efficiency question

[Q]

To all those motor nurds out there (I'm a geek :P)
What would be more efficient?
1) a medium wind (+17T) motor on a 2S pack
2) a very high wind motor on a 4S pack.

So, what would be better, lower voltage + low winds or high voltage + high winds? All other parameters should be considered the same.

[Jay C]

Hard for me to think about it without a sample prop to be honest ... and a target amperage ... but that's just me.

[LBMiller5]

When it comes moving electricity, higher voltage is always more efficient. Losses in a wire are proportional to the square of the current, so if you double the voltage in a motor, and have twice the number of winds in the stator, you will have the same magnetic flux in the poles, but you will have 1/4 the heat loss in the windings. This assumes you are using the same size wire in both motors.

The amount of magnetic flux generated in a stator pole is measured in Amp-Turns. It is equal to the amount of current in the wire times the number of turns of wire around the pole. 10 amps on a 20 turn motor will give you the same magnetic flux as 5 amps on a 40 turn motor.

In order to get more turns of wire to fit, You will probably use a smaller gauge wire to wind the poles. If you do this the resistance will go up, but power loss is directly proportional to resistance, so the loss due to the resistance is less.

In the above motor, lets say that the 20 turn motor is getting 10 volts, while the 40 turn motor is getting 20 volts. Let us also assume that the 20 turn motor has a field resistance of .1 ohms and the 40 turn motor has a resistance of .2 ohms. (This is based on using a wire with half the area and twice the resistance in the 40 turn motor compared to the wire used in the 20 turn motor.)

The loss in the winding is equal to the current squared times the resistance, which is called I2R loss. For the 20 turn motor above, the loss would be equal to:

10 Amps x 10 amps x .1 Ohms = 100 x .1 = 10 watts

For the 40 turn motor the loss would be equal to:

5 Amps x 5 Amp x .2 Ohms = 25 x .2 = 5 watts

Both motors have 200 Amp-Turns of magnetic flux, 10 x 20 for the 20 turn motor, and 5 x 40 for the 40 turn motor, but the 40 turn motor, with double the voltage, has half the power loss.

This is why they run almost 500,000 volts on the power lines coming out of a power plant. It is to keep the current as low as possible. The power company steps this down to about 22,000 volts at the local power station to run to your neighborhood. Then it gets stepped down to 240 volts at the transformer on the pole outside your house.

If they ran 240 volts out of the power plant, the wires would have to be as big around as telephone poles to carry the current, and they would be so heavy you would have to put a tower every 20 feet to support the weight!

Bottom line: Higher voltage = Better efficiency.

Lucien

[Q]

Thanks Lucien,
a little green man whispered in my ear you'd be the one answering the question. LOL

So, the next phase should be to develop a 3D machine for a 4S 500mAh LiPo pack. Might proove to be interesting.
Many electric helicopter pilots are going to a high voltage system using a 20:1 gearing. I just wondered if the higher efficiency at a higher voltage would be applicable for direct drive too.
Thanks for clearing that up!!

[instain]

To all those motor nurds out there (I'm a geek :P)
What would be more efficient?
1) a medium wind (+17T) motor on a 2S pack
2) a very high wind motor on a 4S pack.

So, what would be better, lower voltage + low winds or high voltage + high winds? All other parameters should be considered the same.

The short answer is that, as long as you fit as much copper in the stator as possible (i.e. use thicker wire for lower winds), there will be the *same* amount of heating loss in the motor.

If you draw a schematic of the circuit, you will find that the resistance in the high voltage system is squared compared to that of low voltage system, cancelling the I^2 effect completely.

So, make your decision based on what wire you have, how many winds you can reasonably do, what batteries you have, what the batteries' specifications are, and personal preference :)

Andrew

[Q]

Hmmmmm, lets see here

what wire you have: 26 gauge and 0.3mm. The rest is just an order away
how many winds you can reasonably do: good one, need to test that first
what batteries you have: plenty, I can get more at a very low price too
what the batteries' specifications are: see the above
personal preference: none at this point.

[instain]

Hmmmmm, lets see here

what wire you have: 26 gauge and 0.3mm. The rest is just an order away
how many winds you can reasonably do: good one, need to test that first
what batteries you have: plenty, I can get more at a very low price too
what the batteries' specifications are: see the above
personal preference: none at this point.

