Author Topic: Talking Thermodynamics  (Read 194866 times)

Offline derekwarner

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Re: Talking Thermodynamics
« Reply #795 on: March 25, 2018, 09:50:42 AM »
Sorry all......my words were a little misleading.....'all interconnecting pipework was cast steel' this was true or as intended

'the complete compressors entablature, casings & pipework all painted gloss light cream colour' .......naturally the mass of the entablature was of assumed high quality cast iron, however painted gloss light cream colour

A part of the apprentices tasks was to read the Broom & Wade drawings book.....[text with hand drawn exploded views of great complexity]

This was my first journey into the light hand 'tapping' metallic components with a small hammer

This tapping  :killcomputer: as a maintenance requirement also confirmed the integrity of the cast interconnecting pipespool material as being free of cracking :ThumbsUp:

I am unable to offer comment of cast steel earlier than my 1st experience of the 1890 Broom & Wade compressor name plate

At sometime after the turn of the Century, I believe the original Broom & Wade Engineers, became Broomwade Pty...........

Derek

Derek L Warner - Honorary Secretary [Retired]
Illawarra Live Steamers Co-op - Australia
www.ils.org.au

Offline MJM460

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Re: Talking Thermodynamics
« Reply #796 on: March 26, 2018, 12:56:34 PM »
Hi Derek, it is clear that I am not an expert on metallurgy or history, no surprise there, I have mentioned it before.  I do welcome such comments however, as while they are not thermodynamics, materials science has to advance appropriately to take advantage of the knowledge.  It is no use thermodynamics telling us that higher pressures and temperatures are necessary for increased efficiency if materials science does not at the same time allow us to construct boilers, piping and engines suitable for the higher temperatures and pressures.  Besides, such comments and historical footnotes add interest to a subject that some may otherwise see as somewhat dry.

Not making much progress on the thermodynamic analysis of my recent boiler test runs due to a very hectic home life, but I will get there eventually.

While I am not much on remembering history or metallurgy, I do like topics that involve calculation, especially things involving physics and motion.

I don't know how many others have noticed kvom's thread on the Muncaster Grasshopper engine.  It is going to be another great build.  However, he asked in an early post whether the swinging link was part of a Watts linkage.  I have been interested to notice that the question has not yet been explicitly answered.  It clearly is, but it is not so obvious how it works. 

OK, it is pretty obvious, but the question that attracted me was whether I could demonstrate that  it produced the required vertical linear motion of the top of the piston rod exactly, particularly as the swing link is not fixed on the piston rod centre line as it is on most "conventional" beam engines.  At the same time, it is clear that many successful engines have been built to the design, so how does it work?  Rather than hijack kvom's thread, I thought it might be worth posting the question here.

The Watts motion has at its centre an isosceles triangle, but for this to actually keep the top of the piston rod in a vertical line, the vertex of that triangle must be constrained so that its height above the horizontal line is exactly half of the height of the base of that triangle.  In a conventional beam engine that constraint is provided by the drop link from the intermediate point on the beam.

The link is not necessary on the grasshopper because the beam actually forms not only the top leg of that isosceles triangle, but also constrains the apex of the triangle.  I thought I should be able to calculate the position of a sufficient number of points either prove the design is a perfect watts motion, or alternatively, calculate the probable error.

It turns out that it is a more complex calculation than I anticipated, even for a spreadsheet.  Oh for MathCad or a similar program (and to know how to drive it!).  But I think I have I learned a few things about the Grasshopper design in the process.

If the right hand end of the beam (in the drawings posted) moved in a horizontal straight line the mechanism would be exact with the end of the swing link on the centre line, and the beam end of the link mid way along the beam.  However, there is an error introduced by the rocking column on the right hand end of the beam.  The movement is small, and the error can be reduced by making the column longer.  Muncaster's design has a shorter column than most I have seen.  The error can be further reduced by moving the base of the rocking column towards the cylinder so that it rocks an equal amount each side of the vertical. It also appears that the moving end of the swing link is not at the centre of the beam, but is slightly offset towards the cylinder end.  However, that rocking column moves the end of the beam vertically, by a very small amount admittedly, but it tends to distort that isosceles triangle and pull the top of the piston rod ever so slightly out of line.

