Question

A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that itis 1.20×104 \(\rm Pa\), while in the region of blockage it is 1.15×104 \(rm Pa\). Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 \(rm cm/s\), and the specific gravity of this patient's blood is 1.06. What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

Answer 1

Concepts and reason

The concepts required to solve this problem are Bernoulli's equation and the continuity equation. Initially, write the expression of 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, \(P+\frac{1}{2} \rho v^{2}+\rho g h=\mathrm{constant}\)

Here, \(\mathrm{P}\) is the pressure, \(\rho\) is the density of the fluid, \(\mathrm{v}\) is the speed of the fluid, \(\mathrm{g}\) is the acceleration due to gravity, \(\mathrm{h}\) 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 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 the normal region of the 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 the blocked region of the artery, \(v_{2}\) is the speed of blood at the blocked region, \(h_{2}\) is the height of the blocked region. The arteries are horizontal, so the heights \(h_{1}\) and \(h_{2}\) are the 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, \(\rho=(\) specificgravity \() \rho_{\mathrm{w}}\)

Here, \(\rho_{\mathrm{w}}\) is the density of water. Substitute 1.06 for specific gravity and \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\mathrm{w}}\)

$$ \begin{array}{c} \rho=(1.06)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right) \\ =1060 \mathrm{~kg} / \mathrm{m}^{3} \end{array} $$

Substitute \(1.20 \times 10^{4} \mathrm{~Pa}\) for \(P_{1}, 1.15 \times 10^{4} \mathrm{~Pa}\) for \(P_{2}, 30.0 \mathrm{~cm} / \mathrm{s}\) for \(v_{1},\) and \(1060 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho\) in the equation

\(v_{2}=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}+v_{1}^{2}}\)

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

The density of the blood can be calculated by the product of specific gravity and the density of water. The arteries are horizontal so the height of the normal artery and the blocked artery is same. The speed of the blood from the blocked artery can be calculated using the Bernoulli's equation.

 

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 \(30.0 \mathrm{~cm} / \mathrm{s}\) for \(v_{1}\) and \(1.0166 \mathrm{~m} / \mathrm{s}\) for \(v_{2}\)

$$ \begin{array}{c} \frac{A_{2}}{A 1}=\frac{(30.0 \mathrm{~cm} / \mathrm{s})\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)}{1.0166 \mathrm{~m} / \mathrm{s}} \\ =0.2951 \end{array} $$

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

percentageerror \(=\frac{A 1-A_{2}}{A_{1}} \times 100 \%\)

$$ =\left(1-\frac{A_{2}}{A_{1}}\right) \times 100 \% $$

Substitute 0.2951 for \(\frac{A 2}{A_{1}}\)

$$ \begin{aligned} \text { percentageerror } &=(1-0.2951) \times 100 \% \\ &=70 \% \end{aligned} $$

The percentage of the cross-sectional area of the patient's artery, blocked by the plaque is \(70 \%\)

The continuity equation in case of fluids states that the fluid flowing with speed \(v_{1}\) through a region in the pipe of cross sectional area \(A_{1}\) will be equal to the fluid flowing with speed \(v_{2}\) through another region in the pipe of cross sectional area \(A_{2}\). This can be represented by the following equation: \(A_{1} v_{1}=A_{2} v_{2}\).

 

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