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

[VincentG]

Delta U = Q - W, where delta U is a change in internal energy.

This seems a much more appropriate equation for hot air engines. If you take Q as the "amount of heat withdrawn from the system in a thermodynamic process" (as written in Wiki), the heat "wasted" to the cold sink need not be considered. It is only transferred back to the atmosphere from where it came. Then, after cooling, internal pressure is below 1atm and external atmospheric energy is transferred back to the system on the return stroke(adding to positive work). Only the heat required by the gas to expand against the atmosphere(wneg) and do additional positive work is considered as converted.

[Tom Booth]

The equation W = Qh - Ql, applies to all processes, and the equations: n=1-Qh/Ql=1-Tl/Th only applies to cyclic process.
...

U = Qin + Win + Chemical energy release + etc... Also minus energy out...
...
Mass contains internal energy. That can be kinetic/.5mv^2, potential/mgh, or TPmCvR thermal energy with Temperature being the measure.

...
Linearly dependent means that a dependent variable say, Y, changes with another independent variable say, X, graphed onto a straight line. The equation for it is of the form:

Y = mX + b
Or:
Q=CvT + k

b and the substitute k are the Y intercepts. Meaning when X is Zero, Y will equal b, or k. b in the first equation, k in the second.

...

The Y intercept will be zero, b or k will be zero.

If you look at a chunk of air at 300 K, it will have a starting point internal energy of Ql = 300•Cv + k. If you raise it up to 400 K, it will have an internal energy of Qh = 400•Cv + k.

Qh - Ql = delta Q = (400•Cv + k) - (300•Cv + k)
Delta Q = 400•Cv + k - 300•Cv - k

The k's cancel, Cv is pulled out:

Delta Q = Cv•(400-300)

The starting point doesn't matter it becomes:

Delta Q = Cv•DeltaT

It becomes dependent only on the slope, Cv.
The Y intercept, k equals zero, disappeared too.

One more thing. The total value of internal energy equals the heat added from absolute zero Kelvin so the equation:

Q = Cv•T + k

Reduces to:

Q = Cv•T

Again it is a function of only the Slope. Cv or 1/Cv doesn't matter, as the slope, Cv, cancels out when put into the efficiency equation.
...

Qhin, Qhout. Qhout=-Qhin, so to speak. In engineering a variable is defined as positive in one specific direction, after the analysis is finished, if it has a negative Value, that just means it, a force or motion, is in the opposite direction.
...

...

For the sake of everyone, for general reference, could you please provide a key defining ALL of your variables if you don't mind.

Just for example: you wrote:

equation:

Q = Cv•T + k

Key
Q is blah blah
C is blah blah
v is etc.
etc.

Immediately before or after each new equation introduced would be helpful, especially if the meaning of the variable might be altered in different contexts.

Ps, for the record, your math didn't seem to add up correctly. I'm not going to analyze it. It seemed like you were trying to prove that a single process could be more efficient than for complete cycle. ...

From what you say here, If you read my posts at all, you apparently somehow missed the entire argument or the point(s) being addressed.

Well, it was long and rambling and repetitious and you seem to assume erroneous and misdirected so not worth any serious analysis or thought.

I'm disappointed, you had stated earlier you were giving all this careful thought and consideration over the course of several days.

Anyway, if you could provide that "key" defining the various algebraic symbols or letters including "substitutions" and so forth or variations or changes or potential double meanings or alterations of meaning in different contexts if necessary

I think we have agreed at the beginning that Q represents a quantity of heat transfered in joules, is that correct and are we still adhering to that definition?

Or can Q now somehow represent ALL "internal energy" down to absolute zero?

[matt brown]

Oh! And Matt, if the world is dominated by adiabatic processes, how does heat get into an engine? It doesn't.

Imo, the world is dominated by real processes, neither adiabatic, nor isothermal. But it is limited by temperature and pressure extremes. Those extremes will dictate maximum efficiencies. Those maximums will be further eroded by reality. The processes will be between adiabatic and isothermal and material and design constraints. Smiles

Hey Fool, my previous:

Furthermore, the whole Carnot buzz is a maximum that only directly applies to common isothermal cycles and ignores common adiabatic cycles that dominate the world.

means:

Furthermore, the whole Carnot buzz is a maximum that only directly applies to common isothermal engine cycles and ignores common adiabatic engine cycles that dominate the world.

