# Homework Help: Natural frequency of an Oscillating water column device

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1. Feb 22, 2017

### Dimis414

1. The problem statement, all variables and given/known data
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?

2. Relevant 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).

3. 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.

2. Feb 24, 2017

### haruspex

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. Mar 1, 2017

### Dimis414

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. Mar 1, 2017

### haruspex

That doesn't seem to match any descriptions I can find online. A diagram of the system would help.

5. Mar 3, 2017

### Dimis414

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.

6. Mar 3, 2017

### haruspex

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.
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: Mar 3, 2017
7. Mar 6, 2017

### Dimis414

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).

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. Mar 6, 2017

### haruspex

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.