Talking Thermodynamics

[MJM460]

Hi Captain Jerry.  A very interesting question, as it highlights the procedures necessary to analyse what is going on in our engines.

There are many variables in the problem, but not all are independent.  There are only so many degrees of freedom in a particular problem.  Each variable that we specify reduces the number of degrees of freedom remaining.  We can specify some values but not all.  Once the necessary number of variables has been specified so that there are no more degrees of freedom, all the other variables are dependent on those and can be calculated with an appropriate analysis.

In your particular problem, I am assuming you are referring to a reciprocating engine and you are running it on air.  We can look at steam later if you like.

As you say, if the engine runs nicely with 10 psi, and we try and supply it with 50 psi without increasing the load, it will accelerate until either friction absorbs all the extra power, or something breaks.

To answer the question of why we can’t run it on 50 psi at one fifth of the volume of air, let’s look at those variables.  I feel that in you specification of the problem, you have in mind an operating speed of the engine that allows you to see the motion in action.  The volume in the cylinder of a reciprocating machine is fixed, and hence at fixed rpm, volume of air consumed is fixed, it is no longer an independent variable for us to specify.  Volume and pressure are enough to specify the work done by the engine.  The friction is basically fixed by your construction so that it runs freely, so if you increase the pressure, the speed will increase.  If you increase the pressure, you increase the forces on the piston, so the engine speed will increase, thus increasing the air volume used, or you might increase the load to absorb the extra energy being supplied.

Without a governor, and with sufficiently large pipes and valve ports, you might assume that the cylinder pressure is approximately equal to the supply pressure.  However, if you add that governor to throttle the incoming air, you no longer know the pressure in the the cylinder.  If the engine still runs at the same speed as it did with 10 psi, we can assume that the throttle is actually reducing the supply pressure to about 10 psi.  A much smaller volume of 50 psi air is being drawn from the compressor/accumulator, and this volume of 50 psi air is throttled to 10 psi with a consequent expansion in volume to what your engine was using with the 10 psi supply.

If instead of adding a governor, you put a brake on the engine, and increase the supply pressure to maintain the speed, the engine will be doing more work, and as the volume is the same for the same operating speed, more pressure will be required to supply the required energy to drive the extra load.

Or in principal, you could apply a load as you increase the supply pressure, and allow the engine to slow so that at 50 psi, the engine was running at one fifth of the original speed, so using one fifth of the volume from your air supply.

I say in principal, because I am assuming that in order to see the motion in action, you are probably running near the minimum speed the flywheel can maintain.  Unless you have a multi-cylinder double acting engine, it might not actually run that much slower, as the nature of a reciprocating engine is that the torque from each cylinder is developed in a fluctuating manner.   Continuous running is dependent on the flywheel storing enough energy during each pulse to keep the engine running through the low torque angles.

I wonder if that answers your question, or are you thinking of another configuration?

MJM460

[Captain Jerry]

I am of course thinking of a specific configuration but not all that unusual. Trying to make it as simple as possible, lets take time out of the question by looking at only the power stroke of the piston from TDC to BDC.  Lets assume a cylinder bore of 1.414 inches so that the piston surface is 1 sq. in. and a stroke of 5 inches.  That is a very long stroke but it helps with the math.  And of course we need perfect sealing of the piston and of the valves so that there is no by pass.


Now with the piston at TDC, we open the inlet valve to an adequate supply of air at 50 PSI. but nothing happens because I have a firm grip on the flywheel (the load).


Now I manually allow the flywheel to rotate until the piston makes 20% of its stroke. The resulting swept volume is 1 cu.in. and it fills with air at 50 PSI. and the force on the face of the piston is 50 lbs. so to prevent it from breaking my wrist, I slam the inlet valve closed.  But the force on the piston is still 50 pounds so I let the flywheel rotate a bit more so that the piston moves another inch, doubling the swept volume and reducing the pressure by half.  Now the force on the face of the piston is 25 pounds. 


I let the piston travel another 2 inches, again doubling the volume and halving the pressure so that the force on the piston is now down to 12 pounds. Still a considerable force.  During the last possible travel of the piston the volume is increased by 20% and the pressure decreased at the same ratio so it is down  to about 9.6 psi when the exhaust val opens, still way higher than the 5 psi we were comparing it to.


So if I want to run my engine on 50 psi. air, instead of having the governor throttle the pressure to 5 psi, I have the governor linkage to close the inlet quickly at some fraction of the piston stroke and allow the trapped volume of air apply the appropriate force to balance the load.  What I am describing is of course well known as early cut off ala Corliss, Greene.


Have I described this correctly?

[Captain Jerry]

The above comments really needed to be expanded but the dinner bell rung so I had to wind it up, so this a continuation. Read the previous post first.


