flow rate of blood vs radius - just need explanation?

Question

When the body requires an increased blood flow in a particular organ or muscle, it can accomodate this by increasing the radius of arterioles in that area. What percentage increase in the radius of an arteriole is required to double the flow rate of the blood, all other factors remaining the same?

15.9%
18.9%
41.4%
82.8%
159%


I know that the correct answer is 18.9% but I am not sure how to rationalize this. Can someone explain the logic behind this?
Thank you!

This question was put forth by my physcis teacher and he gave the correct answer as being 18.9% - I just don't understand why. I can't ask him because we only meet once a week.

Answer 1

The flow of a liquid through a tube (volume/unit time) will be given by the cross section of the tube and the linear speed of the fluid through the tube. As the cross section of the tube increases, the volume that can flow through that tube will grow as the square of the radius. This is only half the answer. The outermost layer (lets call it L1) of a fluid flowing through a tube will be slowed down by friction with the wall of the tube which makes this outermost layer travel fractionally slower than the layer immediately inside this outer layer (L2) which will itself be slowed down by friction with L1 and therefore travel fractionally slower than the layer immediately inside it (L3) which, in turn, will be slowed down by friction with L2, etc. all the way down to the innermost layer of the fluid. The internal friction of a fluid is what we call viscosity. If you measure the speeds through the cross section of a fluid flowing through a pipe you will find that the speeds increase from the periphery through the center of the pipe describing a paraboloid (like a bullet point) around the center (that is the center goes fastest). The speed of any given layer is, then, a function of the distance from the edge of the tube and increases as the square of that distance. The overall speed of the fluid is an average (actually an integral but close enough) of all these speeds. As the radius of the tube increases the distance of the central layer to the edge increases in the same proportion and its speed increases by r^2. This phenomenon propagates throughout the entire cross section so that the overall speed also increases a r^2. Put this together with the r^2 increase in the cross sectional area of the tube and you get a flow dependence of the quartic power of r (r^4). An increase of 18.9 percent on the radius of the tube then gives ((1.189)^4 ~ 2) about a two-fold increase in flow. The law describing the flow of a newtonian fluid through a tube is known as Poiseuille's law. This law actually describes blood flow only approximately as blood is not a newtonian fluid and only in small vessels since blood flow in larger vessels is not laminar but rather chaotic.

I hope this explanation helps. (The symbol ^ means "raised to the power of" since this system does not accomodate formatting)

Answer 2

Stop all that jibba jabba. It's FlowRate ~ 1/R
R = 1 / (r^4)
FlowRate = (r^4)
2 = (r^4)
r = 2^(1/4) or 2^0.25
r = 1.189207
%r increase = (r -1.0)*100
=18.9%

Answer 3

Let V be volume Q be volumetric flowrate (∆V/∆dt).
Q = ∆V/∆dt = A∆s/∆t = Av
A = πr^2
Q = πr^2v
2Q/Q = (π(r2)^2v)/(πr^2v)
r2^2 = 2r^2
r2 = r√2
(r2 - r)/r =
(r√2 - r)/r =
(√2 - 1)r/r =
1.414 - 1 =
0.414 =
41.4%