5.2. Low Reynolds number[[Student version, December 8, 2002]] 149
Figure 5.3:(Photograph.) Low Reynolds-number fluid flow past a sphere. The fluid flows from left to right at
R=0.1. The flow lines have been visualized by illuminating tiny suspended metal flakes with a sheet of light
coming from the top. (The black area below the sphere is just its shadow.) Note that the figure is symmetrical; the
time-reversed flow from right to left would look exactly the same. Note also the orderly, laminar character of the
flow. If the sphere were a single-cell organism, a food particle located in its path would simply get carried around it
without ever encountering the cell at all. [From (van Dyke, 1982).] [Copyrighted figure; permission pending.]
a vFigure 5.4: (Schematic.) Motion of a small fluid element, of size,asitimpinges on an obstruction of radiusa
(see Figure 5.3).
Toestimate the frictional force, we first generalize Equation 5.4 to the case where the velocity
of the fluid is not a uniform gradient (as it was in Figure 5.2b). To do so, replace the finite velocity
differencev 0 /dbythe derivative, dv/dx.When a fluid element slides past its neighbor, then, they
exert forces per unit area on each other equal to
f
A
=−η
dv
dx. (5.9)
In the situation sketched in Figure 5.4, the surface areaAof one face of the fluid element^2 is≈^2.
Thenetfrictional forceffrict on the fluid element is the force exerted on it by the one above it,
minus the force it exerts on one below it. We can estimate this difference astimes the derivative
df/dx,orffrict≈η^3 d
(^2) v
dx^2 .Toestimate the derivative, again note thatvchanges appreciably over
distances comparable to the obstruction’s sizea;accordingly we estimate d^2 v/dx^2 ≈v/a^2 .Putting
(^2) Weare still dropping numerical factors to get an estimate; really the area is 6 (^2) .Also, Equation 5.9 sometimes
needs a correction in cases with nonplanar geometry (see Problem 5.9); thus for our example it only gives an estimate
of the viscous force.