Author Topic: Talking Thermodynamics  (Read 194516 times)

Offline MJM460

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Re: Talking Thermodynamics
« Reply #60 on: May 27, 2017, 01:18:24 PM »
Flywheels - a little diversion into physics.

Last time I started on flywheels.  The flywheel accelerates in response to a torque, meaning it spins at an increasing angular velocity, and in so doing stores energy.  When the torque produced by the engine is less than that absorbed by the load, the flywheel returns this energy by slowing down.  This enables a single acting engine to run, and reduces the speed fluctuations of a double acting engine that otherwise result from the normal production of torque.  The physics of this is described by the law of conservation of angular momentum.

We have an intuitive feeling for momentum in linear systems, that "something" which makes a body in motion continue in a straight line unless acted on by an external force (Newtons Law).  It can be quantified as the product of mass and velocity.  It has the units kg.m/s.  (In imperial units ft. lb(mass)/s.)  Velocity is a vector, and when we multiply by mass to get momentum the resulting momentum quantity is also a vector (with the same direction as the velocity).

In spinning objects, such as spinning tops, gyroscopes, bicycle wheels, and ice skaters, the analogous quantity is angular momentum.  It is quantified by multiplying the moment of inertia by the rotational speed.

But what is moment of inertia?  First it must be defined in relation to an axis of rotation.  Then, for a point mass, it is mass times (distance from the axis)^2.  It has units of kg.m^2.  So moment of inertia is proportional to mass, but the distance term means that it is also dependant on the distribution of mass, or shape.  The squared part means that shape is much more important than mass.  We can see that the typical flywheel, which has a heavy rim plus minimal spokes to keep it centred on the axis of rotation, and a small hub to fix it on the shaft, should have a large moment of inertia relative to its mass.  A solid disk would have much more mass (and bearing load), but only slightly larger moment of inertia than a spoked flywheel.

Next, we need to know the rotational speed.  No great mystery in that, but note that the units of angular rotation are radians, not degrees, grad, or even revolutions.   A revolution of 360 degrees is two times Pi radians.  Rotations have no dimensions, but are vectors and have direction.  Hence rotational speed is measured in radians/s, and has the units just "per second".

Now we can multiply moment of inertia (kg.m^2) multiply by angular velocity (per second) to get units of angular momentum as kg.m^2/s.  This is the quantity which tends to average out speed fluctuations.  Like linear momentum it is a vector.  The direction is conventionally defined as coinciding with the axis of rotation.  If you make the typical "thumbs up" sign with your right hand, when your fingers follow the direction of rim rotation, your thumb shows the positive direction.  (This could require a "thumbs down", so don't be insulted, it could be just someone trying to work out the direction of the angular momentum of a spinning top!)

Two interesting points to close on.  Angular momentum and torque are both vectors.  When torques and angular momentum interact, we can predict the resulting vector mathematically by performing a vector product multiplication.  The result has a direction which is determined by a right hand rule.  This vector multiplication predicts all the movements we see when a gyroscope reacts to gravity, or we hold a spinning bicycle wheel by its axle and try to move the axle in various ways.

Second, linear momentum and angular momentum are quite independent quantities with independent conservation laws, and an object can have both at the same time.  (There is a third conservation law which can also be useful to us, but back to that later.)

I hope this answers many common questions about flywheels.  Next time I will try and return to getting more out of our engines.

Thanks for following along.

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

Offline MJM460

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Re: Talking Thermodynamics
« Reply #61 on: May 30, 2017, 12:31:47 PM »
Flywheels (2)

Thinking about my last post, I realised that it might be good to put some numbers to a real flywheel before moving on, just to give a sense of proportion.  I calculated the mass of the flywheel on my little oscillator, nothing complex.  It is machined from a 2 inch bar, cleaned up to 50 mm diameter. 15 mm in the axial direction.  I left the hub and rim full length, and machined a recess in each side to leave a web 6 mm thick.  Nothing fancy, you guys are inspiring me to at least drill some lightening holes or even curved spokes.  But this was a few years ago.

I believe the material is cast iron.  I have used cgs units for such a small flywheel to avoid having too many decimal places in the numbers, but for any other calculations I need to do it in ISO metric or mks system.  I was interested in a few key results.

