15 09 14 sol

Pagina 1

Esercizio 1: Gfumi = 50 kg/s Tfumi = 530 °C cp fumi = 1.1 kJ/kg/K h liq sat = 1154.5 kJ/kg h vap sat = 2794.2 kJ/kg Teva = 263.9 °C h vap SH = 3440.1 kJ/kg h vap RH = 3468.1 kJ/kg Dhis turbina HP = 228 kJ/kg Dhis turbina LP = 1117.8 kJ/kg h ritorno condensa = 1154.5 kJ/kg h condensato = 188.5 kJ/kg G vapore cogenerazione = 2 kg/s pcond = 0.096 bar pinECO = 65 bar DTpp = 10 °C perdite th = 2 % eta is TV = 85 % eta mec-el TV = 98 % eta idr pompa = 75 % a) Tfumi pinch = 273.9 °C Q fumi pinch = 13803.8 kW Dh turbina HP = 193.8 kJ/kg hRHcold = 3246.3 kJ/kg Dh vap HP da rit. cond = 2285.6 kJ/kg Dh vapore HP = 2285.6 kJ/kg Dh vapore IP = 221.8 kJ/kg Q produz vap cogen = 5014.8 kW Q disponobile = 8789.0 kW Gvap HP non cog = 3.51 kg/s Gvap HP tot = 5.51 kg/s Gvap LP = 3.51 kg/s Pel TV = 4309.4 kW b) hinECO = 197.2 kJ/kg DhECO = 957.3 kJ/kg QECO = 3355.7 kW Tcamino = 211.6 °C Qrecuperato = 17159.5 kW Tambiente = 15 °C Q disponibile = 28325.0 kW Eta caldaia = 60.6 % vapore a utenza cogenerativa ritorno condensa Esercizio 2: eccesso aria = 10 % xO2 aria = 0.21 xN2 aria = 0.79 MMO2, kg/kmole 32 MMN2, kg/kmole 28 MMCO, kg/kmole 28 a) CO + 1/2O2 -> CO2 λstechCO = 0.5 moliO2/moleCO N2/O2 aria = 3.76 moliN2/moliO2 λstech = 2.38 moli aria/moleCO λ = 2.62 moli aria/moleCO MMaria = 28.84 kg/kmole α = 2.70 kgaria/kgCO Gfumi = 3.70 kgfumi/kgCO zCO2 = 1 kmoliCO2/kmoleCO zO2 = 0.05 kmoliO2/kmoleCO zN2 = 2.07 kmoliN2/kmoleCO ztot = 3.12 kmoli_fumi/kmoleCO xCO2 = 0.321 kmoliCO2/kmolefumi xO2 = 0.016 kmoliO2/kmolefumi xN2 = 0.663 kmoliN2/kmolefumi b) %CO incombusto = 0.01 % CO incombusto = 1.00E-04 kmoliCO/kmoleCO_comb ECO = 3.21E-05 kmoliCO/kmole_fumi ECO = 32.1 ppm ECO = 40.05 mg/Nm3 ECO = 29.75 ppm @ 3%O2 ECO = 37.17 mg/Nm3 @ 3%O2