Will a funnel in a river pump water?
August 14, 2009 2:57 PM Subscribe
PhysicsFilter: The Setup: a 3 meter hose with a 1 millimeter interior diameter is attached to the small end of a funnel with a 1 meter maximum diameter and a 1 millimeter minimum diameter. This assembly is submerged in a "quick-flowing river" with the large end of the funnel facing into the current (so, "on its side"). The free end of the hose is raised 10 centimeters above the top of the funnel and this happens to be above the water level as well.
The Question: Theoretically, will water come out of the raised end of the hose? If so, what is the equation that will let me input the rate of the "quick-flowing river" get the rate of discharge from the free end of the hose?
I ran this past my engineer father, and he past some of his engineer friends. We have differing opinions.One of the engineers is pretty sure that Bernoulli's Principle will come into play.
Are there examples of this kind of "pump" in the real world? I'd imagine they're VERY inefficient.
I ran this past my engineer father, and he past some of his engineer friends. We have differing opinions.One of the engineers is pretty sure that Bernoulli's Principle will come into play.
Are there examples of this kind of "pump" in the real world? I'd imagine they're VERY inefficient.
I would approach this problem from the point of view of pressure balance. If you knew the speed of the water flowing in the river, you could calculate the pressure that pushes into the funnel. Conversely, air pressure + the column of water between the surface and the small end of the funnel will exert a downward pressure (because of gravity). If the pressure of the flowing river is greater than the “gravitational pressure”, then I would imagine water would come out of the raised end of the hose. That's probably very simplified though, and I don't know how to calculate the “rate of discharge.” The maximum rate of course is however much water enters the large end of the funnel in any second. Complex turbulences would probably prevent all that water from ending up in the small end though.
posted by spaghettification at 3:15 PM on August 14, 2009
posted by spaghettification at 3:15 PM on August 14, 2009
From a first glance, my initial thoughts are that although hypothetically possible, using the "quick flowing river" to push fluid down a 1mm tubing line with the mount above the inlet would NOT allow a discharge.
Imagine it from another perspective: a water fall, with initial velocity of water as zero. Pour it into the funnel at a (theoretical) zero height above the funnel (initial velocity is zero), and have the hose outlet above the source (mouth) of the waterfall. The water will be unable to come out of the hose.
The reason is not the volume of water, but rather the pressure created by the column of water (hydrostatic pressure) which is .433 psi/ft of differential height. Therefore, as long as the outlet of the hose is above the inlet AND the initial velocity of the water is zero, there will be no flow.
Now turn it on its side, and do the same experiment. You have the outlet above the inlet of the funnel. Your result will be the same because the velocity of the water creates a dynamic pressure. At some possible point, the velocity of the water in the stream could create a dynamic pressure great enough to push some water out of the hose, but the start point of the water to create the velocity will still be above the hose outlet. (And we aren't accounting for flow velocity losses due to streambed roughness).
The RAM pump needs to utilize the additional pressure of the closing valve to create the additional force necessary to create the pump action. Additional energy must be input into the system by virtue of a spring or closure mechanism.
It has been many years since I took fluid dynamics though, so my recollection might be a little rusty. But I think the fundamentals are correct.
posted by fox_terrier_guy at 3:21 PM on August 14, 2009
Imagine it from another perspective: a water fall, with initial velocity of water as zero. Pour it into the funnel at a (theoretical) zero height above the funnel (initial velocity is zero), and have the hose outlet above the source (mouth) of the waterfall. The water will be unable to come out of the hose.
The reason is not the volume of water, but rather the pressure created by the column of water (hydrostatic pressure) which is .433 psi/ft of differential height. Therefore, as long as the outlet of the hose is above the inlet AND the initial velocity of the water is zero, there will be no flow.
Now turn it on its side, and do the same experiment. You have the outlet above the inlet of the funnel. Your result will be the same because the velocity of the water creates a dynamic pressure. At some possible point, the velocity of the water in the stream could create a dynamic pressure great enough to push some water out of the hose, but the start point of the water to create the velocity will still be above the hose outlet. (And we aren't accounting for flow velocity losses due to streambed roughness).
The RAM pump needs to utilize the additional pressure of the closing valve to create the additional force necessary to create the pump action. Additional energy must be input into the system by virtue of a spring or closure mechanism.
It has been many years since I took fluid dynamics though, so my recollection might be a little rusty. But I think the fundamentals are correct.
posted by fox_terrier_guy at 3:21 PM on August 14, 2009
The equation you want is listed on that wikipedia page, except what you need to do is apply it at two points on a streamline - one is the entrance to the funnel and the other is the exit of the tubing.
