When physicians diagnose arterial blockages, they quote the reduction in flow rate. If the flow rate in an artery has been reduced to 30% of its normal value due to plaque formation, and the average pressure difference has increased by 40%, by what factor has the plaque reduced the radius of the artery?
% of its normal value
flow rate Q2 =
v2 pi r2^2 = .3 v1 pi r1^2
v2/v1 = .3 (r1/r2)^2
.5 rho v2^2 =P2=1.4 P1=1.4*.5 rho v1^2
v2^2/v1^2 = 1.4
so
v2/v1 = 1.18
so
1.18 = .3(r1/r2)^2
r1/r2 = sqrt (1.18/.3)
= 2
so reduced to half the original radius
To determine the factor by which the plaque has reduced the radius of the artery, we can use the relationship between flow rate and radius in a cylindrical pipe. 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 the radius of the artery before plaque formation is "r." After plaque formation, the flow rate is reduced to 30% of its normal value, and the pressure difference has increased by 40%.
Given that the flow rate is proportional to the fourth power of the radius, we can set up the following equation:
(0.3) = (r_with_plaque / r)^4
To find the factor by which the plaque has reduced the radius, we need to solve for (r_with_plaque / r):
(r_with_plaque / r) = (0.3)^(1/4)
Calculating this expression, we get:
(r_with_plaque / r) ≈ 0.7412
Therefore, the plaque has reduced the radius of the artery by a factor of approximately 0.7412 (or 74.12% of its original value).
To determine the factor by which the plaque has reduced the radius of the artery, we can use the relationship between flow rate, pressure difference, and radius according to the Poiseuille's law. Poiseuille's law states that the flow rate (Q) through a cylindrical tube (like an artery) is directly proportional to the pressure difference (∆P) across the tube and the fourth power of the radius (r) of the tube, and inversely proportional to the viscosity (η) of the fluid.
Mathematically, it can be expressed as:
Q = (∆P * π * r^4) / (8 * η * L)
Where:
Q is the flow rate,
∆P is the pressure difference,
r is the radius,
η is the viscosity of the fluid (assumed to remain constant),
L is the length of the artery (assumed to remain constant),
π is a constant (approximately 3.14159), and
'*' denotes multiplication.
Since we are given that the flow rate has been reduced to 30% of its normal value, we can express it as:
Q' = 0.3Q
We are also given that the pressure difference has increased by 40%, so:
∆P' = ∆P + 0.4∆P = 1.4∆P
Now, let's gather the given information:
Q' = 0.3Q (reduced flow rate)
∆P' = 1.4∆P (increased pressure difference)
We can rearrange Poiseuille's law to find the factor by which the plaque has reduced the radius (r') of the artery.
Q' = (∆P' * π * (r')^4) / (8 * η * L)
Substituting the given expressions for Q' and ∆P', we have:
0.3Q = (1.4∆P * π * (r')^4) / (8 * η * L)
Now, let's solve for the factor by which the plaque has reduced the radius (r'):
(r')^4 = (0.3Q * 8 * η * L) / (1.4∆P * π)
Taking the fourth root of both sides, we get:
r' = (0.3^(1/4) * (Q/(∆P))^(1/4) * (η/(8 * π * L))^(1/4)
Simplifying the equation further, we can express the factor as:
r' = 0.3^(1/4) * (Q/(∆P))^(1/4) * (η/(8 * π * L))^(1/4)
To find the factor, you need to know the values of Q, ∆P, η, and L, and substitute them into the equation.