The moment of inertia turned out to be 537 gm.cm^2.  This is 80% of the moment of inertia for a solid disk, and 436 gm.cm^2 (or 83% ) of this is due to the rim which was 15 mm wide and 6 mm in the radial direction.

What if I had made it of steel?  Well the density of steel is about 7% greater than cast iron, and both mass and moment of inertia increase by 7%.

Brass would make more difference and you might like the appearance or rust resistance.  Brass density is about 17% greater than cast iron, and gives about 17% higher moment of inertia.

What if we want to save some weight?  Perhaps Jim is designing his steam powered plane.  The density of aluminium is only 38% that of cast iron.  So if we use the same design, mass and moment inertia will be 38%.  If the original flywheel was just enough, that would not do.  Can we make a lighter flywheel from aluminium if we modify the design?  The clue is in the R^2 term in moment of inertia.  If we increase the diameter to just 66 mm diameter compared with the original 50 mm) we will have an aluminium flywheel with the same moment of inertia as the original cast iron, but only 60% of the mass.

Now as a practical matter, I found the calculation was quite sensitive to even 0.1 mm change in the overall diameter, and I think we would all agree that flywheel dimensions are not that critical, clearly most are "adequate" rather than bare minimum.  If we want our engine to fly it would be worth experimenting to find the minimum required moment of inertia using cheap steel discs.  Then we could make a lighter flywheel to that moment of inertia by increasing the diameter and using aluminium.

I hope that is of interest.

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 #62 on: June 08, 2017, 12:37:58 PM »
More on producing rotation

Sorry to have been a bit silent for a while.  Family and other activities always need due attention.

I have also decided that this thread needs pictures, surprising that no one has mentioned it, but I think I have overcome some technical challenges, so here goes.

Figure 1 has an overall sketch engine of the simple vertical engine I am thinking of in my terminology and arrangement.  You can see why I was the engineer, not the draughtsman. 

I have included a sketch of the piston showing the important gas pressures on the top and bottom faces.  Remember I am using absolute pressures and I am assuming for the moment an atmospheric exhaust system.  We will discuss condensing later.  Hence the exhaust pressure is a little above atmospheric pressure to push the exhaust gas out of the cylinder.  You can see the relationship between the pressure on the piston faces and the resulting force on the piston.  The exact pressure is not important, and is varying from moment to moment anyway.

Figure 1 also includes sketches showing the forces on the wrist pin, and the triangle of forces which shows the relative size of the forces.  The horizontal force is not due to gas on the sides of the piston, but a result of the angle of the conrod. 

Finally, I have shown the torque calculation.  The force is the conrod load, and the moment arm (a) which causes the torque is shown in red.  Torque varies from zero to a maximum on the power strokes, and for a single acting engine is actually negative during the exhaust stroke.

Figure 2 shows the gas pressure, force triangle and torque for three important cases.

Detail (a) is for the power producing down stroke.  Piston force is downward.  Note that the conrod is in compression.  I have shown the rotation as anticlockwise.  This is the normal trigonometry convention, and means that your calculator gives the correct result for the parts of the circle where trigonometric ratios are negative.  Only important if you put these calculations into a spreadsheet, then increase the crank angle in increments, and have the computer recalculate for each angle through a full revolution.

Detail (b) is for the upstroke of a double acting engine.  Note how the moment arm is shown, and that the piston force is now upward and the conrod is now in tension.

Detail (c) is for the upstroke of a single acting engine.  As there is no steam pressure under the piston, the flywheel provides the necessary energy to push the piston upwards, and expel the exhaust.  Note that the conrod stays in compression for the upstroke of a single acting engine, compared with load reversal for a double acting engine.  This factor is an important consideration in the lubrication of the wrist pin and crank pin journals.  If you are familiar with larger full size engines, the pistons are much heavier in relation to the gas forces, and momentum considerations modify this simple analysis to some extent.

I know I have been labouring these points a bit, but this is the process by which the pressure energy in our gas produces force on the piston, which produces mechanical work when the piston moves in the direction of the gas force.  The force on the piston loads the conrod which gives torque at the output shaft. 

This is the process by which energy in the gas is changed to mechanical work.  This is how our engines really work!

Thank for looking in, more next time.

