Talking Thermodynamics

[MJM460]

Some more diverse issues

Hi Willy,  I have gone back and checked my list of your questions and still have a few from post #92 and Jo's post #93.  I will try and cover those, then hopefully I can get crack to valve events on my little oscillator.

You asked about testing at altitude.  I think it has been indirectly answered, but to be clear, testing is about the strength of the boiler to contain pressure, and not about the boiling point of water.  This means we are only interested in the difference between the inside and outside pressures.  It does not matter whether you test it up the mountain or at the depth of the ocean, so long as you measure the difference between the inside and outside pressures.  We have covered the pressure gauge so you know it measures the difference between the local atmospheric pressure and the measured pressure, so it does just what we want.  You can confidently test up the mountain.

You asked about coal and heat to boil water.  The heat released when burning coal, or any other fuel for that matter, is proportional to the mass of coal. You will find the value called calorific value expressed as kJ/kg of coal.  I haven't spoken about joule as a unit, a bit later, I will.  However, a Joule (J) is a measure of energy equivalent to a Newton meter of mechanical work.  We cannot convert all the energy onto work, or even most of it, but we can use the same units to measure it.  In imperial units, the calorific value of your fuel would be expressed in Btu/lb, a British Thermal Unit being a measure of heat energy.

The calorific value of the coal is not dependant on the pressure, so you need the same mass of coal to get the same heat when you burn it up the mountain.

Now, burning coal requires a certain mass of oxygen to burn a given mass of coal, in the same way as any other fuel.  At sea level, the density of air is well known and depends on temperature.  From density, expressed in kg/m^3, we can calculate the volume of a kg of air and hence the volume required by a mass of coal.  In practice, it will not burn very completely, unless we provide excess air to ensure that every molecule of carbon in the coal has a good chance to meet an available oxygen molecule.  A bit like you need more boys than girls at a dance to ensure that every girl has a good chance of meeting a suitable boy.  About 15% or 20% excess air might be a good place to start.  Too much means heat energy is lost up the stack in the oxygen that was not used and the accompanying nitrogen.

Up the mountain two things happen as you know.  The pressure is lower than at sea level, and so normally is the temperature.  The density depends on both, so both have to be taken into account. The end result is a lower density than sea level, so a greater volume of air is required to carry the same mass of oxygen required for our coal.  Greater volume involves more resistance to flow, so you would need a tall stack, or a very good blower to get enough draft to draw this extra air through the coal bed.  Alternatively you could use a forced draft, perhaps a fan, probably not a compressor, to push the air volume through the bed.  Of course if you are on Mt Everest, you would probably have oxygen bottles in your kit.  If you had enough extra over what you need to get back down, you could get the required mass of oxygen from a much smaller volume of air.  Assuming of course that it is enriched oxygen in your bottles, not just compressed air.  I assume it would be.

I am not a combustion specialist so I do not know how the altered volume of air in the bed would affect combustion.  It could be good or bad, I really don't know, so the combustion efficiency I cannot answer.  I hope someone else can join our conversation to help us both understand that one.  Also, I think there are factors both increasing and reducing heat transfer.  Again I am not sure which one predominates to advise on whether you would need more or less heat transfer area.

You asked about whether it matters if you fill the boiler at sea level (and, I assume tighten the filler connection) then take it up, or fill it at the top.  This is a much more interesting question than I thought when you first asked it. 

Had to attend the grandchildren's school concert tonight.  It is quite late, so this and your last question will have to wait until tomorrow.  Wonderful to see these young people developing their talents.  There will be good musicians for us to listen to well into the future.

I hope this is filling some of the gaps and we are building a firm foundation for the rest of our knowledge base.

Thanks for following along

MJM460

[steam guy willy]

A small point of interest ..the local boiler inspector has said the the regulations state that when testing a boiler for leaks the water must be between 7 and 21 degrees centigrade. this is when the boiler is pumped full of water to check for weeps.  still following along ..interesting stuff ......

[MJM460]

Our mountain top experience.

