Home  5  Books  5  GBEzine  5  News  5  HelpDesk  5  Register  5  GreenBuilding.co.uk
Not signed in (Sign In)

Categories



Green Building Bible, Fourth Edition
Green Building Bible, fourth edition (both books)
These two books are the perfect starting place to help you get to grips with one of the most vitally important aspects of our society - our homes and living environment.

PLEASE NOTE: A download link for Volume 1 will be sent to you by email and Volume 2 will be sent to you by post as a book.

Buy individually or both books together. Delivery is free!


powered by Surfing Waves




Vanilla 1.0.3 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to new Forum Visitors
Join the forum now and benefit from discussions with thousands of other green building fans and discounts on Green Building Press publications: Apply now.




  1.  
    If you have a reservoir full of water and let a tonne of it out, the potential energy released is mass X gravity X the height of the dam above the turbine.

    But releasing that tonne will reduce the depth of water behind the dam a little. So the potential energy from the next tonne of water will be a bit less than the first. And so on, until the reservoir is empty.

    If you integrate over all the tonnes of water in the reservoir, the average energy stored is equivalent to: mass X gravity X the height when the dam is only half full - or a bit more depending on the slope of the banks.

    If the reservoir is located 100s of metres up a hill, losing the effect of half of the dam's height doesn't make much reduction in the amount of energy stored. But for a tidal reservoir only a few metres high, this makes a big impact.

    Worse, for a tidal reservoir, the maximum height is only available for a short period of each tide, when the reservoir is full and the sea level is right out, or v.v.

    So you could generate m.g.h/2 if you quickly emptied the reservoir at that moment, a bit less considering mechanical inefficiencies. But the turbine probably isn't big enough to do that, and you want power output over several hours, so the sea level will have started to rise before you finished emptying the reservoir, and you will lose even more of the height potential energy.

    You might decide to generate during the rising tide as well as the ebb, so you'd ideally keep the level low in the reservoir until the sea level has risen quite high, so there is enough height difference for generating. This might mean you can't get the reservoir completely filled in time for high tide, so can't generate so much in the subsequent falling tide.

    You can do pumped storage with a tidal reservoir. Ideally you pump extra water into the reservoir around high tide time, wait for the tide to go out, then release it, generating more energy than was put in (>100% efficient storage). Obviously this depends on the peak energy demand time aligning with the low tide time, so not possible every day.

    Energy prices can be forecast days ahead, and tides much further, so you can optimise the reservoir levels and times of the generation for best effect.

    A good way to 'store' electricity in any reservoir is to have a big river run into it. During the night you fill the dam with the river, then release at peak electricity time, so 'storing' (time shifting) electricity at 100% efficiency (+- height changes). Some big Scottish hydro schemes do this, and barrages across river estuaries, but not Dinorwig which is too high uphill. Cruachan is reported to deliver 10% of its storage this way.
    • CommentAuthorEd Davies
    • CommentTimeJul 15th 2020
     
    Posted By: WillInAberdeen: “If you integrate over all the tonnes of water in the reservoir, the average energy stored is equivalent to: mass X gravity X the height when the dam is only half full - or a bit more depending on the slope of the banks.”

    Indeed. But the amount of energy needed to pump the water in is, in principle, the same so that wouldn't indicate any difference in efficiency (energy out/energy in).

    However, I have a feeling there's a Betz/thermodynamics-like reason that generating electricity from dropping 10 tonnes of water 10 metres produces less than dropping 1 tonne 100 metres. Or maybe it's just a practical engineering reason. Or maybe I've got the wrong idea completely.

    Posted By: WillInAberdeen: “You can do pumped storage with a tidal reservoir.”

    Yes, Rance does this:

    https://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity#Seawater
  2.  
    That's right, but I was responding to PiMs idea about effectiveness, which I suppose is energy stored per unit area of reservoir. Only very few reservoirs have water pumped into them, but all are concerned about 'effectiveness' in some way. My point is that it's not as simple as area X height, for reasons I mentioned.

    Maybe 'effectiveness' should also refer to the value of the land taken up by the reservoir and the cost of building the dam? It is more effective to store water with a dam across a steep valley or extending a natural lake. Ecologists have views about the habitat value of the land or estuary that is flooded.
  3.  
    I think the Betz limit applies to turbines in a flow where the fluid has mainly velocity (kinetic energy) rather than stored pressure energy, like wind turbines, or those tide turbines that look like them. It states that you can't extract all of the velocity energy, or then the fluid would stop, and clog up your turbine.

