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

[VincentG]

The compressed air thing is interesting in its own right, but my main point is that there seems to be no evidence that heat energy is actually converted to GPE, just that heat energy can be used to gain GPE.

[Tom Booth]

The compressed air thing is interesting in its own right, but my main point is that there seems to be no evidence that heat energy is actually converted to GPE, just that heat energy can be used to gain GPE.

The evidence I see is that heat expands a gas in a Stirling engine that does work, including lifting a weight or whatever. (Gravitational potential energy).

If the heat was not converted to work the gas would remain expanded, and the piston would not be able to return.

But the piston does return. The gas cools and contracts and the piston returns to TDC.

Now, the ongoing debate has.been, does it return due to energy stored in the flywheel pushing it? And/or from "rejecting heat" to a sink?

Or

Due to the heat having been converted to work leaving the working fluid cold so it contracts? (Or put another way, is pushed back by atmospheric pressure.

This has been the focus of many of my own experiments.

If the flywheel is eliminated and conductive cooling is eliminated that only leaves heat conversion into work.

No flywheel, no displacer, no heat sink:

https://youtu.be/LG09AXAjpio

Process of elimination.

[Tom Booth]

Other evidence:

Heat goes in the bottom hot side of the engine but no evidence of heat elevating the temperature of the cold side as logically it should if heat is not being "used up" to run the engine, but is just passing through.

Just for example, I ran this engine for three hours on hot water, the sides insulated so heat could not leave through the sides. Yet the top plate remained at ambient temperature.

https://youtu.be/NtrYSpYD43w

Other than the hot spot around the power piston, likely IMO due to friction.

If it were heat "passing through" the entire top plate would be heating up more evenly, not staying cold everywhere but the power cylinder.

Personally I don't understand how such demonstrations seem to make no impression. Certainly you've seen these videos before. How else can such results be accounted for?

[Fool]

Again, I go back to my very simple question of does the isolated room reach 0k if all the rocks are lifted. Any possible method can be used inside the room to lift the rocks. There are no limitations, let your imagination run wild.

You are still limited by nature. Wild thinking can help by viewing things from different perspectives allowing discovery of creative solutions. Nature still is the ultimate limiter.

Heat engines run on a temperature difference. All the lifting of rocks will be done when all temperature differences are zero. This will be, in a perfectly isolated room, when the temperature is between the highest temperature and lowest temperature. Sort of near the average temperature for the whole room. This will not be zero Kelvin, and will be the same temperature as if the hot and cold just conducted through to each other.

If it is a perfect room, perfect Carnot engine, and perfect Carnot heat pump, driven by perfect rock raising mechanism, driving the heat pump from the rock dropping, will restore the temperature difference back to the starting point. The temperature for repetitive cycles will go from Th and Tc to temperature middle Tm, and back to Th and Tc spread. Always remember a full cycle.

If it is a perfect room and a real engine and a real heat pump cycling it will lead to all the rocks on the ground and all the temperatures in the room will be Tm. This will be the same temperature as if Th conducted to Tc straight across. This is heat death for the perfect room.

If it is a real room, the temperature middle will eventually become the temperature of the surrounding outside. That will happen if the outside mass is sufficiently large enough not to significantly change with the heat input from the room.

The second law doesn't state that 100% is impossible. It just points out that if 100% efficient engine is reversed it will have a COP of 1 (100%).

It also states that if an engine is getting 50% and is reversed the pump will have a COP of 2 (200%). These two combined will return100% of the energy to the same temperature differential.

[VincentG]

Rocket #1 is contained in a room, ignited and set to spin around a pole on a string, flying unloaded until depleted.

Rocket #2 is attached to its maximum payload, ignited and reaches 5ft off the ground before being depleted.

Rocket #3 is anchored to the ground, ignited and sits stationary until depleted.

Which rocket heated the room more with its exhaust?

[Tom Booth]

Rocket #1 is contained in a room, ignited and set to spin around a pole on a string, flying unloaded until depleted.

Rocket #2 is attached to its maximum payload, ignited and reaches 5ft off the ground before being depleted.

Rocket #3 is anchored to the ground, ignited and sits stationary until depleted.

Which rocket heated the room more with its exhaust?

