Several sources seem to assume that the work in this case is ideally the enthalpy gained by
the fluid, but after the compression the compressor must push the fluid out from the compressor? I try to explain this question also in my new blog: http://thermodynamicalmisconceptions.blogspot.com
An application of the first law givesW=h_1 - h_2; The enthalpy used in theSteady flow energy equation automatically takes into account the 'effort' the compressor requires to push the fluid out (this is why you use enthalpy and not internal energy - the pv term accounts for the 'pushing out' across the control volume). The state points are deduced either via the steam tables (for a vapour), or via an explicit formulat (for a gas)
Thanks for the answers so far. I think I was previously suspecting something like this, but then I kind of lead myself astray. So enthalpy somehow (must check this) contains information on also other things than the 'internal' state of the fluid? Should we then say that the enthalpy of the fluid is still rising, or jumps further up, if it is in a situtation where p and t are constant (both are at the maximum) but it is pushed from the compressor to the 'high-pressure side' -- or enthalpy is only defined before and after the compressor ...
Enthalpy does indeed contain the 'something else' beyond internal energy. The 'pv' term in the definition of h=u+pv refers to the pressure work - how much 'work' you do on a parcel fluid to squeeze it into (or out of) your thermodynamic system. If you check out Rogers and Mayhew 'Engineering thermodynamics', p36 onwards, you'll see how the pv term comes about. Most other thermodynamics textbooks that contain analyses of steady flow processes also contain similar derivations.
Another forms . To determine the work isentropico in a compressor with a few such substances as freon, ammonia, between others. You must use the graph P-h, of the respective substance. With the pressure of the evaporator, in saturated and dry steam you look for the enthalpy of entry to the compressor. You move for the line isentropica up to the pressure of the condenser. Where these 2 lines are cut you look for the value of the enthalpy of unload of the compressor. The work isentropico is the difference of the enthalpy of unload fewer enthalpy of entry
The main differences between compression of air and similar gases and refrigerant
compression are:
- technical gases are ideal and have almost constant specific heat (cp = 1 kJ/kgK
for an air) and therefore the enthalpy change during compression can be easily found from the relation between temperature and pressure along isentrope
- refrigerants are usually in the superheated vapor state and as the real gases
can be treated in different way - enthalpy change can be found by using a proper "pressure-enthalpy" chart or tables (and computer procedures)
- a comparable isentropic index for refrigerants is smaller (k=1.1-1.3) than for an air (k=1.4), and the specific work is less than for air.
(Jan Gorski, AGH Uni, Cracow, PL)
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