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Old Sep 15, 2012, 08:57 AM
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How to Calculate Motor Efficiency

The purpose of this thread:

I'm after a set of of formulae that will enable someone to calculate losses firstly, and secondly get efficiency numbers equaling that of simulation programs, i.e. Scorpion_Calc.

The main purpose is to compare motors at similar and different voltages. This can be very valuable when comparing motors of different brands (obviously of the same size and Kv).


For a number of years, I've been building, rewinding, testing & measuring electric motors and I thought that I had a fairly good idea about what's going on inside an electric motor. However, my calculations don't add up when I compare the results to motor simulation programs, i.e. Scorpion_Calc.

My understanding of motor losses are that there are 2 main categories of losses, i.e. copper loss and iron loss. The latter include eddy currents in the stator, hysteresis loss, bearing loss, etc.

1. Copper loss = Rm x IČ , where Rm is internal resistance of one phase of the motor (let's talk about brushless motors for now).

2. Iron loss = Io x volt, where Io is the no-load current measured at 10 volts.

Straight-away I can can see a problem with the formula for iron loss. If Io is measured at 10 volts, then the formula should read: Io x volt/10.

Ok, now let's add copper loss and iron loss to get the total loss in the motor. If we now calculate input Watts (amps x volts), and subtract total loss (previously calculated), we should get output Watts.

Finally, motor efficiency = output Watts / input Watts.

Something is missing... My answers don't agree with simulation programs. Would someone like to comment, please?

Thanks
Christo
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Last edited by Skylar; Sep 17, 2012 at 10:01 AM. Reason: Added the purpose of this thread (in blue)
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Old Sep 15, 2012, 01:36 PM
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Quote:
Originally Posted by Skylar View Post
For a number of years, I've been building, rewinding, testing & measuring electric motors and I thought that I had a fairly good idea about what's going on inside an electric motor. However, my calculations don't add up when I compare the results to motor simulation programs, i.e. Scorpion_Calc.

My understanding of motor losses are that there are 2 main categories of losses, i.e. copper loss and iron loss. The latter include eddy currents in the stator, hysteresis loss, bearing loss, etc.

1. Copper loss = Rm x IČ , where Rm is internal resistance of one phase of the motor (let's talk about brushless motors for now).

2. Iron loss = Io x volt, where Io is the no-load current measured at 10 volts.

Straight-away I can can see a problem with the formula for iron loss. If Io is measured at 10 volts, then the formula should read: Io x volt/10.

Ok, now let's add copper loss and iron loss to get the total loss in the motor. If we now calculate input Watts (amps x volts), and subtract total loss (previously calculated), we should get output Watts.

Finally, motor efficiency = output Watts / input Watts.

Something is missing... My answers don't agree with simulation programs. Would someone like to comment, please?

Thanks
Christo
As you said, motor loss = iron loss + copper loss + mechanical loss (bearing, friction etc).
Iron loss = hysterness loss + eddy current loss.
Hysterness loss is predominant when motor speed is low, while eddy current loss is predominant when motor speed is high. This is derived from the fact that hysterness loss is propotional to "frequency (how many times the current changes its direction)". and eddy current loss is propotional to the square of frequency.

To calculate the total loss of a motor, you need to refer to the original formula of hysterness and eddy current loss calculation. But there is a rough way to estimate the loss.

Following is an example:
If you want to calculate the total loss of a motor running at 20V, driving a certain sized propeller:
First, you need to do a no-load run at 20V, and read the current. If the current is 3A, then 20X3=60W, is roughly iron loss + machnical loss ( to note that this will grow up slightly with the growing of current).
Secondly, you need to measure the Rm. If you have a milliohm meter, you can do a direct measuring of the Rm, the resistance between any two of the three lead of the motor. If you do not have such equipment, you can estimate the Rm by refering to AWG table, and estimate the total length of wire. If you know the length of one of the three wire in the motor, you can get the resistance (r) of that single wire. For Y connection, the resistance of the motor (Rm) = 2r. while for delta connection, Rm = (2/3)r. For other type of connection, like half parallel YY or DD, you can calculate repestively. Let's assume the Rm of the motor is 0.1ohm for now.
Thirdly, you need to do a test run with the propeller. If the voltage still remains to be 20V, current is 20A.

Now, we have the above numbers labelled in Green.
Total loss will be 60W + 20X20 0.1 = 100W.
total input power is 20X20 = 400W.
So the efficiency of this motor, running at 20V, driving such a propeller, is (400-100)/400 = 75%.

This is a very rough way of estimating the efficiency. But it is very easy to get.

Another application of such test is to estimate the best efficiency current range of a motor at certain voltage. This is based on the fact that the motor reaches its highest effiency when iron loss = copper loss. Still, using the above numbers: the best efficiency current will be (60/0.1)^0.5 = 24.5A. It means if the motor is running at 20V, it will have the best efficiency when the working current is 24.5A.

Just to note that, everything is related to voltage. When you talk about effiency, and Io, you should always mention the voltage, to make those number meaningful. They are all related to voltage. So, if you see a motor description, saying the motor's Io is 2A, but without mention the voltage, you should know the man who wrote the description is a "Newbie".
In some motor descriptions, there will be a best efficiency. There is always one best efficiency for a given motor at a given voltage. And the best efficiency for that motor will change with the change of the input voltage. The "best efficiency" is the highest calculated efficiency for such motor.
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Old Sep 15, 2012, 01:59 PM
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Antony (France)
Joined Sep 2003
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Quote:
Originally Posted by Skylar View Post
For a number of years, I've been building, rewinding, testing & measuring electric motors and I thought that I had a fairly good idea about what's going on inside an electric motor. However, my calculations don't add up when I compare the results to motor simulation programs, i.e. Scorpion_Calc.