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 #63 on: June 08, 2017, 12:39:36 PM »
I had no luck attaching two figures, so here is the second

MJM460
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #64 on: June 11, 2017, 02:11:48 PM »
Completing the P-V diagram

I have included a idealised sketch P-V diagram, fig. 3, to illustrate the discussion.  Note I have assumed the pressure during admission to be constant, and I have rounded the corners to acknowledge the point that real valves take a finite time and rotation to open or close.

We have discussed the top boundary of the P-V diagram, shown in blue, admission and expansion of the power stroke, and also the lower boundary, the exhaust stroke, also in blue, but to describe the complete cycle so our engine can run continuously, we need to connect up the ends with the lines shown in red.

At the end of the power stroke, the cylinder pressure is still well above the exhaust pressure when the exhaust valve opens.  The remaining pressure expands into the exhaust system with that familiar chuff or beat of the steam locomotive in heavy load.  Of course when the valve gear is notched up to give earlier cut off, the beat is less distinct as expansion from an earlier cut off may be close to exhaust pressure when the exhaust valve opens.  From the thermodynamics point of view, the significance of this process, called the release, is that all the remaining energy in the steam goes out the exhaust and plays no further part in doing useful work.  When efficiency or water conservation is important, of course we can capture some of this heat in a feed water heater, and return it to the system to use in a later cycle.

At the end of the exhaust stroke, we have to get the cylinder contents back up to the inlet admission pressure.  There are two alternative processes possible, and in practice we probably get a little of both.  In the diagram, I have shown the exhaust valve closing a little before the bottom dead centre, resulting in compression of the steam remaining in the cylinder.  The energy for this compression of course comes from the flywheel, and so subtracts from the net output of the cycle.  Is there an alternative?

The alternative is to close the exhaust at bottom dead centre, and open the inlet valve with the cylinder pressure still at exhaust pressure, so that new steam expands into the cylinder and flows until the pressure equalises at the admission pressure.  We get a little more work output from the cycle, at the expense of using a bit more steam compared with the cycle in which we use compression.

So what is the difference?

The difference comes down to efficiency.  The energy in the steam which expands into the cylinder while it is still at low pressure is downgraded in terms of its ability to produce mechanical work.  It all goes into heat in the exhaust stream.  So there is extra steam consumption but no extra work output.  On the other hand, the energy which goes into compression remains mostly able to be converted back into work.  There are losses, but not total loss.  So a little less work output per cycle, but a steam saving which is greater than the work lost.

There is another more practical benefit to having some compression at the end of the exhaust cycle.  This is often described as cushioning, and is related to the need to stop then reverse the kinetic energy of the moving piston.  It may be even more directly related to softening any noise from slack in crank pins and wrist pins.  I don't know the full explanation, but the observation will have an explanation, whether I understand it or not.

Next time I will look at the more realistic P-V diagram that Maryak provided and how this differs from the idealised ones I have referred to so far.

Thanks for following along

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

Offline MJM460

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Re: Talking Thermodynamics
« Reply #65 on: June 12, 2017, 12:59:10 PM »
Introducing some reality.

Not so many comments these days, I hope I have not turned everyone off after the initial enthusiastic response.  I think we may now be beyond the tedious parts and things should move a bit quicker.

By now, it should be obvious that to get the most power out of our engine, we have to get the highest possible pressure to the face of the piston, and at the same time, the lowest possible exhaust pressure, so that we have the greatest differential pressure difference on the piston.

We have assumed an idealised picture of the p-v diagram.  But what happens in practice?

Maryak has kindly provided a p-v diagram (reply#53, page 4 of this thread) from a text book written for marine engineers, about 1916 I think.  It shows the diagram for both the top and the bottom of the piston.  As the textbook was intended to instruct, I assume it shows features, both good and bad which indicate the health of the cylinder.  So what can we see?  Let's look first at the diagram for the top, starting with admission at the top right, moving anticlockwise around the diagram.

We can see the admission process is not really constant pressure, however a drop from around 110 to 92 is probably a reasonable approximation as we will later see.  Then we appear to have cut-off and expansion, and the diagram again looks reasonable, ending with release, when the exhaust valve opens before bottom dead centre, so the cylinder pressure is right down at exhaust pressure when the return or exhaust stroke begins.  Finally we see compression to perhaps 60% of the inlet pressure.  The vertical section at top dead centre to the start of the down stroke probably implies steam flow in to raise the pressure up to the inlet pressure.  To those of you who have experience with a real indicator diagram, does that sound like a reasonable interpretation.

