Stirlingengineforum.com

So why can an Ultra LTD run on a wet piece of paper with just about 0.5 percent "Carnot efficiency"?

Well, because with that pancake style displacer housing there are a LOT of Joules crossing the system boundary.

What matters is actual Joules because heat is a form of energy. Molecules in motion. Temperature difference CAN make a difference, but not necessarily. What matters is actual Joules entering and leaving as either heat or work. That value can be changed by intelligent engineering. More surface area, more Joules. Higher temperature, more Joules. Pressurize the working fluid, more molecular impacts, more Joules transfered. Switch to a gas with a higher rate of conductivity, more Joules per time interval etc. etc.

The ∆T does not matter if the energy transfer in Joules can be increased in some other way.

Increase the velocity of the molecules impacting the piston and alter the direction of impact with a venturi.

Many many variables have a potential influence on efficiency besides the temperature difference but these areas have hardly been explored.

It's been a few days again, since last hearing from "fool".

Just to reiterate, I do very much appreciate all the time and effort spent, or being spent on this analysis.

I've been waiting to get to the "juicy" part, which I'll highlight in the following repeat quote.

Note, what I'm highlighting is what I consider fallacious for reasons already delineated. Fundamentally it is not a matter of mathematical "error" or false calculations. Rather, it is the fact that our very definition and understanding about the nature of heat has changed over the course of a couple centuries, BUT THE MATH HAS NOT CAUGHT UP with our new, presumably better and more accurate and true understanding about the nature of heat.

We have a new definition of heat as a form of energy, but the math, some of it at least, the Carnot efficiency limit formula in particular, is still based on the old definition.

First question: "How do you define HEAT?"

Second question.

Does all the math conform with or reflect that definition?

Fools detailed analysis has helped me to narrow things down and zero in on the problem area

Qh = Qc + W

Or rearranged:

W = Qh - Qc

Second Law:

Efficiency ratio:

n = W/Qh

Or combining first and second

n = (Qh-Qc)/Qh = 1- Qc/Qh

Qh and Qc have a linear relationship to Temperature. Meaning add 100 Joules get 100 K temperature increase. Add 200 Joules get 200 degrees increase.

So they are of the following form:

Qh = Th x Cv + K
Qc = Tc x Cv + K

Y intercept, constant: K = zero because, at absolute zero Kelvin, there will be no heat. Qh and Qc will be zero at zero Kelvin Temperature.

So those two simplify to:

Qh= CvTh
Qc=CvTc

Putting those into the previous equation for n, efficiency:

n = 1 - Qc/Qh

Or

n = 1 - (CvTc)/(CvTh)

The Cv's cancel, becoming one, which leaves:

n = 1- Tc/Th

K = zero because, at absolute zero Kelvin, there will be no heat

That depends on how heat is defined.

If temperature is viewed as a measure of "heat", well, we have different temperature scales. The Kelvin scale did not exist in Carnot's day so how could an equation that originated with Carnot use a temperature scale that did not yet exist?

But there is a more fundamental problem. The definition of heat itself.

Qh and Qc have a linear relationship to Temperature

While that may in a sense be true, Qh and Qc represent "heat" as energy transfered into or out of the engine. Each Joule, roughly speaking represents or corresponds to one degree increase or decrease in temperature depending on the volume and heat capacity and other complicating factors, and we are talking about gas not water but It takes 4.184 joules, to raise the temperature of one gram of water by 1 degree Kelvin. What that is for a gas, air, helium, etc. depends on the gas, but I think it is fine to use this simplification. Let's say it takes 1 Joule to raise the temperature of one mole of our working fluid 1 degree Kelvin. Whatever. I'm sure that's not accurate but this is a simplification.

That is Qh and Qc

We will say 1 Joule of heat transfered in or out of the engine raises or lowers the temperature of the working fluid 1 degree Kelvin.

What about Th and TC ?

Let's take Tc first.

Just because: "Qh and Qc have a linear relationship to Temperature" can we really just make a substitution: Qc = Tc ?

Recalling our definition of heat:

energy that is transferred from one body to another as the result of a difference in temperature.

Does that definition describe Tc ?

Is TEMPERATURE a transfer of energy from one body to another ?

Does an ambient temperature of 300 K represent 300 Joules (in our simplification) transfered ?

Can we really just swap out Tc and Qc as if these represent equivalent values or even comparable concepts ?