    A hydro dam stores water with pressure rather than kinetic energy. Although there's conversion in the turbine, the Betz limit has less relevance. There just has to be enough potential left over in the water to push it out down the tail race tunnel, which can be short or fat or run downhill.

    If you have a very low dam, you have to let many more tonnes of water out per kWh impounded, so you need bigger diameter tunnels and turbines. There might be an economic point where you choose to build a more restrictive / less efficient tunnel, depending on the costs/ distances/ geology etc I suppose.
    • CommentAuthorEd Davies
    • CommentTimeJul 15th 2020
     
    Posted By: WillInAberdeenI think the Betz limit applies to turbines in a flow where the fluid has mainly velocity (kinetic energy) rather than stored pressure energy, like wind turbines, or those tide turbines that look like them. It states that you can't extract all of the velocity energy, or then the fluid would stop, and clog up your turbine.
    Yes, the Betz limit itself doesn't apply directly to pumped storage and the like.

    AIUI, the Betz limit applies when the fluid is:

    a) non-compressible, as it effectively is for water or air at the sort of pressures and velocities applicable to most turbines of the sort we're discussing here, and, more relevantly,

    b) when the fluid is moving freely, unconstrained by a pipe or tunnel, so that it can flow around the turbine.

    As you say, the limit arises because you can't take all the kinetic energy out of the fluid otherwise it would stop and clog up the output of your turbine. More specifically, the betz limit arises because the back pressure the slowed-down fluid produces or, alternatively, the increased cross-sectional area needed to get rid of the same volume of slower-moving fluid deflects some of the incoming fluid round the turbine.

    If the incoming fluid is constrained in a pipe but you can engineer things to spread it out more as you extract the kinetic energy you can take out more than the 59.3% Dr Betz would keep you to. An example of spreading out like this would be a Pelton wheel where you have a small jet on the input side and the whole size of the buckets or whatever on the output side.
  4.  
    Interesting, but I think it's more general than that, so I am not sure either of those conditions are necessary:

    a) If the fluid were compressible and expanded from a higher pressure through the turbine, there would still need to be enough kinetic or pressure energy left over in the fluid for it to flow out of the turbine. The limit might have a different % value but the principle would be the same.

    b) Imagine building a hypothetical shroud casing around the outside of a wind turbine, shaped exactly the same as a streamline so it didn't interfere at all with the air flow. The flowing air molecules wouldn't be able to tell it was there, so all the aerodynamics would work the same, including the Betz limit.

    You don't have to have a circular (axial) turbine, the same principle would apply to a tangential flow turbine.

    The jets on a Pelton wheel are because it's an impulse turbine - you squeeze the fluid through a narrow jet to make it accelerate (convert its pressure to momentum, Bernoulli style). Then the fluid hits the wheel and transfers the momentum (an impulse, in the mathematical sense). There is still a Betz-ish limit as you need to drain the water out of the turbine, but the % value might be different.

    The nose cone on the front of a wind turbine does the opposite, it makes the flow lines diverge, so the fluid slows down and gains pressure. The stronger the wind, the further upwind of the turbine the divergence begins, so only as much air flows in as can flow out again

    All turbines are preferably designed so that Integral(velocity X density X dArea) stays the same all the way through from inlet to outlet, as mass is always conserved.
    •  
      CommentAuthordjh
    • CommentTimeJul 15th 2020
     
    The wikipedia sketches the applicable conditions and the proof. The Betz limit is precise - 16/27 - and whilst there may be similar limits for circumstances that don't meet the preconditions, I don't think they can be called Betz limits.

    Or even Lanchester limits, since it was he who actually discovered the law.

    https://en.wikipedia.org/wiki/Betz%27s_law
  5.  
    Well to be fair, this was all sorted out by Bernoulli in the 18th Century, for the general case that mass momentum and energy are all conserved.

    Betz has chosen a special case, where he has chosen to neglect the pressure, density and gravity terms in Bernoulli's equation, leading to a particular numerical outcome. But you could pick any other special case you like, such as for a compressible flow down a steep hillside, and solve in the same general way, then name it after yourself! (Or even after somebody else, who you were at war with at the time).

