8.4.21 Describe the principle of operation of an oscillating water column (OWC) ocean-wave energy converter.

Wave energy is one of the ways that we generate electricity today. Waves form through the current of winds that blow on the ocean, and these winds are created by difference in air pressure created by solar energy. Therefore, wave energy is still one of the ways in which we harness solar/wind energy to generate electricity.

Although there are many ways in which wave energy can be used to generate electricity, an oscillating water column is one of the commonly used methods today.
An OWC is built next to a shore, with a wave capture chamber that is used to trap the air and the water of the surface below the walls of the chamber. As the wave pushes a column of water into the wave capture chamber, the water level increases in the chamber.
The excess air occupying the chamber is pushed out of a tunnel in which a Wells turbine is installed.
A wells turbine is designed specially so that it will spin and generate electricity when spun either ways.
As the retreating wave drags the water away from the wave capture chamber, the air pressure inside the chamber significantly decreases and forces a current of air to come back through the tunnel to replace the volume that the water took inside the chamber. As this current of air travels through the tunnel, it spins the wells turbine again, generating electricity.
Therefore, the OWC generates electricity using a wells turbine, by causing a current of air to pass through the tunnel every time a wave pushes water into the chamber and every time the wave retreats.

(Tim Kirk, 2007, p 72)

8.4.22 Determine the power per unit length of a wavefront, assuming a rectangular profile for the wave.

To calculate the power per unit length of a wave front, we need to first find the amount of potential energy that a column of wave loses as it falls down to normal sea level. If we simplify the shape of waves into rectangles like in the diagram above, we can find the amount of potential energy lost when the shaded section of the wave falls back into the "trough" that is below the midline.
The formula to calculate gravitational potential energy is m*g*h, with m=mass, g=gravity, h=change in height
The volume of the shaded region on the wave is A*L* (λ/2)
The mass of the shaded region of the wave is A*L* (λ/2)*p, with p being the density of the water.
Therefore, the transfer of energy with 1 column of wave is A*L* (λ/2)*p*g*A = A^2*L* (λ/2)*p*g

And to find the number of waves that pass a point per unit time f= v/λ, with v representing the velocity of the wave
Therefore the transfer of energy per unit time (power)= A^2*L* (λ/2)*p*g*(v/λ) = 1/2*A^2*L*p*g*v
And so the maximum power per unit length = 1/2*A^2*p*g*v

(Tim Kirk, 2007, p 72)

8.4.23 Solve problems involving wave power.

Calculate the maximum power per unit length available from an oscillating water column under the following conditions.
Average amplitude of waves = 1.50 m
Average velocity of waves = 3.70 m s-1
Density of water = 1.00 x 103 kg m-3

To solve this problem, simply use the formula we have found above.

## Wave power

## 8.4.21 Describe the principle of operation of an oscillating water column (OWC) ocean-wave energy converter.

Wave energy is one of the ways that we generate electricity today. Waves form through the current of winds that blow on the ocean, and these winds are created by difference in air pressure created by solar energy. Therefore, wave energy is still one of the ways in which we harness solar/wind energy to generate electricity.

Although there are many ways in which wave energy can be used to generate electricity, an

oscillating water columnis one of the commonly used methods today.An OWC is built next to a shore, with a

wave capture chamberthat is used to trap the air and the water of the surface below the walls of the chamber. As the wave pushes a column of water into the wave capture chamber, the water level increases in the chamber.The excess air occupying the chamber is pushed out of a tunnel in which a

Wells turbineis installed.A wells turbine is designed specially so that it will spin and generate electricity when spun either ways.

As the retreating wave drags the water away from the wave capture chamber, the air pressure inside the chamber significantly decreases and forces a current of air to come back through the tunnel to replace the volume that the water took inside the chamber. As this current of air travels through the tunnel, it spins the wells turbine again, generating electricity.

Therefore, the OWC generates electricity using a wells turbine, by causing a current of air to pass through the tunnel every time a wave pushes water into the chamber and every time the wave retreats.

(Tim Kirk, 2007, p 72)

## 8.4.22 Determine the power per unit length of a wavefront, assuming a rectangular profile for the wave.

To calculate the power per unit length of a wave front, we need to first find the amount of potential energy that a column of wave loses as it falls down to normal sea level. If we simplify the shape of waves into rectangles like in the diagram above, we can find the amount of potential energy lost when the shaded section of the wave falls back into the "trough" that is below the midline.

The formula to calculate gravitational potential energy is m*g*h, with m=mass, g=gravity, h=change in height

The volume of the shaded region on the wave is A*L*

(λ/2)

The mass of the shaded region of the wave is A*L*

(λ/2)*p, with p being the density of the water.

Therefore, the transfer of energy with 1 column of wave is A*L*

(λ/2)*p*g*A = A^2*L*

(λ/2)*p*g

And to find the number of waves that pass a point per unit time f= v/λ, with v representing the velocity of the wave

Therefore the transfer of energy per unit time (power)= A^2*L*

(λ/2)*p*g*(v/λ) = 1/2*A^2*L*p*g*v

And so the maximum power per unit length =

1/2*A^2*p*g*v(Tim Kirk, 2007, p 72)

## 8.4.23 Solve problems involving wave power.

Calculate the maximum power per unit length available from an oscillating water column under the following conditions.

Average amplitude of waves = 1.50 m

Average velocity of waves = 3.70 m s-1

Density of water = 1.00 x 103 kg m-3

To solve this problem, simply use the formula we have found above.

A=1.50,

p=1.00 x 10^3

g=9.81

v=3.70

1/2*1.50^2*1000*9.81*3.70= 40834 Watts=40.8 kW