The Tidal Resource
In this section we will estimate the total power available to the UK from the tidal resource, then break this down into separate considerations of tidal range and tidal stream technologies - what can they produce, and on what scale would they have to be implemented?
1 Power Available in the Tides
The national power resource from the tides cannot exceed the power delivered to UK waters by the tides, whatever technology is used. The way to find the power delivered by the tides is to remember that they are waves and to work out the energy transported per second per metre length of wave front. When we know this we can multiply by the width of the tidal wavefront to find the total power delivered. This is the same method as used to estimate the energy delivered by ordinary wind generated waves.
To do this, we work out the energy present in one wavelength of the wave and divide by its period. Think about it - to find the speed of a wave you divide its wavelength by its period, so to find the “energy speed” of a wave you divide the energy per wavelength by its period. This is the same as working out the energy density of the wave - the energy per metre in the direction of the wave travel - and multiplying by the wave velocity. Hence the power delivered by a tidal wave (or any other kind of wave) per metre across the wave front is the energy density times the wave velocity.
The energy per wave length is the sum of the potential and kinetic energies of all the water particles in that section. A general result of wave physics is that the average value of the kinetic energy in a wave is equal to the average value of the potential energy. Thus to find the total energy, we can find the potential energy and double it. The potential energy of a wave per wavelength per unit width of wavefront is found to be:
\[\begin{equation} E_\mathrm{P}\ =\ \dfrac{1}{4}\rho gh^2 \lambda \tag{1.1} \end{equation}\]
A proof of this is will be found here (when I have written it - try and prove this yourself, if interested!).
So if we double this and divide by the period T we get
\[\begin{equation} \mathrm{Power}\ =\ \dfrac{1}{2}\rho gh^2 \lambda\ \times\ \dfrac{1}{T} \tag{1.2} \end{equation}\]
But \[\frac{\lambda}{T}\] is the wave speed v. So we can write as
\[\begin{equation} \mathrm{Power}\ =\ \dfrac{1}{2}\rho gh^2 v \tag{1.3} \end{equation}\]
This is generally true for all waves.
We have seen that tidal waves are shallow water waves - their wavelength (half the circumference of the earth) is very definitely much greater than the depth of any water they might travel through. Thus their speed is given by \(v\ =\sqrt{gd}\) where d is the water depth. If we substitute this into (3) in place of \(\frac{\lambda}{T}\) we get
\[\begin{equation} \text{Power} = \dfrac{1}{2}\rho gh^2 \lambda \sqrt{gd} = \dfrac{1}{2}\rho g^\frac{3}{2} \sqrt{d} h^2 \tag{1.4} \end{equation}\]
For wind driven waves, which are deep water waves with wave speed given by \(v\ =\ \sqrt{\frac{g\lambda}{2\pi}}\), the equivalent expression is
\[\begin{equation} \mathrm{Power}\ =\ \dfrac{1}{2}\rho gh^2 \lambda\ \sqrt{\frac{g\lambda}{2\pi}}\ =\ \dfrac{1}{2}\rho g^\frac{3}{2}\sqrt{\frac{\lambda}{2\pi}}h^2 \tag{1.5} \end{equation}\]
What do equations (1.4) and (1.5) tell us?. Try some plausible numbers. In near shore waters we might have depth d = 100m and tidal wave height about 1 -2 m. The wave velocity -the speed at which the tide advances - is then about 30 ms-1. The water particle velocity is then about 0.3 -0.6 ms-1. For a wave height of 1 m, the power per metre width of wave front is, from (1.4) 155 kW. If the tidal wave height is 2m, the power is 600 kW. These powers are much greater than those available from wind driven waves. These may have a wavelength of 70 m or so. Equation (1.5) shows that their power density for wave heights of 1 or 2 m would be 50 kW m-1 or 200 kW m-1. hence tidal waves, which have similar wave heights to wind driven waves, deliver a power per metre length of wavefront which is about three times greater.
This model is simple and has ignored such effects as the Coriolis force, which results from the spin of the earth, or the impact of reflected waves.
However if the power density of 100-600 kW per metre length of tidal wave front is multiplied by the total length of wave front approaching the UK from the south west, north west and north, a total length of some 2000 km, we should not be surprised to find that the total power available has been measured to be 200 - 300 GW (Cartwright et al, 1980).
1.1 Effect of Shoaling
If the water gradually becomes shallower, with minimal reflection or absorption, then the wave power will remain constant. From (1.4) we see that this means that the product \(\sqrt{d}h^2\) remains constant. This means that the tidal wave height increases as the depth reduces, but not rapidly: \(h\propto d^{\frac{1}{4}}%\) to that if the depth reduces by a factor 100, such as on going from the deep ocean to near shore waters, the tidal wave height will increase by a factor 100\(\frac{1}{4}\) \(\approx\) 3.
1.2 Power Density of Tidal Pools
if an artifically created tidal pool is filled rapidly during flood tide and emptied rapidly during ebb tide, power can be generated as the water flows in both directions. The change in potential energy of the water is mgh, where h is the change in height of the centre of mass, which is half the tidal range. This happens every six hours. The mass per unit area of the pool is \(\rho\ \times\ 2h\), so that the power available per unit area of pool is
\[\begin{equation} P\ =\ \dfrac{2 \rho gh^2}{6\ \mathrm{hours}} \tag{1.6} \end{equation}\]
If h = 5 m then we find that P = 23 W m-2. A tidal range of 2 h = 10 m is often exceeded in the Severn Estuary. Thus to generate 1 GW on average the area of tidal pool required is 1 x 109 / 23 \(\approx\) 43 km2, which is about the area of a circular pool of diameter 7.5 km. If power is generated only on ebb or only on flow, then the required area for an average of 1 GW is doubled to 86 km2. The area of the Severn estuary behind the site of the proposed Cardiff-Weston barrage is about 550 km2.
1.3 The UK tidal resource
Areas with good tidal stream resource
Figure 1.1: The UK tidal stream resource (SDC 2007)
Peak tidal flow
Figure 1.2: In dark red, regions with peak tidal flow speeds in excess of 1 ms-1 (Mackay 2008)
Potential barrage sites
Figure 1.3: The most likely tidal barrage sites in the UK