go into this model of tidal power in some detail because most of the offi-

cial estimates of the UK tidal resource have been based on a model that I

believe is incorrect.

Figure G.2 shows a model for a tidal wave travelling across relatively

shallow water. This model is intended as a cartoon, for example, of tidal

crests moving up the English channel or down the North Sea. It’s impor-

tant to distinguish the speed *U* at which the water itself moves (which

might be about 1 mile per hour) from the speed v at which the high tide

moves, which is typically 100 or 200 miles per hour.

The water has depth *d*. Crests and troughs of water are injected from

the left hand side by the 12-hourly ocean tides. The crests and troughs

move with velocity

*v* = √gd.

(G.2)

We assume that the wavelength is much bigger than the depth, and we

neglect details such as Coriolis forces and density variations in the water.

Call the vertical amplitude of the tide *h*. For the standard assumption

of nearly-vorticity-free flow, the horizontal velocity of the water is

near-constant with depth. The horizontal velocity *U* is proportional to the

surface displacement and can be found by conservation of mass:

*U* = *vh*/*d*.

(G.3)

If the depth decreases, the wave velocity *v* reduces (equation (G.2)). For the

present discussion we’ll assume the depth is constant. Energy flows from

left to right at some rate. How should this total tidal power be estimated?

And what’s the *maximum* power that could be extracted?

One suggestion is to choose a cross-section and estimate the average

*flux of kinetic energy* across that plane, then assert that this quantity repre-

sents the power that could be extracted. This kinetic-energy-flux method

was used by consultants Black and Veatch to estimate the UK resource. In

our cartoon model, we can compute the total power by other means. We’ll

see that the kinetic-energy-flux answer is too small by a significant factor.

The peak kinetic-energy flux at any section is

*K*_{BV} = ^{1}⁄_{2}*ρAU*^{3},

(G.4)

where *A* is the cross-sectional area. (This is the formula for kinetic energy

flux, which we encountered in Chapter B.)

The true total incident power is not equal to this kinetic-energy flux.

The true total incident power in a shallow-water wave is a standard textbook

calculation; one way to get it is to find the total energy present in one

wavelength and divide by the period. The total energy per wavelength is

the sum of the potential energy and the kinetic energy. The kinetic energy

happens to be identical to the potential energy. (This is a standard feature

of almost all things that wobble, be they masses on springs or children