Hi Willy,  glad you are still following along.  I really appreciate your insightful questions that prevent me from glossing over details.  I don't know which regulations your inspector is talking about, or what is behind them, other than the fact that pressure will change with temperature, and so a steady temperature is necessary for constant pressure.  In fact, at least a tiny bubble of air is helpful in your boiler when testing, and almost impossible to avoid, otherwise it would be hard to keep the pressure steady through a test.  I do know that in industry, I have been involved in pressure testing in locations where, if we had to wait until the temperature was below 21 C, we would have had to wait until the next ice age.  Well, perhaps at night we would get there, at least in winter, but not for long enough for a big test, and it is very difficult to see any leak at night.  I have also been involved with tests where we would have had to wait until spring if we wanted to test above 7 C, we had to use antifreeze for the test fluid.  In fact, I had to purchase a whole train load, 7 full tanker cars, of antifreeze, and if my memory serves, we still had to get a top up.  I suspect there was a shortage of antifreeze in Canada that year.   Not so far from Brian's territory.  No more war stories, but best to comply with the inspectors requirements.

Back to your question about the difference between filling at sea level or at elevation for our mountain top operation.

Let's first think about the basic difference between the two cases.  The significant issues are the  pressure and temperature.  At sea level, let's assume the pressure is 100 kPa, we have a low pressure system passing over, and the temperature is 30 deg C.  (It is hot at the foot of the mountain).  We fill the boiler to the proper level, and seal the filler plug.  Our steam tables tell us the vapour pressure of the water at 30 C is 4.246 kPa, and of course this is absolute.  The total pressure is 100 kPa, atmospheric pressure is by definition absolute, so the air pressure in our boiler is just under 96 kPa (by subtraction).  We take it up the mountain, the water and trapped air do not know or care what the pressure is outside.  We use an absolute pressure gauge to measure the pressure inside the boiler instead of our standard bourdon tube type, and yes such things exist, so please accept it for now.  What pressure do we expect see on our gauge at the mountain top? 

We might be surprised to find our absolute pressure gauge, when we reach the mountain top, reads only a little below 89 kPa.  Of course our gauge cannot tell whether the water vapour or the air pressure or both have changed, we need our steam tables and some calculations to help with that.  We only know that reading is the total pressure in our sealed boiler.

Our thermometer tells us that the temperature has fallen to 5 deg C.  Just as well it is not below 0!  We do not want freezing to complicate things.  The steam tables tell us that at 5 deg C, the water pressure will be 0.872 kPa.  Near enough to 1 kPa, unless you are Mr. Keenan or Mr. Keyes.  The air mass in the boiler has not changed, and the volume of water has changed but by an insignificant amount.  However the fixed mass of air in the boiler has cooled down along with the water. 

We know that the pressure of a gas at constant volume is proportional to the absolute temperature.  The absolute temperature is temperature in deg C plus 273.  So at sea level, the absolute temperature was 273+30=303 K.  (That is K for Kelvin, and conventionally used without any degree sign.). Up the mountain, the absolute temperature is 273+5=278 K. 

Now we can calculate the air pressure in our boiler as 96*278/303=88 kPa.  The total pressure must therefore be 88 from air plus a touch under 1 for the water, or 89 kPa, just as read on our gauge.  It may even be a little less, as there is still plenty of space between those water molecules, and some of the air molecules which hit the water surface will not bounce back, but will stay in the water.  We find the solubility of air in water is higher at lower temperature. 

Next time we will light the fire, and heat our boiler slowly while we watch what is happening.

Thanks for following along

MJM460

[MJM460]

Following on from my last post.

As we heat the boiler, (it is well insulated, so no miscellaneous heat losses,) we see the pressure and temperature rising.  There is no problem here, as our safety valve, properly set at sea level, looks only at the difference between the internal and external pressure, which is exactly what is required to protect the boiler from over pressure.  We can be confident that it will lift to protect the boiler if required.  (No ice remember.)  It does not matter if it is a spring type, or weight and lever.  (Please ignore the buoyancy effects of the weight in the rarified air if you have a weight type, they will be tiny.)

When it all gets up to 30 deg C, the temperature we had at sea level when we sealed the boiler, we find the pressure is back to 100 kPa, just as it was down there.  The air pressure has increased, and so, independently, has the water vapour pressure.  If we could see in sufficiently clearly, we might see some small bubbles caused by that extra air that dissolved being driven out.

Next we pause at 95 C, the temperature that our water boiled in our earlier saucepan experiment.  The absolute pressure is now about 202 kPa.  This means we have 202 - 84.55 = 117 kPa difference between inside and outside the boiler.  No problem so long as our safety valve setting is above 117 kPa.

