Question 4 of 9: Steady-Flow Air Turbine, Reversibility Check

Question 4: Steady-Flow Air Turbine, Reversibility Check (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Inlet: $P_1=300$ kPa, $T_1=52\ ^\circ$C. Exit: $P_2=100$ kPa, $T_2=12\ ^\circ$C. $\dot m=10$ kg/s. Adiabatic, $\Delta$KE $=\Delta$PE $=0$.

Find. $\dot W$ [kW]; reversibility; $\eta_t$ if irreversible.

Approach

The actual power follows directly from the steady-flow energy balance using the given inlet/exit enthalpies. Reversibility for an adiabatic process is judged purely by the entropy change: compute $s_1$ and $s_2$ from the given $(T,P)$ pairs via the variable-cp air relation and compare. If $s_2>s_1$, the process is irreversible, and the isentropic efficiency follows from a separate isentropic-exit calculation at the same exit pressure.

1. Actual work and power. $h_1=h(325.15\ \text{K})=451.62$ kJ/kg, $h_2=h(285.15\ \text{K})=411.36$ kJ/kg. $$w_{actual}=h_1-h_2=451.62-411.36=40.27\ \text{kJ/kg}.$$ $$\dot W=\dot m\,w_{actual}=10\times40.27=\boxed{402.65\ \text{kW}}.$$
2. Reversibility check. Absolute entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$: $$s_1=3.6563\ \text{kJ/kg}\cdot\text{K},\qquad s_2=3.8394\ \text{kJ/kg}\cdot\text{K}.$$ Since the turbine is adiabatic, any entropy change equals the entropy generated: $$s_{gen}=s_2-s_1=\boxed{0.1832\ \text{kJ/kg}\cdot\text{K}}>0,$$ so the process is irreversible (a reversible adiabatic process would require $s_2=s_1$).
3. Isentropic efficiency. Holding $s_1$ fixed and expanding to $P_2=100$ kPa isentropically gives $T_{2s}=237.67$ K, so $w_s=h_1-h(T_{2s})=451.62-363.61=88.01$ kJ/kg. $$\eta_t=\frac{w_{actual}}{w_s}=\frac{40.27}{88.01}=\boxed{0.4575\ (45.8\%)}.$$