Author Topic: Flywheel calculations  (Read 3021 times)

Offline vtsteam

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Re: Flywheel calculations
« Reply #60 on: March 20, 2023, 12:39:47 PM »
I'll be very interested to see where your explorations go MJM460. What's the worst that can happen in just thinking aloud about something like this?  :Lol:

So, in line with your example, I'm guessing you'll be considering moment of inertia values for internal counterweights, if any.

Interestingly, if small engines of some particular type turn out to need relatively small moments of inertia to run acceptably, and we optimize flywheel size, various masses close to the center of rotation become much more important.
Steve

Online Jasonb

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Re: Flywheel calculations
« Reply #61 on: March 20, 2023, 01:19:20 PM »
looks like I drew the original central web at 3/16" wide not 1/4" once changed we have the same figures.

Yes it's interesting how little moment is needed particularly if the engine is going to spend it's time running at speed. When I did some of those small engines last year I did not think they would run well at low speeds with their tiny flywheels but they did.
« Last Edit: March 20, 2023, 01:24:44 PM by Jasonb »

Offline RReid

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Re: Flywheel calculations
« Reply #62 on: March 20, 2023, 03:33:38 PM »
As one of those who learned much of this stuff many years ago but haven't used it since, I appreciate having this well presented review available without digging out the old texts. Thank you, MJM!

And as an Alibre user and a lazy bones when it comes to hand calcs, I appreciate Jason's contributions as well. Thank you, Jason!
Regards,
Ron

Online MJM460

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Re: Flywheel calculations
« Reply #63 on: March 21, 2023, 10:09:11 AM »
Hi Steve, have to agree.  The worst is usually a spreadsheet of figures that I canít find my way through a year later.  Itís always worth a try, as even finding out I canít get any useful results is still a result, right?

But internal counter weights?  With the formulas discs and for spokes and the parallel axis theorem, most things can be calculated, so it is easy to check if the numbers are significant.  Of course, once the flywheel fell off my little Mamod engine, only the disk and an insignificant big end remained, so those little weights might be more significant with a very small flywheel.

Hi Jason, thank you for checking that.  It is so easily done, but nice to find a reason for discrepancies, itís the value of checking calculations, more important than the small difference in the numbers.  Checking of the computer output is a more difficult issue, but the agreement of those numbers suggests at least we are both calculating the same thing.  Probably the first time anyone has checked the Alibre moment of inertia calculations.

Hi Ron, I think Jason has done us all a favour by showing us that the calculations are there in the program.  I assume you only have to find the right button in one of the tool bars to display the result without further effort.  Iím with you, knowing it is there, I wouldnít bother calculating manually, even with the help of a spreadsheet.  Though I hope that going through the theory has prompted the memory or helped people understand what is behind the number.

When I mention manual calculation, instead of back of an envelope, or even a note book, I actually open a spreadsheet, and put the inputs and formulas into that.  The computer does the arithmetic for me, and I name the range so I can find it again.  And I can easily explore the sensitivity to some of the variables without starting to make mistakes in the calculation.  This way I put many calculations into one sheet instead of ending up with a multitude of files with names I canít associate with the subject.  Surprising how often I come back to the same calculation.  But basically it is still manual input of the required formulas.  And the file is called Scratchpad!

I am making good progress with some flywheel calculations but I need to look out some old files with real run figures for my engines.  Then we can see what we can learn with realistic inputs. 


I hope to have more tomorrow,

MJM460

« Last Edit: March 21, 2023, 10:14:07 AM by MJM460 »
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Online MJM460

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Re: Flywheel calculations
« Reply #64 on: March 22, 2023, 10:31:28 AM »
I learned long ago that flywheels in full size practice are sized on the basis of limiting the instantaneous speed variation within one revolution.  So I did some calculations, just using the basic physics, to get average rotational speed, accelerations, torques and power based on an assumed speed variation.  The number most difficult to find is the actual speed variation for one of our model engines.  I donít have an answer to that yet, which is another issue.  But itís instructive to do the calculations anyway, and then explore the sensitivity of the results to the initial assumption.

I rummaged around my notes and found a number of log sheet records of test runs that I have done on my models.  Unfortunately, most of the ones that would be most useful now, relating to that little single acting oscillating engine, were carried out before I bought the digital tacho.  At the time I was I was most interested in the boiler performance, dictated in part because I had no equipment able to measure the engine output.  But I did find just one result from 2017 that had some engine speed readings, including a maximum reading of 1000 rpm.  And we have already calculated the flywheel inertia.  Should be all we need if we assume a speed variation, just to see what the calculations mean.

