# Head pressure



## BenHugs (Jan 13, 2007)

I just bought a new water pump for my tank and it comes with a graph telling me that at 0 head it pumps 1017gph but at 6'or head it pumps 600gph. How do I calculate this?? my tank has an under gravel filter that this is inline with and both the intake and return enters the bottom of the tank through bulkheads. Do I measure from the pump to the bottom of the tank (outlet) or do I measure to the top of the tank as that is the height that causes the pressure?? Or is it 0 head pressure as gravity is pushing the water into the pump and it just speeds it up and back out???? :roll:


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## Hoosier Tank (May 8, 2007)

So you have a Under Gravel Filter in your tank plumbed out a bulhead at the bottom to an external pump and then back into the tank through another bulkhead at the bottom as well?
Sorry I'm confused.


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## BenHugs (Jan 13, 2007)

Hoosier Tank said:


> So you have a Under Gravel Filter in your tank plumbed out a bulhead at the bottom to an external pump and then back into the tank through another bulkhead at the bottom as well?
> Sorry I'm confused.


That is correct.........I should stop posting at 1am..... then people could understand me :?


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## BenHugs (Jan 13, 2007)

This may be what I was looking for :thumb: 
Still a little confusing though
http://www.aquariumpros.com/articles/headpress.shtml


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## Hoosier Tank (May 8, 2007)

Sorry for the lack of responce after my first reply, but that was a darn good question! I had to mull it over in my brain :lol: 
But your research paid off. 


> If the pump draws water from and returns water to the aquarium through a filter, divide this figure in half as the pump is getting assistance from gravity on the intake sideâ€¦
> 
> If you are returning the water to your aquarium at or above the water surface, do nothing. If however the pump must overcome the pressure of water pushing back from a return line below the surface, you must add some more feet of head to your calculation. For each return, add one feet of head for every foot the return point is located below the water and then add these together. If the return is more than two feet below the water's surface, make that two feet of head for every foot below the surface of your aquarium.


Make sence that you have to add the height of the aquariums water level thats above the return to the head height. After all it is water pressure pushing down on the pump that is trying to push up. 
So yes, add it... I wouldn't even figure in the assistance you get from the intake being below the water level and that way you have a margin in your favor!

Now my concern is, what if you have a leak ANYWHERE in the system, will it empty your tank?


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## Barbie (Feb 13, 2003)

You also have to figure in head pressure that you get when you add any elbows to the system. It can definitely add up quick, but you'll find you've ended up with way more flow than you thought you would in the end. All of the filters and powerheads that we buy for aquarium use show a gph rating, but they neglect to mention that's without media or any resistance such as being attached to a filter . Just make sure you aren't "overpowering" the intake hole size and you should be fine.

Closed loop systems like this are all the rage in sw tanks. Yes, if there's a leak anywhere the whole system will drain to the floor, but they are less prone to leaks as there isn't an open body of water in the sump to have to control, also. Well as long as you buy the right glue and what not for the pressure rating .

Barbie


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## BenHugs (Jan 13, 2007)

Yes my tank would completely drain onto the floor and this is one of the 2 systems the tank came with it also has a drilled overflow into a refrugum then into a wet/dry sump system. This tank was a 255g bargain that used to be saltwater ($500) now it's going to be a hap/peacock tank :thumb:

So if my calculations are correct then my pump sits 2' below the tank but is gravity fed into , so call that 1' but it has to pump into a 3' tall tank and there are no elbows on the output side of the pump so would it be 4' of head pressure which my pumps graph says I will get somewhere aroun 750gph???? sound right???? or would it be the total head pressure divided by 2 so that would be 5' divided by 2 which would then give me around 900gph???
So confusing :lol:


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## kornphlake (Feb 12, 2004)

