Power Required to Drive Large Flywheel

[Rick Rowlands]

I just registered for this forum as I have a problem that needs solved. I am building an industrial museum in Youngstown, Ohio, and the centerpiece of our collection is a 230 ton stationary steam engine which we saved from a local steel mill. I am designing an electric drive for it so that we can roll it over at 15 rpm for demonstration purposes. Before I go any further here is a photo of the engine.
http://inlinethumb19.webshots.com/18130/2908668190033749120S600x600Q85.jpg

The crankshaft and flywheel weigh 230,000 lbs., and rest upon two babbitt bearings, 23" dia. x 40" long. The bearings will have a pressurized lubrication system feeding oil to the underside of the journals before startup, so the coefficient of friction may be around .015 or less. The flywheel rim is 20' dia. and has a 24" square cross section, weighing 102,000 lbs.

I want to run the engine at 15 rpm for the demonstrations.

We have some equipment here which I plan to use for the drive. We have a 20 HP series wound DC mill motor with a full load speed of 725 rpm, drawing 76 amps. The motor has a short term rating of 30 HP, 575 RPM at 116 amps. This motor will drive a planetary gear reducer with an 18.9 to 1 ratio, then from that reducer there would be a double 140 roller chain with a reduction of 2.56 to 1 before finally coupling to the engine crankshaft.

What I am concerned is if the drive has sufficient power to bring the engine up to speed, and how long would it take to accelerate to 15 rpm. Also I am looking to determine what the resistance should be for each step of a five step DC controller, as I will be using control equipment from a DC crane to operate the motor.

For more info. on our industrial museum project visit www.todengine.org or Google Tod Engine.

Thanks

 

[robert2734]

I didn't do any math (or physics). I know of a hundred ton telescope pointed by a twelve horsepower motor. I don't know what the coefficient of friction is. I think you are in the ball park.

 

[dlgoff]

The actual start torque applied should be many times greater than the actual start torque required by the application. The greater difference in torque applied by the motor and the start torque required by the application, the faster the applied acceleration of the electric motor.

http://www.engineersedge.com/motors/torque_electric_motor.htm
Welcome to PF

 

[spacester]

Hi Rick,

Were you looking for guidance in doing the calcs or for someone to do all of it for you?

I'm willing to do the former but not the latter. For something like this, IMO you should know exactly how the answer was found. There is nothing really advanced here, but it isn't exactly trivial either.

 

[Rick Rowlands]

It seems that the engineersedge formulas posted by dlgoff will answer the question regarding acceleration time, as well as torque available at full speed, which is 7000 ft. lbs. So this boils the down to torque required to get the 230,000 lb. mass up to 15 RPM from a dead stop.

From engineersedge:

"The time duration required to accelerate a application from a dead stop to operating speed is given by the following:

T = [ N x WR2 ] / [ Ta x 308 ]

Where:

T = Time ( seconds )
N = Velocity at load ( rpm )
Ta = Average Torque During start ( ft-lbs )
WR2 = Rotating Inertia (lbs-ft3)
W = Weight (lbs)
R = Radius of Gyration (ft2)
308 = Constant derived converting minutes to seconds, mass from weight, and radius to circumference"

What is rotating inertia, radius of gyration and what would be the average torque during start? How do I determine what those are?

 

[berkeman]

It seems that the engineersedge formulas posted by dlgoff will answer the question regarding acceleration time, as well as torque available at full speed, which is 7000 ft. lbs. So this boils the down to torque required to get the 230,000 lb. mass up to 15 RPM from a dead stop.

From engineersedge:

"The time duration required to accelerate a application from a dead stop to operating speed is given by the following:

T = [ N x WR2 ] / [ Ta x 308 ]

Where:

T = Time ( seconds )
N = Velocity at load ( rpm )
Ta = Average Torque During start ( ft-lbs )A
WR2 = Rotating Inertia (lbs-ft3)
W = Weight (lbs)
R = Radius of Gyration (ft2)
308 = Constant derived converting minutes to seconds, mass from weight, and radius to circumference"

What is rotating inertia, radius of gyration and what would be the average torque during start? How do I determine what those are?

Another welcome to the PF.

The correct term is Moment of Inertia:

http://en.wikipedia.org/wiki/Moment_of_inertia

That should give you what you need.

 

[Rick Rowlands]

That gave me a lot of formulas which I don't understand and even more terms which I don't know. Anything that explains it for let's say a practical man who graduated high school with straight Ds in math?

I'm not an idiot when it comes to mechanics, how stuff works, etc., but am clueless when the hieroglyphs start appearing in equations.

 

[spacester]

Hi Rick, good job. I think you're kidding about your math grades.

I do not like that engineersedge formula for two reasons.

