Question

A medical technician is trying to determine what percentage of apatient's artery is blocked by plaque. To do this, she measures theblood pressure just before theregion of blockage and finds that itis 1.20x10⁴ Pa, while in the region of blockage it is1.15x10⁴ Pa. Furthermore, she knows that bloodflowingthrough the normal artery just before the point of blockage istraveling at 30.0 cm/s, and the specific gravity of this patient'sblood is 1.06. What percentageof the cross-sectional area of thepatient's artery is blocked by the plaque?

Answer 1

Concepts and reason

The concepts required to solve this problem are the Bernoulli’s equation and the continuity equation.

Initially, write the expression of the Bernoulli’s equation for the blockage region and the normal artery. Rearrange the expression for the speed of blood at the blockage artery. Then, use the equation of continuity to find the ratio of the areas for both the blockage and the normal artery. Finally, calculate the percentage area of the cross-section of the artery blocked by plaque.

Fundamentals

The expression for the Bernoulli’s equation is,

$P + \frac{1}{2}\rho {v^2} + \rho gh = {\rm{constant}}$

Here, P is the pressure, $\rho$ is the density of the fluid, v is the speed of the fluid, g is the acceleration due to gravity, h is the height.

The expression for the continuity equation is,

${A_1}{v_1} = {A_2}{v_2}$

Here, ${A_1}$ is the cross-sectional area of the region 1 and ${v_1}$ is the speed of the fluid at region 1, ${A_2}$ is the cross-sectional area of the region 2 and ${v_2}$ is the speed of the fluid at region 2.

The expression of the Bernoulli’s equation for the blockage region and the normal artery is,

${P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

Here, ${P_1}$ is the pressure at normal region of artery, ${v_1}$ is the speed of blood at region 1, $\rho$ is the density of the fluid, ${h_1}$ is the height of the region 1, ${P_2}$ is the pressure at blocked region of artery, ${v_2}$ is the speed of blood at blocked region, ${h_2}$ is the height of the blocked region.

The arteries are horizontal so the heights ${h_1}$ and ${h_2}$ are same.

Substitute ${h_1}$ for ${h_2}$ in the above equation and rearrange the equation for ${v_2}$.

$\begin{array}{c}\\{P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_1}\\\\{P_1} + \frac{1}{2}\rho {v_1}^2 = {P_2} + \frac{1}{2}\rho {v_2}^2\\\\{v_2} = \sqrt {\frac{{2\left( {{P_1} - {P_2}} \right)}}{\rho } + {v_1}^2} \\\end{array}$

The density of the blood is,

$\rho = \left( {{\rm{specific gravity}}} \right){\rho _{\rm{w}}}$

Here, ${\rho _{\rm{w}}}$ is the density of water.

Substitute 1.06 for specific gravity and $1000{\rm{ kg/}}{{\rm{m}}^3}$ for ${\rho _{\rm{w}}}$.

$\begin{array}{c}\\\rho = \left( {{\rm{1}}{\rm{.06}}} \right)\left( {1000{\rm{ kg/}}{{\rm{m}}^3}} \right)\\\\ = 1060{\rm{ kg/}}{{\rm{m}}^3}\\\end{array}$

Substitute $1.20 \times {10^4}{\rm{ Pa}}$ for ${P_1}$, $1.15 \times {10^4}{\rm{ Pa}}$ for ${P_2}$, 30.0 cm/s for ${v_1}$, and $1060{\rm{ kg/}}{{\rm{m}}^3}$ for $\rho$ in the equation${v_2} = \sqrt {\frac{{2\left( {{P_1} - {P_2}} \right)}}{\rho } + {v_1}^2}$.

$\begin{array}{c}\\{v_2} = \sqrt {\frac{{2\left( {1.20 \times {{10}^4}{\rm{ Pa}} - 1.15 \times {{10}^4}{\rm{ Pa}}} \right)}}{{1060{\rm{ kg/}}{{\rm{m}}^3}}} + {{\left( {30.0{\rm{ }}\left( {\frac{{{\rm{cm}}}}{{\rm{s}}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}} \\\\ = 1.0166{\rm{ m/s}}\\\end{array}$

The expression of the continuity equation is,

${A_1}{v_1} = {A_2}{v_2}$

Rearrange the expression for $\frac{{{A_2}}}{{{A_1}}}$.

$\frac{{{A_2}}}{{{A_1}}} = \frac{{{v_1}}}{{{v_2}}}$

Substitute 30.0 cm/s for ${v_1}$ and 1.0166 m/s for ${v_2}$.

$\begin{array}{c}\\\frac{{{A_2}}}{{{A_1}}} = \frac{{\left( {30.0{\rm{ cm/s}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}{{1.0166{\rm{ m/s}}}}\\\\ = 0.2951\\\end{array}$

The percentage area of the cross section of the artery blocked by plaque is,

$\begin{array}{c}\\{\rm{percentage error}} = \frac{{{A_1} - {A_2}}}{{{A_1}}} \times 100\% \\\\ = \left( {1 - \frac{{{A_2}}}{{{A_1}}}} \right) \times 100\% \\\end{array}$

Substitute 0.2951 for $\frac{{{A_2}}}{{{A_1}}}$.

$\begin{array}{c}\\{\rm{percentage error}} = \left( {1 - 0.2951} \right) \times 100\% \\\\ = 70\% \\\end{array}$

Ans:

The percentage of the cross-sectional area of the patient’s artery, blocked by the plaque is $70\%$.