Download Calculating The Velocity of Gravity- and Capillarity

Calculating The Velocity of Gravity- and
Capillarity-Driven Flow
A.J. Aranyosi
March 16, 2010
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Problem Description
Imagine we have a microfluidic channel of length L, width w, and height h,
with h << w << L. This device is initially empty, and a drop of liquid
(e.g. blood) is placed at the opening of the device. The resulting flow is driven
by a combination of forces due to gravity and capillarity. We would like to
compute the corresponding velocity of this flow.
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Pressure Due To Gravity
Due to hydrostatic forces, the pressure at the bottom of the droplet of liquid
is given by P = ρgH, where P is the pressure, ρ is the density of the liquid
(roughly 1000 kg/m3 ), g is the gravitational constant (9.8 m/s), and H is the
height of the liquid. For water and similar substances, P ≈ 104 H, with P in
Pascals and H in meters. At atmospheric pressure, P ≈ 105 Pa, so H = 10 m;
that’s a tall column of water! If we replace the water with mercury, at a density
of 13,500 kg/m3 , the height drops to 0.75 m, or about 30 inches (the number
you’re likely to hear in weather reports). For now, though, let’s assume that
the volume of liquid that enters the channel is a negligible fraction of our initial
droplet, so that H is constant.
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Pressure Due To Capillary Action
Capillarity is a force that results from the interaction of cohesion of molecules
of a liquid to each other and adhesion of these molecules to the surrounding
material. For example, water adheres tightly to glass, and so “climbs up” the
walls of a glass tube. As the water on the edges of the tube moves up, the
surface tension resulting from molecular cohesion of water pulls the center up
as well, resulting in a positive capillarity force that moves the water further
along the tube. Mercury, in contrast, is highly cohesive but does not adhere
well to glass, so the molecules near the glass tend to be attracted toward other
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mercury, resulting in a concave downward shape. This shape (which can be
seen in a mercury thermometer), combined with the surface tension of mercury,
results in a negative capillarity force that tends to pull the mercury downwards.
In general, the pressure due to capillarity in a vertical cylindrical tube of radius
r can be described by
2γ cos(θ)
,
(1)
P =
r
where γ is the surface tension between the liquid and air (in J/m2 ; γ = 0.0728J/m2
for water at 20◦ C), θ is the contact angle (in radians, measured from the downward vertical, so water has θ = 20◦ or 0.35 rad while mercury has θ = 140◦ for
contact with glass) between the liquid and wall, and r is the radius of the tube.
For the case of a rectangular channel, the capillary force is dominated by the
smallest dimension, so we can replace r with h in our calculations.
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Resistance To Flow
In lecture we derived an equation describing one-dimensional static Poiseuille
flow,
∂2u
∂P
= µ 2,
(2)
∂x
∂y
where P is the applied pressure, x is the dimension along the length of the
channel, µ is the fluid viscosity, u is the fluid particle velocity (as distinguished
from the volume velocity), and y is the dimension across the channel. With
no-slip boundary conditions (i.e., ∂u
∂y is finite so u(0) = u(h) = 0), it is easy to
show that u(y) = cy(h − y), where c is a constant. The peak velocity umax =
Rh
u(h/2) = ch2 /4, so c = 4umax /h2 . Similarly, u = h1 0 u(y)dy = ch2 /6 = 23 umax ,
so u(y) = 6u
h2 y(h − y). Substituting this back into equation 2 gives
∂P
= −12µu/h2 .
∂x
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(3)
Putting It All Together
Now we’ll combine all of these derivations to determine how flow will be established in our microchannel. We will derive an equation for u as a function of
l, the distance over which fluid has traveled, and use this equation to derive
l(t). Here’s where we get to play fast and loose with physics. We’re going to
assume that the flow in the microchannel is Poiseuille, despite the fact that
capillarity will dominate the shape of the flow near the front. This assumption
will probably make our answer wrong at a detailed level, but we’re not likely
to be off by more than a factor of 2–3, and we should get the scaling factors
correct. If we assume that our applied gravitational and capillary pressure act
over the distance l, then we can substitute these into equation 3 to get
(ρgH +
2γ cos θ
)/l(t) = −12µu/h2 ,
h
2
(4)
which gives
u=
h2
2γ cos θ
(ρgH +
).
8µl(t)
h
(5)
In other words, our fluid velocity will be dominated by the larger of the gravitational and capillary forces, will scale as the square of the smallest dimension h of
the channel, and will be inversely proportional to the distance already traveled
— i.e., the flow will slow down as it moves further along. This result makes
intuitive sense: the driving pressures are essentially constant, but they must act
on an ever-increasing volume of fluid.
∂l
. With this
Finally, we can use equation 5 to solve for l(t), since u = ∂t
substitution,
∂l
h2
2γ cos θ 1
=
(ρgH +
)
,
(6)
∂t
8µ
h
l(t)
or
∂l
(7)
l(t) = C,
∂t
where C =
h2
8µ (ρgH
+
2γ cos θ
)
h
is constant. A simple solution to this nonlinear
√
differential equation is l(t) = At1/2 , so A = 2C, which gives
s
th2
2γ cos θ
l(t) =
(ρgH +
).
(8)
4µ
h
So the length traveled by the fluid will scale with the square root of time and
applied pressure, inversely with the square root of viscosity, and linearly with
the channel height (for gravity-driven flow) or the square root of channel height
(for capillarity-driven flow).
Consequences
Several facts are evident from these equations. First, the force of gravity will
exceed that of capillarity under two conditions. The first condition is when
the channel has already been filled, either due to the capillary force or applied
pressure; in this case, only the pressures due to gravity and applied forces will
cos θ
θ
; i.e., H > 2γρgh
. For
drive fluid flow. The second case is when ρgH > 2γ cos
h
water interfacing with a hydrophilic surface like glass or recently plasma-treated
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. If h = 50 µm,
PDMS, γ = 0.0728J/m2 and θ = 0.35 rad, so H > 1.4×10
h
the gravitational force is less than capillarity for H < 28 cm, and becomes
completely negligible for H < 28 mm. Put another way, the maximum H for a
microfluidic device on a glass slide is 75 mm; for this H, the gravitational force
is less than capillarity for a device height h < 187 µm, and becomes completely
negligible for h < 19 µm. Thus even for the longest device we can make, the
gravitational pressure for a 50 µm high channel will only be 1/3 as big as that
of capillarity — so we can ignore gravity as a driving force in filling the channel.
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The second consequence has to do with the length of the channel. If we want
the device to be completely filled within one minute, for example, equation 8
gives
r
60 ∗ (5 × 10−5 )2 2 × 0.0728 × cos 0.35
l(t) =
= 0.32 m.
(9)
4 × 10−3
5 × 10−5
So for a channel length of 0.2 m, the device will fill in about 48 seconds; for
a channel length of 0.4 m, it will fill in about 67 seconds. In other words, a
reasonable time to fill a reasonable-length channel is about a minute. We can
vary this time somewhat by changing the length and/or the thickness of the
channel, but a minute is a good rough guess.
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