User:PMarmottant/Flow in microchannels

Parallel flows: Poiseuille flows
For parallel flows along axis $$x$$ the velocity writes: $$\mathbf{u}=u_x(y,z) \mathbf{e}_x$$.

The Stokes equation in the permanent regime therefore write, when projected along axis $$x$$
 * $$\mu (\frac{\partial^2 u_x}{\partial y^2}(y,z)+\frac{\partial^2 u_x}{\partial z^2}(y,z))=\frac{\partial p}{\partial x}$$

and along axis $$y$$ and $$z.$$
 * $$0=\frac{\partial p}{\partial y},$$
 * $$0=\frac{\partial p}{\partial z}.$$

These last two equations implie that pressure is a function of $$x$$ only, $$p=p(x)$$. Therefore the right-hand term on the equation along $$x$$ is only a function of $$x$$, and the left hand term is only a function of $$y,z$$, both are equal to a constant.
 * $$\frac{\partial p}{\partial x}=\mathrm{cst}.$$

Circular tube


We consider a tube of radius $$R$$. The boundary condition is $$u_x=0$$ on the tube surface of equation $$1-\frac{y^2}{R^2}+\frac{z^2}{R^2}=0$$. The trial function $$u_x(y,z)=u_0(1-\frac{y^2}{R^2}+\frac{z^2}{R^2})$$ satisfies automatically the boundary condition, and is a solution of Stokes equation, with a constant $$u_0$$ such that the:
 * $$u_x=-\frac{R^2}{4\mu}\frac{\partial p}{\partial x}\left(1-\frac{y^2}{R^2}+\frac{z^2}{R^2}\right)=-\frac{R^2}{4\mu}\frac{\partial p}{\partial x}\left(1-\frac{r^2}{R^2}\right)$$,

with $$r$$ the distance to the tube axis. The flow rate is the integral of the velocity on a cross section $$Q=\int u_x(y,z)dydz$$, and here we obtain:
 * $$Q=-\frac{\pi R^4}{8\mu}\frac{\partial p}{\partial x}.$$

If the pressure drops by $$\Delta P$$ along a tube of length $$L$$, the pressure gradient is $$\partial p/\partial x=-\Delta P/L$$. There is a huge dependance of the flow rate as a function of the tube radius! Reducing by a factor 10 the tube diameter, reduces by a factor 10000 the flow rate for a given pressure gradient.

Rectangular cross-section


The boundary conditions are
 * $$u_x(y=\pm w/2)=0$$
 * $$u_x(y=\pm h/2)=0$$

An expansion in Fourier series along $$y$$ and $$z$$ provides the solution (not derived here) as an infinite sum:
 * $$u_x(y,z)=-\frac{4}{\mu w}\frac{\partial p}{\partial x}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\beta_n^3}\left[ 1-\frac{cosh(\beta_n z)}{cosh(\beta_n z/2)} \right] cosh(\beta_n y), $$

with $$\beta_n=(2n-1)\pi/w$$. The flow rate, following from the integration of the previous flow field along the cross section provides:
 * $$Q=-\frac{8h}{\mu w}\frac{\partial p}{\partial x} \sum_{n=1}^\infty \frac{1}{\beta_n^4}(1-\frac{2}{\beta_n h} tanh(\beta_n \frac{w}{2}) ). $$

Even if this series converges rapidly (it is in $$1/n^5$$ and is calculated with a few terms, an approximate relation is much useful (when $$h< w$$):
 * $$Q\approx \frac{wh^3}{12\mu}\frac{\partial p}{\partial x} \left(1-0.630 \frac{h}{w}\right).$$

It is accurate at 10% when $$w=h$$, and at 0.2% when $$w=h/2$$!

Drag force on a sphere
Assuming a sphere of radius $$R$$ fixed in a flow of velocity $$U$$, it is possible to show that the drag force is exterted along the axis of the flow with strength
 * $$F_{drag}=6\pi \mu R U.$$

Note the linearity of the formula in velocity. This formula is valid for small Reynolds numbers.

This formula is much different at large Reynolds numbers: $$F_{drag}=C_x \pi R^2 \frac{1}{2}\rho U^2, $$ with $$C_x$$ a drag coefficient and $$\pi R^2$$ the apparent surface, while $$\frac{1}{2}\rho U^2$$ is the inertial pressure.

Exercise: compute the free fall velocity of small sphere in liquid.