Can a Flywheel Store 1 MW of Power for 50 Seconds with Only 33% Energy Loss?

[Pepelepoe]

Hi fellow scientists and engineers,

I have a question regarding the design of a flywheel. I know that the actual engineering of a flywheel is very complex and in no way I should expect actual practical answers from this simple question. I'm just trying to get some estimates for a renewable energy project that I'm working on. I'm trying to find out the best way to calculate the dimensions, mass and torque of a flywheel so that it produces 1500 HP during 50 seconds while reducing its kinetic energy to no less than 33%. It's maximum angular velocity should be no more than 4,000 RPMs.

I'm a bit confused with power and the time delta. As you all know Power = (Work) / (delta t). In my calculations I converted 1500 HP into Watts, which is 1.12 MW. I then calculate the Work done by substituting the 1.12 MW for power and assuming that my delta t is 50 seconds. Is this approach correct? What I want to do in essence is be able to store 1 MW of power in a flywheel which feeds a generator during a t(assumed as 50 seconds) second cycle and not loose more than 33% of its speed during this period. Hope that the problem is clear for you guys. Any help is appreciated. Thanks.

 

[billy_joule]

There are probably a few ways to do this, what comes to my mind is as the most straight forward is conservation of energy:

E_rotational = 1/2 I ω²

ΔE = 1.12 MW * 50 s = E_initial - E_final

Where ω_final = 0.67ω_initial
and ω is known

Solve for I (moment of inertia)
Having I you can do some calcs for flywheel size & shape (google will find you a table of I formulas for various simple shapes, disc, ring etc)
You will likely find you need an incredibly large and/or heavy flywheel. Let us know how you get on.