The simple air Brayton cycle shown below produces power (Wet) as well as 10 tons of...

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The simple air Brayton cycle shown below produces power (Wet) as well as 10 tons of refrigeration (ref = 10 tons) with an absorption chiller (COP = 0.7) driven by the waste heat from the turbine. In order to provide the required heat transfer to the chiller the turbine exhaust temperature is T4 = 160°C, and the turbine exhaust air leaving the chiller is Ts = 120°C. Air enters the compressor at T₁ = 27°C and P₁ = 100 kPa, and exhausts the turbine at P4 = 100 kPa. The compressor compresses the air to a pressure of P₂ = 1.5 MPa. The compressor (n.) and turbine (nt) isentropic efficiencies are nc = t = 90%. Treating the Brayton cycle as cold-air standard with specific heats evaluated at 300 K, what is the CHP system total efficiency total, defined as 1 total (Net Power Out) + (Cooling Rate)_ Wnet + Oref Heat Input Rate 2in 0₂ P₂ = 1.5 MPa 2 Compressor Compressor Isentropic Efficiency: nc = 90% (1) T₁ = 27°C P₁ = 100 kPa Combustor 3 Turbine T4 = 160°C, P₁ = 100 kPa 4 Ts=120°C (5) W net Turbine Isentropic Efficiency: nt = 90% Absorption Chiller COP = 0.7 Q = 10 tons Refrigeration Load The simple air Brayton cycle shown below produces power (Wet) as well as 10 tons of refrigeration (ref = 10 tons) with an absorption chiller (COP = 0.7) driven by the waste heat from the turbine. In order to provide the required heat transfer to the chiller the turbine exhaust temperature is T4 = 160°C, and the turbine exhaust air leaving the chiller is Ts = 120°C. Air enters the compressor at T₁ = 27°C and P₁ = 100 kPa, and exhausts the turbine at P4 = 100 kPa. The compressor compresses the air to a pressure of P₂ = 1.5 MPa. The compressor (n.) and turbine (nt) isentropic efficiencies are nc = t = 90%. Treating the Brayton cycle as cold-air standard with specific heats evaluated at 300 K, what is the CHP system total efficiency total, defined as 1 total (Net Power Out) + (Cooling Rate)_ Wnet + Oref Heat Input Rate 2in 0₂ P₂ = 1.5 MPa 2 Compressor Compressor Isentropic Efficiency: nc = 90% (1) T₁ = 27°C P₁ = 100 kPa Combustor 3 Turbine T4 = 160°C, P₁ = 100 kPa 4 Ts=120°C (5) W net Turbine Isentropic Efficiency: nt = 90% Absorption Chiller COP = 0.7 Q = 10 tons Refrigeration Load

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The image you have provided shows a schematic of a modified simple air Brayton cycle combined with an absorption chiller This is a thermodynamic cycle ... View the full answer