With some rearranging,
v12=g(z2-z1)+(p2-p1)/ρ+v22/2
z2-z1 is the height between the centerline of the funnel and the top of the tube. p2-p1 is much trickier - p2 should just be atmospheric pressure, but I think you'd have to measure p1. In fact, the device you describe is essentially a tool to measure p1, given v1 and v2 are known.
fox_terrier_guy - I don't understand why the initial velocity of water would be zero? The device he's talking about is essentially an upside-down, open-ended pitot tube.
posted by muddgirl at 3:37 PM on August 14, 2009 [1 favorite]
With some rearranging,
v12=g(z2-z1)+(p2-p1)/ρ+v22/2
z2-z1 is the height between the centerline of the funnel and the top of the tube. p2-p1 is much trickier - p2 should just be atmospheric pressure, but I think you'd have to measure p1. In fact, the device you describe is essentially a tool to measure p1, given v1 and v2 are known.
fox_terrier_guy - I don't understand why the initial velocity of water would be zero? The device he's talking about is essentially an upside-down, open-ended pitot tube.
posted by muddgirl at 3:37 PM on August 14, 2009 [1 favorite]
Let's see if I can explain my thinking...
The initial velocity of water would be zero, because I was thinking of a "jug" of water. In order to gain velocity, the jug would need to be raised up. It was a convenient starting point to illustrate the idea based on pressure differentials. I think the key to this is that the outlet is above the inlet.
However, I believe you have answered the OP's second question.
posted by fox_terrier_guy at 4:19 PM on August 14, 2009
The initial velocity of water would be zero, because I was thinking of a "jug" of water. In order to gain velocity, the jug would need to be raised up. It was a convenient starting point to illustrate the idea based on pressure differentials. I think the key to this is that the outlet is above the inlet.
However, I believe you have answered the OP's second question.
posted by fox_terrier_guy at 4:19 PM on August 14, 2009
Definitely look up hydraulic rams. Also, you can at least answer the first question on a conservation of energy basis. The water has kinetic energy entering the funnel. That's converted into potential in the hose.
posted by DU at 4:40 PM on August 14, 2009
posted by DU at 4:40 PM on August 14, 2009
While we are waiting for a chemical engineer to solve this (physics - please!) someone could look up the equation for frictional loss in a pipe (DArcy?). Friction goes up with flow rate and I suspect your pipe will just be too long to get the energy balance to work.
posted by Fiery Jack at 5:41 PM on August 14, 2009
posted by Fiery Jack at 5:41 PM on August 14, 2009
My two cents: there is a pressure in the river due to the velocity of the water. This pressure divided by the density of water gives the height of hose to which water will rise. None of the diameters make a difference.
posted by digsrus at 5:48 PM on August 14, 2009
posted by digsrus at 5:48 PM on August 14, 2009
I agree with muddgirl. You want to know if the stagnation pressure (I'm pretty sure that's the right term) at the mouth of the funnel is greater than the pressure needed to support the weight of the column of water it'd be pushing up.
The diameter of the tube is a bit of a red herring I think, if the question is "will water flow" rather than "how much water will flow". (If you want to know the actual flow rate then it gets too complicated for me, what with viscosity of the water, Reynold's numbers, who-knows-what.)
posted by hattifattener at 11:50 PM on August 14, 2009
The diameter of the tube is a bit of a red herring I think, if the question is "will water flow" rather than "how much water will flow". (If you want to know the actual flow rate then it gets too complicated for me, what with viscosity of the water, Reynold's numbers, who-knows-what.)
posted by hattifattener at 11:50 PM on August 14, 2009
Velocity head:
Velocity head can be expressed as
h = v^2 / 2 g
v = sqrt (2gh)
where
v = velocity (ft/s, m/s)
g = acceleration of gravity (32.174 ft/s2, 9.81 m/s2)
(from http://www.engineeringtoolbox.com/velocity-head-d_916.html)
Ignoring the funnel, to get water 0.1m above the surface, you will need ~1.4 m/sec flow. (No friction losses). Through the funky tube arrangement (1mm dia, 3 m long... seriously?) you will get a drip or two. Then crud from the river will plug it up. And you will get no flow, until you clean it again. A cheap and dirty calculator (Hazen Williams, close enough - http://drymine.com.au/dareFX/calc/friction.html gives me 11.75m of water head (umm... about 17PSI) for a 1mm ID at 3m for a flow of 1gallon/hr. Or, to add it all up:
1.4m/s for static head of 0.1m
plus 15m/s for friction loss at 1gph
gives total of 16.4m/s (36mile/hr). I don't know what a quickly flowing river is, but if you get a good number, just run it backwards. You always have the 1.4m/s for the static head, the rest of the flow convert to head (with the equation above), then take that calculated head, and run various flows through the hazen williams friction calculator until you find a flow rate which matches that available head due to velocity. That's your flow (well, within 20%, with the caveats below. This will probably work better with a larger ID)
I'm actually not sure what the funnel will do, if anything at all. First instinct is that the forces all cancel out, except for the hole in the middle... but I'm probably wrong, it's late here.
