Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. What is the fraction of the steam in its initial state?

93%

Explanation:

Hello,

In this case, based on the second state, at 101.33 kPa and 105 °C a saturated vapor is formed, which has an enthalpy of:

Now, since such condition was attained at constant enthalpy, the previous enthalpy equals the mixture's enthalpy at 1100 kPa, thus, for such condition:

In such a way, the quality , or fraction of the steam at the initial condition turns out:

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