Thermodynamics filled in class notes_Part_6

Cycle analysis so t1 t2 v2 v1 1 k 1 100 m3 kg 400

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Unformatted text preview: mber 2011, J. M. Powers. 20 CHAPTER 2. CYCLE ANALYSIS So T1 T2 v2 = v1 1 k −1 = 1.00 m3 kg 400 K 1200 K 1 1.4−1 = 0.06415 m3 . kg (2.22) Note that v2 < v1 as is typical in a compression. The isentropic relation between pressure and volume can be rearranged to give the standard k k P2 v2 = P1 v1 . (2.23) Thus, one finds that P2 = P1 k v1 v2 2 = (1.148 × 10 kP a) m3 kg 3 0.06415 m kg 1.00 1.4 = 5.36866 × 103 kP a. (2.24) Note the pressure has increased in the isentropic compression. Check to see if the ideal gas l...
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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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