To continue, if the piston had been allowed to move only 1/2 inch before the inlet valve closed, the pressure in the cylinder after it had moved 1 inch would still be 25 psi. and even if the valve closed after only 1/4 inch of travel, the pressure would still not be reduced to zero after the full 5 inch stroke. And this is with AIR!  So tell me again, why the expansion of a volume of air cannot drive a model engine?


Oh! Temperature!  Air cools as it expands and steam expands as it cools.  What kind of linguistic mumbo jumbo is that! It is actually the other way around. Air expands as it cools and steam cools as it expands.  What's the difference?  Is that why steam engines have to be run in the back yard? 


It must be getting late. I'm amusing myself with half stated arguments.  The TRUTH is that steam looses it's pressure when it is cooled!


Jerry

[steam guy willy]

Hi Jerry, I think the terms we use have been in place getting on for 300 years .!! the term expansion always confused me until i read in an old book that the elastic nature of steam was a bit like an elastic band pulling the piston down the cylinder !!  also some people say that you don't pull something towards you ...you push it from behind towards you !!  I hope this sheds some light on the matter and it would be much more difficult trying to explain it using German vocabulary !!! with all their multi syllabicular words !!.thinking about the Mallard locomotive that travelled at 126 miles an hour the steam had not much time to cool down whilst passing through the cylinders !!

Willy

[MJM460]

Hi Captain Jerry, that is a different problem that requires a bit more analysis.  But first, let’s back up a bit. 

(Now with time zone issues, I had written all that follows before I saw your follow up post and Willy’s post.  I think those two follow up posts are on an equally interesting but different track, and what follows still applies to your question.  I will then try later in the day to address the additional issues you have raised.)

There are two basic approaches to analysing a reciprocating machine.  First you can treat it as a highly variable process with inlet strokes and exhaust strokes and highly fluctuating torque and speed, a highly unsteady situation which you can analyse moment to moment.  This approach is necessary to closely look at what happens within the machine.

Alternatively you can notice that all those fluctuations are not random, they are cyclic in nature, everything repeats in a predictable manner, so you can take long term averages, and treat it as a steady flow situation.  Perhaps more precisely a quasi-steady state system.  In this case you look at average values over a complete cycle, and don’t worry too much about the moment to moment fluctuations.  This approach is very satisfactory when you are mainly interested in the situation outside the machine, when the machine is only part of the system, and you are really interested in the locomotive,  ship or refrigeration system in which the machine functions.

Yesterday I used that steady flow method.  There is a well known formula for the power produced by a reciprocating engine,
Power = P x L x A x N / 33000.

Where P is the mean effective pressure difference across the piston through the power stroke, L is the length of stroke, A is the area of the piston, and N is the RPM of the engine.  It is commonly used in calculation of engine power from an indicator diagram.  Because the pressure is a differential pressure, atmospheric pressure is not relevant.

The constant 33,000 is an arbitrary constant to account for the curious mish-mash of units that constitutes the imperial system.  I won’t confuse you with the various historical metric systems (which have the same problem) or the SI system (which does not) at this stage.

If you look at that formula, L x A is the swept volume of the engine, or V.  So we have Power = P x V x N / 33000.  With this version you can easily see that for a given power, if you double the pressure you have to halve the swept volume to produce the same power as you proposed.  So an engine with half the swept volume will require double the pressure deliver the same power at the same speed.  But once you have built a reciprocating engine, the swept volume is no longer a variable.  Note swept volume does not include clearance volume which is more easily made changeable.  You can also see that there is an implied assumption that your air pressure is measured and supplied in a manner that makes the assumed pressure a reasonable estimate of the mean effective pressure, which is not totally unreasonable for a simple engine with the valve open for a substantial portion of the stroke in this context.

Your new statement of the problem can still be treated the same way, but now the actual mean effective pressure is very different from your supply pressure.  Without involving calculus, let’s try and estimate it.

First, let’s continue with your 1 square inch piston area, I will leave you to check the piston diameter.  Then we need to use absolute pressure to correctly estimate the change in pressure with volume.  Let’s approximate atmospheric pressure to 15 psi.  (Instead of 14.7) so your initial pressure is 65 psia.  Now, to calculate the change in pressure when you allow the piston to move, you have to allow not only the volume change due to the piston movement, but also the clearance volume.  Again, let’s opt for easy maths and use 10% clearance volume.  It’s a bit high, but will show you the effect more clearly.  With your one square inch piston and five inch stroke, that makes the clearance volume 0.5 cu. in.  When the piston moves 20% of the stroke, or one inch, we now have a volume of 1.5 cu. in. at 65 psia.

Now this is allowed to expand with the inlet cut off.  At the bottom of the stroke, the volume is 5.5 cu. in.  At this volume, the pressure is
65 x 1.5 / 5.5 = 17.7 psia or 2.7 psig.

