Author Topic: Talking Thermodynamics  (Read 105868 times)

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #855 on: April 24, 2018, 01:17:03 AM »
Hi MJM, would the larger flywheel increase the HP on its own ?   old motorbikes always had large flywheels but i think modern ones do not because they need to accelerate very quickly..is this factor taken into account on a beam engine and is there a slight variation in the radian speed of the flywheel during its revolution, however small?.. I have given you the dimensions of the flywheel and bar if you would like to do the calculation adding your own 'adjustments' . If i wanted to increase the MoI with say an inserted band of lead melted into a channel on the outer part would that work and could you work that out.? say a channel 5mm wide and 10mm deep ? sorry for all these questions ,but ,I think they are relevant !!!! The bar should read 170mm btw
Willy.........

Offline derekwarner_decoy

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Re: Talking Thermodynamics
« Reply #856 on: April 24, 2018, 05:32:30 AM »
Just back to the chain drive MJM......yes the paddle shaft is the 4.8:1 reduced speed

I had a 48 tooth S/S chain pinion [that looked :facepalm: terrible] and a second 50 diameter bronze spoked flywheel....when I cross checked dimensions it was a perfect marriage  :Love: ......... [Loctite bonded and bolted together with 6 x M2 HH brass bolts]

The S/S roller chain @ 3.1875 mm pitch so @ 1:16 scale equates to ~~ 2" [51 mm] pitch so is reasonably authentic for size of many of the early Australian and European paddlers

It also provides a very robust engagement between the engine crankshaft & the paddle shaft [although an absolute overkill in strength or breaking load]. With the 4.8:1 reduction, I am expecting a relatively smooth paddle shaft motion at say 25 to 75 RPM, with the engine happily ticking over at the higher speeds

I contemplated how to manufacture resilient couplings for the paddle shafts, but ended up purchasing a pair of commercially manufactured couplings with 73 Duro nitrile joining elements [we can imagine the softness of 70 Duro nitrile O-rings] so I am expecting these will provide a little rotary resilience on startup..... [soft start without the bang & clang]

The 50 diameter spoked wheels each have 4 x M3 tapping's, the 10 tooth crank shaft pinion has 2 x M3 tapping's each for M3 S/S HPGS. Finally, the paddle shafts at the wheel hubs are 7.93 mm and the intention is to secure the wheel hubs to the wheel shaft each with an M3 S/S [standard 50:1] taper pin

Derek
« Last Edit: April 24, 2018, 07:11:49 AM by derekwarner_decoy »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #857 on: April 24, 2018, 11:42:22 AM »
Hi MJM just remembered to weigh the items   flywheel is 530 g  and the bar is 540 g....allmost exactly the same !!
willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #858 on: April 24, 2018, 01:51:01 PM »
Extra flywheels just seem to pop out of the woodwork, like wire coat hangers. I have two more that I did not think of yesterday.  One on a Stewart vertical that I inherited, and one on an oscillator I made after first year of study, over fifty years ago.  But might need some bushings to put them on my 5mm shafts.

Hi Willy, a flywheel is just an inert lump of metal with no way of adding energy to the system, even if it is spinning around.  So it does not increase the power available at all.  Increasing the power requires increasing the rpm, increasing the torque, or both.  However, when it is spinning, it has kinetic energy it gained from the torque that sent it spinning.  The practicality of life on earth is that the spinning flywheel will have some friction drag due to the bearings and air resistance, but with care we can make both relatively small.  So a good flywheel will spin for a long time if we start it spinning then just let it go.  Just like a toy spinning top.  Let's ignore that friction for the moment.

So what is a good flywheel?  It is one with a large moment of inertia.  Mass is part of the equation, but moment of inertia (I) is the property that is significant in any rotating system.  The flywheel stores the energy it gained as it was spun up to speed.  We can quantify the kinetic energy stored by the equation

KE = 1/2 x I x w^2.

Please read that w as the lower case version of the Greek letter, omega, which seems to be universally used for rotational speed.  And as you correctly implied, it is measured in radians per second.  Radians are dimensionless so the units are really "per second" or seconds^-1.  I have no idea how to get Greek letters into the post.  (Dan, do you know how it's done?)

When the flywheel is spinning, it is subject to the rotational version of all Newton's laws of motion.  When we apply more torque (T) than the minimum to overcome friction, the flywheel the flywheel accelerates.  For straight line or linear motion the equation describing this is F = m x a.  For rotational acceleration, the Greek letter alpha is usually used.  The units are radians per second squared, or sec^-2.  Perhaps I will just write alpha in full.  So the rotational analogy for the well known linear motion is T = I x alpha.  The energy is stored in that higher rotational speed as the system accelerates.

