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Questions about Sumps [retitled]

You don't need a sump, but while not having one seems like a time saver initially, in the long run I think you will be much happier if you use one.

Monty, I think what you are talking about will work, though tuning the height of the mast may be iffy. When the power goes out, the siphon will absolutely be broken. I think most people don't do something like that because the alternatives are pretty easy. I think you might lose flow by unpressureising the lines but making the pump work against the head in the mast.
 
magikceph;92759 said:
i think your right cuttlegirl, so far i have been asking way to many questions. im ust gonna take a trip to my LFS and maybe and aqaurium shops in San Francisco. thanks.

If you are that close, check out www.bareefers.org. A great local reef club.
If you go to shops in SF, avoid 6th avenue aquarium, but try to go to lucky ocean.
 
DHyslop;92755 said:
You're right that the height of water in the column would be lower than the pump's theoretical max head pressure because water is being bled off into the tank. The water level will be somewhere between that value and the height of the return. I suspect the water level will be much more than just a few inches above the return due to the degree the pump pressurizes the line (these are centripetal "high-pressure" pumps rather than axial "high-flow" pumps like NASA turbomachinery :smile:. Also given that return outlets are usually pinched off or reduced to some degree to increase velocity for the same discharge at the cost of a little resistance (so in effect a little smaller discharge)--this would push the water level in the vent that much higher.

I think what would end up happening is most of the pump's energy would go into pressurizing this open water column rather than moving water into the aquarium. If ultimately the water level in the vent was, say, two feet above the water level in the tank, the only flow into the tank would be that based on gravity for a certain sized outlet times the hydrostatic pressure (rho*g*2 ft).

I'm starting to hear that music they play in the Warner Brothers cartoons whenever there's a rube goldberg machine :smile:

Dan

This is an aptly named thread...

I'm thinking of this as a steady-state problem (which I'm actually pretending is a statics problem) where once the vent has enough inches of water in it to equalize the pressure, then the water there will just sit there and act like the walls of the pipe in a closed return without the vent, so that after the bootstrap phase when the pump has to get that water up there, the water is just sitting there at some gravitational potential energy and the water's just flowing along happily under it. Since the column of water is a constant mass at a constant height, no work is done by the pump to maintain it past the initial pressurization, when that potential energy is added. Of course, if the pump shuts off, then the pressure in the return goes down, so the force resisting that level seeking the same level as the tank goes away, but in terms of work, I think the pump's energy is going to moving the water from the sump to the tank, against gravity and the resistance of the return line (and nozzle, if any), and the water column just sits there in equilibrium... or, more mathematically, the force is there holding the column up, so F is equal to the weight of the water, but since it's not moving vertically, so ds (or dy, if you will) is zero, so there's no work (integral of F ds ) being done.
 
I think you're making it too complex.

If the water level in the vent is at any point above the water level in the tank--which I think we agree it will be--the pump is expending some amount of energy to hold it there, that is, keeping it from flowing back through the pump into the sump. Just like the lift fan in my hovercraft is eating 6 or 8 hp to keep the air in even after the skirt is pressurized (the bootstrap phase). That's an efficiency loss compared to a normal plumbing setup where the pump is only fighting the head differential from the water level in the tank.
 
DHyslop;92776 said:
I think you're making it too complex.

If the water level in the vent is at any point above the water level in the tank--which I think we agree it will be--the pump is expending some amount of energy to hold it there, that is, keeping it from flowing back through the pump into the sump. Just like the lift fan in my hovercraft is eating 6 or 8 hp to keep the air in even after the skirt is pressurized (the bootstrap phase). That's an efficiency loss compared to a normal plumbing setup where the pump is only fighting the head differential from the water level in the tank.

Dude, you have a hovercraft!

Anyway, that notwithstanding, the hovercraft is a different situation, because there is air escaping from the skirt, reducing the pressure, and the fan has to put energy into moving new air to replace it. The pump is adding new water to compensate for the pressure loss going out into the tank, but once the water level is constant in the vent pipe, no work has to be done to replace water there because none is leaving (or changing height.) It's like if you put a tube attached to a deflated balloon attached to the hovercraft skirt: the fan would have to do more work to get the skirt up to pressure, because it also had to fill the balloon, but once the pressure in the balloon was the same as the pressure the fan was trying to maintain in the skirt, as long as the fan was doing its usual pressure maintenance it wouldn't have to be working any harder to keep the balloon full -- the potential energy that was put in with the extra work to inflate the balloon is exactly enough to assure that air doesn't flow in either direction between the balloon and the skirt (or the vent and the return pipe) -- and where there is no movement of mass, no work is being done. Or, putting it another way, when a brick is resting on a table, the table has to resist the force of gravity wanting the brick to fall to the center of the earth, but the table doesn't have to continually do work to do it... when I set the brick on the table, it compresses the table a bit, just enough that its spring force due to the compression exactly balances the gravitational force of the brick.
 
monty;92785 said:
Dude, you have a hovercraft!

