Dryness Fraction and Entropy Generation in Non-Isentropic Steam Expansion

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The discussion focuses on calculating the dryness fraction and entropy generation during the non-isentropic expansion of superheated steam. The initial calculations for the isentropic case yield a dryness fraction of 88%, using steam tables for enthalpy and entropy values. For the non-isentropic case with an isentropic efficiency of 88%, the actual enthalpy (h2) must be determined first, followed by using the equation h2=hf + X hfg to find the dryness fraction. Additionally, the rate of entropy generation can be calculated by performing an entropy balance, where the change in entropy is multiplied by the mass flow rate. The same values for hf and hfg from the isentropic case are applicable for the non-isentropic calculations.
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Homework Statement


I place the question first:
Superheated steam at an absolute pressure of 10 bar and temperature of 340C expands in a nozzle reversibly and adiabatically until the pressure is 0.1 bar. the mass flow rate is 4.2kg/s.
The questions asking for:
1) Drawing the appropriate diagram for the process.
2) Dryness fraction of the steam ?
3)Dryness fraction if the process is not isentropic and isentropic efficiency is 88% ?

and my particular problem:

I know how to workout dryness fraction when this process is reversible and adiabatic because being isentropic i can use: S1=Sf + XSfg and find the appropriate values off the steam table (found S1 using interpulation between temperatures of 300C and 350C and it was S1= 7.266 kJ/kgK) to eventually find X (dryness fraction which i calculated as 88%). From there i can use X to calculate ideal h2 as well.
However my problem is, when this process is not isentropic, how to we find dryness fraction? i guess i wouldn't be able to use that same equation (used above to find X) as i end up with the same values as before. My second thought is I use the value of isentropic efficiency to find the real value of h2 and then use this real value in equation h2=hf + X hfg (obviously with same hf and hfg that i used in the first part of question) to find X in this case? am i right? any help please?
and One more thing; how do we workout the rate of entropy generation of the steam during expansion?



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The Attempt at a Solution

 
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Any help please?
Its the non-isentropic case with isentropic efficiency of 88% which I'm not sure about. I worked out the dryness fraction for case one (isentropic process) though.
 
kaminho said:
Any help please?
Its the non-isentropic case with isentropic efficiency of 88% which I'm not sure about. I worked out the dryness fraction for case one (isentropic process) though.

If you have the idea h2 which shall be called h2s, then the isentropic efficiency should be

\eta_{isen} = \frac{actual}{ideal}

since the mass flow rate is the same, you will just need the change in enthalpies, so the idea would be h1-h2s. So you just need to get the actual h2 and then use h2=hf+x2hfg.
 
rock.freak667 said:
If you have the idea h2 which shall be called h2s, then the isentropic efficiency should be

\eta_{isen} = \frac{actual}{ideal}

since the mass flow rate is the same, you will just need the change in enthalpies, so the idea would be h1-h2s. So you just need to get the actual h2 and then use h=hf+x2hfg.


Thank you. Can I ask, Just to make sure, in equation h=hf+x2hfg after finding h2, will I be using the same values for hf and hfg which i used in the isentropic case?
Any hints about rate of entropy generation during expansion please?
 
kaminho said:
Thank you. Can I ask, Just to make sure, in equation h=hf+x2hfg after finding h2, will I be using the same values for hf and hfg which i used in the isentropic case?
Any hints about rate of entropy generation during expansion please?


Yes you would just use the hf and hfg at state 2 (the same conditions in the isentropic case)

For the entropy, you can do an entropy balance on the system:

rate of entropy entering + rate of entropy generation = rate of entropy generated

since you have the mass flow rate, calculating rate of entropy entering and rate of entropy generated should be simple.
 
rock.freak667 said:
Yes you would just use the hf and hfg at state 2 (the same conditions in the isentropic case)

For the entropy, you can do an entropy balance on the system:

rate of entropy entering + rate of entropy generation = rate of entropy generated

since you have the mass flow rate, calculating rate of entropy entering and rate of entropy generated should be simple.

would I be doing right if I found S2 and then multiply the \DeltaS by the mass flow rate?
I meant after finding S2 of the irreversible process because S2 of isentropic is obviously same as S1.
 
Last edited:
kaminho said:
would I be doing right if I found S2 and then multiply the \DeltaS by the mass flow rate?
I meant after finding S2 of the irreversible process because S2 of isentropic is obviously same as S1.

I assume you'd want to get the entropy generation rate for the non-isentropic case, else the generation is just zero.

You'd need to get s2 using s2=sf+x2+sfg.

But yes, once you get Δs, just multiply it by the mass flow rate.
 

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