Interpretation:
The final enthalpy of superheated steam as it expands under isentropic conditions should be deduced.
Concept Introduction:
- For a two-phase liquid-vapor equilibrium mixture the specific volume (V), enthalpy (H) and entropy (S) are given as:
- For a process that takes place at constant enthalpy, the change in enthalpy is zero. In other words, the enthalpy in the final state (H2) is equal to that in the initial state (H1). The change in enthalpy is given as:
Final enthalpy of steam = 2597.5 kJ/kg
Given:
Initial pressure of steam, P1 = 500 kPa
Initial Temperature of steam =
Final pressure P2 = 50 kPa
Explanation:
Since this is a constant entropy process, S1 = S2
The initial state enthalpy (H1) and entropy (S1) can be deduced based on the steam table data for superheated steam at 500 kPa and
The final state temperature and enthalpy (H2) can be deduced from steam tables from the calculated initial entropy data.
Calculations:
Step 1:
Calculate the initial state enthalpy (H1) and entropy (S1) at T =
Based on the steam tables at the initial state pressure = 500 kPa we have:
For superheated steam:
At Saturation temperature, T =
Specific enthalpy of vapor, Hg = H1 = 2925.7 kJ/kg
Specific entropy of vapor, Sg = S1= 7.4598 kJ/kg-K
Step 2:
Calculate the final enthalpy (H2)
Since, S2 = S1 We have, S2 = 7.4598 kJ/kg-K
This is wet steam. Hence, for a final pressure P2 = 50 kPa for saturated steam
Specific entropy of vapor, Sg = 7.5939 kJ/kg-K
Specific entropy of liquid, Sf = 1.0910 kJ/kg-K
The steam quality x can be deduced from equation (3) as:
At 50 kPa,
Specific enthalpy of vapor, Hg = 2645.9 kJ/kg
Specific enthalpy of liquid, Hf = 340.47 kJ/kg
Based on equation (2) we have for the initial state enthalpy:
Thus, final enthalpy = 2597.5 kJ/kg
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Introduction to Chemical Engineering Thermodynamics
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