_{1}

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The traditional thermodynamic theory explains the reversible phenomena quite well, except that reversible phenomena are rare or even impossible in practice. Here the purpose is to propose an explanation valid for reversible and also irreversible phenomena, irreversibility being common or realistic. It previously exposed points tricky to grasp, as the sign of the work exchange, the adiabatic expansion in vacuum (free expansion) or the transfer of heat between two bodies at the same temperature (isothermal transfer). After having slightly modified the concepts of heat transfer (each body produces heat according to its own temperature) and work (distinguishing external pressure from internal pressure), the previous points are more easily explained. At last, an engine efficiency in case of irreversible transfer is proposed. This paper is focused on the form of thermodynamics, on “explanations”; it does not question on “results” (except the irreversible free expansion of 1845...) which remain unchanged.

Thermodynamics traditionally [

In the first part, we will check some statements tricky to understand: a mathematical equality considered as a physical principle, an expansion without any work, a negative work represented by a positive area, a transfer of heat between two bodies at the same temperature, and a cycle where heat seems to go from cold to hot! In the second part, we will change the forms of the heat transfer and of the work and consequently of the internal energy. Then in the third part, we will apply these new forms to the first statements which were tricky to understand. Explanations look then more understandable, especially for irreversible phenomena. At last, an efficiency ratio based on differences of temperatures is proposed for irreversible cycles.

Its name or more precisely its numbering—“zero”—is unusual. The reason is historical. The first principles having already been laid, it has nevertheless been considered necessary to add another one prior to the demonstrations. But more than its numbering, it is its content which is surprising as a “principle”. Recall that the principle zero states that if A is in (thermal) equilibrium with B and B with C, then A is with C. This is surprising because that looks like more to a mathematical property (a transitive law, an equivalence relation or an axiom as “two quantities equal to a third are equal”) than to a physical principle. In other branches of physics, it is not stated as a “principle” that if A has the same mass as B and B as C, then A has the same mass as C. This zero principle asks a question: does not it reflect a certain difficulty in characterizing thermal equilibrium?

In 1806, Louis GAY-LUSSAC was the first to experiment expansion in vacuum. Measuring the temperature directly on the air with an alcohol thermometer, he found no variation of temperature [

In 1845, James JOULE did the same experiment, except that it measured the temperature of the surrounding water; he found no variation of temperature either [

In 1865, Gustave-Adolphe HIRN did the same experiment, measuring the temperature through a variation of pressure. He found a light variation of temperature: −0.2˚C [

Since there, the experiment has been done again with carbon dioxide CO_{2} (which is more sensitive than air): it has been noted a decrease of temperature of −0.3˚C [

And today, it is mainly the usual industrial way to cool gases [

And yet History has retained the result of the Joule’s experiment with no variation of temperature!

Let us remind that an ideal gas is a gas that obeys the ideal gas law:

with P: internal pressure of the gas; V: volume of gas; n: amount of substance of the gaz (in moles); R: gaz constant (8.314 J·K^{−}^{1}·mol^{−}^{1}); T: absolute temperature (in Kelvin).

This is a good approximation for gases in usual conditions (when the pressure is relatively low, less than 10 atm.) like the atmospheric pressure. It is often used to describe cycles, compressions and expansions.

There are two laws of Joule: the first one called Joule-Gay Lussac law and the second one called Joule- Thomson law. The first law states that for ideal gas the internal energy depends only on the temperature, the

second one that the enthalpy depends only on its temperature. To check it, these scholars set up an experiment showing that “the gas temperature would remain constant” (which is quite different to the Joule’s law: U depends only on its temperature) with the expansion, due to the formula

The reasoning is as follow: because there is no heat transfer, and because there is no work (

This is why it is still traditionally explained that if the gas cools in experiments, it is that in sudden expansion the gas is far to be ideal [the gas would not obey to Equation (1)]: this is an indirect recognition that traditional theory is not in agreement with experimental results...

What a surprising result: a thermodynamic system which keeps constant its internal energy U and its temperature T, and which does not receive any heat or work but submits an irreversible change! It is the result found by Pr. Joule in 1845.

