Adiabatic Steam Turbine, Efficiency from a Measured Work Output: 04-BS-10, May 2016, Q7

Question 7: Adiabatic Steam Turbine, Efficiency from a Measured Work Output (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: $P_1=1.0$ MPa, $T_1=320\ ^\circ$C. Exit pressure $P_2=20$ kPa. Insulated (adiabatic) turbine, $\dot m=10$ kg/s. Actual work developed $w_{actual}=630$ kJ/kg.

Find. (a) $\eta_t$; (b) $T_2$; (c) $x_2$; (d) reversible or not?

Approach

Fix the inlet state, then find the isentropic exit enthalpy at 20 kPa from the inlet entropy to get the isentropic (maximum possible) work. The isentropic efficiency is the ratio of the actual measured work to that isentropic work. The actual exit enthalpy follows directly from the actual work by an energy balance, which fixes the actual exit state (and its entropy, for the reversibility check).

Inlet state. At 1.0 MPa, 320°C: $h_1=3094.36$ kJ/kg, $s_1=7.1979$ kJ/kg·K.
Isentropic exit and efficiency (part a). At 20 kPa with $s_{2s}=s_1=7.1979$ kJ/kg·K, the exit is wet: $h_{2s}=2372.59$ kJ/kg, giving isentropic work $w_s=h_1-h_{2s}=3094.36-2372.59=721.78$ kJ/kg. $$\eta_t=\frac{w_{actual}}{w_s}=\frac{630}{721.78}=\boxed{0.8728\ (87.3\%)}.$$
Actual exit state (parts b, c). Energy balance on the adiabatic turbine: $$h_2=h_1-w_{actual}=3094.36-630=2464.36\ \text{kJ/kg}.$$ At 20 kPa, $h_f=251.42$ kJ/kg and $h_g=2608.94$ kJ/kg, so $h_2$ lies between them — a two-phase mixture, with $T_2=T_{sat}(20\ \text{kPa})=\boxed{60.06\ ^\circ\text{C}}$ and $$x_2=\frac{h_2-h_f}{h_g-h_f}=\frac{2464.36-251.42}{2608.94-251.42}=\boxed{0.9387}.$$
Reversibility check (part d). Entropy at the actual exit state: $s_2=s_f+x_2 s_{fg}=7.4733$ kJ/kg·K. Since $s_2=7.4733>s_1=7.1979$ kJ/kg·K, entropy increased across this adiabatic turbine, so $$\boxed{\text{the process is irreversible}}$$ (consistent with $\eta_t=87.3\%<100\%$ — a truly reversible adiabatic turbine would be isentropic, $s_2=s_1$, and would achieve $\eta_t=100\%$).

Quantity	Result
(a) $\eta_t$	0.8728 (87.3%)
(b) $T_2$	60.06°C
(c) $x_2$	0.9387
(d)	Irreversible ($s_2>s_1$)

Concept & Theory

Topic. Adiabatic turbine analysis working backward from a measured actual work output to isentropic efficiency, exit state, and reversibility.

Key relations. Isentropic efficiency: $\eta_t=w_{actual}/w_s$ where $w_s=h_1-h_{2s}$ ($s_{2s}=s_1$). Actual exit enthalpy from energy balance: $h_2=h_1-w_{actual}$ (adiabatic, no other energy terms). Second-law check: an adiabatic process is irreversible if and only if $s_2>s_1$.

Why this works. An adiabatic turbine has no heat transfer, so the actual exit enthalpy is pinned directly by the actual (measured) work via the steady-flow energy balance, independent of any efficiency assumption — the efficiency is only needed to characterize how far that actual state falls short of the unattainable isentropic ideal, not to find the actual state itself.

Common pitfall. Assuming the exit must be a two-phase mixture just because the isentropic exit is wet — the actual (irreversible) exit state must be checked independently against $h_f$/$h_g$ at the exit pressure; in this question it does turn out wet (unlike the reheat-cycle case in Question 1, where turbine inefficiency instead pushed an isentropically-wet exit into the superheated region).

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy § Isentropic Efficiencies of Steady-Flow Devices).

Question 7 of 9: Adiabatic Steam Turbine, Efficiency from a Measured Work Output

Question 7: Adiabatic Steam Turbine, Efficiency from a Measured Work Output (15 marks)

Approach