If we then look at the bottom trace, we have a different picture.  The admission starts at a pressure slightly higher pressure than for the top.  Then the expansion section requires some imagination to see, rather the pressure falls quicker than it did on the top diagram down to release.  Then a very flat exhaust pressure until the exhaust valve closes in time to give compression.  The exhaust valve seems to be quite good, as compression is a little better than for the top.  No doubt the student was asked to explain all the departures from the expected trace.  I leave you to think about what faults could cause the pressure trace illustrated, a bit of a detective logic puzzle.

We can see that the engineer did not use a planimeter, but a used simple arithmetic averaging process to find the indicated mean effective (differential) pressure in the cylinder and indicated power.  We can see the top has a higher mean effective pressure, 870, and the area of the diagram shows there are power losses on the underside of the piston which has an average of only 820, presumably psi.

Next time we will start to look at the issues which get in the way of achieving maximum differential pressure across the piston.

Thanks for dropping in.

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

Offline Steamer5

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Re: Talking Thermodynamics
« Reply #66 on: June 12, 2017, 01:24:21 PM »
Hi MJM,
 Still following & enjoying the read & explanations & learning stuff!

Cheers Kerrin
Get excited and make something!

Offline MJM460

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Re: Talking Thermodynamics
« Reply #67 on: June 13, 2017, 01:37:42 PM »
Hi Kerrin,

Thanks for your reply, I am glad you are still finding it worthwhile.  I will assume you speak for many.

I have been doing lots of calculations today and it's getting late but I will have more tomorrow.

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 #68 on: June 14, 2017, 02:10:52 PM »
Supply and exhaust pressure

It is time to ask what pressure we might have at the piston face.  It is obviously less than boiler pressure, but how much less?

On its way from the boiler to the cylinder, the steam must go through pipes, possibly a throttle valve, the steam chest, valves and the cylinder steam passages.  These items all involve pressure loss.  The losses are caused by wall friction in pipes, acceleration losses in entering the pipe end, and more losses when the steam expands into a larger chamber at the end of the pipe.  How can we get an idea of the magnitude of these losses?

If we look at pipe friction, there is a formula known as Darcy's formula which is used in industry to calculate pipe friction.  If we look at my little oscillator again, I have measured the steam flow by measuring the weight of water poured into the boiler before the run, and subtracting the amount I draw out with a syringe at the end.  I also note the time between the engine starting and stopping for the run.  It seems to be about 0.73 kg/hr.  Not much, but I am using a small Meths burner similar to that supplied by Mamod and other small engine manufacturers.  From this, Darcy's formula gives about 1.45 kPa per meter in a 5/32" tube (2.55 mm inside diameter).  As my supply pipe including a superheater is only about 500 mm, I think we can ignore this source of loss.  Alternatively compensate by running the boiler about 1 kPa higher.  Those of you who are familiar with metric units of pressure will appreciate that you would need a very accurate pressure gauge!  Clearly not worth doing a lot more maths to evaluate this more closely, but if you have any doubt, increasing the tube size to 3/16" reduces this loss to about 0.4 kPa/m.

The other losses occur each time the flow path changes cross sectional area.  As the mass flow is the same all the way along the path, changes of flow area involve changes of velocity.  Each time the velocity changes there is a loss of energy. 

To keep this post to a more reasonable length, I will look in more detail at these other losses, next time.

Thanks for dropping 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 #69 on: June 15, 2017, 01:47:32 PM »
Losses along the flow path

Last time I looked at friction losses in the steam line, this time the losses due to velocity changes along the path from the boiler to the cylinder.

We all know about a convergent - divergent Venturi and have probably heard of Bernoulli, and the law which says the pressure energy can basically be converted to velocity energy and vice versa.  However there are some very important qualifications, perhaps not so well known.  First the reduction in area at the entrance must be smooth, not too quick, and the entry must be well rounded.  Not exactly what we see at the boiler outlet or other points in the steam path where the flow area reduces.  Similarly, the exit from a venturi must be very gentle, diverging at about 15 degrees or less with the entry and exit very well rounded to minimise any generation of turbulence.  You see these shapes if you look closely at injectors used on locomotives.  They have to be very carefully made and appear to be a bit of a black art.  Again, nothing like the pipe entrance to the steam chest.  Each of the sudden area reductions or expansions involves an energy loss due to turbulence which dissipates energy in the form of heat, and reduces the ability of the fluid to do work by reducing the pressure.  On a well insulated pipe, (you do insulate your pipes, don't you?) conservation of energy applies and no work is done, but the energy simply goes through to the exhaust as a higher temperature.  So how do we evaluate these losses?