[Tom Booth]

Oh! And Matt, if the world is dominated by adiabatic processes, how does heat get into an engine? It doesn't.

Imo, the world is dominated by real processes, neither adiabatic, nor isothermal. But it is limited by temperature and pressure extremes. Those extremes will dictate maximum efficiencies. Those maximums will be further eroded by reality. The processes will be between adiabatic and isothermal and material and design constraints. Smiles

Hey Fool, my previous:

Furthermore, the whole Carnot buzz is a maximum that only directly applies to common isothermal cycles and ignores common adiabatic cycles that dominate the world.

means:

Furthermore, the whole Carnot buzz is a maximum that only directly applies to common isothermal engine cycles and ignores common adiabatic engine cycles that dominate the world.

What is "Carnot buzz"? I assume you mean the efficiency limit. Isn't that supposed to (allegedly) apply to ALL heat engines?

[matt brown]

Therefore, looking around we find the Carnot Limit to save the day and restore order to the universe.

Therefore we can assume that:

η = 1 – (Qc / Qh). = 1 – (Tc / Th)

Which perhaps it has been noticed is the equation I chose for the title heading for this thread.

But, there is a minor problem with this. Well, a couple problems really, and maybe not so minor.

In the efficiency equation: η = 1 – (Qc / Qh)

Qh and Qc represent quantities of energy or joules of heat (or work) using Qc as the baseline or zero. In other words, starting at 300K (ambient temperature) we have added zero "heat" to Qc to get 300k, that is a given. we add, say, 100 joules so that Qc = 300K and Qh with tbe addition of 100 joules to the given 300 becomes 400K (by the addition of 100 joules to the 300 ambient).

Both above 'equations' are best described as expressions with limited validity. Science is loaded with such expressions and care is required when plugging in real values.

Tom, you're right on my beef with Fool's starting point where work equals the difference between two quantities of temperature. This is so esoteric as to be meaningless.

I wouldn't waste time wading thru the alphabet soup of thermodynamics, since nearly everything has conditions. Here's an interesting example of how careless subbing values into 'equations' can be...

Consider momentum eq. where p=mv
vs kinetic energy eq. where KE=(mv^2)/2

Let's arbitrarily take m=2 whereby p=2v and KE=v^2
If we further arbitrarily take v=1 then p=2 and KE=1 for the same mass or that 1 unit of kinetic energy magically became 2 units of momentum energy. Obviously, total nonsense, and the error is invalid v=1, not m=2.

[matt brown]

What is "Carnot buzz"? I assume you mean the efficiency limit. Isn't that supposed to (allegedly) apply to ALL heat engines?

Carnot limit only applies to compression cycles. My beef with it is that this simple math equation indicating efficiency only works for isothermal compression cycles. The adiabatic compression cycles (Otto, Brayton) have far less efficiency than the ratio of cycle temperature extremes, and the calcs for these (adiabatic cycles) is a tad more complex. So, in a world dominated by ICE adiabatic cycles, who cares about wishy-washy pie-in-the-sky isothermal buzz ???

[Tom Booth]

...The adiabatic compression cycles (Otto, Brayton) have far less efficiency...

If I read you correctly,...

The efficiency "limit" is an upper limit on efficiency. so would apply to cycles that are LESS efficient. as well.

[Tom Booth]

Maybe a graphical depiction will help

Resize_20240108_061624_4202.jpg (150.89 KiB) Viewed 1352 times

Ambient temperature is 300°K
Qc = 300°K
Add 100 joules to get
Qh = 400°K

W = Qh(heat supplied) - Qc(heat "rejected")

Plotting that for a 100% efficient engine:

Resize_20240108_062118_8264.jpg (177.55 KiB) Viewed 1352 times

The second law says (supposedly):


"an efficiency of 1, or 100%, is possible only if there is no heat transfer to the environment (Qc=0)"

The "environment" IS Qc (ambient 300°K)

How can you "transfer" the environment to the environment?