Technically, I have about 3/4 of a hovercraft. The hull is sitting under my parents' deck waiting until I can finish it.

Anyway, that notwithstanding, the hovercraft is a different situation, because there is air escaping from the skirt, reducing the pressure, and the fan has to put energy into moving new air to replace it. The pump is adding new water to compensate for the pressure loss going out into the tank, but once the water level is constant in the vent pipe, no work has to be done to replace water there because none is leaving (or changing height.)

Here's where the assumption is wrong. Yes, some air is escaping under the hovercraft skirt (which is what lubricates it as it travels over a surface), but it really is a small amount. The real factor here is the weight of the craft constantly pressing the skirt and trying to push the air out of the lift fan's hole. Even if the skirt was sealed to the ground the lift fan would have to be running or else the air would escape though it. Likewise a considerable portion of the pump's energy is keeping the column of water from returning to the sump. I believe you are arguing that work has already been done to lift the water up and inertia will make it stay there--on the contrary, force continues to be required to counter gravity, just as a helicopter cannot turn off its engine once it reaches altitude.
 
DHyslop;92786 said:
Technically, I have about 3/4 of a hovercraft. The hull is sitting under my parents' deck waiting until I can finish it.



Here's where the assumption is wrong. Yes, some air is escaping under the hovercraft skirt (which is what lubricates it as it travels over a surface), but it really is a small amount. The real factor here is the weight of the craft constantly pressing the skirt and trying to push the air out of the lift fan's hole. Even if the skirt was sealed to the ground the lift fan would have to be running or else the air would escape though it. Likewise a considerable portion of the pump's energy is keeping the column of water from returning to the sump. I believe you are arguing that work has already been done to lift the water up and inertia will make it stay there--on the contrary, force continues to be required to counter gravity, just as a helicopter cannot turn off its engine once it reaches altitude.

Some of these cases are rather apples-and-oranges comparisons... the helicopter is nowhere near a closed system, so it's rather hard to compare... the hovercraft can push on the ground, while the helicopter has to accelerate air to offset the amount that gravity accelerates it downward.

A very simple proof-of-concept analogy that situations exist where energy is not required to maintain the steady-state is a car jack: lifting the car with a jack requires pumping, to expend energy to do the work of raising the car, but when it's been done, no energy is required to maintain it. I think the column of water is closer to this than the helicopter example once it reaches the steady-state.

In the hovercraft case, I can believe that there's blowback through the fan, and certainly if you turn off the motor, air will go out the fan through the opening, because it's a hole. However, for comparison, if you have an air mattress, you can fill it up, and then shut the valve, and it supports weight. With a big enough pump, you can even fill it up with someone lying on it, if the fan can push air in with greater force than the sleeper's weight. If you shut the valve then, it'll continue to support the sleeper without the pump at all, because the spring force of the compressed air (and the stretching of the mattress walls) is the same as the weight of the sleeper.

My take on the return flow is that this is similar to leaving the pump on the air mattress and punching a hole in the air mattress... now the pump has to move air in at the same rate that air is escaping from the hole. If the pump were turned off, then the weight of the sleeper would force air out the hole and backwards through the pump, but if the pump is up to pushing air in as fast as the air goes out the hole, the pump doesn't have to work any harder than it would to keep the mattress inflated to the same pressure without the sleeper, once it's shoved enough air in to offset the sleeper initially.

I'm not wanting to be argumentative or overly stubborn here, but I'm also usually pretty good at this stuff, as I know you are, Dan, so I'm pretty sure that getting this figured out will be a learning experience... I always want to fine-tune my science and engineering toolbox so I don't misapply the tools.
 
monty;92802 said:
My take on the return flow is that this is similar to leaving the pump on the air mattress and punching a hole in the air mattress... now the pump has to move air in at the same rate that air is escaping from the hole. If the pump were turned off, then the weight of the sleeper would force air out the hole and backwards through the pump, but if the pump is up to pushing air in as fast as the air goes out the hole, the pump doesn't have to work any harder than it would to keep the mattress inflated to the same pressure without the sleeper, once it's shoved enough air in to offset the sleeper initially.

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 :smile:

Dan
 
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 :smile:

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.
 
Ok, so I think the situation in the return line is similar to how I constructed the model of the air mattress. If there's a flow constrictor at the return to the tank, that's like the small hole in the mattress, so the return line, pump, and flow constrictor will be a system that will reach some Peq, where the pump is shoving water into the return pipe, and water's going out of the pipe into the tank, at the same rate, and Peq is where that's all balanced and happy. Let's say I've got the vent pipe, and just for thought experiment, it has a T with a valve that connects it to the return line. We start the system with the valve closed, so it just acts exactly like a normal return line, and after the pump's been running for a while, the return line is pressurized to Peq, and the pump is shoving Meq worth of water into the pipe per second, and Meq water is coming out into the tank (and then Meq is going out the oveflow and back down to the sump, but that's not important for the vent part.)

