Calculating Capacitors for Wave Energy Collectors

[dutchchili]

How can I calculate the capacitor needed for my wave energy collector?

I have have build a wave energy collector. It consists of multiple coils in series with magnets moving up and down, following the waves underneath the unit.
My multi-meter measures anywhere between 60 and 250 milli volts depending on the wave height (0,003 to 0,01 meter).
I would like to attach a LED to the circuit, but the LED requires 1,5V to operate.
So i though to add a capacitor to store the generated energy until it is sufficient to create a 1.5V current.
So far so good?

What capacitor (or other solutions) should i use?
I am aware of the fact that this circuit produces AC power and that i will need to place a diode before the capacitor, and after the LED, to ensure the capacitor is actually charged.

Thanks

 

[anorlunda]

have have build a wave energy collector. It consists of multiple coils in series with magnets moving up and down, following the waves underneath the unit.

What kind of waves? Ocean waves? Radio waves?

So i though to add a capacitor to store the generated energy until it is sufficient to create a 1.5V current.

Create a current at 1.5V for how long a time? As soon as the LED is attached, the energy stored in the capacitor will be depleted and the voltage will go down. Therefore, at best you can get the LED to flash for a short period of time.

A single capacitor won't get you to 1.5 volts. You need a circuit like the ones in this article:https://en.wikipedia.org/wiki/Voltage_multiplier

But even that won't work unless you have enough power. Voltage and power area not the same thing. Power=Voltage*Current. You need to find out how much power your device makes, not just the voltage.

 

[dutchchili]

Thanks Anurlunda,

My device works on water wave in a canal next to my house. The wave height is between 0 and and 10 centimeters (during storm).
The period of the typical wave is 1 second, with a wave height of 1 centimeter.
So all very limited, but fun. The device uses 1,21 T magnets in 7mm diameter coils. I have 9,3 windings covering the entire length of the magnet. I use 4 magnets per coil to maximize the utilization of the entire coil. I measure about 1/3 of the estimated V induced...
I use a 1,5mm diameter wire so very low resistance. I measure A at 0,2 mA.
So...i need the energy storage to get the flash.

Any new thoughts?

I was thinking about compressing water as seen is most wave energy devices, to create a constant flow/pressure.

 

[anorlunda]

So say 200 mV and 0.2 mA. That is 0.2*.0002=0.00004 watts or 40 microwatts. A typical LED needs 150 milliwatts. So you are short by a factor of 0.15/0.00004=3750 from having enough power to light an LED continuously. You need to do much much better to light an LED.

Have you studied this https://en.wikipedia.org/wiki/Wave_power

I was thinking about compressing water as seen is most wave energy devices, to create a constant flow/pressure.

Do you mean focusing? Water is incompressible.

 

[dutchchili]

Yep, it is very clear why the LED won't light.
I did study that article and decided to base the expected output on Faraday's law (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html) as the energy generation comes from magnetic flux.

Using the formula:
-windings * (deltaFlux*area)/deltaTime

The unit should produce 0,94 Volts with 1 cm height waves with a period of 1 second.
Somehow i measure much less. I can image that the multi-meter is part of the problem, as it does not have a fine resolution for AC power. The DCV settings allows for a range of 200 and 2000 mV, and i see it alternating between + and -. Seems trustworthy.

Do you see any err in the below?:

magnet height = 1,4 cm
wire thickness = 1,5 mm
number of windings in 1,4cm heigh coil = 1,4/0,15 = 9,3
(so if i move the magnet, i induce flux in 9 windings)
The magnet is rated a 1,21 Tesla's. It is a cylinder with a 6,4 mm diameter.
The coil has an inner diameter of 7 mm
In the formula i therefore calculated with 0,081 T to compensate for the space between the magnet and the coil.
For the area i used both the 7mm coil diameter and the width of the wire; area = 7+1,5 = 8,5mm -> 3,14 * POWER(8,5/2, 2) = 55,398 mm2 = 0,055 m2
So: 9,3 * (0,081 * 0,055) / 1 = 0,041 V
I have 4 coils in series, so 4 * 0,041V = 0,1657 V
And 4 magnets in each coil , so 4 * 0,1657 = 0,66 V
If the wave is 1 cm high, not only those 9,3 windings get excited by 40% more, leading to 0,94V (i think).

 

[anorlunda]

It's not that simple. You need to know the distance from each turn of wire to each part of each magnet as a function of time.

Google "electric generator design online" It will show a number of calculators you can use to describe your geometry and the trajectories of motion, and then to predict the electric performance.

My guess is that your performance is hurt by too much air distance between the parts. That isn't even mentioned in your calculation.

 

[dutchchili]

Based on all this, and good weather conditions, i tested again. It appears that my multi-meter takes almost a second to take a good reading. I tested this on a battery and it show two to three reading prior to actually indicated 1,5V. That is way to long to measure my wave generator, as the current has already reversed by then, or changed in value.
My LED, which does work when attached to a 1,5V battery, did not light up. Apparently there was insufficient power.

I wonder: if the calculations are accurate and the unit does produce close to 1,5V. Can i simply apply P = V*I where i calculate I using V = I*R?
This would lead about 450 watts of generated power...but the LED doesn't burn.

