24 06 19sol

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Exam Energy Systems 24 June 2019 Problem 1 Data for part A Ambient temperature 20 ambient pressure, bar 1,007 pressure drop of air filter 0,70% Pressure ratio of compressor 13 compressor isentropic efficiency 0,84 pressure drop of combustor 4% heat loss from combustor 0,50%of LHV LHV of fuel, MJ/kg 47 combustor outlet temperature, °C1000 reduced turbine mass flow rate 1500 turbine isentropic efficiency 0,93 pressure drop of turbine exhaust system0,10% mechanical efficiency 0,98 electrical efficiency 0,985 cp/cv of air 1,39 molar mass of air, kg/kmol 28,8 Cp/cv of exaust gases 1,35 molar mass of flue gases, kg/kmol28,4 Data for part B Decrease of air mass flow rate, %30% Results PART A compressor inlet pressure, bar 1,00 pamb*(1 - pressure_drop_air_filer) compressor outlet pressure, bar 13,0 pin_comp * beta combustor outlet pressure, bar 12,48 put_comp*(1 - pressure_drop_comb) turbine inlet mass flow rate, kg/s524,6 from turbine map (m_in_turbo = p_out_comb*1500/(SQRT(Tin_turbo)) Cp or air, kJ/kg-K 1,029 from Cp/Cv or air and molar mass (Cp/Cv)/(Cp/Cv - 1) * (8314/molar mass) Cp of flue gases, kJ/kg-K 1,129 from Cp/Cv or flue gases and molar mass (Cp/Cv)/(Cp/Cv - 1) * (8314/molar mass) Tiso_compressor_outlet, °C 328,9 from isentropic compression of ideal gases T_compressor_oulet, °C 387,7660,7from definition of isentropic efficiency Air/fuel mass ratio of combustor 62,75 from energy balance of combustor: mair * Cpair * (T_comp_out-25)+ mfuel* LHV* (1-heat losses) = mflue_gases * Cpflue * (Tout_comb - 25) fuel mass flow rate, kg/s 8,230 since mflue = mair + mfuel (mass balance) and the definition of air/fuel mass ratio, you can derive that: mfuel = mflue / (alfa + 1) air mass flow rate, kg/s 516,4 turbine outlet pressure, bar 1,008 from ambient pressure and pressure drop…