begin{aligned} = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big) \end{aligned}$$ -Consequently, its gradient $\nabla p$ in spherical coordinates is as follows: +Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows: $$\begin{aligned} \nabla p @@ -140,8 +140,8 @@ $$\begin{aligned} = \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}$$ -According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$. -Let us look at the $r$-component of $\nabla^2 \va{v}$: +According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$. +Let us look at the $$r$$-component of $$\nabla^2 \va{v}$$: $$\begin{aligned} (\nabla^2 \va{v})_r @@ -154,22 +154,22 @@ $$\begin{aligned} &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) \end{aligned}$$ -Substituting $g$ for the expression we found from incompressibility lets us simplify this: +Substituting $$g$$ for the expression we found from incompressibility lets us simplify this: $$\begin{aligned} \eta (\nabla^2 \va{v})_r &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \end{aligned}$$ -The Stokes equation says that this must be equal to the $r$-component of $\nabla p$: +The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$: $$\begin{aligned} \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) \end{aligned}$$ -Where we have inserted $\idv{q}{r}$. -Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$, +Where we have inserted $$\idv{q}{r}$$. +Dividing out $$\eta U \cos\theta$$ leaves an ODE for $$f(r)$$, satisfied by: $$\begin{aligned} @@ -178,17 +178,17 @@ $$\begin{aligned} \end{aligned}$$ Then, thanks to our earlier relation again, -we know that $g(r)$ is as follows: +we know that $$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}$$ -So what about $C_1$, $C_2$, $C_3$ and $C_4$? -For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$, -meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$. -This implies that $C_4 = 0$ and $C_1 = 1$, leaving: +So what about $$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) @@ -199,10 +199,10 @@ $$\begin{aligned} \end{aligned}$$ 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, +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 $C_3 = -3 a / 2$. +yields $$C_2 = a^3 / 2$$ and $$C_3 = -3 a / 2$$. Therefore the full solution is: $$\begin{gathered} @@ -225,8 +225,8 @@ $$\begin{gathered} 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: +described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$. +The drag force $$\va{D}$$ on the surface is: $$\begin{aligned} \va{D} @@ -234,7 +234,7 @@ $$\begin{aligned} = \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}$$ -Where $\vu{e}_r$ is the sphere's surface normal vector. +Where $$\vu{e}_r$$ is the sphere's surface normal vector. The integrand can be expanded as follows: $$\begin{aligned} @@ -242,7 +242,7 @@ $$\begin{aligned} = \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r} \end{aligned}$$ -To calculate this, we start by taking the gradient of the velocity field $\va{v}$: +To calculate this, we start by taking the gradient of the velocity field $$\va{v}$$: $$\begin{aligned} \nabla\va{v} @@ -253,7 +253,7 @@ $$\begin{aligned} + \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big) \end{aligned}$$ -Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$: +Some of these terms are necessary to calculate the stress elements $$\sigma_{rr}$$ and $$\sigma_{\theta r}$$: $$\begin{aligned} \sigma_{rr} @@ -264,6 +264,7 @@ $$\begin{aligned} \\ &= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big) \end{aligned}$$ + $$\begin{aligned} \sigma_{\theta r} &= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big) @@ -276,7 +277,7 @@ $$\begin{aligned} &= - \frac{3 \eta U a^3}{2 r^4} \sin\theta \end{aligned}$$ -At the sphere's surface we set $r = a$, so these expressions reduce to the following: +At the sphere's surface we set $$r = a$$, so these expressions reduce to the following: $$\begin{aligned} \sigma_{rr} @@ -287,7 +288,7 @@ $$\begin{aligned} \end{aligned}$$ 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: +by converting the basis vectors $$\vu{e}_r$$ and $$\vu{e}_\theta$$ to Cartesian coordinates: $$\begin{aligned} \hat{\sigma} \cdot \vu{e}_r @@ -301,11 +302,11 @@ $$\begin{aligned} = \vu{e}_z \frac{3 \eta U}{2 a} \end{aligned}$$ -Remarkably, the stress at every point on the sphere is purely in the $z$-direction! +Remarkably, the stress at every point on the sphere is purely in the $$z$$-direction! This is not entirely unexpected though: symmetry cancels out all other components. -With this, we can do the integrals for $\va{D}$, -which reduce to a surface area factor $4 \pi a^2$: +With this, we can do the integrals for $$\va{D}$$, +which reduce to a surface area factor $$4 \pi a^2$$: $$\begin{aligned} \va{D} @@ -315,7 +316,7 @@ $$\begin{aligned} \end{aligned}$$ At last, we arrive at Stokes' law, -which simply expresses the magnitude of $\va{D}$: +which simply expresses the magnitude of $$\va{D}$$: $$\begin{aligned} \boxed{ @@ -327,8 +328,8 @@ $$\begin{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, +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/): $$\begin{aligned} @@ -336,9 +337,9 @@ $$\begin{aligned} = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 \end{aligned}$$ -Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities, -and $g_0$ is the gravitational acceleration. -Since $D$ acts in the opposite sense of $G$, +Where $$\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} @@ -346,7 +347,7 @@ $$\begin{aligned} = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 \end{aligned}$$ -This is an equation for the **terminal velocity** $U_t$, +This is an equation for the **terminal velocity** $$U_t$$, which we find to be as follows: $$\begin{aligned} @@ -356,7 +357,7 @@ $$\begin{aligned} } \end{aligned}$$ -The falling sphere will accelerate until $U_t$, +The falling sphere will accelerate until $$U_t$$, and then continue falling at constant speed. |