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02 09 13 sol

Study material for Sistemi Energetici L, shared by the Studwiz community and reviewed by moderators.

Sistemi Energetici LFull exam

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Esercizio 1: Degasatore preriscaldatore 5 Vapore da turbina: p, bar 7 39.8 T saturazione, °C 165 250 Dhcond, kJ/kg 2066 1715 cp medio cond, kJ/kg-K 4.76 Acqua da preriscaldare G in, kg/s 150 Tin, °C 130 210 cp medio, kJ/kg-K 4.27 4.47 DT minimo vapore-acqua riscaldata = 3 °C DT minimo condensato-acqua in ingresso = 5 °C c) Degasatore: T acqua in uscita 165 °C DT acqua 35 °C Dh acqua 149.5 kJ/kg Q acqua 22418 kW G vapore 10.85 kg/s G acqua out 160.85 kg/s Preriscaldatore n°5: G acqua in 160.85 kg/s T acqua in uscita 247 °C DT acqua 37 °C Dh acqua 165.39 kJ/kg-K Q acqua 26603 kW T out condensato 215 °C DT condensato 35 °C Dh sottoraffr. condensa 166.6 kJ/kg Dh cond + sottoraffredd. 1881.6 kJ/kg G vapore 14.14 kg/s Esercizio 2: LHV CH4 802 MJ/kmole eccesso aria = 10 % xO2 aria = 0.21 xN2 aria = 0.79 MMCH4, kg/kmole 16 MMO2, kg/kmole 32 MMN2, kg/kmole 28 MMCO2, kg/kmole 44 MMH2O, kg/kmole 18 a) CH4 + 2O2 -> CO2 + 2H2O λstech = 2 moliO2/moleCH4 N2/O2 aria = 3.76 moliN2/moliO2 λstech = 9.52 moli aria/moleCH4 λ = 10.48 moli aria/moleCH4 MMaria = 28.84 kg/kmole α = 18.88 kgaria/kgCH4 Gfumi = 19.88 kgfumi/kgCH4 zCO2 = 1 kmoliCO2/kmoleCH4 165°C 130°C T Q 130°C T Q 250°C zH2O = 2 kmoliH2O/kmoleCH5 zO2 = 0.20 kmoliO2/kmoleCH4 zN2 = 8.28 kmoliN2/kmoleCH4 ztot = 11.48 kmoli_fumi/kmoleCH4 xCO2 = 0.087 kmoliCO2/kmolefumi xH2O = 0.174 kmoliH2O/kmolefumi xO2 = 0.017 kmoliO2/kmolefumi xN2 = 0.721 kmoliN2/kmolefumi MMfumi = 27.7 kg/kmole yCO2 = 0.138 kgCO2/kgfumi yH2O = 0.113 kgH2O/kgfumi yO2 = 0.020 kgO2/kgfumi yN2 = 0.728 kgN2/kgfumi b) cpCO2 = 39.2 kJ/kmole-K cpH2O = 34.3 kJ/kmole-K cpO2 = 30 kJ/kmole-K cpN2 = 29.5 kJ/kmole-K cp, fumi = 31.19 kJ/kmole-K cp, fumi = 1.13 kJ/kg-K Qloss, irr = 1 % Tcamino = 90 °C Tambiente = 25 °C Qcamino = 73 kJ/kgfumi Qcamino = 1454 kJ/kgCH4 LHV CH4 =…

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