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-rw-r--r--source/know/concept/hagen-poiseuille-equation/index.md98
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}