First, the WR^2 term is ambiguous. They use the "X" symbol to multiply by N, but not between 'W' and 'R', and they tell you that the units for WR^2 is lbs*ft^3 which is WRONG - the units are lbs * ft^2

Second, the units of the quantity you find are the inverse of seconds: N is revolutions per minute, and you don't see minutes or seconds anywhere else, so the answer must be per unit time.

Here is my proposed corrected version:

T = Ta / (N * W * R^2)
which is valid for any units, as long as you make the conversions

To make this conversion, you have to realize that W is in lbm ("pounds mass") and T is ft*lbf (ft*pounds force)
The definition of units gives us
lbf = lbm * 32.17
So I can confirm the conversion constant of 308 with this:
2 * pi (radians/revolution) / [ (32.17 lbm / lbf) * [ 60 seconds / minute) ] = 0.003255 = (1 / 307.2)
(with this, all the units cancel except seconds - note that radians is a "dimensionless unit")

So use:
T = 307.2 * Ta / (N * W * R^2)
where
Ta = Ft-lbf
N = rev / minute
W = lbm
R = ft

OK then, to your question:
You need the "Radius of Gyration", and you prolly googled it, and found, as I did, that the definitions seem to be written by mathematicians FOR mathematicians and of little help to us practical types.

But we can find this definition:
"The square root of the ratio of the moment of inertia of a plane figure about a given axis to its area."

So
R = square root of I / A
or
R = (I / A )^(1/2)

Where
R = Radius of Gyration (ft)
I = Moment of Inertia (ft^4)
A = Cross-Sectional Area (ft^2)

We know A, it is the 24 inch square you described
A = 4 ft^2

I is the tricky thing here.
The Wikipedia entry for "Moment of Inertia" is actually for the "MASS moment of Inertia" - they are trying to trick you! :-) What we need is the "AREA Moment of Inertia"
found from
I' = b * h^3 / 12 for a rectangle with the axis running thru the centroid
where
b = base (ft)
h = height (ft)
But this does not account for the distance from the axis of rotation to the centroid of our rectangle, you use the parallel axis theorem for that:
I = I' + A * d^2
where
d = distance between axes

http://www.efunda.com/math/areas/RadiusOfGyrationDef.cfm
http://www.efunda.com/math/areas/ParallelAxisTheorem.cfm

In our case, d is the radius from the center of the wheel to the center of the cross-section
I'm not seeing that number in the thread so far, but that's all we're missing to get this start-up time (which you were very well advised to look at).

I'll post this and let you turn the crank on that much before moving on - are we getting a reasonable answer? (I didn't really check the derivation of my revised version, so maybe I'm missing something.)

 

[Rick Rowlands]

OK it will take a me a couple of rereadings of your reply to fully understand it! :)

One thing I can answer right now is that d would be 19 feet.

 

[spacester]

Cool.

This is a neat project, good for you for taking it on.

d = 19 ft

That is a BIG SUCKER, ain't it?

For some reason the nice photo you provided is no longer showing up in my browser. I'm going to run the numbers and wanted to check that number. I was guessing the diameter was closer to 20 feet than 40 feet . . . but remember that d here is for the radius, not the diameter.

I'll also look at my version of the equation in an effort to verify.

 

[Rick Rowlands]

Oh sorry yes you are right. The wheel is 20' diameter, so d would be 9 feet as the rim is 24" square in cross section.

 

[spacester]

D'oh!

Hang on, I goofed up the lbm to lbf conversion. The result is in seconds, not the inverse of seconds.

More in a few minutes . . .

 

[Rick Rowlands]

If you would like to see more photos of the engine and our project, go here:

http://community.webshots.com/user/todengine

I moved the engine to this site over the last several years, and over the summer erected a building over it, including an 1893 EOT crane which was donated by a local company. Both the crane and the DC motor that will turn the engine will be powered by 250 VDC generated by a diesel locomotive that was donated and recently moved to the site. The project is an attempt to preserve the industrial history of the Youngstown, OH area, by saving and displaying equipment used in local industry. Everything here will operate in some manner, so I am always running into strange problems such as the one I am trying to solve in this thread.

