"Thermoacoustic" Stirling - theory of operation

[Tom Booth]

So, my basic theory of operation is that the wire wool acts as both a heat reserve, similar to a regenerator, but more one-way. It is heated by external application of heat and releases heat internally.

In steady state operation, though, the engine is basically an air spring (at both ends).

Without a load, hardly any heat is used, so hardly any heat enters the engine. The "heat of compression" and the cooling from expansion being roughly equal to the heat sources ( or source and "sink") virtually no heat transfer takes place in or out of the engine.

When a load is applied, however, the "work" output results in additional cooling, which increases the ∆T with compression (that is the working fluid is cooler than the heat input temperature at TDC rather than equal to or hotter), so more heat transfers from the reserve in the wire wool each cycle while a load is being applied.

The greater the work output, the more cooling takes place during the power stroke and the greater the ∆T so the faster the heat can transfer into the engine to compensate.

This explains something I posted many years ago. A write up (VITA technologies: https://pdf.usaid.gov/pdf_docs/pnaas739.pdf ) about a Stirling engine stated that the engine would "grow stronger" as a load was gradually applied.

Increasing the load increases the energy output which results in more cooling which accelerates the heat transfer into the engine due to the increased temperature difference between the source temperature and the working fluid temperature.

So, you could say it is not only a mostly self sustaining "air spring" but also self regulating and self compensating or self governing to some degree.

Long debate on the subject:

viewtopic.php?f=1&t=478&start=15#p1259

[VincentG]

The guys over the pond at stirlingengineboats.com have arguably more hands-on experience than anyone outside of industry with SE's. Here is an alternate take from their website. I tend towards this explanation as it's in line with my observations at face value. However, I suspect there is more at play here, as my engine never seemed to mind an instant application of load.

The load is then released further. The revs jump to 425
And a minute later, the watts and revs fall slightly.

What is this zig-zag pattern about?
With each release of load, the revs soar, the watts rise – then they both fall back.

When the engine is under heavy load, the revs are low. Heat is being poured into the hot cap
at a steady rate by the burner, but at low revs, that heat is being taken out at a slow rate.
This lets the hot cap rise to a higher temperature.

As soon as the load is reduced, that enhanced hot cap temperature make the revs soar.
But those higher revs take heat out of the hot cap more rapidly.
The hot cap temperature drops – and the revs. Until steady state is reached.

We see this repeatedly and reliably when in the boat in a lock.
The engine is sat there – no revs, but with the burner turned down low.
The hot cap temperature rises gently.
When it is time to leave the lock, the hot cap has a higher temperature and there is a surge
of power as the engine crops the extra stored heat in the hot cap.

[Tom Booth]

The guys over the pond at stirlingengineboats.com have arguably more hands-on experience than anyone outside of industry with SE's. Here is an alternate take from their website. I tend towards this explanation as it's in line with my observations at face value. However, I suspect there is more at play here, as my engine never seemed to mind an instant application of load.

The load is then released further. The revs jump to 425
And a minute later, the watts and revs fall slightly.

What is this zig-zag pattern about?
With each release of load, the revs soar, the watts rise – then they both fall back.

When the engine is under heavy load, the revs are low. Heat is being poured into the hot cap
at a steady rate by the burner, but at low revs, that heat is being taken out at a slow rate.
This lets the hot cap rise to a higher temperature.

As soon as the load is reduced, that enhanced hot cap temperature make the revs soar.
But those higher revs take heat out of the hot cap more rapidly.
The hot cap temperature drops – and the revs. Until steady state is reached.

We see this repeatedly and reliably when in the boat in a lock.
The engine is sat there – no revs, but with the burner turned down low.
The hot cap temperature rises gently.
When it is time to leave the lock, the hot cap has a higher temperature and there is a surge
of power as the engine crops the extra stored heat in the hot cap.

I don't disagree, necessarily. The VITA engine was just a burn box using "local" whatever, sticks, leaves, corncobs or whatever. Basically not much control of the rate of combustion.

