A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o

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A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o

f the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized. Determine the entropy change of the steam during this process. Use steam tables.

Engineering

2 answers:

Daniel [329] · Answer 1 · 2026-03-25T11:30:54+03:00

Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

iogann1982 [368] · Answer 2 · 2026-03-25T11:30:44+03:00

The variation in entropy of the steam throughout this process is determined to be 5.55615 KJ/kg. To analyze this, we refer to the steam tables to gather the properties for both states. At State 1, we have a pressure of P1 = 200 KPa, and the quality is calculated as X1 = 1 - 3/4, resulting in a value of 0.25. Consulting the steam table for 200 KPa, we find the specific volume of the liquid Vf = 0.001061 m³/kg and for the gas Vg = 0.88578 m³/kg. Therefore, the difference in specific volumes Vfg comes to Vg - Vf = 0.884719 m³/kg. The entropy values are Sf = 1.5302 KJ/kg.k and Sfg = 5.5968 KJ/kg.k. Subsequently, the specific volume of water at state 1 is computed as V1 = Vf + X1 Vfg = 0.001061 m³/kg + (0.25)(0.884719 m³/kg) leading to a result of V1 = 0.22224075 m³/kg. Correspondingly, the entropy is expressed as S1 = Sf + X1 Sfg = 1.5302 KJ/kg.k + (0.25)(5.5968 KJ/kg.k), resulting in S1 = 2.9294 KJ/kg.k. For State 2, where V2 remains equal to V1 (0.22224075 m³/kg) and X2 = 1 (Saturated Vapor), steam table interpolation yields S2 = 6.6335 KJ/kg.k. The change in entropy, ΔS, is represented by the equation ΔS = m(S2 - S1), calculated as ΔS = (1.5 kg)(6.6335 - 2.9294) KJ/kg.k, yielding ΔS = 5.55615 KJ/K.