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Questions about Sumps [retitled]

Monty, there are actually two other outlets. Oh, actually 3, there is a bypass in the sump as well.

Dan, actually penductors and eductors are kind of out of 'style' because they take a high pressure pump and they are big and ugly. They do a great job though. The thing people are moving towards are prop 'pumps' like the Tunze or the Vortech or DIY maxijet mods. Good stuff if you are looking for flow. Come to RDO dan...you need more things to read!
 
DHyslop;93184 said:
Most everything you typed I agree with. I contend that just by closing the "thought experiment" valve, flow through the return will increase. The bit of water in the vent would strongly prefer to be back down in the sump, just like any one of the many buckets I held high above my head this afternoon would strongly prefer to be on the ground. The only thing that keeps it up there is some exertion on the part of the pump. That is exertion that would be increasing flow if the valve were closed.

I go back to the analogy of a pump with a simple vertical pipe supporting water at maximum head. We agree the water would much prefer to be back in the sump, and only by the great exertion of the pump does it stay up. We both agree this is how it works there. Your contention in this case is that because there's water flowing through the system any bit of water in the vent is immune from having to be held up by the pump's exertion because the system would be in a steady state? In the vein of thought experiments, imagine your vent apparatus, only exaggerated to its logical extreme: The vent tube is tall, and there's only a very small pinhole hole for flow into the aquarium. The water column will be just a hair below max head. There will be a little bit of flow through the pinhole, but the vast majority of the pump's energy is supporting the water column. As we slowly enlarge the pinhole the water level in the vent slowly comes down until we reach the flow and head conditions in your drawing. Yet, at this point--with the water level in the vent still above that in the tank--you believe that the pump is no longer using any of its energy to keep it there, correct?

I'm glad we're converging on thought experiments. I still disagree with your contention in the first paragraph, though. For what it's worth, I've talked to a number of other smart people, all of whom I respect, and found some who agree with each of our points of view.

There is an important principle I'm using, that it sounds like you object to. That is: in an incompressible fluid, if there is no flow, then at that point all the forces are balanced. Another way of looking at that is that if there is no flow somewhere, you can stick a valve at that point, and opening or closing it will not change the behavior of the system at all. In places where there is flow, it is driven by a pressure gradient, and in those places, closing the valve will impede the flow, and the valve will be under load. In the steady state model I'm using, the pressures, the flow magnitude, and flow direction all do not change with time at all, so we can choose any point on the diagram and write down "the pressure here has this value." Closing a valve at any location where the flow is zero will not change the steady state at all. Closing a valve where there is flow will change the system, and the steady state will no longer apply.

When analyzing hydraulic systems like this, pressure and force can be equated, because, while pressure is force per unit area, the force along a pipe is divided by the area of the cross section of the pipe. Also, the pressure difference of a column of water is exactly proportional to the height of the pipe, because again, when you take the cross section out, that removes 2 dimensions, so the weight per unit area is only proportional to the height of the water column. That's the reason why the hydraulic head stuff works, and the reason for Pascal's Principle, and why Pascal could burst a barrel by putting a thin pipe up from it and filling that with a tiny amount of incompressible water.

In any place where there is no net flow, the pressure can be determined by the height of the water above that point, regardless of what may be resisting the pressure. The water doesn't know or care if the pressure at the bottom is maintained by a closed valve, a pump, or something else. Similarly, at the outlet of the pump, the pump only knows about the pressure. At a given pressure, it can move some amount of water. At the pressure at the bottom of its hydraulic head column, the amount of water it can move is zero. The pump doesn't know or care what is causing that pressure. In fact, if you put a valve immediately at the outlet of the pump and shut it, the pump will be just as stressed out over that as if it's pumping against the full hydraulic head column of water. The force holding back the pressure will be in some elasticity in the valve or the pipe walls in that case, rather than in a column of water, but the pump doesn't know the difference... it can't, all it sees is that pressure.

