Chapter 6.7, Problem 16E

To determine

To Show: The given relation between blood pressure in a normal artery and in a constricted artery.

Explanation of Solution

Given Information: The pressure and radius of a normal artery are $P_0$ and $R_0$ . The pressure and radius of a constricted artery are $P$ and $R$ . The blood flow rate in a normal artery and in a constricted artery are to be maintained the same

Formula Used: According to Poiseuille’s Law, the rate of blood flow $F$ in a blood vessel of radius $R$ , length $l$ , viscosity $\text{η}$ and pressure P is given by

  $F=\frac{\text{π}P{R}^4}{8\text{η}l}$

Calculation: Let the blood flow rate in a normal artery be $F_{normal}$ . Then, from Poiseuille’s Law,

  $F_{normal}=\frac{\text{π}{P}_0{R}_0^4}{8\text{η}l}$

Now let the blood flow rate in a constricted artery be $F_{constricted}$ . Then, again from Poiseuille’s Law,

  $F_{constricted}=\frac{\text{π}P{R}^4}{8\text{η}l}$

If the blood flow rate is the same in both the cases, that is $F_{constricted}=F_{normal}$ , then equate the righthand sides of the equations for both the cases to derive the required relation as

  $\frac{\text{π}{P}_0{R}_0^4}{8\text{η}l}=\frac{\text{π}P{R}^4}{8\text{η}l}$

  $\Rightarrow P_0{R}_0^4=P{R}^4$

  $\Rightarrow \frac{P_0}{P}={\left(\frac{R}{R_0}\right)}^4$

If $R$ is three-fourths of the normal value, then put $R=\frac{3}{4}{R}_0$ in the above shown relation to get

  $\frac{P_0}{P}={\left(\frac{\frac{3}{4}{R}_0}{R_0}\right)}^4$

  $\Rightarrow P=P_0\times {\left(\frac{4}{3}\right)}^4$

  $\Rightarrow P=3.16P_0$

Therefore, if the radius of the artery is reduced to three-fourth of its normal value, the pressure in the artery increases by a factor of $3.16$ , that is a factor of more than three.