Questions about Sumps [retitled]

[monty]

Now, every good scientist knows that every model is false. Some models are useful!

And some models are useful in some situations and not in others. Of course, some models aren't useful at all...

I'm starting to think that this is just a terminology problem, in that I'm not disputing that the pump is using energy to do something, just that what it's doing isn't doing work (in the strict physics definition.) Of course, I'm also using this claim to prove that adding the vent pipe doesn't make the pump work any harder, too.

So you're saying that the turbulence created by the impeller would somehow act as a plug? Do you have any evidence this is occuring or do you just think it should?

I'm saying that with respect to a certain model, the whole system of the impeller is equivalent to a plug in the steady-state situation: If there is no flow across the pump, then the system around the pump with no flow doesn't see anything different than anything else that blocks flow in the face of a pressure differential, like a plug or a closed valve. The fact that the turbulence is where the energy driving the pump ends up doesn't have anything to do with the "this is a zero-flow element" aspect... I'm not saying a plug will make turbulence, just like I'm not saying a plug consumes energy. It's just a way of illustrating the point that the pump, in that situation, isn't doing any more work on the column of water than the plug would be.

The only reason I'm bringing up plugs or closed valves it to make the point that zero flow is zero mechanical work... I don't even need to use those (I was just substituting a plug for the pump to make a point)-- If I have a bucket full of water, the bucket doesn't have to do any work to maintain the water line, does it? The walls of the bucket have to resist the force of the water, and if you pour water into the bucket, the sides will bow out a little, which is the water doing work on the bucket until the potential energy stored in the tension of the bucket is the same as the gravitational potential energy of the water relative to the bottom of the bucket.

My position is simple. The pump is doing work to maintain a certain column of water above its natural level in the sump. You*--and every physicist in the country--agree on that because you understand the pump has a maximum head pressure at which the pump is doing work but there is no flow, just water held up. What you're saying is if we punch a hole in that column when the water is at that level we're suddenly going to get free energy: even though the pump is doing all the work it can to hold the water level up, we're somehow going to get 400 gph (what my Mag 7 does) coming out the return line.

*the only way you're not saying this is with your hypothesis about the turbulence of the impeller somehow acting as a plug. If you believe this, then you don't believe the pump has to do work to hold up water and thus you don't believe the pump has a maximum hydraulic head at which there is zero discharge.

My position is also simple: if there is no flow, then no work is being done on the mechanics of the water system. This is in the strictest mechanical definition of work, F ds. Energy is being spent by the pump, but if there's no flow, that energy has to be going somewhere else. Actually, going back to your helicopter example, if the helicopter is hovering, no work (in this sense) is being done on the helicopter, either... the helicopters gravitational potential energy is constant, because it's not moving up or down. The blades are doing work on the air around the helicopter, pushing it downwards, heating it up, and stirring it around, but it's not actually doing work on the helicopter. The effect of this is that it is producing a force that is countering the force of gravity, but "producing a force" is not the same as "doing work."

Here's what would really happen in that situation: the pump is doing work to hold the water up to its max head. We open our return line and the flow coming out of it will be driven by the gravitational potential of the water above it. The traditional formula for maximum flowrate through an aperture at a depth. The water level will come down a bit from the pump's max hydraulic head, but as long as that water level is anywhere above the return height then the flow rate will be less than a convential sealed pipe because some work is always being wasted on holding hydraulic head. I don't see how this is avoidable.

In my model, the pump never gets to max head, because water is flowing out into the tank at some rate that reduces the load on the pump. The max head is only reached if there is no outlet except a vertical run of pipe that's taller than the max head.

Where I disagree with this last paragraph is that I think that the drop in the water level in the vent (above the tanks waterline; where in the tank the outlet is doesn't matter) is exactly the right amount to make the steady-state flow exactly what it would be without the vent at all. While the vent is filling up to that point, some of the pump's energy is doing work to raise that water, but once it's reached that height, no more work is needed to keep it there, and the situation goes back to the same steady-state flow it had without the vent pipe.

Another way to look at this is: if the pump is doing work "to support" the extra column of water, where is that energy going? The water in the vent stays at the same level, so it's not moving-- there's not kinetic or potential energy change there.

I didn't mean to imply that there isn't a max hydraulic head associated with a pump, just that once the water reaches that level, the pump is no longer doing work on the column of water, all of its energy is doing something else (ultimately, making heat). This has the side effect of maintaining the status quo, but it's no longer doing work on the water column, except in the sense of heating up the part of it near the pump.

 

[Thales]

Worriedly sticking my toe in this discussion...
I think there is a lot of turbulence inside the piping, and even more at junctions. I think the water in the vent would not be static, it will keep roiling, moving and flowing trying to get down the vent and out the other exit. The water in the vent will stay in a general range but not at a static level, as the water at the bottom of the vent is drawn along by the water moving out the other exit. The water in the vent would be in flux and would need to be replaced by water from the pump constantly, and therefore the pump would always be doing work.

