Thermal Efficiency of Brayton Cycle
Let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engine. Modern gas turbine engines and airbreathing jet engines also follow the Brayton cycle. This cycle consist of four thermodynamic processes:
isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by WC = H2 – H1.
- isobaric heat addition – the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Qadd = H3 – H2
- isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by WT = H4 – H3
- isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Qre = H4 – H1
As can be seen, we can describe and calculate (e.g. thermodynamic efficiency) such cycles (similarly for Rankine cycle) using enthalpies.
To calculate the thermal efficiency of the Brayton cycle (single compressor and single turbine) engineers use the first law of thermodynamics in terms of enthalpy rather than in terms of internal energy.
The first law in terms of enthalpy is:
dH = dQ + Vdp
In this equation the term Vdp is a flow process work. This work, Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer.
dH = CpdT + V(1-αT)dp
Where Cp is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:
dH = CpdT
At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:
Isobaric process (Vdp = 0):
dH = dQ → Q = H2 – H1 → H2 – H1 = Cp (T2 – T1)
At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:
Isentropic process (dQ = 0):
dH = Vdp → W = H2 – H1 → H2 – H1 = Cp (T2 – T1)
The enthalpy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The thermal efficiency of such simple Brayton cycle, for ideal gas and in terms of specific enthalpies can now be expressed in terms of the temperatures: