The build-up of plaque on the walls of an artery decreases its diameter from 1.1 cm to 0.60 cm. If the speed of the blood flow is 0.15 m/s before reaching the region of plaque build-up, find the speed of blood flow within the plaque region. [3 marks]
According to the continuity equation of steady state incompressible flow,
V*A = constant.
The area ratio A2/A1 = (0.6/1.1)^2
= 0.2975
That is the reciprocal of the speed ratio, V2/V1 = 3.36
Solve for V2. You know what V1 is.
To find the speed of blood flow within the plaque region, we can use the principle of conservation of mass. According to this principle, the mass of the fluid entering a region must be equal to the mass of the fluid leaving that region.
First, we need to calculate the cross-sectional area of the artery before and after the plaque build-up.
The area (A1) before the plaque build-up can be calculated using the formula:
A1 = πr1^2,
where r1 is the radius of the artery before the plaque build-up.
Similarly, the area (A2) after the plaque build-up can be calculated using the formula:
A2 = πr2^2,
where r2 is the radius of the artery after the plaque build-up.
Given that the initial diameter (D1) is 1.1 cm and the final diameter (D2) is 0.60 cm, we can calculate the initial and final radii as follows:
r1 = D1/2 = 1.1 cm/2 = 0.55 cm = 0.0055 m,
r2 = D2/2 = 0.60 cm/2 = 0.30 cm = 0.0030 m.
Now we can calculate the areas:
A1 = π(0.0055 m)^2 ≈ 9.50 x 10^-5 m^2,
A2 = π(0.0030 m)^2 ≈ 2.83 x 10^-5 m^2.
Next, we can use the principle of conservation of mass to find the speed of blood flow within the plaque region. According to this principle, the product of the cross-sectional area and the speed of blood flow should remain constant throughout the flow.
Therefore, we can set up the equation:
A1v1 = A2v2,
where v1 is the speed of blood flow before reaching the region of plaque build-up, and v2 is the speed of blood flow within the plaque region.
Given that v1 = 0.15 m/s, we can rearrange the equation to solve for v2:
v2 = (A1/A2)v1.
Substituting the values obtained earlier, we have:
v2 = [(9.50 x 10^-5 m^2)/(2.83 x 10^-5 m^2)](0.15 m/s)
= 3.34 x 0.15 m/s
≈ 0.50 m/s.
Therefore, the speed of blood flow within the plaque region is approximately 0.50 m/s.