Suppose an obstruction in an artery reduces its radius by 16%, but the volume flow rate of
the blood in the artery is remains the same. By what factor has the pressure drop across the
length of this artery increased?
Q = flow rate = v A
= v * pi * r^2
so
V1 R1^2 = V2 R2^2
V2 = V1 (R1/R2)^2
but we know R2 = (1.-16) R1 = .84 R1
so
V2 = 1.42 V1
then Bernoulii
p + 1/2 rho v^2 = constant
P1 +.5 rho V1^2 = P2 + .5 rho V2^2
P1- P2 = .5 rho (V2^2-V1^2)
so
(P1-P2)/P1 = (V2^2 -V1^1)/V1^2
= (1.4^2 V1^2 - V1^2)/V1^2
=1.4^2-1 = .96
or P2 = 1.96 P1
Thanks a lot.
To determine how the pressure drop across the artery has increased, we can make use of the relationship between flow rate, pressure, and the radius of the artery known as Poiseuille's Law. According to Poiseuille's Law, the flow rate through a cylindrical tube is directly proportional to the fourth power of its radius.
Let's assume that the initial radius of the artery is r, and the obstruction reduces its radius by 16%, resulting in a new radius of (1 - 0.16)r = 0.84r.
Since the volume flow rate remains the same, we can set up the following equation based on Poiseuille's Law:
Flow rate1 = Flow rate2
πr1^4 = π(0.84r)^4
Simplifying and cancelling terms, we have:
r1^4 = (0.84r)^4
Taking the fourth root of both sides, we get:
r1 = 0.84r
We can see that the initial and final radii are different by a factor of 0.84.
Now, let's consider the relationship between pressure and the radius of the artery, keeping in mind that the volume flow rate remains the same. According to Bernoulli's Principle, as the cross-sectional area of a tube decreases, the velocity of the fluid increases, resulting in a decrease in pressure.
As the radius of the artery decreases by a factor of 0.84, we can calculate the corresponding increase in pressure using the inverse relationship:
Pressure1 = 1/(r1)^2
Pressure2 = 1/r^2
Substituting the values, we get:
Pressure1 = 1/(0.84r)^2
Pressure2 = 1/r^2
Simplifying, we find:
Pressure1 = 1/(0.7056r^2)
Pressure2 = 1/r^2
To determine the factor by which the pressure has increased, we can calculate the ratio of Pressure1 to Pressure2:
Pressure1 / Pressure2 = (1/(0.7056r^2)) / (1/r^2) = (r^2) / (0.7056r^2) = 1/0.7056
Calculating the value, we find:
Pressure1 / Pressure2 ≈ 1.417
Therefore, the pressure drop across the length of the artery has increased by a factor of approximately 1.417.