Question

Plaque in an artery: The buildup of plaque on the walls of an artery may decrease its diameter from 0.98 cm to 0.75 cm. If the speed of the blood (rhoblood = 1060 kg/m3 ) flow was 10 cm/s before reaching the region of the plaque buildup, find the speed of the blood flow in the plaque region. Assume both regions have a roughly circular cross-section

Answer 1

To solve this problem it is necessary to apply the concept related to continuity equations and fluid pressure changes.

$A_1v_1 = A_2v_2$

Where

$A_{1,2}$ = Cross-sectional Area at each section

$v_{1,2}$= Velocity at each section

Rearranging to find the value of the velocity at section 2 is

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

$v_2 = \frac{\pi/4 d^2_1}{\pi/4 d^2_2}v_1$

$v_2 = (\frac{d_1}{d_2})^2 v_1$

$v_2 = (\frac{0.98}{0.75})^2 (10)$

$v_2 = 17.0734cm/s$

The velocity at the end is 17.07cm/s

Finally the pressure change can be expressed as:

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

$P_1-P_2 = \frac{1}{2} \rho (v_2^2-v_1^2)$

Replacing we have,

$\Delta P = \frac{1}{2} (1060) ((0.1707)^2-(0.10)^2)$

$\Delta P = 10.14Pa$

Therefore the total change at pressure is 10.14Pa