 

[spacester]

Corrected version:

. . . they tell you that the units for WR^2 is lbs*ft^3 which is WRONG - the units are lbs * ft^2

T = (N * W * R^2) / Ta
which is valid for any units, as long as you make the conversions

To make this conversion, you have to realize that W is in lbm ("pounds mass") and T is ft*lbf (ft*pounds force)
The definition of units gives us a conversion factor called "g-sub-c":
gc= (32.2 ft * lbm) / (lbf * s^2)

So I can confirm the conversion constant of 308 with this:
2 * pi (radians/revolution) / [ [ (32.2 ft*lbm) / (lbf * s^2) ] * [ 60 seconds / minute) ] ] = 0.003255 = (1 / 308)
(with this, all the units cancel except seconds - note that radians is a "dimensionless unit")

So use:
T = (N * W * R^2) / [ 308 * Ta ]
where
Ta = Ft-lbf
N = rev / minute
W = lbm
R = ft

OK let's plug in numbers:
R = (I / A )^(1/2)

We know A, it is the 24 inch square you described
A = 4 ft^2
I' = b * h^3 / 12 for a rectangle with the axis running thru the centroid
I' = 2 * 2^3 / 12 = 1.333 ft^4
I = I' + A * d^2
d = 9 ft

I = 1.333 ft^4 + 4 ft * 9 ft * 9 ft
I = 325.3 ft^4

R = (325.3 / 4)^(1/2)
R = 9.01 ft

Ta = 7000 ft-lbf

T = (N * W * R^2) / [ 308 * Ta ]
T = 15 rpm * 230000 * (9.01 ft)^2 / (308 * 7000 ft-lbf)
T = 129 seconds

2 minutes, 9 seconds to get this sucker going

But let's take a closer look at that 7000 ft-lbf torque. Can you show me where you got that number?

Also, we are assuming that the total mass is in the cross-section of the wheel, but a big chunk is actually in the spokes and the crankshaft, so we are being conservative here. To get a better answer, we're going to need to do a more careful calculation of the Radius of Gyration.

If the cross-section is steel at 0.283 lb/in^3 (a rule of thumb I have memorized), I calculate the wheel itself (no spokes) at 110,600 lbm which would reduce our time to 62 seconds.

edit: "more careful calculation of the Radius of Gyration" should read "more careful calculation of the Rotational Inertia"

 

[Phrak]

What is max pressure of your hydraulic system, and will the crankshafts be moving as well?

 

[sophiecentaur]

Would this simple Measurement procedure help? There seem to be a lot of imponderables if all you are going to do is calculations.
You could try hanging a mass on the outside of the flywheel (or use a belt driven by a falling mass, somewhere else) and see what it takes to get it going. The rate at which the mass accelerates would tell you the effective mass of the system, acting at that point. That would, in effect, be measuring the MI plus Friction of the system in one go - which, I think, is what you need.
Or, if you can translate your requirement into the required peripheral speed and acceleration of the wheel, you could add masses until the acceleration is sufficient. That would tell you the actual Power needed, (Force times Speed or Torque times angular velocity etc.).

The good thing about that method is that it doesn't involve firing up any electric motors and you get a good ball-park idea of what your problem is. It would give you an idea of how long your electric motor might be expected to be 'stalled', for instance. For someone with your facilities and abilities, I am sure you could do this easily and I suspect that is how the original designers may well have worked.

 

[spacester]

@ sophiecentaur: Excellent suggestion!

@ Rick: Let me know if you need any support doing the math for such an experiment.

@ Phrak: Good line of inquiry: It would be a shame if the lubrication was not sufficient and the babbitt bearings wear too rapidly.

 

[Rick Rowlands]

Thanks everyone for the help. I think I have enough information now to go on. My concern was mainly if the DC motor which I have would be sufficient to get the engine turning from a dead stop, and I believe that it will be powerful enough. I've learned a little about power, acceleration, moment of inertia etc. in the process. This entire engine project has been one giant educational adventure for me!

 

[Bob S]

Nice museum piece!

Several things:

1) Make sure the lubrication system will continue running in the event of an electrical power failure. I know of one case during a large power blackout where the journal bearings were ground up, because the generator took ~ 30 min to stop after the power failed. I have seen backup hydraulic pumps that were mechanically coupled to the flywheel.

2) Repeat the calculations in metric units to avoid ambiguities in the English system of units.

3) Brush-type electric motors do not like to "run" with a locked rotor. It will burn the commutator. How do you limit the starting (locked rotor) current? Measure the required static starting torque by tying a rope onto the flywheel and putting half a turn around the rim. How many men does it take to make it turn? Assume 1 man = 300 Newtons of force. Using R = 2.7 meters, the torque would be ~800 to 900 N-m per man.

At 15 RPM, 20 HP = 15,000 watts = torque x 2 pi RPM/60 = 1.57 x torque (in Newtons).

So the available torque (not including drivetrain losses) at 15 RPM is about 9,500 N-m. Use
http://www.unitconversion.org/unit_converter/torque-ex.html
to convert to Lb-ft (I get 7007 lb-ft).

4) Do you have any provision on how to stop the flywheel in case of an emergency?

