Is This Calculation of Steam Expansion and Superheater Efficiency Correct?

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SUMMARY

The discussion focuses on the calculation of steam expansion and superheater efficiency, specifically analyzing steam at 2000 kPa and 250°C expanded isentropically to 550 kPa. The initial state is identified as superheated steam, with the dryness fraction calculated at 95.01% and a change in specific enthalpy of 254.42 kJ/kg. Additionally, the heat energy supplied per kg of steam in the superheater is determined to be 374.07 kJ/kg, with a volume increase of 41.7% due to drying and superheating.

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  • Understanding of thermodynamic properties of steam, including superheated and saturated states.
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  • Knowledge of isentropic processes and their implications in thermodynamics.
  • Ability to perform calculations involving dryness fraction and specific volume.
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  • Investigate the relationship between dryness fraction and specific enthalpy in steam systems.
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rmenon
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Need some help here please,
Steam at 2000 Kpa and 250 degree centigrade is expanded isentropically to a pressure of 550 Kpa.
Determine the
a) dryness fraction of steam
b) change in specific enthalpy of steam
 
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What are your thoughts/ideas on this problem? You need to show your work to get help.
 
This is where I got stuck. I checked the steam tables for h value of superheated steam at 2000 kpa and 250 degree centigrade. Got the values of hf and hfg of wet saturated steam at 2000 kpa, used the formula
x= hsup-hf/hfg and got a value more than 1. This cannot be right because the dryness fraction should be below 100 percent is'nt it. I have'nt reached the b) part yet. I guess I could use s=sf +xsfg to get the specific enthalpy at both initial and the expanded pressures and then get to the change.
Am I getting this right or-------
Thanks and regards
 
First of all, at 2000 Kpa and 250 C, what's the state of water?

To figure this out, look at the saturation temperature at the given pressure of 2000 kpa. It's 212.4 C. Since the initial temperature is 250 C which is greater than the saturation temperature at the given pressure, the initial state is superheated steam. Now, your quality factor will not come into play as it's not saturated vapour.

Since you know the initial pressure and temperature of the superheated vapour, you can look up the corresponding thermodynamic properties (in this case, specific entropy) in the superheated vapour tables.

What's the final state of the vapour? Notice that the question asks for the "dryness fraction of steam". So, the final state is saturated vapour and you have to find the dryness fraction. How will you do this? Can you finish the problem now? Here's a hint. Use the fact that the process is isentropic.
 
Last edited:
Hi,
Think I got it. Can you please double check my answers. I got the dryness fraction 95.01% and the change in enthalpy as 254.42 KJ/Kg

Can you please check if I did this right ?
Q) Steam enters the superheater of a boiler at 2000 kPa and 0.98 dry and leaves at the same pressure and a temperature of 350 degree centigrade. Find
a) heat energy supplied per kg of steam in the super heater
b) percentage increase in volume due to drying and superheating.
Ans-
This is how I approaced the problem.
Orginal volume of 1 kg of steam, I used
v= vf + x(vg - vf) where I got v= 97.66 cm^3 = .097m^3/kg
Then I calculated v for superheated steam at 350 degree centigrade which I got v= 0.1383m^3/kg, the difference in volume being 0.0407m^3/kg which means an increase of 41.7% volume.
Heat supplied -
I calculated the specific entalpies at the two conditions ie. initial and final
inital I used h=hf + hfg where I got 2761.676 kJ/kg
and h at 350 degree centigrade I used the same formula and I got3135.75 kJ/kg
Hence heat energy added in the superheater was 374.07 kJ/kg.
Did I do right ?
Regards
Raj
 

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