For a simple substance, enthalpy --and any other thermodynamic property--depends on two variables, e.g., temperature T and pressure p.

An adiabatic process is a process in which no heat is exchanged. An adiabatic and reversible process has constant entropy s--it is isentropic.

An isenthalpic process has constant enthalpy, and probably there is a myriad ways to realize such a process. However, the most commonly discussed isenthalpic process is the irreversible adiabatic expansion in a throttling device (e.g., flow is pushed through a small orifice or a porous plug).

The process is adiabatic, since the throttle is so small that there is no time for the throughput to exchange heat. The pressure drops due to friction.

First and second law of thermodynamics for such a process reduce to

h(T1,p1) = h(T2,p2)

s(T1,p1) < s(T2,p2)

where indices 1,2 indicate inflow and outflow values.

Only for an ideal gas, where enthalpy depends on temperature only [that is, h(T)], does temperature stays the same. In general, for an arbitrary substance, pressure drops, and temperature can increase, stay the same ,or drop. This depends on the Joule-Thomson coefficient which is defined as J=(dT/dp) for constant h .

Best know use for a throttle is in refrigerator, where they are used to expand liquid cooling fluid at (roughly) room temperature to create the low temperature in the evaporator.

I am from the fluid dynamic field and I differently consider iso- and homo- properties. We consider iso-enthalpic a flow that has total enthalpy constant along a pathline. If the conditions are homogenous upstream, than the process is homentalpic. An homoentalpic flow retains the total temperature constant.

Adiabatic means only that the system does not exchange heat with the environment at the boundaries.

Hello Lothar,

Let me first define them then I give you my opinion about obtaining zeros for both processes.

The adiabatic process occurs when there is no transferable heat or any quantity between the system and its outside surroundings in which the energy transfers to be work solely.

On the other hand, the isenthalpic process occurs when there is no change in enthalpy.

Now, I believe that those processes could be similar or equal to zero when we have a reversible process and the pressure to be constant.

Best regards with your work.

Hayder

Hello Lothar

When a closed system executes a process, it may interact with the surroundings in terms of heat and work. Insulating the system thermally would make the process, whatever, adiabatic. Now, for an ideal gas, system enthalpy is the function only of its thermodynamic temperature by Joule's law. Means to say that a closed system (an ideal gas) undergoing an adiabatic process at constant temperature (absolute) is undergoing an isenthalpic process. Possibility that such a process will occur needs further deliberation.

Dear Khalid,

thank you for the answer. If an adiabatic closed system does not change T during a process it is isenthalpic. OK.

But what is about Joule Thomson effect. This can be made adiacatic by thermal isolation. If a real gas expands, then T changes. This is an isenthalpic process??

Textbook says yes.

Regards

Lothar

Dear Lothar,

I think you can consider the example of a spherical bubble filled with gas that has adiabatic surface so that cannot exchange heat with the environment in which expands. The expansion is a reversible process as it is associated only to normal stress (the pressure) and you can have such variation by mantaining constant the total temperature and total enthalpy (H=cp*T0).

For a simple substance, enthalpy --and any other thermodynamic property--depends on two variables, e.g., temperature T and pressure p.

An adiabatic process is a process in which no heat is exchanged. An adiabatic and reversible process has constant entropy s--it is isentropic.

An isenthalpic process has constant enthalpy, and probably there is a myriad ways to realize such a process. However, the most commonly discussed isenthalpic process is the irreversible adiabatic expansion in a throttling device (e.g., flow is pushed through a small orifice or a porous plug).

The process is adiabatic, since the throttle is so small that there is no time for the throughput to exchange heat. The pressure drops due to friction.

First and second law of thermodynamics for such a process reduce to

h(T1,p1) = h(T2,p2)

s(T1,p1) < s(T2,p2)

where indices 1,2 indicate inflow and outflow values.

Only for an ideal gas, where enthalpy depends on temperature only [that is, h(T)], does temperature stays the same. In general, for an arbitrary substance, pressure drops, and temperature can increase, stay the same ,or drop. This depends on the Joule-Thomson coefficient which is defined as J=(dT/dp) for constant h .

Best know use for a throttle is in refrigerator, where they are used to expand liquid cooling fluid at (roughly) room temperature to create the low temperature in the evaporator.

following

Thanks Gert Van der Zwan

Dear Lothar,

In my understanding, an isenthalpic process is a process that proceeds without any change in the enthalpy. An adiabatic process is one that occurs without transfer of heat or matter between a system and its exterior. In the two processes, energy is transferred only as work.

Following is valid for reversible and irreversible processes: (pvT-Thermodynamics)

Enhalpy: H := U+pV

dotH = dotU + dotp V + pdotV

1st law: dotU = dotQ - pdotV ( closed system, no chemical reactions)

----> dotH = dotQ + dotp V

isenthapic: dotH = 0 ---> dotQ = -Vdotp

isenthalpic + adiabatic ---> dotp =0

Textbook stuff

Best regards to you all

W.M.

Thanks to all of you!!

Lothar

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.