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Narrow tank productivity in the 46 spectra. Henry Young 1982

Explore the potential of narrow tanks in energy production, assessing factors like capacity, demand, and fault analysis. Learn about wave energy challenges, turbine testing, tidal stream design flaws, and marine vegetation impact. Discover methods to enhance energy efficiency, reduce costs, and optimize resources.

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Narrow tank productivity in the 46 spectra. Henry Young 1982

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  1. Narrow tank productivity in the 46 spectra. Henry Young 1982

  2. Narrow tank productivity in the 46 spectra. Henry Young 1982

  3. 35 kW / m x 400 km = 14 GW Installed capacity (2009) = 12.2 GW Planned 19 GW Peak demand ~ 6 GW 14 GW x 8760 hours = 122 TWH Annual Scottish demand = 29 TWH Ratio 4.2

  4. Narrow tank freak wave tests 1978

  5. Inflation from1998 to 2013 is 1.43

  6. From Lords Select Committee on European Communities HL paper 88 May 1988 2013 – 1982 = 31 years Cable length 43 km = 1333 km years. At 10 km years per fault this is 133 faults.

  7. What’s wrong with wave energy? Control by nuclear people >False information Impatient investors >Unreliable components > Ignorance of stresses Waste of sea front by solo devices Inadequate installation equipment Preference for vertical motion.

  8. No names, no pack drill.

  9. Unswept flow passage ≡ leaky pipe

  10. R.A. McAdam , G.T. Houlsby , M.L.G. Oldfield Structural and Hydrodynamic Model Testing of the Transverse Horizontal Axis Water TurbineEWTEC 2011

  11. EWTEC Patras 1998 Edinburgh vertical-axis, variable-pitch with rim power take off.

  12. EWTEC Patras 1998 Edinburgh vertical-axis, variable-pitch with rim power take off.

  13. Cells are 1 minute of arc lat. 1.5 minutes long = 2.617 km2. Power = 6.165 TW x Cf Courtesy Proudman Labs

  14. From Black and Veatch 2011. Using values for the Pentland Firth U = 3m/s, ρ = 1025 kg/m3, channel length = 23 km, channel width = 10 km in combination with a more appropriate bed friction coefficient CD = 0.0015 energy dissipated due to bed friction averaged over a tidal cycle calculated is 4.05 GW.

  15. Laminaria Hyperborea (kelp) are found along the edges of the Pentland Firth at depths up to 30 m. Length can reach 3.5 metres. Cf = ?

  16. 68 mm bob Pentland bed stills. P Hayes. Fisheries Research Aberdeen 2006-8

  17. What’s wrong with tidal stream designs? Ignorance of flow impedance. > Wrong energy input Open flow field equation in a ‘duct’ > Swept area too small Wrong blade support > Bearing load levered up Cramped power compartment Tip vortex losses

  18. Google images

  19. Google images Speed up x 30 Range up x 6000 Payload up x 20,000 Cost per ton-mile down ÷ 100

  20. Something for the simpletons

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