Ok, the whole idea of a sump is something I understand. But what I need to know is what all do I need for a sump? Like, what parts are there to a sump, what makes the sump what it is and what do you need to make it effective and useful?

http://saltaquarium.about.com/cs/bcorsettingup/a/aa020298sumps.htm

:grin:

A sump is simply a smaller tank set below your main tank, with water flowing between the two. The sump is a place equipment that might be unsightly or dangerous to the inhabitants of the display tank.

Installing a sump requires three things:

1) The sump itself: Usually a smaller aquarium that sits in the stand.

2) A pump: sends water from the sump up to the main tank through a vinyl hose or PVC pipe.

3) An overflow: this is installed on the display tank and returns water to the sump via gravity. This typically includes a box inside the display tank who's top edge defines what the water level will be in the display tank. Water "overflows" that edge and is sent down to the sump via a hose or pipe. In this way the flow back down to the sump is always balanced with the flow through the pump.

There are two basic kinds of overflows. The first, and best, involves having holes in the glass of your aquarium and the box installed around them. The other way is a hang-on-back overflow that uses a siphon to transport the water in the box to a second, external box. From there it flows into the sump.

Dan

I know those things, but I need to know what you have to keep inside the sump to keep it running, like, what does the actual filtration. What I'm actually looking for is a list of parts like the protein skimmer and stuff.

I'm not sure what you're looking for here. Obviously you know about protein skimmers. Are you looking for information on biofilters? Most commercial wet/dry filters "replace" the sump in that the water drains directly into it and is pumped directly out of it back to the tank. I personally wanted the best of both worlds so I built a wet/dry inline before the sump.

To be specific, I know that there is the case made of acrylic or glass. But what I need to know is what goes into the plain old glass (or acrylic) box that makes it a filter. What does the filtration? I know that you don't just have a clear box with water running through it and back to the tank. So what makes a sump a sump?

You can use baffles to seperate the sump tank into sections. You could make one section into a wet/dry filter with bio-balls, another into a refugium with live sand, maybe some rock, and macroalgae. This is a great way to maintain a constant pod population in your tank and the macroalgae feeds off of the bad stuff in the water, helping filter the water and oxygenate it.

Check out this link: http://www.melevsreef.com/links.html There are some good examples.

I know that you don't just have a clear box with water running through it and back to the tank.

By definition, this is all a sump is. A sump doesn't need to do any filtration or anything to be a sump. Its just a box with water going through it.

Some things can be added to the sump for filtration. A skimmer and a refugium are the obvious choices, and a wet/dry filter is a good idea but makes it a bit more complex to design because the filter media need to be above the water line. That's why I put mine inline and not in the sump itself. If you do a Google image search for wet/dry filter you'll see most sumps designed for this have the water level low and the media raised up a bit. For this reason its difficult to design a sump that has both a wet/dry filter and a refugium or skimmer.

Dan

Well, I guess what I really need to know is what the stuff does. Here is what I'd actually like to know:

:confused: What is a refugium and what does it do?

:confused: What is a wet/dry filter and what does it do?

:confused: What are these 'bioballs' that I keep hearing about?

:confused: What is the sump actually? Is it the whole filter setup or just the little box?

The word sump has a broad definition. Any tank attached to your main tank can be called a sump. Technically "just the little box." Often the filter is in the sump, more often the refugium is; but neither has to be for it to be called a sump.

A refugium is usually inside a sump. Imagine the sump aquarium with one section partioned off (usually with baffles) with a sandy bottom and some macroalgae. This is a breeding ground for some of the influential little organisms in the tank like amphipods etc. Ideally the pump is in or near the refugium so some of these guys get 'exported' up into the main tank. The pH of your tank water drops a bit at night because photosynthetic bacteria aren't taking up CO2 when the lights are off. If you put a light on the refugium that only comes on at night it will keep the pH more even.

A bioball is part of a wet/dry filter, which I will explain below. It is a little tiny ball that is made up of a bunch of plastic folds or needles so that it has a lot of surface area to it.

A wet/dry filter (also known as a trickle filter) is the queen mother of biological filtration. The bacteria that turn fish pee into more benign substances like oxygen. If you want a whole lot of these bacteria you want to create a lot of surface area for them to grow and make it exposed to air. The best home for them is to have a box full of bioballs and have your tank water trickle through them.

Power filters and canister filters are much less effective because the beneficial bacteria are always underwater and don't get as much oxygen. Some power filters have a "bio-wheel" that gets wet and spins to get some wet/dry action. The downsides of a wet/dry are they take up space and you have to make sure the water that goes into them has a good prefilter because there's a lot of space for particulates to get caught and decompose.

Dan

This is what a bioball looks like. If you can imagine a box full of these things, you can see all the area there is for bacteria to grow.

http://www.marineeco.com/images/bioballs.jpg

Here are two pictures of commercially available sump with an integrated trickle filter. These are the same piece, but one picture shows the bioballs in it and the other doesn't.

http://www.precisionmarine.com/assets/images/T25-2L.jpg

http://www.marinedepot.com/IMD/filter_precision_marine_wet_dry_trickle. jpg

The water flows from the aquarium onto the drip plate with all the holes in it. From here the water trickles through all the bioballs, which rest on the black grid (which is called eggcrate). The middle compartment holds a sponge filter, and then the pump sits on the right.

This is a decent design. The water level is around the top of the sponge, which means only the bottom layers of bioballs are underwater (remember they work better when they're not submerged). Its generally difficult to integrate a wet/dry filter into a sump because you need to have a low water level. Its hard to run a skimmer or do anything else (like a refugium) with a really low water level.

The other alternative for a wet/dry is to put it before the sump, as I have done.

Here are two very good threads (Of course they're good, I started them :lol:). The first is my education in filtration as I was asking questions as I planned my tank:

Tank renovation - Filtration (http://www.tonmo.com/forums/showthread.php?t=5732)

The next thread is photos of my finished setup.

