Narrow tank productivity in the 46 spectra. Henry Young 1982

Narrow tank productivity in the 46 spectra. Henry Young 1982

Narrow tank productivity in the 46 spectra. Henry Young 1982

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

Narrow tank freak wave tests 1978

Inflation from1998 to 2013 is 1.43

2013 – 1982 = 31 years

Cable length 43 km = 1333 km years.

At 10 km years per fault this is 133 faults.

From Lords Select Committee on European Communities HL paper 88 May 1988

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.

No names, no pack drill.

Unswept flow passage ≡ leaky pipe

Open flow field Power1

2 Area Velocity

316

27

Duct Power g Area Velocity Head

R.A. McAdam , G.T. Houlsby , M.L.G. OldfieldStructural and Hydrodynamic Model Testing of the Transverse

Horizontal Axis Water Turbine EWTEC 2011

Edinburgh vertical-axis, variable-pitch with rim power take off.

EWTEC Patras 1998

Edinburgh vertical-axis, variable-pitch with rim power take off.

EWTEC Patras 1998

Cells are 1 minute of arc lat. 1.5 minutes long = 2.617 km2.

Power = 6.165 TW x CfCourtesy Proudman Labs

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.

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 = ?

Pentland bed stills. P Hayes. Fisheries Research Aberdeen 2006-8

68 mm bob

70 60 50 40 30 20 10 0 10 20 30 40 50 60 700

0.2

0.4

0.6

0.8

1

1.2Downstream force on a 140 diameter rotor as a fraction of ideal

1.2

0

fdsi 4

FDSi 4

DR

2

DR2

Xi

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

Google images

Speed up x 30Range up x 6000

Payload up x 20,000Cost per ton-mile down ÷ 100

Google images

Something for the simpletons