Similarly, if there is more torque required by the load, the flywheel experiences a retarding torque, and slows down.  In this case the torque is negative, so alpha also negative.  The energy from the flywheel continues to drive the load, but the system slows in the process, and if more torque is not applied, the system will soon stop.

Your beam engine, and your single cylinder engine, (assuming both double acting) provide a fluctuating torque.  It is zero twice each revolution, at the top and bottom dead centres.  The variation of torque within each revolution is the top half of two sine waves, one 180 degrees after the other so a continuous series of pulses.  The effective average is the RMS value, similar to a RMS voltage with only two diodes for rectification, sound familiar?  So about 70% of the peak value.  Not exactly a sine wave with slider and crank mechanism, but close enough.

When the engine produces that peak torque, (providing it exceeds the load torque) the flywheel absorbs the extra energy, and stores it by accelerating.  When the engine torque is below the load torque, the flywheel gives back the energy, but also slows down.  This happens continuously within each revolution.  If you have a modern electronic governor, it could use a 60 tooth wheel or even more, and measure the speed 60 times each revolution.  If you recorded these readings, I suggest you would see at least 15 different speeds, each one would occur four times in each revolution, depending on just how the torque zeros lined up with the teeth on the measuring wheel.  The difference between the maximum and average, or minimum and average, within each revolution is that speed variation I have been trying to explain.  It is the variation within each revolution, not the variation over several revolutions.

When your engine power rating was increased, that implies higher torque, from higher pressure steam, or higher speed, perhaps both.  To achieve the same speed regulation within each revolution, the flywheel must store and give back more energy in each revolution.  You can see from the equation for kinetic energy, you only have speed and moment of inertia to play with.  So either you allow more speed variation, or you add more moment of inertia.  In reality, the additional moment of inertia will not be the exact requirement, it can't be unless the engine runs at only one average speed, so increasing the moment of inertia will be accompanied by a change in the speed regulation, depending on whether the extra moment of inertia is more or less than required.

Sorry for all the repetition, I guess the issue is whether I made it sufficiently clear that the speed regulation I was talking about, was the speed variation within each revolution.

Your questions about your brass bar and flywheel are totally relevant, my intention is that we should be able to get practical results from all the calculations.  So I will do the calculations, and also for a bar of steel to give say half the moment of inertia of the brass bar.  This will give you a better idea of how much more you need, so we can check alternative flywheel designs, and get rid of that finger whacker.  I wouldn't start machining that channel yet.  If you replace cast iron with lead you are replacing 7.8 relative density for cast iron with 11.3 for lead, so you only benefit from the difference.  It would be better to think in terms of how you could add a "belt" of bronze or steel to the outside of the rim, or even an extra ring each side of the rim to increase the width, if you don't have room for extra diameter.  First, let's see if we can get a better handle on how much moment of inertia you need.

I had a careful look at the drawings for the required dimensions.  All the required information is there and now you have given me the weights we can compare that with calculated weights.  They probably won't match, but is good when they are near the right figure.  It will be tomorrow though.

Hi Derek, that explains the low speed flywheel.  I probably would have done the same in the circumstances, just to get it all connected up and a test run.  Probably even helpful for test running before those paddle wheels are complete.  However, once you have it all in the boat, that flywheel is only ballast due to the slow speed, if you need ballast, it is better to be able to choose the location.  I would suggest that when you have a spare hour, (like the ones all the rest of us have!) make up a simple flanged wheel to support the chain wheel with less weight.  If you need more moment of inertia, it is better to have it on the high speed shaft.  I like the idea of those flexible couplings.  They make the support of the paddle shaft more difficult, but will reduce any sharp impulsive torques from the ripple load of the paddle wheels.

I am most interested to hear how the speeds turn out, as I definitely have a paddler on my ultimate list.  I am working towards a twin cylinder engine.  Would prefer a centre-flue marine boiler for a boat, so may end up using the bought one as I have not been able to work out a suitable Meths burner design for the centre flue.  Got a long way to go on the skills front though.

Well, time flies when you are tending the bar-b-que, roast lamb for 30 guests for a birthday celebration for a niece this evening.  Time to put the chicken on.  Should get this posted after the guests are gone. 

Well, the party went well.  Noisy and chaotic as you might expect with more than half of them children.  All gone now except the two grand children staying overnight.

