Author Topic: Talking Thermodynamics  (Read 84300 times)

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #840 on: April 16, 2018, 02:50:38 PM »
Hi MJM, Interesting questions to answer...... If you are in a room with everything in the room at ambient temperature....if you touch something metal it feels cold ....if  you touch some thing soft and cuddly it feels warm,   so ......at what actual  temp does the metal have to be to not feel cold , is it somewhere between ambient and body temp ??  when i was talking about 'draft' i was thinking about the rush of air causing its own "draft" inside the pipe ! Does the thermocouple act in the same way as you finger ,feeling the metal as being 'cold' ??

willy
« Last Edit: April 16, 2018, 02:57:48 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #841 on: April 17, 2018, 11:42:51 AM »
Hi Willy,

As a furnace draft can be either natural draft, occurring due to the density difference caused by heating of the fluid, or forced draft, driven by a fan, I guess that it is not unreasonable (though not usual) to talk about the draft inside the pipes.  The flow of air (or steam) in the pipe is driven by the compressor or the boiler, and the velocity means that heat transfer is considered forced convection rather than natural convection.  This tends to increase the local temperature gradient near the tube wall, so increases the rate of heat transfer.  As the air compressor discharge is quite hot, in this example the heat transfer is cooling the air.  It would cool quicker in a metal pipe instead of my flexible hose which I assume is mostly rubber, with or without some steel reinforcing mesh.  And certainly, by the time the air reached the connection at the user end of the hose, it was only slightly warm relative to the ambient air temperature.  But further cooling to 14.9 degrees at the engine inlet was a bit of a surprise.

Your comparison of the temperature readings with touching hot or cold surfaces is worth thinking about.  When we touch a metal surface which is say 20 degrees, there is heat transfer, and hence a temperature gradient from our blood supply (about 35 or a touch lower in our fingers) through the flesh and skin to the metal.  Our nerve endings are under our skin, so at an intermediate point on the temperature gradient.  Assuming a large block of metal rather than a thin foil, the heat transfer continues.  The metal,with high conductivity and high specific heat (which determines how much heat is stored in the metal,) the heat transfer continues, but the contact temperature where our skin touches the metal stays close to the bulk metal temperature, the heat transfer continues and the nerve endings can detect the lower temperature.

When we touch something "soft and cuddly", the material is has much lower conductivity than the metal, though similar specific heat.  The lower conductivity means the temperature gradient continues into the bulk material and the temperature in the region of our nerve endings is a little higher so the material feels warmer.  It actually is warmer very close to where we are touching it, heated by transfer from our blood supply.

So the difference in feel is due to the difference in conductivity of the materials and the heat transfer from our blood supply to the object.

So is the situation similar for a thermocouple?  Basically the thermocouple is a welded junction between two different metals that produce a voltage difference in the contact area, and the voltage varies quite predictably with temperature.  You can try it with your digital voltmeter, which will give you something in the range of a few millivolts.

This voltage is measured by a very high impedance circuit, which draws practically no current, perhaps micro amps or less, so the heat generated, volts times amps, is really negligible.

Similarly the mass of a thermocouple is tiny, it is only that little blob where the metals are welded together, mine measure about 1.5 mm in diameter.  If a thermocouple is placed on a flat surface, it sits in that area between the surface and the air, and there is a temperature gradient which affects the thermocouple reading.  However, if the thermocouple is deep in a small diameter hole, or covered with a pad of insulating material, it very quickly arrives at a temperature very close to the material temperature, when there is no further heat transfer.  It takes about 3 - 5 seconds to get within about 0.1 degrees and this time is the reason temperature readings are always quite slow.  However, after this initial period, the the only heat flow to or from the thermocouple is that which occurs along the very thin insulated wires, which are influenced by ambient temperature along most of their length.  I am sure a figure can be calculated, much like the stem correction calculated for an accurate glass thermometer.  But it is small enough that it is usually ignored.

It is hard to believe that my observed temperature difference between the reading with the thermocouple close to the one measuring ambient, when both were about the same, and the temperature reading when I inserted it into the thermowell in the engine inlet piping, which had been sitting with no obvious heating or cooling for most of the day.  Late enough in the day that the night time temperature had gone, and inside my brick garage the temperature reading had been steady for a couple of hours.  No flow, no heat source, long soak time yet nearly two degrees difference in the temperature reading. 

So grasping at straws perhaps, but based on the fact that everything continually looses heat by radiation and at the same time receives heat from everything around it, I am wondering if a difference in emissivity and absorptivity, as on real surfaces, both of these vary with wavelength and direction, and could vary between the brick walls, foil insulated roof, and the brass thermowell and copper piping, which are not polished, but still reasonably shiny.  The third factor in radiation is the reflectivity of a surface, which also varies between materials.  All radiation falling on a solid surface is either absorbed or reflected,  so the difference between brick walls and the metal pipe fittings is a possible explanation. 

