Full tank being emptied - Bernoulli 1

[Peter C]

Hello,

Very commonly people we hear that a tank where column pressure is not big enough to overcome Patm will not empty by itself. In my opinion, most of those who say that forget that once the tank was open to be fulled and, therefore, we must add 1atm pressure to the fluid at any point.

So lets consider the following case: 1) a tank of height H is completly empty, 2) fluid (water in this case) is being pumped in through a valve in the bottom side while air is comming out at the top side, 3) once the tank is completly full, both valves are closed and 4) the valve at the bottom is opened.

When the valve at the bottom is opened, what happens with the fluid inside? Will some of it come out, creating a "void' in the top of the tank (p = pvapor of the fluid)? NOTHING will come out? It will completly empty?
For now, lets ignore that air bubbles may come in from the bottom.

Can you answer using Bernoulli?

Regards

 

[TiCl4]

Are we simply ignoring the fact that due to the dynamics involved, air will gurgle back into the tank through the open valve at the bottom? The tank will eventually empty out, even if the top valve is closed. It simply will go through a constant cycle of spitting water out, drawing vacuum until flow essentially stops, which will allow air back into the vessel, repeating the cycle.

Obviously there are physical constraints - decrease the outlet hole size enough, the surface tension of water will eventually prevail and prevent draining. Similarly, configure the outlet piping in a tortuous path(ala a loop seal) such that air cannot simply flow into the tank, and you will also never drain the tank.

But simply opening a valve (1", 2", 3", etc) on the side of a tank? It'll drain out, and pull a good vacuum on the tank while it does so.

 

[1503-44]

Atmospheric pressure changes and thermal expansion/contraction will too.
No torturous paths needed. We closed the outlet valve.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Peter C]

So just to confirm: if I put a manometer on the topside (Mt) and at the bottom (Mb).

1/ Once all the valves are closed:
Mt = 1bara
Mb = 1 + p*g*h bara

2/ Once the bottom valve is opened
Mt = 1 - p*g*h' (h' measured from the bottom to the top)
Mb = 1bara

where bara is the absolute pressure

DO WE ALL AGREE?

 

[LittleInch]

I think that's good.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

1) Maybe and
2) No

You should know by now the information that is needed to answer your latest question.
What conditions are you talking about now?

Nobody knows the pressure at Mt right now.
We cannot say what the pressure will be until you install the manometer, or provide the appropriate info. All you said is the valve is closed for case1 and open for case 2.

Is there any water in the tank?
What is the water level?
How tall is the tank?
How was it filled?
Is there air inside?
Where exactly will you install Mt?
Is the "top" on the roof?
Is the top somewhere high up on the wall?
Is Mb installed at the outlet elevation?

All we can say for certain is that Mb will be rho g h more than Mt.
And the pressure outside the valve (valve's outlet) is 1 atm.

If somehow you know Mt will read 1 bara, then Mb will read rho g h more than Mt.
Mb need not be 1 bar even when the valve is open. It reads internal tank pressure, as it's installed on the tank wall. That need not the same as the pressure at the valve's outlet, which is 1 atm.

If the tank is full to 10m height and Mt is indeed 1 Bara at the water's surface, when you open the valve, the internal pressure at the valve's inlet will be 1 bara + rho g 10m = 2 bara. The Pressure outside will be 1 atm, say 1 bar. The Pressure difference across the valve will initially be 1 bar. It will decrease as water is expelled. Pressure at the valve's inlet will be 1 bara when the water level finally reaches outlet elevation.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Snickster]

So just to confirm: if I put a manometer on the topside (Mt) and at the bottom (Mb).

1/ Once all the valves are closed:
Mt = 1bara
Mb = 1 + p*g*h bara

Yes agree

2/ Once the bottom valve is opened
Mt = 1 - p*g*h' (h' measured from the bottom to the top)
Mb = 1bara

Yes agree. Case 1 above the pressure at the top plus static pressure equals pressure at bottom which is resisted by the closed valve. For case 2 the pressure at the bottom is greater than what is pushing down from the liquid height so the force of the atmosphere will not allow any liquid to flow out but will compress the liquid until the forces are balanced at the outlet opening of 1 atm. The force at the outlet opening of 1 atm will be balanced by p*g*h' plus the pressure at the top of the tank Pt = Mt = 1 - p*g*h'. Since at the top of tank will be above vapor pressure of the water no vapor will form and liquid will remain in contact with roof and no vapor space will exist.

where bara is the absolute pressure

DO WE ALL AGREE?

 

[Snickster]

Although I do agree with Ticl4 that it seems there would be a tendency for air to bubble up through the inlet like he described above and have intermittent gurgling until tank empties.

 

[1503-44]

No agreement here.

