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+---
+title: "Hagen-Poiseuille equation"
+date: 2021-04-13
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+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 $\ipdv{\va{v}}{t} = 0$,
+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.