Thanks for following along,

MJM460
The more I learn, the more I find that I still have to learn!

Offline MJM460

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Re: Talking Thermodynamics
« Reply #859 on: April 25, 2018, 11:49:44 AM »
Hi Willy, I completed those calculations this afternoon.  I have tried to attach a print of the section of the spreadsheet so you can see the basic steps in the calculation.  My cheap printer makes it quite difficult to get a good clear print from a spreadsheet.

Basically I calculated for a full disk for each of the rim sections, and also for the centre part which is missing.  Using negative density, or subtracting the centre is probably the easiest way to calculate the moment of inertia for the rim.

That flywheel has a moment of inertia of 945,000 x 10-9 kg.m2, approximately 3 times the moment of inertia of my 75 mm flywheel, largely due to the larger diameter.  You can see the spokes and hub make an insignificant contribution, while the majority comes from the outer rim.

The brass bar has 40% higher value, 1,337,871 x 10-9 kg.m2

A steel bar 150 mm long x 25 x 12 mm has a value roughly half of this, 663,000 x 10-9 kg.m2 which would make a convenient half size step up from the flywheel alone.  You can see this would make a very simple temporary flywheel to give you an idea of what you need.  The formula for a bar about its centre of gravity is I = 1/12 x m x R2.

The calculated mass was 567 g for the flywheel and 556 g for the bar, a little higher than the measured weights.  The error is probably density differences for the exact alloy your metal is made from, but also the mass is quite dependent on exact measurements.  I expect the discrepancy is within a reasonable tolerance, and more than satisfactory for the purpose.

But while I was doing the calculations, I could not help but wonder why the extra moment of inertia is required.  Your double acting engine goes through zero torque at the top and bottom dead centre, but for a free running engine it really does not take much of a flywheel to get it through that zero to the point where there is again plenty of torque to overcome friction.  I would expect your flywheel is more than adequate for running, and should even give good speed regulation, even under load.

How does the engine behave without the brass bar?  Does it stop at each dead centre, or just always stop at one end?

I am wondering if you have had the valve chest cover off and carefully checked the eccentric settings.  First I wondered if you just had early cut off which gives a bigger range of rotation at zero torque for starting, but once you spin it over by hand to get it started, there is air pressure on the cylinder during expansion after cut off so most of the range of zero torque disappears.  I don't think this is the issue.

However, if the admission is early, probably depending on the exhaust closure point as well, you could get full steam pressure in the cylinder before top dead centre, which would require much more moment of inertia to turn past the top dead centre and keep it running.  If steam reaches full pressure before top dead centre, the flywheel has to supply enough energy to push some of that steam back against boiler pressure until the piston reaches the top dead centre.  Similarly at the bottom dead centre.

I hesitate to tell you how to set the valve gear, and diagnosis is difficult from such a distance, especially with no information on what was happening that you needed the extra flywheel.  The only obvious thing I can think of is early admission so the steam reaches full pressure before the piston reaches top dead centre, but I am confident that you will be able to find the issue.

In any case, I would not suggest modifying that nice looking flywheel yet.

I hope that is helpful,

MJM460
« Last Edit: April 25, 2018, 11:57:59 AM by MJM460 »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #860 on: April 25, 2018, 10:46:37 PM »
Hi MJM, Thanks for the calculations and the engine is working with its own flywheel now. It was just reluctant to start with. I did some adjustments as well once it got going and it seems fine now !! here is the video.........I was wondering that if the flywheel size makes no difference to increasing the Horsepower did the engine maker need to do this with the Wentworth Engine ?


Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #861 on: April 26, 2018, 01:51:22 PM »
Hi Willy, I was a bit puzzled, as I thought that you had posted the video of the engine running without the brass bar in your engine build thread.  I assume the brass bar convinced you that you had a working engine, and then you did some adjustments, I assume the eccentric settings, and found the bar was no longer necessary, then made the video.  If you have an hour or so, and a 150 mm long piece of 25 x 12 hot rolled steel, you could make up that steel bar and see how you go with only 60% of your current flywheel moment of inertia.  I really expect it should still run well, even if with a few more tweaks to the eccentric positions.

Remember that your engine has two independent speed regulating mechanisms, or three if you count the steam stop valve.  The flywheel regulates the speed within one revolution only.  Well, it would probably keep it running for three or four revolutions, or even more the engine running free, but only by progressively slowing down until it stops.

