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The throttle process in Example  is  an irreversible process. How does a nozzle or sprayhead generate kinetic energy? Find the reversible work and irreversibility assuming an ambient temperature at 25°C. 

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C.V. Throttle. Steady state, adiabatic q = 0 and no shaft work w = 0.

Inlet state:  hi = 346.8 kJ/kg; si = 1.2792 kJ/kg K

Energy Eq  he = hi

Exit state:  P = 291 kPa, he = hi

which is two-phase

se = sf + xsfg = 0.5408 + 0.1638 × 4.9265 = 1.3478 kJ/kg K

The reversible work is the difference in availability also equal to the expression in Eq.

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