 

[anorlunda]

I don't know where you got 450 watts. That sounds too high by a factor of 100K to 1 million.

I already told you how to calculate power in #4.

 

[sophiecentaur]

It consists of multiple coils in series with magnets moving up and down

My multi-meter measures anywhere between 60 and 250 milli volts depending on the wave height

It strikes me that the answer to your problem would probably be to use many more turns of thinner on the same available coil formers. A much higher voltage would reduce the series loss in diodes and improve efficiency.
I agree with @anorlunda that your estimate for power output is probably very optimistic. An important factor which you seem not to be discussing is the Magnetic Circuit. 'Real' alternators are designed with tiny gaps between poles and iron laminations.

 

[dutchchili]

It strikes me that the answer to your problem would probably be to use many more turns of thinner on the same available coil formers. A much higher voltage would reduce the series loss in diodes and improve efficiency.
I agree with @anorlunda that your estimate for power output is probably very optimistic. An important factor which you seem not to be discussing is the Magnetic Circuit. 'Real' alternators are designed with tiny gaps between poles and iron laminations.

I thought about that. From what i found online the resistance of thinner wire would reduce the the output too. And i thought it would be less work making turns with thicker wire. Which was in fact quite heavy work to make a nice coil.
I could not manage multiple layers with this thickness.

 

[dutchchili]

I don't know where you got 450 watts. That sounds too high by a factor of 100K to 1 million.

I already told you how to calculate power in #4.

i calculated the amps first by doing: V(induced)/Resistance of wire: 0,94 / 9,67E-003 resulting in 476 amps...
then V*I = 0,94*476=450 watts.
I did not sound realistic, but what makes the difference between applying these formulas and the actual result?

 

[anorlunda]

Your error is probably that that is not the voltage between the two ends of that wire, it is the voltage across the LED.

The actual voltage difference between the two ends of the wire is probably too small to measure.

 

[dutchchili]

My first attempt was to measure the open circuit voltage. When measuring across the LED i measure the same values. Anywhere between 0 and 200 mV.

 

[russ_watters]

Let's look at the input for a sec and see if we can estimate the energy available to be harnessed.

To make it easier, consider a square wave, with a 1cm amplitude and 1 sec period.

Let's use a 3.8L jug with a flat bottom and surface area of 230 cm2 as our float. That's probably too big, but we'll go with it for now.

Now, to generate a force, the jug has to partially submerge, so figure when submerged 0.5 cm, that's when you get your maximum force and it varies with the wave height. 230*.5*9.8/1000=1.13N max force. If the average is half that, you have 0.0056 Joules per cycle or 5.6 miliwatts.

So, is that in the ballpark of your device? What are you using for a float?

 

[dutchchili]

(https://goo.gl/photos/41hpgzJJS5FAV1VV6)
This the construction. There is a floatation platform made out of cork with 4 sticks on it that move up and down along with the waves. I have created 4 coils that hover above the sticks. The magnets are placed on top of sticks inside the coils. The coils are soldered together to form one big things. (will post a photo).

 

[dutchchili]

Here is the completed prototype:
https://goo.gl/photos/GRXcLuVXU9Z4d3yC9

The tape on the sticks is to prevent it to much amplitude.

 

[sophiecentaur]

I have looked at your images and perhaps you should compare them with the image https://www.researchgate.net/figure/259624017_fig2_Figure-5-3D-model-of-the-six-side-linear-generator. There is a lot in common with the linear alternator in the link and the rotary alternator in this link. Do you notice something that is missing in your design? Those two designs use IRON with the coils wrapped around it and the magnets move across the gaps to give a very high alternating magnetic flux through the coils. I mentioned the Magnetic Circuit requirement in an earlier post. Those images I have posted show the basis of all practical motors and alternators. The induced emf is proportional to the rate of change of magnetic flux and you need to establish a high flux in one direction and a high flux in the opposite direction. Also, the frequency of water waves is very low so you have a lot to make up, compared with a system that changes the polarity many times a second (perhaps 180 times a second, for a simple mains alternator).

 

[dutchchili]

hmm, with nothing rotating it is quite difficult to get something close to those things.
Was my thought to not convert the vertical motion into a rotary motion a mistake?

 

[sophiecentaur]

hmm, with nothing rotating it is quite difficult to get something close to those things.
Was my thought to not convert the vertical motion into a rotary motion a mistake?

The first link shows a linear alternator. The only problem with that design is that the number of gaps traversed by the magnet block is less for small waves.
Using a crank could still be a problem because the length of stroke would need to be matched to the rotation geometry. Some variable lever system could solve that problem.
Do you understand my point about the total need for IRON everywhere, if you want to get an efficiency value that would make the project worth while? You have used thick / stiff wire for you coils which makes them self supporting but that is not important. You can make a cardboard former which can carry ten times as many turns of thinner wire. The resistance would be finite but that loss would be tiny compared with the losses due to the mismatch between the low impedance coil and the necessary diode circuitry.
Actually, you could do a lot worse than get hold of a motor car alternator (from a breaker's yard) and saw the circle of poles and coils apart. Then stick them in a row and do what is done in my first link. You will notice that the thickness of the windings in a car alternator is considerably more than the wire you are using. So resistance should not be a problem.
Please do read a lot of Google hits about alternators before you launch into any more work on this project.