Fluid dynamics problem: Equalize the volume of water in tanks at differing heights

[Hankelec]

TL;DR Summary

Adjustments needed to equalize volume of water in tanks at differing heights.

I am working on a reverse osmosis water system. There are 2 bladder type pressure tanks for storage. One tank is at floor level, the other tank is supported directly above the lower tank 5 feet above the floor. It would be advantageous if these tanks would fill at an equal rate, or approximately equal rate, from a single supply line coming from the RO system. And perhaps they will if I have both tanks pressurized equally and accurately at the same pressure.

The supply is 1/2" plastic. It enters the lower tank at floor level. Just prior to entering the lower tank there is a T junction which travels upward to the upper tank. Let's say the initial pressure of the tanks is set 2 psi below pressure pump kick-in pressure. Kick-in pressure equals 30 psi so empty tank will be pressurized to 28 psi. Kick-out pressure is 50 psi. So I will have a 5' column of water adding a psi or two that will need to be overcome to allow water to enter top tank.

As the top tank begins to accept water, there will be the 5' differential affecting the resistive pressure at its filler inlet. As the top tank's volume increases and its water height increases, will there be developed enough pressure difference relative to the bottom tank that will prevent water from entering top tank.

If supply water tends to oscillate from entering bottom tank then top tank, that is acceptable. What would not be favorable is if bottom tank fills to 30 gallon capacity and top tank is holding only 5 or 10 gallons. So what I am searching for is an offset pressure to best fill each tank to capacity. A 10% error factor is OK.

 

[Baluncore]

A float in each tank, with a connecting rod from the top tank float to the bottom tank, where the inlet to the bottom tank can be restricted if the float levels in their respective tanks differ.

A differential pressure gauge, connected between the tanks, with the control offset by the required pressure difference. The inlet to the lower tank would be restricted if the top tank is too low. There is no restriction into the top tank, apart from the hydrostatic pressure.

The pump, that supplies both tank inlets, turns off when the top tank reaches the high water level.

 

[Lnewqban]

TL;DR Summary: Adjustments needed to equalize volume of water in tanks at differing heights.

I am working on a reverse osmosis water system. There are 2 bladder type pressure tanks for storage. One tank is at floor level, the other tank is supported directly above the lower tank 5 feet above the floor.

I may be wrong, but the difference of heights should not be very critical if both thanks are bladder type.
If both have similar volumes or model number, they should behave just like two springs in a parallel arrangement.

Please, see:
https://inspectapedia.com/water/Water-Tank-Bladders.php

 

[hutchphd]

TL;DR Summary: Adjustments needed to equalize volume of water in tanks at differing heights.

As the top tank begins to accept water, there will be the 5' differential affecting the resistive pressure at its filler inlet. As the top tank's volume increases and its water height increases, will there be developed enough pressure difference relative to the bottom tank that will prevent water from entering top tank.

What??? This makes no sense at all to me (either I don't understand the system or you are imagining nonexistant difficulties) You understand that 5 feet of water gives you 2.5 psi additional pressure head. So the pressure in the upper tank simply be 2.5 psi less. Water will flow in such a way as to maintain that difference. What is the issue ?

 

[erobz]

You understand that 5 feet of water gives you 2.5 psi additional pressure head. So the pressure in the upper tank simply be 2.5 psi less. Water will flow in such a way as to maintain that difference. What is the issue ?

I believe the issue is the flowrate, i.e. they want them to fill at approximately the same rate.

 

[hutchphd]

That makes no sense to me.

1. The system is at equilibrium
2. The pump pressure is increased by $\Delta P$
3. At the mouth of each pressure tank the differential pressure increases by $\Delta P$
4. water flows in according to differential pressure
5. Where is the problem?

Somebody is certainly confused here. (it could be me!)

 

[erobz]

That makes no sense to me.

1. The system is at equilibrium
2. The pump pressure is increased by $\Delta P$
3. At the mouth of each pressure tank the differential pressure increases by $\Delta P$
4. water flows in according to differential pressure
5. Where is the problem?

Somebody is certainly confused here. (it could be me!)

These tanks are effectively parallel branches of a circuit with different resistances. Min. 5 ft of head may or may not be a game changer, but in general think about taking the upper tank to the to top of a mountain.

 

[hutchphd]

Min. 5 ft of head may or may not be a game changer,

He says the system is at 40 psi. Clearly the tank on the mountain top will not be useful to his system.

 

[Hankelec]

What??? This makes no sense at all to me (either I don't understand the system or you are imagining nonexistant difficulties)You understand that 5 feet of water gives you 2.5 psi additional pressure head. So the pressure in the upper tank simply be 2.5 psi less. Water will flow in such a way as to maintain that difference. What is the issue ?

The reason the tanks are pressurized is to maintain water flow to the water outlet, of course, rather than rely on "gravity flow." For example purposes let's say both tanks are above the water outlet. The initial state of the tanks is that one tank is five feet higher than the other tank. Both tanks are pressurized to 28 psi. Now say we open the air filler valve on both tanks until air stop coming out. Both tanks will merely have atmospheric pressure pushing downward upon the bladders. The difference in atmospheric pressure at a 5' difference in height would be neglegable I am assuming. The water inlet to the tank bladders are configured in a "T" configuration. Water start to flow in the supply line, reaches the T intersection, and I have to guess will begin to fill the bottom tank. NOTE: After we lowered the internal pressure in each tank by 28 psi, we closed the air pressure filler valve. As the lower tank begins to fill and compressing the trapped air, you are saying that once the air pressure rises to 2.5 psi, some of the water will be diverted to the upper tank due to a pressure equalibrium. And as long as a 2.5 psi differential is maintained, both tanks will be accepting equal volumes of water from the supply line. I don't see that being the case but that is why I am making this inquiry. How did this scenario sound to you?

 

[erobz]

The reason the tanks are pressurized is to maintain water flow to the water outlet, of course, rather than rely on "gravity flow." For example purposes let's say both tanks are above the water outlet. The initial state of the tanks is that one tank is five feet higher than the other tank. Both tanks are pressurized to 28 psi. Now say we open the air filler valve on both tanks until air stop coming out. Both tanks will merely have atmospheric pressure pushing downward upon the bladders. The difference in atmospheric pressure at a 5' difference in height would be neglegable I am assuming. The water inlet to the tank bladders are configured in a "T" configuration. Water start to flow in the supply line, reaches the T intersection, and I have to guess will begin to fill the bottom tank. NOTE: After we lowered the internal pressure in each tank by 28 psi, we closed the air pressure filler valve. As the lower tank begins to fill and compressing the trapped air, you are saying that once the air pressure rises to 2.5 psi, some of the water will be diverted to the upper tank due to a pressure equalibrium.

There is also dependencies on viscous dissipation in the supply lines to each tank, but that is the general idea.

And as long as a 2.5 psi differential is maintained, both tanks will be accepting equal volumes of water from the supply line. I don't see that being the case but that is why I am making this inquiry. How did this scenario sound to you?

At 2.5 psi differential ( or whatever is required to overcome the height difference ) the flow begins to trickle into the upper tank, then everything becomes complicated to describe analytically. That is my point. as you keep building pressure in the lower, sure...flow is diverting to the upper, but also the upper is pushing back, developing its own pressure at a different rate. Maybe I'm making I'm seeing a mountain in a mole hill, but it doesn't seem immediately reducible at this point (at least to me) without trying some computation.

 

[erobz]

Lets approximate that the flow is steady in each branch over some period of time i.e. in each branch the rate of change in flow in that branch is not changing strongly in time. Also ignore viscous effects in the transmission lines. Furthermore, the top tank is open to atmosphere as a simplified problem, and the total flowrate $Q_T$ is to remain constant in time.