Your quote mentions: "with the burner turned down low.", and; "Heat is being poured into the hot cap... This lets the hot cap rise to a higher temperature. As soon as the load is reduced, that enhanced hot cap temperature make the revs soar." etc

In other words, probably liquid or gas fuel with an adjustable burner that was being turned up and down. That would make it difficult to evaluate the effect of the load alone vs. increasing or decreasing the burner flame (heat input).

It also states: The load is then released further. The revs jump to 425 And a minute later, the watts and revs fall slightly.

This sounds like removing the load, ultimately reduces the engines output in watts as well as RPM, which seems to me in agreement with the VITA statement that increasing the load gradually increases power output.

I think naturally, when a load is first released the engine is already producing more power and will rev, momentarily, "With each release of load, the revs soar, the watts rise – then they both fall back

Some of the statements are questionable IMO like: "But those higher revs take heat out of the hot cap more rapidly.".

I've cited other sources previously that state when the load is reduced on a solar Stirling engine, but the heat input continues the engine quickly overheats.

IMO the higher revs are due to the greater power output that developed while under load. Reducing the load then creates a bottleneck. The excess heat causes reving but also the heat buildup and lack of work output (conversion of heat to work) reduces the ∆T ultimately reducing power.

At any rate, it's not always entirely clear when the burner is turned up or down, also, in both accounts, it's not entirely clear what conclusions are the result of actual measurements, temperature readings, and what is assumed to be true or speculative

Reving or surging or just bogging down in an IC engine is a symptom of some imbalance between the load and the fuel supply. Too rich or too lean, a clogged carb or dirty air filter etc.

I view HEAT as the fuel for a heat engine, so those kind of symptoms, I would say, indicate a load balancing issue.

But, does reducing the load result in a higher or lower temperature?

I would also say reving is increasing the load, or redirecting the heat input from the external load to an internal load, in the form of higher RPM, which is another form of "work".

Does higher RPM cool the engine? "higher revs take heat out of the hot cap more rapidly." OK, but how?

IMO higher RPM is more work.

In a boat, maybe higher RPM means the water cooling circulates faster.

Overall, not really enough detail to draw any hard conclusions, but as I say, I don't necessarily disagree. To me these accounts compliment each other.

Take this statement:

When the engine is under heavy load, the revs are low. Heat is being poured into the hot cap
at a steady rate by the burner, but at low revs, that heat is being taken out at a slow rate.
This lets the hot cap rise to a higher temperature.

Presumably the fuel is turned up high and the burner is pouring heat into the hot cap, but the heavy load keeps the heat input and work output in balance.

Remove or reduce the load and the engine races. Well, was the burner turned down first? It doesn't really say.

Do you have a link to this? I'd be interested in reading the full context as your quote starts out "The load is then released further..."

I'm not really sure the point being made, but again, I don't necessarily see it as an "alternative take" as it seems IMO to complement what I said, though it does not directly relate a higher RPM to "work".

"It does say: "When the engine is under heavy load, the revs are low. Heat is being poured into the hot cap
at a steady rate by the burner, but at low revs, that heat is being taken out at a slow rate.
This lets the hot cap rise to a higher temperature."

I'm not sure by what means or by what logic low RPM with a heavy load = heat being taken out at a slow rate.

IMO the heavy load is removing heat equivalent with the work output or "load". RPM is not necessarily a measure of heat utilization IMO, so, hands on experience or not, I think perhaps too much is being assumed.

Anyway, some experimentation with accurate measurement of heat input vs. work output, temperature RPM etc.would perhaps give a better picture of what's going on.

[VincentG]

http://www.stirlingengineboats.com/Stir ... age23.html

If you click on the index at the top right, there's a whole lot more stuff. Nothing very specific unfortunately.

[Tom Booth]

http://www.stirlingengineboats.com/Stir ... age23.html

If you click on the index at the top right, there's a whole lot more stuff. Nothing very specific unfortunately.

Thanks.

