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 = (A_1)/(A_2)v_1`

`v_2 = (\pi/4 d^2_1)/(\pi/4 d^2_2)v_1`

`v_2 = ((d_1)/(d_2))^2 v_1`

`v_2 = ((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 +(1)/(2)\rho v_1^2 = P_2 +(1)/(2) \rho v_2^2`

`P_1-P_2 = (1)/(2) \rho (v_2^2-v_1^2)`

Replacing we have,

`\Delta P = (1)/(2) (1060) ((0.1707)^2-(0.10)^2)`

`\Delta P = 10.14Pa`

Therefore the total change at pressure is 10.14Pa