It also appears, that the error introduced by the rocking column can be compensated by moving the fixed end of the swing link as Muncaster has done, possibly also involving moving the swing link pin along the beam a little.  I set about to solve the equations by trigonometry, but it's more complicated than it looks.  I tried drawing it out carefully full size, but the various factors are very small and quite difficult to draw sufficiently accurately.  Which of course raises the possibility that the normal tolerances on the linkages and rod packing can actually accommodate the small errors.  I even tried a Meccano model as a 3-D model, but there was too much slack in the pinned joints to be definitive, despite my using bushes at all the critical joints.

I still feel that with more time, I should be able to solve the problem by trigonometry, but it is going to take more time than I have available at the moment.  So I ask, has anyone modelled the engine in a suitable program and been able to prove that Muncaster's design ensures true vertical motion of the top of the piston rod?  Or alternatively worked out just where the fixed end of the swing link should be located to eliminate the error?  Or is there an inevitable small error?

Or perhaps we should just return to Thermodynamics!

Thanks for looking in,

MJM460
« Last Edit: March 26, 2018, 01:02:49 PM by MJM460 »
The more I learn, the more I find that I still have to learn!

Offline paul gough

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Re: Talking Thermodynamics
« Reply #797 on: March 26, 2018, 02:53:36 PM »
Hi MJM, A bit damp up here at the moment, 305mm of rain yesterday, but the frogs like it. I was wondering if you could provide a breakdown of the heat input ascribed to the process of generating steam, as percentages. Let us say we can break the process into the following four parts, 1) heat required for water from 'cold' to 100 C.; 2) heat required for phase transition; 3) heat required to reach operating pressure; 4) heat required to reach desired superheat. This would help me grasp the impact of each stage in packing our molecules with energy. If my thinking is bent, please straighten me out. Regards, Paul Gough.

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #798 on: March 27, 2018, 01:17:09 AM »
HI MJM, i have always wondered why the pivot point for the pivoting radius rod can be in lots of different positions. some below some above  and some quite far away and some quite close to the beam pivot ??

Offline paul gough

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Re: Talking Thermodynamics
« Reply #799 on: March 27, 2018, 09:54:43 AM »
Hi again MJM, A flow on question, sort of connected to the previous is; If one has a boiler being steamed at a given pressure and then it  is steamed at double this pressure, all other things being equal,  does doubling the pressure always require a set percentage more heat to maintain it, or does it vary as pressures rise. Regards, Paul Gough.

Offline MJM460

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Re: Talking Thermodynamics
« Reply #800 on: March 27, 2018, 01:12:19 PM »
Hi Paul, I did hear that you had a shower or two of rain.  Sounds like a lot to most of us, but as the farmers say it takes more than one rainy day to make a wet season.  Still, enough to bring out the frogs for counting.

Thanks for the question, bringing us back to thermodynamics, where I am much more comfortable.  It is a good question, as understanding how much heat is put in to the three stages of producing superheated steam helps us understand many of the issues regarding feed water heating, and why a boiler can not produce as much steam per second when cool water is fed, as opposed to by closed system tests when I measure the steam production "from and at" saturation temperature. 

So the thinking is fine, but before I start quoting figures, let's think about the process we actually follow.  If we fill our boiler with say 20 deg water, the saturation vapour pressure is only 2.34 kPa, that is absolute pressure. But if the air is say 50% humidity, the actual vapour pressure is only 1.17 kPa.  And air has done an amount of work to compress the water to say 100 kPa, or atmospheric pressure.  Now because the change in volume of the water is so small in this compression, even the text books suggest it is close enough to zero.  However, the air really does complicate the calculations.   So let's assume we have previously filled the boiler and run it long enough to expel the air, then allowed it to cool down with the regulator closed so it is really only at 2.34 kPa absolute and no air.  This simplifies the calculations enormously, but if you remember back to Willy's questions about filling the boiler on the mountain, you will remember the rough process.  Of you really want, I can do them again to show the difference, but for tonight, let's just look at the water, and assume the total vapour pressure is only due to water.

