Due to a disease the main artery is enlarged and the area of cross-section bacame 1.6 times the previous value. The s of the blood in unenlarged artery is 0.4 m/s. what is the excess pressure on the walls of the artery due to enlargement. (density of blood is 1060 kg/m^3

To determine the excess pressure on the walls of the artery due to enlargement, we can use the principle of continuity. According to the principle of continuity, the product of the cross-sectional area and the speed of a fluid is constant.

Let's denote the initial cross-sectional area of the unenlarged artery as A1, the final cross-sectional area of the enlarged artery as A2, and the initial speed of blood in the unenlarged artery as v1. We are given that the area of the cross-section became 1.6 times the previous value, so we have:

A2 = 1.6 * A1

We are also given that the speed of blood in the unenlarged artery is 0.4 m/s, so we have:

v1 = 0.4 m/s

Using the principle of continuity, we can set up the following equation:

A1 * v1 = A2 * v2

where v2 is the final speed of blood in the enlarged artery.

Since we know the relationship between A1 and A2, we can substitute the value of A2 in terms of A1:

A1 * v1 = (1.6 * A1) * v2

Simplifying the equation, we get:

v1 = 1.6 * v2

Now, we can solve for v2:

v2 = v1 / 1.6

Plugging in the values, we have:

v2 = 0.4 m/s / 1.6 = 0.25 m/s

Now that we know the final speed of blood in the enlarged artery, we can calculate the excess pressure using Bernoulli's equation, which relates the pressure, speed, and density of a fluid. Bernoulli's equation is given as:

P + (1/2) * ρ * v^2 = constant

where P is the pressure, ρ is the density of the blood (given as 1060 kg/m^3), and v is the speed of the blood.

Let's denote the excess pressure as ΔP, so the equation becomes:

P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2 + ΔP

Since we are interested in the excess pressure, we can rewrite the equation as:

ΔP = P2 - P1 + (1/2) * ρ * (v1^2 - v2^2)

Substituting the known values:

ΔP = 0 - P1 + (1/2) * 1060 kg/m^3 * (0.4 m/s)^2 - (0.25 m/s)^2

Calculating the values, we have:

ΔP = -P1 + 53.2 Pa - 8.85 Pa

Therefore, the excess pressure on the walls of the artery due to enlargement is:

ΔP = -P1 + 44.35 Pa

Note that the negative sign of P1 indicates that the pressure in the enlarged artery is lower than in the unenlarged artery.

Use Bernoulli's equation, after using continuity equation to compute the velocity in the enlarged artery.

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