There is a law of radiation transfer, Kirchoffs law, which says that when a surface emits the same amount of heat as it absorbs, the emissivity and absorptivity are equal.  The obvious corollary of this is that if they are not equal, the heat emitted is different from the heat absorbed, and the temperature will change until the two are equal.

On the other hand, my textbook has no examples which show an object in a room settling at a different temperature than the ambient temperature in the room, so I am not sure if it is considered obvious, or if I am on the wrong track.  So far, I can't think of an alternative explanation.

Getting back to that valve, let's think about the purpose of lap, being the amount by which the valve dimension exceeds that necessary to just cover the steam ports at its central position.  With no lap, the valve just covers the ports and one or the other is open for the full 180 degrees.  With some lap, the valve has to move off its central position by the amount of lap I order to start opening a steam port.  And it means that the port will be open for less than 180 degrees.  If we then rotate the eccentric, so the port for one direction just starts opening at the dead centre, (lead), then there will be admission starting at the dead centre with shut off occurring before 180 degrees, and expansion for the rest of the stroke.  So expansion requires the port to close before it would with no lap.

If we machine the buckle with extra length,so there is a gap. The crank would be rotated to the dead centre, the valve adjusted to be just about to open.  As the crank rotates the port is opened, to end of the valve travel, when the buckle starts moving back.  However, due to the gap, the valve movement does not start until the buckle has travelled the amount of the gap, so later than with a close fitting gap.  However, early cut off requires the valve to move earlier.  So I believe that providing gap in the buckle, (or a wide slot for the nut), is not an alternative to providing lap.  Does that make sense?  Or have I missed something?

That feels like enough to think about for one day,

Oh, by the way, I believe I have now resized those photos I think, so let's see if I have more success today.

Thanks to all 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 #842 on: April 17, 2018, 11:46:05 AM »
My Variation of the Joy valve gear with swing links instead of curved slides (in the third photo) is quite difficult to clearly photograph, bit this one may help a little. 

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 #843 on: April 18, 2018, 01:43:27 PM »
Thinking back to yesterday's dilemma, two unexplained temperature differences, the engine inlet piping compared with atmospheric temperature, then the cooling of the air from the compressor when it reaches the engine inlet. 

I can't add anything to the discussion on radiation heat transfer as a possible explanation of the first, I am by no means sure, but let's think a bit more about that cooling of the inlet air, before it even reached the engine.

First another observation.  Instead of touching various solids, what do we feel if we blow on our finger?  The air definitely feels cool.  So is this due to forced convection cooling of our finger by the air stream?

First try two more experiments.  First with lips pressed together to get a fast flowing air stream, blow on your finger, it feels cool.  Then, with mouth more open, breathe out so the air flow is over your finger.  This time it feels warm.  Not just less cool, but it actually feels warm.  So, is your breath cool or warm?

I suggest the air you breathe out has first filled your lungs, and been in close contact with your blood stream, so I would expect it to be close to blood temperature.  The warm breath is not unexpected.  But if the breath is warm inside the body, why does it feel cool when we purse our lips?

Now remembering that heat only travels from high temperature to low temperature, if the higher velocity air stream feels cooler, it is most likely absorbing heat from your finger, so surely it is actually cooler than your finger.

If you have a thermocouple, try blowing on the thermocouple, first with lips pursed, then with mouth  relaxed open.  The effect of age must be catching up on me, I am sure I used to be able to sustain a breath for much longer than now.  The temperature change, both up and down, seemed too quick to glimpse sensible readings.  My meter has a maximum function so I pressed the "Max" button and the word appeared on the screen, and blew again.  As you would expect, the reading went up and held at the maximum.  Several breaths later, I decided I had found a reliable maximum.  So I reset the meter and tried again.  First noted the room temperature, 20 C.  Sunny day outside.

With my mouth open, the reading reached 34 degrees, seems reasonable for expired air.

When I pursed my lips for the high velocity air stream, the best I could reach was 25 C.

I kept alternating the velocity with quite consistent results.  This time, with the thermocouple, the only heat transfer is the minimal heat required to lift the tiny thermocouple to near the air steam temperature.

Note that the high velocity air, which felt cooler on our finger, registered warmer than room temperature on the thermocouple, but still nearly 9 degrees lower than the reading of the warm breath.

The difference in breath temperatures seems real on the thermocouple, and suggests that while feeling cool on the finger the breath was still 5 degrees above air temperature, so I would expect it to feel warm, but it felt cool.  The high velocity seems to be doing more cooling than you would expect from just considering convection heat transfer.

Apart from heat transfer, the temperature falls if work is done.  But is there any work done when we purse our lips and blow?  If so, where?  Certainly your lungs build pressure, and then contract, so there is pressure causing force, and movement through a distance, but that side is compression and should warm the air, even above the warming by heat transfer in the lungs.

