Magnetic Generator Power Rating

In summary, you would need to push the magnets forward with a force greater than the EMF generated to keep the machine running.Thanks for clarifying that!
  • #1
barendfaber
54
0
Hi,

I am very new to this, so if my questions sound a bit stupid please point me in the right direction.

I want to build my own 2.5 kW magnetic generator, but would like to understand a bit more how this works. Please see attached picture. I will be using the 24 magnet / 18 coil configuration, and the coils will be connected in a star formation. So, I have 18 coils, or 9 pairs. When I move the magnets in a clockwise direction, in one turn, 22 magnets will cause a changing magnetic field on each coil. Let's say I will rotate it 3 times per second.

I hope this does not sound confusing, but which one is my power rating:

1. 18 coils x 1 magnet moving over it (1 changing magnetic field)

2. 18 coils x 22 magnets x 3 revolutions
 

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  • #2
None of these are power ratings. The number of magnets and coils and RPM will give you the frequency of the generator output. The power depends on the EMF generated by coils, and that depends on their construction and the magnetic field in the generator. It also depends on the resistance of the wiring. And, finally, it must be less than the mechanical power fed to the generator's shaft.
 
  • #3
Thanks for the reply. I understand everything you say, however I am still a bit unclear here on how to determine / calculate my power rating. Let's say the EMF induced in each coil every time there is a change in magnetic field is (purely example) 1 volt, my wire gives a resistance of 3 ohm, and my magnets has a magnetic flux of 1.4 Tesla (this will change depending in the size of my magnets, Neodymium magnets). What would be my power rating at 180 rpm?
 
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  • #4
I am not sure what you mean by this: "EMF induced in each coil every time there is a change in magnetic field is (purely example) 1 volt".

Much depends on the geometry of the machine, too. I'd suggest you find a decent book on the theory of electrical machines, otherwise it is unlikely you will succeed.
 
  • #5
I think you are trying to go to deep with this. What I am after is a simple direction - if I have 18 coils, and I move the magnets once in a clockwise direction, so each coil "experiences" 1 magnetic field change and an EMF is induced, is this my power rating (and yes, I understand there are many variables)? Or is my power rating 18 coils x 22 magnetic field changes with an EMF induced (1 rotation)? Or is my power rating 18 coils x 22 magnetic field changes x 3 rotations with an EMF induced?

I am after what I should aim for with induced EMF per coil per magnetic field change to achieve a power rating of 2.5 kW.
 
  • #6
The EMF of one coil is about ## 2fn NB_0S ##, where ## f ## is the number of revolutions per second, ## n ## is the number of magnetic poles, ## N ## is the number of turns in the coil, ## B_0 ## is the max. magnetic flux density, and ## S ## is the crosssection of the coil.

You have got 3 phases, so each phase must generate 2.5 kW / 3 = 830 W. Total resistance is 3 Ohm, thus 1 Ohm per phase, so total EMF per phase must be about 30 V, or 5 V per coil, if they are connected in series.

Since N and S are not specified, I cannot determine if your setup will really generate 5 V EMF per coil. But you could use the formula above to figure out N and S required for that.

Note this is a very rough estimate, because much depends on the true geometry of the magnetic field.
 
  • #7
Hi voko,

Perfect, thank you, this is exactly what I was after! With regards to poles, I will use the layout in the picture in my first post. Blue will be North and Red will be South.

I do apologise but have 1 last question, again this might be a bit up in the air. When there is movement, the magnet will induce an EMF in the copper wire coil, which in turn will generate a magnetic field equal in size to the EMF but opposite to the magnet's field. At what kg force will I need to push the magnets forward to generate continuous EMF? Or is it a bit more complicated than that, and what factors must I keep in mind?
 
  • #8
Well, this is actually quite simple. The mechanical power fed to the generator must be somewhat greater than power produced by it. ## P = \tau \cdot \omega ##, where ## P ## is power, ## \tau ## is torque, and ## \omega ## is angular velocity. Given the required power and frequency, you can obtain the torque.
 
  • #9
  • #10
The EMF frequency is the mechanical frequency multiplied by the number of magnets. With 24 magnets and 3 revolutions per second, you get 72 Hz.

Angular velocity (of the rotor) is the mechanical frequency multiplied by 2 pi. For 3 Hz, that is about 20 rad/second.
 
