Thermodynamics filled in class notes_Part_6

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

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Unformatted text preview: 19 2.1. CARNOT So then T ds = cv P dv + vdP R +P dv, (2.10) =dT cv cv + 1 P dv + vdP, T ds = R R cv cv ∂P 0= +1 P + v , R R ∂v s cv +1 P ∂P = − R cv , ∂v s v R cv +1 P c P −c v , =− cv v c P −c v cv + cP − cv P =− , cv v cP P , =− cv v P = −k . v Take ds = 0, (2.11) (2.12) (2.13) (2.14) (2.15) (2.16) (2.17) Since k > 1, the magnitude of the slope of the isentrope is greater than the magnitude of the slope of the isotherm: ∂P ∂P > . (2.18) ∂v s ∂v T 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 m3 /kg before the isentropic compression. Analyze. Take state 1 to be the state before the compression. Then T1 = 400 K, v1 = 1.00 m3 . kg (2.19) By the ideal gas law P1 = kJ 0.287 kg K (400 K ) RT1 = = 1.148 × 102 kP a. 3 v1 1.00 m kg (2.20) Now isentropically compress to state 2. By the standard relations for a CPIG, one finds T2 = T1 v1 v2 k −1 = P2 P1 k −1 k . (2.21) CC BY-NC-ND. 18 November 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 law 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 v3 v4 1 k −1 = m3 0.19245 kg k 3 = (1.78955 × 10 kP a) 1200 K 400 K m3 kg m3 kg 0.19245 3.000 1 1.4−1 = 3.000 m3 . kg (2.29) 1.4 = 3.82667 × 101 kP a. (2.30) Check: P4 = kJ 0.287 kg K (400 K ) RT4 = 3.82667 × 101 kP a. = 3 v4 3.000 m kg (2.31) A summary of the states is given in Table 2.1. Now calculate the work, heat transfer and efficiency. Take the adiabatic exponent for air to be k = 1.4. Now since cP , cP − cv = R, (2.32) k= cv 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.

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