Calculating Flywheel Mass and Diameter for Efficient Generator Drive

[donblaho]

Hi guys! I have a real life physical problem. I need to calculate the mass and diameter of a flywheel made of steel to run generator with parameters: 123 kVA - 98 kWe; Frequency 50 Hz, 400 V, Class F M8 - 674 nM.

The better question is, how much (kinetic) energy do I have to input into the generator to run it? I can do the calculations, i just don't understand to generator parameters well.

Any tips?

 

[jrmichler]

1) Why do you want a flywheel? If you understand that, then you should be able to size it.

2) The energy input to the generator is equal to the energy out divided by the generator efficiency. If your generator is putting out 80 kW at 90% efficiency, then the energy input is 89 kW.

Keep in mind that the energy input is based on the actual generator energy output, and that the actual generator energy output must be equal to or less than the generator nameplate rating.

 

[berkeman]

I need to calculate the mass and diameter of a flywheel made of steel to run generator with parameters: 123 kVA - 98 kWe

Welcome to PF.

How long do you want your generator to run? The flywheel will obviously slow down as you extract energy from it. And since it will be slowing down, it seems like you will want to generate DC from the flywheel and then follow that with a DC-AC Inverter to give you the steady 50Hz power and output voltage until the flywheel slows too much to generate enough DC power to keep the system going.

 

[donblaho]

Thank you both!

I have to use flywheel, because its "contractor's" request. It is a part of the bigger project. However.

The idea is to spin up the flywheel to f.e. 1500 rpm, let it produce energy to empower the generator for like 10 minutes and at some point (where as you mentioned the generator will not have enough energy to generate the voltage) bring it back to the original rpm and repeat.

I am able to calculate the coefficient of fluctuation (eq. 1)

(eq.1),

but now i need kinetic energy for mass moment (eq. 2). So the question is, am i able to get the energy from the generator parameters?

(eq.2).

Then I will be able to get mass/diameter from its value.

Or am I missing something? Is there a better way to proceed?

 

[berkeman]

(leaving aside for the moment the "contractor requirement")...

Is the energy source that spins up the flywheel only available intermittently? If it is continuously available, why not use it directly? If it is only available intermittently, it seems like pumped hydro might be a better way to do the power smoothing function...

 

[berkeman]

I need to calculate the mass and diameter of a flywheel made of steel to run generator with parameters: 123 kVA - 98 kWe

BTW, if you do end up having to use a flywheel that can store enough energy to generate that power for 10 minutes, what kind of safety precautions are you planning on implementing? How will you build and balance such a flywheel energy storage device? What is your background in large high-power machinery?

 

[jbriggs444]

run generator with parameters: 123 kVA - 98 kWe; Frequency 50 Hz, 400 V, Class F M8 - 674 nM.

The idea is to spin up the flywheel to f.e. 1500 rpm, let it produce energy to empower the generator for like 10 minutes and at some point

So we want 1500 rpm. We want enough energy for 98 kilowatts of power for 10 minutes. We have no figure for how slowly the wheel can be allowed turn. So let's just arbitrarily say 1000 rpm.

This is grist for a first year physics mill. 98 kilowatts for 10 minutes. 98,000 * 10 * 60 = call it 60 megajoules.

We want for this to be the difference in rotational kinetic energy for a flywheel spinning at 1500 rpm versus one spinning at 1000 rpm. Kinetic energy goes as the square of rotation rate. This is a ratio of 1.5 to 1. So 2.25 to 1 for the energy. The total is 2.25. The usable delta is 1.25. We get to utilize 5/9 of the energy stored in the 1500 rpm flywheel. So we need to multiply that 60 megajoules by 9/5 and store 108 megajoules in the full speed flywheel.

Rotational kinetic energy is given by $E=\frac{1}{2}I\omega^2$ where $\omega$ is the rotation rate in radians per second. So we need to convert that 1500 rpm to radians per second. 1500 times 360 gives degrees per minute. Divide by 60 for degrees per second. Multiply by pi/180 to get radians per second. [Or use an online converter]. 157 radians per second. Square this and divide by two and we get 12324.5 radians squared per second squared. We can solve for moment of inertia ($I$) by dividing energy (108 megajoules) by this figure. We need a moment of inertia of 8763 kilogram meter squared. Call it 9000.

If we assume a solid cylinder of mass m and radius r, the moment of inertia we get will be $\frac{1}{2}mr^2$. The mass we need will depend on the radius we choose. Let us arbitrarily suppose a 1 meter radius (6 foot diameter flywheel). Then our multiplier on mass is a simple $\frac{1}{2}$ and we need an 18,000 kilogram wheel. Call it 18 tons.

Assuming that I have not screwed up somewhere along the line.

 

[berkeman]

and store 108 megajoules in the flywheel.

Or roughly equivalent to 100 sticks of dynamite...

Dynamite is usually sold in the form of cardboard cylinders about 200 mm (8 in) long and about 32 mm (1+1⁄4 in) in diameter, with a weight of about 190 grams (1⁄2 troy pound).[6] A stick of dynamite thus produced contains roughly 1 MJ (megajoule) of energy.[7]

https://en.wikipedia.org/wiki/Dynamite

 

[berkeman]

BTW, this looks to be a good reference for FESS:

Critical Review of Flywheel Energy Storage System

https://www.google.com/url?esrc=s&q...QgAN6BAgMEAI&usg=AOvVaw3WvgAdr0gfVEtuUAZd7V0m

 

[donblaho]

Well, contractor made a mistake in given parameters. The duration of spinning flywheel is not 10 minutes, but 10 seconds. Now the mass and energy values are not so crazy.

Thank you all for your help!

Johny