Unformatted text preview: h out,s = 581 kJ/kg, leading to η = ˙ W ← isentropic ˙ W ← = h out,sh in ˙ W ← / ˙ m = 581298 . 44 301 . 68 = 0 . 937 (c) Noting that the compressor boundary temperature at which heat is transferred to the surrounding is unknown, we take both compressor plus surrounding as a control mass and apply the second law of thermodynamics, dS c dt + dS sur dt = ˙ S irr note that compressor is undergoing steady operation, then dS c /dt = 0. The surrounding is a thermal reservoir for which dS sur dT = ˙ S ← heat + ˙ m ( s ins out ) sur ⇒ ˙ S irr = dS sur dT = ˙ Q ← T˙ m ( s ins out ) c ⇒ ˙ S irr ˙ m = ˙ Q ← ˙ mT( s ins out ) c = 0 . 156 kJ/kg K It is assumed that heat is transferred to the surrounding at the inlet temperature. In reality this temperature is higher. 5...
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 Fall '08
 I.L
 Thermodynamics, 1.7 kJ, compressor heat losses, compressor boundary temperature, Sirr dt dt, 0.156 kJ

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