Similar drill, at 95 C the water vapour pressure is 84.55 kPa.  The air pressure that was 96 kPa way back when we sealed the boiler at seal level and 30 C is now 96 x (273+95)/(273+30) = 116 kPa.  So the total is 84.55 + 116, say 202 and remembering that there will be even less air dissolved in the water at this temperature so our calculation might be a little low.

Is it boiling this time?  If we put a screw driver blade to the boiler and the handle against our ear, (being careful not to singe our hair,) it certainly sounds quiet.  If we could see inside sufficiently clearly, we would see some more of those small air bubbles as the air that was dissolved, even at 30 C, is driven out, but no boiling.  Something is different here from when we boiled the kettle with a vented lid.

In case you are worried, we can continue heating until the safety valve lifts without any problems,  good idea to test it while we are watching closely in case it is stuck, and we see the pressure continues to rise, due to both the air increasing in temperature, and due to the water vapour pressure increasing with water  temperature.  Boiling will only start when the safety valve lifts, or we open the stop valve to our engine.  So we have time to think about what is going on.

Remember the description of boiling from our saucepan experiment.  Boiling occurs when the vapour pressure of the water exceeds the total pressure at the liquid surface.  (That should be a bit clearer description of boiling than I used before, all this close looking has helped me clarify things in my own mind, particularly the role of the air in the boiler.)  In our vented kettle, the pressure is fixed by the vent to atmosphere, and the water vapour pressure soon exceeds that when heated.  Any vapour generated expands and results in a large volume of steam, which issues from the vent, or spout whistle, and the air is inevitably entrained in the flow, so is soon lost to the kettle.  So the pressure does not increase.  The turbulent boiling is not anything like equilibrium.

In our sealed boiler, conditions are close enough to equilibrium.  As the water evaporates to keep the vapour in equilibrium with its liquid at the surface, the air pressure, is not only still there, it is also rising with temperature.  This is where Dalton's law of partial pressures comes in.  It says that in a mixture of gasses, each acts on its own, as though the others were not there.  And this is close enough to what happens until very high pressures.  Essentially, while the air is still there, the water vapour pressure cannot exceed the pressure at the surface, and even with out the air, it can only equal, not exceed the surface pressure.  Water cannot undergo that huge expansion into vapour which causes the turbulence we call boiling, the pressure just continues to rise until something changes, preferably nothing more than the safety valve lifting, or the engine throttle opening.

As soon as some mass escapes the fixed volume of our boiler, the pressure must decrease a little.  In a fixed volume, the pressure is proportional to mass.  Some extra water will evaporate to replace the lost steam, but so long as the total pressure is above that equilibrium vapour pressure for the water temperature, no dramatic boiling happens.  Remember we are proceeding slowly here, to trying to stay close enough to equilibrium.  As soon as the pressure drops below the vapour pressure, some water will evaporate which tends to maintain the pressure, but we keep letting some out.   The remaining air is not very important, and some of it is inevitably entrained in the escaping steam, but is not replaced, so soon we are only dealing with water alone, and steam tables tell us the relationship between temperature and pressure, which is now totally water vapour pressure.

A bit of extra heat, and the vapour pressure will exceed the pressure at the surface, and boiling commences.  It is not very dramatic, as we know from every time we start one of our boilers with air and water sealed inside, the same processes happen.  The only difference is the precise starting conditions.  The water which evaporates, absorbs the latent heat from the remaining liquid, cooling it, so only a very limited mass can evaporate, limited by the heat coming in from our fire.

I do seem to use a lot of words in answering these questions.  I hope it is making sense.  But I will wait until tomorrow to talk more about equilibrium, and the other half of this double banger question.

MJM460

[MJM460]

Equilibrium

The issue of equilibrium has been mentioned often, but perhaps its meaning and importance not clear.  Not all processes occur in equilibrium.  As I write this, I am also responsible for supervising the making of soup for lunch.  In the pot, in addition to uninteresting fillers like pumpkin and carrot, etc, there is some water in the bottom, placed there to help conduct the heat so things don't burn, and also some ice, from frozen stock added to help flavour, The impurities in the stock mean I don't really know it's melting temperature, but close enough for our purpose.

