Question

Arteriosclerotic plaques forming on the inner walls of arteries can decrease the effective cross-sectional area of an artery. Even small changes in the effective area of an artery can lead to very large changes in the blood pressure in the artery and possibly to the collapse of the blood vessel.

Imagine a healthy artery, with blood flow velocity ofv0=0.14m/s and mass per unit volume ofρ=1050kg/m3. The kinetic energy per unit volume of blood is given by

K0=12ρv20.

Imagine that plaque has narrowed an artery to one-fifth of its normal cross-sectional area (an 80% blockage).

A) Compared to normal blood flow velocity, v0, what is the velocity of blood as it passes through this blockage? (Show your work)

B) By what factor does the kinetic energy per unit of blood volume change as the blood passes through this blockage?

C) As the blood passes through this blockage, what happens to the blood pressure?

1. It increases by about 41 Pa

2. It increases by about 250 Pa

3. It stays the same

4. It decreases by about 41 Pa

5. It decreases by about 250 Pa

D) Relative to its initial, healthy state, by what factor does the velocity of blood increase as the blood passes through this blockage?

E) By what factor does the kinetic energy per unit of blood volume increase as the blood passes through this blockage?

F) What is the magnitude of the drop in blood pressure, Δp, as the blood passes through this blockage? UseK0 as the normal (i.e., unblocked) kinetic energy per unit volume of the blood. (Show your work)

Answer 1

Concepts and reason

The concepts required to solve this problem are fluid dynamics, Continuity equation, Kinetic energy, and pressure gradient due to velocity.

Initially, use the continuity equation and calculate the velocity of blood as it passes through the blockage. Later, use he kinetic energy per unit volume of fluid to compare the ratio of kinetic energies. Finally, use the pressure gradient equation to calculate the pressure drop and repeat the same for last three parts.

Fundamentals

The continuity equation is since volumetric flow of any incompressible fluid with same density is constant through any cross section that is,

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

Here, $A$ is area of cross section, $v$ is velocity of fluid, $1{\rm{ and 2}}$ are the subscripts that represent two points in a direction of flow.

The kinetic energy per unit volume of a fluid is,

$K = \frac{1}{2}\rho {v^2}$

Here, $\rho$ is the density of fluid, and $v$ is the velocity.

The pressure gradient due to velocity change is,

$\Delta p = \frac{1}{2}\rho \left( {v_1^2 - v_2^2} \right)$

Here, $\rho$ is the density of fluid, $v$ is the velocity and $1{\rm{ and 2}}$ are the subscripts that represent two points in a direction of flow.

A)

Use the continuity equation.

Rearrange the equation of continuity to solve for velocity.

$\begin{array}{c}\\{A_1}{v_1} = {A_2}{v_2}\\\\{v_2} = \left( {\frac{{{A_1}}}{{{A_2}}}} \right){v_1}\\\end{array}$

Substitute $\frac{1}{5}{A_1}$ for ${A_2}$ and $0.14{\rm{ m/s}}$ for ${v_1}$ in the above equation ${v_2} = \left( {\frac{{{A_1}}}{{{A_2}}}} \right){v_1}$.

$\begin{array}{c}\\{v_2} = \left( {\frac{{{A_1}}}{{\left( {\frac{1}{5}} \right){A_1}}}} \right)\left( {0.14{\rm{ m/s}}} \right)\\\\ = 5\left( {0.14{\rm{ m/s}}} \right)\\\\ = 0.70{\rm{ m/s}}\\\end{array}$

B)

Use the equation of Kinetic energy per unit volume.

Substitute ${v_0}$ for $v$, and ${K_0}$ for $K$ in the equation $K = \frac{1}{2}\rho {v^2}$.

${K_0} = \frac{1}{2}\rho v_0^2$

Take the ratio of kinetic energy initially ${K_0}$ and kinetic energy at blockage $K$.

$\begin{array}{c}\\\frac{K}{{{K_0}}} = \frac{{\frac{1}{2}\rho {v^2}}}{{\frac{1}{2}\rho v_0^2}}\\\\\frac{K}{{{K_0}}} = \frac{{{v^2}}}{{v_0^2}}\\\end{array}$

Substitute $0.70{\rm{ m/s}}$ for $v$, and $0.14{\rm{ m/s}}$ for ${v_0}$ in the equation $\frac{K}{{{K_0}}} = \frac{{{v^2}}}{{v_0^2}}$.

$\begin{array}{c}\\\frac{K}{{{K_0}}} = \frac{{{{\left( {0.70{\rm{ m/s}}} \right)}^2}}}{{{{\left( {0.14{\rm{ m/s}}} \right)}^2}}}\\\\ = 25\\\end{array}$

C)

Use the pressure equation.