  • #11
Thanks voko, you are a great help!

I have found this site : http://www.6pie.com/faradayslaw.php, and have used this to calculate how many windings I propose I should have.

N = -1 * (-V/ change in (( tesla * area meters squared)/ seconds))

Volt per coil = 5 V

Magnetic Flux (I want to use Neodymium magnets), so 1.4 Tesla,. and I will divide this by 4 to play safe, giving me 0.35 Tesla.

I propose a magnet size of Width (0.15 m) and Height (0.1 m), giving me an area of 0.015 m2.

Turns per second will be 3, so time will be 0.333333333 s.

Using the above formula, I will need N = -1 * (-5/((0.35*0.015)/0.333333333))) = 317.4603175 windings.

This seems a bit different from the formula you supplied, so is this one correct for me to try and find my windings?

Or if I try yours:

2fnNB0S

2 * 3 revolutions * 22 magnetic poles * 317 windings * 0.35 Tesla * cross sectional area (here I am stuck again, I tried to look at http://en.wikipedia.org/wiki/Cross_section_(geometry [Broken]) but it makes my head spin).

On another point, am I right in saying that if I reduce my resistance, I will need to increase my EMF per coil to achieve my 2.5 kW power rating?
 
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  • #12
The formulas are about the same, but using the web site's formula, you need to factor in the number of magnetic poles into "seconds". This is because the more poles you have, the more frequently the field alternates. Note it is number of "poles", which is twice the number of magnets.

"Cross-section" is the same as "area". I think your coils will be roughly rectangular or perhaps trapezoidal in shape, and it is easy to compute the area of these shapes.

Reducing resistance, you increase the power. The formula is ## P = \frac {E^2} {R} ##. Note this is max power, which is achieved when the generator's output is shorted.
 
  • #13
To the untrained man this can get complicated very quickly!

OK, so using your formula, I need to adjust it to:

2 * 3 revolutions * 22 magnetic poles (is this right?) * 317 windings * 0.35 Tesla * 0.015 m2 (approximate) = 219.681 (this is volt right?).

As per the above, am I right in saying total EMF is 219.681, divided by 3 revolutions, divided by 22 magnetic poles, gives me 3.318 V per coil?
 
  • #14
You have been warned, you need to get a book on some theory!

First of all, is that 22 or 24 magnets? You started with 24. I think this is the correct number for a three-phase generator with 18 coils.

Second, 24 magnets have 48 poles - this is the number you should plug into the formula.

Lastly, 0.015 sq. meters as a coil cross-section is suspicious. This is the area of a square with about 0.12 m a side; you have 18 coils, and you must have gaps between them; so 0.12 m x 35 = 4.3 m; this is the circumference of your stator, meaning the diameter of the generator is greater than 1 meter.

I would assume you want your stator about half a meter wide, then its circumference will be 1.6 meters, dividing by 35 we get 0.05 m the coil diameter, and 0.0025 sq. m its area.

So: 2 * 3 Hz * 48 poles * 317 turns * 0.35 Tesla * 0.0025 sq. m = 80 V per coil. That's 16 times greater than you need, so you need to reduce the number of turns accordingly.
 
  • #15
By the way, assuming your coils are square with 0.05 m a side, and 20 turns per coil, you get 4 m of wire per coil, and 72 m total wiring. That has to be less than 3 Ohm, so that means the wire must have no more than 0.04 Ohms per meter. Assuming you are in the States, that means gauge 20 copper wire.
 
  • #16
Will see what I can do with a book, however working with you is great as you give real time solutions, and if I get stuck you help.

I see where I went wrong. From the picture in my first post, at the top I have two North poles next to each other, and at the bottom I have 2 South poles, so I assumed that these won't induce an EMF as moving 2 north poles over the coil, only 1 would induce an EMF. For my cross section, I just took the area of my magnet, again a bit wrong there!

Awesome, another step closer. I have adjusted the formulation to 2 * 3 Hz * 48 poles * 20 turns * 0.35 Tesla * 0.0025 sq. m = 5.04 V.
 
  • #17
Are your magnets really that huge? 0.1 m x 0.15 m? That's a tad bigger than two iPhone's side by side!
 
