Find the speed of flow in the second pipe of an ideal fluid

[kalisious]

Homework Statement

An ideal fluid with the density of 0.85 x 10³ flows at 7.0 m/s through a level pipe with a radius of 0.50 cm. The pressure in the fluid is 1.4 x 10⁵ N/m³. This pipe connects to a second level pipe with radius of 1.5 cm. Find the speed of flow in the second pipe

Homework Equations

Bernoulli's Equation
P₁+ 1/2density(g)(y₁) + density(g)(y₁) = P₂+ 1/2density(g)(y₂) + density(g)(y₂)
P + 1/2(density)(v) + density(g)(y)

The Attempt at a Solution

Unsure how to find pressure and velocity.

 

[Andrew Mason]

First of all, you have to get the equation right. Wtih Bernouilli's equation stated correctly you will see the solution.

AM

 

[kerimek]

I would approach this problem by using the principles of fluid mechanics and Bernoulli's equation to solve for the speed of flow in the second pipe. Bernoulli's equation states that the total energy of a fluid remains constant along a streamline. In this case, we can use the equation in terms of pressure and velocity:

P1 + 1/2(density)(v1)^2 + density(g)(y1) = P2 + 1/2(density)(v2)^2 + density(g)(y2)

Where P1 and v1 are the pressure and velocity in the first pipe, P2 and v2 are the pressure and velocity in the second pipe, and y1 and y2 are the heights of the fluid in the two pipes.

To solve for the velocity in the second pipe, we need to first find the pressure in the first pipe. We can use the given information to calculate the pressure in the first pipe using the equation P = density x gravity x height:

P1 = (0.85 x 10^3)(9.8)(0.005) = 41.93 N/m^2

We can then plug in the values for P1, v1, and y1 into the Bernoulli's equation and solve for v2:

41.93 + 1/2(0.85 x 10^3)(7)^2 + (0.85 x 10^3)(9.8)(0.005) = P2 + 1/2(0.85 x 10^3)(v2)^2 + (0.85 x 10^3)(9.8)(0.015)

Solving for v2, we get:

v2 = 5.01 m/s

Therefore, the speed of flow in the second pipe is 5.01 m/s.