Thermodynamics filled in class notes_Part_6

287 kg k 1200 k rt2 p2 536866 103 kp a 3 v2 006415

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Unformatted text preview: aw is satisfied at state 2: kJ 0.287 kg K (1200 K ) RT2 = P2 = = 5.36866 × 103 kP a. 3 v2 0.06415 m kg (2.25) This matches. Now the expansion from state 2 to 3 is isothermal. This is the heat addition step in which the volume triples. So one gets v3 = 3v2 = 3 0.06415 m3 kg = 0.19245 m3 . kg (2.26) T3 = T2 = 1200 K. (2.27) The ideal gas law then gives P3 = kJ 0.287 kg K (1200 K ) RT3 = = 1.78955 × 103 kP a. 3 v3 0.19245 m kg (2.28) Process 3 to 4 is an isentropic expansion back to 400 K . Using the isentropic relations for the CPIG, one gets T3 T4 v4 = v3 P4 = P3...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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