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01 07 17 sol

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

Sistemi Energetici LFull exam

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Dati: Valori da tabelle vapore: Assunzioni: Temperatura ingresso olio: TA = 320 °C Temperatura uscita olio: TC = 220 °C Portata olio: mo = 50 kg/s Calore specifico olio: cp,o = 2.9 kJ/kgK peva = 20 bar Tcond = 50 °C Dp in eco = 5 bar DTsc = Teva - T3 = 20 °C a) Calore da raffreddamento olio: Qo=mo*cp,o*(TA-TC) = 14500 kW Pressione di condensazione: pcond = p1 = 0.1235 bar Dp pompa: Dpp = peva + Dpeco - pcond = 24.88 bar Entalpia all'uscita del condensatore: h1 = 209.33 kJ/kg Rendimento idraulico pompa: eta_idr = 70 % Dh in pompa: Dhp = v*Dp/eta_idr. = 3.55 kJ/kg Entalpia ingresso economizzatore: h2 = 212.89 kJ/kg b) Temperatura di evaporazione: Teva = T4 = 212.4 °C Entalpia vapore saturo: h4 = 2798.4 kJ/kg Dh totale in caldaia: Dhtot = h4 - h2 = 2585.5 kJ/kg Portata vapore: mvap = Qo/Dhtot = 5.61 kg/s c) Temperatura uscita economizzatore: T3 = T4-DTsc = 192.4 °C Entalpia liquido saturo nell'evaporatore: hliq,sat = 908.6 kJ/kg Calcore specifico acqua sub-cooling: cp,wsc = 4.5 kJ/kgK Entalpia uscita eco: h3 = hliq,sat - cp,wsc*DTsc = 818.6 kJ/kg Dh in economizzatore: Dheco = h3-h2 = 605.7 kJ/kg Calore in economizzatore: Qeco=m2*(h3-h2) = 3397.1 kW Calore in evaporatore: Qeva = Qoil - Qeco = 11102.9 kW Temp olio uscita eva: TB = TA‐Qeva/(mo*cp,o) = 243.4 °C d) 0 320 212.4 11102.9 243.4 212.4 11102.9 243.4 192.4 14500 220 50 e) DT cold end eva = 31.0 °C DT hot end eva = 107.6 °C LMTDeva = 61.6 K h acqua = 10000 W/m2K h olio = 400 W/m2K Ueva = (1/hsteam + 1/hoil)^(-1) = 384.6 W/m2K Eva area: Aeva = Qeva/(Ueva*LMTDeva) = 468.7 m2 DT hot end eco = 51.0 °C DT cold end eco = 170 °C LMTDeco = 98.87 K h acqua eco = 4000 W/m2K Ueco = 363.6 W/m2K Aeco = 94.48 m2 f) Entalpia ingresso turbina: h5 = h4 = 2798.4 kJ/kg Entropia ingresso in turbina (vap sat): s5 = 6.339 kJ/kgK Pressione…

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