UCCS Renewable Energy PES/ENSC 2500
Chapter 13 Tidal Energy
name: _________________________________
Problem 13.1
3
Calculate the total kinetic energy (in MJ and in kWh) of a 1-m parcel of seawater moving with a velocity of 1
m/s.
The kinetic energy may be written in terms of the density and volume as
3
Using the density of sea water as 1025 kg/m and the values from the problem gives:
In Joules:
or
In kWh:
Problem 13.2
What is the total tidal energy available (at 100% efficiency) during the falling tide from a basin of area 100 km
with a tidal range of 8 m?
2
The potential energy is given as E = mgh/2, where the mass is given in terms of volume as m = ρV and volume is
given in terms of area and height as V = Ah.
Substituting into the expression for energy gives
E = Ah ρg/2 =
2
Problem 13.4
A water turbine 15 m in diameter is placed in a channel with a tidal current moving with a velocity of 3.5 m/s.
Estimate the power produced by the turbine.
The power is given by
P/A = Cρv /2 or P = πr Cρv /2
3
2
3
Using r = 7.5 m, C (typical) = 0.35 and ρ = 1025 kg/m , the power is found to be:
3
Watts:
Problem 13.5
What is the power available from an underwater turbine with a coefficient of performance of 0.4 that is 10 m in
diameter in a current of seawater traveling at a velocity of 2.0 m/s?
Use the values of C = 0.4, ρ = 1025 kg m and v = 2 m/s.
/
3
3
P/A = Cρv =
The total power is, therefore,
P = πr × ( Cρv ) =
2
3
Problem 13.6
For a tidal current with a velocity of 1.5 m/s in a channel that is 0.5 km wide and 28 m deep (on average), calculate
the mass of water per unit time moving through the channel.
Volume per unit time is area multiplied times velocity then
Mass per unit time is given by multiplying this result times density or
Problem 13.7
Calculate the average current velocity that is necessary for an 8-m diameter water turbine with a coefficient of
performance of 0.3 to generate 1 MW.
3
P = CAρv
Solving for velocity gives
Substituting values from the problem