It's also worth noting that the incompressibility constraint causes two other effects that make things simpler: the flow per unit cross sectional area is constant except at 3+ way intersections. Since our hypothetical system has constant diameter pipe everywhere from the pump to the outlet nozzle (where it gets smaller), that means that the flow is exactly the same everywhere in the pipe from the pump to the valve, and from the valve to the outlet. Another effect of the incompressibility is that the resistance to flow can be lumped up for the piping between junctions and treated as a lump sum, which makes things convenient.

(For convenience, I'm going to set atmospheric pressure to zero, since it's a fixed offset and we only care about differences.)

In your initial case of the pump maintaining the pressure against the max head height of water, the analysis is straightforward. None of the water on the outlet side of the pump is moving. All forces are in equilibrium everywhere. Right at the outlet of the pump, the pressure is Pmax. The height of water above that is exactly the amount of water to generate Pmax of pressure, because the pressure is linear in the height of the water. The pressure goes down linearly from Pmax to zero along the column of water, at every point the weight per area of water above that point is exactly equal to the pressure. There is no movement of water into or out of the column. Because of this, if we were to close a valve anywhere between the pump and the high water line, nothing would change-- there's no net force pushing on the water anywhere, so the water doesn't move, so a valve stopping the water from moving doesn't have any flow to resist. This means, counterintuitively, that although the pump is spinning under heavy load to maintain the pressure, it is not moving any water at all; it is running at 0% efficiency, and all its work is going into heat (maybe indirectly by churning water or something). This is why pumps in this situation tend to melt, explode, blow gaskets, stall, or otherwise fail in nasty ways. It's important to notice that the pump doesn't know or care about the column of water; it's just reacting to the pressure Pmax at its outlet, and the column of water doesn't know or care where the Pmax is coming from, whether it's a pump spinning like mad, or a closed valve doing nothing.

Now, in your proposal, at some level in the vertical shaft, we open up some small flow out. This would lead to a transient non-steady-state situation first: at the place where this was opened, some water will start going out the hole. That will mean that the pressure at that point will be a bit lower than it was before, throwing the stability out of whack for a bit. Two things will happen because of this: above the hole, the weight of the column of water will be higher than the pressure, by the amount of pressure relief from the hole. That will mean that, at first, some water will go down from above the hole and out the hole. As that starts to happen, the height of the water column above the pump will go down, and the pressure at the pump will be less than Pmax. Hooray, says the pump, it's efficiency will go up to more than 0%, and it will no longer be churning the local water into heat. The system will reach a steady state again, and that state will have these properties: The pressure at the pump will be exactly the pressure that the pump can move water at the same rate that it's going out the hole. The pressure drop at the hole will be exactly the amount corresponding to the height difference in the water above the hole. The difference in pressure causing the flow of water out the hole is exactly the difference in pressure the pump provides down at the bottom that causes flow out the outlet. The water column above the hole is exactly the height such that no water goes up or down past the hole, it only goes sideways out the hole. The force holding up that water is exactly the pressure that would be there anyway. The column of water above the hole is just like it was before we opened the hole-- it's just a pressure-maintaining system in the steady state, that's not contributing to the flow at all, and it's maintaining the same pressure that would be there if a valve just above the hole were closed.

No matter how big or small that hole is, it's the outflow through the hole that determines the pressure at the pump outlet, and hence the steady-state efficiency of the pump. The water column in the vent will self-correct to the pressure above the hole exactly until it makes no difference at all to the pressure down at the pump, and is providing the exact same pressure containment as a closed valve, or a closed pipe with no vent at all. The only time has an effect is when the system is seeking equilibrium after something changes, like the power goes out to the pump, or someone opens or closes a valve, or the octopus sticks a plug into the outlet hole.

(continued)
 
The only other thing I might add to your response to dw is that returns usually are restricted to some degree. Discharge is area multiplied by velocity and most aquarists prefer a high velocity coming out of the return, so the nozzles are often a bit pinched. The plumbing is otherwise kept wide to reduce frictional losses.

