water through a funnel

March 5, 2007 10:47 AM

Here's a dumb grammar school science question. If you were pour one liter of water through a funnel (say 1 inch) into a one liter container, and you were to do it all at once (so that the water "backed up" in the funnel, would that process be slower to, faster than, or equal to taking another liter of water and pouring it through the same funnel at a slower, steady rate at which hardly any water "backed up." Any references or links are appreciated...
posted by jgballard to science & nature (31 comments total)

Is the container ventilated? You would lift the funnel from the mouth of the bottle, allowing air to escape, for maximum speed when the water "backs up."

Given that, I think pouring as fast as you can would be the quick way. The backed up water would provide pressure, making the flow faster than non-backed up water could ever be.
posted by TheOnlyCoolTim at 10:58 AM on March 5, 2007

It would depend on a number of variables, such as the temperature and the pressure of the water. I would assume you could use the Newtonian fluid equation if memory serves right, as he was using water as a baseline. You'd also have to take into account the friction of the cone and the cone's shape to determine the flow of water. And then you'd have to figure in the flow of water from whatever the non-cone source is and compare the two.

I don't know what grammar school you went to, there'd be significant complex equations involved.
posted by geoff. at 11:07 AM on March 5, 2007

Try it.
posted by aubilenon at 11:09 AM on March 5, 2007 [1 favorite]

You know on second thought it'd be elementary to find the rate of water through a conical structure, but who is to say if it'd be faster than the other method. We'd have to know the rate of flow of the control.
posted by geoff. at 11:09 AM on March 5, 2007

It depends. If the rate of flow through the neck of the funnel is constant, then whether or not the water is "backed up" makes no difference whatsoever. Think about it this way: whether the water is waiting in the top of the funnel or in the bottle makes no difference, as it's still waiting to get to the neck to go through.

In real life, things are more messy, and the rate of flow may not be constant, given force pushing the water through and ventilation of the container. But you *did* say this is a grammar-school level question, so I would just assume constant flow and ventilation.
posted by The Michael The at 11:12 AM on March 5, 2007

I think the tube of the funnel would have to be pretty thin of diameter for you to see a bottleneck—with water, at least. More viscous materials, or something with large particles in it, might jam more easily.

So, assuming a tube size big enough (which I'd say 1" should definitely be) that water tension &c doesn't cause a problem, I don't think you could do a slow-pour that would be as fast as or faster than a filled-up funnel cone.
posted by cortex at 11:15 AM on March 5, 2007

Unfortunately, it is very complicated. There's a yield inversion in the curve. Up to a certain point, more pressure means more flow. But more flow means more turbulence, and the system begins to go chaotic. Above a certain point, more pressure yields less flow because of the turbulence.

Think about traffic on a high way. Up to a certain point, more cars trying to use the highway means more cars pass a certain point per minute. But eventually you start getting turbulence, traffic jams, and the flow rate can crank down drastically. It's essentially the same kind of thing.
posted by Steven C. Den Beste at 11:17 AM on March 5, 2007

Forget about all this turbulent flow jazz. The mass flow rate of the water is the diameter of the funnel, times the length, times the speed of the water falling through. That speed is in turn governed by gravity, which has only accelerated the water from the height of the funnel. Whereas if you pour, gravity has been accelerating the water since it left the lip of the jar. Therefore, as the higher you pour from, the faster the water is moving as it enters the target container. If you could pour a stream exactly the size of the funnel hole, the mass flow rate would therefore be higher. (If you can't pour a stream that steady, pour one smaller but from a little higher.)

Now that's science.
posted by DU at 11:22 AM on March 5, 2007

DU is correct. Another way to think of this is to recall that a stream of water flowing out of a tap gets thinner as it falls (go to the sink and check it out!). If you are pouring from a height the diameter of your stream of water at the source can be *greater* than the diameter of the funnel bore. By the time the water descends to the level of the funnel, the diameter of your stream will have contracted to be less than the diameter of the funnel bore.
posted by Maastrictian at 11:32 AM on March 5, 2007

Above a certain point, more pressure yields less flow because of the turbulence.

Probably. But the point where increasing pressure yields less flow might still be faster flow than no pressure. And after that point, adding more pressure would again increase the water flow.

In reality I guess it would depend on a few things: the diameter of the funnel tube, the width and depth of the top part of the funnel, the height from which you pour the water, the smoothness of the walls of the funnel, the temperature of the water, the length of the funnel, the amount of gravity there is where this experiment takes place, and so on.