To estimate the mean effective pressure over the range 50 psig at the start to 2.7 psig at the end of the stroke, we could take a simple average, about 26 psi.  However the curve of pressure against volume is a hyperbolic shape, and so the mean effective pressure will be something less than this.  We could make a better estimate by calculating the pressure at 50% of stroke, and  calculating an average for the first half and again for the second half, or plot the curve on graph paper and count squares for a better, but still always high estimate.  The repetitive calculations quickly become tedious, so these days we would use a spreadsheet.  Or the integral is probably on your kid’s graphing calculator so you can calculate the area under the curve and use that.  Better to let the kid drive the calculator, I never had a calculator that could do that, and have not learned to drive one, but they are common in secondary school now.  But the answer is probably in the region of 20 to 25 psi.

I think you can see now that your initial 50 psig supply pressure is not the right figure to plug into the P x V calculation.  Of course at the 10 psig supply pressure all the same calculations apply, except that if you cut off at 20% with a 10 psig supply you would expand to less than atmospheric pressure so the pressure difference across the piston will be negative and the piston will probably not travel to the bottom of the stroke without help.

(10 psig = 25 psia.  Expansion from 1.5 cu.in to 5.5 cu. in. by the same calculation as above gives a final pressure 25 x 1.5 / 5.5 = 6.8 psia which is a pretty good vacuum!)

So behind your conundrum is the implied, but perhaps not recognised, requirement to use the mean effective pressure difference in the calculation, along with the effect of that early cut off on that mean effective pressure.

I hope that helps a little.  Back on the additional issues you have raised later.

MJM460

[Captain Jerry]

MJM


I am sorry to say that does not answer my question at all.  The question was why do we run our small engines on 5 psig rather than 50 psig.  and the answer that I expected was that if the speed of the engine is controlled by a throttle type control on top of the steam chest, then the pressure in the cylinder is not at all related to the pressure at the gauge.


In my rush to describe a system before dinner, I incorrectly stated the diameter of the cylinder but after that I was considering the surface area to be 1 sq. in. for simple understanding.  I also chose to ignore what you are calling the clearance volume which I assume to be the space above the piston at TDC because it is arbitrary and only complicates the calculations and because it is possible to conceive of a case where it would be approximately Zero.  Then if I allow the piston to move 1 inch and to hold 1 cu. in. of air at 65 psia, and then close the inlet valve, we will be dealing with the pressure drop that will occur if the piston travels another inch.  I believe that the pressure will drop 50% and be 32.5 psia.  and if the the piston travels another 2 inches, the result will be a volume of 4 cu. in. of air at a pressure of 16.25 psia.  With only another inch of travel available, the volume can increase another 25% of that and at BDC, before the exhaust valve opens, we will have 5 cu. in. of air at 13 psia or -2 psig.  so maybe the exhaust valve should open slightly before BDC.


I chose not to consider average pressure, but it is obviously a positive value, just as I have chosen to ignore the geometry of the crank and the relative torque at the various angles, but it is obvious that an air input pressure of 50 psig. (65 psia) of air with flow cut off at  20% of stroke will be able to turn the crank with a minimal load.  If the load is increased, the cut off point will have to be delayed by some factor and the engine will still turn and if we want the engine to run faster, the cut off point can be delayed further.


I was never expecting the engine to run on 5 psig with an early cut off.  It is also obvious that if an engine with early cut off gear were to be supplied with insufficient air pressure, the governor would be expected to increase the length of time before cut off.


The purpose of my question was to clarify that compressed air does expand and the expansion is sufficient to operate an engine with early cut off gear and to refute the often expressed statement on this and other forums that it is only possible with steam.


I could be wrong as I have mistakenly stated previously.


Jerry

[MJM460]

Hi Captain Jerry, I will try and answer your update while you sleep, as I am sure that some of the points are due to the Apple spell checker which seems to be doing a good job of proving its algorithm wrong these days.

I suspect I have already answered some of the points in the first paragraph or two, but I will emphasise that I have never told you or indeed anyone else that you can’t run a model engine on air.  In fact I have gone so far the other way, that I have actually commissioned a full size steam turbine on air.  We could not make the steam connection for the new turbine until the plant was shut down, and the client did not want to shut down the plant until the system had been proven.  Only a small one, from memory less than 50 kW, but it ran so quietly and smoothly on air that three experts witnessing the test thought it was not running until the exhaust piping started to ice up.  When air expands through a turbine, it does get cool!

So on to the next statements.  First it is necessary to understand that expansion describes a path function, that is, knowing that the air is expanding does not tell you how the air will end up unless you know the conditions under which it is expanding.  Probably not a generally understood concept. 

