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Unformatted text preview: ∂P ∂v v v v v T = − RT 1 v 2 , (2.2) = − P v . (2.3) The slope of the isentrope is found in the following way. Consider ³rst the Gibbs equation, Eq. (1.1): Tds = du + Pdv. (2.4) 17 18 CHAPTER 2. CYCLE ANALYSIS v P T s 1 2 3 4 1 2 3 4 w cycle = ∫ P dv = q cycle = ∫ T ds isentrope isotherm isotherm i s e n t r o p h m is Figure 2.1: Sketch of P − v and T − s planes for a CPIG in a Carnot cycle. Because the gas is calorically perfect, one has du = c v dT, (2.5) so the Gibbs equation, Eq. (2.4), becomes Tds = c v dT + Pdv. (2.6) Now for the ideal gas, one has Pv = RT, (2.7) Pdv + vdP = RdT, (2.8) Pdv + vdP R = dT. (2.9) CC BY-NC-ND. 18 November 2011, J. M. Powers....
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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.
- Fall '11