Many of the descriptions of head pressure I've read range from arbitrary to defying the laws of physics. When a loop of tubing (or rigid pipe) is plumbed from the aquarium back into the aquarium there is no pressure in the loop, if there were a differential pressure from gravity then water would flow through the tubing without a pump. That sounds ridiculous doesn't it, we all know that without a pump water doesn't flow in a loop of tubing the only way to move water in a tube is by having one end open and lower than the water level to create a siphon. Next time you are cleaning your tank start the siphon to drain water out into your bucket and once the siphon has started put your thumb over the bucket end of the hose and put that end in the tank so that both ends are in the aquarium, notice how the water stops flowing through the hose? Water stops flowing because the pressure differential is brought to zero when both ends were placed at the same height. Any differential in pressure no matter how slight would cause water to flow through the hose but it doesn't no matter what you do with the loop of hose as long as the two ends remain submerged in the same aquarium no water will flow through the loop. In a closed loop (as in a canister with both the intake and the return submerged) there is no "pressure" at the lowest point in the loop or anywhere else, there is just as much gravity pushing down on both the intake and the outflow side, the two forces balance each other out and the result is a net pressure of zero. The amount of pressure the pump sees is just the same as if you took that pump and submerged it in the aquarium, it would draw water in the intake and push it out the return with or without any tubing attached. Tubing will cause some friction which does reduce the amount of flow and can be significant but it wouldn't have anything to do with the height of the loop rather the diameter of the tube, the number of bends and the viscosity of the liquid being pumped. Head pressure may be an issue if for some reason the loop had to be primed, until the force is equalized in the hose there will be some pressure differential that the pump has to overcome.

Head pressure really only applies in the case of an open system such as a sump where gravity drains water into a container that is not sealed or pressurized and the pump pushes the water back up into the aquarium, in this case head pressure would be calculated using the distance from water line to water line, regardless of where the intake or outflow might be.


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## BenHugs (Jan 13, 2007)

Kornphlake you should read the link I found after asking this question(lots of varibles).
The reason I have a difference in head pressure is because my tank is drilled in the bottom for both the inlets to the pump and then the water is pumped back in to the tank by another bulkhead "UGJ" so this is where the numbers change. I have the same force acting on both sides of the pump but unlike a canister filter I have to overcome the tank water pressure to circulate the water back into the tank. Yes the water pressure from the inlet does help (found this out in my reading) but the outlet does have head pressure to overcome. I wondered about all this when I decided to switch a high head pump for a more efficient high volume pump.
I do agree with you on the canister filter and open system, but this seems just a little different and from my reading was worth calculating.
Thanks for your info everylittle bit helps :thumb:


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## BenHugs (Jan 13, 2007)

For anyone who's interested the pump I removed from the system was a pentair quiet one 4000hh(high head) capable of pumping up to 13' and pushes 980gph at 120watts. The new pump is also a pentair quiet one 4000 (standard) it pumps upto 10' and pushes 1017gph and only uses 50Watts. Seems to be a no brainer to me. I also needed a new impellor for the 4000hh which is around $20 the new 4000 pump was $60.


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## kornphlake (Feb 12, 2004)

BenHugs said:


> The reason I have a difference in head pressure is because my tank is drilled in the bottom for both the inlets to the pump and then the water is pumped back in to the tank by another bulkhead "UGJ" so this is where the numbers change.


I think we're having a problem with semantics, if you had a difference in pressure water would flow through the loop without a pump. That's not going to happen, the only way to get flow through the loop is to induce a difference in pressure by using a pump. What you have is a restriction on the UGJ side and a free flowing drain on the other side.

There is no difference in pressure, the column of water above the intake is just as heavy as the column of water above the return, even if your intake were at the top of the tank and the return were at the bottom the column of water remains the same. This is difficult for most people to understand, going back to the siphon hose I described in my previous post, once the hose is filled with water plug the open end and put it into the tank. Notice that there is no flow through the hose? Now push one end to the bottom of the aquarium and keep the other end near the top, do you notice any flow through the hose? You shouldn't, if you do it's just your imagination. What is happening is that the column of water is measured from the highest point to the lowest point, in this case the highest point is the water's surface and the lowest point is the lowest point in the loop of tubing. With one end of the hose out of the tank the water column above that end decreases which decreases the pressure on that end which will let water flow through the hose. If you were to plumb that loop so that both ends passed through bulkheads on the bottom of the tank you'd still have the same measurement for the water column, from the water's surface to the lowest point in the loop. There is no differential in pressure or water would flow through the loop without a pump.

