5.3. Biological applications[[Student version, December 8, 2002]] 159
takev≈ 30 μms−^1 to be the known swimming speed anddto be the length of the run, not the
length of the cell. Then we find that to navigate up food gradients a bacterium must swim at least
30 μm,or30bodylengths, before changing direction. And... that’s what they really do.
Attack and escape Takeanother look at Figure 5.3 on page 149. Clearly a solid object, gliding
through a liquid at low Reynolds number,disturbs the fluidout to a distance comparable to its own
diameter. This can be a liability if your livelihood depends on stealth, for example if you need to
grab your dinner before it escapes. Moreover, swimming up to a tasty morsel will actually tend to
push it away, just like your colored blob in the experiment of Section 5.1.3 on page 144. That’s why
many medium-small creatures, not so deeply into the low-Reynolds regime as the bacteria studied
above, put on a burst of speed to push themselves momentarily up to high Reynolds for the kill.
Thus the tiny crustaceanCyclopsmakes its strike by accelerating at up to 12ms−^2 ,briefly hitting
Reynolds numbers as high as 500.
In the same spirit,escapingfrom an attacker will tend just to drag it along with you at low
Reynolds number! Here again, a burst of speed can make all the difference. The sessile protozoan
Vorticella,when threatened, contracts its stalk from 0.2–0.33mmdown to less than half that length
at speeds up to 80mm s−^1 ,the most rapid shortening of any contractile element in any animal. This
impressive performance garners the name “spasmoneme” for the stalk.
5.3.4 Vascular networks
Bacteria can rely on diffusion to feed them, but large organisms need an elaborate infrastructure
of delivery and waste-disposal systems. Virtually every macroscopic creature thus has one or more
vascular networkscarrying blood, sap, air, lymph, and so on. Typically these networks have a
hierarchical, branching, structure: The human aorta splits into the iliac arteries, and so on, down
to the capillary beds that actually nourish tissue. To get a feeling for some of the physical constraints
governing such networks, let’s take a moment to work out one of the simplest fluid-flow problems:
the steady, laminar flow of a simple Newtonian fluid through a straight, cylindrical pipe of radius
R(Figure 5.11a). In this situation the fluid does not accelerate at all, so we can neglect the inertial
term in Newton’s Law even if the Reynolds number is not very small.
Wemust push a fluid to make it travel down a pipe, in order to overcome viscous friction. The
frictional loss occurs throughout the pipe, not just at the walls. Just as in Figure 5.2 on page
145, where each layer of fluid slips on its neighbor, so in the cylindrical geometry the shear will
distribute itself across the whole cross-section of the pipe. Imagine the fluid as a nested set of
cylindrical shells. The shell at distancerfrom the center moves forward at a speedv(r), which we
must find. The unknown functionv(r)interpolates between the stationary walls (withv(R)=0)
and the center (with unknown fluid velocityv(0)).
Tofindv(r)webalance the forces acting on the shell lying betweenrandr+dr.The cross-
sectional area of this shell is 2πrdr,sothe applied pressurepcontributes a force df 1 =2πrpdr
directed along the pipe axis. A viscous force df 2 from the slower-moving fluid at largerrpulls
backward on the shell, while the faster-moving fluid at smallerrdrags it forward with a third force,
df 3 .For a pipe of lengthLthe viscous force rule (Equation 5.9 on page 149) gives
df 3 =−η(2πrL)dv(r)
dr
and df 2 =η(2π(r+dr)L)dv(r+dr)
dr.
Notice thatf 2 is a negative quantity, whilef 3 is positive. Force balance is then the statement that
df 1 +df 2 +df 3 =0.