isoenthalpic vs isoentropy refrigeration process ?

[skan]

Hi.

I'm studying the refrigeration cycle and I know the basic thermodynamics equations (enthalpy, entropy...).
And I have a doubt.

Why is the compression considered approximately isoentropic and the expansion isoenthalpic and no the other way? (the compression isoenthalpic and the expansion isoentropic).
In the basic process (without extra efficiency factors or losses) I think both processes are adiabatic (if done slowly to be reversible).

regards.

[Magoo]

you would be better off studying the Spanish national debt crisis, than this particular subject.

[MikeHolm]

Oh Magoo, thou are indeed a cynic

[skan]

you would be better off studying the Spanish national debt crisis, than this particular subject.

What does it have to do with my question?

One way to get out of the crisis is produce more jobs, for example in refrigeration.

Yo can check that USA and the UK have much more public and external debt than Spain , and they are doing well because they don't care.

[Doug30293]

you would be better off studying the Spanish national debt crisis, than this particular subject.

Economics may seem easier to understand than thermodynamics until one realizes that thermo problems are solvable.

Skan;

Expansion is not considered isentropic because the expansion takes place in a throttling device (TEV or cap tube). Throttles introduce friction to the stream which wastes energy in the form of heat.


Why is the compression considered approximately isoentropic and the expansion isoenthalpic

The key word here is "approximately." Instruction in thermodynamics consists of three years working imaginary problems, followed by more classes where you learn that everything you learned in the first three years was only partially correct.

The friction of pumping losses is not considered in the compression stage because the compressor is imaginary. There is no imaginary throttle.

Doug

[skan]

Economics may seem easier to understand than thermodynamics until one realizes that thermo problems are solvable.

Skan;

Expansion is not considered isentropic because the expansion takes place in a throttling device (TEV or cap tube). Throttles introduce friction to the stream which wastes energy in the form of heat.



The key word here is "approximately." Instruction in thermodynamics consists of three years working imaginary problems, followed by more classes where you learn that everything you learned in the first three years was only partially correct.

The friction of pumping losses is not considered in the compression stage because the compressor is imaginary. There is no imaginary throttle.

Doug

Hi.

In fact I already had that (very) theoretical instruction in thermodynamics many years ago, but I didn't use it in any job, I've been working as a computer programmer instead.
And now the problem is not the theory, I remember most equations and theory, the problem is how to relate that theoretic concepts with true refrigeration devices. Anyway it seems to be easy.

Best regards to everybody except Magoo.

[skan]

I know you mean "irreversible" but I did my best in explaining it in English while keeping the explanation simple.

You say the compressor is imaginary but it could be the "least" imaginary part of the circuit. Isn't it?
You get a very steep and quick change in the conditions (mostly the pressure).

I know that the true diagram doesn't follow the isoentropic lines and it goes up to a little bit higher...

[Doug30293]

Yes, irreversible is probably a better word. I use the cynical word "imaginary" because textbook equations always seem to leave out real world conditions. The irreversible effects of friction and pumping losses in a real compressor are significant. Compare the efficiency of scrolls to reciprocating compressors. The textbook draws both of them as a circle on a diagram.

Compare the textbook PV diagram for a Diesel engine to a real PV diagram. They don't look at all alike. That is because real processes are not adiabatic, isoenthalpic, or isoentropic. It is the same for the refrigeration cycle.

[skan]

What I studied was meteorology, that's why I know about gases and physical processes but I know almost nothing about engines and pumps.
Now I'm starting a course on cryogenics and refrigeration.

[Doug30293]

In thermodynamics, the refrigeration cycle is also called a heat pump cycle. A heat engine is just the opposite of a heat pump. Work goes into a heat pump. Work comes out of a heat engine. The equations are simple. The assumptions used in those equations are often confusing. If you study them long enough you will become a grumpy old cynic like Magoo or myself.

[skan]

In thermodynamics, the refrigeration cycle is also called a heat pump cycle. A heat engine is just the opposite of a heat pump. Work goes into a heat pump. Work comes out of a heat engine. The equations are simple. The assumptions used in those equations are often confusing. If you study them long enough you will become a grumpy old cynic like Magoo or myself.

OK, and why isn't the compression cycle considered isoenthalpic as well?

[skan]

I meant the compression step.

[Doug30293]

OK, and why isn't the compression cycle considered isoenthalpic as well?

The work done by the compressor is added to the fluid primarily in the form of heat. It is called heat of compression and results in an increase in enthalpy. If you look at a PH (Mollier) chart for a refrigeration cycle you will notice the compression is shown as a diagonal line in the superheat region beyond the vapor dome. Temperature and pressure both increase.

It does approximately follow a line of isoentropy which is why the compression is said to be isoentropic. In an actual compressor we have friction which is why it is only an approximation.

It would be helpful if I could include a suitable PH diagram here. I am on a linux machine right now and still learning my way around.

This thread motivated me this week to read the first six chapters of my thermo textbook from college. It is hard to believe I once actually understood some of it.

[skan]

The work done by the compressor is added to the fluid primarily in the form of heat. It is called heat of compression and results in an increase in enthalpy. If you look at a PH (Mollier) chart for a refrigeration cycle you will notice the compression is shown as a diagonal line in the superheat region beyond the vapor dome. Temperature and pressure both increase.

It does approximately follow a line of isoentropy which is why the compression is said to be isoentropic. In an actual compressor we have friction which is why it is only an approximation.

It would be helpful if I could include a suitable PH diagram here. I am on a linux machine right now and still learning my way around.

This thread motivated me this week to read the first six chapters of my thermo textbook from college. It is hard to believe I once actually understood some of it.

Hello.

You don't need to include the Mollier PH diagram, I have it, and I know that the theoretical refreigeration cycle is drawn following an isoentropic line at the compression step.
But I don't know what's the theory behind that, in order to choose an isoentropic and not an isoenthalpic there. Maybe is not theory but experimental data.

[Doug30293]

But I don't know what's the theory behind that, in order to choose an isoentropic and not an isoenthalpic there. Maybe is not theory but experimental data.

Correct. The models are often an attempt to align theory with experimental data. The compression line roughly follows an isotrope so it is called an isoentropic process.

Some areas of thermodynamics, such as the saturation tables and heat transfer analysis, are entirely based on experimental data. The equations are nothing more than curve fit functions.