Chapter 7 – Complementary Experimental Tools 301
Answers
a
The flow rate is given by πa2<v> where a is the pipe radius and <v> is mean speed
of flow. Thus,
v
s
=
×
(
)
×
×
×
=
×
−
−
−
−
−
18 8
10
60
10
9
3
1
6
.
/
/(
(10
10
) m )
1.0
10
2
2
π
3
1
1
m s
1 mm s
−
−
=
The Reynolds number (Re) is given by Equation 6.8, where we approximate the
equivalent length parameter to the diameter of the pipe and the speed to the
mean speed of flow; thus,
Re
3
(1 10 )
(1 10
)
(20
10
)/(1 10
)
0.02
=
×
×
×
×
×
×
=
−
−
−
3
6
3
This is <2100, and so the flow is laminar.
b
Consider a cylindrical pipe shell of radius a with incremental axial length δx. The
cylinder is full of fluid with pressure p at one end and (p + δp) at the other end.
Consider then a solid cylindrical shell of fluid of speed v, which is inside the pipe
and has a radius z, and a shell thickness δz such that z ≤ a. The net force on the
fluid shell is the sum of pressure and viscous drag components, which is zero:
p
p
p
z
z x
v
r
v
r
r
p
x
+
−
(
)
−
(
)
=
∴
=
∂
∂
δ
π
η π δ
η
2
0
2
(2
d
d
d
d
Using boundary conditions of v = 0 at z = a gives the required solution (i.e., a
parabolic profile). The maximum speed occurs when dv/dr = 0, in which z = 0 (i.e.,
along the central axis of the pipe). The volume flow rate is given by
Q
v
r
r
a
=
⋅(
)
∫
0
2π d
which gives the required solution. The mean speed is given by Q/πa2, and it is
easy to then show from the aforementioned that the maximum speed is twice the
mean speed (which here is thus 2 mm s−1).
c
Assume mixing is solely through diffusion across the streamlines. Thus, to be
completely mixed with the fluid in the channel, a molecule needs to have diffused
across the profile of streamlines, that is, to diffuse across the cross-section, a
distance in one dimension equivalent to the diameter or 20 μm. In a time Δt, a
protein molecule with diffusion coefficient D will diffuse a root mean square dis
placement of √(2DΔt) in one dimension (see Equation 2.12). Equating this dis
tance to 20 μm and rearranging indicates that
∆t =
×
×
×
×
(
) =
−
−
−
−
(20
10
2
7
10
10
20
6
2
7
4
2
1
m) /
m s
s.
The mean speed is 1 mm s−1, therefore, the minimum pipe length should be
~20 mm or 2 cm. Comment: this is a surprisingly large length compared to that
of the diameter of, say, a single drop of blood by ca. an order of magnitude. For
proper mixing to occur in lab-on-a-chip devices, it requires either larger channel
lengths (i.e., large chips) or, better, additional methods introduced into the mixing
such as chevron structures to cause turbulence mixing.