Why is it qualified here of surprising result? Because it is unusual to get:

· no work when the volume changes; it would be surprising that the border line, which can be the black top of

· a constant temperature

when in the reversible and in the other irreversible cases, the temperature decreases [

In mathematics, the area is calculated by integrating a function. If the function

In thermodynamics, work is conventionally counted positively when received and negatively when provided by the system. The infinitesimal work exchanged by the system is denoted

The pressure _{2} is higher than V_{1} (that is, when x_{2} higher than x_{1}), the thermodynamics sum W is then negative:

The theorizing of thermodynamics by physicists did not precede but followed or accompanied the development by engineers of thermal engines as the steam engine. They had already realized that the efficiency

done/energy transmitted by the hot source) seemed limited by an unsurpassable value related to the temperatures of the sources.

Was it an impression or a physical principle? Sadi Carnot, who thought the heat was a fluid called caloric, described a clever cycle (not representative of an existing machine) which reaches the performance limit. Let us consider a transformation of this cycle, the isothermal transformation. See

An isothermal transformation is a transformation that takes place at constant temperature, in contact with a thermostat. Take for example a piston air-filled at 20˚ Celsius, in an external medium like water also at 20˚ Celcius. The air volume is adapted to the pressure according to the ideal gas law. This system is in stable equilibrium; it does not change and will not change spontaneously. And there is no reason that for two bodies at the same temperature, heat flows in one direction rather than another. In the case of perfect equality between fluid temperature and thermostat temperature: there is no transfer of heat.

And more generally for isothermal transformation

This cycle cannot work as a thermal engine.

The efficiency

Hence [cf Equation (6)]

The thermal cycle of a motor is traditionally represented as follows in

In motor mode, the heat from the hot source is partly converted in form of work, the balance going to the cold source.

And the thermal cycle of a refrigerating unit is traditionally represented as follows in

In refrigeration mode, the heat from the cold source goes to the hot source and the work too. How is it possible that the heat goes from cold to hot?

The above points will be discussed again after the following assumptions:

Beforehand, what is the internal energy U? “The internal energy of given state cannot be directly measured. (∙∙∙) Though it is a macroscopic quantity, internal energy can be explained in microscopic terms by two theoretical virtual components. One is the microscopic kinetic energy (∙∙∙). The other is the potential energy (∙∙∙). There is no simple universal relation between these quantities of microscopic energy and the quantities of energy gained or lost by the system in work, heat, or matter transfer.” [

with E_{K} for kinetic energy of particles and E_{P} for potential energy

And the main property of U is: “The internal energy is a state function of the system, because its value depends only of the current state of the system and not on the path taken or processes undergone to prepare it.” [

The first new assumption is that all bodies emit heat, and this heat

with:

· s coefficient, unit in [W/(K∙m^{2})]

· A area of the body, in [m^{2}]

· T temperature of the body surface, in [K]

· dt time length of the thermal exchange, in [s]

Remark: if entropy S in [J/K] has the properties of a status function, s is only a coefficient in [W/(K∙m^{2})]. And like C_{p}, s is empirical and should be dependent for example on the temperature.

According to the fundamental principle of dynamics, acceleration undergone by a body is proportional to the net strength it received. The piston is moved because at a time that internal strength was different from the external strength, i.e. the internal pressure was different from the external pressure. See

It is considered here (please read arguments in Appendix) the internal strength or internal pressure, and therefore their work according to the formula:

or

Remark: The fundamental principle of dynamics indicates that if internal pressure equals constantly and exactly to external pressure, then there is no movement (for a body initially at rest): the reversible movement where constantly and exactly

The property to calculate the “variation” of U is traditionally given by

Note than the variation of the “internal” energy would depend from the “external” pressure.

In this essay let us rather have the internal energy to depend from the internal pressure:

or

with

·

· When

· For reversible paths,

· For irreversible paths, we will get a difference in the calculation of_{1} or T_{2}? See

· Then, to calculate the difference of internal energy from the initial point

· It confirms the U value of such defined point depends only of the current state of the system and not on the path taken or processes undergone to prepare it.

