10_55 - h3 - h2 = 0.75 (h3s-h2) therefore: q_12 = mdot...

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Problem 10.55 0.75 eta_s Cold air standard: constant c_p for simplicity. 0.29 = R (kJ/kg,K) 1 = cp (kJ/kg/K) 1 State 2 State 3s State 3 State Dead State (reference values for s and h) 200 p1 (kPa) 200 p2 95 p3s 95 p3 18 p0 (kPa) 380 T1 (K) 320 T2 258.66 T3s 274 T3 233 T0 (K) 147.59 h1 87.35 h2 25.76 h3s 41.16 h3 0 h0 (kJ/kg) -0.2 s1 -0.37 s2 -0.37 s3s -0.31 s3 0 s0 (kJ/kg,K) 194.19 ef1 174.15 ef2 112.56 ef3s 114.49 ef3 s3s - s0 = -0.11043 = cp ln(T3s/T0) - R ln(p3s/p0)
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Unformatted text preview: h3 - h2 = 0.75 (h3s-h2) therefore: q_12 = mdot (h1-h2) 60.24 w_23 = mdot (h2-h3) 46.19 I_12 = mdot (ef1-ef2) 20.04 I_23 = mdot (ef2-af3) - w_23 13.47 The second law efficiencies can be defined as usual: For the heat exchanger: e_X = ef2 / ef1 = 1 - I_12/ef1 0.9 For the turbine: e_t = w_23/(ef2-ef3) 0.77 Overall: e = e_X e_t 0.69...
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