 

[monty]

Worriedly sticking my toe in this discussion...
I think there is a lot of turbulence inside the piping, and even more at junctions. I think the water in the vent would not be static, it will keep roiling, moving and flowing trying to get down the vent and out the other exit. The water in the vent will stay in a general range but not at a static level, as the water at the bottom of the vent is drawn along by the water moving out the other exit. The water in the vent would be in flux and would need to be replaced by water from the pump constantly, and therefore the pump would always be doing work.

There might be some change in the turbulence in the pipe because of the junction. Most of the turbulence and drag and such will be there whether the vent is there or not, though. It's certainly possible that adding the junction for the vent will create an edge that induces a bit more drag, but I was assuming laminar flow for the most part everywhere except the pump.

Anyway, I'm pretty sure that it's not inherent in the system that there will be significant turbulence in the vent pipe. It could even be damped by putting some sort of wadding at the bottom of the vent pipe, or a screen across the opening to the vent... anything that attenuates the high frequencies.

Of course, it's always a bit worrisome to assume that chaotic and resonant effects can be neglected, as in the Tacoma Narrows Bridge, but I think all the significant contributions from that can be lumped into the flow resistance in the pipe. I actually think the water column in the vent will stay relatively still, though, at least if the pump doesn't have any low-frequency periodic surges. And even if it doesn't, although it will add to frictional losses a bit, if it's oscillating around a particular waterline in the vent, the net work is zero, because when it's going down it "pays back" the potential energy it took when it went up.

 

[cuttlegirl]

I think one of you should build this model to test out this hypothetical thread!

 

[Thales]

Monty,

Could you draw a quick diagram of the piping you envision?

 

[DHyslop]

I didn't mean to imply that there isn't a max hydraulic head associated with a pump, just that once the water reaches that level, the pump is no longer doing work on the column of water, all of its energy is doing something else (ultimately, making heat). This has the side effect of maintaining the status quo, but it's no longer doing work on the water column, except in the sense of heating up the part of it near the pump.

You're right that I'm using the term work casually. The pump, like the helicopter, is expending a lot of energy to hold a mass in place. In a geometric sense it is clearly not work because neither the helicopter nor the water are moving. We agree that every pound of helicopter or every inch of hydraulic head is consuming energy just to be held there. Even if the water level in the vent was only an inch above the outlet height, that's an "inch" of energy that cannot be used for flow.

Another way to look at this is: if the pump is doing work "to support" the extra column of water, where is that energy going? The water in the vent stays at the same level, so it's not moving-- there's not kinetic or potential energy change there.

If I lift your bucket of water above my head and hold it there, my arms are expending energy to hold it up, even if no work is being done. There's not any work being done on the water above the outlet in the vent either, but the pump has to do the same thing my arms do. You're arguing that once the water is up there in the vent, no effort needs to be expended to keep it there, its as if I was putting my bucket on a shelf, which I am not (putting the bucket on a shelf would be analagous to putting a plug in the pump aperture).

Dan

 

[monty]

Here's a picture of the vent system I had in mind. Note that the valve (X in a box) at the base of the vent is there more for argument than because it's necessary.



I think the definition of work explains some of the disagreement here, although I still can't find any reason why the pump has to work harder.

I agree that the idea that the pump doesn't care whether there's the column of water there or not is counterintuitive, but I haven't heard anything besides "it's just obvious."

The claims about lifting the bucket seem to be a bit of a tricky issue. There are two cases here. If you lift a bucket of water over your head, it's clear that you have to spend some effort to keep it there. Similarly, if a helicopter is hovering, and the engine cuts out, it's going to fall to the ground.

However, there are other situations where, once the work has been done to lift an object, it doesn't require any work to maintain its being at that height... although it does require force. If you lift the bucket above your head, and put it on a shelf, in addition to not having to do work on the bucket, you no longer have to expend effort (muscle energy) to keep it on the shelf. What's the difference? The shelf doesn't have some magical immunity to gravity, and yet it also doesn't have to be plugged in or anything. Another common situation is a hydraulic lift in a garage: when you drive your car on it, it runs a pump while it's lifting the car, but when it's at the proper height, someone shuts off the pump, and it stays there. Of course, if the pump allowed backflow, the car would sink, so some sort of valve is probably required. In that case, it's possible to get most of the energy the pump used back as the car is lowered, by running the pump backwards as a generator, although real garages don't bother. Then, the only energy lost is the friction in the system on the way up and down.

I think the main thing we're disagreeing on is whether this situation is more like the garage lift, or more like the helicopter... or, analogously, whether it's more like holding the bucket over your head than putting it on the shelf. There was also some discussion of whether the vent would rise to the full hydraulic head of the pump, which I claim would only happen if the pipe is clogged between the vent and the outlet; as long as some water is flowing into the tank, some (hopefully most) of the pump's work is going toward putting water into the tank, not making the vent into a water tower or fountain.

