Encoding and controlling of two droplet trains in a microfluidic network with the loop-like structure

Abstract

A theoretical model is derived mathematically for the encoding and controlling of the navigating of two droplet trains in a microfluidic network with a loop-like structure. The model reveals the relationship between the new outlet droplet train’s arrangement information (output signals) and the parameters including the two droplet trains’ input signals (droplet intervals), tuning flow rates, etc. The theoretical results are compared with the experimental results and they agree with each other. We find that every tuning flow rate corresponds to a certain output signal and a new droplet train can be obtained accurately. The generation orders of the successive droplets of the new droplet train remain unchanged within a certain range of the tuning flow rates. This work can be a useful reference for traffic controlling of two or more droplet trains in many microfluidic networks including the loop structure; the output signal of this work can be the input one for the next level which makes the multilevel studies possible. In addition, this study can help to promote the effective fusion of droplets and further the biological and chemical applications on droplet microfluidics.

Appendices

Appendix 1: Droplet intervals in the dilution or concentration modules

Figure 11 below is a simple network, and it is also a dilution or concentration module as Sessoms et al. (2009). One droplet train moves from channel 1 whose cross section is S ₁ to channel 2 whose cross section is S ₂, the slip factors of droplets equal to β ₁ and β ₂, respectively. The flow rate of channel 1 is Q _c1, and the flow rate of channel 2 is $Q_{c1} + Q_{c2}$ after dilution. Q _c2 can be minus, but all the droplets from channel 1 will go through channel 2.

Fig. 11

A schematic diagram of droplet train dilution or concentration module

For the module, we assume that droplets entering frequency is f ₁ while droplets exit frequency is f ₂ . The velocities of channel 1 and 2 are calculated as below:

$$\left\{ {\begin{array}{*{20}l} {V_{1} = \beta_{1} Q_{c1} /S_{1} ;} \hfill \\ {V_{2} = \beta_{2} \left( {Q_{c1} + Q_{c2} } \right)/S_{2} } \hfill \\ \end{array} } \right..$$

(21)

After Δt time, the entering droplet number is $f_{1} \varDelta t$, and the exit droplet number is $f_{2} \varDelta t$. Meanwhile, the distance of any droplet from channel 1 and 2 is:

$$\left\{ {\begin{array}{*{20}l} {L_{1} = V_{1} \varDelta t = \beta_{1} Q_{c1} \varDelta t/S_{1} ;} \hfill \\ {L_{2} = V_{2} \varDelta t = \beta_{2} \left( {Q_{c1} + Q_{c2} } \right)\varDelta t/S_{2} .} \hfill \\ \end{array} } \right.$$

(22)

If we let λ ₁ and λ ₂ represent the space intervals between droplets which traffic in channel 1 and 2, respectively, then we have:

$$\left\{ {\begin{array}{*{20}l} {\lambda_{1} = L_{1} /\left( {f_{1} \varDelta t} \right) = \beta_{1} Q_{c1} /\left( {S_{1} f_{1} } \right);} \hfill \\ {\lambda_{2} = L_{2} /\left( {f_{2} \varDelta t} \right) = \beta_{2} \left( {Q_{c1} + Q_{c2} } \right)/\left( {S_{2} f_{2} } \right).} \hfill \\ \end{array} } \right.$$

(23)

We can get f ₁ and f ₂ from Eq. (23):

$$\left\{ {\begin{array}{*{20}l} {f_{1} = \beta_{1} Q_{c1} /\left( {S_{1} \lambda_{1} } \right);} \hfill \\ {f_{2} = \beta_{2} \left( {Q_{c1} + Q_{c2} } \right)/\left( {S_{2} \lambda_{2} } \right).} \hfill \\ \end{array} } \right.$$

(24)

