# Magnetic Generator Power Rating

by barendfaber
Tags: generator, magnetic, power, rating
 P: 54 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?
 P: 54 voko, I think I just confused myself again. Lets 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?
 PF Patron Thanks P: 4,448 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.
 PF Patron Thanks P: 4,448 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?
 P: 54 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?
 P: 54 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?
 PF Patron Thanks P: 4,448 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.
 P: 54 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!
 PF Patron Thanks P: 4,448 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.
 P: 54 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?
 PF Patron Thanks P: 4,448 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.
 P: 54 Thank you voko, you have been a great help!
 PF Patron Thanks P: 4,448 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.
 P: 54 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?
 PF Patron Sci Advisor P: 1,956 barendfaber what are you going to use to spin the magnets ? you do realise 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 dont get anything for nothing ;) Dave
 PF Patron Thanks P: 4,448 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.
 P: 54 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.
 P: 54 good morning voko! Sorry for the hassle, maybe that may mean smaller magnets and smaller coils, I had this one in mind originally and changed it, should have maybe stuck with the sandwich design. Yes, I figured, input less inefficiencies = output. If I can just figure out more or less what the set up should be and what my input power is, I will be happy bunny!

 Related Discussions Electrical Engineering 2 Classical Physics 8 Advanced Physics Homework 1 General Engineering 1 Mechanical Engineering 1