How do you model an ideal compression process from saturated vapour at P1 when the entropy of the saturated fluid at P2 (sg2) is more than sg1?
Usually we take s2 = s1 and then find T2 which is in the super-heated region and proceed to find h2 and the compression work = h2-h1.
For some refrigerants, s2 lies in the saturated region, not the super-heated region.
Dear Musadaq
You may refer to:
Hydrocarbon working fluids for a Rankine cycle powered vapor compression refrigeration system using low-grade thermal energy
H Li, X Bu, L Wang, Z Long, Y Lian - Energy and buildings, 2013 - Elsevier
They give a mathematical model for the problem.
"Low values of cpv give a vapour dome such that a compression process starting on the saturated vapour line terminates in the superheated vapour region. With a high value of cpv , however, the vapour dome is ‘undercut’ so that an adiabatic compression process terminates in the two-phase region."
Ref: https://www.ias.ac.in/article/fulltext/reso/017/02/0139-0162
Agree with Mehdi Aliehyaei
According to the above reference:
"Such a ‘wet’ compression must be avoided for most types of compressors. The liquid and vapour sides of the two-phase region are connected and thus high values of also result in greater flash losses (entropy generation) in the expansion device."
Therefore, for such refrigerants there must be some degree of superheat at the compressor inlet.
The deviation rate of the second law of thermodynamics is 14.6%. See the following figure (or annex). It's not a case.
* * * * * * * *
A: Figure 1 is from the authoritative book "Handbook for Estimating Gas-liquid Physical Properties", which calculates enthalpy change H1 by equation of state. From Fig. 1A, the accuracy of the equation of state is very high, but the enthalpy deviation rate derived is very large. The theoretical basis of the derivation is the thermal 2 differential equation.
B. Drawing with the deviation rate of state equation (< 1%) and enthalpy variation deviation rate (Fig. 2), without considering two abnormal points, the deviation rate is regularly arranged. The deviation rate of fitting line and Y axis intersection point C (0,0.146) and thermal 2 differential equation is 14.6%.
The so-called "entropy " doesn't exist at all.
During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
The so-called entropy was such a concept that was derived by mistake in history.
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.
.
.
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Why did the wrong "entropy" appear ?
In summary , this was due to the following two reasons:
1) Physically, people didn't know Q=f(P, V, T).
2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .
If people knew any one of them, the mistake of entropy would not happen in history.
Please read my paper and those answers of the questions related to my paper in my Projects.
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
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