Bernoulli's Implies:

$$ \frac{P(z_1)}{\rho g } + z_1 + \frac{\dot z_1^2}{2g} = \frac{P_{atm}}{\rho g} + z_2 + H + \frac{\dot z_2^2}{2g}$$

Continuity Implies:

$$ Q_T = A \left( \dot z_1 + \dot z_2\right) \implies \dot z_2 = \frac{Q_T}{A} -\dot z_1$$

From integrating continuity we find that:

$$ z_2 = z_{2_o} + \int \left( \frac{Q_T}{A} -\dot z_1 \right) dt $$

Lets eliminate $z_2, \dot z_2$:

$$ \frac{P(z_1)}{\rho g } + z_1 + \frac{\dot z_1^2}{2g} = \frac{P_{atm}}{\rho g}+ z_{2_o} + \int \left( \frac{Q_T}{A} -\dot z_1 \right) dt + H + \frac{\left( \frac{Q_T}{A} -\dot z_1 \right)^2}{2g}$$

Now, take the derivative w.r.t time $t$, to eliminate that integral from the equation:

$$ \frac{1}{\rho g}\frac{dP(z_1)}{dz_1 } \dot z_1 + \dot z_1 + \frac{2\dot z_1 \ddot z_1}{2g} = \cancel{\frac{P_{atm}}{\rho g}}^0+ \cancel{z_{2_o} }^0+ \frac{Q_T}{A} -\dot z_1 + \cancel{H}^0 - \frac{2 \left( \frac{Q_T}{A} -\dot z_1 \right) \ddot z_1}{2g}$$

Where $P(z_1)$ is found from the ideal gas law ( isothermal ):

$$ P(z_1) = \frac{P_o V_o}{V_o - A z_1} $$

$$ \frac{1}{\rho g}\frac{dP(z_1)}{dz_1 } \dot z_1 + 2\dot z_1- \frac{Q_T}{A} + \frac{Q_T}{gA} \ddot z_1 = 0 $$

Given the form of this simplified equation, and the many assumptions that don't strictly hold for all time. I think active control ( like @Baluncore proposes ), or scaled experimentation are going to be desirable approach (maybe it looks worse than it is in actuality). If you try to account for the fact that the flowrates are actually not steady, the upper tank builds pressure, and transmission losses it only gets mathematically worse as far as I can tell.

 

[hutchphd]

I am totally mystified by the analysis. Although (perhaps) true, most of it seems totally irrelevant to me.
Perhaps it would be good to state the engineering requirements here. Why do you care about pressure and flow and what are the requirement for each? Why is there more than one tank?? Why are they at different altitudes?
The pressure and flow in my upstairs bathroom differs slightly from the downstairs one .......I don't really fret about this.

 

[jrmichler]

TL;DR Summary: Adjustments needed to equalize volume of water in tanks at differing heights.

What would not be favorable is if bottom tank fills to 30 gallon capacity and top tank is holding only 5 or 10 gallons. So what I am searching for is an offset pressure to best fill each tank to capacity. A 10% error factor is OK.

Kick-in pressure equals 30 psi so empty tank will be pressurized to 28 psi. Kick-out pressure is 50 psi.

When the tanks are empty, the pressure is 28 PSI.
At kick-in, the pressure in a tank is 30 PSI, and the tank is almost empty.
At kick-out, the pressure in a tank is 50 PSI, and the tank is as full as it gets.
The pressure in a tank is proportional to the amount of water in that tank as shown in the graph below:

Five feet of head pressure difference between the two tanks is 2.2 PSI. The pressure difference between tank empty and tank full is 20 PSI, so the 2.2 PSI of pressure difference is 2.2/20 = 11% of the tank capacity. The bottom tank will be 11% full before the top tank starts to fill, then there will be equal flow into both tanks until the bottom tank is full at 50 PSI. The top tank will be 89% full at that point because the pressure in it will be 47.8 PSI.

In the real world, these differences are so small as to be not noticeable.

 

[Hankelec]

At 2.5 psi differential ( or whatever is required to overcome the height difference ) the flow begins to trickle into the upper tank, then everything becomes complicated to describe analytically. That is my point. as you keep building pressure in the lower, sure...flow is diverting to the upper, but also the upper is pushing back, developing its own pressure at a different rate.

This was also my perspective on the situation. When I came upon this project, the bottom tank was at least 90% full and the top tank virtually empty. Due to slow leak? Maybe. Due to what you describe above? Maybe. IDK. Experimentation is possible but quite problematic since it is desired to not shut system down for any length of time.

 

[erobz]

IDK. Experimentation is possible but quite problematic since it is desired to not shut system down for any length of time.

I meant experiment at small scale. You can buy some relatively cheap bladder tanks used in residential potable water systems at the local hardware store, or online.

Possibly it's just malfunction that is causing the issue, but the math surrounding the problem is complicated enough IMO not to be absolutely sure of the scale of effects.

If the others have firsthand experience with the exact scenario and they say it's always small, then the math reduces, and the problem goes away.

Are there check valves in each branch? If not, is the top tank to slowly draining into the bottom tank after the valve on the main fill line is closed? Lets say the effects during filling are small as suggested. Then at the end of filling you have a head imbalance of 5 ft which will initiate flow from the top to the bottom tank. Calculating equilibrium of the system should be far more accessible than the dynamics of filling the system.

 

[jrmichler]

When I came upon this project, the bottom tank was at least 90% full and the top tank virtually empty.

This does not agree with the tank theory shown in Post #13. If the air bladder of a tank was properly precharged to 28 PSI, then a full tank must be at about 50 PSI, and and empty tank about 30 PSI. But both tanks are connected with a pipe, so they must be at the same pressure. Something is not right, which raises some questions:

1) How do you know the water level in each tank?
2) Did you check the precharge pressure in each tank with the system empty?
3) What was the pressure in each tank at the time mentioned in Post #14? Measured at the air filler.
4) Have you confirmed that each tank has a properly functioning bladder? Bladders can fail by leaking or sticking.

General observation: Whenever two observations disagree, or theory disagrees with observations, it is necessary to learn more about what is really happening.

 

[erobz]

I used work-energy to find the equilibrium state of two identical tanks with identical initial conditions:

$P_o$ is the initial pressure inside each tank
$V_o$ is the initial volume of ideal gas at $P_o$ in each tank
$z$ is the initial height of water in each tank relative to the bottom of each tank.
$A$ is the cross-sectional area
$H$ is the elevation difference of the tanks
$\rho$ the density of water

At time $t=0$ the hypothetical valve isolating them is opened and the system is allowed to establish equilibrium. For the arbitrary parameters I chose this resulted in the lower tank ( beginning at $z = 0.5 \rm{m}$ ) having a final height of nearly $0.7 \rm{m}$, and the upper ( also beginning at $z = 0.5 \rm{m}$ ) having a final height of $0.3 \rm{m}$.

The intersection of $F(z_1)$ with the abscissa is the final height for $z_1$.

If you give the parameters for your actual system of tanks, I can check what the theoretical final state is?

 

[erobz]

Just to be more descriptive about the method, the system is taken to be the mass of fluid in each tank. The pressure in each tank is responsible for the external work on the system:

Referencing the diagram in #11, defining $\uparrow^+$, apply Work -Energy:

$$ -A \left( \int P(z_1)dz_1 + \int P(z_2)dz_2 \right) = \rho A g z_2 ( H+\frac{z_2}{2} ) - \rho A g z ( H+\frac{z}{2} ) + \frac{1}{2} \rho A g z_1^2 - \frac{1}{2} \rho A g z^2 $$

Expanding the RHS:

$$ -A \left( \int P(z_1)dz_1 + \int P(z_2)dz_2 \right) = \rho A g \left( H z_2 + \frac{z_2^2}{2} + \frac{z_1^2}{2} - \overbrace{ z \left( H+ \frac{z}{2} \right) + \frac{z^2}{2} }^{\text{const.}} \right) $$

From Continuity:

$$A( z- z_2) = A( z_1-z )$$

$$ \implies z_2 = 2 z - z_1 $$

$$ \implies dz_2 = -dz_1 $$

Substitute to get everything in terms of $z_1$

$$ -A \left( \int P(z_1) dz_1 - \int P( 2 z - z_1 ) dz_1 \right) = \rho A g \left( H (2 z - z_1)+ \frac{ (2 z - z_1)^2}{2} + \frac{z_1^2}{2} - \overbrace{ z \left( H+ \frac{z}{2} \right) + \frac{z^2}{2} }^{ \text{const.}} \right) $$

Take the derivative of both sides w.r.t. $z_1$:

$$ \frac{d}{dz_1}\left[-A \left( \int P(z_1)dz_1 - \int P( 2 z - z_1 )dz_1 \right) \right] = \frac{d}{dz_1}\left[\rho A g \left( H (2 z - z_1)+ \frac{(2 z - z_1)^2}{2} + \frac{z_1^2}{2} - \overbrace{ z \left( H+ \frac{z}{2} \right) + \frac{z^2}{2} }^{\text{const.}} \right) \right] $$

$$ \implies -A ( P(z_1) - P(2z-z_1) ) = \rho g ( -H -( 2z -z_1) + z_1) = \rho g ( 2( z_1-z) - H )$$

Dividing by the area of the tank $A$ and taking the negative of both sides:

$$ P(z_1) - P(2z -z_1 ) = \rho g (2(z-z_1)+ H ) $$

Assuming isothermal process to equilibrium (after a long time the temperature returns to ambient in each tank) we have that generally:

$$ P(x) = \frac{P_o V_o}{V_o - Ax}$$

Subbing in that result:

$$ \frac{P_o V_o}{V_o - Az_1} - \frac{P_o V_o}{V_o - 2Az + Az_1} = \rho g ( 2( z-z_1) + H ) $$

Then just pull everything to the LHS to graphically find $z_1$ which satisfies:

$$ \frac{P_o V_o}{V_o - Az_1} - \frac{P_o V_o}{V_o - 2Az + Az_1} -\rho g ( 2( z-z_1) + H ) = 0 $$

Given the unsymmetric state of equilibrium, what is the reasoning behind proposing that the tanks will receive equal flowrates in the dynamic case? It seems clear to me that there would be significant bias toward the lower tank filling at a greater rate, because that is where they are going to end up regardless. The friction present in the piping would also drive the bias further in the dynamic case.

 

[Juanda]

Given the unsymmetric state of equilibrium, what is the reasoning behind proposing that the tanks will receive equal flowrates in the dynamic case? It seems clear to me that there would be significant bias toward the lower tank filling at a greater rate, because that is where they are going to end up regardless. The friction present in the piping would also drive the bias further in the dynamic case.

I didn't follow the math you posted step by step so I can't argue whether it's correct or not. I'll try to revisit the thread when I get the chance to check the math. Still, I believe I can answer your question (marked in bold) at least qualitatively. I think I understood the situation when I read post #13 by @jrmichler.

I see it like this:

1. The flow will be towards the tank with the lowest pressure at its intake because the mass flow will follow the path of least resistance.
   NOTE: I consider the intake to be the T fitting where the flow separates.
2. The mass flow towards that tank will increase its pressure.
3. The situation will continue until the pressure at the intake in the lowest pressure tank equalizes with the intake pressure at the other tank.
4. At that point, the mass flow will divide equally between both tanks.

By the way, I think that implies the pressure at the tanks changes linearly with the mass flow. If that is not the case, then the mass flows will keep oscillating around the equal mass flow point repeating the described cycle (1→4). Again, I'm only tackling the qualitative aspect of this situation so I don't have the math to back that claim at the moment but I'm fairly confident about this.

I guess a somewhat valid analogy is to imagine a simple DC circuit powered by a battery and two resistances and two condensers in parallel. Make the condensers equal but have different resistance values. In the dynamic analysis, we would see how initially most of the current flows through the smaller resistance but then the condenser charges and pushes back so more current will start flowing through the other one.
A possible error in the described cycle (1→4) is that I said ALL the fluid goes through the smallest resistance path. I'm pretty sure that doesn't happen in the electrical analogy. Only MOST of the current goes through the least resistance path so I feel like the same would happen in the fluid case although I can't really tell without doing the math.

Feel free to point out any errors in what I posted. It seems the OP had enough already and it's possible there are no truly practical applications (changes too small to be problematic or even noticeable) for what we're discussing. Still, the problem is interesting and poses a chance to learn more about this kind of system I have no experience in.

I'll try to post a diagram describing the tanks filling up and the DC circuit analogy so that it's easier to visualize what I'm seeing in my head but drawings take a time I don't have right now. Maybe I can do it before hitting the bed.

 

[erobz]

I didn't follow the math you posted step by step so I can't argue whether it's correct or not. I'll try to revisit the thread when I get the chance to check the math.

If I need to clarify more let me know.

I guess a somewhat valid analogy is to imagine a simple DC circuit powered by a battery and two resistances and two condensers in parallel. Make the condensers equal but have different resistance values. In the dynamic analysis, we would see how initially most of the current flows through the smaller resistance but then the condenser charges and pushes back so more current will start flowing through the other one.

I think the effective circuit is not so straightforward as that.

At a minimum we need resistances from the pipe proportional to the current in each branch with different proportionalities representing the vicious effects. The liquid head accumulating is not a resistor, it is a gravitational capacitor, and the gas is a capacitor of different characteristics. The liquid capacitor of the upper branch is initially charged to a different value than the liquid capacitor of the lower branch. This is due to the static liquid head differential of $H$. The gas capacitors in each branch are initially equivalent in charge. Maybe someone could figure out the necessary characteristics of each component and model it in SPICE or something. I don't have that skill.

I still don't think the branches get equal flow at any point in time, and when the feed voltage to the circuit is switched off the parallel branches which form a closed loop (having differing accumulated charge in each of the capacitors) will equilibrate.

 

[Juanda]

I'll try to post a diagram describing the tanks filling up and the DC circuit analogy so that it's easier to visualize what I'm seeing in my head but drawings take a time I don't have right now. Maybe I can do it before hitting the bed.

If I need to clarify more let me know.

Let's hope I can track the math by myself. I just need some free time first so I can focus on it. I'll let you know if I get lost at some point.

I think the effective circuit is not so straightforward as that.

At a minimum we need resistances from the pipe proportional to the current in each branch with different proportionalities representing the vicious effects. The liquid head accumulating is not a resistor, it is a gravitational capacitor, and the gas is a capacitor of different characteristics. The liquid capacitor of the upper branch is initially charged to a different value than the liquid capacitor of the lower branch. This is due to the static liquid head differential of $H$. The gas capacitors in each branch are initially equivalent in charge. Maybe someone could figure out the necessary characteristics of each component and model it in SPICE or something. I don't have that skill.

Maybe using "analogy" was taking it too far. I was just trying to show an example of something with similar behavior where previous knowledge could help to have an intuition. (I assume this kind of electrical circuit is something more common in universities than the filling tanks under study)

I still don't think the branches get equal flow at any point in time.

Well, they will tend to have the same flow. We know for sure at t→INF the flow will be the same in both branches because it stops in both when equilibrium is reached. It happens when the pump can't keep pushing fluid in.
The take-home message is that both branches approach that state and the tank with the lower intake pressure will approach it faster also increasing its pressure faster up to the point where they're both filling up at about the same rate because they have about the same pressure at the intake.

and when the feed voltage to the circuit is switched off the parallel branches which form a closed loop (having differing accumulated charge in each of the capacitors) will equilibrate.

