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Narrowing stream of water

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RoyalCat
#1
Jun28-09, 03:12 PM
P: 671
1. The problem statement, all variables and given/known data

A. Explain why a stream of water coming out of a faucet gradually grows more and more narrow as you descend in height.

B. At a certain height below the faucet, the area of the slice of water is A0 and its speed of descent is V0. Calculate the area of the slice of water positioned h below the A0 slice.

C. What is the shape of the curve describing the envelope of the liquid (I think I'm asked for the 2d envelope, rather than the 3d envelope.

2. Relevant equations

Well, this appeared in the kinematics part of my textbook, so any hydrodynamic considerations are most likely irrelevant.
More notably, this appears among a group of questions that solely deal with ballistic trajectories*.

3. The attempt at a solution

*Haha, just as I wrote that line, a possible answer to A popped into my mind:
The individual water molecules have overall random velocity vectors (Though almost all of them point downwards to an extent because of the external water pressure, correct?).
Though assuming they all have the same average magnitude is safe, given the incredible number of particles, this also means they are pointed in different directions.

At the upper slices, we have all sorts of velocity vectors, but the molecules in the lower slices, got there faster because they had a more significant initial velocity downwards, so they got there faster. The most narrow part of the stream is consists of the 'fastest downward' molecules, with each slice being comprised of water molecules of a certain initial downward component to their velocity.

Is this the correct explanation?

This makes sense on its own, but doesn't amount to a hill of beans when I try to approach B with it. A nudge in the right direction would be greatly appreciated.

EDIT:
Ah, Gotta love the Similar Threads for: Narrowing stream of water
I think I'll be all set from here on out. Sorry for the thread clutter.
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tiny-tim
#2
Jun28-09, 03:23 PM
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tiny-tim's Avatar
P: 26,160
Hi RoyalCat!
Quote Quote by RoyalCat View Post
… The individual water molecules have overall random velocity vectors (Though almost all of them point downwards to an extent because of the external water pressure, correct?).
Though assuming they all have the same average magnitude is safe, given the incredible number of particles, this also means they are pointed in different directions.

At the upper slices, we have all sorts of velocity vectors, but the molecules in the lower slices, got there faster because they had a more significant initial velocity downwards, so they got there faster …
you have to treat water as a wood, not trees!
A nudge in the right direction would be greatly appreciated.
Nudge nudge … how does velocity depend on height? and how does area depend on velocity?
Nick89
#3
Jun28-09, 03:24 PM
P: 550
Think about conservation of momentum, possibly in combination with conservation of mass. Conservation of mass tells you the relation between the surface area of the faucet opening and the volume of water, and the surface A0 and the volume of water there. Conservation of momentum then tells you the speed of the water.

RoyalCat
#4
Jun28-09, 03:31 PM
P: 671
Narrowing stream of water

Quote Quote by tiny-tim View Post
Hi RoyalCat!


you have to treat water as a wood, not trees!


Nudge nudge … how does velocity depend on height? and how does area depend on velocity?
Hehe, yeah. The book illustrated the water continuity equation in one example (Ref'd in the question), but didn't quite put it into words. Seeing this one post sorted everything for me in my head. My explanation is completely irrelevant.

Heh, I liked the analogy. :) It's very fitting.

Thanks for all the help. :)

Just curious about your approach here, Nick89, I know you were trying to steer me to the continuity equation, but how exactly does conservation of momentum help me here? Seeing how momentum is not conserved? :3
Nick89
#5
Jun28-09, 04:00 PM
P: 550
It is if you take gravity into account.