As far as I'm concerned, my experiments and observations stem from and are applicable to Stirling engines where heat enters the engine as heat alone, without mass flow in or out of or through the engine and no combustion or chemical changes or reactions.

Combustion of a fuel does not just expand air with heat but also consumes constituents of the air.

I therefore would not speculate on the scenario as the question is irrelevant to my interests and as far as I'm concerned, apples to oranges, whatever might be the outcome.

You are not talking about simply adding heat to expand a gas but a combustion of and chemical changes in the gas.

[VincentG]

I understand your position Tom. I am looking for an academic to show me some math outlining the direct conversion of heat to work.

If it can't be mathematical shown that heat is converted, then how can efficiency be calculated based on the percentage of heat not converted.

[Tom Booth]

I understand your position Tom. I am looking for an academic to show me some math outlining the direct conversion of heat to work.

If it can't be mathematical shown that heat is converted, then how can efficiency be calculated based on the percentage of heat not converted.

The math is in the first part of the title of the thread:

η = 1 – (Qc / Qh)

Also:

The efficiency of a heat engine can be calculated using the formulas e = W/QH and e = 1 - QL/QH, where e is the efficiency, W is the work, QH is the heat input, and QL is the heat output.

https://study.com/learn/lesson/heat-eng ... mples.html

If you supply 100 joules (QH) that all go into the engine and no heat leaves (QL) the engine;

e = 1 - QL/QH
e = 1 - 0/100
e = 1

Note that Q is the actual heat in joules NOT temperature.

Let's say for simplicity; If the ambient air is 300°K and the hot side is 400°K input = 100 joules

If the cold side remains at 300°k the same as ambient QL (the heat leaving the engine) is 0 joules.

Heat does not transfer without a temperature difference.

This is all very straightforward according to the 1st law of thermodynamics or conservation of energy.

It's when T temperature is inserted in place of W joules that the simple equation becomes corrupted and meaningless, as temperature does not actually represent a definable or measurable quantity of heat being transfered.

[Tom Booth]

...
Heat engines run on a temperature difference. All the lifting of rocks will be done when all temperature differences are zero. This will be, in a perfectly isolated room, when the temperature is between the highest temperature and lowest temperature. Sort of near the average temperature for the whole room. This will not be zero Kelvin, and will be the same temperature as if the hot and cold just conducted through to each other.
...

Just FYI I, of course, do not agree with this as there is no accounting for heat as energy being converted to another form, in this case the gravitational potential energy of the lifted rocks.

If the heat energy goes into GPE and also simply spreads out evenly to equalize the temperature, there is again, a doubling of the total energy of the system or room.

A violation of conservation of energy.

The work of lifting the rocks has been accomplished and GPE created without the consumption or transformation of the original thermal energy.

[Fool]

VincentG, you already know the following, so it is just for collaboration.

The first law of thermodynamics, conservation of energy, and the constraints of a complete cycle lead to the following equation:

W=Qh-Qc

This literally and mathematically.means that if work is performed by a device/engine, the heat rejected will be less than the heat absorbed. This is your "colloquial" mathematical proof that heat is converted to work.

It is misleading because heat isn't necessary for the conversion, as is demonstrated in every adiabatic expansion. No heat is conducted. Internal energy is converted to mechanical motion. Pressure times a change-in-volume. W=P•∆V. One stroke only.

To save some of that work for output, heat must be rejected during the return stroke, so the return stroke is easier. A cycle made up entirely of adiabatic processes will have zero work performed.

That is the consequence of Qc being zero, Qh then will also become zero. Expansion temperature drop always equals compression temperature rise, adiabatically. Temperature of the gas now equals the temperature of the hot reservoir. No heat transfer can take place between identical temperatures. Adiabatic bounce only, Zeroth law of thermodynamics. Remember to always consider a full cycle.

All three rockets will heat the room equally.

Rocket number one will convert some exhaust heat to momentum. That kinetic energy will bleed off until all of it is converted to air friction, which is internal energy measured by temperature. It will defuse into a middle room temperature.

Rocket number two will convert some of its exhaust energy to height and velocity. That energy will become internal energy only after it reaches apogee returns to the floor, crashes, and the debri cools to a middle room temperature.

