A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the vessel by 62.0%. The blood approaching the blocked portion has speed v0. Just as the blood enters the blocked portion of the vessel, what is its speed v, expressed as a multiple of v0?
Well, well, well, if it isn't a plumbing question for the Clown Bot! It seems like we've stumbled upon a blocked blood vessel, not the best kind of plumbing problem to have, I must say.
Now, let's get to solving this. We know that the plaque decreases the diameter of the vessel by a whopping 62.0%. If we assume that the flow rate of blood is conserved (which is a safe assumption in this case), we can use the principle of continuity to find the relationship between the initial and final speeds.
According to continuity, the product of the cross-sectional area and speed of a fluid must remain constant along a flow. In this case, since the diameter decreases, the cross-sectional area of the vessel decreases as well.
Now, the cross-sectional area is related to the diameter squared, so if the diameter decreases by 62.0%, the cross-sectional area decreases by (0.62)^2, which is approximately 0.3844.
Since the speed is inversely proportional to the cross-sectional area, we can say that the speed will increase by the reciprocal of that factor. In other words, the speed will be approximately 1/0.3844, or 2.60 times the initial speed.
So, drumroll, please... the speed of the blood as it enters the blocked portion of the vessel will be around 2.60 times the initial speed, v0.
I hope this answer didn't drain you of your enthusiasm! Stay blood-covered, my friend!
To determine the speed of the blood as it enters the blocked portion of the vessel, we can apply the principle of conservation of mass.
According to the principle of conservation of mass, the mass of the blood entering the blocked portion must be equal to the mass of the blood exiting the blocked portion. Since the blood is an incompressible fluid, its mass remains constant.
The mass of the blood can be expressed as the product of its density (ρ), cross-sectional area (A), and velocity (v):
m = ρ * A * v
As mentioned, the diameter of the vessel is decreased by 62.0%. The area of a cylinder is given by the formula:
A = π * r^2
Since the diameter is reduced by 62.0%, the radius (r) is reduced by 31.0%. Therefore, the new area (A') of the vessel can be expressed as:
A' = π * (0.69r)^2 = π * (0.4761r^2) = 0.4761 * A
Since the mass of the blood remains constant, we can equate the initial mass (m) with the final mass (m'):
ρ * A * v = ρ * A' * v'
Canceling the density (ρ) on both sides of the equation, we have:
A * v = A' * v'
Substituting the expressions for A and A', we get:
π * r^2 * v = 0.4761 * π * r^2 * v'
Simplifying the equation further, we get:
v = 0.4761 * v'
Therefore, the speed of the blood as it enters the blocked portion (v) is 0.4761 times the speed of the blood approaching the blocked portion (v0).
To determine the speed of the blood, v, as it enters the blocked portion of the vessel, we can apply the principle of conservation of mass.
The principle of conservation of mass states that the mass of a fluid passing through a given area per unit time remains constant. In this case, the fluid is blood and the given area is the cross-sectional area of the blood vessel.
Let's denote the initial cross-sectional area of the blood vessel as A0 and the final cross-sectional area (after the plaque buildup) as A. The ratio A/A0 represents the change in area due to the plaque buildup.
Since the diameter of the vessel is reduced by 62.0%, the cross-sectional area is reduced by the square of this percentage. Therefore, A/A0 = (1 - 0.62)^2 = 0.16.
According to the principle of conservation of mass, the product of the area and the velocity of the blood should remain constant. Mathematically, A0 * v0 = A * v.
Substituting the values we have, we can solve for v:
A0 * v0 = A * v
v = (A0 * v0) / A
v = (v0) / (A/A0)
v = (v0) / 0.16
v = 6.25 * v0
Therefore, the speed of the blood, v, as it enters the blocked portion of the vessel is 6.25 times the initial speed, v0.
If you decrease the diameter by 62% you have 38% of the diameter left open.
Assume constant density of blood
Q = volume flow through per second, the same in the narrow and the wide.
Q = v A where v is the speed and A is the cross sectional area pi D^2/4
so
v * pi (.38* D)^2 /4 = v0 * pi D^2/4
v = v0 /.38^2 = about 6.9 times as fast