The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[matt brown]

Fool, we cross-posted, but your last post is more like what's req'd. Years ago, I gave up using numbers your way 'cause it looses most guys quickly. The only problem with proving Carnot via internal energy is that few know this relationship. We grow up in a world of adiabatic engines - ICE, steam, whatever - where simple high and low internal energy in intuitive. Venturing from this comfort zone, most guys end up lost in the weeds like Tom and his air compressor tank. I'm a lucky dog, I grew up with all this stuff (loads of various tech).

[Tom Booth]

Qcz is the internal energy of the mass of gas that is in the engine.

It is equivalent to heat that would be added if heating that mass from zero K to Tc K. It is also the amount of heat energy that would come out if cooled from Tc to zero Kelvin.

It is automatically supplied by the ambient temperature. ...

A better way of phrasing it... "...would be added if..." -"...that would come out if..."

Big big "ifs" IMO.

To say "It is automatically supplied by the ambient temperature. ... (to the "internal energy") However is FALSE!

The internal energy is already filled up to that degree, so it is not "supplied". It was not "added". It cannot even be legitimately refered to as "heat".

If you have a jug of water that is already 4/5ths full and top it off with 1/5 you have not "supplied" a full jug of water. The 4/5th were already present.

[matt brown]

Tom, Fool said this perfectly correct. The change from U=0 when K=0 to U="300" when K=300 was 'heat' added/supplied from "surroundings" to "system" via Mother Nature or some lab guys during research; U didn't change via bunny farts.

In your jug analogy, let's consider 0/5ths=0K and 4/5ths=ambient K. First you need the 0/5ths jug (no small feat) but once this is acquired it's easy=peasy to acquire 4/5ths jug from ambient heat without human interaction. Fool is merely quantifying this value, not qualifying this value (how it got there).

[Tom Booth]

Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?

Answer:

Let Th be the temperature of the hot plate/reservoir.
Let Tc be the temperature of the cold plate.

Let Qh be Delta Qh = DQh, the same definition and value, the heat added per cycle.

Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible, for 100%.

Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.

The heat rejected from zero K all the way to Tc will be Qcz

Qhz will become the heat added to get to Qcz plus the extra heat added or DQh to get to Th.

Qhz = M•Cv•Th = Qcz + DQh
Qcz = M•Cv•Tc

Note: Qhz-Qcz = DQh-DQc

Where M is the mass of the working gas, and Cv the heat capacity. Cv example units: Joules per Kelvin per Kilogram, or J/(K•Kg)

W = Work output from the working gas.
W = DQh - DQc
n = efficiency = W/DQh = (DQh - DQc) / DQh

Taking the above equation Qhz = Qcz + DQh and subtracting Qcz from both sides and rearranging to get:
DQh = Qhz -Qcz
Multiplying both sides by n:
N•DQh = n•(Qhz-Qcz)
Dividing both sides by DQh:

n = n•(Qhz - Qcz) / DQh #1

The term n applies to the engine regardless of reference point. It will have the same efficiency burning the same energy at the same temperatures but different units. That allows us to look at the total energy in the system, Delta Qh added to ambient. So Qhz should have the same relation ship to Qcz that DQh has to DQc, in other words, "n" is the same. Basically this says that Qcz can be calculated from n
Qcz = Qhz•(1-n)

Substituting that into the above equation for Qcz:
Qhz = Qcz + DQh
Substituting and distributing:
Qhz = Qhz(1-n) + DQh
Qhz = Qhz - n•Qhz + DQh
Subtracting Qhz - n•Qhz from both sides:
Qhz - (Qhz - n•Qhz) = DQh
Subtracting:
n•Qhz = DQh
Rearranging:

DQh = n•Qhz #2

Combining #1 and #2

n = n•(Qhz - Qcz) / DQh #1
DQh = n•Qhz #2

n = n•(Qhz - Qcz) / n•Qhz
The n's on the right side cancel becoming one, rewriting:

n = (Qhz - Qcz) / Qhz #3

Equation #3 shows that DQh and DQc have now become Qhz and Qcz. In retrospect, it seems logical that the efficiency profile should be the same regardless of absolute scale. Substituting in Qhz and Qcz straight across makes sense, and would have been faster. Note also, that it tends to maximize efficiency by equating DQc with Zero. It's not zero is just the amount left over after inevitable heat rejection and maximum work.