Once the boiler is sealed, with the regulator closed, 100 degrees C is no longer a relevant special point.  When heat is applied to the boiler, the pressure will rise, and the first law of thermodynamics says for heating at constant volume, which also means that no work is done during heating, the heat input is equal to the change in enthalpy.  The answer is different if you apply heat with the boiler outlet open as some of the heat goes into the work done by the expanding steam.

The model of steam behaviour for this and all similar problems is the steam tables.  You may have downloaded these when I provided a reference way back.  Again I will look it out again for you if necessary.

The steam tables tell us the enthalpy of saturated water at 20 degrees C is 83.96 kJ/kg.  This is listed in the tables in the column labelled hf.  This is the starting point.  Probably a bit cool for your location, but equally, too warm for many of our fellow forum members.

Let's assume we will produce steam at 500 kPa.  As the tables list absolute pressures, that is equivalent to 400 kPa gauge or about 57 psig.  I have chosen that pressure because it also appears in the superheat tables without complicating the answers by having to interpolate the tables.  At 500 kPa, (or 0.5 MPa, whichever is listed in your copy of the tables) we look up four figures. We find the saturation temperature is a touch under 152 deg C.  This is the temperature at which the water will start to boil at this pressure.  We also look up hf, 640.23 kJ/kg, so the heat added to this point is 640.23 - 83.96 = 556.27 kJ/kg.

The next column, headed hfg, is the heat necessary to evaporate the water at that pressure.  We read 2108.5 kJ/kg as the latent heat.  And the last figure in the column headed hg is the enthalpy of  the steam without any superheat.  We read hg = 2748.7 which is the same as hf + hfg, which the the enthalpy of steam before any superheat is added.

Now we look at the superheat section of the tables.  Basically a collection of small tables, one for each pressure.  In the table for 500 kPa, the saturation temperature is shown in brackets as 152 as before.  We can go down to say 250 C, or approximately 100 degrees of superheat, and across to the column headed h, where we find the enthalpy as 2960.7 kJ/kg.  Now we can subtract the enthalpy of the saturated steam 2960.7 - 2748.7 = 212 kJ/kg which is the heat added by superheater.  This does not look like a lot in the overall scheme, but it makes a big difference to the engine output as I demonstrated earlier in the engine discussion.  I can look up the post number or revisit that again if you prefer.

So now we have the total enthalpy of the superheated steam, and we can subtract the enthalpy of that cold water to see the total added is 2960.7 - 83.96 = 2876.74 kJ/kg, and we can answer your three questions.

Sensible heat to reach boiling point is 556.27 / 2876.74 = 19.3%

Latent heat to boil the steam is 2108.5 / 2876.74 = 73.3%

Heat added to superheat to 250 C is 212 / 2876.74 = 7.4%

Now the second question, what if we double the pressure, yields a surprising result.

So far I have assumed an absolute pressure of 500 kPa.  So double that is 1000 kPa, (or 1 MPa), a convenient figure which is also directly listed in the superheat tables, equivalent to about 128 psig.

Of we assume the same superheater outlet temperature of 250 degrees C, we get the surprising result that h = 2942.6 kPa which is slightly less than the enthalpy of 2960.7 at the lower pressure.  If we left the regulator closed, and just continued heating to the higher pressure, and still controlled the superheater outlet temperature to 250 C.  Because the saturation pressure is 180 C, this is only 70 degrees of superheat, and we don't have to superheat much further to have equal enthalpy to the lower pressure.  I will try and attach a pressure enthalpy diagram for steam, basically the steam tables presented in graphical form, and you will see from the diagram why this result occurs.  In the table, pressure is in MPa, so look at 0.5 MPa and 1 MPa.  Unfortunately, the temperature lines are absolute so 250 is 250 + 273 = 523 which is about half way between the 500 and 550 degree lines.  I think the picture will be clear enough for you to find the appropriate lines.