Dan

I certainly will admit that if the discharge outlet is a small nozzle to encourage high velocity at the expense of flow, that will make the water level rise higher in the vent, for the same reason that if you have a hose with a nozzle, it gets a lot stiffer from the pressure. And that may make my vent design impractical, if it makes the height particularly high. Although if it's a major flow constriction that causes the water to get close to the max head of the pump, that will make the pump run at a very low efficiency, and will both not move much water into the tank (so very little will go out the overflow through the filtration) and will also make the pump very stressed and likely to burn out. The vent will neither help nor hinder this condition, it's all about the nozzle (and anything else that impedes flow, like barnacles growing inside the pipes and such.)
 
Thales;93222 said:
Monty, there are actually two other outlets. Oh, actually 3, there is a bypass in the sump as well.

Ok, so regardless of whether Dan or I is right in theory, it sounds like the reason this doesn't work in practice is that the vent would have to be really tall... yay, empiricism!
 
monty;93225 said:
That will mean that, at first, some water will go down from above the hole and out the hole. As that starts to happen, the height of the water column above the pump will go down, and the pressure at the pump will be less than Pmax. Hooray, says the pump, it's efficiency will go up to more than 0%

Monty--It feels to me that you've planted all the trees, but can't see the forest. You've got a nice description of everything that happens when you prick the hole with transient non-steady-state equilibrating periods, etc, etc etc--I understand and agree with all of that. Here are the end members: The pump is operating at 0% efficiency when its just holding up water. Its at its maximum efficiency when its just moving water. Thus, when it is doing both (moving water out the outlet and holding some volume of water above that in the vent) it cannot be at its maximum efficiency.

I don't know how else I can say it, but since I'm closing on a house next week maybe I can make an exact model and measure the flow rates with a bucket.

Dan
 
DHyslop;93287 said:
Monty--It feels to me that you've planted all the trees, but can't see the forest. You've got a nice description of everything that happens when you prick the hole with transient non-steady-state equilibrating periods, etc, etc etc--I understand and agree with all of that. Here are the end members: The pump is operating at 0% efficiency when its just holding up water. Its at its maximum efficiency when its just moving water. Thus, when it is doing both (moving water out the outlet and holding some volume of water above that in the vent) it cannot be at its maximum efficiency.

I don't know how else I can say it, but since I'm closing on a house next week maybe I can make an exact model and measure the flow rates with a bucket.

Dan

I'm all for experiments.

I realized in the shower this morning that one simple way of describing where I disagree is that when you say "the bucket of water wants to come down" or whatever, I agree that it has a downward force of gravity, so it needs a force to keep it up, but not that it requires energy to keep it up. It so happens that spinning pumps and animal muscles are systems that use energy to generate forces, but there are other thing, like shut valves, that can happily provide a force to keep something up without using energy at all.

In terms of the pump efficiency/ pinhole thing, I think that the system is such that the water column will go down as a function of the pump's efficiency... in fact, the same pressure that's holding up the water column results from the pump working against the resistance in flow, the rise to the hole, etc. If you think of the water in the vent, rather than something that needs to be maintained, as just a springy energy reservoir that will equalize the pressure so that it's the same as it was when the sealed pipe with no vent was there (or the valve was closed with no water above it), that's pretty much how I'm looking at it... or, another way of looking at it is that the height of the water column is controlled by the flow resistance, and self-adjusts to the level where it is neither helping nor hindering what was going on without the vent.

I'm also not sure about "maximum efficiency"-- it still has to fight the rise to the upper tank and the resistance in the pipe and the nozzle and stuff. If it's pressurizing Thales' return line enough to get more than a foot of rise, then his pump must be sized such that it can push water at a rate that makes the resistance of the outflow a pretty significant flow impedence (otherwise, the flow would be high but the pressure at the top would be fairly low). I'll take "optimal efficiency," but I bet the maximum efficiency for the pump is when you disconnect it so it's pumping water from the sump to the sump, with no rise and no pipe resistance.

