Brayton Cycle

Cycle data

Known properties
  • Air is used as working fluid;
  • Pressure at the compressor inlet 100 kPa;
  • Temperature at the compressor inlet 20 °C;
  • Pressure ratio equal to 7;
  • Temperature at the turbine inlet 500 °C;
  • The turbine and compressor have an isentropic efficiency of 90%.
Input code
using CycleSolver

@solve begin
        st2.p / st1.p = 7
        st1.T = 20 + 273
        st1.p = 100
        st3.T = 500 + 273  
        newCycle[Air] 
            compressor(st1, st2, 90)
            combustion_chamber(st2, st3)
            turbine(st3, st4, 90)
            combustion_chamber(st4, st1)
end

PrintResults()

Output:

1- CYCLE [Air]

State
Name		 T [K] P [kPa] h [kJ/kg] s [kJ/kg.K] x ṁ [kg/s] Mass-flux
fraction
st1 293.0 100.0 419.257 3.8667 1.0
st2 532.122 700.0 662.359 3.9135 1.0
st3 773.0 700.0 918.995 4.3107 1.0
st4 487.236 100.0 616.266 4.3821 1.0

Cycle Properties:

Total Component Value
qin
Q̇in		 256.6365 kJ/kg
0.0 kW		 combustion_chamber:
st2 >> st3		 256.6365 kJ/kg
qout
Q̇out		 197.0086 kJ/kg
0.0 kW		 combustion_chamber:
st4 >> st1		 197.0086 kJ/kg
win
Ẇin		 243.1014 kJ/kg
0.0 kW		 compressor:
st1 >> st2		 243.1014 kJ/kg
wout
Ẇout		 302.7293 kJ/kg
0.0 kW		 turbine:
st3 >> st4		 302.7293 kJ/kg

Thermal efficiency (n) = 23.2344 %