Perhaps I should mention that while the calculation is a constant volume calculation, it would be necessary to have Willy's electric superheater we discussed at one stage.  It is normally necessary to maintain a flow through a conventional superheater to prevent it becoming too hot, though in a small Meths fired boiler, it is not really necessary, though it is advisable to avoid joins in a copper tube superheater.

Hi Willy, different beam engine designs involve slightly different versions of the Watts linkage, but if we stay with the grasshopper for the purposes of your question, an ideal Watts motion has an isosceles triangle on its side with the piston rod as the base.  The apex of that triangle has to be constrained to move exactly half the distance the top of the piston rod moves.  Then the swinging link is fixed on the vertical centreline of the piston rod, and attached to the midpoint of the beam.  The other end of the beam has to move in a horizontal path at the same elevation as the fixed end of the swinging link. 

In a real engine, with a rocking column at the far end of the beam, so the end of the beam moves in an arc instead of a straight line, that isosceles triangle is "pulled over", just a fraction.  Also you will note the swinging link is not fixed at the beam centre.  It is possible by trigonometry or graphically to find a position displaced from the piston rod centre line where the attachment point on the beam is equidistant from the fixed point at top, middle and bottom of the stroke and so constraining the beam to the correct position at these three points.  So the different positions of the ends of the link are selected to compensate for the arc at the other end of the beam.  The fixed end position might not be at the same elevation as the top of the column at the other end, and some designs seem to have a provision which may be intended to make a small vertical adjustment possible. 

But three points, while they completely define the arc swept by the link, do not necessarily mean that the top of the piston rod moves in a straight line between these end points, it may move in a slight curve.  Some trigonometry is required to check the beam position at intermediate points to see if the linkage is correct or has a small error.  I am still looking at it when I get a chance, perhaps over Easter, as I have not been able to produce a satisfactory calculation so I really don't yet know the answer.  I will let you know if I get a satisfactory answer.  But to be fair, if there is an error in the linkage, it is very small and probably accepted by the inevitable tolerances in the pin linkages and rod packing.  I may have to resort to building an engine to see if I can measure any discrepancy for myself.  I really think the issue is quite academic, and any error is not enough to spoil the operation of the published designs that are well proven, but I have read of others who claim there is a necessary small error, I would like to be able to demonstrate it for myself.  I hope that is enough for now.

Thanks for looking in,

MJM460


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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #801 on: March 28, 2018, 02:08:36 AM »
Hi MJM, Thanks for that ...can i ask you about compound,  triple expansion , Quadruple expansion engines ....is there an exact figure for determining the cylinder bores  assuming all the other dimensions are equal ?? also with a beam engine where the inside cylinder is shorter (closer to the beam) is there also a constant for the bore sizes ?? and is this worked out with a formulae ??...thanks...
Willy

Offline paul gough

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Re: Talking Thermodynamics
« Reply #802 on: March 28, 2018, 03:53:02 AM »
Thanks MJM for working through and clarifying the proportions of heat input. The chart certainly shows the 'surprising result', my untutored eye has never noticed this before. Demonstrates the power of visuals over lists of numbers.I wonder how many other surprises might be revealed if I was graced with enough knowledge to know where to look and what to look for! Regards, Paul Gough.

Offline paul gough

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Re: Talking Thermodynamics
« Reply #803 on: March 28, 2018, 09:58:32 AM »
Seems at least one other person in Nth Queensland has /had an interest in thermodynamics, so thought maybe there is someone on the forum who might like to know of the books existence. Found this book in the Op Shop for $1. Maybe its one of Click Springs old books?? Regards Paul Gough.

Offline MJM460

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Re: Talking Thermodynamics
« Reply #804 on: March 28, 2018, 12:33:17 PM »
Hi Willy, that's a bit out of left field, it sounds like you are hatching something! 

