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22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2015

Question 2 of 8: Reheat Rankine Power Plant

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2015. Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); a complete paper is any five questions — three from one part and two from the other, all of equal value. Full worked solutions to all eight questions are given below.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power cycles and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU / LMTD-correction heat-exchanger methods. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and water data from standard tables.

Question 2: Reheat Rankine Power Plant (equal value)

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. Reheat Rankine cycle. Turbine isentropic efficiency $\eta_t=0.85$, pump efficiency $\eta_p=0.80$; total turbine power $\dot W_t=1000\ \text{kW}$.

StateCondition$h$ (kJ/kg)$s$ (kJ/kg·K)
1 — HP inlet3.5 MPa, 350 °C3104.96.6579
3 — LP inlet (reheat)0.5 MPa, 350 °C3167.77.6329
Condensate30 °C liquid, 7.5 kPa125.74—

Find. (a) steam mass flow, (b) pump power, (c) thermal efficiency.

024680100200300400entropy s (kJ/kg·K)T (°C)saturation dome12345,6boiler, 3.5 MPacondenser, 7.5 kPa
Figure 1 — T–s schematic of the reheat cycle: 1→2 HP expansion to 0.5 MPa, 2→3 constant-pressure reheat to 350 °C, 3→4 LP expansion to the 7.5 kPa condenser, condensation to state 5 (30 °C liquid), 5→6 pumping to 3.5 MPa. Every labelled state is plotted at its tabulated $(s,T)$ against the axes shown, so both expansions carry the small entropy increase of the 85 %-efficient stages and the 2→3 reheat runs to the right at constant pressure.

Approach. Find each isentropic turbine exit from the entropy, apply the 85 % efficiency to get the real work, sum the two stage works to fix the mass flow from the 1000 kW output, then size the pump and close the boiler heat balance for the efficiency.

  1. HP turbine (3.5 MPa → 0.5 MPa). At 0.5 MPa ($s_f=1.8607$, $s_{fg}=4.9606$), $x_{2s}=(6.6579-1.8607)/4.9606=0.967$, so $h_{2s}=640.1+0.967(2108.0)=2678.7$ kJ/kg and$$w_{HP}=\eta_t(h_1-h_{2s})=0.85(3104.9-2678.7)=\boxed{362.3\ \text{kJ/kg}},\quad h_2=2742.6\ \text{kJ/kg}.$$
  2. LP turbine (0.5 MPa → 7.5 kPa). At 7.5 kPa ($s_f=0.5764$, $s_{fg}=7.6750$), $x_{4s}=(7.6329-0.5764)/7.6750=0.919$, so $h_{4s}=168.8+0.919(2406.0)=2380.9$ kJ/kg and$$w_{LP}=0.85(3167.7-2380.9)=\boxed{668.8\ \text{kJ/kg}}.$$
  3. (a) Steam mass flow. Total specific turbine work $w_t=362.3+668.8=1031.1$ kJ/kg, so$$\dot m=\frac{\dot W_t}{w_t}=\frac{1000}{1031.1}=\boxed{0.970\ \text{kg/s}}.$$
  4. (b) Pump work and power. Pumping saturated liquid ($v_f=0.001004\ \text{m}^3/\text{kg}$) from 7.5 kPa to 3.5 MPa,$$w_p=\frac{v_f(P_2-P_1)}{\eta_p}=\frac{0.001004(3500-7.5)}{0.80}=4.38\ \text{kJ/kg},\quad \dot W_p=\dot m\,w_p=\boxed{4.25\ \text{kW}}.$$
  5. (c) Thermal efficiency. Boiler adds $q_\text{in}=(h_1-h_6)+(h_3-h_2)=(3104.9-130.1)+(3167.7-2742.6)=2974.8+425.1=3399.9$ kJ/kg; net work $w_\text{net}=1031.1-4.4=1026.7$ kJ/kg. Hence$$\eta_\text{th}=\frac{w_\text{net}}{q_\text{in}}=\frac{1026.7}{3399.9}=\boxed{0.302\ (30.2\%)}.$$
QuantityResult
HP / LP specific work362.3 / 668.8 kJ/kg
Steam mass flow≈ 0.970 kg/s
Pump power≈ 4.25 kW
Thermal efficiency≈ 30.2 %
Check
The net plant output is $1000-4.25\approx996$ kW; the pump consumes under 0.5 % of the turbine output, as expected for a steam cycle (the "back-work ratio" of a vapour cycle is tiny). The condensate is subcooled to 30 °C (below the 40.3 °C saturation at 7.5 kPa), so its enthalpy is taken as $h_f(30\,{}^\circ\text{C})$. The HP exit lands at $h_2=2742.6$ kJ/kg, i.e. $x_2=0.997$ — the steam sits essentially on the saturated-vapour line after a single stage, which is exactly why the cycle reheats before the LP turbine.