Improve knowledge base

1 files changed, 21 insertions, 24 deletions

diff --git a/source/know/concept/hagen-poiseuille-equation/index.md b/source/know/concept/hagen-poiseuille-equation/index.md
index 6484631..52d3ce8 100644
--- a/source/know/concept/hagen-poiseuille-equation/index.md
+++ b/source/know/concept/hagen-poiseuille-equation/index.md
@@ -11,9 +11,8 @@ 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 $$R$$.
+through a cylindrical pipe: the fluid clings to the sides,
+limiting the amount that can pass through per unit time.
 
 Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
 of an incompressible fluid with spatially uniform density $$\rho$$.
@@ -27,13 +26,12 @@ $$\begin{aligned}
     \nabla \cdot \va{v} = 0
 \end{aligned}$$
 
-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 $$z$$.
-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,
+Let the pipe have radius $$R$$, and be infinitely long and parallel to the $$z$$-axis.
+We insert the ansatz $$\va{v} = \vu{e}_z \: v_z(r)$$,
+where $$\vu{e}_z$$ is the $$z$$-axis' unit vector,
+and we are assuming that the flow depends only on $$r$$, not on $$\phi$$ or $$z$$.
+With this, $$\nabla \cdot \va{v}$$ trivially vanishes,
+and in the main equation multiplying out $$\rho$$ yields this,
 where $$\eta = \rho \nu$$ is the dynamic viscosity:
 
 $$\begin{aligned}
@@ -56,7 +54,7 @@ $$\begin{aligned}
     = - G
 \end{aligned}$$
 
-The former equation, for $$p(z)$$, is easy to solve.
+The former equation for $$p(z)$$ is easy to solve.
 We get an integration constant $$p(0)$$:
 
 $$\begin{aligned}
@@ -64,13 +62,12 @@ $$\begin{aligned}
     = 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. $$G = \Delta p / L$$
+This gives meaning to $$G$$: it is the **pressure gradient**,
+which for a pipe of length $$L$$
+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 $$v_z(r)$$,
+As for the latter equation for $$v_z(r)$$,
 we start by integrating it once, introducing a constant $$A$$:
 
 $$\begin{aligned}
@@ -148,8 +145,8 @@ $$\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:
+$$G$$ is an inconvenient quantity here, so we remove it
+by substituting $$R^2 G = 8 \eta \Expval{v_z}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -159,8 +156,8 @@ $$\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 $$P$$,
-since power equals force (i.e. pressure times area) times velocity:
+does work on the fluid at a rate $$P$$.
+Since power equals force (i.e. pressure times area) times velocity:
 
 $$\begin{aligned}
     P
@@ -179,14 +176,14 @@ $$\begin{aligned}
     = D \Expval{v_z}
 \end{aligned}$$
 
-In conclusion, the power $$P$$,
-needed to drive a fluid through the pipe at a rate $$Q$$,
-is given by:
+In conclusion, the power $$P$$ needed to drive a fluid
+through the pipe at a rate $$Q$$ is given by:
 
 $$\begin{aligned}
     \boxed{
         P
         = 8 \pi \eta L \Expval{v_z}^2
+        = \frac{8 \eta L}{\pi R^4} Q^2
     }
 \end{aligned}$$