Dec 18, 2008, 11:02 PM
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# Using motor curves to predict generator performance

If I have the performance curves for a permanent magnet motor, how can I use these curves to predict generator/alternator performance if this motor was used as a generator or alternator? I know that kv can be used to predict voltage output, but how can I use the motor curves (or constants) to predict required torque and current output?
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Engineer for Christ
Quote:
 Originally Posted by bash11 If I have the performance curves for a permanent magnet motor, how can I use these curves to predict generator/alternator performance if this motor was used as a generator or alternator? I know that kv can be used to predict voltage output, but how can I use the motor curves (or constants) to predict required torque and current output?
The generation graph is a mirror image of the motor graphs up until the point of magnetic saturation for a brushed or BLDC machine. This is NOT the case for an inductive machine.

-Alex
 Dec 20, 2008, 08:18 PM Registered User Thread OP Thanks for the answer. Could you explain what you mean by a mirror image? If, for instance, a motor requires 10 amps at 10 volts to produce "x" torque at "y" rpm, does this mean that if you turn the motor at "y" rpms and electrically load it down until the torque is "x", it will produce 10 volts at 10 amps (assuming 100% efficiency)?
 Dec 21, 2008, 01:20 PM Engineer for Christ Yes, that is correct. The efficiency of the motor will not change on behalf of power flow direction.
Dec 21, 2008, 04:30 PM
Registered User
Quote:
 Originally Posted by bash11 Thanks for the answer. Could you explain what you mean by a mirror image? If, for instance, a motor requires 10 amps at 10 volts to produce "x" torque at "y" rpm, does this mean that if you turn the motor at "y" rpms and electrically load it down until the torque is "x", it will produce 10 volts at 10 amps (assuming 100% efficiency)?
I was just about to say �No� but then I read (assuming 100% efficiency) bit. If you are assuming 100% efficiency then electrical power equals mechanical power, and you are not using the motor curves.

The motor constant Kv is all you need. Once you know the rpm then you know the voltage or vice versa.

If you don�t assume 100% efficiency then it is a different story. As Alex said efficiency does not change just because the direction of energy flow changed.

Lets look at a little example.

You have a motor that is 80% efficient. You put 100 Watts in, say 10 volts and 10 amps. You get 80 watts out at some rpm and torque.

Now drive the motor as a generator and electrically load it so you have the same torque and rpm. The mechanical power is the same 80 watts. Do you think you will get 10 volts and 10 amps out? You can�t because then you would be getting 100 watts out with only 80 watts in. You could sell this generator for a LOT of money if it worked that way. Lets say that the efficiency stays at 80%, the electrical power you will get out is 64 watts. The exact voltage and current depend on the Rm of the motor. The voltage will be near 10 volts and the current will be near 6.4 amps.

Also if you are using a brushless motor you will have voltage drop across the rectifier.

