Can isentropic process also be called an adiabatic process? And is the vice verse true?
A simple example of isentropic but non-adiabatic and irreversible process: a working isothermal damper (loose piston in a cylinder with a viscous liquid) which is in perfect thermal contact with a big heat reservoir at temperature T. If the piston moves at speed V overcoming viscous resistance kV (k is the damping factor) it produce the heating rate q= kV^2 due to the viscous dissipation of the piston work and the entropy production q/T. Thus due to the heat flux from the damper to the heat reservoir the damper holds equal in magnitude but negative flow of entropy. The total entropy change in it is zero. But obviously the dumper is neither adiabatic nor reversible.
"is-entropic" simply means constant entropy. "a-diabatic" means without heat exchange. A reversible or frictionless adiabatic process is isentropic.
Isentropic is alway adiabatic, too. But the opposit is wrong. Thus, a adiabatic process must not at all be isentropic.
Think of a standard example: the isenthalpic throtteling, as it is typically implemented to small domestic refrigerators.
Please refer to
-> Baehr, H.-D. Thermodynamik Springer-Verlag, 2002, ISBN 3-540-43256-6-X, (book in german, which can be taken as a "standard" or "reference")
or more common in the english speaking or educational world (but I won't recommend it as a reference):
-> Moran, M. J. & Shapiro, H. N. Fundamentals of Engineering Thermodynamics John Wiley & Sons, 1998, ISBN 0-471-97960-0
adiabatic process occurs without heat transfer with its surrounding.In isentropic process entropy remains constant,it is known as reversible adiabatic process.
adiabatic process occurs without heat transfer with its surrounding.In isentropic process entropy remains constant,it is known as reversible adiabatic process.
adiabatic process occurs without heat transfer with its surrounding.In isentropic process entropy remains constant,it is known as reversible adiabatic process.
Any reversible adiabatic process is isentropic because Entropy change (Delta S) = integral of (Q_rev)/T. Here Q_rev is the heat transferred in an equivalent reversible process. Since Q_rev is 0 for reversible adiabatic process, it is isentropic.
On the other hand, every isentropic process need not necessarily be reversible adiabatic. To find entropy change in any process, all you need to do is to chalk out an equivalent REVERSIBLE path for the process, and then calculate the heat change (dQ/T) for that path. If integral of that (dQ/T) is 0, then and only then, the process is isentropic.
An isentropic process is a reversible and adiabatic process but an adiabatic process is not necessarly reversible and therefore is not isentropic
Isentropic process implies constant entropy. The summation of two quantities may alter entropy; those are heat transfer divided by temperature and the entropy generation. The later is always positive quantity, however, heat could be negative and the summation could be zero, in other words irreversible process with heat rejection could be isentropic process when the summation of heat divided by temperature and the entropy generation is zero.
If the process is adiabatic, then it must be reversible to call it isentropic.
A simple example of isentropic but non-adiabatic and irreversible process: a working isothermal damper (loose piston in a cylinder with a viscous liquid) which is in perfect thermal contact with a big heat reservoir at temperature T. If the piston moves at speed V overcoming viscous resistance kV (k is the damping factor) it produce the heating rate q= kV^2 due to the viscous dissipation of the piston work and the entropy production q/T. Thus due to the heat flux from the damper to the heat reservoir the damper holds equal in magnitude but negative flow of entropy. The total entropy change in it is zero. But obviously the dumper is neither adiabatic nor reversible.
Reversible adiabatic process and isentropic process are the same
Simplest formal explanation: dS=dQ/T+dR, reversibility means that entropy production dR=0, so in that case dS=0 dQ=0
When dR is not equal to zero Shnip's example is exellent.
This question has produced a lot of interest. If we really want to understand the question we need to first understand what entropy is and what heat is. Entropy is a concept that is well recognized for its difficulty. My own belief is that neither concept can be well understood unless we study thermal energy at the molecular level.
Imagine an ideal gas expanding in a perfectly insulated piston-cylinder device. The molecules have kinetic energy and as the expansion takes place they impart this by doing work on the piston. No heat is transferred to/from the molecules from a source/sink. This is an adiabatic process. The temperature of the gas is directly related to the average kinetic energy of the molecules by the Boltzmann constant and is only defined when the gas is at equilibrium. However we have a process which is not an equilibrium situation so we have to idealize the process by imagining it passing through a series of equilibrium points - we call the process quasi-static. In this way it can the be regarded as reversible and therefore isentropic, i.e. the gas has constant entropy. If friction occurs during the expansion process, as in the case of a turbine expansion, where high gas velocity is present, the process deviates from the ideal isentropic process and is not reversible and therefor not isentropic, even though it may be adiabatic.
