For such a process, the final state is the same as the initial state, so the total internal energy change must be zero. Steam (water) that circulates through a closed cooling loop undergoes a cycle. The first law of thermodynamics is then:
dEint = 0, dQ = dW
Thus, the process’s network must equal the net amount of energy transferred as heat. It must be noted, according to the second law of thermodynamics, not all heat provided to a cycle can be transformed into an equal amount of work. Some heat rejection must take place.
Example of Cyclic Process – 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:
The ideal Brayton cycle consists of four thermodynamic processes. Two isentropic processes and two isobaric processes.
- Isentropic compression – ambient air is drawn into the compressor, 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, burning fuel, 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 a turbine, gives up its energy. The work done by the turbine is given by WT = H4 – H3
- Isobaric heat rejection – the residual heat must be rejected to close the cycle. The net heat rejected is given by Qre = H4 – H1
See also: Thermal Efficiency of Brayton Cycle.