The governor regulates the speed by controlling the steam pressure, so actually changes the amount of energy available for conversion to work by adjusting that throttle valve at the steam chest inlet.  The governor should have enough damping, and be slow enough acting, that it only responds to a longer term change in the average speed.  It should not respond to those changes within each revolution, there would be unacceptable wear in the pins of the linkages if it did.  But, if the engine keeps getting faster each revolution, the weights move outwards and the linkage to close the throttle valve a little.  Alternatively, if the engine slows, due to an increase in load, the  governor weights collapse inward, and the linkage opens the throttle to admit more steam.  The spring tension on the linkage determines the speed the engine settles at in the longer term. 

The big trick to governor performance is that it must be slow enough not to respond to those torque fluctuations within each revolution, but fast enough to prevent the engine over-speeding if the load is suddenly lost, for example the belt could come off, or even break on a belt driven load.  Industrial engines have a separate independent speed trip mechanism and a very fast acting throttle valve, called a trip valve, that slams shut if required.

Let's return to the flywheel again for a moment.  The kinetic energy stored is determined by the flywheel moment of inertia and the rotational speed squared.

K. E. = 1/2 x I x (omega)2.

The units of energy are the same as the units work, so the same as the units of torque times rotational angle (T x angle) with the angle measured in radians, which you will remember have no units.  It is also the same as work in a linear system, force times distance.  The key to getting the correct dimensions for force is Newton's equation, F =m x a, so kg.m/s2.

For a rotational system, torque = force times distance, (the distance being the moment arm), so work in both linear and rotational systems has the units of kg.m/s2 x m, or kg.m2/s2 which is the same as the units of K. E. In the equation above.

When the speed of the flywheel slows, the stored energy is reduced.  Conservation of energy says that energy went somewhere, and the answer is that it was returned to the rotating system as work  to help driving the load.  When the torque from the engine is greater than the load, the system speeds up, and the work not absorbed by the load is absorbed by the flywheel, as an increase in speed.  And this cycle continues while the engine runs.

You can see from all of this that the flywheel does not add to the energy of the rotating system, it simply soaks up the excess when it is available, and returns it when there is a deficiency, but both only in very short term.  So the answer to your question on the Wentworth engine is that it was not necessary to modify the flywheel just to keep the engine running, it was included simply to keep the speed fluctuations within each revolution within acceptable limits.  With the increased power available, the load would also have been increased, so more stored energy was required to limit the slowing as the engine passed through the top and bottom dead centres.

I have finally found an Ap that lets me make a quick freehand sketch, as I would with a pencil if we were talking face to face.  They say an Italian needs his hands to talk, I need a pencil!  So let me see if I can include a sketch to illustrate what I have been saying.  Please excuse the shaky freehand experience, the Ap does not seem to include a ruler, let alone a sine function.  I don't know the file size, Apple is quite sure I don't need to know, so refuses to tell me.  If it is too big, I will come back and post them later, when I have access to the computer to resize it.

The first sketch shows the torque in black on the vertical axis and degrees of rotation on the horizontal.  The speed is shown in red on the vertical axis.

The second sketch shows the torque fluctuations for several revolutions, again black.  The red crosses are the rotational speed, measured only once per revolution, for an engine at steady speed.  Measuring speed once per revolution does not see those variations within the revolution.

The blue crosses are the speed, again measured for an engine with steadily increasing speed.  It is the governor which must respond to this steady increase, and shut the throttle plate a little to slow the engine back to the required average speed.  I hope that helps clarify what I have been trying to say.

Thanks for looking in,

MJM460
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #862 on: April 26, 2018, 02:50:50 PM »
HI MJM, Yes that makes things a lot clearer. The Governor on the Beeleigh engine has a 4 stepped pulley on the crankshaft and a 3 step pulley on the Governor side  so there is a lot of options with the speed control. The engine provides the power for up to 5 corn grinding large stones so quite a variation in the load !
. The starting procedure is to pull the handle on the governor valve down to admit steam and let the engine run up to speed. If the rope breaks then the governor will drop and cut off the steam quickly. here are some pics of the arrangement...The governor valve is inside the main steam valve. This model has been a fascinating engine to make and the restoration of the original is well under way!!
« Last Edit: April 26, 2018, 02:58:33 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #863 on: April 27, 2018, 02:52:17 PM »
Hi Willy, it certainly is an interesting engine to be able to look at closely.  You didn't say how the drive is transmitted to the mill-stones, but quite a range, from one or even zero, through to all five, but I assume the speed would be similar, regardless of number of stones being driven.  However, if the engine was powered up, driving all five, and some part of the drive broke, the engine would accelerate quite quickly, and this is what the governor, or these days the separate trip valve, must catch.

