How to calculate torque and power for self starter and flywheel

[Sudhakar Naidu]

How to calculate following input and output parameters-
1. Input at starter motor - Drive torque and power as shown below
2. Output at flywheel axle - torque, power, Kinetic energy, centrifugal force as shown below

Input parameters: starter motor rpm -3600, Gear ratio- 18: 1
if any parameter missed please consider some sample value because i am electrical background, so later i will find correct value.

Thanks in Advance.

 

[russ_watters]

There is no image.

 

[Sudhakar Naidu]

 

[Sudhakar Naidu]

Thanks for quick look.

 

[CWatters]

Assume constant velocity and apply conservation of energy...

power in + power out + losses in gearbox = 0

If you ignore losses in gearbox then

Power out = power in

 

[CWatters]

Also...

Torque out = 18 * Torque in

RPM out = RPM in / 18

 

[jack action]

For power $P$:
$P_{out} = P_{in}$

For Torque $T$, assuming a gear ratio of 18:1:
$T_{out} = 18 T_{in}$

For rpm $\omega$ in rad/s, assuming a gear ratio of 18:1:
$\omega_{out} = \frac{\omega_{in}}{18}$

For the kinetic energy $E$ needed for the system considering the mass moment of inertia ($I$):
$E = \frac{1}{2}I_{out} \omega_{out}^2 + \frac{1}{2}I_{in} \omega_{in}^2$

For the centrifugal force, I'm not sure what you are looking for, since it usually applies to a particle in rotation rather than an object rotating about its center of gravity.

 

[Sudhakar Naidu]

Hi CWatters and Jack,
Thanks for quick solution.

This calculation is for feasible check on 'free energy generators', which are in 'YouTube'. They are really getting that much output or not(not sure).

One doubt on Power, like-
Power output = Power input --> really is this correct at this situation?, because in output power there is 'moment of inertia' this parameter will increases output power right?, not sure just am asking you( am electrical back ground, sorry if am wrong).

One Practical example to be more clear on this:
Consider 1 meter dia of flywheel(Rim type)setup which is ready to rotate, apply 10 - 20N force via bare hand then flywheel will rotate at certain speed. In next moment(after 2- 3 sec) stop suddenly the rotation of the flywheel in 'bare hand', then we feel more force is applied to stop.

My understanding,
Power out == Power in --> is correct, when Flywheel is replaced with same dimension gear wheel. Kindly let me know your thought on this...

Thanks for the understanding.

 

[jack action]

Power in = power out is always correct. When you add inertia (i.e. a flywheel), it will store energy (which will not get out initially). Once 'filled', the then energy in = energy out (or power). Once there will be no energy in, the stored energy will get out. Overall, energy in will always be equal to energy in.

Think of it as an empty water tank with an input hose and an output hose. The water coming in will always equal the water coming out. But the tank will filled before the water began to get out and, once the water will stop coming in, the tank will empty through the output hose.

 

[CWatters]

One doubt on Power, like-
Power output = Power input --> really is this correct at this situation?, because in output power there is 'moment of inertia' this parameter will increases output power right?, not sure just am asking you( am electrical back ground, sorry if am wrong).

A flywheel is a bit like a capacitor. It's an energy store not an energy source. You only get power going in or out of a capacitor when the voltage changes. Likewise, energy or power only flows into or out of a flywheel when the angular velocity changes (eg when the flywheel is accelerating or decelerating).

One Practical example to be more clear on this:
Consider 1 meter dia of flywheel(Rim type)setup which is ready to rotate, apply 10 - 20N force via bare hand then flywheel will rotate at certain speed.
In next moment(after 2- 3 sec) stop suddenly the rotation of the flywheel in 'bare hand', then we feel more force is applied to stop.

Correct. For a brief period you can get it to produce more force (torque). However it doesn't produce any more energy then you put in.

This is exactly the same as charging a battery or a capacitor slowly and then discharging it quickly. If you put 1kW into a battery for 1 Hour it contains 1kWH of energy. You can then take 10kW out but only for 6 mins.

You took more power out (10kW) than you put in (1kW) but you only got the same energy out (1kWH) as you put in (1kWH).

Sorry I had to edit this post to correct mistakes.

 

[Sudhakar Naidu]

Hi CWatters and Jack,
Thanks for very clear examples. Even i believe strongly with you both.

Because of following video(providing i/p 15A and getting output 27A, it is near to 2 times), not satisfied my self... some thing is missed some where... my feeling says, do we missed gravity factor(9.8) in 'flywheel design calculation'?(am not very sure, but output of following video triggers me like that). if any thought on this please...

 

[jim mcnamara]

Let's wrap up this thread. The basic question has been resolved= you cannot create extra energy above what you start with. Period. The End.