· For ideal gas, the property that the macroscopic internal energy U depending only on the temperature (the Joule’s law) remains unchanged.

· Due to Equation (10) and Equation (15bis), the link between the macroscopic value U and the virtual microscopic terms could be changed.

To sum up into a chart (Chart 1).

In function of these hypotheses, points of Paragraph 2 are now discussed again.

Suppose two identical bodies, except for temperature, in a fully insulated enclosure: all the heat emitted from one body A is received by the other body B (and vice versa). According to the first hypothesis of Equation (11) where

· the body A at the temperature T_{A} emits the heat

· the body B at the temperature T_{B} emits the heat

· For A, the balance between the emitted heat and the received heat is the heat

· For B, the balance between the emitted heat and the received heat is the heat

This heat

· We can note in the case of this single two bodies that

This is in agreement with energy conservation, what is lost from one side is gained from the other side.

· If for example

and so

The (hot) body A emits more heat than it receives.

By the same reasoning

Chart 1. Similarities and differences about internal energy.

The (cold) body B receives more heat than it emits.

Thus, the difference in heat is transferred from A to B, or in other words:

The heat “difference” goes from hot to cold.

Result: That prohibits the perpetual motion machine of the second kind where heat is directly removed from a body: for there to be heat transfer we need a heat differential that goes from hot to cold.

When

When heat balance is nil, the bodies remain in thermal equilibrium. But that does not mean the heat exchanged is non-existent, which means the exchange is balanced. That is why it is distinguished. So we will distinguish the “balance heat”

Using this argument on three bodies, mathematics proved that two temperatures equal to a third one are equal, and so the three bodies are in stable equilibrium; it would not be longer a thermodynamical principle, it would be a mathematical property.

For radiative exchange, according to Stéfan-Boltzmann’s law, the emittance M of a black body is:

with

Let us have the hypothesis where the variable s would be equal to kT^{3}:

Note in this hypothesis the variable s increases as a function of temperature.

For a given area,

For T_{A} and T_{B} of the same order of magnitude, let us have

_{A} and T_{B}.

_{B},

and with

and

On a low temperature range and for a given duration t,

First we have to distinguish

· the Joule’s law: the internal energy depends only of the internal temperature for an ideal gas, from

· the so-called experiment of Joule: the adiabatic expansion in vacuum. It is a specific experiment because it is done against vacuum, and so it is irreversible. Let us remind experiments and industrial practice indicate that without heat transfer

In the case of the expansion against vacuum, there is no heat exchange or external work exchange. Thus, according to Equation (14)

Another explanation is to assert absolute vacuum does not exist (even in the intergalactic space, there is one particle per m^{3}), there is always some molecules and thus a (tiny) level of external pressure

In this case, the gas should hardly cool (whereas in practice, the gas really cools).

According to Equation (16):

work is not nil,

A more precise calculation will be to use the formula from the Appendix:

Because

The experimental results are in agreement with the Joule’s law: with the expansion work the temperature is decreasing; and so there is a loss of internal energy (gained by the expansion work).

Conversely, in case of compression, the temperature increases and the internal energy consequently.

During an expansion the volume increases, and so according to Equation (13) where

When the abscise increases for a positive function, the mathematical integral represented by its area is positive

The same reasoning can be applied to the volume restriction and to the negative integral.

There is total correlation between the thermo dynamical work of the internal pressure

“A transformation is reversible if it can return to its initial position by a series of infinitesimal changes. It is only a thought experiment: the actual transformations are all irreversible” [

The

In refrigeration mode, heat of the cold source goes towards the colder fluid. Then the cycle fluid is compressed by the work done up to a temperature above the heat source for the heat dissipation of the fluid. And to close the cycle, the fluid is relaxed and cooled itself to return to the initial conditions.

The

In motor mode, heat from the heat source goes to the cooler fluid, the fluid expansion provides work (and to complete the cycle, the fluid transfers the heat balance to the cold source).

