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.

 

[sreid]

It looks correct to me.

 

[Cockroach]

I'm not to sure. You have several key quantities correctly computed but I can't follow the logic.

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

 

[MintJulep]

I think that you cannot assume constant angular acceleration for this problem.

 

[MintJulep]

Ok, a little bit of web research turned up these:

http://www.diracdelta.co.uk/science/source/f/l/flywheel/source.html

http://science.radioelectronics.biz/Acceleration/angularAcceleration.html

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.

 

[cutinoak]

A steel flywheel 12" in diameter , 6" thick, weighing 196.79 Lbs.,

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

 

[MintJulep]

Cutinoak,

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.

 

[cutinoak]

A 25 hp engine with a 60 tooth gear pulley, and the belt runs to a 24 tooth gear pulley. And that shaft turns the flywheel

engine rpm 4000
pulley ratio's 2.5:1
flywheel rpm 10,000

Anthony

 

[cutinoak]

Let's go this way

How many horsepower would it take to accelerate the flywheel(197 lb) 0-10,000rpm in 60 seconds.

Anthony

 

[MintJulep]

Ok, I will assume a gas engine.

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.

 

[MintJulep]

Horsepower does not produce angular acceleration.

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.

 

[zekeman]

The fundamental problem here is that all motors have Torque/speed charcterustics and to truly do this problem you must have this relationship, viz
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.

 

[Barry1961]

The math looks good to me. The thickness of the flywheel compared to the OD makes the values seem odd. I guess the small OD is due to the high rpm.

Is this a real application? Sounds interesting.

There will be many other factors to make this a functioning device.

Barry1961

 

[zekeman]

Your Math is OK but your assumption of constant torque may not be,as I have remarked, and Cockroach's problem is that w is in rad/sec, not Rev/sec, so it remains to find the torque as a function of speed to finally put this thing to bed.

 

[cutinoak]

Garden tractor pulling:
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

 

[Fabrico]

I hope your gears and axles are up to the flywheel effect! And don't forget to allow for the couter affects. You may let the clutch out and have the tractor start spinning Speaking of the clutch...
scattershield...
gyroscopic effects...
...

 

[BobM3]

Since speed varies you would have to integrate torque x speed to get the power. If you assume the speed varies linearly from 0 to 10,000 RPM and the torque is constant, then the power = 1/2 x torque x max speed. I come up with 12.3 HP.

 

[BobM3]

Not a good answer on my part. 12.3 HP would be the average power needed. If you are sizing a motor rated constant torque up to its base speed then the motor would need to be rated 25.6 HP.

 

[Cockroach]

I believe Zekeman has found it. In my calculation, I mistakingly used "rev/s" and it should properly be "rad/sec".

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

 

[cutinoak]

I want to thank everyone for taking the time to check my figuring and others as well. This weekend I will fire up the lathe and make my flywheel.

Fabrico,
All rotating parts will be sheilded properly. The transmission will be the first thing to fail.

Anthony