If we look at the ideal engine concept, which tells us the maximum amount of work that could be produced from the steam conditions, the thermodynamics tells us we should balance the stages so each stage expands the steam through an equal pressure ratio.  Remembering that the pressures have to be absolute pressures, for a compound engine, we are looking the intermediate stage pressure (Pi),so that P1/Pi = Pi/P2 = sqrt P1/P2, where P1 is the supply pressure and P2 is the exhaust pressure.  To achieve this, we would probably make the ratio of the swept volume of the stages about equal to this same ratio, but it might need a bit of refinement using the specific volume column of the steam tables, probably in practice not a very important factor.  For a triple, the ratio would be the cube root relationship, and four a quad expansion, a one fourth root.

Of course in a real engine, at the start of the expansion in the HP cylinder, we only have the clearance volume at steam pressure when the expansion starts, so the quantity of steam would severely limit the amount of work that could be done by the engine.  We normally leave the inlet valve open for and admission period, then close the inlet valve partway through the stroke and the expansion begins.  The ratio of the volume at cutoff is then expanded to the volume at release, and the pressure ratio for the expansion is determined by that volume ratio.  Then the steam in a compound engine is opened into the second stage.  So the volume of the HP cylinder is then expanded into the increasing volume of the second cylinder during its admission stroke, thus completing the first stage expansion.  When the second stage inlet valve is closed, the expansion continues until the second stage release to the exhaust pressure for a compound, or to the lp cylinder in a triple expansion engine.  Hopefully the pattern is clear to continue to a fourth stage in a four stage expansion.

The situation is complicated by the need to look at the position of the various throws around the crank shaft and the actual valve timing.  Some articles I have read suggest that the valve timing between the cylinders can be made a little more independent by a steam "volume bottle" between the stages, so the HP cylinder exhausts into this volume, while the next stage admits steam from this volume.  It would be particularly important in a reheat cycle where the first stage exhaust is sent back to the boiler for extra heat input, which significantly increases the pipe volume between the stages.

The crank angle and the valve timing give the designer considerable latitude in the best way to control, the expansion and divide the work between the cylinders.  I would need to do a lot more reading to be confident to actually design an engine, but in looking at existing proven designs, I would be looking for that equal volume ratio between the stages.  For equal stroke on each cylinder, it means the same ratio applies to the cross-sectional area so there would be a square root relationship between cylinder diameters on a compound and so on.  And I would be looking closely at the recommended valve timing.

For a beam engine, where the stroke of the cylinders are not equal, the relative length of the strokes would also have to be taken into account in making those equal volume ratios, resulting in a slightly different diameter ratio.  Some of our other members might have more information on this topic.

I am sure the pioneers would have developed rules of thumb for the cylinder diameters, and these rules are probably included in some of your historical books, but I suspect these rules lead to a similar result.  I hope that is enough answer for your purpose, as I don't even have access to my copy of Jamison's book at the moment to see if he says anything helpful on the subject.

Hi Paul, the graphical presentation certainly clarifies the direction of property changes in a way that is difficult to appreciate from the tables.  But the graph is hard to read to sufficient accuracy when the answer to a calculation involves a small difference between two large quantities.  So I find it useful to use the tables to calculate precise answers, but use the diagram alongside to make it clearer just what is happening.  And of course the laws of thermodynamics to tell us which properties are changing in a specific problem.

As an example of something else you might be able to see from the diagram, if we continue to use your example from yesterday, and assume we have kept the regulator closed until 1 MPa but find the load on the engine only requires 0.5 MPa to run at the required speed, so we throttle the steam from 1 MPa to 0.5 MPa.  The first law says that at the throttle valve, there is no heat input or output, and no work done, so the change in enthalpy is determined only by the change in velocity.  This is quite small if the pipe diameter is increased on the low pressure side, so the velocity change is negligible.  So we effectively have constant enthalpy, a vertical line on that diagram.  If you follow that vertical line and see how it compares with the constant temperature lines, you will see that the steam cools just a very little during throttling.

Now remember when we discussed adiabatic expansion, the second law tells us that the work done is the change in enthalpy, and that during the ideal expansion, the entropy is constant.  So, if you follow the slope of the constant entropy line from 1 MPa at the temperature you are considering (either superheated or perhaps just saturated) to the exhaust pressure, the vertical lines show you the change in enthalpy for an ideal engine.  Similarly, if you follow from the conditions at 0.5 MPa, you can see the change to the same exhaust pressure is much less than for the higher pressure.  And of course a real engine does less work in each case, probably quite a similar percentage of the ideal engine enthalpy change.  So you can see that throttling to the lower pressure reduces the potential work output of the engine, as you might expect.