My understanding of motor losses are that there are 2 main categories of losses, i.e. copper loss and iron loss. The latter include eddy currents in the stator, hysteresis loss, bearing loss, etc.

1. Copper loss = Rm x IČ , where Rm is internal resistance of one phase of the motor (let's talk about brushless motors for now).

2. Iron loss = Io x volt, where Io is the no-load current measured at 10 volts.

Straight-away I can can see a problem with the formula for iron loss. If Io is measured at 10 volts, then the formula should read: Io x volt/10.

Ok, now let's add copper loss and iron loss to get the total loss in the motor. If we now calculate input Watts (amps x volts), and subtract total loss (previously calculated), we should get output Watts.

Finally, motor efficiency = output Watts / input Watts.

Something is missing... My answers don't agree with simulation programs. Would someone like to comment, please?

Thanks
Christo
Hello Christo
You have also
a) ESC losses
b) Temperature effect on copper losses (Rm) and iron losses
Louis
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Old Sep 15, 2012, 02:24 PM
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mechanical loss
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Old Sep 15, 2012, 03:37 PM
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South Africa, GP, Pretoria
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Quote:
Originally Posted by Fourdan View Post
Hello Christo
You have also
a) ESC losses
b) Temperature effect on copper losses (Rm) and iron losses
Louis
Thanks for the replies so far, guys.

@Louis: Do I understand you correctly? If I add ESC losses and the temperature losses to copper and iron losses, I'll get to your efficiency figures in Scorpion_Calc?

Why do you include ESC losses? I always thought that motor efficiency gets calculated on it's own.

Christo
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Old Sep 15, 2012, 03:39 PM
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Quote:
Originally Posted by erikc641 View Post
mechanical loss
Hi Erikc641

Mechanical loss is already included in iron loss when measuring Io.

Christo
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Old Sep 15, 2012, 03:53 PM
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If you include the ESC, don't you have to include a prop?
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Old Sep 15, 2012, 04:04 PM
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Originally Posted by zeroback View Post
If you include the ESC, don't you have to include a prop?
Zeroback

I'm not sure what you mean by that. A prop is always included because every prop will cause a motor to run at a certain efficiency. The bigger the prop, the harder the motor works and therefore the lower the efficiency becomes.

By efficiency, I don't mean "maximum efficiency" of the motor.

Christo
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Old Sep 16, 2012, 01:13 AM
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Antony (France)
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Quote:
Originally Posted by Skylar View Post
Thanks for the replies so far, guys.

@Louis: Do I understand you correctly? If I add ESC losses and the temperature losses to copper and iron losses, I'll get to your efficiency figures in Scorpion_Calc?

Why do you include ESC losses? I always thought that motor efficiency gets calculated on it's own.

Christo
Hi Christo
Temperature is acting on Rm (so on copper loss) and eddy currents (part of iron losses)
ESC losses are very difficult to modelize
ESC + motor PMBLDC is a couple
ESC timing and algorithms have also an influence
And finally the prop loading (vs rpm) has to be modelized
Scorpion Calc (SC) is only trying to do its best ..but is not perfect
I repeat again, SC is only created to help to choose before buying and trying.
Just a good order of magnitude to predict static thrust and pitch speed ..
that are the keys of a plane power train.
Regards
Louis
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Old Sep 16, 2012, 02:22 AM
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Quote:
Originally Posted by Skylar View Post
The bigger the prop, the harder the motor works and therefore the lower the efficiency becomes.
Christo
This is not right.
Best efficiency is achieved when copper loss = iron loss. In other words, it is when the current reaches a certain value.
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Old Sep 16, 2012, 04:17 AM
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Originally Posted by modisc View Post
This is not right.
Best efficiency is achieved when copper loss = iron loss. In other words, it is when the current reaches a certain value.
Modisc

I think we have to differentiate between "maximum efficiency" and "better" efficiency" here. I was comparing efficiencies with 2 different sized propellers.

Christo
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Old Sep 16, 2012, 07:24 AM
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Motor efficiency, power system efficiency or propulsive efficiency?

Must be careful what you're asking...
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Old Sep 16, 2012, 01:18 PM
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How to Calculate Motor Efficiency

Quote:
Originally Posted by Odysis View Post
Motor efficiency, power system efficiency or propulsive efficiency?

Must be careful what you're asking...
Quite correct. I've changed the title of this thread to "How to Calculate Motor Efficiency".

Thanks
Christo
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Old Sep 16, 2012, 02:23 PM
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Quote:
Originally Posted by Skylar View Post
Thanks for the replies so far, guys.

@Louis: Do I understand you correctly? If I add ESC losses and the temperature losses to copper and iron losses, I'll get to your efficiency figures in Scorpion_Calc?

Why do you include ESC losses? I always thought that motor efficiency gets calculated on it's own.

Christo
I was referring to this statement. I understood you were calculating the motor efficiency as a stand alone. As soon as an ESC is added it changes electrical values but the motor is useless as until there is a load.
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Old Sep 16, 2012, 05:53 PM
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And the prop has no effect (large vs small). As far as the motor's concerned, it's just a load.

Could be egg beaters in jelly for all it knows!

I don't know how to extract ESC efficiency out - all the models I've seen incorporate it, as it's necessary.

You could try putting the ESC in a well insulated box of known volume, and keeping track of the temperature rise? That'll give you waste heat?
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