Tank photos - Finally! (http://www.tonmo.com/forums/showthread.php?t=6460)

You'll learn a lot from these threads. They should be required reading for ceph-keeping; if I do say so myself :)

Dan

is a sump neccassary for a bandensis cuttleifsh???

...along with the disaster that can occur during a power failure !

Having read all the linked material, I can honestly say "my head hurts!".

Some incredible info there!

The book I bought relegates the use of a sump to "more advanced" aquarists, so I've been searching for this info.

Thanks!!!!!!

The book I bought relegates the use of a sump to "more advanced" aquarists, so I've been searching for this info.

That's fair. Especially given that we relegate the keeping of cephalopods to more advanced aquarists :)

I hope my threads are valuable to newbies. The level of experience I have--paltry by comparison to a storied reefkeeper like Thales--was earned slowly through multiple iterations of equipment and costly mistakes. And hundreds of hours pouring through old threads on sites like Reef Central. Many people come to this site looking for a how-to on setting up a complex system and to some degree I see the lack of discipline or interest in finding the details on their own as a proxy for a lack of the requisite attention to keep these animals properly.

Above all, I think the threads linked above provide a logical narrative of sumps and filtration that are more conducive to learning than any syllabus or article I could ever write.

Dan

Aww, shucks Dan! :grin: Your threads are great!

I tend to avoid system questions on TONMO because there are tons of sites out there dedicated to such questions - I like using TONMO for ceph talk. :grin:

FWIW, I don't care all that much for RC - its too big to be focused on any issue. I like Reefs.org. But, any of the reefing sites have 'New Reefkeepers' forums that are great for browsing to pick up general info. But nothing is going to be better than a couple of good books - I really recommend the complete idiots guide to saltwater aquariums for a general overview, and The Reef Aquarium vol 1 for a more detailed overview. I even more highly recommend seeing if there is a local reef club in your area, joining it, and going around to visit some peoples set ups. Seeing what is going on under the hood is easier than reading it.

I don't think sumps are for more advanced aquarists, but I guess I think saltwater setups themselves require understanding of a whole lot of issues. Without an overall understanding of the systems in general, an aquarist is asking for trouble. There are a bunch of concepts to grok, but once you get a handle on them, its really pretty straightforward - there are just a lot of things to pay attention to. Oh yeah, everyone has their own opinion and there are a billion ways to skin a tank. People with a good understanding of a saltwater system 'get away' with stuff because they know what any omission leads to, and they know what to be on the lookout for and how to compensate. A lot of stuff recommended to newbies is recommended to compensate for the general lack of understanding. Ha! Weird. Smaller tanks aren't really any easier or harder than larger tanks because they all have their own issues based on the same basic principals. Once you understand the idea of a sump, it seems so obvious, and in fact, I don't think you would want to run a tank without one.
The thing that I think gets in most peoples way is they want to set up the tank right now, so they get an unsound understanding of the basics, then, after they get frustrated and kill a bunch of stuff, either quit the hobby, or spend even more money on stuff they should have bought in the first place. A learning system is a good thing, and will even help you out in the long term, but almost no one wants to do it - even though most end up upgrading everything within a year.

Wow. Late night rambly wall of text. Cool.

All excellent points.

I've also started avoiding equipment questions for the same reasons--they've all been answered before on this site, and almost certainly in better detail on others. When I start typing an answer to one of those questions, I invariably end up searching through other sites to find relevant threads. At that point I realize I'm just searching for info for those who are for whatever reason unwilling to do it themselves and I abandon the post.

I have a love-hate relationship with RC. The depth of knowledge that's been posted there--even in a single week--is amazing. Albeit most of the content is from the 'peanut gallery,' its instantly recognizable when someone knows what they're talking about. The problem is there is no organization at all. You can have a half dozen fascinating threads posted every day, but each is buried under hundreds of worthless others never to be seen again. Search isn't effective because there's a hundred go-nowhere threads on a topic, say, drilling a tank, for every one important treatise. You can only find the thing you're looking for if you remember a particular phrase or username.

Thanks guys. Your points, particularly regarding newbies, is exactly what I haven't started a "is there a step-by-step guide to setting up a reef tank" thread. I know all the pieces are here, or linked here from other sites, I've just got to ask the magic 8 ball that is TONMO the right questions and all will be revealed.

I know I want to setup a complete and correct system from day 1, and not short-step anything, I know I want to properly cycle my system for as long as necessary, even though I also know that can mean a relatively non-rewarding period of time...The reward will come when I put in my first few "test" species and have great success with them because I was willing to do my homework and be patient and do it right the first time.

I've been wanting to keep Cephs for many, many years now, and have had tremendous success with tropicals in the meantime, waiting to find a site like this where people like Rich and Jen, etc. were having luck with Cephs. I've bought a few books regarding new marine reef setup and have been reading them non-stop, when not researching points the book skips over online...

I still have a lot to learn regarding the external filtration systems available and their benefits and detriments to our specific application here. Things like what makes a good protein skimmer, how to properly setup the sump, whether to use a trickle filter or a canister (likely both), and how to make that work properly with a large refugium, as well as having space for hatching babies, staging them before release into the (likely 90G bow) main tank, quarantine, as well as extra space for last minute visitors.

Just a while longer, just a few more weeks' researching, and I'll be ready to start my build. Hopefully by next year's crop of babies I'll have a ground-up custom (and cycled) system ready and waiting to fulfill what now seems almost like a lifelong dream.

I've never wanted a sump because all the horror stories I've read about when a power failure occurs like the main tank water being siphoned down to the sump and overflowing it all over the floor. I'm not extremely knowledgable on sumps but how do you avoid the power failure overflow nightmare when the return pump has no power?

I've never wanted a sump because all the horror stories I've read about when a power failure occurs like the main tank water being siphoned down to the sump and overflowing it all over the floor. I'm not extremely knowledgable on sumps but how do you avoid the power failure overflow nightmare when the return pump has no power?