  • #18
Great minds think alike! You replied just as I typed mine. I am based in the UK, but will check USA gauge and find one similar this side. Now I have my coils, windings and size, pretty chuffed, thank you!

Last one, now I need the torque. As per your comments "Well, this is actually quite simple. The mechanical power fed to the generator must be somewhat greater than power produced by it. P=τ⋅ω, where P is power, τ is torque, and ω is angular velocity. Given the required power and frequency, you can obtain the torque.", P is 2.5 kW, right? Struggling a bit to get my head around angular velocity. "Mechanical frequency multiplied by 2 pi. For 3 Hz, that is about 20 rad/second." What will my angular velocity be?
 
  • #19
Magnets are still in the development stage. Wow, your right, two iPhones together might be a bit much, will adjust the settings to see if I can get a smaller area with more turns?
 
  • #20
voko, I think I just confused myself again. Let's say I get Neodymium magnets, magnetized to 1.4 Tesla. If I change my magnet size to 8 cm x 10 cm as an example, what Tesla can I use in my formulation? Should I use 0.08 m x 0.1 m = 0.008 sq. m x 1.4 Tesla, or have I got the wrong end of the stick again?
 
  • #21
Magnets and coils should be similar in size. Yes, you can always trade area for turns, but keep in mind that more turns means more wire, which needs more volume - and increases resistance, unless you use thicker wire, which means even more volume.

20 rad/s is the angular velocity that corresponds to 3 revs per second. More accurately, it is 3 * 2 * 3.14 = 18.9. For 2.5 kW, then, the torque is 132 N*m.
 
  • #22
Tesla is a unit for magnetic flux DENSITY, which means it is magnetic flux PER area. You get the magnetic flux by multiplying the Tesla rating and the area.

8 cm x 10 cm again looks quite big. I am not sure how you are going to arrange your magnets, but in any case that means the circumference of your rotor will be no less than 47 * 8 cm = 376 cm, which again means over one meter for its diameter.

What dimensions do you have in mind for your generator?
 
  • #23
I am unsure about that one. I would like to build a 2.5 kW generator, this will be my first one, but as you most probably have seen from my posts and ramblings, I am very much a newby and am starting from scratch. It seems you are very familiar with this, could I ask what you would recommend then as a starting point to determine magnet size and windings for such a generator?

Awesome for torque. I had a quick look at http://en.wikipedia.org/wiki/Newton_metre, do I need to look at the second conversion factor which is 1 kilogram-force metre = 9.80665 Nm, i.e. 13.46 kilogram-force metre?
 
  • #24
Voko, just to give you an overview of where I am. I have been trying to research magnetic generators on the internet for some time now, and have about 10 PDF's with different magnetic generators, ranging from 1 kW to 1.5 Mw. It always fails to stipulate magnet size, coil gauge and turns, this is why I have turned to the forum to try and get to a formulation that will help me to determine the set up for my generator. I am pretrty good with Excel and am trying to set up a calculator that will help me to try different set ups for my needs.

From your post re the magnetic flux per area, say I have a magnet in size 8 cm high and 4 cm width, this gives me an area of 0.0032 sq. m, if my magnet is magnetized to 1.4 Tesla, does this mean the most I can get from this magnet is 0.0048 Tesla?
 
  • #25
I have never built a generator. What I say here is based purely on my knowledge of physics.

The best recommendation I could give regarding building a generator, is to find a proven design. Then you could either just copy it, or study the design, understand why certain design decisions were taken, and think how you could adapt that to whatever objective you have.

Does any of those PDFs you have found describe an ACTUALLY WORKING generator? Even with some parameters omitted.
 
  • #26
Yes, all the generators are working. I saw one for a 2.5 kW 180 rpm generator (they had output at various rpm's, the main one being 180 rpm), but no specs on magnet size or strength, or copper wire gauge, length or windings. So good paper but not helpful if you want to replicate it.

I do not mean to waste your time, but could I please ask if you could help me with the maths on my proposed generator, so I can have a clear idea of magnet size and strength, and copper wire? I know for a fact as soon as I leave here I will mess the calculations up again!
 
  • #27
I suggest that you take that 2.5 kW working generator as a starting point and fill in the gaps. If you do want to follow this route, please post a link to its design.

