## Squirting

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++\$_
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### Squirting

If you point a garden hose straight up, water comes out, but not very high. If you put your thumb over the end, the water will squirt much higher.

On the other hand, if you have two hoses, one wide and one narrow, and you point them both straight up, the water will squirt higher from the wider hose.

This is confusing to me, probably because my understanding of fluid mechanics is limited to the analogy with electricity. It seems to me that the narrow hose should be equivalent to (say) a 10 ohm resistor, while the wide hose is equivalent to a 1 ohm resistor. Then, the wide hose with your finger over the tip is like a 1 ohm resistor in series with a variable resistor (depending on how much of the pipe is covered by your finger). Now, if we set the variable resistor to 9 ohms by covering most of the mouth we get a total resistance of 10 ohms, so the water should squirt as far as it does with the narrow hose; that is, not as far as with the wide hose.

This is obviously wrong.

Here is an electrical component that may work the same way as the hose. To the negative terminal of the battery, connect first a 1 ohm resistor and then a hot cathode (heated by a separate heater). The hot cathode emits electrons, which strike the anode, which is connected to the positive terminal of the battery. Now, if the cathode is heated to a very high temperature, a lot of electrons are going to come out of it, and as a result the potential difference between the cathode and the anode will be nearly zero. If we cool the cathode, fewer electrons will come out, and this will increase the potential difference, causing the electrons to go faster.

Now, it is easy to see what is going on here. The heat is needed to dislodge the electrons from their comfortable attachments to the positively-charged nuclei in the cathode before they can be attracted to the anode. This local force is the reason that Ohm's law cannot be used to understand the electron gun.

In the case of the water, something similar is obviously happening. Can someone please explain what is going on, in terms of local forces, that causes the water to squirt in violation of Ohm's Law? I suspect it has something to do with the incompressibility of water, which suggests that it should be much harder to squirt a gas than it is to squirt a liquid. Is this true? Am I totally wrong?

thoughtfully
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### Re: Squirting

One thing I see right away is that your narrow hose is not equivalent to a large hose with some portion covered. The walls of the hose are a drag on the flow of water (the narrow hose provides more drag upon a given volume/flow rate of water that is not present when the stream is flowing freely through the air.

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++\$_
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### Re: Squirting

Yes, I agree with that, but it doesn't explain to me why the water shoots further when the thumb is in place. Putting the thumb over the end surely increases the drag forces, even if it increases them less than in the case of the long narrow hose.

I am envisioning my thumb as forming a sort of nozzle, like this:
nozzle.png (1.99 KiB) Viewed 2913 times
If the fact that the thumb is not actually a nozzle is getting in the way, though, then just imagine attaching a real nozzle.

The narrowed portion is of course very short in length. Once the water gets past the nozzle and is out in the air, I agree that the forces are very different from those experienced in the small pipe. But once the water has come out of the tube, I am done observing the forces. All I care about is the amount of energy that the water has when it comes out of the end of the tube -- after that, I know what happens to it.

What I am saying is that in my simplistic "electricity" model, it should be possible to model the nozzle as a resistor. Apparently, though, this is not the case, because it doesn't obey the additive law of resistors.

ajd007
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### Re: Squirting

Let's assume that your water source provides a supply of water at constant pressure. Thus, no matter what type of hose you attach, it will have the same pressure difference through it.
If we have one wide and one narrow hose, the narrow hose will have more drag on the water due to the larger surface area to volume ratio. Thus, the wider hose will have water that shoots out higher.

When you have an initially wide hose, the water reaches a certain velocity in the wide section (the same as the hose that doesnt have a nozzle). Then when it reaches the nozzle, the water speeds up because it is incompressible. The same amount of water entering the nozzle has to exit at any given time. Thus (AV)_1= (AV)_2
Since the cross-sectional area goes down, the velocity of the water goes up and the water shoots out higher. In this case, since the nozzle is short, the additional frictional losses due to the narrowing of the pipe are negligible.

The actual velocity of the water shooting out will depend on many things including the roughness of the pipe, the temperature of the water, the dimensions of the pipe, and the pressure of the incoming water.