The other thing that you might be able to see from the chart, if you compare the constant entropy expansion from saturated steam and compare it with superheated steam at the same pressure.  You might have to calculate exact values to see in this pressure range, that the extra work obtained from superheated steam is produced at higher efficiency than the work produced from saturated steam.  The reason for this, is that the main reason the efficiency of the steam cycle is so low is the high proportion of latent heat required.  All this latent heat is lost with the exhaust.  The extra heat put into superheating does not involve any extra latent heat, so the resultant extra work from superheated steam is obtained at much higher efficiency than the work from saturated steam at the same pressure.

You might also try following the temperature changes as your water is first boiled (at constant temperature and pressure) then further heated at the same pressure.  The diagram clearly shows how the temperature enthalpy and entropy all change as you add superheat at constant pressure (a horizontal line on the pressure enthalpy diagram).

You will find that for some problems the pressure-enthalpy diagram is most helpful, while for others the temperature-entropy diagram is more useful.  If you look around a bit you will even find an enthalpy-entropy diagram, the one to use is generally the one where you make most use of the vertical and horizontal lines rather than needing the curves.  But they are all presenting the same data from the tables.

I think that book is one of a series on many topics, possibly intended for students or those who find a need to understand more of the topic in their work.

Thanks everyone for looking in,

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 #805 on: March 29, 2018, 12:54:12 PM »
Hi Willy,  I was just looking over my post yesterday in response to your problem, and I can see that I did not explain the steps from the volume ratio to the diameter ratio very well.  So let me give you an example.

Let's assume steam supply at 400 kPa and exhaust at 100 kPa, so no condensing.  Remember, absolute pressures so 100 kPa is also atmospheric pressure, just a little lower than standard, for ease of calculation, but realistic in a low pressure weather system.  In imperial units, you might assume exhaust at 14.7 psia, a slight high pressure system and steam supply 58.8 psia, very similar pressures, but because the ratio is the same, supply is four times exhaust, the same relationship between cylinder volumes applies.

For a compound engine, the pressure ratio between the hp and the intermediate pressure should be the square root of the pressure ratio from hp to exhaust, so square root of 4 or 2 and this figure should be used for the cylinder volume ratios.  So V2/V1 = 2, so we can say (d2/d1)^2 = 2 so d2/d1 = 1.414.  The intermediate pressure will be 200 kPa.  If the hp cylinder is 1" diameter, then d2 = 1.414"

For a triple expansion, the pressure ratio should be the cube root (= ratio^1/3) of the overall pressure ratio.  Assuming the same supply and exhaust pressures, the cube root of the pressure ratio, 4^(1/3) = 1.59.  So the lower intermediate pressure = 100 x 1.59 = 159 kPa.

The higher intermediate pressure will be 159 x 1.59 = 253 kPa, and of course 253 x 1.59 = 400.

For our 1" dia hp cylinder, (p2/p1)^1/3 = V2/Vi = (d2/di)^2, so d2/di = (V2/Vi)^(1/3))^1/2 = V2/Vi^(1/6) or 1.26.

For the same 1" hp cylinder intermediate cylinder will have a diameter of 1.26" and the lp cylinder is 1.26 x 1.26 = 1.59" diameter.

In the above calculations I have assumed the stroke of all cylinders is the same.  For your beam engine, the statements about pressure ratio and volume ratio still hold, but the swept volume of the cylinders is a function of d^2x s, so for a compound engine, (p2/p1)^1/2 = V2/V1 = (d2^2 x s2)/(d1^2 x s1).  We can multiply each side by s1/s2 to give V2.s1/V1.s2 = d2^2/d1^2, so d2/d1 = (V2.s1/V1.s2)^1/2.  In other words, we adjust the volume ratio by the ratio of stroke lengths before taking the square root.

