Ram pump -- is water compressibility really relevant?

[Swamp Thing]

Is it really necessary to consider the compressibility of water in order to understand a ram pump?

Steve usually does an excellent job of explaining this kind of thing, but this time I'm not so sure...

Note: Despite the thumbnail, he is not proposing a perpetual motion device.

 

[Ken Fabian]

I built and used one for a time - a variant that included a pressure vessel packed with closed cell rubber rather than just an air chamber - without that there was too much backflow from the non-return valve. It was a design intended for low water 'head' (or drop) and it did work but my site had low water levels and was near the lower limit for the design - only about 750mm - and sand was a problem but a screen filter just slowed the flow too much.

Lack of water compressibility was critical to a strong 'water hammer' - initially I used a polyurethane washer in the 'slam valve' (not sure about correct terminology) and just that bit of compressibility prevented it working. I had to use different materials than the initial design to get it to work.

Even a bit of expansion each 'hammer' from using less rigid materials - a different kind of 'compressibility' - was enough to prevent it working properly.

High density polypropylene worked for the valve washer but the valve seat in the main body - cast iron in place of the original PVC because it was not rigid enough and the hammer impulse was reduced - rapidly rusted and pitted, no doubt accelerated by the sand, sigh. (If I had enough money at the time to use stainless steel I probably could have bought a commercially available variant ram pump by Glockemann. That sort uses a diaphram that pushes a piston.)

It was not very reliable and didn't last beyond a year and a half. In deeper clear water with plenty of drop it would have worked much better and I could have used the more durable PVC instead of cast iron.

 

[Baluncore]

Is it really necessary to consider the compressibility of water in order to understand a ram pump?

Yes. Water is effectively not compressible, when compared to the elasticity of a steel pipe. If the reverse was true, the ram pump could not function.

Ram pumps are easier to understand and work with, if you make the wise assumption that water is incompressible. What is unfortunately elastic, are the pipes, which explains why the twelve or so meters of fixed diameter induction pipe, closest to the pump must be made from a metal such as steel, not from plastic. The remaining upstream part of the induction pipe can be made from cheap and flexible "polypipe" without problems. The steel pipe ensures that a significant pressure pulse is generated and maintained, which will only very slightly expand the elastic steel pipe.

The steel pipe must withstand the water hammer pulse when the waste valve slams shut. That pressure wave travels away from the waste valve, pushing a small amount of water through the check valve, into the bottom of the air reservoir and outlet chamber, as it passes, on its way back up the induction pipe.

The pressure wave generated by the waste valve closing, moves upstream through the induction pipe to the inlet, where it is inverted by the reflection from the open-ended induction pipe, to become a depression wave that runs back down to reopen the waste valve for the next cycle. There is a convenient complication here, in that the pressure inversion due to reflection, may be from the junction between the steel and the plastic induction pipes, since the significantly greater elasticity of the plastic can appear to be an open-end. The same inversion may be obtained from a step increase in pipe diameter at that junction.

The check valve below the air reservoir prevents the pumped head from affecting the lower pressure at the waste valve due to the lower head of water in the induction pipe. The water outlet line from the bottom of the air chamber can be small in diameter, and several kilometres long. That is because the flow through the outlet line is steady, driven continuously by air pressure, and so the flow does not pulse.

There must be air in the chamber to continuously oppose the outlet head, but that air also allows the check valve to operate. It is sometimes necessary to induce a few small bubbles of air into the pump body, to maintain sufficient air in the outlet chamber. Excess air will be vented, through the outlet line as compressed bubbles in the water. The air chamber pressure and volume is therefor self-regulating.

 

[Swamp Thing]

I may not have worded my question in the best way.

Steve Mould's explanation is based on the idea that you can't make complete sense of the device unless you assume that water does compress a little during the pressure spike, sending a shock wave upstream all the way to the reservoir.

I was skeptical about the need to invoke such ideas, and the two replies above seem to confirm my suspicion.

Also, since posting my question it occurred to me that the presence of that air chamber almost guarantees that the water isn't going to compress or swell the pipes to any meaningful extent. I mean, why would the water "bother" to compress when there is a compliant cushion of air to take the shock?

Thanks for the detailed and informative answers.

 

[Baluncore]

If Steve Mould was a perfectionist, he would never commit a video to YouTube. Like every publication, it can only be third-rate. If a few mistakes were fixed it could be imagined as being second-rate, until it was published, when it would naturally revert to being third-rate. First-rate is perfection, and is unobtainable. A publication must be designed for a particular educational level. Maybe we need seven different videos about the hydraulic ram pump, each targeted to a particular audience. The fundamentals need to be correct, but the hand waving will be different.

Steve Mould's explanation is based on the idea that you can't make complete sense of the device unless you assume that water does compress a little during the pressure spike, sending a shock wave upstream all the way to the reservoir.

That is strictly not a shock wave, it is a sound wave in a liquid, the sound of a step function.
A shock wave is a self-maintaining pressure and temperature step in a gas.

The presence of sound waves tells us that water is compressible, but that is where it ends. The "pressure spike" can also be seen as a transient wave travelling along a transmission line, or as a bulge in the pipe that travels with the sound pressure inside.

Also, since posting my question it occurred to me that the presence of that air chamber almost guarantees that the water isn't going to compress or swell the pipes to any meaningful extent. I mean, why would the water "bother" to compress when there is a compliant cushion of air to take the shock?

The system is dynamic, not static. Unlike hydrostatic pressure, the step function is so short that it is not everywhere the same at once.

As the waste valve closes with a snap, a pressure wave travels away from the valve. That forms a wavefront across the induction pipe. As that front passes the check valve, the peak pressure is clamped by the exit of water to the air chamber on one side, but only part of the energy pushes water through the valve, the rest of the wavefront continues to swell the pipe as it travels on towards the induction pipe inlet, where it is reflected. The peak pressure of that travelling step pulse, was limited to some extent, while passing the small aperture of the check valve, to the pressure in the air chamber.

The reflected wave is inverted and of similar but negative amplitude to what remained after passing the air chamber check valve earlier. Note also, there is a negative pressure limit due to cavitation at the reflector. That reflected depression pulse has no problem overcoming the static pressure head in the pump chamber, so the waste valve is opened by atmospheric pressure without any problem.

The forward step-up pressure wave, will be partly cancelled by its inverse on reflection. The pressure pulse width will be determined to some extent by the return transit time of the induction pipe. It will also be limited by the harmonic content of the waste valve closure.

 

[A.T.]

I mean, why would the water "bother" to compress when there is a compliant cushion of air to take the shock?

The compressibility of water is the reason why there is a wave, rather than instantaneous pressure change propagation. Without that delay, the system would settle into some steady flow state, rather than alternating valve operation.

 

[Swamp Thing]

I am still not sure whether finite non-zero compressibility is a key feature, or merely a secondary / tertiary modifying effect. If we compare two simulations, one with finite and realistic compressibility and the other with incompressible fluid, and if we include the air chamber, will there be a huge difference in performance?

If I have understood @Ken Fabian correctly, he seems to believe that compressibility is a minor secondary factor if at all. When I read @Baluncore 's first post I thought he believed the same, but from his second post I think he is making the case that the thing wouldn't work with an incompressible fluid? And @A.T. is unambiguously for the idea that a simulation with incompressible fluid would malfunction.

 

[Swamp Thing]

I decided to start with the simplest toy problem and work it up step by step into a full ram pump simulation. Gemini gives me this diff. eqn. for the simple problem I have sketched below. Does it look right?

$\frac{dQ}{dt} = \frac{\pi d^2}{4\sqrt{X^2+Y^2}} \left( \frac{P}{\rho} + gY \right) - \frac{2Q^2}{\pi d^2 \sqrt{X^2+Y^2}}$

X is the horizontal span and Y the drop. P is the inlet pressure.

 

[A.T.]

And @A.T. is unambiguously for the idea that a simulation with incompressible fluid would malfunction.

Any analysis that directly or indirectly assumes a finite speed of sound in the water, is also implicitly assuming some compressibility. As @Baluncore wrote:

The presence of sound waves tells us that water is compressible, ...

And you need that finite speed wave, for the alternating valve operation.

But as @Baluncore also wrote:

... but that is where it ends.

For example, you don't need to consider things like the change of the total volume of water here.

 

[erobz]

$\frac{dQ}{dt} = \frac{\pi d^2}{4\sqrt{X^2+Y^2}} \left( \frac{P}{\rho} + gY \right) - \frac{2Q^2}{\pi d^2 \sqrt{X^2+Y^2}}$

Dimensionally it's inconsistent. So back to the drawing board. I guess its dimensionally ok. On second thought it appears to be the unsteady Bernoulli equation. But something isn't quite correct because it should be like in the following paper:

https://ocw.mit.edu/courses/2-25-ad...edabdfd4b95c1792_MIT2_25F13_Unstea_Bernou.pdf

 

[Ken Fabian]

Steve Mould's explanation is based on the idea that you can't make complete sense of the device unless you assume that water does compress a little during the pressure spike, sending a shock wave upstream all the way to the reservoir.

Whilst every material outside of a black hole is compressible I do think very low compressibility of water is a crucial factor for ram pumps to work. Loose terminology to say 'lack' of compressibility rather than 'very low' perhaps.

I think a compression wave passing up to the reservoir is NOT a necessary element for it to work, it is just a loss of efficiency. Less water compressibility, less energy will be lost to a sound wave passed to the reservoir to dissipate - and more of that is probably from the pipe expanding and contracting than the water.

I expect the expansion/contraction of the pipework and pump body to dissipate energy; if the materials were rigid enough ram pumps would be less audible and a bit more efficient. I think absolute lack of water compressibility and absolute rigidity of the pipe and pump will improve performance and make it silent.

If we are getting very nitpicky there is probably some energy lost to heat as well.

Seems to me a ram-pump works with a conversion of momentum - the rapid stop of a column of water converts momentum to pressure that is higher enough to overcome the delivery line (head) pressure. Any water compression or expansions of the pipe and pump lowers the peak pressure the stopped momentum makes.

There is a cushioning effect of the air in the vessel which is at the delivery line pressure between pulses; instead of pressure pushing directly against the delivery line as a short pulse it seems to allow the pulse to be absorbed pulse and rebound more slowly to give time for the non-return valve to close. And do compression/expansion internally and usefully rather than produce sound that dissipates energy outside the system.

(In the version I made the added compressible rubber slowed that rebound further. Or else too much water flowed back down before the non-return valve could function. Stronger spring in the non-return valve reduces backflow but resists the upflow.)

 

[A.T.]

I think a compression wave passing up to the reservoir is NOT a necessary element for it to work
...
I think absolute lack of water compressibility and absolute rigidity of the pipe and pump will improve performance and make it silent.

To operate the valves, you need the water pressure to rise and fall in a cyclic manner. Where would that periodic pressure variation come from, without a wave that propagates with finite speed?

 

[Ken Fabian]

To operate the valves, you need the water pressure to rise and fall in a cyclic manner. Where would that periodic pressure variation come from, without a wave that propagates with finite speed?

That is what the main 'slam valve' does (Not sure of correct terminology).

At the start of each cycle water flows down the main water feed pipe and spills out freely, back to the river - effectively a pipe open at both ends - until it is flowing fast enough to slam the valve closed. Most of the water that flows down that first pipe doesn't get pumped, it is released. Which is why it defies no physics; water flowing downhill powers it but only a fraction of it gets pumped.

When that valve slams closed it makes a pressure spike from the momentum of the column of water being stopped. A bit gets pushed past a non-return valve - the bit that gets pumped - whilst the slam valve releases (because the flow has stopped) and water can flow freely down the pipe again. It keeps repeating that cycle.

 

[Baluncore]

I think a compression wave passing up to the reservoir is NOT a necessary element for it to work, it is just a loss of efficiency.

I disagree. A ram pump works poorly if it does not have a straight metal induction pipe. The waste valve does not work reliably without the inverted reflection from the open-ended induction pipe. Note that from the instant the waste valve closes, the pressure in the pump chamber is high, like the head, and that it should not fall again until the reflected wave returns. During that double transit period of the induction line, water will flow up through the check valve into the air reservoir.

Without the induction pipe, which is a transmission line, there would only be a short inductive spike in pressure due to the waste valve closing. Inertia would reduce the flow of water through the check valve. Maybe look at the electronic analog, Time Domain Reflectometry, where a transmission line maintains the voltage until the reflection returns. Note also that an inductor generates only a very short TDR positive pressure pulse. Maybe it's time for a comparative SPICE model of a voltage boost converter using an inductor, and one using a transmission line.

If HDPE is used for the entire induction pipe, the ram pump will fail, or pump poorly. That is because the polypipe expands too much, which reduces the peak pressure, and wastes energy, attenuating the wavefront and so prevents the inverted reflection from reopening the waste valve. When a ram pump with a short or thin walled induction pipe does not work well, replacing it with a thicker walled pipe, or a longer pipe improves things.

Any water compression or expansions of the pipe and pump lowers the peak pressure the stopped momentum makes.

Volume for volume, liquid water is significantly more compressible than solid steel. But a steel pipe has a small sectional wall area compared with the section filled with water, so it is both the pipe that expands and the water that compresses. It does not matter which expands or compresses most. You cannot ignore water compression because that dictates the speed of sound in water, and the reflected wave, which is a fact.

Stronger spring in the non-return valve reduces backflow but resists the upflow.

The check valve, where water from the pressurised chamber enters the air reservoir, does not have a spring. It is a poppet valve or a ball, that relies on gravity until the pressure head builds in the air reservoir.

The check valve must be at the top of the chamber because air is needed in the reservoir, and that air is induced through a pinhole into the chamber of the pump, close to the induction pipe connection. The pinhole usually has a loose fitting pin that can rattle, to keep it from blocking with debris. The pinhole weeps and sucks on each cycle, while being swept clean each time by the waste flow.

I was once given a new ram pump that did not work. It had been replaced with a second hand cast iron classic. Once I removed the manufacturers paint from the blocked pinhole, it worked well and went to a new home. Without air in the reservoir, very little water can enter past the check valve, and the long outlet line will pulse inefficiently, rather than flow smoothly and continuously.

That is what the main 'slam valve' does (Not sure of correct terminology).

That is the "waste valve" that remains open while momentum builds in the induction pipe. It is sometimes called the "clacker valve", because that is the sound it makes. To turn off a ram pump, wedge a stick in the waste valve outlet, so it cannot close.

 

[Ken Fabian]

@Baluncore - My experience was with a variant type, so details vary. My non-return valve had a spring and had other differences but principles are the same.

Why a straight pipe appears to deliver a better pressure pulse seems a different question to how the 'waste valve' re-opens. On mine the momentary lack of flow and gravity did that. As it would, I believe, in the 'traditional' type you describe. You may be overthinking it with inverted reflections; the water stops flowing, the valve re-opens.

I will think about it some more but need something more compelling to change my conclusion that compressibility of water is not necessary for a ram pump to function. For this purpose it effectively un-compressible; even peak pressures are not extreme. And I still think fluid compressibility and/or lack of rigidity of the materials reduces efficiency.

 

[Baluncore]

Why a straight pipe appears to deliver a better pressure pulse seems a different question to how the 'waste valve' re-opens.

It is not. There is a partial reflection at any and every bend in the induction pipe, and again at the open-end of the induction pipe. Optimum valve opening is when there is one clean reflection.

You may be overthinking it with inverted reflections; the water stops flowing, the valve re-opens.

Without an inverted reflection, for the waste valve to open it must be spring-loaded to overcome the head of the induction pipe. The selection of that spring would be a difficult adjustment to make. Yet gravity works well without a spring when there is an inverted reflection.

https://en.wikipedia.org/wiki/Hydraulic_ram#Sequence_of_operation

Meanwhile, the water hammer from the closing of the waste valve also produces a pressure pulse which propagates back up the inlet pipe [17] to the source where it converts to a suction pulse that propagates back down the inlet pipe.[18]

Take a look at references [17] and [18, fig 8].

 

[erobz]

The water stops flowing because someone slammed the gate closed. The pressure spike ( post sound wave beginning to travel toward the sump) work to open the valve back up. Then the system returns to its lower state of pressure once the wave has made it to the inlet of the sump(river). The valve has to now be able to fall again under its own weight (fighting viscous drag as well) carrying enough momentum to slam itself shut again. We need a pressure drop below initial state to see this occur. I believe that pressure drop below initial state is characterized by the reflecting waves @Baluncore brings up, as is shown in the image below.

Image from Engineering Fluid Mechanics, Ninth Edition. Authors Crowe,Elger,Williams,Roberson