Question

An aortic aneurysm is a cardiovascular abnormality where the primary artery supplying blood from the heart irregularly increases in size, usually in the abdomen region. Because of the aneurysm, the radius of the aorta of a patient has increased in size as compared to a healthy aorta. The speed of blood through the healthy portion of the aorta 0.34 m/s when the patient is lying flat. The pressure difference due to the increase in artery is 30 Pa Consider the case where the blood has an average density of 1060 kg/m^3 and is treated as an ideal fluid in steady flow. Consider the case where the artery starts at healthy size, increases size in the aneurysm, then decreases back to the healthy size.

A) (3 pts) Is the pressure higher or lower in the aneurysm as compared to the healthy aorta?

B) (12 pts) What is the blood flow velocity in the aneurysm region? Ignore any dissipative effects from viscosity.

C) (15 pts) If the radius of the healthy aorta is 1.3 cm, what is the radius of the aneurysm?

Answer 1

as it is an ideal fluid in steady flow,

flow rate is constant

==>area*speed of flow is constant

so for the case of aneurysm, area increases and hence speed of flow decreases

now as per bernoulli’s equation:

P1+rho*g*h1+0.5*rho*v1^2=P2+rho*g*h2+0.5*rho*v2^2

where P1=pressure in healthy aorta

P2=pressure in aneurysm

h1=h2=0 (as the person is lying flat)

v1=speed of flow in healthy aorta=0.34 m/s

v2=speed of flow in aneurysm

rho=density of blood=1060 kg/m^3

hence P2-P1=0.5*rho*(v1^2-v2^2)

as v1 > v2, P2-P1>0

so pressure in aneurysm is higher as compared to healthy aorta.

part b:

given P2-P1=30 Pa

==>0.5*1060*(0.34^2-v2^2)=30

==>v2=0.2429 m/s

part c:

area of healthy aorta*speed of flow in healthy aorta=area of aneurysm*speed of flow in aneurysm

==>pi*(0.013^2)*0.34=pi*radius^2*0.2429

==>radius=0.01538 m

=1.538 cm