Thermodynamics filled in class notes_Part_6

Thermodynamics filled in class notes_Part_6 - 19 2.1 CARNOT...

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2.1. CARNOT 19 So then Tds = c v p Pdv + vdP R P b ² = dT + Pdv, (2.10) = ³ c v R + 1 ´ + c v R vdP, Take ds = 0 , (2.11) 0 = ³ c v R + 1 ´ P + c v R v ∂P ∂v v v v v s , (2.12) v v v v s = c v R + 1 c v R P v , (2.13) = c v c P c v + 1 c v c P c v P v , (2.14) = c v + c P c v c v P v , (2.15) = c P c v P v , (2.16) = k P v . (2.17) Since k > 1, the magnitude of the slope of the isentrope is greater than the magnitude of the slope of the isotherm: v v v v v v v v s v v v v > v v v v v v v v T v v v v . (2.18) Example 2.1 (Adapted from BS, 7.94, p. 274) Consider an ideal gas Carnot cycle with air in a piston cylinder with a high temperature of 1200 K and a heat rejection at 400 K . During the heat addition, the volume triples. The gas is at 1 . 00 m 3 /kg before the isentropic compression. Analyze. Take state 1 to be the state before the compression. Then T 1 = 400 K, v 1 = 1 . 00 m 3 kg . (2.19) By the ideal gas law P 1 = RT 1 v 1 = ³ 0 . 287 kJ kg K ´ (400 K ) 1 . 00 m 3 kg = 1 . 148 × 10 2 kPa. (2.20) Now isentropically compress to state 2. By the standard relations for a CPIG, one Fnds T 2 T 1 = p v 1 v 2 P k 1 = p P 2 P
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Thermodynamics filled in class notes_Part_6 - 19 2.1 CARNOT...

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