By Kent W. Mayhew

Carnot cycle: A New thermodynamics perspective

In my previous blogs (see Lost work blog) I discussed how traditional
thermodynamics was a complication of the simple. The real issue being that we wrongly used entropy to explain lost work, when in fact
lost work (W=PdV) is simply the energy required to upwardly displace out atmosphere. This has ramifications throughout thermodynamics
and all of the sciences.

As a continuation I present this blog, which will show why the currently accepted theory behind the
Carnot cycle is misleading, and present a better simpler understanding for the cycle.

I also write this blog in hopes that you will:

1)
understand why traditional thermodynamics needs an overhaul

2) Hopefully you will spot any errors in my analysis so that I may make
corrections, as I am only human.

Two good on-line examples demonstrating the traditionally accepted analysis for the Carnot cycle
are:

A)http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/Thermodynamic_Cycles/Carnot_Cycle

The above
webpage gives the currently accepted/used equations. This is the site whose diagrams I shall basically follow in this discussion,
that being the diagram used herein (below).

B) https://en.wikipedia. org/wiki/Carnot_cycle

One of the simplest types of engine
is a *heat engine* wherein power is obtained by the extraction of thermal energy from a heat reservoir. Consider an engine wherein heat
(dQin) is required for an isobaric expansion process, e.g. the isobaric expansion of a piston-cylinder apparatus filled with an ideal
gas. The issue becomes how can this be done at 100% efficiency? It would need to expand due to a heat source and more importantly
any realistic system would have to displace our atmosphere, resulting in lost work!

Step 1). “Reversible isothermal expansion”; From points 1 to 2: Put
the engine in contact with the hotter heat reservoir at *T1*, and then allow it to isothermally expand from: V1* *to V2. If
it were not in contact with the hot reservoir, then the gas’s temperature would have to decrease as it expands, hence does work. The
gas’s temperature also increases and this to requires energy. Accordingly, isothermal expansion requires heat in [Q(in)].

Step 2). “Isentropic expansion” as an adiabatic process; From points 2 to 3: The hot engine is thermally re-isolated and its volume
is allowed to expand from V2 to V3, until its temperature decreases back to *T. *Herein the engine cools as it does work in displacing
its surroundings i.e. atmosphere.

Step 3). “Reversible isothermal compression”; From points 3 to 4: The engine is now
put into thermal contact with a cooler heat reservoir at *T *' and the gas is now compressed: V3 to V4. In traditional conceptualization
the gas gives up energy [Q(out)], i.e. the system’s total energy has decreased.

Step 4). “Isentropic compression” as an adiabatic
process; From points 4 to 1: The engine/cycle starts at a colder temperature *T *'and is thermally isolated (100% insulated). The gas
is slowly compressed hence its volume decreases from; *V4 *to *V1* and its pressure increases; *P4 *to *P1*, which causes
to the gas’s temperature increasing to back to T1. And thus completes the cycle.

All the steps of Carnot’s cycle
are discussed in terms of theoretically being reversible, hence the cycle is deemed reversible. The cycle’s conceptualizations have
been reworked numerous times since and often starts at another steps other than step 1) as described herein.

Contemporary contemplations
of the cycle discuss the exchanges of energy (Qin, Qout) with the surroundings but fail to properly describe/define the surroundings,
which is a critical error. Rather then adhere to traditional dogma let us now investigate the Carnot cycle employing our new understandings
and perspectives.

* Our new analysis of the steps*

Step 1). “Reversible isothermal expansion”. We realize that the system does
work onto its surroundings. Of course if we do clearly not define our surroundings then there is no way to quantify the amount of
heat/energy that must enter the system in order for it to remain isothermal. If the surroundings are our atmosphere (as would be for
most applications) then insulated expanding systems will cool down, as they upwardly displaces our atmosphere. Logic now dictates
that this step is isothermal because the energy in [Qin] equates to lost work [Wlost], then in terms of the atmosphere’s pressure
[Patm], we can now write:

Wlost =PatmdV eqn (1)

Eqn (1) assumes that our atmosphere is the surroundings, and that this is at 1 atm pressure.

What if the surroundings were a gas filled isometric box. Now there would be no displacement of our atmosphere and the pressure in
the surrounding isometric box would increase. Would it increase isothermally, well that depends upon whether or not, the surrounding
isometric box was also in thermal contact with the same high temperature isothermal heat source/bath/reservoir (T1=T2=Th). If it was
then the work done could be in terms of number of gaseous molecules (N) becomes:

W = -(NkT1)In(V2/V1) = Qin eqn (2a)

Or if you prefer in terms of number of mols (n):

W = -(nRT1)In(V2/V1) = Qin eqn (2b)

So herein we have just seen that the work required depends upon the system’s surrounding. Accordingly by
simply saying surroundings, one cannot expect to know how much work is required. Yet the Carnot cycle as described in contemporary
literature i.e. both our on-line examples, seemingly does just that! Interestingly, the on-line example A) uses eqn (2b) to describe
the energy requirements, even though nothing is stated about what the surroundings are.