Now the equations above for Qhz and Qcz:

Qhz = M•Cv•Th
Qcz = M•Cv•Tc
Substituting into equation #3:

n = (Qhz - Qcz) / Qhz #3

n = (M•Cv•Th - M•Cv•Tc) / (M•Cv•Th)
Rearranging M•Cv
n=M•Cv•(Th-Tc) / (M•Cv•Th)
Canceling M•Cv top and bottom because they become one:

n=(Th-Tc)/Th The final solution.

That shows the logical progression of the mathematical proof.

It's logical to think that higher temperature differences produce higher pressure differences that make greater power to weight and size ratios, ease of construction, operation, and efficiency increases.

1/10 degree temperature, would be like trying to move a piston with 1/10 of a psi. How can that possibly be as efficient as trying to move a piston with 100 psi.

It would take a piston 1000 times larger in area to develop the same force. Power and efficiency sucking bad bulk would be the down fall. I hope this is enough said about the Carnot Theorem.

Overlooking the various objections already addressed and a few more I'll set aside for the moment, the "note" I've highlighted above is just too illogical, twisted and convoluted to ignore.

Note: Qhz-Qcz = DQh-DQc

You've stated:

"Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th"

Forgetting about where the heat did or did not come from or how it got there Qhz represents ALL the "internal energy"

Correct?

Qcz you said:

The heat rejected from zero K all the way to Tc will be Qcz

And also:

Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible, for 100%.

First of all, let's just forget 0°K and put a realistic number on Tc. Let's just say Tc is ambient at 300°K (or about 80°F)

What exactly does "The heat rejected from zero K all the way to Tc will be Qcz" mean anyway?

The heat rejected from the engine depends entirely on the efficiency of the engine and the work output. What is not converted to work is "rejected" to the "cold reservoir". (Or so the theory goes.)

If, for example, Qh is 100 joules and engine efficiency is 50% then Qc (or your DQc) will be 50 joules.

You also show the equation:

Qcz = M•Cv•Tc

It seems you are trying to shoehorn all the heat of the ambient environment, from "zero K all the way to Tc" into the paltry 50 joules or whatever heat is rejected by the engine after some percentage is converted to work.

Your "Qcz" appears to have nothing to do with any heat that ever entered or ever left the engine. It is apparently the ambient environment itself: "The heat (rejected) from zero K all the way to Tc will be Qcz"

Leave out the word "rejected" and we have "The heat... from zero K all the way to Tc..."

This is in some way included or part of the 50 or whatever joules going out from the cold side of the engine to the ambient environment.(?)

How does a variable quantity that is completely dependent on the amount of heat added, the efficiency of the engine and the work output, a quantity which is only whatever heat is left over after the heat added is converted to work equate with the all pervading heat of the ambient environment down to absolute zero?

So "Qhz-Qcz" which is apparently all the heat from Th to 0°k minus all the heat from Tc to 0°K is equivalent to DQh-DQc or the heat supplied to the engine, (or actual joules of energy added) minus the joules left after some of those added joules are converted to work output.

Again, this appears to be an attempt to stuff all the heat of the environment into Qc the heat "rejected" by the engine, whatever that may be, regardless of the actual heat input to the engine or the engine efficiency, that "Qcz" including all the heat down to absolute zero has to be wedged in there one way or another.

Your "note" looks to me like an arbitrary insertion of completely unproven, unsupported, irrational nonsense that doesn't really even make any sense at all on its face.

Nevertheless, it is an accurate representation of the "Carnot efficiency limit" illogical, nonsensical "LAW"

Anyway, I'll try to strain my brain and extend my credulity as far as possible, pressing ever forward line by line, doubtful of my ability to make it all the way to the end. All the way down to the Holy Grail

n=(Th-Tc)/Th The final solution.

Which after all this, by some coincidence, just happens to be the temperature difference.

[Tom Booth]

Taking a steam engine or turbine, where you have actual MASS flow through the engine, then yes, I"d say of course ALL the "internal energy" included in that mass is drawn or flows into and through (and back out of) the steam engine or turbine.

With an external combustion Stirling engine the heat that expands the working fluid is not any physical matter entering and passing through and back out of the engine. So none of the "internal energy" goes in and none gets "rejected".

The only thing "crossing the system boundary" is energy itself.

With a solar Stirling engine the combustion takes place 92 million miles away. Are we supposed to include the "internal energy" of the solar mass as having been "added" and "rejected" in our calculations of a solar powered Stirling engine's efficiency?

Heat is energy not matter, not a "fluid" (caloric).

