Categories: Fluid dynamics, Fluid mechanics, Physics.

# 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 $\mathrm{Re} \approx U L / \nu$.

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.

Let $\mathrm{Re}_\mathrm{gap} \ll 1$.
We are thus dealing with *Stokes flow*, in which case
the Navier-Stokes equations
can be reduced to the following *Stokes equations*:

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:

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

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.