Natural frequency of an Oscillating water column device

In summary, the heave response of an OWC device containing a cylindrical body which is connected to the OWC chamber with brackets and moving as a unit is plotted in Fig.2. Assuming an array of OWC devices floating as a platform then the platform's heave response is plotted in Fig.3. It is observed that there are three maxima near 1.3, 1.8, and 0.9 rad/s, which are probably due to the resonance of the air pressure inside the OWC. However, it is not possible to verify this by maths.
  • #1
Dimis414
6
0

Homework Statement


Assuming that an OWC device is floating/moored in monochromatic waves. When there isn't air pressure distribution inside (atmospheric condition) the device behaves as an undambed body (with the oscillating chamber open to the atmosphere). When there is air pressure oscillation inside the OWC which is the natural frequency of the device?

Homework Equations


For an undamped moored body, the equation of heave motion is:

(-iw(M+A33)+B33+i/w(C33+K33))*U=Fz

where c33 is the restoring spring coeff equals ρgAwl; K33 is the vertical mooring stiffness; M is the body's mass; A33 and B33 is the heave added mass and damping coeff; U is the vertical velocity; Fz is the exciting wave forces and w the wave frequency. Thus the natural frequency in heave motion equals to w=sqrt[(c33+K33)/(M+A33)]

For a moored OWC device the equation of motion equals to:

(-iw(M+A33)+B33+i/w(C33+K33))*U+F*P=Fz

Here F is the complex damping coeff due to air pressure oscillation inside the OWC and P is the air pressure head.

The air pressure term, P, from equation of volume flows, can be written as:

P=(Qd+Qr*U )/(Λ-Qp), where Qd is the exciting volume flow (due to diffraction problem); Qr is the motion radiation volume flow (due to motion radiation problem) and Qp is the pressure radiation volume flow (due to pressure radiation problem or 7th degrees of freedom radiation problem).

The Attempt at a Solution


Assuming that the air inside the OWC is incompressible, which is the natural frequency of the OWC device? It should be different from the undamped body, but can't figure out how to calculate it.
 
Physics news on Phys.org
  • #2
Dimis414 said:
P=(Qd+Qr*U )/(Λ-Qp), where Qd is the exciting volume flow (due to diffraction problem); Qr is the motion radiation volume flow (due to motion radiation problem) and Qp is the pressure radiation volume flow (due to pressure radiation problem or 7th degrees of freedom radiation problem).
I don't begin to understand any of that, so I'm probably out of my depth here, but why is it not simply a matter of treating the airspace as adiabatically compressed? It does not make sense to me to treat it as incompressible, whereas its inertia is so small that I do not think you have to worry about its flow characteristics.
 
  • #3
The issue that we are facing is that when we plot the heave response as a function of the wave frequency, when the oscillating chamber is closed in the upper (no air is going out or in), we observe three maxima. The first is near 0.9 rad/s and it probably occurs due to the first anti-symmetric mode of resonant motions of the water trapped in the annulus between the internal and external cylinders of the OWC’s. The third one is near 1.8 rad/s (its natural frequency). The problem is with the second peak, near 1.3 rad/s, where is probably the resonance of the air pressure inside the OWC. But we can't verify this by maths.
 
  • #4
Dimis414 said:
annulus between the internal and external cylinders
That doesn't seem to match any descriptions I can find online. A diagram of the system would help.
 
  • #5
About natural resonance frequencies relevant articles are: Lateral sloshing in mooving containers by Sandor Silverman and Norman Abramson (where it is described how to obtain the natural frequencies of an open chamber OWC) and among others, Hydrodynamic analysis of three–unit arrays of floating annular oscillating–water–column wave energy converters; Konispoliatis et al. APOR 61 (2016).Assuming that we are examine an OWC device containing a coaxial cylindrical body which is connected to the OWC chamber with brackets and moving as a unit as in Figure 1 (with the air chamber closed no air in and out). The device is moored with TLP tethers. Then its heave response is plotted in Fig.2. Assuming an array of i.e. the above three OWC devices floating as a platform then the platform's heave response is plotted in Fig.3.In the latter figure you can observe the three pro-mentioned maxima. The problem is with the second peak, near 1.3 rad/s, where is probably the resonance of the air pressure inside the OWC. But we can't verify this by maths.
1.jpg
2.jpg
3.jpg
 
  • #6
This situation strikes me as analogous to partial reflections as light goes through media of different indices.
The turbine blades are a partial barrier. There will be a natural frequency of the airspace below the blades, and another for the airspace above the blades, and a third for the combination, maybe.

But let's start with something much simpler. Consider a half closed cylinder inverted, partially submerged and held fixed.
A= cross-sectional area
ρ= density of water
H= cylinder length (height)
x(t)= airspace height
P(t)= pressure in airspace.
Assume that at equilibrium x=x0=height of cylinder top above surrounding water, P=P0=atmospheric.
Assume adiabatic compression/expansion.
Crudely, I treat the oscillating mass of water as being that of the volume currently inside the cylinder. I get
##\rho(h-x)\ddot x+\rho g(x-x_0)=P_0\left(\left(\frac{x_0}{x}\right)^{\gamma}-1\right)##
For small perturbations:
##\omega^2=\frac 1{h-x_0}\left(g+\frac{P_0\gamma}{x_0\rho}\right)##
With e.g. P0=105, γ=1.4, ρ=1000, g=10, h=2, x0=1, that gives ω≈12 rad/s. (But that seems far too much.)

What are the dimensions of the device?
 
Last edited:
  • #7
The device's dimensions are:

Radius of the coaxial cylindrical body: 7m

Draught of the coaxial cylindrical body: 20m

Radius of the oscillating chamber: 14m

Thickness of the oscillating chamber: 1.5m

Draught of the oscillating chamber: 8m

Considered your suggestion for half closed air duct, and assuming that the air turbine characteristics equals to optimum value as in Evans & Porter 1996 () work the vertical response of the platform does not seem to appear the second maxima. Thus we suppose? that the second maxima (near 1.3 rad/s) occurs due to air pressure oscillation when the air chamber is closed (no air goes inside or outside).

4.jpg

D.V. Evans, R. Porter, Efficient calculation of hydrodynamic properties of O: W. C. type devices, J. Offshore Mech. Arct. Eng. 119 (4) (1996) 210–218.
 
  • #8
Dimis414 said:
The device's dimensions are:

Radius of the coaxial cylindrical body: 7m

Draught of the coaxial cylindrical body: 20m

Radius of the oscillating chamber: 14m

Thickness of the oscillating chamber: 1.5m

Draught of the oscillating chamber: 8m

Considered your suggestion for half closed air duct, and assuming that the air turbine characteristics equals to optimum value as in Evans & Porter 1996 () work the vertical response of the platform does not seem to appear the second maxima. Thus we suppose? that the second maxima (near 1.3 rad/s) occurs due to air pressure oscillation when the air chamber is closed (no air goes inside or outside).

View attachment 114163
D.V. Evans, R. Porter, Efficient calculation of hydrodynamic properties of O: W. C. type devices, J. Offshore Mech. Arct. Eng. 119 (4) (1996) 210–218.
Not entirely sure of your terminology, but I think you are saying that in mine H=28, x0=20. For that, my formula gives ω=√2.1 rad/s.
 

Related to Natural frequency of an Oscillating water column device

1. What is the natural frequency of an oscillating water column device?

The natural frequency of an oscillating water column device refers to the frequency at which the device oscillates back and forth due to the movement of waves. It is dependent on the size, shape, and design of the device, as well as the characteristics of the waves in the specific location where the device is installed.

2. How is the natural frequency of an oscillating water column device calculated?

The natural frequency of an oscillating water column device can be calculated using a mathematical formula that takes into account the device's dimensions, the density of the water, and the acceleration due to gravity. However, this calculation can be complex and is usually performed by engineers or researchers using specialized software.

3. Why is the natural frequency important for the performance of an oscillating water column device?

The natural frequency is important because it determines the efficiency of the device in converting the energy of the waves into usable electricity. If the natural frequency of the device does not match the frequency of the waves, the device will not be able to effectively capture the energy and convert it into electricity.

4. How can the natural frequency of an oscillating water column device be optimized?

The natural frequency of an oscillating water column device can be optimized by adjusting the dimensions and design of the device to match the frequency of the waves in the specific location where it will be installed. This can be achieved through computer simulations and testing, as well as on-site adjustments and modifications.

5. What are the factors that can affect the natural frequency of an oscillating water column device?

The natural frequency of an oscillating water column device can be affected by various factors, including the size and shape of the device, the depth and density of the water, and the characteristics of the waves in the specific location. Other factors such as changes in water level, wind, and the presence of other structures or obstacles can also impact the natural frequency of the device.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
19
Views
3K
Replies
25
Views
3K
  • Classical Physics
4
Replies
131
Views
5K
Replies
1
Views
1K
  • General Discussion
Replies
2
Views
3K
  • Other Physics Topics
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
3K
Back
Top