Long Discussion on Torque

Question:

Sirs:

I must determine the exact steps in solving a basic torque/acceleration problem. I hope you will assist in this project.

Given:

A large disc 84" in diameter, weighing 12,167 lbs., WK^2 = 74,470 Lb.-Ft.^2, a flywheel which has a shaft through the center with each end of the shaft to be secured in the appropriate heavy-duty oil-lubricated ball bearing assemblies. The bearing journals for this disc will be dimensioned near the 9.875" dia. This assembly will have an appropriately sized oil cooling and circulation pump to serve both bearings. The unit will rotate between 890 and 900 RPM. Temperature is not a factor because the application is inside and less than 2000' altitude. The proposed centerline distance between bearing assemblies will be 60" and the type of bearing in view would be Pillow Block bearings of your recommendation. The bearing assemblies will be bolted to a large, 4" thick bed plate 48" long by 24" wide solidly secured to large anchor bolts embedded within the concrete columns on which the assembly will rest. There will be a 3/8" heavy duty mesh guard surrounding the flywheel.

I realize that we must insert and use a value corresponding to the friction present at each bearing. The bearing manufacturers will give me the values to use. There will also be some opposition due to windage between the disc and cover but I think it would be insignificant. I also realize that the mounting and base plate data does not really apply but I gave the information to illustrate that the mechanism is secure.

I need help with the basic mathematical steps to accomplish the following:
These applications will have an electric motor in place and will have a double shaft - one of which will direct couple to the flywheel shaft. The other will direct couple to the load but for now we will disregard that issue. My fist task is to couple to the "flywheel" shaft and accelerate the assembly to 900 RPM. I have to generate a basic speed/torque curve for the flywheel and I need help from you.

I have to generate the required mathematical steps and then set the procedure up in an Excel spreadsheet. My technical discipline is in a totally different field but this is a task I must accomplish. I realize that two important issues are friction and windage and I seem to recall something from physics which suggests those two items are the major factors or players in this scenario.

It is not too difficult to visualize a disc with a shaft through its exact center and each shaft end captured inside of an oil-lubricated ball bearing assembly. Applying "X" pounds of force on the end of the 12" lever attached to one end of the shaft - the 12" lever extends from the center of the shaft - initiates rotational movement. To accelerate the rotation from standstill to a given speed in a given time frame - requires "XX" pounds of force on the end of the lever. I tend to believe that the "XX" force amount would change as acceleration goes from zero to a point equal to 90 R.P.M. - perhaps not.

Two oil-lubricated ball bearing assemblies sufficiently large to accommodate a 9.875" shaft diameter, will have some type of friction factor which relates to the portion of weight it is carrying. As the mass accelerates a windage factor must be considered but I don't see that as much of an issue in comparison to a large fan or other object with surfaces sufficient to move air.

How do I solve for the following:
1. I need to determine the required force to be applied to initiate rotation. Assume that I have a 12" lever attached to shaft and extending radially from the center of the shaft to the 3 o'clock position and we are looking at an end view - how do I determine the approximate force at that point to initiate rotation with a view or objective to accelerate the mass to 90 RPM.
2. Assume I initiate rotation and bring it to 90 RPM. What torque value would be required at that point to continue the acceleration to 180 RPM, and then to 360 RPM, etc., up to 900 RPM - total time for the task in case one would be 30 sec.?

Second Question:
Is there a mathematical procedure to determine a torque value required to maintain a given rotation - 900 RPM for example - given the following information to work with?
Total Load WK^2 of 126,683 Lb.-Ft.^2 [Includes load and motor]
Speed to maintain: 900 RPM approximately.
Mass weight of 15,895 pounds.
Diameter of mass: 96"
Disc Thickness 8"
The shaft diameter will also be at the 9.875" point.
Same type of application - what bearings do you suggest here?

I also have to calculate speed/torque curves for other upcoming units of this nature. My objective is to set up in Excel the mathematical steps to calculate the force [ lb.- ft.] required to initiate rotation and accelerate to the next 10% speed increment, then to 20%, 30% etc. - and on to 100% speed - the speed/torque curve.

I hope you will assist with this total process. Our function is to provide the flywheel and shaft assembly complete with bearings and oil pump - plus, provide the large electric motor and switch gear.

