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eski thermo (2)

# eski thermo (2) - 10-1 CHAPTER 10 The correspondence...

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Unformatted text preview: 10-1 CHAPTER 10 The correspondence between the new problem set and the previous 4th edition chapter 8 problem set. New Old New Old New Old 1 new 21 new 41 35 2 new 22 new 42 41 3 1 23 18 43 42 4 2 24 new 44 46 5 3 25 19 45 54 6 5 26 20 46 56 7 6 27 21 47 39 8 7 28 new 48 55 9 9 29 new 49 53 10 15 30 23 50 new 11 16 31 24 51 43 12 10 32 new 52 51 13 11 33 new 53 17 14 49 34 25 54 29 15 47 35 26 55 30 16 13 36 27 56 31 17 14 37 32 57 36 18 38 38 28 58 40 19 52 39 33 59 44 20 8 40 34 60 57 The problems that are labeled advanced starts at number 53. The English unit problems are: New Old New Old New Old 61 new 71 67 81 75 62 58 72 new 82 83 63 60 73 68 83 77 64 63 74 new 84 64 65 61 75 69 85 71 66 80 76 70 86 73 67 62 77 72 87 78 68 new 78 new 69 65 79 79 70 66 80 84 10-2 10.1 Calculate the reversible work and irreversibility for the process described in Problem 5.18, assuming that the heat transfer is with the surroundings at 20 ° C. C.V.: A + B. This is a control mass. Continuity equation: m 2- (m A1 + m B1 ) = 0 ; Energy: m 2 u 2- m A1 u A1- m B1 u B1 = 1 Q 2- 1 W 2 System: if V B ≥ 0 piston floats ⇒ P B = P B1 = const. if V B = 0 then P 2 < P B1 and v = V A /m tot see P-V diagram State A1: Table B.1.1, x = 1 v A1 = 1.694 m 3 /kg, u A1 = 2506.1 kJ/kg m A1 = V A /v A1 = 0.5903 kg State B1: Table B.1.2 sup. vapor v B1 = 1.0315 m 3 /kg, u B1 = 2965.5 kJ/kg V P 2 a P B1 2 m B1 = V B1 /v B1 = 0.9695 kg => m 2 = m TOT = 1.56 kg At (T 2 , P B1 ) v 2 = 0.7163 > v a = V A /m tot = 0.641 so V B2 > 0 so now state 2: P 2 = P B1 = 300 kPa, T 2 = 200 ° C => u 2 = 2650.7 kJ/kg and V 2 = m 2 v 2 = 1.56 × 0.7163 = 1.117 m 3 (we could also have checked T a at: 300 kPa, 0.641 m 3 /kg => T = 155 ° C) 1 W ac 2 = ⌡ ⌠ P B dV B = P B1 (V 2- V 1 ) B = P B1 (V 2- V 1 ) tot = -264.82 kJ 1 Q 2 = m 2 u 2- m A1 u A1- m B1 u B1 + 1 W 2 = -484.7 kJ From the results above we have : s A1 = 7.3593, s B1 = 8.0329, s 2 = 7.3115 kJ/kg K 1 W rev 2 = T o (S 2- S 1 ) - (U 2- U 1 ) + 1 Q 2 (1 - T o /T H ) = T o (m 2 s 2- m A1 s A1- m B1 s B1 ) + 1 W ac 2- 1 Q 2 T o /T H = 293.15 (1.5598 × 7.3115 - 0.5903 × 7.3593 - 0.9695 × 8.0329) + (-264.82) - (-484.7) × 293.15 / 293.15 = -213.3 - 264.82 + 484.7 = 6.6 kJ 1 I 2 = 1 W rev 2- 1 W ac 2 = 6.6 - (-264.82) = 271.4 kJ 10-3 10.2 Calculate the reversible work and irreversibility for the process described in Problem 5.65, assuming that the heat transfer is with the surroundings at 20 ° C. P v 2 1 Linear spring gives 1 W 2 = ⌡ ⌠ PdV = 1 2 (P 1 + P 2 )(V 2- V 1 ) 1 Q 2 = m(u 2- u 1 ) + 1 W 2 Equation of state: PV = mRT State 1: V 1 = mRT 1 /P 1 = 2 x 0.18892 x 673.15 /500 = 0.5087 m 3 State 2: V 2 = mRT 2 /P 2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m 3 1 W 2 = 1 2 (500 + 300)(0.3944 - 0.5087) = -45.72 kJ From Figure 5.10: C p (T avg ) = 45/44 = 1.023 ⇒ C v = 0.83 = C p- R For comparison the value from Table A.5 at 300 K is C v = 0.653 kJ/kg K 1 Q 2 = mC v (T 2- T 1 ) + 1 W 2 = 2 x 0.83(40 - 400) - 45.72 = -643.3 kJ 1 W rev 2 = T...
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