The printed result is attached below.  I have assumed a constant torque and hence acceleration.  Not true of course, but the total kinetic energy into the flywheel for a given speed change is the same whether the torque is constant or variable, so the result is adequate for our purposes.

The average speed is 1000 rpm as measured by the digital tacho, and converted to radians/sec.

With the assumed speed variation, the increase in speed from minimum to maximum occurs over half a revolution for this engine, and we can calculate the time (t) for half a revolution in seconds.

With a little algebra, the angular acceleration is calculated using w2 = w1 + alpha x t where w2 and w1 are the initial and final speeds and alpha is the angular acceleration in radians/sec^2.  (As usual, on my iPad I am using w instead of the text book lower case omega)

The the average torque is I x alpha, note that I have switched the units for I to Kg.m^2.  This avoids the need to keep track of multiple powers of 10 so that torque is in Newton metres, or N.m

Similarly Power for this half revolution is torque times angular velocity.

Note the parallels between those formulas for rotational motion compared with the equivalent ones for linear motion.  Torque and angular velocity and acceleration replace force and linear velocity and acceleration, otherwise the formula is the same.  Resulting directly on Newtons laws.

The result of those simple calculations is interesting!  The engine is running at constant average speed, so it must slow as much during the exhaust cycle as it speeds up during the steam inlet, hence the flywheel gives the same energy back during the exhaust half of the revolution.

The engine is running free with no connected load, so the same power is given back during the exhaust, which means that is the power required to turn the engine at that speed.  The engine has to supply that average power, so the peak power from the engine is about twice that.  About but not exactly, as it takes power to push the exhaust steam out of the cylinder, which is not exactly the same as the inlet half of the revolution.  However, if the power absorbed in the exhaust stroke was less that supplied during steam inlet, the engine would speed up.  As it runs at constant speed, at least until the burner starts running out of fuel, that maximum speed provides a reasonable estimate of the engine power, providing of course that our assumed speed variation is correct.  I hope that logic is clear, but donít hesitate to ask for clarification, or if you disagree, please add to the discussion.

(Note, it is ok for my little meths powered plant and similar low power plants, but not a good idea to run a larger plant unloaded at full power.  You can see from the above calculation that the speed will increase until all the power is absorbed and this is potentially dangerously destructive.  Make sure you do have a suitable load coupled up before running full throttle.)

We can return to that later, but while we are looking at the calculation, there is more we can learn from that calculation.

Time for a break while that much gets digested, then a little more from this exercise tomorrow.

MJM460

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Online MJM460

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Re: Flywheel calculations
« Reply #65 on: March 23, 2023, 10:17:39 AM »
Continuing from yesterday, letís explore what else we can learn from that calculation.

How important is that initial assumption of 5% speed variation?  A spreadsheet is great for exploring ďWhat if?Ē.  I changed the input cell to 2.5%, which. with all else constant, reduced the  power  to 0.75 W.  (Some might argue that even this is optimistic for such an engine.) 

Then, back to the initial 5% speed variation, and reduced the flywheel inertia to 21 kg.mm^2, half the original value.  Again, the power reduced to 0.75 Watts.  The same engine and steam conditions, so the engine is presumably still capable of the original 1.5 Watts, so reducing the flywheel inertia leaves more capacity to drive an external load.  An extra 0.75 over the original 1.5 is a very significant difference.  A heavier than necessary flywheel gives improved speed regulation, but absorbs power that could otherwise drive the external load.

Finally, with the reduced flywheel inertia of 20.5 kg.mm^2, I increased the number in the speed cell until I matched the original power of 1.5 Watts.  The new speed was 1260, so the increase in rpm for half the flywheel inertia was about 1/4 of the original speed.

If we want to run an engine very slowly for display purposes, it is probably preferred to have a very even speed so it is less likely to stop at the slow point of the revolution.  You can see from these trials that a larger moment of inertia reduces the speed variation, but at the same time, increases the power required to accelerate the flywheel during the steam inlet.  This implies more pressure required. 

On the other hand, if you want to run at the lowest possible pressure, a good measure of a well built engine, a smaller flywheel inertia requires less power for the acceleration part of the revolution, but the resulting greater speed variation makes it more likely that the engine will stop at the slow point.

At higher speed for a working engine, the risk of stopping is less, and a lower moment of inertia requires less power for acceleration, and leaves more power available to drive the load.

As always, you have to find the balance between competing requirements to optimise the flywheel size.

These explorations are easier on a real computer, where a spread sheet normally has a goal seeking tool which finds the answer for you, however that function is not available on my iPad so I have had to do it manually by trial and error.