Thank you very much.

Answer:

Here are some thoughts on your problem. First let's consider the no friction/no windage case. My understanding is that you have a flywheel 7 feet in diameter with a total rotating weight of 12,167 lb.. This corresponds to a mass of 3802 slugs in English units. We get the mass by dividing the weight by the acceleration due to gravity, which is 32.2 ft/sec^2.

The moment of inertia of a disk of any thickness is 1/2*mass*r^2 where r is the radius. Since the thickness of the disk does not affect the moment of inertia, the shaft of the flywheel may be included in the disk radius as long as the mass of the shaft is included in the total rotating weight. In our case the moment of inertia, I, is 1/2*3802*3.5^2 = 23,290 slug ft^2

Now if we want to accelerate the flywheel from zero to 90 RPM in 30 seconds, we need to convert the RPM to radians per second to get the angular velocity. 90 RPM is 90*2*pi = 565.5 radians per minute or 9.425 radians per second. The angular velocity at t=30 seconds then should be 9.425 radians per second.

To accelerate from zero radians per second to 9.425 radians per second in 30 seconds, the average angular acceleration must be 9.425/30 = 0.3142 radians per second^2.

The torque required to produce a given acceleration is the desired acceleration times the moment of inertia. This is the rotational equivalent of Newton's second law. In our case the torque would be T=0.3142*23.290 = 7317 slug ft^2/sec^2. The slug though has units of lb*sec^2/ft so we may express the torque as 7317 lb ft.

With the specified moment arm of 12 inches the required tangential force to accelerate a frictionless version of your flywheel to 90 RPM in 30 seconds would be 7317 pounds. In the absence of friction, this same 7317 pound force would increase the speed of the flywheel 90 RPM every 30 seconds up to the speed where the centripetal forces would tear the assembly apart.

If my understanding of the given problem was wrong, you should be able to apply the same logic to any rotating disk, adjusting the mass, radius or time to reach a given speed to fit the specific situation.

Next let's consider the friction effects. You say this wheel will be supported by ball bearings. In my experience loads of this size usually are supported by journal bearings. There is no theoretical reason why balls could not work but the load on the weight bearing balls is going to be large. If you can find a ball bearing that will carry the weight I suspect you will have wear problems.

In any event a properly lubricated journal or ball bearing will require so much less force to overcome friction than is required to accelerate the wheel, that you can probably neglect the effect of bearing friction. Do not forget to take into account the work required to run the lubrication system. If it is to be driven from the main shaft the prime mover will have to supply this force in addition to that required for acceleration.

Windage is a more complicated issue. The wheel is going to drag a sheath of air around with it. This layer of air for a smooth wheel will be quite thin but except for very slow speeds the flow will become turbulent due to a radial component of force arising from the circular motion. With turbulent friction, the frictional force may be approximated as proportional to the square of the tangential speed of the wheel. Even with a fairly low coefficient of turbulent friction (smooth wheel), the energy lost through this mechanism may be significant.

The torque required to overcome friction in any event will be considerably less than that required to meet your acceleration requirements. If you size your prime mover to handle the acceleration, it can run at less than full load once up to speed.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.

JDJ

Question:

Date: Monday, March 20, 2000 7:25 AM

Mr. Jones,

Thank you very much - this is a great help.

In the opening paragraphs you provide 3802 slugs as the mass for a weight of 12,167 lbs., obtained by dividing the weight by the gravity factor of 32.2. 12,167/32.2 = 377.86 but you present a value of 3802 slugs which is then carried forward in the remaining steps. Is this an error or am I overlooking some obvious function?

I am very grateful for your assistance. Subsequent to carefully thinking through the procedure, am I correct in assuming that I can apply the steps to accomplish the following?

I have a machine with a rotating mass weighing 15,895 lbs., 96" in diameter, disc thickness of 8". Using a generalized equation I obtain a WK^2 value of 126,683 Lb.Ft^2 (does this value appear correct to you?). I want to accelerate this rotor to 890 RPM in 30 seconds or less. Would it be correct to divide the problem into 10 parts: 0-90 RPM, 90-180 RPM, 180-270 RPM, etc. - and calculate the torque for each increment using the steps outlined in your letter? This would yield a speed/torque curve for the load .

