Calculating Power & Torque for Flywheel Drives

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To calculate the appropriate motor power and torque for driving a flywheel at 2000 rpm, one must consider the effects of friction, as real systems cannot achieve ideal conditions. In an ideal scenario with zero friction, the power output at 750 rpm equals the power input at 3000 rpm, but this is not achievable in practice due to losses from friction and windage. The discussion emphasizes that the flywheel does not amplify power; it merely stores energy, and the output cannot exceed the input. Calculations for motor size should account for these real-world losses to determine the actual power needed. Understanding these principles is crucial for accurate design and performance expectations in flywheel applications.
Bassam Salman
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Hi
I have a personal project, In link bellow in figure A I need to calculate appropriate power motor (HP) and torque (N.M) to drive flywheel at maximum speed 2000 rpm and what is value of resulting power on 750 rpm pulley.
In figure B if we use two flywheel the weight for each half weight of flywheel in figure A do motor power and resulting power equal to figure A

http://img3.file-upload.com/i/00209/utjydznopqfe.png
 
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At steady state, in an ideal (lossless) system, the power out at 750 rpm will be equal to the power in at 3000 rpm.

Since no real system can be built without losses, you will need to account for the losses in the system to see how much power can be drawn out. Much depends on the quality of your bearings, the alignment of shafting, and the windage of the several elements.
 
:welcome:

Your answer depends on friction. The information you provided says nothing about friction.

With zero friction it takes zero power to maintain a rotation at constant speed.

Are you sure that the question is related to constant speed rather than acceleration?
 
Real life flywheels have friction and thus power loss.
 
Dr.D said:
At steady state, in an ideal (lossless) system, the power out at 750 rpm will be equal to the power in at 3000 rpm.

Since no real system can be built without losses, you will need to account for the losses in the system to see how much power can be drawn out. Much depends on the quality of your bearings, the alignment of shafting, and the windage of the several elements.
Is that mean no power loss on flywheel
 
anorlunda said:
:welcome:

Your answer depends on friction. The information you provided says nothing about friction.

With zero friction it takes zero power to maintain a rotation at constant speed.

Are you sure that the question is related to constant speed rather than acceleration?

Now I need calculate without friction and at constant speed
 
An ideal flywheel requires no power to run at constant rotational speed. Real flywheels (and every other machine part as well) always involve friction at bearings and windage. How much these friction losses are depends upon the shapes, the speeds involved, and the surrounding fluid medium.
 
Bassam Salman said:
Now I need calculate without friction and at constant speed

You are not understanding because I already gave you the answer.

anorlunda said:
With zero friction it takes zero power to maintain a rotation at constant speed.

@Dr.D is telling you the same thing I am.

Edit: rearranged the above.
 
Ok thank you Dr. D and anorlunda
I need value of motor size (calculation) and also calculate out power
 
  • #10
Bassam Salman said:
I need value of motor size (calculation) and also calculate out power
If there is no friction, the output power is equal to the input power; the flywheel doesn't change that.

You aren't thinking that the flywheel let's you get more power out than you put in, are you? That would violate the laws of physics and therefore the rules of this forum...
 
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