You have me confused about the governor though.  When the engine runs, the governor weights fly outwards, and the collar position controls the valve position.  When the engine is running too fast, the weights fly out further, and move the linkage to close the steam valve.  When the engine slows, the weights drop back in and open the valve to restore the set speed.  If the governor drive belt drops off, or breaks, the governor rotates slower and stops, the weights fall inwards, as it would if the engine had slowed, which should move the valve to the full open position.  Not what you need when the governor no longer "knows" the engine speed.

I have given some thought to how to get a feel for how much energy is stored in the flywheel, and what is the impact of the degree of regulation allowed.  I have already mentioned that there are rotational analogies for all the normal laws of motion we learn at high school for linear systems.

I will use your flywheel with a moment of inertia of 935,300E-9 kg.m^2, or 0.0009353 kg.m^2.  Then I will assume you want the engine to run at an average of 500 rpm, and allow a speed variation of +/-10%.  This means the speed varies from 550 rpm maximum to 450 rpm at the slowest point in each revolution.  We have to change these to radians per second for use in the calculations, 57.6 max to 47.1 minimum all while the flywheel rotates 1 revolution, or 2 x Pi radians.

We can use the rotational forms of the equations of motion to calculate the acceleration, and the steady torque necessary to cause this acceleration.  For slowing, acceleration is negative, 0.082 N.m.  We can also calculate the time for one revolution as 0.12 seconds, or an average speed of 500 rpm.  Finally, Power, assuming steady retarding torque is about 4 watts.

Now this does not mean that your engine is producing that power, we have no measurements of what the engine is producing, or how much the speed changes within the revolution.  The flywheel stores energy by speeding up during more than half of the revolution, then returns this stored energy by slowing down during the remainder of the revolution.  It is quite hard to work out what this figure really means as the total energy returned is returned in less than half of the revolution, so the actual load the engine is driving to give this speed variation is probably more than double.  It suggests that this flywheel could be suitable for an engine driving a load of around 10 watts.  That reasoning is pretty obtuse, so don't worry if you can't follow it.  The main conclusion, with a pile of judgement/wild guessing thrown in, is that all the energy figures and rotational speeds are the right order of magnitude, and depending on your steam supply pressure, or probably a bit high, meaning a smaller flywheel would be adequate.

It is interesting to do the same calculations with an average speed of say 1000 rpm.  Double the speed means four times the energy, however, speed regulation based on the same percentage means a wider speed range.  The corresponding "power" is then 34 watts, much higher than I would guess is possible with reasonable steam pressure.  If this is the speed you are aiming for under full load, a considerably smaller flywheel would be adequate.  But you can see a slow rotating beam engine needs a bigger flywheel.

I think the real value in these calculations is not in the actual figures, but in understanding the implications of those equations of motion, and the effect of moment of inertia.  At the end of the day, given that most of us don't have the ability to measure the speed variation within each revolution, the best way to determine the right size for a flywheel is trial and error under the proposed operating conditions.  However, by calculating the moment of inertia, we can compare flywheels, and even use square bars as a temporary flywheel.  We can then select a more conventional wheel with similar moment of inertia.

Thanks for looking in,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #864 on: April 28, 2018, 12:49:28 AM »
Hi MJM, The engine is in bits at the moment as i am making the new bed to accommodate the dynamo, so i will delay the air test for the moment The governor valve on the Beeleigh engine is cunningly incorporated inside the main steam valve, rather like a butterfly valve. I have made a drawing to show the parts in cross section also some photos.They did manage to break the end off when removing it as well.Is it possible to have too large a flywheel on an engine? and with a boiler supplying steam to an engine is it a good idea to have the boiler producing a higher pressure than the engine requires  as a form of backup to keep the piston moving ? all these questions are possibly of a more practical thinking than a technical mind so i think i should do a lot more reading !!!
Willy
« Last Edit: April 28, 2018, 12:52:45 AM by steam guy willy »

Offline derekwarner_decoy

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Re: Talking Thermodynamics
« Reply #865 on: April 28, 2018, 10:48:20 AM »
MJM...when you mentioned....'I like the idea of those flexible couplings.  They make the support of the paddle shaft more difficult'

To aide this, the paddle shaft will be supported by six [6] 1/4" ID Grade 304 ZZ shielded ball bearings in miniature plummer blocks

Only the outer [4] will be splashed with water....& hopefully never immersed  :facepalm:

I am enjoying the continued TT thread, however question in my mind that the brass or steel bar mounted on the crank [as opposed to a uniform disk shaped flywheel] would produce horrendous mechanical pulsations x 2 at each revolution that would result in premature bearing/bush failure due to ovality?

