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Diffstat (limited to 'source/know/concept/lubrication-theory')
| -rw-r--r-- | source/know/concept/lubrication-theory/index.md | 70 |
1 files changed, 35 insertions, 35 deletions
diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md index d2a615e..4015526 100644 --- a/source/know/concept/lubrication-theory/index.md +++ b/source/know/concept/lubrication-theory/index.md @@ -16,15 +16,15 @@ 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. +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/) $\mathrm{Re} \approx U L / \nu$. +[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). +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: $$\begin{aligned} @@ -34,13 +34,13 @@ $$\begin{aligned} \approx \frac{d^2}{L^2} \mathrm{Re} \end{aligned}$$ -If $d$ is small enough compared to $L$, -then $\mathrm{Re}_\mathrm{gap} \ll 1$. -More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$, +If $$d$$ is small enough compared to $$L$$, +then $$\mathrm{Re}_\mathrm{gap} \ll 1$$. +More formally, we need $$d \ll L / \sqrt{\mathrm{Re}}$$, so we are inside the boundary layer, in the realm of the [Prandtl equations](/know/concept/prandtl-equations/). -Let $\mathrm{Re}_\mathrm{gap} \ll 1$. +Let $$\mathrm{Re}_\mathrm{gap} \ll 1$$. We are thus dealing with *Stokes flow*, in which case the [Navier-Stokes equations](/know/concept/navier-stokes/equations/) can be reduced to the following *Stokes equations*: @@ -53,16 +53,16 @@ $$\begin{aligned} = \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big) \end{aligned}$$ -Let the $y = 0$ plane be an infinite flat surface, -sliding in the positive $x$-direction at a constant velocity $U$. +Let the $$y = 0$$ plane be an infinite flat surface, +sliding in the positive $$x$$-direction at a constant velocity $$U$$. On the other side of the gap, -an arbitrary surface is described by $h(x)$. +an arbitrary surface is described by $$h(x)$$. Since the gap is so narrow, and the surfaces' movements cause large shear stresses inside, -$v_y$ is negligible compared to $v_x$. +$$v_y$$ is negligible compared to $$v_x$$. Furthermore, because the gap is so long, -we assume that $\ipdv{v_x}{x}$ is negligible compared to $\ipdv{v_x}{y}$. +we assume that $$\ipdv{v_x}{x}$$ is negligible compared to $$\ipdv{v_x}{y}$$. This reduces the Stokes equations to: $$\begin{aligned} @@ -74,7 +74,7 @@ $$\begin{aligned} \end{aligned}$$ This result could also be derived from the Prandtl equations. -In any case, it tells us that $p$ only depends on $x$, +In any case, it tells us that $$p$$ only depends on $$x$$, allowing us to integrate the former equation: $$\begin{aligned} @@ -82,17 +82,17 @@ $$\begin{aligned} = \frac{p'}{2 \eta} y^2 + C_1 y + C_2 \end{aligned}$$ -Where $C_1$ and $C_2$ are integration constants. -At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$. -Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to: +Where $$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}$$ -The moving bottom surface drags fluid in the $x$-direction -at a volumetric rate $Q$, given by: +The moving bottom surface drags fluid in the $$x$$-direction +at a volumetric rate $$Q$$, given by: $$\begin{aligned} Q @@ -103,9 +103,9 @@ $$\begin{aligned} Assuming that the lubricant is incompressible, meaning that the same volume of fluid must be leaving a point as is entering it. -In other words, $Q$ is independent of $x$, -which allows us to write $p'(x)$ in terms of -measurable constants and the known function $h(x)$: +In other words, $$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{ @@ -114,7 +114,7 @@ $$\begin{aligned} } \end{aligned}$$ -Then we insert this into our earlier expression for $v_x$, yielding: +Then we insert this into our earlier expression for $$v_x$$, yielding: $$\begin{aligned} v_x @@ -130,7 +130,7 @@ $$\begin{aligned} } \end{aligned}$$ -With this, we can find $v_y$ by exploiting incompressibility, +With this, we can find $$v_y$$ by exploiting incompressibility, i.e. the continuity equation states: $$\begin{aligned} @@ -139,7 +139,7 @@ $$\begin{aligned} = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big) \end{aligned}$$ -Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$: +Integrating with respect to $$y$$ thus leads to the following transverse velocity $$v_y$$: $$\begin{aligned} \boxed{ @@ -150,7 +150,7 @@ $$\begin{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, +Although the pressure gradient $$p'$$ need not be zero, we therefore expect that its integral vanishes: $$\begin{aligned} @@ -159,7 +159,7 @@ $$\begin{aligned} = 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}$$ -Isolating this for $Q$, and defining $q$ as below, yields a simple equation: +Isolating this for $$Q$$, and defining $$q$$ as below, yields a simple equation: $$\begin{aligned} Q @@ -169,7 +169,7 @@ $$\begin{aligned} \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}} \end{aligned}$$ -We substitute this into $v_x$ and rearrange to get an interesting expression: +We substitute this into $$v_x$$ and rearrange to get an interesting expression: $$\begin{aligned} v_x @@ -180,7 +180,7 @@ $$\begin{aligned} The first factor is always positive, but the second can be negative, -if for some $y$-values: +if for some $$y$$-values: $$\begin{aligned} h^2 < 3 y (h - q) @@ -188,8 +188,8 @@ $$\begin{aligned} y > \frac{h^2}{3 (h - q)} \end{aligned}$$ -Since $h > y$, such $y$-values will only exist -if $h$ is larger than some threshold: +Since $$h > y$$, such $$y$$-values will only exist +if $$h$$ is larger than some threshold: $$\begin{aligned} 3 (h - q) > h @@ -201,7 +201,7 @@ 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 $v_y$: +This is confirmed by $$v_y$$: $$\begin{aligned} v_y |