But the basic idea that more water in the reservoir will generally make it flow faster can be made obvious by imagining what would happen if you put a whole ocean on top of it. That is, put a big room at normal surface pressure at the bottom of the ocean, and stick a funnel through the roof. The flow of water through the funnel would be very fast.
posted by sfenders at 11:39 AM on March 5, 2007

Seriously, just try it and time yourself. I don't know that you'll let a conclusive answer here; trying it for yourself will provide a quick and easy answer.
posted by Aloysius Bear at 11:40 AM on March 5, 2007 [1 favorite]

Assuming that the container receiving the water is ventilated, the "backed up" condition would probably flow faster than pouring slowly (that is what my experience designing storm drainage systems tells me). However, there are a lot of variables to be considered.

If you want to search for the answer in an engineering or fluid mechanics textbook, your keyword will be "orifice."

Quantifying your answer is a non-trivial exercise in fluid mechanics. This site looks like a decent starting point. If you are starting with zero fluid mechanics knowledge, it will probably be easier to actually run the experiment than to work through the equations.
posted by Uncle Jimmy at 11:57 AM on March 5, 2007

What determines the filling rate of the receptacle is the flow rate of the water exiting the bottom of the funnel. This is the only variable that matters, and it is not constant as some others have speculated. It will increase if more water is backed up in the funnel because the pressure of the water on top will push the water through the funnel bottom more rapidly. Therefore, if you go back and read the exact wording of the original question, the answer is clear-cut. Water will fill up a container faster if the funnel is backed up than if water is poured slowly into the funnel (all else being equal).

Turbulence doesn't enter into it at all. The onset of turbulence occurs when the speed of a fluid is sufficient to overcome internal friction (viscocity). This competition of effects is measured by the Reynolds number, and until a certain Reynold's number is reached (which occurs when a fluid's velocity reaches some threshhold level), the flow is characterized as "laminar." After this point, the flow is turbulent (there is often a transition regime too). This transition can be most easily seen in smoke. Light a stick of incense or a cigarette and watch the smoke come up off of it. As the smoke rises from the tip, it is moving fairly slowly and rises in smooth, regular flow lines. As it acceleartes upward, the inertial forces overcome the cohesive forces and you can observe turbulence in the chaotic and unpredictable flow of the smoke. The transition is fairly obvious. Obviously different fluids will acheive turbulence at different speeds. For example, honey is highly viscous, so you would have to move honey incredibly fast to induce turbulence. Water is less viscous than honey, but still has substantial viscosity. The speeds required to induce turbulence in water are far greater than will occur in any setup involving any reasonably sized funnel and container.
posted by SBMike at 12:20 PM on March 5, 2007

I think the simple answer is that they're equal. The funnel is the rate-limiting step; it sets the rate at which water enters the bottle. Pouring the water faster doesn't override that. Period.

I'm sure all the fluid mechanics cited above is correct, but this was presented as a "dumb grammar school science question."
posted by futility closet at 12:50 PM on March 5, 2007

This question does remind me of a problem we had in fluid mecanics in college. The problem was to determine the contours of a water clock. The clock was funnel-shaped and water was allowed to flow out of a small hole in the bottom. Because of the pressure, the flow rate from the bottom would vary with the amount of water in the container. Given this relationship, it was possible to shape the walls of the container in such a way that the level of the water in the funnel would decrease at a constant rate.

So when the clock was very full, the flow rate out the bottom was very fast. When it was nearly empty, the flow rate out the bottom was very slow. But because the funnel is wider at the top than at the bottom, the rate at which the top surface would fall was constant and could be used to measure time if markings were placed regularly in the funnel. Kind of cool that you could make a clock with no moving parts or circuitry.
posted by SBMike at 1:24 PM on March 5, 2007

DU has it (really nice analysis, btw.) As long as the flow from a full funnel is limited by the nozzle diameter, a stream just wide enough to not back up poured from a higher point will give more flow.
posted by Opposite George at 2:04 PM on March 5, 2007

I disagree with most of the above. Pouring the water as fast as you can without causing a back-up in the funnel will be faster than the backed-up funnel.

