Thermodynamics, isentropic and reversible processes

[erezap]

Hi, I have some questions regarding thermo:
1) Does an isentropic process have to be a quasistatic one? What is the relation between equilibrium and reversibilty?

2) Naturally, there are revesible processes involving control volumes. A gas can go through an isentropic process in a reversible turbine, but as we know, turbines in a steady state have siginficant gradient of pressure inside them- the properties of the gas inside the turbine are not uniform.
How can the gas go through a reversible process in the turbine if it is not in equilibrium? (the gas in the turbine is not in equilibrium because it's properties are not unfiorm)

Also, it is said that the gas goes through an isentropic prcess, and let's say it goes through a turbine, what system is regarded? Is it the gas inside the control volume (always looking at the gas in the control volume- open system) ? is it a closed system that goes through the turbine, exits it and than moves on (following the gas- closed system)?

For example: if you have an ideal gas going through a quasistatic adiabatic process, than the process can be described by the equation: (pV)^k=const
if you have an ideal gas going through a revesible adiabatic process (isentropic)- the process can be described by the equation: (pV)^k=const
The same equation. So does this mean any isentropic process is quasistatic? If so how can the process in the turbine can be quasistatic if the properties of the gas in the trubine are not uniform. If not any isentropic process is quasistatic ,than how can both be equations be the same?

Generaly, how can a process in a control volume be reversible and adiabatic, if the properties of the gas are not uniform?

I must say I am very confused about this, and it's not the first time I took this course.
Thanks, Erez.

 

[h2oski1326]

1) Not really.

Any real process is not isentropic. When modeling thermodynamic systems it is often convenient to use the assumption of an isentropic process as a 'first-cut' model. This is similar to ignoring friction in a mechanical model. This works well in combination with the assumption that a process is quasi-static because increases in entropy can be considered small in some situations.

2) Being uncomfortable with the idea of an 'isentropic' turbine is well founded and that shows some insight that you have realized that, because these devices are just not isentropic. However, it is often a good place to start and is certainly a good place to start when learning the material. This is very similar to physics I when you learn about a block sliding down a frictionless inclined surface. You know perfectly well that there isn't such a thing as a frictionless surface but it is a good place to start, the same applies here.

2a) The equation you have displayed is for a polytropic process of a gas and can actually represent a lot of different situations depending on the value of 'k'. The exponent can be one value if one is modeling the situation as isentropic (e.g. 1.4 for air). Another situation is an isothermal process, where the value of k=1.

Just keep in mind that these are just mathematical models which represent what is actually going on and in a lot of situations it is much more complicated.

 

[Blueshift5]

Hi Erez,

Thank you for your questions regarding thermodynamics, isentropic and reversible processes. I can understand your confusion and I will do my best to clarify these concepts for you.

1) An isentropic process does not necessarily have to be a quasistatic one. Quasistatic processes are those in which the system remains in equilibrium throughout the entire process. On the other hand, an isentropic process is one in which the entropy of the system remains constant. So, while a quasistatic process is always isentropic, an isentropic process may or may not be quasistatic. The relation between equilibrium and reversibility is that a reversible process is one in which the system can be returned to its initial state without any change in the surroundings. In order for a process to be reversible, it must also be quasistatic. So, while equilibrium is a necessary condition for reversibility, it is not a sufficient condition.

2) In the case of a gas going through a reversible turbine, the gas inside the turbine is not in equilibrium because there is a pressure gradient. However, the process is still considered reversible because the gas can be returned to its initial state without any change in the surroundings. This is because the gas inside the turbine is still undergoing an isentropic process, meaning its entropy remains constant. The system being regarded is the gas inside the control volume, as it is the gas that is undergoing the process.

To answer your question about the equations being the same, it is important to note that the equations (pV)^k=const represent different types of processes. The equation for a quasistatic adiabatic process is for a closed system, while the equation for a reversible adiabatic process is for an open system. So, the equations may look the same, but they are describing different processes.

In general, a process in a control volume can be reversible and adiabatic if the properties of the gas are not uniform because the process is still isentropic, meaning the entropy remains constant. The non-uniform properties of the gas do not affect the reversibility of the process.

I hope this helps to clarify some of your confusion about these concepts. If you have any further questions, please do not hesitate to ask. Good luck with your studies!

Best,