Microfluidics/Hydraulic resistance and capacity

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

Contents

[edit] 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 = RhQ,

with Rh 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)
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 Rh = Rh1 + Rh2,
  • two channels in parallel have a resistance 1 / Rh = 1 / Rh1 + 1 / Rh2.

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

[edit] 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 Ch the hydrodynamic capacitance. It is the microfluidic analog of the electrokinetic law I=C\,dU/dt.

[edit] 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:

Ch = KfluidV

[edit] 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 Ktube

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

Ch = KtubeV.

[edit] Modelisation 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.


[edit] 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

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 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 = Kwater, while it takes up to 1000 seconds if the syringe is in plastic K = Kplastic!

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.
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