Derek

PS1...the mock-up image looks a little like spaghetti or a dogs breakfast of bits on a shaft including 1x resilient coupling...[these couplings will look a little too current Century, :stickpoke:  so plan to cover them on the paddle shaft .....so out of sight]

PS2... Willy....one must wonder how that tapered plug valve spool ever sealed in the valve body.... or is it only after years of non use then,,,, stripdown and de-rusting preservation?
« Last Edit: April 28, 2018, 11:19:18 AM by derekwarner_decoy »
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #866 on: April 28, 2018, 01:25:18 PM »
Hi Willy, that is a very clever steam valve design with the governor valve inside the stem.  The combined unit has fewer flanges which are always potential leakage points.  It is a pity it was broken.  I guess someone will work out how to fix it, although if the seal area on the two stems is corroded like the rest of the shaft, it may be preferable to replace one or the other or both.  Or would that be anathema?

Basically, the optimum size for a flywheel is a very flat plateau, and a very large range of flywheels will be satisfactory.  If you have a flywheel that works to your satisfaction, a larger one would do the job with less speed variation within each revolution.  A smaller one would mean the speed would vary more.  As the flywheel gets bigger in moment of inertia, generally the mass will get bigger, and also the diameter.  In most engine plants, mass or diameter, or often both will become a problem, probably before the inevitable air resistance and bearing friction really become significant.  But a bigger flywheel always increases the difficulty of manufacturing, installing and maintaining the plant.

Extra pressure in the boiler does not do much to keep the piston moving through those two zero torque points at top and bottom dead centres.  I suggest that the flywheel is the item which keeps the piston moving through the low torque points in each revolution.  You need to tap into stored energy, or momentum, to carry the engine through those zero torque points.  However, I would expect that extra pressure in the boiler may help the engine respond to sudden load increases, when the governor will quickly open the valve to maintain speed.  But, the extra pressure also has a disadvantage in responding to a sudden load decrease.  I suspect the boiler pressure and also water levels would be operated to optimise boiler operation rather than the engine.  As you say, this is a practical question, and the theory can help predict the effect of those variables, but I suspect that practical boiler operation experience would better inform you of the best way to go.

More reading?  With your library, you certainly have the material.  I don't know about you, but I find that by the time I read all the posts on this forum and spend some time writing this one, I barely have time to keep up with the news, eating and sleeping, and any remaining time seems to get allocated to washing the dishes.  I would have to join Chris's universe to fit in more reading, and he is a bit secretive about how to get there. 

Hi Derek,  glad you are still enjoying the thread.  I like those flexible couplings for the flexibility, and damping they provide.  They look excellent, but I was referring to the difficulty of getting so many solidly mounted bearings lined up.  Sorry I was not more clear.  You are addressing the issue they had on full size paddlers, though your hull will be relatively more stiff so less problematic.  They certainly look ideal for the job.  Ideally you would use only four bearings, with the two couplings supporting a floating section.  As the gear wheel position is determined by your engine layout and would need to be on a solidly supported section, you would need an asymmetrical arrangement, but it is all looking very neat and well thought out.  I am sure you will sort out the issues.

The solid brass bar, drilled and reamed on its centre of mass is perfectly balanced will spin perfectly freely with no pulsations, despite its appearance.  It will stop at any position, just like the round version, but will churn up a bit more air, rather like a badly shaped two blade aeroplane propellor.  If the drill misses the centre line there will be the normal out of balance forces, but no worse than those due to the big end and crank throws.  A small out of balance at suitable orientation can help offset the crankshaft.  The centre of mass is found in the normal way by the intersection of the two corner to corner diagonals drawn on the face to be drilled.  The bar should be straight, and the hole really should be perpendicular to the length of the bar, or you will get a rotating couple, like a two cylinder engine.  You can try it by just drilling a similar shaped rectangular bit of steel and spinning it on a rod of silver steel, use a piece that will need a hole in the centre for its intended use.

I have given a bit more thought to just what those calculations based on the laws of motion are actually telling us.  It is not much use calculating as though the flywheel was returning energy to the system for the full revolution,  because, as we know, the flywheel has to soak up excess energy from the engine for part of the revolution, in order to have energy to return to the system during the rest of the revolution.

