Q2. A piston-cylinder contains 1.5 kg of saturated liquid water at 3 MPa. The water is heated at constant pressure until it reaches becomes a saturated vapour. The piston expands quasi-statically. Assume there is no entropy generated in the water (Sgen=0). a) What is the change in entropy of the water? b) How much heat was added? (Hint: use entropy balance.) c) Why would we assume Sgen=0?
Question
Q2. A piston-cylinder contains 1.5 kg of saturated liquid water at 3 MPa. The water is heated at constant pressure until it reaches becomes a saturated vapour. The piston expands quasi-statically. Assume there is no entropy generated in the water (Sgen=0). a) What is the change in entropy of the water? b) How much heat was added? (Hint: use entropy balance.) c) Why would we assume Sgen=0?
Solution
a) The change in entropy of the water can be calculated using the formula:
ΔS = S2 - S1
Where: S1 is the initial entropy, S2 is the final entropy.
Since the process is isentropic (Sgen=0), the entropy of the system remains constant. Therefore, the change in entropy ΔS = 0.
b) The heat added to the system can be calculated using the entropy balance equation:
Q = T * ΔS
Where: Q is the heat added, T is the absolute temperature, ΔS is the change in entropy.
Since ΔS = 0 (as calculated in part a), the heat added Q = 0. This means no heat was added to the system.
c) We assume Sgen=0 because the process is quasi-static and reversible. In a reversible process, there is no generation of entropy within the system. This is an idealized assumption used to simplify the analysis of the process. In reality, all real processes generate some entropy due to irreversibilities such as friction, heat transfer across a finite temperature difference, etc.
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