Anyway, I'd actually (perversely) appreciate it if you'd be as pedantic as possible in pointing out where the "trees" lead to your "forest," 'cause I'm either missing it, or you have an invalid assumption. I'm most frustrated by knowing I either am wrong and don't understand why, or I'm right but I can't articulate why. (I'd be frustrated by the third possibility that we're both wrong, too, but that seems unlikely.) Although I'll admit to having a stubborn streak, I also am usually not so thick I can't see why I'm wrong...
 
monty;93292 said:
...when you say "the bucket of water wants to come down"...I agree that it has a downward force of gravity, so it needs a force to keep it up, but not that it requires energy to keep it up...


How can a pump provide force without using energy?
 
monty;93292 said:
I'm also not sure about "maximum efficiency"-- it still has to fight the rise to the upper tank and the resistance in the pipe and the nozzle and stuff

This is kind of what I mean with the forest and the trees thing. Those factors are the same whether there's a vent or not so they cancel out--by even bringing them up you're pulling out the hand lens and looking at the tree bark. At best a distraction.

I'll be pedantic--apology in advance: :biggrin2:

1. The pump has a maximum efficiency where as much of its energy as possible is being used to move water. There's an ultimate maximum efficiency if it was moving water horizontally without friction. That's not meaningful to our discussion at all, we care about a relative max efficiency which is the maximum efficiency it can have at the identical given conditions of both our scenarios ("the fight to rise to the upper tank and resistance in the pipe..."). I assumed this was tacit to the discussion so I believed the term "maximum" unmodified would be adequate.

2. As you pointed out, the pump has a minimum efficiency of zero where none of its energy is being used to move water. All of the energy is being spent to hold that water. We know that the energy is holding the water because if we cut off the supply of energy the water is no longer held. A plug is in no way an analogy because a plug does not hold the water only if it is consuming energy.

3. The pump has that minimum efficiency (one extreme end) when it is holding water at its maximum hydraulic head. All the energy is spent supporting the water, none is spent moving water.

We're just looking at steady states--any transition periods between these end-members (say the first few seconds after opening a hole) are ephemeral and their discussion offers no clarity.

4. We now have two end-member scenarios: in one, all the pump's energy is holding up water. The other, all the pump's energy is moving water. Lets put this together in a continuum of infinitessimal steady state conditions between the two:

a. All the energy is holding water up, none to flow. The water level in the vent is at max pump head.

b. Most of the energy is holding water up, a little to flow. This is the pinhole, and the water level in the vent is near max pump head.

c. Half the energy is holding water up, half to flow. For the sake of argument lets say the water level is halfway up the vent.

d. A little of the energy is holding water up, most to flow. The water level in the vent is just a little bit above the outlet.

e. None of the energy is holding water up, all to flow. The water level in the vent is at the same level as the outlet, no higher.

Your drawing of what you take as maximum efficiency physically looks like d. There is a few inches of water above the outlet. I believe your argument is that once the energy was used to lift the water there it doesn't need to be used again and the pump is working at max efficiency to move water. If that were correct, then wouldn't the pump also be working at maximum efficiency in b., the pinhole? Would it be at maximum efficiency regardless of the height of water in the vent? Do you believe the pinhole has the same efficiency?

The logical conclusion to this line of reasoning is this: If we agree the pinhole is less efficient than your drawing, then isn't any water level below your drawing going to have a higher efficiency?

Dan
 
Michael Blue;93296 said:
How can a pump provide force without using energy?

Your chair is providing a "normal" force equal to but in opposite direction to your the force of your weight to keep your bum from falling to the ground, and it does this without using energy. Monty thinks the water in the vent is also being held up by a normal force like this, the force of a shelf keeping a bucket up, a plug keeping water in or a ratchet holding a jack up. I contend that the scenario requires dynamic support, like the helicopter or holding a bucket of water over my head.