Case 1) Valve closed
If h = 10m (that is not specified, but I need a number to do the math) and Mt = 1 Bara, as stated, then
Mb = the 1 bara at Mt + [ρ] * g * 10m * 1 Bar/10m = 2 Bara

Case2) Valve opened. Mb is initially 2 bara.
Pressure differential [Δ]P across the valve is initially =2-1 = 1 bar
Exterior pressure is 1 atm.
As the tank empties, [Δ]P decreases.
After the tank empties, only then is Mb finally = 1 atm

Air will only have the possibility of entering through the valve when the valve does not have a full cross sectional flow. That's going to take some time to get to that point.

If the pressure at the Mt happens to be near 0 and the valve is opened, gravity draws the water down as space above is filled by water vapor, say at Pv = 0.5 psi. Again, if the tank is 10m high, the outlet valve inlet pressure is 0.5 psia + 14.67 psi. 15.17 psia. Close to atm pressure but enough to start outflow at full cross sectional flow. No way for air to get in yet. There is a wall of water in the valve starting outward movement. I'll give it one or two bubbles at most. Or maybe not. Then full outflow at V soon to reach sqrt(2gh). How does air get through the valve into the tank?

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Snickster]

Case 1) Valve closed
If h = 10m (that is not specified, but I need a number to do the math) and Mt = 1 Bara, as stated, then
Mb = the 1 bara at Mt + ρ * g * 10m * 1 Bar/10m = 2 Bara

I work in psi units so ok if the tank is exactly 10 meters high and water is completely to the top with no air space and the vent is then closed and the valve at bottom is then closed the pressure at the bottom at the closed valve is atm plus ρ * g * h = 2 atm = 29.4 psia.

Case2) Valve opened. Mb is initially 2 bara.
Pressure differential ΔP across the valve is initially =2-1 = 1 bar
Exterior pressure is 1 atm.
As the tank empties, ΔP decreases.
After the tank empties, only then is Mb finally = 1 atm

Valve is opened, at interface of the liquid to atmosphere crossection there is an instantaneous imbalance of forces. This causes liquid to flow out of the tank. As the liquid flows a complete vaccuum attemps to form at the very top of the tank. Since a complete vacuum would be below the vapor pressure of liquid then a water vapor space immediately forms above the liquid at the liquid vapor pressure of about 0.5 psia. This causes the forces to balance again at 0.5 (due to vapor pressure) + 14.2 (due to liquid height of 10 meters) to equal atm again at about 14.7, so if everything proceeds quasi statically very little liquid exits the tank and the height is still 10 meters a with very thin vapor space at top.

Air will only have the possibility of entering through the valve when the valve does not have a full cross sectional flow. That's going to take some time to get to that point.

I am not sure about that. If you turn a jug over won't it gurgle out as TiCl4 indicated. If you have a pipe exiting the tank such as in this case even with equillibrim of forces with 0.5 psia vapor pressure plus 14.2 psia of static height to balance with the atm pressure. Within the pipe there is a differential pressure across the crossection due to the height of liquid in the pipe. Wouldn't there be higher pressure at the bottom that would make it imbalanced with atm pressure at the bottom that would want to make the liquid at the bottom of the pipe flow out with air flowing in at the top of the crossection causing intermittent gurgling of air into the vessel. I know if you dip a straw into water and hold a finger at the top and then remove from water the water will not drop out of the straw but there is a lot of surface tension holding the water in preventing bubbling up but I don't know if this will be the case with a large tank and an outlet pipe that is opened to the atmosphere

If the pressure at the Mt happens to be near 0 and the valve is opened, gravity draws the water down as space above is filled by water vapor, say at Pv = 0.5 psi. Again, if the tank is 10m high, the outlet valve inlet pressure is 0.5 psia + 14.67 psi. 15.17 psia. Close to atm pressure but enough to start outflow at full cross sectional flow. No way for air to get in yet. There is a wall of water in the valve starting outward movement. I'll give it one or two bubbles at most. Or maybe not. Then full outflow at V soon to reach sqrt(2gh). How does air get through the valve into the tank?

This is the case described above when the intial pressure was 1 atm and after just a little flow out the pressure at top will immediately drop to vapor pressure of the liquid. Then flow will stop if height of liquid was 10 meters exactly. If initial height was above 10 meters then flow will commence until resuting height is 10 meters. If the static liquid height is below 10 meters to start then no flow will exit the tank at all except for perhaps the gurgling effect, and the resulting pressure at the top of the tank will be Mt = 14.7 - H*(62.4/144)

 

[1503-44]

I see. Right, If the tank is less than 10m of water, maybe I should have picked 5m height, the pressure at the top may not be near vapor pressure, so the tank collapses, or air must get in so more water can come out. I already said that way up above, the 11m tall tank will drain to 10m. I guess I forgot that. So if h < 10m, air can get in when the flow is stopping and restarting, as long as water does not pool around the tank bottom and seal the exit, like the barometer.

However, there is still a mystery here. In this latest problem we don't know how it happens to be, as is stated, that there is initially 1 Bara at the top of the tank. That's why I asked those questions. The 1 bara at Mt. If there is any water in the tank at all and no pressurised air, Mt would be atm - [&rho;] gh, and that is less than 1 bar. So, it must be from from trapped pressurized air during filling, OR h=0, i.e. NO WATER in the tank. And if there is air in the tank, a lot of water, maybe all of it, exits quickly, no matter how tall it is.

