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I've been working some Thermodynamics problems on Power Cycles and I have noticed that if I'm given a specific quality at the exit of the turbine and considering the adiabatic process I will have two properties (s,x) but I will not be able to define the state.

When I say i will not be able to define the state I mean I cannot compute the other state properties. I've been trying to use several software and they scream at when I input these two values. Math also doesn't help. Only intersecting lines in an h-s diagram helps (approximately).

Furthermore, studying the math behind it seems that these two values cannot be given since it is possible to define two specific values of entropy and quality that do not define an existing(real) pressure, so the illusion I get from intersecting values in the graph and approximating the pressure as a solution is only an illusion.

My question is, am I right in my previous claim, or is there a deep connection that I could use to compute such a combination of properties?

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Certain values are indeed invalid, but that usually just results in software complaining it could not find a solution. I believe the reason the software refuses to solve, is that there may be multiple solutions to a (s,x) pair.

For example tracing the 50% line on this graph from wikicommons: http://en.wikipedia.org/wiki/Phase_diagram#mediaviewer/File:Mollier_enthalpy_entropy_chart_for_steam_-_US_units.svg (apparently stackechange doesn't support svgs)
shows that it wiggles back and forth across 1.04 a couple times, so there are four solutions to (1.04,0.5)
This could be especially problematic in the near vertical sections as any tiny variation in entropy would result in a huge variation in output state. The resulting system would thus either be highly unstable or the defining conditions aren't correct. Perhaps the problems you're looking at are meant to be solved in a manner other than using an adiabatic relation or specifying the output quality was just extra information that was not needed to solve the problem.

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  • $\begingroup$ Ah yes! This at least relieves me from the horror of being no values at all since the line is continuous after all. I was given a Rankine Cycle problem and the output from the turbine was fixed by the quality and the adiabatic condition for the turbine hence I complained to the TA who suggested I should use the graph to intersect lines. Neither did I have any parameters for the condensate, so the only values were the (s,x) pair. [It seems i need 15 rep to vote up, sorry for that. Will someone do it for me please.] $\endgroup$
    – bromega
    Commented Feb 2, 2015 at 22:08

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