Longpig,
Good point, it'd be difficult to stop the working gas from escaping unless you could make a properly sealed cylinder.
Tom,
I've had another read through of your first post & the wikipedia entry on Stirling engines and in the cold light of day here's what I'm thinking. Let me know if I'm tripping up along the way.
First of all, I believe that it is theoretically possible for a gas to compress or expand without a change in temperature. In order to achieve this, you have to feed in or take out an amount of heat energy at the same time. In an idealised Stirling engine the expansion and compression phases are, in fact, exactly this: Isothermal compression and isothermal expansion. The expansion takes place inside the hot cylinder, which transfers heat energy into the gas. The compression takes place in the cold cylinder, which absorbs some heat energy from the gas.
In one of your posts you quote someone saying 'Any gas when compressed rises in temperature. Conversely, if it is made to do work whilst expanding, the temperature will drop'. I think the guy is talking about a situation where the compression or expansion happens without much heat energy transfer from the surroundings. For example, when you let off a fire extinguisher the contents expand quickly and there is little heat transfer into the gas, which therefore gets cold.
Leaving that behind, I believe that you were right in thinking that the Stirling engine (or any other kind of heat engine) functions by means of a heat differential between the source and sink. In a heat engine heat-energy flows from the hot source to the cold sink and we divert a portion of that heat out as useful work. The thermodynamic efficiency of a heat engine is the proportion of the heat flowing in from the source that we have diverted out as useful work. It is, sadly, impossible to divert all of the heat-energy flowing through the engine out as work (a 100% efficient engine). In fact the best we can do is a Carnot or Stirling cycle.
I can see the rationale behind what you propose in your posts: If we apply a greater load to the engine and lag the heat sink then two things should happen:
1) More of the heat energy will come out as useful work
2) We block more of the heat energy from being lost into the cold sink
which will cause the engine to run cooler and more efficiently.
I think that you're right about the load application, but not about lagging the cold sink. I think the best way to explain my reasoning is to refer to pressure-volume and temperature-entropy graphs for the cycle. I've put a scan of p-V and T-s sketches highlighting each of the phases here:
http://i1009.photobucket.com/albums/af2 ... gcycle.jpg.
Say we
start looking at the cycle where the gas is a minimum volume and maximum temperature (which looks like this:
http://en.wikipedia.org/wiki/File:Alpha ... ame_8.png), which I've labelled as 'A' on my graphs. At this point the gas is mainly in the hot cyclinder and is at low volume, high temperature and high pressure.
Phase 1 of the cycle, going from 'A' to 'B', is an isothermal expansion: The gas stays at the same temperature, but increases in volume and entropy and decreases in pressure. As you point out in your post, if the gas is expanding at pressure within our cylinder then it is doing work. This is where the p-V graph is useful:
The work done by the gas is the shaded area under the line on the p-V graph, which I've labelled W1. The work done by the gas is transferred to the fly wheel (or some similar energy store). The work going to the fly wheel is important, as we'll need it back later.
This is only half the story for phase 1. As I mentioned above, in order to have an isothermal expansion we need to put heat energy into the gas.
The energy being put into the gas is the shaded area under the line on the T-s graph, which I've labelled Q1.
After phase 1 the engine looks a bit like this:
http://en.wikipedia.org/wiki/File:Alpha ... ame_12.png, and is at the point I've labelled B. The gas is now at high volume, high temperature and high entropy, but lowish pressure.
In
Phase 2, between points labelled 'B' and 'C' of the cycle, the gas is transferred through the regenerator. In doing so it undergoes constant volume (isochoric) cooling. The volume stays constant, but the temperature and the pressure drop. The entropy also drops slightly. Because the volume of the gas stays constant the line on the p-V graph is vertical and there is no area under it; no work is done in this phase (we're assuming no friction in the mechanism and no viscosity in the gas).
The T-s graph is a different matter. The area under the line, labelled Q2, is the heat energy that is transferred from the gas into the regenerator.
After phase 2 the engine looks a bit like this:
http://en.wikipedia.org/wiki/File:Alpha ... ame_16.png, and is at the point I've labelled C. The gas is now at the cold temperature and low pressure, but the volume is still high and the entropy is still highish.
