Suppose an occlusion in an artery reduces
its diameter by 22%, but the volume flow rate of blood in the artery remains the
same. By what factor has the pressure drop
across the length of this artery increased?
To find the factor by which the pressure drop across the length of the artery has increased, we can use the continuity equation and the Bernoulli's equation.
Step 1: Start with the continuity equation:
A1 * V1 = A2 * V2
Where,
A1 = initial cross-sectional area of the artery
V1 = initial velocity of blood flow in the artery
A2 = final cross-sectional area of the artery, reduced by 22%
V2 = final velocity of blood flow in the artery, which remains the same as the volume flow rate is constant.
Step 2: Use the definition of volume flow rate, Q:
Q = A * V
Where,
Q = volume flow rate
A = cross-sectional area of the artery
V = velocity of blood flow in the artery
As given in the problem, the volume flow rate of blood in the artery remains the same. Therefore,
Q = A1 * V1 = A2 * V2
Step 3: Rearrange the equation to solve for V2:
V2 = (A1/A2) * V1
Step 4: Now consider the Bernoulli's equation, which states:
P + (½ * ρ * V^2) + (ρ * g * h) = constant
Where,
P = pressure in the artery
ρ = density of blood
V = velocity of blood flow in the artery
g = acceleration due to gravity
h = height above a reference point
Step 5: Since we are interested in the pressure drop across the length of the artery, we can ignore the gravitational terms and consider only the pressure and velocity terms. The Bernoulli's equation can be simplified to:
P1 + (½ * ρ * V1^2) = P2 + (½ * ρ * V2^2)
Step 6: Rearrange the equation to solve for the pressure difference, ΔP = P2 - P1:
ΔP = (½ * ρ * V1^2) - (½ * ρ * V2^2)
Step 7: Substitute V2 from Step 3 into the equation:
ΔP = (½ * ρ * V1^2) - (½ * ρ * [(A1/A2) * V1]^2)
= (½ * ρ * V1^2) - (½ * ρ * (A1^2/A2^2) * V1^2)
= (½ * ρ * V1^2) * (1 - (A1^2/A2^2))
Step 8: Substitute Q = A1 * V1 (from Step 2) into the equation:
ΔP = (½ * ρ * (Q/A1)^2) * (1 - (A1^2/A2^2))
Step 9: Simplify the equation further:
ΔP = (½ * ρ * Q^2) * (1/A1^2 - 1/A2^2)
Step 10: Substitute the given information that the cross-sectional area of the artery is reduced by 22%:
ΔP = (½ * ρ * Q^2) * (1/A1^2 - 1/(0.78 * A1)^2)
Step 11: Simplify the equation:
ΔP = (½ * ρ * Q^2) * (1/A1^2 - 1/(0.78^2 * A1^2))
= (½ * ρ * Q^2) * (1 - 1/0.78^2)
= (½ * ρ * Q^2) * (1 - 1/0.6084)
= (½ * ρ * Q^2) * (1 - 1.6436)
Finally, we have the factor by which the pressure drop across the length of the artery has increased:
Factor = ΔP / P1
= (½ * ρ * Q^2) * (1 - 1.6436) / P1
To determine the factor by which the pressure drop across the length of the artery has increased, we need to understand the relationship between the diameter of an artery, the flow rate of blood, and the pressure drop.
The flow rate of blood through a pipe (or artery in this case) can be described by the equation of continuity:
A1 * v1 = A2 * v2
where A1 and A2 are the cross-sectional areas of the artery at two different points, and v1 and v2 are the corresponding velocities of blood flow.
Since the volume flow rate of blood remains the same, we can denote it as Q1 = Q2.
Q1 = Q2
A1 * v1 = A2 * v2
We know that the occlusion reduces the diameter of the artery by 22%, which means the new diameter, D2, is 0.78 times the original diameter, D1.
D2 = 0.78 * D1
From the equation of continuity, we can rewrite it using the relationship between the diameter and cross-sectional area of a pipe:
A1 * v1 = A2 * v2
(pi * (D1/2)^2) * v1 = (pi * (D2/2)^2) * v2
(pi * (D1/2)^2) * v1 = (pi * ((0.78 * D1)/2)^2) * v2
By rearranging the equation, we can isolate v2/v1:
v2/v1 = [(D1/2)^2] / [((0.78 * D1)/2)^2]
v2/v1 = (D1/0.78D1)^2
v2/v1 = (1/0.78)^2
v2/v1 = 1.689
Since the pressure drop across a pipe is directly proportional to the square of the flow velocity, the factor by which the pressure drop increased is the square of the ratio of velocities:
((v2)^2)/((v1)^2) = (1.689)^2 = 2.85
Therefore, the pressure drop across the length of the artery has increased by a factor of approximately 2.85.