But some interesting conclusions that hold, even if the real speed variation is different from that initial guess.  It would be really useful if we could actually measure that variation, but that would be a project for another thread.

We can also use the information so far to get a little more information about that engine.  Perhaps I should do a test run or two first to check those figures.

Thanks for looking in,

MJM460


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Online MJM460

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Re: Flywheel calculations
« Reply #66 on: March 24, 2023, 10:44:06 AM »
As an aside, there is some more really cool maths/physics around flywheels.  We are all familiar with linear motion, and the associated forces and velocities.  We know that forces and velocities are vectors which simply means that to describe them fully, you have to include not only magnitude, but also direction.

Also that to add the effect of two vector quantities, we generally visualise this graphically by drawing a triangle, where two sides of the triangle represent two forces or velocities, and the third side is the resultant of the two.  This helps us solve problems like adding our boat speed to the wind speed to get the apparent wind, and similarly for aeroplanes and cars.  (As we generally know the apparent wind, we actually subtract the vectors to get the true wind).  We also know that we can divide a vector into two components at right angles to each other and that a vector has no effect at right angles to itself.

So what about rotational motion.  Well, rotations also involve vectors.  Torques and angular velocities are vectors, so to describe them we need to define not only the magnitude, but also the direction.  The direction of a torque is defined by a ďrule of thumbĒ, if you curl the fingers of your right hand around to make a thumbs up sign, with fingers going the direction of rotation which results from the torque, the direction of the torque is defined by the direction pointed by your thumb.  Now I am sure this definition is a bit arbitrary.  In a left handed world, we might use the left hand and the maths would still work, but it would be very confusing if there were some using each.  So in this world, the mathematical definition used the right hand.  Angular velocity is also a vector.

We see a very interesting result of this vector behaviour in those little gyroscopes that we often see in a good toy shop, or in your local science museum shop.  Basically a flywheel suspended on two bearings and set spinning by winding a string around the shaft, then pulling the string as hard as you are able to set the wheel spinning.  The cage has non-rotating extensions to the shaft, often one with a point, and the other end with a little ball shape.

If you set the wheel spinning and stand it with the shaft vertical on the pointed end, and you set it perfectly vertical is sits spinning without falling over, but more likely, it will be not quite vertical, which results in a torque due to the weight not being quite in the same vertical line as the equal and opposite supporting force on the point.  If the flywheel was stopped it would just fall over.  But when it is spinning, instead of falling, the flywheel axis will progress with the top of the shaft tracing out a small circle.  And this continues for quite a long time.  Just like a spinning top.

We can use that rule of thumb to determine the direction of the torque due to the support point and the weight of the wheel, and sure enough, it is at right angles to the direction of the spin.  Being at right angles, it has no effect on the magnitude of the spin, but simply changes its direction.  So that torque causes a change in the direction of the spin in the  horizontal direction, indicated by that rule of thumb, causing that progression.

Those little gyroscopes are usually supplied with a little stand with a little hollow spherical depression on top.  You can stand the point in that depression and it will balance there as it progresses.  An extreme of this situation is if you put the spherical end of of the cage on top of the stand in that hollow, with the spinning shaft horizontal, and the other end unsupportd.  Even then, the gyroscope does not immediately fall as you might expect, but will just progress around its support. 

Of course, it eventually falls, but the manner of falling is also interesting, but we will get back to that.  That gyroscope action is behind the safety advice for portable circular saws, to line up the saw with the cut before switching on, and wait for it to stop before you move it away.  If you have a job with a large number of successive cuts you might be tempted to move the saw before it stops.  The unexpected direction of movement in response can result in nasty accidents.  If you must move the saw while still rotating, make sure you are holding the body of the saw firmly with both hands and move it slowly while you get to understand the motion.

On the other hand, if the sphere is installed in a ball and set floating so there is no torque resulting from the support, it will just keep spinning with the axis maintaining a constant direction, except for the motion imparted from the rotation of the earth.  If a motor is provided to maintain the rotation against friction, we have the makings of a gyroscopic compass.

All based on the same vector physics as our flywheels.

Thanks for looking in.

MJM460

« Last Edit: March 25, 2023, 06:27:53 AM by MJM460 »
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Online Kim

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Re: Flywheel calculations
« Reply #67 on: March 24, 2023, 04:48:10 PM »
Thanks, MJM for this fun little physics refresher!  I've really enjoyed following along with this thread!

Kim

Online MJM460

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Re: Flywheel calculations
« Reply #68 on: March 25, 2023, 09:37:53 AM »
Thanks Kim, Iím glad you are enjoying it.  I hope there is also something there for those seeing it for the first time.