So I guess its mainly the length of the thin pipe that matters, since it adds additional drag to the water. If you had a short, thin pipe, I would imagine that the water would shoot up higher than the wide pipe.

You could also use Bernoulli's law and some empirical fluid mechanics formulas for frictional loss to calculate the actual velocity in each case.
This site has some good info and a calculator on pipe friction losses:
http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm

jmorgan3
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### Re: Squirting

Let's start off with some definitions:

Static Pressure: This is the pressure of the water in its own reference frame. Imagine tossing a pressure gauge into the flow, and letting it come to rest relative to the water. The pressure it would measure is the local static pressure of the water.

Dynamic Pressure: A quantity equal to (1/2)*density*speed^2. This is the increase in static pressure that the water would experience if brought to a halt (assuming you kept it at the same level, so that there was no effect from gravity and without viscosity). It is sometimes called kinetic head.

Total Pressure: This is the sum of static and dynamic pressures. It is the static pressure that water would attain if brought to rest (with the same caveats as above).

Mass flow: the rate that water is flowing through your hose. Units would be kg/s or pound-mass/s. It is given by density*area*velocity. Mass flow will be constant all the way through your hose. Because density is constant for water, this means that, for a given mass flow, area*velocity is constant. Therefore, as area decreases, velocity will increase.

In an ideal flow, without losses (from viscosity, think skin friction at the hose wall) the total pressure of a flow will be constant. In the real world, however, friction will cause a decrease in total pressure. This decrease is called pressure drop or head loss. Head loss will increase (for a given flow system, e.g. your house's plumbing) with increasing mass flow. This makes sense, because more mass flow means a higher velocity, which means more friction (for fluids, anyway).

The height of the "squirt" will scale with the velocity at the end of the hose. The higher the velocity, the higher the squirt.

Now, some boundary conditions.
-The total pressure upstream of the hose (at the spigot) will be roughly constant. If this doesn't seem right to you, imagine plugging the hose directly into the water tower. The pressure there will be a constant as long as the water level doesn't change much.
-The static pressure at the end of the hose is the same as atmospheric pressure. If this is not exactly true right at the end, it will be true a little after the end of the hose as the stream contracts or expands to come into pressure equilibrium with its surroundings.

Now back to imagining the hose. There is a constant total pressure of water going into the hose. The water then travels at constant velocity (because the hose is constant-area) through the hose, resulting in some head loss. The dynamic pressure is the same at both ends of the hose, because the velocity is the same. The total pressure is lower, due to head loss. Therefore, the static pressure must be lower at the end than at the beginning.

This leads to a bit of a conundrum. It seems that the static pressure has been uniquely determined just by the flow conditions leading up to the exit, but I said above that the static pressure is equal to atmospheric pressure. The answer is that the flow conditions adjust until they result in an exit static pressure that is equal to atmospheric.

When you put your thumb over the end of the hose, you change the area at the end, so the equilibrium conditions change. Keeping mass flow the same would mean a higher velocity at the exit, which (keeping static pressure the same) requires a higher total pressure at the end, which requires a smaller pressure drop, which is requires a smaller mass flow, violating our assumption of constant mass flow. So, mass flow decreases.

With smaller mass flow, there is less pressure drop through the hose, which leads to a higher total pressure at the end. With constant static pressure at the end, this means a higher velocity at the end and, therefore, a higher squirt.

That's half your question, but doesn't address the wider hose. Why does increasing the area also increase end velocity (squirt distance)? The answer is that pressure drop depends not just on mass flow, but also on hose area (and shape and roughness, but that's another story). When you double the hose area (throughout the entire hose, not just the end) and double the flow velocity while keeping hose length the same, you end up with a smaller pressure drop. You can think of it as "the ratio of wall area (where friction happens) to mass flow has decreased." If the pressure drop through the bigger hose is smaller, and the beginning total and ending static pressures are the same, then the bigger hose will have a higher flow velocity than the thinner one, and therefore will squirt higher at the end.

