I How to Calculate Power of a Turning Shaft?

AI Thread Summary
To calculate the power of a turning shaft, it's important to note that if the mass is rotating at a stable speed of 100 rpm without any change in kinetic or potential energy, the power output is effectively zero. However, losses due to friction or other dissipative forces could affect this calculation, although they cannot be estimated with the provided data. The discussion also highlights the concept of "instantaneous power," particularly in applications like steam traction engines, where energy stored in flywheels can provide short bursts of power to overcome temporary loads. This principle is relevant in various mechanical systems, emphasizing the difference between average power and peak power available during specific conditions. Understanding these dynamics is crucial for accurately assessing the performance of rotating systems.
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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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.
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
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