For example, it can be expanding at constant temperature, which requires external heat input.  It can be expanding without heat transfer (adiabatic expansion), it will cool.  it can be expanding at constant pressure.  In this case it will be doing work but the energy expended is being replaced by the necessary fluid input and I suggest it will neither increase or decrease in temperature unless there is also heat transfer.  This one is more tricky, and I might have been wiser to leave it out.  And the expansion could involve more than one of those conditions.  I am sure that you can think of other possibilities, but in each case, the end result will be different and determined by the path.  After the expansion, you know the volume, but you don’t know the pressure.

So, “Air cools as it expands”.  Generally true, unless there is heat input during the expansion.  Try blowing out through tight lips as if blowing out birthday candles, with your finger, you can feel that the air is cool.  If you blow out with your mouth wide open, the air will be warm, close to blood temperature.  Try it!  The cool air for the candle also started at blood temperature before the expansion as it leaves your lips.

“Steam expands as it cools”.  The statement suggests the expansion is the result of the cooling.  I would prefer to say the same as for air, steam cools as it expands.  If you are cooling the steam or air by heat transfer, they both will be reducing in volume not expanding.

With steam there is the additional complication if you are allowing heat transfer, in that the steam can condense in full or in part, which further reduces the specific volume, and the effect may be more or less than the effect of the volume increase.  In fact, even without heat transfer, if the steam expands over an appropriate pressure range while doing work, it will cool enough to partially condense in the cylinder, reducing its specific volume and hence pressure even further, so reducing the work output that might be expected for that degree of expansion, basically by reducing the pressure more than you would expect for the volume change.

Finally, “steam looses its pressure when cooled”.  Well remember that path function again, you have not sufficiently defined the problem.  Imagine an inverted cylinder with a piston making the top end closure.  Assuming the normal perfect seal, no friction.  And a brick or two sitting on top of the piston.  To admit steam, the pressure has to be sufficient to lift the bricks.  Before the piston reaches the top of the cylinder, shut off the steam.  As the cylinder walls transfer the heat from the steam to the atmosphere, the steam reduces in volume as before, but this time the bricks on the piston will maintain the pressure.  We have a case of cooling at constant pressure.

So with statements like expansion, remember to define the path.  Just for interest, work is also a path function.  The end point properties of steam or air in a cylinder doing work depend on the path taken.

You are right about the linguistic mambo jumbo, but I hope I have clarified some of it.

Hi Willy, it is to remove that rubber band analogy that I so often refer back to the molecular theory of matter, and the collisions of those atoms with the vessel or piston walls that is responsible for the pressure.  There is no negative pressure in terms of absolute pressure.  The atoms do not pull the piston, rather, when the pressure is sufficiently low on one side of the piston, the pressure on the other side of the piston is pushing it.  They are always there on each side pushing it, it is simply a matter of pressure difference that determines the direction of the resultant force.  The net force can come from the other side to where you are looking.

Your point about the Mallard cylinders not having much time to loose heat is quite valid, hence so much analysis done on the assumption of adiabatic expansion, i.e. no heat transfer.  But we must always remember that is an approximation to simplify the analysis, and if you can burn your finger on the outside of the cylinder, there is heat transfer.  You would have to run pretty fast to do the experiment.

Hi again Jerry, you are coming back faster than I can answer.  Just as well I have an unusually quiet schedule today.

I think we are getting confused by the terminology here, at least I am.  I am a fairly literal kind of beast.  When I talk about running the engine at a certain pressure, I mean the pressure at the steam chest, which with sufficient size valve ports and passages, is some sort of approximation to the pressure at the piston face while the valves are open.  You are quite right that the supply pressure tells us nothing about the pressure at the engine if we have a manual or governor controlled valve in between.

So if your question is why we run our engines with 5 psi instead of fifty, it is because we make them so large, and apply so little load.  If you want to run at 50 psig with no load other than the inherent friction in the engine, you need to make a much smaller engine.  George B can do it, but most of us seem to find it easier to make larger components, so they are vastly oversized for the work output we demand.

If your question is why we use a 50 psig supply when we only need 5 psig at the engine, the reason depends on whether we are running on air or steam. 

With an air compressor, we tend to like to shut off the compressor to stop the noise.  Then we are running on the energy stored in the accumulator.  And higher pressure means more stored energy so we can run our engine longer before the noise returns.  If like me, your compressor has no accumulator, then it’s a matter of balance between the volume of air supplied by the compressor and the volume of air consumed by your engine.  The compressor throughput actually reduces as the discharge pressure increases.  If your engine is using less air than the compressor provides, the pressure will rise.  In practice it will rise more than we want, so the compressor is equipped with an adjustable relief valve to control the pressure at the compressor, by venting some of the air to the atmosphere.  Our throttle valve if we have one then further reduces the pressure at the engine to control the speed.  Unless we put a pressure gauge on the steam chest (air chest?) we don’t know what the pressure at the engine really is.  I usually set my relief valve to the minimum, barely readable on the gauge, and the engine speed is what it results.  Probably higher than you are trying to achieve.