Regarding your proposed generator, I am not sure how I could help. You should start off by defining its basic parameters, such as its dimension, the rotor-stator layout, etc. Again, this is best done by having in mind something similar that is known to work.
 
  • #28
OK, I will look into it and also work with what you gave me. Just a few points more:

1. If I make my magnets smaller, the cross sectional area of my coils will also become smaller, and I will need to adjust the formulation to keep this in mind?

2. You mentioned "so that means the wire must have no more than 0.04 Ohms per meter. Assuming you are in the States, that means gauge 20 copper wire.", is there a site that will show me what ohm resistance a specific gauge of copper wire will have?

3. For torque, am I right in my conversation of 132 N*m to 13.46 kilogram-force metre?
 
  • #29
The flux density (Tesla) and the area enter separately in the EMF formula. The density value does not change as you vary the size of magnets, but the area does. Again, the formula assumes that coils and magnets are about the same size (area).

Ignore the wire gauge for now, it is simple to determine once you have settled on the geometry of coils and the number of turns.

132 N.m is 13.46 kgf.
 
  • #30
Thank you voko, you have been a great help!
 
  • #31
I have reviewed the discussion and it seems that we have completely neglected the question of efficiency, which has a major impact on the design, and even its feasibility.

1. Determination of the required EMF per coil. For any desired power rating ## P ## (of the entire generator), the power at the per-phase load is ##P_p = P / 3 ##. Let ## r_p ## be the resistance of one phase's internal wiring, ## R_p ## be the resistance of one phase's load, ## E_p ## the phase EMF, and ## I_p ## the phase current. Then, from Ohm's law, ## E_p = (r_p + R_p)I_p = r_pI_p + R_pI_p ##. Multiplying this with ## I_p ##, we have ## E_pI_p = r_pI_p^2 + R_pI_p^2 = r_pI_p^2 + P_p ##. The ## r_pI_p^2 = W_p ## term is the power wasted on heating the internal wiring, so we want to minimize that. Let's say we want ## W_p = wP_p ##, where ## w ## is the "waste factor": this is how much waste heat is produced for the rated power.

Since ## W_p = r_p I_p^2 = w P_p ##, we can determine the phase current from it: ## I_p = \sqrt { \frac {w P_p } {r_p} } ##, and the formula for the phase EMF then becomes: ## \displaystyle E_p = \frac {1 + w } {\sqrt {w}} \sqrt {P_p r_p} = \frac {1 + w } {\sqrt {3w}} \sqrt {P r_p}##.

Taking ## P = 2500 \ W ##, ## w = 0.03 ## (3% waste) and ## r_p = 1 \ Ω ##, we end up with ## E_p = 172 \ V ##. Since there are 6 coils per phase, the required EMF per coil ## E_c = 29 \ V ##. Not five volts as estimated originally!

2. Windings per coil. A more accurate formula for a coil's induced EMF is ## E_c = \frac {\pi N B_s S n f} {\sqrt {2}} ##, where ## N ## is the number of turns per coil, ## B_s ## is the average magnetic flux density in the air gap, ## S ## is the area of the coil's cross-section (and of the magnet at the same time), ## n ## is the number of magnets, ## f ## is the rotational frequency (Hz, revolutions per second). To find out the number of turns: ## \displaystyle N = \frac {\sqrt {2} E_c} {\pi B_s S n f} ##.

## B_s ## could be taken to be 0.5 T. Assuming 0.0025 mm2 square coils (and magnets, meaning 5 cm x 5 cm), ## N = 145 ##.

3. Wire gauge. Each coil has 145 turns, and each turn is 4*0.05 m, thus one coil has 29 m of wire, and one phase has 174 m. Thus the wire's resistance must be no greater than 1/174 = 0.0057 Ω/m = 5.7 mΩ/m. And here comes the catch: that corresponds to the 4 mm2 standard copper wire. That means the diameter of the conductor is 2.25 mm, and with insulation it is probably around 2.5 mm. If you arrange 145 turns of such wire in a single layer, you will end up with a coil over a foot long. That's clearly unacceptable. You could wind that in 7 layers, but then the thickness of the winding will be 35 mm, which is comparable to the size of the coil, which means the the formula we used to get EMF and the number of coils becomes invalid.

