Expansion Tank Mystery?

Empire_2

The only problem is the water would move very slowly with a total 1 psi diff. at suction of pump. Lets say the pump delivered 6 psi, and 2 tanks cancel out the pumps ability to show any increase towards the left tank and show little or no Decrease to produce a differential to flow on the right tank. The only thing in this case that will determine flow is the system loss. In this case it is 5 psi.. The pump will only see a 1 psi diff to flow. If system loss was at 6psi or higher,=no flow....In addition the pump would be spinning its wheeles and cavitating big time...

Mike T.

Dave_12

Expansion Tank Mystery?

We are trying to figure out what happens when a system has multiple expansion tanks in different locations. (We know this is bad design, however it happens and we want to know how to properly analyze it.) We've read through Pumping Away and Primary Secondary Pumping Made Easy.

We've simplified the problem as illustrated on the attached sketch.

The question is - If the expansion tank tie-in represents the point of no pressure change, can this pump move water around the system? Alternatively, what would the pressure be before and after each of the two expansion tanks shown?

gerry gill

awe....

now your making my brain hurt...

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thp_8

:}

If you have a circ. that is rated for 20 gpm @ 5 psi of head your going to get 20gpm in the loop... As far as psi readings, if the gauges are directly in front of the circ. and after it, the static fill is 12 psi, and the piping looks just like that, the discharge gauge will read 14.5 psi and 9.5 psi on the suction gauge...

Empire_2

I respect your opinion

But,...If the tank on the discharge is at 12 psi sp. how are you going to raise your discharge pressure to 14? 12psi is the point of no pressure change. When pumping directly at an expansion tank the pump has to show it diff pressure in the suction side, but there is a tank there to. Also at 12Psi.

Mike T. I think Dan would have the final answer.

thp_8

Halonized

Because the circ., provided that there is close to air free water in the system will establish 20gpm at 5 psi differential on that loop. After start-up the tanks are just along for the ride. You need to look at the tanks as being there to average out the pressure of the system. The tanks main job is to maintain an acceptable pressure from a cold 50*F start-up to a hot 200*F run condition. Thats why we call them expansion tanks. Yes I know if the tanks were Halonized they would be call Point of no pressure change tanks. :-}

thp_8

Halonized

Because the circ., provided that there is close to air free water in the system will establish 20gpm at 5 psi differential on that loop. After start-up the tanks are just along for the ride. You need to look at the tanks as being there to average out the pressure of the system. The tanks main job is to maintain an acceptable pressure from a cold 50*F start-up to a hot 200*F run condition. Thats why we call them expansion tanks. Edit my dad still calls them ballast tanks. Yes I know if the tanks were Halonized they would be call Point of no pressure change tanks. :-}

Dave_12

I thing I may have figured it out

This is a great and simple example of "what happens" when you have several expansion tanks in different parts of a system (a no, no but common).

When you only have one point of tie-in for the expansion tank or tanks, this is the point of no pressure change. As Dan says, you cannot add or take water from the expansion tank because "where would it go."

However in this simple example, we have two expansion tanks that tie-in at two different places in the system. After pondering on this for awhile, I agree with the pressure values offered by thp above (14.5 PSI at the pump discharge and 9.5 PSI at the pump suction), but want to add why this happens. Frankly this stumped me for awhile.

In my judgment, there is no point of no pressure change when you have multiple expansion tanks in different parts of a system. What happens is that one tank will "give up" some water to the system and the other tank will "take some" water from the system. The exchange would be exactly the same amount of water. In this simple example, the pump will do what it has to do and develop 5 PSI across it, and it can happen because each expansion tank accomodates it by transfering a little bit of water to and from the sytem during the pumping cycle. Not a good design, but this is what happens to the system shown in this example.

If anyone sees this from a different point of view technically, please jump in.

thp_8

Pressure Change

Yes there is a point of pressure change it is the circulator.
The point of pressure change that relates to the tank only lasts for about 1 second that when the circ. starts. After that your P.D. is set and the rest is history.

Jed_2

Seems to me

the system will suffer in that that differential battle.

Jed

Jamie_5

pressure change

I think thp is right that the point of pressure change is the circulator, that the pressures at the inlet and outlet of the circ will 9.5 and 14.5 psi, respectively, and that there will be a momentary adjustment in pressures around the system as the pump starts.

Imagine the loop without any expansion tank, but pressurized to 12 psi. With the pump off, there is 12 psi at every point in the system. Although liquids are roughly incompressible, it seems to me that the pump couldn't do anything if it didn't affect the water density locally, i.e. at the inlet and outlet. If it couldn't do that, the pressure would be equal at inlet and outlet and no water would move. Turn on the circulator, the pressure differential is created, and the water flows. I believe the pressure at the inlet and outlet will be 9.5 and 14.5, reasoned as follows.

Assume momentarily that the 5 psi differential could be realized solely as an increase in pressure at the outlet, that is, the outlet pressure would rise to 17 psi while the inlet remained at 12 psi. I believe this is impossible; it would imply that the average pressure in the loop was 14.5 psi without any explanation as to how the water became less dense or water was added to the system. The only way for the average system pressure to remain at 12 psi is for the difference between inlet and outlet to be split as equal amounts of pressure drop at the inlet and pressure rise at the outlet, or 9.5 and 14.5 psi.

