Using instantaneous torque to determine energy

  • #1
Hi all - This may seem like a very simple exercise, but for some reason it's giving me trouble...

I am trying to determine the difference in energy (ultimately looking for kWh) used between two different electric motors by utilizing instantaneous torque measurements I have. I would like to do this so that I can chart energy consumption instantaneously as well across different motor speeds. Here's what my thought process is:

- Integrate torque signal in N-m: This gives me N-m-sec so that I have cumulative torque over a fixed duration (sec)
- But a Watt = 1 Joule/sec...?

Again... Silly question, but somehow i'm stumped
 

Answers and Replies

  • #2
SteamKing
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Boy, I'll say you're stumped. But posting your question in Astronomy and Astrophysics is unlikely to attract many helpful responses.
 
  • #3
CWatters
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Power = Torque * angular velocity.
where
Power is in Watts
Torque is in Nm
Angular Velocity in Rad/s.

To convert rpm to Rad/s multiply the rpm by 2Pi/60.

To convert W to kWh just multiply by (60*60)/1000.

PS: You need both the torque and rpm/angular velocity. You can't do it with just the torque.
 
  • #5
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You stated you were trying to "chart energy consumption instantaneously", but measuring the torque and angular velocity of the motor output won't give you that. It will give you mechanical energy output which will only be a lower bound on energy consumption. Consumption and output will always differ by the efficiency of the motor, which could vary widely depending on speed and load.

Why not measure the electrical energy directly?
 
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  • #6
jim hardy
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CWatters answered in his PS, maybe it could use some emphasis ?

Torque has the units force X distance but is not work.
You can prove that to yourself - put a wrench on a tight nut and push gently. No movement = no work done.


Work has units force X distance moved

that's why you must multiply torque by angular displacement to get work

i'm old,
in the English units of my day
horsepower = 2 X π X Torque(in foot-pounds_RPM /33,000

- Integrate torque signal in N-m: This gives me N-m-sec so that I have cumulative torque over a fixed duration (sec)
- But a Watt = 1 Joule/sec...?
integrate product of torque and ω
∫ nm X radians/sec = nm = joules ,,, since radians are unit-less

any help ?
 
Last edited:

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