If I have a WK^2 value and a corresponding RPM value, 74,470 Lb.-ft.^2, 720 RPM, is it possible to generate a required torque value at 0-72, 72-144, 144-216, etc. and on to the 10th point of 720 RPM? Total acceleration time to be 30 seconds or less.

The clarification of these issues will certainly take care of my problem.

Answer:

Date: Monday, March 20, 2000 9:48 AM

You will learn never to trust my arithmetic. I am currently clueless about how I came to 3802.

I am not sure what WK^2 represents. Please explain.

The procedure I used in the first example should work for the larger machine. Calculate the moment of inertia, calculate the final angular velocity, divide the angular velocity by the time allowed to reach that speed to get the average angular acceleration, multiply the angular acceleration by the moment of inertia to get the average torque required.

The notion of a torque-speed curve implies that required torque is dependent on speed. If you consider only the flywheel in doing the required torque calculation, the speed dependence is probably negligible. I have to say probably because I am not sure what the windage effect is. My intuition is that for a smooth flywheel, loosely shrouded, the windage effects are going to be so small relative to the torque required to accelerate the wheel from rest to 890 RPM in 30 seconds that windage losses may be neglected. You will have to try to find a table of windage losses for various geometries to confirm my intuition.

If in fact the windage loss can be neglected the torque-speed curve is a flat line at constant torque. This means that you do not have to break the speed range up into segments. One value of torque will get the wheel up to speed in the required time.

I am concerned that you are neglecting the effect of the other load attached to the shaft. That load may in fact be quite dependent on speed. Pumps, fans etc all have their own torque-speed characteristics. My bold assertion that the torque will be flat with speed certainly does not hold for loads of that nature.

I am on the road at the moment and will be out of touch until some time tomorrow. If I do not respond in a timely fashion, that is the reason.

Question:

Date: Monday, March 20, 2000 6:49 PM

Greetings,

That is a problem I face continuously. I keep all calculation routines and programs within Excel or I put then in VB.

Here is my understanding of your presentation. I carried it forward adding an extra step or two - check me out on the correctness of the steps.

---------------
Problem:
A steel flywheel 7' in diameter (84"), 8" thick, weighing 9348 Lbs., with a horizontal shaft running through the center and captured within oil-lubricated anti-friction bearings on each end.
1. Determine Mass: Weight/Gravity Acceleration = 9348.7/32.2 = 290.33
2. Determine Moment of Inertia: x mass x radius2 = .5x290.33x3.52 =1778.28 Slugs ft2.
3. To accelerate the flywheel from zero to 90 RPM in 3 seconds or to 890 RPM in 30 seconds, we need to convert the RPM to radians per second to get the angular velocity.
Radians: 90 x 2 x pi = 565.5 radians per minute / 60 = 9.425 radians/sec.
Angular velocity: with t=3 seconds: 9.425 / 3 = 3.142 radians/sec2
4. Torque Required: Desired Acceleration x Moment of Inertia = 3.142 x 1778.28 = 5586.63 Slug ft2/sec2. The Slug though has units of lb*sec2/ft so we may express the torque as 5587 lb.ft.
5. Determine Revolutions/Sec: (S1 - S0)/60 = (90 - 0)/60 = 1.5 Rev per Sec. S1 = 1st Speed Increment = 90, S0 = Previous Speed Increment = 0
6. Determine Power: Torque x Rev per Sec x 6.28 = 5587 x 1.5 x 6.28 = 52626
7. HP required for the increment: Power / 550 = HP = 52626 / 550 = 95.7 HP

Also, if we simply alter the steps to calculate straight from zero to 890 RPM, the following results exist:
Radians Per Sec. = 93.2
Angular Acceleration = 3.11 Radians/Sec
Slug Ft2/Sec2 = 5524
Power = 566556
HP = 1030

