Question 5 of 9: Air Turbine — Power, Reversibility Check, Isentropic Efficiency

Question 5: Air Turbine — Power, Reversibility Check, Isentropic Efficiency (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: $T_1=52\ ^\circ$C, $P_1=300$ kPa. Exit: $T_2=12\ ^\circ$C, $P_2=100$ kPa. $\dot m=10$ kg/s. Adiabatic, $\Delta$KE $=\Delta$PE $=0$.

Find. $\dot W$; reversibility; if not, $\eta_t$.

Approach

The turbine power follows directly from the steady-flow energy balance using the actual inlet and exit enthalpies. To check reversibility, compare the actual specific entropy change (using the full ideal-gas entropy, since inlet and exit are at different pressures) against the $\Delta s=0$ criterion for an adiabatic reversible process; if $\Delta s>0$, find the isentropic efficiency by comparing actual work to the work of an isentropic process reaching the same exit pressure.

1. Actual states and power developed. $h_1=451.62$ kJ/kg ($T_1=325.15$ K), $h_2=411.36$ kJ/kg ($T_2=285.15$ K): $$\dot W=\dot m(h_1-h_2)=10\times(451.62-411.36)=\boxed{402.65\ \text{kW}}.$$
2. Reversibility check. Full ideal-gas entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$ at each state's own pressure gives $s_1=3.6563$ kJ/kg·K, $s_2=3.8394$ kJ/kg·K: $$\Delta s=s_2-s_1=3.8394-3.6563=\boxed{+0.1831\ \text{kJ/kg}\cdot\text{K}}\ (>0).$$ Since the turbine is adiabatic ($q=0$) and $\Delta s>0$, entropy is generated internally ($\dot S_{gen}=\dot m\,\Delta s>0$) — the process is not reversible.
3. Isentropic exit state. Following $s=s_1$ down to $P_2=100$ kPa: $T_{2s}=-35.48\ ^\circ$C, $h_{2s}=363.61$ kJ/kg, giving isentropic work $w_s=h_1-h_{2s}=451.62-363.61=\boxed{88.01\ \text{kJ/kg}}$.
4. Isentropic efficiency. Actual specific work $w_a=h_1-h_2=451.62-411.36=40.27$ kJ/kg: $$\eta_t=\frac{w_a}{w_s}=\frac{40.27}{88.01}=\boxed{0.4575\ (45.7\%)}.$$