Put a ball valve on one side of that loop to act as a restriction to simulate your setup and you still don't get current through the loop because you have not put any pressure differential in the loop, only a restriction. A restriction can not be overcome with head pressure, it is just something that will always be there. An ideal remedy would be to relieve the restriction by using a larger diameter pipe, reducing bends or add more branches to increase the volume of the return.


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## BenHugs (Jan 13, 2007)

Kornphlake I do agree with you. My thoughts are on what would my new pump gph be in this set up. Now I know that the water won't run through the system without a pump. This head pressure question revolves around the fact that water gains roughly a 1/2 psi per foot of elevation. My pumps sits about 5' below the tanks surface so the psi acting on it should be around 2.5 psi of course that psi is acting on both sides of the pump. So how would that equate when trying to figure out what the pump would put out on the chart included on the box???? I'm probably making this more complicated than it is and it was just a nice to know kinda question anyways :roll:


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## kornphlake (Feb 12, 2004)

Contrary to much of the information out there I don't believe any of the .5 PSI per foot mumbo jumbo. Nobody seems to be able to explain how this mystical pressure exists without a differential in pressure. I will agree that there is pressure caused by gravity acting on the column of water above any given point in an aquarium, but the pressure is only relative to the air outside the aquarium, not relative to the water surrounding the "given point" with exactly the same column of water above it. A 5 gallon bucket holds 5 gallons of air at sea level just as it hold 5 gallons of air at the top of mount Everest, when you dump the bucket out you spill 5 gallons of air on the ground rather it's a sandy beach or a snowy peak. Maybe that's a lame analogy but really that's what a pump is doing, it puts water in a bucket then moves the bucket from its left hand to its right hand and dumps it out, the size of the bucket doesn't change no matter what the pressure is so the flow rate doesn't change, you always get one bucket of water when you dump one bucket of water.

If there is any head loss I suggest it is because of a restriction, friction within the line, excessive bends, etc. The suggestion that head pressure increases with water column height would mean that a power head placed just below the surface would flow more water than a power head near the bottom of the aquarium where a larger column of water is above it. That just doesn't make any sense, pressure on the inlet and return are equal so there are no head losses caused by the water column and flow rates will be the same.

Calculating the flow of your pump should be done using the suggestions for flow reduction caused by friction in the tubing, bends or elbows and some margin of error in the constants used as they were most likely developed for salt water systems. I'd ignore the idea of head pressure caused by the water column above the pump. If the pump is rated for 1017 GPH at a depth of 2" of water I assume it will flow 1017 GPH at a depth of 24" of water or 60 feet of water minus the frictional losses already mentioned.

I've read the "Aquatic Systems Engineering" book mentioned in the link and strongly recommend it to any one interested in real theory and derived calculations for water flow through an aquarium. To the best of my recollection it doesn't touch how to calculate head losses in a closed loop, which was something I was hoping would be explained to confirm the information contained on sites like aquariumpros. Unfortunately the link for spa systems is a dead link, I'd expect swimming pool plumbing to be closer to an engineered system rather than the one size fit's some equipment available for aquaria.

I'm not saying I'm right here, I'm just saying that nobody on the forums has been able to explain in an academic way exactly how the definition of head pressure is derived, I've followed links in the past and found none really address how pressure constants and equations were created or how they have been measured to confirm the calculations are correct. They all seem right on the money for an open system like a sump, but in a closed loop things start to become questionable.


