Find the flows and fluxes of exergy in the condenser of Problem 11.32. Use those

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Explanation / Answer

For this case we select To = 12°C = 285 K, the ocean water temperature. The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, 3: 3.0 MPa, 600oC: s3 = 7.5084 kJ/kg K 5641cb C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 7.5084 = 0.6386 + x4 (7.5261) => x4 = 0.9128 => h4 = 188.42 + 0.9128 (2394.77) = 2374.4 kJ/kg C.V. Condenser : qL = h4 - h1 = 2374.4 - 188.42 = 2186 kJ/kg Q.L = m.qL = 25 × 2186 = 54.65 MW = m.ocean Cp ?T m.ocean = Q.L / Cp ?T = 54 650 / (4.18 × 3) = 4358 kg/s The net drop in exergy of the water is F.water = m.water [h4 – h1 – To(s4 – s1)] = 25 [ 2374.4 – 188.4 – 285 (7.5084 – 0.6386)] = 54 650 – 48 947 = 5703 kW The net gain in exergy of the ocean water is F.ocean = m.ocean[h6 – h5 – To(s6 – s5)] = m.ocean[Cp(T6 – T5) – ToCp ln(T6T5) ] = 4358 [ 4.18(15 – 12) – 285 × 4.18 ln 273 + 15273 + 12 ] = 54 650 – 54 364 = 286 kW The second law efficiency is ?II = F.ocean / F.water = 2865703 = 0.05

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