Reversible adiabatic implies isentropic behaviour, but the converse is not true. To see this, recall that for a cycle, dQ_R/T
Every reversible adiabatic process is Isotropic but every Isotropic process in not adiabatic.We know change in entropy for any reversible process is given by dS=dQ/t.Now if process is adiabatic then dQ=0 i.e. dS=0, so every reversible adiabatic is isotropic. But for irreversible process
dS=(dQ/t) + $(entropy generation (always positive)).Now here if suppose dQ=-3(for heat rejection) and $=+3 then dS=0 i.e. isotropic but still the process is not reversible adiabatic
Isentropic process means a constant entropy process. Now entropy is a system property. A reversible process means that there is not net generation of entropy (This where there is an issue with Alexander Schnip's example since it assumes no entropy generation in a piston moving through a viscous liquid). Thus the entropy of the universe has not changed due to the process. The entropy generation can be obtained from a balance equation. But simply we can state that the entropy generated is the sum of the change of entropy of the system and the change of entropy of the surrounding.
delta S(system) + delta S (surrounding) = entropy generated.
The entropy of the surrounding can change only if heat is added or removed from the surrounding. SO if there is no heat exchange between the system and the surrounding (adiabatic) or in other words the there is no entropy change of the surrounding. If it is reversible, then there is no entropy generated since there is no useless energy that is dissipated. So if we go back to the equation
delta S(system) + delta S (surrounding) = entropy generated.
We have delta S (surrounding) as zero due to adiabatic process and entropy generated as zero due to reversible process.
Thus delta S (system) is zero or in other words the entropy of the system remains constant through the process and is thus isentropic.
adiabatic process is not always isentropic in nature .their will be some change in entropy due irreversibility associated but if it reversible the both are same.
reversible adiabatic process is isentropic. Isentropic is ideal process where there is no irreversibility but adiabatic process encounter irreversibility. All isentropic processes are Adiabatic and not vice versa.
All these long-winded answers (some even irrelevant) are unnecessary. The question was simple and straight forward: What is the difference between reversible adiabatic process and isentropic process?
The simple, straight answer is 'none'! I was pained to see some answers here that could not even differentiate between isentropic and isotropic.
A reversible adiabatic process is always isentropic. The reverse is not true.
Entropy in a closed system can change because of two contributions:
1) Entropy fluxes at the system boundaries which are associated to heat fluxes. If the latter do not exist, the process is adiabatic.
2) Entropy production, within the system volume, which is associated to non-reversible processes. If entropy does not change the process is isentropic.
So, you can have an isentropic process, where you extract all the entropy that is produced (with heat fluxes going out of the system), which is not reversible, neither adiabatic.
Of the three conditions adiabatic, reversible and isentropic, if any two of them are valid, also the third one is true.
Sorry, ai the end of the sentence relative to point2) you should read reversible and not isentropic. and instead of does not change you must read is not produced.
"Adiabatic process" refers to a processes that take place in a closed system with no heat interaction with it's surroundings.
"Isentropic process" refers to a processes that take place in a closed system with no heat interaction with the surroundings (adiabatic process) and internally reversible means there is no internal generation of entropy, entropy stays constant, which is what is meant by "isentropic".
Isoentropic, or more briefly isentropic, means that the entropy does not change (iso), and not that the process is adiabatic.
As I already said:
You can have an isentropic process, where you extract all the entropy that is produced (with heat fluxes going out of the system), which is not reversible, neither adiabatic, but the entropy of the system does not change..
Dear colleagues,
We can summarize the difference in the following points:
- Adiabatic process is the process in which there is absolutely no heat loss and gain in the fluid being worked on whereas isentropic process is still an adiabatic process (there’s no heat energy transfer) and is the reversible type (no entropy change).
- Adiabatic process can either be reversible and irreversible but isentropic process is specifically a reversible adiabatic process itself.
I found this definition in the link:
http://www.differencebetween.net/science/difference-between-adiabatic-and-isentropic/
It is true!!???????????????
Best regards
Fateh
Dear Mr. Fateh,
A reversible adiabatic process is always isentropic. The reverse is not true.
Entropy in a closed system can change because of two contributions:
1) Entropy fluxes at the system boundaries which are associated to heat fluxes. If the latter do not exist, the process is adiabatic.