If you consider the DC circuit is not useful to transmit the concept I feel we should drop it altogether and focus back on the tanks. I didn't intend it to be a 1:1 translation where mathematical tools can be easily reused but a mere example of a similar case for a conceptual understanding. I think the confusion comes from my use of "analogy" too lightly.
Regarding the "switched off" circuit you mentioned, the similarities with the hydraulic case break apart even further since the circuit is capable to consume the potential with the resistances but the tanks can't really get rid of the built-up pressure to the same degree.

To conclude, I believe both branches will approach the same rate but I'd need to first run the math myself and check the expressions and plots of the mass flow in each branch to judge how true is that claim.

 

[erobz]

Well, they will tend to have the same flow. We know for sure at t→INF the flow will be the same in both branches because it stops in both when equilibrium is reached. It happens when the pump can't keep pushing fluid in.
The take-home message is that both branches approach that state and the tank with the lower intake pressure will approach it faster also increasing its pressure faster up to the point where they're both filling up at about the same rate because they have about the same pressure at the intake.

With a real pump, the flowrates tend to zero in each branch as time increases, sure... However, that is not the same as saying:

$$ \int_0^{t} Q_{l} ~dt = \int_0^{t} Q_{u} ~dt $$

Anyhow, unless my physics in #18 is incorrect, if they have the same conditions the instant the pump is turned off, they will equalize to have differing volumes of water in each tank.

 

[Juanda]

$$ \int_0^{t} Q_{l} ~dt = \int_0^{t} Q_{u} ~dt $$

Just to be clear, that's not my claim. What I'm saying is that in the beginning the flow rates will differ significantly and they will tend to equalize as time goes on.
At the very limit (t→+INF), they will be the same (zero) but from the very beginning they will approach one another because a greater rate will flow into the tank with smaller pressure which will increase its pressure. It's a kind of feedback cycle that tends to equalize the flow rates.
Only after solving the math I will be able to assess at which point in time the flow rates will be almost equal. Right now this is just a hypothesis.
Actually, even after solving the mass I guess we could still call it hypothesis since we didn't perform any test to verify it but that's out of the scope.
Anyways, I'll try to solve it with some code in the following days and maybe we can arrive at some conclusion after discussing the results we get.

 

[Joe591]

Maybe look up municipal water flow dividers. I think they sound like what you are looking for.

 

[erobz]

@Juanda Putting the exact math in #18 aside for a moment, I'm trying to figure out whether or not you agree with me that the following system cannot possibly be in a state of equilibrium with the valve shut?

 

[Juanda]

@Juanda Putting the exact math in #18 aside for a moment, I'm trying to figure out whether or not you agree with me that the following system cannot possibly be in a state of equilibrium with the valve shut?

View attachment 330318

Yep. It can't be in the way you show it but it can be in equilibrium.
The water level in the upper tank will be lower and the pressurized air in it will have a lower pressure too compared with the tank below.
I couldn't yet get my hands on this. Time keeps rolling and I'm struggling to keep up with it. There are actually a few posts on my waitlist in the same situation. I should have some free time next week so I'll try to invest some time in this.
I plan to start this by defining an equilibrium point by doing Bernoulli on the different paths (A→j and B→j) so that I know the water levels ($z_{1i} \ z_{2i}$) and the air pressures ($P_{1i} \ P_{2i}$) at the beginning. Then, open the valve with an arbitrary pump and see how the system evolves with time.

I'll probably need to check some exercises related to pumps to remind myself how to do this kind of thing but I believe I should be able to handle it.
By the way, I plan to ignore losses in pipes due to friction. I feel it's already hard the way it is and having to deal with Reynolds in a time-varying problem is something I'd rather avoid. We can consider the pipes to be pretty short and wide to consider the losses to be negligible anyways and still get some knowledge out of the situation.
NOTE: I just realized that if I consider the pipes to be very wide I would need to consider the volume they can store so let's just say I ignore Reynolds because of magic instead of the reasoning previously explained. I don't think it's relevant for the thing we are interested in anyways.

 

[Juanda]

I plan to start this by defining an equilibrium point by doing Bernoulli on the different paths (A→j and B→j) so that I know the water levels ($z_{1i} \ z_{2i}$) and the air pressures ($P_{1i} \ P_{2i}$) at the beginning. Then, open the valve with an arbitrary pump and see how the system evolves with time.

So I could solve this part which is the part that requires no extra research from me. If you ( @erobz ) have doubts about equilibrium with the valve being closed being possible as you posted in #25 I believe this should clarify it.

First, let's define some constraints.

• The valve is shut.
• The mass of water is constant and known → the mass that's not in one tank must be in the other. (The pipes must be full at all times for that last statement to be true which depends on the geometry so I'm forcing it to be true with this statement. You can imagine pipes being very small in volume compared with the tanks.)
• Water is incompressible. (I'm fairly certain it's a necessary assumption but not 100% sure about it. I'll throw it in there just in case.)
• Both tanks are the same size. Lenght $L$ and cross-section $A$.
• Both tanks contain the same amount of air which is constant and known.
• Temperature is constant and known.
• Ideal gasses.

There are 4 things we wish to know about the equilibrium point. The pressure of air in both tanks and their water levels. Therefore, we need 4 equations. I'll be referencing this diagram:

View attachment 330401

Eq 1: Bernoulli from A→B
$$P_2 + \rho g(z_2+H)=P_1+\rho gz_1$$

Eq 2: Ideal gas law in Tank 1
$$P_1V_1=mRT \ \left \{ V_1=A(L-z_1) \right \}$$

Eq 3: Ideal gas law in Tank 2
$$P_2V_2=mRT \ \left \{ V_2=A(L-z_2) \right \}$$

Eq 4: Constant mass of water and pipes being full means the water levels are related since water is incompressible and the pipes are always full
$$z_1+z_2=K$$

With those 4 equations, the equilibrium point for $P_1 \ P_2 \ z_1 \ z_2$ can be found for the valve being shut.

Solving the equations shouldn't be too hard but I didn't do it yet.
From the physical perspective $K$ must be smaller than $L$. I'm curious about what will happen if I make $K$ greater than that limit but I guess some values will go negative/complex (either the heights or pressures). Since I'm planning to solve the system numerically I'm also wondering if there's only one solution. I'm pretty sure there's only ONE valid solution but having $P(A(L-z))$ makes me think that there'll be more mathematical solutions either with complex numbers or negative solutions without physical meaning.

Anyways, let me know if you don't agree with this equilibrium point. If we agree, we can move on to the "pumping water in" scenario.

 

[erobz]

So I could solve this part which is the part that requires no extra research from me. If you ( @erobz ) have doubts about equilibrium with the valve being closed being possible as you posted in #25 I believe this should clarify it.

I think you are confused about my doubts...

My doubt is not that an equilibrium exists ( I solved for it myself in #18 ), if you set $K = 2 z_o$, and use my numbers I'm fairly confident you are going to get the same result. That being said, yours is much more concise. My doubts were that the OP could "have their cake and eat it too". Meaning... depending on the exact meaning of "approximately" in the OP, they can't have both the same volume of water and the same pressure; that is not a valid point of equilibrium.

The question really is...can we manufacture a solution where both the pressure and volume vary less than (stated in the OP) 10% from each other, or some nominal desired value for each quantity. The next question is whether or not we can get to the required initial conditions dynamically.

 

[Juanda]

I think you are confused about my doubts...

Seems like that. From #25 I got a different idea.

My doubt is not that an equilibrium exists ( I solved for it myself in #18 ), if you set $K = 2 z_o$, and use my numbers I'm fairly confident you are going to get the same result. That being said, yours is much more concise.

Maybe. I'd need to go into the details of what you posted to be sure. Anyways, since you're OK with the approach I posed I'd try to keep working on top of that when analyzing the dynamic case if I manage to put my head on it.

My doubts were that the OP could "have their cake and eat it too". Meaning... depending on the exact meaning of "approximately" in the OP, they can't have both the same volume of water and the same pressure; that is not a valid point of equilibrium.

The question really is...can we manufacture a solution where both the pressure and volume vary less than (stated in the OP) 10% from each other, or some nominal desired value for each quantity. The next question is whether or not we can get to the required initial conditions dynamically.

Just with passive elements there always be a % difference between tanks. If that's within the target 10% difference you mentioned (and from post #13 by jrmichlel it could be a realistic scenario) then there's nothing extra to manufacture. Just put together the pipes and tanks and it'll be ready to go.
If the height difference between tanks were much larger and you are interested in them being the same at all times I believe you'd need a second pump to feed the higher tank. Maybe active valves with feedback from the tanks' state could do the same but I'm not sure.

 

[erobz]

Maybe. I'd need to go into the details of what you posted to be sure.

I just used work-energy; I effectively derived a reduced form of Bernoulli (hydrostatic) from scratch.

$$ \frac{P_1}{\rho g } + z_1 = \frac{P_2}{\rho g } + z_2 + H $$

If $z_1 + z_2 = 2z \implies z_2 = 2z - z_1$:

$$ \frac{P_1}{\rho g } + z_1 = \frac{P_2}{\rho g } + 2z - z_1 + H $$

$$ \frac{P_1}{\rho g } -\frac{P_2}{\rho g } = 2( z - z_1) + H $$

$$P_1 - P_2 =\rho g ( 2(z -z_1) + H)$$

Subbing in the relationship for the pressures in each tank:

$$P_1 = \frac{P_oV_o}{V_o - Az_1}$$

$$P_2 = \frac{P_oV_o}{V_o - Az_2} = \frac{P_oV_o}{V_o - A( 2z- z_1 )} =\frac{P_oV_o}{V_o - 2Az + Az_1 } $$

That is Identical to the result in #18

$$ \frac{P_o V_o}{V_o - Az_1} - \frac{P_o V_o}{V_o - 2Az + Az_1} = \rho g ( 2( z-z_1) + H ) $$

Anyways, since you're OK with the approach I posed I'd try to keep working on top of that when analyzing the dynamic case if I manage to put my head on it.

You shouldn't use Bernoulli's for what you want to do with a pump model. The flow will not be steady, Bernoulli's is valid for steady flow. What I was proposing in #11 was for constant total volumetric flowrate, where the transient effects were expected to be small because they were tied only to the change in flow to each tank. You are talking about transient total volumetric flowrate on top of that. I don't trust that for what you wish to solve.

If you were solving this for steady flow using Bernoulli's it results in a system of non-linear equations that is usually solved by linearization and iteration ( i.e. The Hardy-Cross Method ).

To be honest, I would nix the method I'm proposing in #11 in its entirety. The method I'm using in #18 does not have steady flow assumption built in. It's readily amenable to include both the work from the pump and the changing kinetic energies of the mass both contained inside the tanks and any piping as the flow slows down. It will amount to solving a single second order non-linear ODE(I think).

 

[erobz]

To be honest, I would nix the method I'm proposing in #11 in its entirety. The method I'm using in #18 does not have steady flow assumption built in. It's readily amenable to include both the work from the pump and the changing kinetic energies of the mass both contained inside the tanks and any piping as the flow slows down. It will amount to solving a single second order non-linear ODE(I think).

I was wrong about this, it gets messy.

We need to be using "Unsteady -Bernoulli"¹ which says in general:

$$ P_1 + \rho g z_1 + \frac{1}{2}\rho v_1^2 = \int_1^2 \rho \frac{\partial v_s}{\partial t} ds + P_2 + \rho g z_2 + \frac{1}{2}\rho v_2^2 \tag{1}$$

If we assume uniformly distributed properties, the streamlines in a particular section of the flow are straight. As a result, the partial derivative of the velocity of the flow in that section does not depend on its position, and can be taken outside the integral. Basically... this ignores the flow transition between tanks and pipes.

Apply $(1)$ between the junction and tank liquid surface in each tank. Splitting the integral at the transition we have that:

$$\begin{aligned} P_j + \frac{1}{2}\rho \left(\frac{A}{A_p}\right)^2 \dot z_1^2 &= \int_j^a \rho \frac{\partial v_s}{\partial t} ds + \int_a^1 \rho \frac{\partial v_s}{\partial t} ds + P_1(z_1) + \rho g z_1 + \frac{1}{2}\rho \dot z_1^2 \\ \quad \\ &= \rho \frac{A}{A_p} \ddot z_1 \int_j^a ds + \rho \ddot z_1 \int_a^1 ds + P_1(z_1) + \rho g z_1 + \frac{1}{2}\rho \dot z_1^2 \\ \quad \\ &= \rho \frac{A}{A_p} \ddot z_1 l + \rho \ddot z_1 z_1 + P_1(z_1) + \rho g z_1 + \frac{1}{2}\rho \dot z_1^2 \\ \quad \\ &= \rho \left( \frac{A}{A_p} l + z_1 \right)\ddot z_1 + P_1(z_1) + \rho g z_1 + \frac{1}{2}\rho \dot z_1^2 \end{aligned}$$

$$ \tag{2} $$

$l$ is just an arbitrary length of pipe joining the junction to tank 1
$A_p$ is the area of the branch.

Applying $(1)$ from the junction to tank 2 we get:

$$ P_j + \frac{1}{2}\rho \left(\frac{A}{A_p}\right)^2 \dot z_2^2 = \rho \left( \frac{A}{A_p} H + z_2 \right)\ddot z_2 + P_2(z_2) + \rho g \left(H+z_2\right) + \frac{1}{2}\rho \dot z_2^2 \tag{3}$$

We also have continuity at the junction:

$$ Q = A \dot z_1 + A \dot z_2 \tag{4}$$

Then assuming the pump is drawing from a very large reservoir at atmospheric pressure:

$$ P_{atm} + \rho g h_p(Q) = P_j + \frac{1}{2}\rho \frac{Q^2}{A_m^2} + \rho L \dot Q \tag{5} $$

Where $A_m$ is the area of the main feeding the system. We can use $(5)$ to eliminate $P_j$ in each of the equations.

For the pump curve we can say generally:

$$ h_p(Q) = C - \beta Q^2 $$Using what we figured out earlier, set the tanks in an initial state of equilibrium at time $t = 0$, pick the thermodynamics for the ideal gas compression (adiabatic?). Turn on the pump, simultaneously solve $(2)$ through $(5)$ under the initial conditions:

$z_1(0) = z_{1,o}$
$z_2(0) = z_{2,o}$
$\dot z_1(0) = 0$
$\dot z_2(0) = 0$

Reference:
[1] https://ocw.mit.edu/courses/2-25-ad...edabdfd4b95c1792_MIT2_25F13_Unstea_Bernou.pdf

 

[Hankelec]

O

I meant experiment at small scale. You can buy some relatively cheap bladder tanks used in residential potable water systems at the local hardware store, or online.

Possibly it's just malfunction that is causing the issue, but the math surrounding the problem is complicated enough IMO not to be absolutely sure of the scale of effects.

If the others have firsthand experience with the exact scenario and they say it's always small, then the math reduces, and the problem goes away.

Are there check valves in each branch? If not, is the top tank to slowly draining into the bottom tank after the valve on the main fill line is closed? Lets say the effects during filling are small as suggested. Then at the end of filling you have a head imbalance of 5 ft which will initiate flow from the top to the bottom tank. Calculating equilibrium of the system should be far more accessible than the dynamics of filling the system.

 

[Hankelec]

erobz. I appreciate the suggestions and interest generated on this question.I have gone ahead with the process of lowering the static start pressure in the upper tank by 3 psi. Fill rate seemed acceptable for both tanks. When initially seeing this system, the lower tank appeared 100% full and the upper tank sounded virtually empty. And like you mentioned, after pressure pump stops, it might be the case that a gravity flow is initiated. I will have an opportunity to check on these tanks and if that is a problem, I will create a thread. One tank is on a stand above the other due to space constraints. It was added after initial installation of system. So at present the system is running a producing a satisfactory amount of RO water. Thanks to all who replied. I am uncertain if this thread will be closed and archived but I will occasionally visit to see if it has or if other conclusions have been drawn. I am also uncertain if the electronics analogy is useful with this problem as I don't know what would represent gravity in the pump off state.

 

[erobz]

I am also uncertain if the electronics analogy is useful with this problem as I don't know what would represent gravity in the pump off state.

Gravitational potential is a capacitor of some variety. While the pump is on and there is flow in the system gravitational potential is accumulating. It is capable of being returned to the lower tank at later time. This is indeed what happens when the pump is switched off, unless the tanks end the cycle in equilibrium.

Anyhow the analogy is not so important at this point. I believe I have captured the necessary system of equations in #31. I doubt turning them into electrical components makes the system any easier to physically solve.

 

[erobz]

P.s. I would not make a new thread about the same topic. The mods don’t like that. They probably wont shut it down to new replies for some time, so I wouldn’t worry about that too much.

 

[Juanda]

I was wrong about this, it gets messy.

We need to be using "Unsteady -Bernoulli"¹ which says in general:

$$ P_1 + \rho g z_1 + \frac{1}{2}\rho v_1^2 = \int_1^2 \rho \frac{\partial v_s}{\partial t} ds + P_2 + \rho g z_2 + \frac{1}{2}\rho v_2^2 \tag{1}$$

Messy is an understatement for me. I had never seen this version of Bernoulli. Fluids are hard... Anyways, this just proves this problem is a chance to learn some new things for me.

After reading the PDF you shared here's my attempt to show the equations before moving into coding where traceability will be much harder. I want to ensure we agree on the methodology before crunching numbers and plotting graphs. I'll express everything in terms of flow instead of velocities which I believe is more closely related to what we're interested in.
I'll need to redraw the diagram and rename a few things. I hate doing it at this stage but I need to do it to keep track of the relevant points and variables assigned to them.

Unsteady Bernoulli $(1)$ from the PDF you shared.

Using $(1)$ between the points $P$ and $D$ gives $(2)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +P_D + \frac{1}{2}\rho v_D^2+\rho g (H+z_D) - P_P-\frac{1}{2}\rho v_P^2 - \rho g z_P = 0 \tag{2}$$

Taking into account the height of $P$ is zero and writing the velocities in terms of flow yields $(3)$
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +P_D + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+z_D) - P_P-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{3}$$

Considering ideal gas law and constant temperature then the pressure of the air inside tank 2 can be expressed in relation to its water level. That it'll be equation $4$.
$$P_DV_2=m_2RT \ \left \{ V_2=S_T(H_T-z_D) \right \} \rightarrow P_D= \frac{m_2RT}{S_T(H_T-z_D)} \tag{4}$$

The water level $z_D$ is related to the water flow $(5)$.
$$z_D = z_{D_0}+\frac{Q_2}{S_T}t \tag{5}$$

Combining $(4)$ and $(5)$, $(6)$ is obtained.
$$P_D= \frac{m_2RT}{S_T(H_T-\left (z_{D_0}+\frac{Q_2}{S_T}t \right ))} \tag{6}$$

The equation $(5)$ can also be used on $(3)$. The results from $(6)$ will be implemented too to give $(7)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +\left (\frac{m_2RT}{S_T(H_T-\left (z_{D_0}+\frac{Q_2}{S_T}t \right ))} \right ) + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+\left (z_{D_0}+\frac{Q_2}{S_T}t \right )) - P_P-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{7}$$

Pressure at P ($P_P$) is an input value for the problem so I'll consider it known. Its expression is $(8)$.
$$P_P = P_{atm}+\rho g h_{pump}(Q)\rightarrow P_P = P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \tag{8}$$

Using the information from $(8)$ into $(7)$ gives $(9)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +\left (\frac{m_2RT}{S_T(H_T-\left (z_{D_0}+\frac{Q_2}{S_T}t \right ))} \right ) + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+\left (z_{D_0}+\frac{Q_2}{S_T}t \right )) - \left (P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \right )-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{9}$$

Finally, it is necessary to evaluate the path integral in $(9)$ that goes from $P$ to $D$. Before doing it I want to clarify the change of variable to have it in terms of the flow instead of the velocity $(10)$.
$$\rho\frac{\partial v}{\partial t}ds \rightarrow \left \{ v=\frac{Q}{S} \right \} \rightarrow \frac{\rho}{S}\frac{\partial Q}{\partial t}ds \tag{10}$$

Here is something that mathematically might not be expressed correctly because I'll go from a general variable $Q$ between the points $P$ and $D$ to the specific flows $Q_2$ and $Q$ in the pipes. I'll write the "generic" flow with $Q_g$ and the generic section with $S_g$ to be able to tell the difference. This will be the equation $(11)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds = \int_{P}^{D} \frac{\rho}{S_g}\frac{\partial Q_g}{\partial t}ds \tag{11}$$

According to the document from MIT, that integral from $P$ to $D$ can be simplified to the integral from $P$ to $F$ because $S_T>>S_P$ which only considers what's going on in the pipes. In this case, it is necessary to note the distinction in the flow from $P$ to $A$ and from $B$ to $F$ because the velocities/flow rates will not be the same. Equation $(12)$.
$$\int_{P}^{D} \frac{\rho}{S_g}\frac{\partial Q_g}{\partial t}ds = \int_{P}^{A} \frac{\rho}{S_p}\frac{\partial Q}{\partial t}ds+\int_{B}^{F} \frac{\rho}{S_p}\frac{\partial Q_2}{\partial t}ds=\frac{\rho}{S_p}\left ( \frac{\partial Q}{\partial t}L_{PA} + \frac{\partial Q_2}{\partial t}(H+L_{JF})\right ) \tag{12}$$

The expression from $(12)$ can be plugged in $(9)$ to obtain the final expression of Unsteady Bernoulli from $(2)$ for Tank 2. Equation $(13)$.
$$\frac{\rho}{S_p}\left ( \frac{\partial Q}{\partial t}L_{PA} + \frac{\partial Q_2}{\partial t}(H+L_{JF})\right ) +\left (\frac{m_2RT}{S_T(H_T-\left (z_{D_0}+\frac{Q_2}{S_T}t \right ))} \right ) + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+\left (z_{D_0}+\frac{Q_2}{S_T}t \right )) - \left (P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \right )-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{13}$$

Following the same logic it is possible to obtain the expression for Unsteady Bernoulli applied to Tank 1.
Finally, with the conservation of mass $Q = Q_1+Q_2$, I feel there is enough information to solve the system and obtain the functions $Q(t)$, $Q_1(t)$ and $Q_2(t)$ although I'd love to have confirmation about it if you know it for sure.

If we are on the same page about the procedure then I'd be ready to put it into Python to try solving it which will probably be a challenge in itself because I don't have a lot of experience with it.

 

[erobz]

Messy is an understatement for me. I had never seen this version of Bernoulli. Fluids are hard... Anyways, this just proves this problem is a chance to learn some new things for me.