If you are running on steam, there is a good reason to run the boiler at much higher pressure than we need at the engine.  The volume of the steam produced by a given heat input obviously depends on the pressure.  If the volume is larger, a greater disengagement space is needed above the water level to minimise the water carryover.  If you run the boiler at higher pressure then throttle the steam to the engine, you get less carryover for a given boiler size.  That said, with my little boilers, I don’t have a throttle valve, I have relatively large piping, and with my little meths burners, the steam chest pressure is not much less than the boiler pressure, and providing I take care not to overfill the boiler, I don’t get excessive carryover operating at quite low pressure.

Jumping to your last paragraph, air under pressure is definitely able to expand and drive a model or indeed a full size engine.  Many of the statements of the type you mention may be in relation to compound engines.  Whether full size or model, you need enough pressure at the inlet that the exhaust after expansion is still above the exhaust pressure.  With air, in practical terms, this exhaust pressure must be above atmospheric, and with zero load on the engine and the necessary inlet pressure, it will not run slowly.  And the exhaust temperature will be quite chilly.  With steam, providing you have an air pump and a condenser, you can run with a lower exhaust pressure, but again unless you put an appropriate load on the engine, it will probably run faster than you want.  Even if you use atmospheric steam pressure inlet, you could still exceed your five psig differential pressure with an effective condenser.

If you expand to a pressure lower than the pressure in the exhaust system, air or steam from the exhaust system will rush into the cylinder when the exhaust valve opens.  You could open the exhaust early, and hence balance the cylinder pressure to atmospheric for the remainder of the piston travel.

The energy in the steam or air is proportional to the mass of air or steam admitted in each revolution.  The mass is proportional to pressure with other factors constant, so if you admit your motive fluid at higher pressure, there is more mass, therefore more energy, and under higher pressure, the engine will run faster unless you increase the load.  It is all about energy balance.  You can’t defeat conservation of energy. 

While you are dealing with work and energy and not concerned with torque fluctuations you don’t need to consider the conrod geometry.  You have defined piston travel which is sufficient.  Note that the time for each increment of piston travel and angle of rotation of the crank shaft will be different for each increment.   If you want to delve further into torque and torque fluctuations, then you will need to consider the con rod and crank geometry.

It is difficult to achieve zero clearance volume, but you can get it close enough to ignore if you want. So long as you are aware of the effect this has on the results of your calculation.

There is one more area for confusion in this discussion.  We are actually using that P x V product in two very different ways.  And it is very easy to unconsciously switch without noticing.

The product of P x V appears in the calculation of the power output of the engine.  In that formula, you need to use the mean effective pressure, otherwise you get the wrong answer for the power developed.

When you use the P x V product to calculate the pressure after a volume change, you are assuming the gas behaviour is the same as a theoretical ideal gas.  Then you use the actual absolute pressures and the volumes as you have done.  When you are using air, this is a reasonable assumption.  However there is a trap if you are using steam.  Steam at the pressure and temperatures we are usually involved with in model engines, is very close to that two phase region of the property map.  Under these conditions steam behaviour is not nearly as close to the ideal gas behaviour, and it is best to use the steam tables to determine the pressure from any change resulting from the expansion process.  As we don’t accurately know the path of the expansion, we have to determine the necessary two independent properties from the conditions we do know at the exhaust condition, and use these properties with the steam tables to determine the final pressure.

I hope I have not introduced too many confusing statements.  And I hope that by continuing the conversation, we are increasing understanding of how the engines work.

Thanks to everyone who is still following.  I hope it’s not getting too heavy.

MJM460

[Zephyrin]

Of course air expands, as steam does, and in my opinion, early cutoff or full gear are functional with air too.
However, the main drawback using compressed air is the excessive cooling (which may freeze the engine) that occurs upon detent as compared to steam, which contains much more "heat"; cooling while steam expands uses the latent heat of evaporation, a large amount of energy gained by the steam during the change of physical state liquid-gas.

[steam guy willy]

Hi Zephyrin,  Your comment  (which may freeze the engine) reminded me about my model of the Beeleigh , Woolf compound  steam engine running on air at the forncett steam museum that after about 30 minutes running just decided to stop. It didn't seize up however it just wouldn't run ??!!!.

