Hi,

to know the thermodynamic , you have to calculate free energy of formation from elements delta(g) or the solubility product log K .For a given substance let us assume Delta(h) as enthalpy of formation from standard elements at 298 and 1 bar and s the entropy of the substance at 298 and 1 bar. We use the fundamental relationships:

Delta(g)= Delta(h) - T*Delta (s)

with delta(s) as s-Sum( ni*si)

ni is the number of elements and si the standard entropy of the element in the standard states.

For example let us calculate Delta(g) of water liquid from Delta(h) = -285,83 kJ/mol and s= 69,95 J/K/Mole

Let us calculate delta (s) of H2O with s(H2)g = 130,68 j/K/Mole and s(O2)g =205,152 J/K/mole these quantities are standard entropy od elements in the standard state .Let us remember that Delta(h) and Delta(g) are equal to 0 fro all standard elements at all temperature and pressure.

considering the reaction of formation

H2(g) + 1/2 (O2)(g) = H2O(l)

Delta(s) = 69,95- (130,68 + 0,5* 205,152) = -163,306 J/K/Mole

Delta (g) = -285,83 - 298,15* -163,306/1000=-237,14 k/J/mole

Be careful that you have to divide the term T*Delta(s) by 1000 because s is J/K/Mole and h and g are in kJ/Mole

I hope you followed me.

For calculation the solubility product ,you have to know the free energy of formation of aqueous ions in the dissociation of a substance in water.

If you have some questions, please let me know.

Ph. Vieillard

University of Poitiers

Dear Ph. Vieillard

Many thanks for your answer.

However, I think my question is not so clear.

Consider water with known enthalpy (h) and entropy (s).

I want to determine its temperature and pressure using the attached thermodynamic table.

From 'Thermodynamic state' I mean 'Super-heated vapor or saturated or sub-cold liquid'.

Hello,

I check computation from table you sent me last day. In fact the internal energy is calculated by the following expression :

(u) = h - P*V

this mean:

u(f)= h(f)-P*v(f)

u(fg)=h(fg)-P*v(fg)

u(g)= h(g)- P*v(g)

In order to know the state of water, plot P versus T with values of table B.1.1 This curve represent the PT equilibrium between Liquid and Vapour

(see the paper Wagner and Pruss, 2002) Above the curve, this is the liquid state and under the curve the is the vapor state.

Did I answer to your question? If no please let me know what is your problem.

Best regards

Ph. Vieillard

Thanks for you attention.

No, this is not my answer.

Consider for example we know that h=400 kJ/kg and s=6.2 kj/kg.K.

The question is: What is the temperature, pressure and specific internal energy?

Thank you for your precise question but the ability to determine the nature of water, temperature, pressure and internal energy from only two parameters h(f) ,s(f) , h(v), s(v), h(fv) and s(fv) (and not from density , v) seems very feasible only by interpolation. This implies that one have to plot each of the six parameters versus P and T . A possibility to express each of the six parameters as a mathematical function of Pressure and T.

Give me one week to study the relationships of h and s versus P and T.

Thank you Viellard.

It is possible to express h and s as a function of P and T.

However, it means we cannot easily determine P and T from the attached table, when h and s are known.

Hello,

The initial data are in pdf file; I would like to have an excel version of all tables because I think that there is a possibility to determine virial equations between these parameters. But I have not the adobe version that transform a pdf file into an excel file . Did you have an excel version ?

Thanks you very much for your help

Ph. Vieillard

Dear Vieillard,

Unfortunately, I don't have these tables in an Excel file.

These table are commonly used by students and engineers to determine the thermodynamic properties of a substance when two independent properties are known.

For example, if P and T are known, other properties can be easily determined using the tables. However, when the two known properties are h and s, it is not so easy and this is my problem.

Hello

This is not a problem. I 'll see to morrow in my lab, somebody which have a program that convert pdf file into excel file .

Bye bye

Hi,

Using tables may not be the best option for this situation. For example, the pdf file you sent corresponds to saturated water, which is already setting one variable and partially specifying the system. For something more general you could try Mollier diagrams. Visually that would probably be easier than inspecting a table. However, I am not sure of how easy is to obtain these for different substances. Otherwise you could do as Vieillard suggests, and basically construct your own chart.

