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26 09 16 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: Temp. aria ingresso compr. = 25 °C Rapporto compressione totale = 10 Temp. aria uscita intercooler = 40 °C Portata aria = 10 kg/s TIT = 950 °C eta,is_c = 0.80 eta,is_t = 0.85 eta,mecc = 0.96 eta,el = 0.97 PCI GN = 47 MJ/kg cp,aria = 1 kJ/kg/K MM,aria = 28.8 kg/kmole cp,gas combusti = 1.1 kJ/kg/K MMgas combusti = 28.4 kg/kmole Temp. acqua in intercooler = 25 °C Temp. acqua out intercooler = 50 °C h aria intercooler = 100 W/m2K h acqua intercooler = 5000 W/m2K Se/Si intercooler = 15 efficienza aletta = 0.8 Temp. acqua refrigerata in interc.= 5 °C b) cv,aria = cp - R/MM = 0.711 kJ/kg/K gamma,aria = 1.406 beta sezione compressore = 3.162 T1 = 298.15 K T2,is = T1 * beta^((gamma-1)/gamma) = 415.70 K = 142.55 °C T2 = T1 + (T2is - T1) / eta,is_c = 445.08 K = 171.93 °C Wpale_c1 = cp * (T2 - T1) = 146.9 kJ/kg T3 = 313.15 K T4,is = T3 * beta^((gamma-1)/gamma) = 436.61 K = 163.46 °C T4 = T3 + (T4,is - T3) / eta,is_c = 467.47 K = 194.32 °C Wpale_c2 = cp * (T4 - T3) = 154.3 kJ/kg Potenza pale compressore = 3012.6 kW c) Q intercooler = 1319.3 kW U = 1/(1/h_acqua + 1/(h_aria*Se/Si*eta_al)) = 967.7 W/m2K DTce intercooler = 15 °C DThe intercooler = 121.9 °C LMTD = 51.03 K A intercooler = 26.7 m2 d) efficacia intercooler = (T2-T3)/(T2-Tacquain) = 0.898 Funzionamento con acqua refrigerata: Cmin/Cmax=cost, UA assunto costante -> efficacia costante DTaria,max con acqua refrigerata = 166.93 °C DTaria con acqua refrigerata = 149.89 °C Taria uscita inter-cooler = 22.04 °C e) Bilancio energia combustore: m_aria*cp_aria*(T2-Tref) + m_GN*PCI = (m_aria+m_GN)*cp_gas*(TIT-Tref) m_GN = 0.19 kg/s f) cv,gas = cp - R/MM = 0.807 kJ/kg/K gamma,gas = 1.363 TIT = 1223.15 K TOT,is = TIT * beta^((1-gamma)/gamma) = 662.75 K = 389.60 °C TOT = TIT - (TIT - TOT,is) * eta,is_t = 746.81 K = 473.66 °C Wpale_t…

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