As evidence for my point of view, I submit that the water behaves like the helicopter or my arm and not like the jack or the shelf. When the pump is turned off the water rushes back down.
 
Michael Blue;93296 said:
How can a pump provide force without using energy?

Some pumps, like a reciprocating pump, can provide a force against backflow without using energy, because they include one-way valves, and there are other things, like any solid object, that can generate force without using energy-- if I set a book on a table, I don't have to have the table plugged in to the wall to keep the book from falling through to the floor. But the table is also not going to lift the book higher into the air: that requires work, which requires energy.

But that's not what I'm trying to get at. The pump is using energy to move the water through the pipe, which pressurizes the water. That pressure is there whether it's in a closed pipe, or whether it's in an open pipe with a column of water pushing down, because the column of water is supplying the same force role as the pressure-containment of the pipe was in the closed system.

I'm not actually saying that the pump is doing anything different in the two (closed and vented) systems, I'm saying that adding the column of water removes the exact same amount of strain energy in the pipe; they're equal forces containing the pressure generated by the pump, so the pump is generating the same flow against the same pressure in both situations. And the water height in the pipe will naturally seek the level where this is the case, if the vent pipe is high enough to accommodate it, anyway.
 
It occurs to me that your question may have led to the core of my disagreement with Dan on this, Michael.

My analysis says that because the top of the vent is open to the air, the column of water provides the same containment force for the pressure in the pipe that the pipe itself did before. The pump has to do work to pressurize the system, which is more visible when there's a column of water, but there is energy stored in either the water column or the stretching of the pipe which provides a force that balances things out in the steady state.

And once the pump has pressurized the system, either by distorting the pipe or raising the water, it no longer has to provide "maintenance" energy, although if it stops, then the system will have a path to depressurize, which is much more obvious when you see a bunch of water drain out, but is almost invisible in the case of the pipe, which is probably rigid enough that it only has expanded in the tenths of millimeters, but it's stiff enough that that deflection provides as much force as a big column of water.
 
monty;93305 said:
My analysis says that because the top of the vent is open to the air, the column of water provides the same containment force for the pressure in the pipe that the pipe itself did before. The pump has to do work to pressurize the system, which is more visible when there's a column of water, but there is energy stored in either the water column or the stretching of the pipe which provides a force that balances things out in the steady state.

Now I see where you're coming from, but I still think its going to take energy to keep that water up there, which is going to inherently reduce the efficiency (a la my pedantic other post).
 
I can't argue the work point as my education in this area is limited, however, I have an idea that might suggest an answer.

Build a simple set up with a tiny pump (it could just flow back into the pump bucket) initially eliminating the vent pipe. Run the system for say 15 minutes and record the water temp before and after the trial.

For the second test, dump the water and refill the bucket with water at the same beginning temperature, add the vent pipe and rerun the test.

If there is no difference in temperature at the end of the two experiments, it would seem that the pump worked no harder with the siphon breaker.

If I was not behind in some coding I would try it just for fun myself.
 
DHyslop;93301 said:
This is kind of what I mean with the forest and the trees thing. Those factors are the same whether there's a vent or not so they cancel out--by even bringing them up you're pulling out the hand lens and looking at the tree bark. At best a distraction.

As opposed to hovercrafts and helicopters, which are directly relevant :roll:

I'll be pedantic--apology in advance: :biggrin2:

hey, I like pedantry... so much I'll be pedantic back. But anyway, when 2 people smart enough to be able to understand something are at odds, it seems like pendantry and experiment are all that's left. Er, unless we prefer a "hearts and minds" campaign to getting to the actual answer...

1. The pump has a maximum efficiency where as much of its energy as possible is being used to move water. There's an ultimate maximum efficiency if it was moving water horizontally without friction. That's not meaningful to our discussion at all, we care about a relative max efficiency which is the maximum efficiency it can have at the identical given conditions of both our scenarios ("the fight to rise to the upper tank and resistance in the pipe..."). I assumed this was tacit to the discussion so I believed the term "maximum" unmodified would be adequate.

