Insulated Steam Turbine — Isentropic Efficiency from Given Actual Work: 04-BS-10, May 2014, Q7

Question 7: Insulated Steam Turbine — Isentropic Efficiency from Given Actual Work (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. $P_1=1.0$ MPa, $T_1=320\ ^\circ$C, $\dot m=10$ kg/s, $P_2=20$ kPa, $w_{actual}=630$ kJ/kg. Adiabatic (insulated), steady-state, steady-flow.

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

Approach

Fix the inlet state, follow the isentrope to the exit pressure to get the ideal exit enthalpy and isentropic work, then compare with the given actual work to get the isentropic efficiency directly. Back out the actual exit enthalpy from the actual work, locate it on the 20 kPa isobar (checking whether it falls in the two-phase dome), and compare actual and inlet entropy to assess reversibility.

Inlet state. $P_1=1.0$ MPa, $T_1=320\ ^\circ$C (superheated): $$h_1=3094.36\ \text{kJ/kg},\qquad s_1=7.1979\ \text{kJ/kg}\cdot\text{K}.$$
Isentropic exit state and isentropic work. At $P_2=20$ kPa with $s_{2s}=s_1$: $$h_{2s}=2372.59\ \text{kJ/kg},\qquad w_s=h_1-h_{2s}=3094.36-2372.59=721.78\ \text{kJ/kg}.$$
Isentropic efficiency (part a). $$\eta_t=\frac{w_{actual}}{w_s}=\frac{630}{721.78}=\boxed{0.8728\ (87.3\%)}.$$
Actual exit state. $h_2=h_1-w_{actual}=3094.36-630=2464.36$ kJ/kg. At 20 kPa, $h_f=251.42$ kJ/kg and $h_g=2608.94$ kJ/kg, so $h_f
Exit quality (part c). $$x_2=\frac{h_2-h_f}{h_g-h_f}=\frac{2464.36-251.42}{2608.94-251.42}=\boxed{0.9387\ (93.87\%)}.$$
Reversibility (part d). Exit entropy at $(P_2,x_2)$: $s_2=7.4733$ kJ/kg·K. Since the turbine is adiabatic ($q=0$), any entropy generation shows up directly as $\Delta s>0$: $$s_2-s_1=7.4733-7.1979=0.2754\ \text{kJ/kg}\cdot\text{K}>0\ \Rightarrow\ \boxed{\text{not reversible}}.$$ This is consistent with $\eta_t=87.3\%<100\%$: any isentropic efficiency below unity in an adiabatic device is itself proof of internal irreversibility (friction, flow separation in the blading, etc.).

Quantity	Result
(a) $\eta_t$	0.8728 (87.3%)
(b) $T_2$	60.06°C (saturation temp. at 20 kPa)
(c) $x_2$	0.9387 (93.87%)
(d) Reversible?	No — $s_2 > s_1$ ($\Delta s=0.2754$ kJ/kg·K)

Concept & Theory

Topic. Adiabatic steam-turbine performance from measured actual work; isentropic efficiency, quality, and the entropy criterion for reversibility.

Key relations. Isentropic efficiency of a turbine: $\eta_t=w_{actual}/w_s=(h_1-h_2)/(h_1-h_{2s})$. Two-phase quality: $x=(h-h_f)/(h_g-h_f)$. For any adiabatic process, $\Delta s\ge0$ with equality only for a reversible (isentropic) process.

Why this works. Working backward from the given actual work rather than an assumed efficiency is the same calculation run in reverse: the isentropic exit state is still needed as the reference against which "actual" is measured, so it must be computed first regardless of which quantity (efficiency vs. actual work) the problem happens to supply.

Common pitfall. Reporting $T_2$ from an enthalpy-only interpolation on the superheated tables. Once $h_2$ is confirmed to fall between $h_f$ and $h_g$ at $P_2$, the exit is two-phase and $T_2$ must equal the saturation temperature at that pressure — it cannot be recovered by treating $h_2$ as if it were a superheated-region enthalpy.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy) & Ch. 5 (Mass and Energy Analysis of Control Volumes).

Question 7 of 9: Insulated Steam Turbine — Isentropic Efficiency from Given Actual Work

Question 7: Insulated Steam Turbine — Isentropic Efficiency from Given Actual Work (15 marks)

Approach