Tom Booth
Registered Senior Member
All off topic as usual, but I could care less if YOU or "FOOL" take me seriously or think I'm "credible" or drop dead. Your a couple of idiots.
Origin and verification of e=(Th-Tc)/Th
- Thread starter Tom Booth
- Start date
exchemist
Valued Senior Member
Really? You could care less means you care a lot. I think you mean you couldn’t care less.
But that proves it: your abject refusal to address the key issue can only be because the truth is too uncomfortable for you to face.
Your experiments are worthless, because the amount of heat predicted by thermodynamics to be emitted is below the detection limit of your experiment. So it’s no wonder you don’t see anything, and the fact you don’t tells you nothing at all about the thermodynamic efficiency of your engine. You are wasting your time.
Tom Booth
Registered Senior Member
Your entire argument is dumb, irrational, ignorant, unreasoning, moronic, and stupid on its face.
That "I could care less" is a common expression. Your chronic "word policing" is another symptom of your mental condition. I don't care at all, but I really COULD care less, but I certainly DON'T care "a lot" that's for certain.
As I already knew from my own painstaking research as well as more than 10 years of unanswered questions, there is no, and never has been any experimental verification of the Carnot limit.
Probably because anyone who ever tried came up against the same negative results. Of course it is "very difficult" to validate a falsehood experimentally, try and try as you might.
Case closed.
The "Carnot limit" can go the way of other myths and fairy tales.
exchemist
Valued Senior Member
Among the ignorant, yes.
There is always the following thread in the Stirling Engine Forum:
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["Tom Booth"]
["Fool"]
["Tom Booth"]The "first law" thermal efficiency equation is:
Qh=Ql+W
Which I don't have a problem with. It does not imply that work can exceed the heat input, but neither does it suggest that the work output in Joules cannot match the heat input in Joules.
The first law does not forbid the existence of an engine more efficient than the Carnot limit, not at all.["/Tom Booth]
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
That is Carnot's Theorem derived from the first and second laws. The first law, conservation of energy, manifests itself into a great many following areas. Maximum efficiency is just one of those.
Here is a more rigorous proof:
Relationship to ideal gas law:
en.m.wikipedia.org
Scroll down to the appropriate section. It uses a slightly higher level of mathematics.
There are other more complicated proofs should you care for them. It has been proven many different ways.
The Carnot Theorem is just for thermodynamic efficiency. Kinematic efficiency reduces the efficiency even further. And not following the ideal engine cycles, also reduces the efficiency.
Efficiency is a valent goal to pursue, but mostly a waste of effort for a starting concern. Power to weight ratio is a much better pursuit. Think American Drag Racer. 1/4 mile per 12 gallons. 48 gpm. Excellent.
["Fool"]
["Tom Booth"]
["Fool"]
["Tom Booth"]The "first law" thermal efficiency equation is:
Qh=Ql+W
Which I don't have a problem with. It does not imply that work can exceed the heat input, but neither does it suggest that the work output in Joules cannot match the heat input in Joules.
The first law does not forbid the existence of an engine more efficient than the Carnot limit, not at all.["/Tom Booth]
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
That is Carnot's Theorem derived from the first and second laws. The first law, conservation of energy, manifests itself into a great many following areas. Maximum efficiency is just one of those.
Here is a more rigorous proof:
Relationship to ideal gas law:
Thermodynamic temperature - Wikipedia
Scroll down to the appropriate section. It uses a slightly higher level of mathematics.
There are other more complicated proofs should you care for them. It has been proven many different ways.
The Carnot Theorem is just for thermodynamic efficiency. Kinematic efficiency reduces the efficiency even further. And not following the ideal engine cycles, also reduces the efficiency.
Efficiency is a valent goal to pursue, but mostly a waste of effort for a starting concern. Power to weight ratio is a much better pursuit. Think American Drag Racer. 1/4 mile per 12 gallons. 48 gpm. Excellent.
["Fool"]
["Tom Booth"]
Last edited:
The following had a great graphic from Wikipedia. I don't know how to insert it yet. After the graphic is my derivation. It got misunderstood in the Stirling Engine Forum Site. Equivalence of Ideal Carnot with Ideal Stirling, cycles:
["Fool"]
From Wikipedia:
Stirling_cycle_pV.svg.png
This line was added to separate graphic from my derivation that follows this line:
Work Energy coming out of the engine is positive.
Heat energy going in is positive.
Vt = Volume top dead center.