I'm not sure what type of pump you mean. What are you trying to do? I just can't get over the 1mm ID. That's really small compared to anything I've dealt with. I don't even know if the equation above is valid for openings that small. Surface tension may be a force that comes into play...
posted by defcom1 at 7:30 AM on August 15, 2009
Velocity head can be expressed as
h = v^2 / 2 g
v = sqrt (2gh)
where
v = velocity (ft/s, m/s)
g = acceleration of gravity (32.174 ft/s2, 9.81 m/s2)
(from http://www.engineeringtoolbox.com/velocity-head-d_916.html)
Ignoring the funnel, to get water 0.1m above the surface, you will need ~1.4 m/sec flow. (No friction losses). Through the funky tube arrangement (1mm dia, 3 m long... seriously?) you will get a drip or two. Then crud from the river will plug it up. And you will get no flow, until you clean it again. A cheap and dirty calculator (Hazen Williams, close enough - http://drymine.com.au/dareFX/calc/friction.html gives me 11.75m of water head (umm... about 17PSI) for a 1mm ID at 3m for a flow of 1gallon/hr. Or, to add it all up:
1.4m/s for static head of 0.1m
plus 15m/s for friction loss at 1gph
gives total of 16.4m/s (36mile/hr). I don't know what a quickly flowing river is, but if you get a good number, just run it backwards. You always have the 1.4m/s for the static head, the rest of the flow convert to head (with the equation above), then take that calculated head, and run various flows through the hazen williams friction calculator until you find a flow rate which matches that available head due to velocity. That's your flow (well, within 20%, with the caveats below. This will probably work better with a larger ID)
I'm actually not sure what the funnel will do, if anything at all. First instinct is that the forces all cancel out, except for the hole in the middle... but I'm probably wrong, it's late here.
I'm not sure what type of pump you mean. What are you trying to do? I just can't get over the 1mm ID. That's really small compared to anything I've dealt with. I don't even know if the equation above is valid for openings that small. Surface tension may be a force that comes into play...
posted by defcom1 at 7:30 AM on August 15, 2009
The flow friction in a capillary (that 1mm ID qualifies) is related to the fourth power of the diameter. The flow in it is almost certainly laminar and the friction loss factor is 64/(Reynolds number).
As pointed out above this is a Pitot tube or piezometer but the 1mm capillary is a bit weird.
posted by jet_silver at 4:51 PM on August 15, 2009
As pointed out above this is a Pitot tube or piezometer but the 1mm capillary is a bit weird.
posted by jet_silver at 4:51 PM on August 15, 2009
Agree with above answers and have included some other considerations:
The solution begins with applying Bernoulli's principle. The sum of pressure, velocity, and gravitational potential energy should be conserved at the inlet of the tube and the outlet of the tube, so you are converting velocity at the inlet to gravitational potential energy at the outlet. That would give you an answer, as others have written above.
Regarding the funnel: a funnel that is placed inside a river (as in your problem) differs from a funnel attached to an end of a hose. In the latter case, you can assume that the water has no where else to go, so the funnel "channels" the flow into a jet. But in this problem, the funnel is just sitting inside the river, so water could just go around the funnel instead of inside it. As an early approximation, you could perhaps ignore the funnel, but I'll get into the nitty-gritty later.
To add one level of sophistication, you would factor in frictional losses at the tube itself. This website has a built in calculator, and as you can see, in addition to the velocity /pressure which we've talked about, the density and viscosity of the water, and the material of the pipe matters. Since your problem is talking about a 1mm tube diameter (relatively small diameter), pipe smoothness actually does matter as you can see in the "Moody diagram." Also, if you want to have an equation for ANY river flow velocity (including extremes) you have to consider that as velocity increases flow goes from laminar to turbulent so it sort of switches the equation you use. If you did not include a funnel in this problem, it's worth mentioning that there is also friction associated with water entering the tube at its inlet, which can be calculated by the entrance loss coefficient. For a 1mm diameter inlet, that frictional loss could be HUGE and more important of a consideration than anything else.
Lastly, the funnel is probably the most complicated part of the problem. So the situation involved is sort of like calculating the friction forces of an object flying through the air (also a fluid), except this time our reference frame places the funnel as stationary and the water as moving. A lot of this would depend on the shape of the funnel, the material it is made out of, the thickness of the front edge of the funnel and whether that edge is round or sharp.