When droplets traffic in a microfluidic network, all the droplets number in the network will be stable under a stable traffic state. That is to say: f ₁ = f ₂. Combining with Eq. (24), we have:

$$\lambda_{2} = S_{1} \beta_{2} \left( {Q_{c1} + Q_{c2} } \right)\lambda_{1} /\left( {S_{2} \beta_{1} Q_{c1} } \right)$$

(25)

If the microchannels’ cross sections are equal to S and the slip factors of droplets in the microchannels are equal to β (Bithi and Vanapalli 2010; Jeanneret et al. 2012), we have:

$$\lambda_{2} = \left( {Q_{c1} + Q_{c2} } \right)\lambda_{1} /Q_{c1}$$

(26)

Appendix 2: Droplet number in branch A and branch B

The total hydrodynamic resistances for branches A, B and C are (n _C  = 0):

$$\left\{ {\begin{array}{*{20}l} {R_{A} = \overline{{R_{A} }} + n_{A} R_{d1} ;} \hfill \\ {R_{B} = \overline{{R_{B} }} + n_{B} R_{d2} ;} \hfill \\ {R_{C} = \overline{{R_{C} }} .} \hfill \\ \end{array} } \right.$$

(27)

We also know that:

$$\left\{ {\begin{array}{*{20}l} {n_{A} = L_{A} /\lambda_{1}^{\text{III}} ;} \hfill \\ {n_{B} = L_{B} /\lambda_{2}^{\text{III}} .} \hfill \\ \end{array} } \right.$$

(28)

Combining with Eqs. (8), (10) and (27), we have:

$$\left\{ {\begin{array}{*{20}l} {\lambda_{1}^{\text{III}} = c_{1} \left( {c_{3} + c_{4} R_{d2} n_{B} } \right)/\left( {c_{5} + R_{d1} n_{A} + R_{d2} n_{B} } \right);} \hfill \\ {\lambda_{2}^{\text{III}} = c_{2} \left( {c_{6} + c_{4} R_{d1} n_{A} } \right)/\left( {c_{5} + R_{d1} n_{A} + R_{d2} n_{B} } \right).} \hfill \\ \end{array} } \right.$$

(29)

where the coefficients above are as follows:

$$\left\{ {\begin{array}{*{20}l} {c_{1} = \lambda_{1}^{\text{I}} /Q_{1} ;} \hfill \\ {c_{2} = \lambda_{2}^{\text{I}} /Q_{2} ;} \hfill \\ {c_{3} = \left( {Q_{1} + Q_{1T} } \right)\left( {\overline{{R_{B} }} + \overline{{R_{C} }} } \right) + \left( {Q_{2} + Q_{2T} } \right)\overline{{R_{B} }} ;} \hfill \\ {c_{4} = Q_{1} + Q_{2} + Q_{1T} + Q_{2T} ;} \hfill \\ {c_{5} = \overline{{R_{A} }} + \overline{{R_{B} }} + \overline{{R_{C} }} ;} \hfill \\ {c_{6} = \left( {Q_{1} + Q_{1T} } \right)\overline{{R_{A} }} + \left( {Q_{2} + Q_{2T} } \right)\left( {\overline{{R_{A} }} + \overline{{R_{C} }} } \right).} \hfill \\ \end{array} } \right.$$

(30)

Substituting Eq. (29) into Eq. (28) yields

$$\left\{ {\begin{array}{*{20}l} {n_{A} = \left( {c_{5} + R_{d1} n_{A} + R_{d2} n_{B} } \right)L_{A} /\left[ {c_{1} \left( {c_{3} + c_{4} R_{d2} n_{B} } \right)} \right];} \hfill \\ {n_{B} = \left( {c_{5} + R_{d1} n_{A} + R_{d2} n_{B} } \right)L_{B} /\left[ {c_{2} \left( {c_{6} + c_{4} R_{d1} n_{A} } \right)} \right].} \hfill \\ \end{array} } \right.$$

(31)

Finally, n _A and n _B can be got by solving Eq. (31).