25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016
Question 2 of 8: Reheat–Regeneration Rankine Cycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2016 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).
Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.
It is solved as the thermodynamics/heat-transfer exam it actually is.
Given. Steam Rankine cycle with reheat and two open feedwater heaters (FWHs).
Given data
Location
Pressure
Temperature
HP turbine inlet (state 1)
3.5 MPa
350°C
HP turbine exit / FWH1 extraction (state 2)
0.8 MPa
—
Reheat exit / LP turbine inlet (state 3)
0.8 MPa
350°C
FWH2 extraction (state 4)
0.2 MPa
—
Condenser (state 5)
10 kPa
—
Find. Thermal efficiency $\eta_{th}$ and net work output per kilogram of throttle-flow steam.
Reheat–regeneration Rankine cycle: 1→2 HP turbine (to first extraction), 2→3 reheat, 3→4 LP turbine (to second extraction), 4→5 LP turbine (to condenser); FWH mixing and pump legs omitted for clarity.
Approach. Fix all six state enthalpies from the steam tables, find the extraction fractions $y_1,y_2$ from open-FWH energy balances (exit = saturated liquid), then sum turbine work minus pump work over net work, and boiler-plus-reheat heat input.
State 1 & state 2 (HP turbine, isentropic). At 3.5 MPa, 350°C: $h_1=3104.0\text{ kJ/kg}$, $s_1=6.4502\text{ kJ/kg}\cdot\text{K}$. At 0.8 MPa, $s_2=s_1$ falls inside the dome ($s_f=2.0462$, $s_{fg}=4.6166$): quality $x_2=\dfrac{6.4502-2.0462}{4.6166}=0.9539$, so
$$h_2=h_f+x_2h_{fg}=721.11+0.9539(2048.0)=2674.8\text{ kJ/kg}$$
State 3 & state 4 (reheat, LP turbine). Reheated to 0.8 MPa, 350°C: $h_3=3161.7\text{ kJ/kg}$, $s_3=7.4089\text{ kJ/kg}\cdot\text{K}$. At 0.2 MPa, interpolating the superheated table between 150°C and 200°C for $s=s_3$ gives $h_4=2826.7\text{ kJ/kg}$.
State 5 (condenser). At 10 kPa, $s_5=s_3$ is wet: $x_5=\dfrac{7.4089-0.6492}{7.4996}=0.9013$, so $h_5=191.81+0.9013(2392.1)=2347.9\text{ kJ/kg}$.
Pump & feedwater-heater states. $h_6=h_f(10\text{ kPa})=191.81$; pump 1 to 0.2 MPa: $h_7=h_6+v_{f}\Delta P=192.00$. $h_8=h_f(0.2\text{ MPa})=504.71$; pump 2 to 0.8 MPa: $h_9=505.35$. $h_{10}=h_f(0.8\text{ MPa})=721.11$; pump 3 to 3.5 MPa: $h_{11}=724.12$.
Extraction fractions (open-FWH energy balances).
$$y_1=\frac{h_{10}-h_9}{h_2-h_9}=\frac{721.11-505.35}{2674.8-505.35}=0.0995,\qquad y_2=\frac{(1-y_1)(h_8-h_7)}{h_4-h_7}=0.1069$$
Turbine, pump, and net work. $W_{HP}=h_1-h_2=429.2$; $W_{LP1}=(1-y_1)(h_3-h_4)=301.6$; $W_{LP2}=(1-y_1-y_2)(h_4-h_5)=380.3$ — total turbine work $1111.1\text{ kJ/kg}$. Pump work sums to $3.7\text{ kJ/kg}$ (dominated by the boiler feed pump), so
$$\boxed{W_{net}=1111.1-3.7 \approx 1107.1\text{ kJ/kg}}$$
Heat input and efficiency. $q_{boiler}=h_1-h_{11}=2379.9$, $q_{reheat}=(1-y_1)(h_3-h_2)=438.5$, so $q_{in}=2818.4\text{ kJ/kg}$ and
$$\boxed{\eta_{th}=\frac{W_{net}}{q_{in}}=\frac{1107.1}{2818.4}\approx 0.393=39.3\%}$$