Anyway, I think this is the framework for how I would solve the problem without going through the number crunching. Fluid mechanics textbooks actually have tables addressing friction based on all sorts of materials, shapes, and situations, but even those are approximations. For those who really care, there are also alternative approaches to arrive at a solution with a simulation based on constructing 3d models on a computer and using the finite element method to solve for flow.
posted by alex3005 at 5:57 PM on August 15, 2009
The solution begins with applying Bernoulli's principle. The sum of pressure, velocity, and gravitational potential energy should be conserved at the inlet of the tube and the outlet of the tube, so you are converting velocity at the inlet to gravitational potential energy at the outlet. That would give you an answer, as others have written above.
Regarding the funnel: a funnel that is placed inside a river (as in your problem) differs from a funnel attached to an end of a hose. In the latter case, you can assume that the water has no where else to go, so the funnel "channels" the flow into a jet. But in this problem, the funnel is just sitting inside the river, so water could just go around the funnel instead of inside it. As an early approximation, you could perhaps ignore the funnel, but I'll get into the nitty-gritty later.
To add one level of sophistication, you would factor in frictional losses at the tube itself. This website has a built in calculator, and as you can see, in addition to the velocity /pressure which we've talked about, the density and viscosity of the water, and the material of the pipe matters. Since your problem is talking about a 1mm tube diameter (relatively small diameter), pipe smoothness actually does matter as you can see in the "Moody diagram." Also, if you want to have an equation for ANY river flow velocity (including extremes) you have to consider that as velocity increases flow goes from laminar to turbulent so it sort of switches the equation you use. If you did not include a funnel in this problem, it's worth mentioning that there is also friction associated with water entering the tube at its inlet, which can be calculated by the entrance loss coefficient. For a 1mm diameter inlet, that frictional loss could be HUGE and more important of a consideration than anything else.
Lastly, the funnel is probably the most complicated part of the problem. So the situation involved is sort of like calculating the friction forces of an object flying through the air (also a fluid), except this time our reference frame places the funnel as stationary and the water as moving. A lot of this would depend on the shape of the funnel, the material it is made out of, the thickness of the front edge of the funnel and whether that edge is round or sharp.
Anyway, I think this is the framework for how I would solve the problem without going through the number crunching. Fluid mechanics textbooks actually have tables addressing friction based on all sorts of materials, shapes, and situations, but even those are approximations. For those who really care, there are also alternative approaches to arrive at a solution with a simulation based on constructing 3d models on a computer and using the finite element method to solve for flow.
posted by alex3005 at 5:57 PM on August 15, 2009
If you want to account for the friction within the tube, you will need to consult a Moody Chart.
Stagnation pressure, pitotube, head loss, entrance loss coefficient... these are all exactly the correct terms to be using to research this. In reality, you would need to know the speed of the river flow, as this enters the problem in multiple terms (stagnation pressure and reynolds #). Without doing any calcs, I agree with defcom1 in that you will probably get 1-2 drips. 1mm is really small (think of how hard it is to blow air with your mouth into a really skinny tube, water is much harder to move). I have a masters in Mechanical Engineering. Finite element methods are probably overkill for something this straightforward, just look at the chart for varying roughness walls and you will get an idea (unless you actually feel like crunching numbers) of the losses involved. You can probably ignore the pressure of the atmosphere since the difference between the entrance and exit is only 10cm, and the atmosphere weighs on the river water as well as the pipe exit (10cm is peanuts in the absolute terms of 1 atm).
alex3005 pretty much nailed it.
posted by kenbennedy at 7:27 AM on August 17, 2009
Stagnation pressure, pitotube, head loss, entrance loss coefficient... these are all exactly the correct terms to be using to research this. In reality, you would need to know the speed of the river flow, as this enters the problem in multiple terms (stagnation pressure and reynolds #). Without doing any calcs, I agree with defcom1 in that you will probably get 1-2 drips. 1mm is really small (think of how hard it is to blow air with your mouth into a really skinny tube, water is much harder to move). I have a masters in Mechanical Engineering. Finite element methods are probably overkill for something this straightforward, just look at the chart for varying roughness walls and you will get an idea (unless you actually feel like crunching numbers) of the losses involved. You can probably ignore the pressure of the atmosphere since the difference between the entrance and exit is only 10cm, and the atmosphere weighs on the river water as well as the pipe exit (10cm is peanuts in the absolute terms of 1 atm).
alex3005 pretty much nailed it.
posted by kenbennedy at 7:27 AM on August 17, 2009
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posted by Solomon at 3:01 PM on August 14, 2009 [1 favorite]