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Diffstat (limited to 'source/know/concept/stokes-law/index.md')
| -rw-r--r-- | source/know/concept/stokes-law/index.md | 33 |
1 files changed, 17 insertions, 16 deletions
diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md index 3a02a83..40212ef 100644 --- a/source/know/concept/stokes-law/index.md +++ b/source/know/concept/stokes-law/index.md @@ -50,10 +50,10 @@ $$\begin{gathered} \\ v_r = U f(r) \cos\theta - \qquad + \qquad \quad v_\theta = - U g(r) \sin\theta - \qquad + \qquad \quad v_\phi = 0 \end{gathered}$$ @@ -76,7 +76,7 @@ $$\begin{aligned} \\ &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta \\ - &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) + &= U \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) \cos\theta \end{aligned}$$ The parenthesized expression must be zero for all $$r$$, @@ -103,16 +103,16 @@ which is as follows for our ansatz $$p(r, \theta)$$: $$\begin{aligned} 0 = \nabla^2 p - &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{p}{\theta} \Big) + &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \pdv{p}{\theta} \sin\theta \Big) \\ 0 &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big) - \frac{\eta U q}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin^2\theta \Big) \\ - &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big) + &= \frac{\eta U}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big) \cos\theta - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta \\ - &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) + &= \eta U \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) \cos\theta \end{aligned}$$ Again, the parenthesized expression must be zero for all $$r$$, @@ -129,7 +129,7 @@ The pressure is therefore: $$\begin{aligned} p - = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big) + = \eta U \Big( \frac{C_3}{r^2} + C_4 r \Big) \cos\theta \end{aligned}$$ Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows: @@ -137,7 +137,7 @@ Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows: $$\begin{aligned} \nabla p = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta} - = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big) + = \vu{e}_r \Big( \eta U \dv{q}{r} \cos\theta \Big) - \vu{e}_\theta \Big( \eta U \frac{q}{r} \sin\theta \Big) \end{aligned}$$ According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$. @@ -148,24 +148,24 @@ $$\begin{aligned} &= \pdvn{2}{v_r}{r} + \frac{1}{r^2} \pdvn{2}{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r} + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta \\ - &= U \cos\theta \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f - + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) + &= U \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f + + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) \cos\theta \\ - &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) + &= U \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) \cos\theta \end{aligned}$$ Substituting $$g$$ for the expression we found from incompressibility lets us simplify this: $$\begin{aligned} \eta (\nabla^2 \va{v})_r - &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) + &= \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta \end{aligned}$$ The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$: $$\begin{aligned} - \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) - = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) + \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta + = \eta U \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) \cos\theta \end{aligned}$$ Where we have inserted $$\idv{q}{r}$$. @@ -213,14 +213,15 @@ $$\begin{gathered} \\ \boxed{ v_r - = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) + = U \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) \cos\theta \qquad v_\theta - = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) + = - U \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) \sin\theta } \end{gathered}$$ + ## Drag force From the definition of [viscosity](/know/concept/viscosity/), |