A transfer of 1 Joule of heat out of the engine or "rejected" from the engine as "waste hear' at 300° K is not equivalent to 300 Joules rejected, is it ?

It seems like when we add 1 Joule (Qh) at say 600°K we then have to either convert that 600 Joules into "work' or else '"reject" whatever portion of the 600 Joules is not converted to work to the "cold reservoir".

So if we add one Joule at a temperature of 600°K and our engine can convert 200 Joules of that 600 Joules into work then the engine (our Carnot engine that is) will need to "reject" 400 Joules.

If the engine only converts the 1 Joule supplied to it, then it will have to "reject" the other 599 of the 600 Joules.

I have an idea!!!!

Suppose we take a very inefficient engine. We supply it with 1 Joule at 600°K, then when it "rejects" 599 Joules of heat, we can use all that "waste heat" to run a much much much more powerful engine!!!

Now, can it be seen where this fallacious substitution is being made?

https://youtu.be/_n3Z_YBzvDQ?si=JK7tIrCaX6lP634j


As well as a maligning of the accepted definition of heat?

The heat input in Joules: Qh is suddenly transformed into "all the energy" of the temperature at which Qh enters the engine, all the way down to absolute zero.

How does this pass muster with ANYONE ?

I'm anxious to see how "fool" or anyone else can make sense of this.

It looks like total bunk to me.


I think it's safe to say that if your cold reservoir is 300°K and the cold plate of your engine is 300°K then Qc will be zero.

No heat will be transfered if the temperature of the.cold side is equivalent to Tc.

You don't have to go to absolute zero to make Qc = 0 you just have to not have any heat transfer and you don't have heat transfer when two things are at thermal equilibrium.

In Caloric theory TEMPERATURE represents the VOLUME of the FLUID, Caloric.

Carnot published Reflections on the Motive Power of Fire in 1824.

Joules experiments were published about 20 years later in 1843.

No such thing as a "Joule" of heat ENERGY in 1824. No equivalence of heat and work, no conservation of energy.

Nobody knew heat was not a substance but only a form of energy. All that was known is that heat was measured by temperature and was considered to be a fluid, so temperature could only represent a volume of heat, the totality of a quantity of heat, the totality of a quantity of a LIQUID.

So naturally Qh, instead of representing, let's say, 100 Joules at 600K can only represent 600... 600....600 .... 600 what? Ja ja ja ja Joules???

1 – (Qc / Qh) is conservation of energy

1 – (Tc / Th) is Caloric theory

Two entirely different conceptual frameworks. Two entirely different ideas about what a quantity of heat consists of and how to measure it. Two entirely different world views, two different realities.

In the first 1 – (Qc / Qh) , Qh - heat is that quantity of an immaterial ENERGY transfered to the engine.

In the second 1 – (Tc / Th) , The - heat is that total measure of a FLUID, Caloric, represented by the temperature.

What these professors or instructors in the videos are doing is adding some Joules of heat to the engine using the modern conception of heat as energy transfered, to bring the temperature up from 300° to. 600° degrees absolute, then measuring the "heat" according to the archaic and obsolete Caloric theory where temperature is a measure of heat as a fluid so that the 300 joules supplied becomes 600 JOULES total VOLUME of a fluid.

The math looks good. You can't argue with the math, it's the flip flopping between two entirely different definitions of heat.

If heat is energy, then the measure of heat is the energy transfered between 300°K and 600°K or 300 Joules

If heat is a fluid, then the measure of ALL the heat is 600°K or 600 Ja ja ja ja Joules ????

No no no no no.

Here is another video on this very subject including all the math in a lengthy derivation where

η = 1 – (Qc / Qh)

The efficiency formula, Is step by step transformed into:

η = 1 – (Tc / Th)

The "Carnot" efficiency formula.


https://youtu.be/M_5KYncYNyc?si=xt8q4waeeycQ0EyL


Perhaps transformed is the wrong word.

It seems more a convoluted substitution.

The ideal gas law, PV=nRT is, I think, arbitrarily introduced without explanation.

PV=nRT however, and please someone correct me if I'm wrong, in this context, has to do with the totality of the internal energy of a gas.

