Thermodynamics filled in class notes_Part_3

Thermodynamics filled in class notes_Part_3 - (1.40 1 q 2 =...

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13 Since the gas is ideal: du = c v dT. (1.16) Since our ideal gas is also calorically perfect, c v is a constant, and we get i u 1 u 2 du = c v i T 1 T 2 dT, (1.17) u 2 u 1 = c v ( T 2 T 1 ) , (1.18) = p 0 . 7175 kJ kg K P (400 K 300 K ) , (1.19) = 71 . 750 kJ kg . (1.20) Also we have Tds = du + Pdv, (1.21) Tds = c v dT + Pdv, (1.22) from ideal gas : v = RT P : dv = R P dT RT P 2 dP, (1.23) Tds = c v dT + RdT RT P dP, (1.24) ds = ( c v + R ) dT T R dP P , (1.25) = ( c v + c P c v ) dT T R dP P , (1.26) = c P dT T R dP P , (1.27) i s 2 s 1 ds = c P i T 2 T 1 dT T R i P 2 P 1 dP P , (1.28) s 2 s 1 = c P ln p T 2 T 1 P R ln p P 2 P 1 P . (1.29) (1.30) Now since P is a constant, s 2 s 1 = c P ln p T 2 T 1 P , (1.31) = p 1 . 0045 kJ kg K P ln p 400 K 300 K P , (1.32) = 0 . 2890 kJ kg K . (1.33) Then one Fnds 1 w 2 = i v 2 v 1 Pdv = P i v 2 v 1 dv, (1.34) = P ( v 2 v 1 ) , (1.35) = (100 kPa ) p 1 . 148 m 3 kg 0 . 861 m 3 kg P , (1.36) = 28 . 700 kJ kg . (1.37) CC BY-NC-ND. 18 November 2011, J. M. Powers.

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14 CHAPTER 1. REVIEW Now du = δq δw, (1.38) δq = du + δw, (1.39) i 2 1 δq = i 2 1 du + i 2 1 δw,
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Unformatted text preview: (1.40) 1 q 2 = ( u 2 − u 1 ) + 1 w 2 , (1.41) 1 q 2 = 71 . 750 kJ kg + 28 . 700 kJ kg , (1.42) 1 q 2 = 100 . 450 kJ kg . (1.43) Now in this process the gas is heated from 300 K to 400 K . One would expect at a minimum that the surroundings were at 400 K . Check for second law satisfaction. s 2 − s 1 ≥ 1 q 2 T surr ? (1.44) . 2890 kJ kg K ≥ 100 . 450 kJ/kg 400 K ? (1.45) . 2890 kJ kg K ≥ . 2511 kJ kg K , yes . (1.46) 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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Thermodynamics filled in class notes_Part_3 - (1.40 1 q 2 =...

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