    Most of the conditions stated in the wiki article are irrelevant. The only ones relevant to wind turbines are that changes in air pressure and density are chosen to be neglected. The rest are just there to avoid them needing calculus in their example 1D derivation.

    The important (and generally applicable) bit of Betz's concept is that as much fluid must flow out of the turbine as flows in, so you cannot extract 100% of the energy. Again to be fair, this was already well understood by mediaeval waterwheel builders, but Betz is a snappier name for the concept.
    • CommentAuthorEd Davies
    • CommentTimeJul 16th 2020
     
    Posted By: WillInAberdeenb) Imagine building a hypothetical shroud casing around the outside of a wind turbine, shaped exactly the same as a streamline so it didn't interfere at all with the air flow. The flowing air molecules wouldn't be able to tell it was there, so all the aerodynamics would work the same, including the Betz limit.
    Yes, it would work the same but only because the hypothetical shroud just moves the problem of air flow out to the sides forward to the front of the shroud. You'd still have higher pressure there than in the free air so air would be deflected sideways.

    Posted By: WillInAberdeenBetz has chosen a special case, …
    Absolutely, though it's an interesting and useful special case but using his rule outside that special case is wrong.

    Obviously you can't extract all the potential and kinetic energy from a fluid but for practical cases where you have a high pressure or high speed flow I think this limitation is trivial compared with friction and other losses. E.g., for a Pelton wheel at the bottom of a 100 metre-high pipe (the sort of thing Paul at the end of the road has) you then have to let the water drop another 500 mm or whatever to get clear of the wheel so you can “only” take 99.5% of the original potential energy. So, yes, there's a theoretical limit but it's way more than Betz's 16/27 and irrelevant for practical purposes.

    On the other hand, when the head available is only a few metres I suspect this sort of limitation becomes more significant hence my vague intuition that low-head pumped storage is intrinsically limited in efficiency.
  6.  
    Derivation of the Edz-Willz Limit for Efficiency of Hydro Generators:

    Consider a fluid of density rho in a reservoir. The height from the tailrace to halfway up the dam is H. (See above for why halfway)

    Consider a flow rate of M kilogrammes/s of fluid, which flows from the surface of the reservoir through an ideal frictionless turbine with an outlet duct of area A.

    Bernoulli: the energy per unit mass = gh + P/rho + 0.5v^2
    (g=gravity, P=pressure, v=velocity)


    At the reservoir surface, P = v = 0 so power available = MgH

    At the outlet, P = h = 0 so unrecoverable power = 0.5 M.v^2

    Continuity: v = M/rho.A

    The Edz-Willz Limit:
    Fraction of unrecoverable energy = out/in
    = v^2/2gH
    = M^2/(2gH.rho^2.A^2)

    where v and A are at the outlet.


    Observations:
    1 fraction of unrecoverable energy is inversely proportional to H, as expected by Ed;

    2 unrecoverable energy is inversely proportional to A^2, so bigger turbines are more efficient, but more expensive --> economic trade off;

    3 the size of the inlet pipe doesn't appear as its kinetic energy is recoverable in the turbine;

    4 the shape or style of the turbine doesn't appear, could even be a waterwheel.

    5 As with the Betz limit, real world hydro plants will always have greater losses than the Edz-Willz limit;

    6 Pumping into storage has the same losses, all signs reversed.


    Examples:
    100m water to Pelton turbine, assume MgH= 1kW and 2" pipe:
    -> M= 1kg/s, A = 0.002m2, v =0.5m/s
    Unrecoverable energy= 0.5^2/(2.10.100) = 0.0125%

    Tidal barrage, height of reservoir above sea at low tide = 2H = 6m, assume MgH=10MW turbine in 10m dia duct
    -> M= 333333kg/s, A= 78m2, v= 4.2m/s
    Unrecoverable energy = 4.2^2/(2.10.3) = 30%


    A century from now, people will call this "The DJH Law".....
Add your comments

    Username Password
  • Format comments as
 
   
The Ecobuilding Buzz
Site Map    |   Home    |   View Cart    |   Pressroom   |   Business   |   Links   
Logout    

© Green Building Press