The human abdominal aorta branches symmetrically near the pelvis into the left a
ID: 583089 • Letter: T
Question
The human abdominal aorta branches symmetrically near the pelvis into the left and right common iliac arteries. The typical flow rate and diameter of the abdominal aorta are 5 L/min and 2 cm. The diameter of both iliac arteries is 1 cm. What is the pressure difference between the abdominal aorta and either iliac artery near the bifurcation? Which vessel has the smaller pressure? Assume flow is uniformly distributed across the cross-section of an artery. Let's say one of the iliac arteries occludes suddenly. Now what is the pressure difference between the abdominal aorta and the remaining unoccluded iliac artery?Explanation / Answer
d1 = 2cm = 0.02 m
A1 = pi*r1^2 = pi*d1^2/4
d2 = 1 cm = 0.01 m
d3 = 0.01 m
A2 = A3 = pi*0.01^2/4
given A1*v1 = 5L/min = (5*10^-3)/60
pi*0.02^2/4*v1 = (5*10^-3)/60
v1 = 0.26 m/s
from equation of continuity
A1*V1 = A2*v2 + A3*v3
v2 = v3 = v
A1*v1 = 2*pi*0.01^2/4*v
A1*v1 = 2*A2*v2
(5*10^-3)/60 = pi*0.01^2/4*v2
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v2 = v3 = 0.53 m/s
from Bernoullis principle
P1 + 0.5*rho*v1^2 = P2 + 0.5*rho*v2^2
P1 - P2 = 0.5*rho*(v2^2 - v1^2)
P1 - P2 = 0.5*1060*(0.53^2-0.26^2)
P1 - P2 =113.05 Pa
+++++++++++
P1 > P2
iliac vessel is at smaller pressure
++++++
B)
A1*v2 = A2*V2
v2 = 1.06 m/s
P1 - P2 = 0.5*1060*(1.06^2-0.26^2)
P1 - P2 = 560 Pa
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