A correctly setup sump will never overflow. People who report horror stories almost always have done something incorrectly.

Remember that a sump typically isn't running with the water level near the top. More often its about halfway full. To set up a sump correctly you need to calculate the volume of water above your overflow and return outlet in the display tank (ie, all of the water that will go into the sump when the pump is off), and then just make sure you keep the water level in the sump low enough so it can hold all that. That's all it takes.

People who don't do that simple planning step will flood. People who install check valves in their return plumbing to try and stop the problem will flood.

Dan

I've never wanted a sump because all the horror stories I've read about when a power failure occurs like the main tank water being siphoned down to the sump and overflowing it all over the floor. I'm not extremely knowledgable on sumps but how do you avoid the power failure overflow nightmare when the return pump has no power?

Funny, I just revamped a system at a wholesaler because they were getting floods when they turned off their main return pump!
I think floods are actually easy to avoid, and any horror stories are most likely from people who didn't really understand the whys or hows of sumps in the first place. Its just one of those things I was talking about that once understood, seems obvious. :grin:
As sump is a volume of water external from the main tank. Usually it is under the tank. You put all your filtration, heaters, carbon, skimmers and whatnot into the sump so you don't have to look at them in the main tank, and you get a larger water volume in the system which give you a more stable system. For a ceph, you also get the added benefit of having less equipment going into the tank that needs to be octo proofed.
So, you have a line from the tank to the sump (either drilled into the side or back, drilled into the bottom with a stand pipe, drilled into the bottom but in an overflow, or in the form of an external hang on overflow), and a return pump in the sump (or external, depending on choice) that pumps water back up to the tank. Generally these are called a return, as in return to tank, and overflow because the water level needs to reach a certain height for the water to overflow into that line to the sump.
Placement of the return and 'overflow' are important. When the power goes out, water will siphon from the tank into the sump via both the return and the overflow, and will drain as much water as their placement allows. If they have been placed as to allow more water to siphon than the sump will allow - boom, wet floors.
Most often this problem is not from the overflow because the overflow needs water pumping into the tank to, well, overflow. The flooding is almost always from the return being plumbed low in the tank. If the return is 2 inches below the surface of the water in the tank, in a power outage, it will drain the tank to that level. If the return is 1/4 inch below the surface of the water in the tank, in a power outage it will drain the tank to that level. Obviously, the return that is two inches below the water line will drain more water than the return that is 1/4 inch below the water line. In the power outage, the return that drains the tank two inches down also exposes to air anything in the top two inches of the tank - with a ceph not so bad, with corals really bad.
I don't trust things like check valves or anti siphon holes on the return lines. Check valves almost always fail because stuff grows on the surface of anything in saltwater, which will make the check valve not seal and the water will drain. Anti siphon holes get overgrown with muck or coraline algae and need to be tended regularly.
I like to idiot proof my systems so, the returns generally go way up high near the surface of the water. I also use the largest sumps I can and generally run them at a maximum of 1/2 volume so even if some jinky return I set up on the fly slips and is two or three inches below the waterline, the sump can handle it. And, I make sure to power outage test the system to make sure the sump can handle the volume drained into it in an outage.

Whew!

This kind of talk is so much more fun/rewarding with people who have done some legwork on their own! :D

Thanks guys. Your points, particularly regarding newbies, is exactly what I haven't started a "is there a step-by-step guide to setting up a reef tank" thread. I know all the pieces are here, or linked here from other sites, I've just got to ask the magic 8 ball that is TONMO the right questions and all will be revealed.

Whoo hoo!

I know I want to setup a complete and correct system from day 1, and not short-step anything, I know I want to properly cycle my system for as long as necessary, even though I also know that can mean a relatively non-rewarding period of time...The reward will come when I put in my first few "test" species and have great success with them because I was willing to do my homework and be patient and do it right the first time.

Thats fantastic. However, what you will find frustrating is there is no correct system. There are a bagillion ways to get any of this done and everyone has an opinion as to why to do what they do. You have to decide what you think will work best for you, and try to leave room for changes down the road.

I've been wanting to keep Cephs for many, many years now, and have had tremendous success with tropicals in the meantime, waiting to find a site like this where people like Rich and Jen, etc. were having luck with Cephs. I've bought a few books regarding new marine reef setup and have been reading them non-stop, when not researching points the book skips over online...

I think that experience puts you in really good shape.

I still have a lot to learn regarding the external filtration systems available and their benefits and detriments to our specific application here. Things like what makes a good protein skimmer,
Much debated topic. I like Euro Reefs or ASM's (ASM's are a Euro Reef knock off and are much cheaper)
how to properly setup the sump, [quote]
:grin:
[quote]whether to use a trickle filter or a canister (likely both),
I run neither!
and how to make that work properly with a large refugium,
Why do you want a fuge? As an algae filter? I haven't found them necessary for ceph tanks or reef tanks - though they are in vogue.
as well as having space for hatching babies
Maybe in the fuge?
, staging them before release into the (likely 90G bow) main tank, quarantine, as well as extra space for last minute visitors.
All good stuff if you have the space.

Just a while longer, just a few more weeks' researching, and I'll be ready to start my build. Hopefully by next year's crop of babies I'll have a ground-up custom (and cycled) system ready and waiting to fulfill what now seems almost like a lifelong dream.

You are well on your way, and I can't wait to see what you come up with!

Hey, as long as you guys are writing awesome sump articles, I have what's probably a really silly question:

why don't people just plumb a big vertical vent pipe into the return line? Then there will be no siphon vacuum if the return pump fails, and water in that vent pipe will rise to the same level as the tank water, so it can never overflow the vertical pipe (assuming all pipes are big enough diameter compared to the pump's max flow rate that resistance is negligible) but there can never be a siphon on the return. Of course, having a big vertical mast next to your tank may be kinda ugly.

