Microfluidics/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 ΔP. This can be summarized in

ΔP = R_hQ,

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

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, expression valid when h < w)

$R_h\approx \frac{12 \mu L}{w h^3(1-0.630 h/w)}$

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 either due to fluid compressibility or 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 dΔP / 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_fluidV

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

	Ktube
steel	$\scriptstyle 0.5\times 10^{-6}$/bar
plastic	0.01/bar
rubber	0.1/bar

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_tubeV.

Modelling of an elastic long tube with a substantial pressure drop

We consider a tube of length L on which a pressure difference ΔP is applied. In the tube, the pressure decreases along the tube coordinate x as P(x) = P₀ + (1 − x / L)ΔP. The dilatation is therefore not homogeneous: larger near the entrance. The volume increase of the tube (compared to the rest situation at pressure P₀) 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 $dx/L\times V_{tube}$.

Equivalent circuit to a long elastic 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 may be considered as a tube of diameter R and volume V, while the cylindrical microchannel has a diameter $r\ll R$ , a volume $v\ll V$, and a length l. The resistance of the microchannel R_v = 8μl / πr⁴ is much larger than that of the syringe: $R_v \gg R_V$. However the capacitance of the syringe is usually much larger: $C_V \gg C_v$, due to a large surface area of walls (if the walls are elastic) and due to much larger volume of the syringe (if the liquid is compressible).

The equivalent circuit is shown in the picture.

Equivalent circuit to a long elastic tube

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 approaches its stationary state assymptotically, with the characteristic time given as follows:

$\tau=R_v \,C_V=\frac{8\mu l}{\pi r^4} V K,$

where K is the effective compressibility (both from the compressibility of liquid and from the elasticity of the container).

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 made from plastic K = K_plastic !

As a conclusion, for practical realization of microfluidic networks:

• avoid elastic tubes and prefer metallic tubes for a faster equilibration
• avoid elastic glues in contact with the liquid: they will compress
• avoid bubbles in the system, the compressibility of any gas 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.