Buoyant force and a water column

[KuriousKid]

TL;DR Summary

How to calculate the amount of water pushed in vertical column

Recently I viewed the Searaser video at
After watching it, I feel the numbers given in the video could be wrong but not sure. So I'd like to know how I can calculate the amount of water pushed by the Searaser device at some height in one wave shot e.g. at 100 ft, 200 ft and 300 ft above sea level.

Let's assume the Buoy cube size is 2 m x 2 m x 2 m square. Assume the wave height is 1 m. So every 9 seconds, one wave comes and pushes the buoy 1 m up with a force of B = rho * V * g = 40000 N or 4000 kg approximately. (If 2 m cube is too big for average wave size, what could be best size in real life?)

Based on this, how can I calculate the amount of water, this force can pump, at height of 100, 200, 300 ft?
I'm taking Pipe diameter of 2.55 inch, with this value, 1 ft water level in the pipe weighs 1 kg. So 100 ft = 100 kg of water in pipe.
If pipe is 3.6 inch then per ft, it weighs 2kg. 100 ft pipe = 200 kg
If pipe is 4.415 inch then per ft, it weighs 3 kg. 100 ft pipe = 300 kg

Since the 4000 kg force is applied only for a second for 1 meter length, how can I calculate how much water will be pumped from the pipe to 100 ft height from sea level and at 1000 ft for 2.55 inch diameter pipe in one wave force?

 

[Baluncore]

It seems all the numbers are based on the initial wrong 1.6m³ per second.

Waves have a frequency and a height. Bigger waves tend to have longer periods.
The ratio of float area to cylinder area reduces the volume pumped, while it increases the pressure.
He wrongly assumes the pressure and volume can both be maximised at the same time.

Pressure * flow rate = power.
1.6 m³ going up and down by 1 metre has a peak hydrostatic pressure difference of 0.1 bar = 10 kPa. Maximum energy would be 1.6 m³ * 10 kPa = 16 kJ per wave.
One cycle per second makes 16 kW maximum, being generous.
But the assumed 10 bar = 10 * 100 kPa is not possible without reducing the pumped volume to about; 1.6 / 100 = 0.016 m³ per second. Which obviously gives the same power.

The float must wait until the hydrostatic pressure has risen before it begins to rise. So there must be a hysteresis in the float to wave surface difference as energy is drawn.
If the surface wave follows a cosine, I believe the float must follow in quadrature, as a sinewave. The product of height difference by volume is rectified by the flap valves.

 

[KuriousKid]

Thank you for your comments.
So how can I solve this problem?

In simple way, what I'm thinking is, if wave displacement is 1 meter, then 2.55 inch diameter pipe can pump 3 kg of water up to 4000 ft.

If I use 4.14 inch diameter pipe, I can pump approximately 9 kg of water up to 1333 ft above sea level?

 

[Baluncore]

So how can I solve this problem?

Ignore the pressure multiplication due to the piston and cylinder. It is not relevant to energy extraction, it only represents a later transformation. It makes the problem too complex to analyse simply.

Analyse only the float in the water, restrained by some mechanism.

The float must rise and fall in the water relative to the water surface, as the waves pass.
There will be no force if the float height does not lag the wave height.
The energy extracted, or work done, is the force on the float, multiplied by the distance moved by the float, integrated over the whole wave cycle.

You could write the equations of the wave and float, then solve it from the math point of view.
But the flap valves needed will make the restraint non-linear;
So it might be better to come up with a numerical model that follows the wave and float position through the wave cycle, computing the integral of energy over the cycle time.

Avoid inches, feet and pounds.
Keep to SI units, metres, kilogram, seconds and pascals.

 

[Baluncore]

Here is a worked example of a wave buoy pump system.

Consider a double acting piston pump. It will have four flap valves that allow water to be pumped from the sea level to a fixed height reservoir. Once primed and operating, the hydrostatic pressure of the column of water to the reservoir will be more or less constant. From that we can identify the piston rod force that must be applied before the piston will move, to lift water to the reservoir.

Consider now the buoy. We make it 2 metre high with a cross-section area of 1 square metre. It is made from concrete, so it will not rust, and weighs 1000 kg. It will float half-in and half-out of the water. That defines the reference line on the buoy. We assume the buoy is constrained by the piston connecting rod to remain upright and in place.

We now set the force needed to move the piston rod to be 500 kg force. 500 * 9.8 = 4900 Newton. That will occur when the buoy is 0.5 metre above or below the neutral reference line. When the wave rises by 0.5 metre above the buoy reference line, the buoy will follow it up, remaining 0.5 metre below the rising wave. When the wave falls by 0.5 metre below the buoy reference line, the buoy will follow it down, but remain 0.5 above the water.

If the waves are less than 1 metre from peak to trough, there will not be sufficient force to operate the pump.

Now increase the wave height to have peaks 1 metre up and troughs 1 metre down. As each wave passes, the buoy will rise and fall by only one metre, that makes a total movement of 2 metres per wave. The force when the buoy moves will be 4900 Newton, so each wave generates 2 m * 4900 N = 9800 joule.

The wave period is likely to be about 10 seconds, so we get 980 J/sec = 980 watt.
A 2 metre wave is certainly not common. The piston force will need to be reduced, or the section of the buoy increased to operate when the waves are smaller.

 

[256bits]

Excellent description of how the system works.

 

[KuriousKid]

Here is sample setup of the problem I was trying to solve. Got this from Youtube video at

Assuming every 10 seconds we get 5000 N force of 1 meter wave [up + down = 1 m], this force lasts only 1 second. So in this 1 second, how much water we can pump and how high? My initial calculations were based on pipe length and weight of water in that pipe column.

 

[Baluncore]

The power in watts generated by the buoy is the energy in joules per second available to lift water. Potential energy, E = m·g·h;
Where h, is the lift height in metres, and g = 9.8, the acceleration due to gravity.
From that calculate the mass of water; m = E / ( g·h ); in kg per second.

I really don't want to watch any videos.

 

[Baluncore]

Here is a graphical model of how a simple double acting buoy pump operates. As a sufficiently large wave passes, the float (magenta) follows the wave (cyan), but with a lag because the buoy must apply sufficient buoyancy force to move the pump. The height difference between the wave surface and buoy is plotted (red). Notice that energy (yellow) is only generated while the buoy is displaced by the fixed offset, and that it is proportional to the slope of the wave. No energy is generated while the wave height is crossing the buoy height as the buoy does not move.

In order to generate energy continuously it would be necessary to employ a minimum of four buoys, separated by 1/8 of a wavelength.

There is an interesting parallel here with the “Peter Principle”. Wave buoy energy generator designs progressively increase in complexity, until the inventor is unable to analyse the system. At that point the design becomes irrational and unreliable. When key information is not provided (because it's relevance is not recognised) by the inventor, it becomes impossible to analyse the complex system.

Wave power also has the challenge of a statistical distribution. Sometimes the waves are big, sometimes they are small. The generator must be optimised to the wide range of wave heights and wavelengths encountered, while not being damaged by huge seas.

Reliability is very important in wave energy generation. The cost of repairing damage after a failure can exceed the cost of installing the system. Future insurance premiums will probably be a critical factor in the design and implementation.

 

[KuriousKid]

This is good graph and information. What I learned from it is, about 4 buoys can be used to continuously pump the water at some level.
I'll perform the simulation of this sometime soon and see how it turns out.

 

[Baluncore]

The simple step by step simulation I used to generate the graph is attached as a text file.

 

[KuriousKid]

Wow, that's genius. I thought some Simulation software will be required to do this.
I should learn the Basic for simulation...