But food!  That should provoke some interest.  I notice that the water has begun to boil, (the lid lifts as necessary to prevent any pressure increase) so there is steam (water vapour), water and ice all in the pot, at the same time, along with some unknown quantity of air.  Now that interests all thermodynamicists, as we all know it is only possible to have water, ice and water vapour at the same time in equilibrium at a specific water vapour pressure and temperature, called the triple point, about 0.01 deg C and 0.6113 kPa, (if that number of decimal places is consistent with the term "about").  Now my thermometer tells me the temperature is hovering around 99 C, (you do monitor your soup temperature don't you?) and my steam tables tell me therefore, the vapour pressure should be close to 100 kPa, well above that triple point pressure.  Of course, there is also air, as my pot is not sealed.  This all  tells me that the system is not in equilibrium.  Equilibrium requires no turbulence and also that the temperature be uniform throughout, which it cannot be with ice, water and water vapour all at the same time, except at that one special temperature and pressure.  Any two, but not the whole three.  And equilibrium is a necessary condition if we are comparing our process with those ideal processes such as adiabatic, isothermal, and similar.  If we are nowhere near equilibrium, our actual results may be very different from our calculations.

But mmmm! That soup smells good.  To make the soup a bit quicker, I started with the stove on high (setting was 9) and watched carefully while everything heated up and the water started boiling.  Then I turned the stove down to only 2, which was enough to maintain the temperature just simmering, with a little excess heat to just evaporate a small amount of steam to cook and soften the vegetables which were not submerged in the water.  No wild boiling or drama, just a wisp of steam from the lid vent, probably not enough to reliably remove all the air, just enough to limit the steam lifting the lid to an amount I could accept.  This keeps the temperature relatively uniform, rather than burning the vegetables where they touch the pot.  Very efficient use of steam to carry a lot of heat from the bottom of the pot to where it is required for the cooking, unlike our small engines which convert only a tiny portion of the heat to the work that we require.  Of course, the rest of the heat is mostly still contained in the exhaust, and could very well be used to cook the vegetables, if you don't mind a bit of oil flavouring.  This is called combined heat and power, or sometimes cogeneration.  And of course some of the heat is lost to the atmosphere through our cladding.

So after a little thermodynamics lesson in the soup, it's  time for a break for lunch.  Next time, I will think about what happens in Willy's alternative scenario, carrying our boiler empty, and filling it at the top of the mountain from a handy stream.

I hope everyone is still with me,

MJM460

[MJM460]

I had hoped to continue today, but my post is not quite ready, so I hope to continue tomorrow.

MJM460

[MJM460]

We fill our boiler at the top of the mountain.

A bad day yesterday.  A beautiful page of writing, full of calculations, all carefully checked, but in the end it did not clearly answer the question.  The knowledge base equivalent of the beautifully machined part that does not fit.  Cutting a second time did not help, and it was all consigned to the bin.  So here goes with a second attempt.

I believe I have covered the first part of Willy's question, what if we fill (and seal) our boiler at sea level, then carry it up the mountain to operate, even if with a few too many calculations.  Now the other part, is it different if we carry the boiler up empty and fill it from a convenient stream at the top? 

We begin by looking at our starting conditions for this case.  It is a way back, so I will remind you that at the top of the mountain, the atmospheric pressure was only 84.55 kPa, remembering that atmospheric pressure is always expressed in absolute pressure units, whether it be measured as 14.7 psi or 101.3 kPa, or some other value as in this case.  Also  we found that water in our vented kettle boiled at 95 deg C.   More recently, we decided the air temperature at the mountain top was only 5 deg C, definitely a bit chilly unless you are in Barrie in early spring, when any plus temperature means it is time to put away the scarves and heavy jackets.

When we fill our boiler, both the water and air will be at 5 deg C, and we assume the metal boiler shell, no longer in our pack, has also cooled to 5 deg C.  The absolute pressure will be 84.55 kPa as before, then we install and tighten the plug and connect our absolute pressure gauge.  At 5 deg C, the water vapour pressure is 0.872 kPa as before, so the air pressure in our boiler is 84.55-0.87= 83.68 kPa.

