a tunnel was constructed with a very large pipe with a cross-sectional area of 1.5m^2. it is at a given height descending to 5m^2 and then narrows to 0.500m^2, where it terminates in a valve. if the pressure at point 2 is the atmospheric pressure and the valve is opened wide wat3er allowed to flow freely, find the speed of the water leaving the pipe.

Question

a tunnel was constructed with a very large pipe with a cross-sectional area of 1.5m^2. it is at a given height descending to 5m^2 and then narrows to 0.500m^2, where it terminates in a valve. if the pressure at point 2 is the atmospheric pressure and the valve is opened wide wat3er allowed to flow freely, find the speed of the water leaving the pipe.

Answer 1

To find the speed of the water leaving the pipe, we can apply the principle of conservation of mass. According to this principle, the mass flow rate will remain constant throughout the pipe.

First, let's find the mass flow rate at point 1 and point 3, and then equate them to find the speed at point 3.

At point 1:
The cross-sectional area is 1.5 m^2.

Let's assume the speed at point 1 is v1, and the density of water is ρ (approximately 1000 kg/m^3, if we consider freshwater).

Since mass flow rate (ṁ) = density (ρ) * velocity (v) * cross-sectional area (A), we can write:

ṁ1 = ρ * v1 * A1

At point 3:
The cross-sectional area is 0.500 m^2.

Let's assume the speed at point 3 is v3.

ṁ3 = ρ * v3 * A3

Since the mass flow rate remains constant, we have:

ṁ1 = ṁ3

ρ * v1 * A1 = ρ * v3 * A3

Since the density of water (ρ) is common on both sides, we can cancel it:

v1 * A1 = v3 * A3

Now let's substitute the given values:

A1 = 1.5 m^2
A3 = 0.500 m^2

v1 * 1.5 = v3 * 0.500

Simplifying the equation:

v1 = (v3 * 0.500) / 1.5

Now we need to find the value of v3. At point 2, the pressure is atmospheric pressure, which can be considered as constant. Therefore, we can apply Bernoulli’s equation between points 2 and 3:

P2 + 0.5 * ρ * v2^2 + ρ * g * h2 = P3 + 0.5 * ρ * v3^2 + ρ * g * h3

Since the pressure at point 2 is atmospheric pressure (P2) and there is no change in height (h2 = h3 = 0), the equation becomes:

P3 + 0.5 * ρ * v2^2 = P3 + 0.5 * ρ * v3^2

Since the pressure difference between point 3 and point 2 is zero (P3 - P2 = 0) as the valve is opened wide, we have:

0.5 * ρ * v2^2 = 0.5 * ρ * v3^2

Canceling the common factor (ρ), we get:

v2^2 = v3^2

v2 = v3

This means the speed at point 2 (v2) is equal to the speed at point 3 (v3). Therefore, we can substitute v2 for v3 in the previous equation:

v1 = (v2 * 0.500) / 1.5

Now you can calculate the speed of the water leaving the pipe by substituting the values you know into this equation and solving for v1.