A horizontal pipe 15 cm in diameter has a smooth reduction to a pipe 8 cm in diameter. If the pressure of the water in the larger pipe is 10 "104 Pa and the pressure in the smaller pipe is 6" 104 Pa, at what rate does water flow through the pipes?
To find the rate at which water flows through the pipes, we can use the principle of continuity, which states that the volume flow rate of an incompressible fluid (like water) is constant at any point in a pipe.
The volume flow rate can be calculated using the equation:
Q = A1 * v1 = A2 * v2
where Q is the volume flow rate, A1 and A2 are the cross-sectional areas of the pipes, and v1 and v2 are the velocities of the water in the pipes.
To find the cross-sectional areas, we can use the formula for the area of a circle:
A = π * r^2
where r is the radius of the pipe.
Given that the diameter of the larger pipe is 15 cm, the radius of the larger pipe is 15/2 = 7.5 cm = 0.075 m.
Similarly, the diameter of the smaller pipe is 8 cm, so the radius of the smaller pipe is 8/2 = 4 cm = 0.04 m.
Using the formula for area, we can calculate the cross-sectional areas:
A1 = π * (0.075)^2
A2 = π * (0.04)^2
Next, we need to find the velocities v1 and v2. We can use Bernoulli's equation for incompressible fluid flow, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant at any point along a streamline.
Applying Bernoulli's equation to the two pipes, we get:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
where P1 and P2 are the pressures at the respective points, and ρ is the density of water.
Given that P1 = 10 * 10^4 Pa, P2 = 6 * 10^4 Pa, and ρ = 1000 kg/m^3, we can rearrange the equation to solve for v2:
v2 = sqrt((P1 - P2) / ρ + v1^2)
After finding v2, we can substitute it back into the continuity equation to solve for the volume flow rate Q:
Q = A1 * v1 = A2 * v2
By plugging in the values for A1, A2, v1, and v2, we can find the rate at which water flows through the pipes.