Essentially the losses are proportional to the velocity squared.  So we need to know the velocity in our steam pipe.  In metric units, the energy due to velocity is just v^2/2, to give N.m/kg, but as we conventionally use kJ, kilojoules as the unit of energy we need to further divide by 1000 to give kJ/kg.  Pressure energy can be found by dividing the pressure in Pascals by the density to give N.m/kg.  Because a pascal is such a small unit, (atmospheric pressure is about 101000 Pa) we generally use kPa, so dividing by density then gives kJ/kg which is directly comparable with the velocity energy.  Then by a little mathematical manipulation, we can show a pressure drop for a change of velocity.

I don't propose to provide all the maths though it is simple enough, as the result is enough to identify some conclusions.  For an exhaust system without condensing, the pressure is close to atmospheric, and steam density about 1.7 kg/m^3.    We can then calculate the pressure loss for a change in velocity.

I tried to insert a table, but it did not come across, but basically for velocity of 10, 30, 100 and 200 m/s, the corresponding velocity pressure is 0.085, 0.765, 8.5 and 34 kPa

Each time our steam enters a pipe, we loose pressure equivalent to the change in velocity and again when it exits into a larger flow area such as the steam chest.  Some extra losses around the valve, and through the valve, entry to the steam passages and again on exit into the cylinder.  Say at least six times the velocity pressure.  Clearly providing the valves open quickly to at least
 area of the steam pipe, these losses would indicate a velocity of around 30 m/s might be acceptable, but losses increase very rapidly above that.

I calculated the velocity for my little oscillator, and found 15 m/s for 5/32" tube, 8 m/s for 3/16" and 5 m/s for 1/4 " tube for the steam supply and about double that for the exhaust.

 I was expecting worse,  but these figures certainly seem to support using 5/32" tube as is typical for supply to these small engines, with a size larger, say 3/16" for the exhaust.  Of course the flow is doubled for a double acting engine, and my little gas burner provides about twice the heat of my Meths burners, so larger tubes are a good idea.  Personally I use 3/16" tube for steam supply and 1/4" exhaust but I have only built small engines so far.  For larger engines it is a matter of calculating the velocity for your steam supply and exhaust, using density from the steam tables for the pressure you intend, and keep the openings in your fittings and steam passages close as practical to or larger than the steam pipe internal diameter.  Ideally aim for less than 20 m/s.

The item I have skipped through quickly is the valves, so I will look at this in more detail next time.

Thanks for dropping in

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

Offline derekwarner

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Re: Talking Thermodynamics
« Reply #70 on: June 16, 2017, 04:52:37 AM »
Back on May the 15th of May...I offered

'I agree understanding what is happening with/within our model steam plants is essential' and spoke of using an inexpensive digital laser pyrometer to help achieve this

So all the lagged lines and temperatures seemed understandable or as expected until I got to the de-oiler  :Mad: ......

This proprietary built de-oiler unit steam inlet is a single 5/32" OD copper tube.......[disregard the second 5/32" inlet installed  for the blowdown of the displacement lubricator] .....the discharge from the de-oiler again a single 5/32" OD copper tube]

I adapted further 5/32" x 0.014 wall brass tubing interconnected with K&S telescoping brass 1/8" x 0.014 wall internal joiners and a lot of bends ....all the way to the chimney connection to atmosphere

The end result [pressure drop caused by my tube work ID and unintentional internal orifice plates :Doh: ] was causing the steam discharge to condense in the de-oiler as opposed to exit the chimney as steam.....

