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Diffstat (limited to 'source/know/concept/hagen-poiseuille-equation')
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1 files changed, 49 insertions, 49 deletions
diff --git a/source/know/concept/hagen-poiseuille-equation/index.md b/source/know/concept/hagen-poiseuille-equation/index.md index d59ee9a..6484631 100644 --- a/source/know/concept/hagen-poiseuille-equation/index.md +++ b/source/know/concept/hagen-poiseuille-equation/index.md @@ -13,12 +13,12 @@ The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**, describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/) through a cylindrical pipe. Due to its viscosity, the fluid clings to the sides, -limiting the amount that can pass through, for a pipe with radius $R$. +limiting the amount that can pass through, for a pipe with radius $$R$$. Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/) -of an incompressible fluid with spatially uniform density $\rho$. -Assuming that the flow is steady $\ipdv{\va{v}}{t} = 0$, -and that gravity is negligible $\va{g} = 0$, we get: +of an incompressible fluid with spatially uniform density $$\rho$$. +Assuming that the flow is steady $$\ipdv{\va{v}}{t} = 0$$, +and that gravity is negligible $$\va{g} = 0$$, we get: $$\begin{aligned} (\va{v} \cdot \nabla) \va{v} @@ -27,26 +27,26 @@ $$\begin{aligned} \nabla \cdot \va{v} = 0 \end{aligned}$$ -Into this, we insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$, -where $\vu{e}_z$ is the $z$-axis' unit vector. -In other words, we assume that the flow velocity depends only on $r$; -not on $\phi$ or $z$. +Into this, we insert the ansatz $$\va{v} = \vu{e}_z \: v_z(r)$$, +where $$\vu{e}_z$$ is the $$z$$-axis' unit vector. +In other words, we assume that the flow velocity depends only on $$r$$; +not on $$\phi$$ or $$z$$. Plugging this into the Navier-Stokes equations, -$\nabla \cdot \va{v}$ is trivially zero, -and in the other equation we multiply out $\rho$, yielding this, -where $\eta = \rho \nu$ is the dynamic viscosity: +$$\nabla \cdot \va{v}$$ is trivially zero, +and in the other equation we multiply out $$\rho$$, yielding this, +where $$\eta = \rho \nu$$ is the dynamic viscosity: $$\begin{aligned} \nabla p = \vu{e}_z \: \eta \nabla^2 v_z \end{aligned}$$ -Because only $\vu{e}_z$ appears on the right-hand side, -only the $z$-component of $\nabla p$ can be nonzero. -However, $v_z(r)$ is a function of $r$, not $z$! -The left thus only depends on $z$, and the right only on $r$, +Because only $$\vu{e}_z$$ appears on the right-hand side, +only the $$z$$-component of $$\nabla p$$ can be nonzero. +However, $$v_z(r)$$ is a function of $$r$$, not $$z$$! +The left thus only depends on $$z$$, and the right only on $$r$$, meaning that both sides must equal a constant, -which we call $-G$: +which we call $$-G$$: $$\begin{aligned} \dv{p}{z} @@ -56,22 +56,22 @@ $$\begin{aligned} = - G \end{aligned}$$ -The former equation, for $p(z)$, is easy to solve. -We get an integration constant $p(0)$: +The former equation, for $$p(z)$$, is easy to solve. +We get an integration constant $$p(0)$$: $$\begin{aligned} p(z) = p(0) - G z \end{aligned}$$ -This gives meaning to the **pressure gradient** $G$: -for a pipe of length $L$, -it describes the pressure difference $\Delta p = p(0) - p(L)$ +This gives meaning to the **pressure gradient** $$G$$: +for a pipe of length $$L$$, +it describes the pressure difference $$\Delta p = p(0) - p(L)$$ that is driving the fluid, -i.e. $G = \Delta p / L$ +i.e. $$G = \Delta p / L$$ -As for the latter equation, for $v_z(r)$, -we start by integrating it once, introducing a constant $A$: +As for the latter equation, for $$v_z(r)$$, +we start by integrating it once, introducing a constant $$A$$: $$\begin{aligned} \dv{}{r}\Big( r \dv{v_z}{r} \Big) @@ -82,7 +82,7 @@ $$\begin{aligned} \end{aligned}$$ Integrating this one more