When Qh (100 joules) is supplied at 400°K Qc at 300°K is a pre-existing baseline or ground state.

Qh is not 400 joules it is 100 joules added ON TOP OF ambient at 300°K

There is NO SUCH THING as a "cold reservoir".

Qc is 300 joules of HEAT
Qh is 100 joules ON TOP OF Qc

How can you say that if 100 joules are supplied 100 joules of work output is not enough to have 100% efficiency?

No, supposedly for 100% efficiency for every 100 joules supplied as heat 400 joules needs to be converted to work.

The ∆T is between 300°K and 400°K

If you want to conceive Qc as being included or "contained within" Qh so 100 joules at 400°K is "really" 400 joules because Qh is 100 joules ON TOP OF the pre-existing 300 (ambient) and 300 joules are "rejected" for each 100 joules converted to work, then Qc just cancels out.

"Carnot efficiency" is representing 100% thermal efficiency (100 joules transfered into the engine with 100 joules work output each cycle) as only 25% "Carnot efficiency".

Carnot efficiency = 1-Tc/Th = 1-300/400 = 1-3/4 = 1/4 = 25%

25% of 400 joules is 100 joules = Qh = W

Qc is a null value. It cancels out.

Effectively 300°K is zero. The pre-existent ambient "baseline" thermal bath of HEAT we live in, not a "cold reservoir".

On an absolute scale, there is no "cold reservoir".just heat, from 0°K on up.

The 400° K "hot reservoir" is just an additional 100 joules ON TOP OF the 300°K reservoir of ambient heat.

So 25% Carnot efficiency = 100% thermal efficiency (100 joules of heat in 100 joules of work output each cycle, cycling between 300°K and 400°K, not between 400°K and zero°K)

[Tom Booth]

Now, of course, if A Stirling engine is behaving during part of the cycle as a heat pump and so stealing even a fraction of a Joule from "the cold side", what would the chart (second one above) look like?

The sinusoidal S curve would dip slightly below 300°K during the "work out" phase and the "Carnot efficiency" would be slightly higher.

Carnot efficiency = 1-Tc/Th = 1-299/400 = 1-0.7475 = 25.25%

By diverting just a small fraction of the work output into refrigeration. A 25.25% Carnot efficiency would result.

25.25% efficiency doesn't sound like any threat to the established order of the universe to me. We still have another 299 joules of ambient heat in the surrounding atmosphere that has not yet been touched at all

viewtopic.php?f=1&t=5410


Of course the heading should have been:

100% efficiency (+) is it possible?

I hate when a typo gets into a topic heading and can't be edited, but oh well

[Tom Booth]

Delta U = Q - W, where delta U is a change in internal energy.

This seems a much more appropriate equation for hot air engines. If you take Q as the "amount of heat withdrawn from the system in a thermodynamic process" (as written in Wiki), the heat "wasted" to the cold sink need not be considered. It is only transferred back to the atmosphere from where it came. Then, after cooling, internal pressure is below 1atm and external atmospheric energy is transferred back to the system on the return stroke(adding to positive work). Only the heat required by the gas to expand against the atmosphere(wneg) and do additional positive work is considered as converted.

I think we are in agreement here?

Resize_20240108_203556_6261.jpg (174.43 KiB) Viewed 1218 times

Though with a few provisos:

Delta U = Q - W, where delta U is a change in internal energy.

I would say the S curve in the above diagram shows the change in internal energy in the course of the cycles.

This seems a much more appropriate equation for hot air engines. If you take Q as the "amount of heat withdrawn from the system in a thermodynamic process" (as written in Wiki)

Not really sure what this is saying, I'd have to see the context, in my diagram Qh is the heat supplied to the engine at the high temperature.

Qh is not Q general or heat generally nor is it temperature. It is the thermal energy in Joules actually entering into the working fluid of the engine to raise the temperature from 300°K to 400°K (which is certainly NOT the same thing as raising the temperature from 0°K!)

the heat "wasted" to the cold sink need not be considered. It is only transferred back to the atmosphere from where it came.

