Questions about Sumps [retitled]

[monty]

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.

 

[monty]

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.

 

[DHyslop]

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).

 

[DWhatley]

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.

 

[monty]

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

I'll be pedantic--apology in advance:

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)

 

[monty]

part 2

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.

Nope, it's all about pressure, not about energy. When you open the pinhole, there is some drain that reduces the pressure. This leads to two things: the water level goes down, and the pump sees lower pressure at its outlet. As soon as the pump sees lower pressure at its outlet, its efficiency goes non-zero by a bit, so it starts to generate flow. The heat it's putting out goes down a bit. When we reach our steady state, the water in the vent has gone down a bit, so there's less force (and less potential energy) coming from above, just enough to equalize the pressure to the pressure down by the pump that will cause it to have the same flow into the system as is going out of the system at the pinhole. If the pressure from the water above the pinhole were any less, then some water flow from the pump would raise it; if the pressure from the water were any more, then the pump wouldn't be able to move enough water to go out the pinhole, so some of the water from the column above would go out the pinhole, in addition to what's coming up from the pump, so there is a feedback system that makes the height of the water above the hole just the perfect amount given the amount of flow the pinhole allows. (the amount of flow the pinhole allows will jitter around a bit as all this settles out, because the pressure at the pinhole will change as the height of the water seeks the appropriate place, but when it's all settled that won't matter.)

This will pretty much extend to your cases b through e-- as the hole resists the flow less, there is less need for backpressure to keep stuff from going up the vent, 'cause most of it's going out the hole eventually. If the hole is so big that water goes out of it faster than the pump can pump against the rise and the resistance from the pipes, then it'll all go out the hole (and any excess capacity of the hole will have let out any water that was up in the vent before we opened the hole, but that's part of the transient behavior we're ignoring).

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?

It's certainly like d, although from Thales' experiment, it sounds like real sump systems have more flow restriction than I realized, so maybe it's more like c. And yeah, in b. the pump is working at the max efficiency it can given the limit of the pinhole. That's part of why I have a problem with "max efficiency"-- if you restrict the outlet of the closed system to a pinhole, with no vent at all, the pressure will go up because of the increased resistance, and the pump will work at almost zero efficiency there, too.

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?

I'm arguing that the water level will seek the appropriate height where it doesn't change the efficiency from whatever efficiency the closed system had, so sure, the pinhole will be less efficient, and the water will accommodate that.

Here's another interesting thought: if you put a plug with a pinhole in my drawing's line before the vent, then the pump will be doing the same amount of work, fighting the same amount of backpressure, and getting the same amount of flow, but there will be no water in the vent, and the water will drain to the octo tank faster than it can get through the pinhole. In that case, the pump is running at the same efficiency, but it's not holding up any water at all. The stored energy that has been bugging you about the water column has now been pushed back to deformation of the pipes before the plug, or similar. But the pump experiences all the same physics down at its outlet, but none of it is doing anything to the vent....

And by the way, if I seem overconfident, it's because I spent the last few years building up an immunity to iocane powder.

Anyway, whaddaya think? I don't have a sense that I was grasping at straws anywhere there, although I have to admit that it's counterintuitive that the pump running at low or zero efficiency is not actually putting energy into fighting the resistance in the steady state, I believe it's one of those counterintuitive-but-true things...

 

[Michael Blue]

I "kind of" see the argument, but I still disagree that without some energy being expended (since it is a pump and not a table or a shelf) there would not be the force to keep the water at a given level.

We're discussing two very different types of force, and the kind created by the pump requires energy.

 

[DHyslop]

Anyway, whaddaya think? I don't have a sense that I was grasping at straws anywhere there, although I have to admit that it's counterintuitive that the pump running at low or zero efficiency is not actually putting energy into fighting the resistance in the steady state, I believe it's one of those counterintuitive-but-true things...

Monty, the only disagreement I have is the one you're touching on here in this paragraph, which is the keystone to the entire discussion. I believe the pump is putting energy into fighting the resistance (Hearts and minds?), and to support that I submit the empirical observation that the water behaves like the helicopter, the hovercraft or the bucket of water over my head: turn off the energy, and the mass comes back down.

If we're at an impass, I should be able to perform the experiment within a couple weeks, and yes, you'll have input on the design

Dan

 

[monty]

Monty, the only disagreement I have is the one you're touching on here in this paragraph, which is the keystone to the entire discussion. I believe the pump is putting energy into fighting the resistance (Hearts and minds?), and to support that I submit the empirical observation that the water behaves like the helicopter, the hovercraft or the bucket of water over my head: turn off the energy, and the mass comes back down.

If we're at an impass, I should be able to perform the experiment within a couple weeks, and yes, you'll have input on the design

Dan

Sounds good to me... I'm picturing the unveiling of a complex Rube Goldberg demonstration at TONMOCON II involving a lot of people in the audience getting soaked. Because, well, that's what science should be like, right?

Of course, I need to figure out if I can actually get to Florida for it, still.

 

[DHyslop]

I'm all for Rube Goldberg devices, but as I previously pointed out it would need the requisite 'Rube Goldberg music.' The Truffle Shuffle may suffice, however.

I won't be at TONMOCON, so the demonstration will probably occur within my soon-to-be garage.