Substitute $0.70{\rm{ m/s}}$ for ${v_2}$, $0.14{\rm{ m/s}}$ for ${v_1}$ and $1050{\rm{ kg}}$ for $\rho$ in the equation $\Delta p = \frac{1}{2}\rho \left( {v_1^2 - v_2^2} \right)$.

$\begin{array}{c}\\\Delta p = \frac{1}{2}\left( {1050{\rm{ kg}}} \right)\left( {{{\left( {0.14{\rm{ m/s}}} \right)}^2} - {{\left( {0.70{\rm{ m/s}}} \right)}^2}} \right)\\\\ \approx - 250{\rm{ Pa}}\\\end{array}$

The negative sign indicates decrease in the pressure.

D)

Use the continuity equation.

Rearrange the equation of continuity to solve for ratio of velocities.

$\begin{array}{c}\\{A_1}{v_1} = {A_2}{v_2}\\\\\frac{{{v_2}}}{{{v_1}}} = \frac{{{A_1}}}{{{A_2}}}\\\end{array}$

Substitute $0.1{A_1}$ for ${A_2}$, $v$ for ${v_2}$, and ${v_0}$ for ${v_1}$ in the above equation$\frac{{{v_2}}}{{{v_1}}} = \frac{{{A_1}}}{{{A_2}}}$.

$\begin{array}{c}\\\frac{v}{{{v_0}}} = \frac{{{A_1}}}{{0.1{A_1}}}\\\\ = 10\\\end{array}$

E)

Use the equation of Kinetic energy per unit volume.

Substitute ${v_0}$ for $v$, and ${K_0}$ for $K$ in the equation $K = \frac{1}{2}\rho {v^2}$.

${K_0} = \frac{1}{2}\rho v_0^2$

Take the ratio of kinetic energy initially ${K_0}$ and kinetic energy at blockage $K$.

$\begin{array}{c}\\\frac{K}{{{K_0}}} = \frac{{\frac{1}{2}\rho {v^2}}}{{\frac{1}{2}\rho v_0^2}}\\\\\frac{K}{{{K_0}}} = {\left( {\frac{v}{{{v_0}}}} \right)^2}\\\end{array}$

Substitute $10$ for $\frac{v}{{{v_0}}}$ in the equation$\frac{K}{{{K_0}}} = {\left( {\frac{v}{{{v_0}}}} \right)^2}$.

$\begin{array}{c}\\\frac{K}{{{K_0}}} = {\left( {10} \right)^2}\\\\ = 100\\\end{array}$

F)

Use the pressure equation.

Substitute ${v_0}$ for ${v_1}$, and $10{v_0}$ for ${v_2}$ in the equation $\Delta p = \frac{1}{2}\rho \left( {v_1^2 - v_2^2} \right)$.

$\begin{array}{c}\\\Delta p = \frac{1}{2}\rho \left( {v_0^2 - {{\left( {10{v_0}} \right)}^2}} \right)\\\\ = - \frac{1}{2}\rho \left( {99v_0^2} \right)\\\end{array}$

Substitute $0.14{\rm{ m/s}}$ for ${v_0}$ and $1050{\rm{ kg}}$ for $\rho$ in the equation $\Delta p = - \frac{1}{2}\rho \left( {99v_0^2} \right)$.

$\begin{array}{c}\\\Delta p = - \frac{1}{2}\left( {1050{\rm{ kg}}} \right)\left( {99{{\left( {0.14{\rm{ m/s}}} \right)}^2}} \right)\\\\ \approx - 1000{\rm{ Pa}}\\\end{array}$

The negative sign indicates decrease in the pressure. The magnitude of pressure drop is 1000 Pa.

Ans: Part A

The velocity of blood as it passes through this blockage is, $0.70{\rm{ m/s}}$.

Part B

The factor by which the kinetic energy per unit of blood volume change as the blood passes through this blockage is 25.

Part C

As the blood passes through this blockage, pressure decreases by about 250 Pa.

Part D

The factor by which the velocity of blood increase as it passes through this blockage is, $10$.

Part E

The factor by which the kinetic energy per unit of blood volume change as the blood passes through this blockage is 100.

Part F

The magnitude of the drop-in blood pressure, as the blood passes through this blockage is $1000{\rm{ Pa}}$.

Answer 2

It is5Vo


if u do v2 = (A1/A2) V1

where V1= initial velocity of theblood
V2= final velocity of theblood

A1= Initial crosssectionalareas
A2= Final crosssectionalareas