Now add an expansion tank, pre-charged at 12 psi, just before the circulator inlet. Again, at rest, the pressure is 12 psi everywhere in the system. When the circulator starts, however, it lowers the pressure on its inlet side, just as it did before. The difference between this and the previous situation is that water from the expansion tank flows from the tank into the piping, due to the pressure difference (created by the pump) between the water in the tank and the water in the pipe, until the pressure is the same at both places. In essence, then, water is added to the piping, though the volume of the piping is the same, implying that the average pressure around the circuit must rise, and the pressure at the inlet to the pump must be higher than 9.5 psi. Hence, the advantage of pumping away.

Now add an expansion tank, precharged at 12 psi, to the outlet side. When the circulator starts, the pressure differential is created, but now the bladder on the expansion tank at the outlet must give way, because it is only under 12 psi pressure and the water at the outlet of the circulator is higher than that. So water moves out of the piping and into the expansion tank until the pressure in the tank is equal to that in the piping. I suggest that the amount of water that moves from the piping into the tank on the outlet side will be equal to the volume that moves out of the piping and into the tank on the inlet side, and that the average pressure around the circuit will again be 12 psi. Hence, inlet and outlet pressures will be 9.5 and 14.5 psi respectively, as thp stated.

If anyone can tell me what's wrong with this reasoning, I would appreciate it.

Empire_2

OK,..OK You guy's got me out #'ed

Let me think about it for a while and I will chime in again. 3 against 1, and I read all the thoughts presented,But in my mind, I still have issues.....Wow now I'm starting to sound like my sister ;-)

Mike T

thp_8

Issues

Glad we could help.

Empire_2

After further review, I am sticking to my Guns...

This or any pump near or at the 5psi loop drop cannot produce any "Delta P" because you cannot raise the pressure at the PONPC of 12psi. The pump in and of itself is rated at whatever performance but it is the system that dictates what happens, in this case. The pump cannot make the pressure go any higher than 12psi due to the tank on the discharge and is also slaved to the 12psi at the tank on the right @ 12psi. Because of this tank(s) set up the perssure drop in the system will dictated by the "Delta P" of this particular system 5psi loss.....

I know that's gonna ruffle feathers, but I am open and see what everyone is saying,...Perhaps I am incorrect, But Damn it feels so good to work the Synaps gaps for a while...

Mike T.

Empire_2

I need a beer ;--)

Damn, now I am starting to wonder.... I thought I had it but Ya All make good points....

Almost time for the "Buffalo Sabres" to take control of the Eastern Division Lead, so excuse me..............

:-----)

thp_8

You can stick to your guns

But as in any gun fight some one has gotta lose. We got a few different laws of physics on are side. If you want to understand this, look at the tank as an expansion tank. Nothing more nothing less.

Empire_2

Hands up,....Dont Shoot ;-)

I am trying to understand and will monitor this topic till I get it thru my thick skull ;-) Just don't see it YET!!! ;0) I'm gonna go back into the saloon and have a drink and ponder all thoughts....................If a am wrong, I apologise, but Damn, thought I had an argument....I guess it just goes to show all dogs can learn new tricks;-)

Mike T.GO "BUFFALO SABRES"

P.S. I gave incorrect opinionated advice on another topic...Boy do I feel like an ****........

Humbled to say so Mike T.;-(

Dave_12

Mike T -- Keep thinging about it

Mike T:

You are right that the pressure can't change at the expansion tank. But when you have more than one tank at different locations, this changes EVERYTHING.

Remember the reason why the pressure can't change--because if water goes into or out of the expansion tank, where would it go?? But when we have two or more expansion tanks on different sides of the pump, water can leave one expansion tank and go into another because the pressures at the expansion tanks WILL BE DIFFERENT when the pump is running.

Keep thinking it through and I believe that the light will come on.

Dave

thp_8

Hello Walls

The point that is not sticking here is that it's just a tank................ The circulator is the point of pressure change............ If the circ. will create 20 gpm @ 5 psi and the loop has 5 psi drop acrost it, the rest is just math. The circ. is going to move 20 gpm with a 5 psi drop, it has to. Where you put the gauges in the system will yield the gauge reading in direct relationship to the piping flow resistance at that point. Nothing more and nothing less.

Empire_2

Got it Guy's

And thanks for clearing that up. Makes sense now. One thing I love about this sight is that you can allways get multiple answers to any question. Geat work

Mike T.

Ron Schroeder

Are these diaphragm type tanks? If the fill pressure is the same as the pre-charge on the tanks (assuming they are diaphram tanks) the pressure at the inlet of the circ will drop by about 5psi because you can't suck the diaphram out of the inlet side tank and the circ. outlet pressure will stay at about 12psi.

If the fill pressure is at least half of the system loss above the pre-charge then with equal sized tanks, the inlet side will drop by about 1/2 the 5psi system loss and the outlet side pressure will rise by about 1/2 of the 5psi system loss once some water gets shifted from one tank to the other.

If the tanks are different sizes, the amount of pressure change will be inversly proportional to the relative size of the tanks.

Imagine a golf ball size tank on the circ outlet and a 55 gallon tank on the circ inlet. Then the pressure on the inlet will hardly change at all but the pressure on the outlet will go up by almost 5 psi.

Ron