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diff --git a/content/know/concept/hagen-poiseuille-equation/index.pdc b/content/know/concept/hagen-poiseuille-equation/index.pdc new file mode 100644 index 0000000..b33e085 --- /dev/null +++ b/content/know/concept/hagen-poiseuille-equation/index.pdc @@ -0,0 +1,203 @@ +--- +title: "Hagen-Poiseuille equation" +firstLetter: "H" +publishDate: 2021-04-13 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-04-13T10:42:46+02:00 +draft: false +markup: pandoc +--- + +# Hagen-Poiseuille equation + +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 $R$. + +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 $\pdv*{\va{v}}{t} = 0$, +and that gravity is negligible $\va{g} = 0$, we get: + +$$\begin{aligned} + (\va{v} \cdot \nabla) \va{v} + = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} + \qquad \quad + \nabla \cdot \va{v} = 0 +\end{aligned}$$ + +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 $z$. +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: + +$$\begin{aligned} + \nabla p + = \vu{e}_z \: \eta \nabla^2 v_z +\end{aligned}$$ + +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 $r$, +meaning that both sides must equal a constant, +which we call $-G$: + +$$\begin{aligned} + \dv{p}{z} + = -G + \qquad \quad + \eta \frac{1}{r} \dv{r} \Big( r \dv{v_z}{r} \Big) + = - G +\end{aligned}$$ + +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 $\Delta p = p(0) - p(L)$ +that is driving the fluid, +i.e. $G = \Delta p / L$ + +As for the latter equation, for $v_z(r)$, +we start by integrating it once, introducing a constant $A$: + +$$\begin{aligned} + \dv{r} \Big( r \dv{v_z}{r} \Big) + = - \frac{G}{\eta} r + \quad \implies \quad + \dv{v_z}{r} + = - \frac{G}{2 \eta} r + \frac{A}{r} +\end{aligned}$$ + +Integrating this one more time, +thereby introducing another constant $B$, +we arrive at: + +$$\begin{aligned} + v_z + = - \frac{G}{4 \eta} r^2 + A \ln{r} + B +\end{aligned}$$ + +The velocity must be finite at $r = 0$, so we set $A = 0$. +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 $v_z(r)$: + +$$\begin{aligned} + \boxed{ + v_z(r) + = \frac{G}{4 \eta} (R^2 - r^2) + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + Q + = 2 \pi \int_0^R v_z(r) \: r \dd{r} + = \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r} + = \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R +\end{aligned}$$ + +We thus arrive at the main Hagen-Poiseuille equation, +which predicts $Q$ for a given setup: + +$$\begin{aligned} + \boxed{ + Q + = \frac{\pi G R^4}{8 \eta} + } +\end{aligned}$$ + +Consequently, the average flow velocity $\expval{v_z}$ +is simply $Q$ divided by the cross-sectional area: + +$$\begin{aligned} + \expval{v_z} + = \frac{Q}{\pi R^2} + = \frac{G R^2}{8 \eta} +\end{aligned}$$ + +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 $2 \pi R L$ +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 $- \sigma_{zr}$): + +$$\begin{aligned} + D + = - 2 \pi R L \: \sigma_{zr} \big|_{r = R} + = - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R} + = 2 \pi R L \eta \frac{G R}{2 \eta} + = \pi R^2 L G +\end{aligned}$$ + +We would like to get rid of $G$ for being impractical, +so we substitute $R^2 G = 8 \eta \expval{v_z}$, yielding: + +$$\begin{aligned} + \boxed{ + D + = 8 \pi \eta L \expval{v_z} + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + P + = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r} +\end{aligned}$$ + +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: + +$$\begin{aligned} + P + = \Delta p \: Q + = G L \pi R^2 \expval{v_z} + = D \expval{v_z} +\end{aligned}$$ + +In conclusion, the power $P$, +needed to drive a fluid through the pipe at a rate $Q$, +is given by: + +$$\begin{aligned} + \boxed{ + P + = 8 \pi \eta L \expval{v_z}^2 + } +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |