Can a siphon be used to generate more power than a conventional penstock?

Discussion in 'General Science & Technology' started by GoTesla, Oct 10, 2020.

  1. GoTesla Registered Member

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    2 days ago I came across patent CA2978935C, which seems to confirm something I have been thinking about for a couple of years – is it possible to extract more power from water using a siphon, than it is using a conventional penstock design?

    In a siphon, because of energy conservation laws, the velocity in the pipe is uniform throughout it’s length. Meaning the speed of water entering the pipe will be the same as that leaving it, so what if the generator is placed at, or near the input, as in the patent above. This still means, as long as the generator does not break the siphon, that there will still be a stream of water exiting the siphon, which could be used to generate, albeit reduced, power using say a pelton or similar turbine.

    For example, if we had a pipe of cross sectional area 1 m2, flow rate of 10 m/s (easy maths), using

    P = (mv2)/2 ( where m = density(in kg)*area*v) we get:

    at the input - (1000*1*10*10*10)/2 = 500,000 watts

    assuming the velocity has now been halved (from Betz’s law) we get:

    at the output – (1000*1*5*5*5)/2 = 62,500 watts

    Does this mean we have gained 62,500 watts, because using a penstock would only generate the same as the input? (I know these figures don't take into account any generator inefficiencies)

    If this is correct I think it may be possible to increase it further!
     
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  3. origin Heading towards oblivion Valued Senior Member

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    That is always true at steady state for any pipe of constant diameter with water flow. How could it not be true? Where would the water go if it the velocity was higher at the pipe inlet? Where would the water come from if it the velocity was higher at the pipe exit?

    Overriding concept: The goes ins equals the goes outs.
     
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  5. exchemist Valued Senior Member

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    Are you sure about this? My understanding is that the water in a downwardly inclined pipe accelerates as expected under gravity, with the result that it occupies a progressively smaller cross-section as it accelerates. This allows the mass flow rate to remain constant, as it must, while permitting the water to flower faster and faster as it must due to gravitational acceleration.


    The water may fill the pipe at the top, but it won't fill it lower down.
     
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  7. sculptor Valued Senior Member

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    It seems that you are describing a penstock system.

    from my perspective,
    I would start by thinking pump.
    A water pump pumps water uphill more efficiently/effectively than is sucks water.
    A siphon sucks ergo if the turbines are located at the top of the siphon, one would want the diameter of the pipe to increase as it goes downhill ---thereby creating more sucking power and increasing the speed of the flow at the inlet.
    Conversely, one would want the diameter of the pipe in a penstock setup to decrease as it approached the turbines, thereby increasing the speed of the flow at the turbines.
     
  8. exchemist Valued Senior Member

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    10,097
    No, I'm just describing water flowing downward freely, through an inclined pipe, i.e. with no restriction.
     
  9. billvon Valued Senior Member

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    19,051
    Yep. A caveat here - in most systems the pipe diameter is constant until the last few feet. In well designed systems the nozzle narrows gradually before the turbine. Many systems ignore the "gradual" part and just have a large feedstock pipe going straight to small nozzles. It still works.

    In such a system you can ignore what happens in between the two ends. You have a large intake and a small nozzle, and since the total mass flow has to be the same, the water at the nozzle is much higher velocity.
     
  10. GoTesla Registered Member

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    I think everyone is describing a penstock system! The point I’m trying to make is, can maximum power in a siphon be generated at the input, which is not true with a penstock? Ignore large or small inlets or outlets for the time being to increase velocity, I’m sure I can overcome this later!
     
  11. origin Heading towards oblivion Valued Senior Member

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    11,133
    I was just pointing out the obvious, same diameter means the same velocity.
     
  12. exchemist Valued Senior Member

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    10,097
    When you say the "input" do you mean the generator turbine is at the inlet to the pipe? If so, then you are limited to a maximum pressure differential across the turbine of 1 bar, because the minimum pressure downstream from it will be zero, i.e. a vacuum. This constrains the power you can get out of it, compared to a turbine located at the bottom of a 100m drop (say), which can have a pressure of 10bar.

    I must admit I don't see the point of this at the moment. The available gravitational potential is simply a function of the head between upper reservoir and the outlet at the bottom. Interposing a siphon does not change that. But I have not looked up the patent. Can you provide a link to it, so we can read what it says and see any diagrams that may help?
     
  13. exchemist Valued Senior Member

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    10,097
    If the pipe is filled, yes.
     
  14. billvon Valued Senior Member

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    19,051
    Nope. Same-same. It should be noted that in the real world siphons will always have slightly lower efficiency due to the need to pump them out periodically. (They will eventually fill with air.)
     
  15. Michael 345 New year. PRESENT is 71 years old Valued Senior Member

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    https://wiki.openstreetmap.org/wiki/Proposed_features/Hydropower_water_supplies

    Oct 13, 2017 — Penstock: A conduit taking water from a river, lake or ... The most the elevation difference is high, the most will be the pressure. ... The siphon is a pipe flow waterway.

    What am I missing

    From a quick read, the article does not go into much explanation, but I gather
    Penstock = water is under power and pushed through the pipes
    Syphon = water is not under power except gravity

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  16. GoTesla Registered Member

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    In a penstock system, energy at the input is potential energy, it is only when the water picks up velocity down the pipe it changes to kinetic energy, which is what we need to move a turbine. The amount of power that could be extracted would depend on where the turbine is placed in the pipe. Obviously, maximum power is at the end of the pipe. With a siphon, the kinetic energy, I believe, is constant throughout the pipe, therefore the turbine could be anywhere in the pipe and still produce maximum power.

    Some years ago, I watched a documentary on how steam powered Atlantic passenger ships utilised almost all the energy in the steam. It came out the boiler at 150 psi and entered the first piston, exited that piston at 100 psi and into a second piston which was 1.5 times the area of the first, exited that piston at 50 psi and into a third which was twice the area of the second, therefore producing 3 pistons providing the same pressure and getting maximum output from the steam.

    What I thought at first was, by moving the maximum amount of kinetic energy to the start of the pipe, like the pistons above, could there be multiple turbines down the length of the pipe each, obviously, producing less power than the previous? I have no idea about fluid dynamics, but I’m pretty sure they would interfere with each other and if nothing else break the siphon. So I thought, one in the pipe one out.

    I don’t think a conventional bladed turbine could be used, as I believe the flow would effectively be

    chopped up by the blades, potentially breaking the siphon. I was thinking more along the lines of Viktor Schauberger’s jet turbine:

    https://hermit-hideaway.blogspot.com/2010/02/viktor-schaubergers-jet-turbine-patent.html

    as this should minimise disruption to the flow. Ignore most of the right up, look at the picture of the turbine and imagine the twisted blade inside the siphon pipe. As long as the water is made to rotate inside the pipe then flow over the blade, water velocity will increase while rotating the blade as it progresses down the cone, as the cross sectional area inside the pipe is reduced, because the cone fills more of the pipe! (half the area would double the speed) In the second world war, the allies proved this design produced about a third more power than they thought was possible. I’m pretty sure someone with a Computational Fluid Dynamics (CFD) Software package could come up with a variation on this design to improve output.

    The reason for a siphon, I thought, was it could be applied retrospectively to existing dams and because water is sucked through the pipe, doubling the speed to increase power is more likely to be achievable than if it is pushed through the pipe, as is the case with a penstock (less resistance!?). Half the cross sectional area, doubles the speed, which quadruples the power output.

    The link to the patent I referenced is:

    https://patents.google.com/patent/CA2978935C/en?oq=CA 2978935
     
  17. exchemist Valued Senior Member

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    10,097
    Thanks for the patent. I'm surprised this was granted, as it seems to be based on some misconceptions. The two most glaring are:

    1) The notion that water turbines work by converting kinetic energy into electrical. Most of them are reaction turbines, which convert gravitational potential energy directly into work, by getting the pressurised water to push on the blades through a pressure drop.

    2) The idea that "molecular cohesion" is significant enough to allow the downhill water to "pull" water through the turbine. The tensile strength of water is only about 0.3 atm, too small to have a material effect. If you have more than a drop of ~13m downstream of the turbine, the column of water will break, leaving a vacuum. So, once a fall of this magnitude is exceeded, it doesn't matter if it is 20 m or 200m, the power output of a turbine above will the be same.

    I don't follow what your idea is, but basically you are limited in any hydro system to the available gravitational potential energy, mgh, in other words the mass x g x the height the mass falls. The maximum theoretical power output will therefore be given by mass flow rate x g x h.

    This is true regardless of the system you use to extract the energy.
     
  18. Michael 345 New year. PRESENT is 71 years old Valued Senior Member

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    Not done the mathes (no need)

    If you think a flow of water being syphoned down a pipe, by gravity, has the same kinetic energy after 1 metre as it has at a 100 metres down you are clearly not all up with gravity or physics in general

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  19. billvon Valued Senior Member

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    Short answer - no.

    The reason that steam engines work well with multiple stages is that the gas is cooling, contracting and changing temperature as it passes through. Thus the ideal gas for a high pressure displacement engine (or turbine) is not the same as the ideal gas for a lower pressure engine. Multistage engines take advantage of this to maximize energy extracted. If the gas that exits the engine is room temperature and at zero pressure you have extracted all the possible energy you can. You can't get there of course but you can come close. (All gas turbines do this.)

    Water doesn't work like that. Once you have harvested all the kinetic energy from a volume of water you are done. No further extraction is possible. So more stages don't help.

    Siphons can be used to deal with difficult penstock routes (i.e. by a microhydro system where the owner doesn't want to dig through solid rock) but they don't increase energy. In fact, compared to a straight run, they decrease it due to increased friction.
     
  20. GoTesla Registered Member

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    Yep, I figured that, that's why I said one in the pipe, one out.
    True, but are we harvesting ALL the kinetic energy. If we did, the patent I referenced clearly would not work, as the siphon would stop!
    Again I realised the main problem in a siphon would be bends, depending on how tight the angle it can drop the flow rate by half. My son was a gas fitter and he had to take into account the length of the pipe, every bend and joint to calculate the flow rate at the boiler.
    Not quite right. A siphon requires both gravity and cohesion to work, for example you cannot siphon sand!
    Where did this come from? Who mentioned 1 metre?

    Just to get things right, a siphon with a head of 5.1 metres will produce a velocity of just over 10 m/s, which is exactly the same as a penstock with a 5.1 m head (except for bend losses). From the formula p=(m*(v*v))/2 it is clear the velocity is significant in how much power is generated, so if this can be increased then more power will be generated. I know I can double the speed in a siphon without significantly affecting the flow rate, I've done it, but can it be done when there is a turbine there, I don't know! I do think however, it is more likely to work in a siphon than conventional penstock, due to water being sucked through the pipe as opposed to pushed through. If it can be made to work, using the same figures I used in post one we get the following:
    cross sectional area of pipe = 1 square metre
    velocity in pipe = 10 m/s, but halving the area doubles the speed to 20 m/s
    density of water = 1000 kg / cubic metre
    (1000*0.5*20*20*20)/2 = 2,000,000 watts
    This is four times the amount calculated in post 1!
     
  21. billvon Valued Senior Member

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    Well but that's not "molecular cohesion" - it's pressure. If the sand provided a pressure seal such that air could not work its way into the pipe you could siphon it. To use a counterexample you can't make a siphon work even with water higher than about 30 feet. Above that height you would need a pressure less than zero (which is impossible) to maintain flow; 'molecular cohesion' does not help you.
     
  22. exchemist Valued Senior Member

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    10,097
    You are thinking about this the wrong way. Velocity is not the key to how much energy can be extracted. As I pointed out before, the energy available for conversion to electricity is related solely to the head and the mass flow rate, i.e. mgh/t.

    You can extract this energy either by conversion to kinetic energy first and then using an impulse turbine (such as a Pelton wheel), or by means of a reaction turbine (such as a Francis turbine) which does not require any particular velocity in the water flow, because it uses the pressure due to the head directly. The efficiency of a Francis turbine is typically >90%, leaving not much scope for further improvements. Most hydroelectric plants use this type of turbine.
     
  23. GoTesla Registered Member

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    Got it! In this formula increasing the velocity would not affect the outcome. I thought power was calculated by 0.5*m*v squared (haven't got a superscript). Increasing velocity in this formula gives the figures I calculated before! However, this creates another dilemma - how does Viktor Schauberger’s jet produce more power (assuming this is true)?
     

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