If nothing else, your BEC will be more efficient at lower voltages (unless it's a switcher).

[Q]

at 4S I'm not using a BEC anyways! :P

[LBMiller5]

Instain,

You are correct in your argument. If you use twice as much wire with twice the resistance per foot, it will completely cancel out the gain due to the lower I2R losses. Perhaps I should have been a little more specific in my post. If you keep the wire size constant, and you want to deliver a specific amount of power, say 100 watts, to a load, if you double the voltage, and cut the current in half, you will have 1/4 the I2R loss in the wire.

A little side note: In the AWG wire size system, every time you go 3 gauge sizes smaller, you double the current carrying capacity of a wire. Using accepted industry safety guidelines, a 14ga wire is rated for 15 amps. An 11ga wire will carry 30 amps, and a 8ga wire will carry 60 amps.

Going the other way it works the same. Every time you increase the wire by 3 gauge sizes, you cut the current capacity in half. For example, a 12ga wire will safely carry 20 amps, a 15ga wire will carry 10 amps, and an 18ga wire will carry 5 amps.

Going back to the 2 motors I described earlier. I will re-run the calculations with real numbers to get an accurate comparison. Let us assume that the 20 turn motor was wound with 26ga wire. In order to keep nice even numbers, let us also assume that it takes 1 inch of wire per turn around the stator pole. (This is fairly close to the actual number, and if we stay consistant throughout this discussion, the actual length is irrelevent as it will cancel out when you do the math)

Winding this motor will take 20 turns per pole times 3 poles, equaling 60 turns of wire. If we add an extra inch to jump the gap between the 3 poles, and add 4 more inches to account for the start and end of the wind, we would end up with a total of 65 inches of wire per phase.

Wire resistances in the AWG Standard Charts are listed in Ohms per 1000 feet of wire. 26ga wire has a resistance of 41.02 Ohms per 1000 feet. Since there are 12 inches to the foot, if we divide 41.02 by 12,000, we will get the resistance of the wire per inch, which for 26ga wire is .003418 Ohms. Taking this value and multiplying it by 65 inches we get a resistance of .2222 Ohms per phase.

Now going back to the 40 turn motor we discussed earlier. If you changed the wire size to 29ga wire, since you only needed to carry half as much current, we can use the above example to calculate the resistance per phase. Since we are winding 40 turns per pole, it would take 120 total turns per phase or 120 inches of wire if each turn takes 1 inch of wire. Adding an inch to connect the phases and two 2 inch start and end tails brings the total length to 125 inches. From the wire charts, 29ga wire has a resistance rating of 82.04 Ohms per foot (Twice that of 26ga wire). Converting this to Ohms per inch we get .006836 Ohms.

The 29ga wire has twice the resistance of the 26ga wire, and since we need about twice as much wire, the total resistance will be about 4 times as large. In our case here, if we multiply .006836 times 125 inches we get .8545 Ohms, compared to .2222 Ohms for the 26ga wire. Now you would expect the resistance of the 29ga wire to be .8888 ohms, 4 times the value for the 26ga wire which is .2222 ohms. Where is the difference coming from?

The length of the wire around the poles does double when you double the number of turns, but it takes the same length of wire to connect the poles and the same length of wire to make the start and end wires. That is why we ended up using 65 inches of wire per phase on the 20 turn motor, and only 125 inches on the 40 turn motor (not 130 inches, which is double the 65 inch used in the 20 turn motor.) Because of this small difference in the total length of the wire, the 40 turn motor does end up having a slight edge on efficiency. If we take the above resistance values, and divide .8545 by .2222 we get a value of 3.8456. In this case, the current has been cut in half, and our resistance went up by a factor of 3.8456. If we calculate the I2R loss for the 40 turn motor it equals (.5 x .5) x 3.8456 = .9614. This makes the increase in efficiency equal to (1-.9614)/1 = 3.86%.