My experience suggests that windage is a small consideration in this scenario. I am keeping a keen eye on the main load conditions. I have available to me all of the main load characteristics, speed/torque curves, type of electric motor starter in use - plus, we require sufficient data from the power utility and from the consulting project electrical engineer (power distribution data, transformers, cable size and distance, etc.) - to model the complete application in terms of starting impact on the system, voltage drop at each buss location, acceleration time, and the temperature rise on the motor's rotor bars, end rings, rotor assembly, and stator windings. We design the electric motor torque and performance characteristics to fit the model - altering the type of starter (reduced voltage tap settings, etc.). My engineering design program will produce up to twenty points of speed/torque/current data for the given motor design. I have to plug in at least ten points of speed/torque data for the driven machine - now, I must also add speed/torque data for the flywheel. These two "load curves" will subtract from the motor speed/torque capacity (curve) yielding "acceleration torque". My program "slices" this up into ten or more points and calculates acceleration time from point to point - and the kilowatt/sec value at each point for the rotor bars, end rings, rotor assembly, and stator winding. This is summed to yield a total temperature rise for each of the items. This is the major point to evaluate from the design perspective.

I am unclear about the flywheel acceleration issue in terms of:
1. It appears that the calculation yields specific data for zero to 90 rpm.
2. We obtain essentially the same information if we then go from 90 to 180 rpm, etc.
3. This tends to suggest we have a straight line torque.
4. If we run the numbers 0 to 890 rpm you will notice that the power results in close to 1000 horsepower which then suggests that the total power consumed is additive - 96 HP x 10 points = 960 HP.
5. This is an area I must be clear in my thinking and calculations. Recently I overheard a "technical " discussion between two "experts" and involving this very same issue - except their machine was a wood chipper which essentially is similar to a flywheel - mostly mass. On an X - Y axis graph with speed left to right and torque bottom to top, one person argued that the straight line torque would start low on the torque side (lower left of the graph) and rise to a high position as the speed approached full speed. His graph indicated torque at start somewhere around 10% of the motor full load torque and reached approximately 80% of motor full load torque at full speed - 720 RPM in that case. He argued that "pure mass" will most generally have that type of curve. The other person argued that the mass would show a torque requirement at start much higher than 10% with an abrupt drop to some level and then follow mostly a straight line going from zero speed to 720 rpm - emphasizing that the torque magnitude would remain fairly static through acceleration. I must have correct data to work with - i.e., at 10% speed I must know what the torque requirement is - same at 20%, 30% etc. - matching the load point and motor points.

Am I out to lunch here or what?

Answer:

Date: Tuesday, March 21, 2000 9:43 AM

Your calculations appear to be correct.

In calculating the flywheel torque we made the assumption that we were going to accelerate the flywheel at a constant rate from zero to 900 RPM in 30 seconds. The flywheel torque load under that assumption is going to be a constant 5587 ft lb, neglecting bearing friction and flywheel windage. The horsepower requirements to maintain a constant torque on the flywheel will be proportional to the instantaneous angular velocity, just as you show with the formula...horsepower=torque*2*pi*RPM/60. The faster the wheel is spinning, the greater horsepower is required to increase its speed through the next speed increment, even when the torque is constant. A curve of power vs RPM for the flywheel would be a straight line with a slope of torque*2*pi/60.

It sounds like one of your experts is looking at the torque vs speed curve for the load and the other is looking at the torque vs speed curve for a typical induction motor.

If you drive this thing with a simple induction motor, the torque characteristics of the motor will be governed by the following ideas. At low speed the counter EMF due to the generator action in the motor will be small, resulting in a large initial torque from the initial inrush of current through the windings. As the apparatus comes up to speed, the current through the rotor windings produce a rotating magnetic field that cuts the windings of the stator, inducing a voltage that opposes the applied voltage. This "counter electro-motive force" limits the current through the motor and therefore reduces the torque available. It is possible to manipulate the current through the motor through external controls as you describe, using some sort of current limiting device during startup. This is necessary to reduce the heat load the motor must dissipate. Or you can simply limit the number of startups per unit time procedurally.

It seems to me that you have worked out the load torque vs speed curve. Just add a torque equal to the moment of inertia times the rate of change of angular velocity to account for the flywheel, to whatever the torque vs speed curve is for the working load. Now you should size the motor so that its decreasing torque vs speed curve intersects the load's increasing torque vs speed curve at 900 RPM.

Regards,

JDJ