Flywheel help 1

[cutinoak]

Please check my flywheel figure.

I have a steel flywheel 12" diameter 6" thick, and want to turn it 0-10,000 rpm in 60 seconds. Please eliminate bearing and air drag.

weight (12 / 2)^2) x 3.1416 x 6 x .29 = 196.79lbs
mass 196.79 x 32.2 = 6.11
slugs ft^2 .5 x 6.11 x ((12 / 2) / 12)^2 = .76
radians/ minute 10000 x 2 x 3.1416 = 62832
radians/second 62832/60 = 1047.20
radians/sec^2 1047.20 x 60 = 17.45
slug ft^2/sec^2 17.45 x .76 = 13.33
same as lb.ft 13.33
power 10000 / 60 x 13.33 x 6.28 = 13955
HP 13955 / 550 = 25.37


So it would take a 25 hp engine to run the flywheel from 0-10,000 rpm 60 seconds.

 

[BobM3]

Don't forget that some of the engine's power will be needed to accelerate its own inertia.

 

[zekeman]

Good point BobM3. In fact it may be high compared with the the flywheel when referred to the engine shaft . Then the general way to do this is to reflect all the inertia to the engine shaft , so the "sum" of the inertias are Isum where
Isum=Ifl/n^2+...Iothers, n being the gear ratio between the engine and the flywheel shaft. And then from the Newton differential equation
Isum*dw/dt=T(w)
w= engine speed, rad/sec
T(w)= the torque at the engine as a function of w
dt= differential time
dw=differential speed

From this (recharacterizing what I had previously shown for motors) solve for dt, the differential of t, the time.
Isum/T(w)*dw=dt
To get the answer you integrate both sides, the integral of the right side being exactly t, the time we are seeking and the integral of the left side between the limits of zero and the final speed,wfinal/n. The method of solving this is simple.
From the known torque speed curve of a selected engine, you plot the reciprocal of the torque, 1/T(w) against the engine rad/sec, and to get that integral (remember this from your first course in calculus?), you obtain the area under that curve between 0 and wfinal and multiply by Isum, a rather staightforward process so you can finally write that

t=area under the 1/T(w) vs w curve between 0 and wfinal mutiplied by Isum

Hope this clarifies (or does it complicate?) matters.

 

[desertfox]

Hi cutinoak

I based my calculations on energy and to answer your last question ie:- "how many horse power would it take to accelerate the flywheel to 10000rpm in 60 secounds"

kinetic energy
of flywheel = 1/2 * I * w^2= 620255.6401 joules

I= 1.13121kgm^2

w= (10000/60)* 2*pi

power= kinetic energy/ time

power= 620255.6401/60 = 10337.594 watts

10337.594/745.7= 13.863HP

regards desertfox

 

[Darken99]

So how did everything turn out? I was thinking you may want to have a way to release the flywheel after the inital acceleration. That way you do not have to use any energy to maintain the rpm of the flywheel.

 

[cutinoak]

I had so much OT at work I have not had the time. I will reply when my tractor is tested.


Anthony

 

[Fabrico]

Re: "I was thinking you may want to have a way to release the flywheel after the inital acceleration."

That might also come in handy if a problem develops in the engine or elsewhere. If you have a rod knock, or worse, turning off the key won't do much good.

Most pullers have a "deadman" system in the ignition or fuel delivery. The super-flywheel would do an excellent job of circumventing that too.

I wouldn't spend too much on the project. It may get banned after the first run or mishap; which ever occurs first

 

[cutinoak]

There is a clutch on both sides of the flywheel. A brake band will wrap around the the diameter of the flywheel.

Anthony