Dryness fraction of saturated steam

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SUMMARY

The discussion centers on the unexpected increase in the dryness fraction of saturated steam when expanded isentropically through a turbine. The initial conditions involve a saturated steam mixture at 330°C with a vapor mixture fraction of 0.5, flowing at 6 kg/s, and expanding to 150°C. The calculations show that the dryness fraction increases to 0.53151, contradicting the expectation of a decrease. This anomaly is attributed to the assumptions made regarding isentropic efficiency and the low initial dryness fraction.

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  • Understanding of thermodynamic principles, specifically isentropic processes.
  • Familiarity with steam tables and enthalpy calculations.
  • Knowledge of the concept of dryness fraction in steam systems.
  • Basic proficiency in performing thermodynamic calculations involving turbines.
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  • Learn how to utilize steam tables for enthalpy and entropy calculations.
  • Investigate the effects of varying the initial dryness fraction on turbine performance.
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Thermodynamic engineers, mechanical engineers, and students studying steam turbine operations and thermodynamic cycles will benefit from this discussion.

maserati1969
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Hi all, any thoughts on the following problem please?

When saturated steam is expanded isentropically across a turbine, the dryness fraction should decrease. However, when I calculated the answer to the following question, it actually increased slightly. Any ideas where I went wrong?

Question:
Saturated steam mixture at temperature 330°C with vapour mixture fraction of 0.5 is flowing through a turbine at a flow rate of 6 kg/s. Steam is expanded isentropically to temperature 150°C at the turbine exit. Assume that the turbine external walls are fully insulated (adiabatic turbine). Calculate enthalpy at exit.

My solution:
Using enthalpy formula at inlet (3300C) and data from table: h1 = hf +x.hfg
h1 = 2095.95 kJ/kg
Is isentropic so s1 = s2 So entropy at inlet: s1 = sf +x.sfg
s1 = 4.4969 kJ/kg.K also = s2
Therefore at 150 degrees C: entropy: 4.4969 = 1.8418 + (x.4.9953)
Therefore x = 0.53151 (This should decrease, not increase)
Now use new ‘x’ value for h2:
h2 = 632.18 + (0.53151.2113.8)
h2 = 1755.7 kJ/kg

Many thanks in advance.
Regards
 
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Hi,

I get the same result. Perhaps the exercise composer didn't worry about reality. 100% isentropic efficiency ?
And ##x_{in}=0.5## is low: for ##x_{in}>0.55## it doesn't increase any more.

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Thank you BvU
 
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