Vb = Volume bottom dead center.
M Mass of the working gas.
R gas constant for working gas
ln() = Natural log function.
Th and Tc Temperatures hot an cold
W12 work for each process, respectively. Four total. 2 isothermal. Two constant volume.
Q12, Heat transfered for each of four process respectively.
The following is for the ideal Stirling Cycle as depicted in the above graphic.
W12 = M•R•Th•ln(Vb/Vt) positive because work is coming out.
W23 = 0. Zero volume change.
W34 = M•R•Tc•ln(Vt/Vb) negative because work is going in.
W41 = 0. Zero volume change.
Q12 = W12 = M•R•Th•ln(Vb/Vt) positive because the energy is going in.
Q23 = -M•Cv•(Th-Tc) negative because the energy is coming out.
Q34 = W34 = M•R•Th•ln(Vt/Vb) negative because the energy is coming out. Vt smaller than Vb, ln() function negative for values less than one.
Q41 = -M•Cv•(Tc-Th) positive because the energy is going in. Tc smaller than Th, two negatives combined to become a positive.
Q23 and Q41 cancel each other, being an equal and opposite regenerator processes. Ideally.
n = Wout/Qin = (W12 + W34) / (Q12)
Substituting:
n={(M•R•Th•ln(Vb/Vt))+(M•R•Tc•ln(Vt/Vb)}/{M•R•Th•ln(Vb/Vt)}
Using the log identity ln(x) = -ln(1/x), for ln(Vt/Vb) = -ln(Vb/Vt), the equation becomes:
n={(M•R•Th•ln(Vb/Vt))-(M•R•Tc•ln(Vb/Vt)}/{M•R•Th•ln(Vb/Vt)}
Canceling M•R•ln(Vb/Vt) top and bottom:
n=(Th-Tc)/Th
Ideally and a maximum for the temperatures given.
It is direct observable science from the observed relationship between temperature, pressure, volume, energy, mass and the linear coefficients of heat Cv, and ideal gas constant R for the real gas, Rn for Nitrogen, or whatever.
Again this doesn't equate Q to T. The differences just cancel out in the equations of n=W/Q, PV=MRT, and Q=MCvT, for a full cycle. Doesn't work for single strokes. Real engines will be worse.
["Fool"]
Those are verifications of the Carnot Theorem.
["Fool"]
From Wikipedia:
Stirling_cycle_pV.svg.png
This line was added to separate graphic from my derivation that follows this line:
Work Energy coming out of the engine is positive.
Heat energy going in is positive.
Vt = Volume top dead center.
Vb = Volume bottom dead center.
M Mass of the working gas.
R gas constant for working gas
ln() = Natural log function.
Th and Tc Temperatures hot an cold
W12 work for each process, respectively. Four total. 2 isothermal. Two constant volume.
Q12, Heat transfered for each of four process respectively.
The following is for the ideal Stirling Cycle as depicted in the above graphic.
W12 = M•R•Th•ln(Vb/Vt) positive because work is coming out.
W23 = 0. Zero volume change.
W34 = M•R•Tc•ln(Vt/Vb) negative because work is going in.
W41 = 0. Zero volume change.
Q12 = W12 = M•R•Th•ln(Vb/Vt) positive because the energy is going in.
Q23 = -M•Cv•(Th-Tc) negative because the energy is coming out.
Q34 = W34 = M•R•Th•ln(Vt/Vb) negative because the energy is coming out. Vt smaller than Vb, ln() function negative for values less than one.
Q41 = -M•Cv•(Tc-Th) positive because the energy is going in. Tc smaller than Th, two negatives combined to become a positive.
Q23 and Q41 cancel each other, being an equal and opposite regenerator processes. Ideally.
n = Wout/Qin = (W12 + W34) / (Q12)
Substituting:
n={(M•R•Th•ln(Vb/Vt))+(M•R•Tc•ln(Vt/Vb)}/{M•R•Th•ln(Vb/Vt)}
Using the log identity ln(x) = -ln(1/x), for ln(Vt/Vb) = -ln(Vb/Vt), the equation becomes:
n={(M•R•Th•ln(Vb/Vt))-(M•R•Tc•ln(Vb/Vt)}/{M•R•Th•ln(Vb/Vt)}
Canceling M•R•ln(Vb/Vt) top and bottom:
n=(Th-Tc)/Th
Ideally and a maximum for the temperatures given.