I have to go now. Later, I'll try to post some graphs to make my point clearer. If I have made any sense to anyone, I might try to explain how gases behave in nozzles. For now, http://en.wikipedia.org/wiki/De_Laval_nozzle should give you some idea.
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gmalivuk
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### Re: Squirting

++\$_ wrote:On the other hand, if you have two hoses, one wide and one narrow, and you point them both straight up, the water will squirt higher from the wider hose.
Will it? If so, I agree that it's because of the increased friction along the entire length of the narrow hose. But I've never actually done a comparison with a pair of garden hoses, since I've never had multiple sizes of hosing on hand. As a thought experiment, though, I'd actually have expected it to come out the narrower hose faster. (For example, if you hook a 4" fire hose to your garden spigot, it's not going to fountain out the end at all, but rather just gently flow over the edge of the hose.)
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++\$_
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### Re: Squirting

jmorgan3 wrote:<snip>
That makes sense. So essentially, by putting my thumb over the end, I limit the mass flow of the water, which results in decreased head loss, and this accounts for the increased dynamic pressure at the end of the hose.

I will insist on trying to make an analogy to electrical phenomena. If I take my hot cathode and cool it down, I limit the current, which results in a smaller voltage drop across the resistor, and this accounts for the increased energy of each individual electron emitted from the CRT.

I think this makes sense.

thoughtfully
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### Re: Squirting

++\$_ wrote:
jmorgan3 wrote:<snip>
That makes sense. So essentially, by putting my thumb over the end, I limit the mass flow of the water, which results in decreased head loss, and this accounts for the increased dynamic pressure at the end of the hose.

I will insist on trying to make an analogy to electrical phenomena. If I take my hot cathode and cool it down, I limit the current, which results in a smaller voltage drop across the resistor, and this accounts for the increased energy of each individual electron emitted from the CRT.

I think this makes sense.

The trouble with electrical analogies is that the cause of potential difference is just EM, whereas with the hydraulics, you have that screwball gravity with its charge being equivalent to its resistance to acceleration and such. The incompressibility doesn't translate well either, I'd imagine.

Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away.
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++\$_
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### Re: Squirting

Yeah, I think this illustrates that compressibility is an issue. Gravity, not so much, since you can just assume that your system doesn't have much of a vertical extent.

The other real problem, in my opinion, is the complete inability to explain transformers or, in general, most electromagnetic phenomena. (You can get away with a freely-rotating paddle wheel as an inductor analogy.)

Mighty Jalapeno
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### Re: Squirting

My kingdom for my old Fluid Dynamics 363 binder.

thoughtfully
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### Re: Squirting

++\$_ wrote:Yeah, I think this illustrates that compressibility is an issue. Gravity, not so much, since you can just assume that your system doesn't have much of a vertical extent.

The other real problem, in my opinion, is the complete inability to explain transformers or, in general, most electromagnetic phenomena. (You can get away with a freely-rotating paddle wheel as an inductor analogy.)

Wait.. you can have inductors, but not transformers?
If you need magnetic effects, use relativistically flowing water

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Zamfir
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### Re: Squirting

++\$_ wrote:Yeah, I think this illustrates that compressibility is an issue. Gravity, not so much, since you can just assume that your system doesn't have much of a vertical extent.

The other real problem, in my opinion, is the complete inability to explain transformers or, in general, most electromagnetic phenomena. (You can get away with a freely-rotating paddle wheel as an inductor analogy.)

Doesn't that make transformers two coupled paddle wheels in two streams?

gorcee
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### Re: Squirting

You can also show how/why this works by computing the momentum integral for the fluid.

Fluids, just like rigid bodies, enforce a conservation of momentum criterion. Putting your thumb over the end of a hose only constrains the area at the tip of the hose, so naturally we have a higher velocity and a higher squirt (just like jmorgan explained). With a wider hose, he is also right, you have a greater mass flux through an area.

Considering the inviscid case for a moment (because it's accurate enough for our discussion), the Euler equations can be shown to satisfy a Hamiltonian condition, and you can treat them as a symplectic system. That is, you have some area-preserving property w/r.t. the mass flow rate. This holds in reverse, too. Of course, I typed all this up thinking I had my aero books here in the office, but I don't. I can actually spit out the equations this afternoon once I get a chance to double-check to make sure I'm not crazy. This should clarify what I'm trying to say.

