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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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This note was uploaded on 03/15/2010 for the course ENGRD 221 taught by Professor Staff during the Fall '08 term at Cornell.
- Fall '08