The reason I'm not too keen on the name "maximum" it's the maximum given a bunch of stuff, and we seem to disagree on whether messing that stuff will make a difference in the efficiency of the pump. But I'm ok with thinking of it as the same efficiency it has in the closed system with no vent at all (or with the valve closed, which I think we agree is the same thing, right?) A more direct reason I care, though, is that your pinhole is changing the flow resistance to something that'll effect the pump's efficiency.


2. As you pointed out, the pump has a minimum efficiency of zero where none of its energy is being used to move water. All of the energy is being spent to hold that water. We know that the energy is holding the water because if we cut off the supply of energy the water is no longer held. A plug is in no way an analogy because a plug does not hold the water only if it is consuming energy.

In my worldview, this description has a subtle problem in it. When the pump is at zero efficiency, it is pumping against a pressure that is causing it to be unable to move any water. Therefore, all of the energy going into the pump is being turned into something other than flow (which moves water, and does mechanical work). Ultimately, this is heat, although it may stir up the sump or something. The pump really doesn't see anything beyond the pressure it's fighting against at its outlet, so it isn't doing anything different when that pressure is caused by a column of water than it is when it's caused by a cork. Although the cork has to be able to hold back the same pressure as that column of water, which is what I was getting at in the other post; the cork or outlet pipe or something will deform, and store some mechanical potential energy to respond passively to the pressure the pump generates until the pressure either overwhelms the pump down to zero efficiency, or overwhelms the cork. Or melts the windings in the pump motor or blows a gasket or whatever.

3. The pump has that minimum efficiency (one extreme end) when it is holding water at its maximum hydraulic head. All the energy is spent supporting the water, none is spent moving water.

Nope, all the energy is spent pushing on some pressure that's pushing back, which leads to a lot of heat generation but no mechanical work. The fact that the pressure is caused by there being a column of water there is immaterial; it could be a piston and a big rock, it could be a big spring, it could be shock absorber from a 74 trans am, it just has to be something that, when pushed, wants to push back. And if the pusher can push back hard enough without failing, the pump is turned into a heater, pretty much.

However, it is true that without the pump churning away, something other than the pump would have to hold back that force. But that could be a passive thing like a shut valve. In fact, if there were a valve right at the pump outlet, shutting the valve wouldn't change the pump's view of things anyway, because it's maintaining the max pressure it can with no flow, and it wouldn't change the column of water/spring/whatever side, because all it knows is that it's pushing with pressure Pmax on something that isn't yielding.

I know this seems extra bonus pedantic, but I think you're confusing "turning energy into heat with the side effect of maintaining pressure" with "spending energy on the mechanical work process of moving water." It's certainly valid to say that the pump running at zero efficiency is causal to the pressure being maintained, but when talking about conservation of energy, or mechanical work, or conservation of momentum, the pump is doing zero work, nothing has any net momentum, the pump's energy draw is being converted into heat, and past the pressure being Pmax at the pump outlet, the rest of the system is just sitting there.

We're just looking at steady states--any transition periods between these end-members (say the first few seconds after opening a hole) are ephemeral and their discussion offers no clarity.

sounds good to me

4. We now have two end-member scenarios: in one, all the pump's energy is holding up water. The other, all the pump's energy is moving water. Lets put this together in a continuum of infinitessimal steady state conditions between the two:

repeat complaint that the pumps energy is applying force to counter a pressure, which is only indirectly holding up water. The actual energy is going to heat, no energy is going into the changing the pressure.

a. All the energy is holding water up, none to flow. The water level in the vent is at max pump head.

As weird as it seems, none of the energy is actually going into holding the water up. The pump is in an equilibrium where, because of the opposing force/pressure, the pump can't put any more energy into the column of water, but the column of water pushing back can't put any of its potential energy back into the pump, it's a Mexican standoff. (you may have gotten the point for being first to mention hovercraft, but I clearly score for being the first to invoke an image from a spaghetti western!)

(continued)
 

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