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Date: Thu, 3 Jun 2021 19:30:38 +0200
Subject: Expand knowledge base

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+---
+title: "Lubrication theory"
+firstLetter: "L"
+publishDate: 2021-06-03
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-05-11T14:34:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Lubrication theory
+
+**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.
+
+For simplicity, we limit ourselves to 2D
+by assuming that everything is constant along the $z$-axis.
+Consider a gap of width $d$ (along $y$) and length $L$ (along $x$),
+where $d \ll L$, containing the fluid.
+Outside the gap, the lubricant has a
+[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$.
+
+Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different.
+This is because advection will dominate along the $x$-axis (gap length),
+and viscosity along the $y$-axis (gap width).
+Therefore:
+
+$$\begin{aligned}
+    \mathrm{Re}_\mathrm{gap}
+    \approx \frac{|(\va{v} \cdot \nabla) \va{v}|}{|\nu \nabla^2 \va{v}|}
+    \approx \frac{U^2 / L}{\nu U / d^2}
+    \approx \frac{d^2}{L^2} \mathrm{Re}
+\end{aligned}$$
+
+If $d$ is small enough compared to $L$,
+then $\mathrm{Re}_\mathrm{gap} \ll 1$.
+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
+the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
+can be reduced to the following *Stokes equations*:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \: \Big( \pdv[2]{v_x}{x} + \pdv[2]{v_x}{y} \Big)
+    \qquad \quad
+    \pdv{p}{y}
+    = \eta \: \Big( \pdv[2]{v_y}{x} + \pdv[2]{v_y}{y} \Big)
+\end{aligned}$$
+
+Let the $y = 0$ plane be an infinite flat surface,
+sliding in the positive $x$-direction at a constant velocity $U$.
+On the other side of the gap,
+an arbitrary surface is described by $h(x)$.
+
+Since the gap is so narrow,
+and the surfaces' movements cause large shear stresses inside,
+$v_y$ is negligible compared to $v_x$.
+Furthermore, because the gap is so long,
+we assume that $\pdv*{v_x}{x}$ is negligible compared to $\pdv*{v_x}{y}$.
+This reduces the Stokes equations to:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \pdv[2]{v_x}{y}
+    \qquad \quad
+    \pdv{p}{y}
+    = 0
+\end{aligned}$$
+
+This result could also be derived from the Prandtl equations.
+In any case, it tells us that $p$ only depends on $x$,
+allowing us to integrate the former equation:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 + C_1 y + C_2
+\end{aligned}$$
+
+Where $C_1$ and $C_2$ are integration constants.
+At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$.
+Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U
+\end{aligned}$$
+
+The moving bottom surface drags fluid in the $x$-direction
+at a volumetric rate $Q$, given by:
+
+$$\begin{aligned}
+    Q
+    = \int_0^{h(x)} v_x(x, y) \dd{y}
+    = \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h}
+    = - \frac{p'}{12 \eta} h^3 + \frac{U}{2} h
+\end{aligned}$$
+
+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 $x$,
+which allows us to write $p'(x)$ in terms of
+measurable constants and the known function $h(x)$:
+
+$$\begin{aligned}
+    \boxed{
+        p'
+        = 6 \eta \: \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big)
+    }
+\end{aligned}$$
+
+Then we insert this into our earlier expression for $v_x$, yielding:
+
+$$\begin{aligned}
+    v_x
+    &= 3 y (y - h) \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) - \frac{U h}{h^2} y + \frac{U h^2}{h^2}
+    %\\
+    %&= U \frac{3 y (y - h) - h y + h^2}{h^2} - Q \frac{6 y (y - h)}{h^3}
+\end{aligned}$$
+
+Which, after some rearranging, can be written in the following form:
+
+$$\begin{aligned}
+    \boxed{
+        v_x
+        = U \frac{(3 y - h) (y - h)}{h^2} - Q \frac{6 y (y - h)}{h^3}
+    }
+\end{aligned}$$
+
+With this, we can find $v_y$ by exploiting incompressibility,
+i.e. the continuity equation states:
+
+$$\begin{aligned}
+    \pdv{v_y}{y}
+    = - \pdv{v_x}{x}
+    = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
+\end{aligned}$$
+
+Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$:
+
+$$\begin{aligned}
+    \boxed{
+        v_y
+        = - 2 h' \frac{U h - 3 Q}{h^4} y^2 (h - y)
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    0
+    = \int_L p'(x) \dd{x}
+    = 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x}
+\end{aligned}$$
+
+Isolating this for $Q$, and defining $q$ as below, yields a simple equation:
+
+$$\begin{aligned}
+    Q
+    = \frac{1}{2} U q
+    \qquad \quad
+    q
+    \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
+\end{aligned}$$
+
+We substitute this into $v_x$ and rearrange to get an interesting expression:
+
+$$\begin{aligned}
+    v_x
+    &= U \frac{3 y^2 - h y - 3 h y + h^2}{h^2} - U q \frac{3 y^2 - 3 h y}{h^3}
+    \\
+    &= U \Big( 1 - \frac{y}{h} \Big) \Big( 1 - \frac{3 y (h - q)}{h^2} \Big)
+\end{aligned}$$
+
+The first factor is always positive,
+but the second can be negative,
+if for some $y$-values:
+
+$$\begin{aligned}
+    h^2 < 3 y (h - q)
+    \quad \implies \quad
+    y > \frac{h^2}{3 (h - q)}
+\end{aligned}$$
+
+Since $h > y$, such $y$-values will only exist
+if $h$ is larger than some threshold:
+
+$$\begin{aligned}
+    3 (h - q) > h
+    \quad \implies \quad
+    h > \frac{3}{2} q
+\end{aligned}$$
+
+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 $v_y$:
+
+$$\begin{aligned}
+    v_y
+    = - U h' \frac{2 h - 3 q}{h^4} y^2 (h - y)
+\end{aligned}$$
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
-- 
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