BL motors and load

Ok, so this just crossed my mind. Do brushless motors really run that much less efficient with a lesser load (IE- being "undergeared" in a truck)? As evidence, I always read posts about people being undergeared on other forums, and see that their motors are running obscenely hot.

The reason I ask this is because I just took the plunge and ordered that 4800kv V2 4 pole motor, and I remember back when people were running pletty motors (which some were also 4 poles, as you know) on Uforces, they'd have to use some crazy pinions to keep the motor temperature in check.

But anyways, my original question; how come such a load is needed on the motor in order for it to operate more efficiently? Does it have to do with it having to expend energy one way or another, as in, if X amount of watts goes in the motor, X-Y must come out of the motor (Y being heat, X being work done to move the gears and wheels and stuff). So unloaded, you have more Y than X due to conservation of energy (same amount of energy goes in, but some is burned off as heat, and since you're undergeared in this scenario, not near as much wattage is going to be needed to perform work, so that large amount of wattage leftover is heat).

If anyone has any links or explanations, please post them. I'm curious. This'll give me a good starting point for gearing the V2 motor as well.

[BrianG]

This is a good question.

With no load, a motor's current will be its "Io" rating (for the rated voltage of course), which is usually under 5A for all but the extremely low turn motors. So, with no load, the power dissipated is the voltage X this Io rating. For example: if a motor has an Io rating of 2A @ 14v, that's 28w dissipated with no load.

When we put a load on the motor, it pulls more current, so it would stand to reason that more current = higher power.

I've done a little reading on this and there are a couple explanations, each of which doesn't sound quite right.

1) An unloaded motor doesn't direct any of the input power to "work" so all the power is wasted as heat. Load a motor and more power goes to making it move. Kinda makes sense at first glance, but there's something wrong - too simplistic in this theory.

2) A conflicting answer says this isn't true. An unloaded motor heats up because it generally is not moving (or not as fast) so there is less airflow to cool the motor. This makes more sense to me.

I've run an underloaded motor and it does get hotter than a "properly" loaded motor. Is this because the vehicle is not moving as fast to cool the motor with airflow? I would think that the back-EMF factor would have something to do with it as well...

[othello]

Good question.

I think it all comes down to operate a motor in it's best efficiency range.

When looking for example at one of Lehners motor diagramms i think it gets more understandable.

7,13A -> 39,9W heat
10,18A -> 37W heat
20,54A -> 36,6W heat
29,26A -> 40,8W heat
40,59A -> 52W heat
50,65A -> 66,4w heat
60,7A -> 89,1W heat
69,94A -> 112,6W heat

Let's start with the easyest two states.

Unloaded state:
When unloaded (low amps) efficiency of a motor is bad. In this case Lehner starts its test at 7A where efficiency hovers around 60%. So 40% is wasted in heat. 100 Watt input equals to 60 Watt output and 40 Watt of generated heat. Typicaly a motor is not moving when unloaded and there is no airflow around the motor. Exchange of heat between motor and air is at it's lowest while it is in a state where it generates quite huge amounts of heat (due too bad efficiency). Now a good motor design comes into play: heat from the windings (where most of the heat is generated due to copper resistance) must be transfered as quickly as possible to the motor housing so that heat exchange with the surrounding (cooler air) can be achieved.

Too much load (overgeared):
In this state you will see Amp spikes which exceeds 50A easely. Starting from this amperage level efficiency starts to drop. A vicious circle begins. The more power you will require (more Amps) the more Watt in heat is generated. But not linearly but more like exponentialy. At 70A the motor will already generate 112W in heat (now imagine holding a 110W+ light bulb in your hand). Motor mass is not enough to take this amount of heat for a long time and air flow is not sufficient any more to dissipate heat. Your motor will heat up. If you're lucky your battery will be drained so fast that your motor will not reach a critical point where the insulation of your whindings starts to melt and will result in burnt up windings.

If you operate your motor in its best efficiency range with proper airflow dissipating 40-60 Watt is way more easy then if your motor is not moving and almost generates the same amount of heat (unloaded state). I think this is the easyest part to acknowledge.

Undergeared for me is operating your motor somewhat below it's best efficiency range where amps may be low but heat build up is almost the same as if running your motor in its best efficiency range but with lesser airflow. Resulting in a motor which heats up faster.

I hope this makes sense.

Thank you Brian for your explanations. I see what you mean about all the energy having to be dissipated (the Io) under no load. Only a small percentage of the watts in is used to actually turn the rotor, so that other large portion of energy can't just go poof; law of conservation of energy says it's gotta do something or go somewhere, so it turns into heat.

I'm leaning towards explanation number 1, just because it seems to me that it would affect the motor's temperature far more than speed (especially if the motor is mounted inboard like on a 4wd touring car, and air flow is limited). Airflow is a huge factor in cooling though, I can't doubt that (though air is supposed to be a pretty crappy fluid for heat transfer I suppose).

I understand what you're saying othello, but that raises another question. Why do brushless motors have a best efficiency range under a given load?

Or furthermore; what is affecting the efficiency throughout the RPM and power band?

My guess is that startup of a brushless motor means a high current draw to initially get moving. Once you are already moving though, it takes a fraction of the current to keep the motor going, and it also takes a fraction of the startup current to go from say 30,000 RPM to 40,000 RPM. Am I on to it or no?

[othello]

>Why do brushless motors have a best efficiency range under a given load?

The number of turns for a given motor will influence how thick the copper wires can be. A smaller copper wire has more resistance (resulting in a bigger voltage drop) and will heat up quicker at a certain amperage when compared to a thicker wire. While the wire heats up resistance also rises. So for a given copper wire you will only be able to flow a certain current before heating it up too much. This is one part of the equation why a motor might be more efficient at 50A then at 100A.

A Neu 1515/2Y (1100kv) for example has thinner copper windings then a Neu 1515/1Y (2200kv). Optimum efficiency for the 2Y (2 turns) might be reached between 10-50A. When flowing more then 50A over those thin windings, those windings will heat up too fast (resistance is 0.019 ohm). The 1Y motor only has a resistance of 0.006 ohm. This is why you can flow a lot more current over its windings before they will heat up too much. And this is also the reason why it will suck more current then the 1Y motor.

If you would apply say 30V to those motors while you block their rotor then theoreticaly they will suck up to:

Neu 1515/2Y (0,019ohm): 30V/0,019ohm=1578A
Neu 1512/1Y (0,006ohm): 30V/0,006ohm=5000A

Also explaining why you will need much better batteries with a higher C rate for a low turn motor.

Other efficiency losses do come from bearing friction, hystheresis ... all this will be part of the overall efficiency at a given load.

>My guess is that startup of a brushless motor means a high current draw to initially get moving. Once you are already moving though, it takes a fraction of the current to keep the motor going, and it also takes a fraction of the startup current to go from say 30,000 RPM to 40,000 RPM. Am I on to it or no?

Normally, with perfect grip under constant acceleration amp draw will get higher and higher while revving up. Shortly before top speed when acceleration gets slower and slower your amp draw will get less. While holding top speed you normally will need less Amps when compared to accelerating.

Ah.. Thanks for shedding some light on that.

[GriffinRU]

Inductive losses would be load and RPM dependent. (Core properties, windings. PWM...)
And not to forget about inertia and rotor weight.
All the rest is very simple to calculate.

[zeropointbug]

One of many things that can be said about the Neu motors, is that they do not suffer from no/low load efficiency losses. You can rev the thing for hours with no load and it won't get hot. A feigao will literally get hot within seconds.