It brought back a memory of a science programme I heard on the radio some time ago.  The speaker explained that the air you breathe out is not entering a vacuum, it has to displace the air already filling the space and push it away.  This is consistent with a statement in the thermodynamics text book that it is not always easy to see if work is being done, or which is the boundary which moves to produce the work.  I was sceptical at the time, but this little experiment seems to support the theory.

The relevance to this thread, is that it returns us to the concept of enthalpy.  You will notice that I always use the enthalpy columns of the steam tables, and don't refer to the internal energy columns.  The definition of enthalpy is internal energy plus pressure times the specific volume.  The pressure times volume term is often called flow work.  It accounts for the work at the system boundary of a system with continuous flow.

When you breathe out with your mouth relaxed open, there is very little resistance to the out breath, so minimal pressure build up.

When you purse your lips, you provide resistance to the outgoing breath which requires your lungs to provide extra pressure to expel the air.  Not much, but it is enough to have a significant effect.

It is the flow work associated with the expansion of that slightly pressurised air that makes the high velocity jet not only feel cooler, but actually is cooler than the air in your lungs.

I hope that helps with understanding of the concept of enthalpy, and also adequately explains just why that high velocity breath out both feels, and is, cooler than the more gentle stream from your partly open mouth.  However, if you have an alternative explanation, please, do tell.

Thanks for looking in,

MJM460
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Offline Admiral_dk

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Re: Talking Thermodynamics
« Reply #844 on: April 18, 2018, 09:03:44 PM »
I'm probably telling you something you know more about than me, but it just hit me, that a freezer has a thin orifice where the gas expands after and that lowers the temperature - could a similar restriction or even turbulence somewhere in your setup have a similar effect ?

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #845 on: April 18, 2018, 11:30:54 PM »
Hi MJM , more fascinating stuff..and i guess Robert Stirling was having similar thoughts !!   Also after i washed my hair recently i was using the hair dryer at full blast and it felt hot on top of my head but i could distinctly feel a cold draft lower down at the sides  ? !! i am becoming more and more aware of thermodynamics around the home and when i fill the sink with hot water, instead of adding cold if it is too hot i have a metal tub that i put in the sink and swirl the water around ,and this cools it and saves money on the water rates !! and when i have a bath i use my temp gauge  to adjust it to about 106 fareinhieght. !  Also i was at the car  boot today and there was an inlet manifold gauge scaled in   " Inches of mercury absolute "   , this was reading at 28. he said it came off a Lancaster bomber so i don't know how accurate it was and the needle was about 1/3rd of the way round the scale....so if it was measuring vacuum why was it not calibrated in Lbs square"?? Another experiment for you.....you could suspend two apple with thermocouples attached and blow between them and see what readings you get as they swing towards each other ...or use the compressor, not forgetting to video it of course !!!!
Willy....
« Last Edit: April 18, 2018, 11:37:35 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #846 on: April 19, 2018, 01:13:07 PM »
Hi Admiral DK.  That is a good question, and it lets me get back to a topic where I am quite comfortable.  However I am always open to learning something new, there are some very knowledgable people on this forum.

The degree of cooling over the pressure drop at an orifice depends on the condition of the fluid at the upstream side of the orifice.  If it is a superheated vapour, well away from the two phase region and also well above the fluid freezing temperature, there is a degree of cooling determined by the Joule Thompson coefficient.  The is coefficient is yet another property of the fluid, and may be positive or negative.  If the vapour is right on the saturation line, it might result in a wet vapour, or a slightly superheated vapour depending on the shape of that p-h curve.

However, if the fluid is either saturated or sub-cooled liquid, as in a refrigeration system, as the pressure drops, some of the liquid evaporates, and as there is usually no source of heat at the orifice, the necessary energy comes from the sensible heat in the liquid, resulting in a large temperature drop.  The portion of the fluid which evaporates flows through the evaporator to the compressor suction, while the liquid is boiled off by heat absorbed in the evaporator, which is at a pressure with a corresponding boiling temperature lower than the contents of the fridge, which in turn are cooled by the loss of heat to the evaporator.

Just how much of the liquid is evaporated can be worked out by remembering that for an orifice or similar throttling process, there is no work done and no heat transfer in or out.  So the first law says the enthalpy does not change.  Then on the pressure-enthalpy diagram, or a property chart similar to the steam tables but for the refrigerant, you can easily see the proportion of the fluid which evaporates.  It varies quite a bit with different refrigerants.

In a large refrigeration system, the pressure drop occurs at a pressure reducing valve and the no heat transfer assumption is pretty good.  In a small domestic refrigerator, the orifice would be very small and liable to blockage.  Instead, a laminar flow orifice is used.  A fancy name for a very long, small bore tube, through which the velocity is small enough to be below the critical Reynolds number, so laminar flow, under the prevailing pressure difference.  You can see it as a long loose coil if you look around the motor compartment of your fridge.  The length of the tube is such that, as the fluid cools, there would be some heat flow in, which results in a small increase in enthalpy, but the advantage of a larger bore (larger than an orifice would be) being less liable to blockage, outweighs the small unintended heat gain.

