Example 12.7 from College Physics, 12.4 Viscosity and Laminar Flow; Poiseuille’s Law

Using Flow Rate: Plaque Deposits Reduce Blood Flow

Suppose the flow rate of blood in a coronary artery has been reduced to half its normal value by plaque deposits. By what factor has the radius of the artery been reduced, assuming no turbulence occurs?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Assuming laminar flow, Poiseuille’s law states that

$Q = \frac{(P_{2} - P_{1}) {πr}^{4}}{8 η l} .$
12.46

We need to compare the artery radius before and after the flow rate reduction.
Solution
With a constant pressure difference assumed and the same length and viscosity, along the artery we have

$\frac{Q_{1}}{r_{1}^{4}} = \frac{Q_{2}}{r_{2}^{4}} .$
12.47

So, given that $Q_{2} = 0 . 5 Q_{1}$ , we find that $r_{2}^{4} = 0 . {5 r}_{1}^{4}$ .
Therefore, $r_{2} = {(0.5)}^{0.25} r_{1} = 0 . 841 r_{1}$ , a decrease in the artery radius of 16%.
Discussion
This decrease in radius is surprisingly small for this situation. To restore the blood flow in spite of this buildup would require an increase in the pressure difference $(P_{2} - P_{1})$ of a factor of two, with subsequent strain on the heart.

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