The engine torque has a sinusoidal form.  Well, a scotch crank torque characteristic is sinusoidal, a cross head and crank gives a bit more complex form, but a sine form is close enough approximation.  As you know from working with AC electric power, the steady output equivalent to the work done by the sine form is the r.m.s value, which is 1/square root of 2 or 0.707 times the peak value.  If we look up the sine tables (or use trigonometry) and find the angle whose sine is 0.707, we find it is 45 degrees.  So for the first 45 degrees, the engine output is less than the average, while the second 45 degrees, it is greater.  For the next 45 degrees, it is still above the average, but decreasing and for the last 45 degrees of the half revolution it is again less.  Similarly for the remaining half of a full revolution.  So for half the revolution, the engine power is more than average, and for the other half it is less.  For half of the revolution, the flywheel soaks up the excess energy by speeding up, then for the second half, the flywheel returns this energy to the system while slowing down.

If we redo the calculations based on the speed reduction occurring in only half of the revolution, and increasing during the other half, we find the engine supplies the equivalent of 17 watts excess power during the high torque half of the cycle, while the flywheel returns the equivalent of 17 watts during the low torque part of the cycle. 

The steady torque required to slow or accelerate the flywheel within half a revolution is 0.3 N.m.  You could calculate whether your engine is able to produce torque of this order or more by calculating the force on the piston at your supply pressure and multiplying by the crank throw.

Remember these calculations are not based on any engine measurements, but simply the values at which the speed variation equals the assumed 10% with the given moment of inertia.  But I believe it means the flywheel would achieve this performance for an engine with an average power considerably larger than that 17 watts, as the figure is based on a speed variation of only +/-10%.

I suppose it might be possible to estimate the engine power from these calculations, provided we increased the moment of inertia to account for the effect of the engine rotating parts.  But the difficulty of measuring the speed variation within the revolution means it is probably not practical  for most of us.  However, it seems to demonstrate that the flywheel should be adequate for your engine.

One more factor influencing the required size of the flywheel is the load characteristic.  If the load imposes a constant torque throughout each revolution, such as a winch, or loaded conveyor, or your mill stones, the load exceeds the engine output by a considerable amount at some points of the revolution.  If however the load is say a reciprocating pump, the load torque is also zero twice per revolution.  If this characteristic is connected with the zeros at the same orientation, and compared with the engine torque, there will be hardly any deficiency all around, and the flywheel will only have to overcome friction to keep the system moving.

On the other hand, a load which varies with speed such as a centrifugal pump or fan will impose less torque during the slower parts of the revolution and more torque when the speed is faster.  This characteristic will tend to reduce the required flywheel inertia a little.

Well, some quite theoretical calculations, but only reliant on Newton's laws of motion, which in turn rely on the fundamental conservation laws of physics.  I can give more details of the calculations if anyone is interested, but the main purpose in presenting this is to give a better understanding of how the flywheel does its job.

Thanks for looking in,

MJM460
« Last Edit: April 28, 2018, 01:32:32 PM by MJM460 »
The more I learn, the more I find that I still have to learn!

Offline derekwarner_decoy

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Re: Talking Thermodynamics
« Reply #867 on: April 29, 2018, 10:40:53 AM »
Thanks MJM....."Ideally you would use only four bearings, with the two couplings supporting a floating section"

Just a point of clarification, this is the case...two [2] plummer blocks are to be the output shaft restraint elements on either side of the engine....or essentially a part of the engine

The paddle wheels will be supported on either side of each wheel...[so these are the four [4] bearings you suggest]...then the resilient coupling on each side of the chain drive shaft ......[remembering this is not the engine crankshaft]

At this stage in time, it is little difficult to setup and photograph the final drive....

I take your point thankyou ....so for the final alignment of the shaft, I don't believe rigidity of the hull will be the issue, however I can see me asking a colleague with the facility of a milling machine table large enough to accept & setup the hull & then take a witness mill cut  :hammerbash: over the four [4] plummer block mounting surfaces 

Derek
__________________________

PS1......can you imagine our Shipwrights 100 years ago on the Murray Riverbanks with 30ft of rubber hose with a glass tube in each end [under the hull of the a Paddler being built] and comparing the level of the water at each bearing support? .....I cannot think of any alternate method!

