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

Coordinates on a 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

Channel of 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

$F_{drag}=6\pi \mu RU.$

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.