Question 1 of 6: Combined Cycle Plant

Question 1: 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. Gas mass flow \(G = 100\ \text{kg/s}\); cold-air-standard properties from page 14, \(c_p = 1.005\ \text{kJ/kg}\cdot\text{K}\), \(k = 1.4\). The state points printed on page 9 are:

Find. The ten quantities (a)–(j): the combustor duty, the two steam flows, the two machine outputs, three machine efficiencies, the pump work and the overall plant efficiency.

Approach. Apply the steady-flow energy equation to one component at a time — combustor, heat-recovery steam generator, feedwater heater, then each machine — using the printed state points, and finish by summing the shaft work against the fuel heat.

1. Part (a) — Heat input to the combustion chamber. The combustor raises the compressed air from state 2 to state 3 at constant pressure, so on the cold-air-standard basis $$\dot Q_{cc} = G\,c_p (T_3 - T_2) = 100 \times 1.005 \times (1000 - 422)$$ $$\boxed{\dot Q_{cc} = 58\,089\ \text{kW} = 58.09\ \text{MW}}$$ This is the total fuel heat released, and it is the denominator of every efficiency asked for later.
2. Part (b) — Main steam flow from the heat-recovery steam generator. The HRSG is adiabatic to the surroundings, so the heat given up by the gas between 4 and 5 equals the heat taken up by the water between 9 and 10: $$G\,c_p (T_4 - T_5) = M (h_{10} - h_9)$$ $$100 \times 1.005 \times (418 - 159) = 26\,029.5\ \text{kW}$$ Dividing by the enthalpy rise of the water, $$M = \frac{26\,029.5}{3196 - 610} = \frac{26\,029.5}{2586}$$ $$\boxed{M = 10.07\ \text{kg/s}}$$ So roughly one tenth of a kilogram of steam is raised per kilogram of turbine exhaust — the usual order for an unfired HRSG.
3. Part (c) — Bled steam flow from the direct-contact heater. In an open heater the bled steam \(m\) at state 11 mixes directly with the condensate \((M-m)\) at state 7 and leaves as saturated water at state 8. Energy and mass balances give $$m\,h_{11} + (M - m)\,h_7 = M\,h_8 \quad\Longrightarrow\quad m = M\,\frac{h_8 - h_7}{h_{11} - h_7}$$ $$m = 10.066 \times \frac{605 - 136}{2719 - 136} = 10.066 \times \frac{469}{2583}$$ $$\boxed{m = 1.828\ \text{kg/s}}$$ which is 18.2 % of the main steam flow. The balance of the flow, \(M - m = 8.238\ \text{kg/s}\), passes on to the condenser.
4. Part (d) — Net power of the gas turbine. The compressor is on the same shaft as the turbine, so the net output is the turbine expansion work less the compression work: $$\dot W_{GT} = G\,c_p\big[(T_3 - T_4) - (T_2 - T_1)\big] = 100 \times 1.005 \times \big[(1000-418) - (422-30)\big]$$ $$= 100.5 \times (582 - 392) = 100.5 \times 190$$ $$\boxed{\dot W_{GT} = 19\,095\ \text{kW} = 19.10\ \text{MW}}$$ The back-work ratio is \(392/582 = 0.67\) — two thirds of the turbine's gross work is consumed by its own compressor, which is characteristic of a gas turbine.
5. Part (e) — Power of the steam turbine. The machine expands the full flow \(M\) from state 10 to the bleed point 11, and only the remainder \((M-m)\) from 11 to the exhaust 12: $$\dot W_{ST} = M(h_{10} - h_{11}) + (M-m)(h_{11} - h_{12})$$ $$= 10.066(3196 - 2719) + 8.238(2719 - 2201) = 4801 + 4267$$ $$\boxed{\dot W_{ST} = 9068\ \text{kW} = 9.07\ \text{MW}}$$ The bottoming cycle therefore contributes about a third of the plant output, the classic split for a combined cycle.
