Improve knowledge base

1 files changed, 34 insertions, 33 deletions

diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md
index 4015526..54200d3 100644
--- a/source/know/concept/lubrication-theory/index.md
+++ b/source/know/concept/lubrication-theory/index.md
@@ -9,8 +9,7 @@ categories:
 layout: "concept"
 ---
 
-**Lubricants** are widely used
-to reduce friction between two moving surfaces.
+**Lubricants** are widely used to reduce friction between two moving surfaces.
 In fluid mechanics, **lubrication theory**
 is the study of fluids that are tightly constrained in one dimension,
 especially those in small gaps between moving surfaces.
@@ -34,32 +33,32 @@ $$\begin{aligned}
     \approx \frac{d^2}{L^2} \mathrm{Re}
 \end{aligned}$$
 
-If $$d$$ is small enough compared to $$L$$,
-then $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
+If $$d$$ is small enough compared to $$L$$, then $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
 More formally, we need $$d \ll L / \sqrt{\mathrm{Re}}$$,
 so we are inside the boundary layer,
 in the realm of the [Prandtl equations](/know/concept/prandtl-equations/).
 
 Let $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
-We are thus dealing with *Stokes flow*, in which case
+We are then dealing with *Stokes flow*, in which case
 the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
 can be reduced to the following *Stokes equations*:
 
 $$\begin{aligned}
     \pdv{p}{x}
     = \eta \: \Big( \pdvn{2}{v_x}{x} + \pdvn{2}{v_x}{y} \Big)
-    \qquad \quad
+    \qquad \qquad
     \pdv{p}{y}
     = \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big)
 \end{aligned}$$
 
-Let the $$y = 0$$ plane be an infinite flat surface,
+Let the $$y = 0$$ plane be an infinite flat surface
+(a good approximation because $$d \ll L$$),
 sliding in the positive $$x$$-direction at a constant velocity $$U$$.
-On the other side of the gap,
-an arbitrary surface is described by $$h(x)$$.
+On the other side of the gap, an arbitrary surface
+is described by a height function $$h(x)$$.
 
 Since the gap is so narrow,
-and the surfaces' movements cause large shear stresses inside,
+and the surfaces' movements cause large shear stresses inside it,
 $$v_y$$ is negligible compared to $$v_x$$.
 Furthermore, because the gap is so long,
 we assume that $$\ipdv{v_x}{x}$$ is negligible compared to $$\ipdv{v_x}{y}$$.
@@ -68,7 +67,7 @@ This reduces the Stokes equations to:
 $$\begin{aligned}
     \pdv{p}{x}
     = \eta \pdvn{2}{v_x}{y}
-    \qquad \quad
+    \qquad \qquad
     \pdv{p}{y}
     = 0
 \end{aligned}$$
@@ -92,18 +91,20 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The moving bottom surface drags fluid in the $$x$$-direction
-at a volumetric rate $$Q$$, given by:
+at a volumetric rate $$Q(x)$$, given by:
 
 $$\begin{aligned}
-    Q
-    = \int_0^{h(x)} v_x(x, y) \dd{y}
-    = \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h}
-    = - \frac{p'}{12 \eta} h^3 + \frac{U}{2} h
+    Q(x)
+    &= \int_0^{h(x)} v_x(x, y) \dd{y}
+    \\
+    &= \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h}
+    \\
+    &= - \frac{p'}{12 \eta} h^3(x) + \frac{U}{2} h(x)
 \end{aligned}$$
 
-Assuming that the lubricant is incompressible,
-meaning that the same volume of fluid must be leaving a point as is entering it.
-In other words, $$Q$$ is independent of $$x$$,
+Let us assume that the lubricant is incompressible,
+meaning that the same volume of fluid must be both leaving and entering the gap.
+In that case, $$Q$$ is independent of $$x$$,
 which allows us to write $$p'(x)$$ in terms of
 measurable constants and the known function $$h(x)$$:
 
@@ -139,7 +140,8 @@ $$\begin{aligned}
     = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
 \end{aligned}$$
 
-Integrating with respect to $$y$$ thus leads to the following transverse velocity $$v_y$$:
+Integrating with respect to $$y$$ therefore leads to
+the following transverse velocity $$v_y$$:
 
 $$\begin{aligned}
     \boxed{
@@ -148,10 +150,11 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Typically, the lubricant is not in a preexisting pressure differential,
-i.e it is not getting pumped through the system.
-Although the pressure gradient $$p'$$ need not be zero,
-we therefore expect that its integral vanishes:
+Usually, the lubricant is not getting pumped through the system.
+In that case, although the pressure gradient $$p'$$ need not be zero in all points
+(i.e. there may be complex dynamics inside the gap),
+we do expect that its integral across the gap vanishes
+(because both sides are at the same pressure):
 
 $$\begin{aligned}
     0
@@ -164,7 +167,7 @@ Isolating this for $$Q$$, and defining $$q$$ as below, yields a simple equation:
 $$\begin{aligned}
     Q
     = \frac{1}{2} U q
-    \qquad \quad
+    \qquad \qquad
     q
     \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
 \end{aligned}$$
@@ -178,27 +181,25 @@ $$\begin{aligned}
     &= U \Big( 1 - \frac{y}{h} \Big) \Big( 1 - \frac{3 y (h - q)}{h^2} \Big)
 \end{aligned}$$
 
-The first factor is always positive,
-but the second can be negative,
-if for some $$y$$-values:
+The first factors are always positive, but the last one can be negative,
+if any $$y$$-values satisfy:
 
 $$\begin{aligned}
     h^2 < 3 y (h - q)
-    \quad \implies \quad
+    \qquad \implies \qquad
     y > \frac{h^2}{3 (h - q)}
 \end{aligned}$$
 
-Since $$h > y$$, such $$y$$-values will only exist
+Since $$h \le y$$, such $$y$$-values will only exist
 if $$h$$ is larger than some threshold:
 
 $$\begin{aligned}
     3 (h - q) > h
-    \quad \implies \quad
+    \qquad \implies \qquad
     h > \frac{3}{2} q
 \end{aligned}$$
 
-If this condition is satisfied,
-there will be some flow reversal:
+If this condition is satisfied, there will be some flow reversal:
 rather than just getting dragged by the shearing motion,
 the lubricant instead "rolls" inside the gap.
 This is confirmed by $$v_y$$: