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My question really pertains to improving the oscillating water column device to generate greater electricity. The way the current system works, when a wave with a crest hits the wall, the water inside also increases in height and expels air through a Wells turbine. However, what if instead of just having a straight inclined plane wall, the wall was enveloped in a V-shape so that the wave can be directed to the center? Would that create a wave with a greater height, and thus also increase the height in the inside and expel more air?

For instance, what if this enter image description here had a V-shaped opening to in a way horizontally compress and merge the waves together before they hit the wall?

enter image description here

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  • $\begingroup$ it is unclear which V shape you are referring to in the picture $\endgroup$
    – jsotola
    Aug 31, 2022 at 4:15
  • $\begingroup$ @jsotola I've made it clearer. $\endgroup$
    – ARJ
    Aug 31, 2022 at 19:19
  • $\begingroup$ now the inlet looks blocked ... are you trying to draw a funnel? ... is it a funnel if viewed from above? $\endgroup$
    – jsotola
    Aug 31, 2022 at 19:41
  • $\begingroup$ @jsotola The inlet is at the bottom and its still open. Those are two rigid flaps that serve a similar purpose to a funnel. However, funnels are usually circular while these are V-shaped. $\endgroup$
    – ARJ
    Aug 31, 2022 at 19:52

2 Answers 2

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Your idea has merit, and the fundamental principle involved is called impedance matching. Here is how it works:

For any power source we wish to harness, that source can be characterized by two variables: an effort variable in conjunction with a flow variable. For example, for electric power the effort is voltage and the flow is current: volts x amps = power, in watts. For rotating machinery, the effort is torque and the flow is RPM: torque x RPM = power. For hydraulic systems, the effort variable is pressure head and the flow variable is (for example) cubic inches per second.

We want power to flow out of the source at the maximum possible rate, and then to capture it without undue losses. If the source consists of a huge pressure at a tiny flow rate and the capture element wants to see a huge flow rate being pushed by a tiny pressure, then most of the energy trying to get out of the source will bounce off the capture element and the rate of power transfer will be small.

The ratio of the effort versus the flow in a system element establishes its characteristic impedance. A high-impedance source for example pushes a tiny flow with a huge effort and a low-impedance source does the opposite: it conducts a huge flow that is pushed by a small effort. Maximum power transfer occurs when the source impedance is matched to the capture impedance.

If the source and sink are not matched, we can fix this by inserting a transformer between them. For example, a small loudspeaker is not well-matched to the air surrounding it and we fix that by inserting a horn between the loudspeaker and the ambient air, which acts as an impedance-matching transformer.

For wave power, the waves have a characteristic impedance and to match them to a capture element may require inserting a funnel-like tube between them that transforms a lot of flow with a little pressure behind it into a smaller flow with more pressure behind it.

The economics of source-to-sink matching is the most fundamental issue in capturing wave power, and a textbook on wave power engineering will explain this in detail for you.

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  • $\begingroup$ Sorry, but I'm a bit confused by your response. My end goal is to direct more water to the center of the wall in hopes to reduce cost of building a wider device. My thinking is that if a V-shaped funnel is used, the water depth at the wall would be higher, and so more water would rise. $\endgroup$
    – ARJ
    Aug 31, 2022 at 19:39
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This is already being done, you just need to rotate your reference frame 90 degrees. The concentrator is the sea bottom beneath the device. A gradually sloping seabed (a beach, in technical terms) will do all kinds of wonderful things to straighten out and condition an unruly wind-driven gravitational free surface. It costs nothing and there are square miles of it in play. You can't compete with that. Instead, you optimize the device for the conditioned seaway as you find it. It will have a tidal profile, a characteristic energy profile, a characteristic wave orbital shape, and characteristic net momentum that is a function of the beach geometry and the impinging deep-water seaway. The basic idea is to optimize the sea floor overpressure to amplify the vertical component of the wave orbitals. This is the impedance matching that matters, to put it in the terms of Neils' answer.

Analyzing these things is tricky. You are basically going to be pumping the same water in and out over and over again, so the issue is what sort of energy is being transported through the water. Driven free surface waves have three energy components, and it turns out they all transport at different speeds, and the group velocity of the wave train works out to be the weighted average of the transport speed of these different components. But you aren't going to extract these components equally with your machine, so you need to calculate the transport velocity, power, and spectrum of the stuff you can actually extract. And that is, as they say, a nontrivial problem.

http://web.mit.edu/flowlab/NewmanBook/Tulin.pdf

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  • $\begingroup$ Just to clarify, are you implying that the most optimal place to generate electricity from ocean waves is near the beach right before the wave begins to break? Also, I understand how the wave refracts and straightens as it gets closer to the shore, but at that point, is there a way to converge the waves into a single spot as to get a large wave height? In other words, if there is a 1 ft. high, 10 ft. long wave approaching the shore, can the length be mushed together into a 1 ft. long space with the water having a height of 10 ft? $\endgroup$
    – ARJ
    Sep 6, 2022 at 21:27

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