Flywheel & Energy: Calculating Efficiency

In summary, the conversation is about a flywheel experiment in which energy is transferred from a falling mass to the flywheel. The question is whether the energy is fully absorbed by the flywheel when it reaches its maximum angular acceleration or its maximum angular velocity. There is also a discussion about the calculation of power and the understanding of the mechanism involved in the experiment. Suggestions are made to measure with a smaller mass to better understand the behavior of the system.
  • #1
joe_leen
3
0
Hi all,

I have a flywheel which I'm putting energy into from the rest position. The energy is been put in from a mechanism attached to a falling 30 kg mass.

Now the mass falls for 0.392 seconds before it hits the ground. The flywheel accelerates and reaches a maximum acceleration at 0.434 seconds. The maximum angular velocity is reached at 0.857 seconds. I have this set up in a lab and indirectly measuring the data from a hall sensor and magnet attached to the flywheel.

The question is this:

The falling mass puts in an energy of 567 watts. I want to calcuate the efficiency of the mechanism that transfers the energy into the flywheel. So is all the energy from the falling mass absorbed by the flywheel when the flywheel reaches its maximum Angular Acceleration or it's maximum Angular Velocity ?

I think maximum acceleration but open to discussion.


Thanks for any help !

Joe
Ireland
 
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  • #2
I don't understand why the acceleration of the rotational flywheel motion reaches its maximum after the driving weight has reached the ground. And why is the angular velocity still accelerating after the weight has reached the ground?

I guess there is something about the mechanism I don't understand.

You write "The falling mass puts in an energy of 567 watts". But energy is not measured in Watts (but Joules). A quantity in Watts is a power, and indicates change of energy per unit time in your system of weight+flywheel. But the power generated by the action of gravity on your system should be time-dependent in this case, since the force is constant but the speed is not.

The kinetic energy of the weight is not transferred to the flywheel at the moment the weight hits the ground. It is rather converted to heat and deformation at the instant of collision between it and the ground.

Maybe there are some things about this contraption that I didn't understand :-)
 
  • #3
Sorry, I ment to say the power the falling mass is putting into the flywheel.

The Total Energy the mass has before it's released is given by the potental energy equation: PE = mass(kg)*gravity(m/s^2)*height(m) - giving units in joules.

From the kinematic equations of motion for a body moving linearly at a constant rate of acceleration the time for the mass to hit the ground can be calculated (V = U + at).

Then divide the Total PE by the time taken for the mass to fall and you get the Total Power of the falling mass. This power is transferred into the flywheel. As the mass is moving at 3.85 m/s before it hits the ground, as you say it transfers it's kinetic energy into the ground, resulting in heat, sound and deformation.

Well think about this, at the point of maximum angular velocity (the point when it just starts to slow down) the angular acceleration must be zero and changing to a negative value. This ties in with reaching the maximum angular acceleration before the maximum angular velocity.

When the maximum acceleration is reached the flywheel doesn't instantaneously stop increasing in angular velocity, remember the angular acceleration at this point is still positive but reducing, hence the velocity can still increase, just at a lower rate.

Added a graph of the data, should help.

Still questioning when to equate when is the power absorbed by the flywheel equivalent to the power of the falling mass.
 

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  • #4
What are the dimensions of the flywheel, of the acceleration of gravity applied to a 30kg gives 294N of tension force applied to the rope(assuming this device uses rope to connect the mass?) or whatever, then that force, depending on where it is applied, gives the torque, so I am no expert but I think I need to know the mass, and moment of inertia of the flywheel to really know what going on here.

At how many seconds does the angular acceleration become 0? It should be pretty close to .392 since that is when gravity is balanced by the normal force of the ground Oh. .857seconds... So it accelerates for over twice as long as the force is applied...I think it depends on the mechanism perhaps... But I would go with angular velocity, as when there is still angular acceleration, energy must have been stored in your mechanism and is being dispenesed still. until it is 0 and the sum of forces is then 0, so work is no longer being done on the wheel by your mechanism..at least not positive work?
 
  • #5
joe_leen said:
Then divide the Total PE by the time taken for the mass to fall and you get the Total Power of the falling mass.

This is what I would call "average power".

This power is transferred into the flywheel.

Yes, but not in its entirety. This potential energy is converted into kinetic energy of the falling mass, as well as kinetic energy in the flywheel. And when the first part is lost upon impact, your are left with the kinetic energy in the flywheel.

Well think about this, at the point of maximum angular velocity (the point when it just starts to slow down) the angular acceleration must be zero and changing to a negative value. This ties in with reaching the maximum angular acceleration before the maximum angular velocity.

I think the angular acceleration should be constant from the time the weight is released until it hits the ground, since the force of gravity is constant in this case, and the mass and inertia of your system is constant. The acceleration value will depend on the moment of inertia of the flywheel and the mass of the falling object, as well as the torque of your "rope" upon the flywheel.