I certainly will admit that the energy balance in the helicopter/ bucket holding situation is a lot harder to be clear about. I think we agree that no work is being done on the bucket/helicopter. However, because the thing isn't falling to the ground, something is exerting a force on it. In the shelf case, the shelf is exerting a force, which is in some compression of the shelf or bending of the brackets or some such. In the case of muscles, in order to resist significant forces by applying forces, muscles need to expend energy. That's more because of how muscles work, though, and because they have more versatility than the shelf. In fact, if you have a heavy backpack on, and go from sitting to standing, once you lock your knees, you don't have to put any significant muscle force into standing, because your knee joints are acting like the shelf.

In the helicopter case, one can make some weird arguments about the flow of things... in the pedantic definition of work, no work is done on the helicopter at all; it's not moving up or down, so it's subject to the force of gravity and an equal and opposite force of lift at the rotor hub. That force in turn comes from lift in the blades, which are rotating, and pushing on the air, which is being accelerated downwards. The process of accelerating the air downwards does require work, and the energy loss in this causes the blades to be subject to drag that tries to slow down their rotation rate. What the engine is spending energy on (and doing work towards) is fighting the slowdown of the blades that the air friction is causing.

I don't think the column of water is particularly analogous to either of these situations, by the way. The pump running at full bore holding up the full hydraulic head is a lot like the hovering helicopter, but I think once you get the outlet, some aspects change to be closer to the hydraulic lift...

One of several approaches I've used to model this is actually a DC circuit, say a battery (the pump) and a resistor (the tank outlet & other flow resistance) -- the return goes back to the negative terminal of the battery, and is assumed to be negligible resistance (or it's lumped into the resistor, whatever). The vent is (I claim) like adding a capacitor before the resistor, with the other side of the capacitor going to ground/negative. When that's added, the current through the resistor will drop for a bit while some current goes to charging the capacitor, but once enough charge is stored that the voltage across the capacitor is the same as the voltage drop across the resistor, all the current will go back through the resistor as before and the load on the battery will be exactly the same, and the capacitor will do nothing and won't add any load on the battery until something happens to change the voltage. It's not exactly the same, because this model mashes the water pressure and the height of the water in a gravitational field into one thing, but by Pascal's Principle the height and pressure are tied, so I think the mathematics turn out the same. Of course, I eschew electrical stuff that's got transients and inductance and resonance just like I hate hydrodynamics that has oscillations and vortices and such... but I think I we can characterize this system without needing that stuff.

 

[DWhatley]

Monty,
I don't understand why the water won't just fountain out of the top of the vent pipe if your experimental valve is open. You show reduced flow into the tank so isn't that a resistor and won't the water follow the path of least resistence or am I forgetting that the lift of the pump will be exceeded at the line drawn for water height?

 

[monty]

Monty,
I don't understand why the water won't just fountain out of the top of the vent pipe if your experimental valve is open. You show reduced flow into the tank so isn't that a resistor and won't the water follow the path of least resistence or am I forgetting that the lift of the pump will be exceeded at the line drawn for water height?

As water goes up the vent pipe, the weight of the water pushes back down, so there's some limit to how high it will go up. If the octopus decided to block the outlet to the tank completely, then the water would rise to the hydraulic max head of the pump, so the water can never go out of the vent higher than that. However, presumably you want good flow into the tank, so effectively the resistance of the outlet into the tank should be low enough in practice that the vent pipe level never gets anywhere near that high. If there's a concern that the outlet might get blocked completely, having a properly tuned relief valve down by the pump can prevent both the water going too high in the vent, and also would make the pump less inclined to melt down or blow gaskets if the plumbing gets clogged.

Of course, Dan thinks I'm wrong on some of this, so it might be unwise to build a tank based on my sketch until we work out the details... I think we both agree that water won't go any higher than the max head of the pump, and that if the outlet isn't blocked it won't get all that high in normal operations (because the closer the water gets to the max hydraulic head level, the higher the backpressure there has to be reducing the flow)

 

[Michael Blue]

From what little I think I understand of your diagram here, wouldn't it be difficult to "balance" the restriction of flow back into the main tank with the desired height of the water in the vent?

Or am I missing something?

 

[monty]

From what little I think I understand of your diagram here, wouldn't it be difficult to "balance" the restriction of flow back into the main tank with the desired height of the water in the vent?

Or am I missing something?

I was assuming the restriction of flow should be minimized, except as needed to get a bit of velocity for circulation in the tank. There's no particular "desired" height of water in the vent, except that it's strongly desired not to overflow, but if there's a lot of height there, it probably means that the pump is doing a lot of work to fight the flow restriction, while its main goal should be circulation through the filtration, so that would probably be a sign that the outlet into the tank should be bigger... a high water line in the vent is a sign that the plumbing is keeping the pump to below its capacity for circulation through the overflow/filter setup.

 

[DHyslop]

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?

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

 

[Thales]

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall.

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

 

[DHyslop]

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall.

Makes sense--to get any considerable discharge through small diameter pipe means our pumps are designed for high pressure!

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

This is where penductors, etc come in, right? Taking some energy from the return pump discharge and distributing it into a greater area of somewhat milder current? I should spend more time on your precious reefs.org

Dan

 

[monty]

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall.

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

I'm confused-- was there any outlet other than up the vertical pipe, or was this just testing the hydraulic head of the pump?