Dan
 Dec 22, 2008, 08:53 AM Engineer for Christ And following the example of Dan and the 80% efficient motor, to get 10 Volts at 10 Amps out you would need y *1.125 rpm and x *1.125 torque. Thus 125% of you previous output yeilds you original input. Good point on the efficiency, Dan.
 Dec 22, 2008, 01:18 PM Registered User Thread OP Thanks for your answers. This was a good discussion. Instead of 1.25x torque and rpm, ins't it 1.12x? I would think that it would be the sqrt of the reciprical of .8. I am just making sure I understand this.
 Dec 22, 2008, 01:47 PM Engineer for Christ Yes assuming that rpm and torque increase linearly together a 1.125 multiplier must be used for each. If only one quantity such as torque increases you could also state that y rpm and x*1.25 torque yields 10 Amps and 10 volts. In any case: Power out = efficiency * Power in This is regardless of power flow direction. However, the efficiency will change with loading. So in the case of volts, Amps, torque and rpm we can write: Volts * Amps = efficiency * torque * rpm - or - Volts * Amps * efficiency = torque * rpm - for a motor application And also: Torque * rpm = efficiency * Volts * Amps - or - Torque * rpm * efficiency = Volts * Amps - for a generator application. -Alex
 Mar 11, 2009, 07:31 PM Registered User Hello, i have some question also since i am also doing the reserach by hand cranking the DC geared motor to generate electricity. Now i am using a DC Geared motor 12V, Max current 0.3A. Rated speed 56rpm rated torque 593mN.m That means if i crank with 56rpm and 593mN.m it will simply generate 12V with 0.3A if there is 100% efficiency. But is it possible to generate larger current by faster cranking? Let say if i crank at 120rpm, can i generate more than 12V and more than 0.3A?
 Mar 12, 2009, 07:49 AM Engineer for Christ In short, yes you could exceed the 12V and .3A by cranking faster. However the 100% efficiency is not a real world assumption. I'd guess you setup is more like 60% efficient. To make the 12V, .3A, you are probably going to need about 800mNm torque and somewhere around 70 rpm. So if you are able to crank 120 rpm, you'll probably get about 20-22 volts or so with the terminals unloaded. The current and voltage will vary with load. At .3A, you'll probably see about 20V. Higher currents will have a voltage drop.
 Mar 12, 2009, 12:05 PM Dieselized User The actual voltages and currents in the generator circuit will depend mostly on the load. If you short the motor terminals you will see much more current and much less voltage. You have to fully calculate the load in order to see what will really happen. Even then if you use a power point tracking device you can make it operate in a different RPM or current range and still run the load. It all depends on what you need. Greg
 Mar 13, 2009, 05:15 AM Registered User Thank you. There is a very usefull information for me. About my circuit. My motor actually connect to a LM7805 +5V Voltage regulator and then to charge a mobile phone. When i use a 12V power supply instead the motor to connect to circuit, output voltage is 5V and i get constantly 0.5A to charging the phone. The motor specification is when cranked at 57rpm, will get 12V and 0.3A. Let say the motor has 100% efficiency (I ignore lost for more easy understanding), If i crank at 57rpm and connected to the phone, will i still get 0.3A only? or 0.5A and voltage will drop from 12V? And 1 more things i found that when i instead the motor to a 12V battery and then 9V battery, the output voltage is the same, 5V, but the current is much different, for 12V is 0.48A and for 9V is 0.18A, i think should not like this if the output current is depend on load. Anyone know where is the problem?
 Mar 13, 2009, 09:12 AM Dieselized User 9V batteries in general have very high resistance. This is probably what is limiting your current in that case. Measure the actual battery voltage under load to verify. This brings up another issue. Is the motor resistance is high, then you may have trouble with low voltage under high current load. They way it works is that the circuit should pull .5A when the voltage from the regulator is 5V. So as long as you have enough voltage from the motor (5V+ dropout voltage) it's going to pull .5A from the motor. If motor resistance is high, you may need to generate a lot of voltage that turns into heat within the motor. The result is that you will need to crank harder (more torque), but maybe not need to turn 57RPM. If the motor resistance is high, and it's not capable of supplying 0.5A, even when cranking at 57RPM you'll have trouble. A DC-DC converter could help allowing you to crank faster and convert some of that voltage to current in the converter. Without having detailed information about the motor it's hard to say what will actually happen. Greg
 Mar 13, 2009, 02:55 PM Engineer for Christ I believe what Greg is trying to tell you is somewhere in the system is going to be your "bottleneck" limiter. But that's all dependent on the system. You cannot ignore efficiency because that is precisely what is going to limit you. Let me explain this way: Say you have an infinite source at you 5V, how much amperage will you load draw now assuming your power source is infinite. Now you take your input and figure can it source the amount (and type) of power the load is taking? One of these will be your limiting factor. I'm going to assume your 12V battery was far more powerful than what the load required, so there's your infinite source model. Your load draws .5A @ 5V. Now we'll factor in the minimum drop of 1.1 V (know as drop out voltage) across the regulator (this is a voltage drop there that cannot be ignored). So now you have [email protected] This is 3.05 Watts. That's your load profile. Now you need your source profile. We'll assume a fixed motor efficiency of 65% (eventhough it's not really fixed, we need a starting point). At 57 rpm the motor produces 12V * .3A *.65 eff = 2.34 Watts We know the load is capable of accepting 3.05 Watts, but our motor is only capable of 2.34, so the motor is our limiter. We know that our load is going to operate at a fixed voltage so then we can calculate our expected current from there: 2.34 Watts/6.1 Volts = .383 Amps - So at 57 rpm Expect about .383 Amps Now where will the motor produce the maximum power to our load? Simple. Again assume efficiency (I'm again guessing at 65%). 3.05/2.34 = 1.31 motor rpm increase needed 57*1.31 = 75 rpm or more to produce the loads full potential of 6.1V @ .5A This is not exact, but it should be close enough. -Alex
 Mar 13, 2009, 04:35 PM Dieselized User Alex. I don't think it works that way. If we had motor constants we could determine if the motor is or isn't capable of supplying the current. Load is 6.1V at 0.5A. The motor voltage is dependent on RPM and current is dependent on input torque if we assume 100% efficiency. Now resistance of the motor comes into play. This has an effect on maximum current and voltage when load is applied. Output current will be dependent simply on the load circuit, if the resistance of the motor is low enough to make more than 6.1V @ 0.5A. That would be 29 RPM and ~1Nm torque with no losses. If the motor can't make more than 6.1V at 0.5A due to losses, RPM must be raised in order to make enough voltage, required torque does not change. Any voltage above 6.1V is wasted so it does not make sense to turn any faster than required. It seems there is a current limiter in the phone, so we can assume current will never exceed 0.5A. A DC-DC converter helps because if the motor can't supply more than 0.3A we can convert the [email protected] (3.6W) to [email protected] (3.05W). Greg Last edited by gkamysz; Mar 13, 2009 at 04:42 PM.

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