How Does Load Impact Electric Motor Efficiency and Speed?

In summary: M Motors have a lower efficiency rating at full load due to the use of commutators. Commutators allow the motor to run at a higher speed with less current, so the rated efficiency at full load may be lower.
  • #1
tyquestions
5
0
Does anybody know of an equation or rule of thumb for power decrease as load increases?
As an example an electric golf cart will go say 15mph with one 100lb driver. Put 4, 200lb people in it and it will go much slower, say 9mph. Is there a way to determine what that loss will be other than simple observation?
Does an electric motor's efficiency rating play into it? One motor I am lookng at is 86% efficient. Does that mean 86% of the incoming electrical power is converted into mechanical power? If so, is that under no load conditons or some sort of industry standard load?
Thanks
Tyler
 
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  • #2
tyquestions said:
Does anybody know of an equation or rule of thumb for power decrease as load increases?
As an example an electric golf cart will go say 15mph with one 100lb driver. Put 4, 200lb people in it and it will go much slower, say 9mph. Is there a way to determine what that loss will be other than simple observation?
Does an electric motor's efficiency rating play into it? One motor I am lookng at is 86% efficient. Does that mean 86% of the incoming electrical power is converted into mechanical power? If so, is that under no load conditons or some sort of industry standard load?
Thanks
Tyler

I don't think it's the power decreasing, it's that given a fixed amount of power, with a higher load, you go slower.

Power is Work/Time, or Force*Distance/Time. So if the required force is going up, the time takes longer.

Don't know what load the 86% is at -- it doesn't give some curve or more test details in the datasheet? You could call the manufacturer to find out, perhaps.
 
  • #3
I’m going to make some assumptions because this problem can’t be solved without them. To the extent my assumptions are wrong, the answer will be wrong.

The first assumption is that the power out (angular velocity x torque) of the motor is constant. This won’t be strictly true because as the output shifts from higher angular velocity and lower torque to lower angular velocity and higher torque the windings and battery will carry more current, heat up and lose efficiency due to heat.

The second assumption is that the torque required is directly proportional to the weight of the golf cart and cargo.

From the problem description I don’t know if the four 200 lb people replace the driver or if they are in addition to the driver. I will assume the latter.

So, if the cart plus 100 lbs. goes 15 mph and the cart plus 900 lbs. goes 9 mph, I think we can make a proportion where c is the weight of the empty cart.

15/9 = (c + 900)/(c + 100)
c = 1100 lbs.
top speed = (12 * 15)/11 = 16.36 mph
speed = (16.36 * 1100) / (load + 1100)

Is this what you were looking for?
 
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  • #4
tyquestions said:
Does an electric motor's efficiency rating play into it?
Yes
One motor I am lookng at is 86% efficient. Does that mean 86% of the incoming electrical power is converted into mechanical power?
If so, is that under no load conditons or some sort of industry standard load?
Thanks
Tyler
Yes, its the conversion from electric to mechanical power. The efficiency is of course zero at no load (no output / a non zero current). The rated efficiency will peak at a given load. See the attached typical DC more operating curves.

Good mfn data sheet:
http://hades.mech.northwestern.edu/wiki/images/e/ee/Maxon_Motor_Guide.pdf
 

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  • #5
mheslep said:
Yes
Yes, its the conversion from electric to mechanical power. The efficiency is of course zero at no load (no output / a non zero current). The rated efficiency will peak at a given load. See the attached typical DC more operating curves.

Good mfn data sheet:
http://hades.mech.northwestern.edu/wiki/images/e/ee/Maxon_Motor_Guide.pdf

NEMA motors typically give full-load efficiency, not no-load. Most motors are designed to operate more efficiently at their designed load, so it makes sense to rate them based on full-load instead of no-load.

Although they may give nominal efficiency as well for worst case scenarios.

CS
 
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What is DC efficiency % and why is it important?

DC efficiency % is a measure of how effectively a direct current (DC) power supply converts input power into usable output power. It is important because it indicates how much energy is being wasted during the conversion process.

How is DC efficiency % calculated?

DC efficiency % is calculated by dividing the output power by the input power and multiplying by 100%. The formula is: (Output Power / Input Power) * 100%

What is the ideal DC efficiency %?

The ideal DC efficiency % is 100%, meaning that all of the input power is converted into usable output power with no energy wasted.

How does load affect DC efficiency %?

The load, or the amount of power being drawn from the power supply, can affect the DC efficiency %. The efficiency tends to decrease as the load increases. This is because more energy is required to power the load, resulting in more energy being wasted during the conversion process.

What factors can impact DC efficiency %?

There are several factors that can impact DC efficiency %, including the quality of components used in the power supply, the design and topology of the power supply, and the operating conditions such as temperature and voltage. Other external factors like power source fluctuations and load variations can also affect efficiency.

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