Taylor scraping flow

In fluid dynamics, Taylor scraping flow is a type of two-dimensional corner flow when one of the wall is sliding over the other with constant velocity, named after G. I. Taylor.

Flow description

Consider a plane wall located at $\theta =0$ in the cylindrical coordinates $(r,\theta )$, moving with a constant velocity $U$ towards the left. Consider an another plane wall(scraper) , at an inclined position, making an angle $\alpha$ from the positive $x$ direction and let the point of intersection be at $r=0$. This description is equivalent to moving the scraper towards right with velocity $U$. It should be noted that the problem is singular at $r=0$ because at the origin, the velocities are discontinuous, thus the velocity gradient is infinite there.

Taylor noticed that the inertial terms are negligible as long as the region of interest is within $r<<\nu /U$, thus within the region the flow is essentially a Stokes flow. George Batchelor gives a typical value for lubricating oil with velocity $U=10{\text{cm/s}}$ as $r<<0.4{\text{cm}}$. Then for two-dimensional planar problem, the equation is

$\nabla ^{4}\psi =0,\quad u_{r}={\frac {1}{r}}{\frac {\partial \psi }{\partial \theta }},\quad u_{r}=-{\frac {\partial \psi }{\partial r}}$

where $\mathbf {v} =(u_{r},u_{\theta })$ is the velocity field and $\psi$ is the stream function. The boundary conditions are

${\begin{aligned}r>0,\ \theta =0:&\quad u_{r}=-U,\ u_{\theta }=0\\r>0,\ \theta =\alpha :&\quad u_{r}=0,\ u_{\theta }=0\end{aligned}}$

Solution

Attempting a separable solution of the form $\psi =Urf(\theta )$ reduces the problem to

$f^{iv}+2f''+f=0$

with boundary conditions

$f(0)=0,\ f'(0)=-1,\ f(\alpha )=0,\ f'(\alpha )=0$

The solution is

$f(\theta )={\frac {1}{\alpha ^{2}-\sin ^{2}\alpha }}[\theta \sin \alpha \sin(\alpha -\theta )-\alpha (\alpha -\theta )\sin \theta ]$

Therefore the velocity field is

${\begin{aligned}u_{r}&={\frac {U}{\alpha ^{2}-\sin ^{2}\alpha }}\{\sin \alpha [\sin(\alpha -\theta )-\theta \cos(\alpha -\theta )]+\alpha [\sin \theta -(\alpha -\theta )\cos \theta ]\}\\u_{\theta }&=-{\frac {U}{\alpha ^{2}-\sin ^{2}\alpha }}[\theta \sin \alpha \sin(\alpha -\theta )-\alpha (\alpha -\theta )\sin \theta ]\end{aligned}}$

Pressure can be obtained through integration of the momentum equation

$\nabla p=\mu \nabla ^{2}\mathbf {v} ,\quad p(r,\infty )=p_{\infty }$

which gives,

$p(r,\theta )-p_{\infty }={\frac {2\mu U}{r}}{\frac {\alpha \sin \theta +\sin \alpha \sin(\alpha -\theta )}{\alpha ^{2}-\sin ^{2}\alpha }}$

Forces on the scraper

The tangential force and the normal force on the scraper due to pressure forces and viscous forces are

$F_{t}={\frac {2\mu U}{r}}{\frac {\sin \alpha -\alpha \cos \alpha }{\alpha ^{2}-\sin ^{2}\alpha }},\quad F_{n}={\frac {2\mu U}{r}}{\frac {\alpha \sin \alpha }{\alpha ^{2}-\sin ^{2}\alpha }}$

The same scraper force if resolved according to Cartesian coordinates (parallel and perpendicular to the lower plate i.e. $F_{x}=F_{t}\sin \alpha +F_{n}\cos \alpha ,\ F_{y}=-F_{t}\cos \alpha +F_{n}\sin \alpha$) are

$F_{x}={\frac {2\mu U}{r}}{\frac {\alpha -\sin \alpha \cos \alpha }{\alpha ^{2}-\sin ^{2}\alpha }},\quad F_{y}={\frac {2\mu U}{r}}{\frac {\sin ^{2}\alpha }{\alpha ^{2}-\sin ^{2}\alpha }}$

As noted earlier, all the forces become infinite at $r=0$, because the velocity gradient is infinite there. In real life, there will be a huge pressure at the point of point, which depends on the geometry of the contact.