When is Steam not an Ideal gas?

I am trying to get some help on this question. it is directly from the NCEES ME practice exam TFS 2019.

At first glance I used the constant entropy process for ideal gas and I got a negative outlet temperature in F. How can I tell if it is not an ideal gas process from the question?  

The solution used saturated condition at 3-psia to solve the question and I am not sure where that came from.

Question: Steam (600F, 300 psia) is expanded in an 80% adiabatic efficiency turbine to 3 psia. The enthalpy (Btu/lbm) of the resulting 3-psia steam is most nearly:

A  880

B  970

C  1,040

D 1,120

Thank you.

Dense gases such as steam in power plants and refrigerants in refrigeration cycles in general cannot be modeled as ideal gases. Only when the pressure is VERY low can steam be treated as an ideal gas (this is done in psychrometrics, where the partial pressure of water vapor is in the order of less than 2 psia or so). 
 

You have comprehensive steam tables available. Use them and avoid catastrophic mistakes like the one you described.

ICM said:

The solution used saturated condition at 3-psia to solve the question and I am not sure where that came from.

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1. With the inlet conditions, find the enthalpy at the inlet, h1, and entropy at the inlet, s1

2. The entropy at the discharge for the isentropic process (s2s) is the same as s1, so for the isentropic process you have s2s and pressure.

3. Is s2s < sg at 3 psia? If yes, you have a saturated mixture at the end of the isentropic process. Calculate quality at the end of the isentropic process, x2s

4. With x2s find the enthalpy at the exit of the isentropic process h2s

5. With h2s and the definition of isentropic efficiency find the enthalpy at the discharge of the actual process.

Slay the P.E. said:

Dense gases such as steam in power plants and refrigerants in refrigeration cycles in general cannot be modeled as ideal gases. Only when the pressure is VERY low can steam be treated as an ideal gas (this is done in psychrometrics, where the partial pressure of water vapor is in the order of less than 2 psia or so). 
 

You have comprehensive steam tables available. Use them and avoid catastrophic mistakes like the one you described.

Click to expand...

Thank you for the explanation. The use of psychometrics to check for ideal gas is new to me, thanks. 

If I swap the inlet/outlet conditions and replace the turbine with a compressor, will the constant entropy for ideal gas model work?. Since the inlet condition can now be approximated as an ideal gas?

I guess my follow-up question is, if the inlet condition can be modeled as an ideal gas, then I can use the constant entropy equations for ideal gas to calculate the outlet condition.

ICM said:

Thank you for the explanation. The use of psychometrics to check for ideal gas is new to me, thanks. 

If I swap the inlet/outlet conditions and replace the turbine with a compressor, will the constant entropy for ideal gas model work?. Since the inlet condition can now be approximated as an ideal gas?

I guess my follow-up question is, if the inlet condition can be modeled as an ideal gas, then I can use the constant entropy equations for ideal gas to calculate the outlet condition.

Click to expand...

No. Just never use ideal gas assumptions for steam. It’s safer that way. Use tables or a Mollier diagram.
 

For this particular problem: If you start at 3 psia and compress, there is some liquid present at the “inlet” therefore that liquid-vapor mixture is far from being an ideal gas. So that’s a huge no-no.

Even if you start with saturated or slightly superheated vapor at 3 psia (a condition in which the steam arguably behaves like an ideal gas) you are taking it to a condition of pressure high enough so the deviation from ideal gas behavior is significant, so whatever it is you’re calculating is likely to be wrong. 
 

Just please drop this idea of wanting to use Pv=RT for steam in power problems. 
 

Regarding psychrometrics: in typical HVAC applications, the partial pressure of water vapor is very small so moist air is modeled as a mixture of two ideal gases: dry air and water vapor. If you read in a thermo book the derivation of the equations used to plot the psychrometric chart you will see they model the water vapor as an ideal gas. But, you don’t have to worry about this, just use your psychrometric chart.

So let me repeat: become familiar with the Mollier diagram and the steam tables. Use those for steam power problems.

The better question would be, when is STEAM ever an ideal gas. It's quite rare that you can treat it as one. It really only ever acts as an ideal gas when it is in a highly superheated vapor state.