To see this, imagine pouring at very near that fastest rate, then increase the speed just the tiniest amount; as soon as you exceed that fastest rate, the funnel will fill to a considerable depth even though you are pouring only the tiniest bit faster than you were just the instant before. During the time the funnel is filling to the depth it will reach at the new pouring speed, the output at the bottom of the funnel will be less than the amount you are pouring in by a considerable margin (otherwise the funnel would not fill at all), but the amount per second you are pouring in has barely increased. Now as you continue to pour at the barely faster rate, the container you are pouring from will abruptly run out, but the funnel will still be backed up, and it will take some little time for that water to drain from the funnel, within fairly wide limits.

That is the extra amount of time pouring into a backed up funnel takes, the time it takes the funnel to drain while the pouring container is empty, because at the old pouring speed, which is only infinitesimally less than the new speed (so that the amount of time you spend pouring is almost identical in the two cases), all the water would have been in the receptacle immediately after the pouring container was empty.

Why? In dynamic terms, it's because some of the kinetic energy of the poured stream of water is dissipated in the backed up funnel, but when the funnel is not backed up, none of that kinetic energy is lost until the water is in the ultimate receptacle, so it must arrive there more quickly.
posted by jamjam at 2:12 PM on March 5, 2007

"within fairly wide limits" was meant to be at the end of the first paragraph, not the second.
posted by jamjam at 2:15 PM on March 5, 2007

For a grammar school answer it would be faster if the fluid backed up. From a fluid dynamics standpoint it is faster because the volume of water is using the full diameter of the funnel neck as opposed to just the interior surface area as it drains the water. Discounting friction which will be the same in both cases and turbulence which will be minimal, a higher volume relates directly to quicker drainage.

As others have said, the pressure of the head will also increase the speed of the water through the funnel neck. I think jamjam is assuming that you slosh the water into the funnel but kinetic energy losses would be slight IMO.
posted by JJ86 at 2:25 PM on March 5, 2007

Ok, I didn't want to break out the math here, but I think we're losing some clarity without it, so here goes:

Bernoulli's equation says: (1/2)*rho*v^2 + rho*g*h + p = constant.

Where:
rho = density of the fluid
v = fluid velocity
g = accel. due to gravity
h = height of the fluid
p = pressure

So compare the top surface of the water in the funnel (surf. 1) to the bottom surface of water in the funnel at the exit hole (surf. 2)

We know that rho doesn't change. Neither does g.

We can call the height of the bottom surface 0 and the height of the top surface remains simply h.

v1 could be approximated to be zero if the top part of the funnel is wide enough.

The pressure on both surfaces should be equal because both surfaces are touching air. Therefore the pressure on both surfaces should only be air pressure.

Simplifying this equation for these assumptions gives:

v2 = (2*g*h)^1/2

What does this mean? That the exit velocity can be uniquely determined from the height of the water in the funnel.

This relationship is steady-state also. As long as the height of water in the funnel is maintained at that height, the exit velocity will also remain constant. If the water level is increased, the exit velocity will increase as well. Of course, to maintain this new height with the increased exit velocity, it is also required that the input pour rate is increased as well (steady-state implies that things are entering and leaving at equal rates). Likewise, pouring water into the funnel more slowly will decrease both the water level in the funnel and the exit rate of the water.

Think about this for a second. The only way to decrease the water level in the funnel (your second condition where "hardly any water backed up") would be to pour slower from your original container, which would also slow the exit rate from the funnel. High water level in the funnel implies fast exit from the funnel as well as fast input. Jamjam is wrong in this point of his analysis. A marginal increase in input pouring would only give a marginal increase in funnel depth (the height would not keep rising indefinitely)

In some sense the water level in the funnel is a direct indication of the total flow rate.

In the case where you pour quickly, there will be slow flow rates at the beginning when the funnel is filling up and at the end when the funnel is emptying, but the net result will always be faster than the case where the low funnel level is maintained throughout.

Everybody arguing that it would actually be slower with a full funnel are essentially saying that the funnel empties faster when empty than when full, which violates Bernoulli's law. It doesn't work that way.
posted by SBMike at 3:08 PM on March 5, 2007

SBmike, you fail to grasp that a positive feedback loop develops the moment any water backs up into the funnel; it does so because, as soon as any water pools in the funnel the kinetic energy of the water falling from the pouring container is partially dissipated in the funnel pool. Before that, all the kinetic energy of the poured water carried it down into the the receptacle. Only when the water reaches a certain depth within the funnel will it reattain the flow rate from the bottom of the funnel it had before any backing up happened in the first place. Above that depth in the funnel, the water will flow from the bottom of the funnel faster than the fastest rate when there is no back-up, but at that point it has considerable catching up to do, which it generally will not be able to do before the Tortoise crosses the finish line. That is why I included my misplaced proviso 'within broad limits.'
posted by jamjam at 3:40 PM on March 5, 2007

