NivaarExam PrepOfficial exam papers ↗

04-BS-10 · December 2017

Question 1 of 9: Reheat Rankine Cycle 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 2017. 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, interpolation not required). 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. All state properties (water/steam, R-134a, air, N₂, CO₂, H₂) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric result.

Question 1: Reheat Rankine Cycle with Second-Law Efficiency (20 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. Turbine-1 inlet (state C): $P=8$ MPa, $T=480\ ^\circ$C. Reheat to 480 °C at 0.7 MPa (state E) before turbine-2, which expands to the condenser pressure of 8 kPa (states A/F). $\dot m=2.63\times10^5$ kg/h $=73.056$ kg/s. $\eta_{t1}=\eta_{t2}=0.88$, $\eta_p=0.80$. Source $T_H=1000$ K, sink $T_0=288$ K.

StateDescriptionPh (kJ/kg)s (kJ/kg·K)
ACondenser exit, sat. liquid8 kPa173.840.5925
BPump exit (actual)8 MPa183.900.5989
CTurbine-1 inlet8 MPa, 480°C3349.656.6613
DTurbine-1 exit (actual)0.7 MPa2815.526.8246
EReheat exit / turbine-2 inlet0.7 MPa, 480°C3439.317.8755
FTurbine-2 exit (actual)8 kPa2582.368.2468

Find. (a) $\dot Q_{in}$ [kW]; (b) $\dot Q_{out}$ [kW]; (c) $\dot W_{net}$ [kW]; (d) $\eta_{th}$; (e) $\eta_{II}$.

Entropy s (kJ/kg·K)Temperature TQ1 — Reheat Rankine cycle, non-isentropic pump & both turbine stages (T–s)saturation domeBCDEFA
Fig. Q1 — T–s state points for the reheat Rankine cycle (A→B actual pump, B→C boiler, C→D actual turbine 1, D→E reheater, E→F actual turbine 2, F→A condenser). State F lands just outside the dome (superheated) despite D landing well inside it — see Step 3.

Approach

Fix the condenser-exit and both turbine-inlet states directly from the given pressures and temperatures, then run the pump and each turbine stage isentropically to get the ideal work, divide/multiply by the stated efficiency to get the actual work, and back out each actual exit enthalpy. Summing the two boiler-side heat additions (main boiler and reheater) and the single condenser heat rejection, weighted by the given mass flow rate, gives rates (a)–(d); part (e) treats the given source/sink temperatures as the exergy reference to size the exergy supplied by the heat input.

  1. Condenser exit and actual pump. Saturated liquid at 8 kPa: $h_A=173.84$ kJ/kg, $s_A=0.5925$ kJ/kg·K ($T_{sat}=41.51\ ^\circ$C). Isentropic pump exit at 8 MPa gives $w_{p,s}=8.046$ kJ/kg; actual: $$w_{p,a}=\frac{w_{p,s}}{\eta_p}=\frac{8.046}{0.80}=\boxed{10.058\ \text{kJ/kg}},\qquad h_B=183.90\ \text{kJ/kg}.$$
  2. Turbine 1 (C→D), 8→0.7 MPa. At 8 MPa, 480°C: $h_C=3349.65$ kJ/kg, $s_C=6.6613$ kJ/kg·K. Isentropic exit at 0.7 MPa gives $w_{t1,s}=606.96$ kJ/kg; actual: $$w_{t1,a}=\eta_{t1}\,w_{t1,s}=0.88\times606.96=\boxed{534.12\ \text{kJ/kg}},\qquad h_D=2815.52\ \text{kJ/kg}.$$
  3. Reheat and turbine 2 (E→F), 0.7→0.008 MPa. At 0.7 MPa, 480°C: $h_E=3439.31$ kJ/kg, $s_E=7.8755$ kJ/kg·K. Isentropic exit at 8 kPa is wet, giving $w_{t2,s}=973.80$ kJ/kg; actual: $$w_{t2,a}=\eta_{t2}\,w_{t2,s}=0.88\times973.80=\boxed{856.95\ \text{kJ/kg}},\qquad h_F=2582.36\ \text{kJ/kg}\ (T_F=44.70\ ^\circ\text{C, superheated}).$$ Turbine inefficiency has pushed the actual exit just past the saturated-vapor line even though the isentropic exit would have landed well inside the dome.
  4. Heat input rate (part a). Two heat-addition legs: the boiler ($B\to C$) and the reheater ($D\to E$). $$q_{in}=(h_C-h_B)+(h_E-h_D)=(3349.65-183.90)+(3439.31-2815.52)=3165.75+623.79=3789.54\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=73.056\times3789.54=\boxed{276{,}846\ \text{kW}}\ (276.8\ \text{MW}).$$
  5. Heat rejected rate (part b). Single condenser leg, $F\to A$: $$q_{out}=h_F-h_A=2582.36-173.84=2408.52\ \text{kJ/kg}.$$ $$\dot Q_{out}=\dot m\,q_{out}=73.056\times2408.52=\boxed{175{,}956\ \text{kW}}\ (176.0\ \text{MW}).$$
  6. Net power output (part c). $$w_{net}=w_{t1,a}+w_{t2,a}-w_{p,a}=534.12+856.95-10.06=1381.01\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=73.056\times1381.01=\boxed{100{,}890\ \text{kW}}\ (100.9\ \text{MW}).$$ Cross-check: $\dot Q_{in}-\dot Q_{out}=276{,}846-175{,}956=100{,}890$ kW $=\dot W_{net}$ ✓.
  7. Thermal efficiency (part d). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1381.01}{3789.54}=\boxed{0.3644\ (36.4\%)}.$$
  8. Second-law efficiency (part e). Treat the given source temperature as the reservoir supplying $\dot Q_{in}$ and the sink temperature as the dead-state reference $T_0$: the maximum (exergy) work obtainable from that heat rate is $\dot X_{in}=\dot Q_{in}(1-T_0/T_H)$. $$\dot X_{in}=276{,}846\times\left(1-\frac{288}{1000}\right)=276{,}846\times0.712=\boxed{197{,}115\ \text{kW}}.$$ $$\eta_{II}=\frac{\dot W_{net}}{\dot X_{in}}=\frac{100{,}890}{197{,}115}=\boxed{0.5118\ (51.2\%)}.$$
QuantityResult
(a) $\dot Q_{in}$276,846 kW (276.8 MW)
(b) $\dot Q_{out}$175,956 kW (176.0 MW)
(c) $\dot W_{net}$100,890 kW (100.9 MW)
(d) $\eta_{th}$0.3644 (36.4%)
(e) $\eta_{II}$0.5118 (51.2%)
← Paper overview