Download UCCS PES/ENSC2500 Chapter 12 Wave Energy name: ______________________________

UCCS PES/ENSC2500
Chapter 12 Wave Energy
name: ______________________________
Problem 12.2
A small boat with two occupants and a 20-hp outboard motor (total mass = 500 kg) moves up and down in a wave
with a height of 1 m and a period of 10 seconds.
Calculate the ratio of the average wave power lifting the boat from the trough of the wave to the crest of the wave
to the power available from the motor.
The energy difference between the top and the bottom of the wave is:
E = mgh =
This gives an equivalent power of:
P = E/t =
Converting this to horsepower gives:
Problem 12.3
An ocean wave has a velocity of 8.5 m/s.
What are its period and wavelength?
Solving for T in terms of v give:s
The wavelength of the wave is related to its period as:
Problem 12.4
An ocean wave has a height of 2.5 m and a period of 10 s.
What is the total power available from 20 m of wave front for this wave?
The height and period are related to the power per unit length as
For 20 m of wave front this gives a total power available of
Problem 12.5
The average wave power available in Hawaii is 100 kW/m.
For waves with a period of 10 s, what is the average wave height?
The height and period are related to the power per unit length as
Where A is the amplitude (A = H/2) and ρ = 1025 kg/m .
3
Then solving for the amplitude gives:
5
Using P/l = 10 W/m for a period of T = 10 s, the amplitude is found to be
2
–3
Note that W = kg m s so the above units give amplitude in m.
From this result, the height is
H=
Problem 12.7
In deep-water, tsunamis have a relatively small height (typically 2 m) and a very long period (typically 30
minutes).
The amplitude becomes larger as they pile up when they reach shallow water.
For a tsunami in deep water, calculate the energy per meter of wave front (in J/m).
Compare this with the energy per meter of width of a 2-m wide, 1500-kg automobile traveling at 120 km/h.
This comparison emphasizes the damage that can be caused by a tsunami.
From equation (12.6)
Using the values above gives (noting that A = H/2)
For the vehicle the kinetic energy is
2
E = mv /2
Converting 120 km/h to m/s gives
Calculating the energy gives
For the 2 m wide vehicle this is E /l = _________________________J/m.
The tsunami provides about _________________ times as much energy per unit length/width.
Problem 12.8
A wave travels at 10 m/s and has a height of 2 m.
If it is incident on a wave energy device that generates electricity from wave energy with an average efficiency
of 20%, how large would the device have to be to generate 1 MWe?
The height and period are related to the power per unit length as
Where A is height/2.
so
Then using the values above gives
At 20% efficiency this is
Therefore 1 MWe requires