Obviously it depends on which cylinder has the shorter stroke.

That's some fairly heavy maths, I hope I got it correct, it has taken a bit of work on dealing with exponents, but I hope it better answers your question for the compound and triple expansion engine.  I hope you can follow this through to a quadruple expansion, the diameter ratio is 1.19).  I will give everyone a chance to absorb that, and if necessary, point out any errors in the maths, and work the example of the beam engine to see the difference it makes whether the hP or lp cylinder has the shorter stroke.

Thanks for looking in,

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

Offline Dan Rowe

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Re: Talking Thermodynamics
« Reply #806 on: March 29, 2018, 02:03:50 PM »
MJM, just so you know there is a "sup" and "sub" button in the text editor to allow superscript and subscript text which is more work for the author but makes reading formulas easier. All you have to do is select the text to modify and then hit the button.

Cheers Dan

ShaylocoDan

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #807 on: March 30, 2018, 12:05:40 AM »
Hi MJM , I have taken the measurements from the Beeleigh and the cylinder bores are   HP 270 mm and the LP 440 mm the shorter HP cylinders is 83% of the LP cylinder . The exhaust is fed directly to the LP steam chest  So i was wondering how this configures with modern calculations This engine was made about 1830 so how developed were Mr Woolfs understanding of currant thermodynamic laws then ?? On a triple expansion the throws of the cranks are  equal. Is this just for aesthetics or could the throws be different to save metal and weight ??
 just thinking about these things as there is no imminent self designed triple on the way.....
Willy.
.

Offline MJM460

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Re: Talking Thermodynamics
« Reply #808 on: March 30, 2018, 11:04:33 AM »
Hi Dan, good to hear from you again.  Thanks for the tip about subscripts and superscripts, I will try them this time.  On that topic, do you have a preference for "." or "x" for multiply?  And are you happy with the "^" for "raised to the power of"?  I suppose the alternative is the whole exponent in superscript.  Also are you comfortable with the "E" notation, as in 2E6 for 2 x 10^6.  The iPad does not like it but many spreadsheets and scientific calculators offer it as an option.  These conventions can vary between countries, so not always easy to know which an international membership would find easiest.

Hi Willy, with those diameters and stroke of the hp cylinder = 0.83 times the stroke of the lp cylinder, we can put the pressure ratio = volume ratio.  Rather simpler maths than knowing the pressures and calculating the relative diameters.  Of course the stroke ratio comes from the beam dimensions.

Remember the volume of each cylinder is proportional to the diameter squared times the stroke.  I have used the subscript 1 for the first or hp cylinder, and 2 for the second, or lp cylinder.  As a formula we have

Rp= (V2/V1) = (d22 x s2)/(d12x s1) = (d22/d12)x s2/s1

When we substitute those measurements, d1 = 270, d2=  440, we find Rp= 4402 / 2702 x 1 / 0.83 = 3.2

(Wow! That was difficult.  My ipad will not allow me to select the text for some reason, I seem to have to delete and retype.  Not easy.  I have to post to see how I went on the process.  Works better if I retype first then delete!)
 
So I would conclude the engine might have been designed for a pressure ratio of 3.2.  The square root of this is 1.79 so the intermediate pressure would be about 1.79 x exhaust pressure.

If the exhaust pressure was atmospheric pressure, say 100 kPa, then the supply pressure would be 100 x 3.2 = 320 kPa.  This is 220 kPa gauge pressure or about 31 psig.  I would expect the boiler to be a operated at a bit higher pressure, to allow for throttling and piping losses, but does that sound something like the name plate pressure? 

The intermediate pressure would be 1.79 x 100 = 179 kPa.  Again we can calculate gauge pressure of 79 kPa, or about 11 psig.  It would be interesting to know whether Mr Woolf was aiming for that square root relationship, or had some other basis for sizing those cylinders.  (Perhaps it will be in that haul of model engineer magazines.  Great score!) In any case the fluid mechanics is such that the intermediate pressure tends to settle at a level that works with the actual steam pressures cylinder volumes and valve timing.