On re-reading my last post, I notice some missing words when talking about the gyroscope.  I intended to say put the ball end of the cage on top of the stand with the spinning shaft horizontal, and the other end unsupported.  I have corrected that though it was probably obvious to those familiar with a gyroscope.

The rotational equivalent of the linear F = m x a is similar in appearance,

Torque = moment of inertia x angular acceleration

It is a mathematical expression of  Newtonís second law and a consequence of the conservation of angular momentum.

Now for that gyroscope sitting horizontal with one end sitting on the support, we have already worked out the direction of movement, which makes a horizontal circle.  The torque is equal to the weight of the wheel times the distance to the support, so that formula gives us the acceleration.  However, in this case, the rotation is around the vertical axis (plus it is also spinning around its horizontal axis) so for the progression calculation, we need the moment if inertia about that vertical axis.  Remember the additional ones in the result of Jasonís calculation?  It is a little more than half the moment about the primary axis.

And the gyroscope does not continue to accelerate but quickly reaches a steady angular velocity of progression.  Basically this is due to the friction at the support point, which results in a torque that by the rule of thumb is at right angles to the axis of spin and resists the progression.  In turn, this friction torque changes the direction of the axis of spin downwards resulting in the graceful slow tilting down of the gyroscope until it falls off the tower.  Nothing like the motion you would expect if you tried to support it by one end only when it is not spinning.

The purpose of this little side track is to illustrate the differences and similarities between linear and rotational motion.  Besides, gyroscopes are fascinating things

Thanks for looking in

MJM460

The more I learn, the more I find that I still have to learn!

Online MJM460

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Re: Flywheel calculations
« Reply #69 on: March 26, 2023, 12:29:51 PM »
Back in post no 65 I used the flywheel inertia and an assumed speed variation to calculate the power into the system from the flywheel during the exhaust half of the revolution of my little single acting oscillating engine.  It obviously takes the same power during the steam inlet half of the revolution.

 With the engine running unloaded at constant speed, the energy from the flywheel is the total energy into that system during the exhaust cycle.  With the engine running at constant speed, all that energy is absorbed in friction and in pushing out the exhaust steam.  This is approximately the same as the energy needed to run the engine for the inlet part of the cycle.  The steam also has to accelerate the engine during that steam inlet.  Hence the total energy input from the steam is twice the energy returned by the flywheel during the exhaust cycle.

If we use the standard formulas relating power, torque and Brake Mean Effective Pressure, with a little algebra, with the torque already calculated, we can arrive at a figure for the BMEP.  Remember that BMEP is an average pressure for the power stroke, and is the mean differential between the two sides pressure of the piston. 

Just for interest, I did the calculation, and for my little single acting oscillating engine.

The calculated power is 0.56 watts.  Then I calculated a BMEP of 26 kPa.   This is based on an average speed of 720 rpm, a speed variation of +/-2.5%, flywheel inertia of 41 kg.mm^2.  The engine is 12 mm bore and 16 mm stroke.  The peak pressure in the cylinder will be somewhat higher than the Mean, but itís hard to guess the pressure variation in an oscillating engine.  If it is approximately sinusoidal, the peak would be about 40% higher than the mean.

For reference, the boiler temperature for that test run was 110 C which means a saturation pressure of 143 kPa absolute.  My steam plant has a 3/16 tube to connect the engine to the boiler compared with 5/32 which is possibly more common, so there might be minimal pressure loss between the boiler and the engine inlet.  With atmospheric pressure on the other side of the piston, about 43 kPa differential.   This is only about 60% higher than the calculated BMEP.

Unfortunately, I have no way of measuring the actual speed variation, it is just a guess.  For a relatively high inertia flywheel on such a small engine, it is reasonable to expect the speed variation to be small.  All things considered that BMEP estimate is surprisingly close to the boiler pressure, but I can make no claim to the accuracy other than that the peak pressure cannot be higher than the boiler pressure.  Similarly that calculated power is no more accurate than the guessed speed variation, similarly it is an upper limit.

I think the main conclusion is that the actual speed variation, if we could measure it, would be something less than the +/- 2.5% I have assumed here.  However, if we could measure it ..Ö

 I think that is about as far as I can take those calculations.  We started with calculating the moment of inertia of a flywheel, and how it is more than just mass.  Did a little exploration of rotational motion and linear motion principles, side tracked into the vector maths associated with rotational motion and gyroscopes, and finally showed how flywheel inertia and instantaneous speed variation are related to torque and power of an engine running unloaded at constant speed. 

I do hope that is of interest.

MJM460



« Last Edit: March 26, 2023, 12:33:58 PM by MJM460 »
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