Measuring the "efficiency" of an external combustion engine is not like measuring tidal flow of the ocean through a turbine where water is flowing through. It's more like the energy of waves on the surface..

Waves are energy. They have no mass or volume, no "internal energy". They are the energy. Likewise with heat passing into a Stirling engine. There is no MASS to exhaust, along with its "internal energy".

[Fool]

Qhz = M•Cv•Th = Qcz + DQh
Qcz = M•Cv•Tc

Note: Qhz-Qcz = DQh-DQc

Good pickup, it was a typographical error. Not a logic error. The correction follows:

"Qhz = M•Cv•Th = Qcz + DQh
Qcz = M•Cv•Tc

Note: Qhz-Qcz = DQh "


Taking:
Qhz = M•Cv•Th = Qcz + DQh
Qhz = Qcz + DQh
Subtracting Qcz from both sides:
Qhz - Qcz = DQh
100 = 400 - 300

Clarification:

Qcz is the energy stored in the engine's gas mass at Tc. It is the same amount as if it were all converted to heat and conducted out of the engine to a cold reservoir, leaving it at zero K. The example I use here is 300 K for 300 Joules.

At the above derivation point we do not know what DQc is. If it were zero, that equation would be true and n would be 100%, always. So indeed we are striving to maximize "n".

A similar equation that is derived from the first ones is :
n•(Qhz-Qcz) = (DQh-Dqc)
0.25•(400-300) = (100-75)

Look for that derivation. I'm sorry that this a brain burner. Thanks for helping to straighten it out. I spent weeks going over this, wasn't real easy for me either. Looking forward to hammering it out straighter.

[Tom Booth]

OK, much better.

So far then, with that little glitch ironed out we have the typical heat/work diagram.

heat-work-100.jpg (147.51 KiB) Viewed 389 times

What we know is (using the diagram's variables)

Th = 400°K
Tc = 300°K
Qh = 100 joules

We don't yet know the work output therefore
We don't know Qc

Though I'm pretty sure this is not exactly accurate, for the sake of simplicity and ease of calculations we are going to say that for each degree of temperature up or down one joule is added or removed.

per/what?

The engines working fluid volume?

Per mole? Cubic centimeter?

I'm having some issues/difficulty with what represents what quantities of what in your analysis at times.

Qhz for example
Qcz
DQh ?

These are aspects of what volume(?) of what(?)

The entire volume of working fluid, each mole, each cubic centimeter ?

Qcz in particularly, seems like a bit of a wildcard.

Qc is heat "rejected" into(?) Qcz which is ?

Ummm .. All the heat between 0°K and Tc (?) I think ?

If the heat Qc (or your DQc ?) leaves (is "rejected" from) the engine, goes out of the working fluid into the environment...

Let's just say for example the engine is 50% efficient. 100 joules are added, 50 are converted to work output so 50 joules are "rejected".

Those 50 joules dissipate into the surroundings and are no longer attached to or associated with any volume of working fluid

So Qcz it would appear, must represent the "cold reservoir" and/or ambient environment at a temperature of 300°K

This is problematic in my mind because I think we are trying to maintain some 1 to 1 correspondence between degrees of temperature and joules (per some specific volume of gas/substance)

We now have 50 joules "rejected" from the working fluid into the surrounding, virtually "infinite" ambient surroundings.

So when you say:

The heat rejected from zero K all the way to Tc will be Qcz

Then proceed to include Qcz in various calculations;

Such as: Qcz = M•Cv•Tc .... etc. etc.

What mass is represented here?

The "heat" has left the working fluid and dispersed into the vast unknown or 300°K "cold reservoir" or the ambient environment "outside" of the "system" boundary.

Basically the nice neat 1 to 1 correspondence between joules an degrees C or K per quantity of "working fluid" has evaporated once the heat leaves the "system".

So in what way it can be stated that: Qcz = M•Cv•Tc or "The heat rejected from zero K all the way to Tc will be Qcz".

Qcz is a mist in the wind. A wisp of 50 joules of heat forever lost to the environment, commingled with the vast, "infinite" cold "reservoir".

Again, in the case of a turbine or steam engine where the "working fluid" is represented by an actual mass of fluid that passes into, through and out of the engine, we could follow each "charge" of intake and exhaust. Maybe even collect the exhaust gases but Qc has no mass.

