Chapter 9, Problem 85P

(a)

To determine

By what percentage the blood pressure difference between the ends of artery increased.

Answer to Problem 85P

The percentage that the blood pressure difference between the ends of artery increased is $220\%$.

Explanation of Solution

Write the expression to find the percentage of increase of blood pressure difference between the ends of artery.

$P_B=\frac{\Delta P_a-\Delta P_u}{\Delta P_u}\times 100\%$ (I)

Here, $P_B$ is the percentage of increase of blood pressure difference between the ends of artery, $\Delta P_a$ is the pressure difference between the ends when arteriosclerosis is present, $\Delta P_u$ is the pressure difference between the ends of the artery when it is undisturbed.

Write the expression for Poiseuille’s law for viscous flow.

$\Delta P=\frac{8\eta L}{\pi r^4}\left(\frac{\Delta V}{\Delta t}\right)$

Here, $\Delta P$ is the pressure difference between the ends, $\frac{\Delta V}{\Delta t}$ is the volume flow rate, $r$ is the inner radius of the artery, $L$ is the length of the artery, $\eta$ is the viscosity of the fluid.

Re-write the equation (I) using above Law to find the percentage of increase of blood pressure difference between the ends of artery.

$\begin{array}{c}{P}_B=\frac{\frac{8\eta L}{\pi r_a{}^4}\left(\frac{\Delta V}{\Delta t}\right)-\frac{8\eta L}{\pi r_u{}^4}\left(\frac{\Delta V}{\Delta t}\right)}{\frac{8\eta L}{\pi r_u{}^4}\left(\frac{\Delta V}{\Delta t}\right)}\times 100\%\\ =\frac{\frac{1}{r_a^4}-\frac{1}{r_u^4}}{\frac{1}{r_u^4}}\times 100\%\\ =\left({\left(\frac{r_u}{r_a}\right)}^2-1\right)\times 100\%\end{array}$

Conclusion:

Substitute $1$ for $r_u$ and $0.75$ for $r_a$ to find the percentage of increase of blood pressure difference between the ends of artery.

$\begin{array}{c}{P}_B=\left({\left(\frac{1}{0.75}\right)}^2-1\right)\times 100\%\\ =220\%\end{array}$

(b)

To determine

By what factor the blood flow rate through the artery decrease.

Answer to Problem 85P

The factor at which the blood flow rate through the artery decrease is $0.68$.

Explanation of Solution

Write the expression for Poiseuille’s law for viscous flow.

$\frac{\Delta V}{\Delta t}=\frac{\pi r^4\Delta P}{8\eta L}$ (II)

Here, $\Delta P$ is the pressure difference between the ends, $\frac{\Delta V}{\Delta t}$ is the volume flow rate, $r$ is the inner radius of the artery, $L$ is the length of the artery, $\eta$ is the viscosity of the fluid.

Write the expression for the factor of blood flow decreases in artery.

$\begin{array}{c}\left|\frac{changeinflowrate}{originalflowrate}\right|=\frac{{\frac{\Delta V}{\Delta t}}_u-{\frac{\Delta V}{\Delta t}}_a}{{\frac{\Delta V}{\Delta t}}_u}\\ =1-\frac{{\frac{\Delta V}{\Delta t}}_a}{{\frac{\Delta V}{\Delta t}}_u}\end{array}$

Substitute equation (II) in expression to find the factor of blood flow decreases in artery.

$\begin{array}{c}\left|\frac{changeinflowrate}{originalflowrate}\right|=1-\frac{\frac{\pi r_a{}^4\Delta P}{8\eta L}}{\frac{\pi r_u{}^4\Delta P}{8\eta L}}\\ =1-{\left(\frac{r_a}{r_u}\right)}^4\end{array}$

Conclusion:

Substitute $1$ for $r_u$ and $0.75$ for $r_a$ to find the factor of blood flow decreased in artery.

$\begin{array}{c}\left|\frac{changeinflowrate}{originalflowrate}\right|=1-{\left(\frac{0.75}{1}\right)}^4\\ =0.68\end{array}$