Question 1 of 9: Reheat Rankine Cycle — Non-Ideal Turbines and Pump, with Second-Law Efficiency
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — December 2014. 3 hours, Closed-Book Exam
(approved calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and
charts supplied in an appendix). Part A: answer 2 of Questions 1-3 (20 marks each). Part B: answer
4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the first two Part-A and first four
Part-B questions as they appear in the answer book are marked. All nine questions (Part A complete,
Part B complete) are solved below for completeness.
Reference texts: Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering
Thermodynamics, 8th ed. State properties for the real-fluid questions (R-134a, water/steam,
moist air, air) were computed from high-accuracy equations of state in place of printed property-table
interpolation. Question 5 is the exception: it directs that the mixture be treated as an ideal gas,
so its N₂ and O₂ enthalpies are taken from the ideal-gas tables supplied with the paper.
Question 1: Reheat Rankine Cycle — Non-Ideal Turbines and Pump, with Second-Law Efficiency (20 marks)
Fig. Q1 — T–s state points for the reheat Rankine cycle (1→2
pump, 2→3 boiler, 3→4 turbine 1, 4→5 reheater, 5→6 turbine 2, 6→1 condenser).
Approach
Fix the condenser-exit and boiler-exit states directly from the given pressures/temperatures, run
the isentropic pump and both isentropic turbine legs to get the ideal exit enthalpies, then apply
each component's isentropic efficiency to get the actual states. Net specific work and total heat
input follow directly, and the rates use the given mass flow rate; the second-law efficiency compares
actual net work to the maximum (exergy) work obtainable from the same heat input.
Condenser exit and Pump (1→2). Saturated liquid at 7.5 kPa: $h_1=168.75$
kJ/kg, $s_1=0.5763$ kJ/kg·K. Isentropic pump exit at 8 MPa: $h_{2s}=176.79$ kJ/kg, so
$w_{p,s}=h_{2s}-h_1=8.04$ kJ/kg. Actual pump work
$$w_{p,a}=\frac{w_{p,s}}{\eta_p}=\frac{8.04}{0.80}=\boxed{10.05\ \text{kJ/kg}},\qquad h_2=h_1+w_{p,a}=178.80\ \text{kJ/kg}.$$
Turbine 1 (3→4), isentropic then actual. Following $s=s_3$ down to the
reheat pressure (0.8 MPa): $h_{4s}=2760.72$ kJ/kg, so $w_{t1,s}=h_3-h_{4s}=576.35$ kJ/kg. Actual:
$$w_{t1,a}=\eta_t\,w_{t1,s}=0.88\times576.35=\boxed{507.19\ \text{kJ/kg}},\qquad h_4=h_3-w_{t1,a}=2829.88\ \text{kJ/kg}.$$
Turbine 2 (5→6), isentropic then actual. Following $s=s_5$ down to the
condenser pressure (7.5 kPa): $h_{6s}=2432.46$ kJ/kg, so $w_{t2,s}=h_5-h_{6s}=994.95$ kJ/kg. Actual:
$$w_{t2,a}=\eta_t\,w_{t2,s}=0.88\times994.95=\boxed{875.56\ \text{kJ/kg}},\qquad h_6=h_5-w_{t2,a}=2551.85\ \text{kJ/kg}.$$
Net specific work and total heat input.
$$w_{net}=(w_{t1,a}+w_{t2,a})-w_{p,a}=1382.75-10.05=\boxed{1372.69\ \text{kJ/kg}}.$$
$$q_{in}=(h_3-h_2)+(h_5-h_4)=3158.27+597.53=\boxed{3755.80\ \text{kJ/kg}}.$$
Net power output and boiler heat-input rate (parts a, b).
$$\dot W_{net}=\dot m\,w_{net}=20\times1372.69=\boxed{27{,}454\ \text{kW}}.$$
$$\dot Q_{in}=\dot m\,q_{in}=20\times3755.80=\boxed{75{,}116\ \text{kJ/s}}.$$
Second-law (exergy) efficiency (part d). Modeling the boiler/reheater heat
addition as supplied from a source at $T_H=1500$ K:
$$x_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=3755.80\times\left(1-\frac{300}{1500}\right)=3004.64\ \text{kJ/kg}.$$
$$\eta_{II}=\frac{w_{net}}{x_{in}}=\frac{1372.69}{3004.64}=\boxed{0.4569\ (45.7\%)}.$$