So, the fastest way to pour water through a funnel is to climb up a very tall ladder directly above it, and pour the water so that it passes straight through without losing its kinetic energy? It'd have to be a large quantity of water to make up for the time spent climbing the ladder, but given that it sounds possible. Of course, it kind of misses the point of using a funnel.
posted by sfenders at 4:01 PM on March 5, 2007

Jamjam, it seems like you're comparing the case with a funnel to one without a funnel. If the funnel doesn't slow down the kinetic energy of the water, the two aren't in contact, and you might as well hypothesize the funnel out of the problem. The original question seemed to presume that there would be some level of water in the funnel, and whether keeping this level high or low would affect the flow rate. You're saying you could fill up the final container faster by pouring such that the funnel stays empty, but you'd have to pour incredibly slowly to keep the funnel from filling. You could always increase the exit rate by maintaining a steady-state pour with a higher funnel depth.

To put it another way, as long as the funnel depth is constant (whether it's constantly 0 or some other number), the pour rate is equal to the fill rate. So imagine increasing the pour rate. When the pour rate is zero, the fill rate is zero. If the pour rate is very small (so small that the funnel stays empty), the fill rate will be very small also. At some point as you pour faster, water will "back up" into the funnel as you say, but if it's steady state (funnel depth not changing), the exit rate will be this faster pour rate. It will not slow down once the funnel starts filling.

I'm certain of this. Put some money on it?
posted by SBMike at 4:24 PM on March 5, 2007

SBMike is correct. If you read the question, the OP asked whether the flow would be faster with the funnel a little backed up or a lot backed up. Turbulence has nothing to do with it. Simply, higher water level equals higher pressure and higher pressure equals higher flow rate as any five-year-old with a soda straw knows. This ain't rocket science, although SBMike nearly got us there. (By the way, I like your water clock problem. I might try it on some calculus students I'm tutoring.)

Jamjam and sfenders are talking about a the case of no funnel and no backup, which is different. I think everyone could agree with the trivial case that if there were no funnel in the way, the flow would be fastest.
posted by JackFlash at 5:54 PM on March 5, 2007

The way I read it, the question is, essentially, "a funnel gravity drains at X rate. Can you jam water down it faster by pouring from some height (i.e. without the intermediate storage in the funnel?)" I say no. The funnel acts as an orifice and limits the mass flow rate. The best you could do pouring is to break even.

But, I often have to pour a five gallon jug of almost boiling water with dissolved chemicals into a ridiculously small funnel. The strategy there is to pour, because the funnel overflows quickly, and I want to get all the chemicals in there. No spillage allowed. Bonus: the chemical dust makes you want to sneeze, while you're carefully holding up to 40 pounds of hot water jug at a precise angle.
posted by ctmf at 7:13 PM on March 5, 2007

I'm sorry to say this, JackFlash and SBMike, but I remain convinced that your understanding of this problem is deficient-- but even if so, I'm afraid I am also convinced I am unlikely to suceed in explaining it to either one of you.
posted by jamjam at 7:17 PM on March 5, 2007 [1 favorite]

If the funnel doesn't slow down the kinetic energy of the water, the two aren't in contact, and you might as well hypothesize the funnel out of the problem.

Ah, but you are over-simplifying here! The funnel might slow down the water a little, catching drops that aren't aimed exactly right, fill up a little bit, but still leave most of the water going through with most of its kinetic energy intact. I know this because I just tried it.

Lacking a proper funnel, I used a styrofoam cup with a 3/8" hole in the bottom. Held it a few inches beneath the water faucet in the kitchen sink. Less than ideal for the "aim at the hole" method of getting water through, since the aerated stream was quite turbulent at the speed required. Less than half an inch of water pooled up. The pressure it was under from the tap seemed like it would've been very easy to accomplish by pouring. Consistently it got enough water through the cup to fill up a half-litre container about 25% more quickly than adjusting the flow to keep the "funnel" close to full.

I guess the success of this will depend on the shape of the funnel. If its tube is long and tapered, it might be faster to fill the funnel to the top. But a 1" small diameter is pretty big for one litre of water as specified in the question, so I strongly suspect it'd be faster to do like jamjam suggests.