Now you might have noticed that I have jumped around a bit on the number of significant figures and accuracy.  A barometer is in fact an absolute pressure gauge and is calibrated in hectopascals, (10^2 Pa, or hPa, it seems Mr. Apple does not know they exist!)  Standard atmospheric pressure is 1013 hPa, and that fourth significant figure is quite significant in determining the weather we will experience as the highs and lows of our weather pattern pass over.

You will also notice that I have not bothered to look at humidity.  Despite its "couldn't care less" attitude to air molecules, water is very inclusive about water molecules.  They are all treated equally, whether they arrived with the air as humidity, or they evaporated from the water.  The vapour pressure of water is only dependant on the total number of molecules in the space, which in turns depends on the temperature of the liquid at the surface.

Our operating conditions are dictated by the boiler design which is determined by the difference between inside and outside pressures.  Now, our standard pressure gauge measures just this very difference.  So up the mountain, we still operate to the gauge pressure, not the absolute pressure.  The same gauge pressure at the top of the mountain, where the atmospheric pressure is lower than at sea level, means the absolute pressure in the boiler is lower, and hence the boiling temperature of the water (once all the air is expelled) will be lower.

The other difference will be the starting conditions when we are up the mountain.  At sea level the air density is a little more, so there is a little more mass of air in the boiler which will add to the water vapour pressure so changing the total pressure in the boiler until it has all gone out mixed with steam production.  Also, air, water and the boiler mass will be sea level temperature of 30 deg C.

At the mountain top, the air density is less, and the temperature is less.

Because of the lower density, if we fill the boiler at the mountain top, we will trap a smaller mass of air when we insert the plug.  This will always exert a lower partial pressure contribution to the total pressure than the greater mass we trapped at sea level.  However, once we reach our operating pressure and allow a little steam to escape, the air will be mixed with the steam and soon gone.  Then both cases are the same.  Boiling commences as soon as the vapour pressure of the liquid exceeds the total pressure at the liquid surface.

The temperature effect is unchanged between the two cases, as we have already noted the boiler and its contents will be at the same temperature whether we fill at sea level or at the top. So our starting conditions at the top of the mountain are 5 deg C for both cases and slightly less air partial pressure for the case when we fill at the top.

I hope that answers that question, so next time on to the issue of heat.

MJM460

[MJM460]

A question of heat

Hi Willy,  now for the final question to the list I made from your earlier posts, the question of heat.

 It was only one word in your original question, I almost missed it.  I am assuming the implied question was, "Is there a difference in the heat required when we operate the boiler up the mountain compared with sea level?"

I will work with the operating conditions we have already chosen, 275 kPa(g) as normal operating pressure.  This means a slightly lower absolute pressure on the mountain than at sea level, but the boiler has been designed for a difference between inside and outside pressure, not an absolute pressure.  Our standard pressure gauge provides the right measure of operating pressure, both on the plain and on the mountain.

We can determine the required heat input from the starting conditions and our operating conditions, using the law of conservation of energy and the steam tables.  Conservation of energy is the generalised form of the first law of thermodynamics.  I am only considering the boiler, no superheater for simplicity.  It will adequately illustrate the principle. 

Heating the water from our starting conditions until it reaches our operating pressure is a constant volume process in our sealed boiler.  A feature of this process is that no external work is done.

Careful application of the first law of thermodynamics reveals that for a constant volume process, the heat input required is equal to the change in internal energy.  Now the steam tables conveniently tabulate values of internal energy over the range we are interested in, so perhaps I had better briefly explain a bit more about steam tables.  You might have a copy in a thermodynamics text book, or you can purchase them as a separate booklet.  Or you can, as usual, ask Google.

Steam tables have two basic sections.  The first is just a solid block of figures that apply to the two phase range of conditions, that is when you have both liquid and vapour in equilibrium, and is the section used for boiling and condensing conditions.  Even this is divided into two parts.  One part has the temperature in the first column, and corresponding equilibrium pressure in the second.  The other part has pressure in the first column and the corresponding equilibrium temperature in the second.  Use the one that best suits your particular conditions, in particular whether you know the pressure, and want to know the temperature, or do you know the temperature.

The other part of the steam tables looks like a number of separate small tables.  This section covers the superheat range.  More about that section when we talk about boilers.

