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- Mar 8, 2004
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part 2
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).
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.
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...
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).
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.
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...
