Dear researchers
I want to use commercial oils like Thermiol-VP1 as working fluid in solar collectors and insufficient data are available for these oils in the references. How can I calculate exergy and entropy properties?
Dear sir
chemical exergy zigma(xiechi)
For physical exergy (h-h0)-t0(s-s0)
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.
Dear Mr. Shufeng Zhang
Thank you for adding your answer. Some point should be considered which was not included in your definition, as can be seen. You are using the ordinary derivative instead of the partial one for Q. In fact, Q is a function of (X, Y, Z, ..), as you mentioned, consequently its derivative must be of the partial type. This fact has been obviously represented in all reference books. In my opinion, there is a problem in your definition. If not, do not hesitate to aware me.
Yours,
N.Mahdavi
Dear Navid Mahdavi,
It seemed that you are not familiar with thermodynamics.
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.
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
Maybe, you can understand.
Please carefully study thermodynamics before you comment.
Dear Mr. Shufeng Zhang ,
How did you act to reject one of the most important thermodynamic relationships?!
You probably have no adequate insight into the thermodynamic cycles.
In dQ/T, the relationship of Q and T is not the ratio of dQ and T or the product of 1/T and dQ, but the relationship to Find the Original Function of 1/T in dQ/T=1/TdQ
For ΔQ/T=1/T *ΔQ, so,
in ΔQ/T, the relationship of Q and T is the ratio of ΔQ and T or the product of 1/T and ΔQ.
But in dQ/T, the relationship of Q and T is not the ratio of dQ and T or the product of 1/T and dQ, but the relationship to Find the Original Function of 1/T in dQ/T=1/TdQ.
For we know Q is not a single valued-function of T, in fact, Q=f(P, V, T), so, ΔQ/T can NOT turn into dQ/T.
All this time, people thought that dQ was not a complete differential, but dQ/T was considered as a complete differential.
However, this is wrong.
The facts are just the opposite: dQ is actually a complete differential, but dQ/T is meaningless.
in ΔQ/T, the relationship of Q and T is the ratio of ΔQ and T or the product of 1/T and ΔQ, so, in ΔQ/T, Q and T can be any relationship.
But in dQ/T, the relationship of Q and T is not the ratio of dQ and T or the product of 1/T and dQ, but the relationship to Find the Original Function of 1/T in dQ/T=1/TdQ.
For we know Q is not a single-valued function of T, (in fact, Q=f(P, V, T) ), so, ΔQ/T can NOT turn into dQ/T, or say, dQ/T is meaningless.
The problem is not whether dQ is meaningful or not here, it is 1/TdQ is meaningless !
Q = f(T, V, P) is a process quantity which varies with path, it has innumerable forms between the same original and terminal states, and has a unique form for fixed reversible process path. When the given path is fixed, Q = f(T, V, P) is the system state variable.
So, dQ=df(T, V, P) is a perfect differential, it is meaningful;
but the integral variable of 1/TdQ is self-contradictory (T and T, V, P), so, 1/TdQ=1/Tdf(T, V, P) is meaningless, that is ∫T 1/TdQ = ∫T 1/Tdf(T, V, P) is not a meaningful integral, or say, it is not a integral at all.
That is, ΔQ/T can NOT turn into dQ/T,
there is not the result∮1/TdQ=0 at all.
The so-called "entropy" does NOT exist at all.
When A and B are any two quantities, you CAN NOT turn a product of any two quantities AΔB into AdB directely, then said B=f(A) !
e.g., when L is the Cable Length, I is the current intensity of the Cable, could you turn LΔI into LdI ?
NO, you certainly can’t ! for there is not the function relationship as I=f(L) or L=f(I), so, Lid is meaningless.
Similely, could you turn 1/TΔQ into 1/TdQ directely then said Q=f(T) ?
NO, you certainly can’t, for there is not the function relationship as Q=f(T) at all, in fact we know Q≠f(T).
And once we know Q≠f(T), then we immediately know that 1/TΔQ can’t be turned into 1/TdQ.
In fact, Q=f(P, V, T), generally speaking, two of them(P, V, T) are independent.
The premise of differential is the existence of a differentiable function !
The premise of integral is the existence of a function !
Dear Mr. Shufeng Zhang ,
Further to your answer which was removed/edited, maybe I couldn’t describe my points well to you so that you can understand them. If Clausius, such a great scientist, could not understand both thermodynamic and mathematic (based on your previous statements), it doesn’t matter that I can’t either.
Good luck
I don't think a researcher should have such prejudice as "If Clausius, such a great scientist, could not understand both of thermodynamic and mathematic, …… ", what a researcher follows should not be authority but be objective facts, a researcher should think independently, but not simply obey authoritative conclusion.
The premise of differential is the existence of a differentiable function !
The premise of integral is the existence of a function !
To a single integral, first of all, there must be a Function y=f(x), then, ∫ ydx=∫x f(x)dx may be meaningful, similarly,
∫ xdy=∫x xdf(x) =∫x xf'(x)dx=∫x G(x)dfx=∫x dF(x) is meaningful.
But, as we know, Q is NOT a single valued function of T, In fact, as I pointed out in my paper, Q=f(T, V, P), so, 1/TdQ=1/Tdf(T, V, P) is meaningless, that is ∫T 1/TdQ = ∫T 1/Tdf(T, V, P) is not a meaningful integral, or say, it is not a integral at all.
Dear Navid Mahdavi,
I didn't humiliate and scoff you or anyone, I'm just stating the fact that the Predecessor ( not you) really made a mistake, because the predecessor was an authority, so, the descendants believed in that authority.
Dear Navid Mahdavi,
I have deleted another answer that may led to misunderstanding, you see, there is no such answers that may lead to misunderstanding.
Dear Navid Mahdavi,
Do you know how this formula ∮1/TdQ=0 was derived ?
Do you remember the definition of the integral ?
Did you read my paper ?
Sorry, I really don't think there is any point in going on with the discussion. So much for this.
Thank you for your discussion, enjoy yourself !
Best regards.
Dear Mr. Shufeng Zhang ,
Yes, I do know how this formula ∮1/TdQ=0 was derived. It seems to me that you do not have enough knowledge of partial differential equations according to what you have presented in this discussion and also your paper.
Good luck
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