So getting back to the question, in my mashup air test system, air is clearly a superheated gas, and the Joule Thompson coefficient means it cools a bit as the pressure drops in an orifice.  So this could explain some or all of the temperature drop I observed.  On the other hand, in a refrigeration system, it is liquid being throttled, and the amount of liquid which evaporates at constant enthalpy ensures considerable cooling.

I have had a look for values for the J-T coefficient, and while most Google references seem to come up with all the theory, the ones giving an actual value are hard to find.  It looks like about 0.25  C per bar pressure drop seems in the right order.  My compressor was set at 30 psig, say 2 bar gauge, so around 0.5 degrees could be explained by Joule-Thomson cooling, but not the observed temperature difference.  It can be moor useful cooling when larger pressure drops are available, and used in liquid air plants and similar.

Hi Willy, the air from the hair dryer is quite hot, test it with your thermocouple, so if you are thinning on top, like most of us, you will feel the heat, intensified by the good heat transfer resulting from forced convection.  When this air, which starts with low humidity, evaporates the moisture in your damp hair, the heat necessary to supply the latent heat to evaporate the water comes from the sensible heat in the moisture, which is cooled, and some comes from your head so it even feels cool as well.  When all the moisture is gone, you will feel the heat of the hair dryer. 

All that, so long as the humidity in the room stays low.  The evaporating moisture raises the room humidity.  If you have a lot of damp hair, in a small room, humidity gets high enough to condense on the mirror, and there is no more drying.  I was in a shop one day, before I started thinking too deeply about all this, when the lady ahead of us was complaining that her hair dryer was too small as it would not dry her hair.  Caught by surprise, and none of my business really, I did not think quick enough to ask if the bathroom had a fan, and did she switch it on.  The fan replaces the humid air from the drier with fresh less moisture laden hair, and the drying continues.  The salesman was obviously not into thermodynamics,and duly sold her a drier with a bigger fan and heating element.  Which no doubt did not solve the problem.

When you think of all those applications around the home, it is quite a significant subject in daily life.  Sometimes the effect is quite small, but when you wonder which action to take, it gives a sensible direction to try.

Before aneroid barometers, atmospheric pressure was measured by a mercury in glass manometer.  Still is when accurate readings are required.  But in the days before SI and even calculators, the conversion to psi or bar was pretty awkward, so the pressure was quoted in inches of mercury, or mm in metric countries.  Standard atmospheric pressure is 101.325 kPa, which is  29.92 inches of mercury or 760.002 mm of mercury.  Pressures below standard meant a slight vacuum, but there was confusion as to whether to quote how much below standard as inches of vacuum, or absolute pressure.

Of course, actual atmospheric pressure various as the various weather patterns progress over us, but if the pressure was really only 28 inches, I suggest you might be preparing for a cyclone.  One inch of mercury is equivalent to 0.49 psi, so nearly 2" low.  But check pressure with the weather bureau, to reset the gauge at the correct calibration.

Now, apples, thermocouples, and air jets!  Newton, Bernoulli, and temperature?  Surely this time you are being silly.  Must be time to learn how to insert those poke-poke emojis....  No, I'm still not ready for that!

Thanks for following along,

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 #847 on: April 20, 2018, 02:35:03 AM »
Hi MJM, Thanks for the explanations...  actually It was the warmest april day for 70 years according to the weather man  don't know about emojis ,but the clues are in the number of exclamation marks !!!!!

Offline MJM460

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Re: Talking Thermodynamics
« Reply #848 on: April 20, 2018, 02:04:39 PM »
Hi Willy, I assume the warm day was the day you were using the hair drier?  The electric element of the hair drier gets hot enough to dissipate all the electrical power input by heating air, just like your electric boiler elements dissipate all the heat by heating water.  So it will always feel hot on dry skin, whatever the ambient temperature.  But in raising the temperature, the drier lowers humidity, so the hot air stream is able to absorb more moisture from the damp hair.

Sorry, I did not notice the importance of the number of !'s.  Never mind.

A few days back, in post #825, I finally returned to the question of power from air compared with power from steam.  I did the calculations for air at 27 degrees C which seemed a reasonable estimate for the air you might be using at a show.  While the answer was that expansion of air gave less work then expansion of steam on a volumetric basis, it left open the question of whether this difference was simply due to the low temperature of the air compared with the steam.

I was distracted by other issues but tonight, I finally went back and redid the calculations for air at 127 degrees.  Not exactly equal to the steam temperature, but close enough, and as usual, selected because it appears directly in the air tables I am using.  It is also enough above the temperature I used last time to indicate the effect of the initial air temperature.

