Biological Physics: Energy, Information, Life

160 Chapter 5. Life in the slow lane: the low Reynolds-number world[[Student version, December 8, 2002]]

flow

v(r)

v(r+dr)

r

R

ω(r)

ω 0

R

r

a b

Figure 5.11:(Sketches.) (a)Inlaminar pipe flow, the inner fluid moves faster than the outer fluid, which must be
motionless at the pipe wall (the no-slip boundary condition). We imagine concentric cylindrical layers of fluid sliding
overeach other. (b)The torsional drag on a spinning rod in viscous fluid. This time the inner fluidrotatesfaster
than the outer fluid, which must be at rest farawayfrom the rod. Again we imagine concentric cylindrical layers of
fluid sliding over each other, since the angular velocityω(r)isnot constant but rather decreases withr.

Since dris very small we can evaluate dv/drat the point (r+dr)using a Taylor series, dropping
terms with more than one power of dr:

dv(r+dr)

dr

=

dv(r)

dr

+dr×

d^2 v

dr^2

+···.

Thus adding df 2 to df 3 gives 2πηL(ddvr+rd

(^2) v
dr^2 ). Adding df^1 and requiring the sum to be zero gives
rp
Lη

+

dv

dr

+r

d^2 v

d^2 r

=0.

This is a differential equation for the unknown functionv(r). You can check that its general solution
isv(r)=A+Blnr−r^2 p/ 4 Lη,whereAandBare constants. We had better chooseB=0,since
the velocity cannot be infinite at the center of the pipe. And we need to takeA=R^2 p/ 4 Lηin order
to get the fluid to be stationary at the stationary walls. This gives our solution, the flow profile for
laminar flow in a cylindrical pipe:

v(r)=(R

(^2) −r (^2) )p
4 Lη

. (5.16)

Your Turn 5f

After going through the math to check the solution (Equation 5.16), explain in words why every

factor (except the 4) “had” to be there.

Now we can see how well the pipe transports fluid. The velocityvcan be thought of as theflux
ofvolumejv,orthe volume per area per time transported by the pipe. The total flow rateQ,with
the dimension of volume per time, is then the volume fluxjv=vfrom Equation 5.16, integrated
overthe cross-sectional area of the pipe:

Q=

∫R

0

2 πrdrv(r)=

πR^4

8 Lη

p. (5.17)