Again, that is ALL the energy at a given temperature, NOT merely the Joules supplied at T1 and the Joules "Rejected" at T2 but the totality of internal energy of the gas at the isotherms T1 and T2

But, of course, the word "heat" is a slippery one. It is pretty natural for most people to refer to heat as the total content or internal thermal energy, and forget the technical scientific or thermodynamics(?) definition that heat is only that portion of energy transfered.

Calculating the state variables for internal energy or archaic "heat content" at T1 and T2 is not the same thing as calculating the heat supplied to the engine at T1 and T2. Is it????

Is the total internal energy level at the isotherms T1 and T2 the same as the Joules supplied to or removed from the engine at the temperatures T1 and T2 ???

All of the "HEAT"

Again, what is "heat" ?

Again, the definition of heat is not being adhered to.

"Heat" as energy in Joules transfered

Vs

"Heat" as total internal energy. (archaic "heat content).

Is it then actually legitimate to take the efficiency formula η = 1 – (Qc / Qh) and Shoe in the "Carnot" efficiency formula via the Ideal gas law.?

Apples to Oranges IMO

It is making a false equivalency between the heat supplied and "rejected" at T1 and T2 and the total "heat content" at T1 and T2.

Even if there was any such thing as "heat content" it is still just tossing an equal sign between two dissimilar values or unrelated concepts.

Energy supplied vs. Total internal energy.

BTW, I pointed out this apparent mathematical discrepancy on the Physics forum back in January of 2013.

I never received any response then either, other than to have the thread locked and to be banned from the forum, almost immediately.


https://www.physicsforums.com/threads/a ... st-4247448

Jan 28, 2013 #79
Tom Booth
DaleSpam said:

The calculations are a correct analysis of the scenario. The fact that you don't like the answer doesn't make a bit of difference.
________

I like the answers fine. If it's understood what it means. most people don't so its misleading.
_________

Then please post a scientific reference that says that you can use another temperature scale for these equations.
__________

It isn't the scale. It's that actual utilization of supplied heat to such a heat engine in practical terms. The 100% of all the heat collected by the solar panels for example is something different from ALL THE HEAT down to absolute zero. Carnot efficiency does not describe practical utilization of the supplied heat, it describes how much of all the heat. ALL of it. That means if you are talking about a hot air engine running on heat in the air, the engine has to cool the air to absolute zero to be 100% "efficient". The engine could be working great making all the energy anyone could ever need but the Carnot efficiency numbers would make it look like a dog. A waste of time.

Reference: https://www.physicsforums.com/threads/a ... st-4247448

So, what has happened to "fool"??

After more than ten years I was really looking forward to some kind of answer or resolution.

Carnot efficiency appears to have nothing to do with utilization of the actual "heat" supplied to the engine, as heat has been defined.

You supply 1000 Joules of actual heat/energy and suddenly, to utilize that 1000 Joules of heat you have to use "all the heat" down to absolute zero?

All the "heat"?

Please someone, explain to me how this actually makes sense.

Tom Booth wrote:You supply 1000 Joules of actual heat/energy and suddenly, to utilize that 1000 Joules of heat you have to use "all the heat" down to absolute zero?

There are simple answers and always PHD level answers. The simple answer is I've already answered it earlier in this thread.

Another simple way is to ask the question, if you start at zero K and add your 1000 Joules to a work conversion machine arriving at 100K, then, after conversion, the temperature becomes zero K, how much energy would be left to be ejected at Ql?

That's right zero. All of it would be converted to work. 100% efficiency when Tl = zero K. 1000 Joules of work.

Starting the temperature of Ql at 300 K, and add 1000 Joules, Th now become 400 K. Qh is now 4000 Joules, and Ql is 3000 Joules. ∆Q is still Qh-Ql=1000 Joules. Converting 4000 Joules all the way to zero K would convert 4000 Joules to work, 3000 Joules of free ambient heat for a conversation of 100%. However, stopping at 300 Kelvin would reduce that to:

Work = Qh - Ql = 4000-3000=1000

Efficiency for the work and total heat "used" by the machine is:

Efficiency = n = (Qh-Ql)/Qh = (4000-3000)/4000 = 0.25
This "sets" "defines" "describes" the maximum efficiency at 25%. It won't/can't change. It is now an absolute.

For 100% efficiency the machine would need to have a Ql temperature of zero K and the work out would have to be 4000 Joules. Not just converting 1000 Joules to work.

