Good timing! I am just composing an article about the entropy extrema.
Total entropy is the sum of an “intrinsic contribution”, which is Integral(Cvmid/T) dT
(Cvmid: ideal gas isochoric heat capacity), R ln Vm, the residual entropy, and
some constant (reference) terms. If you move away from the critical point along the vapour pressure curve, the residual term vanishes.
If your fluid has a small Cvmid, the resulting entropy curve is monotonous (your fluid is a “wet medium” in the language of ORC processes); if it has a large Cvmid, you will see an entropy maximum (“dry medium”). As a rule of thumb, you will see a maximum if Cpmid/R(0.7 Tc) [ideal-gas isobaric heat capacity at 70% of the critical temperature is above 11.
OK, thank you. I will find a similar criterion as the ¨retrogradicity¨ parameter corresponding to smax on the saturated vapor line, but relating to the conditions of
saturated vapor enthalpy hmax, see any pressure-enthalpy chart for a pure fluid.
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
Dear Friend,
I hope that if you deepen your knowledge of classical and statistical thermodynamics you will be able to understand the essence of a number of concepts and laws of heat physics. First of all, I recommend the following publications: http://web.mit.edu/beretta/www/Beretta-IAP-Thermodynamics.htm I wish you continued enthusiasm for work, but also greater self-control regarding formulated hypotheses and scientific theses. Jan Gorski
Calculus is not "take for granted", ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
It is well known that calculus has a definition.
Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.
Based on the definition of calculus, we know:
to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.
As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,
∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.
1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.
2) On the other hand, In fact, Q=f(P, V, T), then
∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).
We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.
that is, the so-called "entropy " doesn't exist at all.
Dear Friend,
you probably have no proper backgrounds in thermal sciences and such discussion only based on the mathematical analysis have no sense. Study, please, more deaply the proper topics in advanced thermodynamics and then formulate such 'thesis'. Jan
- Badges
- Science topic