In
Phase 3, between points 'C' and 'D' of the cycle, is an isothermal compression: The gas stays at the same temperature, but decreases in volume and entropy. The pressure of the gas rises. In order to make this happen two things must occur:
1) The fly-wheel (or similar) must do work on the gas. The amount of work needed is equal to the area under the p-V line, labelled W3. The fly-wheel should have this energy stored up, as we put W1 energy into it in phase 1, and W1 is greater than W3. However, imagine if we were using our Stirling engine to power a electricity generator. Imagine if, between phases 1 and 3 we had sucked more energy out than the difference between W1 and W3 as electricity. There would now be less than W3 energy left in the fly-wheel. We would not have enough energy left to compress the gas and the engine would stall.
2) Heat energy is lost to the cold sink. The quantity lost is equal to the shaded area under the line on the T-s graph. You can see that you can reduce the amount of energy lost by reducing the temperature at the cold sink. In order to eliminate the energy lost you'd have to have your cold sink at absolute zero. Good luck finding a gas that will work there.
As you probably noticed, phase 3 is a real bummer. We have to put work back into the gas and we lose energy to the cold sink. I guess you're trying to think of a way to eliminate phase 3. The trouble is that for this to be a cycle you have to get the gas back to the starting point - point A.
So why do we bother having a cycle? Well, if you can find an large supply of hot, compressed gas (like we have at point A) then you're onto a winner: You can do just phase 1 and let it all out through a turbine and get lots of work. You can sell that and be really rich. Sadly, there aren't many large supplies of hot, compressed gases just sitting around for the taking. Before you ask, Congress doesn't count - neither does parliament, as they're not sitting at the moment. If you want to be rich you'd probably be better off looking for the chemical or nuclear equivalent of a supply of hot, compressed gas. I guess the petrochemical and uranium industries got there first, though. The sun, perhaps? They can't monopolise the sun, can they? The trouble with energy from petrochemicals, uranium and the sun is that we have to turn it into useful work somehow....
Anyhow, I digress. The gas is now at point D. It is back at the starting volume, but it is still at the cold temperature and low pressure. The engine will look a bit like this:
http://en.wikipedia.org/wiki/File:Alpha ... rame_4.png.
Phase 4 is the reverse of phase 2: The gas passes back through the regenerator, undergoing isochoric heating. The gas goes from point D to point A. The volume stays the same, so there's no work done, but the heat energy the gas lost to the regenerator in phase 2 is reabsorbed (so area Q4=Q2, assuming the regenerator is perfectly conservative and doesn't lose any energy down the back of the couch). This raises the temperature and pressure back up to the values the gas started with, at point A.
Sorry if you already know all that and I'm wasting your time, but hopefully it'll help to explain the path of my thinking. You said:
"the 'cold' end of a displacer chamber in a Stirling Engine at ambient temperature is perhaps not cooling the air in the chamber at all but on the contrary actually heating it and reducing efficiency. If this is true then efficiency could be increased by insulating the cold end of the chamber against the heat intrusions from the external ambient air."
As I've noted above, I think that in a Stirling engine the compression and expansion strokes happen at constant temperature. I think what you're describing would be a different kind of engine; lets call it the Booth Engine. The p-V and T-s diagrams for the Booth cycle are here:
http://i1009.photobucket.com/albums/af2 ... hcycle.jpg.
Starting at the same point, with the same gas as we did with the Sterling cycle: The gas has a high temperature and pressure, but low volume and entropy.
Phase 1 is an expansion where the gas increases in volume (by definition) and entropy, but decreases in temperature and pressure.
The temperature of the working gas dips below that of the cold sink. As with the Stirling engine, heat energy is absorbed from the hot source and work is done by the gas.
Regenerators improve the efficiency of engines. However, a regenerator will be difficult to incorporate into the Booth engine, for the same reason that phase 3 will be tricky. We will miss out on
phase 2.
In
phase 3, we would like to reduce the entropy so that we can return the gas to its original state in phase 4. In order to do that we need the gas to lose some heat energy, as it does in phase 3 of the Stirling cycle. However, heat energy flows from hot to cold. If our working gas is colder than the sink then heat will flow from the sink into the gas. The entropy of the gas will increase and get even further away from the starting point.
