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Water vapor is cooled in a closed, rigid tank from 520°C and 100 bar to a final temperature of 270°C. Determine the final pressure, in bar, and sketch the process on T–y and p–y diagrams.
Water vapor is cooled in a closed, rigid tank from 520°C and 100 bar to a final temperature of 270°C. Determine the final pressure, in bar, and sketch the process on T–y and p–y diagrams.
Final answer:In this thermodynamics problem, a high-temperature and high-pressure system is cooled at a constant volume. Using the ideal gas law, it can be shown that if the temperature decreases at a constant volume, then pressure also decreases proportionally. In these conditions, the final pressure is calculated to be 68.5 bar.Explanation:The subject in focus here deals withthermodynamics, specifically property relations and phase transitions. We start with a system at high temperature (520°C = 793K) and high pressure (100 bar = 100 x 105 Pa) which is allowed to cool at a constant volume (definition of a rigid tank) to a lower temperature (270°C=543K).Since the process happens in a closed, rigid tank, the volume remains constant. For the ideal gases, if the temperature decreases at a constant volume, then pressure will also decrease proportional to the temperature by the ideal gas law (pV = nRT, in which p=pressure, V=volume, n=number of moles, R=gas constant, T=temperature).Therefore, the final pressure can be determined using the initial and final temperature and the initial pressure. The equation will be in the form of:p2 = p1*(T2/T1)This gives final pressure as: p2 = 100 x (543/793) =68.5 bar.For T–y and p–y diagrams, because the volume and hence the mass of water vapor remains constant, y (quality of steam) remains same through the process. Hence, straight horizontal lines in both diagrams represent the process.Learn more about Thermodynamics here:brainly.com/question/34045237#SPJ11...