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## kornphlake (Feb 12, 2004)

I guess I should edit my bucket analogy, with equal water column height you fill the bucket on the left side and dump the bucket on the right side. With unequal water column height you fill the bucket on the left side and dump it on the right side but before you can tip the bucket back up gravity reverses itself and 1/4 of the water rushes back in the bucket. That would be an open system, which is something that can't be described very well with the bucket model.


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## BenHugs (Jan 13, 2007)

Kornphlake you've done a great job explaining this to me. I thought that it worked much as you put it but then went reading and thats where I started getting confused. I do know through work that yes you do get .5psi per foot of elevation but like you say that wouldn't apply in a closed loop system.
I'm gonna say that my pump should get very close to the 1017gph that it's rated for as the pump discharge goes straight into the tank with no bends. All the bends are before the pump and shouldn't affect performance (correct me if I'm wrong).
Kornphlake thanks for the insite :thumb:


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## kornphlake (Feb 12, 2004)

Like I said there are a lot of websites that claim head pressure exists in a closed loop because of gravity. It's easy to get confused, you have to remember that the internet doesn't have any standards for publication, anyone can publish anything they want on a website rather it is right or wrong. A lot of websites have dozens of pages of great information with one or two items that may be half-truths that people don't recognize as incorrect because they are buried in pages and pages of credible information. I'll bet you could find a website with lots of credible information for the first time hobbyist that says you only need to set up a new tank and let the filter run for a couple days before adding fish, you could probably find a couple websites that claim you only need to do water changes once every couple months... I've got a couple books published by Tetra Press and TFH Publications that make those exact claims, how could the two most prominent publishers in the hobby be wrong?

I'm actually thinking about putting together an article with some diagrams and equations to describe head pressure and show why it's irrelevant in a closed system. I don't have any empirical evidence that my assumptions are correct, equipment for measuring flow isn't cheap and I'm afraid a home built flowmeter would be incredibly inaccurate. I consider myself more of a mentor than a teacher, I'm better at showing you how to do something then supervise you while you do it yourself rather than explain everything you need to know and expect you to be capable of making critical decisions on your own. The ink may never reach the paper on this topic.

As far as losses due to restriction (bends, reduced pipe diameter, friction, etc) it doesn't matter if the restriction is before or after the pump, it will all affect the flow rate of the pump. You have the freedom to put the pump anywhere you want in the loop, if it is 2" from the bulkhead for the drain the losses will be equal to the losses of the entire loop, if the pump is placed 2" from the return bulkhead the losses are the same the only difference would be rather the loop has a relative negative pressure or a relative positive pressure. Neither is more or less important it just is what it is.

One thing to consider is that pumps are much more efficient at pushing water out the outlet side than they are at sucking water in the inlet side. Think of it this way, air pressure at sea level is about 15 psi, a pure vacuum at sea level would be 15 psi of "suck," that is the most pressure a pump could possibly "suck." Once pressure reaches zero there is no more work that can be done, there is no such thing as a negative absolute pressure. On the other hand it's very easy to compress air to hundreds of psi, a pump is simply more efficient at creating pressure than it is taking pressure away. You are going to have some restriction in the inlet side of the pump there is no way to avoid that, but I would suggest that you do all you can to move the restriction from the inlet side to the outlet side and add a valve to throttle the flow on the return side of the pump if necessary.


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## BenHugs (Jan 13, 2007)

Kornphlake 
Thanks for the insite once again.
I can't change my system as it is now up and running, but the way it is set up I'm sure I don't have any negative pressure acting on my pump and this is my reasoning. I have two bulkheads that put water into a ocean clear red sea canister filter the water then goes to the pump which is sitting at the lowest part of my stand and then straight back into the tank. I feel that I do indeed have gravity on my side as my red sea canister filter has a gauge on it that is always reading a positive pressure. With the system shut off it reads close to 2psi and when running still has over .5psi. I know it's not a flow gauge but these ocean clear psi gauges are really cheap. maybe you could make one of them work for your testing. :thumb:


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