What we need in order to complete the Booth cycle is something that will absorb the heat energy from the gas. Unfortunately that has to be a sink that is colder than the gas. I believe it is impossible to have a heat engine in which the working gas falls to a temperature below that of the coldest sink in the engine. If it does, it will be impossible to complete that cycle.
In any case. you'll notice that we absorbed less energy and got less work in phase 1 of the Booth cycle than we did from the Stirling cycle. If we did think of a phase 3 that made the Booth cycle work then I still think it would be less efficient than the Stirling cycle.
So, what would happen if we kept with the Stirling cycle (ie isothermal expansion in phase 1), but lagged the heat sink? In our theoretical engine the sink has infinite thermal mass: it can absorb any amount of heat energy and not get hotter. Real heat sinks aren't like this. The cold piston absorbs heat energy from the gas and then passes it out to its surroundings. If more heat comes into the piston than it it can pass to the surroundings then it will begin to warm up. If it warms up then the temperature that the gas is reduced to (Tcold) will increase.
If Tcold increases then the line for process 3 on our p-V and T-s graphs will rise. The work needed to compress the gas and the heat rejected to the cold sink will increase. We will get less useful work out and the efficiency of the engine will drop.
The efficiency of the engine depends on the difference in temperature between the hot source and the cold sink.
So, that's what I reckon about lagging the cold cylinder. How about the question of differing loads? Imagine you've set up your Stirling engine to lift a light weight or low gearing and you apply a source of heat to the hot cylinder. The engine starts off turning slowly. The engine goes through a cycle and a certain amount of work is left in the fly-wheel (remember that the amount of work got from a cycle was W1 - W3).
In turning over the engine also lifts the weight a distance. The work needed to do this is the product of the weight of the load and the distance it gets raised in one turn of the engine. However, what happens if this is less work than was left in fly-wheel? The energy stored in a fly wheel is related to the speed at which it is spinning, so if there is left over work, then the fly-wheel spins faster. If the fly-wheel is spinning faster then the engine is also turning over faster, so the engine also begins to speed up.
As the engine speeds up any mechanical inefficiencies will grow - greater energy will be lost due to friction in the working fluid and in the mechanism. More importantly the working gas will be in contact with the hot source, cold sink and regenerator for shorter periods each cycle. As the engine spins faster the working gas will cease to reach the temperature of the hot source when heated, or the cold sink when cooled. The efficiency of an engine is related to the temperature difference between the hottest temperature the gas reaches at one end of the cycle and the coolest temperature the gas reaches at the other end of the cycle, so as the engine spins faster it will become less efficient.
At some point the engine will have sped up to the point where the work put out in a cycle matches the work needed to lift the weight the distance it is set to go in one cycle. At that point it will settle at that speed.
If you put a heavier weight on the string then the same sort of process will occur, but the engine will settle at a lower speed (assuming the extra load doesn't stall it). Because it is running slower the gas will have more time to reach the higher and lower temperatures and it will run more efficiently.
That's all theory, though. Being from a country that has tended to favour empiricism over that nasty continental synthesis, I encourage people to test it. We can measure the work being output by lifting a weight. However, can anyone think of a way to measure the amount of heat going through? Perhaps we could measure the rate of ice melting at the cold end?
Bear in mind that this is more efficiently thermodynamically speaking. ie. You are getting more of the energy that passes through the cycled gas out as work. However, most of the engines we're looking at here are fueled by candles or suchlike. The heat put out by your fuel candle is unrelated to how fast your engine is running. Any heat that isn't taken up by the engine is just lost around outside into the atmosphere. You may get more useful work done in this situation by putting a lighter weight on. The engine will run fast an thermodynamically less efficiently, but the work will be done quicker and you can then snuff out the candle. Whilst the engine has been thermodynamically less efficient, you may have made a more efficient use of your fuel. That's why most engines have a device like Watt's governor (
http://en.wikipedia.org/wiki/Centrifugal_governor) or a hit-and-miss mechanism (
http://en.wikipedia.org/wiki/Hit-and-miss_engine) to match the fuel being used to the load applied and to stop the engine spinning too fast.
As I say, that's my tuppence-worth. Feel free to pick holes.
Please forgive me, I didn't mean to be rude.