2) Entropy production, within the system volume, which is associated to non-reversible processes. If entropy does not change the process is isentropic.
So, you can have an isentropic process, where you extract all the entropy that is produced (with heat fluxes going out of the system), which is not reversible, neither adiabatic.
Of the three conditions adiabatic, reversible and isentropic, if any two of them are valid, also the third one is true.
The reversible adiabatic process is an isentropic( constant entropy) process but an isentropic process might be an irreversible and not adiabatic process.
But what if in a non-adiabatic irreversible process, the amount of entropy generation (dR) is higher or lower than the entropy extracted through the thermal flux (dQ/T). This will render a positive or negative dS value which is not isentropic. So we can say that non-adiabatic, irreversible process may be isoentropic but an adiabatic reversible process will necessarily be isoentropic.
Entropy is a thermodynamic state function that characterizes any type of transformation experienced by a working substance or a system or both in the different thermodynamic states that are carried out. By definition the entropy can not take the value zero and its value does not change during the course of an isentropic process or it is equal reversible adiabatic.
Dear Mr. Molina,
the prefix iso- means that the quantity remains constant (in old greek isos means equal; e.g.: isotherm, isobaric, isovolume, isotropic, etc.).
So isentropic means that the entropy remains constant (we do not say isoentropic since it is cacophonous). But this can happen also if the process is not reversible (please see my previous anwer). Therefore, the answer 2.1) is not correct.
Regards
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
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.
Calculus is not "take for granted".
In fact, 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.
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 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.
A detailed explanation of that ΔQ/T can NOT turned into dQ/T:
In thermodynamics, because 1/TdQ is used to integrate∮1/TdQ, and 1/T is a single function of T, that is 1/T=f1(T) , so, only when Q is also a single function of T, Q=f2(T), and Q=f2(T) is a derivable function ( obviously, f1(T)=1/T is a continuous function), then, the original function of 1/T exist: 1/TdQ=1/Tdf2(T) =(1/T)f′2(T)dT =f3(T)dT =Cdf4(T) , C is a constant. and, ∮1/TdQ=∮T (1/T)df2(T) =∮T Cd f4(T) ( ∫T f(T)dF(T) is a Stieltjes integral, namely, ∮T f(T)dF(T) is also a Stieltjes integral. ) ( in ∮T and ∫T , T is a subscript ) then, ∮T (1/T)dQ is measingful. However, because we have known that Q is NOT a single-valued function of T, in fact, Q=f (T, V, P), generally speaking, two of these three variables(T, V, P) are independent. So, ∮1/TdQ=∮T (1/T)df2( T, V, P) is meaningless, the reason is that the integrand is a function of one variable and there is only one integral sign∮, but there are several variables after the differential sign d. Compare it with the definition of integral, obviously, ∮1/TdQ=∮T (1/T)df2( T, V, P) is neither a single integral nor a multiple integral, in fact, there's no such integral as ∮T (1/T)df2( T, V, P) at all ( there's no such integral as ∫x f(x)dF(x, y, ...) at all ). Or say, it is not a integral or it is a wrong integral. No doubt, the "integral" ∮(1/T)dQ is meaningless. So, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
Dear Mr.Shufeng Zhang,
I would suggest that before you criticize the entropy concept, you should read:
Thermodynamics and an Introduction to Thermostatistics, by Herbert B. Callen and the work of Lars Onsager, Nobel Prize in Chemistry in 1968.
Entropy is a function of the state of a system which can vary because of production inside the system or fluxes at the system boundaries.
The total internal energy is again a function of the thermodynamic state of a system which can vary because of fluxes at the system boundaries. No production is involved.
The quantity ∮T (1/T)dQ represents, in an elementary form, the entropy variation due to fluxes.
Please, apply yourself to a deeper study of thermodynamics.
Regards and Good luck.
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.
Dear Giovanni Maria Carlomagno,
Please read my paper and those answers of those questions on my paper in my Projects.
The understanding(NOT definition) on heat and work( or say, process quantity) of the standard textbook IS WRONG !
The standard textbook says that
1) heat Q and work W are process quantities. This is right.
2)heat Q and work W(or say, process quantities) can not be the functions of system state variables. However, this is WRONG !
In fact, when the process path is fixed, the process variable is ALSO a state variable.
So,
on the one hand, heat Q and work W are process quantities.
on the other hand, when the process path is fixed, to the fixed process path, heat Q and work W are state quantities.
