Question

Water flows vertically out of a circular faucet that has a diamter d=0.050m, height = 0.55m, speed after falling that height = 3.58 m/s.

What is the volume of water that exits the faucet in 2.0 seconds? Please write formula used to solve this eq.

Answer 1

Let us consider two points of the flow of water vertically. One is the highest point 1 and the other the lower point 2 which is on the ground.

Using Bernoullis equation,

Here, density of water is r, acceleration due to gravity is g, speed at the lower point is 2, speed at the highest point is v₁ and lower point is v₂, pressure at point 1 is P₁ and at 2 is P₂.

As the water comes out from faucet, the pressure at the point 1 and 2 is atmospheric pressure.

Thus,

P₁ = P₂

Also, the water comes out from the faucet near to ground.

Thus,

h₂ =0

It follows that

Substitute 3.58 m/s for v₁, 9.8 m/s² for g and 0.55 m for h₁ in the above equation

This is the speed at which the water comes out from the faucet.

Now, the radius of the circular faucet is

r = diameter /s

= 0.050 m / 2

= 0.025 m

Now, the area of the circular faucet is

From equation of continuity, the rate of volume flow of water is

Substitute 0.025 m for r and 4.8576 m/s for v₂ in the above equation is

Thus, the volume of the water flow in 2.0 seconds is

Rounding off to three significant figures, the volume of water that exits from the faucet in 2.0 seconds is 0.0191 m³.