I started reading the "book" from the begining. The first thing I notice is this description of how a Stirling engine works:

Page 3 – Comparison: The Four Stroke and the Stirling

Imagine a single cylinder, 4 stroke, motorcycle engine. Block off the inlet and outlet valves.
Remove the spark plug and block off its hole.

Air is trapped inside the cylinder.

Imagine that we could heat the gas inside the cylinder – above the piston.
As the temperature of the gas rises, pressure is created. In just the same way as a petrol explosion – but far, far weaker and slower. That pressure pushes the piston - the shaft turns.

Now imagine that we can cool the gas above the piston.
It shrinks. The pressure reduces and the piston now returns.

Repeat the process of heating and cooling. The piston pushes and then returns. Repeatedly.

This is the Stirling engine.

There is no mention of any temperature reduction resulting from the conversion of heat into work.

That data presented later is also interpreted in the same way is not surprising.

This external heating and cooling is a view I held myself for some number of years while first starting out with an interest in Stirling engines.

It was not until happening upon some texts relating to methods for liquifying gases in expansion engines that I began to realize that the "work" output itself can be a method or means for reducing temperature. As a gas expands in an engine and does "work" the "heat" is converted to "work" and a cooling or refrigeration effect is the result.

Classical heat engine theory, being largely derived from the Carnot theory that ALL the heat entering a heat engine must be removed to a cold sink or "reservoir", does not really recognize "work" output as any kind of means for reducing the temperature of the engine.

IMO, if heat is simply a form of energy, then such a scenario, where heat is added to the engine and then ALL the heat removed while ALSO outputting work, which is another form of energy, violates conservation of energy.

You have energy going in AS HEAT and then ALL that heat going out AS HEAT to the sink, and then ALSO, energy going out as WORK. That would be a fiat creation of energy from nothing.

So, the later view was that SOME energy was converted to work output, but it still didn't seem to be recognized that this resulted in cooling so less heat needed to be removed by transfer to a sink.

The second "law" of thermodynamics absolutely forbids ALL the heat being converted to work, but IMO there is no empirical basis for that. The simplest "toy" Stirling engine violates the second law with complete impunity.

[Tom Booth]

This seems representative of the prevailing viewpoint:

What is the theoretical maximum efficiency of a heat engine and how does it relate to the second law of thermodynamics?

Ans :- The theoretical maximum efficiency of a heat engine is related to the second law of thermodynamics, which states that heat cannot be converted completely into work. This means that in any heat engine, there will always be some amount of heat energy that is not converted into work.

The theoretical maximum efficiency of a heat engine is defined as the maximum fraction of heat energy that can be converted into work. This maximum efficiency is given by the Carnot efficiency, which is calculated as follows:

Carnot Efficiency = 1 - T_C / T_H

where T_C is the temperature of the cold reservoir and T_H is the temperature of the hot reservoir. The Carnot efficiency is the upper limit of efficiency that can be achieved by any heat engine, regardless of its design.

In other words, no heat engine can have an efficiency greater than the Carnot efficiency because the second law of thermodynamics sets a limit on the maximum amount of heat energy that can be converted into work. The Carnot efficiency represents the maximum possible efficiency of a heat engine operating between two temperatures.

https://www.quora.com/What-is-the-theor ... modynamics

And yet an LTD Stirling can run on a temperature difference of less than 1 degree.

To me, that demonstrates the incredible efficiency with which these engines are able to convert even a nearly imperceptible amount of heat into mechanical motion or "work".

Yet the second law aka Carnot limit would mathematically place the efficiency of such an engine at near zero.

[Jack]

I think there's a difference between work and useable work output.
"Any" work output has a much lower mechanical threshold to overcome friction and gravity than "useable" work output. Because for it to be useable to us more contraptions need to be bolted on which result in extra friction.
At some point you go so far back in what you consider work output that it is still there, but can't even overcome the lightest friction. Then it goes so far back that it's barely measurable. But even then it's still there. And even further back to venture into the theoretical realm, which is what the carnot cycle is.