PS2.....yes, but I still need to close my eyes to accept the brass bar [true & = & on axis] provides that uniform harmonic rotational weight
« Last Edit: April 30, 2018, 02:52:30 AM by derekwarner_decoy »
Derek Warner - Honorary Secretary [Retired]
Illawarra Live Steamers Co-op - Australia
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #868 on: April 29, 2018, 01:39:46 PM »
Hi Derek, I am most interested to understand your drive train and how you are using those couplings, so I will be keenly following your progress.  So, as I understand it, the paddle wheels are each supported by a bearing each side, and the chain wheel is on the floating section between them, supported only by the two flexible couplings.   Are the flexible couplings suitable for the side loading by the chain and that extra flywheel weight?

In addition to the hose and water, I assume that those shipwrights would have used a tight wire stretched across between the outer bearings and measured the centre ones from that.  I once had a boss who was involved in alignment of the machinery in our Snowy Mountains hydro project, (many years earlier) and he often told us how the turbines were all aligned by plumb bob and tight wires.  It would have involved considerable skill.  However, I wonder if your model bearings can be aligned by threading the four bearings on a straight rod just enough under diameter to allow the bearings to slide into place, rather than needing that milling machine.

There is no need to close your eyes imagining that bar.  Perhaps the best way to demonstrate the situation is by carefully marking and drilling a bar at its centroid, a piece of steel say 150 mm long, or even a piece of 4 x 2, so you can suspend it on a piece of rod held horizontal, or even the parallel part of the drill shank.  If you have drilled it at the right place, it will balance on that rod, otherwise balance it up with a touch of the sander.  It will balance with the long dimension vertical or horizontal, or anywhere in between.  If you give it a spin you will feel no rotating force, and it will even demonstrate gyroscopic effects, so long as it is not too long for your arms.  I suggest you just try it.  A more practical example is a two bladed aeroplane propellor, which has to be well balanced on a high speed engine.  Similarly, a conventional ships propellor.

You don't in fact need a regular steel bar, any shape will balance at its centroid, and spin without any rotating force, it is a very interesting feature of the centre of mass and rotational motion.  Even your engineers square has a centroid, and will spin around that.  Of course in this case, the centroid is not on either of the arms, but in between them, so you would need some of Willy's perspex to make a brace between the arms to support a bearing at the centroid.  You may need to break out the old Meccano set to make up that one.

The big advantage of the conventional circular flywheel with spokes and a heavy rim is that it is the optimum shape to provide the largest moment of inertia from a given mass of material.  Alternatively it requires the smallest mass to provide a given moment of inertia.  In addition, it experiences less air resistance than other shapes, due to relatively streamlined shape.  Even with a circular flywheel, the energy used to create the turbulence can produce quite a bit of heat if the guard does not allow sufficient ventilation.  This is a significant issue on large, high speed couplings, despite provisions to fill the space around bolts to minimise turbulence.  (Interesting that that line of thought brought us back to couplings, which have many of the properties of small flywheels, I must still be thinking about them in the back of my mind!)

Obviously a conventional flywheel is more appropriate on our models for many reasons, but it is more difficult to mock up on an experimental basis.  The idea of calculating the moment of inertia of a bar, which can easily be adjusted, and using this to find a suitable moment of inertia for a new engine was simply to provide a method of quickly and cheaply arriving at a suitable flywheel size that can be compared with available flywheel sizes before purchase.  The number does not have to be very precise, the hub and spokes contribute minimal moment of inertia, so just calculate the the moment of inertia of the rim of the available flywheels and compare with the moment of inertia of the bar.  Remember recently, Brian was considering whether to make his flywheels four inch diameter or four and a half from material at hand before he finally purchased two.  I had not thought of using a bar at the time, we can thank Willy for his thinking outside the box, so had not worked it through to suggest it at the time.

A bit shorter post tonight to make up for yesterday,

Thanks for looking in

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #869 on: April 30, 2018, 02:45:24 AM »
Hi MJM, Thanks for the latest posts and answers to my questions.. I suppose these are the same questions that people have asked in the past and then did lots of experiments to find the answers. !! every time i ask you something you provide the answer which i am grateful for as it would take a lot of reading to find out myself !! the info you provide would be quite difficult to find out from a book as you cannot ask a book questions unfortunately .....I was having a bath tonight and i filled it a bit too full and a bit too hot ,so i was wondering if you have a quantity of water at a certain temperature ,it would take a length of time to cool down. but if you had half the amount of water at the same temp with the same ambient conditions would it take half the time to cool down the same amount ??
The question being if i emptied the water to the correct level would i not have to wait so long for it to reach the more comfortable level..?? Also if you used a bar to get a good performance ,could you use that info to design an adequate flywheel to give the same result and is there a formulae for this ??
Willy
« Last Edit: April 30, 2018, 02:51:05 AM by steam guy willy »