Form Joule Brayton

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Brayton cycle Δp% 0-1 % Pressure losses due to air filter, combustor and exhaust system Δp% 2-3 % Δp% 4-5 % T0 K p0 bar T1 K p1 bar T2 K p2 bar T3 K p3 bar T4 K p4 bar T5 K p5 bar θ air - cp air kJ/kgK R 8'314 J/kmolK θ flue gas - R* is the gas constant divided by cp flue gas kJ/kgK the molar mass of the molecule We assume that T0 and p0, i.e. ambient condition, are given T is always expressed in Kelvin degrees p is expressed in bar T3, temperature at the combustor outlet is equal to the temperature at the turbine inlet and is assumed to be known βc is a known value. βt must be computed as p3/p4 Compressor work: Turbine work: Electric power: Efficiency: mfuel kg/s Lc kW mair kg/s Lt kW βc - Pel kWe βt - η % NB formulas are coherent and valid provided that Δp losses are expressed as a percentage of inlet values T [K] s [kJ/kgK] p0 p1 p2 p3 p4 0 1 2 3 4 5 cp = R∗ γ γ − 1 cp − cv = R∗ γ = cp cv θ = γ − 1 γ T1 = T0 p1 = 1 − Δp01 ∙ p0 T2 = T1 ∙ 1 + βc θair − 1 ηc p2 = βc ∙ p1 T3 = TIT = COT p3 = 1 − Δp23 ∙ p2 T4 = T3 ∙ 1 + βt −θFG − 1 ηt p4 = p5 1 − Δp45 T5 = T4 p5 = p0 Lc = mair ∙ cp,air ∙ T2 − T1 Lt = mFG ∙ cp,FG ∙ T3 − T4 Pel = Lt − Lc ∙ ηel ∙ ηmec η = Pel mfLHV Energy balance of the combustor (from which missing data can be computed): α kgAir/kgFuel cp air kJ/kgK Tin air K cp fuel kJ/kgK Tin fuel K cp flue gas kJ/kgK COT K mfuel kg/s ξ*100 % mair kg/s LHV kJ/kg COT = combustor outlet temperature Tin air = T end of compression stage (T2) Solving for α: Reduced mass flow rate: Free space for additional computations: T3, temperature at the combustor outlet is equal to the temperature at the COMBUSTOR Air, Tin air Fuel, Tin fuel Flue gasses, COT Thermal losses, ξ α = mair mfuel mair + mfuel = mFG TR = 298.15 K α ∙ cp,air ∙ Tin,air − TR + cp,fuel ∙ Tin,fuel − TR + 1 − ξ ∙…