Flywheel help
Flywheel help
(OP)
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.
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.





RE: Flywheel help
RE: Flywheel help
Using energy concepts, angular impulse equals the change in angular momemtum. Let T=torsion, t=time, J=polar moment of mass inertia and w=angular momentum. Then:
T t = J (wf - wi) with wi=0 starting from rest.
But horsepower is simply torque times RPM adjusted for units [T]=ft lbf, [N]=RPM for power to be [P]=HP. So this would imply:
P = T N / 5252 or T = (25 HP)5252/10000 RPM = 13.13 ft lbf
I prefer metric, so 13.13 ft lbf = 17.802 m N. So computing the mass moment of polar inertia as J = 0.5 m r^2 for m=mass, r=flywheel radius,
J = 0.5 (89.32 kg)(0.15240 m)^2 = 1.03726 kg m^2
Assuming your mass is correct, namely 197 lbf is 876.3 N or dividing by g=9.81 m/s^2, I get 89.32 kg. Of course, the 12 inch diameter wheel is 0.15240 m radius.
Manipulating the main equation for "t", you would get:
t = J wf/T = (1.03761 kg m^2)(166.66667 rev/s)/17.802 m N
t = 9.714 s
Checking dimensional consistancy of the equation:
t = (kg m^2)(1/s)/(m N) = m/(m/s) = s, as expected.
So I got just under ten (10) seconds, roughly one-sixth of your estimate at sixty (60) seconds.
Originally I thought maybe I goofed a unit conversion or dropped a "g" factor in the imperial computations. I can't find an obvious error, maybe a fresh set of eyes can.
I got 9.7 seconds.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Flywheel help
RE: Flywheel help
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It is not clear from the OP if the flywheel needs to be held at 10,000 rpm once it gets there, nor do we know the characteristics of the drive. Without this, we can't go too much further towards the solution.
RE: Flywheel help
1. Determine Mass: Weight/Gravity Acceleration = 196.79/32.2 = 6.11
2. Determine Moment of Inertia: ½ x mass x radius^2 = .5x6.11x.5^2 =0.7637 Slugs ft2.
3. To accelerate the flywheel from zero to 1000 RPM in 6 seconds or to 10000 RPM in 60 seconds,
we need to convert the RPM to radians per second to get the angular velocity.
Radians: 1000 x 2 x pi = 6283.18 radians per minute / 60 = 104.72 radians/sec.
Angular velocity: with t=6 seconds: 104.72 / 6 = 17.45 radians/sec2
4. Torque Required: Desired Acceleration x Moment of Inertia = 17.45 x 0.7637 = 13.33 Slug ft2/sec2.
The Slug though has units of lb*sec2/ft so we may express the torque as 13.33 lb.ft.
5. Determine Revolutions/Sec: (S1 - S0)/60 = (1000 - 0)/60 = 16.67 Rev per Sec. S1 = 1st Speed Increment = 1000,
S0 = Previous Speed Increment = 0
6. Determine Power: Torque x Rev per Sec x 6.28 = 13.33 x 16.67 x 6.28 = 1395.21
7. HP required for the increment: Power / 550 = HP = 1395.21 / 550 = 2.53 HP
It would take a 2.53hp engine 6 seconds to accelerate the flywheel to 1000 rpm
or a 253hp engine 6 seconds to accelerate it to 10000 rpm
or a 25.3hp engine 60 seconds to accelerate it to 10000 rpm
There seems like something is still wrong. 25 hp should be able to turn the flywheel 10,000rpm quicker than 60 seconds. More like what Cockroach said 9.7 But I can't follow his rythym.
Anthony
RE: Flywheel help
In both your posts you use the word "engine". Does this mean internal combustion engine?
An internal combusion engine will not be constant torque or constant power over a range of 10,000 rpm.
An electric motor is approximately constant power.
A hydraulic motor could be approximately constant torque.
RE: Flywheel help
engine rpm 4000
pulley ratio's 2.5:1
flywheel rpm 10,000
Anthony
RE: Flywheel help
How many horsepower would it take to accelerate the flywheel(197 lb) 0-10,000rpm in 60 seconds.
Anthony
RE: Flywheel help
A 25 HP gas engine produces 25 HP only at some specific rpm. If we assume that your engine makes 25 HP at 4000 rpm then you have a maximum of about 33 lb ft of torque to work with. I have never seen a gas engine with a perfectly flat torque poer curve, so in all likelihood you have less torque at lower speed. You need the torque vs rpm curve for your specific engine to solve this problem.
Once you know the torque available at any instant in time you can calculate the acceleration for that instant, and thus the delta-omega for that instant. Keep going until the sum of the delta-omegas equals 10,000. However many instants you needed to calculate is how long it will take.
RE: Flywheel help
Torque produces angular acceleration.
For ANY drive, horsepower is the produce of torque X RPM X a constant to get whatever units you need.
You need to know torque vs rpm, or torque vs time.
RE: Flywheel help
T=k*f(v) which states that torque is a function of velocity (angular velocity here) for a class of motors and k is a constant that reflects the size. Then you can write
I(dv/dt)=k*f(v).This is then solved as follows
I/k*dv/f(v)=dt
integrating these we get
I/k*integral(dv/f(v)--limits vfinal to 0=tfinal=60 seconds
The integral on the left side can be solved numerically .
From this eq, you can get k from which you determine the motor and thus its hoorsepower.
RE: Flywheel help
Is this a real application? Sounds interesting.
There will be many other factors to make this a functioning device.
Barry1961
RE: Flywheel help
RE: Flywheel help
It's a sport/competition where a garden tractor hook up to a weighted sled. winner is determined by distance sled is pulled.
Most people add (dead)weight to their tractor to weigh in class 1050 lbs or other classes. My weight is added in a kinetic energy flywheel. A stock GT (garden tractor) has 18-25hp with 20-33ft.lbs torque. My tractor will have the engine torque plus torque saved up in the form of kinetic energy.
I have the engine clutched to the flywheel, clutched to the trasmission. A pull of the sled will only last 12-20 seconds. If I take 60 seconds to build up rpm of the flyweel then the ft.lb of torque multiplies. So if I use it in all in ten seconds I will have 80 ft.lb of toque plus that of my engine.
Sneaky huh!
Anthony
RE: Flywheel help
scattershield...
gyroscopic effects...
...
RE: Flywheel help
RE: Flywheel help
RE: Flywheel help
So everything is good up to the line:
t = J wf/T = (1.03761 kg m^2)(166.66667 rev/s)/17.802 m N
which should read:
t = J wf/T = (1.03761 kg m^2)(1047.2 rad/s)/17.802 m N
t = 61.04 s
Therefore my answer was out by 2 X pi. I believe this puts us all on the same page.
Thanks Zekeman.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Flywheel help
Fabrico,
All rotating parts will be sheilded properly. The transmission will be the first thing to fail.
Anthony
RE: Flywheel help
RE: Flywheel help
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.
RE: Flywheel help
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
RE: Flywheel help
RE: Flywheel help
Anthony
RE: Flywheel help
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
RE: Flywheel help
Anthony