Better to get all those questions answered.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[pierreick]

Hi,
How can you get one bar absolute at the top of the tank when the valve is closed?
Are you compressing air trapped in your system when you fill your tank from the bottom?

Pierre

 

[Snickster]

The OP original post shows there is a bleed valve at top of tank that is open during filling. I assumme a pump fills the tank up which is capable of providing the required head due to liquid level with atm acting at top, and vent valve is closed after all air is out of tank, then the lower block valve is closed as OP indicated. In this case the pressure at the bottom of the tank will be atmospheric plus static head. What I stated in my previous posts is still the way I see it.

 

[Peter C]

@snickster, when you say that in scenario 2/ one would have a pressure difference generating flow (1atm + p.g.h - 1atm = pressure difference generating flow). In that case, if we just see the equation, ALL liquid would come out, not mattering the height of the tank...
that's something I can't understand. Could you please clarify?

Thank you

 

[Snickster]

1atm + p.g.h - 1atm = pressure difference generating flow

Yes when the lower valve is first opened but pressure at top of tank gradually but quickly reduces to vapor pressure of water at 0.5 psia and then forces are balanced and flow stops and then:

0.5 psia + p.g.h - 1atm = 0

 

[Peter C]

@snickster, thank you. But solve your second equation... you have a p.g.h = 1atm - 0.5 psi

this can only be true if h is big enough to almost reach atm pressure (about 10m). If my tank is a 5m tank, this can never be true... but in no moment we have add values so far, so how can the equations be true for a high tank but not for a tank with 5m?

 

[Snickster]

@snickster, thank you. But solve your second equation... you have a p.g.h = 1atm - 0.5 psi

this can only be true if h is big enough to almost reach atm pressure (about 10m).

Correct, the assumption was that the tank was equal to or greater than 10 m. If greater than 10 m flow will exit the tank until p.g.h = 1atm - 0.5 psi

If my tank is a 5m tank, this can never be true... but in no moment we have add values so far, so how can the equations be true for a high tank but not for a tank with 5m?

So for a tank 5 m high with full liquid level then vent closed then lower valve closed the pressure at bottom of tank will be:

14.7 psia (atm) + 7.1 psi (p.g.h) = 21.8 psia

Now when the lower valve is opened the liqiud will want to flow out of tank due to the unbalaced force. The only way this can happen is the liquid level at the top to drop. If the liquid level at top drops whatsoever a vacuum will need to form since there will then be a seperation of the liqiud surface from the underside of the tank roof with no air in it and therefore zero pressure absolute. However since this absolute vacuum will then be below the vapor pressure of the water of about 0.5 psia then any void developed at top will be filled with the vapor pressure of the water. However if the space did fill with the vapor pressure of the water the the net pressure at the bottom of the tank will be 0.5 + 7.1 = 7.6 psia. But the pressure of the atmosphere at the open end of the pipe is 14.7 psia and since this is greater than the pressure pushing down in the tank the extenal outside pressure wins and compresses the liquid so that the static liquid pressure is 7.1 psi and the resulting pressure at top of liquid is 7.6 psia - the force of the liquid pushing on the underside of the roof from the external atm pressure minus liquid static pressure. Since this is above vapor pressure then no water vapor space will exist. This must be the resulting equillibrium condtion as I see it. In fact it will never reach vacuum at top of tank and water will stop flowing as soon as pressure above liquid dropped to 7.6 psia which will basically be instantly and no water will really flow per se.

 

[1503-44]

I'm good with that. Nicely done. Thanks.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[TiCl4]

If you turn a jug over won't it gurgle out as TiCl4 indicated.

Except it does. Take any milk jug or water bottle with a spout, fill it to the top, and turn it over, spout pointing downwards. Water will come out, the jug will flex inwards as pressure drops, and flow will temporarily stop as air enters the jug, temporarily reinflating it slightly. The cycle then repeats. This produces the classic "glug glug glug" effect. Flow may not stop completely, depending on size of the spout and bottle. Sometimes it reduces flow and you have simultaneous water flow down and air flow upwards.

Edit: Your example with the straw is indeed what you indicated - surface tensions. I acknowledged this in my original post.

Either way, water empties out and air flows it. This is simple stuff.

 

[Compositepro]

The air gurgling into the tank is a trivial issue that simply distracts us from the main topic of discussion. It can be prevented with a U-bend (or P-trap) in the drain line. The way to develop understanding of fundamentals is to simplify things to the very basics. Then add complexity as needed. In this case even the effect of the vapor pressure of water is an unnecessary complication, although It may be important in real life. Too many engineers think their job is to consider every possible situation or complication. Ultimately it is, but that is not how you start to develop an understanding of basic principles, which I think this thread is about. This thread should have ended with the barometer analogy.

 

[1503-44]

So right.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."