Prelim_2_problem_3

Prelim_2_problem_3 - the speciﬁc internal energy of state...

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Prelim II - Problem 3: Free Expansion of Water Vapor Given : An insulated container filled with saturated water vapor at an unknown temperature expands to fill the container and winds up vapor at a known temperature and pressure. Find : The specific entropy and specific internal energy after filling the container, the initial temperature and pressure of the water, and the entropy produced by the process. Assumptions 1. Insulated container Δ Q = 0. 2. No work is done on the system. 3. System is drawn as shown in the figure. 4. Closed system. The specific entropy and specific internal energy of the water vapor at state 2 can found from the tables by interpolation: u = 2529 . 7 kJ / kg s = 7 . 44045 kJ / kg · K We first apply the First Law: Δ U = Δ Q - Δ W (1) Because of the first two assumptions, we can say that Δ U = 0. Because the system is closed, we know that

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Unformatted text preview: the speciﬁc internal energy of state 1 is the same as the speciﬁc internal energy of state 2. By looking at the values for speciﬁc internal energy of a saturated vapor in Table A-2, we ﬁnd that: T = 120 ◦ C p = 1 . 985 bar s = 7 . 1296 kJ / kg · K 1 To ﬁnd the entropy produced by the process, we use the Second Law: Δ S = Z 2 1 δQ T + σ (2) Because of the ﬁrst assumption, there is no heat transfer, so the entropy produced during the process is: m ( s 2-s 1 ) = σ σ/m = 0 . 31085 kJ / kg · K σ = 3 . 1085 kJ / K This process is not reversible because in order for a process to be reversible, there must be no entropy generation. This is an irreversible process because σ > 0, where as if σ < 0 the process would be impossible. 2...
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