Document information
- University
- Politecnico di Milano
- Degree programme
- Mechanical Engineering
- Subject
- Sistemi Energetici L
- Academic year
- 2014-2015
- Classification
- Exam · Full exam
- Content
- Exam paper only
- Original format
- Text
- Searchable text
Study material for Sistemi Energetici L, shared by the Studwiz community and reviewed by moderators.
Study material for Sistemi Energetici L, shared by the Studwiz community and reviewed by moderators.
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Dati Assunzioni Fumi turbina a gas: Potenza = 15 MW rendimento = 35 % Gfumi = 50 kg/s cp,f = 1.03 kJ/kg-K TOT = 450 °C Composizione volumetrica: x O2 = 12.5 % x CO2 = 4 % x H2O = 8 % x N2 = 75.5 % MMO2 = 32 kg/kmole MMCO2 = 44 kg/kmole MMH2O = 18 kg/kmole MMN2 = 28 kg/kmole ENOx = 50 mg/Nm3 dry gas @ 15%O2 p steam = 4 bar Tsat = 143.6 °C (from steam tab.) hsat steam = 2738.1 kJ/kg (from steam tab.) hsat liquid = 604.66 kJ/kg (from steam tab.) DTpp = 10 °C Heat losses: ξ =1 % Post-combustione: T comb = 550 °C LHV CH4 = 50 MJ/kg TCH4 = 25 °C MMCH4 = 16 kg/kmole Riferimento cogenerazione: eta,el-ref = 55 % eta,th-ref = 90 % ESERCIZIO 1: a) MM gas combusti: MMg = Σxi*MMi = 28.34 kg/kmole Portata molare gas: Mg = G/MMg = 1.764 kmole/s Portata volumetrica gas: Vg = 39.545 Nm3/s xO2 fumi secchi: xO2,fs = xO2/(1-xH2O) = 12.5 / (1 - 0.08) = 13.59 % ENOx @ xO2,fs = 50 * (21 - 13.59) / (21 - 15) = 61.78 mg/Nm3 fumi secchi @ concentrazione O2 reale ENOx in fumi reali umidi = 61.78 * (1 - 0.08) = 56.83 mg/Nm3 in gas reali G NOx = ENOx reali * Vg = 56.83* 39.545 / 1000 = 2.25 g/s b) Risolvendo il seguente bilancio di energia: Gg*cp,g*(TOT-Tref) + Gf*LHVf = (Gg+Gf)*cp,g*(Tcomb-Tref): Portata massica CH4: Gf = 0.1041 kg/s Portata molare CH4: Mf = Gf / MMCH4 = 0.00651 kmole/s Heat input as LHV: Qf = Gf * LHVCH4 = 5.21 MW c) Portata molare componente i nei gas combusti: Mi,fg MO2,fg = xO2_GTg*M_GTg - 2*MCH4 = 0.20752 kmol/s MCO2,fg = xCO2_GTg*M_GTg + MCH4 = 0.07708 kmol/s MH2O,fg = xH2O_GTg*M_GTg + 2*MH2O = 0.15416 kmol/s MN2,fg = xN2_GTg*M_GTg = 1.33204 kmol/s Portata molare totale: Mtot,fg = ΣMi,fg = 1.771 kmol/s Composizione gas combusti: xi = Mi,fg / Mtot,fg x O2 = 11.72 % x CO2 = 4.35 % x H2O = 8.71 % x N2 = 75.22 % ESERCIZIO 2: a) Q da gas = mg*cp,g*(TOT-Teva-DTpp)*(1-ξ) = 15112…
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