Talk:Tidal barrage

Efficiency in Energy Example
Why is the estimated efficiency of the power generation system in the worked example so low?

A typical efficiency for a Hydro-Turbine-Generator system would be in the region of 75% I suspect this is a transcription error, and the LOSSES were meant to be 30%.

Ref: Energy Studies, Shepherd & Shepherd, Chapter 9.

86.179.143.74 (talk) 17:48, 18 February 2012 (UTC)


 * The head is too variable. --Robertiki (talk) 10:59, 6 June 2014 (UTC)
 * I agree. The example is poorly worded and doesn't explain.  It should start by considering a more general form for the potential energy that takes account of the bathymetry of the enclosed basin, so we get surface areas S(h) at heights h above Chart Datum (the lowest low tide) that can vary because of sloping shores, mudflats etc. Then from E=mgh we get m is S(h) times density times an infinitesimal layer height of dh, multiplied by gh to get potential energy, and the overall potential energy is the sum of all these layers from Chart Datum to highest high tide, H - or integral of S(h)ρh dh from 0 to H.  Next introduce the idea that tides vary (reference article on tides), and that therefore the potential energy from a given tide will also vary according to the range between high and low tide (the integral runs between high and low tide for the cycle with h corrected for the low tide - S(h)ρ(h-L) dh from L to H1 where L is low tide and h1 high tide for the cycle, not between highest and lowest possible tides).
 * The next step is to point out that potential energy could only be realised if the water could be held back behind the barrage until low tide, when it is instantaneously released, generating a massive power spike. In reality, that is not possible nor desirable given a need for reasonably continuous power output: water flows through turbines approximately in accordance with the Orifice Equation (derived e.g. in Chapter 40 of Feynman Lectures on Physics) that limits the flow rate (m^3/sec) to CdA.sqrt(2gh) where Cd is an empirical coefficient that depends on details of design and performance of the turbines and their feed channels, A is the area of the turbine orifices, g is acceleration due to gravity, and h is now the head - the difference in water levels between the basin and the sea. While the water is flowing the tide level is also changing, affecting the head.
 * Energy from a particular tide is maximised by ensuring that as much water as possible has flowed out of the basin through the turbines before the head becomes too low to provide significant output. That involves matching the number and size of turbines (and hence the overall area of their orifices) with the size of the basin in an optimisation (since the highest tides only occur a small proportion of the time it is not necessarily worthwhile investing in extra capacity).  The result is that only a fraction of the simple potential energy is actually available.  This fraction has little to do with turbine efficiency.  Prandle derived some stylised simplified examples in his much cited paper, available e.g. here: https://doi.org/10.1680/maen.2009.162.4.147
 * Design of tidal barrage power schemes | Proceedings of the Institution of Civil Engineers - Maritime Engineering (icevirtuallibrary.com)
 * Later work showed that Prandle's assumptions tended to overestimate the energy available, particularly from flood mode operation, because turbine efficiency falls when used for reverse turbining, and because sloping bathymetry means that the water height inside the basin rises rapidly when the surface area is small, reducing the available head quickly from its starting value. Also, real world tides and water flows need more sophisticated modelling that reflects that water levels are uneven, and tides are influenced by the existence of a barrage among other factors. 92.18.134.237 (talk) 19:22, 25 September 2023 (UTC)