Question

A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20×10⁴ Pa , while in the region of blockage it is 1.15×10⁴ Pa . Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 27.5 cm/s , and the specific gravity of this patient's blood is 1.06.

Part A: What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

Answer 1

Specific gravity of blood = 1.06

Density of blood = = 1.06 x 1000 kg/m³ = 1060 kg/m³

Pressure of the blood in the unblocked area = P₁ = 1.2 x 10⁴ Pa

Speed of flow of blood in the unblocked area = V₁ = 27.5 cm/s = 0.275 m/s

Cross-sectional area of the artery in the unblocked area = A₁

Pressure of the blood in the blocked area = P₂ = 1.15 x 10⁴ Pa

Speed of flow of blood in the blocked area = V₂

Cross-sectional area of the artery in the blocked area = A₂

Assuming the artery is horizontal.

By bernoulli's equation,

V₂ = 1.01 m/s

By continuity equation,

A₁V₁ = A₂V₂

A₁(0.275) = A₂(1.01)

A₂ = 0.2722A₁

The cross-sectional area of the artery in the blocked area is 27.22% of the unblocked area, therefore the plaque blocks 72.78% of the cross-sectional area.

A) Percentage of the cross-sectional area of the artery blocked by plaque = 72.78%