Attempting some calculations, leads me to understand that my engine steam discharge line sizes are also too small  :facepalm:

So the first modification is to increase the discharge tube size from the de-oiler top plate to the chimney top to 1/4" OD x 0.014" wall without any internal reduction influence....and then trial & measure for the reduction in condensate to water to steam ratio

Dependent on that trial, a second possibility is to increase the actual engine steam discharge tube line sizes and a new manufactured squareish box type de-oiler with similar 1/4" full flow inlet/s

[I have previously confirmed that lagged steam tube to the engine provide a ~~ 3 degree C + gradient over non-lagged tubes] 

Thanks MJM460.......a great thought provoking thread :ThumbsUp:

Derek
« Last Edit: June 16, 2017, 10:40:26 PM by derekwarner_decoy »
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 #71 on: June 16, 2017, 01:40:24 PM »
Thanks Derek for your feedback and for telling us about your adventures.  It is sad to find after all the work to make a neat installation as you have, that it does not work the way you expected due to something no one ever talks about.  It is the sort of issue that prompted me to start this thread.  We need to talk about thermodynamics and other obscure engineering subjects to avoid some of the issues you have found.

You did not say what size engine, but what I can see in the photo suggests it may be two cylinder with cranks at 90 degrees which suggests double acting.  Unless you are into real watchmaking, I have no doubt that 1/8" is too small anywhere in the exhaust steam system.  Probably OK for a water outlet, but then surface tension can trip you up in those sizes.

Instead of building a new separator, how about taking the exhaust from that "filler" plug, is it 1/4" or perhaps 5/16"?  I assume it is actually for emptying, so a removable section of the exhaust would give you access after a day's run.

Of course the practicalities of connecting the exhaust to an engine cylinder often preclude using a reasonable size pipe.  But a fitting which screws into the block on one end can be larger at the other.  The loss from one small passage right at the engine is probably acceptable.  Then continue in 1/4" or, for a bigger engine, perhaps 5/16" to the separator.  And even bigger if your model permits it from the separator to the stack.  Again, you can get away with smaller entry to the separator especially if it is tangential.  The velocity will give a swirl which helps separate the condensate.  Perhaps a topic for another time.

I remember your point about the radiation temperature instrument, and they certainly have their uses, especially for temperature difference comparisons.  I will put a little post on temperature measuring a little later.  (It is on my list.)

I probably put the emphasis on the wrong part of the maths in my last post and glossed over the useful bit.  You can calculate the pressure loss for each change in velocity using

Pressure loss = density x velocity squared / 2000

Pressure will be in kPa when you use m/s for velocity, and kg / m^3 for density.  The two is a constant in the energy equation what ever units are chosen, while the divide by 1000 converts pressure in Pascals to kPa. 

You need to devise a test run generally involving weights to determine your approximate flow rate.  For those less familiar with steam tables, they do not appear to contain density, but instead have columns for specific volume which is 1/density.  With flow rate, and density, you can easily calculate velocity for each tube size.  Not mental arithmetic, for me anyway, but computers are magnificent for such tasks.

I know the Bernoulli equation implies that this velocity pressure is converted back to static pressure  when the velocity reduces, but in practice this only happens in a well made  venturi, and you can estimate that you loose a velocity pressure at each entrance and each exit for your tubes.

I hope this lets you get things working well without too much trial and error.  Look for a velocity less than 20 m/s.

Back to valves next time.

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 #72 on: June 16, 2017, 02:08:15 PM »
still following along .....with the help of my scientific dictionary !!...interesting info here.....in the old days engines only had two pipes inlet and exhaust. this made model engines look less unclutered with sleek lines and uninterrupted views. later on with all the thermodynamic knowledge they began to become more of a plumbers nightmare !!....A comparison can be made with an old Morriss minor A series engine and the modern car engine ....lots of room in the engine compartment then  and now there is hardly any room to get a spanner in .......
.
« Last Edit: June 16, 2017, 02:15:02 PM by steam guy willy »

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #73 on: June 16, 2017, 02:19:36 PM »
Also can we talk about sound    ?  My morris Minor is very loud compared to a Rolls Royce so is some of the energy converted to sound...............?

Offline derekwarner

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Re: Talking Thermodynamics
« Reply #74 on: June 16, 2017, 11:24:26 PM »
OK thanks MJM460....

I am awaiting the supply of 1/4" female copper long radius bends for the new exhaust trunking exiting the de-olier........[1/4" OD x 0.014" wall full flow]

Will report back on physical results with temperatures...however the final result  = volume of water consumed : to exhaust steam condensed & retained as water in the de-oiler......

Derek
« Last Edit: June 17, 2017, 10:49:30 AM by derekwarner_decoy »
Derek L Warner - Honorary Secretary [Retired]
Illawarra Live Steamers Co-op - Australia
www.ils.org.au

 

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