Actually it never leaves, so cannot be "transfered back" the internal energy fluctuates following the S curve "above" it.

Starting at 300°k (per mole or some fixed quantity of the working fluid) 100 Joules as thermal energy are added raising the temperature 100 degrees to 400°K with 100 Joules work output the temperature falls back down to 300°K

The "wasted" heat is not so much "wasted" as just never touched at all. It did not enter into the engine. It is represented by the 300°K which is simply a given or the baseline or starting point before additional heat was added.

Then, after cooling, internal pressure is below 1atm and external atmospheric energy is transferred back to the system on the return stroke(adding to positive work).

IMO for this to take place there would have to be some refrigerating effect that brings the pressure down below atmospheric. The rubber band car rolling back FURTER than pushed. The car pushed up the hill rolls back down to the bottom of the hill and once it reaches the bottom KEEPS GOING until it hits something that stops it.

Only the heat required by the gas to expand against the atmosphere(wneg) and do additional positive work is considered as converted.

I would say Qh is "the heat required by the gas to expand against the atmosphere(wneg) and do additional positive work"

Qh is converted to work.

My S curve separates the increase in U (internal energy) due to Qh (heat addition) from the decrease due to Work output in time, but heat input and work output often happen simultaneously.

However, the modern school of thought taught in academia is that only 25% of Qh can be converted into work.

Of the 100 Joules represented by Qh, according to the modern interpretation of the Carnot limit, 75% absolutely MUST be 'rejected" back to the "sink" or back to atmosphere so that only 25% or 25 Joules At AN ABSOLUTE MAXIMUM can be converted to work.

"Real engines are much LESS efficient, real engines have friction, heat losses, vibration etc

Of Qh which is 25% of "the heat" down to ABSOLUTE ZERO the academic claims we can only utilize 25% of that 25% or 25% of Qh.

75% or more of Qh or 75 joules of the heat entering the engine to expand the working fluid MUST flow through to the 'cold reservoirs'.

Why?

Because?

Because why?

Well, "entropy'? Otherwise you would have perpetual motion, over unity, the creation of energy out of nothing, that would violate conservation of energy.

But I added 100 joules, how is converting more than 25% of that a violation of conservation of energy???

Carnot rules!!

You can't beat Carnot!!!

Go ahead and try building an engine that violates the Carnot limit. No one has done so in 200 years!!!

Personally I think the academics, Kelvin, Clausius or one of them made a slight mathematical slipup, due to conceptualizing "All the HEAT" as all the internal energy, all the caloric fluid, down to absolute zero rather than all the HEAT transfered or Qh.

For heat to be Transfered there needs to be a temperature difference.

At 300°K everything is in equilibrium. You don't include the thermal energy in equilibrium at and below 300°K in the equation of heat transfered. You can't say after adding 100 joules that you need to "reject". 375 joules to convert 25 Joules into work.

But that is what the academics are teaching their students. They are using math that was derived from Caloric theory.

Once an idea has been put into a mathematical formula it is set in stone. But if heat is energy then the math being used, or the way it is being interpreted is WRONG! Horrifically wrong. Absolutely, insanely wrong. It's a mathematical clusterfuck.

[Tom Booth]

Just FYI, I put the Wikipedia quote in Google and word for word got zero exact matches.

The original passage reads:

The first law of thermodynamics is often formulated as the sum of contributions to the internal energy (U) from all work (W), done on or by the system, and the amount of heat (Q) supplied or withdrawn from the system in a thermodynamic process.[1]

The word "supplied" from "supplied or withdrawn" had been, I assume, inadvertantly left out, but harmonizes better. Qh is heat supplied Qc however is NOT "heat "wasted" to the cold sink.

This is a mathematical gaffe.


Earlier in the thread "fool" wrote:

One more thing. The total value of internal energy equals the heat added from absolute zero Kelvin...

In context here: viewtopic.php?f=1&t=5601&start=15#p21199


"....the heat added...." ?????

