How Does Pipe Expansion Affect Water Pressure?

[omgitsmonica]

Homework Statement

Water flows at 6.1 m/s in a horizontal pipe with a pressure of 3 ´ 104 N/m2. If the pipe widens to twice its original radius, what is the pressure in the wider section? Give answer in Pascals.

Homework Equations

volume = length x area
rate = volume / time
rate = volume x area

The Attempt at a Solution

Again, my professor's equations leave a bit to be desired. If anyone has a better equation (or is willing to help me solve this problem) please let me know. Sorry to post twice in a row!

 

[noumed]

Since the question is asking for Pressure, try to find an equation that includes Pressure. Also, give this a thought, would the pressure increase or decrease if the area increases?

 

[thomate1]

I would approach this problem by using the fundamental equation of fluid dynamics, the Bernoulli's equation. This equation states that the total energy of a fluid remains constant along a streamline. In other words, the sum of the kinetic energy, potential energy, and pressure energy of the fluid remains constant.

In this problem, we can use the Bernoulli's equation to determine the pressure in the wider section of the pipe. The equation is as follows:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Where P1 and P2 are the pressures at points 1 and 2, ρ is the density of water, v1 and v2 are the velocities at points 1 and 2, g is the acceleration due to gravity, and h1 and h2 are the heights at points 1 and 2.

In this problem, we are given the velocity at point 1 (6.1 m/s), the pressure at point 1 (3 x 10^4 N/m^2), and the fact that the pipe widens to twice its original radius. We can use this information to determine the velocity at point 2 and the height difference between points 1 and 2.

First, we can use the continuity equation (rate = volume x area) to determine the velocity at point 2. Since the pipe is twice its original radius, the area at point 2 will be four times the area at point 1. Therefore, the velocity at point 2 will be half the velocity at point 1, or 3.05 m/s.

Next, we can use the equation for volume (volume = length x area) to determine the height difference between points 1 and 2. Since the area at point 2 is four times the area at point 1, the height difference between points 1 and 2 will be one-fourth the length of the pipe.

Now, we can plug these values into the Bernoulli's equation and solve for P2, the pressure at point 2. The result will be in Pascals, as requested.

In summary, as a scientist, I would use the fundamental equation of fluid dynamics, the Bernoulli's equation, to solve this problem. This approach allows us to take into account the different variables involved and determine