Am I missing something painfully obvious?

Monty,

I'm not quite sure what you're asking. The return line is pressurized, so the water level in a vent pipe would try to rise to the hydraulic head of the pump, usually >15 ft. If the vent pipe was lower than that water would spray out. If the vent pipe was higher than that, it probably has more volume of water to go into the sump than the tank.

Dan

I guess if you made the mast tall enough it would work, but I think you would then have the pump fighting against the pressure the mast was creating.
People drill 'anti siphon' holes just over the waterline of the return, so if the power goes out, they suck air and break the siphon. They really only work with short runs of tubing though.

Monty,

I'm not quite sure what you're asking. The return line is pressurized, so the water level in a vent pipe would try to rise to the hydraulic head of the pump, usually >15 ft. If the vent pipe was lower than that water would spray out. If the vent pipe was higher than that, it probably has more volume of water to go into the sump than the tank.

Dan

I think you got the question right-- I didn't think the pressure had to be that high in the return line. My thinking (which, like I said, is probably wrong) was that the water in the vertical vent pipe would seek the same level as in the main tank, because it's connected with a pretty large (hence low resistance) pipe. Isn't the hydraulic head the height that the pump could reach if it were just pumping into a vertical pipe that has no outlet except at the top, before the backpressure from the weight of the water offsets the pump's capacity, or am I misunderstanding the term? I'd think that the flow from the vent pipe into the tank into the overflow would relieve pressure far before it could build up a head of more than a few inches unless the return pump was a NASA surplus or something. Pascal's Principle (and his barrel experiment) show that the pressure in a vertical standpipe can be enormous even in a narrow pipe, so I'd think that the pipe between the vent and the tank would relieve the pressure far before a vertical head built up, unless that section were clogged (in which case I guess a pipe less than 15' would become a fountain, which is not so hot, but in the non-vented system it would build up a huge amount of backpressure and probably either stress the pump to death or burst some component, wouldn't it?)

Of course, despite having some confidence in my engineering reasoning, I also assume people don't do this because my reasoning doesn't apply somehow, so I'm not sure how to say the above without sounding irrationally overconfident... I'm probably just thinking about the problem wrong somehow, and maybe I'm just not getting your explanation because I'm thick.

is a sump neccassary for a tank of bandensis?

No, a sump is not necessary, but it does make things easier. Do you have access to a local fish store that has salt water aquariums in your area? I think if you could go to a store and see what we are talking about you might be able to understand the basic setup of a salt water tank.

Also, is there a public aquarium in your area? You could learn sooo much by volunteering at an aquarium. This is how I started learning about salt water aquariums, the staff have a wealth of information and tips about caring for salt water animals.

I'd think that the flow from the vent pipe into the tank into the overflow would relieve pressure far before it could build up a head of more than a few inches unless the return pump was a NASA surplus or something. Pascal's Principle (and his barrel experiment) show that the pressure in a vertical standpipe can be enormous even in a narrow pipe, so I'd think that the pipe between the vent and the tank would relieve the pressure far before a vertical head built up, unless that section were clogged (in which case I guess a pipe less than 15' would become a fountain, which is not so hot, but in the non-vented system it would build up a huge amount of backpressure and probably either stress the pump to death or burst some component, wouldn't it?)

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 :). 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 :)

Dan

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.

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.

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.

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 :). 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 :)

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.

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.

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.

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.

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 :)

Dan

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.

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.

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

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.

oh, sorry, that was not a sump, that was only a bioball chamber set up.

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? :)

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.

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.

Worriedly sticking my toe in this discussion...
I think there is a lot of turbulence inside the piping, and even more at junctions. I think the water in the vent would not be static, it will keep roiling, moving and flowing trying to get down the vent and out the other exit. The water in the vent will stay in a general range but not at a static level, as the water at the bottom of the vent is drawn along by the water moving out the other exit. The water in the vent would be in flux and would need to be replaced by water from the pump constantly, and therefore the pump would always be doing work. :grin:

Worriedly sticking my toe in this discussion...
I think there is a lot of turbulence inside the piping, and even more at junctions. I think the water in the vent would not be static, it will keep roiling, moving and flowing trying to get down the vent and out the other exit. The water in the vent will stay in a general range but not at a static level, as the water at the bottom of the vent is drawn along by the water moving out the other exit. The water in the vent would be in flux and would need to be replaced by water from the pump constantly, and therefore the pump would always be doing work. :grin:

There might be some change in the turbulence in the pipe because of the junction. Most of the turbulence and drag and such will be there whether the vent is there or not, though. It's certainly possible that adding the junction for the vent will create an edge that induces a bit more drag, but I was assuming laminar flow for the most part everywhere except the pump.

Anyway, I'm pretty sure that it's not inherent in the system that there will be significant turbulence in the vent pipe. It could even be damped by putting some sort of wadding at the bottom of the vent pipe, or a screen across the opening to the vent... anything that attenuates the high frequencies.

Of course, it's always a bit worrisome to assume that chaotic and resonant effects can be neglected, as in the Tacoma Narrows Bridge, but I think all the significant contributions from that can be lumped into the flow resistance in the pipe. I actually think the water column in the vent will stay relatively still, though, at least if the pump doesn't have any low-frequency periodic surges. And even if it doesn't, although it will add to frictional losses a bit, if it's oscillating around a particular waterline in the vent, the net work is zero, because when it's going down it "pays back" the potential energy it took when it went up.

I think one of you should build this model to test out this hypothetical thread!

Monty,

Could you draw a quick diagram of the piping you envision?

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.

You're right that I'm using the term work casually. The pump, like the helicopter, is expending a lot of energy to hold a mass in place. In a geometric sense it is clearly not work because neither the helicopter nor the water are moving. We agree that every pound of helicopter or every inch of hydraulic head is consuming energy just to be held there. Even if the water level in the vent was only an inch above the outlet height, that's an "inch" of energy that cannot be used for flow.