time, -thereby introducing another constant $B$, +thereby introducing another constant $$B$$, we arrive at: $$\begin{aligned} @@ -90,11 +90,11 @@ $$\begin{aligned} = - \frac{G}{4 \eta} r^2 + A \ln{r} + B \end{aligned}$$ -The velocity must be finite at $r = 0$, so we set $A = 0$. +The velocity must be finite at $$r = 0$$, so we set $$A = 0$$. Furthermore, the Navier-Stokes equation's *no-slip* condition -demands that $v_z = 0$ at the boundary $r = R$, -so $B = G R^2 / (4 \eta)$. -This brings us to the **Poiseuille solution** for $v_z(r)$: +demands that $$v_z = 0$$ at the boundary $$r = R$$, +so $$B = G R^2 / (4 \eta)$$. +This brings us to the **Poiseuille solution** for $$v_z(r)$$: $$\begin{aligned} \boxed{ @@ -104,8 +104,8 @@ $$\begin{aligned} \end{aligned}$$ How much fluid can pass through the pipe per unit time? -This is denoted by the **volumetric flow rate** $Q$, -which is the integral of $v_z$ over the circular cross-section: +This is denoted by the **volumetric flow rate** $$Q$$, +which is the integral of $$v_z$$ over the circular cross-section: $$\begin{aligned} Q @@ -115,7 +115,7 @@ $$\begin{aligned} \end{aligned}$$ We thus arrive at the main Hagen-Poiseuille equation, -which predicts $Q$ for a given setup: +which predicts $$Q$$ for a given setup: $$\begin{aligned} \boxed{ @@ -124,8 +124,8 @@ $$\begin{aligned} } \end{aligned}$$ -Consequently, the average flow velocity $\Expval{v_z}$ -is simply $Q$ divided by the cross-sectional area: +Consequently, the average flow velocity $$\Expval{v_z}$$ +is simply $$Q$$ divided by the cross-sectional area: $$\begin{aligned} \Expval{v_z} @@ -133,12 +133,12 @@ $$\begin{aligned} = \frac{G R^2}{8 \eta} \end{aligned}$$ -The fluid's viscous stickiness means it exerts a drag force $D$ -on the pipe as it flows. For a pipe of length $L$ and radius $R$, -we calculate $D$ by multiplying the internal area $2 \pi R L$ +The fluid's viscous stickiness means it exerts a drag force $$D$$ +on the pipe as it flows. For a pipe of length $$L$$ and radius $$R$$, +we calculate $$D$$ by multiplying the internal area $$2 \pi R L$$ by the [shear stress](/know/concept/cauchy-stress-tensor/) -$-\sigma_{zr}$ on the wall -(i.e. the wall applies $\sigma_{zr}$, the fluid responds with $- \sigma_{zr}$): +$$-\sigma_{zr}$$ on the wall +(i.e. the wall applies $$\sigma_{zr}$$, the fluid responds with $$- \sigma_{zr}$$): $$\begin{aligned} D @@ -148,8 +148,8 @@ $$\begin{aligned} = \pi R^2 L G \end{aligned}$$ -We would like to get rid of $G$ for being impractical, -so we substitute $R^2 G = 8 \eta \Expval{v_z}$, yielding: +We would like to get rid of $$G$$ for being impractical, +so we substitute $$R^2 G = 8 \eta \Expval{v_z}$$, yielding: $$\begin{aligned} \boxed{ @@ -158,8 +158,8 @@ $$\begin{aligned} } \end{aligned}$$ -Due to this drag, the pressure difference $\Delta p = p(0) - p(L)$ -does work on the fluid, at a rate $P$, +Due to this drag, the pressure difference $$\Delta p = p(0) - p(L)$$ +does work on the fluid, at a rate $$P$$, since power equals force (i.e. pressure times area) times velocity: $$\begin{aligned} @@ -167,10 +167,10 @@ $$\begin{aligned} = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r} \end{aligned}$$ -Because $\Delta p$ is independent of $r$, -we get the same integral we used to calculate $Q$. -Then, thanks to the fact that $\Delta p = G L$ -and $Q = \pi R^2 \Expval{v_z}$, it follows that: +Because $$\Delta p$$ is independent of $$r$$, +we get the same integral we used to calculate $$Q$$. +Then, thanks to the fact that $$\Delta p = G L$$ +and $$Q = \pi R^2 \Expval{v_z}$$, it follows that: $$\begin{aligned} P @@ -179,8 +179,8 @@ $$\begin{aligned} = D \Expval{v_z} \end{aligned}$$ -In conclusion, the power $P$, -needed to drive a fluid through the pipe at a rate $Q$, +In conclusion, the power $$P$$, +needed to drive a fluid through the pipe at a rate $$Q$$, is given by: $$\begin{aligned} |