After reading the PDF you shared here's my attempt to show the equations before moving into coding where traceability will be much harder. I want to ensure we agree on the methodology before crunching numbers and plotting graphs. I'll express everything in terms of flow instead of velocities which I believe is more closely related to what we're interested in.
I'll need to redraw the diagram and rename a few things. I hate doing it at this stage but I need to do it to keep track of the relevant points and variables assigned to them.
View attachment 330727Unsteady Bernoulli $(1)$ from the PDF you shared.
View attachment 330728Using $(1)$ between the points $P$ and $D$ gives $(2)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +P_D + \frac{1}{2}\rho v_D^2+\rho g (H+z_D) - P_P-\frac{1}{2}\rho v_P^2 - \rho g z_P = 0 \tag{2}$$

Taking into account the height of $P$ is zero and writing the velocities in terms of flow yields $(3)$
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +P_D + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+z_D) - P_P-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{3}$$

I believe the last term on the right should be:

$$\frac{1}{2}\rho \left ( \frac{Q_2}{S_p} \right ) ^2 $$

We are summing along a single streamline, what you have written is the kinetic energy along both streamlines.

The water level $z_D$ is related to the water flow $(5)$.
$$z_D = z_{D_0}+\frac{Q_2}{S_T}t \tag{5}$$

That is only if $Q,Q_2,Q_1$ are constant in time, I think it should be:

$$z_D = z_{D_0}+ \int \frac{Q_2}{S_T} dt \tag{5}$$

That propagates into a few other equations.

The equation $(5)$ can also be used on $(3)$. The results from $(6)$ will be implemented too to give $(7)$.
$$\int_{P}^{D}\rho\frac{\partial v}{\partial t}ds +\left (\frac{m_2RT}{S_T(H_T-\left (z_{D_0}+\frac{Q_2}{S_T}t \right ))} \right ) + \frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2+\rho g (H+\left (z_{D_0}+\frac{Q_2}{S_T}t \right )) - P_P-\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2 = 0 \tag{7}$$

Pressure at P ($P_P$) is an input value for the problem so I'll consider it known. Its expression is $(8)$.
$$P_P = P_{atm}+\rho g h_{pump}(Q)\rightarrow P_P = P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \tag{8}$$

You could just pretend $P_p$ is a constant for now for slightly more simple expressions, but that is up to you. It's going to be a beast to solve either way.

 

[Juanda]

I believe the last term on the right should be:

$$\frac{1}{2}\rho \left ( \frac{Q_2}{S_p} \right ) ^2 $$

We are summing along a single streamline, what you have written is the kinetic energy along both streamlines.

I'm integrating along a single streamline in this case. However, along that streamline, the velocity is not constant because the flow $Q$ must divide into $Q_1$ and $Q_2$. Since the cross-section of the pipes is constant, the velocity of the fluid after the separation must be smaller. At the junction, there's a discontinuity I'm choosing to ignore which is why I created the points $A$, $B$, and $C$ around it and "infinitely" close to it. That discontinuity would of course disappear with a more detailed analysis such as CFD but this already is complex enough the way it is.
At the point $P$, we have $v_P$ which in terms of the flow it is $Q/S_p$ ($S_p$ is the cross-section of the pipe) and at the point $D$ we have $v_D$ which in terms of the flow is $Q_2/S_T$ ($S_T$ is the cross-section of the tank).
Therefore, I think the equation $(3)$ is correct.

That is only if $Q,Q_2,Q_1$ are constant in time, I think it should be:

$$z_D = z_{D_0}+ \int \frac{Q_2}{S_T} dt \tag{5}$$

That propagates into a few other equations.

You're absolutely right here. I'll need to update that when I try to solve it.

You could just pretend $P_P$ is a constant for now for slightly more simple expressions, but that is up to you. It's going to be a beast to solve either way.

It's true I could. However, I'll have to discretize the rest of the problem anyways so maybe there's not much extra effort needed to describe the pump as a function of the flow. I just need to be careful when choosing the values of $\alpha$ and $\beta$ to make sure there'll be no backflow. I think the equations should predict that scenario without needing modifications (assuming the pipes are always full) but I'm interested in seeing first the "simple" case.

 

[erobz]

I'm integrating along a single streamline in this case. However, along that streamline, the velocity is not constant because the flow $Q$ must divide into $Q_1$ and $Q_2$. Since the cross-section of the pipes is constant, the velocity of the fluid after the separation must be smaller. At the junction, there's a discontinuity I'm choosing to ignore which is why I created the points $A$, $B$, and $C$ around it and "infinitely" close to it. That discontinuity would of course disappear with a more detailed analysis such as CFD but this already is complex enough the way it is.
At the point $P$, we have $v_P$ which in terms of the flow it is $Q/S_p$ ($S_p$ is the cross-section of the pipe) and at the point $D$ we have $v_D$ which in terms of the flow is $Q_2/S_T$ ($S_T$ is the cross-section of the tank).
Therefore, I think the equation $(3)$ is correct.

I don't think that a streamline begins until after we are outside the infinitesimal ball we're calling the junction. You should be going from $B$ to $D$, and $C$ to $E$ IMO.

Imagine a cylinder is rolling down a hill of mass $2M$, and it splits. On the path where the object is becomes the object, to where that particular object is going, I'm not going to uses kinetic energies involving $2M$ to describe a particular portion of its mechanical energy just after it became an object of mass $M$. I think the same is true here.

But I'm not going to argue further. You're modeling it, do whatever you want.

 

[Juanda]

I don't think that a streamline begins until after we are outside the infinitesimal ball we're calling the junction. You should be going from $B$ to $D$, and $C$ to $E$ IMO.

Why would you think that though? The flow starts at $P$ and goes to both $D$ and $E$ according to the diagram.

It'd be similar to something like this. Would you have trouble applying Bernoulli in this situation to find the state at A, B, C and D? With Bernoulli and mass conservation is possible. The only difference when compared with the problem we're discussing is that the flow isn't reuniting again after the junction but accumulating in the tanks.

Besides, in the problem with the tanks, it is necessary to use Bernoulli P→E and P→D so that the equations are related and can be solved simultaneosly in a system.

I'm not trying to argue for the sake of arguing but I'd prefer we both agree on how to model the problem before trying to implement it in the simulation.

 

[erobz]

If we go directly from A to D in the side problem you propose we learn nothing about the division of flow. We must visit the junctions in our equations.

We go from $P$ ( Which is just outside the pump) to the junction, then from the junction to each of the endpoints independently.

 

[erobz]

If we go directly from A to D in the side problem you propose we learn nothing about the division of flow. We must visit the junctions in our equations.

We go from $P$ ( Which is just outside the pump) to the junction, then from the junction to each of the endpoints independently.

Ok, after a quick example mimicking the problem at hand a bit more( one were the streamlines don't reunite), I convinced myself that we don't always have to visit nodes (the problem you chose as an instructive example, we do have to visit the nodes if we are to learn anything of the flow division - and we also would need to consider head loss to solve it). It could be that I just like visiting nodes out of habit, and may not be the most "streamlined" approach. I'm not 100% sure at the moment if what we are discussing isn't just going to give the same results in the end (but I believe they will)...have to think on that a bit. But I see now that (at least) your equations are good.

 

[erobz]

This is my take on it. on the incoming line where it splits you put a flow divider. there whole purpose is to split the flow to near 50/50.

We are basically just pontificating on the possibility of this at this point to try to learn something.

 

[Joe591]

We are basically just pontificating on the possibility of this at this point.

I'm not sure what you mean

 

[Juanda]

Ok, after a quick example mimicking the problem at hand a bit more( one were the streamlines don't reunite), I convinced myself that we don't always have to visit nodes (the problem you chose as an instructive example, we do have to visit the nodes if we are to learn anything of the flow division - and we also would need to consider head loss to solve it). It could be that I just like visiting nodes out of habit, and may not be the most "streamlined" approach. I'm not 100% sure at the moment if what we are discussing isn't just going to give the same results in the end (but I believe they will)...have to think on that a bit. But I see now that (at least) your equations are good.