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.

If I lift your bucket of water above my head and hold it there, my arms are expending energy to hold it up, even if no work is being done. There's not any work being done on the water above the outlet in the vent either, but the pump has to do the same thing my arms do. You're arguing that once the water is up there in the vent, no effort needs to be expended to keep it there, its as if I was putting my bucket on a shelf, which I am not (putting the bucket on a shelf would be analagous to putting a plug in the pump aperture).

Dan

Here's a picture of the vent system I had in mind. Note that the valve (X in a box) at the base of the vent is there more for argument than because it's necessary.

4637

I think the definition of work explains some of the disagreement here, although I still can't find any reason why the pump has to work harder.

I agree that the idea that the pump doesn't care whether there's the column of water there or not is counterintuitive, but I haven't heard anything besides "it's just obvious."

The claims about lifting the bucket seem to be a bit of a tricky issue. There are two cases here. If you lift a bucket of water over your head, it's clear that you have to spend some effort to keep it there. Similarly, if a helicopter is hovering, and the engine cuts out, it's going to fall to the ground.

However, there are other situations where, once the work has been done to lift an object, it doesn't require any work to maintain its being at that height... although it does require force. If you lift the bucket above your head, and put it on a shelf, in addition to not having to do work on the bucket, you no longer have to expend effort (muscle energy) to keep it on the shelf. What's the difference? The shelf doesn't have some magical immunity to gravity, and yet it also doesn't have to be plugged in or anything. Another common situation is a hydraulic lift in a garage: when you drive your car on it, it runs a pump while it's lifting the car, but when it's at the proper height, someone shuts off the pump, and it stays there. Of course, if the pump allowed backflow, the car would sink, so some sort of valve is probably required. In that case, it's possible to get most of the energy the pump used back as the car is lowered, by running the pump backwards as a generator, although real garages don't bother. Then, the only energy lost is the friction in the system on the way up and down.

I think the main thing we're disagreeing on is whether this situation is more like the garage lift, or more like the helicopter... or, analogously, whether it's more like holding the bucket over your head than putting it on the shelf. There was also some discussion of whether the vent would rise to the full hydraulic head of the pump, which I claim would only happen if the pipe is clogged between the vent and the outlet; as long as some water is flowing into the tank, some (hopefully most) of the pump's work is going toward putting water into the tank, not making the vent into a water tower or fountain.

I certainly will admit that the energy balance in the helicopter/ bucket holding situation is a lot harder to be clear about. I think we agree that no work is being done on the bucket/helicopter. However, because the thing isn't falling to the ground, something is exerting a force on it. In the shelf case, the shelf is exerting a force, which is in some compression of the shelf or bending of the brackets or some such. In the case of muscles, in order to resist significant forces by applying forces, muscles need to expend energy. That's more because of how muscles work, though, and because they have more versatility than the shelf. In fact, if you have a heavy backpack on, and go from sitting to standing, once you lock your knees, you don't have to put any significant muscle force into standing, because your knee joints are acting like the shelf.

In the helicopter case, one can make some weird arguments about the flow of things... in the pedantic definition of work, no work is done on the helicopter at all; it's not moving up or down, so it's subject to the force of gravity and an equal and opposite force of lift at the rotor hub. That force in turn comes from lift in the blades, which are rotating, and pushing on the air, which is being accelerated downwards. The process of accelerating the air downwards does require work, and the energy loss in this causes the blades to be subject to drag that tries to slow down their rotation rate. What the engine is spending energy on (and doing work towards) is fighting the slowdown of the blades that the air friction is causing.

I don't think the column of water is particularly analogous to either of these situations, by the way. The pump running at full bore holding up the full hydraulic head is a lot like the hovering helicopter, but I think once you get the outlet, some aspects change to be closer to the hydraulic lift...

One of several approaches I've used to model this is actually a DC circuit, say a battery (the pump) and a resistor (the tank outlet & other flow resistance) -- the return goes back to the negative terminal of the battery, and is assumed to be negligible resistance (or it's lumped into the resistor, whatever). The vent is (I claim) like adding a capacitor before the resistor, with the other side of the capacitor going to ground/negative. When that's added, the current through the resistor will drop for a bit while some current goes to charging the capacitor, but once enough charge is stored that the voltage across the capacitor is the same as the voltage drop across the resistor, all the current will go back through the resistor as before and the load on the battery will be exactly the same, and the capacitor will do nothing and won't add any load on the battery until something happens to change the voltage. It's not exactly the same, because this model mashes the water pressure and the height of the water in a gravitational field into one thing, but by Pascal's Principle the height and pressure are tied, so I think the mathematics turn out the same. Of course, I eschew electrical stuff that's got transients and inductance and resonance just like I hate hydrodynamics that has oscillations and vortices and such... but I think I we can characterize this system without needing that stuff.

Monty,
I don't understand why the water won't just fountain out of the top of the vent pipe if your experimental valve is open. You show reduced flow into the tank so isn't that a resistor and won't the water follow the path of least resistence or am I forgetting that the lift of the pump will be exceeded at the line drawn for water height?

Monty,
I don't understand why the water won't just fountain out of the top of the vent pipe if your experimental valve is open. You show reduced flow into the tank so isn't that a resistor and won't the water follow the path of least resistence or am I forgetting that the lift of the pump will be exceeded at the line drawn for water height?

As water goes up the vent pipe, the weight of the water pushes back down, so there's some limit to how high it will go up. If the octopus decided to block the outlet to the tank completely, then the water would rise to the hydraulic max head of the pump, so the water can never go out of the vent higher than that. However, presumably you want good flow into the tank, so effectively the resistance of the outlet into the tank should be low enough in practice that the vent pipe level never gets anywhere near that high. If there's a concern that the outlet might get blocked completely, having a properly tuned relief valve down by the pump can prevent both the water going too high in the vent, and also would make the pump less inclined to melt down or blow gaskets if the plumbing gets clogged.

