5.3. Biological applications[[Student version, December 8, 2002]] 161
Equation 5.17 is theHagen–Poiseuille relationfor laminar pipe flow. Its applicability extends
beyond the low-Reynolds regime studied in most of this chapter: As mentioned earlier, the fluid
doesn’t accelerate at all in laminar pipe flow. Thus we can use Equation 5.17 as long as the Reynolds
number is less than a thousand or so. This regime includes all but the largest veins and arteries in
the human body (or the entire circulatory system of a mouse).
The general form of Equation 5.17 can be expressed asQ=p/Z,where thehydrodynamic re-
sistanceZ=8ηL/πR^4 .The choice of the word “resistance” is no accident. The Hagen–Poiseuille
relation says that the rate of transport of some conserved quantity (volume) is proportional to a
driving force (pressure drop), just as Ohm’s law says that the rate of transport of charge is propor-
tional to a driving force (potential drop). In each case the constant is called “resistance.” In the
context of low Reynolds-number fluid flow, transport rules of the formQ=p/Zare quite common
and are collectively calledDarcy’s law.(Athigh Reynolds number turbulence complicates matters,
and no such simple rule holds.) Another example is the passage of fluid across a membrane (see
Problem 4.10). In this context we writeZ=1/(ALp)for the resistance, whereAis the mem-
brane area andLpis called the “filtration coefficient” (some authors use the synonym “hydraulic
permeability”).
Asurprising feature of the Hagen–Poiseuille relation is the very rapid decrease of resistance as
the pipe radiusRincreases. Two pipes in parallel will transport twice as much fluid at a given
pressure as will one. But a single pipe with twice the area will transportfourtimes as much, because
πR^4 =(1/π)(πR^2 )^2 ,andπR^2 has doubled. This exquisite sensitivity is what lets our blood vessels
regulate flow with only small dilations or contractions:
Example Find the change in radius needed to increase the hydrodynamic resistance of a blood
vessel by 30%, other things being equal. (Idealize the situation as laminar flow of a
Newtonian fluid.)
Solution: We wantp/Qto increase to 1.3 times its previous value. Equation 5.17
says that this happens when (R′)−^4 /R−^4 =1.3, orR′/R=(1.3)−^1 /^4 ≈ 0 .94. Thus
the vessel need only change its radius by about 6%.5.3.5 Viscous drag at the DNA replication fork
Tofinish the chapter, let us descend from physiology to the realm of molecular biology, which will
occupy much of the rest of this book.
Amajor theme of the chapters to come will be that DNA is not just a database of disembodied
information, but aphysical objectimmersed in the riotous thermal environment of the nanoworld.
This is not a new observation. As soon as Watson and Crick announced their double-helix model of
DNA structure, others asked: How do the two strands separate for replication, when they’re wound
around each other? One solution is shown in Figure 5.12. The figure shows aY-shaped junction
where the original strand (top) is being disassembled into two single strands. Since the two single
strands cannot pass through each other, the original must continually rotate (arrow).
The problem with the mechanism sketched in the figure is that the upper strand extends for a
great distance (DNA is long). If one end of this strand rotates, then it would seem that the whole
thing must also rotate. Some people worried that the frictional drag resisting this rotation would
beenormous. Following C. Levinthal and H. Crane, we can estimate this drag and show that on
the contrary it’s negligible.