Water Flow from 17.0mm Faucet: When Does it Narrow to 10mm?

[jkb]

Water flows out of a 17.0 mm-diameter sink faucet at 2.20 m/s.

At what distance below the faucet has the water stream narrowed to 10 mm diameter?

I am confused on how to approach this question as it is vertical and therefore using the bernoulli's eqn. for this makes no sense? or at least i think it doesnt? thanks for any help!

 

[civil_dude]

Conservation of mass is required. The water will gain velocity as it leaves the faucet, and thus as the velocity increase due to gravity, the cross section must decrease. Find what velocity is required for the same flow to flow through the smaller cross section. Then use kinematics to find the distance it takes to reach that velocity.

 

[kyle8921]

I would approach this question by first understanding the physical properties of water flow. The Bernoulli's equation is applicable for horizontal flow, but we can still use the principles of fluid dynamics to answer this question.

The flow of water through a faucet is a combination of laminar and turbulent flow. As the water exits the faucet, it will initially have a cylindrical shape with a diameter of 17.0 mm. However, as it travels through the air, it will experience friction and turbulence, causing it to gradually narrow in diameter.

To determine the distance at which the water stream narrows to 10 mm diameter, we can use the concept of the continuity equation. This equation states that the volume flow rate of a fluid remains constant throughout the flow. In other words, the volume of water that passes through the faucet per second will be the same as the volume of water that passes through the narrowed section per second.

We can use the equation Q = Av, where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity. We know that the volume flow rate is the same for both the initial 17.0 mm diameter and the narrowed 10 mm diameter sections. We also know the velocity of the water at the faucet (2.20 m/s). By rearranging the equation, we can solve for the cross-sectional area (A) at the narrowed section.

A = Q/v = (17.0 mm)^2 x (2.20 m/s) / (10 mm)^2 = 6.38 mm^2

Now, we can use the formula for the area of a circle (A = πr^2) to solve for the radius (r) of the narrowed section.

r = √(A/π) = √(6.38 mm^2 / π) = 1.28 mm

This means that the radius of the water stream at the narrowed section is 1.28 mm. To find the distance below the faucet, we can use the equation for the circumference of a circle (C = 2πr).

Distance = C/2 = (2π x 1.28 mm)/2 = 4.02 mm

Therefore, the water stream will narrow to 10 mm diameter at a distance of 4.02 mm below the faucet. It is important to note that this calculation is based on ideal conditions and there may be some