Whether this is directly relevant to the original question is unclear, I think. But maybe it's a good way to win bets with passing physicists.

Put some money on it?

Let me select the funnel, and you're on.
posted by sfenders at 8:10 PM on March 5, 2007

There are two quite different scenarios being considered here, with two different sets of reasoning required.

The first case is a stream of water being poured from above the funnel, that either doesn't hit the walls of the neck on the way down, or powers its way on through by way of extreme kinetic energy at delivery. With a suitable nozzle above the funnel to direct the flow, and suitable pressure behind that nozzle, you could fill a container pretty much arbitrarily fast this way; it will be faster than any variation of the second scenario, which is where water sits in the funnel on the way through and drains from the bottom by gravity alone.

Unfortunately for all the fans of the high-powered first scenario, it's not what the question is about:

would that process be slower to, faster than, or equal to taking another liter of water and pouring it through the same funnel at a slower, steady rate at which hardly any water "backed up."

Both these cases are variants of the second scenario, and the question explicitly states that the supply flow in the steady-flow case is slower than the flow in the "dump it all in and wait" case.

The first case (dumping all the water in at once) will in fact get the water through faster, because in this scenario the head of water in the funnel is what determines the pressure inside its neck, which in turn determines the flow rate out of the neck into the receiving container; for any given diameter and length of funnel neck, the higher the head, the faster the flow.

All of this assumes that the receiving container is more like a bucket than a bottle, so that the air displaced by the incoming water can escape. If it can't - if the receiver is in fact bottle-shaped, and the funnel blocks the escape of air so that air and water must flow in opposite directions through the funnel neck - then the fastest way to get the water down the funnel is to pour it in at such a rate and in such a way that it swirls down, making a vortex inside the neck with a narrow air passage up the middle. If water completely blocks the neck and the air has to bubble to get out, flow is slowed dramatically.

This is very similar to the case of the fastest way to empty a bottle of water. If you just turn a bottle upside down so the water gloggles and blups out, it will take longer to empty than if you give the bottle a bit of a swirl to get a vortex going in the neck.
posted by flabdablet at 8:12 PM on March 5, 2007

Jamjam, you might want to look up Torricelli's Law. It says that the velocity of water from an orifice is given by:

v = (2gh)^1/2

where g is the acceleration due to gravity and h is the height of the water. If you remember your physics that is exactly the same equation for the velocity of a falling object. This is not surprising since they both can be derived from the same law of conservation of energy.

So if you somehow manage to pour water from a height at the top edge of the funnel so that it just barely misses the wall of the funnel it will exit with exactly the same velocity as if you filled the funnel to the brim. So that gives the theoretical maximum velocity for either method (unless you are going to claim to be able to pour fluid down a funnel from a height of, say, 5 feet without touching the funnel walls.)

Any other condition other than your theoretical optimum will give a slower flow. It's very straight forward physics. I doubt your claim that you can pour a liquid at exactly your optimum rate and accuracy, neither too fast nor too slow. On the other hand it is very easy to keep the funnel full to the brim which is the theoretical maximum flow rate. Even if you could miraculously maintain your optimum flow, you still can't beat the full funnel rate.

Oh, and the sfenders case is using a pressurized water jet to accelerate the water. The OP is talking about only using gravity. If you allow other external forces then I'll just use an electric pump.
posted by JackFlash at 8:50 PM on March 5, 2007

My five-year old son and I did a quarter-scale version of this experiment earlier tonight. Our setup: 9-oz conical funnel with 0.5-inch diameter opening, timed using an iPod nano in stopwatch mode. We timed three different methods of draining water from the funnel: a) stopping the funnel with a finger, filling the funnel completely, then removing finger; b) "fast pouring" the water into the open funnel from a height of about an inch above the funnel; c) "slow pouring" the water into the open funnel to "minimize turbulence" from the same height.

For each method we performed three trials. The results were fairly conclusive:

• a) fill, then release: 2.43 sec
• b) "fast pour": 2.96 sec
• c) "slow pour": 3.79 sec

Stdevs are a) 0.08 sec, b) 0.16 sec, c) 0.02 sec.

I'll leave the modeling of this to the theoriticians!
posted by ldenneau at 9:57 PM on March 5, 2007 [2 favorites]

Sounds like fun getting your son involved. Maybe he'll be a famous scientist some day.
posted by JackFlash at 10:02 PM on March 5, 2007