It is direct observable science from the observed relationship between temperature, pressure, volume, energy, mass and the linear coefficients of heat Cv, and ideal gas constant R for the real gas, Rn for Nitrogen, or whatever.
Again this doesn't equate Q to T. The differences just cancel out in the equations of n=W/Q, PV=MRT, and Q=MCvT, for a full cycle. Doesn't work for single strokes. Real engines will be worse.
["Fool"]
Those are verifications of the Carnot Theorem.
The question is why are both Carnot and Tom correct? The answer is buffer pressure. The calculation I just provided works for both an engine running in a vacuum or with a buffer pressure. Tom is working exclusively with a buffer pressure. Although the efficiency equations are not affected. the heat gets rejected in different ways depending on having a buffer pressure or not.
The fact that an isothermal power stroke is 100% efficient. Heat in equals work out only applies to zero buffer pressure. The derivation, using calculus, assumes integration to zero pressure. And or zero Kelvin. Running an atmosphere or buffer pressure, the zero in calculus is now the buffer pressure.
A nonzero buffer means the power stroke is no longer 100%. Example, say 30%. 70% of the heat coming in goes out as work through the piston into the cold sink known as the atmospheric pressure. This energy is lost directly to the outside air. It is gone. It doesn't get rejected as heat through the cold plate. In fact the cold plate gets colder during this stroke. The gas, by definition, isothermal, stays hot.
The regenerator and displacer change the gas temperature to cold Tc, while saving that energy difference and using zero work (constant volume).
It gets worse for the return stroke. The work lost during the power stroke, is returned by atmospheric pressure. The gas adiabatically heats from the retuned work. It heats less than the work from the power stroke, because it is colder than for the expansion stroke. The cold plate might be slightly colder than the atmosphere. The air rushing back into the piston cylinder is also slightly colder. That is called hysteresis. That cold air is pulled into Tom's insulation, further cooling the cold side of the engine and reducing the work returned.
The displacer and regenerator return the energy back to hot Th.
In other words, for a 80% Carnot rule engine, the buffer pressure eats 70% of the rejected heat as atmospheric work. Leaving 10% or less to come out of the returning piston and cold plate. 10W Qh, 2W power out, 7W heat removed by working the atmosphere directly. 1W removed by the cold plate, which is fanned by the piston pushing air in and out
Through the insulation.
Now just imagine how those figures would decrease if the engine could only absorb 1/10 or 1/100 that amount of power?
It may very well reject more heat with insulation than not, for atmospheric LTD engines, running unloaded, only.
If anything, Tom's experiment verified Carnot. It still needs measurement of work output to confirm that theory. Thanks Tom.
The fact that an isothermal power stroke is 100% efficient. Heat in equals work out only applies to zero buffer pressure. The derivation, using calculus, assumes integration to zero pressure. And or zero Kelvin. Running an atmosphere or buffer pressure, the zero in calculus is now the buffer pressure.
A nonzero buffer means the power stroke is no longer 100%. Example, say 30%. 70% of the heat coming in goes out as work through the piston into the cold sink known as the atmospheric pressure. This energy is lost directly to the outside air. It is gone. It doesn't get rejected as heat through the cold plate. In fact the cold plate gets colder during this stroke. The gas, by definition, isothermal, stays hot.
The regenerator and displacer change the gas temperature to cold Tc, while saving that energy difference and using zero work (constant volume).
It gets worse for the return stroke. The work lost during the power stroke, is returned by atmospheric pressure. The gas adiabatically heats from the retuned work. It heats less than the work from the power stroke, because it is colder than for the expansion stroke. The cold plate might be slightly colder than the atmosphere. The air rushing back into the piston cylinder is also slightly colder. That is called hysteresis. That cold air is pulled into Tom's insulation, further cooling the cold side of the engine and reducing the work returned.
The displacer and regenerator return the energy back to hot Th.
In other words, for a 80% Carnot rule engine, the buffer pressure eats 70% of the rejected heat as atmospheric work. Leaving 10% or less to come out of the returning piston and cold plate. 10W Qh, 2W power out, 7W heat removed by working the atmosphere directly. 1W removed by the cold plate, which is fanned by the piston pushing air in and out
Through the insulation.
Now just imagine how those figures would decrease if the engine could only absorb 1/10 or 1/100 that amount of power?
It may very well reject more heat with insulation than not, for atmospheric LTD engines, running unloaded, only.
If anything, Tom's experiment verified Carnot. It still needs measurement of work output to confirm that theory. Thanks Tom.
He hasn't been banned.