How to Calculate Power of a Turning Shaft?

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Discussion Overview

The discussion revolves around calculating the power of a turning shaft, specifically in the context of a mass rotating at a stable speed of 100 rpm. Participants explore the implications of power in relation to kinetic and potential energy, as well as the effects of external forces and energy storage systems like flywheels.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests a formula for calculating power in relation to a mass turning at a stable speed, indicating a desire for clarity on the topic.
  • Another participant argues that if there is no change in kinetic or potential energy, the power is effectively zero, although they acknowledge potential losses due to friction or other dissipative forces.
  • A different participant introduces the concept of instantaneous power, referencing a historical example of steam traction engines and their ability to deliver power boosts when needed, despite the average power rating being lower.
  • Further clarification is provided regarding the operation of flywheels, emphasizing that they can deliver short bursts of power while slowing down, which is relevant to specific applications.
  • Another participant reiterates the idea of instantaneous power in the context of flywheels and compares it to the specifications of other systems, such as a motor cruiser's radar system.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of power, with some asserting that power is zero under certain conditions, while others highlight the relevance of instantaneous power and energy storage systems. The discussion remains unresolved regarding the calculation and implications of power in this context.

Contextual Notes

Participants note that the discussion is limited by the lack of information regarding dissipative forces and specific conditions affecting the power calculation.

Soko
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Hello, can someone tell me how, or provide me a formula, to calculate the power of the turning shaft? in Nm or Watts, I don't mind. The mass (23Kgr) is turning at stable speed of 100 rpm by external independent force. The shaft is supposed to be on bearings both sides..
I made a picture to save words..
powerofturningmass.png

Thank you
 

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Soko said:
The mass (23Kgr) is turning at stable speed of 100 rpm
So the power is 0. Or in other words there is no change in KE or PE so no work is being done.

The one exception would be any losses due to friction or other dissipative forces. But there is no way to estimate those with the given information.
 
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Likes   Reactions: CWatters and russ_watters
Although the Energy in the rotating system is the most obvious measure, Power can also be relevant.
Many years ago, I was at a steam traction engine rally and there was a demonstration of a traction engine used as a 'ploughing engine', using a cable around a wheel, slung under the body of the engine. The engines operated in pairs and used the cable to pull a plough back and forth across a field. There could be large instantaneous loads when the plough came across a boulder. The horsepower rating would be a humble 35hp, which would not deal with this load. But the flywheel could overcome the problem over a short time.
The 'Instantaneous Power' available from the flywheel was, apparently, part of the spec of these engines. Sloppy terminology? Not really because it was relevant to the particular application. It's similar to comparing the weight of a hammer with the damage it can do.
 
Just in case you think Sophiecentaur's contradicts Dale's post that's not the case. The flywheels Sophiecentaur mentions slow down while delivering the power boost needed to get the plough past the boulder. Once past the boulder the steam engines have to bring the flywheel back up to full speed ready for the next one.
 
CWatters said:
Just in case you think Sophiecentaur's contradicts Dale's post that's not the case. The flywheels Sophiecentaur mentions slow down while delivering the power boost needed to get the plough past the boulder. Once past the boulder the steam engines have to bring the flywheel back up to full speed ready for the next one.
Yes. The Power that is quoted is available for a short time from Energy stored in the flywheel. When the man on the tannoy used the phrase "Instantaneous power" I did a double take but, of course the same approach is used with the spec of a motor cruiser's Radar system. 4kW pulses and the boat uses 24W of electrical power from the boat's system.
 

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