It must be emphasized that in
order for any PV engine to be useful to us here on Earth then it must be at some point in its cycle, expand and displace our atmosphere.
Any cycle, or step thereof which does not, is only useful when the isometric box acts as the surroundings, hence is limited to motions
by devices inside that isometric box. Which is not particularly useful to any of us standing outside the box, here on Earth’s surface.

Now ask is this process reversible. If this is a useful Step in a cycle, therefore it displaces our atmosphere, and then no
it is not a reversible step, because lost work is passed onto our atmosphere. Conversely, if it were not a useful step in the cycle
then it could be reversible, as could be the case if an isometric box were its surroundings attached to the same hot heat bath. Obviously,
there are inherent problems with Step 1) of this Carnot cycle being reversible, if we want to extract any useful from it.

Up to this point we treated the system as being isothermal as the name commonly used for this step implies. But the step is not isothermal,
because the system’s/engine’s temperature also increased. Specifically the system was put into contact with a higher temperature heat
bath/thermal reservoir to initiate this step.

So we must also consider the amount of thermal energy that has increased
inside the system/engine. In terms of the molar specific heat (Cv) and temperature change this energy is:

Q(tempincrease) = nCvdT eqn (3)

Hence the total input of energy required for this non-isothermal step is:

Qin = nCv(T’-T) + Patm(Vf-Vi) eqn (4)

Eqn (4) assumes that the surrondings are the atmosphere.

Step 2). “Isentropic expansion”
as an adiabatic process. This step is deemed adiabatic because it is insulated. Herein the gas continues to expand hence cools; therefore
the expanding gas must be doing work onto something. Again there is no clear clarification as to what the surroundings are.

Let us assume that the surroundings are our atmosphere. If this step 1) is cooling down because it is expanding, thus it is displacing
our atmosphere, hence it is doing work as would be defined by eqn 1).

It sounds fine until you ask; how does this step
actually do any mechanical work? If the system expanded in step 1) and its surroundings were the atmosphere, then logic dictates that
the expansion in Step 1) ended at 1 atm pressure. Just because the system is at a higher temperature does not give us the right to
think that it is capable of doing any more work. In order for a system to do work onto its surroundings that system must be at a higher
pressure. The fact that the system/engine is now at a higher temperature and equal pressure would mean that all it can do is radiate
heat. If this heat can be extracted, controlled and then utilized to do work, then so be it, but it still has to be transformed into
mechanical energy by some device whose efficiency is 100%.

What if the surroundings were the isometric box? Well now the system/engine
can expand only if its pressure is higher than that inside the isometric box. First of all this implies that this is not a useful
step in a useful process to us here on Earth’s surface. Secondly; would the expansion in Step 1) not have stopped at the pressure
inside the isometric box? If yes then we are back; at how does the system even do any mechanical work?

If there was any real
logic behind the expansion described in Step 2) then the energy lost could be calculated using the molar specific heat and temperature
change. i.e.:

Qlost = nCv(T’-T) eqn (5)

Now ask is this process adiabatic. Does not adiabatic means no exchange of energy whatsoever with its surrounding? Since
in Step 2) the system/engine supposedly cools because it is supposedly doing work onto its surroundings. So by definition it is not
really an adiabatic process!

Step 3) “Reversible isothermal compression”. Since the compression is isothermal at the colder
temperature (Tc=T3=T4), then the ability of the gas to do work does not change I.e. d(PV)=0. The compression of the gas does however
require work as defined by either:

W = -(NkT3)In(V4/V3) eqn (6a)

Or:

W = -(nRT3)In(V4/V3) eqn
(6b)

Eqn 6a and 6b define the work that results in a pressure increase, hence an increase in the potential to do
work, where the potential to do work is with respect to the surroundings. Note that herein we assume that the compression results
in a pressure increase with an accompanying molecular volume decrease and that the gas obeys the ideal gas law (PV=NkT).

Since temperature increases occur with compression, then in step 3) there must be heat flowing out of the system in order for the
system/engine to be isothermal. Therefore in terms of heat out [Qout] we would write

W = -(NkT3)In(V4/V3) = Qout eqn (7a)

Or:

W = -(nRT3)In(V4/V3) = Qout eqn (7b)

Is step 3) reversible? Yes it seems to be a reversible step in this Carnot cycle.