Now look at converting just 1000 Joules at a time to work and heat out, (not 4000) ∆Q = 1000. n is still set at 0.25. just the starting point has changed. Not starting at 4000 Joules, starting at 1000 Joules, nor are we starting at 100 K, starting at 400 K.

Work out = ∆Q•n = n(Qh-Ql) = n•∆Qh = 1000 x .25 = 250, per cycle. ( No longer W=Qh-Ql.) <<< This is the main difference.

Heat out at Tl = ∆Ql = 750.

The difference is between used heat and converted heat. To convert heat to work, it takes a full cycle, thus, removing "back work" from gained work. It is also know as the work of compression, or rejection of heat.

Your question revolves around used heat meaning 100% accountability. It needs to be about how much can be converted to work. You seem to already agree that real engines are not perfect. You need to learn why "perfect" ideal cycles also don't convert 100% of the heat to work. They all require "back work". Question converted to o work and residual heat rejection.

If you use ∆Q for ∆T, the equation for work becomes W=n(Qh-Ql) not W = (Qh-Ql) {for absolute Temperature}.

If I go any further it will be proceeding into the area of a PHD level of explanation. I do not have a PHD so would expect one or more of the following, as it will be thesis quality material:

1: One or more universities to contact me and offer a honorary PHD-ME

2 : At least an honorary MS-ME degree.

3: A full scholarship for the purpose of earning an honorary PHD, to be completed on line (I'm not moving out of the PNW.).

That last one would have to be honorary because I'm 65 and any new degrees wouldn't pay society back for very long, although very worth it.

It would involve several proofs of the Carnot efficiency formula, and Stirling equivalent. Also equivalent electrical motor, and possibly mechanical formulas. It would also require several uninterrupted posts here and weeks of time, perhaps months. It already tooks weeks to respond to this thread, sorry for the delay, I was cogitating and researching before answering and this was a simple answer.

Ps I hope you all are coping with the weather okay. It's been harsh lately for many and you too Tom.

Fool wrote: ↑Fri Jan 19, 2024 5:58 am

Tom Booth wrote:You supply 1000 Joules of actual heat/energy and suddenly, to utilize that 1000 Joules of heat you have to use "all the heat" down to absolute zero?

There are simple answers and always PHD level answers. The simple answer is I've already answered it earlier in this thread.

Another simple way is to ask the question, if you start at zero K and add your 1000 Joules to a work conversion machine arriving at 100K, then, after conversion, the temperature becomes zero K, how much energy would be left to be ejected at Ql?

That's right zero. All of it would be converted to work. 100% efficiency when Tl = zero K. 1000 Joules of work.

Starting the temperature of Ql at 300 K, and add 1000 Joules, Th now become 400 K. Qh is now 4000 Joules, and Ql is 3000 Joules. ∆Q is still Qh-Ql=1000 Joules. Converting 4000 Joules all the way to zero K would convert 4000 Joules to work, 3000 Joules of free ambient heat for a conversation of 100%. However, stopping at 300 Kelvin would reduce that to:

Work = Qh - Ql = 4000-3000=1000

Efficiency for the work and total heat "used" by the machine is:

Efficiency = n = (Qh-Ql)/Qh = (4000-3000)/4000 = 0.25
This "sets" "defines" "describes" the maximum efficiency at 25%. It won't/can't change. It is now an absolute.

For 100% efficiency the machine would need to have a Ql temperature of zero K and the work out would have to be 4000 Joules. Not just converting 1000 Joules to work.

....

Maybe we can pause there and mull that over a bit.

You certainly provide here an admirable rendition of the reasoning (?) process used by academics generally to arrive at the "Carnot efficiency".

I have to ask again, however, how do we define "HEAT".

If at 300°K everything is in thermal equilibrium, then there is no "heat".

You wrote above:

Starting the temperature of Ql at 300 K, and add 1000 Joules, Th now become 400 K. Qh is now 4000 Joules, and Ql is 3000 Joules.

Thank you for proving my point!

Qh is the heat transfered into the engine

That was/is 1000 Joules of HEAT transfered.

Add that to the 3000 Joules of "heat" (archaic def. Internal energy) that were already present while everything existed in a state of thermal equilibrium brings the temperature up to 400°K

The amount of "heat" transfered (Qh) you say now amounts to 4000 Joules.

Like I said, this is confounding an obsolete, invalid definition of heat with the well established modern definition.

How can 1000 Joules just "become" 4000 Joules?

