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?
Strategy
Assuming laminar flow, Poiseuilleâs law states that
Q = (p2 - p1)pir^4/8nl.We need to compare the artery radius before and after the flow rate reduction.

Respuesta :

Answer:

1.18

Explanation:

The flow rate of blood is proportional to the fourth power of its radius as given the Poiseuille's law.

The law is :

[tex]$Q \propto r^4$[/tex]

It is given here that the flood flow rate is been reduced to half its normal value. Therefore, [tex]$Q_1 = \frac{1}{2}Q_2$[/tex]

So, for the radius [tex]$r_1$[/tex] and [tex]$r_2$[/tex], the ratios of their flow rates are :

[tex]$\frac{Q_1}{Q_2}=\frac{r_1^4}{r_2^4}$[/tex]

It is given that the flow rate is reduced to half. So we have,

[tex]$\frac{Q_1}{2Q_1}=\frac{r_1^4}{r_2^4}$[/tex]

or [tex]$r_2=2^{1/4}{r_1}$[/tex]

[tex]$r_2=1.18 \ r_1}$[/tex]

So the radius changes by a factor of 1.18