Therefore, the angular velocity will reach its maximum at the instant the weight reaches the ground.

Are you sure the time-resolution of your measurement apparatus is good enough for the very rapid acceleration you seem to get from the 30kg mass? How about trying the experiment with a 5kg mass first so that the acceleration phase takes a longer time so that the behaviour is easier to study?
 
  • #6
The angular acceleration should stop as soon as the weight impacts the ground, unless - the mechanism has some spring or rubber band like effect that stores energy while the weight is descending, in which case angular acceleration will continue until the tension in the mechansism goes to zero. If there's a rope of significant mass supporting the weight, then the weight of the rope could also be a factor.
 
  • #7
Your rigth about the resolution of the measurement. I currently only have two magnets on the outer edge of the flywheel which the hall sensor detects. The outer diameter of the flywheel is 310mm. So going to make a high resolution sprocket to capture detailed data. Flywheel mass is 23.4 kg and moment of Inertia is 0.2589 kg m2.

I agree that the maximum angular acceleration of the flywheel should coinside with the instant the mass hits the ground. My data set is too sparse to detect this exact instant in time.

Ok here is the fun bit. Their appears to be too "power" balancing points. The first is as follows:

A) The sum of the Positive Instantaneous Power measurements is 813 watts (this is summed during the period the angular acceleration is positive).

The sum of the negative Instantaneoust Power measurements is -793 watts (this is summed during the period the angular acceleration is negative).

This leaves a difference of 20 watts, which I attribute to windage and also friction in the device.

But this implies a total input of "power" of 813 watts. As the falling mass input an average of 566 watts of power, something not correct with this analysis.

B) If you sum the positive Instantaneous Power upto the point of maximum angular acceleration you get 396 watts.

If you sum the remaining Instantaneous Power, that is from the point of maximum angular acceleration to the time the flywheel stops rotating you get a power measurement of -376 watts.

So here you still have a 20 watt "loss" in the system but crucially there now is a difference of 170 watts (566 - 396) between what the falling mass inputs and what's apparently input to the flywheel.

I'm more inclined to believe that there is a further "loss" or dissipation of energy to the tune of 170 watts somewhere, which gives an efficiency of the device of 70% (396/566).

The way I'm inputting the power of the falling mass to the flywheel involves gears.

So what I'm thinking is that the 30% loss is attributed to losses resulting from the input device / method. And the 20 watt loss is primarily due to the bearings and windage of the flywheel accelerating from zero to a maximum and back to rest. At this point the input mechanism of gears is de-coupled from the system.

Look forward to your comments on this !
The primary eqations I'm using are

Instantaneous Power = Torque * Angular Velocity

Instantaneous Torque = Moment of Inertia * Angular Acceleration

Hence:

Instantaneous Power = Moment of Inertia * Angular Acceleration * Angular Velocity

Does the instantaneous power measurement incorporate the rotational kinetic energy of the flywheel ( 0.5 * I * Angular Velocity2 )

Also the mechanism joining the falling mass to the flywheel is "solid" as in it offers no energy storage capacity. So no rubber band effect going on.
 
Last edited:

1. What is a flywheel and how does it work?

A flywheel is a mechanical device that stores rotational energy. It consists of a heavy disc or wheel with a shaft through its center. When a force is applied to the wheel, it rotates and stores the energy as rotational kinetic energy. The wheel continues to spin until the energy is needed, at which point it can be used to power a machine or perform work.

2. How is the efficiency of a flywheel calculated?

The efficiency of a flywheel is calculated by dividing the output energy by the input energy and multiplying by 100%. The output energy is the work done by the flywheel, while the input energy is the energy put into the flywheel to make it rotate. The formula for efficiency is: Efficiency = (Output Energy / Input Energy) x 100%. A higher efficiency indicates that less energy is lost in the conversion process.

3. What factors affect the efficiency of a flywheel?

The efficiency of a flywheel can be affected by several factors, including the material and weight of the flywheel, the friction between the flywheel and its bearings, and the speed of rotation. Generally, a heavier and smoother flywheel made of a durable material such as steel will have a higher efficiency.

4. Can the efficiency of a flywheel be improved?

Yes, the efficiency of a flywheel can be improved by reducing the friction between the flywheel and its bearings, using a more efficient material for the flywheel, and increasing the speed of rotation. Additionally, regular maintenance and lubrication of the flywheel can help maintain its efficiency over time.

5. What are the practical applications of flywheels?

Flywheels have many practical applications, including energy storage for renewable energy sources, such as wind and solar power. They are also used in vehicles, such as hybrid cars and buses, to store and release energy during acceleration and deceleration. Flywheels are also used in machinery and industrial equipment to store and release energy, providing a smooth and continuous power source.

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