22-Mec-B3 Energy Conversion and Power Generation · May 2018
Question 2 of 8: Combined Cycle Steam Turbine
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mec-B3
Energy Conversion and Power Generation, National Examinations, May 2018.
Three hours, closed book. Two sections: Section A is calculative
(Questions 1–5) and Section B is descriptive (Questions 6–8).
Candidates answer four questions from Section A and two from Section B; six
questions of 10 marks each constitute a complete paper (60 marks). Reference
data for individual questions are bound in as attachments on pages 9–15,
reference formulae and constants on pages 16–19, and Granet &
Bluestein steam tables are supplied. All eight questions are solved
below, because the set is a study resource rather than a timed attempt.
Reference texts.
Granet, I. and Bluestein, M., Thermodynamics and Heat Power, 6th ed.
— the steam tables issued with this paper, and the vapour-cycle and
gas-cycle chapters.
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering
Approach, 9th ed. — Brayton and Rankine cycles, combined cycles,
isentropic component efficiencies.
El-Wakil, M. M., Powerplant Technology — heat recovery steam
generators and pinch behaviour, condensers and their off-design response,
boiler heat-absorption surfaces, load scheduling.
Lamarsh, J. R. and Baratta, A. J., Introduction to Nuclear
Engineering, 4th ed. — the pressurised water reactor, primary loop
heat removal, once-through steam generators.
Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of
Turbomachinery, 7th ed. — hydraulic turbines, net head and the energy
accounting of a waterway.
Wherever the paper's own attachments carry a value that duplicates a computed
result — the Koeberg terminal temperature difference and back pressure on
page 10, the gas-turbine output quoted in the preamble to Question 2,
the published rating of the Oconee unit — that printed value is used as an
independent check and the agreement is quoted in the answer.
Given. An unfired single-pressure heat recovery steam generator takes 145 kg/s of turbine exhaust from 560 °C down to 180 °C, raising steam at 1.40 MPa and 540 °C from feedwater at 30 °C; the steam expands to a 30 °C exhaust with an internal efficiency of 85 %.
$42.5\ \text{MW}$ from $2.69\ \text{kg/s}$ at $40\,000\ \text{kJ/kg}$
Find. The terminal temperatures on a temperature against path-length diagram, the steam mass flow raised, the power the bottoming turbine develops and the overall efficiency of the combined plant.
Question 2(a) — temperature against path length through the heat recovery steam generator. The gas (red) falls linearly from 560 °C to 180 °C; the water and steam (blue) flow counter to it, entering the economiser at 30 °C, evaporating at the 1.40 MPa saturation temperature of 195.0 °C and leaving the superheater at 540 °C. The closest approach, the pinch, is 62.8 K at the evaporator inlet.
Approach. Take the gas-side duty from a sensible-heat balance, divide it by the enthalpy rise of the water to get the steam flow, expand the steam from the superheater state to the condenser saturation pressure using the internal efficiency, then add the two shaft outputs and divide by the single fuel input.
Part (a) — establish the terminal points and check the profile is feasible. At 1.40 MPa the steam tables give a saturation temperature of 195.0 °C with $h_f = 830.0$ and $h_{fg} = 1958.9\ \text{kJ/kg}$, so the water is heated from 30 °C to 195.0 °C in the economiser, evaporated at constant temperature, then superheated to 540 °C. The four terminal temperatures shown on the diagram are gas 560 °C in and 180 °C out, water 30 °C in and steam 540 °C out.
Part (b) — heat recovered from the gas. Treating the products as an ideal gas of constant specific heat $c_p = 1.148\ \text{kJ/kg K}$ (the value the paper assigns to hot gas), $$\dot{Q} = M_{\text{gas}}\,c_p (T_{\text{in}} - T_{\text{out}}) = 145 \times 1.148 \times (560-180) = \boxed{63\,250\ \text{kW}} .$$
Steam mass flow from the water-side enthalpy rise. The superheated table at 1.40 MPa gives $h = 3474.8$ at 500 °C and $3695.4\ \text{kJ/kg}$ at 600 °C, so at 540 °C linear interpolation gives $h_{\text{steam}} = 3562.4\ \text{kJ/kg}$; feedwater at 30 °C and 1.40 MPa carries $h_{fw} = 127.0\ \text{kJ/kg}$. Hence $$M_{\text{steam}} = \frac{\dot{Q}}{h_{\text{steam}} - h_{fw}} = \frac{63\,250}{3562.4 - 127.0} = \boxed{18.4\ \text{kg/s}} .$$ The gas-to-steam mass ratio of about 7.9 is typical of an unfired heat recovery boiler on a machine of this size.
Confirm the profile does not cross. The economiser takes $18.41 \times (830.0 - 127.0) = 12\,940\ \text{kW}$, which is 20.5 % of the total duty. Because the gas cools linearly, its temperature where the water reaches saturation is $$T_{\text{pinch}} = 180 + 0.2046 \times (560-180) = 257.8\ ^\circ\text{C},$$ which is 62.8 K above the 195.0 °C boiling water. The profile is therefore feasible, and the pinch is generous — the low steam pressure chosen for this plant is what makes such a deep gas cooling possible.
Part (c) — the end states of the expansion. The turbine exhausts at 30 °C, so the condenser sits at the corresponding saturation pressure $p_{\text{cond}} = 4.25\ \text{kPa}$, where $h_f = 125.7$, $h_{fg} = 2429.8\ \text{kJ/kg}$, $s_f = 0.4368$ and $s_{fg} = 8.0152\ \text{kJ/kg K}$. Interpolating entropy in the same superheat table gives the inlet state as $s_1 = 7.712\ \text{kJ/kg K}$, so an isentropic expansion ends wet at $$x_{2s} = \frac{s_1 - s_f}{s_{fg}} = \frac{7.712-0.4368}{8.0152} = 0.908,$$ $$h_{2s} = 125.7 + 0.908 \times 2429.8 = 2332.4\ \text{kJ/kg} .$$
Actual enthalpy drop and shaft power. The internal efficiency scales the ideal drop: $$\Delta h_s = 3562.4 - 2332.4 = 1230.0\ \text{kJ/kg},\qquad \Delta h = \eta_i \Delta h_s = 0.85 \times 1230.0 = 1045.5\ \text{kJ/kg},$$ $$P_{\text{steam}} = M_{\text{steam}} \Delta h = 18.41 \times 1045.5 = \boxed{19\,250\ \text{kW} = 19.3\ \text{MW}} .$$ The real exhaust enthalpy is $3562.4 - 1045.5 = 2516.9\ \text{kJ/kg}$, a quality of 0.984, so the last stage is barely wet — a consequence of the very high superheat carried at only 1.40 MPa, and welcome news for blade erosion.
Part (d) — overall combined-cycle efficiency. The two shafts share a single fuel input, so $$\dot{Q}_{\text{fuel}} = 2.69 \times 40\,000 = 107\,600\ \text{kW},$$ $$\eta_{cc} = \frac{P_{\text{gas}} + P_{\text{steam}}}{\dot{Q}_{\text{fuel}}} = \frac{42\,500 + 19\,250}{107\,600} = \boxed{0.574 = 57.4\ \%} .$$ The gas turbine alone would return $42\,500/107\,600 = 39.5$ %, so the bottoming cycle adds nearly 18 percentage points for no extra fuel at all — the whole commercial case for the combined cycle in one line.