I'm glad we agree. Knowing information about all the nodes within a streamline would be possible by applying Bernoulli to them but I chose the points I considered most relevant (exiting the pump and at the top of the tanks).
I'm then ready to try to write all this in code to solve it. Let's see if I can actually do it and get some physically possible results that make any sense.

 

[erobz]

@Joe591 Some say the flow just splits 50/50 without intervention, other suggest control ( like yourself). @Juanda and I are trying to tease something out of underlying physics for "fun".

 

[Joe591]

Some say the flow just splits 50/50 without intervention, other suggest control ( like yourself). @Juanda and I are trying to tease something out of underlying physics for "fun".

oh I see. yeah, I'm reasonably convinced it will require intervention. I tend to solve my problems with technology rather than math though, lol...

I must admit I don't have the faintest idea about the math that your using, haha

 

[erobz]

I'm glad we agree. Knowing information about all the nodes within a streamline would be possible by applying Bernoulli to them but I chose the points I considered most relevant (exiting the pump and at the top of the tanks).
I'm then ready to try to write all this in code to solve it. Let's see if I can actually do it and get some physically possible results that make any sense.

Can you write up the system of equations you are going to solve in final reduced form if you get a chance and talk about the numerical approach? I think this is going to be challenging system to solve (even programmatically).

 

[Juanda]

Can you write up the system of equations you are going to solve in final reduced form if you get a chance and talk about the numerical approach? I think this is going to be challenging system to solve (even programmatically).

I still owe you this. It's not that I have forgotten it but I know it's going to take at the very least a couple of hours being focussed to be able to do it and I can't get to it now. I'll try next week though.

 

[Juanda]

@erobz I think I got the system of equations you were talking about in post #48.
Equations will be referred to this diagram.

View attachment 330727

The dynamic state when filling it up can be defined by the equations $(1)$, $(2)$ and $(3)$.
Equation $(1)$ is unsteady Bernoullin from $P$ to $D$.
$$
\overbrace{\frac{\rho}{S_p}\left ( \frac{\partial Q}{\partial t}L_{PA} + \frac{\partial Q_2}{\partial t}(H+L_{JF})\right )}^{\int_{P}^{D} \frac{\rho}{S_g}\frac{\partial Q_g}{\partial t}ds}
+
\overbrace{\left (\frac{m_2RT}{S_T(H_T-\left (z_{D_0}+ \int \frac{Q_2}{S_T} dt \right ))} \right )}^{P_D}
+
\overbrace{\frac{1}{2}\rho \left ( \frac{Q_2}{S_T} \right ) ^2}^{\frac{1}{2}\rho v_D^2}
+
\overbrace{\rho g (H+\left (z_{D_0}+ \int \frac{Q_2}{S_T} dt \right ))}^{\rho g (H+z_D)}
-
\overbrace{\left (P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \right )}^{P_P}
-
\overbrace{\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2}^{\frac{1}{2}\rho v_P^2}
=
0 \tag{1}
$$

Equation $(2)$ is unsteady Bernoulli from $P$ to $E$.
$$
\overbrace{\frac{\rho}{S_p}\left ( \frac{\partial Q}{\partial t}L_{PA} + \frac{\partial Q_1}{\partial t}L_{CG}\right )}^{\int_{P}^{E} \frac{\rho}{S_g}\frac{\partial Q_g}{\partial t}ds}
+
\overbrace{\left (\frac{m_1RT}{S_T(H_T-\left (z_{E_0}+ \int \frac{Q_1}{S_T} dt \right ))} \right )}^{P_E}
+
\overbrace{\frac{1}{2}\rho \left ( \frac{Q_1}{S_T} \right ) ^2}^{\frac{1}{2}\rho v_E^2}
+
\overbrace{\rho g \left (z_{E_0}+ \int \frac{Q_1}{S_T} dt \right )}^{\rho g z_E}
-
\overbrace{\left (P_{atm}+ \rho g \left ( \alpha - \beta Q^2 \right ) \right )}^{P_P}
-
\overbrace{\frac{1}{2}\rho \left ( \frac{Q}{S_p} \right ) ^2}^{\frac{1}{2}\rho v_P^2}
=
0 \tag{2}
$$

Equation $(3)$ is mass conservation.
$$Q=Q_1+Q_2 \tag{3}$$

From those 3 equations, I'm hoping it's possible to obtain the functions $Q(t)$, $Q_1(t)$ and $Q_3(t)$ from which it would also be possible to obtain the functions $z_E(t)$ and $z_D(t)$.

It'd be necessary to define the geometry of the problem (pipe lengths, tank's cross-section, tank's height, tanks' height difference), the amount of air in each tank, the system's temperature, pump's properties, gravity, fluid's density, and the initial conditions (initial heights and flows).

To set up the initial conditions $(t=0)$ I'd prefer to start the system from an equilibrium point. I feel like it's not necessary but I think it's convenient because it's more realistic and the oscillatory nature of the result will be suppressed (or so I think).

To find the initial conditions there are different ways of doing it. Previously in the thread (post #27), we discussed using Bernoulli from $E$ to $D$ and the conservation of mass. I'll change the method to using Bernoulli from $P$ to $E$ and Bernoulli from $P$ to $D$.
$$
\overbrace{\frac{m_2RT}{S_T(H_T-z_{D_0})}}^{P_{D_0}}
+
\rho g(H+z_{D_0})
=
P_{P_0} \tag{4}
$$
$$\overbrace{\frac{m_1RT}{S_T(H_T-z_{E_0})}}^{P_{E_0}}
+
\rho gz_{E_0}
=
P_{P_0} \tag{5}
$$

The change is because of two main reasons. First, the initially proposed method required to know the initial water in the system which is not realistic. It's easier to install a manometer at $P$ and read the pressure when the valve is closed. Secondly, doing it this way is more consistent with the previous set of equations $(1)$, $(2)$ and $(3)$ and the equations $(6)$ and $(7)$ that will be written later and describe the new equilibrium at $t \rightarrow \infty$ when the amount of mass that entered the system is unknown but the pressure at $P$ is known.

So, for the initial conditions, we have $(4)$ and $(5)$.
During the fill-up process the equations are $(1)$, $(2)$ and $(3)$.
Finally, for $t \rightarrow \infty$ we know the flow tends to 0 so we can use "normal" Bernoulli again just like it was done to set up the equations $(4)$ and $(5)$. Also, there will be no difference between having the inlet valve open or closed. That defines the equations $(6)$ and $(7)$
$$\overbrace{\frac{m_2RT}{S_T(H_T-z_{D_f})}}^{P_{D_f}}
+
\rho g(H+z_{D_f})
=
P_{atm}+\rho g \alpha \tag{6}
$$
$$\overbrace{\frac{m_1RT}{S_T(H_T-z_{E_f})}}^{P_{E_f}}
+
\rho gz_{E_f}
=
P_{atm}+\rho g \alpha \tag{7}
$$

It's useful to define some "checkpoints" to make sure the results are correct. Some of them are:

• Functions for flow and height should not oscillate since we started from an equilibrium point with zero velocity.
• We know that the evolution described by $(1)$, $(2)$ and $(3)$ when starting from $(4)$ and $(5)$ with zero velocity must tend towards the result from $(6)$ and $(7)$ with zero velocity as time goes on.
• The functions $z_E$ and $z_D$ can never be negative. It'd imply the tanks have no water and from the beginning, we said the pipes are always full so the very minimum is that the height is zero. From the previous points, it can also be understood that their slopes can't be negative either.
• Since I'll be working with absolute pressure. No pressures can be negative either.

I think from there the last point would be to try to code it to find a numerical solution. I'm not too experienced in this but by looking at the presented math I feel it should be possible.