It also depends on what pressure he expected to achieve in that condenser.  If he expected to achieve say 50 kPa (7.35 psia) then he would only require an inlet pressure of 3.2 x 50 = 160 kPa.  That is only 60 kPag or 8.5 psig, may be a boiler operating at 10 psig, which sounds a bit low.  It is possible that not much vacuum was expected or achieved.

The test of the theory is whether the answers are somewhere close to the real answer, so putting myself on the line here, and I hope I have the maths right.

I would think that Mr Woolf would have been right up on the latest theory at the time, and he would have had a finely tuned intuition as to how it all worked.  However he had to rely on less accurate data than we have available these days.  (What was the footnote on that book you posted a while back?)  He may have also had ideas on valve timing that affected relative cylinder volumes.

In an engine with a conventional crank shaft, the throws are normally equal.  This gives the same piston speed for each cylinder which is an important figure in control of wear rates.  I suspect, but don't really know, that this may also reduce balancing problems.  I don't really know if there is any reason why an engine could not be made with unequal stroke lengths.  Perhaps others may like to comment.

Thanks for looking in,

MJM460
« Last Edit: March 30, 2018, 11:11:37 AM by 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 #809 on: March 31, 2018, 11:13:01 AM »
Hi Willy, I hope that you are going to let us all down from the suspense by telling us what the real steam conditions were for your engine.  Calculations are ok but only really informative if they are helpful in understanding reality.  I am most interested to know whether Mr Woolf used a similar logic to design his engine for the available steam conditions or had other ideas.

Just as a little side note, I have been following Brian's trials and tribulations with fuel for his Poppin.    He seems to have tried three fuels, methanol, ethanol and propanol.  It is hard to know from the chemistry how each would burn.  The chemistry is simple enough.  Methanol has one carbon, three hydrogens and an OH group.  It is usually used as antifreeze and quite poisonous.  I don't know the composition in the form it is normally sold, pure, or with water or other substances in a mixture.  I have no experience of burning it, but in a chemical plant it needs good fire protection.

Ethanol is the normal drinking alcohol.  Two carbons plus five hydrogens and an OH in the arrangement CH3 - CH2 - OH.  Apart from alcoholic drinks is is commonly sold as methylated spirits, a name that has always confused me as it is chemically ethyl alcohol and normally has five percent water.  It is sometimes called denatured alcohol, and I believe the denaturing agent is methanol, which makes it poisonous, possibly explains the name and sometimes it is coloured blue, both to discourage drinking.  Presumably only a small addition, as it does not appear in the list of ingredients on the label.  It is used as a cleaning agent and a stove fuel.  It is the one I use for my boilers, and also in my stove when I am camping.  My wife uses it as a window cleaner.  In all, we use quite a lot of it.  It seems to burn without much residue, though the mixture sold in these parts is water clear.  Ethanol mixes with water in all proportions, and that gives it a great safety advantage in our use of it for fuel.  Water will extinguish any ethyl alcohol fire, so you only need some water at hand in case any spilled fuel catches fire.  Not quite intuitive, as water should not be used to extinguish other hydrocarbon fires.

Isopropyl alcohol has three carbons and seven hydrogens plus that OH group that makes it an alcohol.  Usually sold as rubbing alcohol and as others have mentioned contains some oil additives  so as not to dry the skin.  I imagine these might be left behind when it is burned, so might foul wicks or fine jets, but again I have no experience of burning it.

Many of these differences have been pointed out by those commenting on Brian's thread.

I see no reason the basic carbon- hydrogen- OH compounds should not be quite similar in the burning characteristics, with the carbon hydrogen ratios causing variations in heat output and the CO2 and water content of the combustion gases.  But without direct experience, I don't know if they actually burn in a similar manner.  I would suspect the main reason for any difference might be the various additives included to make them better for the specific applications they are usually used for.  Another case where theory has to be backed up by practice.

Does anyone else have more information on any differences in the burning of these compounds?  Just something to think about while Willy gets back to us on those steam conditions.

Thanks for looking in,

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

 

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