Question

A patient with an aneurysm (a widening of the artery due to a weakness in the arterial wall) in their aorta is lying horizontally. The radius of the healthy part of the aorta is 1.20 cm and in the aneurysm it is 3.50 cm. The normal speed of blood in the aorta is 0.400 m/s.

a) What is the speed of the blood in the aneurysm?

b)The gauge systolic pressure of blood leaving the heart and entering the aorta is 120 mmHg. What is the absolute pressure in the aneurysm? (pblood= 1060 kg/m^3)

Answer 1

a)

r₁ = radius of healthy part = 1.20 cm

r₂ = radius of other part = 3.50 cm

v₁ = speed in healthy part = 0.4 m/s

v₂ = speed in other part

using equation of continuity

A₁ v₁ = A₂ v₂

(r₁²) v₁ = (r₂²) v₂

(1.20)² (0.4) = (3.50)² v₂

v₂ = 0.22 m/s

b)

P₁ = pressure in aorta = 120 mm Hg = 1.6 x 10⁴ pa

P₂ = pressure in aneurysm = ?

using equation of bernoulli

P₁ + (0.5) v₁² = P₂ + (0.5) v₂²

(1.6 x 10⁴) + (0.5) (1060) (0.4)² = P₂ + (0.5) (1060) (0.22)²

P₂ = 16059.1 Pa = 120.5 mm Hg