travelling waves in water of depth *d* that is shallow compared to the wave-

length of the waves (figure G.2). The power per unit length of wavecrest

of shallow-water tidal waves is

*ρg*^{3/2}√*d**h*^{2}/2.

(G.1)

Table G.3 shows the power per unit length of wave crest for some plausible

figures. If *d* = 100 m, and *h* = 1 or 2 m, the power per unit length of wave

crest is 150 kW/m or 600 kW/m respectively. These figures are impressive

compared with the raw power per unit length of ordinary Atlantic deep-

water waves, 40 kW/m (Chapter F). Atlantic waves and the Atlantic tide

have similar vertical amplitudes (about 1 m), but the raw power in tides is

roughly 10 times bigger than that of ordinary wind-driven waves.

Taylor (1920) worked out a more detailed model of tidal power that

includes important details such as the Coriolis effect (the effect produced

by the earth’s daily rotation), the existence of tidal waves travelling in the

opposite direction, and the direct effect of the moon on the energy flow in

the Irish Sea. Since then, experimental measurements and computer models

have verified and extended Taylor’s analysis. Flather (1976) built a

detailed numerical model of the lunar tide, chopping the continental shelf

around the British Isles into roughly 1000 square cells. Flather estimated

that the total average power entering this region is 215 GW. According

to his model, 180 GW enters the gap between France and Ireland. From

Northern Ireland round to Shetland, the incoming power is 49 GW. Be-

tween Shetland and Norway there is a net loss of 5 GW. As shown in

figure G.4, Cartwright et al. (1980) found experimentally that the average

power transmission was 60 GW between Malin Head (Ireland) and Florø

(Norway) and 190 GW between Valentia (Ireland) and the Brittany coast

near Ouessant. The power entering the Irish Sea was found to be 45 GW,

and entering the North Sea via the Dover Straits, 16.7 GW.

This section, which can safely be skipped, provides more details behind

the formula for tidal power used in the previous section. I’m going to

h(m) |
ρg^{3/2}√dh^{2}/2(kW/m) |
---|---|

0.9 | 125 |

1.0 | 155 |

1.2 | 220 |

1.5 | 345 |

1.75 | 470 |

2.0 | 600 |

2.25 | 780 |