User:PMarmottant/Hydraulic resistance and capacity

We present here simple tools to compute the flow in complex network of channels, just knowing the applied pressure.

Hydrodynamic resistance
We have seen in the previous chapter that flow rate $$ Q$$ in a channel is proportional to the applied pressure drop $$\Delta P$$. This can be summarized in
 * $$\Delta P=R_h Q,$$

with $$R_h$$ the hydrodynamic resistance. This expression is formally the analog of the electrokinetic law between voltage difference and current, $$U=R I$$.

The expression for the hydraulic resistance is:
 * channel of circular cross-section (total length $$L$$, radius $$R$$):
 * $$R_h=\frac{8\mu L}{\pi R^4}$$


 * rectangular cross-section (width $$w$$ and height $$h$$)
 * $$R_h\approx \frac{12 \mu L}{w h^3(1-0.630 h/w)}$$

In a network of channels, equivalent resistances can be computed (as in electrokinetics):
 * two channels in series have a resistance $$R_h=R_{h1}+R_{h2}$$,
 * two channels in parallel have a resistance $$1/R_h=1/R_{h1}+1/R_{h2}.$$

These laws provide useful tools for the design of complex networks. Actually Kirchhoff's laws for electric circuits apply, being modified in:
 * the sum of flow rates on a node of the circuit is zero
 * the sum of pressure differences on a loop is zero

Hydrodynamic capacitance
The volume of fluid in a channel can change just because of a change in pressure: this is due either to fluid compressibility or either channel elasticity. This behavior can be summarized with
 * $$Q=C_h \frac{d\Delta P}{dt}$$

with $$C_h$$ the hydrodynamic capacitance. It is the microfluidic analog of the electrokinetic law $$I=C\,dU/dt$$.

Compressible fluid in a container
A pressure increase can compress the fluid in a container. The compressibility is measured by
 * $$K_{fluid}=-\frac{1}{V}\frac{\partial V}{\partial P}.$$

For water its value is $$K=4.6\times 10^{-5}/bar$$ which is usually negligible since pressure are usually less than a bar. For air it is $$K=1/P=1/bar$$ which considerable if pressure attain a bar.

The flow rate entering a tube of volume $$V$$, because of fluid compression $$d\Delta P/dt$$ is:
 * $$Q=-\frac{dV}{dt}=V\left(-\frac{1}{V}\frac{\partial V}{\partial P}\right) \frac{d\Delta P}{dt}=K_{fluid} V \frac{d\Delta P}{dt}$$

The hydrodynamic capacitance is therefore:
 * $$C_h=K_{fluid} V$$

Elastic tubes
We define the tube dilatability as
 * $$K_{tube}=\frac{1}{V_{tube}}\frac{\partial V_{tube}}{\partial P}$$

It has a positive sign, since the tube volume increases with pressure.

The tube dilatability is approximately the inverse of the Young modulus $$K\simeq 1/E.$$ The following table gives order of magnitude of this dilatability for different materials This value can be interpreted in the following way: if the pressure is increased by 1 bar the relative volume increase is $$K_{tube}$$

Assuming a uniform pressure in the tube (which is not true in long tube where pressure decreases subtantially) , we find a flow rate entering the tube to inflate to be
 * $$Q=\frac{dV_{tube}}{dt}=K_{tube}V\frac{\Delta P}{dt}$$

The hydrodynamic capacitance is therefore
 * $$C_{h}=K_{tube}V.$$

Modelisation of an elastic long tube with a substantial pressure drop
We consider a tube of length $$L$$ on which a pressure difference $$\Delta P$$ is applied. In the tube, the pressure decreases along the tube coordinate $$x$$ as $$P(x)=P_0+(1-x/L)\Delta P$$. The dilatation is therefore not homogeneous: larger near the entrance. The volume increase of the tube (compared to the rest situation at pressure $$P_0$$) is
 * $$\Delta V=\int_0^L \left(1-\frac{x}{L}\right)\Delta P \; K_{tube} \frac{dx}{L} V_{tube}=\frac{\Delta P}{2}K_{tube} V.$$

We have integrated the inflation of small volumes $$dx/L\times V_{tube}$$.

We obtain that the flow due to dilatation is
 * $$Q=\frac{d\Delta V}{dt}=C_{tube}\frac{d\Delta P/2}{dt}$$

meaning that only half the pressure difference loads the volume capacitor. The capacitor is placed in the middle of the channel, where the overpressure is half, see figure.

Application: syringe injection in a microchannel
The syringe is has a tube diameter $$R$$ and a volume $$V$$, while the (cylindrical) microchannel has a diameter $$r\ll R$$ and a volume $$v\ll V$$, and a length $$l$$. The resistance of the microchannel is much larger than that of the syringe $$R_v=8\mu l/\pi r^4 \gg R_V$$. However the capacitance of the syringe is much larger $$C_V \gg C_v$$, because of the larger volume.

The equivalent circuit is therefore The total flow is distributed in the microchannel branch and the capacitor branch:
 * $$Q_{piston}=\frac{1}{R_v}\Delta P+C_V\frac{d\Delta P}{dt}$$

If the piston is suddenly started, initially water or tube elasticity will absorb the flow, and the flow is stationary only for time larger tha a characteristic transient time
 * $$\tau=R_v \,C_V=\frac{8\mu l}{\pi r^4} V K,$$

with $$K$$ the compressibility of either water or the syringe tube.

As an example, we take a microchannel of radius 10 micrometers, length 1 cm and a syringe of volume 1cc: the characteristic time is 10 seconds, if the syringe is rigid (glass) and $$K=K_{water}$$, while it takes up to 1000 seconds if the syringe is in plastic $$K=K_{plastic}$$!

As a conclusion, for practical realization of microfluidic networks:
 * avoid flexible tubes and prefer metallic tubes for a faster equilibration
 * avoid flexible glues in contact with the liquid: they will compress
 * avoid bubbles in the system, their compressibility is extremely high compared to plastic!
 * impose pressure with a valve, instead of piston velocity: the pressure equilibrates at the speed of sound in the liquid and changes in pressure are very rapidly applied to the whole system.