6. Part (f) — Isentropic efficiency of the air compressor. For compression the ideal work is the smaller of the two, so the isentropic temperature rise goes on top: $$\eta_{C} = \frac{T_{2S} - T_1}{T_2 - T_1} = \frac{344 - 30}{422 - 30} = \frac{314}{392}$$ $$\boxed{\eta_{C} = 0.801 = 80.1\ \%}$$
7. Part (g) — Isentropic efficiency of the gas turbine. For expansion the ideal work is the larger, so the actual temperature drop goes on top: $$\eta_{T} = \frac{T_3 - T_4}{T_3 - T_{4S}} = \frac{1000 - 418}{1000 - 353} = \frac{582}{647}$$ $$\boxed{\eta_{T} = 0.900 = 90.0\ \%}$$ Both figures are typical of a modern industrial machine; note the turbine is the better of the two, as it always is, because the expanding flow is stabilised by a favourable pressure gradient.
8. Part (h) — Internal efficiency of the steam turbine. The internal (isentropic) efficiency compares the actual work of Step 5 with the work the same two flows would do expanding reversibly along the isentrope through state 10, i.e. through 11S and on to 12S: $$\dot W_{ST,s} = M(h_{10} - h_{11S}) + (M-m)(h_{11S} - h_{12S})$$ $$= 10.066(3196-2634) + 8.238(2634-2025) = 5657 + 5017 = 10\,674\ \text{kW}$$ $$\eta_{ST} = \frac{9068}{10\,674}$$ $$\boxed{\eta_{ST} = 0.850 = 85.0\ \%}$$ Taken section by section the same data give \(\eta_{HP} = (3196-2719)/(3196-2634) = 84.9\ \%\) for the expansion down to the bleed point. The lower section must be referred to the isentrope through the actual state 11 — this is the state left blank as 12SS on page 9; entering the steam tables at 0.4 MPa with \(h_{11} = 2719\ \text{kJ/kg}\) gives \(s_{11} = 6.852\ \text{kJ/kg}\cdot\text{K}\), and expanding at that entropy to 0.005 MPa gives \(h_{12SS} = 2088\ \text{kJ/kg}\), so \(\eta_{LP} = (2719-2201)/(2719-2088) = 82.1\ \%\).
9. Part (i) — Work done by the pumps. Both pumps are read straight off the table as enthalpy rises. The feed pump handles the full flow from 8 to 9: $$\dot W_{FP} = M (h_9 - h_8) = 10.066 \times (610 - 605) = 50.3\ \text{kW}$$ The condensate pump handles \((M-m)\) from 6 to 7, and the table rounds both enthalpies to 136 kJ/kg, so its work is below the resolution of the printed data. Estimating it independently from \(w = v_f\,\Delta p = 0.001005 \times (400-5) = 0.40\ \text{kJ/kg}\) gives only 3.3 kW. Taking the table at face value, $$\boxed{\dot W_{\text{pumps}} = 50.3\ \text{kW}}$$ which is 0.55 % of the steam-turbine output — the reason pump work is routinely neglected in a steam cycle but never in a gas cycle.
10. Part (j) — Overall efficiency of the plant. The pumps are driven from the steam-turbine output, so the saleable power is $$\dot W_{net} = \dot W_{GT} + \dot W_{ST} - \dot W_{\text{pumps}} = 19\,095 + 9068 - 50 = 28\,113\ \text{kW}$$ $$\eta_{\text{plant}} = \frac{\dot W_{net}}{\dot Q_{cc}} = \frac{28\,113}{58\,089}$$ $$\boxed{\eta_{\text{plant}} = 0.484 = 48.4\ \%}$$ Compare the gas turbine on its own: \(19\,095/58\,089 = 32.9\ \%\). Bolting a modest steam cycle onto its exhaust lifts the plant by 15.5 percentage points without burning another gram of fuel, which is the whole argument for the combined cycle.