We will use the first section, called the saturated steam table, as we have both liquid and vapour in our boiler.  After pressure and temperature, the  next two columns show the specific volume of liquid and the specific volume of saturated vapour.  Specific volume is simply the reciprocal of density, and I assume the reason for the convention is because when calculations were done by hand, specific volume involved less division, so was easier to deal with.  Specific volume is the volume in cubic meters occupied by 1 kg of fluid.  You can see the specific volume of the liquid is very close to 0.001 m^3 or 1 litre per kg as we all know, while as vapour, it occupies a much larger volume and you can see the expansion ratio in these two columns.

Then we have normally have three columns for internal energy.  The first, Uf, is for water just about to boil, referred to as saturated liquid.  The third, Ug, is for saturated vapour, vapour when the last of the liquid has just evaporated, before superheat starts.  The second column, Ufg, is the difference between the two, which saves a step in many calculations, again for the days when calculations were done by hand.

At sea level, we sealed our boiler at 30 deg C which is our starting condition.  We could look up the internal energy of water at 30 deg C,  and again at our operating pressure, but this is where things get complicated!  Some of the operating pressure is due to the remaining air, so the water is not yet up to operating temperature.  It will not be until we let the air out, and we will also let out steam in the process.  Once we let some steam out, we no longer have a constant volume process, so internal energy no longer tells us what we need to know.

It all makes a precise calculation quite difficult. So I will try and answer your question a different way that I hope will tell you what you want to know.

I think that is enough for one session, so back to this question again tomorrow.

I hope it is all making sense so far.

MJM460

[MJM460]

Continuing with the question of how much heat is required to boil water.

My first application of conservation of energy to this problem led to a further difficulties when my initial assumption of a constant volume process fell down.

Let's start again, this time looking only at the water.  Starting at 30 deg C, we gently heat our boiler until the pressure reaches our operating pressure of 275 kPa(g).  Now we let out a little steam air mixture to control the pressure at 275 kPa(g).  We no longer have a constant volume process, as our steam is expanding into our outlet pipe.  If we hold it all at our selected operating pressure, we have changed to a constant pressure process.  We can no longer rely on the internal energy to solve our problem, we have to go back to the first law of thermodynamics and apply it to a constant pressure process. 

This time our text book introduces the concept of enthalpy, one of your magic words that so far I have managed to avoid.  The first law analysis tells us that for a constant pressure process, the heat input equals the change in enthalpy!  Again, in most thermodynamics texts if you want to look at exactly how that is determined from conservation of energy.

So I am going to have to try and explain enthalpy in one easy lesson.  That will test me, but let's have a go.

Many of the thermodynamic problems associated with both gas and liquid are solved by applying those basics laws of physics, conservation of energy, and conservation of momentum.  I have referred to those previously.  We start with the three basic properties that we can measure, pressure, temperature and specific volume (or density if you prefer), then apply conservation of energy and/or conservation of momentum to find the solution.  When we do this, some combinations of the measured properties come up often.  Specifically, the combination U + p x v.  Internal energy plus pressure times specific volume.  It turns out that this combination behaves very like another property, but we have no gauge to measure it directly.  Steam can have the same value for U+pv for the same mass when evaluated at different pressure or temperature.  Some processes can proceed along a path on a pressure-temperature diagram, along which the value of U+pv is the same all the way.  Just as we can proceed along a path where the pressure, or the temperature (but not both) are the same all the way.  Hence this combination of U+pv behaves like a property, and is given the name enthalpy, and the symbol, h.  Enthalpy, like internal energy, has the units of kJ/kg and looks rather like a particular sort of energy.  It occurs so often, and is so useful that the next three columns of the steam tables list the value of enthalpy in a similar format to the columns for internal energy.  In case you are wondering, it occurs frequently in processes involving work at an external boundary, such as in our engines.  I hope that is sufficient for the moment.

So back to the question.  If we control the pressure in the boiler as the steam escapes, it becomes a constant pressure process which we can analyse using the enthalpy columns of the steam tables.  To find out how much heat is required to heat our water from the cusp of boiling until all the liquid is evaporated into steam while we let out some steam and air to hold the pressure constant, we can simply look up the change of enthalpy.  Furthermore, it turns out that we can use change of enthalpy right back to our starting point, even though the first part of our heating is really constant volume.  This conveniently avoids the issue of just how we changed from constant volume to constant pressure.

If we continue just looking at the water, we can now start at 30 deg C, and look up the enthalpy for water, hf = 125.79 kJ/kg.  We can then look at the enthalpy of the saturated vapour at our operating absolute pressure of 375 kPa, (275 kPa(g)) and find hg = 2735.6 kJ/kg.  Then the heat input required by the water is 2735.6 - 125.79 = 2609.8 kJ/kg.

Now if we go to the mountain top we do the same analysis with two differences.  First the starting temperature, and hence the enthalpy of the water, is lower at only 5 deg C, and second our absolute pressure at 275 kPa(g) is only 359.55 kPa absolute compared with 375 kPa at the foot of the mountain.

Now starting at 5 deg C, we can see the enthalpy of the saturated liquid is just 20.98 kJ/kg.   Our operating pressure of 359.55, say 360, is close enough to the saturation pressure for 140 deg C.  You can see hg for the steam released from the boiler will be 2733.9 kJ/kg.  When we insert the figures and do the subtraction, we see that we need 2733.9 - 20.98 = 2712.9 to boil our water on the mountain.

Comparing the figures, we see that we need 103.1 kJ/kg extra to boil our water at the mountain top.  That is interesting, but how did it come about?

If we look closely at the figures, we see that we needed 104.8 extra to heat the water from 5 deg C to our mountain top boiling point of 140 deg C, while the enthalpy of the steam at the lower absolute pressure on the mountain is only 1.7 less than it was at the higher absolute pressure on the plain.

My conclusion is therefore, that we will need more heat to produce our steam at the mountain top, and the reason is almost entirely due to the lower starting temperature.  The slightly lower absolute pressure at our operating gauge pressure makes a difference of only 1 part in about 2600, which is well beyond the accuracy of most of our experiments, while the extra heat to get from 5 to 30 degrees, (104.8 kJ/kg) is about 4% of the heat required to boil from 30 deg C.

A small boiler like we might use for these experiments might contain about 1 kg of copper.  This would require 42 kJ/kg to heat from 30 to 140 deg and about 52  to heat from 5 to 140, thus adding about 12 kJ/kg to our extra heat requirement, and we need to allow a little for the lagging.  The other losses will be somewhere near proportional to the heat required by the water.  On this basis the total heat required on the mountain would be not much more than 5% more than on the plain.

The differences we have found may not seem very much. They could have been made to stand out a bit better had I continued assuming you might want to climb Mt Everest.  But I am trying to stay practical, and let's face it, it's crowded up there these days, unlike when Sir Edmond Hillary did the climb.  It is unlikely that the crowd would be willing to hang around, stamping their feet to keep warm, while you carefully measure your boiler temperature!

Next time, back to my little oscillator and its exhaust port, I hope it will have something for everyone.

Thanks for hanging in there,

MJM460

[derekwarner]

.....'Next time, back to my little oscillator and its exhaust port, I hope it will have something for everyone'

Looking forward to the exhaust lesson MJM.....this is where I am currently chasing my temperatures OK ...however without clear resolution 

Derek

[MJM460]

Returning to talking about engines.
 
Hi Derek, glad to have you still looking in.  In your previous posts, you were recognising that your exhaust piping was too small and were awaiting parts, which I assume have now arrived.  I assume that "still chasing temperatures" means you might be trying to condense the steam.  Or are you just trying to collect the oil?   Or are you trying to achieve vacuum conditions for your exhaust?  Please let me know which.  First, two posts on oscillating engines, then I will go on to exhaust.  I am sure that through that discussion, together we will be able to solve the problem.

 Those questions and comments by Willy and Jo, were deceptive in their depth, and quite a challenge to answer clearly.  I hope that I have done them justice.  Some of the issues will be discussed further when we come to boilers, and will also help our discussion of condensers.  They have given me the prompt to deal with some basic issues before I go too far.  Thank you both, and thank you to the others who have contributed by asking questions or commenting so far.

Back in post #90, I had been talking about valves and their part in conducting the sequence of processes that enable our engines to run in a continuous manner.  I had looked at the requirements for the valve opening to match the volume swept by the piston.  It turns out that the inlet valve opening matches quite closely the opening required, based on looking at required velocity in the port to maintain pressure as the piston goes down from top dead centre.