From "absolute zero" ????


This appears to be the source of the error, or one of the errors.

Heat was "added" to a pre-existing ambient thermal bath at 300°K in complete thermal equilibrium. Heat was not "added from absolute zero", heat was added from 300°K as a starting point.

If you view heat as a fluid filling a container and the container is already full to the 300 milliliter mark and you add 100 milliliters you then have 400 milliliters of 'HEAT' fluid.

This is why I asked for clarification at the start of this thread. viewtopic.php?f=1&t=5601#p21059

"How do you define "heat"?

Is heat energy TRANSFERED? The 100 milliliters?

OR

Is heat the whole 400 milliliters?

The current Carnot "Buzz" has it that heat added is the 100 milliliters transfered into the engine. But when it comes to converting the heat into work....

Now, all of a sudden "heat" is defined as the entire 400 milliliters, the 100 added AND the 300 that was already in there!!!!

Now to use up the 100 milliliters (or J oules) I have to convert all 400 (300 given PLUS 100 added) into heat. On top of that only 25% of the 100 added can be converted to work. 75% of the heat added PLUS the 300 pre-existing Joules must be "rejected" all the way down to absolute zero if you want to convert the 100 Joules ALL into work, or even 25% of the 100 joules supplied.

100 Joules added = Qh
25 Joules converted to work
375 Joules "rejected" to the sink = Qc (+75% of Qh)
???

You can't fully utilize even 25% of the 100 Joules you added without sending "the other" 375 Joules of "heat", or the totality of the remaining internal energy (down to absolute zero) through to the "cold reservoir".

You can't define heat as energy transfered and then turn around and define heat as "all the heat" down to absolute zero, 300 joules of which has just been sitting there at thermal equilibrium and NOT EVER TRANSFERED AT ALL!!!!


I really don't know how scientists and mathematician of any kind of sound mind could look at this for 200 years and continue considering it a "LAW of the universe".

It's pure lunacy!

[Tom Booth]

Well, I said: You can't define heat as energy transfered and then turn around and define heat as "all the heat" down to absolute zero.


But that is exactly what is being done on a routine basis, day after day, year after year:

https://youtu.be/V3nNgygrmsI?si=i-UqFOzWlsUNID-o


So if I run my LTD engine on a cup of coffee, it cannot be 100% efficient at utilizing the "heat" transfered from that cup of coffee unless the engine can take the heat from the coffee and somehow blow solid chunks of dry ice out the exhaust, (if it had an exhaust).

It would have to refrigerate the cup of coffee down to absolute zero.

I guess that sounds reasonable.

[Tom Booth]

So if anyone tells you they have an engine that exceeds the Carnot limit " YOU KNOW THAT THEY ARE LYING!"

Let's say I supply 100 Joules of heat Qh at 500°K

Taking ambient, at 300°K as our "cold reservoir"

The Carnot efficiency limit is calculated:

Efficiency= 1-Tc/Th
1-300/500
1-3/5
1-0.6
0.4 = 40%

Now on an absolute scale the ∆T 500-300=200

That 40% is the ∆T

The Carnot efficiency IS the ∆T always

Carnot said that the power of a heat engine results from the "fall" in temperature and Clausius, Kelvin or somebody put that in mathematical terms and put it in all the textbooks as the gospel of Carnot.

So naturally the Carnot limit equation is simply a mathematical representation of the temperature difference - always.

The difference between T1 and T2 on the absolute temperature scale is always the Carnot efficiency. Always.

500-300=200
200 is 40% of 500

So, the Carnot efficiency is 40%

O.40×500=200

Now since in someone's mind Qh, the heat supplied, is no different than "all the heat" all the way down to absolute zero we can now apply that 40% to our 100 Joules of heat input because if 500 Joules can equal 100 Joules than the reverse must also be true, right?

So, you can only convert, AT MOST 40 Joules of heat out of every 100 supplied into work per cycle and must "reject" the other 60 Joules out of every 100 supplied.

If anyone tries to tell you otherwise, that they can get 41% efficiency from their engine, "You KNOW" for a fact "they are LYING".