Ok, so now I open the valve. Now there's this vertical pipe sticking up. On the bottom of the pipe, the pressure is Peq + 1, and at the top, it's one atmosphere. naturally, water is going to want to rise vertically in the pipe at first. So, water starts going up the pipe a bit, relieving the pressure to less than Peq. The pump sees this and starts moving more than Meq into the system, because as the pressure goes down, its capacity goes up. Less water will be going out into the tank, too, since the flow that way is driven by Peq. So, more water is going in than coming out, so the water level continues to rise in the vent pipe. As the water level goes up, its weight pushes back down on the water in the main branch of the return pipe, proportionally to how high the water is in the pipe above the level in the tank (because the amount of water in the tank above where the vent causes it to want to backflow a bit to raise it to that level even if the pump isn't doing anything). So more water goes into the vent pipe until enough water is there that it pushes down at Peq on the return pipe. Now, the outlet, the pump, and the bottom of the vent all see Peq. In the vent, Peq is the same as the weight of the water column (assuming the vent is tall enough that this fits!) so water neither goes in nor out of the vent. At the pump, just as before, the pressure it sees is Peq, so it pumps at a flow rate of Meq, just as before. At the outlet, the pressure is also Peq, so it also has a flow of Meq into the tank, just as before. Since the pressure is the same and the flow rate is the same at the pump, the pump is doing the same work that it was before the valve was opened, now that we're back to the stable state.

Of course, to actually break the siphon, there should be a loop of return pipe that's raised above the waterline in the tank, and the vent should be installed there, so that if the pump power goes out, when the column of water falls (because the pressure drops from Peq down to zero, or one, since I forget if I'm talking absolute or relative) some water will go back to the tank, some will go into the sump, but the return line won't be pulling down to the sump because the vent breaks the vacuum.

I reiterate that I'm not sure there isn't some sneaky conceptual bug in this, but I can't spot a flaw in this reasoning.
 
monty;92818 said:
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.

Bingo.

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.

The pump is doing work--it is consuming energy and that energy is spent to move air, not generate heat. The impeller of the pump is not a plug that fills the entire aperture--air is rushing out as best it can and work needs to be done to replace it. Something I learned about hovercraft lift fans is it is useful to conceptualize a fan filling a reservoir as two seperate holes--an in hole and an out hole. In reality they're the same physical hole, but that makes it easier to model what's happening.

I believe your device is a flavor of perpetual motion machine. Its doing work to keep the sleeper up but doesn't have to expend energy to do it.

Dan
 
DHyslop;92832 said:
The pump is doing work--it is consuming energy and that energy is spent to move air, not generate heat. The impeller of the pump is not a plug that fills the entire aperture--air is rushing out as best it can and work needs to be done to replace it. Something I learned about hovercraft lift fans is it is useful to conceptualize a fan filling a reservoir as two seperate holes--an in hole and an out hole. In reality they're the same physical hole, but that makes it easier to model what's happening.

I believe your device is a flavor of perpetual motion machine. Its doing work to keep the sleeper up but doesn't have to expend energy to do it.

Dan

I certainly agree that the pump is doing some work on something. However, I think that the point at which the pump cannot fight the pressure (what I called Pmax) is exactly the point where all of the work the pump is doing goes into entropy.

I'm not really understanding the perpetual motion machine analogy, since nothing in the system is in motion except the pump, and it also seems like the perpetual motion misconception is the opposite of what I'm saying: the fallacy there is that people assume that no work needs to go to entropy at all.

Maybe both the "the pump is a single hole" vs "the pump is to holes" models are both silly... they're certainly both conceptual and don't reflect reality. In a fan pump at Pmax, there is pressure trying to go out past the fan blades, and lift from the blades pushing against the pressure. The lift will be higher at the outside edge of the blades, because that part's moving faster (although prop designers have all these "blade shape changes with radius" tricks, I think they can't get around this too much). So in the inside of the fan, the air will be going out, and at the outer edge air will be pushed in to compensate. So really, what the pump is expending energy on is recirculating air locally.

My claim is that at Pmax this is a local phenomenon around the region of the pump, so if it were in a tube that had the internal pressure on one side and the external pressure on the other, all the churning would be happening inside the tube, and to the rest of the system, it would be more or less equivalent to a shut valve; no air would flow through the tube, and all the churning going on inside the tube is invisible to the rest of the system as long as the external system keeps Pmax on one side and 1 atmosphere on the other.
 
HOLY JEEZ, Dhyslop, you just explained to me what i could not get. People have been practically screaming at me to stop posting questions and you just answered it in one post. But there is only one question can you purcahse a sump like the one you took a picture of, you know the one below the picture of the bioballs.
 
monty;92846 said:
Maybe both the "the pump is a single hole" vs "the pump is to holes" models are both silly... they're certainly both conceptual and don't reflect reality.

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

all the churning would be happening inside the tube, and to the rest of the system, it would be more or less equivalent to a shut valve

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? :smile:

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.

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.

Dan

*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.
 

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