Question 2 of 6: Combined Cycle Plant

Question 2: Combined Cycle Plant (15 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. The Question 1 machine with a single-pressure heat recovery boiler and steam turbine added:

Find. The gas and steam temperature profiles, every enthalpy round the steam circuit, both machine outputs and the combined cycle efficiency, the pinch-point temperature difference, and a discussion of what the pinch means for the boiler design.

Approach. The gas side sets how much heat is available; the steam side sets how much enthalpy each kilogram of steam must absorb. Dividing one by the other gives the steam flow, after which both powers follow directly. The pinch is then found by walking back up the gas line from the boiler inlet through the superheater and evaporator duties.

1. Part (b) — the low-pressure end of the steam cycle. Condensate leaves the condenser saturated at 40 °C, so from the steam tables $h_1 = h_f = 167.53\ \text{kJ/kg}$, $v_f = 0.001008\ \text{m}^3\text{/kg}$. The feed pump raises it to 1.6 MPa: $$w_p = v_f\,(p_2 - p_1) = 0.001008 \times (1600 - 7.4) = 1.61\ \text{kJ/kg}$$ so the water entering the economiser carries $h_2 = 167.53 + 1.61 = 169.1\ \text{kJ/kg}$. The pump work is under one part in two thousand of the turbine work but it is retained because the boiler duty is measured from state 2, not state 1.
2. Part (b) — the boiler states. At 1.6 MPa the saturation temperature is 201.37 °C with $h_f = 858.6\ \text{kJ/kg}$ and $h_g = 2794.0\ \text{kJ/kg}$; superheated to 350 °C at the same pressure the steam tables give $h_3 = 3146.0\ \text{kJ/kg}$ and $s_3 = 7.0713\ \text{kJ/kg}\cdot\text{K}$. Each kilogram therefore absorbs $h_3 - h_2 = 2976.9\ \text{kJ/kg}$ in the boiler, split 689 / 1935 / 352 kJ/kg between economiser, evaporator and superheater.
3. Part (b) — the turbine exhaust state. An isentropic expansion to the condenser holds $s = 7.0713$, which at 40 °C (where $s_f = 0.5724$ and $s_{fg} = 7.6832$) implies a quality $$x_{4s} = \frac{7.0713 - 0.5724}{7.6832} = 0.8459 \quad\Rightarrow\quad h_{4s} = 167.53 + 0.8459 \times 2406.0 = 2202.7\ \text{kJ/kg}$$ The isentropic drop is $3146.0 - 2202.7 = 943.3\ \text{kJ/kg}$, so the actual drop is $0.88 \times 943.3 = 830.1\ \text{kJ/kg}$ and $$h_4 = 3146.0 - 830.1 = \boxed{2315.9\ \text{kJ/kg}}$$ corresponding to a wetness of only 10.7 % at exhaust, comfortably inside the erosion limit.
4. Part (c) — gas turbine output. The fuel supplies $$\dot Q_{fuel} = \dot m_f \times \mathrm{HV} = 13 \times 40 = 520\ \text{MW}$$ and at the stated 28 % simple-cycle efficiency the gas turbine delivers $$P_{gt} = 0.28 \times 520 = \boxed{145.6\ \text{MW}}$$ which reproduces the 145.1 MW computed independently in Question 1.
5. Part (c) — steam flow and steam turbine output. Treating the exhaust as cold-standard air with $c_p = 1.005$ and taking no casing loss, the boiler recovers $$\dot Q_{hrb} = \dot m_g c_p (t_{g,in} - t_{g,out}) = 1000 \times 1.005 \times (392 - 177) = 216.1\ \text{MW}$$ and since each kilogram of steam takes 2976.9 kJ, $$\dot m_s = \frac{216075}{2976.9} = \boxed{72.58\ \text{kg/s}}$$ $$P_{st} = \dot m_s\,\Delta h_{act} = 72.58 \times 830.1 = \boxed{60.3\ \text{MW}}$$ against a feed-pump demand of only 117 kW.
6. Part (c) — combined cycle efficiency. The fuel input is unchanged, because the bottoming cycle burns nothing: $$\eta_{cc} = \frac{P_{gt} + P_{st} - P_{pump}}{\dot Q_{fuel}} = \frac{145.6 + 60.3 - 0.1}{520} = \boxed{39.6\ \%}$$ The steam tail adds 11.6 percentage points, lifting the plant from 28 % to nearly 40 % — a 41 % relative gain, and the reason essentially every new gas-fired station is a combined cycle.
7. Part (d) — the pinch-point temperature difference. The pinch sits where evaporation begins, because from there onwards the water temperature is held at $t_{sat}$ while the gas keeps falling. Walking down the gas line from the boiler inlet through the superheater and evaporator duties, $$\dot Q_{sup} + \dot Q_{evap} = \dot m_s (h_3 - h_f) = 72.58 \times 2287.4 = 166.0\ \text{MW}$$ $$\Delta t_{gas} = \frac{166030}{1000 \times 1.005} = 165.2\ \text{K} \quad\Rightarrow\quad t_{g,pinch} = 392 - 165.2 = 226.8\ ^\circ\text{C}$$ $$\Delta T_{pinch} = 226.8 - 201.37 = \boxed{25.4\ \text{K}}$$ The economiser then removes a further 50.0 MW, cooling the gas by 49.8 K to 177 °C — exactly the boiler gas outlet temperature printed in the question, which closes the whole steam side independently.
8. Part (e) — significance of the pinch, and how to improve the boiler. Both alternatives are answered. Heat transfer and capital cost: the pinch is the smallest temperature difference anywhere in the boiler and therefore the point that governs surface area, since $\dot Q = UA\,\Delta T_{lm}$ and it is the pinch that drags the log-mean difference down. Choosing a small pinch recovers more heat and raises steam production, but the evaporator area needed grows roughly as the reciprocal of the pinch, so area and cost rise steeply as the pinch is squeezed; below about 8–10 K the exchanger becomes uneconomic and control becomes fussy, while a pinch much above 25 K wastes recoverable heat up the stack. The 25 K found here is generous, which is why the stack is still at 177 °C. Modification: the thermodynamic loss in a single-pressure boiler is the large temperature gap between the gas line and the flat evaporation line. Dual- or triple-pressure steam generation, with a supplementary low-pressure evaporator fed from the tail of the gas, folds the water line closer to the gas line and drops the stack temperature; raising the main steam pressure and adding reheat lifts the mean temperature of heat addition in the steam cycle itself. A lower condenser pressure would also help, though it is limited by cooling-water temperature.