The third rocket will convert some of its exhaust to heating the rocket motor housing and into an uneven room internal energy, as all three do. All that energy will be dissipated into a middle room temperature. It will ultimately be the same middle temperature as for all three rockets.

This only applies to a perfect-closed-isolated-room that is identical for all three experiments. And the complete cycle, of rocket and temperature life, is considered.

Work doesn't disappear unless it is converted to heat in a non reversible process such as friction. Then the energy is just dispersed not gone. Some can be stored temporarily in a reversible, or partially reversible mechanical or other, process.

I've demonstrated several times in the distant pages of this thread that the first, and zeroth law, combined with the third law, heat spontaneously flows only from hot to cold never from cold to hot (Third Law), and also considering a complete cycle, that the Carnot theorem is easily derived. It is a continuation of the colloquial but misleading "heat gets converted to work", mathematical proof.

It is less misleading to say, "Only Some 'Thermal Energy' gets converted to work, the rest gets dispersed.". Leave out the word, "heat". All that is needed is P•∆V. Remember it is only for a complete cycle.

In other words to challenge these laws and theorems a measurement of work output must be made. As well as other factors to be measured, such as Th, Tc, inside and out. Pressure and Volume. Time. Room temperature. Energy supplied, Amps, Volts. Etc... Be meticulous.

This equation says it all:
W=Qh-Qc

[Fool]

The work of lifting the rocks has been accomplished and GPE created without the consumption or transformation of the original thermal energy.

The room temperature will only be slightly cooler than if the hot and cold conducted, depending on how massive the room is compared with the stored energy in the rocks. Yes some difference. Perhaps negligible. It will be far from zero Kelvin. It will still be between the two.

Energy from a hot torch applied to a cold anvil will transfer heat to the anvil. The temperature of the anvil will not change much. The energy gained by the anvil will dissipate into the atmosphere with some, but very little, temperature difference. It might be so small as to not be measurable.

[Tom Booth]

...
W=Qh-Qc

...

It is misleading because heat isn't necessary for the conversion, as is demonstrated in every adiabatic expansion. No heat is conducted. Internal energy is converted to mechanical motion. ...

In that case:

∆U=Q+W

Though given the equivalence of heat and work the distinction is moot IMO.

Adding heat increases internal energy, work decreases internal energy.

In reality for transfers to or from a gas, work or heat are just different words to describe a transfer of kinetic energy in or out of the gas. Actually identical on a molecular or atomic level.

Fool likes to imagine some significant distinction between heat and internal energy. For a gas, there really isn't any.

The "vibrations" of the hot plate transfer energy to the gas which increases the ",vibration" of the gas. The gas transfers the same to the piston.

A simple transfer of kinetic energy

Hot plate molecule to gas molecule, gas to piston.

[Fool]

Temperature equates to internal energy through the following equation:

Q=M•Cv•T=U

Heat is:
∆Q=M•Cv•∆T not equal to U

The equation:
n = (Qh-Qc)/Qh

Could be rewritten:
n = (Uh-Uc)/Uh

Which has nothing to do with heat, only the currently contained internal energy when at the hottest and coldest. For a full cycle:

But Tom is assuming it is written:
n = (∆Qh-∆Qc)/∆Qh

I hope this points out how they are different, but can be interchanged in a ratio equation, a fraction.

n=(Uh-Uc)/Uh

Uh = M•Cv•Th
Uc = M•Cv•Tc

Mathematically showing some thermal energy can be converted to work, but typically not all.

Tom likes to equate an energy river to an energy lake. Not the same but can be interchanged in some equations.

[Tom Booth]

You seem to be assuming that heat going into a gas can "disappear" becoming some form of "internal energy" as in a chemical reaction or some rearrangement of internal molecular structure.

Gases however are in their simplest possible structure, so there is no such internal rearrangement possible, unless undergoing a phase change.

Therefore, in a gas, in a Stirling engine, heat work and "internal energy" are all identical. There is no "latent heat" as with steam.

In a gas, the only measure of its internal energy is it's temperature.

[VincentG]

Work is a function of pressure x volume.

Pressure is a function of heat applied to a gas.

But that does not mean that heat converts to work. Work is just a consequence of heat. Heat creates pressure and pressure converts to work as the heat does what heat does, attempt to reach equilibrium.