One comment, though. The de Laval nozzle isn't the greatest example for this phenomenon, because it's designed specifically with compressibility in mind. In fact, that makes it so most of the fluid mechanics behind a de Laval nozzle are misleading for our discussion.

Zamfir
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### Re: Squirting

gorcee wrote:Considering the inviscid case for a moment (because it's accurate enough for our discussion),

No, in the inviscid case the exhaust velocity stays constant when you squirt, and the mass flow simply goes down. Without friction, the kinetic energy of the water is only a function of the height, not of the geometry.

If you attach a frictionless fountain to a water tower filled with frictionless water in frictionless air, the fountain water will always spray up to the height of the water level in the tower.

gorcee
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### Re: Squirting

Zamfir wrote:
gorcee wrote:Considering the inviscid case for a moment (because it's accurate enough for our discussion),

No, in the inviscid case the exhaust velocity stays constant when you squirt, and the mass flow simply goes down. Without friction, the kinetic energy of the water is only a function of the height, not of the geometry.

If you attach a frictionless fountain to a water tower filled with frictionless water in frictionless air, the fountain water will always spray up to the height of the water level in the tower.

I don't debate this, but that's not what I was referring to with respect to the inviscid assumption. I was more referring to the symplectic property generated by the Euler equations, which govern inviscid flow. In any case, I was being unclear, but I was discussing the effect created by suddenly reducing the cross-sectional area of a fluid flow, not the wider vs narrow hose part of the problem. Also, I should have been a little more clear: I wasn't describing a perfectly inviscid fluid, but rather, was discussing the mathematical properties of the Euler equations, which are a decent approximation of flow phenomena for scales in which viscosity can be considered negligible.

bentheimmigrant
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### Re: Squirting

++\$_ wrote:If you point a garden hose straight up, water comes out, but not very high. If you put your thumb over the end, the water will squirt much higher.

On the other hand, if you have two hoses, one wide and one narrow, and you point them both straight up, the water will squirt higher from the wider hose.

I'm still struggling to understand your system. Do the two hoses diverge from the same source? If that's the case, then the height between the hoses will not follow a simple rule of "wider one shoots higher." If they both have the same flow rate, then the narrow hose will by nature have a higher exit velocity and higher squirt. If they share a source, then obviously the majority of the water would have follow the path of least resistance, i.e. will flow out the wider hose. I'd imagine someone could put together a fairly simple set of equations to estimate the flow in the two hoses based on bernoulli's equation, but it's been at least 2 years since I did that kind of thing, and the starting point is not obvious to me at the moment.
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FrancovS
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### Re: Squirting

++\$_ wrote:
jmorgan3 wrote:<snip>

I will insist on trying to make an analogy to electrical phenomena. If I take my hot cathode and cool it down, I limit the current, which results in a smaller voltage drop across the resistor, and this accounts for the increased energy of each individual electron emitted from the CRT.

I have trouble with this analogy because I would rather model the nozzle as a resistor too, because it is essentially a small hose. In the fat hose->nozzle->fat hose->end of the hose arrangement, water would have it's speed reduced by the nozzle. The real trouble is with the end of the hose, which would "dissipate" more pressure if water is flowing faster. The way it acts depends on the last "component". Maybe you could model it as a voltage source with voltage equal to the voltage of the last component attached.

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### Re: Squirting

Am I the only one who clicked on this thread thinking it would be sexual oriented?
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gmalivuk
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### Re: Squirting

I think that for a second every time I see it in my egosearch or when I search for unread posts in all the places I moderate. (Of course, for me that's compounded by the fact that I also moderate LSR, where such a sexual topic would be pretty unremarkable.)
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### Re: Squirting

big boss wrote:Am I the only one who clicked on this thread thinking it would be sexual oriented?

No. Lol.

My first thought upon opening this topic was a disappointed "Man, this thread is *so* not what I thought it was gonna be about..."