The real issue comes with the exhaust valve opening.  While the piston is slowing in its decent, the piston is full of steam which has done its work and must be expelled in the exhaust stroke and this is the most demanding event from the capacity point of view.  The slide valve is not too bad, with wide ports with sides parallel to the edge of the valve plate, so opening occurs quite quickly, and allows rapid exhaust of the pressure in the cylinder.  But looking so closely does show the advantage of those wide straight sided ports typical of a slide valve engine, in case you are tempted to just drill round ports.  A piston valve has a similar characteristic, with the ports wrapping around the valve.

That brings me back to my little oscillating engine.  Most of us have built, or contemplated building at least one of those, or at least seen them in action in toy engines like the Mamod range and others, excellent models which gave many of us our first hands on experience of steam.  I still have the one so much enjoyed by my brother and I.  Of course in those days, real trains were hauled by those magnificent steam locomotives, with the associated sound and billowing steam.  And we got coal specs in our eye when we put our heads out of the windows, which actually opened.

In an oscillating engine, also known as a wobbler, the valve port is formed by the overlapping portion of the drilled ports, one in the cylinder and one in the port block on the engine stand.  They are generally located so the openings do not overlap at all at top and bottom dead centre.  As the engine rotates, the cylinder port starts to overlap the port in the block on the stand to start admitting steam.  A typical design has them finally fully overlapping at the maximum cylinder movement, about 80 deg of rotation, after which they start closing again.  There is some variation, sometimes planned and sometimes due to accuracy of drilling in the required location.  There is only a momentary complete circular opening.  For most of the revolution, the opening is only the overlapping section of the drilled cylinder and steam ports.  In case the words are not clear, I have attached two sketches.  One shows the general port layout, the other, four steps in the port opening, the open section filled in red.  Once maximum opening is reached, they progressively close until bottom dead centre.  The exhaust port opens in a similar manner on the upstroke.  With a little trigonometry and some circle geometry in a spreadsheet, we can calculate the actual opening at each point of the engine revolution.

It should be obvious however, even without calculation, that just when the engine needs a large exhaust opening to quickly exhaust the steam, the port is nearly closed, and only slowly opens as the exhaust stroke progresses.  Consequently, the piston starts its return stroke for the exhaust with significant pressure in the cylinder, and the port is not fully open until the piston is over half way back to the top.

We have already seen that pressure in the cylinder on the exhaust stroke actually works against the piston movement, subtracting from the work done in the power stroke and reducing the output of the engine.  In a single acting engine, the work necessary to drive the exhaust stroke comes from the flywheel, and in the process, the engine slows during each exhaust stroke.  Despite this, the engines do run, and quite well, though they do not handle much load.  If we could allow the cylinder to exhaust freely, the engine would be considerably more capable.

Next time, I will look at two possible solutions that have been proposed at various times to overcome this problem.

MJM460

[steam guy willy]

Hi, still following along........one of my favourite sayings is : the adiabatic enthalpy : principal  but i don't know if you can use the two words together !! It does however impress lesser mortals !! One question on a more practical note is ..Where is the best place to introduce the feed water into the boiler ??
Or is this a topic all in itself ?   also rather then a round hole for the Wobbler port would it be beneficial to have an oblong hole so as to introduce cut off  ?? this is also a whole new subject possibly !!

[Maryak]

.Where is the best place to introduce the feed water into the boiler ??

In full size practice, feed water is introduced via internal feed pipes which evenly distribute the water along the length of the steam drum in watertube boilers. In in firetube boilers there is a distribution box fitted in the middle of the shell which acts as a form of feed heater. The idea is to reduce the effect of the difference in temperatures between the boiler water temperature and the feed water temperature.

Regards
Bob

[derekwarner]

....."Or are you trying to achieve vacuum conditions for your exhaust?  Please let me know which"

MJM.....explanatory PM sent so as not to discolour  or cloud your thread......

Looking forward to the next instalment & reading with interest  ....Derek   

[Zephyrin]

the port is nearly closed, and only slowly opens as the exhaust stroke progresses.  Consequently, the piston starts its return stroke for the exhaust with significant pressure in the cylinder, and the port is not fully open until the piston is over half way back to the top.

as opening and closure takes time, to get widely open port when needed, we need advance in the steam distribution, but a simple wobbler cannot cope to that.
And there no doubt that straight edges are preferable for the steam ports, as the rate of increase of the opened area is much higher than with circular holes for an identical valve displacement.