I hope that this helps.

Best,

Oscar

Hello,

I just finished to plot H and S versus T and different constant Pressure ( see Excel files and Word files) All figures are given excel files, sheets "données P T H" and "données P T S" for enthalpy and entropy. There are two cases of plots One for vapor (figures up of the sheet) and one for liquid in compressed state (down of the sheet) . In my opinion , the most interesting state is the temperature range (40°C to 800°c) Files in word presents the zoomed figures for H and S in this range.

So it is possible for a definite range of T to expressed the enthalpy (or the entropy) as a mathematical equation ( not strictly linear) of Temperature and pressure .

For a given value of H and S you have to resolve the two mathematical equation with the two unknown values which are T and P.

Do you follow me in my teaching.

If you need some helps , do not hesitate.

Best regards

Dear Vieillard,

I greatly appreciated your effort in preparing these files.

We have to generate h=function(T, P) and s=function(T, P), and then, solve two equations for the two unknowns. However, it is not so easy to determine a functional form which can capture all the points with reasonable accuracy. If we can do so, it can be replaced with these big tables.

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.

I've been doing some light research in the area of your original question. Due to the length of your previous conversation, I've only reviewed the first few entries, and then jumped to this. Consider this a fresh start.

Many fluids have empirical formulas that have been developed to describe the thermodynamic properties of the fluid. (Water will be discussed specifically below.) For a long time, these equations were in a cubic form. More recently, these have been updated to exist in the form of a Reduced Helmholtz Free Energy. 'F' is the symbol used to represent Helmholtz Free Energy. The formula is in the form: F/(RT) = phi(delta, tau) ; where 'R' is the ideal gas constant, T is the temperate on an absolute scale, delta is the reduce density, and tau is the inverse reduced temperature. Delta is defined as density divided by critical density. Tau is the critical temperature divided by temperature, both on an absolute scale. Using various combinations of the partial derivatives of phi(delta, tau), all of the thermodynamic quantities of a fluid can be described. Because of this property, the empirical formulas have become the state-of-the-art in thermodynamic calculation.

In 1995, the International Association on the Properties of Water & Steam (IAPWS) adopted a Helmholtz explicit equation of state for ordinary water for general and scientific use (The IAPWS-95 formulation). This can be found on their webside, www.iapws.org . This paper contains the all information necessary to calculate water's thermodynamic properties based on the temperature and density of water. In 1997, IAPWS adopted an industrial formulation, IAPWS-IF97. This is a slightly less accurate formulation, but it utilizes a variety of "backward equations" to calculate properties based on input other than temperature and pressure. The IAPWS-IF97 formulation can directly calculate pressure as a function of specific enthalpy and specific entropy, which can then calculate temperature based on pressure and specific entropy or pressure and specific enthalpy.

I'm currently working on my own software library to calculate the properties of water using the IAPWS-95 formulation, using iterative techniques to calculate temperature and density from any other combination of independent intrinsic properties (P&T, T&s, T&h, P&h, h&s, etc.).

I'm having difficulty in calculating temperature and density from enthalpy and entropy, but so far I've been successful in finding iterative solutions to the other combinations I've tried. I've been specifically avoiding the IAPWS-IF97 formulation. The framework I'm developing should be translatable to all the various other fluids for which Helmholtz equations of state have been developed.

Thanks for your explanations. I hope the software work well.

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

I'm not sure whether we can specify the state of the substance using these two properties unless we have at least one more datum to add, like the state of the substance before or after a process in which it is involved, for example throttling, vaporization,condensation or so.

To take it more easily, I can propose you some simple software and apps using which you're able to get the results much more quickly than ever:

"CAT" (Computer-Aided Thermodynamics) for Windows, and "ConvertPad" for Android.

Good luck!

M Reza Moharreri Thanks for your comment.

It is not possible to determine the state of the fluid using h and s, even with the CATT software.

Checkout my software I'm working on. coolingpower.sourceforge.io It will calculate the state of water given enthalpy and entropy of a single phase fluid. It can't do two-phase; I'm still working on a useful algorithm and logic to correctly detect thesatuayed mixture.