Bob S

 

[sophiecentaur]

"2) Repeat the calculations in metric units to avoid ambiguities in the English system of units."
Oy Oy. It's decades since the 'English' used those old fashioned units! It's only the uncivilised part of the world that uses pints for anything other than Beer.

The lubrication pump has to be driven off the flywheel itself, surely. Wouldn't the original system have been like that?

 

[Bob S]

"2) Repeat the calculations in metric units to avoid ambiguities in the English system of units."
Oy Oy. It's decades since the 'English' used those old fashioned units! It's only the uncivilised part of the world that uses pints for anything other than Beer.

I was referring to pounds, poundals, and pound-force. I once used English ball bearings that had static torque measured in stone-barleycorns. Honest.

The lubrication pump has to be driven off the flywheel itself, surely. Wouldn't the original system have been like that?

From the NY Times, referring to the Big Blackout of 1965:
In New York City, as monitoring dials went wild, a crucial operator at the main power control station made frantic phone calls upstate instead of preemptively shutting down the city's system to protect it. Automatic shutoffs took over and suddenly closed down the Con Ed system for the first time in its history, causing damage to the power system. A 1,000 megawatt steam-turbine generator nicknamed "Big Allis" burned up its bearings."
Rumor has it that the bearing lubrication system on Big Allis (from Allis Chalmers) failed because it ran on electric power.
Bob S

 

[sophiecentaur]

Well well. And I thought it was only modern equipment that was thrown together!

 

[Rick Rowlands]

A reserve oil pump operated from the drive system itself. I had not thought about that but it makes perfect sense. I would still use electric pumps for the prelube and circulating system, but this third pump would protect against a power failure.

Originally there was a 1000 gallon oil reservoir in the roof trusses and a duplex booster pump next to the engine. So as long as there was oil in the reservoir the engine had lubrication.

The system will have a low oil pressure cutoff anyways, so that if we lost high pressure lube to the bottom of the main bearings the drive would shut off.

As for an emergency brake, maybe set the motor up so that it could also become a dynamic brake. Not sure of the advantage would be in having an emergency stop ability, these engines never had them when run on steam.

 

[Phrak]

A reserve oil pump operated from the drive system itself. I had not thought about that but it makes perfect sense. I would still use electric pumps for the prelube and circulating system, but this third pump would protect against a power failure.

Yes, this really is your first concern. First lift-off the shaft from the journal bearings with an auxilary pump as well as possible. Otherwise you're going to gall the bearings. Second, calculate or discover by experimentation the torque required to overcome the remaining stiction. Third, discover the required HP to turn it at the target speed.

In an ideal world the pressure required, in psi, is the weight of the wheel in pounds divided by D*L. D is the shaft diameter in inches. L is the sum of the lengths of the journals in inches. I'd double or tripple that.

Edit: Is 375 psi unusually for an haudralic system?

 

[sophiecentaur]

I hope you plan to take out a hefty third party insurance - or at least notify your insurers. Just imagine the havoc if the flywheel jumped off at full speed - it would demolish the whole high street! The total stored energy would be a lot of MJ (or BThUs or G ft pdls etc.)

 

[Bob S]

In the early 1960's, cracks were discovered emanating from the keyways in both 70-ton Bevatron (particle accelerator at LBL Berleley) alternator flywheels, that spun at ~1200 RPM, and spun down to ~800 RPM every time the machine pulsed. It was calculated that if a flywheel split into 4 equal pieces, one piece would probably land on the U.C. Berkeley campus. So they were replaced ASAP.
Bob S

 

[SystemTheory]

Does this differential equation for the mechanical system offer you some insight? I use it to study the spin-up characteristics of tiny permantent magnet motors. The principles are the same but the actual model you use may need some refinements:

[tex]T_{IN} = J_{EQ}\frac{d \omega}{dt} + D\omega + T_{C}[/tex]

where the input torque goes into accelerating the equivalent moment of inertia [itex]J_{EQ}[/itex] up from 0{rad/s} to final speed [itex]\omega_{F} = 1.57[/itex]{rad/s} and the equation includes a damping factor D that is proportional to shaft speed and a constant counter-torque factor on the right side.

The equivalent flywheel intertia as "seen by the drive motor" is a value reflected through the gearbox:

[tex]J_{EQ} = \frac{J}{G^{2}}[/itex]

where G is the speed reduction ratio of the gearbox and J is the flywheel moment of inertia.

Notice the starting torque is high while accelerating the large flywheel mass up to speed, and this draws a large current for the period of acceleration, therefore the current flow in the motor, its duration of flow during the transient response (acceleration), and the gearbox ratio are related factors of significance in the design. Chances are the factor [itex]D \omega[/itex] is negligible in this low rpm system with oil under pressure in the bearings.