Water is flowing through a horizontal pipe of varying cross section. At section 1,the diameter is 12cm at a pressure of 80000 Pa. At section 2, the pipe has a diameter of 6cm and the pressure is 60000 Pa. Find the speed of water through sections 1 and 2.
To find the speed of water through the sections 1 and 2, you can use Bernoulli's principle, which relates the pressure of a fluid to its velocity.
Bernoulli's principle states that the total energy per unit mass of a fluid remains constant along a streamline. This principle can be expressed by the equation:
P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2
Where:
P1 and P2 are the pressures at sections 1 and 2, respectively.
ρ is the density of water.
v1 and v2 are the velocities of water at sections 1 and 2, respectively.
g is the acceleration due to gravity.
h1 and h2 are the elevations of sections 1 and 2 relative to a reference point.
Since both sections are horizontal, the elevations (h1 and h2) can be considered the same, and therefore, the terms ρ * g * h1 and ρ * g * h2 cancel out.
Our goal is to find v1 and v2. Now, let's rewrite the equation by neglecting the elevation terms:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Given:
P1 = 80000 Pa
P2 = 60000 Pa
diameter1 = 12 cm = 0.12 m
diameter2 = 6 cm = 0.06 m
First, we need to calculate the velocities v1 and v2. To do that, we need to find the cross-sectional areas of the two sections:
A1 = π * (diameter1/2)^2
A2 = π * (diameter2/2)^2
Let's calculate the areas:
A1 = π * (0.12/2)^2
A2 = π * (0.06/2)^2
Now, let's compute the areas:
A1 = π * 0.06^2
A2 = π * 0.03^2
Next, we can rewrite the equation using the calculated areas:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Since the pressures and densities are known, and we have the areas, we can solve for v1 and v2.