Interestingly, the difference in answer from expansion of air at 27 C was only 3 kJ/m3 which, in a total of 530 kJ/kg, is about 0.5% different, so not significant.  It is definitely not worth constructing an air heater to increase the work from your engines.  So the answer is still that expansion of air gives about 85% of the work from expansion of steam, but 92% of the work from expansion of superheated steam.

The difference when the air supply was at 127 degrees was that the exhaust would be -13 C instead of -78 C for expansion of 27 C air through the same pressure ratio, while the steam exhaust was wet at 100 kPa, so 99.6 C

All the calculations were on the basis of an ideal adiabatic expansion of steam.  A real engine might give around 70% of the work output of the adiabatic engine, so the corresponding temperatures would be -46 C for expansion of 27 degree air, and +29 C for expansion of the 127 degree air.

And above all, remember that while the inlet valve is open, there is no expansion in the cylinder, the potential work output is the same whether the fluid is air or steam.  The remaining expansion after the inlet valve closes, in a single cylinder engine, is much less than the expansion in an adiabatic engine for the same inlet and exhaust pressures, so the differences in work output and the temperature effect on the exhaust will both be much less pronounced.  It would probably take a compound or even a triple expansion engine to actually achieve the assumed expansion and see the maximum potential cooling of the air.

I do hope that this, at last, answers the question to everyone's satisfaction.

Thanks for looking in,

MJM460

« Last Edit: April 22, 2018, 11:19:04 AM by MJM460 »
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #849 on: April 21, 2018, 12:25:30 PM »
No comments on yesterday's thoughts, so I thought I would post the formula for calculating the moment of inertia for a flywheel to add to the comments I posted about Brian's new flywheels on his thread about the air motor.

Basically, the important property of a flywheel is not mass, but moment of inertia.  Each element of a flywheel contributes to the moment of inertia by an amount equal to its mass times the distance from the axis of rotation squared.  All such contributions can be added or subtracted to make up the total moment of inertia of a complex part, just the same as you can estimate the mass of a complex component by adding and subtracting smaller elements.

The result of this is that really only a few simple formulae are required to cover most imaginable shapes for a flywheel.

First we need the formula for the moment of inertia of a circular disk.

Moment of inertia, I = 1/2 x M x R2

M = the mass of the disk, measured in kg.  To calculate this, first Calculate its volume, using diameter and thickness Pi x d2 / 4 x t

Diameter thickness and radius are all measured in meters

Then multiply by its density.  Steel is 7800 kg/m3, aluminium 2700 kg/m3,  bronze 8666 kg/m3, cast iron 7270 kg/m3 should cover most requirements.

If you have a simple disk flywheel like some low power Stirling engines, this will be all you need.

If you want a flywheel with a rim and spokes, you can subtract the moment of inertia of a disk equal in diameter to the inner diameter of the rim, then add a small diameter disk the diameter and thickness of the hub.

The spokes can each be approximated to a rod, rotated about its end, with the end at the centre of the flywheel and length equal to the inner radius of the rim.

The formula for this component is I = 1/3 x M x L2

You can ignore the tiny contribution of the end of the rod which is double counted in the hub, or you can reduce the length of the hub by the thickness of the spokes to avoid the double counting.

Sometimes, instead of spokes, I have seen flywheels with a thin disk supporting the rim, with holes bored in that disk for both ornamental and weight reduction reasons.  To allow for these, you first calculate the moment of inertia for the thin disk supporting the rim, then calculate the MoI for the material in the hole about its centre, and make allowance for the fact that it is not rotating about its own axis, but an axis displaced from its centre by a radial distance s, also measured in meters.  Then the formula for the moment of inertia for the metal you remove to make the hole is I = 1/2 x M x R2 + M x s2

Multiply this by the number of holes and subtract the total reduction in moment of inertia due to the holes.

This procedure, calculating the MoI of a component about its centre, then adding M x s^2 to account for the axis of rotation displaced from its centre, is quite general.  For example, the MoI of a rectangular bar rotated about its centre of mass is calculated by I = 1/12 x M x L^2.  If you want to rotate it about its end, you add the impact of moving the axis by L/2, M x (L/2)^2. 

If you do the maths you will see that 1/12 x M x L^2 + M x (L/2)^2 = 1/3 x M x L^2.

Of course, one spoke rotating about its end would result in a rotating out of balance force, but unless you go out of your way to make a flywheel with only one spoke, the out of balance force is balanced by the other spokes for any equally distributed number of spokes.

With these few simple formula, by adding and subtracting the various shapes, you can calculate the moment of inertia of most common flywheel shapes.

Using a spreadsheet, you can easily do some trial and error calculations to arrive at the necessary adjustments to a flywheel dimensions to make one with a different material, or to accommodate different space requirements.

