Im a bit confused as to the efficiency curve; your theory says max power should be at 50% efficiency, but i guess i dont understand the graphs for certain motors then because according to the pmac-ds graphic http://www.electricmotorsport.com/store/pdf-downloads/PMAC-DS.pdf i found here, it looks like at 50% efficiency its only producing 15 pound inches of torque.
Motor efficiency
- Thread starter muffildy
- Start date
liveforphysics
100 TW
muffildy said:
All motors start from 0% efficiency, head towards peak efficiency at the point where copper losses equal all other losses, and then taper down towards 0% efficiency again (no-load RPM is always 0% efficient, as power in equals parasitic loss or it would be increasing RPM still).
You're looking at the 50% efficiency on the front side of the curve. Peak power happens at the back-end 50% on the curve, which rarely is displayed on graphs.
how would i run this motor at its highest efficiency?
You say it starts at 0%, goes to max, and then back down to 0%? So how can you keep it out of the low efficiency best?
If im reading the chart correctly, then it looks like
60-200 pound inches
4000 rpm 3200 rpm
So for every 5.7 rpm lost i gain 1 pound inch of torque while maintaining 80% or better efficiency?
Now of course 3200-4000 rpm isnt really a usable rpm range; This motor takes anywhere from 24-96v i think, and changing the volts changes the rpm by a constant for this kind of motor right?
Would that mean then to stay at 80% efficiency throughout most of the rpm range i would need to vary the volts and keep the torque required above a certain minimum level?
You say it starts at 0%, goes to max, and then back down to 0%? So how can you keep it out of the low efficiency best?
If im reading the chart correctly, then it looks like
60-200 pound inches
4000 rpm 3200 rpm
So for every 5.7 rpm lost i gain 1 pound inch of torque while maintaining 80% or better efficiency?
Now of course 3200-4000 rpm isnt really a usable rpm range; This motor takes anywhere from 24-96v i think, and changing the volts changes the rpm by a constant for this kind of motor right?
Would that mean then to stay at 80% efficiency throughout most of the rpm range i would need to vary the volts and keep the torque required above a certain minimum level?
an electric hybrid lightweight vehicle; eg 1400# car.
Is efficiency related to the rpm as well as the torque?
I found a document for a brushless DC motor (not sure if its perm magnet or not)
www.powertecmotors.com/aeffe3.doc
this seems to indicate that the percent of efficiency is worst at low rpm low torque, and best at max rpm max torque? If we keep torque above 50% and rpm above 50% its efficiency is actually fairly good, but the question is if this chart is what i can expect out of the motor were discussing.
Is efficiency related to the rpm as well as the torque?
I found a document for a brushless DC motor (not sure if its perm magnet or not)
www.powertecmotors.com/aeffe3.doc
this seems to indicate that the percent of efficiency is worst at low rpm low torque, and best at max rpm max torque? If we keep torque above 50% and rpm above 50% its efficiency is actually fairly good, but the question is if this chart is what i can expect out of the motor were discussing.
Ok, found an interesting equation:
Efficiency = {(torque*rpm)/9549} kw / {volts*amps/1000} kw
Using this equation plugging in numbers from that chart i get some interesting results...
For example:
at the 80 pound inches mark (9.03 Nm) it looks to be about 3800 rpm 74 v 60 amps
So (9.03*3800)/9549 = 3.59 kw output
with (74*60)/1000 = 4.44 kw input
3.59/4.44 = 80.85% efficiency which is fairly close to the efficiency that they have in the chart....but
for some reason motenergy says i need 180 amps DC to the controller in order to get 125 amps Ac to the motor. Is there a difference between AC amps and DC amps? If there is none, then wouldnt the real efficiency be (74*(180/125)*60)/1000 = 6.39 and 3.59/6.39 = 56% efficiency? so very confused
Efficiency = {(torque*rpm)/9549} kw / {volts*amps/1000} kw
Using this equation plugging in numbers from that chart i get some interesting results...
For example:
at the 80 pound inches mark (9.03 Nm) it looks to be about 3800 rpm 74 v 60 amps
So (9.03*3800)/9549 = 3.59 kw output
with (74*60)/1000 = 4.44 kw input
3.59/4.44 = 80.85% efficiency which is fairly close to the efficiency that they have in the chart....but
for some reason motenergy says i need 180 amps DC to the controller in order to get 125 amps Ac to the motor. Is there a difference between AC amps and DC amps? If there is none, then wouldnt the real efficiency be (74*(180/125)*60)/1000 = 6.39 and 3.59/6.39 = 56% efficiency? so very confused
Yes, efficiency is power out divided by power in. The 9549 converts rpm to radians per second.
The first chapter of this book is a good place to start learning motor theory:
http://books.google.co.uk/books?id=gbIDM60AvGAC&printsec=frontcover&dq=austin+hughes&hl=en&ei=S0GCTooTy5g6_q2FIg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC0Q6AEwAA#v=onepage&q&f=false
See also:
http://lancet.mit.edu/motors/motors3.html
WRT your car, you need to start by working out the maximum torque you require at the driving wheel.
The first chapter of this book is a good place to start learning motor theory:
http://books.google.co.uk/books?id=gbIDM60AvGAC&printsec=frontcover&dq=austin+hughes&hl=en&ei=S0GCTooTy5g6_q2FIg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC0Q6AEwAA#v=onepage&q&f=false
See also:
http://lancet.mit.edu/motors/motors3.html
WRT your car, you need to start by working out the maximum torque you require at the driving wheel.
those links dont seem to explain why i need 180 amps to go into the controller in order to get 125 amps into the motor. Wouldnt this mean im using 44% more DC amps in order to get the AC amps, therefore reducing the efficiency?
edit, also:
in the link i found:
V = E + IR = 8 + (4 x .5) = 10 volts
using data i found in http://www.electricmotorsport.com/store/pdf-downloads/PMAC-DS.pdf
I see a phase resistance of .0125 ohms
So, would this mean the equation above changes to 8+(4x.0125) = 8.05? doesnt seem right - would be around 99% efficiency if that was true.
edit, also:
in the link i found:
V = E + IR = 8 + (4 x .5) = 10 volts
using data i found in http://www.electricmotorsport.com/store/pdf-downloads/PMAC-DS.pdf
I see a phase resistance of .0125 ohms
So, would this mean the equation above changes to 8+(4x.0125) = 8.05? doesnt seem right - would be around 99% efficiency if that was true.
Similar threads