The reason we end up in this situation is because we set a very high standard for efficiency: just 3% losses. Let's see what we get if we make that 10%. Then ## E_p = 100 \ V, E_c = 16.7 \ V, \ N = 83 ##. This results in about 100 meters of wire per phase, thus requiring its resistance to be no greater than 10 mΩ/m, and that means the 2.5 mm2 standard copper wire. Its diameter is 1.8 mm; with insulation, 2 mm. Thus the length of a single-layer coil becomes 166 mm; a four-layer coil is then under 5 cm length, with the thickness of the winding at 16 mm, which will probably work.
 
  • #32
Hi voko,

Thank you for the extra time you spent on this, and thanks for laying out the formulation sin such detail. I ahve been able to easily do the calculations following your instructions.

I have 2 more questions:

1. If we had to change the design to 2 magnets per coil instead of 1, so have an outer ring of magnets, then the copper wire coils, then another ring of magnets, will this mean we can reduce the number of windings in the coil, and could this help?

2. Based on the recalculations above, would my kg force to continue spinning the generator change, or is it still as calculated previously?
 
  • #33
barendfaber

what are you going to use to spin the magnets ?
you do realize that to generate 2.5kW out and even if that were possible your input power to spin the magnets is likely to well exceed 3kW ??

you don't get anything for nothing ;)

Dave
 
  • #34
If you have a sandwich rotor, then you should expect the flux density will be higher, probably much higher. So yes, you should expect smaller coils. How much smaller, however, is a complex question, I need to think more about it.

The total mechanical power required at the max rated power is ## P_m = (1 + w)P ##, where ## w ## is the waste factor introduced previously. Then ## \tau = (1 + w) \frac P \omega ##. As you can see, you need more input power and more torque when you account for the inefficiency.
 
  • #35
hi dave,

yes, absolutely aware of this. i want to build my first wind turbine, however need to figure out what my generator needs to look like in order for me to proceed.
 
<h2>1. What is a magnetic generator power rating?</h2><p>A magnetic generator power rating is a measurement of the amount of electrical power that can be produced by a magnetic generator. It indicates the maximum output of the generator in terms of watts or kilowatts.</p><h2>2. How is the power rating determined for a magnetic generator?</h2><p>The power rating of a magnetic generator is determined by several factors, including the strength of the magnets used, the size and design of the generator, and the efficiency of the conversion of magnetic energy into electrical energy.</p><h2>3. What is the typical power rating for a magnetic generator?</h2><p>The power rating of a magnetic generator can vary greatly depending on its design and intended use. However, most magnetic generators have a power rating in the range of 1-10 kilowatts.</p><h2>4. Can the power rating of a magnetic generator be increased?</h2><p>Yes, the power rating of a magnetic generator can be increased by using stronger magnets, improving the design and efficiency of the generator, and increasing the size of the generator. However, there may be practical limitations to how much the power rating can be increased.</p><h2>5. Is the power rating the only important factor when considering a magnetic generator?</h2><p>No, the power rating is just one aspect to consider when evaluating a magnetic generator. Other important factors include the cost, durability, and reliability of the generator, as well as its compatibility with your power needs and existing electrical system.</p>

1. What is a magnetic generator power rating?

A magnetic generator power rating is a measurement of the amount of electrical power that can be produced by a magnetic generator. It indicates the maximum output of the generator in terms of watts or kilowatts.

2. How is the power rating determined for a magnetic generator?

The power rating of a magnetic generator is determined by several factors, including the strength of the magnets used, the size and design of the generator, and the efficiency of the conversion of magnetic energy into electrical energy.

3. What is the typical power rating for a magnetic generator?

The power rating of a magnetic generator can vary greatly depending on its design and intended use. However, most magnetic generators have a power rating in the range of 1-10 kilowatts.

4. Can the power rating of a magnetic generator be increased?

Yes, the power rating of a magnetic generator can be increased by using stronger magnets, improving the design and efficiency of the generator, and increasing the size of the generator. However, there may be practical limitations to how much the power rating can be increased.

5. Is the power rating the only important factor when considering a magnetic generator?

No, the power rating is just one aspect to consider when evaluating a magnetic generator. Other important factors include the cost, durability, and reliability of the generator, as well as its compatibility with your power needs and existing electrical system.

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