Flywheel Energy Storage and small engine

[grey]

I posted this on another section, but got no replies in three days...so thought maybe I'd had posted in the wrong section...apologies.

I have an idea pertaining to energy storage in flywheels, something like they do in the KERS, and need assistance here...

Ok, my idea involves using a flywheel of around 10 kg (steel) with I~1 kg.m^2, to be charged to an RPM of around 7,000 RPM.
This allows E = 0.5 x 1 x (7000x2x pi /60)^2 = 268 KJ of energy to be stored.

I want to charge it with a small engine (something like a Honda GX50) which, according to the charts will give max torque of 3 Nm at 4000 rpm and max power of 1.6 kW at 7000 rpm (approx).

I want to discharge it to run a light vehicle (170 kg including driver) such that the discharge takes place at one wheel only (a three wheeler). I want discharge in pulses.

Now, the questions:
1. How do I calculate the time it takes to charge the flywheel to 7000 RPM? Can I do it without using some intermediate device like a CVT or motor/generator?
2. How do I calculate the discharge time (and characteristics)?

Thanks in advance.

 

[Hologram0110]

I think that calculating an accurate time would be difficult because the numbers on the torque/RPM curve are for steady-state operation at the given RPM value and not for rapdily increasing RPM values where the inertia of the engine would be come significant. To get a realistic number, I would think you need much more information. It would likely be simpler measure it on the device in question.

The second question, would depend greatly on how you couple the fly wheel to the drive train.

 

[cjl]

Hologram: I would guess that the flywheel would have significantly more inertia than the engine, which means that you could safely ignore the engine's own inertia and simply use the power curve as given.

1) You probably could do it without any intermediate devices, but it might be difficult to start the engine with a heavy flywheel. If you have a torque curve for the engine, then it's fairly easy to calculate the acceleration of the flywheel using \tau = I\alpha

2) Discharge characteristics depend heavily on how you discharge it.

 

[pallidin]

I love flywheels! So much potential energy storage.(Or would it be more proper to say kinetic energy storage?)
A word of caution, though. Anything spinning at 7,000 rpm's is potentialy extremely dangerous.
Always kept safety first and foremost! Best of luck.

 

[grey]

Thanks for the reply guys.
Ok, back to the issues:
1. Charging. I have the engine curve available.
http://www.honda-engines.com/engines/gxh50.htm
how do i calculate it? asking because the torque keeps on changing (in a function which i don't know) so that i don't have an expression for T (torque)

And let's say, I have a diesel engine (with an almost constant torque of 7 Nm), does this make the job of calculating this easier? As in:
T = I x a
T = I x dw/dt
T dt = I dw
integrating (as everything is almost constant)
w = Tt/I
?

2. Discharging. I was thinking of discharging it by directly coupling with the axle (which spins on bearings with the wheel). Does the same expression ( T=Ia) suffice?
Also, in fundamental of vehicle dynamics, they say that to account for rotational objects in the car, you multiply the mass of the vehicle by a factor. If i want to do the opposite (account for linear movement for a rotational reference) what do i do?

 

[RonL]

Thanks for the reply guys.
Ok, back to the issues:
1. Charging. I have the engine curve available.
http://www.honda-engines.com/engines/gxh50.htm
how do i calculate it? asking because the torque keeps on changing (in a function which i don't know) so that i don't have an expression for T (torque)

And let's say, I have a diesel engine (with an almost constant torque of 7 Nm), does this make the job of calculating this easier? As in:
T = I x a
T = I x dw/dt
T dt = I dw
integrating (as everything is almost constant)
w = Tt/I
?

2. Discharging. I was thinking of discharging it by directly coupling with the axle (which spins on bearings with the wheel). Does the same expression ( T=Ia) suffice?
Also, in fundamental of vehicle dynamics, they say that to account for rotational objects in the car, you multiply the mass of the vehicle by a factor. If i want to do the opposite (account for linear movement for a rotational reference) what do i do?

Because I suck at math and equations, I'll put the idea out and let you work the numbers.
My thought is to find the sweet spot based on max engine torque of 4,000 rpm and add weight to the flywheel to compensate for lower speed storage, then spin it closer to the torque speed of your engine, (I'm thinking somewhere between 4 and 5 thousand) keep cycles as short as possible for quicker charge and discharge of storage.

I think I'm on thin ice with the forum, so send a PM to me if you can and I'll give you an idea about how to make use of some of the wasted heat off the engine.

Ron

 

[uart]

Thanks for the reply guys.
Ok, back to the issues:
1. Charging. I have the engine curve available.
http://www.honda-engines.com/engines/gxh50.htm
how do i calculate it? asking because the torque keeps on changing (in a function which i don't know) so that i don't have an expression for T (torque)

And let's say, I have a diesel engine (with an almost constant torque of 7 Nm), does this make the job of calculating this easier? As in:
T = I x a
T = I x dw/dt
T dt = I dw
integrating (as everything is almost constant)
w = Tt/I
?

2. Discharging. I was thinking of discharging it by directly coupling with the axle (which spins on bearings with the wheel). Does the same expression ( T=Ia) suffice?
Also, in fundamental of vehicle dynamics, they say that to account for rotational objects in the car, you multiply the mass of the vehicle by a factor. If i want to do the opposite (account for linear movement for a rotational reference) what do i do?

Grey, assuming you can get the thing started it will take roughly 5 to 6 minutes to spin it up from 1000 RPM to 7000 RPM with that torque/speed characteristic.

Your equations are ok. Just make sure you use radian/sec for "w" (2 Pi / 60 * revs/min). You can get a simple estimate by guessing the "average" torque, say 2.3 to 2.4 Nm, and using: t = Iw/T.

The problem with "average" torque is that it's not really a normal average, since it spends more time in T(w) regions where the torque is small and less time where the torque is large. You can handle this properly however by just taking reciprocals. As in :

dw/dt = T(w)/I

dt/dw = I/T(w)

\Delta\,t = I \, \int \frac{1}{T(w)} \, dw

I hope that makes sense. Basically all it means is that you should take the area under the reciprocal of the torque speed curve (dont forget to change the x-axis to radians/sec first) and then multiple this area by "I" to get the time in seconds. Make some allowance for windage and coupling losses and the moment of inertia of engine itself, otherwise you'll under estimate the time.

 

[grey]

Thanks uart

assuming you can get the thing started

This is because the inertia of the flywheel opposes engine operation, right? So, practically (without experimenting) the engine might not start at all?

In this case, suppose I have the engine already running at 4000-5000 rpm (max torque)...then how do i proceed with the same routine (of calculating the time to charge it).