Hi MJM, thanks for this exhaustive synopsis and could you suggest perhaps a different method of constructing small engines to run on air successfully ,with perhaps different clearances  etc etc . Other peoples engines seem to run OK for the whole day. Was it just because my engine was a compound ??  Perhaps it should have had an oil feed , or even a candle under the engine ??!! (silly).    ?? With refrigerator pumps are there different clearances to cope with the 'coldness' ?? Keep reading through your posts and coming up with questions  ........

Willy

[Captain Jerry]

Hi, MJM,


I don't mean to draw to fine a line here and I certainly did not interpret anything that you said to mean that an engine would not run on compressed air.  It is also clear that you have real work to do and that I am fully retired.  My only real work is to keep the grass cut and with 8+ acres to maintain that can be a full time job, but it is raining and has been for the last several days so I have time on my hands.


However, we seem to be on different pages.  I am trying to simplify a complicated system with multiple interacting factors to make it possible to understand just one of those factors and I am talking only theoretically. In doing so, I may have failed to complete specified the conditions of the experiment.  I think I said that I was assuming a constant temperature, not because I don't understand that it is a factor but because there is no way to specify temperature changes easily.  I should have said that the cylinder block is a large block of brass with resistance heating precisely regulated to hold the system at a constant 75 deg. F. which is the temp in my shop.


My comments on linguistic mumbo jumbo was to emphasize that it is meaningless to describe a three factor system by describing only two of them.  Such statements are neither true or false, they are incomplete. Almost political in the effort to make a point.


I also hope it is clear that when I mention early cut off, I am talking about a system that is unrelated to an eccentric driven system but mean to describe a governor driven system where inlet events are unrelated to exhaust events and which react to speed/load changes rather than control lever settings.


Clearly, ideal conditions can never be achieved but they provide a  point of reference so that when actual results are not the same as ideal, it is possible to understand the effects of changes to the real system.  Such factors as friction, leaks, valve adjustment, etc., can be better evaluated and methods to measure the other factors can be considered.


I jumped into this discussion without having read all of the preceding posts so I think I will take some time off to allow this thread to resume it's original direction.

[AVTUR]

MJM, Willy and others

I have been trying to follow Willy's question and replies but the thread is moving very quickly, far too quickly for me.

As I understand it, Willy wants to run a steam engine which has variable cut-off on air. The question is what pressure does he need?

I may have got this wrong but I will try to answer it!

First, forget Boyle's and Charles's laws. They are only of use in school exams. The two equations required are the perfect gas equation [PV = mRT] and, importantly, the equation for expansion and compression [PV^n = constant]. For brevity I will not re-introduce the gas equation. n in the expansion/compression equation equals 1.4 when the process is adiabatic (no heat is lost to the outside world) and the gas is diatomic (the gas molecules have two atoms each). Since the process will not be adiabatic a lower value for n is usually used (something like 1.35). Air is essentially a diatomic gas.

Taking the liberty of making up figures I will try to do a sum
Taking the total volume of cylinder  as 5 in^3 with the air being admitted and expelled from the cylinder during the first and last 20% of the stroke. The piston area is taken as 1 in^2.
The cylinder pressure at the opening of the exhaust will be at atmosphere (this gives the minimum of wasted energy), P2, = 0 lbf/in^2(gauge) = 14.7 lbf/in^2(absolute). The volume of the air in the cylinder will be, V2, = 5 x (1 – 0.2) = 4 in^3.
Considering the expansion of high pressure air to get this, saying n = 1.35, our expansion constant, C, = P2 x V2^n = 14.7 x 4^1.35 = 14.7 x 6.50 = 95.5
At the start of the expansion, when the inlet closes, the volume of air will be, V1, = 5 x 0.2 = 1 in^3. Our expansion equation will give an air supply pressure, P1, = C /V1^n = 95.5/1^1.35 = 95.5/1 =95.5 lbf/in^2(absolute) = 80.8 lbf/in^2 (gauge).

[For curiosity we can calculate the mass of air required and the exhaust air temperature. Taking the gas constant for air, R, as 1151, (trust me) and an air supply temperature, T1, = 15°C = 288 K, the mass of air, m, = P1 x V1 / (R x T1) = 95.5 x 1 / (1151 x 288) = 0.000288 lb. The exhaust air temperature, T2, = P2 x V2 / (m x R) = 14.7 x 4 / (0.000288 x 1151) = 177 K = -96°C.]

The amount of power that will be produced can be calculated by using a “simplified” version of the equation that MJM gave (at Tech we knew it as “no pale ale” – nPALE). We can draw a PV diagram for the device, attached. The area under the thick black lines is the expansion which includes atmospheric pressure. We can get a mean pressure, say 60 lbf/in^2 (my brain rebelled when it was asked to do the calculus) from which we can calculate the mean force on the piston which will (60 - 14.7) x piston area (it is assumed that the atmosphere is acting on the other side of the piston) = 45.3 lbf. The work produced by this expansion is force x distance = force x (cylinder volume / piston area) = 45.3 x 5 / 1 = 226.5 lbf.in  = 18.88 lbf.ft. To get the power we just multiply this work by a speed, dreaming up a figure 60 rev/min. I am saying that the expansion only takes place once a revolution (single acting) so we get 18.88 x 60 = 1133 lbf.ft/min = 0.034 hp.

