To calculate the ideal/maximum amount of work that can be obtained from a open, steady-state fluid stream, one first performs a first law analysis:

$\Delta \dot H = \dot Q + \dot W_s$

The maximum work will be obtained from a reversible process, where we know the following will be true:

$\Delta \dot S = \frac{\dot Q_{rev}}{T_{surr}}$

or

$Q_{rev} = T_{surr}\Delta \dot S$

so

$\dot W_{s_{ideal}} = \Delta \dot H - T_{surr}\Delta \dot S$

The **isentropic efficiency** is the ratio of the
actual power obtained relative to the power that would be obtained
in a reversible process.

$\eta_{isetropic} = \frac{\dot W_{actual}}{\dot W_{rev}} = \frac{\dot W_{actual}}{\dot W_{ideal}}$

Correct reversible calculations for real systems using isentropic efficiencies

The isentropic efficiency of a power cycle (like a Rankine cycle) is the ratio of the actual power obtained in the cycle, relative to the power that would be obtained in a reversible power cycle.

Calculate the maximum amount of work (i.e., the "ideal work") that can be obtained from a steady-state flow of $N_2$ at 800K and 50bar whose surroundings are 300K and 1bar. If a real process operates between these limits with an isentropic efficiency of 55%, what is the real work obtained?