DHyslop;92805 said:
Without the hole the pump needs x amount of energy to prevent backflow and support the sleeper. A certain discharge of air is flowing back out the pump, and the pump needs x amount of energy to push an equivalent discharge of air in. Without a sleeper the static pressure is lower and there's a lesser value for x.
When we cut the hole the pump has to use x plus y amount of energy (y being that required to add enough air to match air lost through the hole). A pump with a sleeper needs x. A pump with a sleeper and a hole needs x + y.
What we care about is y: the amount of energy used to move the air which is analagous to the water we want to move through the return pump. If we just have a water pump in a standard return line it only uses y--there is no sleeper. If we have an open vent that has any hydraulic head above the return, that's our sleeper and we need to spend x to support him.
I think you are arguing that if the pump can keep up then it doesn't matter if there's a sleeper there or not. An easy way to test this would be to inflate an air mattress, get on top of it and cut a small hole in it with the pump still running. We don't agree that if you get off the bed the pump has an easier time of it?
I'm not trying to be argumentative or stubborn either. I just think its an interesting thought experiment
Dan
One of my working assumptions is that the pump has no way of knowing whether there is a sleeper or a hole at all-- the only thing that impacts the pump is the pressure difference across the pump.
I'm assuming it's OK to disregard stuff like turbulence, pressure waves, transients, and stuff like that, and pretty much assuming that the air in the mattress is a uniform pressure everywhere, so there's really just a value of the pressure inside, and atmospheric pressure on the outside, so, if the internal pressure is P atmospheres and the outside is 1 atmosphere, the pressure gradient across the pump is P - 1.
The pump has a few properties that are probably important. One is what the biggest pressure difference is that it can fight. Let's call that Pmax. If it's filling the mattress without a hole or a sleeper, then it'll fill it until P - 1 = Pmax, and then it'll be in a steady state, where the backflow is equal to the flow, and the mattress will be at a constant pressure, and really the air flow across the pump is zero.
From an energy balance standpoint, this may be exactly where we're at odds. I claim that in some sense, the pump is now doing zero work on the air in the mattress. Obviously, the pump is expending energy, and that energy is going somewhere, probably heat in the form of friction in the vorticity or some kooky hydrodynamic thing like that. As proof that the pump is doing no work, though, I point out that at that point you can replace the pump with a plug, and then no work is required to keep the mattress at a constant pressure.
Before pressure in the mattress got to Pmax + 1, air was flowing into the mattress. As the pressure built up, the mattress was stretched, and the air was compressed, so there was potential energy stored in the system in the form of air pressure and mattress wall stretching. Also, it's worth noting that the amount of air the pump moves per second is somehow related to the pressure differential dP = P - 1. Even when dP = 0, the pump will only move some particular mass (volume?) M of air per second. as dP rises from zero to Pmax, the amount of air moved will decrease monotonically from M down to zero, but I suspect it stays close to M for a while and then falls off to zero pretty steeply as you get close to Pmax (but my argument doesn't depend particularly on this)
Staying with no holes, if the mattress is up at Pmax and the sleeper gets on it, then the pressure will go up momentarily. When dP is larger than Pmax, the pump can't fight against the pressure, and backflow causes air to go out of the mattress. As that happens, the stretching of the walls goes down enough that the pressure is reduced (because there's less mass of air in the mattress at the same pressure) until enough air has escaped that it compensates for supporting the sleeper. Then dP will be at Pmax exactly, and the air moved by the pump will be zero again.
Note that this is a different situation than if the hole were plugged; in that case, the pressure in the mattress would go up and the weight of the sleeper would be supported by the compression of the air and the stretching of the mattress.
Ok, so let's get the lazy bum off of the mattress again, and get the pump running maintaining the mattress at Pmax again. Now, I take an ice pick and poke a hole in the mattress. The amount of air that goes out the hole will depend on the pressure, too. In fact, it's got the same dP as the pump, but it's got some flow rate based on P and the size of the hole. If it's a huge hole, then it'll let air out even faster than M, the max capacity of the pump at zero dP, and then the pump is just doomed, and the mattress will deflate. If it's a small enough hole, though, there will be some equilibrium pressure, a little lower than Pmax, where the pump can shove replacement air in at the same rate it goes out the hole at that pressure. I'll call that Peq, since Pequilibrium is too hard to type. In this state, the pump is doing work moving the air through the mattress, and also losing a bunch of energy to the friction stuff like it was in the earlier steady-state fighting the backflow, churning up a bunch of air and heating it, but not doing any mechanical work... the only mechanical work is the flow replacing the air going out the hole. Another way of looking at this is that the pump sees P going down, and just moves enough air in to replace it; its entire view of the universe is dP, so it doesn't know there's a hole, it just knows dP is low enough that it can shove air in at some rate, and yet dP never seems to go up. (the pump is probably pretty frustrated at this point.)
Ok, now the sleeper gets back on the mattress. The instantaneous effect is just like before: the pressure inside goes up, so dP across both the hole and the pump is bigger. It was steady at Peq rather than Pmax, but besides that it's the same as without the hole. Now, air escapes from both the hole and the backflow from the pump, but the effect is the same-- air goes out of the mattress until enough is gone that the system is back to Peq. The air going out is exactly enough that the stretching/compression is reduced in potential energy enough to support the sleeper. Then, we're back at Peq, and the flow is exactly the same: the pump pumps some air in at the rate it does faced with dP = Peq, which is the same rate that the air goes out the hole, so the flow-through in the steady state is the same with or without the sleeper there. As far as the pump and the hole are concerned, the dP is just Peq, and it has no way of knowing that the sleeper is there, once the transients are over and it's back to equilibrium. Since it's moving the same amount of air across the same amount of pressure difference, I claim that there is no way it could be expending a different amount of energy in this situation than the steady-state without the sleeper.
I'm going to stop here lest I exceed the size one post is allowed... applying this to the vented return line is left to the next post.