Air Turbine — Reversibility Check: 04-BS-10, May 2018, Q4

Question 4: 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, $\Delta ke=\Delta pe=0$. Variable specific heats (air tables).

Find. Power developed; whether the process is reversible; if not, $\eta_t$.

Approach

The actual power follows directly from the steady-flow energy balance using variable-specific-heat air enthalpies. Reversibility for an adiabatic, steady-flow device is checked via the entropy balance ($\dot S_{gen}=\dot m(s_2-s_1)$ since $\dot Q=0$); if $s_2>s_1$ the process is irreversible, and the isentropic efficiency compares the actual work to the work of an isentropic expansion between the same pressures.

Actual turbine work and power. From air tables (variable $c_p$): $h_1=451.62$ kJ/kg (at 325.15 K), $h_2=411.36$ kJ/kg (at 285.15 K). $$w_{actual}=h_1-h_2=451.62-411.36=40.27\ \text{kJ/kg}.$$ $$\dot W=\dot m\,w_{actual}=10\times40.27=\boxed{402.7\ \text{kW}}.$$
Reversibility check. For an adiabatic steady-flow device, entropy generation equals the exit-minus-inlet specific entropy (no heat-transfer term): $$s_1=s_1^\circ-R\ln\frac{P_1}{P_{ref}},\qquad s_2=s_2^\circ-R\ln\frac{P_2}{P_{ref}}.$$ Evaluating from the air table's standard-state entropy function gives $s_1=3.6563$ and $s_2=3.8394$ kJ/kg·K, so $$s_{gen}=s_2-s_1=3.8394-3.6563=\boxed{0.1832\ \text{kJ/kg}\cdot\text{K}}>0.$$ Since $s_{gen}>0$, the process is irreversible (a reversible adiabatic process would require $s_2=s_1$).
Isentropic efficiency. Find the isentropic exit temperature $T_{2s}$ by matching the standard-state entropy function at $P_2$ to the inlet's ($s_1^\circ - s_{2s}^\circ = R\ln(P_2/P_1)$), giving $T_{2s}=237.67$ K $=-35.48\ ^\circ$C and $h_{2s}=363.61$ kJ/kg. The isentropic work is $$w_s=h_1-h_{2s}=451.62-363.61=88.01\ \text{kJ/kg}.$$ $$\eta_t=\frac{w_{actual}}{w_s}=\frac{40.27}{88.01}=\boxed{0.4575\ (45.8\%)}.$$

Quantity	Result
Power developed	402.7 kW
Reversible?	No ($s_{gen}=0.1832$ kJ/kg·K $>0$)
$\eta_t$	0.4575 (45.8%)

Concept & Theory

Topic. Adiabatic steady-flow turbine reversibility check via the entropy balance, followed by an isentropic-efficiency calculation when the process proves irreversible.

Key relations. Steady-flow energy balance: $w=h_1-h_2$ (adiabatic, no KE/PE). Entropy balance for an adiabatic control volume: $\dot S_{gen}=\dot m(s_2-s_1)$. Isentropic efficiency: $\eta_t=w_{actual}/w_s$, with $w_s=h_1-h_{2s}$ found from matching entropies at the exit pressure.

Why this works. An adiabatic process is reversible only if it is also isentropic ($s_2=s_1$); any measured entropy increase across an adiabatic device is unambiguous proof of internal irreversibility (friction, turbulence, etc.), independent of the sign or magnitude of the actual work produced.

Common pitfall. Checking "reversibility" by comparing the actual exit temperature to some assumed isentropic value without first computing entropies directly — the direct entropy-balance test ($s_{gen}\gtrless0$) is unambiguous and doesn't require solving for $T_{2s}$ first.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy) & Ch. 8 (Isentropic Efficiencies).

Question 4 of 9: Air Turbine — Reversibility Check

Question 4: Air Turbine — Reversibility Check (15 marks)

Approach