Discussion
The inlet and outlet temperatures of the flowing fluid during a steady state flow process are the same as that of the surroundings. If the changes in kinetic and potential energies are neglected, the maximum power that can be obtained is equal to
Answer: C
The maximum power equals the rate of decrease in Gibbs free energy of the flowing fluid.
For a steady flow exchanging heat only with surroundings at \(T_{0}\), with kinetic and potential energy changes neglected, the maximum (reversible) work per unit mass is:
\[ w_{max} = (h_{1} - T_{0}s_{1}) - (h_{2} - T_{0}s_{2}) \]
When the fluid enters and leaves at the surroundings temperature, \(T_{1} = T_{2} = T_{0}\), so \(h - T_{0}s\) is exactly the Gibbs function \(g = h - Ts\) at each end. The maximum power is therefore the mass flow rate times the drop in g.
The Helmholtz function \(a = u - Ts\) plays this role for a closed system, not a flow. An increase in enthalpy would mean energy going into the fluid, and the drop in internal energy ignores both the flow work and the entropy term.
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