Question

A simple ideal Brayton cycle uses argon as the working fluid. At the beginning of the compression, P1 = 15 psia and T1 = 70°F, the maximum cycle temperature is 1220°F, and the pressure in the combustion chamber is 150 psia. The argon enters the compressor through a 3 ft2 opening with a velocity of 200 ft/s. Determine the rate at which entropy is generated by the cycle. The temperature of the source is the same as the maximum cycle temperature, and the temperature of the sink is the same as the minimum cycle temperature.

Answer 1

Entropy generation in an ideal Brayton cycle is zero since all processes are reversible and there are no irreversibilities within the system to generate entropy.

Step-by-step explanation:

The student's question relates to finding the rate of entropy generation for an ideal Brayton cycle using argon as the working fluid. However, the Brayton cycle, as an ideal cycle, does not lead to entropy generation within the system because all processes are reversible. Entropy may change as heat crosses the system boundaries, but this is not the same as entropy generation due to irreversibilities. Calculating entropy changes in an actual Brayton cycle requires knowledge of each process's irreversibilities, which cannot be determined without additional data on the components' efficiencies and the specific entropy values of argon at the given states. Further calculations for ideal processes can be simplified using the assumption of adiabatic and isentropic compression and expansion.