How does the finite quantity of heat added suddenly quadruple from 1000 to 4000?

Well, going in heat was defined as the actual heat added in Joules, but once inside heat is redefined as all the thermal energy content. The 3000 that existed already at thermal equilibrium, before another 1000 was added plus the 1000 added becomes 4000 total "HEAT".

But the 3000 pre-existing units of "internal energy" does not meet the modern definition of "heat" as it already existed, it was NOT transfered into the engine. It was not supplied.

It's like I hold up a shot glass and ask you, "can you drink this ounce of brandy?" You agree.

Then I pour the ounce of brandy into another gallon jug of brandy and say, "there ya go".

It is like that. Only it's more like having three shots, adding a fourth ounce. You tell me you will drink the one ounce. You start drinking and pour three drops into a brandy cake, and one drop into you. Explaining that at 25% you will consume a total of one ounce, and the cake will get three. You are forced to stop at one shot gone, three left. You now have drunk .25 ounce, and there is .75 ounces in the cake, which you cook evaporating it to be gone forever, cold plate heat rejection.

Cold heat is useless and amounts to 3000 Joules, in reference to zero.

Please understand that ∆Q is what you add, Qh is the sum total of the heat capacity in respect to the temperature. That is why Ql is 3000, Qh is 4000 and ∆Q is 1000. Work is equal to n•∆Q.

Ql equals the internal energy at Tl, if there is no other energy stored in the engine. It equals the heat that would be added to bring the gas up to ambient temperature.

Fool wrote: ↑Fri Jan 19, 2024 2:00 pm It is like that. Only it's more like having three shots, adding a fourth ounce. You tell me you will drink the one ounce. You start drinking and pour three drops into a brandy cake, and one drop into you. Explaining that at 25% you will consume a total of one ounce, and the cake will get three. You are forced to stop at one shot gone, three left. You now have drunk .25 ounce, and there is .75 ounces in the cake, which you cook evaporating it to be gone forever, cold plate heat rejection.

Cold heat is useless and amounts to 3000 Joules, in reference to zero.

Please understand that ∆Q is what you add, Qh is the sum total of the heat capacity in respect to the temperature. That is why Ql is 3000, Qh is 4000 and ∆Q is 1000. Work is equal to n•∆Q.

Ql equals the internal energy at Tl, if there is no other energy stored in the engine. It equals the heat that would be added to bring the gas up to ambient temperature.

Please understand that ∆Q is what you add, Qh is the sum total of the heat capacity in respect to the temperature.

That is ABSOLUTELY NOT how these problems are presented or described in textbooks and online instructions/videos etc.

Qh is the heat added in ALL such examples problems I've ever looked at and definitely NOT "the sum total of the heat capacity in respect to the temperature".

Well, I take that back.

It is treated as the heat added, when the heat is added. Qh is the heat ENERGY transfered into the engine from the heat source (in Joules).

Once inside the engine, then suddenly it becomes, (via the Caloric "Carnot Limit" based theory of heat as a fluid substance measured by temperature) as you say: 'the sum total of the heat capacity in respect to the temperature".

You can't just transform 1000 Joules of heat ENERGY added into 4000 Joules once it is inside the engine.

That's OK as a fraternity prank but it isn't any kind of valid science, IMO. How anybody could ever have imagined that is beyond me.

Anyway, let's see a textbook problem where it says: Qh is "the sum total of the heat capacity in respect to the temperature". Let's get down to some brass tacks.

Maybe this one:


https://youtu.be/V3nNgygrmsI?si=ONZ4-3saP8yBkoCS


He clearly states, as stated in any and all such examples, that Qh is the heat received from the hot reservoir.

If you can find something somewhere that states otherwise, please do so.

That is ABSOLUTELY NOT how these problems are presented or described in textbooks and online instructions/videos etc.

Thanks. Yes, the text books mix that up. That is why I have strived to make it more clear than the current text books. I don't have a PHD so I tend to need it simplified. You will need to weigh it on your own without the ability of verifying it by some authoritative text.

I define my terms and use them correctly. If there is a flaw in my mathematics, please point it out, or have someone else. If my math appears solid let me know, I may send it to a university. If you will prefer, I can choose different symbols for those concepts.

Fool wrote: ↑Fri Jan 19, 2024 7:10 pm

That is ABSOLUTELY NOT how these problems are presented or described in textbooks and online instructions/videos etc.