I suggest that model flywheels are rarely designed by a complete theoretical analysis of the torque curve and careful selection of the flywheel moment of inertia to control the engine speed in strict limits as a full size often is.  This probably means that the flywheels are rarely just optimal in terms of dimensions and mass.  More likely, they look about right, and are, like those Rolls Royce engines, adequate.  There is no reason not to try a smaller flywheel, if you have one, or a larger one, than called for by the design, although it's another variable you may not want to play with, if you have a new engine which may be difficult to start.  But if you would like a different size flywheel for a running engine, a few calculations and trying a different flywheel you have access to, might give you a good idea of how much leeway you have to reduce weight.  A good reason to standardise on shaft dimensions where practical for similar size engines.

I hope this little diversion into mechanics is of interest.

MJM460

Edited due to minor errors and omissions which could cause confusion.
« Last Edit: May 02, 2018, 06:02:51 AM by MJM460 »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #850 on: April 21, 2018, 02:15:28 PM »
Hi MJM,  Thanks for the info about flywheels    When i first attempted to start my new freelance engine on air it did not seem to want to operate !  I thought the flywheel was too small ( i used one that i had in store) so to give it more moment of inertia ( aka "welly" ) i attached this large chunk of brass to it.  It then sprang into life as if by magic !!! Intuitively ,one might think that if the  flywheel is balanced then there would be no advantage as one side has to lift against gravity ,whilst the other side is helped by gravity, or ,is that the answer !!!!!!!

« Last Edit: April 21, 2018, 02:23:05 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #851 on: April 22, 2018, 11:05:00 AM »
Hi Willy, I really like your thinking.  We may not have a box of spare flywheels to try but we all have bars of metal.  Steel is as good as anything, and moment of inertia does not require any particular shape.  Your rectangular bar is as good a shape as any for experimentation, though like a propellor, it can give your fingers a whack of you are not careful.  Gives new meaning to the term flywheel made from bar stock!

If the moment of inertia is not enough, instead of using a heavier bar section, you can simply bolt a short piece at each end for much more inertia in proportion to the extra mass.  But do use through bolts in shear with nuts.  If one comes loose, it will not be good.  Mass at a larger distance from the axis is much more effective than mass near the axis, due to that r2 term.

To calculate the moment of inertia of your bar stock flywheel about its centre of mass is simple enough.

I = 1/12 x M x L2

If you add weights to the ends (while being careful to keep it all in balance), the additional moment of inertia due to each weight equals the mass times radius squared.  The radius is the distance from the centre of the added mass to the axis of rotation.  Strictly you should add the moment of inertia about its own centre as well, but in practice, this is small enough to ignore without significant error.

Perhaps you could do the calculation and let us know the answer.

When you are satisfied you have enough inertia, you can then simply check the flywheel you want to use by calculation.  If it is a bit low, adjust your temporary set up to match, and try it.  Will save spending money on a flywheel that is too small.

Perhaps I should say a bit more about how the "correct" size of flywheel is determined.

When the engine is running, we know for a single cylinder engine that the torque goes through zero and a maximum twice each revolution.  For a single acting engine, it still goes through zero twice, but also a maximum on the power stroke, and a negative maximum on the exhaust stroke each revolution.  A multi-cylinder engine goes through a number of maximums and minimum torques each revolution, though the minimums are still usually positive unless all the cranks are on the zero and 180 degree points.

The load the engine drives also has a torque characteristic that may vary throughout each revolution, like a reciprocating pump or compressor, or may be constant throughout each revolution  like a winch.  When the engine torque is greater than the required torque, the whole lot speeds up, when it is less, it all slows down, and this all happens each revolution.

The flywheel inertia means the flywheel speeds up when there is excess torque, storing rotational kinetic energy in the process, and when the torque is negative, it slows down, returning energy to the system.  The necessary flywheel inertia is that required to keep that speed fluctuation within acceptable limits.  To give you an idea, speed limits of +/-7% within each revolution is considered a quite stringent limit for a 400 rpm system.  (I have been up against this limit on an electrically driven reciprocating compressor, where those fluctuations affected the motor current and were potentially enough to upset the city power system.  Fortunately the engineering was all well done, the flywheel was adequate and the power supply authority happy.)

My non-contacting tachometer only detects the flywheel position once per revolution, so cannot detect the speed variation within the revolution.  The electronics these days is fast enough that if you put the reflectors at multiple points about the flywheel, it is possible to measure the speed multiple times per revolution.  This is done routinely in modern digital governors instead of flyweights.  If you are then able to record these readings and graph them on a suitable time scale, you can easily see the variation.  I have commonly used both 24 and 60 tooth wheels on the shaft.   The digital governor calculates the speed multiple times per revolution even on a high speed turbine, and does not wait for a full revolution to respond to a speed change.  But not so fascinating to watch as your flyweight governor.

The smallest useful moment of inertia for a flywheel has to keep the engine turning, but I suggest that visibly nearly stopping twice each revolution would not meet the expectation, it needs to look reasonably steady even at the lowest speed you want to run, when the flywheel stores the least energy.