Of course, Dan thinks I'm wrong on some of this, so it might be unwise to build a tank based on my sketch until we work out the details... I think we both agree that water won't go any higher than the max head of the pump, and that if the outlet isn't blocked it won't get all that high in normal operations (because the closer the water gets to the max hydraulic head level, the higher the backpressure there has to be reducing the flow)

From what little I think I understand of your diagram here, wouldn't it be difficult to "balance" the restriction of flow back into the main tank with the desired height of the water in the vent?

Or am I missing something?

From what little I think I understand of your diagram here, wouldn't it be difficult to "balance" the restriction of flow back into the main tank with the desired height of the water in the vent?

Or am I missing something?

I was assuming the restriction of flow should be minimized, except as needed to get a bit of velocity for circulation in the tank. There's no particular "desired" height of water in the vent, except that it's strongly desired not to overflow, but if there's a lot of height there, it probably means that the pump is doing a lot of work to fight the flow restriction, while its main goal should be circulation through the filtration, so that would probably be a sign that the outlet into the tank should be bigger... a high water line in the vent is a sign that the plumbing is keeping the pump to below its capacity for circulation through the overflow/filter setup.

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?

The only other thing I might add to your response to dw is that returns usually are restricted to some degree. Discharge is area multiplied by velocity and most aquarists prefer a high velocity coming out of the return, so the nozzles are often a bit pinched. The plumbing is otherwise kept wide to reduce frictional losses.

Dan

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall. :grin:

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall. :grin:

Makes sense--to get any considerable discharge through small diameter pipe means our pumps are designed for high pressure!

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

This is where penductors, etc come in, right? Taking some energy from the return pump discharge and distributing it into a greater area of somewhat milder current? I should spend more time on your precious reefs.org :grin:

Dan

Interestingly enough I have a output from a return pump that isn't being used. I added a foot of tubing to it above the water line, pointed it up and opened the valve a little bit. The tube filled up very quickly and overflowed, so I shut the valve. So, I think the height of the vent needed to not overflow would be unusefully tall. :grin:

Oh- there has also been movement away from high velocity to higher flow among reefers in the past year or two. The higher velocity tends to damage coral tissues, so more flow is preferable.

I'm confused-- was there any outlet other than up the vertical pipe, or was this just testing the hydraulic head of the pump?

Monty, there are actually two other outlets. Oh, actually 3, there is a bypass in the sump as well.

Dan, actually penductors and eductors are kind of out of 'style' because they take a high pressure pump and they are big and ugly. They do a great job though. The thing people are moving towards are prop 'pumps' like the Tunze or the Vortech or DIY maxijet mods. Good stuff if you are looking for flow. Come to RDO dan...you need more things to read!

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.

(continued)

The only other thing I might add to your response to dw is that returns usually are restricted to some degree. Discharge is area multiplied by velocity and most aquarists prefer a high velocity coming out of the return, so the nozzles are often a bit pinched. The plumbing is otherwise kept wide to reduce frictional losses.

Dan

I certainly will admit that if the discharge outlet is a small nozzle to encourage high velocity at the expense of flow, that will make the water level rise higher in the vent, for the same reason that if you have a hose with a nozzle, it gets a lot stiffer from the pressure. And that may make my vent design impractical, if it makes the height particularly high. Although if it's a major flow constriction that causes the water to get close to the max head of the pump, that will make the pump run at a very low efficiency, and will both not move much water into the tank (so very little will go out the overflow through the filtration) and will also make the pump very stressed and likely to burn out. The vent will neither help nor hinder this condition, it's all about the nozzle (and anything else that impedes flow, like barnacles growing inside the pipes and such.)

Monty, there are actually two other outlets. Oh, actually 3, there is a bypass in the sump as well.


Ok, so regardless of whether Dan or I is right in theory, it sounds like the reason this doesn't work in practice is that the vent would have to be really tall... yay, empiricism!

:grin: !

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%

Monty--It feels to me that you've planted all the trees, but can't see the forest. You've got a nice description of everything that happens when you prick the hole with transient non-steady-state equilibrating periods, etc, etc etc--I understand and agree with all of that. Here are the end members: The pump is operating at 0% efficiency when its just holding up water. Its at its maximum efficiency when its just moving water. Thus, when it is doing both (moving water out the outlet and holding some volume of water above that in the vent) it cannot be at its maximum efficiency.

I don't know how else I can say it, but since I'm closing on a house next week maybe I can make an exact model and measure the flow rates with a bucket.

Dan

Monty--It feels to me that you've planted all the trees, but can't see the forest. You've got a nice description of everything that happens when you prick the hole with transient non-steady-state equilibrating periods, etc, etc etc--I understand and agree with all of that. Here are the end members: The pump is operating at 0% efficiency when its just holding up water. Its at its maximum efficiency when its just moving water. Thus, when it is doing both (moving water out the outlet and holding some volume of water above that in the vent) it cannot be at its maximum efficiency.

I don't know how else I can say it, but since I'm closing on a house next week maybe I can make an exact model and measure the flow rates with a bucket.

Dan

I'm all for experiments.

I realized in the shower this morning that one simple way of describing where I disagree is that when you say "the bucket of water wants to come down" or whatever, I agree that it has a downward force of gravity, so it needs a force to keep it up, but not that it requires energy to keep it up. It so happens that spinning pumps and animal muscles are systems that use energy to generate forces, but there are other thing, like shut valves, that can happily provide a force to keep something up without using energy at all.

In terms of the pump efficiency/ pinhole thing, I think that the system is such that the water column will go down as a function of the pump's efficiency... in fact, the same pressure that's holding up the water column results from the pump working against the resistance in flow, the rise to the hole, etc. If you think of the water in the vent, rather than something that needs to be maintained, as just a springy energy reservoir that will equalize the pressure so that it's the same as it was when the sealed pipe with no vent was there (or the valve was closed with no water above it), that's pretty much how I'm looking at it... or, another way of looking at it is that the height of the water column is controlled by the flow resistance, and self-adjusts to the level where it is neither helping nor hindering what was going on without the vent.

