Question

The buildup of plaque on the walls of an artery may decrease its

diameter from 1.0 cm to 0.90 cm. The speed of blood flow was 17 cm/s before
reaching the region of plaque buildup. (Density of blood at 37°C 1060 kg/m3)

(1) Find the speed of blood flow within the plaque region.
(2) Find the pressure drop within the plaque region.

Answer 1

(1) The speed of blood flow within the plaque region is approximately
`\(21.0 \ \text{cm/s}\)`. (2) The pressure drop within the plaque region is approximately
`\(21.2 \ \text{kPa}\)`.

(1) Speed of Blood Flow within the Plaque Region:

Given:

- Initial diameter
`\(D_1 = 1.0 \ \text{cm}\)`

- Speed of blood flow before plaque buildup
`\(v_1 = 17 \ \text{cm/s}\)`

- Reduced diameter within the plaque region
`\(D_2 = 0.90 \ \text{cm}\)`

Using the continuity equation,
`\(A_1v_1 = A_2v_2\)`, where
`\(A_1 = (\pi D_1^2)/(4)\) and \(A_2 = (\pi D_2^2)/(4)\)`:

`\[v_2 = (A_1)/(A_2)v_1\]`

Substitute the values:

`\[v_2 = ((\pi (1.0)^2)/(4))/((\pi (0.90)^2)/(4)) * 17 \approx (1.0)/(0.81) * 17 \approx 21.0 \ \text{cm/s}\]`

(2) Pressure Drop within the Plaque Region:

Using Bernoulli's equation:

`\[P_1 + (1)/(2)\rho v_1^2 = P_2 + (1)/(2)\rho v_2^2\]`

The pressure drop
`(\(\Delta P\))` is given by
`\(\Delta P = P_2 - P_1\)`.

Substitute the values:

`\[\Delta P = (1)/(2)\rho(v_2^2 - v_1^2) = (1)/(2) * 1060 \left((21.0)^2 - (17)^2\right) \approx 2122 \ \text{Pa} \approx 21.2 \ \text{kPa}\]`

The complete question is:

(attached)