It should be stressed that the above power is the theoretical produced by the expansion. Quite a bit of it will be used to pump the air out of the cylinder (and perhaps into the cylinder), overcoming friction and other losses. Also the above IS NOT valid for steam.

I hope I have answered the correct question. I am happy to be quizzed.

And I could have been carving some metal. Now my brain hurts.

AVTUR

[steam guy willy]

Hi MJM  et al,  Thinking on the practical side was the Newcomen engine 100% efficient as over the compleat cycle the hot steam returned to water at ambient temp   so gave up all its thermalicity ?    Unlike the Mallard engine that did not really lose its temperature during the engine cycle ?? Can steam at pressure only do work by losing  its temperature

Willy

[MJM460]

Hi Zephyrin, thanks for joining in.  I suspect the real issue with air is not the temperature drop on expansion, but the fact that it starts at a much lower temperature, usually close to ambient, so the end temperature is below the freezing point of the water which is always present in the air as humidity.   With steam, it starts at a much higher temperature than the usual air supply temperature, so the end temperature is more moderate.  And as you point out, when the expansion continues into the wet region, the steam condenses at least partly, thus giving up some of that latent heat.  This holds the temperature instead of requiring extra cooling to supply the energy for the work done.

Hi Willy, you mentioned that your beautiful Beenleigh engine had stopped and I was hoping that you would tell us more about it.  If it did not seize, I assume it will now turn over ok, and even run if you give it a short burst of air?

I suggest that the problem is mostly due to the fact that it is a compound engine, so involves expansion of air to do work, and gets quite cold in the process.  That little turbine that I ran on air for the client produced frost on the exhaust pipes so quickly that it must have been very cold inside the pipes.   Now compressed air contains all the water from the humidity of the supply air when it leaves the compressor hot.  Usually the compressor has an after cooler to reduce the air temperature to ambient, during which much, though not all of the water drops out in a trap that hopefully has been provided.  But the remaining water is enough to cause problems, and when your engine cools the air I would expect ice would form inside the engine or exhaust, possibly as snow or that rhyme you get when fog freezes, and lodge in some place that locks up the engine.  When the whole engines warms back to ambient temperature the ice melts and with luck is free again and undamaged.

I think lubrication is the other problem with running on air.  With  a simple engine with minimal early cut off, there is minimal cooling, and I understand that they run for considerable periods at your various shows.  Obviously the displacement lubricator used for steam engine do not work, it would be worth talking to other exhibitors about what they do.  Do they squirt in a little oil at the start, does the water in the air lubricate them enough?  Or do they use a low temperature lubricant that does not get too viscous in the engine?  Unfortunately I don’t know the answers, but with all the work you put into that beautiful engine, it is sure worth asking the questions.  It would be great to see it running again.

Certainly different components of engines and compressors operate at different temperatures, so I am sure that manufactures of equipment calculate and allow for different amounts of thermal expansion in components where it might change critical clearances.  The biggest problem occurs when different materials with different coefficients of expansion are used.  If all the materials are the same or similar in coefficient of expansion then the clearances only change when there is a differential temperature between components. On our small engines the differences are surely very small, and if we follow one of our colleagues’ mantra, and have a little clearance, it will never get in the way.  But I believe we have others on the forum with more expertise in this area, perhaps they will come in with more suggestions.

Regarding the Newcomen engine, I could say that it is definitely not 100% efficient because the second law of thermodynamics says that is not possible, but that would be a cop out, or begging the question.  First we should say what we mean by efficiency of the engine.  The definition I prefer is the fraction of the input energy that is converted to work.  Unfortunately a significant portion of the input energy to any engine appears in the exhaust, and in a steam engine, the normally very low efficiency figures are because the exhaust steam is usually mostly still in the vapour phase, so still has all that latent heat.  Sometimes it is easier to look at the losses, which for a 100% efficient engine should be zero.  So look at where energy is lost rather than being converted to work.  If I remember rightly, in that engine, the exhaust is condensed by a water spray.  Then the warm water produced is discharged to the river.   If you do an energy balance you will see that it takes a large quantity of water to condense a small quantity of steam, so there is a lot of heat lost in that discharge stream.  Does that answer the question? Or am I thinking of the wrong engine?