Step 4). “Isentropic compression”
as an adiabatic process. Since the compression is not isothermal, then the ability of the gas to do work does change I.e. d(PV) does
not equal 0. The compression of the gas does however require work. If the compression was isothermal then we could write either in
terms of the colder temperature (Tc=T3=T4):

W = -(NkT4)In(V1/V4) eqn (8a)

Or:

W = -(nRT4)In(V1/V4) eqn (8b)

However, the compression is not isothermal as described by eqn 8 (a or b). Since the temperature of the system/engine
also increased and the thermal energy associated with this temperature increase could be expressed in terms of molecular heat capacity
(cv) and number of molecules (N), then:

Q = Ncv(T4-T1) eqn (9a)

Or if you prefer in terms of the molar heat capacity (Cv).

Q=nCv(T4-T1) eqn (9b)

Is this step 4) reversible? Yes it is one of the reversible steps in this Carnot cycle.

*Traditional Demise*

In part due to the lost work by useful systems employing the Carnot cycle, both the second law of thermodynamics and its
correlation to entropy were conceived. Traditional conceptions were based upon a false pretext for Steps 2 & 3) discussed herein.
The pretense that isobaric volume increase (PdV) can be contemplated in terms of an isothermal entropy increase (TdS), became universally
accepted:

W = TdS = dE + PdV eqn (10)

Eqn (10) is really part
of the basis for entropy’s traditional consideration that related randomness to work. And from here developed the second law concept
that such entropy increases could not be reversed without an input of energy.

It is interesting that in some traditional
analysis of the Carnot cycle that entropy change is only considered in terms of energy in (Qin) or out (Qout), herein entropy is used
in place of heat capacity (see: __https://en.wikipedia.org/wiki/Carnot_cycle__).

TdS = Qin eqn (11)

Of course in considering entropy in terms of heat capacities could make one ponder what does this have to do
with randomness?

No matter which version of entropy you prefer, we can see why the Carnot cycle is void of true logic,
and our new understanding was readily accomplished without even considering entropy.

The traditional demise actually goes
further. Once work was equated to entropy change [eqn(10)], then it seemingly became necessary to think in terms that all forms of
work should be expressed in terms of isothermal entropy change. This indoctrinated confusion is now so engrained that the science
community now thinks that all work involves entropy changes, rather then limiting such entropy change to lost work as defined by:

Wlost =
PatmdV eqn (12)

Yet the sad reality is that
all this is nothing more then a continuation of circular logic. Think about it, you design a math to explain empirical data for lost
work, and then you exclaim that the empirical data proves your mathematical based theory and then you universally apply it.

To further exasperate the situation, Boltzmann designs his brilliant mathematical conceptualizations (basis of statistical thermodynamics),
and determines his constant (*k*) so that it equates to lost work here on Earth [eqn(12)] for gases. The logic now has been reinforced
because now statistical thermodynamics seemingly now explains what we witness, hence is claimed to be inarguable proof, when in actuality
it very inception means that it was equated to what was empirically known, hence is a continuation of circular logic.

Ultimately the perception of work in terms of isothermal entropy change, and the proof provided by Boltzmann’s eloquent math are all
based upon circular logic, yet this remains fundamental to traditional thermodynamics. Even so, there exists no real readily envisioned
explanation for lost work in traditional thermodynamics beyond isothermal entropy change. This theoretical blunder has implications
endeared by all the realms of science, often complicating the simple.

Hopefully I have not only demonstrated that both
the Carnot cycle and thermodynamics have been poorly conceived over the last century and a half. It is time to end the indoctrination
and simplify the science once and for all.

Sincerely Kent Mayhew (copyright of)

Commentary:

It should be stated that any of the
equations wherein the work is defined in terms of natural logarithmic function of volume change, can equally be written in terms of
natural logarithmic functions of pressure. For example eqn 2a can equally be written in terms of pressure change that being:

W = -(NkT1)In(V2/V1) = (NkT)In(P2/P1)

The above is based upon the realization that work as defined in eqn 2a is based upon the assumption that the ideal gas law
is valid, hence pressure multiplied by volume is directly proportional to temperature when defining the gas’s energy.

Understanding some unusual terms used herein.

1) Ability to do work: d(PV). This is really based upon my new understanding that
work should not be limited to isobaric isothermal processes as eqn (10) does. Specifically we must start thinking in terms of

W = dE + d(PV) eqn 13

For the isobaric case we could write:

W= dE + PdV eqn 14

The ability to do work has nothing to do with whether or not a system will do work. In order for a system to do work, then its pressure
must be greater than the pressure of its surroundings. So you may consider the ability to do work in terms of a system being able
to do work against a pressure approaching zero.

2) The potential to do work increases as the pressure increases. I consider the
potential to be:

W = VdP eqn 15

3)
Useful system is a system that can provide useful energy or power man and/or machine here on Earth’s surface.

It must be emphasized
that the above are my terminology based upon my thoughts, which are still in their infancy stage. Accordingly, the terminology
may change the future after input is obtained from others. Input from others is always appreciated.

KWM

Blog: Carnot cycle