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- Mar 8, 2004
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DHyslop;92888 said: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.