Thanks. Yes, the text books mix that up. That is why I have strived to make it more clear than the current text books.

Wow!

Well, OK I think maybe that is actually my point. Thanks for the confirmation.

However, the textbooks and online sources etc. are pretty consistent in HOW they mix that up.

They use one definition for heat when the heat is being supplied to the engine but then use the old Caloric theory once the heat is inside the engine.

The math is really not at all complicated. The Carnot formula is ridiculously overly simplistic, actually.

Efficiency = 1 - Tc/Th

That's nothing more and nothing less than the temperature difference.

That it kinda, sorta resembles this other equation:

Efficiency = 1 – (Qc / Qh) does not justify actually equating the two.

So now, maybe you could expound upon the "Fool's" version of how Qh was actually always Th all along, or whatever it is your saying. Perhaps it is at least more consistent in regard how "heat" is defined, I hope.

Anyway, I don't think it could be any worse, or much more mixed up than the official doctrine.

That you are able to rationalize all this in some way, at least in your own mind is interesting.

Personally, I think a Joule is a Joule is a Joule. A fixed quantity of energy does not magically transform from 1000 Joules into 4000 joules just because it crosses the system boundary into the engine.

Heat is either energy, or it is a quantity of some fluid substance measured on the temperature scale like milliliters or quarts, but ALL of these sources engage in the same switcheroo.

First heat is Qh, the energy in Joules added to the system, then once inside, when it comes to utilization of that same energy by the engine, Qh and Qc are equated with Th and Tc and "heat" suddenly becomes a state variable representing ALL the internal energy, including the general ambient heat that was already there, that was never added along with Qh, that was not "supplied", transfered, or added at all.

This is mixing up and confusing two entirely different definitions of heat, two entirely different methods of measuring heat. If you define heat as energy TRANSFERED, that definition completely excludes the "internal energy" that was already inside the engine as it sat in thermal equilibrium with the ambient surroundings.

I can find no justification for closely associating the temperature difference with the actual quantity of energy transfered in Joules.

One Joule at 300°K or at 600°K or at any other temperature is still just 1 Joule.

1000 Joule transfered into the engine cannot be transformed into 4000 Joules with a little number juggling here and there.

I don't have a PHD so I tend to need it simplified. You will need to weigh it on your own without the ability of verifying it by some authoritative text.

I define my terms and use them correctly. If there is a flaw in my mathematics, please point it out, or have someone else. If my math appears solid let me know, I may send it to a university. If you will prefer, I can choose different symbols for those concepts.

Very generally, I can see how the air inside the engine is limited to a certain small volume.

The heat capacity of that volume of air is limited to a certain quantity of heat per cubic centimeter per degree Kelvin.

So, the greater the ∆T, you could say, the bigger the "gulp" of heat the engine can take in each cycle. For every degree of "cooling" (by whatever means) there will be that much more heat capacity, the engine can potentially transform that much more heat into work

So, with zero ∆T the air in the engine cannot take in any heat. With a small temperature difference, the engine can take in a small amount of heat each cycle.

With a big temperature difference, the engine can take in that much more heat, there can be more expansion per cycle and more heat transformed into work per cycle.

So, I'm not denying there is a relationship between ∆T and potential work output.

Much of thermodynamics has been translated from other languages, so I've often thought, perhaps this all has something to do with a misinterpretation of the word "rejected".

If I'm trying to get into a club, but I don't have a pass, I'll be "rejected' at the door and I won't get into the club.

When I'm "rejected" I'm not let in, allowed to hang around for a while, and then "rejected" and thrown out the back door.

I can see how given a limited heat capacity and low ∆T, heat applied to the engine on the HOT side could be "rejected". It never gets inside the engine.

This is very different from heat getting in, mixing with the air inside the engine and then being "rejected" back out through the engine, through the working fluid to the cold side and back out.

Because of the limited heat capacity, there is no use in blasting the engine with more heat than it could possibly take in. Most of the heat will be "rejected" and not be able to enter the engine at all. Cool down the cold side and the heat capacity per cycle can be increased. That let's more heat in to be converted. Less heat is "rejected".

Mathematically, it may not seem to make a difference. Some heat is converted to work and the rest is "rejected" back to ambient.

Well, I think it makes a HUGE difference.

I think it makes all the difference in the world, in terms of how an engine is designed and operated.