tidal amplitude
(half-range) h
optimal boost
height b
with pumping
without pumping
1.0 6.5 3.5 0.8
2.0 13 14 3.3
3.0 20 31 7.4
4.0 26 56 13

that generation has an efficiency of εg = 0.9 and that pumping has an
efficiency of εp = 0.85. Let the tidal range be 2h. I’ll assume for simplicity
that the prices of buying and selling electricity are the same at all times, so
that the optimal height boost b to which the pool is pumped above high
water is given by (marginal cost of extra pumping = marginal return of
extra water):

Defining the round-trip efficiency ε = εgεp, we have

For example, with a tidal range of 2h = 4 m, and a round-trip efficiency of
ε = 76%, the optimal boost is b = 13 m. This is the maximum height to
which pumping can be justified if the price of electricity is constant.

Let’s assume the complementary trick is used at low tide. (This requires
the basin to have a vertical range of 30 m!) The delivered power per unit
area is then

where T is the time from high tide to low tide. We can express this as the
maximum possible power density without pumping, εg2ρgh2/T, scaled up
by a boost factor

which is roughly a factor of 4. Table G.10 shows the theoretical power
density that pumping could deliver. Unfortunately, this pumping trick
will rarely be exploited to the full because of the economics of basin con-
struction: full exploitation of pumping requires the total height of the pool
to be roughly 4 times the tidal range, and increases the delivered power
four-fold. But the amount of material in a sea-wall of height H scales as
H2, so the cost of constructing a wall four times as high will be more than
four times as big. Extra cash would probably be better spent on enlarging
a tidal pool horizontally rather than vertically.

The pumping trick can nevertheless be used for free on any day when
the range of natural tides is smaller than the maximum range: the water

Table G.10. Theoretical power density from tidal power using the pumping trick, assuming no constraint on the height of the basin’s walls.