The equation I mean is this. I'm not sure if you know this one though, it's what we used for basic fluid dynamics:
[tex]\frac{\partial}{\partial t} \iiint_V \rho \mathbf{v} \, dV + \iint_S \rho \mathbf{v} (\mathbf{v} \cdot \mathbf{n}) \, dS = - \iint_S p \mathbf{n}\, dS + \iiint_V \rho \mathbf{g} \, dV[/tex]

Looks difficult, but basically it tells you that the momentum that goes in should equal the momentum that goes out. Since the momentum at the bottom is larger (the same volume of water, with a higher velocity (due to gravity!), the area of the stream should become smaller to compensate.
RoyalCat
#6
Jun28-09, 04:14 PM
P: 671
Quote Quote by Nick89 View Post
It is if you take gravity into account.

The equation I mean is this. I'm not sure if you know this one though, it's what we used for basic fluid dynamics:
[tex]\frac{\partial}{\partial t} \iiint_V \rho \mathbf{v} \, dV + \iint_S \rho \mathbf{v} (\mathbf{v} \cdot \mathbf{n}) \, dS = - \iint_S p \mathbf{n}\, dS + \iiint_V \rho \mathbf{g} \, dV[/tex]

Looks difficult, but basically it tells you that the momentum that goes in should equal the momentum that goes out. Since the momentum at the bottom is larger (the same volume of water, with a higher velocity (due to gravity!), the area of the stream should become smaller to compensate.
Yeah, I think I follow you. Fluid dynamics looks like a daunting subject though, I reckon I'll be coming back here once I start studying it, haha.
It all simplifies greatly when you're dealing with uniformly spread and uniformly dense liquids, right?

What I got for the stream of water is the following:
At a certain height below the faucet, the area of the slice is A0, and the velocity is V0, and at a certain height h below that, the area of the slice is A1 and the velocity is V1.

At every point along the stream, the same mass of water must pass at a given time interval.

In our case, it simplifies to the expression:
A0*V0 = A1*V1
Since the mass of the water passing through the certain height is only changed by the surface of the slice (Seeing how everything else would be equal on both sides of the equation).

V1 = √(V0+2gh)
(Kinematics, or energy, if you feel like multiplying things by m)

And from there on it's just pure algebraic manipulation.
Nick89
#7
Jun28-09, 04:20 PM
P: 550
I don't know how you got to your final equation so quickly (although I am pretty sure it's correct), but yes, that is the basic idea. (You probably skipped some steps).

So yes, conservation of mass (or the continuity equation) tell you that A0 V0 = A1 V1.

Actually, you don't even need conservation of momentum for question A (but you probably do for B and C), since you know that V1 > V0, because the water is falling, accelerating, due to gravity. So, A1 must decrease to compensate for the increase in V1.
RoyalCat
#8
Jun28-09, 04:28 PM
P: 671
Quote Quote by Nick89 View Post
I don't know how you got to your final equation so quickly (although I am pretty sure it's correct), but yes, that is the basic idea. (You probably skipped some steps).

So yes, conservation of mass (or the continuity equation) tell you that A0 V0 = A1 V1.

Actually, you don't even need conservation of momentum for question A (but you probably do for B and C), since you know that V1 > V0, because the water is falling, accelerating, due to gravity. So, A1 must decrease to compensate for the increase in V1.
Yes, the question referenced an earlier example given in the book in regards to a syringe squeezing out water through a cylinder at a constant velocity, with the water being ejected through the tip of the needle at some other velocity. The question asked for the height of the spurt of water, but once you know the nozzle velocity that really is trivial.

I'd have to be crazy to try and use that many integrals at once without knowing what they're even referring to, haha!

The A0 V0 = A1 V1 equation appeared there in its entirety, it was just a matter of putting all the pieces together. Conservation of mass, the continuity equation (In not as many mathematical terms, heh), really was the peg to hold it all together.

It seems quite trivial to me though, that since the same 'amount' of fluid has to be passing through every point along the axis of movement (Intuitively, any other situation would not provide a flow, or result in a build-up of fluids at a certain point), and the only thing affecting the 'amount' of fluid for an incompressible fluid that spreads out evenly along the path that it travels, is what determines the 'size' of it at a certain height. In our case, it's the surface of the slice. Naturally, everything else is irrelevant as it makes no difference.

Thanks again. :)


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