Hydropower - Environmental Impacts of Renewable Energy - page 127

Reservoir 1000 × 1000 m

Mean depth
--------------- 10 m ---------------

300 m (984.25 ft)

Penstock

Power station

FIGURE 4.4 Schematic representation of a hydropower station. (Adapted from Tovey, N.K.,
ENV-2E02 Energy Resources (2004-2005) , University of East Anglia, Norwich, UK, 2005.)

Because the potential energy stored in the reservoir is converted into kinetic
energy at the inlet to the water turbine, we can equate:

m × g × h = mV 2 (4.18)

where:
m = Mass of water.
g = Acceleration due to gravity (can be taken as 10 m/s 2 in most applications).
h = Head difference.
V = Velocity of water at the inlet.

■ Example 4.13 *
Problem: A reservoir has an area of 1 km 2 , and the difference between the crest of
the dam and the inlet to a hydro station is 10 m (see Figure 4.4). The station runs at an
overall efficiency of 80% and is situated 305 m below the crest of the dam. The rain-
fall is 1000 mm per annum, the catchment area of the reservoir is 10 times the area
of the reservoir, and the run is 50%. What should the rated output of the turbine be if
its maximum output is designed to be 5 times the mean output at the site? What is the
maximum time the station could operate at full power during a sustained drought?

Solution: Mean head between maximum and minimum levels = 305 - 10/2 = 300 m.
Average annual flow into the reservoir is equal to 50% of 10 times the area multiplied
by the rainfall:

Average annual flow = 0.5 × 10 × 1000 × 1000 × 1 = 5,000,000 m 3

* Adapted from Tovey, N.K., ENV-2E02 Energy Resources (2004-2005) , University of East Anglia,
Norwich, UK, 2005, p. 109.