But can steam only do work by losing its temperature?  Certainly some of the energy in steam is is converted to work, so when work is done, some energy is transferred across the boundary of the system, and the fluid in the system then has less energy.  In all the cases I can think of that means it will have a lower temperature.  The energy is not lost, it has just moved outside the system we are looking at, usually to the surrounding atmosphere.

Hi Captain Jerry, I can see that we all need the rain to stop.  But if you could send some here, most of our country would welcome it.  Don’t worry about the real work, I have been retired for ten years now, but as others have found, I have less spare time now than when I was working.  I think it was Dr. Parkinson who had something to say on that matter.  But I would like to get on the same page with you.

Simplifying a complicated system to understand the effect of one factor is a good idea, but it would help to start by identifying or describing the factor you are trying to understand.  I don’t think I have caught on to that yet.  When simplifying the problem, it is a good idea to first identify all the variables involved.  Then it is important to identify which of these variables are independent so you can try and specify them, and which are dependent, meaning they result from the process, and can’t be separately controlled.  Then we can move on to see what simplifying assumptions we should make.  Consider what conditions you know at the start of the process, and what conditions you know at the end, and what process is occurring.  I am guessing that you are talking about an expansion process.  In this case, temperature changes are often the result of the process and if you try and independently alter the temperature, you are almost certainly introduce a second process of heat transfer, so making the issue more complicated.  I would suggest leaving the temperature changes to the thermodynamics.

I would certainly encourage you to engage theory to help guide your experiments.  To do this, I would suggest you first try and define the process you are trying to understand.  Then try and clearly define the theory you intend to use.  Before proceeding further, it is a good idea to do a  check and make sure the theory is sound, accepted physics.  Unfortunately, as you have found, much of the so called theory expounded around the club tea rooms is not so sound, and often fails when put to the test. 

Using an ideal process to provide a point of reference is a good idea, used across all kinds of engines.  The key requirement, is that the ideal process selected can be analysed by calculation.

So to help us all get on the same page, how about starting by defining or at least describing the process that you are trying to understand.  It is even possible that the answer is already known, which will simplify the discussion somewhat.

Hi Avtur, thank you for joining in and especially for providing such a clear and useful description of the ideal gas and the compression /expansion equations.  We all need to print that paragraph and stick it in the front of our preferred reference books.  It will be pinned above my desk for sure.  Polytropic never quite cut it for me.  You were just in time to save me from getting deeper into trouble.

Normally I only put up one post each day, but yesterday I was having a quiet day, and the thread did move on quite fast.  I have had much less spare time today.  I am not surprised that it was a bit hard to follow.  Just to clarify, while Willy has asked most of the questions to date, this one came from Captain Jerry.  You may have noticed his other thread on a quick release valve mechanism, and I suspect the question has arisen out of that.  But I am sure we will get to it soon.  In the mean time I hope you have been able to get back to cutting some metal.

Achieved a milestone today, I backed the car in through our gate and went inside to sit in my chair for a cup of coffee.

What’s so special about that you might be thinking?  Well, I last pulled out of that gate three months ago and have been on a little road trip.  Sixteen thousand, eight hundred km later, I have kept the water on the right, and completed the loop around the country, even if I did cut the odd corner, as time did not allow the coastal road the whole way.  Crossing the Nullarbor is one of those rights of passage.  And now you can see why I have made the odd mention of holes in the net.  It claims to cover over 95% of the population, but the other 5% are spread out over a very large area.  The misconceptions they can create with statistics!

Thanks for looking in,

MJM460

[Captain Jerry]

MJM,


Had I known that you were on such an epic trip, I would not have bothered you with poorly stated questions or examples. If I were to take such a trip, I think I would enjoy it more without the intrusion of electronics and other peoples thoughts. Maybe a very simple GPS for safety. Open windows and paper maps. Did you take a good supply of vegamite? 


Land cruises of that are interesting but my preference is the deep blue water and endless skies.  When I first started cruising, it was primarily magnetic compass and charts.  GPS was available but the readout was position only and sometimes quite slow to resolve.  Communication was short wave.


It won't stop raining here for a few more days and might get much more intense as hurricane Dorian passes. 


Jerry

[steam guy willy]

Hi MJM, more thoughts on heat engines...a steam engine running on compressed air (cold) is still capable of doing useful work..An engine using compressed air turning over a diesel engine for instance  or a turbine that you made ,  however the heat is used in producing the compressed air perhaps ?  unless it is supplied by a water wheel at the bottom of a Dam supplied by rain !! So are these still thermodynamic heat engines or are there different equations for working out the formulae ??  Hope you had a loyely trip.

I have been to Norway , Hong Kong ,Germany, Europe Turkey, Greece,The Med etc and think everybody should travel just to see that other people are basically just like us with the same dreams and aspirations for peace and love !!

Willy