When we discuss the heat and work, what we discuss is just the heat and work of ONE process path, although it can be any one process path, but it is just ONE given(fixed) process.
A state quantity depends only on the thermodynamic state of the system.
Now you say:
2)heat Q and work W(or say, process quantities) can not be the functions of system state variables. However, this is WRONG !
In fact, when the process path is fixed, the process variable is ALSO a state variable.
So,
on the one hand, heat Q and work W are process quantities.
on the other hand, when the process path is fixed, to the fixed process path, heat Q and work W are state quantities.
When we discuss the heat and work, what we discuss is just the heat and work of ONE process path, although it can be any one process path, but it is just ONE given(fixed) process.
Probably you seem to ignore that your process paths from one thermodynamic state to another can be infinite, each one with a different Q and a different W.
In any case, please invite me when you receive your Noble prize for your great discovery.
Regards
Dear Giovanni Maria Carlomagno,
you said:
"your process paths from one thermodynamic state to another can be infinite, each one with a different Q and a different W."
Right, you are right.
this is just what I said in my paper "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 a fixed(given) reversible process path."
So are W and E.
(Above all, all reversible process paths are distributed on a surface determined by the system state equation. ANY path in this surface is a path that can exist.)
Now, in a surface, there are two points: point1 and point 2 ( corresponding to state 1 and state 2 ), then,
how many curves can you draw that connect point 1 to point 2 ? obviously, no doubt, you can draw innumerable ( countless, infinite ) curves, each curve is a process path, each process path corresponds to a function of Q( and W, E), so, 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 a fixed(given) reversible process path.
A detailed explanation of that ΔQ/T can NOT turned into dQ/T:
In thermodynamics, because 1/TdQ is used to integrate∮1/TdQ, and 1/T is a single function of T, that is 1/T=f1(T) , so, only when Q is also a single function of T, Q=f2(T), and Q=f2(T) is a derivable function ( obviously,
f1(T)=1/T is a continuous function), then, the original function of 1/T exist:
1/TdQ=1/Tdf2(T)
=(1/T)f′2(T)dT
=f3(T)dT
=Cdf4(T) , C is a constant.
and,
∮1/TdQ=∮T (1/T)df2(T)
=∮T Cd f4(T)
( ∫T f(T)dF(T) is a Stieltjes integral, namely,
∮T f(T)dF(T) is also a Stieltjes integral. )
( in ∮T and ∫T , T is a subscript )
then, ∮T (1/T)dQ is measingful.
However, because we have known that Q is NOT a single-valued function of T, in fact, Q=f (T, V, P), generally speaking, two of these three variables(T, V, P) are independent.
So, ∮1/TdQ=∮T (1/T)df2( T, V, P) is meaningless, the reason is that the integrand is a function of one variable and there is only one integral sign∮, but there are several variables after the differential sign d.
Compare it with the definition of integral, obviously, ∮1/TdQ=∮T (1/T)df2( T, V, P) is neither a single integral nor a multiple integral, in fact, there's no such integral as ∮T (1/T)df2( T, V, P) at all ( there's no such integral as ∫x f(x)dF(x, y, ...) at all ). Or say, it is not a integral or it is a wrong integral.
No doubt, the "integral" ∮(1/T)dQ is meaningless.
So, ΔQ/T can not be turned into dQ/T.
That is, the so-called "entropy " doesn't exist at all.
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.
When you don't know the function relationship of B and A, could you turn AΔB into AdB directely ?
NO !
In AΔB, the relationship of A and ΔB is the product of A and ΔB, so, A and B can be any two quantities and can be any relationship, that is there can have function relationship or have NO function relationship between A and B.
But, in AdB, the relationship of A and dB is not the product of A and dB, but the relationship to Find the Original Function of A in AdB.
As we know, only when we knew A=f(B), namely, the relationship of A and B is single valued function relationship, and A=f(B) is a continuous function, B is a derivable function( include Independent Variable Function), you can turn AΔB into AdB, otherwise, you can not turn AΔB into AdB directely.
That is, 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).
Similely, could you turn 1/TΔQ into 1/TdQ directely ?
NO, you certainly CAN NOT, for there is not the function relationship as Q=f(1/T) (namely Q=f(T) ), let alone we knew Q≠f(T).
And once we know Q≠f(T), then we immediately know that 1/TΔQ CAN NOT turned into 1/TdQ.
In any case, please invite me when you receive your Noble prize for your great discovery.
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