How much is too much moment of inertia?  In principle, bigger always gives better speed regulation.   However, bigger usually means more friction in the bearings, and more air friction resistance, so it is definitely too big if the friction forces absorb the entire output of the engine.  Besides, there is usually some weight or dimensional constraint, while our speed regulation requirements are really not very stringent.  So once the engine appears to run at a reasonably steady speed, the smallest and lightest flywheel that looks appropriate is just right, but a bit more never hurts.  Remember that you can make it lighter by making it larger diameter, material density is another factor to play with in balancing performance, diameter and mass.  The D2 term means that a small change in diameter makes a relatively large change in inertia, even with the same mass, so you can slim down a heavy looking rim.

Your comment about gravity not making any difference if the flywheel is balanced is quite correct.  It does not matter if the flywheel is round, or bar shaped, or even propellor shaped, bit must be reasonably balanced.

All this stuff about flywheels and moment of inertia might be easier to understand if you compare it with linear systems.  Most of us have a fair understanding of the place of mass in a linear system, moving through distance, at velocity, force and acceleration.  Associated with linear systems is kinetic energy and linear momentum.  And I hope you have noticed that conservation of energy,  and linear momentum are absolute laws of physics, while conservation of mass is near enough for all our practical purposes, but not absolutely correct.

Rotational systems have exact analogies for all these quantities.  Angle of rotation is the rotational analogy of distance, angular velocity the analogy of linear velocity, torque the analogy of force, rate of change of angular velocity the analogy of acceleration, and moment of inertia the analogy of mass.  Conservation of angular momentum is an absolute law of physics along side conservation of linear momentum, but there is no analogy to conservation of mass.  This is because moment of inertia and hence rotational kinetic energy are not only mass dependant but also dependent on the distribution of that mass.  By moving the mass on a spinning object, you change the moment of inertia, however, as conservation of angular momentum still applies, if the moment of inertia changes, the angular velocity must change to maintain the angular momentum.  The classic example is a skater spinning with arms outstretched, then drawing them in close to the body to spin faster.  This applies to all systems with variable geometry.  With the analogous properties substituted in all of the linear laws of motion, the exact same laws apply to rotational motion.

I hope that helps a little with understanding flywheels and how they work.

Thanks for looking in,

MJM460
« Last Edit: April 22, 2018, 11:17:25 AM by MJM460 »
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Offline derekwarner_decoy

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Re: Talking Thermodynamics
« Reply #852 on: April 22, 2018, 11:53:55 AM »
Hullo Willy......I too was a little concerned for your hands or fingers with that rotation a hunk of brass  :facepalm:

I found the text from MJM very interesting as I am now just going thru the mass of inertia in my build

50 diameter x 96gm flywheel on the shaft 4 diameter
18 diameter x 10gm chain pinion on axis on the shaft 4 diameter [not shown]
4.8:1 chain reduction
68 diameter x 120gm flywheel/chain pinion x 6.35 shaft diameter
2 x 25 diameter x 35gm shaft couplings x 6.35 shaft diameter
2 x 130 diameter x 300gm paddle wheels on 6.35 shaft diameter

So my main concern is to ensure there is adequate mechanical security  :killcomputer: of the drive within it's individual components

Derek
« Last Edit: April 22, 2018, 01:30:27 PM by derekwarner_decoy »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #853 on: April 22, 2018, 01:02:32 PM »
Hi MJM, Just a little bit more about flywheels .... this is about a similar engine to the Beeleigh mill  to raise the 'horses power' !! I don't know if the thinking was %100 correct but possibly was as they did know something about this in 1868...So Question,, is it advisable and proper to increase pressure and flywheel weight together ,and are there tables for this ??

Offline MJM460

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Re: Talking Thermodynamics
« Reply #854 on: April 23, 2018, 01:36:20 PM »
Hi Derek, that is a very interesting and compact arrangement.  You have put a lot of work into that.  I like the small chain and wheel.

And you have added another critical variable into the rotational equations for a flywheel, the rotational speed.  With the speed reduced by 1:4.8, the energy stored is proportional to 1/4.82 or 0.04 times the energy at the higher speed.  Is there a reason for putting a flywheel on the low speed shaft?

I presume that the paddle wheels are on that same slow speed shaft.  Their moment of inertia adds to the flywheel on that shaft, and of course they are also the load for the engine.  The torque required by the paddle wheels is likely to be fluctuating as each float dips into the water.  Change in momentum involves a torque, just like change in linear momentum involves a force, the rotational analogy of F = M x a.  The change of flywheel angular momentum with each ripple in the torque from the paddles and from the engine, results in reversing torques, which will test your wheel attachment to the shafts.  I presume that is what you are thinking about with the attachment considerations.  Are you thinking in terms of taper locks, or perhaps keys for reliable attachments?  I will be most interested in seeing further progress.

