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 it is $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$, while in the region of blockage it is role="math" localid="1668168100834" $\mathbf{1}\mathbf{.}\mathbf{15}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 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?

Short answer

The percentage of the cross-sectional area of the patient’s artery blocked by plague is 71%.

Step-by-step solution

Given data

• The blood pressure before the blockage is ${\mathrm{P}}_1=1.20\times {10}^4\mathrm{Pa}$.

• The blood pressure in the region of blockage is ${\mathrm{P}}_2=1.15\times {10}^4\mathrm{Pa}$.

• The velocity of blood flow is $v_1=30cm/s$.

• The specific gravity is n = 1.06.

Understanding the Bernoulli’s principle

In this problem, Bernoulli’s equation will be applied before the blockage region and after the blockage region to find the velocity of blood in the blockage region.

Determination of the velocity of blood in blockage region

The relation obtained from Bernoulli’s equation can be written as:$\begin{array}{c}{P}_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2\\ P_1+\frac{1}{2}\left(\eta \times {\rho }_w\right)v_1^2=P_2+\frac{1}{2}\left(\eta \times {\rho }_w\right)v_2^2\end{array}$

Here, pis the density of block, $p_w$is the density of water and is the velocity of the blood flow in blockage region.

Substitute 1.06 for n , 30 cm/s for $v_1$, 1000 kg/$m^{\mathit{3}}$for $p_w$, $1.20\times {10}^{4}Pa$for $P_1$and $1.15\times {10}^{4}Pa$for $P_2$in the above relation.

$\begin{array}{r}\left(1.20\times {10}^4\mathrm{Pa}\right)+\left[\begin{array}{c}\frac{1}{2}\left(1.06\times 1000\mathrm{kg}/{\mathrm{m}}^3\right)\times \\ {\left(30\mathrm{cm}/\mathrm{s}\times \frac{1\mathrm{m}/\mathrm{s}}{100\mathrm{cm}/\mathrm{s}}\right)}^2\end{array}\right]=\left(1.15\times {10}^4\mathrm{Pa}\right)+\frac{1}{2}\left(1.06\times 1000\mathrm{kg}/{\mathrm{m}}^3\right)v_2^2\\ v_2=1.01\mathrm{m}/\mathrm{s}\end{array}$

Determination of the percentage blockage

The relation to find percentage blockage can be written as:

$A_1{v}_1=A_2{V}_2$

Here, $A_1$and $A_2$are the area before and after the blockage region.

Substitute 1.01 m/s for $v_2$and30 cm/s for$v_1$in the above relation.

$\begin{array}{r}{\mathrm{A}}_1\left(30\mathrm{cm}/\mathrm{s}\times \frac{1\mathrm{m}/\mathrm{s}}{100\mathrm{cm}/\mathrm{s}}\right)={\mathrm{A}}_2(1.01\mathrm{m}/\mathrm{s})\\ {\mathrm{A}}_2=0.29{\mathrm{A}}_1\\ \mathrm{The}\mathrm{percentage}\mathrm{blockage}\mathrm{is}\mathrm{calculated}\mathrm{as}:\\ \mathrm{\%}\mathrm{B}=\frac{{\mathrm{A}}_1-{\mathrm{A}}_2}{{\mathrm{A}}_1}\times 100\mathrm{\%}\\ \mathrm{\%}\mathrm{B}=\frac{{\mathrm{A}}_1-0.29{\mathrm{A}}_1}{{\mathrm{A}}_1}\times 100\mathrm{\%}\\ \mathrm{\%}\mathrm{B}=\frac{0.71{\mathrm{A}}_1}{{\mathrm{A}}_1}\times 100\mathrm{\%}\\ \mathrm{\%}\mathrm{B}=71\mathrm{\%}\end{array}$

Thus, the percentage blockage is 71%.