And

If this cycle is more realistic, it will generate complicated equations to calculate variable temperatures. That is why it is suggested here the following simplification, knowing that t_{H} and t_{C} are approximately (not exactly) constant. Let us characterize it with equations:

The heat transferred to the (relative) hot fluid from the Hot source is (cf Equations (11) and (18)):

Note: the Hot source and the (relative) hot fluid being substantially at the same temperature, we consider firstly that they have substantially the same s_{H} (idem for s_{C}).

For writing formulas reason, let us simplify the expression: _{H} (and _{C}).

And the heat transferred from the (relative) cold fluid to the Cold source is [cf Equation (11) and Equation (19)]:

Preliminary remark: according that we adhere or not to the hypothesis expressed by Equation (15) taking into account internal work, demonstration hereafter does not change, it is why it is denoted by

For a motor cycle:

Motor efficiency

or according to Equation (44):

and using equalities (40) and (42):

On the same range of temperature,

which, for effective engine, is more accurate than the usual limit inequality

To get a better understanding, let us apply to a theoretical example, the irreversible Carnot engine, as described in

The effective efficiency is according to definition Equation (45):

If we use the efficiency of irreversible cycle based on temperatures as seen Equation (48):

which has the same value than the effective efficiency,.

Thus the Carnot efficiency based on sources temperatures is, according to Equation (5):

which is slightly higher than the effective efficiency.

We can of course find the efficiency of 60% if we consider the cycle temperatures instead of the sources temperatures. The trouble is Carnot “engine” with isothermal evolution is unrealistic as explained §2.4. So, let us check another simple example.

Then we will apply this property of the efficiency to a non-biphasic cycle: the very simple but more realistic elevator engine given by Pr. Richard Taillet [

In this non-biphasic cycle, the net work is −45 kJ for a heat transfer of (122 + 868 =) 990 kJ. The cycle effective efficiency is very low, (45/990 =) 5%.

When we apply the Carnot efficiency, with a 746 K hot temperature and a 300 K cold temperature, the maximum efficiency is according to Equation (5):

The maximum efficiency (60%) based on sources temperatures is very far of this cycle efficiency (5%).

To apply the new efficiency formula, we strictly should have used the instantaneous temperatures and integrate them, which would be very difficult here. So, only to give an order of magnitude, we will take in this example an average temperature of ((373 + 746)/2 =) 560 K for the hot side, and an average temperature of ((600 + 300)/2 =) 450 K for the cold side. An irreversible efficiency should be closed to

An irreversible efficiency (19%) based on differences of temperatures is closer to the effective efficiency (5%).

Previous calculation (19%) was done with a very approximative but simple average temperature only to explain. It has also been taken final fluid temperatures equal to the source temperature, which would mean a perfect heat exchanger. In practice, we will get for example 4 to 5 K between the final fluid temperature and the

Chart 2. Cycle results.

source temperature:

which would give

and, because

so

which is in good correlation with the 5% of the effective efficiency.

Slightly modifying the concepts of heat transfer and work, we have given different explanations, and hopefully clearer on:

· the thermal balance and the heat transfer direction;

· the effects of expansion in the so-called experiments of Joule;

· the direct reading of the work in the Clapeyron diagram;

Chart 3. Comparison chart.

· a more accurate use of diagram of thermal cycles, especially for irreversible transformations in the motor cycle and the refrigeration cycle;

· a new property for irreversible engine efficiency:

You will find a chart (Chart 3) to sum up the proposed formulas vs. the traditional formulas.

Without questioning the experimental results and the reversible predictions, the purpose of this essay is contributing to evolve the presentation of the thermodynamics traditional explanations, taking better account of Irreversibility.

I would like to thank Daniel Mandineau for his teaching of traditional thermodynamics.

1) Work of external pressure

Because

this means

b) Work of internal pressure

More than the work supplied by the external, it would be more rigorous to take in count the work received (or supplied) by the (internal) system:

Due to the conservation of energy, the trouble is why the supplied work

To get a movement, it is necessary to have

3) Effective work

Effective work, supplied and received, is in fact

this will be difficult to integrate

So previous equation can be approximated by

See

For pragmatic reasons (for example Clapeyron diagram, cf §4.3), it has been chosen here to represent