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.
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