11 02 19sol

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Exam Energy Systems proff. Consonni, Martelli, Romano - 11 February 2019 Problem 1 natural gas composition mol fraction LHV [kJ/kg] molar mass LHV [kJ/kmol] CH4 0,90 50,1 16 801,6 C2H6 0,07 47,8 30 1434 N2 0,03 0 28 0 total 1,00 47394 17,34 821,82 fuel temperature 15,00 °C air temperature 15,00 °C cp_fuel 2,20 kJ/kgK cp_air 1,05 kJ/kgK air composition mol fraction LHV molar mass N2 0,79 0 28 O2 0,21 0 32 total 1 0 28,84 mass flow rate of natural gas 0,02 kg/s heat losses from wall 0,005 incomplete combustion losses 0 stack temperature 140 °C cp_flue gases 1,12 kJ/kgK xO2 in dry flue gases 0,04 Stoichiometric reaction betaO2 stoichiom CH4 (mol O2/mol CH4) 2 using direct eq: nC+nH/4-nO/2+nS betaO2 stoichiom CH2H6 (mol O2/mol C2H6) 3,5 using direct eq: nC+nH/4-nO/2+nS betaO2 stoichiom mol O2/mol fuel 2,045 beta_O2_CH4*xCH4+beta_C2H6*xC2H6 air/fuel stoichiom mass ratio (kgair/kgfuel) 16,196 beta_O2_stoich*(MM_air/MM_fuel)/xair_O2 Combustion reaction with excess of air FUEL + betaO2 (O2 + 3.76 N2) --> A CO2 + B O2 + C N2+ D H2O write linear system of equations with atomic balance eq. and condition on 4% O2; unknowns are A, B, C, D, beta_O2 A coeff stoichiom CO2 (mol CO2/mol fuel) = xCH4,ng + 2*xC2H6,ng1,040 D coeff stoichiom H2O = 2*xCH4,ng + 3*xC2H6,ng 2,010 to calculate betaO2, the following equations can be used: B = betaO2 - A - D/2 B/(A+B+C) = xO2,gas = 0.04 C = xN2,ng + betaO2*3.76 --> betaO2 (moli O2/mole fuel) = [A+D/2+0.04*(xN2,ng-D/2)] / (1-0.04*4.76) betaO2 (moli O2/mole fuel) = 2,478 B coeff stoichiom O2 (mol O2/mol fuel) = betaO2 - A - D/2 0,433 C coeff stoichiom N2 (mol N2/mol fuel) = xN2,ng + betaO2*3.769,351 air / fuel molar ratio 11,799 beta_O2/xair_O2 air/fuel mass ratio = alfa = 19,624 air/fuel_molar*(MM_air/MM_fuel) excess of air 0,212…