You appear to be assigning some mass to your Qcz with Qcz = M•Cv•Tc but it is unclear to me where this is coming from or what mass is being represented by M

Since Qc is no longer associated with the engines working fluid but was "rejected" to the outside environment, it would seem to be M in association with Qcz in the above equation can not represent any volume of working fluid, per cubic centimeter or otherwise, so, what then?

Qcz seems like it has become some kind of orphaned mathematical abstraction not associated with any concrete physical reality and I'm having a hard time bringing it down to earth trying to figure out what it might represent in some real world scenario involving a Stirling engine, as in our example, with an input of 100 joules of heat.

We have 100 joules going in

50 say, converted to work

50 "rejected". A finite quantity of energy

Then I'm confronted with Qcz

The heat rejected from zero K all the way to Tc ...

Logically as here expressed Qcz would be 0°K to 300°K and with 1°K = 1 Joule

Qcz = 300 joules of "heat rejected".

Do you see my dilemma?

Qh, the heat supplied was 100 joules

We now, with Qcz, apparently, have 300 joules being rejected

100 joules - 50 joules = 300 joules

Quite the magic trick.

[Tom Booth]

Tom, I really admire your patience with this "Fool".

But like Don Quixote, it is . . . .

In my (practical) world, efficiency is the relation between what you put in, versus what you get out.

Please, read again : What YOU put in !

So if the input is 1 Kelvin, where does "the rest" of the heat comes from, Fool ?
...

Indeed.

As the input is reduced and the efficiency of the engine is increased the discrepancy with where all this heat being "rejected" actually comes from becomes more and more of a glaring problem as the value of Qcz or "All the heat between 0°K and Tc" at any real achievable ambient environment temperature cannot be reduced.

It will always, in a 300°K environment be 300 joules regardless if the input is only 1 joule.


So with my little LTD experiments.

So many joules supplied, so many out as "work". At a supposedly very low efficiency I'm not seeing the tremendous "waste heat" output predicted by the Carnot limit formula.

I do very much value and appreciate "fool's" grappling with this and putting so much sincere time and effort into it while allowing me the opportunity to pick it apart.

[Fool]

M is the Mass of the gas inside the engine.

Qcz = MCvTc applies to the energy in the engine's gas mass. It is such that Cv, specific heat coefficient of that gas, times the mass equals one.

That is, one Joule per degree temperature increase for the entire volume/mass inside the engine.

When the heat is outside the engine, it is no longer part of that equation.

When the volume of the engine is being reduced, the gas is being worked on. The amount of work is related to the absolute pressure, as well as other things. That gas temperature would increase if heat weren't rejected. The work would also increase. So that return stroke is a combination of heat rejection and expenditure of work. The two added together becomes DQc.

That work and heat expenditure must be smaller than the produced work during the expansion stroke.

I like to think heat transfer suffers from a dilution factor. Put 100 Joules into a tiny pebble or BB and it would be at 1000 degrees. Put that same 100 J into a giant boulder or large anvil, and the temperature would hardly increase. An engine running from that 1/2 degree hotter than ambient anvil, would have the same, or larger, volume of gas at the same ambient pressure to be compressed to the same level and have way less forward work produced. That is because Delta T corresponds to Delta Pressure. Probably wouldn't run let alone be as efficient.

The dilution factor I mention is called Entropy. Also known as, heat delivered at a temperature. Same amount of energy, at a lower temperature/pressure. Way less usable.

[Tom Booth]

M is the Mass of the gas inside the engine.

Qcz = MCvTc applies to the energy in the engine's gas mass. It is such that Cv, specific heat coefficient of that gas, times the mass equals one.

That is, one Joule per degree temperature increase for the entire volume/mass inside the engine.

When the heat is outside the engine, it is no longer part of that equation.

...

??????

Your definition of Qcz was as follows:

The heat rejected from zero K all the way to Tc will be Qcz

So,...

We have 300 joules of internal energy in the engine to begin with.

We add 100 joules to "the system" inside the engine.

This raises the temperature from 300°K to 400°K (inside the engine)

We do 50 joules of work

We reject 50 joules of "waste heat"

The working fluid inside the engine should now be what?

400 less 50 work, less 50 rejected.

Back to 300°K

Is this not correct?

The internal energy is the same as before we added the 100 joules

Correct?

Now, I still have the problem of Qcz

"The heat rejected from zero K all the way to Tc will be Qcz"

And you now say:

"When the heat is outside the engine, it is no longer part of that equation"

The 300 joules we started with before adding 100 additional joules and then deducting 50+50=100 joules hasn't moved.