Now I realize that this is splitting hairs, but every bit counts, and the 40 turn motor, wound with 29ga wire, running at twice the voltage, will be 3.86% more efficient than the 20 turn motor wound with 26ga wire. (IE 3.86% less heat loss)

Now if we step out of the box of pure theory and look at this from a practical standpoint, you are going to be very hard pressed trying to find 29ga wire. The only sizes of magnet wire that are commonly stocked by most vendors are the even numbered sizes: 22ga, 24ga, 26ga, 28ga, 30ga etc. Because we are winding such tiny stators, wire gauge size becomes a big issue, and fitting extra turns in can present a problem with the larger wires such as 20, 22 and 24 gauge sizes.

Theoretically, you should be able to wind twice as many turns of 29 gauge wire on a stator pole as you can with 26ga wire. In actual practice, since the smaller wire lays neater and pulls together tighter, you can actually put more than twice as much wire on. Somewhere between 250% and 275% more is easily do-able. So where am I going with this?

In "The Real World" we would probably use 28ga wire to wind the 40 turn motor, not 29ga wire since 29ga wire is almost impossible to purchase. Now if we run the numbers as before using 125 inches of 28ga wire to wind our 40 turn stator we get significantly different efficiency values.

28ga wire has a resistance of 65.31 Ohms per 1000 feet, which equals .005443 Ohms per inch. Taking this number times 125 inches gives a resistance of .6803 Ohms per phase. When we compare this to the value given earlier for the 20 turn motor of .2222 Ohms, the ratio of the resistance is now .6803/.2222 which equals 3.062.

Now if we calculate the I2R losses compared to the 20 turn motor we get (.5 x.5) x 3.062 = .7654. Now if we compare this to the 20 turn motor and calculate the rise in efficiency we get (1-.7654)/1 = 23.46%.

So now we have 2 comparisons. If we use 29ga wire and wind a 40 turn motor, it will be 3.86% more efficient than a 26ga 20 turn motor. (IE 3.86% less internal heat loss).

If we wind the 40 turn motor with 28ga wire, it will be 23.46% more efficient. (IE 23.46% less internal heat loss).

Well, this post ended up being a little longer than I thought it would, but hopefully everyone can agree with the numbers presented here. As always, I am open to any and all comments or other points of view concerning these assumptions.

Keep Winding!!!

Lucien

[Q]

Lucien, I was hoping and expecting you would come back with numbers like you did. The one thing you forgot to mention is my age and blood type. LOOOOL

Hey, you know what? After I've built the motor test stand I'll do a test of both types of motors and get some real numbers.

[instain]

LBMiller5,

That sounds like a better argument. I would counter by suggesting that you could parallel smaller strands of wire to get equal efficiency out of a motor. i.e. 20T 26AWG on two LiPo cells or 2x10T 26AWG on one LiPo. They now have the same amount of copper and the one with twice the current has 1/4 the resistance so the efficiency is equal.

If you want to try comparing two different gauges of wire, it gets more complicated and the results will change depending on the motor and wire dimensions. Not that your calculations aren't helpful and valid, but you could probably show just about anything that way :). I do agree that for these small motors it does help to use thinner wire for the reasons you listed.

In the end, it comes down to doing whatever you can to get as much copper as possible into the stator. Whether this is done by careful selection of wire thickness, parallel smaller strands, or adjusting the voltage to match your winding, you will come out the same in the end.

Andrew

[Q]

Now, let me throw in a, possibly stupid, extra factor.

Considering the facts above, would the high voltage motor have more torque, or would the torque be the same for both motors?

[instain]

Now, let me throw in a, possibly stupid, extra factor.

Considering the facts above, would the high voltage motor have more torque, or would the torque be the same for both motors?

As Lucien said, ampere*turns makes the flux so they should have the same torque either way.

Andrew

[LBMiller5]

I concur with Instain. The torque is directly proportional to the number of amp-turns per pole, so both motors would have essentially the same torque.

Lucien

[Q]

OK, so the only advantage between both motors is that I can fly twice as long on twice the battery size if the high voltage motor uses twice the number of winds and I don't consider the motor leads a factor.

This is not entirely true if you take the extra weight of the battery into consideration of course.


I do get why more and more people go to a high voltage system in their helicopters. They use the SAME motor on a higher voltage and step down the rpm with a gearing adjustment. The Ri is the same, yet the amps go down, so the efficiency of the motor goes up. The people using the system say brushless sytems are more efficient at higher RPMs. I guess they 'd better say it's more efficient at a lower amp and higher voltage for the same power output.