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diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md new file mode 100644 index 0000000..04acd30 --- /dev/null +++ b/source/know/concept/lubrication-theory/index.md @@ -0,0 +1,215 @@ +--- +title: "Lubrication theory" +date: 2021-06-03 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics +layout: "concept" +--- + +**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*: + +$$\begin{aligned} + \pdv{p}{x} + = \eta \: \Big( \pdvn{2}{v_x}{x} + \pdvn{2}{v_x}{y} \Big) + \qquad \quad + \pdv{p}{y} + = \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big) +\end{aligned}$$ + +Let the $y = 0$ plane be an infinite flat surface, +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 $\ipdv{v_x}{x}$ is negligible compared to $\ipdv{v_x}{y}$. +This reduces the Stokes equations to: + +$$\begin{aligned} + \pdv{p}{x} + = \eta \pdvn{2}{v_x}{y} + \qquad \quad + \pdv{p}{y} + = 0 +\end{aligned}$$ + +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: + +$$\begin{aligned} + v_x + &= 3 y (y - h) \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) - \frac{U h}{h^2} y + \frac{U h^2}{h^2} +\end{aligned}$$ + +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. |