I'm also not sure about "maximum efficiency"-- it still has to fight the rise to the upper tank and the resistance in the pipe and the nozzle and stuff. If it's pressurizing Thales' return line enough to get more than a foot of rise, then his pump must be sized such that it can push water at a rate that makes the resistance of the outflow a pretty significant flow impedence (otherwise, the flow would be high but the pressure at the top would be fairly low). I'll take "optimal efficiency," but I bet the maximum efficiency for the pump is when you disconnect it so it's pumping water from the sump to the sump, with no rise and no pipe resistance.

Anyway, I'd actually (perversely) appreciate it if you'd be as pedantic as possible in pointing out where the "trees" lead to your "forest," 'cause I'm either missing it, or you have an invalid assumption. I'm most frustrated by knowing I either am wrong and don't understand why, or I'm right but I can't articulate why. (I'd be frustrated by the third possibility that we're both wrong, too, but that seems unlikely.) Although I'll admit to having a stubborn streak, I also am usually not so thick I can't see why I'm wrong...

...when you say "the bucket of water wants to come down"...I agree that it has a downward force of gravity, so it needs a force to keep it up, but not that it requires energy to keep it up...


How can a pump provide force without using energy?

I'm also not sure about "maximum efficiency"-- it still has to fight the rise to the upper tank and the resistance in the pipe and the nozzle and stuff

This is kind of what I mean with the forest and the trees thing. Those factors are the same whether there's a vent or not so they cancel out--by even bringing them up you're pulling out the hand lens and looking at the tree bark. At best a distraction.

I'll be pedantic--apology in advance: :grin:

1. The pump has a maximum efficiency where as much of its energy as possible is being used to move water. There's an ultimate maximum efficiency if it was moving water horizontally without friction. That's not meaningful to our discussion at all, we care about a relative max efficiency which is the maximum efficiency it can have at the identical given conditions of both our scenarios ("the fight to rise to the upper tank and resistance in the pipe..."). I assumed this was tacit to the discussion so I believed the term "maximum" unmodified would be adequate.

2. As you pointed out, the pump has a minimum efficiency of zero where none of its energy is being used to move water. All of the energy is being spent to hold that water. We know that the energy is holding the water because if we cut off the supply of energy the water is no longer held. A plug is in no way an analogy because a plug does not hold the water only if it is consuming energy.

3. The pump has that minimum efficiency (one extreme end) when it is holding water at its maximum hydraulic head. All the energy is spent supporting the water, none is spent moving water.

We're just looking at steady states--any transition periods between these end-members (say the first few seconds after opening a hole) are ephemeral and their discussion offers no clarity.

4. We now have two end-member scenarios: in one, all the pump's energy is holding up water. The other, all the pump's energy is moving water. Lets put this together in a continuum of infinitessimal steady state conditions between the two:

a. All the energy is holding water up, none to flow. The water level in the vent is at max pump head.

b. Most of the energy is holding water up, a little to flow. This is the pinhole, and the water level in the vent is near max pump head.

c. Half the energy is holding water up, half to flow. For the sake of argument lets say the water level is halfway up the vent.

d. A little of the energy is holding water up, most to flow. The water level in the vent is just a little bit above the outlet.

e. None of the energy is holding water up, all to flow. The water level in the vent is at the same level as the outlet, no higher.

Your drawing of what you take as maximum efficiency physically looks like d. There is a few inches of water above the outlet. I believe your argument is that once the energy was used to lift the water there it doesn't need to be used again and the pump is working at max efficiency to move water. If that were correct, then wouldn't the pump also be working at maximum efficiency in b., the pinhole? Would it be at maximum efficiency regardless of the height of water in the vent? Do you believe the pinhole has the same efficiency?

The logical conclusion to this line of reasoning is this: If we agree the pinhole is less efficient than your drawing, then isn't any water level below your drawing going to have a higher efficiency?

Dan

How can a pump provide force without using energy?

Your chair is providing a "normal" force equal to but in opposite direction to your the force of your weight to keep your bum from falling to the ground, and it does this without using energy. Monty thinks the water in the vent is also being held up by a normal force like this, the force of a shelf keeping a bucket up, a plug keeping water in or a ratchet holding a jack up. I contend that the scenario requires dynamic support, like the helicopter or holding a bucket of water over my head.

As evidence for my point of view, I submit that the water behaves like the helicopter or my arm and not like the jack or the shelf. When the pump is turned off the water rushes back down.

How can a pump provide force without using energy?

Some pumps, like a reciprocating pump, can provide a force against backflow without using energy, because they include one-way valves, and there are other things, like any solid object, that can generate force without using energy-- if I set a book on a table, I don't have to have the table plugged in to the wall to keep the book from falling through to the floor. But the table is also not going to lift the book higher into the air: that requires work, which requires energy.

But that's not what I'm trying to get at. The pump is using energy to move the water through the pipe, which pressurizes the water. That pressure is there whether it's in a closed pipe, or whether it's in an open pipe with a column of water pushing down, because the column of water is supplying the same force role as the pressure-containment of the pipe was in the closed system.

I'm not actually saying that the pump is doing anything different in the two (closed and vented) systems, I'm saying that adding the column of water removes the exact same amount of strain energy in the pipe; they're equal forces containing the pressure generated by the pump, so the pump is generating the same flow against the same pressure in both situations. And the water height in the pipe will naturally seek the level where this is the case, if the vent pipe is high enough to accommodate it, anyway.

It occurs to me that your question may have led to the core of my disagreement with Dan on this, Michael.

My analysis says that because the top of the vent is open to the air, the column of water provides the same containment force for the pressure in the pipe that the pipe itself did before. The pump has to do work to pressurize the system, which is more visible when there's a column of water, but there is energy stored in either the water column or the stretching of the pipe which provides a force that balances things out in the steady state.

