Dryness Fraction and Entropy Generation in Non-Isentropic Steam Expansion

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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?



Homework Equations





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

[tex]\eta_{isen} = \frac{actual}{ideal}[/tex]

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

[tex]\eta_{isen} = \frac{actual}{ideal}[/tex]

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 [tex]\Delta[/tex]S 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 [tex]\Delta[/tex]S 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.