So if you want to build an engine that's higher in efficiency you'd have to build it very light so it can even be powered by the smaller temperature differences. But there's a point where that no longer results in a useable engine. Depending on the goal of course.

[VincentG]

And yet an LTD Stirling can run on a temperature difference of less than 1 degree.

To me, that demonstrates the incredible efficiency with which these engines are able to convert even a nearly imperceptible amount of heat into mechanical motion or "work".

Yet the second law aka Carnot limit would mathematically place the efficiency of such an engine at near zero.

IMO the Carnot limit is biased towards reducing heat to absolute zero. A more realistic equation would factor in the ambient conditions as a starting point. In other words, an internal combustion engine that ingests ambient temperature air, and exhausts ambient temperature air after complete heat to power utilization(assuming no other thermal losses) should be at 100% efficiency. But the equation does not allow for that.

[Jack]

That still assumes you can extract every little bit of extra heat and convert it into work. Which might be theoretically possible, but in the real world it's not. Not as long as we're using moving parts. Because every moving part will need a minimum of heat to be able to be pushed into movement. Due to friction, gravity and inertia. So all the heat below that threshold will be lost.
And this is all assuming you can build a fully insulated machine and no heat is lost through conduction.

[VincentG]

Agreed Jack, but the Carnot limit is theoretical anyway, so I was assuming a theoretical engine with 100% mechanical efficiency in order to isolate the equation.

[Jack]

In that case, I don't think the carnot is trying to reduce heat to absolute zero. It is merely the most efficient path between any two temperatures.

I'm only into this stuff for a few weeks now, but I see how it pulls you in. This constant "fight" between power and efficiency. It's never ending haha.

[VincentG]

Never ending is an understatement haha.

Thats a good way to put it, the most efficient path. But why cant the shorter path, say down to 300k not 0k, be as efficient?

[Jack]

It can be just as efficient. But not as powerful.

[Tom Booth]

Personally I think the Carnot efficiency equation has been misinterpreted.

If the ∆T is between 300° K and 400° K the old idea was that temperature is a measure of a THING, a substance, not a measure of energy.

So, temperature was interpreted as a kind of volume, like 400 milliliters of heat.

If you put heat in at 400°K viewed as a volume of some actual substance, like 400 milliliters of a fluid and your engine exhaust is 300°K, (ambient) the logical conclusion is that only 100 milliliters of this HEAT fluid could be utilized and the other 300 went to "entropy" or "waste heat". The portion of heat that could not be utilized.

In that context the equation Efficiency = 1 - Tc/Th ,makes sense. But only if heat is an actual substance.

Naturally, if you put superheated steam into a steam engine at 400°K and the exhaust is 300°K (back down to ambient) a quantity of heat FLUID equal to the 300°K volume of this heat substance has to have leaked away somewhere.

IMO the Carnot limit is biased towards reducing heat to absolute zero. A more realistic equation would factor in the ambient conditions as a starting point. In other words, an internal combustion engine that ingests ambient temperature air, and exhausts ambient temperature air after complete heat to power utilization(assuming no other thermal losses) should be at 100% efficiency. But the equation does not allow for that.

Actually, it does.

Efficiency = 1 - Tc/Th

Efficiency = 1 - 300K/400K

Efficiency = 25%

75% of the heat juice is going to waste!

From 0°K to 300°K is 75% of 400°K

From 300°K to 400°K is only 25% of the heat that was put into the engine at 400°K

This only makes sense if heat is a substance and 400°K represents a quantity of that substance.

As ENERGY temperature is not even a measure of a quantity of energy.

How many joules are in 400°K air?

The question is meaningless.

The Carnot efficiency equation is just as meaningless outside the context of Caloric theory where heat was thought of as an actual substance of some sort.

[Jack]

While heat is not a substance, if we're working with fluid, we better equate it to one. All matter reacts in a predictable way to heat input. So whether Carnot meant pure energy or its effect on matter, they are the same thing.