And once the pump has pressurized the system, either by distorting the pipe or raising the water, it no longer has to provide "maintenance" energy, although if it stops, then the system will have a path to depressurize, which is much more obvious when you see a bunch of water drain out, but is almost invisible in the case of the pipe, which is probably rigid enough that it only has expanded in the tenths of millimeters, but it's stiff enough that that deflection provides as much force as a big column of water.

My analysis says that because the top of the vent is open to the air, the column of water provides the same containment force for the pressure in the pipe that the pipe itself did before. The pump has to do work to pressurize the system, which is more visible when there's a column of water, but there is energy stored in either the water column or the stretching of the pipe which provides a force that balances things out in the steady state.

Now I see where you're coming from, but I still think its going to take energy to keep that water up there, which is going to inherently reduce the efficiency (a la my pedantic other post).

I can't argue the work point as my education in this area is limited, however, I have an idea that might suggest an answer.

Build a simple set up with a tiny pump (it could just flow back into the pump bucket) initially eliminating the vent pipe. Run the system for say 15 minutes and record the water temp before and after the trial.

For the second test, dump the water and refill the bucket with water at the same beginning temperature, add the vent pipe and rerun the test.

If there is no difference in temperature at the end of the two experiments, it would seem that the pump worked no harder with the siphon breaker.

If I was not behind in some coding I would try it just for fun myself.

This is kind of what I mean with the forest and the trees thing. Those factors are the same whether there's a vent or not so they cancel out--by even bringing them up you're pulling out the hand lens and looking at the tree bark. At best a distraction.


As opposed to hovercrafts and helicopters, which are directly relevant :roll:


I'll be pedantic--apology in advance: :grin:


hey, I like pedantry... so much I'll be pedantic back. But anyway, when 2 people smart enough to be able to understand something are at odds, it seems like pendantry and experiment are all that's left. Er, unless we prefer a "hearts and minds" campaign to getting to the actual answer...


1. The pump has a maximum efficiency where as much of its energy as possible is being used to move water. There's an ultimate maximum efficiency if it was moving water horizontally without friction. That's not meaningful to our discussion at all, we care about a relative max efficiency which is the maximum efficiency it can have at the identical given conditions of both our scenarios ("the fight to rise to the upper tank and resistance in the pipe..."). I assumed this was tacit to the discussion so I believed the term "maximum" unmodified would be adequate.


The reason I'm not too keen on the name "maximum" it's the maximum given a bunch of stuff, and we seem to disagree on whether messing that stuff will make a difference in the efficiency of the pump. But I'm ok with thinking of it as the same efficiency it has in the closed system with no vent at all (or with the valve closed, which I think we agree is the same thing, right?) A more direct reason I care, though, is that your pinhole is changing the flow resistance to something that'll effect the pump's efficiency.



2. As you pointed out, the pump has a minimum efficiency of zero where none of its energy is being used to move water. All of the energy is being spent to hold that water. We know that the energy is holding the water because if we cut off the supply of energy the water is no longer held. A plug is in no way an analogy because a plug does not hold the water only if it is consuming energy.


In my worldview, this description has a subtle problem in it. When the pump is at zero efficiency, it is pumping against a pressure that is causing it to be unable to move any water. Therefore, all of the energy going into the pump is being turned into something other than flow (which moves water, and does mechanical work). Ultimately, this is heat, although it may stir up the sump or something. The pump really doesn't see anything beyond the pressure it's fighting against at its outlet, so it isn't doing anything different when that pressure is caused by a column of water than it is when it's caused by a cork. Although the cork has to be able to hold back the same pressure as that column of water, which is what I was getting at in the other post; the cork or outlet pipe or something will deform, and store some mechanical potential energy to respond passively to the pressure the pump generates until the pressure either overwhelms the pump down to zero efficiency, or overwhelms the cork. Or melts the windings in the pump motor or blows a gasket or whatever.


3. The pump has that minimum efficiency (one extreme end) when it is holding water at its maximum hydraulic head. All the energy is spent supporting the water, none is spent moving water.


Nope, all the energy is spent pushing on some pressure that's pushing back, which leads to a lot of heat generation but no mechanical work. The fact that the pressure is caused by there being a column of water there is immaterial; it could be a piston and a big rock, it could be a big spring, it could be shock absorber from a 74 trans am, it just has to be something that, when pushed, wants to push back. And if the pusher can push back hard enough without failing, the pump is turned into a heater, pretty much.

However, it is true that without the pump churning away, something other than the pump would have to hold back that force. But that could be a passive thing like a shut valve. In fact, if there were a valve right at the pump outlet, shutting the valve wouldn't change the pump's view of things anyway, because it's maintaining the max pressure it can with no flow, and it wouldn't change the column of water/spring/whatever side, because all it knows is that it's pushing with pressure Pmax on something that isn't yielding.

I know this seems extra bonus pedantic, but I think you're confusing "turning energy into heat with the side effect of maintaining pressure" with "spending energy on the mechanical work process of moving water." It's certainly valid to say that the pump running at zero efficiency is causal to the pressure being maintained, but when talking about conservation of energy, or mechanical work, or conservation of momentum, the pump is doing zero work, nothing has any net momentum, the pump's energy draw is being converted into heat, and past the pressure being Pmax at the pump outlet, the rest of the system is just sitting there.


We're just looking at steady states--any transition periods between these end-members (say the first few seconds after opening a hole) are ephemeral and their discussion offers no clarity.


sounds good to me


4. We now have two end-member scenarios: in one, all the pump's energy is holding up water. The other, all the pump's energy is moving water. Lets put this together in a continuum of infinitessimal steady state conditions between the two:


repeat complaint that the pumps energy is applying force to counter a pressure, which is only indirectly holding up water. The actual energy is going to heat, no energy is going into the changing the pressure.