Remember for rotational systems, it is the moment of inertia, or more correctly, the second moment of mass that is important, it takes the place of mass in a linear system.  For the paddle wheels you need to calculate the moment of inertia of each component, and add them all together.  Fortunately you have many identical except components and you only have to calculate the value for one set and multiply by the number of identical sets.  You can calculate those rings from the formulae I have already calculated, but you may prefer to use the direct formula for a ring, I = M x r2

Hi Willy, your beam engines have hp and lp cylinders, but equivalent to 180 degree crank locations, so two zero torque points each revolution.  The flywheel must store the excess energy by accelerating during the power stroke, and return this energy to the system when the engine torque is less than the load by slowing down.  Increasing the power rating implies increasing the load, perhaps pumping to a greater height, or larger volume, by working at higher speed.  The extra power implies either extra torque from the engine, or higher speed, so the flywheel has extra energy to store each power stroke, and has to return the greater amount of energy to drive the load as the engine torque goes down to zero at the top and bottom dead centres.  Either the engine can be allowed a greater speed fluctuation, or the degree of regulation can be maintained by adding extra moment of inertia to the flywheel.  A band around the outer diameter is a very efficient way of adding moment of inertia as the large radius ensures the maximum increase in moment of inertia for the weight added.  So it makes sense that the two were linked.  Did they have tables?  Or did they perhaps have a rule of thumb for the moment of inertia required for a given power of engine?  I don't know.  These days, a computer is used to calculate the moment of inertia of the whole machinery string, and is combined with a detailed analysis of the torque characteristics of the engine and load, and these figures are used to determine the extra moment of inertia required to ensure the necessary speed regulation is achieved.  I think you mentioned this once before, but I have probably been able to give a more detailed comment this time.  I hope it helps.

I mentioned yesterday that most of us don't have a stock of spare flywheels available to try, however when I looked around, even I had the four shown in the attached picture.  In fact I also have a second very similar to the small one, and even two commercial ones,  one a Mamod and the other a similar model, though those two have different shaft sizes.  All of mine have been drilled and reamed 5 mm diameter, so are interchangeable if they fit on the base.

This afternoon, I put all the dimensions in a spreadsheet and calculated the moment of inertia for each.

The little casting is yet to be machined, but it will be made for the same size shaft as the others.  I don't think that it has had enough fondling to meet the required high standard required on this forum before I start making swarf.  Consequently I had to estimate the finished dimensions.

The calculations are very sensitive to the actual dimensions, (as I found when I tried to match calculated mass to the actual measured weight.  So I assumed the same material density for each which at least makes them all comparable on a dimensional basis.

The smallest is 45 mm diameter, 14 mm wide and weighing 157 g.  The moment of inertia was 20.16E-9 kg.m2.  The E-9, or x 10^-9 is necessary to make the units correct for use in other equations of motion.

The next is 65 mm diameter, 17.4 mm wide and weighs 310 g.  The moment of inertia was 75.53E-9 kg.m2.

The third turned from solid is 75 mm diameter, 16.4 wide and weighs 447 g.  The moment of inertia  was 148.06E-9 kg.m2.

Finally the cast flywheel looks like it will yield a flywheel 58 mm in diameter, 12 mm wide and weighing 129 g with a moment of inertia of 14.95E-9 kg.m2.

The cast one looks anomalous.  Careful examination of all the dimensions shows that the cast flywheel rim is only 11 mm thick in the radial direction, compared with 14 mm on the 45 mm turned wheel.  When I increased the dimension to 14 on both, the larger spoked flywheel had the same moment of inertia with less weight.  That r2 term is so significant that quite small differences have a large impact.

You can see that the larger ones have significantly more moment of inertia than in proportion to their extra mass.  The calculations also show that more than half the moment of inertia comes from the rim, over 70% for the larger ones.  I was a bit timid machining out those webs, but the spokes contribute only 10% of the moment of inertia of the cast example.  The hubs were also relatively insignificant.

I guess I now have some known flywheels to try on my next engine.  But in fact they are all relatively large for my small engines.  (Jason might call them series 12).  Even the smallest is adequate for my single acting oscillator, so they are all more than adequate for the double acting engines, and even more so for a multi-cylinder which might be my next.  Perhaps I should steal Willy's idea, and make a bar stock mock up to see how small I can go on each.  This thread is producing more ideas for projects than I can keep up with.  At least I know I can make more flywheels from that bar I used for the 45 mm one, make them thinner, and perhaps machine out some spokes.

I have been a bit slack not doing those calculations long ago, but you have all prompted me to at last open a spreadsheet, measure my flywheels and put some real figures into the discussion.  I hope you have found it interesting.

MJM460

« Last Edit: April 23, 2018, 01:55:52 PM by MJM460 »
The more I learn, the more I find that I still have to learn!