8.3 Verification
117
Table 8.3 Droplet sequence attempts
First attempt:
t c 1 c 2 m 1 c 4
c 6
c 3 m 2 c 5 c 7 c 8
1
2
3
4
5
6
7
8
Second attempt:
t c 1 c 2 m 1 c 4
c 6
c 3 m 2 c 5 c 7 c 8 m 3 c 9 c 10 c 11 c 12 m 4 c 13
1
2
3
4
5
6
7
8
9
10
11
12
13
a module or parts of a channel contain a header droplet, contain a payload droplet,
or does not contain a droplet at all, is represented by an empty circle ( ), a filled
circle ( ), or an empty cell ( ), respectively.
In the first droplet sequence (at the top of Table 8.3), a header (denoted as )
is injected in time step t = 1. At the first bifurcation, this header takes the
channel c 2 because c 2 ’s hydraulic resistance is lower than those of c 3 (i.e., Table 8.3
shows that c 2 requires 2 time steps compared to 3 time steps of c 3 ). In time
step t = 2, the payload (denoted as
) gets injected. Since the header is still
in channel c 2 (and, therefore, the hydraulic resistance of this channel is increased),
the payload intentionally flows into channel c 3 . After the payload passed channel c 3 ,
the operation of module m 2 is executed in time step t = 6. However, in the chosen
droplet sequence, the two droplets would coalesce afterwards in channel c 8 in time
step t = 8. Therefore, the chosen sequence is invalid.
In the second droplet sequence (at the bottom of Table 8.3), the header is
injected in time step t = 1 and the payload is injected in time step t = 3.
Therefore, it is avoided that the droplets would again coalesce in time step t = 8
as in the first sequence. When further considering the droplet’s flow, the header