The 300 joules is never "rejected", never "outside the engine".

So, what do you mean by: "When the heat is outside the engine"

Are we relying on quantum teleportation?

50 joules -> out (conducted)
50 joules -> out (work)
300 joules

First the 300 joules are in the engine.

"Qcz = MCvTc applies to the energy in the engine's gas mass"

Then we "do work and reject 50 joules and

The same 300 joules are outside the engine

"

The heat rejected from zero K all the way to Tc will be Qcz"

Qcz still appears to be some mathematical abstract with magical powers of teleportation.

[Tom Booth]

Now, again, take a steam engine.

The WATER we make the steam from is outside the engine.

Let's say that water contains 300 joules per cylinder charge.

We add 100 joules to boil the water to make steam under pressure.

The steam is introduced to the cylinder.

50 joules work

Now, this makes sense.

The water vapor (mass) is "rejected", along with "all the heat". All of the internal energy of that fluid mass. The "latent" 300 joules and the added, but rejected 50 are together exhausted with the spent water vapor.

Is this applicable when evaluating the efficiency of a Stirling engine?

I'm straining to see how.

[Tom Booth]

Well, it looks like teleportation of the 300 joules from inside to outside the engine, or...

The same 300 joules exists BOTH inside and outside?

Of course, after work and heat "rejection" the TEMPERATURE is 300°K inside and outside the engine.

This is the problem with drawing an equivalence between "heat" and temperature and then tossing around equations using abstract variables that don't appear to consistently represent anything in the real world

[Fool]

Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.

It becomes the unused residual that must be recompressed after the expansion stroke. It takes an equivalence of DQc, 75 J, of energy to recompress the gas, but only if kept at Tc. If the temperature is allowed to rise during the recompression, the energy needed for compression, will be grater, a combination of temperature rise and Work.

If Tc is much lower, that compression stroke will require less work input. Hence more efficient.

If expansion stroke is at a temperature higher than Th the power stroke will produce greater energy, and the efficiency will be greater.

The recompression energy is colloquially called the "heat" of compression. It is a combination of the heat rejected, so as to not get hotter than Tc, and the 'Pressure times Delta Volume' Work, that is put back into the gas mass, to return to initial ambient starting conditions. Matt calls it back work.

Starting with:
n=(DQh-DQc)/DQh
and Th and Tc
We don't know what DQh or DQc are. We can calculate Qhz and Qcz from Cv, M (mass), and Th, and Tc. From those, we can calculate:
DQh=Qhz-Qcz.

Calculating DQc is more difficult. It would be:
DQc=DQh•(1-n)

Since 'n' isn't known, we derive an expression using the known values. That turns out to be:
n=(Qhz-Qcz)/Qhz

The derivation is in an earlier post. I may repost for convenience after all corrections are hammered out. With added explanations.

It might help to think of Qcz as a dilution factor. 100 Joules is being mixed in with 300 Joules. It makes it harder to convert 100 Joules fully into work, because recompression is fighting against the residual 300 Joules.

If it went all the way to zero Kelvin, you would convert 400 Joules to Work, cycle back for free, for 100% efficiency. It would however, require Tc to be zero, and that 400 Joules be added each cycle. In other words, the only free return stroke is at zero Kelvin.

[VincentG]

n=(Th-Tc)/Th The final solution.

That shows the logical progression of the mathematical proof.

It's logical to think that higher temperature differences produce higher pressure differences that make greater power to weight and size ratios, ease of construction, operation, and efficiency increases.

1/10 degree temperature, would be like trying to move a piston with 1/10 of a psi. How can that possibly be as efficient as trying to move a piston with 100 psi.

It would take a piston 1000 times larger in area to develop the same force. Power and efficiency sucking bad bulk would be the down fall. I hope this is enough said about the Carnot Theorem.

Does all this math boil down to an opinion here? I can think of a lot of practical ways that lower temperature and pressure would be more efficient than higher. What does power to weight have to do with efficiency?

This formula has no business being applied to a closed cycle Stirling engine that is capable of lowering internal temperature to atmospheric or below. The Stirling cycle is a scavenging cycle that recovers the energy normally lost to open cycles.

[Fool]

Best definition of Qcz:

Qcz=M•Cv•Tc

M = Total Mass of the gas inside the engine.
Cv = Specific heat Coefficient.
Tc = Temperature of the gas when at Tc, and approximately ambient.

'M' is chosen in above example for a one to one relationship with temperature, for convenience.