It is "impossible" to utilize even one more Joule out of the 100 with that 200°∆T

If they say their engine can utilize 41 joules of every 200 supplied they are SCAM artists trying to steal money from investors, or just ignorant fools! Don't believe it!

https://youtu.be/LUoUb4hGMH8?si=pGY4s56WN-V57o7k


They deserve everything they have coming to them, the dirty filthy LIARS!!!

[Tom Booth]

Does everyone (or anyone at all) see what's going on here?

At a ∆T of 300°K and 500°K
Qh represents 40% of ALL the "HEAT" down to absolute zero

But how is "HEAT" being defined?

"ALL" the "internal energy" down to absolute zero is not the same thing as Qh or the heat that was actually transfered into the engine, but they are being treated as equivalent!

Or, not exactly equivalent but the 40% value which applies to all the heat (sic. Not transfered so not "heat") on the absolute scale is used in the efficiency calculation, but NOW instead of applying it to ALL the "heat" (sic.) down to absolute zero, we are going to apply it to the finite number of Joules transfered into the engine, whatever that happens to be.

Resize_20240109_015054_4444.jpg (266.18 KiB) Viewed 1168 times

I'm hoping that "fool" will be able to explain how and why such a substitution is justified.

How does a finite number of Joules supplied suddenly become ALL the internal energy down to absolute zero?

[Tom Booth]

OK, let me get this straight.

Qh, the heat added or supplied at some arbitrary temperature, as a ratio of the absolute temperature is some percentage of "all the heat" down to absolute zero.

For example:

Qh may be 10,000 joules supplied to the engine.

If supplied at a temperature of 600°K and ambient happens to be 300°K the difference is 300°K

300 is 50% of 600

Therefore only 50% of the 10,000 Joules can be utilized for "work" output

Therefore half the heat supplied can be converted to work and half the heat or more MUST absolutely be "rejected", and go out of the engine to the "sink".


Now what if we use different temperatures?


Qh is the same 10,000 joules supplied to the engine.

If Qh is supplied at a temperature of just 400°K and ambient is still 300°K the difference is 100°K

100 is 25% of 400

Therefore only 25% of the 10,000 Joules can be utilized for "work" output

Therefore one quarter of the heat supplied can be converted to work and three quarters of the heat MUST absolutely be "rejected", and go out of the engine to the "sink".

2,500 Joules work output
7,500 Joules "waste heat"

Let's reduce the ∆T more, like our little LTD engine.

Qh is still 10,000 joules supplied to the engine.

If supplied at a temperature of 310°K, just 10°K above ambient. This is a "heat of your hand" supper efficient Kontex LTD

10 is about 3% of 310 (3.226% to be exact)

Therefore only 3% of the 10,000 Joules supplied by the heat of your hand can be utilized for "work" output

97% of every 10,000 Joules MUST absolutely be "rejected", and go out of the engine to the "sink".

Seems like the cold side should be nearly as hot as your hand since 97% of the heat is going straight through. Right?

How far do we need to go before cognitive dissonance starts to tell us something is wrong with this picture?

Let's try an "Ultra LTD" that can run on just a 2°C ∆T

This is the most amazingly efficient Stirling engine ever built, it can run on a wet piece of paper using environmental heat.

That calculates to LESS THAN 1% efficiency.(0.66%) but, ignoring that "fact", this is arguably the most efficient engine ever built.


But nearly all of the 10,000 Joules must be rejected. (According to the Carnot limit).

About 9,300 of every 10,000 Joules supplied goes towards "waste heat" so the Ultra efficient "Ultra LTD" can only utilize about 600 of every 10,000 Joules of heat supplied.

This is hopeless. The harder I try and the more efficient I make my engine the LESS efficient it becomes.

I give up.

Stirling engines suck. These little LTD engines that can run on a whisper of heat are horribly inefficient. No point in trying really.

So what if it can run on ambient heat using only a wet piece of paper.

Carnot rules. You can't beat the Second Law of Thermodynamics.

https://youtu.be/ARD3ctp80ac?si=oMnHrkIHWZ4-IN3A